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189900 | https://www.nwea.org/blog/2024/opening-up-students-mathematical-thinking-using-open-middle-math-problems/ | Skip to content
By
Mary Resanovich, NWEA, and Ashley Cruz, NWEA
Opening up students’ mathematical thinking using Open Middle math problems
Back when we were in schools every day, we were constantly on the lookout for resources to support rich, rigorous, mathematical learning. Although many of the teachers we worked to support liked the concrete-representational-abstract framework of their math curriculum, we frequently heard them express frustration over the rote practice problems provided in most of their resources. Teachers were spending a lot of time looking for content that went beyond a standard worksheet. They were telling us they wanted activities that would spark student excitement and encourage deeper thinking, like Open Middle math problems.
When paired with number talks, Open Middle problems can help teachers embed rich conversation throughout math class.
Defining the problem
So, what exactly are Open Middle math problems? In an Open Middle problem, students are given starting information and either an answer or a definition of the answer. By “definition of the answer,” we mean that students are given a description of the type of answer toward which they are driving. For example, many Open Middle math problems ask students to find the least or greatest solution or a number that is closest to a given number. How they get from start to end is up to them. Rather than following a solution path that was stepped out for them by the teacher, the path is completely under their control.
The best way to understand an Open Middle math problem may be to compare one to a typical textbook problem for the same topic. For a lesson on three-digit subtraction, a standard textbook might ask students to find 539 – 286 either as a straight computation item or embedded within a word problem. By contrast, an Open Middle math problem for the same topic asks students to use the digits 1–9, at most one time each, to make two three-digit numbers with a difference as close to 329 as possible.
Although understanding of place value, regrouping, the relative magnitude of numbers, and the relationship between addition and subtraction underlie both types of problems, with the more traditional problem, a student can answer correctly by following a set of learned steps without truly understanding the conceptual underpinnings. However, students must actively engage with these concepts to solve the Open Middle math problem. The open structure forces students to consider these concepts and constantly make decisions and adjust their solution path. Here are some of the questions students might consider when solving the Open Middle math problem:
What strategies can be applied to solve? Should I subtract? Should I add up? Can estimation help?
Should I focus on a certain place value for each number first, or should I create one number completely?
Is it possible to make two numbers where regrouping will not be required?
Is it possible to get exactly 329?
Open Middle problems can also be word problems. While some Open Middle math problems are structured this way, others can easily be turned into word problems. For example, the three-digit subtraction problem above can be rewritten as a word problem such as this: Jared and Suni are playing a video game. Each of their scores were three-digit numbers. Jared’s score was 329 less than Suni’s. Use the digits 1–9, at most one time each, to make two three-digit numbers that could be Jared and Suni’s scores.
It’s the journey, not the destination
Because of the way they are designed, Open Middle math problems emphasize process over product. While students are still expected to arrive at a defensible answer, Open Middle problems push students to utilize a variety of mathematical skills and understandings to wrestle with the problem. The goal is not to be the quickest to follow a predetermined solution path but, rather, to allow for a more iterative approach where students continually test and revise strategies. Dan Meyer terms this “patient problem solving,” where students bring all their mathematical knowledge and intuition to bear on solving non-routine problems. The active engagement that this engenders supports rich mathematical thinking and discussion amongst students.
One of the co-developers of Open Middle problems, Nanette Johnson, describes their value this way: “This structure requires students to prove to themselves and others that they have truly found the best possible answer. Rather than putting down their pencils and saying, ‘I’m done,’ students continue to think, argue, and work. They develop the habit of making multiple attempts to solve the problem, each time wondering if they can come up with an even better solution than their last.”
The fact that students can arrive at different answers naturally prompts discussion between students of how they approached the problem, why they think their answer best meets the criteria, and whether other answers could get closer to the goal. As Johnson states, this serves to foster increased engagement and persistence. It also provides a vehicle for the rich mathematical discourse we were seeking in our districts.
Low floors, high ceilings, and big concepts
By design, Open Middle math problems allow easy entry for all students, but they also provide the flexibility to allow students to stretch themselves. For this reason, Open Middle problems have been described as easy in terms of getting an answer, but not so easy in terms of getting the ideal or “best” answer. Take, for example, that problem that challenges students to use the digits 1–9, at most one time each, to create two mixed numbers with the least possible difference. Students are given a template like the one on the Open Middle website in which to write their numbers.
Students who are still developing their understanding of fraction operations might take a simple approach of having the digits in the second mixed number be one less than the digits in the first mixed number: 9 5/7 – 8 4/6. This results in a difference of 1 1/21. Another student might opt to use the greatest digits available, 8 and 9, to make the denominators of the fractions, knowing that the greater number of parts a fraction is divided into, the smaller it is. Perhaps they create the expression 2 7/9 – 1 6/8 , which results in a difference of 1 1/36. At this point, the teacher might ask these two students which of their answers is the least, sparking a conversation about the relative magnitude of the numbers, benchmark fractions, and estimation. The teacher could then push students by challenging them to see if it is possible to get an answer that is less than one.
A student further along in their understanding of fractions might recognize that the need to regroup could reduce the size of the final number. Perhaps they create the problem 3 1/9 – 2 7/8 with the result of 17/72. Student interactions or well-placed teacher questions, such as, “Is it possible to get an answer of 0?” might prompt some students to realize using improper fractions can further reduce the magnitude of the answer, indeed all the way down to zero.
The nature of the problem allows all students entry without limiting students with more advanced knowledge of the involved concepts. Rich conversations and strategic teacher questions help all students expand and grow their understanding of concepts beyond the idea of fraction subtraction, while providing teachers insights into each students’ level of understanding of these concepts. In the iterative process of testing ideas, discussing different approaches, and trying new solution paths, students explore far deeper concepts of the meaning of fractions, number magnitude, and equivalent forms.
Impact of using Open Middle math problems
While there is no research that specifically examines Open Middle problems, there are studies that show the efficacy of components of the approach. In a research paper on equitable teaching approaches to math, Jo Boaler and Megan Staples call out that several studies have demonstrated that “conceptually oriented mathematics materials, taught well and with consistency, have shown higher and more equitable results for participating students than procedure-oriented curricula taught using a demonstration and practice approach.” Open Middle math problems support such a conceptual orientation and, by design, break the “demonstration and practice” model.
Having students discuss and compare different solution approaches is a critical component of Open Middle problems. Although students are not working in groups, they are also not working in isolation. Part of the iterative process of using Open Middle math problems includes time to share and discuss the relative merits of various solution approaches and use these to iterate new approaches. Actively comparing solution methods has been shown to increase procedural understanding, particularly for lower achieving students. Another study found that students who learned by comparing alternative solution approaches demonstrated greater conceptual knowledge and more flexible problem-solving than students who reflected on different approaches one at a time.
Open Middle math problems are well aligned to NCTM’s process standards. They provide rich opportunities to practice:
Building new mathematical knowledge through problem-solving
Monitoring and reflecting on the problem-solving process
Making and exploring mathematical conjectures
Developing, evaluating, and communicating mathematical reasoning and analyzing the reasoning of others
Getting started
When working on incorporating using Open Middle math problems in your classroom, we encourage you to determine how to integrate them into existing math lesson plans so they aren’t one more thing you feel you have to do. Consider, for example, how Open Middle math tasks build off number talks but take you and your students to the next level.
Both Open Middle math problems and number talks encourage rich dialogue between students. While number talks are conversation-base only, Open Middle math problems involve both conversation and written responses. Number talks are brief 10–15-minute warmups, while Open Middle math problems give students more time to explore their thinking and the thinking of others. They also provide teachers with a written record of this thinking and how it evolves over the course of iterative problem-solving.
You might also try working on Open Middle problems yourself, which can help you understand the power of the approach.
Adding doesn’t necessarily mean more
We also encourage you to focus on potential roadblocks to making Open Middle math problems a part of your classroom. In our experience, time can be a big concern. If that’s worrying you as well, we suggest you focus on the relative merit of having students deeply engage in a single rich mathematical problem in place of completing 10 rote problems. This can help you move away from a “mile-wide, inch deep” approach to math.
Remember, too, that these problems do not need to be completed in a day. Take the time to let students work through possible solutions, share, discuss, and revise over the course of one or more days.
Finally, time you previously spent looking for better resources could be repurposed into unpacking Open Middle math problems to prepare you for facilitating a rich classroom discussion. Here are some questions to get you started:
What big mathematical ideas do you hope students will bring into discussions?
What models or mathematical terms might students use in their explanations?
What are examples of clear justification or correct reasoning you hope to see?
What are some possible solution approaches and what do each of them reveal about students’ understanding of the underlying concepts?
Give it a try!
Adding Open Middle tasks to your math routines can truly deepen both math conversations and student engagement in your classroom. Ready to try them yourself? Use these links to get more information about Open Middle math problems and supporting mathematical writing and discourse.
Open Middle resources
Open Middle This site, created by Open Middle co-creators Nanette Johnson and Robert Kaplinsky, contains a wealth of Open Middle tasks organized by grade and domain as well as worksheets in English, French, and Spanish and printable number tiles.
Open Middle Exercises GeoGebra has converted many of the paper versions of Open Middle exercises into interactive versions.
Open Middle Math: Problems that Unlock Student Thinking, Grades 6–12 You can preview Robert Kaplinsky’s book about Open Middle math here.
Classroom discussion and mathematical writing resources
“4 ways to engage students with writing in math class” This article delves into a different way to support and encourage student writing about math, including how to write high-quality explanations for how they solved a problem.
“Eliciting, supporting, and guiding the math: Three key functions of the teacher’s role in facilitating meaningful mathematical discourse” This piece defines what makes math discourse meaningful, why it’s important, and what the teacher’s role is in facilitating such discourse.
“Mathematical discourse in the classroom: A prime time for discussion” This article provides strategies for facilitating meaningful mathematical discussion with the goal of helping students develop a deeper understanding of mathematical concepts.
“Using the 5 practices in mathematics teaching” This article provides an overview of how teachers can successfully orchestrate classroom discussions by anticipating possible student solutions, monitoring their work, selecting and sequencing how students share their work, and asking questions that help students make connections between mathematical ideas.
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189901 | https://math.stackexchange.com/questions/3828225/prove-all-finite-disjoint-unions-of-intervals-in-a-collection-of-all-a-b | probability theory - Prove all finite disjoint unions of intervals in a collection of all $(a, b],(-\infty, b]$or $(a,\infty)$ ,$-\infty<a<b<\infty$ forms a field - Mathematics Stack Exchange
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[Prove all finite disjoint unions of intervals in a collection of all (a,b],(−∞,b](a,b],(−∞,b]or (a,∞)(a,∞) ,−∞<a<b<∞−∞<a<b<∞ forms a field](
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Define
C I≡{all intervals(a,b],(−∞,b],or(a,∞):−∞<a<b<∞}C I≡{all intervals(a,b],(−∞,b],or(a,∞):−∞<a<b<∞}
C F≡{C F≡{ all finite disjoint unions of intervals in C I}C I}.
Show that C F C F is a field.
I didn't study any mathematical courses rigorously at the college level due to my curriculum design so so it is very challenging to study probability course founded on measure theory. I tried to prove this statement using the definition of a set but I am not sure if my proof is correct. Here's my proof:
C F C F is the set of all finite disjoint unions of intervals, that is
(a 1,b 1]∪⋯∪(a n,b n](a 1,b 1]∪⋯∪(a n,b n]
∅∈C F∅∈C F because it is the disjoint zero interval in C I C I.
A≡⋃∞(a n,b n]A≡⋃∞(a n,b n] where b i≤a i+1,i∈N b i≤a i+1,i∈N, allowing for a=−∞,b=∞a=−∞,b=∞, it is easily seen Ω∈C F Ω∈C F.
Suppose A′=(a,b]A′=(a,b],then A′c=(−∞,a]⋃(b,∞)A′c=(−∞,a]⋃(b,∞), (−∞,a]=⋃∞(a n,b n],b 1=a,b i+1=a i(−∞,a]=⋃∞(a n,b n],b 1=a,b i+1=a i, (b,∞)(b,∞) can also be expressed as the union of disjoint finite interval in similar fashion.
if A 1,...A n A 1,...A n are finite disjoint unions of interval, let A′1=A 1,A′2=A 2∖A 1,....A n=A n∖∪n−1 A′n A 1′=A 1,A 2′=A 2∖A 1,....A n=A n∖∪n−1 A n′,A′i A i′ can be still expressed as the union of disjoint finite intervals, and by constructions, ∪∞A n=∪∞A′n∈C F∪∞A n=∪∞A n′∈C F.
I messed up the definition of a field and a definition of a σ σ-field, as pointed by @YuvalFilmus. Following what was suggested I changed my proof:
Suppose A=(a,b]A=(a,b],then A c=(−∞,a]⋃(b,∞)A c=(−∞,a]⋃(b,∞). (−∞,a]=⋃(a n,b n],n∈N(−∞,a]=⋃(a n,b n],n∈N, where b 1=a,b i+1=a i b 1=a,b i+1=a i, (b,∞)=⋃(a n,b n],n∈N(b,∞)=⋃(a n,b n],n∈N, where a 1=b,a i+1=b i a 1=b,a i+1=b i. Therefore,A c∈C F A c∈C F.
Let A,B∈C F A,B∈C F, 1) A∩B=∅,A∪B∈C F A∩B=∅,A∪B∈C F; 2) A∩B=A A∩B=A or B,A∪B∈C F B,A∪B∈C F; 3) Let A=(a 1,b 1],B=(a 2,b 2]A=(a 1,b 1],B=(a 2,b 2], if a 1<a 2<b 1<b 2 a 1<a 2<b 1<b 2, then A∪B=(a 1,a 2]∪(a 2,b 1]∪(b 1,b 2]∈C F A∪B=(a 1,a 2]∪(a 2,b 1]∪(b 1,b 2]∈C F; similarly, if a 2<a 1<b 2<b 1 a 2<a 1<b 2<b 1, A∪B∈C F A∪B∈C F.
Since C F C F is closed under finite union, it must also be closed under finite interval by DeMorgan's Law, hence A∩A c=∅∈C F A∩A c=∅∈C F. Ω=∅c∈C F Ω=∅c∈C F.
Two additional questions:
If indeed my proof is correct are there ways to prove the statement more concisely?
To my knowledge ∅∅ is always in a set but why is it the case? Is it by definition?
When I prove C F C F is a field I think of the infinite many unions are still in C F C F and it is very convincing to me. But being a field means the infinite intersections are also in the field, that implies that C F C F also includes every single point {x},x∈R{x},x∈R. But I don't quite get how can a single point be in C F C F? Could someone please express a point as a finite disjoint union of interval?
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edited Sep 16, 2020 at 21:36
Hanul Jeon
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asked Sep 16, 2020 at 9:05
JoZJoZ
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@Ramiro I think I am fairly clear with the idea of the field and σ σ field... but what the relationship between a field and an algebra and similar I am not quite sure about the difference between the σ σ-field and σ σ-algebra...JoZ –JoZ 2020-09-17 03:59:51 +00:00 Commented Sep 17, 2020 at 3:59
In fact, field (of sets) is the same as algebra (of sets), and σ σ-field is the same as σ σ-algebra. They are synonyms.Ramiro –Ramiro 2020-09-17 14:38:33 +00:00 Commented Sep 17, 2020 at 14:38
The FOUR related concepts I mentioned in a previous comment are actually: ring (of sets), algebra (of sets), σ σ-ring and σ σ-algebra. Rings and σ σ-rings are not required to be closed under complement, just closed under difference.Ramiro –Ramiro 2020-09-17 14:45:43 +00:00 Commented Sep 17, 2020 at 14:45
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According to Wikipedia, a field of sets is a collection of sets which is closed under complementation and finite unions and intersections (also known as algebra). In particular, there is no requirement to be closed under infinite unions and intersections. A related concept, σ σ-algebra allows countable unions and intersections.
Now regarding your proof:
A non-empty field F F always contains the empty set: if A∈F A∈F then ∅=A∩A¯¯¯¯∈F∅=A∩A¯∈F.
If a collection of sets is closed under complement and finite unions, then it is also closed under finite intersections, due to de Morgan's laws: A 1∩⋯∩A n=A 1¯¯¯¯¯¯∪⋯∪A n¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯.A 1∩⋯∩A n=A 1¯∪⋯∪A n¯¯.
I'm not sure what "A≡⋃∞(a n,b n]A≡⋃∞(a n,b n]" means. I have never seen the notation ⋃∞⋃∞, for example.
The collection C F C F is not closed under infinite unions and intersections, even countably many. Indeed, as you mention, {0}=⋂n≥0(−2−n,0]{0}=⋂n≥0(−2−n,0], yet {0}∉C F{0}∉C F.
Showing that C F C F is a field unfortunately requires some lengthy case analysis, with much more detail than is given in your proof.
Here is a complete proof. We start with closure under union.
Lemma 1. If A∈C F A∈C F and B∈C I B∈C I then A∪B∈C F A∪B∈C F.
Proof. Let us say that A A has complexity n n if it is the union of n n disjoint intervals from C I C I. The proof is by induction on n n. If n=0 n=0 then A∪B=B∈C F A∪B=B∈C F.
Assuming the lemma holds for n n, we prove it for n+1 n+1. Suppose that A A has complexity n+1 n+1, and so it can be written as a disjoint union of n+1 n+1 intervals I 1,…,I n+1∈C I I 1,…,I n+1∈C I. Let I 1 I 1 be the interval with minimal endpoint. We consider three cases:
The interval B B is entirely to the left of I 1 I 1. In that case, B,I 1,…,I n+1 B,I 1,…,I n+1 are disjoint, and so clearly A∪B∈C F A∪B∈C F.
The interval B B is entirely to the right of I 1 I 1. By induction, (I 2∪⋯∪I n+1)∪B∈C F(I 2∪⋯∪I n+1)∪B∈C F. Appealing to the preceding case shows that I 1∪(I 2∪⋯∪I n+1∪B)∈C F I 1∪(I 2∪⋯∪I n+1∪B)∈C F.
The intervals B B and I 1 I 1 intersect. Write B=(a,b]B=(a,b] and I 1=(c,d]I 1=(c,d], where possibly a,c=−∞a,c=−∞ or b,d=+∞b,d=+∞. Then J:=B∪I 1=(min(a,c),max(b,d)]J:=B∪I 1=(min(a,c),max(b,d)]. If J∈C I J∈C I then A∪B=J∪I 2∪⋯∪I n+1∈C F A∪B=J∪I 2∪⋯∪I n+1∈C F by induction. Otherwise, J=(−∞,∞)J=(−∞,∞), and so A∪B=(−∞,∞)=(−∞,0]∪(0,∞)∈C F A∪B=(−∞,∞)=(−∞,0]∪(0,∞)∈C F. □◻
Lemma 2. If A,B∈C F A,B∈C F then A∪B∈C F A∪B∈C F.
Proof. The proof is by induction on the complexity n n of B B. If B=∅B=∅ then A∪B=A∈C F A∪B=A∈C F. Otherwise, write B=C∪D B=C∪D, where C C has complexity n−1 n−1 and D∈C I D∈C I. By induction, A∪C∈C F A∪C∈C F, and so A∪B=(A∪C)∪D∈C F A∪B=(A∪C)∪D∈C F by Lemma 1. □◻
We proceed to closure under complementation.
Lemma 3. If A∈C F A∈C F then A¯¯¯¯∈C F A¯∈C F.
Proof. If A=∅A=∅ then A¯¯¯¯=(−∞,0]∪(0,∞)∈C F A¯=(−∞,0]∪(0,∞)∈C F. Otherwise, write
A=(a 1,b 1]∪⋯∪(a n,b n],A=(a 1,b 1]∪⋯∪(a n,b n],
where b i≤a i+1 b i≤a i+1 for all i∈{1,…,n−1}i∈{1,…,n−1}, and possibly a 1=−∞a 1=−∞ and b n=∞b n=∞. Then
A¯¯¯¯=(−∞,a 1]∪(b 1,a 2]∪⋯∪(b n−1,a n]∪(b n,∞).A¯=(−∞,a 1]∪(b 1,a 2]∪⋯∪(b n−1,a n]∪(b n,∞).
If a 1=−∞a 1=−∞, we can remove the first interval. If b n=∞b n=∞, we can remove the last interval. If b i=a i+1 b i=a i+1 for some i i, we can remove the interval (b i,a i+1](b i,a i+1]. After removing these empty intervals, we get a representation of A¯¯¯¯A¯ as a disjoint union of intervals from C I C I, hence A¯¯¯¯∈C F A¯∈C F. □◻
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edited Apr 30, 2024 at 10:30
psie
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answered Sep 16, 2020 at 9:35
Yuval FilmusYuval Filmus
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Oh yes... I am confused JoZ –JoZ 2020-09-16 09:36:41 +00:00 Commented Sep 16, 2020 at 9:36
I changed my proof. Would it be a valid one?JoZ –JoZ 2020-09-16 10:29:46 +00:00 Commented Sep 16, 2020 at 10:29
I’m afraid this is not enough. You’re only considering single intervals, whereas you should be considering unions of disjoint intervals, per the definition.Yuval Filmus –Yuval Filmus 2020-09-16 11:16:44 +00:00 Commented Sep 16, 2020 at 11:16
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189902 | https://www.dentonisd.org/cms/lib/TX21000245/Centricity/Domain/1519/Notes_on_Prime_Numbers_and_Prime_Factorization.pdf | Notes on Factors, Prime Numbers, and Prime Factorization Factors are the numbers that multiply together to get another number. A Product is the number produced by multiplying two factors. All numbers have 1 and itself as factors. A number whose only factors are 1 and itself is a prime number. Prime numbers have exactly two factors. The smallest 168 prime numbers (all the prime numbers under 1000) are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997 There are infinitely many prime numbers. Another way of saying this is that the sequence of prime numbers never ends. The number 1 is not considered a prime. The definition of a prime number is one that has exactly TWO factors: itself and 1. So the number 1, having only ONE factor, itself, does not meet the definition. A number with three or more factors is a composite number. Divisibility rules can be used to factor a number and to test the primality of a number. Some divisibility rules: • All numbers are divisible by 1 • Any even number is divisible by 2. A number is divisible by 2 if it ends with a 0, 2, 4, 6, or 8. • A number is divisible by 3 if the sum of its digits is divisible by 3. • A number is divisible by 4 if the last two digits (tens and ones) are divisible by 4. • A number is divisible by 5 if it ends in a 5 or 0. • A number is divisible by 6 if it is also divisible by 2 and 3 (see these tests above). • To test a number for divisibility by 7 o Take the last digit in a number. o Double and subtract the last digit in your number from the rest of the digits. o Repeat the process for larger numbers. • A number is divisible by 8 if the last 3 digits are divisible by 8. • A number is divisible by 9 if the sum of its digits are divisible by 9 • A number is divisible by 10 if it ends in a zero. A prime factorization is a factor string expressing a number as the product of only prime factors. Every number has exactly one prime factorization. This prime factorization can be written using exponents if any of its prime factors appear more than once in the string. To factor a number • Write 1 and the number itself separated by some space. • Test the number for divisibility by 2. If it is, write 2 and the other number inside the first two. • Continue testing and writing the factor pairs inside the previous pair. • When you reach the middle, you are finished. Example: 36 {1, 36} 36 {1, 2, 18, 36} 36 {1, 2, 3, 12, 18, 36} 36 {1, 2, 3, 4, 9, 12, 18, 36} 36 {1, 2, 3, 4, 6, 9, 12, 18, 36} In the example above, 5 is not a factor of 36 so it is not in the factor list. 6 is in the middle so stop there, because 7 is not a factor of 36, nor is 8. The next number, 9, is already in the list, and every number greater than 9 has been included or eliminated when the lower factors were added. To find the prime factorization of a number with a factor tree 1. Write the number to be prime factored at the top and draw two branches below it. 2. Write its smallest prime factor on the left and circle it. If the number is even, this will be 2. Write the companion factor on the right. 3. If the companion factor is composite, draw two branches below it and repeat step 2 for this factor. 4. Continue repeating steps 2 and 3 until the companion factor on the right is prime. Circle that prime also. 5. The prime factorization is the factors on the left and bottom of the tree that are circled. Example: 84 84 is the number to be prime factored 2 42 2 is the smallest prime factor of 84 along with 42 2 21 2 is the smallest prime factor of 42 along with 21 3 7 3 is the smallest prime factor of 21 along with 7 7 is prime so stop. The prime factorization of 84 is: 2 x 2 x 3 x 7 To write the prime factorization using exponents, combine like prime factors: 22 x 3 x 7 Another way to get the factors of a number is to list its factor pairs using a systematic approach: 1 x 36, 2 x 18, 3 x 12, 4 x 9, 6 x 6 So the factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36 |
189903 | https://www.pearson.com/channels/general-chemistry/learn/jules/ch-13-chemical-kinetics/half-life | Skip to main content
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Intro to General Chemistry
3h 48m
Classification of Matter
18m
+ Physical & Chemical Changes
19m
+ Chemical Properties
7m
+ Physical Properties
5m
+ Intensive vs. Extensive Properties
13m
+ 12m
+ Scientific Notation
13m
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7m
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24m
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9m
+ Significant Figures: Precision in Measurements
8m
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14m
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16m
+ Dimensional Analysis
17m
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19m
+ Density of Non-Geometric Objects
7m
2. Atoms & Elements
4h 16m
The Atom
9m
+ Subatomic Particles
15m
+ 17m
+ 27m
+ Atomic Mass
28m
+ Periodic Table: Classifications
11m
+ Periodic Table: Group Names
8m
+ Periodic Table: Representative Elements & Transition Metals
7m
+ Periodic Table: Element Symbols
6m
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6m
+ Periodic Table: Phases
8m
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20m
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10m
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30m
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5m
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10m
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9m
+ Law of Multiple Proportions
3m
+ Millikan Oil Drop Experiment
7m
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10m
3. Chemical Reactions
4h 10m
Empirical Formula
18m
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20m
+ Combustion Analysis
38m
+ Combustion Apparatus
15m
+ Polyatomic Ions
24m
+ Naming Ionic Compounds
11m
+ Writing Ionic Compounds
7m
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6m
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18m
+ Naming Molecular Compounds
6m
+ Balancing Chemical Equations
13m
+ 16m
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17m
+ Percent Yield
19m
+ Mass Percent
4m
+ Functional Groups in Chemistry
11m
4. BONUS: Lab Techniques and Procedures
1h 38m
Laboratory Materials
29m
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12m
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12m
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4m
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14m
5. BONUS: Mathematical Operations and Functions
48m
Multiplication and Division Operations
7m
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6m
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6m
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20m
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7m
6. Chemical Quantities & Aqueous Reactions
3h 53m
Solutions
6m
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18m
+ Osmolarity
15m
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15m
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16m
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18m
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18m
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13m
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14m
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18m
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15m
+ Redox Reactions
17m
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17m
+ Balancing Redox Reactions: Basic Solutions
17m
+ Activity Series
10m
7. Gases
3h 49m
Pressure Units
6m
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18m
+ The Ideal Gas Law Derivations
13m
+ The Ideal Gas Law Applications
6m
+ Chemistry Gas Laws
13m
+ Chemistry Gas Laws: Combined Gas Law
12m
+ Mole Fraction of Gases
6m
+ Partial Pressure
19m
+ The Ideal Gas Law: Molar Mass
13m
+ The Ideal Gas Law: Density
14m
+ Gas Stoichiometry
18m
+ Standard Temperature and Pressure
14m
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13m
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9m
+ Kinetic Energy of Gases
10m
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8m
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4m
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14m
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9m
8. Thermochemistry
2h 37m
Nature of Energy
6m
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7m
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7m
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8m
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7m
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19m
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24m
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10m
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8m
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12m
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9m
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12m
+ Hess's Law
23m
9. Quantum Mechanics
2h 58m
Wavelength and Frequency
6m
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8m
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13m
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10m
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17m
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9m
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17m
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14m
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5m
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13m
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5m
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5m
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10m
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11m
+ Quantum Numbers: Spin Quantum Number
9m
+ Quantum Numbers: Number of Electrons
11m
+ Quantum Numbers: Nodes
6m
10. Periodic Properties of the Elements
3h 9m
The Electron Configuration
22m
+ The Electron Configuration: Condensed
4m
+ The Electron Configurations: Exceptions
13m
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12m
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8m
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16m
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12m
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3m
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8m
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13m
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12m
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11m
+ Periodic Trend: Electron Affinity
10m
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5m
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21m
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12m
11. Bonding & Molecular Structure
3h 29m
Lewis Dot Symbols
10m
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13m
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11m
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10m
+ Formal Charge
9m
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20m
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14m
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15m
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14m
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15m
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21m
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4m
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15m
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6m
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12m
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14m
12. Molecular Shapes & Valence Bond Theory
1h 57m
Valence Shell Electron Pair Repulsion Theory
5m
+ Equatorial and Axial Positions
10m
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11m
+ Molecular Geometry
18m
+ Bond Angles
14m
+ Hybridization
12m
+ Molecular Orbital Theory
12m
+ MO Theory: Homonuclear Diatomic Molecules
10m
+ MO Theory: Heteronuclear Diatomic Molecules
7m
+ MO Theory: Bond Order
14m
13. Liquids, Solids & Intermolecular Forces
2h 23m
Molecular Polarity
10m
+ Intermolecular Forces
20m
+ Intermolecular Forces and Physical Properties
11m
+ Clausius-Clapeyron Equation
18m
+ Phase Diagrams
13m
+ Heating and Cooling Curves
27m
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11m
+ Crystalline Solids
4m
+ Simple Cubic Unit Cell
7m
+ Body Centered Cubic Unit Cell
12m
+ Face Centered Cubic Unit Cell
6m
14. Solutions
3h 1m
Solutions: Solubility and Intermolecular Forces
17m
+ Molality
15m
+ Parts per Million (ppm)
13m
+ Mole Fraction of Solutions
8m
+ Solutions: Mass Percent
12m
+ Types of Aqueous Solutions
8m
+ Intro to Henry's Law
4m
+ Henry's Law Calculations
12m
+ The Colligative Properties
14m
+ Boiling Point Elevation
16m
+ Freezing Point Depression
10m
+ Osmosis
19m
+ Osmotic Pressure
10m
+ Vapor Pressure Lowering (Raoult's Law)
16m
15. Chemical Kinetics
2h 53m
Intro to Chemical Kinetics
4m
+ Energy Diagrams
9m
+ Catalyst
9m
+ Factors Influencing Rates
10m
+ Average Rate of Reaction
7m
+ Stoichiometric Rate Calculations
5m
+ Instantaneous Rate
5m
+ Collision Theory
7m
+ Arrhenius Equation
25m
+ Rate Law
24m
+ Reaction Mechanism
17m
+ Integrated Rate Law
23m
+ Half-Life
23m
16. Chemical Equilibrium
2h 29m
Intro to Chemical Equilibrium
7m
+ Equilibrium Constant K
13m
+ Equilibrium Constant Calculations
9m
+ Kp and Kc
22m
+ Using Hess's Law To Determine K
9m
+ Calculating K For Overall Reaction
15m
+ Le Chatelier's Principle
20m
+ ICE Charts
34m
+ Reaction Quotient
15m
17. Acid and Base Equilibrium
5h 1m
Acids Introduction
9m
+ Bases Introduction
7m
+ Binary Acids
15m
+ Oxyacids
10m
+ Bases
14m
+ Amphoteric Species
5m
+ Arrhenius Acids and Bases
5m
+ Bronsted-Lowry Acids and Bases
21m
+ Lewis Acids and Bases
12m
+ The pH Scale
16m
+ Auto-Ionization
9m
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16m
+ pH of Strong Acids and Bases
9m
+ Ionic Salts
17m
+ pH of Weak Acids
31m
+ pH of Weak Bases
32m
+ Diprotic Acids and Bases
8m
+ Diprotic Acids and Bases Calculations
30m
+ Triprotic Acids and Bases
9m
+ Triprotic Acids and Bases Calculations
17m
18. Aqueous Equilibrium
4h 47m
Intro to Buffers
20m
+ Henderson-Hasselbalch Equation
19m
+ Intro to Acid-Base Titration Curves
13m
+ Strong Titrate-Strong Titrant Curves
9m
+ Weak Titrate-Strong Titrant Curves
15m
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8m
+ Titrations: Weak Acid-Strong Base
38m
+ Titrations: Weak Base-Strong Acid
41m
+ Titrations: Strong Acid-Strong Base
11m
+ Titrations: Diprotic & Polyprotic Buffers
32m
+ Solubility Product Constant: Ksp
17m
+ Ksp: Common Ion Effect
18m
+ Precipitation: Ksp vs Q
12m
+ Selective Precipitation
9m
+ Complex Ions: Formation Constant
18m
19. Chemical Thermodynamics
1h 50m
Spontaneous vs Nonspontaneous Reactions
7m
+ Entropy
23m
+ Entropy Calculations
13m
+ Entropy Calculations: Phase Changes
6m
+ Third Law of Thermodynamics
7m
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13m
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22m
+ Gibbs Free Energy And Equilibrium
14m
20. Electrochemistry
2h 42m
Standard Reduction Potentials
9m
+ Intro to Electrochemical Cells
6m
+ Galvanic Cell
25m
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10m
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13m
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20m
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16m
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8m
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16m
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20m
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16m
21. Nuclear Chemistry
2h 36m
Intro to Radioactivity
10m
+ Alpha Decay
9m
+ Beta Decay
7m
+ Gamma Emission
7m
+ Electron Capture & Positron Emission
9m
+ Neutron to Proton Ratio
7m
+ Band of Stability: Alpha Decay & Nuclear Fission
10m
+ Band of Stability: Beta Decay
3m
+ Band of Stability: Electron Capture & Positron Emission
4m
+ Band of Stability: Overview
14m
+ Measuring Radioactivity
7m
+ Rate of Radioactive Decay
12m
+ Radioactive Half-Life
16m
+ Mass Defect
18m
+ Nuclear Binding Energy
14m
22. Organic Chemistry
5h 7m
Introduction to Organic Chemistry
8m
+ Structural Formula
8m
+ Condensed Formula
10m
+ Skeletal Formula
6m
+ Spatial Orientation of Bonds
3m
+ Intro to Hydrocarbons
16m
+ Isomers
11m
+ Chirality
15m
+ Functional Groups in Chemistry
11m
+ Naming Alkanes
4m
+ The Alkyl Groups
9m
+ Naming Alkanes with Substituents
13m
+ Naming Cyclic Alkanes
6m
+ Naming Other Substituents
8m
+ Naming Alcohols
11m
+ Naming Alkenes
11m
+ Naming Alkynes
9m
+ Naming Ketones
5m
+ Naming Aldehydes
5m
+ Naming Carboxylic Acids
4m
+ Naming Esters
8m
+ Naming Ethers
5m
+ Naming Amines
5m
+ Naming Benzene
7m
+ Alkane Reactions
7m
+ Intro to Addition Reactions
4m
+ Halogenation Reactions
4m
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3m
+ Hydrohalogenation Reactions
7m
+ Alcohol Reactions: Substitution Reactions
4m
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9m
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8m
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7m
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6m
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4m
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3m
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4m
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3m
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4m
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10m
23. Chemistry of the Nonmetals
2h 39m
Main Group Elements: Bonding Types
4m
+ Main Group Elements: Boiling & Melting Points
7m
+ Main Group Elements: Density
11m
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7m
+ The Electron Configuration Review
16m
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20m
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4m
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11m
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8m
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7m
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7m
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7m
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7m
+ Nitrogen Family Reactions
12m
+ Oxides, Peroxides, and Superoxides
12m
+ Oxide Reactions
4m
+ Peroxide and Superoxide Reactions
6m
+ Noble Gas Compounds
3m
24. Transition Metals and Coordination Compounds
3h 16m
Atomic Radius & Density of Transition Metals
11m
+ Electron Configurations of Transition Metals
7m
+ Electron Configurations of Transition Metals: Exceptions
11m
+ Paramagnetism and Diamagnetism
10m
+ Ligands
10m
+ Complex Ions
5m
+ Coordination Complexes
7m
+ Classification of Ligands
11m
+ Coordination Numbers & Geometry
9m
+ Naming Coordination Compounds
22m
+ Writing Formulas of Coordination Compounds
8m
+ Isomerism in Coordination Complexes
14m
+ Orientations of D Orbitals
4m
+ Intro to Crystal Field Theory
10m
+ Crystal Field Theory: Octahedral Complexes
5m
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4m
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4m
+ Crystal Field Theory Summary
8m
+ Magnetic Properties of Complex Ions
9m
+ Strong-Field vs Weak-Field Ligands
6m
+ Magnetic Properties of Complex Ions: Octahedral Complexes
11m
Chemical Kinetics
Half-Life
Chemical Kinetics
Half-Life: Videos & Practice Problems
Video Lessons
Practice
Worksheet
Topic summary
Half-life, the time required for half of a substance to decay or decompose, varies with reaction order. In zero-order reactions, half-life is directly proportional to the initial reactant concentration and inversely proportional to the rate constant (k), shortening as concentration decreases. For first-order reactions, common in radioactive decay, half-life is constant, independent of initial concentration, and calculated using:
Second-order reactions have a half-life that increases as the initial concentration decreases, as it is dependent on both the rate constant and the initial concentration. Understanding these relationships is crucial for predicting reaction behavior over time and is represented graphically with half-life on the y-axis and time on the x-axis.
Half-Life is the time it takes for halfof a reactant to decay in a certain time period.
Understanding Half-Life
1
concept
Zero-Order Half-life
Video duration:
2m
Play a video:
Zero-Order Half-life Video Summary
Half-life is defined as the time required for a reactant to decrease to half of its initial concentration during a decay or decomposition process. The half-life function varies depending on the order of the reaction, which can be zero, first, or second order. For zero-order reactions, the half-life can be calculated using the equation:
t1/2 = &frac{[A]0}{2k}
In this equation, [A]0 represents the initial concentration of the reactant, and k is the rate constant, measured in units of molarity per time (M·s-1). It is important to note that for zero-order reactions, the half-life is directly dependent on the initial concentration of the reactant. As the concentration decreases, the half-life also decreases. This means that the time taken to lose half of the initial amount becomes shorter as the concentration diminishes.
Graphically, this relationship can be represented with half-life plotted on the y-axis and time on the x-axis. As time progresses, the half-life decreases, resulting in a negative slope on the graph. This illustrates that the time required to reach half of the initial concentration shortens over time, reflecting the nature of zero-order kinetics.
2
example
Zero-Order Half-life Example
Video duration:
1m
Play a video:
Zero-Order Half-life Example Video Summary
The reverse Haber reaction is characterized as a zero-order reaction, which is indicated by its rate constant units of molarity per second (M/s). To determine the half-life of this reaction, we can use the formula for the half-life of a zero-order reaction, which is given by:
In this equation, represents the half-life, is the initial concentration of the reactant (in this case, ammonia), and is the rate constant. Given that the initial concentration of ammonia is moles per liter and the rate constant is M/s, we can substitute these values into the equation.
Substituting the values, we have:
Calculating this gives:
Thus, the half-life for the reverse Haber reaction at the specified conditions is approximately seconds. This calculation illustrates the relationship between the initial concentration of a reactant and the rate constant in determining the time required for half of the reactant to be consumed in a zero-order reaction.
3
Problem
Decomposition of a certain substance Y at 45°C was found to be zero order. What is the half-life of substance Y if it took 15.5 minutes to decompose 67% of this substance? [Y]0 = 0.25 M.
A
0.058 min
B
0.029 min
C
12 min
D
23.5 min
4
concept
First-Order Half-Life
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First-Order Half-Life Video Summary
In the study of radioactive processes, it is essential to understand that they adhere to a first-order rate law. The half-life of a first-order reaction can be calculated using the equation:
t1/2 = \frac{\ln{2}}{k}
Here, t1/2 represents the half-life, k is the rate constant, and ln 2 is a constant value approximately equal to 0.693. This means that when you see 0.693 on a formula sheet, it is derived from the natural logarithm of 2. The units of the rate constant k are time-1, indicating that it is inversely related to time.
One of the key insights regarding the half-life equation is that it does not include the initial concentration of the reactant. This indicates that the half-life remains constant throughout the reaction, independent of the concentration of the reactant. As a result, the half-life will not change over time, making it a reliable measure for first-order reactions.
When graphing this relationship, the half-life can be plotted against time, where the y-axis represents the half-life and the x-axis represents time. This visualization reinforces the concept that the half-life remains consistent, as both ln 2 and k are constants in this context. Keeping these principles in mind will aid in understanding the behavior of first-order reactions and their half-lives.
5
example
First-Order Half-Life Example
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First-Order Half-Life Example Video Summary
In chemical kinetics, the half-life of a reaction is a crucial concept, particularly for first-order reactions. The half-life is defined as the time required for the concentration of a reactant to decrease to half of its initial value. For first-order reactions, the half-life can be calculated using the formula:
In this case, the rate constant is given as at a temperature of 35 degrees Celsius. To find the half-life, we substitute the value of into the formula:
Calculating this gives:
It is important to note that while the initial concentration of nitrogen dioxide is provided as , it does not affect the calculation of the half-life for a first-order reaction. The half-life is solely dependent on the rate constant . This illustrates that in some problems, not all provided information is necessary for solving the question at hand.
6
Problem
Radioactive plutonium-239 (t1/2 = 2.41 × 105 yr) is used in nuclear reactors and atomic bombs. If there are 5.70 × 102 g of plutonium isotope in a small atomic bomb, how long will it take for the substance to decay to 3.00 × 102 g?
A
2.23×105 yr
B
2.60×105 yr
C
517 yr
D
600 yr
7
Problem
Which of the following statements is False?
A
The average rate of a reaction decreases during a reaction.
B
The rate of zero order reactions are not dependent on concentrations.
C
The rate of a first order reaction is dependent on concentrations.
D
The half-life of a first order reaction is dependent on the initial concentration of reactant.
concept
Second-Order Half-Life
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Second-Order Half-Life Video Summary
In second order reactions, the relationship between the rate of reaction and the concentration of reactants is defined by a specific rate law. The half-life () for these reactions can be expressed with the formula:
Here, represents the rate constant, and is the initial concentration of the reactant. The units for the rate constant in second order reactions are typically expressed as molarity-1 time-1.
One key aspect of second order reactions is that the half-life is inversely related to the initial concentration. As the initial concentration decreases, the half-life increases. This occurs because the formula indicates that as becomes smaller, the value of becomes larger. Essentially, the half-life extends over time, meaning it takes progressively longer to reduce the concentration of the reactant by half.
This characteristic of second order reactions contrasts with first order reactions, where the half-life remains constant regardless of the concentration. This consistency in first order processes is why they are often preferred in radioactive decay scenarios. Understanding how the half-life varies with concentration in second order reactions is crucial for predicting the behavior of chemical systems over time.
example
Second-Order Half-Life Example
Video duration:
1m
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Second-Order Half-Life Example Video Summary
In a second-order reaction, the relationship between half-life and initial concentration is defined by the formula:
Where is the half-life, is the rate constant, and is the initial concentration of the reactant. To find the initial concentration, we can rearrange this equation:
Given that the half-life is 0.45 seconds, we need to determine the rate constant . In this case, is provided as . Plugging in the values, we can calculate the initial concentration:
Calculating this gives:
Considering significant figures, the initial concentration of the reactant is approximately:
This calculation illustrates the relationship between the half-life of a second-order reaction and the initial concentration, emphasizing the importance of the rate constant in determining the concentration of reactants at the start of the reaction.
10
Problem
Use the data below to determine the half-life of decomposition of NOCl reaction which follows 2nd order kinetics.
A
1.00×104 s
B
1.80×105 s
C
2.02×104 s
D
1.022×103 s
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The expression that relates h (the number of half-lives) and y (the number of years) is derived from the definition of a half-life, which is the time required for half of a sample of a radioactive substance to decay. The expression can be written as:
Here, t1/2 represents the half-life of the radioactive substance in years. This equation shows that the number of half-lives (h) is equal to the total time elapsed (y) divided by the half-life of the substance. If you know the half-life of a substance and the amount of time that has passed, you can use this expression to calculate how many half-lives have occurred during that period. This is particularly useful in fields like geology, archaeology, and physics, where understanding the decay of radioactive isotopes is important for dating materials and studying processes over time.
To find the number of half-lives that have elapsed, you need to know two key pieces of information: the total time that has passed and the half-life duration of the substance in question. The half-life is the time it takes for half of the original amount of a radioactive substance to decay.
Here's a step-by-step method to calculate the number of half-lives:
Determine the half-life of the substance. This information is usually provided in the problem or can be found in scientific literature.
Measure or find out the total time that has passed since the substance began to decay.
Divide the total time by the half-life of the substance to find the number of half-lives.
For example, if the half-life of a substance is 5 years and 20 years have passed, you would calculate the number of half-lives as follows:
So, four half-lives have
To calculate the number of half-lives that have passed for a sample of a radioactive substance, you'll use the formula:
Here's how you do it step by step:
Determine the total time that has elapsed since the substance began to decay.
Find out the half-life of the substance, which is the time it takes for half of the substance to decay. This information is typically provided in textbooks or reference materials.
Divide the total time elapsed by the half-life of the substance.
For example, if you have a substance with a half-life of 4 years and you want to know how many half-lives have passed in 16 years, you would calculate it as follows:
This means that in 16 years, the substance has gone through 4 half-lives.
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189904 | https://mathoverflow.net/questions/489209/binomial-probability-distribution-inequality | Skip to main content
Binomial probability distribution inequality
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I have a binomial probability question. I have run simulations up to 1,000,000 and it seems to hold true, but the inequality seems quite delicate.
Let k,l be positive integers s.t. k>l and define n=k+l
Let p be the success probability of a binomial distribution which solves the following equation:
p2(1−p)2=kPr(≤k successes from n draws)lPr(≥k successes from n draws)()
Goal: We wish to show that
Pr(≤k−1 successes from n−1 draws)≥p
To verbalize, one makes n draws from a binomial distribution, it could be that there are less than k successes, in which case the right term's numerator counts the number of missing successes. Likewise, it could be that there are more than k successes, in which case the right term's denominator counts the excess successes.
Motivation: We are considering a version of the coupon collector's problem. In the classic coupon collector problem, a child has a box of cereal each day, and there are two prizes. Prize 1 is in the box with probability p and Prize 2 is in the box with probability 1−p. The probability that each prize is in each box is i.i.d. In the classical problem, the collection is complete once one of each prize is obtained.
In our formulation, we consider the problem where the collector wishes to get k of the first prize and l of the second prize. Moreover, the distribution of prizes is ideal. Condition () identifies the p which minimizes the expected number of draws needed to complete the collection.
One of several questions that we are investigating is: if the collector could get one prize for free, before beginning her collection procedure, which prize would it be better for her to obtain? Our conjecture is that she (the collector) would be best off if she could obtain the more common prize.
Note: We have obtained this result, but I am leaving this up in case anyone is interested.
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edited Jul 16 at 18:05
MDR
asked Mar 10 at 23:30
MDRMDR
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I think it could help if you disclose how the conjectured inequality arose.
– Iosif Pinelis
Commented
Mar 11 at 13:29
Dear Iosif, Of course, I will add it to the question.
– MDR
Commented
Mar 11 at 17:38
What is meant by "third term" and "fourth term", please?
– Gerry Myerson
Commented
May 20 at 2:40
1
Dear Gerry, thanks for taking a look at the problem and sorry for the delay. You are right, I had formulated it as a different condition earlier and didn't fix the description. I've updated it now.
– MDR
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Jul 16 at 18:07
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Thanks for considering our inequalities. We have managed to solve it, and I thought that it could be helpful to post the answer for the community.
Let q=1−p. Notice that:
Pr(≤ k successes from n draws)1p=∑i=0∞(n+ik)pkqi+l
The formulation on the right assumes that you make a sequence of iid draws and keeps track of when you have encountered exactly k successes and at least l failures. The multiplier on the left is because, if any path which has ≤k successes from n draws will definitely end up in the right hand sum and will stay there for 1p periods on average.
Using this observation, condition () becomes
p2lq∑i=0∞(n+il)pk+iql=q2kp∑i=0∞(n+ik)pkqi+l
This reduces to
∑i=0∞(n+i−1l−1)(n+i)pk+iql−1=∑i=0∞(n+i−1k−1)(n+i)pk−1qi+l(1)
Assume the contrapositive of our goal which is:
p>Pr(≤k−1 successes from n−1 draws)
which can be rewritten as:
pPr(≤l−1 failures from n−1 draws)>Pr(≤k−1 successes from n−1 draws)q
Using the same reduction technique as before, it becomes:
∑i=0∞(n+i−1l−1)pk+iql−1>∑i=0∞(n+i−1k−1)pk−1qi+l
Next, notice that the inside terms satisfy the single-crossing property. They are equal when i=−1, then initially the left hand side increases more quickly than the right, then eventually the opposite happens. This means that for any truncation at t:
∑i=t∞(n+i−1l−1)pk+iql−1>∑i=t∞(n+i−1k−1)pk−1qi+l(2t)
This is a contradiction because (1) is an affine combination of (2t).
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answered Apr 7 at 15:01
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189905 | https://pmc.ncbi.nlm.nih.gov/articles/PMC5561167/ | The Conserved Sonic Hedgehog Limb Enhancer Consists of Discrete Functional Elements that Regulate Precise Spatial Expression - PMC
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. 2017 Aug 8;20(6):1396–1408. doi: 10.1016/j.celrep.2017.07.037
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The Conserved Sonic Hedgehog Limb Enhancer Consists of Discrete Functional Elements that Regulate Precise Spatial Expression
Laura A Lettice
Laura A Lettice
1 MRC Human Genetics Unit, MRC Institute of Genetics and Molecular Medicine, University of Edinburgh, Edinburgh EH4 2XU, UK
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1, Paul Devenney
Paul Devenney
1 MRC Human Genetics Unit, MRC Institute of Genetics and Molecular Medicine, University of Edinburgh, Edinburgh EH4 2XU, UK
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1, Carlo De Angelis
Carlo De Angelis
1 MRC Human Genetics Unit, MRC Institute of Genetics and Molecular Medicine, University of Edinburgh, Edinburgh EH4 2XU, UK
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1, Robert E Hill
Robert E Hill
1 MRC Human Genetics Unit, MRC Institute of Genetics and Molecular Medicine, University of Edinburgh, Edinburgh EH4 2XU, UK
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1 MRC Human Genetics Unit, MRC Institute of Genetics and Molecular Medicine, University of Edinburgh, Edinburgh EH4 2XU, UK
∗
Corresponding author bob.hill@igmm.ed.ac.uk
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Received 2016 Jun 26; Revised 2017 May 17; Accepted 2017 Jul 13; Collection date 2017 Aug 8.
© 2017 The Authors
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PMCID: PMC5561167 PMID: 28793263
Summary
Expression of sonic hedgehog (Shh) in the limb bud is regulated by an enhancer called the zone of polarizing activity regulatory sequence (ZRS), which, in evolution, belongs to an ancient group of highly conserved cis regulators found in all classes of vertebrates. Here, we examined the endogenous ZRS in mice, using genome editing to establish the relationship between enhancer composition and embryonic phenotype. We show that enhancer activity is a consolidation of distinct activity domains. Spatial restriction of Shh expression is mediated by a discrete repressor module, whereas levels of gene expression are controlled by large overlapping domains containing varying numbers of HOXD binding sites. The number of HOXD binding sites regulates expression levels incrementally. Substantial portions of conserved sequence are dispensable, indicating the presence of sequence redundancy. We propose a collective model for enhancer activity in which function is an integration of discrete expression activities and redundant components that drive robust expression.
Keywords:Shh expression, limb development, ZRS, enhancer, HoxD genes, Werner mesomelic syndrome, polydactyly, genome editing, phenotype
Graphical Abstract
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Highlights
•
The ancient vertebrate enhancer, the ZRS, shows sequence plasticity
•
Discrete regulatory activities are assigned to specific sites in the enhancer
•
The number of HOXD binding sites determines the level of Shh expression
•
Robust expression is a collective of regulatory and redundant information
Lettice et al. examine the composition of a highly conserved limb-specific enhancer, the ZRS, by dissecting the endogenous sequence using genome editing. Analysis of the resulting phenotype gives insights into the complex composition of the enhancer, which integrates discrete expression activities and redundant elements to drive accurate spatiotemporal gene expression.
Introduction
The basis of embryonic development lies in the spatiotemporal control of gene expression, which is mediated by remote cis-regulatory elements. These cis-acting elements, or enhancers, are fundamental to evolution and disease. Despite these important roles, major unanswered questions remain about the information encoded by the enhancer sequence and the importance of the overall structural architecture to enhancer activity. One class of enhancers that operates during embryogenesis are those that are highly conserved, acting at long distances from their target genes (Visel et al., 2009). Here we focus on a highly conserved element called the zone of polarizing activity regulatory sequence (ZRS) that is responsible for the spatiotemporal expression of Shh during limb bud development (Lettice et al., 2003, Sagai et al., 2005) and is essential for specifying digit identity and number. This enhancer is ∼770 bp in length and shows a high degree of similarity in vertebrates across a lengthy evolutionary timescale, including sharks and rays (Dahn et al., 2007), and, in accord, the mouse shows >70% similarity with the coelacanth (lobe-finned fish) sequence (Figure S1). Hence, the ZRS has remained highly invariant against a backdrop of major evolutionary changes to the anatomy of the appendicular skeleton, which includes the transition of fish fins to tetrapod limbs (Gehrke and Shubin, 2016). The structural organization of this class of deeply conserved vertebrate enhancers is under strong selective constraints, and, even in light of binding site redundancies exhibited by transcription factors, few sequence changes are present.
The ZRS is located 800–1,000 kb away from the Shh promoter in the mouse and human and is necessary and sufficient for accurately activating and maintaining Shh expression in the limb (Lettice et al., 2003, Sagai et al., 2005). An enhancer evolves not simply as a regulator that switches gene expression on or off but must also solve the challenges of regulating expression from a distance (Lettice et al., 2014) while controlling gene activity accurately in space and time and at the appropriate levels. Based on the evolutionary stasis of the ZRS, it is reasonable to expect that the sequence was finely honed during evolution so that that there is little tolerance for sequence change. Indeed, point mutations in and duplications of the ZRS result in a spectrum of appendicular skeletal defects (Anderson et al., 2012). Point mutations in well over 20 different positions scattered across the ZRS cause autosomal dominant limb defects, called “ZRS-associated syndromes” (Wieczorek et al., 2010). Some of the conditions associated with ZRS mutations include preaxial polydactyly type 2, triphalangeal thumb polysyndactyly, syndactyly type 4, and Werner mesomelic syndrome (WMS).
To investigate the structural composition of this highly conserved vertebrate enhancer, we used genome editing technology (Dow, 2015) to target deletions in three regions within the ZRS. Because ZRS activity is limb-specific, the phenotypes were expected to be overt, accessible, and nonlethal. The regions that were targeted contain the 5-bp site responsible for WMS (Anderson et al., 2012), the single mutation responsible for hemimelic extra toes (Hx) (Lettice et al., 2008) in the mouse, and a previously identified site for binding the HAND2 transcription factor (Osterwalder et al., 2014). This approach generated an overlapping series of mutations and deletions that scan across 250 bp of the endogenous ZRS. Here we show that the ZRS encodes multiple, diverse functions that contribute to the enhancer activity. Spatial restriction of expression is, in part, controlled by a small repressor domain that confines Shh expression to the posterior limb bud margin. In contrast, large overlapping domains regulate expression levels contingent on the number of HOXD binding sites. In addition, in response to insertion mutations, cryptic, unique phenotypes were generated that revealed the functional plasticity potentially encoded in an enhancer. Mutational analysis, however, also showed that, even though the enhancer is highly conserved, it could still tolerate quite substantial losses of sequence information without causing an abnormal phenotype. We propose a collective model for enhancer composition in which discrete activities and redundant sequences in the ZRS accrue to provide a robust regulatory response during development.
Results
ZRS Mutations in the Mouse Mimic WMS
WMS is associated with point mutations in a single, short, 5-bp stretch of the ZRS (green box, Figure 1A) that results in preaxial polydactyly of the hands and feet but is uniquely associated with short limb dwarfism because of tibial hypoplasia. WMS results from any heterozygous point mutation at position 404 (in humans) (Figure 1A, Lettice et al., 2008), a heterozygous A-to-G change two nucleotides downstream at position 406 (Norbnop et al., 2014), and a homozygous C-to-T change at position 402 (VanderMeer et al., 2014; green bases, Figure 1A). These three nucleotide positions lie within a highly conserved site and, to date, are the only sites associated with this syndrome.
Figure 1.
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Mutational Analysis of the WMS Site in the ZRS
(A) The position of the three sites within the ZRS that were targeted for mutation analysis; the WMS site is boxed. The conservation of the region containing the Werner mesomelic syndrome site is shown. The 3 nucleotides (green) mutated in WMS are shown in a green box and are labeled WMSΔ5, the Cu mutation is highlighted by the red box, and the position of the gRNA is contained in the black box (the PAM site is underlined and in italics). The position of the+A and+AA insertions is also shown. The wild-type and mutant allele sequences are shown at the top.
(B, G, and H) The wild-type hind limb (B) and expression patterns of Shh (G) and Ptc (H) at E11.5 hind limb buds are shown for comparison.
(C and D) The hindlimbs of the Cuban mutation (a G > A point change) shows an extra anterior digit in the heterozygote (C), a polydactylous hindlimb, and the hypoplastic tibia in the homozygote (D).
(I–L) Shh expression in the Cu homozygote at E11.5 (I) and E12.5 (J) and Ptc in the heterozygote at E11.5 (K) and E12.5 (L). Ectopic expression is highlighted by the black arrows.
(E, F, M, and N) The heterozygous WMSΔ5 deletion mutants. The hindlimb shows the absence of the tibia and polydactyly (E) and, unlike in the Cu mutation, polydactyly on the forelimb (F). Strong ectopic expression of Shh (M) and Ptc (N) is observed (highlighted by arrows).
Scale bars, 500 μm (B and C), 1 mm (D–F), and 100 μm (G–N).
Initially, to examine the nature of these mutations in the mouse and to ensure that it is possible to recreate the human abnormality, a G-to-A replacement in position 404 originally reported in a Cuban family (labeled Cu, Figure 1A; Lettice et al., 2003) was generated in mice using conventional “knockin” technology (Lettice et al., 2014). The resulting heterozygous mice exhibited extra preaxial digits on the hindlimbs (Figure 1C), whereas homozygotes, in addition, had bent legs because of tibial dysplasia (Figure 1D). Bone stains confirmed the loss of the terminal portion of each tibia (Figure 1D), which copied the dysplastic tibiae of WMS patients; however, unlike the patients, the forelimbs in mice were unaffected, and tibia dysplasia only occurred in the homozygous mutant.
To further investigate the nature of the dominant mutations at the WMS position, we targeted deletions using CRISPR/Cas9. A guide RNA (gRNA) targeted to this region (black box, Figure 1A) resulted in a number of different deletions and insertions. The most common mutation that was recovered was the precise removal of the five base pairs (called the WMSΔ5 deletion) (green box, Figure 1A) implicated as the site of WMS. The hindlimb phenotype of the WMSΔ5 mutant mice is similar to that observed in the homozygous Cu mutant mice in that the hindlimbs show extra preaxial digits and the tibia is hypoplastic, ranging from partial loss of the distal portion of the bone to its complete absence (Figure 1E). In contrast to the point mutation, these phenotypes occurred in both heterozygous and homozygous WMSΔ5 mice, and both genotypes exhibit preaxial polydactyly (PPD) in the forelimbs (Figure 1F). Thus, the strength of the WMSΔ5 allele is similar to the point mutation in humans. No differences in the severity of the phenotypes were observed between heterozygous and homozygous mice.
Analysis of Shh expression in the developing limb buds in the homozygous Cu mutant embryos showed normal expression at embryonic day 10.5 (E10.5), whereas, by E11.5, ectopic anterior expression of Shh (Figure 1L) was observed in approximately half of the embryos examined (3 of 7 mice). By E12.5, ectopic Shh occurred at the anterior margin (Figure 1J) in an outgrowth of limb tissue in all embryos examined. Heterozygous embryos showed normal Shh expression at all stages examined, but analysis of Ptc1, a sensitive readout of Shh signaling, showed ectopic, anterior expression at both E11.5 and E12.5 (Figures 1K and 1L), showing that low but sufficient levels of ectopic Shh were present in all mutant limb buds at these stages. Both heterozygous and homozygous WMSΔ5 mutant embryos showed appreciably more Shh and Ptc expression at the ectopic site of the hindlimb bud on E11.5 (Figures 1M and 1N) than what was detected in the homozygous Cu embryos, with some also showing ectopic expression in the forelimbs. Thus, the levels of ectopic Shh signaling detected reflected the final phenotype, with long-bone abnormalities arising in limb buds expressing higher levels of ectopic Shh earlier in development. The clustering of the human mutations within a short 5-bp sequence causing WMS indicates that this is a single important site for transcription factor binding, whereas the deletions confirm that WMS is due to the loss of binding of a repressor that actively represses ectopic expression.
Small Insertions Extend the Limb Phenotypic Spectrum
A second set of mutations arose adjacent to the 5-bp WMS site, resulting in the insertion of either one or two additional adenosines (called+A and+AA in Figure 2A). The mutant phenotype generated in the WMS+A mutant heterozygotes was a lengthening of the first digit and, sometimes, the addition of an extra terminal phalange on digit 1 of the hindlimbs (Figure 2B) with normal forelimbs. Shh expression appears normal in+A mutant limb buds at E11.5 (Figure 2C); however, the phenotype suggests a low level of expression at the ectopic, anterior margin of the limb bud. Insertion of+AA resulted in a more severe phenotype that has not been previously described for ZRS-associated mutations. This dinucleotide insertion caused typical PPD in the forelimbs (Figure 2D), but, in the hindlimbs, extra digits occurred centrally in the digital array (Figure 2E), occasionally in conjunction with long-bone anomalies (1 in 7 heterozygotes) (Figure 2F). The+AA hind limb buds showed an extended pattern of ectopic expression ranging from the posterior margin all around the distal edge of the limb bud (Figure 2G). The plasticity of a developmental enhancer in producing morphological changes has been investigated in Drosophila (Swanson et al., 2010) and in the mouse; it is clear that point mutations in the ZRS give rise to additional preaxial digits and homeotic transformations of the thumb to a finger (Anderson et al., 2012). The+AA mutant embryo presents an unusual skeletal configuration in the digital ray, indicating that further cryptic plasticity is uncovered by mutational events that disrupt the enhancer’s organization.
Figure 2.
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Insertion Mutations Disrupt Limb Development to Generate an Unusual Skeletal Phenotype
(A) The position of the adenosine insertions are shown adjacent to the WMS site to create the+A and the+AA mutations.
(B) The skeletal features of the+A hindlimb show a triphalangeal digit 1 (arrowhead).
(C) Expression of Shh on E11.5 in the hindlimb of the+A mutant.
(D and E) Forelimb of the+AA mutant (D) showing fusion and duplication of internal digits (asterisks), whereas the hindlimb (E) shows bifurcation at the tip of the extra preaxial digit (arrowheads) and the centrally located extra digit (arrow).
(F) Bending of the hindlimb caused by a shortening of the tibia is shown.
(G) E11.5 WMS+AA hindlimb bud, showing expression of Shh along the entire distal edge (black arrowheads).
Scale bars, 500 μm (B, D, and E), 100 μm (C and G), 1 mm (F).
Large Regions within the ZRS Are Dispensable
We next focused our mutation analysis on two highly conserved regions 3′ of the WMS site (Figure 3A; see Figure S1 for sequence comparison) to delve into the function of previously identified sites that are putatively important for Shh gene regulation. Corresponding gRNAs were designed that overlapped these sites (sequences shown as boxes in Figures 3B and 3J, with the protospacer adjacent motif [PAM] sites in italics). One region contains the conserved Ebox that binds the transcription factor HAND2 (Osterwalder et al., 2014; Figure 3A; blue nucleotides, Figure 3B) crucial to the spatial specific activation of Shh in the posterior margin of the limb bud. The second region contains the Hx mouse point mutation that lies at position 553 (Figure 3A; red nucleotide, Figure 3J), shown by transgenic analysis to operate as a dominant gain-of-function mutation and to encode important structural features crucial for enhancer activity (Lettice et al., 2014). In addition, the Hx site is embedded in a large region of the enhancer that is crucial for the long-range activity of the enhancer.
Figure 3.
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Mutational Analysis of the Ebox and the Hx Sites in the ZRS
(A) The ZRS (yellow rectangle) and the relative locations of the WMS site, the Ebox, and the Hx mutation are indicated. Boxes highlight the relative positions of the sequences shown in (B) (Ebox) and (J) (Hx), respectively. Linking these two regions, the position of the 3′Δ127 deletion is also shown. The gRNA sequences are boxed in the wild-type sequences in (B) and (J) with the PAM site (italics).
(B) The EBox (highlighted in blue font) and the deleted nucleotides for each mutation. The numbers of the homozygous animals analyzed are indicated below each mutation (as n= ).
(C and D) Representative forelimb (C) and hindlimb (D) from an EboxΔ17 homozygote demonstrate no detectable deviation from the wild-type.
(E–I) Shh expression in hindlimbs on E11.5 for the wild-type (E) and EboxΔ3 (F), EboxΔ17 (G), EboxΔ16 (H), and EboxΔ8 (I) homozygotes, showing a normal pattern of expression.
(J–L) The mutant sequence affected by the 3′ deletions near Hx (J). The wild-type sequence with the position of the Hx mutation (red base and box) is indicated. The position of Hoxsite 4 is highlighted in orange. The sequences of all the deletions are shown below, and the numbers of the homozygous animals analyzed are indicated below each mutation (as n= ). The apparent unaffected forelimb (K) and hindlimb (L) of the large 3′Δ127deletion are shown.
(M and N) The levels of expression of Shh at E11.5 hindlimb buds, shown by in situ hybridization (M) and by quantification by qRT-PCR (N) in 3′Δ127 homozygous embryos.
Scale bars, 500 μm (C, D, K, and L) and 100 μm (C–I and M). Error bars indicate ± SEM.
A series of overlapping deletions targeting the Ebox were identified (Figure 3B); two of these disrupted the Ebox: EboxΔ3, which was Ebox-specific, removing the 3 nucleotides from the middle, and EboxΔ17, which deleted the Ebox and surrounding nucleotides. Two other deletions, EboxΔ8 and EboxΔ16, removed nucleotides at the 3′ side of the Ebox. None of these four deletions showed a phenotype either as heterozygotes or homozygotes (the number of homozygotes analyzed is shown as n= in Figure 3B). For example, the largest deletion, EboxΔ17, which disrupts the Ebox and removes surrounding nucleotides, showed wild-type skeletal patterns in both the fore- and hindlimbs (Figures 3C and 3D, respectively). Shh expression in the limb buds for all four deletions showed the normal posterior pattern in homozygous embryos (Figures 3F–3I). The EboxΔ3 and EboxΔ17 deletions suggest that removal of a single Ebox site has no detectable effect on Shh expression. Possibly, a second conserved Ebox site downstream that has a lower affinity for HAND2 (Osterwalder et al., 2014) may compensate for this loss. The EboxΔ8, EboxΔ16, and EboxΔ17 mutations overlap in a conserved region (Figure 3B; Figure S1), deleting a total of 24 bp. No deletions in this region affected the limb phenotype, showing that a substantial region of conserved information can be disrupted.
Using gRNA targeted to the Hx mutation, we identified four deletions, 3′Δ42, 3′Δ11, 3′Δ12, and 3′Δ8, all of which are encompassed in 56 bp, including the Hx site (Figure 3J), and none of these had an effect on limb phenotype. Similar to the deletions created for the Ebox, these removed highly conserved nucleotide stretches: the 3′Δ8 and Δ12 deletion disrupting a HOXD binding site (Hoxsite 4; orange nucleotides, Figure 3J; Figure 5) (see below). 3′Δ11 and 3′Δ42 remove the Hx mutant site, and no polydactylous phenotype is detected in the heterozygotes, confirming that, unlike the WMS mutations, the Hx point change is a gain-of-function mutation (Lettice et al., 2014). The two other deletions, 3′Δ8 and 3′Δ12, do not contain the Hx mutant site but do remove the adjacent, highly conserved sequences containing the Hoxsite4, and these do not show a heterozygous phenotype. Homozygous mutants were made for all of these deletions, and no phenotype was detected (numbers are shown in Figure 3J). The larger, 127-bp deletion (3′Δ127) (Figure 3A; Figure S1) confirmed this tolerance for loss of conserved sequence. The large 3′Δ127 allele showed no dominant effect on digit number, and, in the homozygous state, there was no influence on the limb phenotype (n= 7) (Figures 3K and 3L), whereas both in situ hybridization and qRT-PCR showed no appreciable change in the expression profile or levels (Figures 3M and 3N). This deletion showed that a large region of conserved sequence can be deleted from this enhancer. Because the mutational analysis was performed at the endogenous locus, the lack of a phenotype suggests that the loss of the 3′Δ127 sequence is compensated for, thus indicating that there is encoded redundancy within the enhancer.
Figure 5.
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HoxD Binding to Conserved Motifs in the ZRS
The top line shows the genomic sequence around the 4 Hox binding sites (designated Hoxsites 1–4), with the position of the WMS region indicated. The relative positions of the WMSΔ20, WMSΔ48, and WMSΔ110 deletions are indicated. Below, the sequences of the 4 Hoxsites are shown in the same orientation. The consensus binding sites for each of the proteins (HOXD9, D10, and D11) are shown as position weight matrices under their gene names. For each triplet of EMSAs, the lanes are shown as binding to a labeled Hoxsite oligonucleotide with no competition, with excess of the wild-type oligo as competitor and with the mutated Hoxsite oligo in competition to show specificity of binding to the Hoxsite. Specific binding is indicated by the arrowheads. In the case of the E11.5 limb bud extract binding to Hoxsite 2, a higher mobility shift is observed, indicated by the asterisk. The non-specific band (arrow) is marked as a comparison with Figure S2.
Large Deletions Encompassing the WMS Site Incrementally Affect Expression Levels
Three other deletions were generated (Figure 4A; Figure S1) when making the WMS mutations: a 20-bp deletion, WMSΔ20, which included the 5-bp site of the WMSΔ5 and two other deletions, WMSΔ48 and WMSΔ110, both of which lost 21 bp on the 3′ side of the WMS site, removing the Ebox element but extending to different positions at the 5′ end. The WMSΔ20 deletion, unexpectedly, showed no observable limb phenotype, neither a dominant phenotype displaying extra toes nor, in the WMSΔ20/WMSΔ20 homozygote, a loss of activity phenotype displaying skeletal deficiencies (n= 5) (Figures 4B and 4C). The WMSΔ20 mouse was further crossed to the Shh-null mutation to make the WMSΔ20/Shh null compound heterozygote to expose any subtle loss of activity, but these, again, showed no abnormal phenotype (n= 5). Analysis of Shh expression in WMSΔ20 homozygotes showed little observable differences in expression pattern (Figure 4H) compared with the wild-type (Figure 4K), and levels of expression measured by qRT-PCR were not affected significantly (Figure 4L). Thus, the deleterious phenotypic effects of the WMSΔ5 mutations were lost in the larger WMSΔ20 deletion.
Figure 4.
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Deletions Near the WMS Site Reduce the Levels of Shh Expression
(A) The three deletions WMSΔ20, WMSΔ48, and WMSΔ110 are shown relative to the position of the WMS (green line), Ebox sites (blue line), and Hx mutation (red) within the ZRS (yellow rectangle).
(B, C, and H) The homozygous WMSΔ20 mutants. No limb abnormalities are detected in the forelimb (B) or the hind limb (C) of the homozygous WMSΔ48.
(D–G) WMSΔ48 limbs (D) and (E). The forelimb (D) shows loss of a digit and two terminal phalanges on the adjacent digit (arrowheads), and the hindlimb in (E)shows partial loss of digit 3 (arrow). Loss of the middle digit in the forelimb (F) and hind limb (G) are shown in the WMSΔ110 deletion.
(H–K) Shh expression on E11.5 in the hindlimbs of WMSΔ20 (H), WMSΔ48 (I), WMSΔ110 (J), and wild-type (K) embryos.
(L) The outcome of the quantification by qRT-PCR of Shh expression in E11.5 limb buds from wild-type and WMSΔ20, WMSΔ48, and WMSΔ110 homozygous embryos. WMSΔ48 and WMS110 expression levels are significantly (p< 0.001) lower than wild-type levels (Kruskal-Wallis test with Dunn’s multiple comparisons). Error bars indicate ± SEM.
Scale bars, 1 mm (B and C), 500 μm (D–G), and 100 μm (H–K).
Two deletions, WMSΔ48 and WMSΔ110 (Figure 4A), were examined, and, in the homozygote, removal of these sequences resulted in loss of digits. The WMSΔ48 mutation showed loss of up to one digit on each on the forepaws (Figure 4D), with some elements being retained and soft and hard tissue syndactyly and fusion being observed. The hindpaws were mildly affected, some showing only partial loss of a single digit (digit 3) (Figure 4E). The WMSΔ110 embryos showed a precise loss of one digit on all four paws (Figures 4F and 4G), with the rest of the digits apparently unaffected. Because these phenotypes were seen only in the homozygous state, these were loss-of-activity mutations resulting in a decrease in enhancer activity. Indeed, Shh expression was lower but was retained at the posterior margin of the limb bud at E11.5, but by in situ hybridization, levels in Δ48 (Figure 4I) appeared to be appreciably lower than in the wild-type (Figure 4K), with further reductions in WMSΔ110 (Figure 4J). Levels of RNA measured by qRT-PCR showed a significant reduction in Shh, in both WMSΔ110 and WMSΔ48, compared with the wild-type (Figure 4L).
Multiple Conserved HOX Binding Sites Control the Expression Levels of Shh
Within the ZRS, a highly conserved 6-bp element composed of the sequence CATAAA was detected at four positions (boxed in Figure 5 and Figure S1). This 6-bp sequence is embedded in sites that compare well with the consensus motif established for the 5′ Hoxd genes (the motif for HOXD9–11 is shown in Figure 5), and these were numbered Hoxsites 1–4. Genetic analysis of the HOX complexes previously demonstrated that the 5′ Hoxd genes (Hoxd 10–13) and that their counterparts in the Hoxa locus regulate Shh expression in the limb (Tarchini et al., 2006), and chromatin immunoprecipitation (ChIP) showed that at least two Hox proteins, HOXD10 and 13, directly bind to the ZRS (Capellini et al., 2006). Three of these identified sites (Hoxsites 1, 2, and 3) are contained within the Δ110 deletion, the Δ48 contained two sites (Hoxsites 2 and 3), and the Δ20 contained only Hoxsite 3(Figure 5; Figure S1). Hoxsite 4 was deleted in the series that included the Hx site (Figure 3J) discussed above. Each of the 5′ Hoxd genes was cloned into the vector pT7CFE1-CHis for subsequent expression in the human in vitro expression system (1-Step Human Coupled IVT Kit, Thermo Fisher Scientific). The 5′HOXD proteins were synthesized (Figure S2) and used in an electromobility shift assay (EMSA) to establish binding to double-stranded oligonucleotides containing one of these four sites (Table S1). The in vitro-synthesized HOXD9, 10, and 11 proteins showed the highest binding activity with these sites (Figure 5), whereas HOXD12 and 13 showed lower activity across all oligos (Figure S2B). The HOXD proteins showed different preferences for these sites; HOXD9 and 11 bound all four sites, with D9 showing a preference for sites 3 and 4, whereas D11 favored Hoxsites 1, 2, and 3. HOXD10 bound site 3 but bound weakly to Hoxsites 1, 2, and 4 (Figure 5). Specificity of binding was shown using competitor oligonucleotides with either wild-type sequence or Hox binding site mutations (Figure 5; Table S1). Nuclear extracts from E11.5 limb buds were also used in the EMSA (Figure 5) and were found to bind to all sites, and the binding was specific for the putative HOX binding motif.
The role played by the three HOX sites (Hoxsites 1–3) contained in the Δ110 deletion on ZRS activity was assayed in a series of transgenic embryos. Each site was mutated by replacing three bases in the CATAAA element (the mutations for each site are shown in Figure 6A) in a construct carrying the mutated, full-length ZRS driving the expression of a LacZ reporter gene. Mutations in each of these HOXD binding sites were made individually or in combination, and expression was examined on E11.5 in each injected transgenic embryo (the transient G 0 embryo). As a measure of the relative extent of expression in each transgenic embryo, the width of expression as a percentage of limb bud width was plotted to show the trends (individual limbs are represented by dots in Figure 6P). Mutations in individual Hoxsites had no observable effects on transgenic expression in the limb bud (compare Figure 6B with Figures 6C–6E; Figure 5P); however, mutations in the two sites (Hoxsites 2 and 3) that were contained in the Δ48 deletion or mutations in Hoxsites 1 and 3 showed detectably decreased expression (Figures 6F, 6G, and 6P). Mutation of all three sites (Hoxsites 1-–3) showed even further decreases (Figures 6H and 6P) comparable with the ZRS carrying the Δ110 deletion (Figures 6I and 6P). The accumulative decrease in expression of the endogenous Shh in the Δ48 and Δ110 deletions correlates with the progressive loss of the HOXD binding sites Hoxsites 1–3.
Figure 6.
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Transgenic Analysis of Embryos Carrying Mutant ZRS Sequences
(A) The sequences of the wild-type Hoxsites 1–3 and the mutated sequences (designated MutHoxsite) that were used in the transgenic constructs.
(B–E) Limb buds from transgenic embryos (E11.5) carrying the following ZRS sequences driving LacZ expression: the wild-type ZRS sequences (B); MutHoxsite 1(C), MutHoxsite 2 (D), and MutHoxsite 3 (E). (C)–(E) show no effect on expression of mutating single Hoxsites.
(F–H) Mutating combinations of sites results in lower LacZ expression: MutHoxsite2+3 (F), MutHoxsite 1+3 (G), and MutHoxsite 1,2+3 (H).
(I) The low level of expression in MutHoxsite 1,2+3 is reproduced in the WMSΔ110 construct.
(J–L) Addition of the Cu point mutation (J) or deletion of WMSΔ5 (K) results in distal and ectopic anterior expression, whereas deletion of WMSΔ20 (L) returns expression to wild-type levels.
(M–O) The Cu mutation in combination with mutant Hox sites: MutHoxsite 3+Cu (M), MutHoxsite 2+Cu (N), and MutHoxsite 2+3+Cu (O).
(P) Graphical representation of the LacZ expression patterns resulting from mutations within the ZRS. The width of the expression domain was divided by the width of the limb and expressed as a percentage. One spot represents the extent of reporter gene (LacZ) expression for each individual limb from a set of transient transgenic embryos. Data were subjected to one-way ANOVA and Tukey HSD test, and those that differ significantly from the wild-type are indicted (∗p ≤ 0.05, ∗∗∗p ≤ 0.001, ∗∗∗∗p ≤ 0.0001).
Scale bars, 100 μm.
Deletions of the WMS Domain Restore the Wild-Type Phenotype
The deletion in WMSΔ20 removes the WMS repressor site but also includes Hoxsite 3. Transgenics carrying either the Cu point mutation (Lettice et al., 2003; Figure 6J) or the WMSΔ5 deletion (Figure 6K) drives reporter gene expression to an elevated level in the posterior margin of the limb bud (Figure 6P) with appreciable ectopic expression. The WMSΔ20 deletion appears to return transgenic expression to wild-type levels (Figures 6L and 6P). Loss of the WMS repressor in combination with the mutant Hoxsite 3 binding site may be sufficient to nullify the increased and ectopic expression by the WMS mutations. To examine this possibility, transgenic mice carrying the Cu mutation in the presence of the 3-bp replacement (see above) that disrupts Hoxsite 3 was used in the transgenic assay and showed no detectable upregulation of the reporter on E11.5 and, importantly, no ectopic expression (Figures 6M and 6P). To show that the lack of ZRS upregulation was due to the independent action of the WMS mutations and loss of Hoxsite binding, transgenics carrying the Cu change and a different Hoxsite mutation (Hoxsite 2) also predominantly showed the wild-type pattern of expression (one of five G 0 embryos retained ectopic expression) (Figures 6N and 6P). Reductions in expression were shown in the presence of the WMS point mutation when two Hoxsites were mutated (Figures 6O and 6P). The transgenic expression reflects the WMSΔ20 deletion, suggesting that when the WMS mutation, which affects the binding of a repressor, is in the presence of mutations that disrupt binding of an activator, the mutations effectively cancel each other out, giving rise to wild-type expression levels. The independent action of these opposing activities emphasizes the combinatorial nature of elements that operate in the ZRS.
Discussion
The aim of this study was to investigate the composition of a vertebrate enhancer that falls into the highly conserved class of elements (Ovcharenko et al., 2004). These vertebrate enhancers represent a class in which the structural architecture is under selective constraints, resulting in apparent structural inflexibility in both the redundancy and the positioning of transcription factor binding motifs. These enhancers, which range in size from 100 bp to >1 kb, have the capacity to bind a substantial number of transcription factors, arguing that, within a single functional element, there is also a degree of structural complexity. This structural complexity has enabled the dissection of the ZRS into discrete regulatory activities. The expression pattern of the Shh gene in the limb bud is a consolidation of activities that control restriction of expression to the posterior margin, spatial and temporal expression, levels of expression, and long-range promoter activation (summarized in Figure 7A).
Figure 7.
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Schematic of the Composition of the ZRS and Its Role in the Shh and HoxD Regulatory Loop
(A) A representation of different functional regions and sites established for the ZRS. The ZRS is represented by the yellow rectangle, and the positions of the WMS 5bp site (green), the Ebox (blue), the Hx mutations (red), and Hoxsites (orange) are indicated in the ZRS. The positions of the 5 ETS sites that control the position of the expression boundary are represented by the red ovals. The region that contributes to regulating levels is in the blue box, and the deletions that revealed this activity are shown. The region that mediates long-range activity is shown in the black box. The large region of this domain that is redundant is shown by the gray shading. The two systems that control posterior restriction are shown below the ZRS rectangle, indicating the position of the two ETV binding sites and the position of the WMS 5-bp site.
(B) Summary of the positive feedback loop between the 5′ HoxD genes and Shh to reinforce expression of Shh. The ZRS (yellow box) and its position relative to Shh is shown on the left, whereas a schematic of the HoxD complex, including the two flanking regulatory domains (the early enhancer and the late enhancer), is depicted by green boxes on the right. Early-expressing 5′ HOXD proteins bind Hoxsites 1–3 within the ZRS to establish the levels of Shh expression in the initial stages of limb development (arrow 1). The levels of Shh expression are dependent on the number of Hox sites occupied. SHH, in turn, is crucial for the shift in HoxD gene expression to the later genes, in particular Hoxd13 (arrow 2). HOXD13 subsequently binds to sites at the 5′ end of the ZRS, as established by Leal and Cohn (2016) (arrow 3), to maintain Shh expression later in limb development.
Other examples of a complex arrangement of components have been reported, including an elegant analysis of the Drosophila spa enhancer that showed that structural organization underlies correct developmental gene expression (Swanson et al., 2010); for instance, sequence elements were defined that regulate long-range activity and others that repress expression in the wrong cell type. Our analysis surveying deletions further showed a complex organization that included unexpected redundancy incorporated into the enhancer. The model for enhancer action we propose here is one that relies on consolidation of discrete, discernible activities acting as a collective. This collective model suggests an integration of these discrete activities and redundant elements in delivering robust spatiotemporal developmental expression.
Hox Genes Function at the ZRS to Regulate Levels of Expression
We show the homotypic clustering of conserved HOXD binding sites (Hoxsites 1–4) in the ZRS. At least three of these sites (Hoxsites 1–3) are clustered in a 110-bp domain of the ZRS and regulate levels of Shh expression. The 5′ Hoxd genes, which include Hoxd 9–13, are fundamental to limb patterning and are expressed in a temporal collinear fashion, with the Hoxd9 gene expressing earliest in the limb bud, followed in sequence by Hoxd13 being expressed latest (Tarchini and Duboule, 2006). A clustering of highly conserved sites that contain the core motif for binding the 5′ HOXD proteins operates in an accumulative manner to regulate the activity levels of the ZRS enhancer. In an in vitro assay, we showed that the early 5′ HOXD proteins (HOXD9–11) bind this motif, suggesting that these play an initial role in establishing the activity levels of the ZRS. The region of the ZRS that contains three of these HOXD motifs is crucial for activity, and deletions show decreasing Shh expression corresponding to the number of Hox binding sites lost. In addition, loss of a Hox binding site counterbalances the increased and ectopic expression generated by the loss of the WMS repressor site. Thus, multiple HOXD factors coordinate, through binding at multiple sites, the expression levels of Shh.
Additional HOXD binding sites have been identified near the 5′end of the ZRS that have a preference for binding HOXD13 (Leal and Cohn, 2016; Figure 7B). Two sets of HOX sites, therefore, regulate gene expression, reacting to the temporal changes in the expression of the 5′ Hoxd genes. We suggest that the sites we identified play a role in establishing the levels of Shh expression in the initial stages of limb development by binding the early-expressing 5′ HOXD proteins (HOXD10 and 11) but adjust to the changing embryonic environment within the developing limb by also interacting with the later-expressed HOXD13 at different sites.
This establishes a regulatory loop that operates by positive feedback, reinforcing the expression of the Shh gene by the 5′Hoxd genes (Figure 7B). The early 5′ HOXD proteins interact with the ZRS at the HOX binding sites examined in this study to establish the levels of Shh expression based on the sum of the sites occupied (arrow 1, Figure 7B). SHH, in turn, is crucial for the shift in the regulation of HoxD gene expression from a set of early-acting enhancers to the enhancers at the 5′ end of the gene cluster (arrow 2, Figure 7B) that regulate the late-expressing genes, in particular Hoxd13 (Zákány et al., 2004). HOXD13 subsequently binds to sites at the 5′ end of the ZRS, as established by Leal and Cohn (2016) (arrow 3, Figure 7B). We suggest that a temporal response to HoxD genes is important for continued Shh expression as the regulatory environment in the limb bud changes over the 2 days when Shh is expressed in the mouse limb.
In accord, python and boa snakes, which have lost the HOXD13 binding sites in the ZRS, initially express Shh in the rudimentary limb buds, presumably dependent on the early 5′ HOXD protein binding sites we established; however, Shh expression is lost later, and limb development is prematurely terminated. This loss of the HOXD13 binding sites, in combination with loss of an ETS binding site (Leal and Cohn, 2016, Kvon et al., 2016), is responsible for the loss of limbs in these snakes.
Homotypic clustering of binding sites in the ZRS appears to play a number of roles in determining the spatial expression pattern of Shh expression in the embryonic limb bud. We previously showed multiple binding sites for the ETS factors, ETS1 and GABPα (Lettice et al., 2012). Multiple occupancy of these sites determines the extent of the boundary of Shh expression. Mutations in the human ZRS that generate an extra ETS site result in the extension of this expression boundary and ectopic expression in the limb bud, resulting in preaxial polydactyly (Lettice et al., 2012, Laurell et al., 2012). Here, occupancy of multiple HOXD binding sites regulates the levels of expression, and sequential loss of these sites results in a gradual decrease in expression levels. Homotypic clustering of binding sites in the ZRS, therefore, operates to incrementally adjust expression of the Shh gene and is therefore a fundamental mechanism for fine-tuning the regulatory activity of the enhancer.
ZRS Activity and Congenital Abnormalities
Mutations in the human ZRS cause skeletal abnormalities (Anderson et al., 2012). The point mutations act in a dominant fashion to cause digital abnormalities, and, presumably, most operate by switching restricted posterior expression to expression at both the posterior margin and an ectopic site at the anterior margin. One set of point mutations generates additional binding sites for ETS1/GABPα transcription factors, acting as dominant gain-of-activity mutations (Lettice et al., 2012). WMS, on the other hand, is highlighted by point mutations in three distinct positions in a single 5-bp site. The action of these point mutations, confirmed by the WMSΔ5 deletion data, is consistent with loss of binding of a repressor and, thus, an overall loss of functional activity. Hence, point mutations in the ZRS have two modes of action, operating as both gain and loss of activity, but both result in dominant genetic effects on the phenotype.
Insertions Reveal Cryptic Phenotypes
The WMS+AA insertional mutation reveals an unusual phenotype, showing the latent capacity for phenotypic innovation carried by this enhancer. The potential for appreciable morphological change shows that developmental enhancers may have the capacity for change without undergoing large sequence and structural changes in evolution. Selection against such substantial morphological changes may be one of the evolutionary constraints operating on the ZRS, but, in contrast, this also highlights the capacity for appreciable change in vertebrate evolution. These additions reveal the plasticity that is potentially hidden within an enhancer in controlling the phenotype and highlights mechanisms that may be available for phenotypic change during the evolution of an enhancer.
Evolution of the ZRS
The function of a cis regulator is encoded in its molecular architecture. Overlapping deletions in the ZRS that would predictably disrupt this architecture were made near and encompassing the proposed Ebox binding site and the WMS site that removed a total of 44 bp of highly conserved sequence, and these do not affect the limb phenotype. Moreover, the large 3′Δ127 mutation removes the conserved sequence from the 3′ half of the ZRS, which overlaps these 44 bp and also displays neither a limb phenotype nor a detectable reduction in expression. The ability to compensate for loss of sequence information suggests that there is encoded redundancy within the enhancer. This seemingly redundant activity may contribute to phenotypic robustness during development. Robustness is deemed important to buffer developmental processes from environmental and genetic perturbations and was proposed as canalization by Waddington (1942). For enhancers, such redundancy is widespread in Drosophila (Cannavò et al., 2016). Secondary or “shadow enhancers” in Drosophila provide redundant activity for the primary enhancer, and analyses of specific examples show that these can buffer a developmental process against environmental perturbations (Frankel et al., 2010, Perry et al., 2010). It is clear that the ZRS is able to tolerate losses of a highly conserved sequence without affecting the phenotype under ideal breeding conditions and a defined genetic background. In contrast with shadow enhancers, the robustness apparent in the ZRS is encoded within a single enhancer element because no compensatory activity is apparent in ZRS deletions. Hence, redundancy is an important characteristic of enhancers, whether this is encoded in secondary enhancers or contained within a single element, such as in the ZRS.
The evolutionary stability of the ZRS sequence raises a number of questions about the evolvability of this and perhaps other highly conserved enhancers. In addition, this stability occurs in light of the major morphological changes that have occurred to the limb during vertebrate evolution. Thus, the ZRS displays low sequence variability in a morphologically plastic developmental system. The recurrent role the ZRS plays in the diverse species analyzed so far is to ensure that Shh is expressed specifically along the posterior margin of the developing appendage, whether it is an embryonic fin (Dahn et al., 2007) or a limb bud. Many of the genes and signaling pathways known to regulate Shh in the mouse, such as the HoxD complex, Hand2, Gli3, and the fibroblast growth factor (FGF) pathway, are implicated in chick and fish, suggesting that the gene network responsible for Shh activation is also conserved (Gehrke and Shubin, 2016). For vertebrate enhancers, which are found in all vertebrate classes from cold-blooded fish to warm-blooded mammals, it is unlikely that the apparent robustness is a response to environmental factors because these insults would be different for each species. The genetic network of transcription factors and signaling pathways that converge at the ZRS is complex, and we suggest that the regulatory robustness observed for the ZRS buffers against variability and perturbations in this genetic network. This network, which converges at the ZRS, would therefore have evolved early in vertebrates, operating relatively unchanged in the appendicular skeleton in all classes of vertebrates. The conserved enhancer architecture is a response to this complex network and would be a constant factor that pervades species evolution during the morphological changes that have occurred during the fin-to-limb transition.
Experimental Procedures
Production and Analysis of CRISPR Mice
gRNAs were designed using the Optimized CRISPR design site ( and the exact guides were chosen on the basis of their precise location relative to the desired sites in the ZRS (The selected oligonucleotides are listed in Table S1). Oligos were cloned into the px330 vector (Addgene) (Cong et al., 2013), and DNA was prepared using the QIAGEN Plasmid Maxi kit according to the manufacturer’s protocol.
Transgenic mice were made by pronuclear injection of plasmid DNA at a concentration of 5 ng/μL. All resulting pups were screened phenotypically and had their ZRS sequence amplified by PCR and sequenced. All genotyping was performed by direct sequencing.
Skeletal preparations were stained simultaneously with alizarin red and Alcian blue (Nagy et al., 2009a, Nagy et al., 2009b). Whole-mount in situ hybridization was performed as described previously (Hecksher-Sørensen et al., 1998) using probes for Shh (Echelard et al., 1993) (a kind gift from Andy McMahon) and Ptc (Hayes et al., 1998) (a kind gift from Chris Hayes). qRT-PCR for Shh expression was performed on individual pairs of limb buds as described by Lettice et al. (2014). Expression was normalized within a litter to the wild-type level, and statistical significance was calculated by Prism using Kruskal-Wallis test with Dunn’s multiple comparisons. Mouse studies were approved by the University of Edinburgh animal welfare and ethical review board (AWERB) and carried out under the auspices of the United Kingdom Home Office.
EMSAs/In Vitro Translated Proteins
The coding regions of mouse Hoxd 9–13 were amplified by PCR using KOD polymerase (Merck Millipore). The primers used are listed in Table S1. Products were cloned into the expression vector pT7CFE1-CHis for subsequent expression in the human in vitro expression system (1-Step Human Coupled IVT Kit, Thermo Fisher Scientific) following the manufacturer’s instructions. Synthesis of each of the HOXD proteins was verified on a western blot using a rabbit anti-His tag antibody (2365, Cell Signaling Technology) (Figure S2) before the protein was used in an EMSA. The double-stranded oligonucleotides were biotin-labeled by the manufacturer (Sigma) and assayed to ensure that each was labeled to a similar specific activity. EMSAs were conducted as described previously (Lettice et al., 2012), and we used either 2 μL of a 1/25 dilution of protein from the in vitro translated (IVT) reaction or 4 μg of limb bud extract (prepared using the NE-PER Nuclear and Cytoplasmic Extraction Reagent Kit, Thermo Scientific). The specificity of binding was confirmed by competition with 100× excess of either unlabeled wild-type or mutant (mut) Hoxsite oligonucleotides. (Table S1).
Mutant ZRS Transgenic Constructs
Reporter gene transgenic analyses were made as described previously (Lettice et al., 2012). The mutant ZRS deletions (WMS Δ5, Δ20, and Δ110) used were generated by PCR using primers ZRSF and R (Table S1) from the appropriate mutant DNA. The Cu point mutation and MutHoxsite constructs were created using the primers listed in Table S1 and the QuikChange II Site-Directed Mutagenesis Kit (Agilent). For combinations of sites, multiple rounds of mutagenesis were conducted, and the correctly mutated ZRS was subsequently cloned into fresh lacZ-containing vector.
Author Contributions
Conceptualization, L.A.L.; Methodology, L.A.L.; Validation, L.A.L.; Investigation, L.A.L., P.D., and C.D.A.; Writing – Original Draft, R.E.H.; Writing – Review & Editing, L.A.L. and R.E.H.; Visualization, L.A.L.; Supervision, L.A.L. and R.E.H.; Funding Acquisition, R.E.H.
Acknowledgments
We would like to thank Richard Mort for help with the statistical analysis and MRC Central Services for providing DNA sequence support. We would also like to thank the staff at the Evans Building for expert technical assistance. Invaluable advice on the manuscript was given by David Fitzpatrick, Nick Hastie, and Wendy Bickmore. This work was supported by an MRC core program grant.
Published: August 8, 2017
Footnotes
Supplemental Information includes two figures and one table and can be found with this article online at
Supplemental Information
Document S1. Figures S1 and S2 and Table S1
mmc1.pdf (632.4KB, pdf)
Document S2. Article plus Supplemental Information
mmc2.pdf (5.8MB, pdf)
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189906 | https://www.reddit.com/r/learnmath/comments/1ahtq8g/what_is_the_proof_that_if_ab0_either_a_or_has_to/ | What is the Proof that if ab=0 either a or has to be 0? : r/learnmath
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Long bones
Long bones are hard, dense bones that provide strength, structure, and mobility. The thigh bone (femur) is a long bone. A long bone has a shaft and two ends.
Some bones in the fingers are classified as long bones, even though they are short in length. This is due to the structure and shape of the bones, not their size.
Long bones contain yellow bone marrow and red bone marrow, which produce blood cells.
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Linda J. Vorvick, MD, Clinical Professor, Department of Family Medicine, UW Medicine, School of Medicine, University of Washington, Seattle, WA. Also reviewed by David C. Dugdale, MD, Medical Director, Brenda Conaway, Editorial Director, and the A.D.A.M. Editorial team.
Silverstein JA, Moeller JL, Hutchinson MR. Common issues in orthopedics. In: Rakel RE, Rakel DP, eds. Textbook of Family Medicine. 9th ed. Philadelphia, PA: Elsevier; 2016:chap 30.
Standring S. Functional anatomy of the musculoskeletal system. In: Standring S, ed. Gray's Anatomy: The Anatomical Basis of Clinical Practice. 42nd ed. Philadelphia, PA: Elsevier; 2021:chap 5.
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Long bones
A long bone is a bone that has a shaft and 2 ends and is longer than it is wide. Long bones have a thick outside layer of compact bone and an inner medullary cavity containing bone marrow. The ends of a long bone contain spongy bone and an epiphyseal line. The epiphyseal line is a remnant of an area that contained hyaline cartilage that grew during childhood to lengthen the bone. All of the bones in the arms and legs, except the patella, and bones of the wrist, and ankle, are long bones.
Illustration
Long bones
A long bone is a bone that has a shaft and 2 ends and is longer than it is wide. Long bones have a thick outside layer of compact bone and an inner medullary cavity containing bone marrow. The ends of a long bone contain spongy bone and an epiphyseal line. The epiphyseal line is a remnant of an area that contained hyaline cartilage that grew during childhood to lengthen the bone. All of the bones in the arms and legs, except the patella, and bones of the wrist, and ankle, are long bones.
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189908 | https://www.expii.com/t/thermochemical-equation-overview-examples-11112 | Expii
Thermochemical Equation — Overview & Examples - Expii
A thermochemical equation is a balanced equation which includes the physical states of the reactants and products and the enthalpy change for the reaction.
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Thermochemistry and Thermodynamics
Thermochemical Equation — Overview & Examples
A thermochemical equation is a balanced equation which includes the physical states of the reactants and products and the enthalpy change for the reaction.
Thermochemical Equation — Overview & Examples
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Eric Sears
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As we've studied thermodynamics, we've seen a lot of different chemical reactions. Our goal was to calculate the enthalpy or internal energy. We had multiple calculation methods. For example, we've studied Hess's law and heats of formation. But why did we do all those calculations? We wanted to find the reaction enthalpy. It tells us the energy transferred during a chemical reaction. If we know it, we can write a thermochemical equation.
What is a thermochemical equation?
A thermochemical equation is a balanced chemical equation with its heat of reaction. They look like, A+B→CΔH=Xjoules. You already know how to write and balance chemical equations. So, all you have to do is add the ΔH or ΔU values.
So, why do we care about thermochemical equations? They give us important information. For example, consider the combustion of gasoline. We can compare the calorimetry value to an engine. That helps determine an engine's efficiency. All car companies are trying to improve fuel efficiency. They compare their cars' engines to the thermochemical data.
Solving Thermochemical Equation Problems
Most thermochemical equation questions are stoichiometry problems. You'll use the mole ratios, molar mass, and density to solve the questions. But, you'll need to consider the energy component.
A new type of problem is efficiency problems. For example, the question might say a process is only sixty percent efficient. So, you'd have to account for the forty percent loss. We'll look at some examples below. As a side note, efficiency problems are an application of the first law of thermodynamics. Energy may be lost from the system to the surroundings. But, the total energy is conserved.
Thermochemical Equation Example Problems
I drive a 2007 ford focus. Its tank holds 14.0 gallons. Let's do some calculations based on my car. The primary fuel in gasoline is octane. It's an organic molecule. Octane has eight carbons and eighteen hydrogens. So, its molecular formula is C8H18.
Car engines combust octane. The combustions do work by moving the engine's pistons. Through a series of gears, the piston's movement rotates the tires. Unfortunately, internal combustion engines are only about 35% efficient.
Octane's enthalpy of combustion is −10942.0KJ. Its density is 0.703gml.
Write the thermochemical equation.
Calculate the enthalpy if I use half a tank of gas.
Find the energy used and lost by the engine from consuming a full tank.
To write the thermochemical equation, balance the equation for octane combustion. Add the enthalpy of combustion to the end.
2C8H18(g)+25O2(g)→16CO2(g)+18H2O(g)ΔH=−10942.0KJ
To calculate the enthalpy for half a tank of gas, we'll need to use the density and the mole ratios. We'll use the density to find the grams of octane burned. Next, we can convert grams to moles. Finally, we can use the mole ratios to solve for the enthalpy. But first, we have to use dimensional analysis to convert gallons to SI units.
7.0gallons×3785.41mlgallon×0.703gml×1114.23gmol=163.1mols.
Now that we found the moles, we can use our thermochemical equation. We see that every two moles of octane release −10942.0KJ.
Our molar enthalpy is −10942KJ2mols=−5471KJmol.
163.1mols×−5471KJmol=−892401.7KJ
A full tank's energy is double the half tank. So, it releases 1784803.3KJ. We said an engine is only 35% efficient. So, the percentage of energy lost is 65%.
1784803.3KJ×0.35=−624681.2KJused
1784803.3KJ×0.65=−1160122.1KJlost
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Reaction at Constant Pressure vs. Constant Volume — Overview
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(Video) Thermochemical Equations Practice Problems
by Tyler DeWitt
In this video, Tyler DeWitt shows you four thermochemical equation example problems. First, you'll learn how to find the total heat released. In the other two examples, you have to do the opposite! You're given the heat and have to find the starting mass. Both types of problems could show up on an exam. So, this video will help you master thermochemical equations.
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189909 | https://www.youtube.com/watch?v=AS8PVnG1GIQ&pp=0gcJCfwAo7VqN5tD | 2 Tangent Lines to a Circle Which Intersect at an External Point
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in this lesson we're going to take a look at what happens when two tangent lines uh intersect at a common external point so for example here's our Circle it has Center o we'll make an external point we'll call this point p and we'll draw some tangent lines to the circle that intersect that P so start here we have a tangent line and again another point Point another line starting at p and tangent to the circle so we have two tangent lines drawn uh what's important about these is that these are both congruent to each other so let's say this uh Point here where it intersects the circle is a and here where it's tangent is B so we have AP is congruent to BP we could also draw the radius here and a radius from o to B as well so we know that a tangent line and a radius are perpendicular to each other so this is a right angle and this is also a right angle right now now we have a a quadrilateral two congruent sides and two congruent angles we were to draw another line from point P to the center of the circle we now have two triangles and these two triangles are actually congruent okay they both have these sides congruent they have this angle congruent and they have these sides congruent because they're both radiuses and radiuses of a of the same Circle are congruent and actually we also know that this side op is congruent to itself this side op is in both triangles see op is part of this triangle and op is also part of this triangle so because the same exact side is in both triangles we know that this side is congruent in both triangles it's the same exact side that's called the reflexive theorem so since we know that these two triangles are congruent we also know that this angle angle APO is congruent to angle BPO so we have angle Apo is congruent to angle BPO and of course this angle a o p is congruent to angle b o p so we have two tangent lines when we uh also draw a line from the external point to the center and draw the radiuses went over with these two congruent triangles which just happen to be right triangles because of this right angle so if we had to find any one of these sides in here uh we can actually use Pythagorean theorem we just have that this side is a this leg is B hypotenuse is C so we do a 2+ B s+ c^2 we can find the missing sides if we knew two of them so if we knew this tangent line and we knew the radius we could find this hypotenuse see |
189910 | https://www.ucd.ie/mathstat/t4media/Polynomial%20UCD.pdf | UCD Senior Maths Enrichment: Polynomials Tianyiwa Xie October 19th 2024 1 Basics What does a polynomial look like?
P(x) = anxn + an−1xn−1 + ... + a0 We specify that an ̸= 0. The degree of the polynomial is n.
It is also important to specify where does the coefficients live. If the coefficients are in Z, we say P(x) ∈Z[x]. Similarly, if coefficients are in Q or C, P(x) ∈Q[x] or C[x].
1.1 Identities Let’s look at some basic polynomial identities.
(x + y)n = n X i=0 n i xnyn−i xn −yn = (x −y)(xn−1 + xn−2y + xn−3y2 + ... + xyn−2 + yn−1) And when n is odd, we can also substitute y for −y 1 in the previous expression to get xn + yn = (x + y)(xn−1 −xn−2y + xn−3y2 −... −xyn−2 + yn−1) The second identity above immediately gives us this theorem (try proving it!): Theorem 1. Take P(x) ∈Z[x] and a, b ∈Z. Then (a −b) | P(a) −P(b) Corollary 1. Suppose p is a prime and a ≡b (mod p). Take P(x) ∈Z[x]. Then P(a) ≡P(b) (mod p).
2 Roots What happens to the graphs of the polynomial when the degree n is odd or even?
To guess the position of the root, we can make use of special values. For instance, suppose that P(x) ∈Z[x], P(a) < 0 and P(b) > 0, then there must be a zero between a and b. Be careful though, since this doesn’t tell us anything about the number of zeros in this interval.
Remember that if r is a root of P(x), then x −r is a factor of p, and we can write p = (x −r)p′ for some other polynomial p′.
Theorem 2 (Fundamental Theorem of Algebra). Every polynomial P(x) ∈C[x] with degree n has n complex roots, counting multiplicity. That is, is can be factorised into P(x) = c(x −r1)(x −r2) · · · (x −rn) for some c ∈C.
Corollary 2. Every polynomial of degree n has at most n distinct roots.
But be careful! This factorization is done in C. If we have a polynomial in Z, say, it doesn’t always have a root in Z, or even Q. Consider x2 + 1, for instance.
Indeed, x2 + 1 is irreducible in Z[x].
1Careful! This only works when n is odd 1 3 Degree Let’s talk a bit more about the degree of the polynomial. In P(x) = anxn + · · · + a0, the leading term anxn is the most important. It completely dominates the polynomial’s behaviour as |x| goes to infinity.
So when studying asymptotic behaviour, in complexity theory, for example, we often only care about the leading term.
3.1 Lagrange interpolation polynomial Another important question. We are given n data points. Can we find a polynomial that passes through all of them?
Theorem 3. Suppose we are given n + 1 points x1, x2, . . . , xn+1, such that no two lie on a vertical line.
Then there is exactly one polynomial of degree n that passes through them.
There is good reason to believe this theorem is true, and that is the idea of degrees of freedom.
Here, “degree of freedom” means how many free variables there are that governs the behaviour of an object. For example, for an unknown polynomial P(x) = anxn + · · · + a0, the degree of freedom is n + 1, because the behaviour of P(x) is completely determined by the n + 1 unknown coefficients.
In general, If the degree of freedom is n, you need n data points to determine the polynomial. This concept also shows up in linear equations: If you want to solve for one unknown x, you need one equation (e.g. 3x = 5).
If you want to solve for two unknowns, you need two equations to do that. Just knowing x + y = 1 wouldn’t tell you what are x and y. But once you have another equation, for example, 3x + 2y = 3, you can solve for x and y.
In general, if you want to solve for n unknowns, then you need n equations to do thata.
aTo be more precise you need at least n equations, to account for “useless” equations; it is also possible to have unsolvable equations.
Sketch proof. Suppose that xi = (αi, βi). Take a polynomial P(x) = anxn + an−1xn−1 + · · · + a1x + a0, we try to find the coefficients by substituting the values in: anαn 1 + an−1αn−1 1 + · · · + a1α1 + a0 = β1 anαn 2 + an−1αn−1 2 + · · · + a1α2 + a0 = β2 · · · anαn n+1 + an−1αn−1 n+1 + · · · + a1αn+1 + a0 = βn+1 This now becomes a linear system with a0, a1, ..., an as the unknown variables. We can now solve by systematically multiplying each line by certain factors and cancelling out terms.
We have n + 1 variables and n + 1 equations, so in general the equation can be solve uniquely2.
The polynomial that fits all the points are called Lagrange interpolation polynomial.
3.1.1 Polynomial fitting A very basic problem in Machine Learning is polynomial fitting. Given a set of data, how to predict what comes next using this data? One way is to model this set of data using polynomials.
1. There is a set of training data, we use this to train the model; Then there’s the test data, we use this to test how good is the model.
2. Fix the degree of the polynomial that you want to approximate with.
3. Get the polynomial that minimalises the root mean error 3.
4. Use test set to calculate test error, to see how good is the fitting.
2There are some special cases, but as long as the points doesn’t lie on vertical lines the system of equations can be solved.
3This is just a way to measure error, using the sum of the square of the error of the model at each point.
2 Minimizing the data set error is not always the best idea, because as we have seen, it is always possible4 to find a polynomial that fits all the data points perfectly.
To fit all the data, your polynomial has to become very squiggly, and that causes overfitting. It makes worse predictions, because we are learning too much from the noise of the data. As you can see in the following graph, when overfitting occurs, the training error goes to zero, but the test error becomes large5.
4 Irreducibility Definition 1. P(x) is irredicible in Z[x] if it cannot be written as P(x) = P1(x)P2(x), for non-trivial polynomials P1 and P2. Similarly for P(x) ∈Q.
As it happens, irreducibility in Z[x] is already very powerful: Theorem 4 (Gauss). If a polynomial with integer coefficients is reducible over Q, then it is reducible over Z.
Theorem 5 (Eisenstein). Let P(x) = anxn +an−1xn−1 +...+a0 be a polynomial with integer coefficients such that p | ai for 0 ≤i ≤n −1, p ∤an and p2 ∤a0. Then P(x) is irreducible over Z.
4as long as there are no two points on the same vertical line 5Image credit: Pattern Recognition and Machine Learning by Christopher M. Bishop.
3 Proof. Prove this by contradiction. Suppose not. Then we can write P(x) = g(x)h(x), where g(x) = bmxm + bm−1xm−1 + · · · + b1x1 + b0 h(x) = cn−mxn−m + cn−m−1xn−m−1 + · · · + c1x1 + c0 and m < n. This is to ensure the factorization is not trivial, and is important later on. Also, we may assume that m ≤(n −m).
Note: a0 = b0c0 a1 = b0c1 + b1c0 a2 = b0c2 + b1c1 + b2c0 · · · ak = bkc0 + bk−1c1 + bk−2c2 + ... + b1ck−1 + b0ck Then P(x) = g(x)h(x) gives us that p|a0 = b0c0. Without loss of generality, suppose that p|b0, then p ∤c0 because p2 ∤a0.
Now we can do this one coefficient at a time.
p|a0 = b0c0. Suppose that p|b0.
p|a1 = b0c1 + b1c0 = ⇒p|b1c0 = ⇒p|b1 p|a2 = b0c2 + b1c1 + b2c0 = ⇒p|b2c0 = ⇒p|b2 · · · p|am = bmc0 + bm−1c1 + · · · + b0cm = ⇒p|bmc0 = ⇒p|bm But then, p|bmcn−m = am, which is a contradiction, because p ∤am.
Exercise 1. Let p be a prime number. Show that P(x) = xp−1 + xp−2 + · · · + x + 1 is irreducible.
Proof. First, notice that P(x) = xp−1 x−1 . Let’s change x to y + 1. Then: P(y + 1) = (y + 1)p −1 y = yp + p 1 yp−1 + · · · + p 1 y y = yp−1 + p 1 yp−2 + p 2 yp−3 + · · · + p 1 Eisenstein criterion applies, so P(y + 1) is irreducible, so P(x) is.
5 Newton-Raphson In numerical analysis, the Newton–Raphson method, also known simply as Newton’s method, is a root-finding algorithm. The most basic version starts with a real-valued function f, its derivative f ′, and an initial guess x0 for a root of f. If f satisfies certain assumptions and the initial guess is close, then x1 = x0 −f(x0) f ′(x0) is a better approximation of the root than x0. Geometrically, this is the intersection on the x axis by the tangent at point x0, see graph 6.
6Image credit: wikipedia 4 The process is then repeated xn+1 = xn −f(xn) f ′(xn) until a close enough approximation is reached.
6 Homework 1. Prove that there is no polynomial P(x) with integral coefficients with the property P(7) = 5 and P(15) = 9.
2. Prove that there is no polynomial which has the property that P(k) = 2k for all positive integers k.
3. P(x) = an + an−1xn−1 + · · · + a0 is a polynomial with integer coefficients and ana0 ̸= 0. Prove that if r = p q (in lowest terms) is a rational root of P(x) then p is a divisor of a0 and q is a divisor of an.
4. Let P(z) and Q(z) be polynomials with complex coefficients of degree greater than or equal to 1 with the property that P(z) = 0 if and only if Q(z) = 0 and P(z) = 1 if and only if Q(z) = 1.
Prove that the polynomials are equal.
5. Find all polynomials P such that P(x)P(x + 2) = P(x2).
6. (IMO 1993) Let f(x) = xn + 5xn−1 + 3, where n > 1 is an integer. Prove that f(x) cannot be expressed as the product of two non-constant polynomials with integer coefficients.
7. Bonus: try to implement Newton-Raphson yourself using a computer program.7 7This is the basis of the Cambridge introductory computational project.
5 |
189911 | https://surferhelp.goldensoftware.com/vec_shd_img/IDD_Vector_Scaling.htm | Topic path:
Vector Map Scaling Properties
To edit a 1-grid vector map or 2-grid vector map, click once on the vector map to select it. The properties for the vector map are displayed in the Properties window. The Scaling page contains placement and sizing options for vector symbols. Each symbol is located at a grid node.
| |
| |
| Change vector scaling properties in the Properties window on the Scaling page. |
Symbol Origin
The symbol is located on the grid node with the Symbol origin options. To change the Symbol origin, click on the existing option and select a new option from the list. The symbol can be placed at the Tail, at the Center, or at the Head. At the Tail, places the end of the vector symbol on the grid node. At the Head , places the tip of the arrow symbol at the grid node. At the Center places the center of the vector at the grid node.
Scaling Method
There are three ways to scale vectors listed in the Scaling method option. You can scale the symbols between the minimum and maximum data values linearly, logarithmically, or by square root. Linear scaling provides a better visual representation of symbols that, for the most part, are scaled in one dimension (such as arrows with varying shaft length). When scaling the arrows in two dimensions (symbol width and shaft length), Square root or Logarithmic scaling displays the arrows more effectively. To change the Scaling method , click the current option. A list is displayed. Click on the new option and the map automatically updates.
Reverse Vector Orientation
The direction the arrowhead points can be reversed by checking the Reverse vector orientation check box. By default, vectors point in the "downhill" direction. Checking the Reverse vector orientation points the vectors in the "uphill" direction.
Vector Magnitude (data)
Set the Minimum and Maximum data values for the vectors in the Magnitude (data) section. Check the Use data limits box to set the vector minimum and maximum to the grid minimum and maximum values. Setting a new Minimum and Maximum is useful in displaying a series of maps in which you would like all of the vectors to be scaled the same even though the data minimum and maximum may differ.
To set the Minimum or Maximum value, highlight the existing value and type a new value. Values are in Z magnitude units.
Shaft Length
Set the range of the arrow shaft length in the Shaft Length section. The Shaft Length is the length of the arrow symbol from the tip of the tail to the tip of the arrow head. The Minimum value is the smallest shaft length displayed in the map at the Minimum value specified by the Magnitude (data) section. The Maximum value is the longest shaft length displayed in the map at the Maximum value specified by the Magnitude (data) section. Enter a new value into the boxes to change the length. Values are in page units. Values can be between 0 and 10 inches (0 and 25.4 centimeters).
Head Length
Set the range of the head length in the Head Length section. The Head Length is the length of the arrow head portion of the arrow symbol. The Minimum value is the smallest arrow head displayed in the map at the Minimum value specified by the Magnitude (data) section. The Maximum value is the longest arrow head displayed in the map at the Maximum value specified by the Magnitude (data) section. Enter a new value into the boxes to change the length. Values are in page units. Values can be between 0 and 10 inches (0 and 25.4 centimeters). If the Head Length values match the Shaft Length values, the arrow heads will be the entire length of the vector map symbol.
Symbol Width
Set the range of the symbol width in the Symbol Width section. The Symbol Width is the width of the arrow head portion of the arrow symbol at the widest point. The Minimum value is the smallest arrow head displayed in the map at the Minimum value specified by the Magnitude (data) section. The Maximum value is the longest arrow head displayed in the map at the Maximum value specified by the Magnitude (data) section. Enter a new value into the boxes to change the length. Values are in page units. Values can be between 0 and 5 inches (0 and 12.7 centimeters).
See Also
1-Grid Vector Map
2-Grid Vector Map
Data Page - 1-Grid Vector Map
Data Page - 2-Grid Vector Map
Symbol Page - Vector Map
Clipping Symbols on Vector Maps |
189912 | https://artofproblemsolving.com/wiki/index.php/Classical_Mechanics?srsltid=AfmBOopTw2ZU3zOzz6wiUlRpbwU__KNyTT5ecxkCu-c7F4V6y0PDud8Y | Art of Problem Solving
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Classical Mechanics
Overview
Mechanics is the study of movement. Kinematics, mechanical forces, work, power, energy, and matter are all part of mechanics.
Kinematics is the study of (relative) motion - displacement, velocity, acceleration etc. The two relations at the heart of kinematics are: and where is displacement at time , is velocity, is acceleration, and is time. Uniform rectilinear motion, projectile motion, uniform circular motion, and simple harmonic motion are some of the types of problems studied in kinematics.
The rules of physics are almost fully summarized by the three famous laws of motion formulated by Isaac Newton:
A body continues to be in its state of uniform rectilinear motion until it is disturbed by an external force. This property is known as inertia.
The rate of change of momentum of a body with respect to time is directly proportional to the force acting on it.
Every action has an equal and opposite reaction.
Mass is one of the two most basic intrinsic properties of a body. It is a measure of its inertia. Momentum is defined as the product of the mass and velocity of a body. Force is something that changes or tends to change the momentum of a body, or, informally, "a push or pull".
Mechanical work is defined by the relation where is work done, is force, is displacement, and subscripts and denote the initial and final states respectively. Similarly, mechanical power is defined as where is power delivered and is velocity. Energy is the other basic intrinsic property of a body. Mechanical energy is simply the capacity of a body to do mechanical work.
Among the various properties of matter are elasticity, surface tension, and viscosity. The most important one is gravity. Gravity is indeed considered one of the most mysterious things not only in physics but in science as a whole.
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189913 | https://www.physicsforums.com/threads/convert-cm-1-to-ev-solve-with-solution.71529/ | Convert cm^-1 to eV: Solve with Solution • Physics Forums
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Convert cm^-1 to eV: Solve with Solution
Thread starter Fowler_NottinghamUni
Start date Apr 14, 2005
TagsFactors
Apr 14, 2005
1
Fowler_NottinghamUni
7 0
Apparently In optical spectroscopy it is common to use the unit cm^-1 tro define the energy (or frequency of a photon that is emitted or absorbed in a transition between electronic states of an atom. This is derived from the reciprocal of the wavelength of the photon (wavelength in cm). Determine the conversion factor that converts cm^-1 to eV. I have no idea how to do this in the problem sheet that he gave us the solution to this particular problem means that we can use the result to complete the rest of the problem.
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New adaptive optics system promises sharper gravitational-wave observations
Physics-informed AI learns local rules behind flocking and collective motion behaviors
New perspectives on light-matter interaction: How virtual charges influence material responses
Apr 14, 2005
2
inha
576 1
E=hc/lambda. plug in the constants and see what's the energy of a 1cm wavelength photon. I don't think there's really anything more to it.
Apr 21, 2005
3
R0man
509 0
The conversion factor to convert cm^-1 to eV is 1 cm^-1 = 0.00012398 eV. This can be derived from the relationship between energy and wavelength, where energy is equal to Planck's constant (h) times the speed of light (c) divided by the wavelength. In this case, we can use the fact that the wavelength is in cm and the energy is in eV to determine the conversion factor.
First, we need to convert the wavelength from cm to meters, as the speed of light is typically given in meters per second. This can be done by multiplying the wavelength in cm by 0.01 (since there are 100 cm in 1 meter).
Next, we can rearrange the equation to solve for energy, which gives us energy = (hc)/wavelength. Plugging in the values for Planck's constant (6.626 x 10^-34 Js) and the speed of light (3.00 x 10^8 m/s), we get energy = (6.626 x 10^-34 Js 3.00 x 10^8 m/s)/wavelength.
Since we want the energy in eV, we can divide the energy by the conversion factor for Joules to eV, which is 6.24 x 10^18. This gives us energy = (6.626 x 10^-34 Js 3.00 x 10^8 m/s)/wavelength (1 eV/6.24 x 10^18 J). Simplifying this, we get energy = 1.06 x 10^-5/wavelength eV.
Finally, to convert from cm^-1 to eV, we need to take the reciprocal of the wavelength (since cm^-1 is the reciprocal of wavelength in cm). This gives us the conversion factor of 1 cm^-1 = 1/wavelength eV. Plugging in the value for wavelength (in meters), we get 1 cm^-1 = 1/(wavelength in cm 0.01) eV. Simplifying this, we get 1 cm^-1 = 0.00012398 eV.
Therefore, to convert from cm^-1 to eV, we simply need to multiply the value in cm^-1 by 0.00012398. For example, if we have a value of 500 cm
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Thread 'Schrödinger equation and classical wave equation'
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
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M
By Meden Agan
Wednesday, 6:58 AM
Replies: 39
Thread 'Is a table levitating possible, even at a minuscule probability?'
I am not sure if this falls under classical physics or quantum physics or somewhere else (so feel free to put it in the right section), but is there any micro state of the universe one can think of which if evolved under the current laws of nature, inevitably results in outcomes such as a table levitating? That example is just a random one I decided to choose but I'm really asking about any event that would seem like a "miracle" to the ordinary person (i.e. any event that doesn't seem to...
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S
By syed
Sep 16, 2025
Replies: 32
Thread 'Energytime uncertainty for particle decay'
If a particle decays into two gamma rays, what is the energy time uncertainty? ΔEΔt = h bar or ΔEΔt = h bar/2 Please explain.
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By jjson775
Sep 15, 2025
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189914 | https://www.quora.com/How-do-you-calculate-all-three-unknown-angles-given-two-sides-and-an-included-angle | Something went wrong. Wait a moment and try again.
Congruent Triangles
Law of Cosines
Classical Mathematics
Geometric Mathematics
5
How do you calculate all three unknown angles given two sides and an included angle?
Alexander Mathey
Former Chemical Engineer, retired, lives in Athens, GR
·
Author has 5.6K answers and 10.9M answer views
·
1y
·
First you find the third side by the ‘Law of Cosines’:
c = √(a² + b² - 2ab cosC)
Then, knowing all three sides and one angle, you find the sines of the other angles by the ‘Law of Sines’:
sinA / a = sinB / b = sinC / c
a, b, c are the lengths of the sides and A, B, C are the measures of the angles opposite to each side.
Last, you find the ‘arcsin’ corresponding to each angle’s sine.
353 views
·
Related questions
How can we find an unknown angle in a triangle given two known sides and one known angle? How can we find a missing side in a triangle given the other two sides and one angle?
How do I find angles and sides with only 1 angle given?
What is the most efficient way to solve for all three angles when given two sides and an included angle in a triangle?
How do you get all three sides and angles when given any two sides and an angle?
What is the method for finding an unknown length in trigonometry when given two sides and their included angle?
Richard Kleinschmidt
Math Teacher (1970–present)
·
Author has 113 answers and 110.5K answer views
·
2y
Related
What is the formula for finding unknown angles in triangles when two sides and an included angle are known?
It turns out that the easiest way to find the unknown angles is to find the unknown side first (using the Law of Cosines). For example, suppose the two sides are 3 and 9, and the angle between them is 20 degrees. (Put your calculator in degree mode for this example.)
Step 1 find unknown side:
x=sqrt(3^2 + 9^2 - 3 9 cos 20)
You'll get an ugly irrational number tha starts with 6.2655…. . Save that number in your calculator.
Step 2 find the angles:
The angle opposite 9 will be
A=InvCos((x^2 + 3^2 - 9^2)/(23x))
Note InvCos might be alt-cos or something like that on your calculator. I get an irratio
It turns out that the easiest way to find the unknown angles is to find the unknown side first (using the Law of Cosines). For example, suppose the two sides are 3 and 9, and the angle between them is 20 degrees. (Put your calculator in degree mode for this example.)
Step 1 find unknown side:
x=sqrt(3^2 + 9^2 - 3 9 cos 20)
You'll get an ugly irrational number tha starts with 6.2655…. . Save that number in your calculator.
Step 2 find the angles:
The angle opposite 9 will be
A=InvCos((x^2 + 3^2 - 9^2)/(23x))
Note InvCos might be alt-cos or something like that on your calculator. I get an irrational angle that starts with 150.57… .
At this point you could find the 3rd angle by subtracting from 180
180 - 20 - 157.57… = 9.425…
I prefer calculating the angle with the Law of Cosines again and then checking that the 3 angles sum to 180. So
B= InvCos((x^2 + 9^2 - 3^2)/(29x))
Note. You may have wondered why I didn't use the Law of Sines. You could but then you need to deal with the problem of obtuse angles. Using Law of Cosines is safer in this time when calculators are easily available. I'm old enough that I could do this by hand with a book of tables, but why?
Frank Jackson
BSc(Hons) in Mathematics, University of Bristol (Graduated 1967)
·
Author has 128 answers and 81.5K answer views
·
3y
Related
How do you find an unknown side in a triangle when two sides and one angle or two angles and one side are given?
There are a number of variants to this question. Let the sides be , , and the opposite internal angles be , , .
For two angles and one side, there is only one variant as given two angles the third angle is easily found because the sum of the three angles equals 180 degs. Hence by sine law if , say, is the given side we get
where is the circumradius of triangle ABC. Consequently, and are easily calculated.
For two sides and one angle, there are two variants. These are:
Given , , we can simply apply cosine rule to find b
There are a number of variants to this question. Let the sides be , , and the opposite internal angles be , , .
For two angles and one side, there is only one variant as given two angles the third angle is easily found because the sum of the three angles equals 180 degs. Hence by sine law if , say, is the given side we get
where is the circumradius of triangle ABC. Consequently, and are easily calculated.
For two sides and one angle, there are two variants. These are:
Given , , we can simply apply cosine rule to find because
.
Given , , we can apply sine rule to find angle B because
.
However here we have two solutions because we can calculate , but to determine the angle we can have or 180 degs.
We must next compare and .
If
we have two solutions for the angle that can form a triangle. Otherwise, if we wish to form a triangle and if
we have only one solution for angle B.
In both of these cases we have two angles and two sides so we can proceed as above for two angles and one side to find .
The drawings below demonstrate the conditions for two, one or no solutions. Also note that the circumradius of triangle ABC or A’BC is and
.
When
we have two solutions
When we have a unique solution
When
we cannot form a triangle, hence no solution.
Rohan Kumar
5y
Related
How do you find an angle when you have all three sides?
To solve an SSS triangle:
use The Law of Cosines first to calculate one of the angles
then use The Law of Cosines again to find another angle
and finally use angles of a triangle add to 180° to find the last angle.
We use the "angle" version of the Law of Cosines:
[x2 means x squared]
cos(C) = [a2 + b2 − c2]/2ab
cos(A) = [b2 + c2 − a2]/2bc
cos(B) = [c2 + a2 − b2]/2ca
(they are all the same formula, just different labels)
Example:-
In this triangle we know the three sides:
[opposite side of angle a,b,c]
a = 8,
b = 6 and
c = 7.
Use The Law of Cosines first to find one of the angles. It doesn't matter which one.
To solve an SSS triangle:
use The Law of Cosines first to calculate one of the angles
then use The Law of Cosines again to find another angle
and finally use angles of a triangle add to 180° to find the last angle.
We use the "angle" version of the Law of Cosines:
[x2 means x squared]
cos(C) = [a2 + b2 − c2]/2ab
cos(A) = [b2 + c2 − a2]/2bc
cos(B) = [c2 + a2 − b2]/2ca
(they are all the same formula, just different labels)
Example:-
In this triangle we know the three sides:
[opposite side of angle a,b,c]
a = 8,
b = 6 and
c = 7.
Use The Law of Cosines first to find one of the angles. It doesn't matter which one. Let's find angle A first:
cos A = (b2+ c2 − a2) / 2bc
cos A = (62+ 72− 82) / (2×6×7)
cos A=(36 + 49 − 64) / 84
cos A=0.25
A=cos−1(0.25)
A = 75.5224...°
A = 75.5° to one decimal place.
Next we will find another side. We use The Law of Cosines again, this time for angle B
cos B = (c
2+ a2− b2)/2ca
cos B = (72+ 82− 62)/(2×7×8)
cos B = (49 + 64 − 36) / 112
cos B = 0.6875
B = cos−1(0.6875)
B = 46.5674...°
B = 46.6° to one decimal place
Finally, we can find angle C by using 'angles of a triangle add to 180°':
C = 180° − 75.5224...° − 46.5674...°
C = 57.9° to one decimal place
Now we have completely solved the triangle i.e. we have found all its angles.
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Related questions
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What is the most efficient way to solve for all three angles when given two sides and an included angle in a triangle?
Find the missing side using the Law of Cosines:
c² = a² + b² - 2ab cosC
Then, knowing all three sides and one angle, apply the Law of Sines:
a/sinA = b/sinB = c/sinC
Supposing the known angle is C, you get:
sinA = (a/c)sinC → A = arcsinA
B = 180 - (A + C)
Alexander Mathey
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How do I find an angle when given 2 sides and 1 angle?
The given angle must be the one between the given sides, otherwise there exist two solutions:
If a, b are the known sides, c the unknown side and C the known angle, then apply the so called ‘Law of cosines’:
c² = a² + b² - 2ab cos(C)
Now, knowing all three sides, use again the low of cosines to solve for any of the angles A or B:
A = arccos [(b² + c² - a²)/2bc]
B = arccos [(a² + c² - b²)/2ac]
The given angle must be the one between the given sides, otherwise there exist two solutions:
If a, b are the known sides, c the unknown side and C the known angle, then apply the so called ‘Law of cosines’:
c² = a² + b² - 2ab cos(C)
Now, knowing all three sides, use again the low of cosines to solve for any of the angles A or B:
A = arccos [(b² + c² - a²)/2bc]
B = arccos [(a² + c² - b²)/2ac]
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Alexander Mathey
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How do you find an unknown angle from three sides of a triangle?
Use the so called ‘Law of cosines’, which states that the square of any side of any triangle equals the sum of the squares of the other two sides minus twice their product multiplied by the cosine of the angle between them:
a² = b² + c² - 2bc cos(α)
Solving for cos(a) you get:
cos(α) = (b² + c² - a²)/2bc
The unknown angle is the arc having that cosine:
α = arccos(α)
Use the so called ‘Law of cosines’, which states that the square of any side of any triangle equals the sum of the squares of the other two sides minus twice their product multiplied by the cosine of the angle between them:
a² = b² + c² - 2bc cos(α)
Solving for cos(a) you get:
cos(α) = (b² + c² - a²)/2bc
The unknown angle is the arc having that cosine:
α = arccos(α)
Umesh Chauhan
Former Studied at A P Shah Institute of Technology (2019–2022)
·
5y
Related
How do I find an angle with 3 sides?
we can only calculate angle , when all three sides are given .
let assume sides are : AB= 15cm, BC=12cm, CA=10cm.
By applying cosine rule (i.e c^2=a^2+b^2–2abcosC),
solution is right here,
This is the only one angle, I calculated, but you can similarly calculate all angle by this same procedure.
I think this is the answer you want , if yes and understood then good, but if not then you need serious practice.
GOOD LUCK AND ENJOY :))))
JAI HIND
we can only calculate angle , when all three sides are given .
let assume sides are : AB= 15cm, BC=12cm, CA=10cm.
By applying cosine rule (i.e c^2=a^2+b^2–2abcosC),
solution is right here,
This is the only one angle, I calculated, but you can similarly calculate all angle by this same procedure.
I think this is the answer you want , if yes and understood then good, but if not then you need serious practice.
GOOD LUCK AND ENJOY :))))
JAI HIND
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What is the formula for finding unknown angles in triangles when two sides and an included angle are known?
Try using the cosine law …. c^2 = a^2 + b^2 - 2 (a)(b) cos (included angle). (a & b are the sides)
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189915 | https://brainly.com/question/13428534 | [FREE] Three isotopes of argon occur in nature – ^{36}{18}\text{Ar}, ^{38}{18}\text{Ar}, - brainly.com
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Chemistry
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Three isotopes of argon occur in nature – 18 36Ar, 18 38Ar, 18 40Ar.
Calculate the average atomic mass of argon to two decimal places, given the following relative atomic masses and abundances of each of the isotopes:
Argon-36: 35.97 amu; 0.337%
Argon-38: 37.96 amu; 0.063%
Argon-40: 39.96 amu; 99.600%
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39.95amu
Explanation
Given parameters:
Relative atomic mass of Ar-36 = 35.97amu abundance = 0.337%
Relative atomic mass of Ar-38 = 37.96amu abundance = 0.063%
Relative atomic mass of Ar-40 = 39.96amu abundance = 99.600%
Unkown:
Relative atomic mass of Ar = ?
Solution:
The relative atomic mass is the average mass of all the isotopes of an element. To solve for the average atomic mass, we use the expression below:
RAM = m₃₆α₃₆ + m₃₈α₃₈ + m₄₀α₄₀
where m is the mass of the isotope
α is the abundance of the isotope
RAM of Argon = (100 0.337 x 35.97) + (100 0.063 x 37.96) +(100 99.6 x 39.96)
RAM of Argon = 39.95amu
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The average atomic mass of argon, calculated from its isotopes and abundances, is 39.98 amu. This was obtained by converting the isotopes' abundances into decimal form and applying the average atomic mass formula. The calculation included argon-36, argon-38, and argon-40, resulting in the average value of 39.98 amu.
Explanation
To calculate the average atomic mass of argon, we need to consider its isotopes and their respective abundances. We are given three isotopes of argon:
Argon-36 with a relative atomic mass of 35.97 amu and an abundance of 0.337%
Argon-38 with a relative atomic mass of 37.96 amu and an abundance of 0.063%
Argon-40 with a relative atomic mass of 39.96 amu and an abundance of 99.600%
Step-by-step calculation:
Convert the percentages into decimal form:
Argon-36: 0.337% = 0.00337
Argon-38: 0.063% = 0.00063
Argon-40: 99.600% = 0.99600
Use the formula for average atomic mass:
Average Atomic Mass=(m 36⋅a 36)+(m 38⋅a 38)+(m 40⋅a 40)
Where:
m 36,m 38,m 40 are the masses of the isotopes, and a 36,a 38,a 40 are their respective abundances in decimal form.
Substitute the values into the formula:
Average Atomic Mass=(35.97⋅0.00337)+(37.96⋅0.00063)+(39.96⋅0.99600)
=0.12119+0.02387+39.83736
=39.98242 amu
When rounded to two decimal places, the average atomic mass of argon is 39.98 amu.
Examples & Evidence
For instance, if we consider chlorine with isotopes 35Cl and 37Cl having abundances of 75.77% and 24.23% respectively, similar calculations yield an average atomic mass of approximately 35.48 amu, illustrating the concept of averaging atomic masses using abundance.
The calculation methods used adhere to standard practices for determining average atomic mass based on isotope abundances, as supported by chemistry textbooks and resources.
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The average atomic mass of Argon is calculated by adding the contributions from all its isotopes, which are based on their relative masses and abundances. In the case of Argon, the average atomic mass calculated is 39.96 amu.
Explanation
The average atomic mass of Argon is calculated by summing the products of the relative atomic masses of each isotope and their respective fractional abundances. Following are the calculations for Argon:
Argon-36 contributes (35.97 amu)(0.00337) = 0.12 amu
Argon-38 contributes (37.96 amu)(0.00063) = 0.02 amu
Argon-40 contributes (39.96 amu)(0.99600) = 39.82 amu
In conclusion, the average atomic mass of Argon is the sum of these three contributions:
0.12 amu + 0.02 amu + 39.82 amu = 39.96 amu (to two decimal places)
Learn more about Average Atomic Mass here:
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### Free Chemistry solutions and answers
Community Answer 7 Calculate the average atomic mass of a sample of a mixture of argon (Ar). The mixture is 90% argon-36 and 10% argon-38. Argon-36 has an atomic mass of 35.968 amu. Argon-38 has an atomic mass of 37.962 amu
Community Answer 5.0 1 (SK015) Naturally occurring argon consists of thre isotopes , argon-36, argon-38, argon-40 in the ratio of 4.86: 1 : 1423 i) Calculate the percentage composition of the three isotopes. ii) Calculate the relative atomic mass of argon to four significant figures.
Community Answer Argon has three naturally-occurring isotopes: 99.6% of 40Ar, with an atomic weight of 39.9624 amu, 0.337% of 36Ar, with an atomic weight of 35.9676 amu, and 0.063% of 38Ar, with an atomic weight of 37.9627 amu. Calculate the average atomic weight of Ar. Round off the answer to six significant figures. Do not include the units.
Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain?
Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized
Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. Massing out120 g of sodium chloride (acbnela)____5. Suspending glassware over the Bunsen burner (rwei zeagu)____6. Used to pour liquids into containers with small openings or to hold filter paper (unfenl)____7. Mixing a small amount of chemicals together (lewl letpa)____8. Heating contents in a test tube (estt ubet smalcp)____9. Holding many test tubes filled with chemicals (estt ubet karc) ____10. Used to clean the inside of test tubes or graduated cylinders (iwer srbuh)____11. Keeping liquid contents in a beaker from splattering (tahcw sgasl)____12. A narrow-mouthed container used to transport, heat or store substances, often used when a stopper is required (ymerereel kslaf)____13. Heating contents in the lab (nuesnb bneurr)____14. Transport a hot beaker (gntos)____15. Protects the eyes from flying objects or chemical splashes(ggloges)____16. Used to grind chemicals to powder (tmraor nda stlepe) __
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Community Answer 8 If the temperature of a reversible reaction in dynamic equilibrium increases, how will the equilibrium change? A. It will shift towards the products. B. It will shift towards the endothermic reaction. C. It will not change. D. It will shift towards the exothermic reaction.
Community Answer 4.8 52 Which statements are TRUE about energy and matter in stars? Select the three correct answers. Al energy is converted into matter in stars Only matter is conserved within stars. Only energy is conserved within stars. Some matter is converted into energy within stars. Energy and matter are both conserved in stars Energy in stars causes the fusion of light elements
Community Answer 4.5 153 The pH of a solution is 2.0. Which statement is correct? Useful formulas include StartBracket upper H subscript 3 upper O superscript plus EndBracket equals 10 superscript negative p H., StartBracket upper O upper H superscript minus EndBracket equals 10 superscript negative p O H., p H plus P O H equals 14., and StartBracket upper H subscript 3 upper O superscript plus EndBracket StartBracket upper O upper H superscript minus EndBracket equals 10 to the negative 14 power.
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Contents
Chapter 1: Algebra Review
Chapter 1.1: Exponents and Scientific Notation
Chapter 1.2: Radicals and Rational Exponents
Chapter 1.3: Dividing Polynomials
Chapter 1.4: Solve Equations with Fraction or Decimal Coefficients
Chapter 1.5: Use a General Strategy to Solve Linear Equations
Chapter 1.6: Solve a Formula for a Specific Variable
Chapter 1.7: Use a Problem-Solving Strategy
Chapter 2: Systems of Equations
Chapter 2.1: Slopes and Their Graphs
Chapter 2.2: Graphing Linear Equations
Chapter 2.3: Perpendicular and Parallel Lines
Chapter 2.4: Introduction to Systems of Equations
Chapter 2.5: Systems of Linear Equations: Two Variables
Chapter 2.6: Systems of Linear Equations: Three Variables
Chapter 2.7: Solving Systems with Cramer's Rule
Chapter 3: Right Angle Trigonometry
Chapter 3.1: Introduction to The Unit Circle: Sine and Cosine Functions
Chapter 3.2: Angles
Chapter 3.3: Right Triangle Trigonometry
Chapter 3.4: Unit Circle
Chapter 3.5: The Other Trigonometric Functions
Chapter 4: Further Applications of Trigonometry
Chapter 4.1: Introduction to Further Applications of Trigonometry
Chapter 4.2: Non-right Triangles: Law of Sines
Chapter 4.3: Non-right Triangles: Law of Cosines
Chapter 4.4: Polar Coordinates
Chapter 4.5: Polar Coordinates: Graphs
Chapter 4.6: Polar Form of Complex Numbers
Chapter 4.7: Parametric Equations
Chapter 4.8: Parametric Equations: Graphs
Chapter 4.9: Vectors
Chapter 4.10: Vector Addition and Subtraction: Analytical Methods
Chapter 5: Factoring and Quadratics
Chapter 5.1: Greatest Common Factor
Chapter 5.2: Factoring by Grouping
Chapter 5.3: Factoring Trinomials where a = 1
Chapter 5.4: Factoring Trinomials where a ≠ 1
Chapter 5.5: Factoring Special Products
Chapter 5.6: Factoring Quadratics of Increasing Difficulty
Chapter 5.7: Choosing the Correct Factoring Strategy
Chapter 5.8: Solving Quadriatic Equations by Factoring
Chapter 5.9: Age Word Problems
Chapter 5.10: Completing the Square
Chapter 5.11: The Quadratic Equation
Chapter 5.12: Solving Quadratic Equations Using Substitution
Chapter 5.13: Graphing Quadratic Equations—Vertex and Intercept Method
Chapter 6: Rational Expressions and Equations
Chapter 6.1: Reducing Rational Expressions
Chapter 6.2: Multiplication and Division of Rational Expressions
Chapter 6.3: Least Common Denominators
Chapter 6.4: Addition and Subtraction of Rational Expressions
Chapter 6.5: Reducing Complex Fractions
Chapter 6.6: Solving Complex Fractions
Chapter 6.7: Solving Rational Equations
Chapter 6.8: Rate Word Problems: Speed, Distance and Time
Chapter 6.9: Rational Functions
Chapter 7: Periodic Functions
Chapter 7.1: Introduction to Periodic Functions
Chapter 7.2: Graphs of the Sine and Cosine Functions
Chapter 7.3: Graphs of the Other Trigonometric Functions
Chapter 7.4: Inverse Trigonometric Functions
Chapter 8: Radical Expressions and Equations
Chapter 8.1: Reducing Square Roots
Chapter 8.2: Reducing Higher Power Roots
Chapter 8.3: Adding and Subtracting Radicals
Chapter 8.4: Multiplication and Division of Radicals
Chapter 8.5: Rationalizing Denominators
Chapter 8.6: Radicals and Rational Exponents
Chapter 8.7: Rational Exponents (Increased Difficulty)
Chapter 8.8: Radicals of Mixed Index
Chapter 8.9 Solving Radical Equations
Chapter 9: Introduction to Exponential and Logarithmic Functions
Chapter 9.1: Introduction to Exponential and Logarithmic Functions
Chapter 9.2: Exponential Functions
Chapter 9.3: Graphs of Exponential Functions
Chapter 9.4: Logarithmic Functions
Chapter 9.5: Graphs of Logarithmic Functions
Chapter 9.6: Logarithmic Properties
Chapter 9.7: Exponential and Logarithmic Equations
Chapter 9.8: Exponential and Logarithmic Models
Chapter 9.9: Fitting Exponential Models to Data
Chapter 10: Complex Numbers
Chapter 10.1: Complex Numbers
Chapter 10.2: Polar Form of Complex Numbers
Reference section
Reference Section
Pre-Calculus
Chapter 10.1: Complex Numbers
Learning Objectives
In this section you will:
Add and subtract complex numbers.
Multiply and divide complex numbers.
Simplify powers of .
Figure 1.
Discovered by Benoit Mandelbrot around 1980, the Mandelbrot Set is one of the most recognizable fractal images. The image is built on the theory of self-similarity and the operation of iteration. Zooming in on a fractal image brings many surprises, particularly in the high level of repetition of detail that appears as magnification increases. The equation that generates this image turns out to be rather simple.
In order to better understand it, we need to become familiar with a new set of numbers. Keep in mind that the study of mathematics continuously builds upon itself. Negative integers, for example, fill a void left by the set of positive integers. The set of rational numbers, in turn, fills a void left by the set of integers. The set of real numbers fills a void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. In this section, we will explore a set of numbers that fills voids in the set of real numbers and find out how to work within it.
Expressing Square Roots of Negative Numbers as Multiples of
We know how to find the square root of any positive real number. In a similar way, we can find the square root of any negative number. The difference is that the root is not real. If the value in the radicand is negative, the root is said to be an imaginary number.The imaginary number is defined as the square root of
So, using properties of radicals,
We can write the square root of any negative number as a multiple of Consider the square root of
We use and not because the principal root of is the positive root.
A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written where is the real part and is the imaginary part. For example, is a complex number. So, too, is
Imaginary numbers differ from real numbers in that a squared imaginary number produces a negative real number. Recall that when a positive real number is squared, the result is a positive real number and when a negative real number is squared, the result is also a positive real number. Complex numbers consist of real and imaginary numbers.
Imaginary and Complex Numbers
A complex number is a number of the form where
is the real part of the complex number.
is the imaginary part of the complex number.
If then is a real number. If and is not equal to 0, the complex number is called a pure imaginary number. An imaginary number is an even root of a negative number.
How To
Given an imaginary number, express it in the standard form of a complex number.
Write as
Express as
Write in simplest form.
Expressing an Imaginary Number in Standard Form
Express in standard form.
Show Solution
In standard form, this is
Try It
Express in standard form.
Show Solution
Plotting a Complex Number on the Complex Plane
We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number, we need to address the two components of the number. We use the complex plane, which is a coordinate system in which the horizontal axis represents the real component and the vertical axis represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs where represents the coordinate for the horizontal axis and represents the coordinate for the vertical axis.
Let’s consider the number The real part of the complex number is and the imaginary part is 3. We plot the ordered pair to represent the complex number as shown in (Figure).
Figure 2.
Complex Plane
In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis, as shown in (Figure).
Figure 3.
How To
Given a complex number, represent its components on the complex plane.
Determine the real part and the imaginary part of the complex number.
Move along the horizontal axis to show the real part of the number.
Move parallel to the vertical axis to show the imaginary part of the number.
Plot the point.
Plotting a Complex Number on the Complex Plane
Plot the complex number on the complex plane.
Show Solution
The real part of the complex number is and the imaginary part is –4. We plot the ordered pair as shown in (Figure).
Figure 4.
Try It
Plot the complex number on the complex plane.
Show Solution
Adding and Subtracting Complex Numbers
Just as with real numbers, we can perform arithmetic operations on complex numbers. To add or subtract complex numbers, we combine the real parts and then combine the imaginary parts.
Complex Numbers: Addition and Subtraction
Adding complex numbers:
Subtracting complex numbers:
How To
Given two complex numbers, find the sum or difference.
Identify the real and imaginary parts of each number.
Add or subtract the real parts.
Add or subtract the imaginary parts.
Adding and Subtracting Complex Numbers
Add or subtract as indicated.
Show Solution
We add the real parts and add the imaginary parts.
Try It
Subtract from
Show Solution
Multiplying Complex Numbers
Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately.
Multiplying a Complex Number by a Real Number
Lets begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. Consider, for example, :
How To
Given a complex number and a real number, multiply to find the product.
Use the distributive property.
Simplify.
Multiplying a Complex Number by a Real Number
Find the product
Show Solution
Distribute the 4.
Try It
Find the product:
Show Solution
Multiplying Complex Numbers Together
Now, let’s multiply two complex numbers. We can use either the distributive property or more specifically the FOIL method because we are dealing with binomials. Recall that FOIL is an acronym for multiplying First, Inner, Outer, and Last terms together. The difference with complex numbers is that when we get a squared term, it equals
How To
Given two complex numbers, multiply to find the product.
Use the distributive property or the FOIL method.
Remember that
Group together the real terms and the imaginary terms
Multiplying a Complex Number by a Complex Number
Multiply:
Show Solution
Try It
Multiply:
Show Solution
Dividing Complex Numbers
Dividing two complex numbers is more complicated than adding, subtracting, or multiplying because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator to write the answer in standard form We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of is For example, the product of and is
The result is a real number.
Note that complex conjugates have an opposite relationship: The complex conjugate of is and the complex conjugate of is Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another.
Suppose we want to divide by where neither nor equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply.
Multiply the numerator and denominator by the complex conjugate of the denominator.
Apply the distributive property.
Simplify, remembering that
The Complex Conjugate
The complex conjugate of a complex number is It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged.
When a complex number is multiplied by its complex conjugate, the result is a real number.
When a complex number is added to its complex conjugate, the result is a real number.
Finding Complex Conjugates
Find the complex conjugate of each number.
Show Solution
1. The number is already in the form The complex conjugate is or
2. We can rewrite this number in the form as The complex conjugate is or This can be written simply as
Analysis
Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by
Try It
Find the complex conjugate of
Show Solution
How To
Given two complex numbers, divide one by the other.
Write the division problem as a fraction.
Determine the complex conjugate of the denominator.
Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.
Simplify.
Dividing Complex Numbers
Divide: by
Show Solution
We begin by writing the problem as a fraction.
Then we multiply the numerator and denominator by the complex conjugate of the denominator.
To multiply two complex numbers, we expand the product as we would with polynomials (using FOIL).
Note that this expresses the quotient in standard form.
Simplifying Powers of i
The powers of are cyclic. Let’s look at what happens when we raise to increasing powers.
We can see that when we get to the fifth power of it is equal to the first power. As we continue to multiply by increasing powers, we will see a cycle of four. Let’s examine the next four powers of
The cycle is repeated continuously: every four powers.
Simplifying Powers of
Evaluate:
Show Solution
Since we can simplify the problem by factoring out as many factors of as possible. To do so, first determine how many times 4 goes into 35:
Try It
Evaluate:
Show Solution
Can we write in other helpful ways?
As we saw in (Figure), we reduced to by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of may be more useful. (Figure) shows some other possible factorizations.
Factorization of
Reduced form
Simplified form
Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method.
Access these online resources for additional instruction and practice with complex numbers.
Adding and Subtracting Complex Numbers
Multiply Complex Numbers
Multiplying Complex Conjugates
Raising i to Powers
Key Concepts
The square root of any negative number can be written as a multiple of See (Figure).
To plot a complex number, we use two number lines, crossed to form the complex plane. The horizontal axis is the real axis, and the vertical axis is the imaginary axis. See (Figure).
Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts. See (Figure).
Complex numbers can be multiplied and divided.
To multiply complex numbers, distribute just as with polynomials. See (Figure) and (Figure).
To divide complex numbers, multiply both numerator and denominator by the complex conjugate of the denominator to eliminate the complex number from the denominator. See (Figure) and (Figure).
The powers of are cyclic, repeating every fourth one. See (Figure).
Section Exercises
Verbal
Explain how to add complex numbers.
Show Solution
Add the real parts together and the imaginary parts together.
What is the basic principle in multiplication of complex numbers?
Give an example to show that the product of two imaginary numbers is not always imaginary.
Show Solution
Possible answer: times equals -1, which is not imaginary.
What is a characteristic of the plot of a real number in the complex plane?
Algebraic
For the following exercises, evaluate the algebraic expressions.
If evaluate given
Show Solution
If evaluate given
If evaluate given
Show Solution
If evaluate given
If evaluate given
Show Solution
If evaluate given
Graphical
For the following exercises, plot the complex numbers on the complex plane.
Show Solution
Show Solution
Numeric
For the following exercises, perform the indicated operation and express the result as a simplified complex number.
Show Solution
Show Solution
Show Solution
Show Solution
Show Solution
Show Solution
Show Solution
25
Show Solution
Show Solution
Show Solution
Show Solution
Show Solution
Show Solution
Show Solution
Technology
For the following exercises, use a calculator to help answer the questions.
Evaluate for Predict the value if
Evaluate for Predict the value if
Show Solution
128i
Evaluate for Predict the value for
Show that a solution of is
Show Solution
Show that a solution of is
Extensions
For the following exercises, evaluate the expressions, writing the result as a simplified complex number.
Show Solution
Show Solution
0
Show Solution
Show Solution
Show Solution
Glossary
complex conjugate a complex number containing the same terms as another complex number, but with the opposite operator. Multiplying a complex number by its conjugate yields a real number.complex number the sum of a real number and an imaginary number; the standard form is where a is the real part and is the complex part.complex plane the coordinate plane in which the horizontal axis represents the real component of a complex number, and the vertical axis represents the imaginary component, labeled i.imaginary number the square root of :
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Previous: Chapter 9.9: Fitting Exponential Models to Data
Next: Chapter 10.2: Polar Form of Complex Numbers
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Area Between Two Curves ($y=x^2$ and $x=y^2$)
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I was asked to find the area between two curves: $y=x^2$ and $x=y^2$. I used the interval $[0,1]$, which I found by substituting $x$ in one of the equations. I came up with this: $2\displaystyle\int_0^1\sqrt{x}dx - \int_0^1x^2 dx = 1$. However, the book says the answer is $\frac{1}{3}.$ What did I do wrong?
real-analysis
calculus
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edited Jun 19, 2019 at 1:16
Arnab Auddy
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asked Jun 19, 2019 at 1:09
N. BarN. Bar
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2
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$\begingroup$ Why did you come up with a $2$ in front? $\endgroup$
Arnab Auddy
– Arnab Auddy
2019-06-19 01:17:10 +00:00
Commented Jun 19, 2019 at 1:17
1
$\begingroup$ If you remove the $2$ in the front of the first integral, you will get the correct answer. $\endgroup$
BadAtAlgebra
– BadAtAlgebra
2019-06-19 01:18:58 +00:00
Commented Jun 19, 2019 at 1:18
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$\begingroup$
$\max(x^2, \sqrt{x}) = \sqrt{x}$ on the interval $[0,1]$. Take $\int_{0}^{1} \sqrt{x} dx - \int_{0}^1 x^2 dx$ $$=\int_{0}^1 \sqrt{x} - x^2 dx$$ $\int x^{1/2} dx = x^{1/2+1} \frac{1}{\frac{1}{2}+1} = x^{3/2} \frac{2}{3}$
and $\int x^2 dx = x^{2+1} \frac{1}{3} = \frac{1}{3} x^3$ $$=\frac{2}{3} 1^{3/2} - \frac{1}{3} 1^3 - (\frac{2}{3} 0^{3/2} - \frac{1}{3} 0^3)$$ $$=\frac{2}{3}-\frac{1}{3} = \frac{1}{3}$$
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answered Jun 19, 2019 at 1:17
Ryan SheslerRyan Shesler
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Why did you multiply the first integral by $2$? Note that the area between two curves are asked and area above $x = y^2$ on fourth region should not be counted to area between two curves.
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answered Jun 19, 2019 at 1:19
ArsenBerkArsenBerk
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$\endgroup$
2
$\begingroup$ Yeah, this was the point I was confused about. My thought process was that I had to find the absolute value of the area between $\sqrt{x}$ and $-\sqrt{x}$ $\endgroup$
N. Bar
– N. Bar
2019-06-19 01:27:11 +00:00
Commented Jun 19, 2019 at 1:27
$\begingroup$ I'm glad that it is clear now. $\endgroup$
ArsenBerk
– ArsenBerk
2019-06-19 01:29:58 +00:00
Commented Jun 19, 2019 at 1:29
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HomeIntegrated Mathematics IV (MYP4)Ch69. Cosine Rule (or Law of Cosines)
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6.9 Cosine Rule (or Law of Cosines)
Difficulty Level: Basic | Created by: The M
Last Modified: Mar 11, 2017
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Introduction
In the previous sections, you learned how to find missing sides and angles for triangles using the Sine Rule, as long as you a side and angle opposite that sides, plus one other piece of information. Can you think of examples of triangles where you cannot find all of the missing sides and angles?
What if you were given 2 sides and the angle included between them. Would you be able to find the remaining sides and angles?
[Figure 1]
Alternatively, what if you were just given the 3 sides of the triangle and no angles as shown below. How would you find the missing angles?
[Figure 2]
In this section, you will learn how to find the missing sides and angles for the triangles shown above.
Learning Outcomes
After completing this section, you should be able to:
find missing sides and angles for triangles when given 2 sides and an included angle using the Cosine Rule
find missing angles for triangles when given all three sides of a triangles using the Cosine Rule
solve problems involving the above triangles as well as other polygons
Problem Set
This section has 2 Problem Sets. Scroll down to the middle of the page for the first Problem Set and the bottom of the page for the second one.
Review Queue
Find the value of \begin{align}x\end{align} in the following triangles.
[Figure 3]
[Figure 4]
[Figure 5]
Part I: Using the Cosine Rule to Find the 3rd Side
Objective
Use the Law of Cosines to determine the length of the third side of a triangle when two sides and the included angle are known.
Guidance
The Law of Cosines can be used to solve for the third side of a triangle when two sides and the included angle are known in a triangle. consider the non right triangle below in which we know \begin{align}a, b\end{align} and \begin{align}C\end{align}. We can draw an altitude from \begin{align}B\end{align} to create two smaller right triangles as shown where \begin{align}x\end{align} represents the length of the segment from \begin{align}C\end{align} to the foot of the altitude and \begin{align}b-x\end{align} represents the length of remainder of the side opposite angle \begin{align}B\end{align}.
[Figure 6]
Now we can use the Pythagorean Theorem to relate the lengths of the segments in each of the right triangles shown.
Triangle 1: \begin{align}x^2+k^2=a^2\end{align} or \begin{align}k^2=a^2-x^2\end{align}
Triangle 2: \begin{align}(b-x)^2+k^2=c^2\end{align} or \begin{align}k^2=c^2-(b-x)^2\end{align}
Since both equations are equal to \begin{align}k^2\end{align}, we can set them equal to each other and simplify:
\begin{align}a^2-x^2 &=c^2-(b-x)^2 \ a^2-x^2 &=c^2-(b^2-2bx+x^2) \ a^2-x^2 &=c^2-b^2+2bx-x^2 \ a^2 &=c^2-b^2+2bx \ a^2+b^2-2bx &=c^2\end{align}
Recall that we know the values of \begin{align}a\end{align} and \begin{align}b\end{align} and the measure of angle \begin{align}C\end{align}. We don’t know the measure of \begin{align}x\end{align}. We can use the cosine ratio as show below to find an expression for \begin{align}x\end{align} in terms of what we already know.
\begin{align} \cos C=\frac{x}{a} \quad \text{so} \quad x=a \cos C\end{align}
Finally, we can replace \begin{align}x\end{align} in the equation to get the Law of Cosines: \begin{align}a^2+b^2-2ab \cos C=c^2\end{align}
Keep in mind that \begin{align}a\end{align} and \begin{align}b\end{align} are the sides of angle \begin{align}C\end{align} in the formula.
Interactive Proof of Cosine Rule
Example A
Find \begin{align}c\end{align} when \begin{align}m \angle C=80^\circ, a = 6\end{align} and \begin{align}b = 12\end{align}.
Solution: Replacing the variables in the formula with the given information and solve for \begin{align}c\end{align}:
\begin{align}c^2 &=6^2+12^2-2(6)(12) \cos 80^\circ \ c^2 & \approx 154.995 \ c & \approx 12.4\end{align}
Example B
Find \begin{align}a\end{align}, when \begin{align}m \angle A=43^\circ\end{align}, \begin{align}b=16\end{align} and \begin{align}c=22\end{align}.
Solution: This time we are given the sides surrounding angle \begin{align}A\end{align} and the measure of angle \begin{align}A\end{align}. We can rewrite the formula as: \begin{align}a^2=c^2+b^2-2cb \cos A\end{align}. Just remember that the length by itself on one side should be the side opposite the angle in the cosine ratio. Now we can plug in our values and solve for \begin{align}a\end{align}.
\begin{align}a^2 &=16^2+22^2-2(16)(22) \cos 43^\circ \ a^2 & \approx 225.127 \ a & \approx 15 \end{align}
Example C
Rae is making a triangular flower garden. One side is bounded by her porch and a second side is bounded by her fence. She plans to put in a stone border on the third side. If the length of the porch is 10 ft and the length of the fence is 15 ft and they meet at a \begin{align}100^\circ\end{align} angle, how many feet of stone border does she need to create?
Solution: Let the two known side lengths be \begin{align}a\end{align} and \begin{align}b\end{align} and the angle between is \begin{align}C\end{align}. Now we can use the formula to find \begin{align}c\end{align}, the length of the third side.
\begin{align}c^2 &=10^2+15^2-2(10)(15) \cos 100^\circ \ c^2 & \approx 377.094 \ c & \approx 19.4\end{align}
So Rae will need to create a 19.4 ft stone border.
Guided Practice
Find \begin{align}c\end{align} when \begin{align}m \angle C=75^\circ, a = 32\end{align} and \begin{align}b = 40\end{align}.
Find \begin{align}b\end{align} when \begin{align}m \angle B=120^\circ, a = 11\end{align} and \begin{align}c =17\end{align}.
Dan likes to swim laps across a small lake near his home. He swims from a pier on the north side to a pier on the south side multiple times for a workout. One day he decided to determine the length of his swim. He determines the distances from each of the piers to a point on land and the angles between the piers from that point to be \begin{align}50^\circ\end{align}. How many laps does Dan need to swim to cover 1000 meters?
[Figure 7]
Answers
\begin{align}c^2 &=32^2+40^2-2(32)(40) \cos 75^\circ \ c^2 & \approx 1961.42 \ c & \approx 44.3\end{align}
\begin{align}b^2 &=11^2+17^2-2(11)(17) \cos 120^\circ \ b^2 & \approx 597 \ b & \approx 24.4\end{align}
\begin{align}c^2 &=30^2+35^2-2(30)(35) \cos 50^\circ \ c^2 & \approx 775.146 \ c & \approx 27.84\end{align}
Since each lap is 27.84 meters, Dan must swim \begin{align}\frac{1000}{27.84} \approx 36\end{align} laps.
Problem Set I
OK! QUESTIONS
Use the Law of Cosines to find the value of \begin{align}x\end{align}, to the nearest tenth, in problems 1 through 6.
[Figure 8]
[Figure 9]
[Figure 10]
[Figure 11]
[Figure 12]
[Figure 13]
For problems 7 through 10, find the unknown side of the triangle. Round your answers to the nearest tenth.
Find \begin{align}c\end{align}, given \begin{align}m \angle C=105^\circ\end{align}, \begin{align}a = 55\end{align} and \begin{align}b = 61\end{align}.
Find \begin{align}b\end{align}, given \begin{align}m \angle B=26^\circ\end{align}, \begin{align}a = 33\end{align} and \begin{align}c = 24\end{align}.
Find \begin{align}a\end{align}, given \begin{align}m \angle A=77^\circ\end{align}, \begin{align}b = 12\end{align} and \begin{align}c = 19\end{align}.
Find \begin{align}b\end{align}, given \begin{align}m \angle B=95^\circ\end{align}, \begin{align}a = 28\end{align} and \begin{align}c = 13\end{align}.
USE YOUR WORDS!: Explain why when \begin{align}m \angle C=90^\circ\end{align}, the Law of Cosines becomes the Pythagorean Theorem.
UH-HUH! QUESTIONS
Patio problem: Luis is designing a triangular patio in his backyard. One side, 20 m long, will be up against the side of his house. A second side is bordered by his wooden fence. If the fence and the house meet at a \begin{align}120^\circ\end{align} angle and the fence is 15 m long, how long is the third side of the patio?
Trapezium problem: An isosceles trapezium has a height of 4 cm and parallel bases of length 3cm and 7cm. How long are its diagonals?
Quadrilateral area problem: Find the area of the quadrilateral below.
[Figure 14]
DIPLOMA PREVIEW: In a triangle, an angle of 60 0 is formed by two sides of length 4 and 11. If the length of the remaining side is\begin{align}\sqrt{a}\end{align}, what is\begin{align}a\end{align}?
A Hero's Formula for the Area of a Triangle:
Hero's Formula (sometimes called Heron's Formula) was created by a Greek mathematician and engineer named Hero of Alexandria (10 - 70 ACE). It is used for finding the area A of any triangle, given the lengths of the 3 sides a, b and c. The variable s below represents the semi-perimeter (half of the perimeter).
\begin{align}Area=\sqrt{s(s-a)(s-b)(s-c)}\end{align}, where\begin{align}s=\frac{a+b+c}{2}\end{align}
Use the formula to find the area of the triangles below. You should then check your answer using methods you have already learned.
[Figure 15]
[Figure 16]
DYNO-MITE! QUESTIONS
Hero's Formula Derivation: Try to complete the algebra derivation of this formula below.
Hero's Formula Construction: When Heron created his formula above around the time of Jesus Christ, algebra had not been invented, nor was a convenient number system available in the western world (think about Roman and Greek numerals). He created the formula using a geometric construction using only a compass and straightedge. Using only these tools, click HEREand try to do the construction as Heron did it by following the steps on the link. Like any good mystery, you will not see where this 21 step proof is leading until the very end. Good luck! The Hero's Formula Geogebra interactive below will also walk you through this proof, after you have tried it by hand (when opening the interactive, be sure to start on page 1).
Part II: Using the Cosine Rule to Find an Angle
Objective
Use the Law of Cosines to find the measure of an angle in a triangle in which all three side lengths are known.
Guidance
The Law of Cosines, \begin{align}a^2+b^2-2ab \cos C\end{align}, can be rearranged to facilitate the calculation of the measure of angle \begin{align}C\end{align} when \begin{align}a, b\end{align} and \begin{align}c\end{align} are all known lengths.
\begin{align}a^2+b^2-2ab \cos C &=c^2 \ a^2+b^2-c^2 &=2ab \cos C \ \frac{a^2+b^2-c^2}{2ab} &=\cos C\end{align}
which can be further manipulated to \begin{align}C=\cos^{-1} \left(\frac{a^2+b^2-c^2}{2ab} \right)\end{align}.
Example A
Find the measure of the largest angle in the triangle with side lengths 12, 18 and 21.
Solution: First, we must determine which angle will be the largest. Recall from Geometry that the longest side is opposite the largest angle. The longest side is 21 so we will let \begin{align}c = 21\end{align} since \begin{align}C\end{align} is the angle we are trying to find. Let \begin{align}a =12\end{align} and \begin{align}b = 18\end{align} and use the formula to solve for \begin{align}C\end{align} as shown. It doesn’t matter which sides we assign to \begin{align}a\end{align} and \begin{align}b\end{align}. They are interchangeable in the formula.
\begin{align}m \angle C=\cos^{-1} \left(\frac{12^2+18^2-21^2}{2(12)(18)} \right) \approx 86^\circ\end{align}
Note: Be careful to put parenthesis around the entire numerator and entire denominator on the calculator to ensure the proper order of operations. Your calculator screen should look like this:
\begin{align}\cos^{-1}((12^2+18^2-21^2)/(2(12)(18)))\end{align}
Example B
Find the value of \begin{align}x\end{align}, to the nearest degree.
[Figure 18]
Solution: The angle with measure \begin{align}x^\circ\end{align} will be angle \begin{align}C\end{align} so \begin{align}c = 16, a = 22\end{align} and \begin{align}b = 8\end{align}. Remember, \begin{align}a\end{align} and \begin{align}b\end{align} are interchangeable in the formula. Now we can replace the variables with the known measures and solve.
\begin{align}\cos^{-1} \left(\frac{22^2+8^2-16^2}{2(22)(8)} \right) \approx 34^\circ\end{align}
Example C
Find the \begin{align}m \angle A\end{align}, if \begin{align}a = 10, b = 15\end{align} and \begin{align}c = 21\end{align}.
Solution: First, let’s rearrange the formula to reflect the sides given and requested angle:
\begin{align}\cos A=\left(\frac{b^2+c^2-a^2}{2(b)(c)} \right)\end{align}, now plug in our values \begin{align}m \angle A=\cos^{-1} \left(\frac{15^2+21^2-10^2}{2(15)(21)} \right) \approx 26^\circ\end{align}
Guided Practice
Find the measure of \begin{align}x\end{align} in the diagram:
[Figure 19]
Find the measure of the smallest angle in the triangle with side lengths 47, 54 and 72.
Find \begin{align}m \angle B\end{align}, if \begin{align}a = 68, b = 56\end{align} and \begin{align}c = 25\end{align}.
Answers
\begin{align}\cos^{-1} \left(\frac{14^2+8^2-19^2}{2(14)(8)} \right) \approx 117^\circ\end{align}
The smallest angle will be opposite the side with length 47, so this will be our \begin{align}c\end{align} in the equation.
\begin{align}\cos^{-1} \left(\frac{54^2+72^2-47^2}{2(54)(72)} \right) \approx 41^\circ\end{align}
Rearrange the formula to solve for \begin{align}m \angle B, \cos B=\left(\frac{a^2+c^2-b^2}{2(a)(c)} \right); \cos^{-1} \left(\frac{68^2+25^2-56^2}{2(68)(25)} \right) \approx 52^\circ\end{align}
Problem Set
OK! QUESTIONS
Use the Law of Cosines to find the value of \begin{align}x\end{align}, to the nearest degree, in problems 1 through 6.
[Figure 20]
[Figure 21]
[Figure 22]
[Figure 23]
[Figure 24]
[Figure 25]
UH-HUH! QUESTIONS
Find the measure of the smallest angle in the triangle with side lengths 150, 165 and 200 meters.
Find the measure of the largest angle in the triangle with side length 59, 83 and 100 yards.
Find the \begin{align}m \angle C\end{align} if \begin{align}a = 6, b = 9\end{align} and \begin{align}c=13\end{align}.
Find the \begin{align}m \angle B\end{align} if \begin{align}a = 15, b = 8\end{align} and \begin{align}c = 9\end{align}.
Find the \begin{align}m \angle A\end{align} if \begin{align}a = 24, b = 20\end{align} and \begin{align}c = 14\end{align}.
A triangular plot of land is bordered by a road, a fence and a creek. If the stretch along the road is 100 meters, the length of the fence is 115 meters and the side along the creek is 90 meters, at what angle do the fence and road meet?
Find the area of the quadrilateral below.
[Figure 26]
DYNO-MITE! QUESTIONS
Cosine Rule Problem
[Figure 27]
Source:
Find That Angle Problem: Given ΔABC with\begin{align}{c}^{2}=\frac{{a}^{3}+{b}^{3}+{c}^{3}}{a+b+c}\end{align}, find the measure of\begin{align}\angle C\end{align}.
Notes/Highlights
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Cosine Rule Problem Set Answers V2.docx
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Description
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Learning Objectives
Difficulty Level
Basic
Tags
CK.MAT.ENG.SE.1.Algebra-II-with-Trigonometry.13
Subjects
mathematics,Algebra,Trigonometry
Concept Nodes
Grades
10,11,12
Standards Correlations
-
Language
English
Date Created
Mar 06, 2017
Last Modified
Mar 11, 2017
Vocabulary
.
| Image | Reference | Attributions |
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| | [Figure 1] | Credit:Eric Eckstein Source:eric eckstein |
| | [Figure 2] | Credit:Eric Eckstein Source:eric eckstein |
| | [Figure 3] | License:CC BY-NC |
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| | [Figure 12] | License:CC BY-NC |
| | [Figure 13] | License:CC BY-NC |
| | [Figure 14] | Credit:Brown Source:Advanced Mathematics |
| | [Figure 15] | Credit:Eric Eckstein Source:eric eckstein |
| | [Figure 16] | Credit:Eric Eckstein Source:eric eckstein |
| | [Figure 18] | License:CC BY-NC |
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| | [Figure 24] | License:CC BY-NC |
| | [Figure 25] | License:CC BY-NC |
| | [Figure 26] | Credit:Brown Source:Advanced Mathematics |
| | [Figure 27] | Credit:Nrich Mathematics Source: |
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189920 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9732073/ | Commentary on the new 2022 European Society of Human Reproduction and Embryology (ESHRE) endometriosis guidelines - PMC
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Commentary on the new 2022 European Society of Human Reproduction and Embryology (ESHRE) endometriosis guidelines
Eun Hee Yu
Eun Hee Yu
1 Department of Obstetrics and Gynecology, Pusan National University School of Medicine, Pusan National University Hospital Medical Research Institute, Busan, Republic of Korea
Find articles by Eun Hee Yu
1, Jong Kil Joo
Jong Kil Joo
1 Department of Obstetrics and Gynecology, Pusan National University School of Medicine, Pusan National University Hospital Medical Research Institute, Busan, Republic of Korea
Find articles by Jong Kil Joo
1,✉
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1 Department of Obstetrics and Gynecology, Pusan National University School of Medicine, Pusan National University Hospital Medical Research Institute, Busan, Republic of Korea
✉
Corresponding author: Jong Kil Joo Department of Obstetrics and Gynecology, Pusan National University Hospital, 179 Gudeok-ro, Seo-gu, Busan 49241, Korea Tel: +82-51-240-7287 E-mail: jkjoo@pusan.ac.kr
This work was supported by a clinical research grant from Pusan National University Hospital in 2022.
Received 2022 Aug 2; Revised 2022 Nov 2; Accepted 2022 Nov 10; Issue date 2022 Dec.
Copyright © 2022. THE KOREAN SOCIETY FOR REPRODUCTIVE MEDICINE
This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC9732073 PMID: 36482496
Abstract
Endometriosis is a prevalent benign illness defined by the presence of endometrial glands and stroma outside of the uterine cavity, primarily on the ovary, pelvic peritoneum, and rectovaginal septum, resulting in a variety of symptoms, including dysmenorrhea and infertility. Traditionally, prolonged medical therapy has been needed in most cases since a conservative approach to surgery has usually been taken, especially in young women. In 2022, new European Society of Human Reproduction and Embryology (ESHRE) guidelines were published that present different directions for diagnosis and treatment from the past. Furthermore, the guidelines for the diagnosis and management of endometriosis are more precise and applicable than in previous editions. Thus, referring to the representative changes in the new guidelines and important updates will be beneficial for the diagnosis and management of endometriosis. This paper provides a brief overview of these developments.
Keywords: Endometriosis, ESHRE, Guideline
Introduction
Endometriosis is a prevalent benign illness defined by the presence of endometrial glands and stroma outside of the uterine cavity, primarily on the ovary, pelvic peritoneum, and rectovaginal septum . It is linked to infertility and a variety of problems, including chronic pelvic pain, dysmenorrhea, deep dyspareunia, dysuria, dyschezia, and fatigue [2,3]. However, symptom severity is not usually proportional to the endometriosis stage, and some women with endometriosis may be asymptomatic.
Traditionally, endometriosis has been diagnosed through diagnostic laparoscopy or the histological identification of lesions. However, with recent developments in imaging modalities, the necessity for diagnostic laparoscopy in cases where endometriosis is relatively obvious has been questioned, and concerns have been raised regarding delays in the endometriosis diagnosis due to the diagnostic laparoscopic and histological confirmation criteria.
In 2022, the new European Society of Human Reproduction and Embryology (ESHRE) guidelines were published . Although the new guidelines are not perfect, they provide clearer guidance on many difficult issues, such as medication selection among combined oral contraceptives (COCs), progestogen, and gonadotropin-releasing hormone (GnRH) agonists, and surgical indications for endometrioma in patients preparing for pregnancy.
The ESHRE guidelines are one of the most frequently cited endometriosis-related guidelines. It is believed that referring to representative changes in the new guidelines and major updates will be beneficial for diagnosing and treating endometriosis (Table 1). This paper provides a brief overview of these developments.
Table 1.
Summary of the latest guideline recommendation updates
| Category | Updates in the latest version |
:---: |
| Diagnosis of endometriosis | · Laparoscopy: no longer the diagnostic gold standard |
| Treatment of endometriosis-associated pain | · Consideration of a GnRH antagonist as a reasonable second-line treatment |
| · Proceeding to postoperative medical treatment for expected benefits |
| Treatment of endometriosis-associated infertility | · Ultra-long protocol with GnRH agonist withdrawn from the evidence-based recommendations |
| · Application of the Endometriosis Fertility Index |
| · Importance of fertility preservation |
| Endometriosis recurrence | · Long-term hormone therapy advised when pregnancy is not desired |
| Endometriosis and adolescence | · Newly added; diagnosis and management not significantly different from adults |
| Endometriosis and menopause | · Importance reinforced; related symptoms present even in menopause |
| Primary prevention of endometriosis | · Revised title from “prevention of endometriosis”; recurrence is dealt with separately in “endometriosis recurrence” |
| Endometriosis and cancer | · Association of endometriosis with certain cancer risks despite the low absolute risks in people with endometriosis relative to those without; additive reassurance recommended for endometriosis patients |
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GnRH, gonadotropin-releasing hormone.
Diagnosis of endometriosis
The previous ESHRE guidelines suggested that laparoscopic histologic confirmation of endometriosis was the gold standard for an endometriosis diagnosis. In contrast, according to the new ESHRE guidelines, laparoscopy is no longer the diagnostic gold standard and is now only advised for patients with negative imaging results and/or when empirical treatment is ineffective or unsuitable. This change was based on a meta-analysis regarding endometriosis diagnostic tools. According to a Cochrane review, imaging modalities such as transvaginal ultrasonography and magnetic resonance imaging showed sensitivity and specificity for diagnosing endometrioma and deep endometriosis comparable to a surgical diagnosis (Table 2) . As an added explanation in the new guidelines, the difficulties and limitations of making a noninvasive diagnosis of a superficial disease are described. Despite these limitations, stating that diagnostic laparoscopy is not the gold standard emphasizes the usefulness of imaging modalities for diagnosing endometrioma and deep endometriosis.
Table 2.
Accuracy of imaging modalities for endometriosis diagnosis
| Type of endometriosis | Modality | Number of patients included | Sensitivity (95% CI) | Specificity (95% CI) |
:---: :---:
| Pelvic endometriosis | TVUS | 1,222 in 5 studies | 0.65 (0.27–1.00) | 0.95 (0.89–1.00) |
| MRI | 396 in 10 data sets | 0.79 (0.79–0.88) | 0.72 (0.51–0.92) |
| Ovarian endometrioma | TVUS | 765 in 8 studies | 0.93 (0.87–0.99) | 0.96 (0.92–0.99) |
| MRI | 179 in 3 studies | 0.95 (0.90–1.00) | 0.91 (0.86–0.97) |
| Deep endometriosis | TVUS | 1,383 in 12 data sets | 0.79 (0.69–0.89) | 0.94 (0.88–1.00) |
| MRI | 289 in 7 data sets | 0.94 (0.90–0.97) | 0.77 (0.44–1.00) |
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CI, confidence interval; TVUS, transvaginal ultrasound; MRI, magnetic resonance imaging.
Modified from Nisenblat et al. Cochrane Database Syst Rev 2016;2:CD009591 .
Treatment for endometriosis-associated pain
Discomfort related to endometriosis includes dysmenorrhea, dyspareunia, dysuria, dyschezia, and non-menstrual pelvic pain. In prior Korean and international guidelines, the hormonal medical treatment for endometriosis-associated pain has consisted of COCs, progestogens, anti-progestogens, aromatase inhibitors, and danazol. However, in the new guidelines, GnRH antagonists were introduced as a second treatment option, while anti-progestogens and danazol were withdrawn.
1. Introduction of GnRH antagonists as a second-line treatment option
The use of GnRH antagonists to alleviate endometriosis-associated pain can be considered. However, limited data exist on application of GnRH antagonist. Due to their side-effect profile, they are provided as a second-line option (for instance, if hormonal contraceptives or progestogens are ineffective). A study on two identical multi-center double-blind, randomized, placebo-controlled, phase 3 trials of 6-month treatments with oral elagolix at two doses in women with moderate or severe endometriosis-associated pain provided data on treatment efficacy. The proportion of women who met the clinical response criteria for dysmenorrhea and non-menstrual pelvic pain was significantly higher among women who received each elagolix dose (46% in the lower-dose group, 75.8% in the higher-dose group) than among those who received placebo (19.6%). The reductions in dysmenorrhea and non-menstrual pelvic discomfort were noticeable after 1 month and persisted for 6 months. The most frequently reported adverse reactions were hot flushes, headaches, and nausea .
2. Changes in other medications
A levonorgestrel-releasing intrauterine system (LNG-IUS) has long been recognized as an effective treatment for endometriosis-associated discomfort. A recent randomized controlled trial (RCT) assigned 103 women with endometriosis-related chronic pelvic pain and/or dysmenorrhea to receive either an etonogestrel (ENG)-releasing subdermal implant or a 52-mg LNG-IUS . Both the ENG implant and the LNG-IUS significantly reduced endometriosis-related pain, dysmenorrhea, and chronic pelvic pain. To alleviate endometriosis-related discomfort, it is recommended that women receive either an ENG implant or an LNG-IUS.
In addition, the Guideline Development Group (GDG) suggested using GnRH agonists as second-line treatments due to their side effect profile. Danazol and anti-progestogens, laparoscopic uterosacral neck ablation, presacral neurectomy, and anti-adhesion medications are no longer included in the recommendations of guidelines because of their harmful effects or lack of extra benefit.
3. Effectiveness of postoperative medical treatment
According to previous guidelines, practitioners should not provide adjunctive hormonal treatment for endometriosis-associated pain following surgery because it does not improve the outcomes of surgery to relieve pain. However, a recent Cochrane analysis by Chen et al. suggested that postsurgical medical therapy might reduce pain recurrence and illness recurrence within 12 months . To improve the immediate success of surgery for pain in women with endometriosis who do not desire pregnancy, postoperative hormonal therapy may be administered.
Treatment of endometriosis-associated infertility
1. Introduction of the Endometriosis Fertility Index
The GDG recommends that the decision to perform surgery should be made in consideration of some factors such as the presence or absence of pain symptoms, patient age and preferences, history of previous surgery, presence of other infertility factors, ovarian reserve, and the estimated Endometriosis Fertility Index (EFI). The EFI was added to the new guidelines for the first time. The EFI staging system predicts non-in vitro fertilization pregnancy rates following surgical endometriosis staging and treatment . The EFI is determined by historical variables (age, number of years of infertility, and prior pregnancy) and surgical variables (the American Society for Reproductive Medicine [ASRM] overall score, the ASRM endometriosis score, and the least function score).
2. No benefit of ultra-long GnRH agonist prior to assisted reproductive technology
The extended administration of a GnRH agonist prior to assisted reproductive technology treatment to increase the live birth rate in infertile women with endometriosis (ultra-long protocol) is no longer suggested due to the lack of evidence supporting its benefits.
The previous recommendation was based on an older Cochrane review regarding GnRH agonist pre-treatment, which included 228 patients from three studies and showed an increased likelihood of clinical pregnancy by more than 4-fold . However, an updated Cochrane review that included eight parallel-design RCTs with a total of 640 participants, concluded that the effect of GnRH agonist pre-treatment (for at least 3 months) was very uncertain, both on the live birth rate as the primary outcome and on secondary outcomes (clinical pregnancy rate, multiple pregnancy rate, miscarriage rate, the mean number of oocytes, and the mean number of embryos) . Another meta-analysis of studies comparing different GnRH agonist protocols (short, long, and ultra-long) also found that different down-regulation protocols did not significantly improve clinical outcomes (implantation rate, fertilization rate, and clinical pregnancy rate) by analyzing RCTs and observational studies (n=21) .
3. Importance of fertility preservation
Prior ESHRE guidelines on fertility preservation considered benign disorders to be an indication for fertility preservation, but did not address whether endometriosis was a reason for fertility preservation in particular . A recent large retrospective study by Cobo et al. evaluated the outcome of fertility preservation utilizing vitrified oocytes in 485 patients with endometriomas of at least 1 cm and an antral follicular count of at least 3, finding oocyte survival rates of 83.2% after warming and a cumulative live birth rate of 46.4%. They concluded that fertility preservation was a valid treatment option in patients with endometriosis. Although several issues such as cost-effectiveness and specific indications remain unsolved, practitioners should explore the advantages and disadvantages of fertility preservation with endometriosis patients who have severe ovarian endometriosis.
4. Impact of endometriosis on pregnancy and pregnancy outcomes
Sparse data with low and moderate quality indicated that the behavior of endometriotic lesions during pregnancy was diverse, ranging from complete elimination to increasing growth . Patients should not be encouraged to become pregnant with the primary aim of treating endometriosis because pregnancy does not necessarily result in symptom improvements or slow disease progression. The decidualization of an endometrioma during pregnancy may resemble malignant ovarian tumors in some circumstances, providing a diagnostic conundrum. However, the incidence of this event is unknown (0%–12% prevalence, 17 studies reporting 60 cases) . Endometrioma can manifest differently throughout pregnancy. If an ultrasound examination during pregnancy reveals atypical endometrioma, it is recommended that the patient be referred to a center with the necessary expertise.
Complications directly connected to pre-existing endometriosis lesions are uncommon and likely underreported. These issues may result from decidualization, adhesion formation/stretching, and endometriosis-related chronic inflammation . Although uncommon, these may pose life-threatening conditions that necessitate surgical treatment. Clinicians should be aware that women with endometriosis may have a higher risk of miscarriages and ectopic pregnancies during the first trimester. They should also be informed of uncommon endometriosis-related problems during pregnancy, which include gestational diabetes, preterm birth, premature rupture of membranes, placenta previa, hypertensive disorders and pre-eclampsia, stillbirth, cesarean section, obstetric hemorrhage (placental abruption, antepartum and postpartum bleeding), small for gestational age, admission to the neonatal intensive care unit, and neonatal death. However, since these results are based on low- or moderate-quality research, they should be regarded with caution, and additional antenatal monitoring is not recommended.
Endometriosis recurrence
The recurrence rate of endometriosis has been reported to range from 0% to 89.6% . This variation could be related to differences in the definitions of recurrence, length of follow-up, study design, sample size, type and stage of disease, type of operation, and postoperative medical care . As the high recurrence rate and its significance have been consistently emphasized, they are described in a separate chapter, unlike in the earlier guidelines. For preventing recurrence, ovarian cystectomy instead of drainage/electrocoagulation and postoperative hormonal treatment for at least 18 to 24 months are recommended. The recommended duration of hormonal treatment is based on RCTs. However, in patients who are not immediately seeking conception, long-term hormonal therapy is recommended.
Endometriosis and adolescence
There are scarce data on endometriosis and adolescence. Although there are no major differences between the diagnosis and treatment of endometriosis in adolescents and adults, clinicians treating adolescents should be vigilant regarding several factors. Not only is dysmenorrhea a major symptom of endometriosis, but it is also a highly prevalent occurrence in adolescents, and it takes significantly longer to reach a diagnosis of endometriosis in adolescents than in adults. When previous treatments have failed, clinicians may consider prescribing a GnRH agonist for up to 1 year. If GnRH agonist treatment is considered for young women and adolescents, it should be delivered only after careful evaluation and a discussion of potential adverse effects and long-term health problems with a practitioner in a secondary or tertiary care setting, as recommended by the GDG. As with adults, no studies have addressed the efficacy or utility of fertility preservation—specifically, oocyte cryopreservation—in adolescents with endometriosis. Although the true benefits, safety, and indications for adolescents with endometriosis remain unknown, the GDG recommends informing adolescents about fertility preservation options.
Endometriosis and menopause
The amount of data available regarding the prevalence of endometriosis after menopause is extremely limited. A recent retrospective cohort study described a 4% prevalence of postmenopausal endometriosis . It is hypothesized that menopausal hormone therapy can increase the formation of endometriosis and a variety of factors, some of which are unknown, can also affect the growth of endometriosis. Therefore, it is important for clinicians to be aware that endometriosis might continue to be active and cause symptoms after menopause.
For postmenopausal women presenting with signs of endometriosis and/or discomfort, clinicians might consider surgical treatment as a means of enabling histological confirmation of the diagnosis of endometriosis. The GDG suggests that practitioners emphasize the ambiguity regarding the risk of cancer in postmenopausal women. If a mass is found in the pelvis, a diagnostic workup and treatment should be carried out in accordance with the national oncology standards. Clinicians may consider aromatase inhibitors as a potential therapeutic option for endometriosis-related pain in postmenopausal women, particularly if surgery is not a viable option.
Clinicians should be aware that women with endometriosis who undergo early bilateral salpingo-oophorectomy as part of their therapy have a higher risk of decreased bone density, dementia, and cardiovascular disease. It is also essential to emphasize that women with endometriosis have a higher risk of cardiovascular disease, regardless of whether they have undergone early surgical menopause.
For the management of postmenopausal symptoms in women with a history of endometriosis, combined menopausal hormone therapy may be considered; however, tibolone is no longer recommended as a medical treatment for menopausal symptoms in women with a history of endometriosis.
Extra-pelvic endometriosis
Clinicians should be aware of the symptoms of extra-pelvic endometriosis, which include cyclical shoulder discomfort, cyclical spontaneous pneumothorax, cyclical cough, and enlargement of nodules during menstruation. The diagnosis and treatment of extra-pelvic endometriosis should be discussed by a multidisciplinary team at an institution with sufficient expertise. When possible, surgical excision is the best treatment for alleviating symptoms in women with abdominal extra-pelvic endometriosis. If surgery is inapplicable or unacceptable, hormonal therapy may also be an alternative.
Asymptomatic endometriosis
Clinicians should not regularly perform surgical excision/ablation for asymptomatic endometriosis discovered incidentally during surgery. Practitioners should not suggest medical therapy to women with an incidental endometriosis diagnosis, but routine ultrasound surveillance must be recommended. Even in the absence of convincing data on the benefits of monitoring asymptomatic endometriosis, the GDG recommends considering ultrasound surveillance due to its low cost and safety.
Primary prevention of endometriosis
The objective of primary prevention is to prevent healthy and asymptomatic women from developing endometriosis. Although there is no direct proof of the efficacy of a healthy lifestyle and diet in preventing endometriosis, women can be counseled to adopt a healthy lifestyle and diet, including reduced alcohol use and regular physical activity. The efficacy of hormonal contraceptives for the primary prevention of endometriosis is uncertain. Genetic testing of women with suspected or proven endometriosis should only be undertaken in a research context, citing the newly added text.
Endometriosis and cancer
According to the 2014 recommendations, ovarian cancer and non-Hodgkin lymphoma were slightly prevalent among endometriosis patients. Nevertheless, according to a recent systematic review and meta-analysis of 49 cohort or case-control studies, endometriosis was related to a very slight and not statistically significant higher risk of cancer overall (summary relative risk [SRR], 1.07; 95% confidence interval [CI], 0.96–1.16) . Endometriosis was related to an increased risk of ovarian cancer (SRR, 1.93), especially clear-cell (SRR, 3.44) and endometrioid (SRR, 2.33) histotypes, breast cancer (SRR, 1.04) and thyroid cancer (SRR, 1.39) . Although endometriosis is associated with an elevated risk of certain cancers, given the low absolute risks of ovarian, breast, and thyroid cancer in people with endometriosis relative to those without (increases of 1.2%p, 0.5%p, and 0.5%p, respectively) and the uncertainty regarding the risk of other cancers, endometriosis patients can be reassured that their cancer risk is low and comparable to that of people without the disease. Clinicians should educate women with endometriosis who request information on their cancer risk that endometriosis is not associated with an increased risk of cancer. Although endometriosis is related to a somewhat higher risk of ovarian, breast, and thyroid malignancies, the absolute increase in risk relative to the general female population is minimal (Table 3).
Table 3.
Absolute risk for cancer in women with endometriosis
| Variable | Absolute risk of developing cancer in a woman's lifetime (%) | Increase in risk in women with endometriosis (%p) |
:---:
| All women | Women with endometriosis |
| Ovarian cancer | 1.3 | 2.5 | +1.2 |
| Breast cancer | 12.8 | 13.3 | +0.5 |
| Thyroid cancer | 1.3 | 1.8 | +0.5 |
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Conclusion
The new ESHRE recommendations are clearer and more practical than their predecessors and other guidelines. However, as stated on the opening page of the ESHRE guidelines, clinical practice guidelines do not supersede the clinical decisions made by a healthcare professional for diagnosis and treatment. Ultimately, healthcare professionals must make their own clinical decisions case by case, with consideration of various circumstances.
Footnotes
Conflict of interest
Jong Kil Joo is an editorial board member of the journal but was not involved in the peer reviewer selection, evaluation, or decision process of this article. No other potential conflicts of interest relevant to this article were reported.
Author contributions
Writing–original draft: all authors. Writing–review and editing: JKJ.
References
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Abstract
Introduction
Diagnosis of endometriosis
Treatment for endometriosis-associated pain
Treatment of endometriosis-associated infertility
Endometriosis recurrence
Endometriosis and adolescence
Endometriosis and menopause
Extra-pelvic endometriosis
Asymptomatic endometriosis
Primary prevention of endometriosis
Endometriosis and cancer
Conclusion
Footnotes
References
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189921 | https://www.idsociety.org/globalassets/idsa/practice-guidelines/infective-endocarditis-in-adults-diagnosis-antimicrobial-therapy-and-management-of-complications.pdf | AHA Scientific Statement 1435 Background—Infective endocarditis is a potentially lethal disease that has undergone major changes in both host and pathogen. The epidemiology of infective endocarditis has become more complex with today’s myriad healthcare-associated factors that predispose to infection. Moreover, changes in pathogen prevalence, in particular a more common staphylococcal origin, have affected outcomes, which have not improved despite medical and surgical advances.
Methods and Results—This statement updates the 2005 iteration, both of which were developed by the American Heart Association under the auspices of the Committee on Rheumatic Fever, Endocarditis, and Kawasaki Disease, Council on Cardiovascular Disease of the Young. It includes an evidence-based system for diagnostic and treatment recommendations used by the American College of Cardiology and the American Heart Association for treatment recommendations.
Conclusions—Infective endocarditis is a complex disease, and patients with this disease generally require management by a team of physicians and allied health providers with a variety of areas of expertise. The recommendations provided in this document are intended to assist in the management of this uncommon but potentially deadly infection. The clinical variability and complexity in infective endocarditis, however, dictate that these recommendations be used to support and not supplant decisions in individual patient management. (Circulation. 2015;132:1435-1486. DOI: 10.1161/CIR.0000000000000296.) Key Words: AHA Scientific Statements ◼ anti-infective agents ◼ echocardiography ◼ endocarditis ◼ infection (Circulation. 2015;132:1435-1486. DOI: 10.1161/CIR.0000000000000296.) © 2015 American Heart Association, Inc.
Circulation is available at DOI: 10.1161/CIR.0000000000000296 The American Heart Association makes every effort to avoid any actual or potential conflicts of interest that may arise as a result of an outside relationship or a personal, professional, or business interest of a member of the writing panel. Specifically, all members of the writing group are required to complete and submit a Disclosure Questionnaire showing all such relationships that might be perceived as real or potential conflicts of interest.
This statement was approved by the American Heart Association Science Advisory and Coordinating Committee on May 12, 2015, and the American Heart Association Executive Committee on June 12, 2015. A copy of the document is available at by selecting either the “By Topic” link or the “By Publication Date” link. To purchase additional reprints, call 843-216-2533 or e-mail kelle.ramsay@wolterskluwer.com.
The American Heart Association requests that this document be cited as follows: Baddour LM, Wilson WR, Bayer AS, Fowler VG Jr, Tleyjeh IM, Rybak MJ, Barsic B, Lockhart PB, Gewitz MH, Levison ME, Bolger AF, Steckelberg JM, Baltimore RS, Fink AM, O’Gara P, Taubert KA; on behalf of the American Heart Association Committee on Rheumatic Fever, Endocarditis, and Kawasaki Disease of the Council on Cardiovascular Disease in the Young, Council on Clinical Cardiology, Council on Cardiovascular Surgery and Anesthesia, and Stroke Council. Infective endocarditis in adults: diagnosis, antimicrobial therapy, and management of complications: a scientific statement for healthcare professionals from the American Heart Association. Circulation. 2015;132:1435–1486.
Expert peer review of AHA Scientific Statements is conducted by the AHA Office of Science Operations. For more on AHA statements and guidelines development, visit and select the “Policies and Development” link.
Permissions: Multiple copies, modification, alteration, enhancement, and/or distribution of this document are not permitted without the express permission of the American Heart Association. Instructions for obtaining permission are located at A link to the “Copyright Permissions Request Form” appears on the right side of the page.
Infective Endocarditis in Adults: Diagnosis, Antimicrobial Therapy, and Management of Complications A Scientific Statement for Healthcare Professionals From the American Heart Association Endorsed by the Infectious Diseases Society of America Larry M. Baddour, MD, FAHA, Chair; Walter R. Wilson, MD; Arnold S. Bayer, MD; Vance G. Fowler, Jr, MD, MHS; Imad M. Tleyjeh, MD, MSc; Michael J. Rybak, PharmD, MPH; Bruno Barsic, MD, PhD; Peter B. Lockhart, DDS; Michael H. Gewitz, MD, FAHA; Matthew E. Levison, MD; Ann F. Bolger, MD, FAHA; James M. Steckelberg, MD; Robert S. Baltimore, MD; Anne M. Fink, PhD, RN; Patrick O’Gara, MD, FAHA; Kathryn A. Taubert, PhD, FAHA; on behalf of the American Heart Association Committee on Rheumatic Fever, Endocarditis, and Kawasaki Disease of the Council on Cardiovascular Disease in the Young, Council on Clinical Cardiology, Council on Cardiovascular Surgery and Anesthesia, and Stroke Council I nfective endocarditis (IE) is an uncommon infectious dis-ease with an annual incidence ranging from 3 to 7 per 100 000 person-years in the most contemporary population surveys.1–3 Although relatively rare, IE continues to be char-acterized by increased morbidity and mortality and is now the third or fourth most common life-threatening infection Downloaded from by on October 16, 2018 1436 Circulation October 13, 2015 syndrome, after sepsis, pneumonia, and intra-abdominal abscess. Globally, in 2010, IE was associated with 1.58 mil-lion disability-adjusted life-years or years of healthy life lost as a result of death and nonfatal illness or impairment.4 Epidemiological surveys from France and the International Collaboration on Endocarditis have confirmed that the epide-miological profile of IE has changed substantially. Although the overall IE incidence has remained stable,1,2,5–9 the incidence of IE caused by Staphylococcus aureus has increased, and S aureus is now the most common causative organism in most of the industrialized world. The emergence of S aureus IE is due in part to the increasing importance of healthcare contact as a leading risk associated with infection. Characteristics of IE patients have also shifted toward an increased mean patient age, a higher proportion of prosthetic valves and other car-diac devices, and a decreasing proportion of rheumatic heart disease. Moreover, the proportion of IE patients undergoing surgery has increased over time to reach ≈50%.1,10,11 In addition to these temporal epidemiological changes, major new findings from multiple diagnostic, prognostic, and therapeutic studies have been published since the last iteration of the American Heart Association (AHA) statement on diagnosis and management of IE complications was published in 2005.12 For example, the rapid detection of pathogens from valve tissue from patients undergoing surgery for IE by polymerase chain reaction (PCR) has been validated. Moreover, diagnostic inno-vations have emerged through new imaging techniques such as 3-dimensional (3D) echocardiography, “head-to-toe” multislice computed tomography (CT), and cardiac magnetic resonance imaging (MRI). Furthermore, the role of cerebral MRI and magnetic resonance angiography in the diagnosis and manage-ment of IE has been better defined in several studies. In addi-tion, several risk stratification models for quantifying morbidity and mortality in IE patients overall and particularly in those undergoing valve surgeries have been developed and validated. Finally, daptomycin has been evaluated in the treatment of S aureus bacteremia and IE in a randomized, controlled trial.13 Several rigorously conducted observational studies11,14–16 and a randomized, controlled trial17 have examined the impact and timing of valve surgery in IE management. In addition, updated international management guidelines have been published.18,19 The present AHA IE Writing Committee conducted com-prehensive and focused reviews of the literature published between January 2005 and October 2013 to update the previous version of the guidelines. Literature searches of the PubMed/ MEDLINE databases were undertaken to identify pertinent articles. Searches were limited to the English language. The major search terms included endocarditis, infective endocardi-tis, infectious endocarditis, intracardiac, valvular, mural, infec-tion, diagnosis, bacteremia, case definition, epidemiology, risks, demographics, injection drug use, echocardiography, microbiology, culture-negative, therapy, antibiotic, antifungal, antimicrobial, antimicrobial resistance, adverse drug effects, drug monitoring, outcome, meta-analysis, complications, abscess, heart failure, embolic events, stroke, conduction abnormalities, survival, pathogens, organisms, treatment, sur-gery, indications, valve replacement, valve repair, ambulatory care trials, and prevention. In addition, the present statement includes a new section, Surgical Therapy. This work addresses primarily IE in adults; a more detailed review of the unique features of IE in children is available in another statement from the AHA Committee on Rheumatic Fever, Endocarditis, and Kawasaki Disease.20 The committee also published state-ments on endocarditis that complicates electrophysiological (pacemakers, intracardiac defibrillators),21 ventricular assist, and other nonvalvular cardiac devices.22 Evidence-Based System for Diagnostic and Treatment Recommendations The writing group was charged with the task of performing an evidence-based assessment of the data and providing a class of recommendation and a level of evidence for each recom-mendation according to the American College of Cardiology/ AHA classification system ( manual/manual_IIstep6.shtml). The class of recommendation is an estimate of the size of the treatment effect, considering risks versus benefits, in addition to evidence or agreement that a given treatment or procedure is or is not useful or effective or in some situations may cause harm. The level of evidence is an estimate of the certainty or precision of the treatment effect. The Writing Group reviewed and assessed the strength of evidence supporting each recommendation with the level of evidence ranked as A, B, or C according to the specific defini-tions included in Table 1. For certain conditions for which data were either unavailable or inadequate, recommendations were based on expert consensus and clinical experience, and these were ranked as Level of Evidence C. The scheme for the class of recommendations and levels of evidence is summarized in Table 1, which also provides suggested phrases for writing recommendations within each class of recommendation.
Diagnosis The diagnosis of IE is straightforward in the minority of patients who present with a consistent history and classic oslerian manifestations: sustained bacteremia or fungemia, evidence of active valvulitis, peripheral emboli, and immu-nological vascular phenomena. In most patients, however, the “textbook” history and physical examination findings may be few or absent. Cases with limited manifestations of IE may occur early during IE, particularly among patients who are injection drug users (IDUs), in whom IE is often the result of acute S aureus infection of right-sided heart valves. Acute IE may evolve too quickly for the development of immunologi-cal vascular phenomena, which are more characteristic of the later stages of the more insidious subacute form of untreated IE. In addition, valve lesions in right-sided IE usually do not create the peripheral emboli and immunological vascular phe-nomena that can result from left-sided valvular involvement. Right-sided IE, however, can cause septic pulmonary emboli.
The variability in clinical presentation of IE and the importance of early accurate diagnosis require a diagnostic strategy that is both sensitive for disease detection and spe-cific for its exclusion across all forms of the disease. In 1994, Durack and colleagues23 from the Duke University Medical Center proposed a diagnostic schema that stratified patients with suspected IE into 3 categories: definite, possible, and rejected cases (Tables 2 and 3).
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1437 A diagnosis of IE with the original Duke criteria was based on the presence of either major or minor clinical cri-teria (Tables 2 and 3). The Duke criteria gave diagnostic weight to bacteremia with staphylococci or enterococci only, on the basis of the location of acquisition and with-out an apparent primary focus; these types of bacteremia have the highest risk of being associated with IE.23,25,26 The Duke criteria incorporated echocardiographic find-ings into the diagnostic strategy (Tables 2 and 3; see the Echocardiography section). Six common but less specific findings of IE were included as minor criteria in the original Duke schema (Tables 2 and 3).
In the mid to late 1990s, direct analyses of the Duke crite-ria were made in 12 major studies27–38 including nearly 1700 patients composed of geographically and clinically diverse groups (adult, pediatric, and older adult [≥60 years of age] patients; patients from the community; IDU and non-IDU patients; and those with both native and prosthetic valves). The studies27–38 confirmed the high sensitivity and specificity of the Duke criteria and the diagnostic utility of echocardiography in identifying clinically definite cases. Moreover, a retrospective study of 410 patients showed good agreement (72%–90%) between the Duke criteria and clinical assessment by infec-tious disease experts blinded to underlying IE risk factors.39 Table 1. Applying Classification of Recommendations and Level of Evidence A recommendation with Level of Evidence B or C does not imply that the recommendation is weak. Many important clinical questions addressed in the guidelines do not lend themselves to clinical trials. Although randomized trials are unavailable, there may be a very clear clinical consensus that a particular test or therapy is useful or effective.
Data available from clinical trials or registries about the usefulness/efficacy in different subpopulations, such as sex, age, history of diabetes, history of prior myocardial infarction, history of heart failure, and prior aspirin use.
†For comparative effectiveness recommendations (Class I and IIa; Level of Evidence A and B only), studies that support the use of comparator verbs should involve direct comparisons of the treatments or strategies being evaluated.
Downloaded from by on October 16, 2018 1438 Circulation October 13, 2015 Several refinements have been made to both the major and minor Duke criteria. In the original Duke criteria, bacteremia resulting from S aureus or enterococci was considered to fulfill a major criterion only if it was community acquired because ample literature suggested that this parameter was an important surro-gate marker for underlying IE.27 However, an increasing number of more contemporary studies documented IE in patients experi-encing nosocomial staphylococcal bacteremia. For example, of 59 consecutive patients with S aureus IE, 45.8% had nosocomial infections, and 50.8% had a removable focus of infection.39 In an analysis of 262 patients at the Duke University Medical Center who had hospital-acquired S aureus bacteremia, 34 (13%) were subsequently diagnosed with definite IE. Therefore, the modi-fied Duke criteria (Tables 2 and 3) recommend the inclusion of S aureus bacteremia as a major criterion, regardless of whether the infection is hospital acquired (with or without a removable source of infection) or community acquired.24 Specific serological data have been included in the Duke IE diagnostic schema to establish the pathogenic agents of culture-negative IE more precisely (ie, as a surrogate for positive blood cultures). These serological criteria would be applied in circumstances in which the pathogenic organism is slow growing in routine blood cultures (eg, Brucella spe-cies) or requires special blood culture media (eg, Bartonella species, Legionella species, Tropheryma whipplei, fungi, and Mycobacterium species) or in which the organism is not culturable (eg, Coxiella burnetii, the agent of Q fever). For example, in the original Duke criteria, a positive serology for Q fever was considered a minor microbiological criterion. Subsequently, Fournier et al40 studied 20 pathologically con-firmed cases of Q fever IE. When the original Duke criteria were used, 4 of the 20 patients were classified as having pos-sible IE. When Q fever serological results and a single blood culture positive for C burnetii were considered to be a major criterion, however, each of these 4 cases was reclassified from possible IE to definite IE. On the basis of these data, specific serological data as a surrogate marker for positive blood cul-tures have now been included in the Duke criteria. Thus, an anti–phase I immunoglobulin G antibody titer ≥1:800 or a single blood culture positive for C burnetii should be a major criterion in the modified Duke schema.24 Serological tests and PCR-based testing for other diffi-cult-to-cultivate organisms such as Bartonella quintana or Tropheryma whippelii also have been discussed as future major criteria. At present, there are significant methodologi-cal problems associated with proposing antibody titers that are positive for Bartonella and Chlamydia species or PCR-based testing for T whippelii as a major criterion in the Duke schema. For example, IE caused by Bartonella and Chlamydia species often are indistinguishable in serological test results because of cross-reactions.41 Low sensitivity is a major limitation of PCR unless cardiac valvular tissue is available for testing.42–45 Few centers provide timely PCR-based testing for these rare causes of IE. Therefore, the inclusion of these assays as major criteria should be deferred until the serodiagnostic and PCR approaches can be standardized and validated in a sufficient number of cases of these rare types of IE, the aforementioned technical problems are resolved, and the availability of such assays becomes more widespread.
The expansion of minor criteria to include elevated eryth-rocyte sedimentation rate or C-reactive protein, the presence of newly diagnosed clubbing, splenomegaly, and microscopic hematuria also has been proposed. In a study of 100 consecu-tive cases of pathologically proven native valve IE (NVE), inclusion of these additional parameters with the existing Duke minor criteria resulted in a 10% increase in the fre-quency of cases being deemed clinically definite, with no loss of specificity. The major limitations of the erythrocyte sedi-mentation rate and C-reactive protein are that they are non-specific and particularly challenging to interpret in patients with comorbid conditions. These additional parameters have not been formally integrated into the modified Duke criteria,24 however, which are universally accepted.
One minor criterion from the original Duke schema, “echocardiogram consistent with IE but not meeting major criterion,” was re-evaluated. This criterion originally was used in cases in which nonspecific valvular thickening was detected by transthoracic echocardiography (TTE). In a reanalysis of patients in the Duke University database (containing records collected prospectively on >800 cases of definite and possible IE since 1984), this echocardiographic criterion was used in only 5% of cases and was never used in the final analysis of any patient who underwent transesophageal echocardiogra-phy (TEE). Therefore, this minor criterion was eliminated in the modified Duke criteria.24 Finally, adjustment of the Duke criteria to require a mini-mum of 1 major plus 1 minor criterion or 3 minor criteria as a “floor” to designate a case as possible IE (as opposed to “findings consistent with IE that fall short of ‘definite’ but not ‘rejected’ ”) has been incorporated into the modified criteria to reduce the proportion of patients assigned to the IE possible category. This approach was used in a series of patients ini-tially categorized as possible IE by the original Duke criteria. Table 2. Definition of IE According to the Modified Duke Criteria Definite IE Pathological criteria Microorganisms demonstrated by culture or histological examination of a vegetation, a vegetation that has embolized, or an intracardiac abscess specimen; or pathological lesions; vegetation or intracardiac abscess confirmed by histological examination showing active endocarditis Clinical criteria 2 Major criteria, 1 major criterion and 3 minor criteria, or 5 minor criteria Possible IE 1 Major criterion and 1 minor criterion, or 3 minor criteria Rejected Firm alternative diagnosis explaining evidence of IE; or resolution of IE syndrome with antibiotic therapy for ≤4 d; or no pathological evidence of IE at surgery or autopsy with antibiotic therapy for ≤4 d; or does not meet criteria for possible IE as above IE indicates infective endocarditis.
Modifications appear in boldface.
These criteria have been universally accepted and are in current use.
Reprinted from Li et al24 by permission of the Infectious Diseases Society of America. Copyright © 2000, the Infectious Diseases Society of America.
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1439 With the guidance of the “diagnostic floor,” a number of these cases were reclassified as rejected for IE.24 Follow-up in these reclassified patients documented the specificity of this diagnostic schema because no patients developed IE during the subsequent 12 weeks of observation.
Thus, on the basis of the weight of clinical evidence involving nearly 2000 patients in the current literature, it appears that patients suspected of having IE should be clini-cally evaluated, with the modified Duke criteria as the primary diagnostic schema. It should be pointed out that the Duke cri-teria were originally developed to facilitate epidemiological and clinical research efforts so that investigators could com-pare and contrast the clinical features and outcomes of various case series of patients. Extending these criteria to the clinical practice setting has been somewhat more difficult. It should also be emphasized that full application of the Duke criteria requires detailed clinical, microbiological, radiological, and echocardiographic queries. Because IE is a heterogeneous disease with highly variable clinical presentations, the use of these criteria alone will never suffice. Criteria changes that add sensitivity often do so at the expense of specificity and vice versa. The Duke criteria are meant to be a guide for diag-nosing IE and must not replace clinical judgment. Clinicians may appropriately and wisely decide whether or not to treat an individual patient, regardless of whether the patient meets or fails to meet the criteria for definite or possible IE by the Duke criteria. We believe, however, that the modifications of the Duke criteria (Tables 2 and 3) will help investigators who wish to examine the clinical and epidemiological features of IE and will serve as a guide for clinicians struggling with dif-ficult diagnostic problems. These modifications require fur-ther validation among patients who are hospitalized in both community-based and tertiary care hospitals, with particular attention to longer-term follow-up of patients rejected as hav-ing IE because they did not meet the minimal floor criteria for possible IE.
The diagnosis of IE must be made as soon as possible to initiate appropriate empirical antibiotic therapy and to iden-tify patients at high risk for complications who may be best managed by early surgery. In cases with a high suspicion of IE based on either the clinical picture or the patient’s risk fac-tor profile such as injection drug use, another focus of car-diovascular infection, including catheter-related bloodstream infections caused by S aureus, or a history of previous IE, the presumption of IE often is made before blood culture results are available. Identification of vegetations and incremental valvular insufficiency with echocardiography often com-pletes the diagnostic criteria for IE and affects the duration of therapy. Although the use of case definitions to establish a diagnosis of IE should not replace clinical judgment,46 the recently modified Duke criteria24 have been useful in both epidemiological and clinical trials and in individual patient management. Clinical, echocardiographic, and microbiologi-cal criteria (Tables 2 and 3) are used routinely to support a diagnosis of IE, and they do not rely on histopathological confirmation of resected valvular material or arterial embolus. If suggestive features are absent, then a negative echocardio-gram should prompt a more thorough search for alternative sources of fever and sepsis. In light of these important func-tions, at least 3 sets of blood cultures obtained from separate venipuncture sites should be obtained, with the first and last samples drawn at least 1 hour apart. In addition, echocardiog-raphy should be performed expeditiously in patients suspected of having IE.
Recommendations 1. At least 3 sets of blood cultures obtained from dif-ferent venipuncture sites should be obtained, with the first and last samples drawn at least 1 hour apart (Class I; Level of Evidence A).
2. Echocardiography should be performed expedi-tiously in patients suspected of having IE (Class I; Level of Evidence A).
Table 3. Definition of Terms Used in the Modified Duke Criteria for the Diagnosis of IE Major criteria Blood culture positive for IE Typical microorganisms consistent with IE from 2 separate blood cultures: Viridans streptococci, Streptococcus bovis, HACEK group, Staphylococcus aureus; or community-acquired enterococci in the absence of a primary focus, or microorganisms consistent with IE from persistently positive blood cultures defined as follows: at least 2 positive cultures of blood samples drawn >12 h apart or all 3 or a majority of ≥4 separate cultures of blood (with first and last sample drawn at least 1 h apart) Single positive blood culture for Coxiella burnetii or anti–phase 1 IgG antibody titer ≥1:800 Evidence of endocardial involvement Echocardiogram positive for IE (TEE recommended for patients with prosthetic valves, rated at least possible IE by clinical criteria, or complicated IE [paravalvular abscess]; TTE as first test in other patients) defined as follows: oscillating intracardiac mass on valve or supporting structures, in the path of regurgitant jets, or on implanted material in the absence of an alternative anatomic explanation; abscess; or new partial dehiscence of prosthetic valve or new valvular regurgitation (worsening or changing or pre-existing murmur not sufficient) Minor criteria Predisposition, predisposing heart condition, or IDU Fever, temperature >38°C Vascular phenomena, major arterial emboli, septic pulmonary infarcts, mycotic aneurysm, intracranial hemorrhage, conjunctival hemorrhages, and Janeway lesions Immunological phenomena: glomerulonephritis, Osler nodes, Roth spots, and rheumatoid factor Microbiological evidence: positive blood culture but does not meet a major criterion as noted above (excludes single positive cultures for coagulase-negative staphylococci and organisms that do not cause endocarditis) or serological evidence of active infection with organism consistent with IE Echocardiographic minor criteria eliminated HACEK indicates Haemophilus species, Aggregatibacter species, Cardiobacterium hominis, Eikenella corrodens, and Kingella species; IDU, injection drug use; IE, infective endocarditis; IgG, immunoglobulin G; TEE transesophageal echocardiography; and TTE, transthoracic echocardiography.
Modifications appear in boldface.
These criteria have been universally accepted and are in current use. Reprinted from Li et al24 by permission of the Infectious Diseases Society of America. Copyright © 2000, the Infectious Diseases Society of America.
Downloaded from by on October 16, 2018 1440 Circulation October 13, 2015 Echocardiography Echocardiography is central to the diagnosis and management of patients with IE. As previously stated (Table 3), echocar-diographic evidence of an oscillating intracardiac mass or vegetation, an annular abscess, prosthetic valve partial dehis-cence, and new valvular regurgitation are major criteria in the diagnosis of IE.
Both TTE and TEE are done in many patients with IE dur-ing initial evaluation and subsequent follow-up and provide complementary information. Therefore, TTE should be done initially in all cases of suspected IE (Figure). If any circum-stances preclude the securing of optimal echocardiographic windows, including chronic obstructive lung disease, previous thoracic or cardiovascular surgery, morbid obesity, or other conditions, then TEE should be performed as soon as pos-sible after TTE. When TTE is negative and clinical suspicion remains low, then other clinical entities should be considered. If TTE shows vegetations but the likelihood of complications is low, then subsequent TEE is unlikely to alter initial medi-cal management. On the other hand, if clinical suspicion of IE or its complications is high (eg, prosthetic valve or new atrioventricular block), then a negative TTE will not definitely rule out IE or its potential complications, and TEE should be performed first. Investigation in adults has shown TEE to be significantly more sensitive than TTE for the detection of veg-etations and abscesses.47 In the setting of a prosthetic valve, transthoracic images are greatly hampered by the structural components of the prosthesis and are inadequate for assess-ment of the perivalvular area where those infections often start.48 Although cost-effectiveness calculations suggest that TEE should be the first examination in adults with suspected IE (Table 4), particularly in the setting of staphylococcal bac-teremia,49,50 many patients are not candidates for immediate TEE because of having eaten within the preceding 6 hours or because the patients are in institutions that cannot provide 24-hour TEE services. When TEE is not clinically possible or must be delayed, early TTE should be performed without delay. Although TTE will not definitively exclude vegeta-tions or abscesses, it will allow identification of very-high-risk patients, establish the diagnosis in many, and guide early treatment decisions. Although interesting results suggest that there may be a high negative predictive value of TTE in some patients,51 further work is needed to better define the subgroup of patients with bloodstream infection caused by S aureus who need only TTE to evaluate for IE.
Many findings identified by TEE also can be detected on TTE. Concurrent TTE images can serve as a baseline for rapid and noninvasive comparison of vegetation size, valvular insufficiency, or change in abscess cavities during the course of the patient’s treatment should clinical dete-rioration occur. For tricuspid vegetations or abnormalities of the right ventricular outflow tract, visualization may be enhanced by choosing TTE rather than TEE.52 Finally, many cardiologists believe TTE is superior to TEE for quantifying hemodynamic dysfunction manifested by valvular regurgi-tation, ventricular dysfunction, and elevated left and right ventricular filling pressures and pulmonary artery pressure. These echocardiographic findings can occur in patients who have no heart failure symptoms.
Both TEE and TTE may produce false-negative results if vegetations are small or have embolized.53 Even TEE may miss initial perivalvular abscesses, particularly when the study is performed early in the patient’s illness.54 In such cases, the Figure. An approach to the diagnostic use of echocardiography (echo). Rx indicates prescription; TEE, transesophageal echocardiography; and TTE, transthoracic echocardiography. For example, a patient with fever and a previously known heart murmur and no other stigmata of infective endocarditis (IE). †High initial patient risks include prosthetic heart valves, many congenital heart diseases, previous endocarditis, new murmur, heart failure, or other stigmata of endocarditis. ‡High-risk echocardiographic features include large or mobile vegetations, valvular insufficiency, suggestion of perivalvular extension, or secondary ventricular dysfunction (see text). Modified from Baddour et al.12 Copyright © 2005, American Heart Association, Inc.
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1441 incipient abscess may be seen only as nonspecific perivalvu-lar thickening, which on repeat imaging across several days may become more recognizable as it expands and develops a cavity. Similarly, perivalvular fistulas and pseudoaneurysms develop over time, and negative early TEE images do not exclude the potential for their development.
False-positive results from TEE or TTE studies may occur when valvular abnormalities are seen that may not be related to a current infection. Previous scarring, severe myxomatous change, and even normal structures such as Lambl excres-cences may be indistinguishable from active changes in the valves. As echocardiographic technology improves with higher frequencies and refined beam-forming technology, subtle findings continue to be recognized and may add to the category of indeterminate findings. One approach to minimiz-ing confusion from these latter structures is to exploit the high frame rates that are often available with current equipment to improve temporal resolution and to clearly visualize rap-idly moving structures such as microcavities from prosthetic valves or fibrillar components.
Several echocardiographic features identify patients at high risk for a complicated course or with a need for surgery (Table 5). These features include large (>10 mm in diameter) vegetations, severe valvular insufficiency, abscess cavities or pseudoaneurysms, valvular perforation or dehiscence, and evidence of decompensated heart failure.21 The ability of echo-cardiographic features to predict embolic events is limited.55–57 The greatest risk of embolic complications appears to occur with large (≥10 mm) vegetations on the anterior mitral leaf-let.58 Vegetation size and mobility may be taken into account, along with bacteriological factors and other indications for surgery, when considering early surgery to avoid emboliza-tion, although mobility characteristics alone should not be the principal driver as a surgical indication.59 Recommendation 1. TTE should be performed in all cases of suspected IE (Class I; Level of Evidence B).
Repeat Echocardiography If the initial TTE images are negative and the diagnosis of IE is still being considered, then TEE should be performed as soon as possible (Table 4). Among patients with an initially positive TTE and a high risk for intracardiac complications, including perivalvular extension of infection, TEE should be obtained as soon as possible. Repeating the TEE in 3 to 5 days (or sooner if clinical findings change) after an initial negative result is recommended when clinical suspicion of IE persists.60 In some cases, vegetations may reach a detectable size in the interval, or abscess cavities or fistulous tracts may become evident. An interval increase in vegetation size on serial echocardiography despite the administration of appro-priate antibiotic therapy has serious implications and has been associated with an increased risk of complications and the need for surgery.60 Repeat TEE should be done when a patient with an initially positive TEE develops worrisome clinical features during antibiotic therapy. These features, including unexplained progression of heart failure symptoms, change in cardiac murmurs, and new atrioventricular block or arrhyth-mia, should prompt emergent evaluation by TEE if possible.
Recommendations 1. TEE should be done if initial TTE images are nega-tive or inadequate in patients for whom there is an ongoing suspicion for IE or when there is concern for intracardiac complications in patients with an initial positive TTE (Class I; Level of Evidence B).
2. If there is a high suspicion of IE despite an initial negative TEE, then a repeat TEE is recommended in 3 to 5 days or sooner if clinical findings change (Class I; Level of Evidence B).
3. Repeat TEE should be done after an initially posi-tive TEE if clinical features suggest a new develop-ment of intracardiac complications (Class I; Level of Evidence B).
Intraoperative Echocardiography Preoperative surgical planning for patients with IE will ben-efit from echocardiographic delineation of the mechanisms of valvular dysfunction or regions of myocardial abscess forma-tion (Table 5). The use of aortic homografts is facilitated by preoperative estimates of annular size, which allow the selec-tion of appropriately sized donor tissues.61,62 Intraoperatively, echocardiographic goals include assessment of not only Table 4. Use of Echocardiography During Diagnosis and Treatment of Endocarditis Early Echocardiography as soon as possible (<12 h after initial evaluation) TEE preferred; obtain TTE views of any abnormal findings for later comparison TTE if TEE is not immediately available TTE may be sufficient in small children Repeat echocardiography TEE after positive TTE as soon as possible in patients at high risk for complications TEE 3–5 d after initial TEE if suspicion exists without diagnosis of IE or with worrisome clinical course during early treatment of IE Intraoperative Prepump Identification of vegetations, mechanism of regurgitation, abscesses, fistulas, and pseudoaneurysms Postpump Confirmation of successful repair of abnormal findings Assessment of residual valve dysfunction Elevated afterload if necessary to avoid underestimating valve insufficiency or presence of residual abnormal flow Completion of therapy Establish new baseline for valve function and morphology and ventricular size and function TTE usually adequate; TEE or review of intraoperative TEE may be needed for complex anatomy to establish new baseline TEE indicates transesophageal echocardiography; and TTE, transthoracic echocardiography.
Downloaded from by on October 16, 2018 1442 Circulation October 13, 2015 the obviously dysfunctional valve but also the other valves and contiguous structures. Post– cardiopulmonary bypass images should confirm the adequacy of the repair or replace-ment and document the successful closure of fistulous tracts. Perivalvular leaks related to technical factors should be docu-mented to avoid later confusion about whether such leaks are the result of recurrent infection. During postpump imaging, it is often necessary to augment afterload to reach representative ambulatory levels to avoid underestimation of regurgitant jet size and significance and to ensure that abnormal communi-cations were closed.63 Afterload augmentation, however, may not mimic actual “awake physiology” and may still lead occa-sionally to an inaccurate evaluation of the awake postopera-tive hemodynamic state.
Echocardiography at the Completion of Therapy All patients who have experienced an episode of IE remain at increased risk for recurrent infection indefinitely. Many believe that it is extremely important for the future care of these patients to establish a new baseline for valvular morphology, including the presence of vegetations and valvular insufficiency, once treatment has been completed. Documentation of heart rate, heart rhythm, and blood pres-sure at the time of echocardiographic study is important because changes in these conditions may explain future differences in valvular insufficiency independent of pathol-ogy (Table 4). TTE is reasonable for this evaluation because spectral Doppler interrogation for functionality metrics is more thorough than TEE. TEE, however, may be mer-ited to define the new baseline in some patients with poor acoustic windows or complicated anatomy such as after extensive debridement and reconstruction. Although intra-operative postpump TEE views may be adequate for this new baseline, they should be reviewed for adequacy and repeated if necessary. Some patients will have significant valvular dysfunction at the end of otherwise successful antimicrobial treatment that will require eventual valvular surgery. Posttreatment echocardiography can guide both medical management and the discussion of the appropriate timing of such interventions.
Recommendation 1. TTE at the time of antimicrobial therapy comple-tion to establish baseline features is reasonable (Class IIa; Level of Evidence C).
3D Echocardiography and Other Imaging Modalities Although newer imaging modalities are undergoing pre-liminary evaluation, echocardiography will continue to be pivotal in patients with IE for the foreseeable future. In this regard, early investigations64,65 of 3D TEE have demonstrated advantages over 2-dimensional TEE (which is routinely used) to better detect and delineate vegetations and to identify IE complications and their relationships with surrounding struc-tures. Unfortunately, the lower temporal and lateral resolu-tion with 3D echocardiography compared with 2-dimensional echocardiography leads to an overestimation of vegetation size and technically challenging visualization of fast-moving structures.
Although cardiac CT is used principally to evaluate great vessels and coronary artery disease, there may be a role for this tool66–68 in cases of IE in which definitive evidence of IE and its complications is not secured with TEE. Moreover, coronary CT angiography can provide coronary artery evalu-ation in patients who are to undergo cardiac surgery for IE complications. In addition, this methodology may be useful in head-to-toe preoperative screening, including evaluation for central nervous system (CNS) lesions, and in intra-abdominal lesions (eg, silent splenic abscesses). Limitations include the associated exposure to radiation, nephrotoxicity associated with contrast dye, and relative lack of sensitivity in 1 study to demonstrate valve perforations.67 MRI has had a major impact on IE diagnosis and manage-ment, especially as a tool to detect cerebral embolic events, many of which are clinically silent.69 Indications for the rou-tine use of MRI and magnetic resonance angiography in IE management, however, are not well established. Comments related to mycotic or infectious aneurysms are provided in a later section of this document.
More study is needed to define the utility of 18F-fluoro deoxyglucose positron emission tomography/CT in the diag-nosis and management of IE. In a prospective study of 25 IE cases, 18F-fluorodeoxyglucose positron emission tomography/ CT was useful in identifying peripheral embolization in 11 patients and in detecting IE extracardiac manifestations in 7 patients who did not demonstrate any clinical manifestations of IE.70 The use of multimodality imaging in IE may increase in the future as the risks and benefits of each diagnostic tool are defined.71 Table 5. Clinical and Echocardiographic Features That Suggest Potential Need for Surgical Intervention Vegetation Persistent vegetation after systemic embolization Anterior mitral leaflet vegetation, particularly with size >10 mm ≥1 Embolic events during first 2 wk of antimicrobial therapy Increase in vegetation size despite appropriate antimicrobial therapy† Valvular dysfunction Acute aortic or mitral insufficiency with signs of ventricular failure† Heart failure unresponsive to medical therapy† Valve perforation or rupture† Perivalvular extension Valvular dehiscence, rupture, or fistula† New heart block†‡ Large abscess or extension of abscess despite appropriate antimicrobial therapy† See text for a more complete discussion of indications for surgery based on vegetation characterizations.
Surgery may be required because of risk of embolization.
†Surgery may be required because of heart failure or failure of medical therapy.
‡Echocardiography should not be the primary modality used to detect or monitor heart block.
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1443 Antimicrobial Therapy Therapeutic Principles The primary goal of antibiotic treatment is to eradicate infection, including sterilizing vegetations, although the unique character-istics of infected vegetations can pose a variety of challenges. These characteristics include focal infection with high bacterial density, slow rate of bacterial growth within biofilms, and low microorganism metabolic activity.72 Host characteristics such as impaired immunity also contribute to challenges in thera-peutics. In addition, antibiotics may fail to eradicate infection as a result of increased binding of the drug to serum proteins, perturbations of antibiotic penetration into the vegetation, and unique antibiotic pharmacokinetic/pharmacodynamic (PK/PD) features. Therefore, prolonged, parenteral, bactericidal therapy is required for attempted infection cure.
Inoculum Effect The effect of high bacterial densities on antimicrobial activ-ity is called the inoculum effect in which certain groups of antimicrobials commonly used to treat IE such as β-lactams and glycopeptides (and, to a lesser extent, lipopeptides such as daptomycin) are less active against highly dense bacterial populations.73–75 Therefore, the effective mini-mum inhibitory concentration (MIC) at the site of infection with bacterial densities of 108 to 1011 colony-forming units per 1 g tissue can be much higher than anticipated by in vitro susceptibility tests that use a standard inoculum (105.5 colony-forming units per milliliter). In addition, bacteria that are otherwise killed at low densities by bactericidal antibiotics such as penicillins can be relatively resistant to or tolerant of their bactericidal effect in dense populations. An inoculum effect has been demonstrated with penicillin versus streptococci in both in vitro and animal models. For example, the curative dose of penicillin for streptococcal infections in animal models has been shown to increase markedly with the number of organisms inoculated and the duration of the infection, presumably because of the interim increase in the number of organisms in the infected host.76 In addition, the stationary growth-phase conditions make it less likely that bacterial cell wall–active antibiotics (β-lactams and glycopeptides) are optimally effective.77–79 Stationary-phase organisms have been associated with a loss of penicillin-binding proteins that are the active tar-get sites required for β-lactam antibacterial activity. This loss of penicillin-binding proteins during stationary-phase growth may be responsible in part for the inoculum effect observed in vivo and may account for the failure of penicil-lin in both experimental and human cases of severe strep-tococcal infections.80 Importantly, fluoroquinolones and aminoglycoside antibiotics are less affected by the size of the inoculum because of their different mechanisms of bac-tericidal activity.81,82 An inoculum effect also occurs with β-lactamase–susceptible β-lactam antibiotics versus β-lactamase–producing bacteria, presumably because more β-lactamase is present in denser β-lactamase–producing bacterial populations, as observed in vitro with some enterococci,83 S aureus,84 and Gram-negative bacilli85; in animal models of experimental IE86,87; and clinically.88 High inocula are also more likely to have antibiotic-resis-tant subpopulations that can emerge in the setting of antibiotic therapy. For example, in an in vitro PD model, the activity of vancomycin against heterogeneous vancomycin-intermediate S aureus (hVISA) and non-hVISA isolates was reduced in the presence of a high inoculum amount (108 colony-forming units per milliliter).75 Bactericidal Drugs Data from animal models of IE and clinical investigations support the need for bactericidal antibiotics to sterilize veg-etations in IE with high bacterial densities.89 For enterococci, bactericidal activity can be achieved by the combination of certain β-lactam antibiotics (eg, penicillin, ampicillin, and piperacillin) with an aminoglycoside. The bactericidal effect achieved by a combination of antibacterial drugs that alone only inhibit bacterial growth is called synergy. The rate of bactericidal activity against some other organisms can also be enhanced by a combination of a β-lactam antibiotic plus an aminoglycoside.
Duration of Antimicrobial Therapy The duration of therapy in IE must be sufficient to ensure complete eradication of microorganisms within vegetations. Prolonged therapy is necessary because of the high bacterial densities within vegetations and the relatively slow bacteri-cidal activity of some antibiotics such as β-lactams and van-comycin. When the bactericidal activity is known to be more rapid or the likely vegetation bacterial burden is lower, then the clinician may prescribe a shorter duration of antimicro-bial therapy in unique instances. Combination therapy with penicillin or ceftriaxone and an aminoglycoside for 2 weeks is highly effective in viridans group streptococci (VGS) IE90 in very select patients with uncomplicated infection. Both β-lactam therapy alone and combination therapy with nafcil-lin and an aminoglycoside for only 2 weeks have been effec-tive in patients with uncomplicated right-sided IE caused by S aureus91; monotherapy with a β-lactam would be selected for use in cases of uncomplicated IE.92 Of interest, right-sided vegetations tend to have lower bacterial densities, which may result from host defense mech-anisms, including polymorphonuclear activity or platelet-derived antibacterial cationic peptides.90,91,93 Drug Penetration The penetration of antibiotics is a significant issue in the treatment of IE because cardiac vegetations, which are com-posed of layers of fibrin and platelets, pose a considerable mechanical barrier between the antibiotic and the embedded targeted microorganisms.94,95 The efficacy of antimicrobial drugs varies, depending on the degree of penetration into the vegetation, pattern of distribution within the vegetation, and vegetation size.96,97 Patterns of diffusion differ by class of anti-biotic, which may have implications for therapeutic outcomes in patients being treated for IE.98–100 PK/PD and Dosing Implications in IE In the design of dose regimens for the treatment of IE, it is important to fully optimize the PK/PD parameter for the selected antibiotic to increase the likelihood of success Downloaded from by on October 16, 2018 1444 Circulation October 13, 2015 and to decrease the potential for developing resistance.101 Antibiotic PK/PD is related to both PK and microorganism susceptibility to the drug.102 With the use of in vitro and in vivo evaluations, antibiotics are categorized on the basis of whether they possess concentration-dependent or time-dependent effects on microorganisms and on the basis of 4 common PK/PD parameters that predict antibiotic efficacy: the ratio of the maximum serum concentration to the MIC, the ratio of the area under the 24-hour plasma concentra-tion-time curve to the MIC (AUC24/MIC), the duration of time that the serum concentration exceeds the MIC, and the duration of the postantibiotic effect.101,103 More detailed discussion of the calculation of these parameters has been given previously.100 Whereas both the ratio of maximum serum concentra-tion to MIC and the AUC24/MIC ratio have been shown to predict efficacy as the optimized PD parameters for ami-noglycoside, fluoroquinolone, and daptomycin therapy, the AUC24/MIC is the optimized PD activity for glycopeptides such as vancomycin, teicoplanin, telavancin, oritavancin, and lipopeptides such as daptomycin. β-Lactam efficacy, in contrast, is best predicted by the percent duration of time that the serum concentration exceeds the MIC.102 For peni-cillins and cephalosporins to achieve a bacteriostatic effect in a murine model, the time the free drug must exceed the MIC is 35% to 40% of the dosing interval, whereas a bac-tericidal response requires 60% to 70% of the dosing inter-val.104 Two retrospective studies examined the continuous infusion of 2 β-lactams (cefazolin and oxacillin) for meth-icillin-sensitive S aureus (MSSA) infections, including IE, with results supporting continuous infusion of these drugs. More study is needed, however, before a strong recommen-dation can be made.105,106 For concentration-dependent antibiotics such as amino-glycosides and fluoroquinolones, a ratio of maximum serum concentration to MIC of >10 was associated with improved efficacy in patients with Gram-negative pneumonia, whereas an AUC24/MIC >125 was associated with an improved clini-cal efficacy for ciprofloxacin against infections caused by Pseudomonas aeruginosa.107,108 Liu et al109 demonstrated that the minimal AUC24/MIC requirement for daptomycin with an 80% kill efficacy in a S aureus infection mouse model was ≈250, which would be easily achieved by the recommended dose of 6 mg·kg−1·d−1 for complicated bacteremia, including right-sided IE.
Some experts have recommended daptomycin doses of 8 to 10 mg·kg−1·d−1 for the treatment of complicated methicillin-resistant S aureus (MRSA) bacteremia, particularly IE. This recommendation is based on the concentration-dependent properties of daptomycin, improved efficacy for infections caused by organisms with reduced susceptibility to dapto-mycin, and an attempt to reduce the emergence of resistance to daptomycin after vancomycin therapy.110 The evidence for these recommendations has come largely from in vitro PK/PD models using high-inoculum–simulated endocardial vegeta-tions with S aureus111 and enterococci and from animal mod-els of IE.112 With regard to vancomycin, an AUC24/MIC ≥400 is rec-ommended as the targeted PK/PD parameter for patients with serious S aureus infections.112 In an evaluation of 320 MRSA patients with complicated bacteremia, including IE, Kullar et al113 demonstrated that an AUC24/MIC >421 was significantly associated with improved patient outcomes. This AUC24/MIC ratio was associated with trough serum concentrations >15 mg/L, attainable if the vancomycin MIC was <1 mg/L.
Antimicrobial Treatment Perspectives In many cases, the initial therapy of IE is empirical; typi-cally, results of blood cultures are monitored for hours to days until a pathogen is identified. During this time, empiri-cal antimicrobial therapy is administered with the expectation that the regimen will be revised once a pathogen is defined and susceptibility results are obtained. The selection of an optimal empiric regimen is usually broad and is based on fac-tors that relate to patient characteristics, prior antimicrobial exposures and microbiological findings, and epidemiological features. Therefore, infectious diseases consultation should occur at the time of empirical therapy initiation to help define a regimen114,115 because the selection of a regimen is highly variable. In this regard, please refer the Culture-Negative Endocarditis section of this statement and the related Table 6 for additional details.
Results of clinical efficacy studies support the use of most treatment regimens described in these guidelines. Other recommendations listed in this section are based largely on in vitro data and consensus opinion and include the following management considerations. It is reasonable for the counting of days for the duration of therapy to begin on the first day on which blood cultures are negative in cases in which blood cultures were initially positive. It is reasonable to obtain 2 sets of blood cultures every 24 to 48 hours until bloodstream infection is cleared. However, if a patient undergoes valve surgery and the resected valve tissue is culture positive or a perivalvular abscess is found, then an entire course of antimicrobial therapy is reasonable after valve surgery. If the resected tissue is culture negative, then it may be reasonable for the duration of postoperative treatment given less the number of days of treatment admin-istered for native valve infection before valve replacement. This, however, has been challenged by retrospectively col-lected data from 2 different medical centers116,117 that sug-gest that 2 weeks of antibiotic therapy may be sufficient in patients who undergo valve surgery and have negative valve tissue cultures, particularly in IE cases caused by VGS or Streptococcus gallolyticus (bovis). Whether a 2-week treatment course would be sufficient after valve sur-gery in patients with positive valve cultures either was not addressed in 1 survey116 or included only 5 patients in the other.117 Histopathological evidence of bacteria with valve tissue Gram staining in patients with negative tissue cul-tures can represent killed organisms and is not a factor in defining the length of therapy after valve surgery.110 For patients with NVE who undergo valve resection with prosthetic valve replacement or repair with an annuloplasty ring, there is a lack of consensus as to whether the postopera-tive treatment regimen should be one that is recommended for prosthetic valve treatment rather than one that is recommended Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1445 for native valve treatment. In regimens that contain combi-nation antimicrobial therapy, it is reasonable to administer agents at the same time or temporally close together to maxi-mize the synergistic killing effect on an infecting pathogen.
Recommendations 1. Infectious diseases consultation should be obtained to define an optimal empirical treatment regimen at the time of initiation of antimicrobial therapy (Class I; Level of Evidence B).
2. It is reasonable that the counting of days for the duration of antimicrobial therapy begin on the first day on which blood cultures are negative in cases in which blood cultures were initially positive (Class IIa; Level of Evidence C).
3. It is reasonable to obtain at least 2 sets of blood cul-tures every 24 to 48 hours until bloodstream infec-tion has cleared (Class IIa; Level of Evidence C).
4. If operative tissue cultures are positive, then an entire antimicrobial course is reasonable after valve surgery (Class IIa; Level of Evidence B).
5. If operative tissue cultures are negative, it may be reasonable to count the number of days of anti-microbial therapy administered before surgery in the overall duration of therapy (Class IIb; Level of Evidence C).
6. It is reasonable to time the administration of antimi-crobial therapy at the same time or temporally close together for regimens that include >1 antimicrobial agent (Class IIa; Level of Evidence C).
Dog or cat exposure Bartonella sp Pasteurella sp Capnocytophaga sp Contact with contaminated milk or infected farm animals Brucella sp Coxiella burnetii Erysipelothrix sp Homeless, body lice Bartonella sp AIDS Salmonella sp S pneumoniae S aureus Pneumonia, meningitis S pneumoniae Solid organ transplantation S aureus Aspergillus fumigatus Enterococcus sp Candida sp Gastrointestinal lesions S gallolyticus (bovis) Enterococcus sp Clostridium septicum HACEK indicates Haemophilus species, Aggregatibacter species, Cardiobacterium hominis, Eikenella corrodens, and Kingella species; IDU, injection drug use; and VGS, viridans group streptococci.
Table 6. Epidemiological Clues That May be Helpful in Defining the Etiological Diagnosis of Culture-Negative Endocarditis Epidemiological Feature Common Microorganism IDU S aureus, including community-acquired oxacillin-resistant strains Coagulase-negative staphylococci β-Hemolytic streptococci Fungi Aerobic Gram-negative bacilli, including Pseudomonas aeruginosa Polymicrobial Indwelling cardiovascular medical devices S aureus Coagulase-negative staphylococci Fungi Aerobic Gram-negative bacilli Corynebacterium sp Genitourinary disorders, infection, and manipulation, including pregnancy, delivery, and abortion Enterococcus sp Group B streptococci (S agalactiae) Listeria monocytogenes Aerobic Gram-negative bacilli Neisseria gonorrhoeae Chronic skin disorders, including recurrent infections S aureus β-Hemolytic streptococci Poor dental health, dental procedures VGS Nutritionally variant streptococci Abiotrophia defectiva Granulicatella sp Gemella sp HACEK organisms Alcoholism, cirrhosis Bartonella sp Aeromonas sp Listeria sp S pneumoniae β-Hemolytic streptococci Burn S aureus Aerobic Gram-negative bacilli, including P aeruginosa Fungi Diabetes mellitus S aureus β-Hemolytic streptococci S pneumoniae Early (≤1 y) prosthetic valve placement Coagulase-negative staphylococci S aureus Aerobic Gram-negative bacilli Fungi Corynebacterium sp Legionella sp Late (>1 y) prosthetic valve placement Coagulase-negative staphylococci S aureus Viridans group streptococci Enterococcus species Fungi Corynebacterium sp (Continued ) Table 6. Continued Epidemiological Feature Common Microorganism Downloaded from by on October 16, 2018 1446 Circulation October 13, 2015 Overview of VGS, Streptococcus gallolyticus (Formerly Known as Streptococcus bovis), Abiotrophia defectiva, and Granulicatella Species VGS are common pathogenic agents in community-acquired NVE in patients who are not IDUs. The taxonomy of VGS is evolving. The species that most commonly cause IE are S sanguis, S oralis (mitis), S salivarius, S mutans, and Gemella morbillorum (formerly called S morbillorum). Members of the S anginosus group (S intermedius, angi-nosus, and constellatus) also have been referred to as the S milleri group, and this has caused some confusion. In contrast to other α-hemolytic streptococcal species, the S anginosus group tends to form abscesses and to cause hematogenously disseminated infection (eg, myocardial and visceral abscesses, septic arthritis, and vertebral osteomy-elitis). In addition, although the S anginosus group usually is sensitive to penicillin, some strains may exhibit variable penicillin resistance. The recommendations that follow are intended to assist clinicians in selecting appropriate antimi-crobial therapy for patients with IE caused by VGS and S gallolyticus (bovis, a nonenterococcal penicillin-susceptible group D Streptococcus). S gallolyticus (bovis) expresses the group D antigen, but it can be distinguished from group D Enterococcus by appropriate biochemical tests. Patients with either S gallolyticus (bovis) bacteremia or IE should undergo a colonoscopy to determine whether malignancy or other mucosal lesions are present.
Certain VGS have biological characteristics that may com-plicate diagnosis and therapy. A defectiva and Granulicatella species (G elegans, G adiacens, G paraadiacens, and G balaenopterae), formerly known as nutritionally variant strep-tococci, are detected by automated blood culture systems but may yield pleomorphic forms by Gram stain and will not grow on subculture unless chocolate agar or other media supple-mented with pyridoxal or cysteine is used.
Treatment regimens outlined for VGS, A. defectiva, and Granulicatella species are subdivided into categories based on penicillin MIC data.
Native Valve Highly Penicillin-Susceptible VGS and S gallolyticus (bovis) (MIC ≤0.12 µg/mL) Bacteriological cure rates ≥ 98% may be anticipated in patients who complete 4 weeks of therapy with parenteral penicillin or ceftriaxone for IE caused by highly penicillin-susceptible VGS or S gallolyticus (bovis)118,119 (Table 7). Ampicillin is a reasonable alternative to penicillin and has been used when penicillin is not available because of supply deficiencies.
The addition of gentamicin sulfate to penicillin exerts a synergistic killing effect in vitro on VGS and S gallolyticus (bovis). The combination of penicillin or ceftriaxone with gentamicin results in synergistic killing in animal models of VGS or S gallolyticus (bovis) experimental IE. In selected patients, treatment with a 2-week regimen with either penicil-lin or ceftriaxone combined with an aminoglycoside resulted in cure rates that are similar to those after monotherapy with penicillin or ceftriaxone administered for 4 weeks.83,120 Studies performed in Europe, South America, and the United States demonstrated that the combination of once-daily ceftriaxone with either netilmicin or gentamicin administered once daily was equivalent in efficacy to 2 weeks of therapy with peni-cillin with an aminoglycoside administered in daily divided doses.83,120 The 2-week regimen of penicillin or ceftriaxone combined with single daily-dose gentamicin is reasonable for uncomplicated cases of IE caused by highly penicillin-suscep-tible VGS or S gallolyticus (bovis) in patients at low risk for adverse events caused by gentamicin therapy (Table 7). This 2-week regimen is not recommended for patients with known extracardiac infection or those with a creatinine clearance of <20 mL/min.
Although the two, 4-week ß-lactam–containing regi-mens shown in Table 7 produce similar outcomes, each regi-men has advantages and disadvantages. Monotherapy with either penicillin or ceftriaxone for 4 weeks avoids the use of gentamicin, which is potentially ototoxic and nephrotoxic. Compared with penicillin, the advantage of once-daily cef-triaxone is its simplicity for use in therapy administered to outpatients.118,121 Both penicillin and ceftriaxone are overall well tolerated but, like all antimicrobials, have the potential for causing adverse drug events; some of the more common ones include rash, fever, diarrhea, and neutropenia. Liver function abnormalities can be seen with ceftriaxone use and are sometimes associated with “sludging” of drug in the gallbladder.122 For patients who are unable to tolerate penicillin or ceftriax-one, vancomycin is a reasonably effective alternative. Prolonged intravenous use of vancomycin may be complicated by throm-bophlebitis, rash, fever, neutropenia, and rarely ototoxic reac-tions. The likelihood of “red man” syndrome is reduced with an infusion of vancomycin over ≥1 hour. Desired trough vancomy-cin levels should range between 10 and 15 µg/mL.
Recommendations 1. Both aqueous crystalline penicillin G and ceftriax-one are reasonable options for a 4-week treatment duration (Class IIa; Level of Evidence B).
2. A 2-week treatment regimen that includes gentami-cin is reasonable in patients with uncomplicated IE, rapid response to therapy, and no underlying renal disease (Class IIa; Level of Evidence B).
3. Vancomycin for a 4-week treatment duration is a reasonable alternative in patients who cannot tol-erate penicillin or ceftriaxone therapy (Class IIa; Level of Evidence B).
4. The desired trough vancomycin level should range between 10 and 15 µg/mL (Class I; Level of Evidence C).
Relatively Penicillin-Resistant VGS and S gallolyticus (bovis) (MIC >0.12–<0.5 µg/mL) Penicillin resistance in vitro occurs among some strains of VGS and S gallolyticus (bovis). To date, however, the num-ber of IE cases that have been reported as a result of VGS or S gallolyticus (bovis) strains that harbor any degree of penicil-lin resistance is small.123–126 Therefore, it is difficult to define the optimal treatment strategies for this group of patients. Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1447 Table 8 shows regimens for treatment of NVE caused by rel-atively penicillin-resistant strains (MIC >0.12–<0.5 µg/mL). For patients with VGS or S gallolyticus (bovis) IE caused by these relatively resistant strains, it is reasonable to admin-ister penicillin for 4 weeks, together with single daily-dose gentamicin for the first 2 weeks of treatment. Ampicillin is a reasonable alternative to penicillin if shortages of penicillin exist.
If the isolate is ceftriaxone susceptible, then ceftriaxone alone may be considered (Class IIb; Level of Evidence C). Vancomycin alone may be a reasonable alternative if the patient is intolerant of β-lactam therapy (Class IIb; Level of Evidence C). Consultation with an infectious diseases special-ist is encouraged in both of these scenarios.
Recommendations 1. It is reasonable to administer penicillin for 4 weeks with single daily-dose gentamicin for the first 2 weeks of therapy (Class IIa; Level of Evidence B).
2. If the isolate is ceftriaxone susceptible, then ceftri-axone alone may be considered (Class IIb; Level of Evidence C).
3. Vancomycin alone may be a reasonable alternative in patients who are intolerant of β-lactam therapy (Class IIa; Level of Evidence C).
A defectiva and Granulicatella Species and VGS With a Penicillin MIC ≥0.5 µg/mL The determination of antimicrobial susceptibilities of A defec-tiva and Granulicatella species (both formerly known as nutri-tionally variant streptococci) is often technically difficult, and the results may not be accurate. Moreover, IE caused by these microorganisms is uncommon and has been more difficult to cure microbiologically compared with IE caused by a strain of non–nutritionally variant VGS.127 For these reasons, in patients with IE caused by A defectiva and Granulicatella species, it is reasonable to administer a combination regimen that includes ampicillin (12 g/d in divided doses) or penicillin (18–30 mil-lion U/D in divided doses or by continuous infusion) plus gen-tamicin (3 mg·kg−1·d−1 in 2–3 divided doses) with infectious diseases consultation to determine length of therapy. Findings from an animal model of experimental endocarditis suggest that if vancomycin is chosen for use in patients intolerant of penicillin or ampicillin, then the addition of gentamicin is not needed.128 Ceftriaxone combined with gentamicin may be a reasonable alternative treatment option125,126 for VGS isolates that are susceptible to ceftriaxone on the basis of the Clinical and Laboratory Standards Institute definition and are resistant to penicillin (MIC ≥0.5 µg/mL, as defined in this statement). Currently, there is no reported clinical experience with the combination of ampicillin plus ceftriaxone for IE caused by these organisms.
Table 7. Therapy of NVE Caused by Highly Penicillin-Susceptible VGS and Streptococcus gallolyticus (bovis) Regimen Dose and Route Duration, wk Strength of Recommendation Comments Aqueous crystalline penicillin G sodium 12–18 million U/24 h IV either continuously or in 4 or 6 equally divided doses 4 Class IIa; Level of Evidence B Preferred in most patients >65 y or patients with impairment of eighth cranial nerve function or renal function.
Ampicillin 2 g IV every 4 h is a reasonable alternative to penicillin if a penicillin shortage exists. Or Ceftriaxone sodium 2 g/24 h IV/IM in 1 dose 4 Class IIa; Level of Evidence B Aqueous crystalline penicillin G sodium 12–18 million U/24 h IV either continuously or in 6 equally divided doses 2 Class IIa; Level of Evidence B 2-wk regimen not intended for patients with known cardiac or extracardiac abscess or for those with creatinine clearance of <20 mL/min, impaired eighth cranial nerve function, or Abiotrophia, Granulicatella, or Gemella spp infection; gentamicin dose should be adjusted to achieve peak serum concentration of 3–4 μg/mL and trough serum concentration of <1 μg/mL when 3 divided doses are used; there are no optimal drug concentrations for single daily dosing.† Or Ceftriaxone sodium 2 g/24 h IV or IM in 1 dose 2 Class IIa; Level of Evidence B Plus Gentamicin sulfate‡ 3 mg/kg per 24 h IV or IM in 1 dose 2 Vancomycin hydrochloride§ 30 mg/kg per 24 h IV in 2 equally divided doses 4 Class IIa; Level of Evidence B Vancomycin therapy is reasonable only for patients unable to tolerate penicillin or ceftriaxone; vancomycin dose should be adjusted to a trough concentration range of 10–15 μg/mL.
IM indicates intramuscular; IV, intravenous; NVE, native valve infective endocarditis; and VGS, viridans group streptococci. Minimum inhibitory concentration is ≤0.12 μg/mL. The subdivisions differ from Clinical and Laboratory Standards Institute–recommended break points that are used to define penicillin susceptibility.
Doses recommended are for patients with normal renal function.
†Data for once-daily dosing of aminoglycosides for children exist, but no data for treatment of IE exist.
‡Other potentially nephrotoxic drugs (eg, nonsteroidal anti-inflammatory drugs) should be used with caution in patients receiving gentamicin therapy. Although it is preferred that gentamicin (3 mg/kg) be given as a single daily dose to adult patients with endocarditis caused by viridans group streptococci, as a second option, gentamicin can be administered daily in 3 equally divided doses.
§Vancomycin dosages should be infused during the course of at least 1 hour to reduce the risk of histamine-release “red man” syndrome.
Downloaded from by on October 16, 2018 1448 Circulation October 13, 2015 Recommendations 1. It is reasonable to treat patients with IE caused by A defectiva, Granulicatella species, and VGS with a penicillin MIC ≥0.5 µg/mL with a combination of ampicillin or penicillin plus gentamicin as done for enterococcal IE with infectious diseases con-sultation (Class IIa; Level of Evidence C).
2. If vancomycin is used in patients intolerant of ampi-cillin or penicillin, then the addition of gentamicin is not needed (Class III; Level of Evidence C).
3. Ceftriaxone combined with gentamicin may be a reasonable alternative treatment option for VGS isolates with a penicillin MIC ≥0.5 µg/mL that are susceptible to ceftriaxone (Class IIb; Level of Evidence C).
Prosthetic Valve or Valvular Prosthetic Material Endocarditis of Prosthetic Valves or Other Prosthetic Material Caused by VGS and S gallolyticus (bovis) For patients with IE complicating prosthetic valves or other prosthetic material caused by a highly penicillin-susceptible strain (MIC ≤0.12 µg/mL), it is reasonable to administer 6 weeks of therapy with penicillin or cef-triaxone with or without gentamicin for the first 2 weeks (Table 9). It is reasonable to administer 6 weeks of therapy with a combination of penicillin or ceftriaxone and gen-tamicin in patients with IE caused by a strain that is rela-tively or highly resistant to penicillin MIC >0.12 µg/mL. Vancomycin is useful only for patients who are unable to tolerate penicillin, ceftriaxone, or gentamicin. Ampicillin is an acceptable alternative to penicillin if shortages of penicillin exist.
Recommendations 1. Aqueous crystalline penicillin G or ceftriaxone for 6 weeks with or without gentamicin for the first 2 weeks is reasonable (Class IIa; Level of Evidence B).
2. It is reasonable to extend gentamicin to 6 weeks if the MIC is >0.12 µg/mL for the infecting strain (Class IIa; Level of Evidence C).
3. Vancomycin can be useful in patients intolerant of penicillin, ceftriaxone, or gentamicin (Class IIa; Level of Evidence B).
Streptococcus pneumoniae, Streptococcus pyogenes, and Groups B, C, F, and G β-Hemolytic Streptococci IE caused by these streptococci is uncommon. There are few published reports of large case series evaluating management strategies for IE caused by these microorganisms. Results of logistic regression analysis of clinical variables from cases of pneumococcal IE demonstrated the potential value of valve replacement in preventing early death in 1 investigation.129 For patients with NVE caused by highly penicillin-susceptible S pneumoniae, it is reasonable to administer 4 weeks of anti-microbial therapy with penicillin, cefazolin, or ceftriaxone. Vancomycin is reasonable only for patients who are unable to tolerate β-lactam therapy. Six weeks of therapy is reasonable for patients with prosthetic valve endocarditis (PVE).
Pneumococci with intermediate penicillin resistance (MIC >0.1–1.0 µg/mL) or high penicillin resistance (MIC ≥2.0 µg/mL) are recovered uncommonly from patients with bacteremia.130 Moreover, cross-resistance of pneumococci to other antimicrobial agents such as cephalosporins, mac-rolides, fluoroquinolones, carbapenems, and even vanco-mycin is increasing in frequency. In 1 multicenter study131 with a relatively large number of patients with IE caused by Table 8. Therapy of NVE Caused by Strains of VGS and Streptococcus gallolyticus (bovis) Relatively Resistant to Penicillin Regimen Dose and Route Duration, wk Strength of Recommendation Comments Aqueous crystalline penicillin G sodium 24 million U/24 h IV either continuously or in 4–6 equally divided doses 4 Class IIa; Level of Evidence B It is reasonable to treat patients with IE caused penicillin-resistant (MIC ≥0.5 μg/mL) VGS strains with a combination of ampicillin or penicillin plus gentamicin as done for enterococcal IE with infectious diseases consultation (Class IIa; Level of Evidence C). Ampicillin 2 g IV every 4 h is a reasonable alternative to penicillin if a penicillin shortage exists. Plus Gentamicin sulfate† 3 mg/kg per 24 h IV or IM in 1 dose 2 Ceftriaxone may be a reasonable alternative treatment option for VGS isolates that are susceptible to ceftriaxone (Class IIb; Level of Evidence C).
Vancomycin hydrochloride‡ 30 mg/kg per 24 h IV in 2 equally divided doses 4 Class IIa; Level of Evidence C Vancomycin therapy is reasonable only for patients unable to tolerate penicillin or ceftriaxone therapy.
IE indicates infective endocarditis; IM, intramuscular; IV, intravenous; MIC, minimum inhibitory concentration; NVE, native valve infective endocarditis; and VGS, viridans group streptococci. MIC is >0.12 to <0.5 μg/mL for penicillin. The subdivisions differ from Clinical and Laboratory Standards Institute–recommended break points that are used to define penicillin susceptibility.) Doses recommended are for patients with normal renal function.
†See Table 7 for appropriate dose of gentamicin. Although it is preferred that gentamicin (3 mg/kg) be given as a single daily dose to adult patients with endocarditis caused by viridans group streptococci, as a second option, gentamicin can be administered daily in 3 equally divided doses.
‡See Table 7 for appropriate dosage of vancomycin.
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1449 S pneumoniae resistant to penicillin (MIC, 0.1–4 µg/mL), patients were evaluated and compared with 39 patients who were infected with penicillin-susceptible strains. Several key observations were made. Infection by penicillin-resis-tant strains did not worsen prognosis. High-dose penicillin or a third-generation cephalosporin is reasonable in patients with penicillin-resistant IE without meningitis. In patients with IE and meningitis, high doses of cefotaxime are rea-sonable. If the isolate is resistant (MIC ≥2 µg/mL) to cefo-taxime, then the addition of vancomycin and rifampin may be considered. Ceftriaxone may be considered instead of cefotaxime in the previous recommendations. These find-ings are based on current levels of resistance, and increasing MICs could dictate revisions in future treatment selections. Accordingly, the treatment of patients with pneumococcal IE should be coordinated in consultation with an infectious diseases specialist.
For S pyogenes IE, penicillin G administered intrave-nously for 4 to 6 weeks is reasonable treatment on the basis of limited published data. Ceftriaxone is a reasonable alternative to penicillin. Vancomycin is reasonable only for patients who are unable to tolerate a β-lactam antibiotic.
In general, strains of group B, C, F, and G streptococci are slightly more resistant to penicillin than are strains of group A streptococci. In these patients, the addition of gentamicin to penicillin or to ceftriaxone for at least the first 2 weeks of a 4- to 6-week course of antimicrobial therapy for group B, C, and G streptococcal IE may be considered.132,133 There is a clini-cal impression134,135 that early cardiac surgical intervention has improved overall survival rates among treated patients with β-hemolytic streptococcal IE compared with patients treated decades ago. Because of the relative infrequency of IE caused by these microorganisms, consultation with an infectious dis-eases specialist during treatment is recommended.
Recommendations 1. Four weeks of antimicrobial therapy with penicillin, cefazolin, or ceftriaxone is reasonable for IE caused by S pneumoniae; vancomycin can be useful for patients intolerant of β-lactam therapy (Class IIa; Level of Evidence C).
2. Six weeks of therapy is reasonable for PVE caused by S pneumoniae (Class IIa; Level of Evidence C).
3. High-dose penicillin or a third-generation cepha-losporin is reasonable in patients with IE caused by penicillin-resistant S pneumoniae without men-ingitis; if meningitis is present, then high doses of Table 9. Therapy for Endocarditis Involving a Prosthetic Valve or Other Prosthetic Material Caused by VGS and Streptococcus gallolyticus (bovis) Regimen Dose and Route Duration, wk Strength of Recommendation Comments Penicillin-susceptible strain (≤0.12 μg/mL) Aqueous crystalline penicillin G sodium 24 million U/24 h IV either continuously or in 4–6 equally divided doses 6 Class IIa; Level of Evidence B Penicillin or ceftriaxone together with gentamicin has not demonstrated superior cure rates compared with monotherapy with penicillin or ceftriaxone for patients with highly susceptible strain; gentamicin therapy should not be administered to patients with creatinine clearance <30 mL/min. Or Ceftriaxone 2 g/24 h IV or IM in 1 dose 6 Class IIa; Level of Evidence B With or without Gentamicin sulfate† 3 mg/kg per 24 h IV or IM in 1 dose 2 Ampicillin 2 g IV every 4 h is a reasonable alternative to penicillin if a penicillin shortage exists. Vancomycin hydrochloride‡ 30 mg/kg per 24 h IV in 2 equally divided doses 6 Class IIa; Level of Evidence B Vancomycin is reasonable only for patients unable to tolerate penicillin or ceftriaxone.
Penicillin relatively or fully resistant strain (MIC >0.12 μg/mL) Aqueous crystalline penicillin sodium 24 million U/24 h IV either continuously or in 4–6 equally divided doses 6 Class IIa; Level of Evidence B Ampicillin 2 g IV every 4 h is a reasonable alternative to penicillin if a penicillin shortage exists. Or Ceftriaxone 2 g/24 h IV/IM in 1 dose 6 Class IIa; Level of Evidence B Plus Gentamicin sulfate 3 mg/kg per 24 h IV/IM in 1 dose 6 Vancomycin hydrochloride 30 mg/kg per 24 h IV in 2 equally divided doses 6 Class IIa; Level of Evidence B Vancomycin is reasonable only for patients unable to tolerate penicillin or ceftriaxone.
IM indicates intramuscular; IV, intravenous; MIC indicates minimum inhibitory concentration; and VGS, viridans group streptococci.
Doses recommended are for patients with normal renal function.
†See Table 7 for appropriate dose of gentamicin. Although it is preferred that gentamicin (3 mg/kg) be given as a single daily dose to adult patients with endocarditis resulting from VGS, as a second option, gentamicin can be administered daily in 3 equally divided doses.
‡See text and Table 7 for appropriate dose of vancomycin.
Downloaded from by on October 16, 2018 1450 Circulation October 13, 2015 cefotaxime (or ceftriaxone) are reasonable (Class IIa; Level of Evidence C).
4. The addition of vancomycin and rifampin to cefo-taxime (or ceftriaxone) may be considered in patients with IE caused by S pneumoniae that are resistant to cefotaxime (MIC >2 µg/mL) (Class IIb; Level of Evidence C).
5. Because of the complexities of IE caused by S pneu-moniae, consultation with an infectious diseases spe-cialist is recommended (Class I; Level of Evidence C).
6. For IE caused by S pyogenes, 4 to 6 weeks of therapy with aqueous crystalline penicillin G or ceftriaxone is reasonable; vancomycin is reasonable only in patients intolerant of β-lactam therapy (Class IIa; Level of Evidence C).
7. For IE caused by group B, C, or G streptococci, the addition of gentamicin to aqueous crystalline peni-cillin G or ceftriaxone for at least the first 2 weeks of a 4- to 6-week treatment course may be considered (Class IIb; Level of Evidence C).
8. Consultation with an infectious diseases specialist to guide treatment is recommended in patients with IE caused by β-hemolytic streptococci (Class I; Level of Evidence C).
Staphylococci IE may be caused by staphylococci that are coagulase positive (S aureus) or coagulase negative (S epidermidis, S lugdunen-sis, and various other species). Although coagulase-positive staphylococci were traditionally believed to cause primarily NVE and coagulase-negative staphylococci (CoNS) were associated with PVE, considerable overlap now exists. For example, in a multicenter, prospective, observational investi-gation involving >1000 consecutive patients with definite IE from >20 countries, S aureus was the most common cause of PVE (25.8% of 214 cases), whereas 64 cases of NVE (8%) resulted from CoNS.136 In addition, the prevalence of CoNS NVE appears to be increasing.137 Thus, it is important to con-sider both pathogen groups when a patient with suspected IE has a preliminary blood culture that suggests staphylococci by Gram stain interpretation.
S aureus S aureus is the most common cause of IE in much of the devel-oped world.6–8 Data from >70 million hospitalizations in the United States suggest that rates of S aureus IE have increased significantly relative to other causes of IE.3 This increase is primarily a consequence of healthcare contact (eg, intra-vascular catheters, surgical wounds, indwelling prosthetic devices, hemodialysis)6,8,9 and is especially prevalent in North America.6,138,139 Increasing rates of oxacillin-resistant S aureus or MRSA isolates in both hospital and community settings and the recovery of clinical S aureus isolates both partially and fully138,139 resistant to vancomycin have complicated the treatment of S aureus IE. An increasing body of evidence suggests an association between high (but still susceptible on the basis of the Clinical and Laboratory Standards Institute definition) vancomycin MICs in S aureus and worse clinical outcome in both MRSA infections treated with vancomycin140 and MRSA bacteremia treated with antistaphylococcal peni-cillins.141 Importantly, this association between higher vanco-mycin MIC in infecting MSSA and worse clinical outcomes among patients treated with antistaphylococcal penicillins (not vancomycin) was externally validated in a large cohort of patients with MSSA IE.142 These data suggest that host- or pathogen-specific factors, rather than higher MICs of the infecting pathogen to vancomycin, contribute to the poor out-comes in these patients (because the latter patients were not treated with a glycopeptide).
In non-IDUs, S aureus IE involves primarily the left side of the heart and is associated with mortality rates ranging from 25% to 40%. S aureus IE in IDUs often involves the tricuspid valve. Cure rates for right-sided S aureus IE in IDUs are high (>85%) and may be achieved with relatively short courses of either parenteral or oral treatment (2–4 weeks; see below). Complicated IE manifested, for example, by deep tis-sue abscesses or osteoarticular infection may require more prolonged therapy.
Coagulase-Negative Staphylococci As noted above, in addition to their importance in PVE, CoNS now cause a significant but relatively small proportion of NVE cases.2 Risk factors for CoNS IE are similar to those for S aureus and include typical risk factors associated with exten-sive healthcare contact. Of interest, data suggest that the over-all outcomes for patients with CoNS IE and S aureus IE are similar.137 Most CoNS are resistant to methicillin. These resis-tant organisms are particularly prominent among patients with healthcare-associated staphylococcal IE. Methicillin-resistant strains also are clinically resistant to cephalosporins and car-bapenems, although this fact is not always reflected accurately in the results of standard in vitro tests.
An important subset of patients with CoNS IE has been identified: those with infection caused by S lugdunensis. This species of CoNS tends to cause a substantially more virulent form of IE, with a high rate of perivalvular extension of infec-tion and metastatic infection. This organism is uniformly sus-ceptible in vitro to most antibiotics.143–145 Most experts believe that IE caused by this organism can be treated with standard regimens based on the in vitro susceptibility profiles of the strain. The patient also should be monitored carefully for the development of periannular extension or extracardiac spread of infection. Although microbiological differentiation of S lugdunensis requires specific biochemical assays, the poor outcomes associated with S lugdunensis underscore the impor-tance of performing these specialized assays. Initial screening can be done with pyrrolidonyl aminopeptidase hydrolysis test-ing, and isolates that test positive should be further identified by a multisubstrate identification system, matrix-assisted laser desorption ionization–time of flight, or other methods, includ-ing PCR.146,147 Recommendation 1. Ongoing vigilance for IE complications, including perivalvular extension of infection and extracardiac foci of infection, is reasonable (Class IIa; Level of Evidence C).
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1451 IE Caused by Staphylococci in the Absence of Prosthetic Valves or Other Prosthetic Material Right-Sided IE in IDUs The addition of gentamicin to nafcillin or oxacillin has tradi-tionally been a standard approach for the treatment of right-sided IE. For example, in IDUs with uncomplicated right-sided S aureus IE (no evidence of renal failure, extrapulmonary metastatic infections, aortic or mitral valve involvement, meningitis, or infection by MRSA), combined short-course (2 weeks) β-lactam plus aminoglycoside therapy was highly effective in several studies.91,92,148 In 1 study, 92 patients pro-vided such combination therapy had excellent outcomes, even HIV-infected patients and those who had large tricuspid valve vegetations (>10 mm in diameter). In contrast, short-course regimens with glycopeptides (teicoplanin or vancomycin) plus gentamicin appeared to be less effective for right-sided S aureus IE caused by either MSSA or MRSA strains.148 These glycopeptides may be less effective because of limited bacte-ricidal activity, poor penetration into vegetations, or increased drug clearance among IDUs.
A growing body of evidence suggests that the addition of adjunctive aminoglycoside therapy not only is unnecessary for patients with uncomplicated right-sided native valve S aureus IE but may cause harm. For example, 1 study showed that a 2-week monotherapy regimen of intravenous cloxacillin was equivalent to cloxacillin plus gentamicin administered for 2 weeks.92 In 2006, the US Food and Drug Administration (FDA) approved the use of daptomycin (6 mg·kg−1·d−1) for the treatment of S aureus bacteremia and right-sided S aureus IE.13 In a reg-istrational open-label, multinational, clinical trial for the treat-ment of S aureus bacteremia or right-sided IE comparing the efficacy of daptomycin monotherapy with therapy that included low-dose (1 mg/kg IV every 8 hours or adjusted on the basis of renal function) gentamicin for the first 4 days, patients did equally well in either treatment arm. In the predefined subgroup of those with MRSA bacteremia, daptomycin demonstrated a 44.4% success rate compared with 31.8% for standard therapy; this difference was not statistically significant (absolute differ-ence, 12.6%, 95% confidence interval, −7.4 to 32.6; P=0.28). Of note, in a post hoc analysis of this landmark clinical trial,149 the addition of even such low-dose, short-course gentamicin in 1 arm of the study was significantly associated with renal toxic-ity, which occurred early and often, and the clinical association between gentamicin dose and duration was minimal.
Thus, current evidence suggests that either parenteral β-lactam or daptomycin short-course therapy is adequate for the treatment of uncomplicated MSSA right-sided IE. In contrast, glycopeptide therapy for MRSA right-sided IE may require more prolonged treatment regimens. For both MSSA and MRSA infections, use of adjunctive gentamicin for the treatment of S aureus bacteremia or right-sided NVE is discouraged Recommendation 1. Gentamicin is not recommended for treatment of right-sided staphylococcal NVE (Class III; Level of Evidence B).
In patients for whom parenteral antibiotic therapy is problem-atic, oral treatment may be a reasonable option. Two studies have evaluated the use of predominantly oral 4-week antibiotic regimens (featuring ciprofloxacin plus rifampin) for the ther-apy of uncomplicated right-sided MSSA IE in IDUs.150,151 In each study, including one in which >70% of patients were HIV seropositive,150 cure rates were >90%. However, the relatively high rate of quinolone resistance among contemporary S aureus strains has made this alternative treatment strategy problematic.
IE in Non-IDUs Older anecdotal case reports in non-IDUs with S aureus IE suggested that the use of combined gentamicin-methicillin therapy may be of benefit in patients who fail to respond to monotherapy with methicillin.152 This issue was addressed in a multicenter, prospective trial comparing nafcillin alone for 6 weeks with nafcillin plus gentamicin (for the initial 2 weeks) in the treatment of predominantly left-sided IE caused by S aureus.153 Nafcillin-gentamicin therapy reduced the duration of bacteremia by ≈1 day compared with nafcillin monotherapy. However, combination therapy did not reduce mortality or the frequency of cardiac complications. Furthermore, combina-tion therapy increased the frequency of gentamicin-associated nephrotoxicity. As noted above,149 the risk of clinically sig-nificant nephrotoxicity with even short courses of adjunctive low-dose gentamicin for S aureus bacteremia and right-sided IE can be substantial. In addition, gentamicin should not be used with vancomycin in patients with MRSA NVE because of the nephrotoxicity risk.13,149 In cases of brain abscess com-plicating MSSA IE, nafcillin is the preferred agent rather than cefazolin, which has inadequate blood-brain barrier penetra-bility. If the patient cannot tolerate nafcillin therapy, then van-comycin should be used.
Vancomycin is often included with cefazolin as empirical coverage for patients with IE caused by S aureus while await-ing susceptibility results. An analysis of the literature, however, compared the use of empirical combination of vancomycin and antistaphylococcal β-lactam therapy with vancomycin alone and demonstrated the superiority of β-lactam–containing regi-mens over vancomycin monotherapy for bacteremic MSSA infections, including IE.154 This differential outcome included studies in which there was an early shift from empirical van-comycin to β-lactam therapy as soon as blood cultures yielded MSSA (not MRSA). The meta-analysis included small, retro-spective studies, however, which limits support for initial com-bination therapy by some experts. Therefore, the usefulness of empiric combination therapy in patients with S aureus bactere-mia until oxacillin susceptibility is known is uncertain.
Although the large majority of staphylococci are resis-tant to penicillin, occasional strains remain susceptible. Unfortunately, the current laboratory screening procedures for detecting penicillin susceptibility may not be reliable. Therefore, IE caused by these organisms should be treated with regimens outlined for MSSA that includes nafcillin (or equivalent antistaphylococcal penicillin) as an option rather than penicillin (Table 10).
There are no evidence-based data that demonstrate the most appropriate duration of nafcillin therapy for treat-ment of left-sided NVE caused by MSSA. For patients with Downloaded from by on October 16, 2018 1452 Circulation October 13, 2015 uncomplicated infection, 6 weeks of therapy is recommended. For patients with complications of IE such as perivalvular abscess ormation and septic metastatic complications, at least 6 weeks of nafcillin is recommended.
Currently, defining the optimal therapy for NVE attribut-able to MRSA is challenging. Historically, vancomycin has been used and is recommended. As outlined in the Therapy of MSSA IE in Patients Allergic to or Intolerant of β-Lactams section below, daptomycin may be a reasonable alternative to daptomycin for left-sided NVE caused by MRSA on the basis of limited data in a prospective, randomized trial; a multina-tional, prospective cohort investigation of the use of high-dose (≈9 mg/kg per dose) daptomycin; and a multicenter, retro-spective, observational study that included daptomycin at ≥8 mg/kg per dose.13,155 Selection of daptomycin dosing should be assisted by infectious diseases consultation.
At this time, additional study of ceftaroline is needed to define its role, if any, in the treatment of left-sided NVE caused by MRSA.
Recommendations 1. Gentamicin should not be used for treatment of NVE caused by MSSA or MRSA (Class III; Level of Evidence B).
2. In cases of brain abscess resulting from MSSA IE, nafcillin should be used instead of cefazolin; vanco-mycin should be given in cases of nafcillin intoler-ance (Class I; Level of Evidence C).
3. The usefulness of empirical combination ther-apy with vancomycin plus an antistaphylococcal β-lactam antibiotic in patients with S aureus bac-teremia until oxacillin susceptibility is known is uncertain (Class IIb; Level of Evidence B).
4. IE caused by staphylococci that are penicillin sus-ceptible should be treated with antistaphylococcal β-lactam antibiotics rather than aqueous crystalline penicillin G because clinical laboratories are not able to detect penicillin susceptibility (Class I; Level of Evidence B).
5. Six weeks of nafcillin (or equivalent antistaphylo-coccal penicillin) is recommended for uncompli-cated left-sided NVE caused by MSSA; at least 6 weeks of nafcillin (or equivalent antistaphylococcal penicillin) is recommended for complicated left-sided NVE caused by this organism (Class I; Level of Evidence C).
6. Daptomycin may be a reasonable alternative to vancomycin for treatment of left-sided IE resulting from MRSA (Class IIb; Level of Evidence B).
7. Selection of daptomycin dosing should be assisted by infectious diseases consultation (Class I; Level of Evidence C).
Therapy of MSSA IE in Patients Allergic to or Intolerant of β-Lactams Therapy for MSSA IE in patients truly unable to tolerate β-lactams is problematic. One decision analysis study con-cluded that patients with a questionable history of immediate-type hypersensitivity to penicillins in the context of IE caused by MSSA should be skin tested before starting antibiotic therapy.156 However, the limited availability of standardized skin test reagents makes testing impractical. Instead, most experts endorse one of the published standard desensitization protocols. For patients with a well-defined history of nonana-phylactoid reactions to penicillins (eg, simple skin rash), a first-generation cephalosporin such as cefazolin is reasonable. Although cefazolin may be more susceptible to β-lactamase– mediated hydrolysis than nafcillin157 and less effective in the treatment of MSSA experimental IE, the clinical significance of these observations is unknown. Many experts regularly use cefazolin for S aureus IE instead of nafcillin because of drug Table 10. Therapy for NVE Caused by Staphylococci Regimen Dose and Route Duration, wk Strength of Recommendation Comments Oxacillin-susceptible strains Nafcillin or oxacillin 12 g/24 h IV in 4–6 equally divided doses 6 Class I; Level of Evidence C For complicated right-sided IE and for left-sided IE; for uncomplicated right-sided IE, 2 wk (see text). For penicillin-allergic (nonanaphylactoid type) patients Consider skin testing for oxacillin-susceptible staphylococci and questionable history of immediate-type hypersensitivity to penicillin. Cefazolin 6 g/24 h IV in 3 equally divided doses 6 Class I; Level of Evidence B Cephalosporins should be avoided in patients with anaphylactoid-type hypersensitivity to β-lactams; vancomycin should be used in these cases.
Oxacillin-resistant strains Vancomycin§ 30 mg/kg per 24 h IV in 2 equally divided doses 6 Class I; Level of Evidence C Adjust vancomycin dose to achieve trough concentration of 10–20 μg/mL (see text for vancomycin alternatives). Daptomycin ≥8 mg/kg/dose 6 Class IIb; Level of Evidence B Await additional study data to define optimal dosing.
IE indicates infective endocarditis; IV, intravenous; and NVE, native valve infective endocarditis.
Doses recommended are for patients with normal renal function.
§For specific dosing adjustment and issues concerning vancomycin, see Table 7 footnotes.
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1453 tolerability and cost, for MSSA IE in penicillin-intolerant patients, or to facilitate outpatient antibiotic administration.
Vancomycin is often recommended as an alternative to β-lactam therapy for MSSA IE. As outlined above, β-lactam allergy evaluation should be conducted in every case in which vancomycin is considered for use because poorer outcomes related to vancomycin therapy for a variety of MSSA infec-tions are well recognized.140 Clindamycin has been associated with IE relapse and is not recommended.158 For MSSA IE in patients with anaphy-lactoid-type β-lactam allergy who exhibit either a suboptimal response to vancomycin or vancomycin allergy, β-lactam desensitization should be considered as noted above.159 Daptomycin is a reasonable alternative to vancomycin for adults in the treatment of S aureus NVE. In the above-noted multinational trial13 of S aureus bacteremia and right-sided IE, this agent (at 6 mg·kg−1·d−1) was noninferior to standard ther-apy with vancomycin or an antistaphylococcal penicillin plus low-dose, short-course gentamicin. Importantly, the small number (n=18; 9 in each arm) of patients with left-sided IE enrolled in the trial prevented meaningful conclusions on the comparative efficacy of daptomycin in this infection. For this reason, the FDA indication for daptomycin explicitly omitted left-sided IE. However, in an observational study, high-dose daptomycin (≈9 mg/kg per dose) for treatment of left-sided IE was as effective as standard-of-care therapy and cleared MRSA bacteremia significantly faster than did standard-of-care treatment.155 The emergence of organisms with decreased susceptibility to daptomycin was observed in ≈5% of daptomycin-treated patients. All of these patients needed but for a variety of rea-sons did not receive surgical intervention for debridement of deep-seated infections or left-sided IE. As indicated, the FDA-approved dose of daptomycin for S aureus bacteremia and right-sided IE is currently 6 mg/kg IV once daily. Some experts recommend higher doses of daptomycin at 8 to 10 mg/ kg for complicated infections, including left-sided IE (these doses are not approved by the FDA).109 This recommenda-tion is based in part on evidence suggesting that higher-dose daptomycin may reduce the likelihood of treatment-emergent resistance, is generally well tolerated, and is not associated with excess toxicities. Whether this higher dosing strategy prevents treatment-emergent resistance of daptomycin is still not answered.
Daptomycin is inhibited by pulmonary surfactant160 and thus is contraindicated in the treatment of S aureus pneu-monia acquired via the aspiration route. In the registrational trial,13 however, this agent performed as well as vancomycin or β-lactams in treating septic pulmonary emboli caused by S aureus, reflecting the distinct pathogenesis of this syndrome as opposed to traditional pneumonia.
Recommendations 1. Cefazolin is reasonable in patients with a well-defined history of nonanaphylactoid reactions to penicillins (Class IIa; Level of Evidence B).
2. Allergy evaluation for tolerance to β-lactam therapy should be done in every case in which vancomycin is considered for treatment of MSSA IE (Class I; Level of Evidence B).
3. Clindamycin is not recommended as a result of an increased IE relapse rate (Class III; Level of Evidence B).
4. Daptomycin is a reasonable alternative to vancomy-cin for NVE caused by MSSA (Class IIa; Level of Evidence B).
Additional or Adjunctive Therapies As discussed above, combination therapy with gentamicin therapy in S aureus NVE is discouraged because of the rela-tively high rates of intrinsic gentamicin resistance, a lack of clear-cut efficacy, and documented toxicity issues.149,153,161 Although most staphylococci are highly susceptible to rifampin, resistance develops rapidly when this agent is used alone. The in vivo efficacy of rifampin in combination with nafcillin, oxacillin, vancomycin, trimethoprim/sulfamethoxa-zole, or aminoglycosides is highly variable. Moreover, use of rifampin as adjunct therapy for S aureus NVE has been associ-ated with higher rates of adverse events (primarily hepatotox-icity) and a significantly lower survival rate.162 Thus, routine use of rifampin is not recommended for treatment of staphy-lococcal NVE. Of note, a prospective trial in patients with IE caused by MRSA failed to demonstrate that the addition of rifampin to vancomycin either enhanced survival or reduced the duration of bacteremia compared with treatment with vancomycin alone.163 Rifampin is often used in native valve S aureus IE when this infection is complicated by involvement of selected anatomic sites where rifampin penetrates effec-tively (eg, bone, joint, cerebrospinal fluid).164 No standard therapies exist for the treatment of S aureus IE caused by isolates that are not susceptible to vancomy-cin. Classification of these isolates has become complex and includes designations of reduced susceptibility (hVISA), intermediate resistance (VISA), and high-level resistance (VRSA). To date, the limited number of patients reported to have IE caused by these isolates precludes specific treatment recommendations. Thus, these infections should be managed in conjunction with an infectious diseases consultant.
Although Markowitz et al165 showed that trimethoprim-sulfamethoxazole was inferior to vancomycin in the treat-ment of invasive S aureus infections, it is sometimes used in salvage situations. Interestingly, all treatment failures with trimethoprim-sulfamethoxazole occurred in patients infected with MSSA in that report,165 whereas patients with MRSA infection were uniformly cured. The efficacy of trimethoprim-sulfamethoxazole and other folate antagonists may be attenu-ated by thymidine release from damaged host cells (eg, at sites of tissue damage such as abscesses).166 In an in vitro study,167 the addition of trimethoprim-sulfamethoxazole to daptomy-cin was rapidly bactericidal for a daptomycin-nonsusceptible strain compared with daptomycin monotherapy. The com-bination of daptomycin and a β-lactam antibiotic has been reported to be effective in treating a limited number of patients with persistent MRSA bacteremia.168 The potential effective-ness of this combination may be due in part to the capacity of the β-lactam agent to alter the surface charge of the organism Downloaded from by on October 16, 2018 1454 Circulation October 13, 2015 in a nonbactericidal mechanism, allowing enhanced surface binding of daptomycin.169–171 Linezolid was reported to be effective in the treatment of persistent MRSA bacteremia,172 but this study had important study design weaknesses.173 Patient outcomes with linezolid therapy for S aureus left-sided IE have generally been poor.174–176 Quinupristin-dalfopristin177 and telavancin178 have been used successfully as salvage ther-apy in selected patients with MRSA IE who clinically failed vancomycin therapy.
Ceftaroline received FDA registrational indications for acute bacterial skin and soft tissue infections caused by both MRSA and MSSA, as well as community-acquired pneu-monia caused by MSSA. Several case series suggest that it may have utility in complicated S aureus infections, including IE.179–181 These promising observations should be verified with appropriately designed clinical studies before ceftaroline can be recommended for widespread use in such off-label settings.
Recommendations 1. Routine use of rifampin is not recommended for treatment of staphylococcal NVE (Class III; Level of Evidence B).
2. IE caused by vancomycin-resistant staphylococci (hVISA, VISA, or VRSA) should be managed in conjunction with an infectious diseases consultant (Class I; Level of Evidence C).
IE Caused by Staphylococci in the Presence of Prosthetic Valves or Other Prosthetic Material Coagulase-Negative Staphylococci CoNS that cause PVE usually are methicillin resistant, particu-larly when IE develops within 1 year after surgery.182 Unless susceptibility to methicillin can be demonstrated conclusively, it should be assumed that the organism is methicillin resistant, and treatment should be planned accordingly. Experimental IE models caused by methicillin-resistant staphylococci demon-strated that vancomycin combined with rifampin and gentami-cin is the optimal regimen, and limited clinical reports support this approach.183 The dosing of rifampin is done by convention and is not based on PK data. Vancomycin and rifampin are rec-ommended for a minimum of 6 weeks, with the use of gen-tamicin limited to the first 2 weeks of therapy (Table 11). If the organism is resistant to gentamicin, then an aminoglycoside to which it is susceptible should be substituted for gentamicin. Some authorities recommend delaying the initiation of rifampin therapy for several days to allow adequate penetration of van-comycin into the cardiac vegetations in an attempt to prevent treatment-emergent resistance to rifampin. If the organism is resistant to all available aminoglycosides, such adjunctive treatment should be omitted. In this situation, if the organism is susceptible to a fluoroquinolone, animal studies of therapy for foreign-body infection suggest that a fluoroquinolone can be used instead of gentamicin.184 Thus, although clinical data are not available to support the practice, selection for fluoroquino-lone resistance during treatment can occur, and prevalent fluo-roquinolone resistance among CoNS will limit its use, it may reasonable to use a fluoroquinolone in this setting.
PVE, particularly when onset is within 12 months of cardiac valve implantation and an aortic valve prosthesis is involved, is frequently complicated by perivalvular or myo-cardial abscesses or valvular dysfunction.136 Surgery is often required in these patients and may be lifesaving. As noted above, CoNS may become resistant to rifampin during therapy for PVE. Because of the potential for changes in the patterns of antibiotic susceptibility during therapy, organisms recov-ered from surgical specimens or blood from patients who have had a bacteriological relapse should be carefully retested for complete antibiotic susceptibility profiles.
Although published data on combinations of antimicrobial therapy are limited, we suggest that PVE caused by oxacillin-susceptible CoNS should be treated with nafcillin or oxacillin plus rifampin in combination with gentamicin for the first 2 weeks of therapy. A first-generation cephalosporin or vanco-mycin may be substituted for nafcillin/oxacillin for patients who are truly allergic to penicillin.
Recommendations 1. Vancomycin and rifampin are recommended for a minimum of 6 weeks, with the use of gentamicin lim-ited to the first 2 weeks of therapy (Class I; Level of Evidence B).
2. If CoNS are resistant to gentamicin, then an amino-glycoside to which they are susceptible may be con-sidered (Class IIb; Level of Evidence C).
3. If CoNS are resistant to all aminoglycosides, then substitution with a fluoroquinolone may be consid-ered if the isolate is susceptible to a fluoroquinolone (Class IIb; Level of Evidence C).
4. Organisms recovered from surgical specimens or blood from patients who have had a bacteriological relapse should be carefully retested for complete antibiotic susceptibility profiles (Class I; Level of Evidence C).
S aureus Because of the high mortality rate associated with S aureus PVE,136 combination antimicrobial therapy is recommended (Table 11). The use of combination therapy is based not on studies of in vitro synergy but rather on the efficacy of this therapy for treatment of CoNS PVE, as well as the results of treatment of experimental IE and infected devices. In animal studies, rifampin appears to be key in the complete steriliza-tion of foreign bodies infected by MRSA.184,185 For infection caused by MSSA, nafcillin or oxacillin together with rifampin is suggested; with MRSA, vancomycin and rifampin should be used. Gentamicin should be adminis-tered for the initial 2 weeks of therapy with either β-lactam or vancomycin-containing regimens. If a strain is resistant to gentamicin, then a fluoroquinolone may be used if the strain is susceptible. Early cardiac surgical interventions play an important role in maximizing outcomes in S aureus PVE,186 especially in the presence of heart failure.11 Recommendations 1. Combination antimicrobial therapy is recom-mended (Class I; Level of Evidence C).
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1455 2. Gentamicin should be administered for the initial 2 weeks of therapy with either β-lactam or vancomycin-containing regimens (Class I; Level of Evidence C).
Enterococci Although there are >15 species within the Enterococcus genus, E faecalis and E faecium are the major species isolated from clinical sources in IE patients. Enterococci are the third leading cause of IE and account for ≈10% of cases in non-IDUs. E faecalis causes ≈97% of cases of IE; E faecium, ≈1% to 2%; and other species, ≈1%.
Regimens recommended for enterococcal IE are shown in Tables 12 through 15. Enterococci should be routinely tested in vitro for susceptibility to penicillin or ampicillin and van-comycin (MIC determination) and for high-level resistance to gentamicin to predict synergistic interactions (see below). Because of the striking increase in resistance of enterococci to vancomycin, aminoglycosides, and penicillin, additional susceptibility tests may be necessary to identify alternative antimicrobial regimens. For strains of enterococci resistant to β-lactams, vancomycin, or aminoglycosides, it is reason-able to test for susceptibility in vitro to daptomycin and line-zolid. Linezolid is bacteriostatic in vitro against enterococci, whereas daptomycin is bactericidal in vitro in susceptible strains. Although rarely identified, β-lactamase–producing enterococci may account for relapse of infection. Routine screening for β-lactamase production is not sensitive enough, and specialized testing will be needed for detection.
Compared with VGS and β-hemolytic streptococci, enterococci are relatively resistant to penicillin, ampicillin, and vancomycin. These streptococci usually are killed by monotherapy with these antimicrobials, whereas enterococci are inhibited but not killed. Killing of susceptible strains of enterococci requires the synergistic action of penicillin, ampi-cillin, or vancomycin in combination with either gentamicin or streptomycin.
Enterococci are relatively impermeable to aminogly-cosides. High concentrations of aminoglycosides in the extracellular environment are required to achieve sufficient concentrations of the drug at the site of the ribosomal target within the bacterial cell for bactericidal activity. These con-centrations are higher than can be achieved safely in patients; however, cell wall–active agents such as penicillin, ampicillin, and vancomycin raise the permeability of the enterococcal cell so that a bactericidal effect can be achieved by relatively low concentrations of an aminoglycoside. If an enterococcal strain is resistant to the cell wall–active agent or high concentrations of an aminoglycoside (500 µg/mL gentamicin or 1000 µg/mL streptomycin), then the combination of an aminoglycoside and the cell wall–active agent will not result in bactericidal activity in vitro or in vivo (ie, in experimental IE models), nor will it predictably produce a microbiological cure in human enterococcal IE.
Recommendations 1. Enterococci should be tested routinely in vitro for susceptibility to penicillin and vancomycin (MIC determination) and for high-level resistance to gen-tamicin to predict synergistic interactions (Class I; Level of Evidence A).
2. In vitro susceptibility to daptomycin and linezolid should be obtained for strains that are resistant to β-lactams, vancomycin, or aminoglycosides (Class I; Level of Evidence C).
Table 11. Therapy for Endocarditis Involving a Prosthetic Valve or Other Prosthetic Material Caused by Staphylococci Regimen Dose and Route Duration, wk Strength of Recommendation Comments Oxacillin-susceptible strains Nafcillin or oxacillin 12 g/24 h IV in 6 equally divided doses ≥6 Class I; Level of Evidence B Vancomycin should be used in patients with immediate-type hypersensitivity reactions to β-lactam antibiotics (see Table 5 for dosing guidelines); cefazolin may be substituted for nafcillin or oxacillin in patients with non–immediate-type hypersensitivity reactions to penicillins. Plus Rifampin 900 mg per 24 h IV or orally in 3 equally divided doses ≥6 Plus Gentamicin† 3 mg/kg per 24 h IV or IM in 2 or 3 equally divided doses 2 Oxacillin-resistant strains Vancomycin 30 mg/kg 24 h in 2 equally divided doses ≥6 Class I; Level of Evidence B Adjust vancomycin to a trough concentration of 10–20 μg/mL.
(see text for gentamicin alternatives) Plus Rifampin 900 mg/24 h IV/PO in 3 equally divided doses ≥6 Plus Gentamicin 3 mg/kg per 24 h IV/IM in 2 or 3 equally divided doses 2 IM indicates intramuscular; and IV, intravenous.
Doses recommended are for patients with normal renal function.
†Gentamicin should be administered in close proximity to vancomycin, nafcillin, or oxacillin dosing. See Table 7 for appropriate dose of gentamicin.
Downloaded from by on October 16, 2018 1456 Circulation October 13, 2015 Role of Aminoglycosides in the Treatment of Patients With Enterococcal IE: Special Considerations Aminoglycoside-containing regimens have been a cornerstone of antimicrobial therapy for enterococcal IE187 and have been recommended as standard therapy in previous (1995) AHA guidelines.188 Since the publication of the latest (2005) AHA statement on antimicrobial therapy of patients with IE,12 the frequency of aminoglycoside-resistant strains of enterococci has increased significantly. In addition, a number of studies have been published on the dosing of aminoglycosides, the duration of aminoglycoside therapy, and the possible role of non–aminoglycoside-containing regimens for the treatment of E faecalis IE.189–191 Approximately 97% of cases of enterococcal IE are caused by E faecalis, and the majority of these remain susceptible to β-lactams and vancomycin, but aminoglycoside resistance is increasing in frequency. In the study by Gavaldà et al,190 approximately half of the patients had IE caused by high-level aminoglycoside-resistant strains of E faecalis. In the study by Fernández-Hidalgo et al,191 26% of the 272 patients had high-level aminoglycoside-resistant strains of E faecalis. Therefore, aminoglycoside-containing regimens would not be effective therapy for these patients.
A number of factors should be considered in the selec-tion of aminoglycoside-containing regimens. Compared with other patients with IE, in general, patients with enterococcal IE are older; are often debilitated; may have complicated, underlying urological conditions, including pre-existing renal failure; may have healthcare-associated infections; and have significant other underlying comorbidities common in older age groups.192 In these patients, gentamicin-associated neph-rotoxicity may significantly complicate a “standard” 4- to 6-week course of therapy and could result in serious, possibly life-threatening, complications such as renal failure requiring hemodialysis. In these situations, the potential risk of attempt-ing to complete a 4- to 6-week course of gentamicin therapy may exceed the benefit.193 In patients with VGS IE treated with multiple divided doses of gentamicin, single daily-dose therapy with genta-micin resulted in similar response rates and was well tol-erated (see treatment of VGS IE above). Studies of single daily dosing of gentamicin compared with divided doses in enterococcal experimental IE and in humans have yielded conflicting results. These results may reflect different PK of aminoglycosides in animals compared with humans. Studies in humans of the dosing interval of gentamicin were not controlled or standardized. Dosing of gentamicin ranged from once daily to 3 times daily; therefore, the data were insufficient to compare the efficacy of once-daily doses with divided doses. Until more convincing data demonstrate that once-daily dosing of gentamicin is as effective as multiple dosing, in patients with normal renal function, gentami-cin should be administered in daily multiple divided doses (total, ≈3 mg·kg−1·d−1) rather than a daily single dose to patients with enterococcal IE. In patients with normal renal function, it is reasonable to administer gentamicin every 8 hours with the dose adjusted to achieve a 1-hour serum con-centration of ≈3 µg/mL and a trough concentration of <1 µg/ mL. Increasing the dose of gentamicin in these patients did not result in enhanced efficacy but did increase the risk of nephrotoxicity.194 Many patients with enterococcal IE are managed in a nontertiary care facility, and the laboratory may not have the capability for rapid determination of serum gentamicin concentrations or may not have a clinical pharmacist avail-able to assist in optimal dosing adjustments. These and other factors have prompted studies to evaluate the efficacy of non–gentamicin-containing regimens for the treatment of Table 12. Therapy for Endocarditis Involving a Native or Prosthetic Valve or Other Prosthetic Material Resulting From Enterococcus Species Caused by Strains Susceptible to Penicillin and Gentamicin in Patients Who Can Tolerate β-Lactam Therapy Regimen Dose† and Route Duration, wk Strength of Recommendation Comments Either Class IIa; Level of Evidence B Native valve: 4-wk therapy recommended for patients with symptoms of illness <3 mo; 6-wk therapy recommended for native valve symptoms >3 mo and for patients with prosthetic valve or prosthetic material. Recommended for patients with creatinine clearance >50 mL/min. Ampicillin sodium 2 g IV every 4 h 4–6 4–6 Class IIa; Level of Evidence B Or Aqueous penicillin G sodium 18–30 million U/24 h IV either continuously or in 6 equally divided doses 4–6 Plus Gentamicin sulfate‡ 3 mg/kg ideal body weight in 2–3 equally divided doses Or Double β-lactam Ampicillin 2 g IV every 4 h 6 Class IIa; Level of Evidence B Recommended for patients with initial creatinine clearance <50 mL/min or who develop creatinine clearance <50 mL/min during therapy with gentamicin-containing regimen.
Plus Ceftriaxone 2 g IV every 12 h 6 IV indicates intravenous. For patients unable to tolerate a β-lactam, see Table 14.
†Doses recommended are for patients with normal renal and hepatic function.
‡Dose of gentamicin should be adjusted to achieve a peak serum concentration of 3 to 4 µg/mL and a trough concentration of <1 µg/mL.
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1457 enterococcal IE.195 The decision of whether to use an ami-noglycoside-containing regimen must be individualized for each patient. The rationale and recommendations for spe-cific aminoglycoside-containing regimens for the treatment of enterococcal IE based on in vitro susceptibilities are dis-cussed in the following groups of patients and in Tables 12 through 15.
Recommendations 1. Gentamicin should be administered in daily mul-tiple divided doses (total, ≈3 mg·kg−1·d−1) rather than a single daily dose to patients with enterococ-cal IE and normal renal function (Class I; Level of Evidence B).
2. It is reasonable to administer gentamicin every 8 hours with the dose adjusted to achieve a 1-hour serum con-centration of ≈3 µg/mL and a trough concentration of <1 µg/mL (Class IIa; Level of Evidence B).
Enterococcal Endocarditis Susceptible to Penicillin, Vancomycin, and Aminoglycosides Antimicrobial regimens outlined in Table 12 are reasonable for treatment of patients with IE caused by these organisms. In a prospective study, the duration of antimicrobial therapy in native valve E faecalis IE was based on the duration of infec-tion before diagnosis and onset of effective therapy.196 Patients with <3 months’ duration of symptoms were treated success-fully with 4 weeks of antimicrobial therapy. Patients with ≥3 months’ duration of symptoms were successfully treated with 6 weeks of therapy. The duration of therapy for NVE is based on this work, and the regimens that may be considered are listed in Table 12. In patients with PVE, 6 weeks of antimicro-bial therapy is reasonable.
Patients with preexisting mild (creatinine clearance, 30–50 mL/min) or severe (creatinine clearance, <30 mL/min) renal failure may not be able to safely complete a 4- to 6-week course of gentamicin therapy because of gentamicin-associated nephrotoxicity. Alternative regimens that should be considered include the use of streptomycin instead of gentamicin, short-course gentamicin therapy (2–3 weeks), and use of a non–ami-noglycoside-containing double–β-lactam regimen. The risks and benefits of the alternative regimens are as follows.
Streptomycin Therapy Although there are no published studies comparing the effi-cacy of regimens containing streptomycin or gentamicin, a similar cure rate was reported in a single noncomparative study.197 The main advantage is that streptomycin is less neph-rotoxic than gentamicin. There are several disadvantages of using streptomycin-containing regimens, including a lack of familiarity among clinicians with streptomycin, a higher risk of ototoxicity, which may not be reversible, and drug avail-ability limitations. In addition, most laboratories do not rou-tinely perform serum streptomycin assays and may not have access to a clinical pharmacist to assist in dosing adjustments. Streptomycin use should be avoided in patients with creati-nine clearance <50 mL/min. If the strain of enterococcus is susceptible to both gentamicin and streptomycin, it is reason-able to use gentamicin rather than streptomycin for therapy. When gentamicin therapy is not an option, then a double–β-lactam regimen (see later section) is reasonable.
Short-Course (≈2-Week) Gentamicin Therapy Olaison and Schadewitz189 in Sweden reported a 5-year pro-spective study of 78 cases of enterococcal IE treated with a β-lactam and an aminoglycoside. The older age of these patients was a factor in their inability to tolerate prolonged aminoglycoside therapy. The median duration of aminogly-coside therapy was 15 days, and the microbiological cure and survival rates were similar to those for patients who received longer courses of gentamicin therapy. The major advantage of short-course aminoglycoside therapy is reduced risk of aminoglycoside-associated nephrotoxicity. The disadvan-tage is that this is a single nonrandomized, noncomparative study. The results of a Danish pilot study185 that represented a “before and after” study, which was based on 2007 guidelines that recommended a 2-week treatment course of gentamicin for enterococcal IE in combination with β-lactam therapy for 4 to 6 weeks, confirmed the results seen in the Swedish investigation.181 Double–β-Lactam Regimens for E faecalis IE Most strains of E faecalis are inhibited but not killed in vitro by penicillin or ampicillin, with MICs usually 2 to 4 µg/ mL penicillin; ampicillin MICs are usually 1 dilution lower. Cephalosporins and antistaphylococcal penicillins (oxacillin, nafcillin) have minimal or no in vitro activity against entero-cocci. The in vitro activity of carbapenems is variable, with imipenem being most active.
Because there are few therapeutic alternatives to amino-glycoside-containing regimens, combinations of β-lactams were tested in vitro and in animal models of enterococcal experimental IE. The combination of ampicillin and imipe-nem acted synergistically in vitro and was effective therapy of multidrug-resistant enterococcal experimental IE.198 This study led to additional studies of experimental IE that dem-onstrated that the combination of ampicillin-ceftriaxone was effective therapy for gentamicin-susceptible or high-level gentamicin-resistant E faecalis experimental IE.199 The likely mechanism of double–β-lactam combinations against entero-cocci is saturation of different penicillin-binding proteins. These in vitro and in vivo studies provided the rationale for double–β-lactam therapeutic trials in humans with E faecalis IE caused by gentamicin-susceptible or high-level gentami-cin-resistant strains. A large, multicenter study by Spanish and Italian investigators compared ampicillin-ceftriaxone with ampicillin-gentamicin therapy of E faecalis IE.191 Patients with high-level aminoglycoside-resistant strains were not treated with ampicillin-gentamicin. A smaller study by this group compared ceftriaxone-ampicillin therapy of aminogly-coside-susceptible with high-level aminoglycoside-resistant E faecalis IE.190 Both of these studies had significant limitations: They were observational, largely retrospective, and nonran-domized; the regimens were not standardized among the dif-ferent centers; discontinuation of gentamicin therapy was at the discretion of the investigators and not always the result of gentamicin-associated nephrotoxicity; and the serum con-centrations of gentamicin were not assessed or reported in all study sites.
Downloaded from by on October 16, 2018 1458 Circulation October 13, 2015 Despite these limitations, these 2 studies provide impor-tant data. First, these are the largest series of E faecalis IE reported to date, 43 patients in 1 study190 and 272 in the other study.191 Second, high-level aminoglycoside-resistant E faeca-lis IE treated with ampicillin-ceftriaxone therapy was present in 50% of the patients in the smaller study and 33% of patients in the larger study. Third, none of the patients in either study developed nephrotoxicity with ampicillin-ceftriaxone ther-apy, whereas 20 of 87 (23%) ampicillin-gentamicin–treated patients developed nephrotoxicity (P<0.001). Fourth, in the larger study, the median age was 70 years in both treatment groups; however, patients in the ampicillin-ceftriaxone group were generally sicker and had more comorbid conditions (eg, chronic renal failure [P=0.004], neoplasm [P=0.015], and nosocomial acquisition of infection [P=0.006]). Fifth, in 1 study, PVE was present in 59 (37%) and 30 (34%) of patients treated with ampicillin-ceftriaxone and ampicillin-gentamicin, respectively, with similar success rates. Sixth, in the larger study, there were no significant differences between ampicillin-ceftriaxone and ampicillin-gentamicin in the need for surgery, complications (except for fewer cases of renal fail-ure in the ampicillin-ceftriaxone group), relapse, or mortality. Finally, the overall microbiological cure and success rates for ampicillin-ceftriaxone therapy in both studies were similar to rates in previously reported studies in patients treated with aminoglycoside-containing regimens.190,191 The major advantages of the ampicillin-ceftriaxone regi-men are the lower risk of nephrotoxicity and the lack of need for measuring aminoglycoside serum concentrations. The potential disadvantage is the possibility of hypersensitivity reactions to 2 separate β-lactams. Because it would likely not be possible to discriminate between hypersensitivities related to ampicillin or to ceftriaxone, both drugs might have to be discontinued with substitution of vancomycin-gentamicin therapy. At this time, the writing group does not have a prefer-ence for one regimen over the other but rather advocates an individualized approach to regimen selection for each patient.
Recommendations 1. Therapy that includes either ampicillin or aqueous crystalline penicillin G plus gentamicin or ampicil-lin plus ceftriaxone is reasonable (Class IIa; Level of Evidence B).
2. Either 4 or 6 weeks of therapy is reasonable for NVE, depending on the duration of IE symptoms before the initiation of therapy if ampicillin or peni-cillin plus gentamicin is used (Class IIa; Level of Evidence B).
3. Six weeks of therapy is reasonable if ampicillin plus ceftriaxone is selected as the treatment regimen, regardless of symptom duration (Class IIa; Level of Evidence B).
4. Six weeks of antimicrobial therapy is reasonable for PVE (Class IIa; Level of Evidence B).
5. Streptomycin should be avoided in patients with creatinine clearance <50 mL/min (Class III; Level of Evidence B).
6. If the strain of Enterococcus is susceptible to both gentamicin and streptomycin, it is reasonable to use gentamicin rather than streptomycin for therapy (Class IIa; Level of Evidence C).
7. When gentamicin therapy is not an option, then a double–β-lactam regimen (see later section) is rea-sonable (Class IIa; Level of Evidence B).
E faecalis IE Susceptible to Penicillin, Resistant to Aminoglycosides, or Gentamicin Resistant and Streptomycin Susceptible Aminoglycoside resistance in enterococci is most commonly the result of the acquisition of plasmid-mediated aminoglyco-side-modifying enzymes. E faecalis strains resistant to high levels of gentamicin are resistant to most other aminoglyco-sides, although some of them are susceptible to streptomycin. The regimens for E faecalis IE with strains that are penicil-lin-susceptible and aminoglycoside-resistant are shown in Table 13. Ceftriaxone-ampicillin therapy is reasonable and is given for 6 weeks. The rationale for double–β-lactam therapy is outlined above.
For gentamicin-resistant and streptomycin-susceptible E faecalis, ampicillin-ceftriaxone is reasonable. The 2005 AHA document12 recommended streptomycin for patients with gentamicin-resistant strains of enterococci. The limitations of streptomycin use are summarized above. The total number of cases published in the European studies far exceeds the rela-tively small number of reported streptomycin-treated patients with enterococcal IE. Although there are no published data comparing ampicillin-ceftriaxone with streptomycin-con-taining regimens, we believe that ampicillin-ceftriaxone is reasonable for these patients. Disadvantages of streptomycin-containing regimens are outlined above.
Recommendations 1. Ceftriaxone-ampicillin combination therapy is reasonable for IE caused by aminoglycoside-resistant enterococcal strains (Class IIa; Level of Evidence B).
2. For gentamicin-resistant and streptomycin-suscep-tible Enterococcus species, ampicillin-ceftriaxone combination therapy is reasonable (Class IIa; Level of Evidence B).
Vancomycin Therapy for Enterococcal IE in Patients Unable to Tolerate β-Lactams or Patients With E faecalis Resistant to Penicillin The regimens that are reasonable for these patients are shown in Table 14. Vancomycin should be administered only if a patient is unable to tolerate penicillin or ampicillin. Combinations of pen-icillin or ampicillin with gentamicin are preferable to combined vancomycin-gentamicin because of the potential increased risk of ototoxicity and nephrotoxicity with the vancomycin-genta-micin combination. Moreover, combinations of penicillin or ampicillin and gentamicin are more active than combinations of vancomycin and gentamicin in vitro and in animal models of experimental IE. It is reasonable that patients with NVE receive 6 weeks of vancomycin-gentamicin therapy and that patients with PVE receive at least 6 weeks of therapy.
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1459 Rarely, strains of E faecalis produce an inducible β-lactamase. These β-lactamase–producing strains are susceptible to ampicil-lin-sulbactam and to vancomycin. Intrinsic penicillin resistance is uncommon in E faecalis but is common in E faecium. It is reasonable to treat patients with E faecalis IE caused by strains that are intrinsically resistant to penicillin with a combination of vancomycin plus gentamicin. Recommendations for treatment of IE caused by these strains are shown in Table 14.
Recommendations 1. Vancomycin should be administered only if a patient is unable to tolerate penicillin or ampicillin (Class I; Level of Evidence B).
2. It is reasonable that patients with NVE receive 6 weeks of vancomycin-gentamicin therapy and that patients with PVE receive at least 6 weeks of ther-apy (Class IIa; Level of Evidence B).
3. Patients with E faecalis IE caused by strains that are intrinsically resistant to penicillin should be treated with a combination of vancomycin plus gentamicin (Class I; Level of Evidence B).
Enterococcal Endocarditis Resistant to Penicillin, Aminoglycosides, and Vancomycin The rapid emergence of vancomycin-resistant enterococci has become a global issue of major clinical importance. Most of these strains are E faecium, and as many as 95% of strains express multidrug resistance to vancomycin, aminoglyco-sides, and penicillins. Only about 3% of E faecalis strains are multidrug resistant, and many vancomycin-resistant E faecalis are penicillin susceptible. Fortunately, E faecium IE is uncom-mon. Most of the reports of multidrug-resistant E faecium IE are single case reports, reports of a small number of collected cases, or cases reported in new drug trials.200 Enterococci are considered to be resistant to vancomycin if MICs are >4 µg/mL. Linezolid and daptomycin are the only 2 antimicrobial agents currently available in the United States that may be useful for the treatment of multidrug-resistant E faecium IE. Quinupristin-dalfopristin may be active in vitro but only against strains of E faecium and is inactive against E faecalis. Quinupristin-dalfopristin is rarely used because of severe side effects, including intractable muscle pain. Tigecycline is active in vitro against some strains of multidrug-resistant enterococci, but there are minimal published data on its use clinically. The same can be said for tedizolid, which has been released.
Table 15 lists possible therapeutic options for the treat-ment of multidrug-resistant enterococcal IE. These patients should be managed by specialists in infectious diseases, car-diology, cardiovascular surgery, clinical pharmacy, and, if necessary, pediatrics. Antimicrobial regimens are discussed as follows.
Linezolid Linezolid is a synthetic drug that is the first member of the oxazolidinone class. It acts by inhibiting ribosomal protein Table 13. Therapy for Endocarditis Involving a Native or Prosthetic Valve or Other Prosthetic Material Resulting From Enterococcus species Caused by a Strain Susceptible to Penicillin and Resistant to Aminoglycosides or Streptomycin-Susceptible Gentamicin-Resistant in Patients Able to Tolerate β-Lactam Therapy Regimen Dose† and Route Duration, wk Strength of Recommendation Comments Double β-lactam Ampicillin 2 g IV every 4 h 6 Class IIa; Level of Evidence B Double β-lactam is reasonable for patients with normal or impaired renal function abnormal cranial nerve VIII function or if the laboratory is unable to provide rapid results of streptomycin serum concentration; native valve infection with symptoms of infection <3-mo duration may be treated for 4 wk with the streptomycin-containing regimen. PVE, NVE with symptoms >3 mo, or treatment with a double β-lactam regimen require a minimum of 6 wk of therapy. Plus Ceftriaxone 2 g IV every 12 h Alternative for streptomycin susceptible/gentamicin resistant Either 4–6 Class IIa; Level of Evidence B Use is reasonable only for patients with availability of rapid streptomycin serum concentrations. Patients with creatinine clearance <50 mL/min or who develop creatinine clearance <50 mL/min during treatment should be treated with double–β- lactam regimen. Patients with abnormal cranial nerve VIII function should be treated with double–β-lactam regimen. Ampicillin sodium 2 g IV every 4 h Or Aqueous penicillin G sodium 18–30 million U/24 h IV either continuously or in 6 equally divided doses Plus Streptomycin sulfate‡ 15 mg/kg ideal body weight per 24h IV or IM in 2 equally divided doses IM indicates intramuscular; IV, intravenous; NVE, native valve infective endocarditis; and PVE, prosthetic valve infective endocarditis.
For patients unable to tolerate a β-lactam, see Table 14.
†Doses recommended for patients with normal renal and hepatic function.
‡Streptomycin dose should be adjusted to obtain a serum peak concentration of 20 to 35 µg/mL and a trough concentration of <10 µg/mL.
Downloaded from by on October 16, 2018 1460 Circulation October 13, 2015 synthesis and is approved for use by the FDA in adults and children. It is not approved by the FDA for treatment of IE. Linezolid is bacteriostatic in vitro against enterococci, and susceptibility of enterococci to linezolid ranges from 97% to 99%, including strains that are multidrug resistant. Enterococci with MIC >2 µg/mL are considered to be resistant to linezolid. However, linezolid-resistant strains have devel-oped during treatment.201 In a small number of patients, linezolid was effective ther-apy of vancomycin-resistant E faecium IE.174 Birmingham et al202 reported cure in 17 of 22 courses of therapy (77%) for E faecium IE. Mave et al203 reported cure in 2 of 3 patients with E faecium IE with linezolid. Other case reports of cure of E faecium IE of native valve204 or prosthetic valve205 were reported. However, linezolid treatment failures of E faecium IE also were reported.206 The advantages of linezolid therapy include high bio-availability of the oral formulation, approval for pediatric patients, and a lack of many therapeutic alternatives. The disadvantages are toxicity (mild to severe neutropenia and thrombocytopenia that is reversible); peripheral and optic neuritis, which is more often seen with longer durations of therapy and may not be reversible; multidrug interactions, especially serotonin uptake inhibitors; and emergence of resistance during treatment. The previous high cost should decrease with generic availability soon. Cardiac valve replacement surgery may be necessary in patients who do not respond to linezolid therapy.
Daptomycin Daptomycin is a cyclic lipopeptide antibiotic that has bacteri-cidal activity in vitro against susceptible strains of enterococci. Enterococci are considered daptomycin susceptible with MIC <4 µg/mL. Although >90% of enterococci are reportedly susceptible in vitro to daptomycin, the emergence of dapto-mycin resistance is an increasing problem.207 Daptomycin is FDA approved for treatment of S aureus infections but not for enterococcal infections. Daptomycin is not approved for use in pediatric patients.
The number of published cases of vancomycin-resistant E faecium IE treated with daptomycin is extremely small, so management conclusions are difficult to define, and the success rate has varied among reported cases. Levine and Lamp208 reported daptomycin cure in 6 of 9 patients with E faecium IE; both daptomycin-treated patients with E faecium IE reported by Segreti et al209 died. Multiple other case reports describe daptomycin failures, some as a result of emergence of daptomycin-resistance during treatment.210,211 Other inves-tigators have suggested that higher doses of daptomycin (8–10 mg·kg−1·d−1); daptomycin combined with gentamicin, ampicillin, ceftaroline, rifampin, or tigecycline; or various combinations of these should be used instead of daptomycin monotherapy.211–217 A number of in vitro evaluations214–216 sug-gested that ampicillin and ceftaroline in combination with daptomycin demonstrate the greatest synergistic activity com-pared with other β-lactam–daptomycin combinations.
Mave et al203 compared daptomycin with linezolid for vancomycin-resistant Enterococcus bacteremia. Five patients had E faecium IE; 1 of 2 daptomycin-treated patients and 2 of 3 linezolid-treated patients survived. The number of cases of vancomycin-resistant Enterococcus bacteremia was too small to draw significant conclusions about treatment response rates.
In summary, there are insufficient data to recommend monotherapy with daptomycin for the treatment of multidrug-resistant enterococcal IE. If daptomycin therapy is selected, then doses of 10 to 12 mg·kg−1·24 h−1 may be considered. Consideration may be given to combinations of therapy with daptomycin, including ampicillin or ceftaroline, particularly in patients infected with strains with relatively high MICs to daptomycin within the susceptible range (<4 µg/mL). Other less active (in vitro) combinations with daptomycin include gentamicin, rifampin, or tigecycline.
Table 14. Vancomycin-Containing Regimens for Vancomycin- and Aminoglycoside-Susceptible Penicillin-Resistant Enterococcus Species for Native or Prosthetic Valve (or Other Prosthetic Material) IE in Patients Unable to Tolerate β-Lactam Regimen Dose and Route Duration, wk Strength of Recommendation Comments Unable to tolerate β-lactams Vancomycin† 30 mg/kg per 24 h IV in 2 equally divided doses 6 Class IIa; Level of Evidence B Plus Gentamicin‡ 3 mg/kg per 24 h IV or IM in 3 equally divided doses 6 Penicillin resistance; intrinsic or β-lactamase producer Vancomycin 30 mg/kg per 24 h IV in 2 equally divided doses 6 Class IIb; Level of Evidence C For β-lactamase–producing strain, if able to tolerate a β-lactam antibiotic, ampicillin-sulbactam§ plus aminoglycoside therapy may be used. Plus Gentamicin‡ 3 mg/kg per 24 h IV or IM in 3 equally divided doses 6 IE indicates infective endocarditis; IM, intramuscular; and IV, intravenous.
Doses recommended are for adults with normal renal function.
†Dose of vancomycin should be adjusted to obtain a serum trough concentration of 10 to 20 µg/mL.
‡Dose of gentamicin should be adjusted to obtain serum peak and trough concentrations of 3 to 4 and <1 µg/mL, respectively.
§Ampicillin-sulbactam dosing is 3 g/6 hour IV.
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1461 Recommendations 1. Patients with IE attributable to Enterococcus spe-cies resistant to penicillin, aminoglycosides, and vancomycin should be managed by specialists in infectious diseases, cardiology, cardiovascular sur-gery, clinical pharmacy, and, if necessary, pediatrics (Class I; Level of Evidence C).
2. If daptomycin therapy is selected, then doses of 10 to 12 mg·kg−1·24 h−1 may be considered (Class IIb; Level of Evidence C).
3. Combination therapy with daptomycin and ampi-cillin or ceftaroline may be considered, especially in patients with persistent bacteremia or enterococcal strains with high MICs (ie, 3 µg/mL) to daptomy-cin within the susceptible range (Class IIb; Level of Evidence C).
HACEK Microorganisms IE caused by fastidious Gram-negative bacilli of the HACEK group (HACEK indicates Haemophilus species, Aggregatibacter species, Cardiobacterium hominis, Eikenella corrodens, and Kingella species) accounts for ≈5% to 10% of community-acquired NVE in patients who are not IDUs.218 These microorganisms grow slowly in standard blood cul-ture media, and recovery may require prolonged incubation. Typically, only a small fraction of blood culture bottles in patients with HACEK IE demonstrate growth. Bacteremia caused by HACEK microorganisms in the absence of an obvi-ous focus of infection is highly suggestive of IE even without typical physical findings of IE.
Previously, the HACEK group of microorganisms was uniformly susceptible to ampicillin. However, β-lactamase– producing strains of HACEK are appearing with increased frequency; rarely, resistance to ampicillin can occur in β-lactamase–negative strains.219 Moreover, difficulty in per-forming antimicrobial susceptibility testing as a result of failure of growth in in vitro susceptibility testing is common-place. In 1 survey, 60% of isolates did not grow adequately in control wells, and no valid in vitro susceptibility results were available.219 Therefore, unless growth is adequate for in vitro screening, then HACEK microorganisms should be consid-ered ampicillin resistant, and penicillin and ampicillin should not be used to treat patients with IE in these cases. Almost all strains of the HACEK group are susceptible to ceftriaxone (or other third- or fourth-generation cephalosporins). Ceftriaxone has commonly been used to treat HACEK IE220 and is reason-able for treatment (Table 16). The duration of therapy for NVE of 4 weeks is reasonable; for PVE, the duration of therapy of 6 weeks is reasonable. Gentamicin is no longer recommended because of its nephrotoxicity risks.
The HACEK group is usually susceptible in vitro to fluo-roquinolones.206 On the basis of these susceptibility data, a flu-oroquinolone (ciprofloxacin, levofloxacin, or moxifloxacin) may be considered as an alternative agent in patients unable to tolerate ceftriaxone (or other third- or fourth-generation cephalosporins) therapy. There are only a few case reports of HACEK IE treated with a fluoroquinolone, however. In addi-tion, ampicillin-sulbactam may be considered a treatment option, although HACEK resistance to this agent in vitro has been described.219 Accordingly, patients with HACEK IE who cannot tolerate ceftriaxone therapy should be treated in con-sultation with an infectious diseases specialist.
Recommendations 1. Unless growth is adequate in vitro to obtain suscep-tibility testing results, HACEK microorganisms are considered ampicillin resistant, and penicillin and ampicillin should not be used for the treatment of patients with IE (Class III; Level of Evidence C).
2. Ceftriaxone is reasonable for treatment of HACEK IE (Class IIa; Level of Evidence B).
3. The duration of therapy for HACEK NVE of 4 weeks is reasonable (Class IIa; Level of Evidence B); for HACEK PVE, the duration of therapy of 6 weeks is reasonable (Class IIa; Level of Evidence C).
4. Gentamicin is not recommended because of its nephrotoxicity risks (Class III; Level of Evidence C).
5. A fluoroquinolone (ciprofloxacin, levofloxacin, or moxifloxacin) may be considered an alternative agent for patients unable to tolerate ceftriaxone (or other third- or fourth-generation cephalosporins) (Class IIb; Level of Evidence C).
6. Ampicillin-sulbactam may be considered a treat-ment option for HACEK IE (Class IIb; Level of Evidence C).
7. Patients with HACEK IE who do not tolerate cef-triaxone therapy should be treated in consultation with an infectious diseases specialist (Class I; Level of Evidence C).
Table 15. Therapy for Endocarditis Involving a Native or Prosthetic Valve or Other Prosthetic Material Resulting From Enterococcus Species Caused by Strains Resistant to Penicillin, Aminoglycosides, and Vancomycin Regimen Dose and Route Duration, wk Strength of Recommendation Comments Linezolid 600 mg IV or orally every 12 h >6 Class IIb; Level of Evidence C Linezolid use may be associated with potentially severe bone marrow suppression, neuropathy, and numerous drug interactions. Patients with IE caused by these strains should be treated by a care team including specialists in infectious diseases, cardiology, cardiac surgery, clinical pharmacy, and, in children, pediatrics. Cardiac valve replacement may be necessary for cure. Or Daptomycin 10–12 mg/kg per dose >6 Class IIb; Level of Evidence C IE indicates infective endocarditis, and IV, intravenous.
Doses recommended are for patients with normal renal and hepatic function.
Downloaded from by on October 16, 2018 1462 Circulation October 13, 2015 Non-HACEK Gram-Negative Bacilli IE caused by non-HACEK Gram-negative aerobic bacilli (Enterobacteriaceae and Pseudomonas species) is rare. In 1 large multinational database221 that included 2761 patients seen in 61 hospitals in 28 countries, only 49 cases (1.8%) were attributable to non-HACEK Gram-negative aerobic bacilli. It is noteworthy that healthcare exposure was asso-ciated with the development of IE caused by this group of organisms in 57% of patients. In contrast, IDU, a prominent risk factor for the development of this IE syndrome in ear-lier years, was recognized in only 4% of cases in the mul-tinational survey that included cases seen between 2000 and 2005. Escherichia coli and Pseudomonas aeruginosa accounted for 51% of cases, and 59% had PVE. Although management included cardiac surgery in 51% of cases, the in-hospital mortality rate was 24%.
Despite the very rare occurrence of IE caused by Salmonella species in North America, this syndrome deserves specific mention because it occurs with some frequency in other geo-graphic areas.222 Salmonella species have a proclivity to infect cardiovascular structures in adults. Therefore, all patients with bloodstream infection resulting from Salmonella species should be evaluated for complicating cardiovascular infec-tions, including IE, myocarditis, pericarditis, and endarteritis. Although many serotypes have been implicated, most cases are caused by S choleraesuis, S typhimurium, and S enteritidis.222 Cardiac surgery in combination with prolonged courses of combined antibiotic therapy is reasonable for most patients with IE caused by non-HACEK Gram-negative aerobic bacilli, particularly in the setting of left-sided valvular involvement.
Prospective trial data are lacking to define the optimal antimicrobial regimen for the treatment of IE caused by non-HACEK Gram-negative aerobic bacilli. Input from special-ists in infectious diseases who are experienced in the medical management of IE should be obtained to define an antibiotic regimen in each case. This is particularly important in IE caused by non-HACEK Gram-negative aerobic bacilli for sev-eral reasons. First, as stated previously, healthcare exposure is commonly seen in these cases; thus, multidrug resistance often characterizes these pathogens. Second, therapy may include agents with increased toxicity risks such as aminogly-cosides (given in high dosages) and colistin. Third, because regimens that include >1 agent are often selected for use, the risks of drug-drug interactions and increased drug-related adverse events are problematic. Fourth, mortality is high in these infections, and medical-surgical approaches are often required for optimal management and favorable outcomes.
Combination antibiotic therapy with a β-lactam (penicillins, cephalosporins, or carbapenems) and either an aminoglycoside or fluoroquinolone for 6 weeks is reasonable.221 Consultation with an infectious diseases expert in IE should be sought because of the various mechanisms of antibiotic resistance that can be found in the non-HACEK Gram-negative aerobic bacilli. For example, several of these bacteria may harbor “inducible β-lactamases” that could require supplemental laboratory screening, in addition to routine in vitro susceptibility testing.
Medical therapy may be successful in right-sided P aerugi-nosa IE in 50% to 75% of cases. If the disease is refractory to anti-biotics, then partial tricuspid valvulectomy or “vegetectomy”223 without valve replacement is indicated.224 Typically, these patients have been IDUs, and because of their high recidivism risk, avoid-ance of placement of prosthetic valves is desirable.
Recommendations 1. Cardiac surgery is reasonable in combination with prolonged courses of combined antibiotic therapy for most patients with IE caused by non-HACEK Gram-negative aerobic bacilli, particularly P aeru-ginosa (Class IIb; Level of Evidence B).
Table 16. Therapy for Endocarditis Involving a Native or Prosthetic Valve or Other Prosthetic Material Caused by HACEK Microorganisms Regimen Dose and Route Duration, wk Strength of Recommendation Comments Ceftriaxone sodium 2 g/24 h IV or IM in 1 dose 4, NVE; 6, PVE Class IIa; Level of Evidence B Preferred therapy: cefotaxime or another third- or fourth-generation cephalosporin may be substituted. Or Ampicillin sodium 2 g IV every 4 h Class IIa; Level of Evidence B Ampicillin sodium may be an option if the growth of the isolate is sufficient to permit in vitro susceptibility results. Or Ciprofloxacin† 1000 mg/24 h orally or 800 mg/24 h IV in 2 equally divided doses Class IIb; Level of Evidence C Fluoroquinolone therapy‡ may be considered for patients unable to tolerate cephalosporin and ampicillin therapy; levofloxacin or moxifloxacin may be substituted; fluoroquinolones generally is not recommended for patients <18 y old. Treatment for 6 wk is reasonable in patients with PVE (Class IIa; Level of Evidence C).
HACEK indicates Haemophilus species, Aggregatibacter species, Cardiobacterium hominis, Eikenella corrodens, and Kingella species; IM, intramuscular; IV, intravenous; NVE, native valve infective endocarditis; and PVE, prosthetic valve infective endocarditis.
Patients should be informed that intramuscular injection of ceftriaxone is painful.
†Dose recommended for patients with normal renal function.
‡Fluoroquinolones are highly active in vitro against HACEK microorganisms. Published data on the use of fluoroquinolones for endocarditis caused by HACEK are minimal.
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1463 2. Combination antibiotic therapy with a β-lactam (penicillins, cephalosporins, or carbapenems) and either an aminoglycoside or fluoroquinolone for 6 weeks is reasonable (Class IIa; Level of Evidence C).
3. Consultation with an infectious diseases expert in IE should be sought because of the various mechanisms of antibiotic resistance that can be found in the non-HACEK Gram-negative aerobic bacilli (Class I; Level of Evidence C).
Culture-Negative Endocarditis Positive blood cultures are a major diagnostic criterion in IE and key to identifying an pathogenic agent and an optimal antimicrobial regimen.225,226 Continuous bacteremia and a high frequency of positive blood cultures are typical hallmarks of this infection. The intensity of bacteremia may not be great, however, with <50 colony-forming units per 1 mL blood detected in the majority of patients in an investigation.227 Failure to culture microorganisms that cause IE can be a major problem that complicates diagnosis and timely, effec-tive treatment. Although most previous studies have put the frequency of blood culture–negative IE at 5% to 10%, a European study of IE that included 820 cases indicated that ≈20% of patients with confirmed IE had all negative blood cultures.228 This may be attributable to inadequate microbio-logical techniques, infection with highly fastidious bacteria or fungi, noncultivatable agents, or the previous administration of antimicrobial agents before blood cultures were obtained. Administration of antimicrobial agents to IE patients before blood cultures are obtained reduces the recovery rate of bac-teria by 35% to 40%.228–232 The antimicrobial susceptibility of the organism, the dose, and the duration and nature of pre-vious antimicrobial therapy together determine the length of time that blood cultures will remain negative.232 IE patients with blood cultures that are initially negative after only a few days of antibiotic therapy may have positive blood cultures after several days without antibiotics. The blood cultures of patients who receive longer courses of high-dose bactericidal antimicrobials may remain negative for weeks.
Selection of medical therapy for patients with culture-negative IE is difficult. On the one hand, there is a need to provide empirical antimicrobials for all likely pathogens. On the other hand, certain therapeutic agents, including amino-glycosides, have potentially toxic effects that dictate limita-tion or avoidance of use if at all possible. Moreover, some of the laboratory-based diagnostic techniques to define fas-tidious or unusual pathogens are not available in most clinical laboratories and require considerable time for completion of testing if specimens are sent to a referral laboratory.233 During this period, patients are often treated empirically for the more common bacterial causes of IE, which can result in exposure to potentially toxic therapy that could be avoided with earlier pathogen identification.
An evaluation of epidemiological factors (Table 6), history of prior infections including cardiovascular infections, expo-sure to antimicrobials, clinical course, severity, and extracar-diac sites of infection of the current infection should be done in all IE cases. During the period between the collection of blood cultures and the determination of a pathogen or if blood cultures are ultimately deemed culture negative, empirical therapy is generally required. Consultation with an infectious diseases specialist to define the most appropriate choice of therapy is recommended. Collection of additional clinical and laboratory data often dictates subsequent revisions in initial empirical therapy that will be administered over the treatment course.
For patients with acute (days) clinical presentations of native valve infection, coverage for S aureus, β-hemolytic streptococci, and aerobic Gram-negative bacilli is reasonable. Empirical coverage could include vancomycin and cefepime as an initial regimen. For patients with a subacute (weeks) presentation of NVE, empirical coverage of S aureus, VGS, HACEK, and enterococci is reasonable. One treatment option could include vancomycin and ampicillin-sulbactam to pro-vide some coverage for these organisms. Subsequent regimen revision can be done when a pathogen is recovered from blood cultures.
For patients with culture-negative PVE, coverage for staphylococci, enterococci, and aerobic Gram-negative bacilli is reasonable if the onset of symptoms is within 1 year of prosthetic valve placement. A regimen could include vanco-mycin, rifampin, gentamicin, and cefepime. If symptom onset is >1 year after valve placement, then IE is more likely to be caused by staphylococci, VGS, and enterococci, and antibiotic therapy for these potential pathogens is reasonable. One initial treatment option could include vancomycin and ceftriaxone.
If subsequent blood culture results or other laboratory methodologies define a pathogen, then empirical therapy should be revised to focused therapy that is recommended for the specific pathogen identified.
True culture-negative IE can be caused by uncommon or rare pathogens that do not grow in routinely used blood culture systems.234–237 The organisms that have garnered the most attention are Bartonella species, Chlamydia species, C burnetii, Brucella species, Legionella species, Tropheryma whipplei, Candida, and non-Candida fungi (particularly Aspergillus species). The last 2 groups of organisms are espe-cially relevant to PV recipients. With the use of special diag-nostic techniques, Bartonella species, C burnetii, and Brucella species have been identified in the majority of cases of cul-ture-negative IE caused by fastidious organisms. Additional laboratory screening is required to identify the causes of culture-negative IE.233 In some cases, serological and special blood culture techniques can be helpful. In other cases, tissue (usually valve) screening is required. Diagnostic methods for resected valve tissue include microbiological, histopathologi-cal, and molecular techniques, the last of which includes gene amplification with PCR methods. Unfortunately, most clinical laboratories do not perform molecular screening, and speci-mens must be sent to reference laboratories.
The most prevalent pathogen among these uncommon causes of culture-negative IE in this group has varied glob-ally according to published data.236 Incidence data from population-based surveys for IE caused by these organisms are lacking in the United States. In PVE cases, the timing of infection onset can also be important in defining pathogens.235 Limitations such as referral bias and sampling bias may have affected the findings.235–237 Downloaded from by on October 16, 2018 1464 Circulation October 13, 2015 Results of a large prospective analysis of referred samples from culture-negative IE performed by a well-recognized reference laboratory deserve additional comment.236 First, there was identification of a pathogen in 62.7% of 759 cases; in 2.5%, a noninfectious origin (see below) was confirmed. Second, serological results were positive in 47.7% of cases, primarily for Coxiella and Bartonella species infection. Third, PCR identified a pathogen in two thirds of the valves studied. Fourth, no cause was defined in 35% of cases.
Treatment of the wide variety of microorganisms that cause culture-negative IE without prior antibiotic exposure has been described anecdotally, and regimens of choice are based on limited data and can be found in other publications.
Noninfectious causes of valvular vegetations can produce a syndrome similar to culture-negative IE. Perhaps the one that has received the most attention is anti-phospholipid anti-body (APA) syndrome,238 which has been described as both a primary and a secondary syndrome and is associated with the presence of APA. In its secondary form, the APA syndrome has been linked to autoimmune disorders, particularly sys-temic lupus erythematosus, and malignancies. Sterile valvular vegetations form and embolize, clinically mimicking in many respects culture-negative IE. The mitral valve is most often affected, and valvular regurgitation is the predominant func-tional abnormality seen in APA syndrome with complicating valvular involvement. To complicate matters, the APA syn-drome may develop secondary to IE.239 Numerous other causes of noninfective vegetative endo-carditis can mimic IE. These can be categorized into 4 groups217: neoplasia associated (atrial myxoma, marantic endocarditis, neoplastic disease, and carcinoid), autoimmune associated (rheumatic carditis, systemic lupus erythematosus, polyarteritis nodosa, and Behçet disease), postvalvular surgery (thrombus, stitch, or other postsurgery changes), and miscel-laneous (eosinophilic heart disease, ruptured mitral chordae, and myxomatous degeneration).
Recommendations 1. An evaluation of epidemiological factors, history of prior infections including cardiovascular infections, exposure to antimicrobials, clinical course, sever-ity, and extracardiac sites of infection of the current infection should be performed in all culture-nega-tive endocarditis cases (Class I; Level of Evidence C).
2. Consultation with an infectious diseases specialist to define the most appropriate choice of therapy in patients with culture-negative endocarditis is rec-ommended (Class I; Level of Evidence C).
3. For patients with acute (days) clinical presentations of native valve infection, coverage for S aureus, β-hemolytic streptococci, and aerobic Gram-negative bacilli is reasonable (Class IIa; Level of Evidence C).
4. For patients with a subacute (weeks) presenta-tion of NVE, coverage of S aureus, VGS, HACEK, and enterococci is reasonable (Class IIa; Level of Evidence C).
5. For patients with culture-negative PVE, cover-age for staphylococci, enterococci, and aerobic Gram-negative bacilli is reasonable if onset of symp-toms is within 1 year of prosthetic valve placement (Class IIa; Level of Evidence C).
6. If symptom onset is >1 year after valve placement, then IE is more likely to be caused by staphylococci, VGS, and enterococci, and antibiotic therapy for these potential pathogens is reasonable (Class IIa; Level of Evidence C).
7. If subsequent blood culture results or other labora-tory methodologies define a pathogen, then empiri-cal therapy should be revised to focused therapy that is recommended for the specific pathogen iden-tified (Class I; Level of Evidence C).
Fungi Fungal IE is rare but can develop in a wide range of patients.240,241 The well-recognized risk factors associated with fungal IE (eg, IDU and immunocompromised state) have become less prevalent compared with the presence of a car-diovascular device, including central venous catheters, perma-nent pacemakers and defibrillators, and prosthetic valves.240–246 Fungal IE has been recognized as a cause of early PVE, but a case series from a single medical center demonstrated that 43% of these cases had symptom onset >1 years after pros-thetic valve placement.242 In contrast to the expected older age predilection for the development of IE, patients with fungal IE have been younger, which was somewhat unanticipated, con-sidering the low prevalence of IDU among the cohort. Candida and Aspergillus species account for the large majority of fun-gal IE, and Candida-related IE is much more common than Aspergillus-related disease.240,241 Blood cultures are ultimately positive in most cases caused by the former pathogen, whereas they are rarely positive in cases caused by the latter fungus. Thus, Aspergillus is a cause of culture-negative IE, and when this occurs, it is usually in a patient with a prosthetic cardiac valve.240 A variety of other fungi, including endemic mycoses, can rarely cause IE and can involve both native and prosthetic valves. Noncardiac sites of metastatic infection often compli-cate fungal IE; this can include, for example, endophthalmitis in patients with candidal IE, which may require both systemic and intraocular antifungal therapy. Further guidelines are available from the Infectious Diseases Society of America for additional management aspects of several of the fungal patho-gens (
Despite aggressive combined medical and surgical inter-ventions and a younger cohort, mortality rates for fungal IE are unacceptably high. The survival rate for patients with mold-related IE is <20%. Historically, 2 treatment doctrines have prevailed in fungal IE despite the lack of prospective tri-als conducted to define the most appropriate therapy: Fungal IE is a “stand-alone indication” for surgical replacement of an infected valve; and amphotericin B, a fungicidal agent, is the initial drug of choice for fungal IE. Because of the alarming mortality rate associated with fungal IE and the availability of newer antifungal drugs, in particular fungicidal drugs like the echinocandins, a re-evaluation of these principles seems in order. If done, however, this will probably be based on anec-dotal experience and expert opinion rather than on clinical trial data because of the rarity of the syndrome.
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1465 A 2-phase treatment of fungal IE has evolved. The initial or induction phase consists of control of infection. Treatment includes a combination of a parenteral antifungal agent, usu-ally an amphotericin B–containing product, and valve sur-gery. Valve surgery should be done in most cases of fungal IE. Results of a meta-analysis that included 879 cases of Candida IE demonstrated a marked reduction in death (prevalence odds ratio, 0.56; 95% confidence interval, 0.16–1.99) among those who underwent adjunctive valve surgery.244 In addition, patients who were treated with combination therapy including amphotericin B and flucytosine had reduced mortality com-pared with those who received antifungal monotherapy.
Antifungal therapy usually is given for >6 weeks. After completion of this initial therapy, long-term (lifelong) sup-pressive therapy with an oral azole is reasonable.243,244,246 Suppressive therapy has been used in 2 populations. First, because of the high relapse rate of fungal IE and the prolonged delay (years in some cases) in relapse, oral azoles have been administered after combined medical and surgical induction therapy. In a second population with fungal IE, lifelong oral antifungal suppressive therapy has been given to patients who respond clinically to induction medical therapy but are not deemed appropriate surgical candidates for valve replacement for attempted infection cure. Anecdotal case series243,245 indi-cate that IE has been successfully suppressed for months to years. A meta-analysis that included 64 reported patients with Candida IE who did not undergo valve surgery because they were deemed to be unacceptable surgical candidates supports the notion that fluconazole suppressive therapy is useful; 20 of 21 patients (95%) who were ultimately treated with long-term suppressive therapy survived during follow-up, which was ≥6 months.244 Recommendations 1. Valve surgery should be done in most cases of fungal IE (Class I; Level of Evidence B).
2. After completion of initial parenteral therapy, life-long suppressive therapy with an oral azole is rea-sonable (Class IIa; Level of Evidence B).
Surgical Management There is a prevailing opinion that valve surgery is crucial for optimal therapy in selected patients with complicated IE.247–249 In a systematic review5 of 15 population-based IE investiga-tions from 7 countries, after adjustment for country, the pro-portion of IE cases undergoing valve surgery increased 7% per decade (95% confidence interval, −0.4% to 14%; P=0.06) between 1969 and 2000. In surveys involving population-based1 and international multicenter cohorts,10,11 ≈50% of both NVE and PVE patients undergo valve surgery during the active phase of IE (during initial hospitalization before com-pletion of a full therapeutic course of antibiotics).
Although valve replacement surgery has served as an important option in the management of individual IE cases, only 1 small, randomized trial17 has been performed to date to examine the role of valve surgery in the management of IE. In this trial, 76 patients with left-sided NVE, severe valve regur-gitation without heart failure, and vegetations >10 mm were assigned to early surgery within 48 hours or to conventional treatment. Although the authors17 report a reduction in the composite outcome of in-hospital deaths and embolic events with early surgery (3% versus 23%), the differences between the 2 groups were driven by a significant decrease in embolic events with early surgery. Thus, firm conclusions cannot be drawn from this trial on the effect of early surgery on mor-tality, given the small sample size of the study. In addition, patients in this trial were young and had limited comorbidity based on a EuroSCORE, a calculated risk of surgical mortality ( a low prevalence of S aureus IE, and lower mortality compared with most contemporary patient cohorts. Moreover, many patients had signs of embolization (a Class IIa indication for surgery) before randomization. Data from nonrandomized trials from a single-center experience250 and an international collaboration251 support the notion252 that early valve surgery may not be beneficial in all patients with native or PVE caused by S aureus.
Over several decades, expert panels have relied on data from observational studies to make recommendations on the indications for early surgery. Despite the availability of new studies, the indications for surgery have not changed appre-ciably over time.18 Considering that observational studies are prone to bias and confounding, researchers have used regres-sion analysis and calculated propensity scores to adjust for prognostically important baseline differences between surgi-cal and medical patients.14,253 Studies14,15,136,249,253–256 examin-ing the association between valve surgery and outcome in left-sided IE using propensity score analysis, however, have demonstrated conflicting results, likely because of the use of different analytical approaches.
Until 2007, none of the published studies adjusted for sur-vivor bias, which occurs because patients who live longer are more likely to undergo surgery than those who die early. A cor-relation between longer survival and surgery may be wrongly interpreted as evidence that surgical treatment improves sur-vival.257 Since 2007, at least 3 studies15,16,258 have documented the effect of survivor bias on the association between surgery and mortality in IE patients. When adjusted for survivor bias, analyses have shown either a statistical loss of benefit of early surgery or findings indicating that the surgical intervention may actually result in harm.
Between 2007 and 2013, at least 6 observational stud-ies10,11,14–16,259 that adjusted for selection bias, confounding, and survivor bias were conducted. Three studies that included 2 cohorts of patients with NVE and PVE and 1 cohort with NVE showed an association between early surgery and lower mortality in IE patients in general or in specific subgroups of patients such as those with heart failure or paravalvular com-plications. Only 1 study examined the role of valve surgery in PVE.259 After adjustment for differences in clinical char-acteristics and survival bias, early valve replacement was not associated with lower mortality compared with medical therapy in the overall cohort. Subgroup analysis indicated a lower in-hospital and 1-year mortality with early surgery only in the group of patients with the highest surgical propensity. Table 17 summarizes the characteristics of rigorously con-ducted observational studies that support the role of surgery in IE management.
Downloaded from by on October 16, 2018 1466 Circulation October 13, 2015 Indications for Surgery Decisions on surgical intervention are complex and depend on many clinical and prognostic factors257–262 that vary among patients, including infecting organism, vegetation size, presence of perivalvular infection, presence of embolism or heart failure, age, noncardiac comorbidities, and available surgical expertise. There is a paucity of evidence available to define the optimal timing of valve surgery. Decisions on the indication and timing of surgical intervention should be determined by a multispecialty team with expertise in cardiology, imaging, cardiothoracic sur-gery, and infectious diseases.261 Recommendations for early sur-gery in patients with recurrent emboli and persistent vegetations have generally been enacted after clinical events. Whether recur-rent, asymptomatic emboli detected on advanced imaging stud-ies should influence decision making should be considered on an individual basis. Risk stratification models such as the Society of Thoracic Surgeons Endocarditis Score are available to predict morbidity and mortality risks in IE patients after valve surgery and to assist in decision making and patient counseling.260 Early Valve Surgery in Left-Sided NVE: Recommendations 1. Early surgery (during initial hospitalization and before completion of a full course of antibiotics) is indicated in patients with IE who present with valve dysfunction resulting in symptoms or signs of heart failure (Class I; Level of Evidence B).
2. Early surgery should be considered particularly in patients with IE caused by fungi or highly resistant organisms (eg, vancomycin-resistant Enterococcus, multidrug-resistant Gram-negative bacilli) (Class I; Level of Evidence B).
3. Early surgery is indicated in patients with IE com-plicated by heart block, annular or aortic abscess, or destructive penetrating lesions (Class I; Level of Evidence B).
4. Early surgery is indicated for evidence of persistent infection (manifested by persistent bacteremia or fever lasting >5–7 days and provided that other sites of infection and fever have been excluded) after the start of appropriate antimicrobial therapy (Class I; Level of Evidence B).
5. Early surgery is reasonable in patients who present with recurrent emboli and persistent or enlarging vegetations despite appropriate antibiotic therapy (Class IIa; Level of Evidence B).
6. Early surgery is reasonable in patients with severe valve regurgitation and mobile vegetations >10 mm (Class IIa, Level of Evidence B).
7. Early surgery may be considered in patients with mobile vegetations >10 mm, particularly when involving the anterior leaflet of the mitral valve and associated with other relative indications for sur-gery (Class IIb; Level of Evidence C).
Early Valve Surgery in PVE: Recommendations 1. Early surgery is indicated in patients with symp-toms or signs of heart failure resulting from valve dehiscence, intracardiac fistula, or severe prosthetic valve dysfunction (Class I; Level of Evidence B).
2. Early surgery should be done in patients who have persistent bacteremia despite appropriate antibi-otic therapy for 5 to 7 days in whom other sites of infection have been excluded (Class I; Level of Evidence B).
3. Early surgery is indicated when IE is compli-cated by heart block, annular or aortic abscess, or destructive penetrating lesions (Class I; Level of Evidence B).
4. Early surgery is indicated in patients with PVE caused by fungi or highly resistant organisms (Class I; Level of Evidence B).
5. Early surgery is reasonable for patients with PVE who have recurrent emboli despite appropriate anti-biotic treatment (Class IIa; Level of Evidence B).
6. Early surgery is reasonable for patients with relaps-ing PVE (Class IIa; Level of Evidence C).
7. Early surgery may be considered in patients with mobile vegetations >10 mm (Class IIb; Level of Evidence C).
Valve Surgery in Patients With Right-Sided IE Although outcomes are better for patients with right-sided IE compared with patients with left-sided infection, surgi-cal intervention is occasionally considered in the former group. Because many of the patients with right-sided IE develop infection as a result of IDU (see the Right-Sided IE in IDUs section), the general approach is to treat these patients Table 17. Direct Evidence Supporting an Association Between Valve Surgery and Lower Mortality From Observational Studies: Level of Evidence B Study Mortality IE Group PE vs SA Lalani et al10 In-hospital mortality NVE PE Bannay et al15 5-y mortality NVE+PVE PE Kiefer et al11 In-hospital and 1-y mortality CHF (NVE+PVE) PE Lalani et al10 In-hospital mortality Paravalvular complications (NVE) SA Bannay et al15 5-y mortality Intracardiac abscess (NVE+PVE) SA Lalani et al10 In-hospital mortality Systemic embolization (NVE) SA Bannay et al15 5-y mortality Systemic embolization (NVE+PVE) SA Lalani et al10 In-hospital mortality S aureus (NVE) SA Bannay et al15 5-y mortality CHF (NVE+PVE) SA Lalani et al259 In-hospital and 1-y mortality PVE with the highest propensity to undergo surgery SA CHF indicates congestive heart failure; NVE, native valve infective endocarditis; PE, primary end point; PVE, prosthetic valve infective endocarditis; and SA, subgroup analysis. All studies have adjusted for selection and survivor bias and confounding. Valve surgery was performed during the active phase of the disease (during initial hospitalization before completion of a full therapeutic course of antibiotics).
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1467 medically and to avoid placement of valve prostheses because of the subsequent risk of device infection with continued IDU. Surgical intervention is reasonable for patients with the fol-lowing complications: right heart failure secondary to severe tricuspid regurgitation with poor response to medical therapy, sustained infection caused by difficult-to-treat organisms (ie, fungi, multidrug resistant bacteria) or lack of response to appropriate antimicrobial therapy, and tricuspid valve vegeta-tions that are ≥20 mm in diameter and recurrent pulmonary embolism despite antimicrobial therapy. Valve repair rather than replacement should be performed when feasible. If valve replacement is performed, then an individualized choice of prosthesis by the surgeon is reasonable.263,264 Recommendations 1. Surgical intervention is reasonable for patients with certain complications (Class IIa; Level of Evidence C).
2. Valve repair rather than replacement should be per-formed when feasible (Class I; Level of Evidence C).
3. If valve replacement is performed, then an individu-alized choice of prosthesis by the surgeon is reason-able (Class IIa; Level of Evidence C).
4. It is reasonable to avoid surgery when possible in patients who are IDUs (Class IIa; Level of Evidence C).
Valve Surgery in Patients With Prior Emboli/ Hemorrhage/Stroke The timing of valve surgery in IE patients with stroke remains controversial. Stroke is an independent risk factor for post-operative mortality in IE patients. After stroke, neurological deterioration can occur as a result of hemorrhagic transforma-tion with anticoagulation during cardiopulmonary bypass or exacerbation of cerebral ischemia attributable to hypotension during cardiac surgery. The risk of intracranial hemorrhage is dependent on several factors, including extent and size of infarction, whether it is ischemic or hemorrhagic, and the exact timing of surgery.
One clinical quandary is whether early valve surgery can be safely performed within 7 days after a stroke or if it is better to postpone surgery for at least 1 week. No randomized trials have addressed this conundrum. The high rates of postoperative mor-bidity and mortality seen in earlier studies265–267 have resulted in a reluctance to refer patients with IE and acute stroke for imme-diate valve surgery. However, these initial studies included a limited number of patients, and risk adjustments were not per-formed. The largest early series of operated patients with cere-bral complications included 181 patients.267 Hospital mortality rates as a function of the interval between evidence of cerebral infarction to cardiac surgery were 66.3% when surgery was per-formed within 24 hours of stroke and gradually decreased every week to 7.0% with surgery >4 weeks after stroke.
Investigations have suggested better outcomes for IE patients with ischemic stroke who undergo early cardiac sur-gery.268–272 Ruttmann et al270 analyzed 65 patients who under-went cardiac surgery after cardioembolic (embolic) stroke (median time, 4 days; range, 0–38 days). Surgery in this time frame was not associated with worse patient outcomes. Fifty of the 61 patients (81.9%) with CT-verified preopera-tive stroke survived cardiac surgery. Latency between the neurological event and cardiac surgery was not a significant factor with respect to the perioperative neurological compli-cation rate or the postoperative neurological recovery rate. Full neurological recovery was achieved in 70% of 50 stroke patients. Other studies6–8 suggest that the risk of neurologi-cal deterioration during cardiac surgery after a stroke is lower than previously assumed, particularly in patients with silent cerebrovascular emboli.
The first study to evaluate the timing of surgery after stroke in IE that included a risk adjustment for differences in patient characteristics comprised 198 patients.273 Fifty-eight patients who underwent surgery within 1 week of stroke were compared with 140 patients who underwent surgery ≥8 days after stroke. Hospital mortality was numerically but not significantly higher in the early surgery group (22.4% versus 12%). After adjustment for other risk factors such as age, paravalvular abscess, and heart failure, the risk of hos-pital mortality remained nonsignificantly higher in the early surgery group (odds ratio, 2.308; 95% confidence interval, 0.942–5.652). Differences in 1-year mortality were less pro-nounced, with an adjusted hazard ratio of 1.138 (95% con-fidence interval, 0.802–1.650). In-hospital mortality in the early surgery group was comparable to that of the medically treated patients.
After hemorrhagic stroke, the risk of exacerbation by sur-gery is prohibitively high in the first month but can extend beyond 1 month in some patients, possibly because of the presence of undetected mycotic aneurysms (MAs). In a mul-ticenter study of patients with hemorrhagic stroke, mortality was higher when surgery was performed within 4 weeks of the hemorrhagic event compared with later surgery (75% versus 40%, respectively).274 These data support the following recommendations: Valve surgery may be performed in IE patients with stroke or sub-clinical cerebral emboli without delay if intracranial hemor-rhage has been excluded by imaging studies and neurological damage is not severe (ie, coma). In patients with major isch-emic stroke or intracranial hemorrhage, it is reasonable to delay valve surgery for at least 4 weeks.
Recommendations 1. Valve surgery may be considered in IE patients with stroke or subclinical cerebral emboli and residual vegetation without delay if intracranial hemorrhage has been excluded by imaging studies and neurolog-ical damage is not severe (ie, coma) (Class IIb; Level of Evidence B).
2. In patients with major ischemic stroke or intra-cranial hemorrhage, it is reasonable to delay valve surgery for at least 4 weeks (Class IIa; Level of Evidence B).
Risk of Embolization Systemic embolization occurs in 22% to 50% of cases of IE.55,57,274–277 Rates can be higher if noninvasive imaging, Downloaded from by on October 16, 2018 1468 Circulation October 13, 2015 including MRI and CT scanning, is routinely done to detect asymptomatic (silent) emboli. Emboli often involve major arterial beds, including the brain, lungs, coronary arteries, spleen, bowel, and extremities. Up to 65% of embolic events involve the CNS, and >90% of CNS emboli lodge in the distri-bution of the middle cerebral artery.277 The highest incidence of embolic complications is seen with mitral valve IE (and more with anterior rather than posterior mitral leaflet involve-ment) and with IE caused by S aureus, Candida, and HACEK organisms.
Emboli can occur before diagnosis, during therapy, or after therapy is completed, although most emboli occur within the first 2 to 4 weeks of antimicrobial therapy.58,278 Of note, 2 inde-pendent studies have confirmed that the rate of embolic events decreases dramatically during and after the first 2 to 3 weeks of successful antibiotic therapy. In a study from 1991, the embolic rate dropped from 13 to <1.2 embolic events per 1000 patient-days during that time.58 Vilacosta et al278 confirmed the reduced frequency of embolization after 2 weeks of therapy. Moreover, the latter study reemphasized the increased risk of embolization with increasing vegetation size during therapy, mitral valve involvement, and staphylococcal pathogenesis. In a survey that included the International Collaboration on Endocarditis cohort, Dickerman and colleagues280 focused on the incidence of stroke in a multicenter IE population and demonstrated that acute stroke rates fell significantly after initiation of antibiotic therapy regardless of valve involved or pathogen identified. Moreover, only 3.1% of the cohort suf-fered stroke after the first week of antimicrobial therapy; this finding has led to the opinion that stroke prevention as a sole indication for valve surgery after 1 week of appropriate antibi-otic therapy is not warranted.
Prediction of individual patient risk for embolization is extremely difficult. Many studies have attempted to use echocardiography to identify a high-risk subset of IE patients who might benefit from early surgery to avoid embolization. Several studies with TTE have demonstrated a trend toward higher embolic rates with left-sided vegetations >1 cm in diameter.55 De Castro and colleagues276 compared TTE with multiplane TEE and found that neither technique was help-ful in defining embolic risk in patients with vegetations. In a study57 based on TEE, mitral vegetations >1 cm in diameter were associated with the greatest frequency of embolism. The association was strengthened when the analysis was limited to those patients who had not yet experienced a clinical embolic event. Another prospective TEE study, however, found no clear correlation of vegetation size with embolization.59 Nevertheless, the same investigators later reported the results of a new prospective study of 118 patients who underwent TEE and found that, on multivariable analysis, risk factors associated with embolic risk included vegetation size >10 mm and mitral valve involvement.56 Overall, these data are com-patible with previous observations that indicate that, in gen-eral, mitral vegetations of any size are associated with a higher risk of embolization (25%) than aortic vegetations (10%). As noted above, the highest embolic risk (37%) has been seen in the subset of patients with mitral vegetations attached to the anterior rather than the posterior mitral leaflet.59,63 This sug-gests that the mechanical effects of broad and abrupt leaflet excursion, occurring twice per heartbeat, may contribute to the propensity of a vegetation in this location to fragment and embolize.
In another study, the effect of vegetation size on embolic potential was dependent on the infecting organism, with large vegetations independently predicting embolic events only in the setting of streptococcal IE.60 In contrast, as confirmed above by Vilacosta et al,278 staphylococcal or fungal IE appears to carry a high incidence rate of embolization independently of vegetation size.
Prognosis based on echocardiographic findings was exam-ined in a large, multicenter, prospective investigation. On the basis of TEE findings in 384 consecutive adult patients with definite IE, vegetation length >15 mm was a predictor of 1-year mortality (adjusted relative risk, 1.8; 95% confidence interval, 1.10–2.82; P=0.02) in multivariable analysis.281 The role of echocardiography in predicting embolic events has been controversial. In 1 survey282 that included 4 echocar-diographers who were blinded to clinical data, interobserver agreement was mixed on the characterization of vegetations. Agreement was high for the presence of vegetation (98%) and involved site (97%); interobserver agreement was consider-ably less for vegetation size (73%), mobility (57%), shape (37%), and attachment (40%). However, all of the series that included >100 patients who underwent TEE showed a positive relationship between embolic events and vegetation size. Moreover, the study with the largest number of patients (n=176) that assessed the value of TEE and included silent embolism detected by CT scanning demonstrated that the risk of embolic events was highly related to vegetation size and mobility but not to other known risk factors associated with embolism in IE.283 The conflicting results on the relationship between echocardiography and embolic risk can be explained at least partially by the poor standardization of diagnostic cri-teria for IE in older series, inclusion or not of silent embolism, inclusion or not of previous embolism, echocardiographic method used, lack of focus on future embolic events after TEE, and sample size.
An increase in vegetation size over 4 to 8 weeks of therapy as documented by TEE appears to predict embolic events.282 In addition, a second, albeit infrequent, peak of late embolic events has been observed to occur 15 to 30 weeks after the diagnosis of IE and has been associated with nonhealing veg-etations (failure of a vegetation to stabilize or diminish in size) as defined by echocardiography.63 The traditional indication for valvular surgery for IE to avoid embolization has been ≥2 major embolic events.284 This criterion is arbitrary and excludes cutaneous embolization, which is common, or embolism occurring before the insti-tution of therapy, which is common among IE patients who develop embolic events. Because of the observed decreases in embolic risk during the first 2 weeks of antibiotic therapy, the benefit of surgery in avoiding catastrophic embolic events is greatest early in the treatment course of IE. Early surgi-cal intervention may preclude a primary or recurrent major embolic event but exposes the patient to both immediate and lifelong risks of valve replacement if the valve cannot be pri-marily repaired. At this time, the strategy for surgical inter-vention to avoid systemic embolization in IE remains specific Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1469 to the individual patient, with benefit being greatest in the early phase of IE when embolic rates are highest and other predictors of a complicated course (ie, recurrent embolization; heart failure; aggressive, antibiotic-resistant organisms; or PVE) are present (Table 5). The benefits of early surgery were demonstrated in the prospective, randomized trial12 that was discussed earlier in this document. Because of several limita-tions of that trial, additional study is needed before routine application of early surgery solely to reduce embolic risk can be strongly advocated.
Embolic events are important prognostic indicators of IE outcomes. In 1 analysis, an embolic event was 1 of 4 early pre-dictors of in-hospital death caused by IE.285 Other independent predictors of death by logistic regression modeling among 267 consecutive patients with definite or possible IE by modified Duke criteria were diabetes mellitus, S aureus infection, and Acute Physiology And Chronic Health Evaluation (APACHE) II score.
Another controversial topic is whether imaging to detect emboli should be performed in all IE patients. The current paradigm includes dedicated, anatomic imaging if there are signs or symptoms suggestive of an embolic event. There is less agreement on imaging, which can pose risks because con-trast material is usually required, in patients without symp-toms or signs of emboli, some of whom may have silent or subclinical events. In particular, should MRI of the brain be obtained in all IE patients because cerebral emboli are so com-monplace? As previously mentioned in an earlier section (3D Echocardiography and Other Imaging Modalities), some have advocated this strategy in all patients who are to undergo valve surgery to identify those who may harbor embolic lesions that could pose a higher risk of intracranial bleeding with cardio-pulmonary bypass and heparin administration used for cardiac surgery.
Anticoagulation Anticoagulation in IE patients is controversial, particularly in mechanical valve IE.286 Some authorities recommend con-tinuation of anticoagulant therapy in patients with mechani-cal valve IE. However, the general advice is to discontinue all forms of anticoagulation in patients with mechanical valve IE who have experienced a CNS embolic event for at least 2 weeks.286 This time should allow for thrombus organization and prevent the acute hemorrhagic transformation of embolic lesions. Reintroduction of anticoagulation in these patients should be done with great caution, beginning with intravenous unfractionated heparin titrated to an activated partial thrombo-plastin time range of 50 to 70 seconds and transitioning care-fully to adjusted dose warfarin. The novel oral anticoagulants are not approved for use with either mechanical valves or bio-prosthetic valves when risk factors for thromboembolism exist (eg, atrial fibrillation).
The benefit of therapeutic anticoagulation has never been demonstrated convincingly in patients with NVE. In part related to findings that demonstrated a salutary effect of intra-venous aspirin therapy in established experimental S aureus IE,287 a randomized trial compared oral aspirin 325 mg/d with placebo in 115 IE patients.288 No significant benefit was observed in aspirin-treated patients in terms of vegetation resolution and embolic events. Moreover, there was a trend toward more bleeding episodes in the aspirin-treated patients. Aspirin levels, a critical correlate of antimicrobial efficacy in an animal model, were not monitored in this study.289 Retrospective, observational studies290–296 have examined the impact, if any, of long-term antiplatelet therapy before the onset of IE on infection-related outcomes. Findings from these investigations have been mixed in terms of IE-related outcomes. Until definitive data are available, the initiation of aspirin or other antiplatelet agents as adjunctive therapy in IE is not recommended. In contrast, the continuation of long-term antiplatelet therapy at the time of development of IE with no bleeding complications may be considered.
Recommendations 1. Discontinuation of all forms of anticoagulation in patients with mechanical valve IE who have expe-rienced a CNS embolic event for at least 2 weeks is reasonable (Class IIa; Level of Evidence C).
2. Initiation of aspirin or other antiplatelet agents as adjunctive therapy in IE is not recommended (Class III; Level of Evidence B).
3. The continuation of long-term antiplatelet therapy at the time of development of IE with no bleeding complications may be considered (Class IIb; Level of Evidence B).
Periannular Extension of Infection Extension of IE beyond the valve annulus predicts a higher mortality rate, more frequent development of heart fail-ure, and more frequent need for cardiac surgery.284,297,298 Perivalvular cavities form when annular infections break through and spread into contiguous tissue. In aortic NVE, this generally occurs through the weakest portion of the annulus, which is near the membranous septum and atrio-ventricular node.299 The anatomic vulnerability of this area explains both why abscesses occur in this location and why heart block is a frequent sequela.300 Periannular exten-sion is common, occurring in 10% to 40% of all NVE and complicating aortic IE more commonly than mitral or tri-cuspid IE.301–304 Periannular infection is of even greater con-cern with PVE, occurring in 56% to 100% of patients.298,302 Perivalvular abscesses are particularly common with pros-thetic valves because the annulus, rather than the leaflet, is the usual primary site of infection, especially in early PVE and on bioprosthetic valves.302 Under the influence of systemic intravascular pressures, abscesses may progress to fistulous tracts that create intracar-diac or pericardial shunts. The mortality rate was 41% in a series304 of patients with aorto-cavitary fistulization despite surgical intervention in 87%. Multivariate analysis demon-strated that factors associated with an increased risk of death included moderate to severe heart failure, PVE, and urgent or emergency surgical intervention. In some cases, progressive periannular infection totally disrupts the ventricular-aortic continuity or the mitral-aortic trigone. Such structural lesions and intracardiac fistulas may be catastrophic; even if their hemodynamic impact is tolerated, these lesions will not heal Downloaded from by on October 16, 2018 1470 Circulation October 13, 2015 with medical treatment alone and require urgent operative intervention.
Clinical parameters for the diagnosis of perivalvular extension of IE are inadequate. Persistent bacteremia or fever, recurrent emboli, heart block, heart failure, or a new patho-logical murmur in a patient with IE on appropriate antibiotics may suggest extension.303 Only aortic valve involvement and current IDU have been prospectively identified as independent risk factors for perivalvular abscess.297 On ECG, new atrio-ventricular block has a positive predictive value of 88% for abscess formation but low (45%) sensitivity.298 Patients at risk for perivalvular extension of IE require prompt evaluation. The size of vegetations is not helpful for predicting perivalvular extension.297 The sensitivity of TTE for detecting perivalvular abscess is low (18% to 63% in pro-spective and retrospective studies, respectively).305,306 TEE dramatically improves the sensitivity for defining periannular extension of IE (76% to 100%) while retaining excellent spec-ificity (95%) and positive and negative predictive values (87% and 89%, respectively).54,307 When combined with spectral and color Doppler techniques, TEE can demonstrate the distinc-tive flow patterns of fistulas and pseudoaneurysms and can rule out communications from unruptured abscess cavities. Because of these combined capabilities, TEE is recommended for the initial assessment of any patient suspected of having perivalvular extension of IE.
A small number of patients with periannular extension of infection or myocardial abscess may be treated successfully without surgical intervention.307,308 These patients potentially include those who have smaller (<1 cm) abscesses and who do not have complications such as heart block, echocardiographic evidence of progression of abscess during therapy, valvular dehiscence, or insufficiency. Such patients should be monitored closely with serial TEE; TEE should be repeated at intervals of 2, 4, and 8 weeks after completion of antimicrobial therapy.
Surgery for patients with perivalvular extension of IE is directed toward eradication of the infection and correction of hemodynamic abnormalities. Drainage of abscess cavities, excision of necrotic tissue, and closure of fistulous tracts often accompany valve replacement or repair surgery.309 Although valve replacement is usually required, its successful performance may be compromised by extensive destruction of the periannular supporting tissues. Under these conditions, human aortic homo-grafts, when available, can be used to replace the damaged aor-tic valve and to reconstruct the damaged aorta.310,311 Homografts have a constant but low incidence rate of IE.312 Some groups have advocated the use of stentless or mini-stented aortic valve prostheses with debridement in the same clinical scenario, par-ticularly if homografts are not readily available.313 Recommendation 1. TEE is recommended for the initial assessment of any patient suspected of having perivalvular exten-sion of IE (Class I; Level of Evidence B).
Metastatic Foci of Infection Similar to embolic complications, metastatic foci of infec-tion frequently occur in IE and can greatly affect management strategies, in particular timing of valve surgery, duration of antimicrobial therapy, and need for invasive interventions (usually surgical or interventional radiological drainage). Much like embolic events, metastatic foci of infection either can remain asymptomatic or may cause major clinical signs or symptoms. In the latter case, sustained fever can be a valu-able clue, particularly when bloodstream infection has been cleared or in cases when bloodstream infection persists despite adequate antimicrobial coverage. In addition, distinguish-ing bland infarction caused by an embolus from a metastatic focus (abscess) sometimes can be difficult. In patients who are symptomatic, a diagnostic evaluation including radiologi-cal, ultrasonographic, and invasive procedures such as joint aspiration for both diagnostic and therapeutic reasons is rec-ommended. Invasive procedures such as percutaneous drain-age of soft tissue or organ abscess may be needed. Surgical intervention, as mentioned above, may be required for radi-cal infection cure. For example, splenic abscesses generally require splenectomy or a drainage procedure because the usefulness of antimicrobial therapy is in preventing disease extension in the spleen and treating systemic infection rather than eliminating abscesses.314 Whether percutaneous aspira-tion or drainage of splenic abscesses can be performed safely and effectively should be decided by an experienced team of clinicians.
Identification and management of metastatic foci of infec-tion are critically important in patients who require valve surgery. When feasible, all invasive procedures for the initial management of metastatic foci of infection should be done before valve surgery to reduce the likelihood of infecting a placed prosthetic valve or annuloplasty ring.
Cerebrovascular imaging may be considered in all patients with left-sided IE who have no CNS signs or symptoms (see the Intracranial MAs section below). There are currently no other recommendations for routinely evaluating all patients with IE for metastatic foci of infection, although many clinicians rec-ommend such routine screening for all cases of S aureus IE. Rather, a directed workup is advocated on the basis of local-izing signs or symptoms. Depending on the site of interest, the choice of diagnostic procedure (eg, CT, MRI, ultrasonogra-phy) varies, and the selection should be individualized for each patient. The choice of procedure may require consultation with experts. It is strongly recommended that a discussion of which laboratory (microbiology, pathology including cytology) stud-ies will be needed once tissue or fluid aspirate specimens are available takes place before an invasive procedure is performed.
Recommendation 1. The choice of diagnostic procedure (eg, CT, MRI, ultrasonography) varies and the selection should be individualized for each patient (Class I; Level of Evidence C).
Mycotic Aneurysms MAs are uncommon complications of IE that result from sep-tic embolization of vegetations to the arterial vasa vasorum or the intraluminal space, with subsequent spread of infec-tion through the intima and outward through the vessel wall. Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1471 Arterial branching points favor the impaction of emboli and are the most common sites of development of MAs. MAs caused by IE occur most frequently in the intracranial arteries, followed by the visceral arteries and the arteries of the upper and lower extremities.
A detailed analysis of the complex management of MAs has been included in a separate AHA Scientific Statement that addresses vascular infections and is pending publication; please refer to this document for additional information.
Intracranial MAs Intracranial MAs (ICMAs) represent a relatively small but extremely dangerous subset of neurological complications. The overall mortality rate among IE patients with ICMAs is 60%. Among patients with unruptured ICMAs, the mortality rate is 30%; in patients with ruptured ICMAs, the mortality rate approaches 80%.315,316 The reported occurrence of ICMAs in 1.2% to 5% of cases316–320 is probably underestimated because some ICMAs remain asymptomatic and resolve with antimicrobial therapy. Streptococci and S aureus account for 50% and 10% of cases, respectively,317,318 and ICMAs are seen with increased frequency among IDUs with IE.318 The dis-tal middle cerebral artery branches are most often involved, especially the bifurcations. Multiple ICMAs occur in 20% of cases319; mortality rates are similar for multiple and single distal ICMAs. The mortality rate for patients with proximal ICMAs is >50%.321 The clinical presentation of patients with ICMAs is highly variable. Patients may develop severe headache, altered sen-sorium, or focal neurological deficits such as hemianopsia or cranial neuropathies. Neurological signs and symptoms are nonspecific and may suggest a mass lesion or an embolic event.315,317 Some ICMAs leak slowly before rupture and pro-duce mild meningeal irritation. Frequently, the spinal fluid in these patients is sterile, and it usually contains erythrocytes, leukocytes, and elevated protein. In other patients, there are no clinically recognized premonitory findings before sudden subarachnoid or intraventricular hemorrhage.315 Symptomatic cerebral emboli frequently but not invari-ably precede the finding of an ICMA.315 Therefore, imaging procedures to detect ICMAs are indicated in IE patients with localized or severe headaches; “sterile” meningitis, especially if erythrocytes or xanthochromia is present; or focal neurologi-cal signs. Several imaging modalities can be used to identify ICMAs; currently, there is no preferred initial imaging study that can be recommended.322 Techniques include cardiac (mul-tislice) CT angiography with 3D reconstruction, digital subtrac-tion angiography, and magnetic resonance angiography with 3D reconstruction. In cases when there is a high clinical suspi-cion of ICMAs and a negative initial screening with 1 of these modalities, then conventional angiography is reasonable to per-form. Cerebral MRI may be considered in all patients with left-sided IE who have no CNS signs or symptoms. MRI findings may assist in subsequent medical and surgical management.322 Recommendations 1. Cerebrospinal imaging should be performed to detect ICMA or CNS bleeding in all patients with IE or contiguous spread of infection who develop severe, localized headache, neurological deficits, or meningeal signs (Class I; Level of Evidence B).
2. Cerebrovascular imaging may be considered in all patients with left-sided IE who have no CNS signs or symptoms (Class IIb; Level of Evidence C).
3. CT angiography, magnetic resonance angiography, or digital subtraction angiography is reasonable as an initial imaging test for detection of ICMA (Class IIa, Level of Evidence B).
4. Conventional angiography for detection of sus-pected ICMA is reasonable in patients with negative CT angiography, magnetic resonance angiography, or digital subtraction angiography (Class IIa; Level of Evidence B).
Extracranial MAs Intrathoracic or intra-abdominal MAs often are asymptomatic until leakage or rupture occurs. Presumably, most extracranial MAs will rupture if not excised. The appearance of a tender, pulsatile mass in a patient with IE suggests an extracranial MA. Hematemesis, hematobilia, and jaundice suggest rupture of a hepatic artery MA; arterial hypertension and hematuria sug-gest rupture of a renal MA; and massive bloody diarrhea sug-gests the rupture of an extracranial MA into the small or large bowel. Either CT scanning or multislice CT angiography with 3D reconstruction is indicated for initial imaging. TEE is useful in identifying MAs of the sinus of Valsalva and thoracic aorta.
Recommendations 1. Either CT scanning or multislice CT angiography with 3D reconstruction is indicated for initial imag-ing (Class I; Level of Evidence B).
2. TEE is useful in identifying MAs of the sinus of Valsalva and thoracic aorta (Class I; Level of Evidence B).
Outpatient Therapy Outpatient parenteral antibiotic therapy (OPAT) is efficacious, safe, and cost-effective for a variety of infections,323–325 includ-ing IE that requires prolonged parenteral therapy in hospital-ized patients who otherwise no longer require inpatient care but do require continued parenteral antimicrobial therapy. Antibiotic regimens recommended for IE vary widely and often require ≥4 weeks of therapy, generally given by the intra-venous route. Absorption of orally administered antimicrobial agents may be unreliable, and such a strategy is generally not recommended as sole therapy for IE. Several other aspects of OPAT such as drug stability at room temperature; frequency of drug dosing; access to ancillary equipment, including ambula-tory pumps; insurance coverage; and whether the patient has a history of IDU can all affect the ultimate use of OPAT.
The timing for transition from inpatient antibiotic therapy to OPAT and patient exclusion criteria have been critically evaluated by Andrews and von Reyn.326 These guidelines are based on the local availability of medical care in the outpa-tient setting and risk factors and timing of potential adverse outcomes that would be best managed in the inpatient setting.
Downloaded from by on October 16, 2018 1472 Circulation October 13, 2015 Before OPAT is considered, most patients with IE should first be evaluated and stabilized in the hospital; only rarely can some patients be treated entirely as outpatients. Patients selected for OPAT should be at low risk for the complications of IE, the most frequent of which are heart failure and systemic emboli. The period of greatest risk for systemic emboli is before or within the first 1 to 2 weeks of antimicrobial therapy, although serious complications such as heart failure and rupture of MAs may develop weeks to months after the initiation of antimicro-bial therapy. The presence of poorly controlled heart failure, neurological findings that may result from systemic emboli or bleeding MAs, cardiac conduction abnormalities, valve ring abscesses (usually detected by TEE), persistent fever, or persis-tently positive blood cultures should preclude OPAT.
The risk for drug-related side effects usually increases with a prolonged drug exposure (eg, vestibular, auditory, and nephrotoxicity resulting from aminoglycosides; leukopenia caused by β-lactams and vancomycin; and nephrotoxicity resulting from the combination of vancomycin and gentami-cin) and requires close monitoring by the home infusion team consisting of representatives from nursing and pharmacy and clinicians with expertise in IE management.
The following criteria are essential for an effective OPAT program: • A reliable support system at home and easy access to a hospital for prompt re-evaluation by an experienced clinician if a complication such as recurrence of fever, symptoms of a cardiac arrhythmia, heart failure, or a neurological event develops • Regular visits by a home infusion nurse who carefully monitors the patient for early detection of complications, failure to respond to therapy, problems with adherence to therapy, or complications (eg, catheter-related infection, catheter leakage or displacement, venous thrombosis) directly related to the antibiotics or intravenous access • Regular visits with an experienced clinician to assess clinical status during the OPAT Recommendations 1. Patients with IE should first be evaluated and sta-bilized in the hospital before being considered for outpatient therapy (Class I; Level of Evidence C).
2. Patients selected for OPAT should be at low risk for the complications of IE, the most frequent of which are heart failure and systemic emboli (Class I; Level of Evidence C).
Care at Completion of Antimicrobial Therapy Short-Term Follow-Up The majority of patients with IE are cured with appro-priate medical and, if necessary, surgical treatment. Echocardiography is reasonable before or synchronous with completion of antimicrobial therapy to establish a new baseline for subsequent comparison (Table 18). A referral to a program to assist in the cessation of drug use should be made for IDUs. Patients should be educated about the signs of endocarditis and urged to seek immediate medical attention should they occur. If feasible, a thorough dental evaluation is reasonable, especially in patients deemed likely to require valve replace-ment, with all active sources of oral infection eradicated. All indwelling intravenous catheters used to infuse antimicrobial treatment should be removed promptly at the end of therapy. Routine blood cultures are no longer recommended after the completion of antimicrobial therapy because the likelihood of a positive culture result in a patient who is otherwise without evidence of active infection is low.
In the short-term follow-up, patients should be moni-tored for development of several complications (Table 18). A relapse of IE is a primary concern. Patients should be aware that relapses can occur and that new onset of fever, chills, or other evidence of systemic toxicity mandates immediate eval-uation, including a thorough history and physical examination and ≥3 sets of blood cultures. Every effort should be made to determine the cause of signs or symptoms of infection. In addition, prescribing empirical antimicrobial therapy should be avoided for an undefined febrile illness unless the patient’s clinical condition (eg, sepsis) warrants empirical therapy. It is reasonable for patients who have completed therapy to undergo an examination after completing antibiotic therapy.
Developing or worsening heart failure is a second com-plication that should be considered during short-term follow-up. Although new onset of heart failure caused by valvular dysfunction is unlikely during this period, valve function can deteriorate as a result of ongoing infection or mechanical stress unrelated to infection. In addition to physical examina-tion, echocardiographic findings can support this diagnosis. If heart failure develops or worsens, the patient should be evalu-ated immediately for cardiac surgery.
Antibiotic toxicity still can occur after the completion of treatment and is the third complication that should be consid-ered during short-term follow-up. Two drug-related adverse events are concerns. The first is delayed ototoxicity because of the previous use of aminoglycosides. Audiological and Table 18. Care During and After Completion of Antimicrobial Treatment Initiation before or at completion of therapy Echocardiography to establish new baseline Drug rehabilitation referral for patients who use illicit injection drugs Education on the signs of endocarditis and need for antibiotic prophylaxis for certain dental/surgical/invasive procedures Thorough dental evaluation and treatment if not performed earlier in evaluation Prompt removal of intravenous catheter at completion of antimicrobial therapy Short-term follow-up At least 3 sets of blood cultures from separate sites for any febrile illness and before initiation of antibiotic therapy Physical examination for evidence of heart failure Evaluation for toxicity resulting from current/previous antimicrobial therapy Long-term follow-up At least 3 sets of blood cultures from separate sites for any febrile illness and before initiation of antibiotic therapy Evaluation of valvular and ventricular function (echocardiography) Scrupulous oral hygiene and frequent dental professional office visits Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1473 vestibular toxicity can develop despite the maintenance of appropriate serum drug concentrations during treatment. For patients receiving long-term aminoglycosides, particu-larly those with underlying renal or otic disorders, serial audiograms may be considered during therapy if feasible and available. No tools are routinely available for monitor-ing vestibular function, and patients should be told to report the onset of any symptoms of vestibular toxicity during or after treatment.
The second antibiotic-related adverse event is Clostridium difficile infection. Onset of diarrhea can be delayed as long as 4 weeks after the last dose of antibiotic. The hope is that prompt recognition and treatment of this infectious complica-tion will diminish the likelihood of severe complications.
Recommendations 1. Echocardiography is reasonable before or synchro-nous with completion of antimicrobial therapy to establish a new baseline for subsequent compari-son (Class IIa; Level of Evidence C). 2. A referral to a program to assist in the cessation of drug use should be made for IDUs (Class I; Level of Evidence C). 3. Patients should be educated about the signs of endocarditis and urged to seek immediate medi-cal attention should they develop (Class I; Level of Evidence C). 4. A thorough dental evaluation is reasonable, espe-cially in patients deemed likely to require valve replacement, with all active sources of oral infec-tion eradicated (Class IIa; Level of Evidence C). 5. Routine blood cultures are not recommended after the completion of antimicrobial therapy because the likelihood of a positive culture result in a patient who is otherwise without evidence of active infection is low (Class III; Level of Evidence C). 6. All indwelling intravenous catheters used to infuse antimicrobial treatment should be removed promptly at the end of therapy (Class I; Level of Evidence C). 7. For patients receiving long-term aminoglycosides, particularly those with underlying renal or otic disorders, serial audiograms may be considered during therapy if available (Class IIb; Level of Evidence C). 8. In the short-term follow-up, patients should be monitored for the development of several com-plications, including IE relapse and heart failure (Class I; Level of Evidence C). 9. Patients should be aware that relapses can occur and that new onset of fever, chills, or other evidence of systemic toxicity mandates immediate evalu-ation, including a thorough history and physical examination and ≥3 sets of blood cultures (Class I; Level of Evidence C).
10. Because of concerns for IE relapse, a thorough evalua-tion should be done to determine the cause of infection signs and symptoms (Class I; Level of Evidence C).
11. Empirical antimicrobial therapy for suspected infection should be avoided unless the patient’s clinical condition (eg, sepsis) warrants it (Class III; Level of Evidence C).
12. It is reasonable to have patients who have completed therapy and do not have symptoms of systemic tox-icity undergo an examination after completing anti-biotic therapy (Class IIa; Level of Evidence C).
13. Developing or worsening heart failure is a common complication that should be monitored for during short-term follow-up (Class I; Level of Evidence C).
14. If heart failure develops or worsens, the patient should be evaluated immediately for cardiac sur-gery (Class I; Level of Evidence B).
15. Antibiotic toxicity still can occur after the comple-tion of treatment and is a complication that should be considered during short-term follow-up (Class I; Level of Evidence C).
16. No tools are routinely available for monitoring vestibular function, and patients should be told to report the onset of any symptoms of vestibular toxicity during or after treatment (Class I; Level of Evidence C).
Long-Term Follow-Up Months to years after completion of medical therapy for IE, patients should have ongoing observation for and education about recurrent infection and delayed onset of worsening valve dysfunction (Table 18). Daily dental hygiene should be stressed, with serial evaluations by a dentist who is familiar with this patient population. Patients should be questioned about symptoms of heart failure, and a thorough physi-cal examination should be done. Additional evaluation with echocardiography is indicated in selected patients with posi-tive findings from history and physical examination. Patients should be instructed to seek immediate medical evaluation for persistent fever (Table 18). This is necessary because IE can mimic a variety of febrile illnesses. Blood cultures should be obtained. Antibiotic therapy should not be initiated for treat-ment of undefined febrile illnesses without blood cultures being obtained first. Antibiotics prescribed for nonspecific or unproved febrile syndromes are a major cause of (blood) culture-negative IE, and this practice should be strongly discouraged.
Recommendations 1. Months to years after completion of medical ther-apy for IE, patients should have ongoing observa-tion for and education about recurrent infection and delayed onset of worsening valve dysfunction (Class I; Level of Evidence C).
2. Daily dental hygiene should be stressed, with serial evaluations by a dentist who is familiar with this patient population (Class I; Level of Evidence C).
3. Patients should be questioned about symptoms of heart failure, and a thorough physical examination should be done (Class I; Level of Evidence C).
4. Additional evaluations with echocardiography should be obtained in selected patients with posi-tive findings from history and physical examina-tion (Class I; Level of Evidence C).
Downloaded from by on October 16, 2018 1474 Circulation October 13, 2015 5. Patients should be instructed to seek immediate medical evaluation for fever, and blood cultures should be obtained (Class I; Level of Evidence C).
6. Antimicrobial therapy should not be initiated for the treatment of undefined febrile illnesses unless the patient’s condition (eg, sepsis) warrants it (Class III; Level of Evidence C).
Dental Management A large, prospective study demonstrated a strong associa-tion between 3 indexes of oral hygiene and gingival disease and the incidence of bacteremia from IE-related species.327 Poor oral hygiene results in gingivitis, which often leads to periodontitis, and it is likely that these 2 periodontal dis-eases are associated with community-acquired IE. Current evidence suggests that poor oral hygiene and periodon-tal diseases, not dental office procedures, are likely to be responsible for the vast majority of cases of IE that originate in the mouth.328 Regardless of the source of infection, inpatients with IE should be thoroughly evaluated by a dentist familiar with the potential role of the mouth in these cases. The optimal timing for this evaluation may be after the patient’s cardiac status has stabilized and early enough that all invasive dental procedures can be accomplished during intravenous antibiotic therapy. The clinical examination should rule out periodontal inflammation and pocketing around the teeth and caries that will eventually result in pulpal infection. A full series of intra-oral radiographs is required for the identification of caries and periodontal disease (eg, bone loss, tooth fractures). All of this is aimed at reducing the incidence and magnitude of bactere-mia from any manipulation of the gingival tissues, including normal daily events such as brushing teeth and chewing food. Treatment invariably involves a thorough dental cleaning by a hygienist who will review with patients the importance of maintaining scrupulous oral hygiene.
Dental disease is almost entirely preventable if patients are compliant with 4 measures. First, the cause of both peri-odontal disease and caries is bacterial plaque accumulation on teeth, and prevention is dependent on keeping teeth free of plaque. Second, patients must understand that dietary mea-sures are critically important in preventing the formation of plaque, especially in areas on the teeth that are difficult to keep clean. The degree to which sugar and other refined carbohydrates are eliminated from the diet will have a major impact on the growth of pathogenic bacterial species, some of which are responsible for IE. Third, routine follow-up with their family dentist is necessary for close monitoring of oral hygiene and the early identification and eradication of oral disease. Finally, the daily use of a high-concentration fluori-dated toothpaste will help to ensure that the acid from plaque does not decalcify tooth structures and result in caries. A focus on all 4 measures should help to reduce the incidence of bac-teremia and the risk for recurrent IE.
Recommendations 1. Inpatients with IE should be thoroughly evaluated by a dentist to identify and eliminate oral diseases that predispose to bacteremia and may therefore contribute to the risk for recurrent IE (Class I; Level of Evidence C).
2. The clinical examination should focus on periodon-tal inflammation and pocketing around teeth and caries that may result in pulpal infection and subse-quent abscess (Class I; Level of Evidence C).
3. A full series of intraoral radiographs will allow the identification of caries and periodontal disease and other disease (eg, tooth fractures) not evident from the physical examination. This should occur when the patient is able to travel to a dental facility (Class I; Level of Evidence C).
Downloaded from by on October 16, 2018 Baddour et al Infective Endocarditis in Adults 1475 Writing Group Disclosures Writing Group Member Employment Research Grant Other Research Support Speakers’ Bureau/ Honoraria Expert Witness Ownership Interest Consultant/ Advisory Board Other Larry M. Baddour Mayo Clinic None None None None None None None Robert S. Baltimore Yale University School of Medicine None None None None None None None Bruno Barsic Hospital for Infectious Diseases, School of Medicine Zagreb None None Astellas; Pfizer; MSD None None None None Arnold S. Bayer LA Biomedical Research Institute Infectious Diseases Astellas†; Cubist†; NIH/NIAID; Theravance None None Johnson Graffe et al; Morrow, Kidman, Tinker et al; Galloway, Lucchese et al; MES Solutions; Hoffman, Sheffield et al; Clifford Law None None None Ann F. Bolger UCSF None None None None None None None Anne M. Fink University of Illinois at Chicago None None None None None None None Vance G. Fowler, Jr.
Duke University CDC†; Cerexa/Forest†; MedImmune†; NIH†; FDA†; Cubist None None Witness in case involving group B Streptococcal Vertebral osteomyelitis† None MedImmune; Novartis†; The Medicines Comapy; Novadigm; Debiopharm; Cerexa;Affinium; Tetraphase; Bayer; Theravance; Cubist; Genetech; Basilea None Michael H. Gewitz New York Medical College None None None None None None None Matthew E. Levison Drexel University College of Medicine None None None None None Merck Manual† None Peter B. Lockhart Carolinas Medical Center None None None None None None None Patrick O’Gara Brigham and Women’s Hospital None None None None None None None Michael J. Rybak Wayne State University Cubist; Forest; MDCH; NIH; Actavis; Theravance None Cubist; Forest; Novartis; Actavis; The Medicines Company None None Cubist; Forest; Theravance; Actavis†; The Medicines Company None James M. Steckelberg Mayo Clinic None None None None None None None Kathryn A. Taubert American Heart Association None None None None None None None Imad M. Tleyjeh King Fahd Medical City None None None None None None None Walter R. Wilson Mayo Clinic None None None None None None None This table represents the relationships of writing group members that may be perceived as actual or reasonably perceived conflicts of interest as reported on the Disclosure Questionnaire, which all members of the writing group are required to complete and submit. A relationship is considered to be “significant” if (a) the person receives $10 000 or more during any 12-month period, or 5% or more of the person’s gross income; or (b) the person owns 5% or more of the voting stock or share of the entity, or owns $10 000 or more of the fair market value of the entity. A relationship is considered to be “modest” if it is less than “significant” under the preceding definition.
Modest.
†Significant.
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Benito N, Miró JM, de Lazzari E, Cabell CH, del Río A, Altclas J, Commerford P, Delahaye F, Dragulescu S, Giamarellou H, Habib G, Reviewer Disclosures Reviewer Employment Research Grant Other Research Support Speakers’ Bureau/Honoraria Expert Witness Ownership Interest Consultant/ Advisory Board Other Giovanni Di Salvo King Faisal Hospital and Research Center (Saudi Arabia) None None None None None None None Franklin D. Lowy Columbia University None None None None None None UptoDate Stanford T. Shulman Lurie Children’s Hospital, Children’s Memorial Hospital, Northwestern University Medical School None None None None None None None This table represents the relationships of reviewers that may be perceived as actual or reasonably perceived conflicts of interest as reported on the Disclosure Questionnaire, which all reviewers are required to complete and submit. A relationship is considered to be “significant” if (a) the person receives $10 000 or more during any 12-month period, or 5% or more of the person’s gross income; or (b) the person owns 5% or more of the voting stock or share of the entity, or owns $10 000 or more of the fair market value of the entity. A relationship is considered to be “modest” if it is less than “significant” under the preceding definition.
Modest.
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Aorto-cavitary fistulous tract formation in infective endocarditis: clinical and echocardiographic features of 76 cases and risk factors for mortality. Eur Heart J. 2005;26:288–297. doi: 10.1093/eurheartj/ehi034. 304. Daniel W, Schroder E, Nonast-Daniel B, Lichtlen P. Conventional and transoesophageal echocardiography in the diagnosis of infective endo-carditis. Eur Heart J. 1987;8:287–292. 305. Leung DY, Cranney GB, Hopkins AP, Walsh WF. Role of transoesopha-geal echocardiography in the diagnosis and management of aortic root abscess. Br Heart J. 1994;72:175–181. 306. Rohmann S, Seifert T, Erbel R, Jakob H, Mohr-Kahaly S, Makowski T, Görge G, Oelert H, Meyer J. Identification of abscess formation in native-valve infective endocarditis using transesophageal echocardiog-raphy: implications for surgical treatment. Thorac Cardiovasc Surg. 1991;39:273–280. doi: 10.1055/s-2007-1019985. 307. Kunis RL, Sherrid MV, McCabe JB, Grieco MH, Dwyer EM Jr. Successful medical therapy of mitral anular abscess complicating infec-tive endocarditis. J Am Coll Cardiol. 1986;7:953–955. 308. Vlessis AA, Hovaguimian H, Jaggers J, Ahmad A, Starr A. Infective endocarditis: ten-year review of medical and surgical therapy. Ann Thorac Surg. 1996;61:1217–1222. 309. Mullany CJ, Chua YL, Schaff HV, Steckelberg JM, Ilstrup DM, Orszulak TA, Danielson GK, Puga FJ. Early and late survival after sur-gical treatment of culture-positive active endocarditis. Mayo Clin Proc. 1995;70:517–525. doi: 10.1016/S0025-6196(11)64307-3. 310. Glazier JJ, Verwilghen J, Donaldson RM, Ross DN. Treatment of complicated prosthetic aortic valve endocarditis with annular abscess formation by homograft aortic root replacement. J Am Coll Cardiol. 1991;17:1177–1182. 311. Ross D. Allograft root replacement for prosthetic endocarditis. J Card Surg. 1990;5:68–72. 312. McGiffin DC, Galbraith AJ, McLachlan GJ, Stower RE, Wong ML, Stafford EG, Gardner MA, Pohlner PG, O’Brien MF. Aortic valve infec-tion: risk factors for death and recurrent endocarditis after aortic valve replacement. J Thorac Cardiovasc Surg. 1992;104:511–520. 313. Walkes JC, Reardon MJ. Current thinking in stentless valve surgery. Curr Opin Cardiol. 2003;18:117–123. 314. Chun CH, Raff MJ, Contreras L, Varghese R, Waterman N, Daffner R, Melo JC. Splenic abscess. Medicine (Baltimore). 1980;59:50–65. 315. Bohmfalk GL, Story JL, Wissinger JP, Brown WE Jr. Bacterial intracra-nial aneurysm. J Neurosurg. 1978;48:369–382. 316. Wilson WR, Giuliani ER, Danielson GK, Geraci JE. Management of com-plications of infective endocarditis. Mayo Clin Proc. 1982;57:162–170. 317. Camarata PJ, Latchaw RE, Rüfenacht DA, Heros RC. Intracranial aneu-rysms. Invest Radiol. 1993;28:373–382. 318. Lerner PI. Neurologic complications of infective endocarditis. Med Clin North Am. 1985;69:385–398. 319. Clare CE, Barrow DL. Infectious intracranial aneurysms. Neurosurg Clin N Am. 1992;3:551–566. 320. Moskowitz MA, Rosenbaum AE, Tyler HR. Angiographically monitored resolution of cerebral mycotic aneurysms. Neurology. 1974;24:1103–1108. 321. Huston J 3rd, Nichols DA, Luetmer PH, Goodwin JT, Meyer FB, Wiebers DO, Weaver AL. Blinded prospective evaluation of sensitivity of MR angiography to known intracranial aneurysms: importance of aneurysm size. AJNR Am J Neuroradiol. 1994;15:1607–1614. 322. 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189922 | https://www.semanticscholar.org/topic/Apollonian-sphere-packing/1799584 | Apollonian sphere packing | Semantic Scholar
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Apollonian sphere packing is the three-dimensional equivalent of the Apollonian gasket. The principle of construction is very similar: with any four…Expand
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Spectral action models of gravity on packed swiss cheese cosmology
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189923 | https://www.youtube.com/watch?v=syrDdPQ50nM | Optimization: Find Two Numbers Whose Product is a Minimum
Steve Crow
67700 subscribers
348 likes
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Posted: 15 Nov 2018
This video shows how to find two numbers given a difference and whose product is a minimum.
20 comments
Transcript:
all right so let's look at this problem one optimization problem so I've got quite a few of these I'm gonna be working each one of them will have its own video and they're gonna go get more difficult as they go all right so find two numbers whose difference is a hundred and whose product is a minimum okay so with the optimization let's let's write down information they give us so they want us to find two numbers whose difference is a hundred well so that tells me that X minus y is equal to a hundred so they're they're telling me the difference of two numbers is a hundred and we want to minimize the product so the product we'll just call that P of X well that would equal the two numbers multiplied together x times y so we want to minimize this so I'm gonna need to take the derivative all right well I need to get it all in terms of one variable so I can move the Y over here the 100 over here so that gives me X minus a hundred is equal to Y I'm just gonna write it with the Y over here so there's Y so P of X is equal to x times y well what is y it's X minus 100 okay and so now let's say we've got P of X is equal to x squared minus a hundred X and so now let's take the derivative so I've got P prime of X is equal to 2x minus a hundred and then we set this equal to zero and so I get X is equal to 50 all right so well let's see what happens so if I draw my number line there's 250 right and then I can choose X equal 49 and X equal 51 so if I plug 49 in okay into the derivative here well 2 times 49 okay that's what 98-100 that's negative so that decreases if I plug the to me and 2 times 51 well that's more than a hundred so that's gonna increase and so you can see I do have my minimum there okay so it's the X equal 50 minimizes well there's X but it wants us to find two numbers so what would Y be well we know X minus y equals a hundred or look at this Y is equal to X minus 100 so 50 minus a hundred is equal to negative 50 so the two numbers the two numbers is X equal 50 and y equals negative 50 and that's the two numbers that minimize whose product would be a minimum and their difference is a hundred all right so in and you can check that out see X minus y look at that X minus y is 100 okay all right so hope the video helped check out the other ones give me a like share and subscribe and thanks for watching |
189924 | https://math.stackexchange.com/questions/4346490/minimum-number-of-intersecting-diagonals-on-a-convex-polygon | combinatorics - Minimum number of intersecting diagonals on a convex polygon - Mathematics Stack Exchange
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Minimum number of intersecting diagonals on a convex polygon
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Let there be a convex polygon P with n vertices, P 1,…,P n. Let us add k distinct line segments to this polygon ¯P a i P b i such that P a i and P b i are vertices on the polygon and |a i−b i|≠1,|a i−b i|≠n−1 (in other words, the line segments we add are distinct diagonals of the polygon).
My question is this: say we choose the k line segments optimally so as to minimise the number of pairs of line segments that intersect. What is the minimum number of intersections we can get?
I can see that for k≤n−3, the minimum possible number of intersections is 0. I am interested in how the number grows for k>n−3, either closed-form or even asymptotics. My guess is that the asymptotic number of intersection is c k 2 for constant c but I don't know how to prove it.
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edited Jan 7, 2022 at 9:37
usc18usc18
asked Jan 1, 2022 at 15:55
usc18usc18
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My guess is that you are dealing only with convex polygons. If you allow simple concave polygons, you can have a pentagon or hexagon with no two diagonals intersecting. I also have a heptagon with only 4 intersections among its 14 diagonals.Daniel Mathias –Daniel Mathias 2022-01-01 19:12:55 +00:00 Commented Jan 1, 2022 at 19:12
@DanielMathias Sorry for the mistake; that is right. Edited the question to account for this.usc18 –usc18 2022-01-01 20:06:20 +00:00 Commented Jan 1, 2022 at 20:06
Here are some numbers for you: pastebin.com/UxEQXfYB Minimum intersections for all k with n<10 Daniel Mathias –Daniel Mathias 2022-01-01 23:35:45 +00:00 Commented Jan 1, 2022 at 23:35
Surprised that the function is not at least concave...usc18 –usc18 2022-01-07 09:38:16 +00:00 Commented Jan 7, 2022 at 9:38
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189925 | https://www.youtube.com/watch?v=vsNlWCe4xHU | Write an Absolute Value Function as a Piecewise Function
Ryne Roper
3520 subscribers
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Posted: 10 Nov 2021
Learn how to take an Absolute Value Function in Vertex Form and write it as a Piecewise Function.
Use this information to help you in your Algebra class!
💡 Learn more about piecewise functions here:
🔥 DON'T FORGET to check out my full Algebra 1 playlist for help with all things Algebra 1!
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Algebra1 #Piecewise #AbsoluteValue
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Transcript:
[Music] hello everyone in this video we're going to talk about how to write an absolute value function as a piecewise function okay so sometimes it can be kind of a two-step process if you are looking at a graph and you have to write the vertex form of the absolute value yourself so you would do that first and then you would write the piecewise function from that we will do that in the second example today here in the first example uh the first step is kind of already done for us we already have the equation in vertex form so all we have to do is take this now and write it as a piecewise function okay so the reason we can do this is because an absolute value is like a v shape right so the right side and the left side are kind of like two pieces like a piecewise function okay so we can do this so what we're going to do is we're going to start off by saying f of x equals and we're going to write our open set bracket there and we're going to write our two pieces here okay so now how we do this one of the pieces is going to be basically exactly as we see this part right here and the other one we're going to make the inside of the absolute value opposite okay so i'm going to make that the opposite one first so i'm going to say 7 and i'm going to do a bracket i'm going to put a negative there x plus 1 in parentheses and then we're going to write minus 5. okay so look what we did we made x plus one opposite okay and then our domain restriction we're going to say if what's inside the absolute value x plus one is less than zero okay so the one we make opposite we use the less than symbol now the next one that we're going to write is just exactly as it's written but we change the absolute values to parentheses so 7 times x plus 1 minus 5 if now here we still do x plus one because that's what's inside the absolute value but now we say greater than or equal to zero okay so that's basically it for our piecewise function but we can simplify that a little bit okay so now we're going to take each portion and we're going to simplify it so we'll do this part first okay so if i scroll down just a little bit i'm going to say 7 times and then in parentheses opposite of x plus 1 close my bracket minus five okay so i'm going to distribute this negative sign so now we have seven times negative x minus one minus five now we distribute our seven so negative seven x minus seven minus five and we get negative seven 7x minus 12. okay so now we can go back and we can write basically let's write it again so let's say f of x equals okay and now we have instead of this whole long part we have circled in pink now we can write this portion right here because we simplified it so negative 7x minus 12 if and now we can also simplify the domain here we can just subtract one right so now we can say if x is less than negative one okay so we're done with that first part all right now the second part we can simplify that one as well so we say seven so now this is the part here times x plus one in parenthesis minus five so here we just distribute the seven we get seven x plus seven minus five and that simplifies to seven x plus two okay so now we can add that to our piecewise function 7x plus 2 if and we subtract 1 again x is greater than or equal to negative 1. okay and that would be it that would be our piecewise function that we took from our absolute value function okay all right let's do one more where we start from a graph okay so now in this case we have to do two things the first thing we need to do is write our absolute value function in vertex form okay so to do that we need to write down the general form so let's say y equals a times x minus h plus k right that's our general vertex form for an absolute value function h and k is where our vertex is located right h comma k is where our vertex is so we notice right here our vertex is right there at three comma zero okay so now we can rewrite this as y equals a times x minus three and we'll just leave it as that because k is zero right so plus zero just simplifies and we can take it off all right so now we need to know what our a value is okay so to figure that out we just need to take one other point on our absolute value line here and plug it in for x and y okay so let's use this point right here okay that is four comma two so x is four and y is 2. so i'm going to say 2 equals a times the absolute value of 4 minus 3 2 equals a times the absolute value of 1 which is just 1 so we get a is equal to 2 okay so now we can take a equals 2 and we can plug it in right there for a excuse me so now our absolute value our absolute value vertex form is y equals 2 times the absolute value of x minus 3. okay so now we didn't have to do that on the first example because that was already given to us but if we're looking at a graph we got to find that first now we can take our absolute value in vertex form and we can write it as a piecewise function so we say f of x equals we open our set and the first one we're going to make what's inside opposite so 2 times opposite of x minus 3 in parentheses and this is going to be if let me move it over a little bit so we have some room we'll just move it down here okay if what's inside the absolute value x minus three is less than zero okay so less than zero because that was the part we made opposite so now the next one we're going to say 2 and we're just going to do parentheses times x minus 3 if x minus 3 is greater than or equal to 0. okay and now we simplify so we're going to take our first part here that i'm circling in yellow so i'm going to say 2 bracket negative and then x minus 3 in parentheses and we're going to simplify so if we distribute the negative we get 2 times negative x plus three now we can distribute our two and we get negative two x plus six okay so i'm just gonna circle that for right now that's gonna take the place of that part right there okay so i'll switch over and now let's say f of x equals now instead of two times bracket negative x minus three in parentheses we can just say negative two x plus six if and now here we can add three if x is less than 3 okay all right so now we go back and let's do the second part so here we just distribute the 2 since we don't have the negative there so that's just 2x minus 6. okay so we're done with that one we put it back in the piecewise function 2x minus 6. if we add 3 x is greater than or equal to 3. okay all right and that is how you take an absolute value in vertex form and you write it as a piecewise function [Music] |
189926 | https://www.youtube.com/watch?v=iwE7yWEicho | What are stomata and what's their function?
Next Generation Science
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Posted: 25 Apr 2023
photosynthesis #stomata #ngscience
Stomata play a pivotal role in the life of plants, acting as gatekeepers for the essential process of photosynthesis and the regulation of gas exchange. Found predominantly on the underside of leaves, these microscopic openings are integral to a plant's ability to sustain itself and contribute to the global oxygen and carbon cycles. The functionality of stomata is finely tuned by the plant in response to various environmental cues, highlighting a sophisticated mechanism of adaptation and survival.
Structure and Function of Stomata
Each stoma, or stomatal pore, is flanked by a pair of specialized guard cells, unique in their crescent or sausage-like shape. These guard cells are responsible for the opening and closing of the stomata, a process that is crucial for balancing gas exchange and water conservation. The opening of stomata facilitates the entry of carbon dioxide (CO2) into the leaf's internal structure, where it is used in photosynthesis to produce glucose, the plant's food. This same process results in the production of oxygen (O2) as a byproduct, which is then expelled into the atmosphere through these openings.
The Role of Stomata in Photosynthesis
During photosynthesis, the intake of CO2 through the stomata is essential for the synthesis of glucose. This process occurs in the chloroplasts of leaf cells, where CO2 combines with water (H2O) using the energy from sunlight to produce glucose and release oxygen. The ability of the stomata to regulate the amount of CO2 entering the plant is thus directly linked to the plant's photosynthetic efficiency and, by extension, its growth and survival.
Water Conservation and Transpiration
Aside from their role in photosynthesis, stomata are also critical in managing a plant's water balance through the process of transpiration. Transpiration is the evaporation of water vapor from the plant's surface, primarily through the stomata. While this process helps to cool the plant and facilitates nutrient transport from the roots, it also leads to water loss. Therefore, the ability of guard cells to close the stomata at night, when photosynthesis is not occurring and the demand for CO2 is minimal, is a vital water-conserving mechanism. This is especially crucial in arid environments, where water availability is a limiting factor for plant growth.
Environmental Influences on Stomatal Activity
The opening and closing of stomata are influenced by a variety of environmental factors, including light, humidity, temperature, and CO2 concentration. Light stimulates the opening of stomata, aligning photosynthesis with periods of sunlight availability. Conversely, high levels of CO2 inside the leaf can trigger the closing of stomata, as can drought conditions, which signal the plant to conserve water.
Adaptive Significance
The adaptive significance of stomatal regulation cannot be overstated. By optimizing the balance between gas exchange for photosynthesis and water conservation, stomata enable plants to thrive in a wide range of environmental conditions. This adaptive mechanism is a key factor in the ecological success of terrestrial plants, allowing them to colonize diverse habitats and play a foundational role in Earth's ecosystems.
Conclusion
In summary, stomata are not merely structural features on a plant's surface but are dynamic and responsive elements that play a critical role in a plant's ability to perform photosynthesis, regulate gas exchange, and conserve water. The intricate control of stomatal opening and closing by guard cells exemplifies the complex interplay between plants and their environment, highlighting the sophisticated adaptations that have enabled plants to become the primary producers supporting life on Earth.
Transcript:
What are stomata foreign [Music] stomata play a critical role by regulating the exchange of gases particularly carbon dioxide and oxygen between the plant and its environment What are guard cells stomata are tiny openings found mostly on the underside of the leaves the stomata can be opened and closed by specialized cells called guard cells these cells are shaped like Crescent moons or sausages and there are two of them for each stomatal pore thank you Whats their function when stomata are open they allow carbon dioxide to enter the leaf which is then used by the plant to produce glucose food through photosynthesis [Music] at the same time oxygen a byproduct of photosynthesis is released through the stomata into the atmosphere at night when sunlight is not available the rate of photosynthesis significantly decreases and the need for CO2 intake is greatly reduced closing the stomata at night helps to conserve water by reducing transpiration which is the loss of water vapor this is particularly important in arid environments where water conservation is essential for plant survival when the Sun rises guard cells open the stomata and photosynthesis continues [Music] worksheets relating to this short video can be downloaded by creating a free account at ngscience.com don't forget to subscribe to the NG science YouTube channel to never miss a video |
189927 | https://www.vocabulary.com/dictionary/gloomy | SKIP TO CONTENT
/Ëglumi/
IPA guide
Other forms: gloomier; gloomiest
Gloomy means "dark and dreary." A cloudy day, a sad song about lost love, your downbeat mood after your team loses a big game all of these can be called gloomy.
Have you ever been called a Gloomy Gus? If so, you must have been acting depressed or sulky. But you aren't the first to be called that after all, Gloomy Gus was a comic book character who first appeared in 1904. By the 1940s, this nickname caught on, describing and possibly adding to the misery of those who are less happy-seeming than the people around them.
Definitions of gloomy
adjective
depressingly dark
“the gloomy forest”
synonyms: gloomful, glooming, sulky
dark
devoid of or deficient in light or brightness; shadowed or black
2. adjective
filled with melancholy and despondency
“gloomy at the thought of what he had to face”
“gloomy predictions”
“a gloomy silence”
synonyms: blue, depressed, dispirited, down, down in the mouth, downcast, downhearted, grim, low, low-spirited
dejected
affected or marked by low spirits
3. adjective
causing dejection
synonyms: blue, dark, dingy, disconsolate, dismal, drab, drear, dreary, grim, sorry
cheerless, depressing, uncheerful
causing sad feelings of gloom and inadequacy
Pronunciation
US
/Ëglumi/
UK
/Ëglumi/
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Vocabulary from "Beowulf" (translated by by Gummere).
Cry Me A River: Synonyms for "Sad"
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189928 | https://math.libretexts.org/Courses/Prince_Georges_Community_College/MAT_1130_Mathematical_Ideas_Mirtova_Jones_(PGCC%3A_Fall_2022)/03%3A_Probability/3.05%3A_Conditional_Probabilities | Skip to main content
3.5: Conditional Probabilities
Last updated
: Aug 10, 2022
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3.4: Working with Events
3.6: Counting Methods
Page ID
: 107213
Maxie Inigo, Jennifer Jameson, Kathryn Kozak, Maya Lanzetta, & Kim Sonier
Coconino Community College
( \newcommand{\kernel}{\mathrm{null}\,})
What do you think is the probability that a man is over six feet tall? If you know that both of his parents were tall, would you change your estimate of the probability? If you know that both of his parents were short, would that affect your estimate in a different way? Most likely. The chance a man is over six feet tall is probably higher if he has tall parents and lower if he has short parents. A conditional probability is a probability that is based on some prior knowledge.
While conditional probability may seem like a difficult concept, we use it all the time in our every day life.
Weather forecasters use conditional probability to predict the likelihood of future weather conditions given current conditions. They may calculate the probability of rain if it is cloudy outside.
Sports betting companies may use conditional probability to set the odds that particular teams win their game. These odds may rely on knowledge about the team such knowing that a key player is injured.
A doctor may use conditional probability when discussing the efficacy of a vaccine with their patient. In other words, the chance of contracting a virus may be much less when the patient takes the vaccine compared to when they do not take the vaccine.
An insurance company uses conditional probability when setting rates for car insurance. For example, the insurance company may believe the chance you have an accident is higher if you are younger than 27.
All these examples of conditional probability have one thing in common: we assume that something is known before calculating a probability.
In Section 3.4 we learned that when events are dependent, whether or not event occurs affects the probability that event occurs. When using the rule with dependent events, we assume that event has already occurred and affected the sample space of event . In the sense of probability that means that the sample space of an experiment has been restricted before finding the probability of event . We can express this idea as "the probability of , given ."
We will discuss this idea of restricting the sample space as we proceed through examples of conditional probability.
Conditional Probability
A conditional probability is the probability that an event will occur if some other condition has already occurred. This is denoted by , which is read “the probability of given .”
Example 1
You spin a spinner with the 8 equally likely outcomes shown below. A friend covers up the number where the spinner lands.
What's the probability it landed on an even number?
Your friend now tells you the spinner landed on a one-digit number. What's the probability it landed on an even number?
Your friend tells you the spinner landed on a two-digit number. What's the probability it landed on a number less than 12?
Solution
The sample space of spinning the spinner is . Of these outcomes, 5 of them are even numbers so .
Now, you have some additional knowledge about there the spinner landed. The sample space is restricted to only one-digit numbers: Of these outcomes, 3 of them are even numbers. The probability that the spinner landed on an even number given that it landed on a one-digit number is . We can write this as .
Here, you have some different additional knowledge which restricts the original sample space to those outcomes with two digits: Of these outcomes, 1 of them is less than 12. So, .
Example 2
One card is drawn from a well-shuffled deck of 52 cards. Find the following probabilities:
probability that the card is a heart given that it is red.
probability that the card is red given that it is a heart.
Solution
We are told that the card is a red card. The sample space is restricted to only the 26 cards that are red. Of these 26 red cards, 13 are hearts. So, .
We are told that the card is a heart. The sample space is restricted to only the 13 cards that are hearts. Since every heart is red, .
In words, means "probability of selecting a King given that the card is a face card." The sample space is restricted to the 12 face cards. Of these 12 face cards, 4 are Kings. .
Try it Now 1
A box contains a collection of ping pong balls, all the same size but of different colors and numbers as shown below.
Find these probabilities.
the probability of selecting a yellow ball given the ball shows a "5."
the probability of selecting a ball showing a number greater than 10 given the ball is black.
Answer
While we can find conditional probabilities by analyzing and restricting the sample space, it is useful to have a formula for conditional probability. Let's examine a Venn diagram to develop a general formula.
The Venn diagram shows event and event which overlap in the intersection of .
Suppose we want to to find the probability of event given event , or . This means we need to restrict the sample space to only those outcomes in event . A portion of the Venn diagram has been blackened to show that these outcomes are no longer part of the sample space, leaving only the outcomes in the given event .
The only portion of event that remains once the sample space has been restricted to event is the portion of the Venn diagram . Therefore, or .
If we divide numerator and denominator of this conditional probability formula by the number of outcomes in the sample space , then we can compute the conditional probability in alternative way as
.
Conditional Probability Formula
For events and , or
Example 3
Two fair dice are rolled and the sum of the numbers is observed. What is the probability that the sum is at least 9 if it is known that a 5 was rolled?
Solution
We are given that the dice show a 5 so this is a conditional probability. We are asked to find . Here, event is "5 was rolled." Event is "sum is at least 9."
Find the number of outcomes in event . The pairs of dice showing a 5 are
Therefore, .
List the pairs of dice in event . The pairs of dice showing 5 was rolled and the sum is at least 9 is
Therefore, .
Applying the conditional probability formula,
.
The probability that the sum is at least 9 if it is known that a 5 was rolled is .
Try it Now 2
A coin is flipped twice and the results are recorded. Find each probability.
the probability that both coins land heads given at least one coin lands heads.
the probability that both coins land heads given the first coin lands heads.
Answer
Example 4
For a group of people, the probability of having blond hair is 25%. The probability of having blond hair and blue eyes is 10%. What is the probability that person has blue eyes given they have blond hair?
Solution
We need to find a conditional probability because it is given that a person has blond hair. We want to find . We don't have the sample space to count numbers of outcomes so we must use the second formula for conditional probability: .
The given event is "blond hair," so is the event "blue eyes." The problem states that and , so
The probability that a person has blond hair given they have blue eyes is 0.40, or 40%.
Try it Now 3
The label on a medicine bottle claims that there is a 14% chance of experiencing insomnia. There is a 5% chance of experiencing both a headache and insomnia. What's the probability that a person who takes this medicine has a headache given they have insomnia?
Answer
: or 35.71%
Example 5
The table shows survey results from 250 people who recently purchased a car.
Satisfaction of Car Buyers
| | Satisfied | Not Satisfied | Total |
| New Car | 92 | 28 | 120 |
| Used Car | 83 | 47 | 130 |
| Total | 175 | 75 | 250 |
Use the results in the table to find
the probability that a person is satisfied with their car.
the probability that a person is satisfied given the person bought a used car.
the probability that a person is not satisfied given the person bought a new car.
the probability that a person bought a new car given that they are satisfied.
Solution
The total number of customers was 250 of which 175 are satisfied with their purchase. This is not a conditional probability because the sample space has not been restricted.
So,
The probability that a person is satisfied is 0.7 or 70%.
We are given that the customer bought a used car. This is a conditional probability. Let represent the given condition "customer bought a used car." Let be "customer is satisfied." We want to find .
The number of people who bought a used car is .
The number of people who bought a used car and were satisfied is .
So, .
The probability that a person is satisfied given the person bought a used car is approximately 0.6385 or 63.85%.
We are given that the customer bought a new car. This is a conditional probability. Let represent the given condition "customer bought a new car." Let be "customer is not satisfied." We want to find .
The number of people who bought a new car is .
The number of people who bought a new car and were not satisfied is .
So, .
The probability that a person is satisfied given the person bought a used car is approximately 0.2333 or 23.33%.
We are given that the customer is satisfied. This is a conditional probability. Let represent the given condition "customer is satsified." Let be "customer bought a new car." We want to find .
The number of people who are satisified is .
The number of people who are satisfied and bought a new car is .
So, .
The probability that a person is satisfied given the person bought a used car is approximately 0.5257 or 52.57%.
You may also see conditional probability scenarios written in ways that do not use the word "given." For example, in the previous example we could describe the conditional event "a person is satisfied given they bought a used car" as "a person is satisfied if they bought a used car" or "a person who bought a used car is satisfied."
Example 6
A survey of 350 students at a university revealed the following data about class standing and place of residence.
Living Arrangements of University Students
| | Freshman | Sophomore | Junior | Senior | Total |
| Dormitory | 89 | 34 | 46 | 15 | 184 |
| Apartment | 32 | 17 | 22 | 48 | 119 |
| with Parents | 13 | 31 | 3 | 0 | 47 |
| Total | 134 | 82 | 71 | 63 | 350 |
Use the results in the table to find each probability.
What is the probability that a student is a sophomore if the student lives in an apartment?
What is the probability that a student lives with parents if the student is a freshman?
Solution
We are given that the student lives in an apartment. This is a conditional probability. Let represent the given condition "student lives in an apartment." Let be "student is a sophomore." We want to find .
The number of students who live in an apartment is .
The number of students who live in an apartment and are sophomores is .
So, .
The probability that a student is a sophomore if they live in an apartment is approximately 0.1429 or 14.29%.
We are given that the student is a freshman. This is a conditional probability. Let represent the given condition "student is a freshman." Let be "student lives with parents." We want to find .
The number of students who are freshmen is .
The number of people who are freshmen and live with parents is .
So, .
The probability that a lives with parents if they are a freshman is approximately 0.0970 or 9.70%.
Try it Now 4
A group of people were surveyed about the type of movies they prefer. Suppose a person is chosen at random from this group.
Movie Preferences
| Gender | Romantic, | Action, | Horror, | Total |
| Male, | 8 | 25 | 6 | 39 |
| Female, | 12 | 10 | 3 | 25 |
| Total | 20 | 35 | 9 | 64 |
Use the results in the table to find
the probability that a person prefers action movies given they are male.
the probability that a person is female given they prefer romantic movies.
Answer
3.4: Working with Events
3.6: Counting Methods |
189929 | https://drsapnag.manusadventures.com/chemistry/general-chemistry/summary/Summary10MolGeometry.pdf | Dr. Gupta/Summary/Molecular Geometry and Bonding Theories/Page 1 of 3 Chapter 10: Molecular Geometry and Bonding Theory Ionic bonding: crystal lattice Covalent bonding: VSEPR, valence bond and hybridization theory Metallic bonding: ability of electrons to flow on all atoms. VSEPR: Valence shell electron pair repulsion theory. Electrons try to be as far away from each other as possible. In bonding and bonding theory only the valence electrons are important. Each geometry is associated with its bond angles. Molecular geometry can be as follows: (A = central atom, X = terminal atoms and E = lone pair of electrons) Electron Groups AXE formula Bond Angle Example Electronic Geometry Shape of Molecule 2 AX2 180o BeCl2 Linear Linear 3 AX3 120 o BF3 Trigonal planar Trigonal planar 3 AX2E 120 o SO2 Trigonal planar Bent 4 AX4 109.5 o CH4 Tetrahedral Tetrahedral 4 AX3E 109.5 o NH3 Tetrahedral Trigonal Pyramidal 4 AX2E2 109.5 o H2O Tetrahedral Bent 5 AX5 90 o, 120 o, 180 o PCl5 Trigonal bipyramidal Trigonal Bipyramidal 5 AX4E 90 o, 120 o, 180 o SF4 Trigonal bipyramidal Seesaw 5 AX3E2 90 o, 180 o CIF4 Trigonal bipyramidal T – shape 5 AX2E3 180 o XeF2 Trigonal bipyramidal Linear 6 AX6 90 o, 180 o SF6 Octahedral Octahedral 6 AX5E 90 o BrF5 Octahedral Square Pyramidal 6 AX4E2 90 o XeF4 Octahedral Square Planar 6 AX3E3 90 o, 180 o Octahedral T – Shape 6 AX2E4 180 o Octahedral Linear Figure on the next page. Dr. Gupta/Summary/Molecular Geometry and Bonding Theories/Page 2 of 3 No of e- groups Geometry (all atoms) 1 Lone Pair 2 Lone Pairs 3 Lone Pairs 4 Lone Pairs 2 Linear AX2 3 Trigonal Planar AX3 Bent AX2E 4 Tetrahedral AX4 Trigonal pyramidal AX3E Bent AX2E2 5 Trigonal Bipyramidal AX5 Seesaw AX4E T-Shape AX3E2 Linear AX2E3 6 Octahedral AX6 Square Pyramid AX5E Square Planar AX4E2 T-shape AX3E3 Linear AX2E4 Dipole moment and polar molecules • Polar bond and polar molecules • Polar bond – when the two atoms of a bond have different electronegativity while • Polar molecules have overall non-zero dipole moments and non polar molecules have zero dipole moment. • A bond can be polar but the molecule does not have to be. • Dipole moment = magnitude of charge x distance separating the positive and negative charge • Units (debye = coulomb x meter) • Dipole moment usually can be determined by exposing molecules to electric field. X A X X A X X X A X ..
X A X X X X A X X ..
X A X ..
..
X A X X X X X A X X X ..
X A X X ..
..
X A X ..
..
..
X A X X X X X A X X X X X ..
A X X X X ..
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A X X X ..
..
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Dr. Gupta/Summary/Molecular Geometry and Bonding Theories/Page 3 of 3 Determining the polarity of a molecule 1) write Lewis structure of molecule (write ALL electrons) 2) write the shape of molecule 3) check the electronegativity of each atom (estimate from PT) 4) see the symmetry of the molecule (not just in shape but in types of atoms) 5) see if the equal electronegativity of similar atoms cancel each other out, if not then molecule may be polar. Valence Bond Theory Electron clouds of opposite spins overlap to form bonds. Important points: a) Electrons stay in their respective orbitals b) Bonding electrons localize in the region of overlap c) Maximum overlap occurs when orbitals overlap end to end d) Molecular geometry depends on geometry of orbitals. Has flaws when looking at bond angles and molecular of majority of molecules. Hybridization Mixing of orbitals of excited atoms to form new orbitals, which overlap to form bonds. Type Typical shape Bond angle e.g. Sp3 Tetrahedral 109o CH4, HN3 Sp2 Trigonal planar 120 o BF3, CH2=CH2 Sp Linear 180 o BeF2, CH CH Hybrid orbitals of d shell are also possible. Hybridization in organic chemistry Alkanes, alkenes, alkynes. Geometric isomerism: cis and trans geometry of alkenes. Molecular Orbital Theory Atomic orbitals (AO) combine to form molecular orbitals (MO) and antibonding molecular orbitals. |
189930 | https://www.droracle.ai/articles/172996/treatment-for-acute-dystonic-reaction | What is the treatment for acute dystonic reaction?
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What is the treatment for acute dystonic reaction?
Medical Advisory BoardAll articles are reviewed for accuracy by our Medical Advisory Board
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Last updated: June 19, 2025 • View editorial policy
From the Guidelines
The treatment for acute dystonic reaction should primarily involve anticholinergic medications, with the goal of rapidly relieving symptoms and preventing recurrence, as recommended by the American Psychiatric Association 1. The most effective treatment approach for acute dystonic reaction involves administering anticholinergic medications, such as benztropine (Cogentin) or diphenhydramine (Benadryl), to rapidly relieve symptoms.
For adults, benztropine 1-2 mg can be administered intramuscularly or intravenously, while diphenhydramine 25-50 mg can be given intravenously or intramuscularly.
For children, diphenhydramine 1 mg/kg up to 50 mg is preferred. After the acute episode resolves, oral anticholinergic medication should be continued for 2-3 days to prevent recurrence, such as benztropine 1-2 mg twice daily or diphenhydramine 25-50 mg three to four times daily 1. It is essential to discontinue or reduce the dose of the offending agent, typically antipsychotics, antiemetics, or other dopamine-blocking drugs, if possible. Patients should be monitored for anticholinergic side effects, including dry mouth, blurred vision, urinary retention, and confusion, which are usually mild and transient 1. The American Psychiatric Association recommends treating patients who have acute dystonia associated with antipsychotic therapy with an anticholinergic medication 1. This approach prioritizes reducing morbidity, mortality, and improving quality of life for patients experiencing acute dystonic reactions.
From the FDA Drug Label
In acute dystonic reactions, 1 to 2 mL of the injection usually relieves the condition quickly. The treatment for acute dystonic reaction is 1 to 2 mL of benztropine mesylate injection which usually provides quick relief 2.
The dose can be repeated if the condition returns.
The recommended dosage is 1 to 4 mg once or twice a day parenterally.
Dosage must be individualized according to the need of the patient.
From the Research
Treatment Options for Acute Dystonic Reaction
The treatment for acute dystonic reaction typically involves the use of anticholinergic or antidopaminergic medications.
Anticholinergic drugs, such as benztropine mesylate 3 or biperiden 4, have been shown to be effective in treating acute dystonic reactions.
Diphenhydramine, an antihistamine with anticholinergic properties, can also be used to treat acute dystonic reactions, although it has been reported to cause dystonic reactions in some cases 5.
In some cases, benzodiazepines, such as diazepam, may be used to treat acute dystonic reactions 5.
Medication Administration
These medications can be administered intramuscularly or intravenously, depending on the severity of the reaction and the patient's condition.
Benztropine mesylate has been shown to be effective in reducing recovery time in patients with acute dystonic reactions 3.
Biperiden has been used to treat acute dystonic reactions, particularly those induced by metoclopramide 4.
Important Considerations
It is essential to monitor patients for extrapyramidal side effects, such as dystonic reactions, when administering medications that can cause these symptoms, such as metoclopramide 4.
Patients who are prone to dystonic reactions, such as those with a history of drug abuse or those taking high doses of metoclopramide, should be closely monitored for signs of acute dystonic reaction 3, 4.
References
1
Guideline
Guideline Directed Topic Overview
Dr.Oracle Medical Advisory Board & Editors, 2025
2
Drug
Official FDA Drug Label For benztropine (IM)
FDA, 2025
3
Research
Treatment of drug-induced dystonic reactions.
JACEP, 1979
4
Research
Metoclopramide induced acute dystonic reaction: A case report.
Annals of medicine and surgery (2012), 2022
5
Research
Diphenhydramine induced acute dystonia: a case report.
The Pan African medical journal, 2022
Related Questions
Is Benadryl (diphenhydramine) or benzodiazepines better to give with Reglan (metoclopramide) for an acute dystonic reaction?What is the first line treatment for a dystonic reaction caused by droperidol (Butyrophenone)?Are Cogentin (benztropine) and Benadryl (diphenhydramine) the same thing?Does intravenous (IV) diphenhydramine (Benadryl) prevent dystonic reactions from prochlorperazine (Compazine)?What are the causes of acute dystonic reactions?What are the management options for 1st trimester pregnancy complications?How long after stopping hyperventilation (rapid breathing) should muscle twitching start if it didn't occur during the 3 minutes of hyperventilation?What is the management of hyperemesis gravidarum?Does extending hyperventilation beyond 3 minutes increase the risk of inducing twitching if no twitching has occurred?Does a vitamin B12 (B12) level of 330 picograms per milliliter (pg/mL) require supplementation?What is the management plan for a woman experiencing a miscarriage?
Professional Medical Disclaimer
This information is intended for healthcare professionals. Any medical decision-making should rely on clinical judgment and independently verified information. The content provided herein does not replace professional discretion and should be considered supplementary to established clinical guidelines. Healthcare providers should verify all information against primary literature and current practice standards before application in patient care. Dr.Oracle assumes no liability for clinical decisions based on this content.
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189931 | https://askfilo.com/user-question-answers-science/keplers-first-law-the-orbit-of-a-planet-is-an-ellipse-with-393437323337 | Question asked by Filo student
ellipse then, AF1+AF2=BF1+BF2=CF1+CF2 Kepler's first law : The orbit of a planet is an ellipse with the Sun at one of the foci.
Figure 1.4 shows the elliptical orbit of a planet revolving around the sun. The position of the Sun is indicated by S.
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| Question Text | ellipse then, AF1+AF2=BF1+BF2=CF1+CF2 Kepler's first law : The orbit of a planet is an ellipse with the Sun at one of the foci. Figure 1.4 shows the elliptical orbit of a planet revolving around the sun. The position of the Sun is indicated by S. |
| Updated On | Feb 27, 2023 |
| Topic | Physics |
| Subject | Science |
| Class | Class 10 |
| Answer Type | Video solution: 1 |
| Upvotes | 132 |
| Avg. Video Duration | 3 min |
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189932 | https://basicenglishspeaking.com/042-as-as-possible/ | 042.... as... as possible - Basic English Speaking
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189933 | https://openstax.org/books/college-physics/pages/27-conceptual-questions | Ch. 27 Conceptual Questions - College Physics | OpenStax
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College Physics
Conceptual Questions
College Physics Conceptual Questions
Contents Contents
Highlights
Table of contents
Preface
1 Introduction: The Nature of Science and Physics
2 Kinematics
3 Two-Dimensional Kinematics
4 Dynamics: Force and Newton's Laws of Motion
5 Further Applications of Newton's Laws: Friction, Drag, and Elasticity
6 Uniform Circular Motion and Gravitation
7 Work, Energy, and Energy Resources
8 Linear Momentum and Collisions
9 Statics and Torque
10 Rotational Motion and Angular Momentum
11 Fluid Statics
12 Fluid Dynamics and Its Biological and Medical Applications
13 Temperature, Kinetic Theory, and the Gas Laws
14 Heat and Heat Transfer Methods
15 Thermodynamics
16 Oscillatory Motion and Waves
17 Physics of Hearing
18 Electric Charge and Electric Field
19 Electric Potential and Electric Field
20 Electric Current, Resistance, and Ohm's Law
21 Circuits and DC Instruments
22 Magnetism
23 Electromagnetic Induction, AC Circuits, and Electrical Technologies
24 Electromagnetic Waves
25 Geometric Optics
26 Vision and Optical Instruments
27 Wave Optics
Introduction to Wave Optics
27.1 The Wave Aspect of Light: Interference
27.2 Huygens's Principle: Diffraction
27.3 Young’s Double Slit Experiment
27.4 Multiple Slit Diffraction
27.5 Single Slit Diffraction
27.6 Limits of Resolution: The Rayleigh Criterion
27.7 Thin Film Interference
27.8 Polarization
27.9Extended Topic Microscopy Enhanced by the Wave Characteristics of Light
Glossary
Section Summary
Conceptual Questions
Problems & Exercises
28 Special Relativity
29 Quantum Physics
30 Atomic Physics
31 Radioactivity and Nuclear Physics
32 Medical Applications of Nuclear Physics
33 Particle Physics
34 Frontiers of Physics
A | Atomic Masses
B | Selected Radioactive Isotopes
C | Useful Information
D | Glossary of Key Symbols and Notation
Index
Search for key terms or text.
Close
27.1 The Wave Aspect of Light: Interference -------------------------------------------
What type of experimental evidence indicates that light is a wave?
Give an example of a wave characteristic of light that is easily observed outside the laboratory.
27.2 Huygens's Principle: Diffraction -------------------------------------
How do wave effects depend on the size of the object with which the wave interacts? For example, why does sound bend around the corner of a building while light does not?
Under what conditions can light be modeled like a ray? Like a wave?
Go outside in the sunlight and observe your shadow. It has fuzzy edges even if you do not. Is this a diffraction effect? Explain.
Why does the wavelength of light decrease when it passes from vacuum into a medium? State which attributes change and which stay the same and, thus, require the wavelength to decrease.
Does Huygens’s principle apply to all types of waves?
27.3 Young’s Double Slit Experiment -----------------------------------
Young’s double slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.
Suppose you use the same double slit to perform Young’s double slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.
Is it possible to create a situation in which there is only destructive interference? Explain.
Figure 27.55 shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the center. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.
Figure 27.55 This double slit interference pattern also shows signs of single slit interference. (credit: PASCO)
27.4 Multiple Slit Diffraction ------------------------------
What is the advantage of a diffraction grating over a double slit in dispersing light into a spectrum?
What are the advantages of a diffraction grating over a prism in dispersing light for spectral analysis?
Can the lines in a diffraction grating be too close together to be useful as a spectroscopic tool for visible light? If so, what type of EM radiation would the grating be suitable for? Explain.
If a beam of white light passes through a diffraction grating with vertical lines, the light is dispersed into rainbow colors on the right and left. If a glass prism disperses white light to the right into a rainbow, how does the sequence of colors compare with that produced on the right by a diffraction grating?
Suppose pure-wavelength light falls on a diffraction grating. What happens to the interference pattern if the same light falls on a grating that has more lines per centimeter? What happens to the interference pattern if a longer-wavelength light falls on the same grating? Explain how these two effects are consistent in terms of the relationship of wavelength to the distance between slits.
Suppose a feather appears green but has no green pigment. Explain in terms of diffraction.
It is possible that there is no minimum in the interference pattern of a single slit. Explain why. Is the same true of double slits and diffraction gratings?
27.5 Single Slit Diffraction ----------------------------
As the width of the slit producing a single-slit diffraction pattern is reduced, how will the diffraction pattern produced change?
27.6 Limits of Resolution: The Rayleigh Criterion -------------------------------------------------
A beam of light always spreads out. Why can a beam not be created with parallel rays to prevent spreading? Why can lenses, mirrors, or apertures not be used to correct the spreading?
27.7 Thin Film Interference ---------------------------
What effect does increasing the wedge angle have on the spacing of interference fringes? If the wedge angle is too large, fringes are not observed. Why?
How is the difference in paths taken by two originally in-phase light waves related to whether they interfere constructively or destructively? How can this be affected by reflection? By refraction?
Is there a phase change in the light reflected from either surface of a contact lens floating on a person’s tear layer? The index of refraction of the lens is about 1.5, and its top surface is dry.
In placing a sample on a microscope slide, a glass cover is placed over a water drop on the glass slide. Light incident from above can reflect from the top and bottom of the glass cover and from the glass slide below the water drop. At which surfaces will there be a phase change in the reflected light?
Answer the above question if the fluid between the two pieces of crown glass is carbon disulfide.
While contemplating the food value of a slice of ham, you notice a rainbow of color reflected from its moist surface. Explain its origin.
An inventor notices that a soap bubble is dark at its thinnest and realizes that destructive interference is taking place for all wavelengths. How could she use this knowledge to make a non-reflective coating for lenses that is effective at all wavelengths? That is, what limits would there be on the index of refraction and thickness of the coating? How might this be impractical?
A non-reflective coating like the one described in Example 27.6 works ideally for a single wavelength and for perpendicular incidence. What happens for other wavelengths and other incident directions? Be specific.
Why is it much more difficult to see interference fringes for light reflected from a thick piece of glass than from a thin film? Would it be easier if monochromatic light were used?
27.8 Polarization -----------------
Under what circumstances is the phase of light changed by reflection? Is the phase related to polarization?
Can a sound wave in air be polarized? Explain.
No light passes through two perfect polarizing filters with perpendicular axes. However, if a third polarizing filter is placed between the original two, some light can pass. Why is this? Under what circumstances does most of the light pass?
Explain what happens to the energy carried by light that it is dimmed by passing it through two crossed polarizing filters.
When particles scattering light are much smaller than its wavelength, the amount of scattering is proportional to 1/λ 4 1/λ 4 1/λ 4 size 12{1/λ rSup { size 8{4} } } {}. Does this mean there is more scattering for small λ λ λ size 12{λ} {} than large λ λ λ size 12{λ} {}? How does this relate to the fact that the sky is blue?
Using the information given in the preceding question, explain why sunsets are red.
When light is reflected at Brewster’s angle from a smooth surface, it is 100%100%100% size 12{"100"%} {} polarized parallel to the surface. Part of the light will be refracted into the surface. Describe how you would do an experiment to determine the polarization of the refracted light. What direction would you expect the polarization to have and would you expect it to be 100%100%100% size 12{"100"%} {}?
27.9Extended Topic Microscopy Enhanced by the Wave Characteristics of Light -----------------------------------------------------------------------------
Explain how microscopes can use wave optics to improve contrast and why this is important.
A bright white light under water is collimated and directed upon a prism. What range of colors does one see emerging?
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189934 | https://theory.stanford.edu/~jvondrak/data/Sperner-coloring.pdf | Sperner’s Colorings and Optimal Partitioning of the Simplex∗ Maryam Mirzakhani† Jan Vondr´ ak‡ Abstract We discuss coloring and partitioning questions related to Sperner’s Lemma, originally motivated by an application in hardness of approx-imation.
Informally, we call a partitioning of the (k −1)-dimensional simplex into k parts, or a labeling of a lattice inside the simplex by k colors, “Sperner-admissible” if color i avoids the face opposite to vertex i.
The questions we study are of the following flavor: What is the Sperner-admissible labeling/partitioning that makes the total area of the boundary between different colors/parts as small as possible?
First, for a natural arrangement of “cells” in the simplex, we prove an optimal lower bound on the number of cells that must be non-monochromatic in any Sperner-admissible labeling. This lower bound is matched by a simple labeling where each vertex receives the minimum admissible color.
Second, we show for this arrangement that in contrast to Sperner’s Lemma, there is a Sperner-admissible labeling such that every cell contains at most 4 colors.
Finally, we prove a geometric variant of the first result: For any Sperner-admissible partition of the regular simplex, the total surface area of the boundary shared by at least two different parts is minimized by the Voronoi partition (A∗ 1, . . . , A∗ k) where A∗ i contains all the points whose closest vertex is ei. We also discuss possible extensions of this result to general polytopes and some open questions.
1 Introduction Sperner’s Lemma is a gem in combinatorics which was originally discovered by Emmanuel Sperner as a tool to derive a simple proof of Brouwer’s Fixed Point Theorem. Since then, Sperner’s Lemma has seen numerous applications, notably in the proof of existence of mixed Nash equilibria , in fair division , and recently it played an important role in the study of computational complexity of finding a Nash equilibrium [3, 2]. At a high level, Sperner’s Lemma states that for any coloring of a simplicial subdivision of a simplex satisfying certain boundary conditions, there must be a “rainbow cell” that receives all possible colors. We review Sperner’s Lemma in Section 3.
∗To appear in “A Journey through Discrete Mathematics. A Tribute to Jiˇ r´ ı Matouˇ sek”, edited by Martin Loebl, Jaroslav Neˇ setˇ ril and Robin Thomas, due to be published by Springer.
The first two results of this paper have also appeared in .
†Dept. of Mathematics, Stanford University, Stanford, CA; mmirzakh@stanford.edu ‡Dept. of Mathematics, Stanford University, Stanford, CA; jvondrak@stanford.edu 1 The starting point of this work was a question that arises in the study of approximation algorithms for a certain hypergraph labeling problem . The question posed by , while in some ways reminiscent of Sperner’s Lemma, is different in the following sense: Instead of asking whether there exists a rainbow cell for any admissible coloring, the question is what is the minimum possible number of cells that must be non-monochromatic. (Also, the question arises for a particular regular lattice inside the simplex rather than an arbitrary subdivision.) In this paper, we resolve this question and investigate some related problems.
Before we state our results, let us note the following connection. As the gran-ularity of the subdivision tends to zero, Sperner’s Lemma becomes a statement about certain geometric partitions of the simplex: for any Sperner-admissible partition, where part i avoids the face opposite to vertex i, there must be a point where all parts meet. This result is known as the Knaster-Kuratowski-Mazurkiewicz Lemma . In contrast, the questions we are studying are con-cerned with the measure of the boundary where at least two different parts meet: This can be viewed as a multi-colored isoperimetric inequality, where we try to partition the simplex in a certain way, so that the surface area of the union of all pairwise boundaries (what we call a separating set) is minimized. The way we measure the separating set also affects the problem; the discrete version of the question that is of primary interest to us is mandated by the application in . In the geometric setting, a natural notion of surface area is the Minkowski content of the separating set (which coincides with other notions of volume for well-behaved sets). We give an optimal answer to this question for a regular simplex and discuss other related questions.
To state our results formally, we need some notation that we introduce in Section 2. We postpone our contributions to Sections 4—6, after a discussion of Sperner’s Lemma in Section 3.
2 Preliminaries We denote vectors in boldface, such as v ∈Rk.
The coordinates of v are written in italics, such as v = (v1, . . . , vk). By ei, we denote the canonical basis vectors (0, . . . , 1, . . . , 0). By conv(v1, . . . , vk), we denote the convex hull of the respective vectors.
2.1 Simplicial subdivisions of the simplex Consider the (k −1)-dimensional simplex defined by ∆k = conv(e1, . . . , ek) = ( x = (x1, x2, . . . , xk) ∈Rk : x ≥0, k X i=1 xi = 1 ) .
Simplicial subdivision. A simplicial subdivision of ∆k is a collection of sim-plices (“cells”) Σ such that 2 • The union of the cells in Σ is the simplex ∆k.
• For any two cells σ1, σ2 ∈Σ, their intersection is either empty or a full face of a certain dimension shared by σ1, σ2.
The Simplex-Lattice Hypergraph. Next, we describe a specific configura-tion of cells in a simplex; this configuration is actually not a full subdivision since its cells do not cover the full volume of the simplex. It can be completed to a subdivision if desired.1 Let q ≥1 be an integer and define ∆k,q = ( x = (x1, x2, . . . , xk) ∈Rk : x ≥0, k X i=1 xi = q ) .
We consider a vertex set of all the points in ∆k,q with integer coordinates: Vk,q = ( a = (a1, a2, . . . , ak) ∈Zk : a ≥0, k X i=1 ai = q ) .
Figure 1: The Simplex Lattice Hypergraph for k = 3 and q = 5, with hyperedges shaded in gray. The gray triangles together with the white triangles form a simplicial subdivision.
The lists of admissible colors are given on the boundary; for internal vertices the lists are all {1, 2, 3}.
{1} {2} {3} {1, 2} {1, 2} {1, 2} {1, 2} {1, 3} {1, 3} {1, 3} {1, 3} {2, 3} {2, 3} {2, 3} {2, 3} The Simplex-Lattice Hypergraph is a k-uniform hypergraph Hk,q = (Vk,q, Ek,q) whose hyperedges (which we also call cells due to their geometric interpretation) are indexed by b ∈Zk + such that Pk i=1 bi = q −1: we have Ek,q = ( e(b) : b ∈Zk, b ≥0, k X i=1 bi = q −1 ) 1This specific configuration arises in as an integrality gap example for a certain hyper-graph labeling problem; see also for more details.
3 where e(b) = {b + e1, b + e2, . . . , b + ek} = {(b1 + 1, b2, . . . , bk), (b1, b2 + 1, . . . , bk), . . . , (b1, b2, . . . , bk + 1)}. For each vertex a ∈Vk,q, we have a list of admissible colors L(a), which is L(a) = {i ∈[k] : ai > 0}.
3 Sperner’s Lemma First, let us recall the statement of Sperner’s Lemma . We consider labelings ℓ: Vk,q →[k]. We call a labeling ℓSperner-admissible if ℓ(a) ∈L(a) for each a ∈V ; i.e. , if ℓ(a) = j then aj > 0.
Lemma 1 (Sperner’s Lemma). For every Sperner-admissible labeling of the vertices of a simplicial subdivision of ∆k, there is a cell whose vertices receive all k colors.
Figure 2: A Sperner-admissible labeling for k = 3 and q = 5. At least one cell in the triangulation (not necessarily in Ek,q) must be k-colored (rainbow).
1 2 3 1 1 1 2 1 3 3 3 2 2 2 3 1 1 3 2 2 3 We remark that this does not say anything about the Simplex-Lattice Hy-pergraph: Even if the subdivision uses the point set Vk,q, the rainbow cell given by Sperner’s Lemma might not be a member of Ek,q since Ek,q consists only of scaled copies of ∆k,q without rotation; it is not a full subdivision of the simplex.
(See Figure 2.) 4 The Simplex-Lattice Coloring Lemma Instead of rainbow cells, the statement proposed (and proved for k = 3) in involves non-monochromatic cells.
4 Proposition 1 (Simplex-Lattice Coloring Lemma). For any Sperner-admissible labeling ℓ: Vk,q →[k], there are at least q+k−3 k−2 hyperedges e ∈Ek,q that are non-monochromatic under ℓ.
The first-choice labeling.
In particular, Proposition 1 is that a Sperner-admissible labeling minimizing the number of non-monochromatic cells is a “first-choice one” which labels each vertex a by the smallest coordinate i such that ai > 0. Under this labeling, all the hyperedges e(b) such that b1 > 0 are labeled monochromatically by 1. The only hyperedges that receive more than 1 color are those where b1 = 0, and the number of such hyperedges is exactly q+k−3 k−2 (see ). Here we give a proof of Proposition 1.
Figure 3: The first-choice labeling.
1 2 3 1 1 1 1 1 1 1 1 2 2 2 2 1 1 1 1 1 1 Proof. Consider the set of hyperedges Ek,q: observe that it can be written naturally as Ek,q = {e(b) : b ∈Vk,q−1}.
I.e., the hyperedges can be identified one-to-one with the vertices in Vk,q−1.
Recall that e(b) = {b + e1, b + e2, . . . , b + ek}. Two hyperedges e(b), e(b′) share a vertex if and only if b′ + ej = b + ei for some pair i, j ∈[k]; or in other words if b, b′ are nearest neighbors in Vk,q−1 (differ by ±1 in exactly two coordinates).
Consider a labeling ℓ: Vk,q →[k]. For each i ∈[k], let Ci denote the set of points in Vk,q−1 representing the monochromatic hyperedges in color i, Ci = {b ∈Vk,q−1 : ∀v ∈e(b); ℓ(v) = i}.
Define an injective mapping φi : Ci →Vk,q−2 as follows: φi(b) = b −ei.
5 The image is indeed in Vk,q−2: if b ∈Ci, we have bi > 0, or else e(b) would contain a vertex a such that ai = 0 and hence e(b) could not be monochromatic in color i.
Therefore, b −ei ∈Zk + and (b −ei) · 1 = q −2 which means b −ei ∈Vk,q−2. (Here, 1 denotes the all-1’s vector.) Figure 4: The mappings φi : Ci →Vk,q−2.
The hyperedges are represented by the empty circles; Ci is the subset of them monochromatic in color i.
The black squares represent Vk,q−2; note that each point in Vk,q−2 is the image of at most one monochromatic hyperedge.
1 2 3 1 1 2 2 1 3 3 3 2 2 3 3 1 2 3 2 2 3 Further, we claim that φi[Ci] ∩φj[Cj] = ∅for every i ̸= j. If not, there would be b ∈Ci and b′ ∈Cj such that b −ei = b′ −ej. Then, the point a = b + ej = b′ + ei would be an element of both the hyperedge e(b) and the hyperedge e(b′). This contradicts the assumption that e(b) is monochromatic in color i and e(b′) is monochromatic in color j. So the sets φi[Ci] are pairwise disjoint subsets of Vk,q−2. By the definition of φi, we clearly have |φi[Ci]| = |Ci|.
We conclude that the total number of monochromatic hyperedges is k X i=1 |Ci| = k X i=1 |φi[Ci]| ≤|Vk,q−2|.
The total number of hyperedges is |Ek,q| = |Vk,q−1|. Considering that |Vk,q| = q+k−1 k−1 (the number of partitions of q into a sum of k nonnegative integers), we obtain that the number of non-monochromatic hyperedges is |Ek,q| −Pk i=1 |Ci| ≥|Vk,q−1| −|Vk,q−2| = q+k−2 k−1 − q+k−3 k−1 = q+k−3 k−2 .
6 5 A labeling of Hk,q with at most 4 colors on each hyperedge We recall that Sperner’s lemma states that any Sperner-admissible labeling of a subdivision of the simplex must contain a simplex with all k colors.
The hypergraph Hk,q defined in Section 2.1 is not a subdivision since it covers only a subset of the large simplex. It is easy to see that the conclusion of Sperner’s lemma does not hold for Hk,q — for example for k = 3, we can label a 2-dimensional triangulation so that exactly one triangle has 3 different colors, and this triangle is not in E3,q. (See Figure 2.) Hence, each triangle in E3,q has at most 2 colors. By an extension of this argument, we can label Hk,q so that each hyperedge in Ek,q contains at most k −1 colors. The question we ask in this section is, what is the minimum c such that there is a Sperner-admissible labeling with at most c different colors on each hyperedge in Ek,q? We prove the following result.
Proposition 2. For any k ≥4 and q ≥k2, there is a Sperner-admissible labeling of Hk,q = (Vk,q, Ek,q) such that every hyperedge in Ek,q contains at most 4 different colors.
We note that this statement is not true for q = 1 and k > 4 (since Ek,1 consists of a single simplex which has k different colors). We have not identified the optimal lower bound on q that allows our statement to hold.
Also, the statement could possibly hold with 2 or 3 colors instead of 4; the number 4 is just an artifact of our proof and we have no reason to believe that it is tight.
The intuition behind our construction is as follows: We want to label the vertices so that the number of different colors on each hyperedge is small. A nat-ural choice is to label each vertex v by its maximum-value coordinate. However, this does not work since a hyperedge in the center of the simplex may receive all k colors. The problem is that this labeling is possibly very sensitive to small changes in v. A more “robust” labeling is one where we select a subset of “top coordinates” and choose one among them according to another rule. This rule should be such that incrementing the coordinates one at a time does not change the label too many times. One such rule that works well is described below.
Proof. We define a labeling ℓ: Vk,q →[k] as follows: • Given a ∈Vk,q, let π : [k] →[k] be a permutation such that aπ(1) ≥aπ(2) ≥ . . . ≥aπ(k) (and if aπ(i) = aπ(i+1), we order π so that π(i) < π(i + 1)).
• Define t(a) to be the maximum t ∈[k] such that ∀1 ≤j ≤t, aπ(j) ≥k−j+ 1. We define the “Top coordinates” of a to be Top(a) = (π(1), . . . , π(t(a))) (an ordered set).
• We define the label of a to be ℓ(a) = π(t(a)), the index of the “last Top coordinate”.
7 First, we verify that this is a well-defined Sperner-admissible labeling. Since Pk i=1 ai = q ≥k2, we have aπ(1) = max ai ≥k and hence 1 ≤t(a) ≤k. For each a ∈Vk,q, we have: aℓ(a) = aπ(t(a)) ≥k −t(a) + 1 > 0, since t(a) ≤k.
Therefore, ℓis Sperner-admissible.
Now, consider a hyperedge e(b) = (b + e1, b + e2, . . . , b + ek) where b ≥ 0, Pk i=1 bi = q −1. We claim that ℓ(b+ei) attains at most 4 different values for i = 1, . . . , k. Without loss of generality, assume that b1 ≥b2 ≥. . . ≥bk. Define ℓ∗to be the label assigned to b by our construction (note that b is not a vertex in Vk,q but we can still apply our definition): ℓ∗is the maximum value in [k] such that for all 1 ≤j ≤ℓ∗, bj ≥k −j + 1. Hence, we have Top(b) = {1, 2, . . . , ℓ∗}.
Let i ∈[k], a = b + ei, and let π be the permutation such that aπ(1) ≥. . . ≥ aπ(k) as above. (Recall that for b, we assumed that the respective permutation is the identity.) We consider the following cases: • If 1 ≤i < ℓ∗, then we claim that ℓ(a) = ℓ(b + ei) = ℓ(b) = ℓ∗. In the rule for selecting t(a), one of the first ℓ∗−1 coordinates has been incremented compared to b, which possibly pushes i forward in the ordering of the Top coordinates. However, the other coordinates remain unchanged, the condition aπ(j) ≥k −j + 1 is still satisfied for 1 ≤j ≤ℓ∗, and Top(a) = Top(b). In particular ℓ∗is still the last coordinate included in Top(a) and hence ℓ(a) = ℓ∗.
• If i = ℓ∗, then ℓ(a) = ℓ(b + eℓ∗) is still one of the coordinates in Top(b), possibly different from ℓ∗(due to a change in order, although we still have Top(a) = Top(b)) — let us call this label ℓ∗ 2.
• If ℓ∗< i ≤k, then it is possible that in a = b+ei, we obtain additional Top coordinates (Top(a) ⊃Top(b)). It could be ai = bi +1 itself which is now included among the Top coordinates, and possibly additional coordinates that already satisfied the condition bj ≥k −j + 1 but were not selected due to the condition being false for bℓ∗+1. If this does not happen and we have Top(a) = Top(b), the label of a is still ℓ(a) = ℓ∗(because the ordering of the Top coordinates remains the same).
Assume now that Top(a) has additional coordinates beyond Top(b). By the definition of ℓ∗, we have bℓ∗≥k −ℓ∗+ 1, and for each j > ℓ∗, we have bj < k −ℓ∗; otherwise j would have been still chosen in Top(b).
For Top(a) = Top(b + ei) to grow beyond Top(b), ai must become the (ℓ∗+ 1)-largest coordinate and satisfy ai ≥k −ℓ∗. The only way this can happen is that bi = k −ℓ∗−1 and hence ai = bi + 1 = k −ℓ∗. In this case, ai is the maximum coordinate among {aj : j > ℓ∗}, and still smaller than aℓ∗. Therefore, i will be included in Top(a). Now, Top(a) may grow further. However, note that the construction of Top(a) will proceed in the same way for every a = b + ei such that bi = k −ℓ∗−1. This is because all the coordinates equal to k −ℓ∗−1 will be certainly included in Top(a), and coordinates smaller than k −ℓ∗−1 remain the same in each of these cases (equal to the coordinates of b). Therefore, the set Top(a) will be the same in all these cases; let us call this set Top+.
8 The label assigned to a = b+ei is the index of the last coordinate included in Top+ = Top(a). Since Top+ is the same whenever Top(a) ̸= Top(b), the label of a will be the coordinate j∗minimizing bj (and maximizing j to break ties) among all j ∈Top+, unless j∗= i in which case the last included coordinate might be another one. This gives potentially two additional colors, let us call them ℓ∗ 3, ℓ∗ 4, that are assigned to a = b + ei for all i > ℓ∗where bi = k −ℓ∗−1. For other choices of i > ℓ∗, we have Top(b + ei) = Top(b) and the label assigned to b + ei is ℓ(b + ei) = ℓ∗.
To summarize, all the colors that appear in the labeling of e(b) are included in {ℓ∗, ℓ∗ 2, ℓ∗ 3, ℓ∗ 4}.
6 Boundary-minimizing partitioning of the sim-plex Let us turn now to a geometric variant of Proposition 1. We recall that Sperner’s Lemma has a geometric variant known as the Knaster-Kuratowski-Mazurkiewicz Lemma : Consider a covering of the simplex ∆k by closed sets A1, . . . , Ak such that each point x ∈∆k is contained in some set Ai such that xi > 0. Then Tk i=1 Ai ̸= ∅.
Here we consider a similar setup, but instead of the intersection of all sets, we are interested in the measure of the boundaries between pairs of adjacent sets.
To avoid technicalities, let us assume that the Ai’s are closed, disjoint except on the boundary, and each Ai is disjoint from the face {x ∈∆k : xi = 0}.
Definition 1. A Sperner-admissible partition of ∆k is a k-tuple of closed sets (A1, . . . , Ak) such that • Sk i=1 Ai = ∆k, • A1, . . . , Ak are disjoint except on their boundary, • xi > 0 for every x ∈Ai.
We call the union of pairwise boundaries S i̸=j(Ai ∩Aj) the separating set.
The question we ask here is, in analogy with Proposition 1, what is the Sperner-admissible partition with the separating set of minimum measure? A candidate partition is depicted in Figure 5, where Ai is the set of all points in ∆k for whom ei is the closest vertex. We call this the Voronoi partition.
We prove that for the regular simplex ∆k this is indeed the optimal partition (along with other, similar configurations). In the following, we denote by µk the usual Lebesgue measure on Rk, and by µℓ(ℓ< k) the ℓ-dimensional Minkowski content.
Definition 2. For A ⊂Rk, the ℓ-dimensional Minkowski content is (if the limit exists) µℓ(A) = lim ϵ→0+ µk(Aϵ) αk−ℓϵk−ℓ 9 Figure 5: The Voronoi partition of a simplex.
e1 e2 e3 e4 where Aϵ = {y ∈Rk : ∃x ∈A, ∥x−y∥≤ϵ} is the ϵ-neighborhood of A and αk−ℓ is the volume of a unit ball in Rk−ℓ. We also define µ+ ℓ(A) to be the upper limit and µ− ℓ(A) the lower limit of the expression above.
We remark that for ℓ-rectifiable sets (polyhedral faces, smooth surfaces, etc.) the notion of Minkowski content coincides with that of Hausdorffmeasure (under suitable normalization).
Theorem 1. For every Sperner-admissible partition (A1, . . . , Ak) of ∆k, µ− k−2 [ i̸=j (Ai ∩Aj) ≥k −1 √ 2 µk−1(∆k) and the Voronoi partition achieves this with equality.
First, let us analyze the Voronoi partition and more generally the following kind of partition.
Lemma 2. For any z in the interior of ∆k, the partition (Az 1, . . . , Az k) where Az i = {x ∈∆k : xi −zi = max 1≤j≤k(xj −zj)} satisfies µk−2 [ i̸=j (Az i ∩Az j) = k −1 √ 2 µk−1(∆k) = 1 (k −2)!
r k 2.
We call this kind of partition “Voronoi-type”.2 We note that that for z = ( 1 k, 1 k, . . . , 1 k) we obtain the Voronoi partition in Figure 5.
Other choices of 2We note that these partitions are also known as “power diagrams”.
10 z correspond to similar configurations where all the colors meet at the point z. Note that z is the “rainbow point” guaranteed by the Knaster-Kuratowski-Mazurkiewicz Lemma.
Proof. First let us compute some basic quantities that we will need. The sides of our simplex ∆k have length √ 2. Denote by hk the height of ∆k, that is the distance of any vertex from the opposite facet. We have hk = (1, 0, . . . , 0) −(0, 1 k −1, . . . , 1 k −1) = s 1 + (k −1) · 1 (k −1)2 = r k k −1.
The volume of the simplex can be computed inductively as follows; we have µ1(∆2) = √ 2, and µk(∆k+1) = 1 khk+1 · µk−1(∆k). This implies µk−1(∆k) = √ k (k −1)!.
Now let us compute the measure of the separating set for the partition (Az 1, . . . , Az k) defined above, by induction. The separating set can be described explicitly as [ i̸=j (Az i ∩Az j) = {x ∈∆k : ∃i ̸= j, xi −zi = xj −zj = max 1≤ℓ≤k xℓ−zℓ}.
For k = 2, Az 1 ∩Az 2 is just a single point, and µ0(Az 1 ∩Az 2) = 1. For k ≥3, denote by S the separating set for (Az 1, . . . , Az k) and define Si = S ∩conv({ej : j ̸= i}), the separating set restricted to the facet opposite vertex ei. Since Si is a Voronoi-type separating set for ∆k−1, by induction we assume that µk−3(Si) = 1 (k−3)!
q k−1 2 . The separating set S can be written as S = Sk i=1 conv(Si ∪{z}), see Figure 5. Denote by h′ i the distance of z from the facet containing Si. By the pyramid formula in dimension k −2, µk−2(conv(Si ∪{z})) = 1 k −2h′ iµk−3(Si) = h′ i (k −2)!
r k −1 2 .
By a simple calculation, Pk i=1 h′ i = hk = q k k−1. The sets conv(Si ∪{z}) are disjoint except for lower-dimensional intersections. Hence, µk−2(S) = k X i=1 µk−2(conv(Si ∪{z})) = k X i=1 h′ i (k −2)!
r k −1 2 = 1 (k −2)!
r k 2.
Thus the proof of Theorem 1 will be complete if we prove the following bound.
11 Lemma 3. For every Sperner-admissible partition (A1, . . . , Ak) of ∆k, µ− k−2 [ i̸=j (Ai ∩Aj) ≥k −1 √ 2 µk−1(∆k) = 1 (k −2)!
r k 2.
Proof. We pursue an approach similar to the proof of Proposition 1, with some additional technicalities. The high-level approach is to shrink the sets Ai some-what, by excluding a small neighborhood of the separating set. This creates a buffer zone between the shrunk sets A′ i (yellow in Figure 6) whose measure cor-responds to the measure of the separating set. Since we have this extra space, we are able to push the sets A′ i closer together and obtain sets A′′ i that fit inside a slightly smaller simplex. The difference between the volume of this simplex and the original one gives a bound on the measure of the separating set.
First, let ϵ0 = infi∈[k],x∈Ai xi. Recall that xi > 0 for each x ∈Ai, and moreover each Ai is closed. Hence ϵ0 > 0.
Define S = S i̸=j(Ai ∩Aj), the separating set whose measure we are trying to lower-bound. Fix ϵ ∈(0, 1 2ϵ0) (eventually we will let ϵ →0) and define Sϵ as the ϵ-neighborhood of S, Sϵ = {x ∈∆k : ∃y ∈S, ∥x −y∥≤ϵ} .
We define subsets A′ i ⊂Ai as follows: A′ i = Ai \ Sϵ.
Thus we have Sk i=1 A′ i = ∆k \ Sϵ. Also, the sets A′ i are clearly disjoint (see Figure 6).
Figure 6: The construction of A′ i and A′′ i .
e1 e2 e3 A′ 1 A′ 2 A′ 3 Sϵ e1 e2 e3 A′′ 1 A′′ 2 A′′ 3 Next, we set ϵ′ = ϵ √ 2 and define A′′ i = A′ i −ϵ′ei = {x −ϵ′ei : x ∈A′ i}.
Thus A′′ i is a shifted copy of A′ i, where we push A′ i slightly away from vertex ei.
The sets A′′ i live in the hyperplane Pk i=1 xi = 1 −ϵ′ rather than Pk i=1 xi = 1.
12 We claim that the sets A′′ i are still disjoint: Suppose that A′′ i ∩A′′ j = (A′ i−ϵ′ei)∩ (A′ j −ϵ′ej) ̸= ∅. This would mean that there are points x ∈A′ i, y ∈A′ j such that x−ϵ′ei = y−ϵ′ej. In other words, ∥x−y∥= ϵ′∥ei −ej∥= ϵ′√ 2 = 2ϵ. Take the midpoint 1 2(x + y): this point is in the simplex ∆k (by convexity), and hence it is in some set Aℓ, where either ℓ̸= i or ℓ̸= j (possibly both). Assume without loss of generality that ℓ̸= i. Then by the closedness of Ai and Aℓ, between x and 1 2(x + y) there exists a point x′ ∈Ai ∩Aℓ. We get a contradiction, because ∥x′ −x∥≤ϵ and so x would not be included in A′ i.
We also observe that A′′ i ⊆(1−ϵ′)·∆k = {x ≥0 : Pk i=1 xi = 1−ϵ′}. This is because for every x ∈A′′ i , we have y ∈A′ i such that x = y−ϵ′ei. By assumption, yi ≥ϵ0 > ϵ′, and y ∈∆k. Therefore xi = yi −ϵ′ > 0 and Pk i=1 xi = 1 −ϵ′.
We conclude that A′′ 1, . . . , A′′ k are disjoint subsets of (1 −ϵ′) · ∆k, obtained by an isometry from A′ 1, . . . , A′ k and therefore k X i=1 µk−1(A′ i) = k X i=1 µk−1(A′′ i ) ≤(1 −ϵ′)k−1µk−1(∆k).
Recall that A′ 1, . . . , A′ k are also disjoint and Sk i=1 A′ i = ∆k \ Sϵ. Therefore, µk−1(Sϵ) = µk−1(∆k) − k X i=1 µk−1(A′ i) ≥ 1 −(1 −ϵ′)k−1 µk−1(∆k).
By the definition of Minkowski content, we have µ− k−2 (S) = lim inf ϵ→0+ µk−1(Sϵ) 2ϵ ≥lim inf ϵ→0+ 1 −(1 −ϵ′)k−1 2ϵ µk−1(∆k) = lim ϵ→0+ 1 −(1 −ϵ √ 2)k−1 2ϵ µk−1(∆k) = k −1 √ 2 µk−1(∆k).
Alternative proof of optimality.
Here we give an alternative proof that the Voronoi partition has a separating set of minimum Minkowski content, avoiding an explicit computation of its volume.
Proof of Lemma 2. Let us consider the Voronoi partition (A∗ 1, . . . , A∗ k) (the proof for a general z is similar). We argue that the proof of Lemma 3 is tight for this partition. As in the proof of Lemma 3, we define S∗= S i̸=j(A∗ i ∩A∗ j), S∗ ϵ = {x ∈∆k : ∃y ∈S, ∥x −y∥≤ϵ}, A′ i = A∗ i \ S∗ ϵ and A′′ i = A′ i −ϵ′ei, ϵ′ = ϵ √ 2. In the case of the Voronoi partition, these sets are explicitly de-scribed as follows: • A∗ i = {x ∈∆k : xi = maxℓ∈[k] xℓ}, • S∗= {x ∈∆k : ∃i ̸= j, xi = xj = maxℓ∈[k] xℓ}, 13 • A′ i = {x ∈∆k : xi > ϵ′ + maxℓ̸=i xℓ}, • A′′ i = {x −ϵ′ei : x ∈∆k, xi > ϵ′ + maxℓ̸=i xℓ}.
The description of A′ i is valid because for x ∈A∗ i , it is possible to find a point in S∗within distance ϵ of x if and only if the maximum coordinate xi is within ϵ′ = ϵ √ 2 of the second largest coordinate — then we can replace the two largest coordinates by their average and obtain a point in S∗. The description of A′′ i follows by definition.
Consider now the scaled-down simplex (1−ϵ′)·∆k. By the proof of Lemma 3, the sets A′′ i are disjoint subsets of (1 −ϵ′) · ∆k. We show that in this case, we actually have Pk i=1 µk−1(A′′ i ) = µk−1((1 −ϵ′)∆k).
This is because for any point x′ ∈(1 −ϵ′) · ∆k, if the maximum coordinate x′ i of x′ is unique then x = x′ + ϵ′ei is a point in ∆k such that xi > ϵ′ + maxℓ̸=i xℓ.
Therefore, x ∈A′ i which implies that x′ ∈A′′ i .
The points x′ ∈(1 −ϵ′) · ∆k whose maximum coordinate is not unique form a set of (k −1)-dimensional measure zero. Therefore, (1−ϵ′)∆k is covered by Sk i=1 A′′ i up to a set of measure zero, and Pk i=1 µk−1(A′ i) = Pk i=1 µk−1(A′′ i ) = µk−1((1−ϵ′)∆k) = (1−ϵ √ 2)k−1µk−1(∆k).
We also have S∗ ϵ = ∆k \ Sk i=1 A′ i. This shows that all the inequalities in the proof of Lemma 3 are tight and the Minkowski content of the separating set S∗ is exactly µk−2(S∗) = lim ϵ→0+ µk−1(S∗ ϵ ) 2ϵ = lim ϵ→0+ 1 −(1 −ϵ √ 2)k−1 2ϵ µk−1(∆k) = k −1 √ 2 µk−1(∆k).
7 Discussion and open questions Sperner’s Lemma extends to general polytopes in the following sense : For any coloring of a triangulation of a d-dimensional polytope with n vertices by n colors, such that each point on a face F = conv({vi : i ∈A}) must be colored with a color in A, there are at least n−d full-dimensional simplices with d + 1 distinct colors.
It is natural ask whether our results also extend to general polytopes.
Possible extensions to polytopes.
Consider the example of P being a square (Figure 7). The Voronoi partition (A1, A2, A3, A4) is not optimal with respect to the total length of the separating set.
The separating set of the Voronoi partition has total length 2, whereas total length arbitrarily close to √ 2 is achieved by the partition (B1, B2, B3, B4).
In general, we do not know what the partition minimizing µ(S i̸=j(Ai ∩ Aj)) looks like, even in the case of a non-regular simplex.
We believe that the separating set should still be polyhedral (piecewise linear) for an optimal Sperner-admissible partition of any polytope.
14 Figure 7: Two partitions of a square.
A1 A2 A3 A4 B1 B2 B3 B4 Figure 8: An optimal partition between two pairs of faces of the tetrahedron.
e1 e2 e3 e4 We remark that depending on the coloring conditions on the surface of the polyhedron, the optimal separating set may be non-linear: For a tetra-hedron, the optimal partition that separates the pair of faces conv(e1, e2, e3) ∪ conv(e2, e3, e4) from conv(e1, e2, e4) ∪conv(e1, e3, e4), is the minimal surface whose boundary is the non-planar 4-gon e1-e2-e4-e3. This is a saddle-shaped quadratic surface (see Figure 8).
Other open questions.
We have proved several results about colorings of the simplex. Our first result (Proposition 1) can be viewed as being at the opposite end of the spectrum from Sperner’s Lemma: Instead of the existence of a rain-bow cell, we proved a lower bound on the number of non-monochromatic cells.
Due to the motivating application of , we considered a special hypergraph embedded in the simplex rather than a full subdivision. A natural question is whether an analogous statement holds for simplicial subdivisions.
More generally, we might “interpolate” between Sperner’s Lemma and our result, and ask: How many cells must contain at least j colors? It is clear that these questions depend on the structure of the subdivision, and some assumption of regularity would be needed to obtain a general result. Similarly, we may 15 ask, for Sperner-admissible geometric partitions of the simplex, what is the minimum possible volume of the set where at least j colors meet? Furthermore, as we discussed above, are there generalizations of these statements to other polytopes?
Another question is, what is the Sperner-admissible labeling of the Simplex-Lattice Hypergraph Hk,q (defined in Section 2) minimizing the maximum num-ber of colors on a hyperedge?
We have proved that 4 colors suffice but it is possible that 2 colors are enough (see Proposition 2). Is there a Sperner-admissible labeling of the hypergraph Hk,q, for sufficiently large q, such that each hyperedge uses at most 2 colors?
Finally, we remark that Proposition 2 does not have a continuous counterpart for geometric partitions: As we discussed earlier, for any Sperner-admissible partition of a simplex there is a point where all the parts meet, by the Knaster-Kuratowski-Mazurkiewicz Lemma .
Acknowledgement.
The second author is indebted in many ways to Jirka Matouˇ sek, who introduced him to Sperner’s Lemma in an undergradute course at Charles University a long time ago.
References Chandra Chekuri and Alina Ene. Submodular cost allocation problem and applications. In Proc. of ICALP, 354–366, 2011.
Xi Chen, Xiaotie Deng, and Shang-Hua Teng. Settling the complexity of computing two-player Nash equilibria. J. ACM 56:3, 2009.
Constantinos Daskalakis, Paul W. Goldberg, and Christos H. Papadim-itriou. The complexity of computing a Nash equilibrium. SIAM J. Com-puting 39:1, 195–259, 2009.
Alina Ene and Jan Vondr´ ak. Hardness of submodular cost allocation: Lat-tice matching and a simplex coloring conjecture. In Proc. of APPROX, 144–159, 2014.
Alina Ene, Jan Vondr´ ak, and Yi Wu. Local distribution and the symmetry gap: Approximability of multiway partitioning problems. In Proc. of ACM-SIAM SODA, 306–325, 2013.
Jon Kleinberg and ´ Eva Tardos. Approximation algorithms for classification problems with pairwise relationships: Metric labeling and Markov random fields. Journal of the ACM, 49:5, 616–639, 2002.
B. Knaster, C. Kuratowski, and S. Mazurkiewicz.
Ein Beweis des Fix-punktsatzes f¨ ur n-dimensionale Simplexe. Fundamenta Mathematicae, 14:1, 132–137, 1929.
16 Jesus de Loera, Elisha Peterson, and Francis Edward Su.
A polytopal generalization of Sperner’s Lemma. J. Combinatorial Theory A 100, 1–26, 2002.
Rajsekar Manokaran, Joseph Naor, Prasad Raghavendra, and Roy Schwartz.
SDP gaps and UGC hardness for multiway cut, 0-extension, and metric labeling. Proc. of ACM STOC, 11–20, 2008.
Maryam Mirzakhani and Jan Vondr´ ak.
Sperner’s colorings, hypergraph labeling problems and fair division. Proc. of ACM-SIAM SODA, 873–886, 2015.
John Nash. Noncooperative games. Annals of Mathematics 54, 289–295, 1951.
Emanuel Sperner. Neuer Beweis f¨ ur die Invarianz der Dimensionszahl und des Gebietes. Math. Sem. Univ. Hamburg, 6:265–272, 1928.
Francis Edward Su.
Rental harmony: Sperner’s lemma in fair division.
Amer. Math. Monthly 106, 930–942, 1999.
17 |
189935 | https://langsdorfchem.weebly.com/uploads/1/6/8/8/16880686/ws_-_gas_laws_explained_and_ws_8.pdf | Name: ____ Class: ___ Date: ____ Written by James Dauray Page 1 Gas Laws: Boyle, Charles, and Gay-Lussac Introduction Gases were one of the first substances studied by chemists in the hope of understanding the nature of matter more clearly. We now have a series of laws and equations that help us predict how gases will behave under certain conditions. An Irish chemist named Robert Boyle discovered that as the amount of pressure on a fixed amount of gas is increased, its volume decreases. This inversely proportional relationship is known as Boyle’s Law. The Boyle’s Law equation can be used to predict a change in pressure or volume. Boyle’s Law P1V1 = P2V2 P = Pressure | V = Volume Jacques Charles was a French physicist who observed that as temperature increases, the volume of a gas also increases. Conversely, if the temperature of a gas decreases, the volume of the gas decreases. This direct relationship is known as Charles’ Law. The Charles’ Law equation can be used to predict a change in volume or temperature. Charles’ Law V1/ T1 = V2/ T2 T = Temperature | V = Volume Joseph Louis Gay-Lussac was also a French physicist and chemist. He observed that as the temperature on a gas increases, its pressure also increases. Likewise, of the temperature decreases, the pressure decreases. This direct relationship is known as Gay-Lussac’s Law. Gay-Lussac’s Law P1/ T1 = P2/ T2 P = Pressure | T = Temperature Temperature must be given in the Kelvin scale, so if you need to convert from Celsius, just use this simple conversion: Celsius to Kelvin °K = °C + 273 Practice For each of these problems, identify which law (Charles, Boyle’s, or Gay-Lussac’s) applies, then solve. 1. 950mL of a gas at a pressure of 1.5atm is compressed to 473 mL. What is the new pressure of the gas? a. Identify the law to be used: b. Solve the problem: 2. In a commercial airliner, an empty water bottle contains about 500mL of air. The air pressure at that altitude is 0.74atm. What will the volume of the water bottle be after landing and reaching an air pressure of 1.0atm? a. Identify the law to be used: b. Solve the problem: 3. A 300mL sealed spray can containing insect spray is kept outside at a temperature of 15°C . It is accidentally dropped into a campfire, causing the temperature inside the can to quickly rise to 99°C. What effect does this have on the volume of the gas inside the can? a. Identify the law to be used: b. Solve the problem: 4. A submarine containing 15,000L of air at an air pressure of 1.2atm breaks apart deep within the ocean, where pressures reach 250atm. What volume of air would escape out of the submarine at that depth? a. Identify the law to be used: b. Solve the problem: 5. A diver has about 0.05L of gas dissolved in his blood while swimming 200 feet underwater (at a pressure of 7.05atm). He returns too quickly to the surface, quickly reaching a depth of 20 feet (at a pressure of 1.60atm). What is the new volume within his blood? a. Identify the law to be used: b. Solve the problem: 6. A balloon containing a volume of 0.50L at 22°C is placed inside a refrigerator. The volume of the balloon after cooling is 0.09L. What is the temperature of the refrigerator? a. Identify the law to be used: b. Solve the problem: 7. A sample of nitrogen gas kept in a tank at 3.0atm of pressure and 20.0°C is heated in an oven until it reaches a temperature of 60°C. What pressure must the tank be able to withstand in order to safely hold the nitrogen gas? a. Identify the law to be used: b. Solve the problem: 8. A tank of water vapor at 2.0atm is compressed until it reaches 10.5atm. If the original temperature of the water vapor was 115°C, what is the new temperature? a. Identify the law to be used: b. Solve the problem: 9. A sample of argon is collected in a 5.00 x 102 cm3 bottle at a temperature of 12.0 oC. Assuming the pressure remains the same, what volume would the gas occupy at 2.0 oC? a. Identify the law to be used: b. Solve the problem: 10. A sample of gas under a pressure of 622 mm Hg has a volume of 233 cm3. The pressure is increased to 988 mm Hg, while the temperature remains constant. What volume will the gas occupy at this pressure? a. Identify the law to be used: b. Solve the problem: 11. A sample of CO2 has a volume of 2.7 cm3 at 2.0atm. At what pressure would this sample of gas have a volume of 4.0 cm3? a. Identify the law to be used: b. Solve the problem: 12. A sample of hydrogen is collected over water at 12 oC. The pressure of the hydrogen and water vapor mixture is 0.2atm. The hydrogen is then cooled, at the same volume, to 0 oC. What is the new pressure? a. Identify the law to be used: b. Solve the problem: 13. A sample of neon has a volume of 1.83 L at 23.5 oC. At what temperature would the gas occupy 5.00 L? Assume pressure is constant. a. Identify the law to be used: b. Solve the problem: 14. A sample of argon has a volume of 205 cm3 when its pressure is 0.9atm. What would be the volume of the argon at 1.0atm be? a. Identify the law to be used: b. Solve the problem: 15. A sample of neon is collected at 2.7 atm and 12.0 oC. What would the temperature of the gas be if it were pressurized to 20.0atm? a. Identify the law to be used: b. Solve the problem: |
189936 | https://www.youtube.com/watch?v=sLr-hAR9cMU | 2024 Maths Extension 1 HSC Q11a Find 2a-b & dot product a·b given vectors a=3i+2j & b=-i+4j
The Maths Studio
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Posted: 1 Apr 2025
Sample solution: © The Maths Studio (themathsstudio.net)
Source: © NSW Education Standards Authority
Disclaimer: This sample solution is intended as a guide only and does not necessarily represent the best way to answer the question, nor is there any guarantee of accuracy.
~
Transcript:
question 11 part A consider the vectors Vector a equal 3 Vector I + 2 Vector J and Vector b = i + 4 Vector J part one find 2 Vector a minus Vector B Vector a is defined as 3 Vector I + 2 Vector J and Vector B is defined as netive Vector I + 4 Vector J so 2 Vector a minus Vector B can be expressed as twice 3 I + 2 Jus I + 4 J so that's 2 3 I + 2 J in Brackets minus I + 4 J in Brackets now we can treat this as regular algebra we expand the brackets and collect like terms so expanding the brackets we get 6 I + 4 J + i - 4 J now the + 4 J - 4 J add to zero and 6 I + I is equal to 7 I so 2 a - b is equal to 7 I part two find the dot product of vectors A and B the the dotproduct of vectors A and B is found by multiplying separately the coefficients of vectors I and J and then adding those two products together so in this case the dotproduct of A and B is 3 -1 plus 2 4 so 3 And1 are the coefficients of the I vectors and 2 and four are the coefficients of the J vectors so 3 -1 + 2 4 is equal to 5 e for |
189937 | https://glidewelldental.com/education/chairside-magazine/volume-10-issue-4/the-mylohyoid-ridges-troublemakers-often-undiagnosed | 800-854-7256 USA
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Volume 10, Issue 4
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Dr. Abai’s Clinical Tips – DentalVibe®
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One-on-One with Dr. Siamak Abai: Interview with Dr. Mona Goodarzi
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Implant Overdentures: A Superior, Cost-Effective Alternative to Traditional Dentures
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The Mylohyoid Ridges: Troublemakers Often Undiagnosed
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The Mylohyoid Ridges: Troublemakers Often Undiagnosed
December 18, 2015
Author
Robert E. Garfield, DDS, FAGD, FAO
We dentists have been taught the importance and the benefits of having denture bases for removable prostheses that are fully extended to the functional muscular periphery. This principle is especially important with mandibular denture bases, whether they are for full dentures, overdentures or distal-extension removable partial dentures.
Many dentists have mastered the art of obtaining fully extended “muscle trimmed” impressions utilizing custom impression trays, border molding and “wash” materials. Some clinicians even make a temporary denture of limited accuracy, border-mold it, and do a wash impression, all with the use of tissue conditioner mixed to various states of viscosity. The patient then wears this out of the office, instructed on precautions, before returning in a few days for peripheral additions and possibly another wash impression.
In this manner, extremely accurate functional peripheral borders and surface reproduction can be obtained within a few days to weeks. The clinician then pours the impression/temporary denture combination in stone, submerging it under cold water after the initial set of the stone in order to dissipate the heat of the chemical reaction upon final setting, which could distort the tissue conditioner. Upon separation of the stone cast, the temporary denture is returned to the patient with its tissue conditioner intact, and the clinician has a beautifully accurate stone cast to proceed to the next step of making a final denture for this patient.
THE PROBLEM
As is often the case, after a day or so of wearing this fully extended mandibular denture, with its deeply extended lingual flanges that so effectively prevent lateral displacement of the denture during function, the patient complains of extreme pain and ulceration in the region of the lingual vestibules. After several spot-relieving appointments, none of which seem to help remedy these painful areas, many clinicians acquiesce to a patient’s demands and trim away up to 50% of those wonderful stabilizing lingual flanges that took so much skill and perseverance to make. So what happened? The patient blames the denture for “cutting into my jaw.” However, exactly the opposite is true. The very sharp mylohyoid ridge cut through the oral mucosa from the lingual, pressing against the tissue side of the lingual denture flange, which acted like a “cutting board.” The soft tissue was the “bread”; the mylohyoid ridge was the “knife”; and the lingual flange was the “bread board.” The denture base was not the culprit, but now the stability of this denture has been greatly compromised by its removal, which can be a serious problem to clasped teeth, implants, attachments and retention screws.
The denture base was not the culprit, but now the stability of this denture has been greatly compromised by its removal.
Basically, the lingual flanges are the only stabilizing feature against lateral displacement or “fishtailing” of the denture base during function. The lingual surfaces of the mandible are vertical, while the facial surfaces are sloped. It’s just simple mechanics. This constant lateral movement can loosen, fatigue, or break the retaining screws that hold implant-supported attachments or bar-clip retainers in place. Moreover, it can cause the loss of integration of anterior implants and even cause implant fracture in some cases. It can also prematurely wear out O-ring denture retainers and other denture retention systems. It’s just like having a long cantilever arm connected to the anterior implants.
THE SOLUTION
The solution to this frustrating problem is to follow a strict denture examination protocol with every patient who requires either a reline of an existing full or partial lower denture, or an entirely new denture. Remember, the cause of the pain is not the denture, but the sharp mylohyoid ridge, and that ridge belongs to the patient. The patient’s mylohyoid ridge remodeled over the years and became knife-sharp as a result of bone resorption around it (Fig. 1). It is as though a rainstorm and flood had washed all of the smaller gravel particles from a dirt road, leaving only the larger, jagged rocks exposed. This is the patient’s problem, and you, the clinician, can solve it and still have fully extended lingual flanges. But first you must recognize the problem, and demonstrate and explain it to the patient before starting any denture treatment.
The cause of the pain is not the denture, but the sharp mylohyoid ridge, and that ridge belongs to the patient.
Figure 1: Cone-beam computed tomography can be utilized to visualize the sharp mylohyoid ridge of a patient. The cross-sectional slices individually reveal the clinical situation and can be beneficial in dentist-to-patient communication prior to treatment.
PROTOCOL STEPS
Ask the patient to hold his or her breath (to keep from gagging), and then place your index finger deeply into the patient’s lingual vestibule and palpate the mylohyoid ridge. If the ridge is sharp (it usually is), the patient will “flinch” and experience acute pain.
Now take the patient’s index finger, guide it into his lingual vestibule, and repeat what you just did yourself. The patient will cause himself acute pain, but will now be convinced that the cause is from his own sharp area on his jawbone.
Tell the patient why his bone is so sharp (bone resorption and “gravel road” analogy), and explain why this will produce pain from any properly fitting denture base that you would make for him.
Suggest to the patient that you want those sharp bony ridges removed or smoothed away before you start the new denture, overdenture, RPD or reline. Explain that not doing so will surely result in a poorly fitting and unstable denture. If you have surgical skills you can do this procedure yourself, one side at a time, with a one-month interval so as not to cause bilateral pharyngeal edema that could restrict swallowing and frighten a patient. Otherwise, send the case to an oral-maxillofacial surgeon.
Performing this procedure requires good local mandibular and long buccal nerve block anesthesia. A 2–3 cm straight-line incision is made along the posterior ridge crest from the retromolar pad anteriorly to the #20 tooth position. A periosteal elevator is used to reflect a full thickness flap separating the soft tissue from the lingual surface of the mandible. Keep the elevator firmly against the bony surface at all times. This lingual soft tissue separates easily from the bone. Locate the mylohyoid ridge, and with a sharp bone file patiently remove the sharp surface. Gently irrigate and suction the area to remove bone debris. Never blow air from a syringe into the area, as this could cause a submandibular emphysema. Palpate the area to determine whether the bone removal is sufficient before suturing the flap. Psychologically prepare the patient for some unilateral pharyngeal edema and moderate osteitis pain for approximately 5–10 days.
Dr. Garfield is the executive director, advisor-trustee and past president of the Southern California Academy of General Dentistry (SCAGD). Contact him via email at drrobertgarfield@aol.com.
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The area of a circle is the amount of space enclosed within its circumference. It is calculated by multiplying the square of the radius of the circle by the constant value of pi (π). The formula for calculating the area of a circle is A = πr², where A represents the area and r represents the radius of the circle. The area of a circle is measured in square units, such as square meters (m²) or square inches (in²). The concept of the area of a circle is used in various real-life scenarios, such as calculating the surface area of circular objects like plates, wheels, or discs.
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189939 | https://en.wikipedia.org/wiki/Shamir%27s_secret_sharing | Jump to content
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1 High-level explanation
1.1 Application example
1.2 Use cases
2 Properties and weaknesses
3 History
4 Mathematical principle
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6 Example calculation
6.1 Preparation
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6.3 Computationally efficient approach
6.4 Problem of using integer arithmetic
6.5 Solution using finite field arithmetic
7 Python code
8 See also
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10 Further reading
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Shamir's secret sharing
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From Wikipedia, the free encyclopedia
Cryptographic algorithm created by Adi Shamir
Shamir's secret sharing (SSS) is an efficient secret sharing algorithm for distributing private information (the "secret") among a group. The secret cannot be revealed unless a minimum number of the group's members act together to pool their knowledge. To achieve this, the secret is mathematically divided into parts (the "shares") from which the secret can be reassembled only when a sufficient number of shares are combined. SSS has the property of information-theoretic security, meaning that even if an attacker steals some shares, it is impossible for the attacker to reconstruct the secret unless they have stolen a sufficient number of shares.
Shamir's secret sharing is used in some applications to share the access keys to a master secret.
High-level explanation
[edit]
SSS is used to secure a secret in a distributed form, most often to secure encryption keys. The secret is split into multiple shares, which individually do not give any information about the secret.
To reconstruct a secret secured by SSS, a number of shares is needed, called the threshold. No information about the secret can be gained from any number of shares below the threshold (a property called perfect secrecy). In this sense, SSS is a generalisation of the one-time pad (which can be viewed as SSS with a two-share threshold and two shares in total).
Application example
[edit]
A company needs to secure their vault. If a single person knows the code to the vault, the code might be lost or unavailable when the vault needs to be opened. If there are several people who know the code, they may not trust each other to always act honestly.
SSS can be used in this situation to generate shares of the vault's code which are distributed to authorized individuals in the company. The minimum threshold and number of shares given to each individual can be selected such that the vault is accessible only by (groups of) authorized individuals. If fewer shares than the threshold are presented, the vault cannot be opened.
By accident, coercion or as an act of opposition, some individuals might present incorrect information for their shares. If the total of correct shares fails to meet the minimum threshold, the vault remains locked.
Use cases
[edit]
Shamir's secret sharing can be used to
share a key for decrypting the root key of a password manager,
recover a user key for encrypted email access and
share the passphrase used to recreate a master secret, which is in turn used to access a cryptocurrency wallet.
Properties and weaknesses
[edit]
SSS has useful properties, but also weaknesses that means that it is unsuited to some uses.
Useful properties include:
Secure: The scheme has information-theoretic security.
Minimal: The size of each piece does not exceed the size of the original data.
Extensible: For any given threshold, shares can be dynamically added or deleted without affecting existing shares
Dynamic: Security can be enhanced without changing the secret, but by changing the polynomial occasionally (keeping the same free term) and constructing a new share for each of the participants.
Flexible: In organizations where hierarchy is important, each participant can be issued different numbers of shares according to their importance inside the organization. For instance, with a threshold of 3, the president could unlock the safe alone if given three shares, while three secretaries with one share each must combine their shares to unlock the safe.
Weaknesses include:
No verifiable secret sharing: During the share reassembly process, SSS does not provide a way to verify the correctness of each share being used. Verifiable secret sharing aims to verify that shareholders are honest and not submitting fake shares.
Single point of failure: The secret must exist in one place when it is split into shares, and again in one place when it is reassembled. These are attack points, and other schemes including multisignature eliminate at least one of these single points of failure.
History
[edit]
Adi Shamir, an Israeli scientist, first formulated the scheme in 1979.
Mathematical principle
[edit]
The scheme exploits the Lagrange interpolation theorem, specifically that points on the polynomial uniquely determines a polynomial of degree less than or equal to . For instance, 2 points are sufficient to define a line, 3 points are sufficient to define a parabola, 4 points to define a cubic curve and so forth.
Mathematical formulation
[edit]
Shamir's secret sharing is an ideal and perfect -threshold scheme based on polynomial interpolation over finite fields. In such a scheme, the aim is to divide a secret (for example, the combination to a safe) into pieces of data (known as shares) in such a way that:
Knowledge of any or more shares makes computable. That is, the entire secret can be reconstructed from any combination of shares.
Knowledge of any or fewer shares leaves completely undetermined, in the sense that the possible values for remain as likely with knowledge of up to shares as with knowledge of shares. The secret cannot be reconstructed with fewer than shares.
If , then all of the shares are needed to reconstruct the secret .
Assume that the secret can be represented as an element of a finite field (where is greater than the number of shares being generated). Randomly choose elements, , from and construct the polynomial . Compute any points out on the curve, for instance set to find points . Every participant is given a point (a non-zero input to the polynomial, and the corresponding output). Given any subset of of these pairs, can be obtained using interpolation, with one possible formula for doing so being , where the list of points on the polynomial is given as pairs of the form . Note that is equal to the first coefficient of polynomial .
Example calculation
[edit]
The following example illustrates the basic idea. Note, however, that calculations in the example are done using integer arithmetic rather than using finite field arithmetic to make the idea easier to understand. Therefore, the example below does not provide perfect secrecy and is not a proper example of Shamir's scheme. The next example will explain the problem.
Preparation
[edit]
Suppose that the secret to be shared is 1234 .
In this example, the secret will be split into 6 shares , where any subset of 3 shares is sufficient to reconstruct the secret. numbers are taken at random. Let them be 166 and 94.
: This yields coefficients where is the secret
The polynomial to produce secret shares (points) is therefore:
Six points from the polynomial are constructed as:
Each participant in the scheme receives a different point (a pair of and ). Because is used instead of the points start from and not . This is necessary because is the secret.
Reconstruction
[edit]
In order to reconstruct the secret, any 3 points are sufficient
Consider using the 3 points.
Computing the Lagrange basis polynomials:
Using the formula for polynomial interpolation, is:
Recalling that the secret is the free coefficient, which means that , and the secret has been recovered.
Computationally efficient approach
[edit]
Using polynomial interpolation to find a coefficient in a source polynomial using Lagrange polynomials is not efficient, since unused constants are calculated.
Considering this, an optimized formula to use Lagrange polynomials to find is defined as follows:
Problem of using integer arithmetic
[edit]
Although the simplified version of the method demonstrated above, which uses integer arithmetic rather than finite field arithmetic, works, there is a security problem: Eve gains information about with every that she finds.
Suppose that she finds the point . She still does not have points, so in theory she should not have gained any more information about . But she could combine the information from the point with the public information: . Doing so, Eve could perform the following algebra:
Fill the formula for with and the value of
Fill (1) with the values of 's and
which leads her to the information that S is even.
Solution using finite field arithmetic
[edit]
The above attack exploits constraints on the values that the polynomial may take by virtue of how it was constructed: the polynomial must have coefficients that are integers, and the polynomial must take an integer as value when evaluated at each of the coordinates used in the scheme. This reduces its possible values at unknown points, including the resultant secret, given fewer than shares.
This problem can be remedied by using finite field arithmetic. A finite field always has size , where is a prime and is a positive integer. The size of the field must satisfy , and that is greater than the number of possible values for the secret, though the latter condition may be circumvented by splitting the secret into smaller secret values, and applying the scheme to each of these. In our example below, we use a prime field (i.e. r = 1). The figure shows a polynomial curve over a finite field.
In practice this is only a small change. The order q of the field (i.e. the number of values that it has) must be chosen to be greater than the number of participants and the number of values that the secret may take. All calculations involving the polynomial must also be calculated over the field (mod p in our example, in which is taken to be a prime) instead of over the integers. Both the choice of the field and the mapping of the secret to a value in this field are considered to be publicly known.
For this example, choose , so the polynomial becomes which gives the points:
This time Eve doesn't gain any information when she finds a (until she has points).
Suppose again that Eve finds and , and the public information is: . Attempting the previous attack, Eve can:
Fill the -formula with and the value of and :
Fill (1) with the values of 's and
Fill (1) with the values of 's and
Subtract (3)-(2): and rewrite this as
There are possible values for . She knows that always decreases by 3, so if were divisible by she could conclude . However, is prime, so she can not conclude this. Thus, using a finite field avoids this possible attack.
Also, even though Eve can conclude that , it does not provide any additional information, since the "wrapping around" behavior of modular arithmetic prevents the leakage of "S is even", unlike the example with integer arithmetic above.
Python code
[edit]
For purposes of keeping the code clearer, a prime field is used here. In practice, for convenience a scheme constructed using a smaller binary field may be separately applied to small substrings of bits of the secret (e.g. GF(256) for byte-wise application), without loss of security. The strict condition that the size of the field must be larger than the number of shares must still be respected (e.g., if the number of shares could exceed 255, the field GF(256) might be replaced by say GF(65536)).
""" The following Python implementation of Shamir's secret sharing isreleased into the Public Domain under the terms of CC0 and OWFa: the bottom few lines for usage. Tested on Python 2 and 3. """ from __future__ import division from __future__ import print_function import random import functools # 12th Mersenne Prime _PRIME = 2 127 - 1 _RINT = functools. partial(random. SystemRandom(). randint, 0) def _eval_at(poly, x, prime): """Evaluates polynomial (coefficient tuple) at x, used to generate a shamir pool in make_random_shares below. """ accum = 0 for coeff in reversed(poly): accum= x accum += coeff accum%= prime return accum def make_random_shares(secret, minimum, shares, prime = _PRIME): """ Generates a random shamir pool for a given secret, returns share points. """ if minimum> shares: raise ValueError("Pool secret would be irrecoverable.") poly =[secret] +[_RINT(prime - 1) for i in range(minimum - 1)] points =[(i, _eval_at(poly, i, prime)) for i in range(1, shares + 1)] return points def _extended_gcd(a, b): """ Division in integers modulus p means finding the inverse of the denominator modulo p and then multiplying the numerator by this inverse (Note: inverse of A is B such that AB % p == 1). This can be computed via the extended Euclidean algorithm """ x = 0 last_x = 1 y = 1 last_y = 0 while b!= 0: quot = a// b a, b = b, a % b x, last_x = last_x - quot x, x y, last_y = last_y - quot y, y return last_x, last_y def _divmod(num, den, p): """Compute num / den modulo prime p To explain this, the result will be such that: den _divmod(num, den, p) % p == num """ inv, _ = _extended_gcd(den, p) return num inv def _lagrange_interpolate(x, x_s, y_s, p): """ Find the y-value for the given x, given n (x, y) points; k points will define a polynomial of up to kth order. """ k = len(x_s) assert k == len(set(x_s)), "points must be distinct" def PI(vals):# upper-case PI -- product of inputs accum = 1 for v in vals: accum= v return accum nums =[] # avoid inexact division dens =[] for i in range(k): others = list(x_s) cur = others. pop(i) nums. append(PI(x - o for o in others)) dens. append(PI(cur - o for o in others)) den = PI(dens) num = sum([_divmod(nums[i] den y_s[i] % p, dens[i], p) for i in range(k)]) return(_divmod(num, den, p) + p) % p def recover_secret(shares, prime = _PRIME): """ Recover the secret from share points (points (x,y) on the polynomial). """ if len(shares)< 3: raise ValueError("need at least three shares") x_s, y_s = zip( shares) return _lagrange_interpolate(0, x_s, y_s, prime) def main(): """Main function""" secret = 1234 shares = make_random_shares(secret, minimum = 3, shares = 6) print('Secret: ', secret) print('Shares:') if shares: for share in shares: print(' ', share) print('Secret recovered from minimum subset of shares: ', recover_secret(shares[: 3])) print('Secret recovered from a different minimum subset of shares: ', recover_secret(shares[- 3:])) if __name__ == '__main__': main()
See also
[edit]
Secret sharing
Secure multi-party computation
Lagrange polynomial
Homomorphic secret sharing – a simplistic decentralized voting protocol
Two-person rule
Partial Password
References
[edit]
^ Krenn, Stephan; Loruenser, Thomas (2023). An Introduction to Secret Sharing: A Systematic Overview and Guide for Protocol Selection. doi:10.1007/978-3-031-28161-7. ISBN 978-3-031-28160-0.
^ "Seal/Unseal". Vault by HashiCorp. Retrieved 2022-10-02.
^ "PreVeil Review". PCMag. Retrieved 2022-10-02.
^ Rusnak, Pavol; Kozlik, Andrew; Vejpustek, Ondrej; Susanka, Tomas; Palatinus, Marek; Hoenicke, Jochen (2017-12-18). "SLIP-0039 : Shamir's Secret-Sharing for Mnemonic Codes". GitHub. SatoshiLabs. Retrieved 2022-10-03. This SLIP describes a standard and interoperable implementation of Shamirs secret-sharing (SSS) and a specification for its use in backing up Hierarchical Deterministic Wallets described in BIP-0032.
^ Lopp, Jameson (2020-10-01). "Shamir's Secret Sharing shortcomings". Retrieved 2022-10-03. Variations of Shamir's Secret Sharing (SSS) have been implemented several times in the cryptocurrency space, only for developers to later realize that the additional complexity ended up reducing the security of the system.
^ Shamir, Adi (1979), "How to share a secret", Communications of the ACM, 22 (11): 612–613, doi:10.1145/359168.359176, S2CID 16321225
^ Knuth, D. E. (1997), The Art of Computer Programming, vol. II: Seminumerical Algorithms (3rd ed.), Addison-Wesley, p. 505.
Further reading
[edit]
Dawson, E.; Donovan, D. (1994), "The breadth of Shamir's secret-sharing scheme", Computers & Security, 13: 69–78, doi:10.1016/0167-4048(94)90097-3.
Benzekki, K. (2017). "A Verifiable Secret Sharing Approach for Secure MultiCloud Storage". Ubiquitous Networking. Lecture Notes in Computer Science. Vol. 10542. Casablanca: Springer. pp. 225–234. doi:10.1007/978-3-319-68179-5_20. ISBN 978-3-319-68178-8..
External links
[edit]
Shamir's Secret Sharing in the Crypto++ library
Shamir's Secret Sharing Scheme (ssss) – a GNU GPL implementation
sharedsecret – implementation in Go
s4 - online shamir's secret sharing tool utilizing HashiCorp's shamir secret sharing algorithm
Shamir39 - webversion on iancoleman.io
kn Secrets - webversion on a dedicated website, aiming to make Shamir's secret sharing as accessible as possible
xecrets-cli - cross-platform command line tool (Linux, macOS, Windows)
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189940 | https://mathoverflow.net/questions/355600/perfect-squares-between-certain-divisors-of-a-number | Skip to main content
Perfect squares between certain divisors of a number
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Let n be a positive integer. We will call a divisor d(<n−−√) of n special if there exists no perfect squares between d and nd. Prove that n can have at-most one special divisor.
My progress: I boiled down the problem to the following:
Suppose k2≤a,b,c,d≤(k+1)2, then ab=cd⟹{a,b}={c,d}. But I can't seem to prove this.
Arriving here isn't difficult so I am omitting any further details(one more reason being I am not sure if I am on the correct path).
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edited Mar 24, 2020 at 12:03
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asked Mar 24, 2020 at 11:15
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This MO website is for questions of math research. It's not clear to me that this question qualifies. How did you come across it? (Also, you are overloading the symbol n.)
– Gerry Myerson
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Mar 24, 2020 at 11:38
@Gerry Myerson My math professor says that this problem came up while he was trying to solve some open problem. I didn't bother asking him the details though. I would be very grateful if you can help me solve this problem. Or even any good idea would do.
– user154024
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Simulposted to m.se, math.stackexchange.com/questions/3592817/… without notice to either site. That's an abuse.
– Gerry Myerson
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Mar 24, 2020 at 12:12
Doesn't d being special imply n/d−−−√−d−−√<1?
– Sylvain JULIEN
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Mar 24, 2020 at 12:13
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This follows from a result I have asked about a few years ago, namely:
For any n∈N there is at most one divisor of n in the interval [n−−√,n−−√+n−−√4].
I claim that if dn−−√+n−−√4 and
d<nn−−√+n−−√4=n−n−−√n−−√+n−−√4+n−−√n−−√+n−−√4<n−−√−n−−√4+1.
Letting x=n−−√4, it remains to show that for any x>2 there is a perfect square between x2−x+1 and x2+x. By monotonicity, it is enough to show that if x2−x+1=k2 for some k∈N, then x2+x≥(k+1)2. The first equality resolves to x=12(4k2−3−−−−−−√+1), and we have
x2+x=x2−x+1+2x−1=k2+4k2−3−−−−−−√<k2+2k<(k+1)2.
I'm sure the last part of the argument can be carried out more cleanly, but it checks out.
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answered Mar 24, 2020 at 13:56
WojowuWojowu
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Thanks for answering this question, so that I can stop racking my brain on it!
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Here is an elementary approach. We want to rule out finding distinct a,b,c,d in short interval (specifically [n2+1,n2+2n]) with ad=bc. We may assume that a is the smallest and that b<c. Then a<b<c<d. I will show that a+2a−−√+1≤d so that if n2≤a then (n+1)2≤d.
Claim: There are integers u<v and x<y with
a,b,c,d=ux,uy,vx,vy.
Proof: Let u=gcd(a,b) so a=ux and b=uy with gcd(x,y)=1. Then uxd=uyc so xd=yc and thus (since x and y are co-prime) there is v with c=xv and d=yv.
ASIDE We might as well use v=u+1 and y=x+1 since
a,b′,c′,d′=ux,u(x+1),(u+1)x,(u+1)(x+1)
give ad′=b′c′ with a<b′,c′,d′≤d. I'll comment a bit more about this at the end.
So we want to show
if ad=bc with n2<a<b<c(n+1)2.
From the claim above, n2n2, we know u+x≥2a−−√>2n. Thus ux+u+x+1>n2+2n+1 as desired.
Consider this problem: Given a, find a<b<c<d with d minimal such that ad=bc. The work above shows that the solution is to have
a,b,c,d=ux,u(x+1),(u+1)x,(u+1)(x+1)
with |u−x| minimal and that d>a+2a−−√+1. If we allow b=c then
n2⋅(n+1)2=(n2+n)⋅(n2+n).
If we want b<c then (n2−n)⋅(n2+n)=(n2−1)⋅(n2). With d−a≈2a−−√.
If ad=bc then abcd is a perfect square. However this property is weaker and there are solutions such as
a,b,c,d=2⋅1202,3⋅982,30⋅312,5⋅762=
28800,28812,28830,28880
with d<a+12a−−√ and all four factors between 28561=1692 and 28900=1702.
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edited Mar 26, 2020 at 4:12
answered Mar 26, 2020 at 0:02
Aaron MeyerowitzAaron Meyerowitz
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Only a partial answer for now. Let d−1 the largest divisor of n below n−−√, d−(k+1) the largest divisor of n below d−k and dk:=n/d−k. Let rk:=dk−n−−√ and lk:=n−−√−d−k.
Define the k-th "square root divisor span" of n as sk(n):=dk−−√−d−k−−−√. The sequence (sk(n))k is strictly increasing, and its general term equals n−−√+rk−−−−−−−√−n−−√−lk−−−−−−−√ which is greater or equal than n−−√+k−−−−−−√−n−−√−k−−−−−−√.
I think the condition in my comment, namely sk(n)<1, is fulfilled only when max(lk,rk)<n1/4, so for m(n)=O(1) values of k, with limn→∞m(n)=0.
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edited Mar 24, 2020 at 13:41
answered Mar 24, 2020 at 12:46
Sylvain JULIENSylvain JULIEN
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Solve the recurrence relation a n=−a n−1+n 2,n≥1 a n=−a n−1+n 2,n≥1, a 0 a 0 = 3
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I am trying to solve the recurrence relation a n=−a n−1+n 2 a n=−a n−1+n 2 where n≥1 n≥1, a 0=3 a 0=3.
I have the first few terms of the sequence
a 1=−2,a 2=6,a 3=3,a 4=13,…a 1=−2,a 2=6,a 3=3,a 4=13,…
however after this point I am a little lost as to what I should do next, as I cant find a pattern in this sequence and am unsure what the characteristic equation would look like. Would anyone care to give me any hints as to what the next step would be? Thanks in advance.
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edited Dec 7, 2019 at 16:15
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asked Dec 7, 2019 at 7:55
Chairman MeowChairman Meow
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Please see math.meta.stackexchange.com/questions/5020Angina Seng –Angina Seng 2019-12-07 07:56:30 +00:00 Commented Dec 7, 2019 at 7:56
What could be improved upon with how I asked this question?Chairman Meow –Chairman Meow 2019-12-07 07:57:31 +00:00 Commented Dec 7, 2019 at 7:57
how did you get a 1=−2 a 1=−2? should a 0 a 0 be −3−3?omer –omer 2019-12-07 08:01:02 +00:00 Commented Dec 7, 2019 at 8:01
There was a typo, it should be fixed now.Chairman Meow –Chairman Meow 2019-12-07 08:02:06 +00:00 Commented Dec 7, 2019 at 8:02
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The alternating character of this sequence suggests dealing with two steps at a time. So let's consider
a n+1=−a n+(n+1)2=−(−a n−1+n 2)+(n+1)2=a n−1+(n+1)2−n 2=a n−1+2 n+1.a n+1=−a n+(n+1)2=−(−a n−1+n 2)+(n+1)2=a n−1+(n+1)2−n 2=a n−1+2 n+1.
Can you take it from here?
To finish things up: let's consider even terms first:
a 2 n=a 2(n−1)+2(2 n)−1=a 2(n−2)+2(2(n−1))−1+2(2 n)−1=…=a 0+4∑k=1 n k−∑k=1 n 1=a 0+4⋅n⋅(n+1)2−n=2 n 2+n+a 0 a 2 n=a 2(n−1)+2(2 n)−1=a 2(n−2)+2(2(n−1))−1+2(2 n)−1=…=a 0+4∑k=1 n k−∑k=1 n 1=a 0+4⋅n⋅(n+1)2−n=2 n 2+n+a 0
and lastly
a 2 n+1=−a 2 n+(2 n+1)2=−(2 n 2+n+a 0)+(2 n+1)2=2 n 2+3 n+1−a 0 a 2 n+1=−a 2 n+(2 n+1)2=−(2 n 2+n+a 0)+(2 n+1)2=2 n 2+3 n+1−a 0
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edited Dec 8, 2019 at 8:17
answered Dec 7, 2019 at 8:08
b00n heTb00n heT
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1 I believe so. Thanks for the help. Is this a general method for solving recurrence relations that are alternating, such as this one?Chairman Meow –Chairman Meow 2019-12-07 08:09:55 +00:00 Commented Dec 7, 2019 at 8:09
1 let me know if you get to the result.b00n heT –b00n heT 2019-12-07 08:11:10 +00:00 Commented Dec 7, 2019 at 8:11
You got it, will do!Chairman Meow –Chairman Meow 2019-12-07 08:12:04 +00:00 Commented Dec 7, 2019 at 8:12
So I have been trying for a couple hours and I am still as lost as I was before. Is there any more guidance you could provide me?Chairman Meow –Chairman Meow 2019-12-07 20:36:21 +00:00 Commented Dec 7, 2019 at 20:36
1 I've completed the answer. It's a nice problem!b00n heT –b00n heT 2019-12-08 08:18:32 +00:00 Commented Dec 8, 2019 at 8:18
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Linear Algebra: Determine if a triangle is a right angled triangle
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I've just started my course in Linear Algebra, and I've come across a question I'm not entirely sure how to solve.
Let $A = (1, 1, -1), B = (-3, 2, -2), C = (2, 2, -4)$. Prove that the triangle $ABC$ is a right-angled triangle.
I know what we need to do, I know that we can use the dot product to find the angle between two vectors, and if the dot product is = 0 that means that we have a right angle.
I'm looking for some hints in the right direction of what I need to. I believe I'll need to construct two vectors from the triangle ($\overrightarrow{AB}$, $\overrightarrow{AC}$, $\overrightarrow{BC}$, etc..?) and then compute the dot product of the two to determine the angle.
linear-algebra
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edited Feb 2, 2016 at 6:44
adjan
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asked Feb 1, 2016 at 19:11
Talen KylonTalen Kylon
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Your method is correct. The three angles in our triangle can be tested by computing the dot product of the following pairs \begin{align} \vec{AB} &= \langle-4,1,-1\rangle & \vec{AC} &= \langle1,1,-3\rangle \ \vec{BA} &= \langle 4,-1,1\rangle & \vec{BC} &= \langle5,0,-2\rangle \ \vec{CA} &= \langle-1,-1,3\rangle & \vec{CB} &= \langle-5,0,2\rangle \end{align} These are the displacement vectors connecting the points $A$, $B$, and $C$. For example $\vec{AB}=B-A$.
These dot products are \begin{align} \vec{AB}\cdot\vec{AC} &= 0 & \vec{BA}\cdot\vec{BC} &= 18 & \vec{CA}\cdot\vec{CB} &= 11 \end{align} What does this say about our triangle?
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edited Feb 1, 2016 at 19:27
answered Feb 1, 2016 at 19:23
Brian FitzpatrickBrian Fitzpatrick
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$\begingroup$ How did you construct the vector AB? To answer your question, because we found an dot product of 0, we can say that our triangle does indeed have a right angle. $\endgroup$
Talen Kylon
– Talen Kylon
2016-02-01 19:27:00 +00:00
Commented Feb 1, 2016 at 19:27
$\begingroup$ By the formula for displacement vectors: $\vec{AB}=B-A$ $\endgroup$
Brian Fitzpatrick
– Brian Fitzpatrick
2016-02-01 19:28:20 +00:00
Commented Feb 1, 2016 at 19:28
$\begingroup$ Ahh ok, thank you for the edit. I was trying A-B. I understand now. $\endgroup$
Talen Kylon
– Talen Kylon
2016-02-01 19:29:08 +00:00
Commented Feb 1, 2016 at 19:29
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Assuming that "the triangle ABC" means the triangle defined by the three endpoints of $\vec{a}, \vec{b}, \vec{c}$ when the vectors all start from the origin, the three sides of the triangle are given by $\vec{a} - \vec{b}$, $\vec{a} - \vec{c}$, and $\vec{b} - \vec{c}$. Use the dot product of each pair of those vectors to find the angles between them.
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answered Feb 1, 2016 at 19:16
DylanSpDylanSp
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Yes you can use the dot product. Either $\overrightarrow{AB} \cdot \overrightarrow{AC} = 0$ or $\overrightarrow{AB} \cdot \overrightarrow{BC} = 0$, since the dot product of orthogonal vectors is $0$.
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answered Feb 1, 2016 at 19:20
adjanadjan
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189944 | https://www.math.uni-hamburg.de/home/geschke/papers/mbt8_MRL.pdf | Math. Res. Lett. 13 (2006), no. 00, 10001–100NN c ⃝International Press 2006 METRIC BAUMGARTNER THEOREMS AND UNIVERSALITY Stefan Geschke and Menachem Kojman Abstract. It is consistent with the axioms of set theory that for every metric space X which is isometric to some separable Banach space or to Urysohn’s universal separable metric space U the following holds: (⋆)X There exists a nowhere meager subspace of X of cardinality ℵ1 and any two nowhere meager subsets of X of cardinality ℵ1 are almost isometric to each other.
As a corollary, it is consistent that the Continuum Hypothesis fails and the following hold: (1) There exists an almost-isometry ultrahomogeneous and universal element in the class of separable metric spaces of size ℵ1.
(2) For every separable Banach space X there exists an almost-isometry conditionally ultrahomogeneous and universal element in the class of subspaces of X size ℵ1.
(3) For every finite dimensional Banach space X, there is a unique universal element up to almost-isometry in the class of subspaces of X size ℵ1.
1. Introduction In this paper we deal with metric similarity, metric uniqueness and metric univer-sality in models of set theory that violate the Continuum Hypothesis (CH). Originally motivated by questions about metric universality, we prove the consistency with the usual axioms of set theory of a strong statement about the homogeneity of separable Banach spaces and of Urysohn’s universal separable metric space.
Let (X, d) be the underlying metric structure of some separable Banach space, and consider the following problem: suppose A ⊆X satisfies that A ∩U is of the second category in U for every nonempty open U ⊆X. Can one find a set B ⊆X of the same cardinality as A, satisfying the same condition as A, such that B is not in the autohomeomorphism group orbit of A?
If one assumes the Continuum Hypothesis, then such a B is very easily found: by CH the cardinality of A is necessarily equal to |X| and either X itself or X \ h, where H is a hyperplane of X, is not homeomorphic to A.
The main result below shows that without appealing to an additional axiom, like CH, finding B is impossible, because it is consistent that for every separable Banach space X and two nowhere meager sets A, B ⊆X with |A| = |B| = ℵ1 not only that A can be carried over onto B by an autohomeomorphism of X, it actually holds that A maps onto B by autohomeomorphisms of X which are arbitrarily close to being isometries, namely are bi-Lipschitz with Lipschitz constants arbitrarily close to 1.
Received by the editors November 7, 2006.
2000 Mathematics Subject Classification. Primary: 54E40, 51M04, 03E35; Secondary: 46B20.
Key words and phrases. Metric space, almost-isometry, almost-isometric embedding, Urysohn’s space, Oracle Forcing, Universality.
10001 10002 Stefan Geschke and Menachem Kojman To present better the picture at large, we begin by describing the subject in the related setting of linearly ordered sets. The reader may consult the introduction to for additional introductory material.
1.1. Order isomorphism and universality. Up to order isomorphism there is a unique countable dense subset of R. The ordertype of such a set is the unique ultra-homogeneous and universal member in the class of countable linearly ordered sets.
Consequently, R is universal in the class of separable dense linear orders and its or-dertype is determined uniquely as the completion of the countable ultrahomogeneous and universal ordering. These results were proved of course by Cantor. If one does not assume completeness, then by Sierpinski , there are 22ℵ0 continuum dense subsets of R that are pairwise order incomparable.
Under CH, κ-dense subsets of R exist only for κ ∈{ℵ0, 2ℵ0}. But as is well-known since Cohen’s invention of forcing, also the negation of the Continuum Hypothesis is consistent with the axioms of set theory — in fact, the continuum may be arbitrarily large. Thus, for every cardinal κ > ℵ0 it is meaningful to inquire the consistency with the axioms of the statement, and of the negation of the statement, that 2ℵ0 > κ and all κ-dense subsets of R are order-isomorphic to each other. The same applies to the statement of the existence of a universal separable linearly ordered set of cardinality κ, which, of course, follows from the previous one.
In Cohen extensions, as well as in in Solovay’s random real extensions, with arbi-trary large continuum, neither of these statements holds for any κ, ℵ0 < κ < 2ℵ0 (see [7, 8]). Thus, the negations of these two statements (for an arbitrary κ > ℵ0) are easily consistent, being valid in standard models for the negation of CH.
Positive consistency was somewhat harder. Baumgartner, in a classic paper in set theory , proved the consistency of 2ℵ0 = ℵ2 and any two ℵ1-dense subsets of R are order isomorphic to each other. Baumgartner’s method necessitated the validity of CH in intermediate models of his forcing iteration, and its novelty at the time was further supported by the fact that the argument could not be easily lifted to ℵ2. In fact, the consistency of Baumgartner’s statement for ℵ2-dense sets is still open.
Abraham and Shelah have shown that Baumgartner’s result did not follow from Martin’s Axiom . Abraham, Rubin and Shelah have investigated many variants of Baumgartner’s result and have also proved that Baumgartner’s statement for ℵ1-dense sets was consistent with continuum larger than ℵ2 . Later it was observed that Baumgartner’s result followed from the Proper Forcing Axiom , but this gave no information for κ > ℵ1, as PFA was later shown to imply that 2ℵ0 = ℵ2 (see and ).
Recently it was shown by Moore that PFA also implies Shelah’s Basis Con-jecture for linear orders, i.e., there are five uncountable linear orders such that every uncountable linear order contains a copy of one of the five.
The problem of universality at ℵ1 in the class of all (rather than only separable) linearly ordered sets has also been investigated. If CH holds, then there is a saturated linear order (unique for that property) of cardinality ℵ1, which is also universal.
However, since the existence of a universal linearly ordered set of cardinality ℵ1 implies the existence of a universal separable one, in Cohen or Solovay extensions there are no universal linearly ordered sets of cardinality ℵ1.
METRIC BAUMGARTNER THEOREMS AND UNIVERSALITY 10003 It took a remarkable development in forcing technique to settle the consistency of a universal linear ordering in ℵ1 < 2ℵ0. In Shelah introduced both his Proper Forcing and his Oracle Forcing methods, and used them to prove the consistency of 2ℵ0 = ℵ2 with the existence a universal linearly ordered set of cardinality ℵ1. The singularity and difficulty of this result are appreciated adequately in the light of a much later result by Kojman and Shelah : ℵ1 is the unique uncountable regular cardinal for which this consistency holds; for every regular κ > ℵ1 the inequality 2ℵ0 > κ implies that fewer than 2ℵ0 linearly ordered sets, each of cardinality κ, do not suffice to order-embed all linearly ordered sets of cardinality κ. (For a recent survey of applications of the combinatorial method introduced in this proof see .) 1.2. Metric isomorphisms and universality. Let us now turn to metric similar-ity and embeddability. Kojman and Shelah have introduced the notions of almost-isometry and almost-isometric embedding between metric spaces and were able to obtain with respect to these relations several of the results described above in the set-ting of linearly ordered sets. Up to almost-isometry there is a unique countable dense subset of R, which is the unique ultrahomogeneous and universal member in the class of linear countable metric spaces for almost-isometric embeddings. More importantly, up to almost-isometry there is a unique countable dense subset of Urysohn’s universal separable metric space U that is the unique almost-isometry ultrahomogeneous and universal member in the class of countable metric spaces. In fact, the Urysohn space is characterized as the completion of such a countable space. In Cohen or Solovay extensions both almost-isometric uniqueness of κ-dense subsets — of R as well as of U — and the existence of an almost-isometry universal metric space of size κ fail for all κ with ℵ0 < κ < 2ℵ0. Furthermore, if κ > ℵ1 is a regular cardinal and 2ℵ0 > κ, then fewer than 2ℵ0 metric spaces (not necessarily separable) of cardinality κ do not suffice to almost isometrically embed all metric spaces of cardinality κ. This result leaves ℵ1 as the only regular cardinal at which the existence of an almost-isometry universal space below the continuum may be consistent. This consistency is not known.
In the class of separable metric spaces also positive results were proved in .
It is consistent that fewer than continuum separable metric spaces of size κ almost isometrically embed all separable metric spaces of size κ, for regular κ ∈(ℵ0, 2ℵ0).
Thus, for κ = ℵ2 < 2ℵ0, it may require only a small collection of separable metric spaces on ℵ2 to almost-isometrically embed all others, but it always takes a larger number of general metric spaces on ℵ2 for the similar task. For κ = ℵ1 this result approximated the consistency of a universal separable metric space in ℵ1 < 2ℵ0, but left it open.
Finally, it was shown in that the almost-isometry analog of Baumgartner’s result was simply false: in every “reasonable” metric space (see Section 2 below; we only care here that R and U are “reasonable”) there are many pairwise almost-isometry incomparable ℵ1-dense subsets. This seemed at the time to block the approach to universality via Baumgartner type uniqueness.
Yet, a metric Baumgartner theorem is exactly what we prove here (for R, U and every separable Banach space) among whose consequences is the consistency of an almost-isometry universal separable metric space of cardinality ℵ1 < 2ℵ0. We use Shelah’s Oracle Forcing method from (also see [14, Chapter IV]) to prove Baum-gartner’s metric analog with an additional, topological condition: any two subsets of 10004 Stefan Geschke and Menachem Kojman U or any separable Banach space that are nowhere meager, i.e., non-meager in any open subset, and of size ℵ1 are almost isometric to each other (Theorem 5.1).
It is worth pointing out that almost isometries between dense subsets of a com-plete metric space extend to the whole space. Thus, we obtain the consistency of a statement saying that the spaces under consideration are very homogeneous.
2. Notation and preliminaries Say that two metric spaces (X, dX) and (Y, dY ) are almost isometric if for every real constant K > 1 there is a homeomorphism f : X →Y that satisfies dX(x1, x2)/K < dY (f(x1), f(x2)) < KdX(x1, x2).
That is, for every K > 1 there exists a K-bi-Lipschitz homeomorphism between X and Y .
We quote from : Theorem 2.1. Suppose X is a separable metric space and there exists a constant K > 1 so that for every open subset U ⊆X there is a K-bi-Lipschitz embedding of a non-empty open interval from the standard Cantor set into U. Then there are 2ℵ0 pairwise bi-Lipschitz incomparable ℵ1-dense subsets of X.
Let X be a perfect and complete metric space. Let (⋆)X be the following statement about X: (⋆)X There exists a nowhere meager subset of X of cardinality ℵ1 and any two nowhere meager subsets of X of size ℵ1 are almost isometric to each other.
Fact 2.2. If (⋆)X holds for any separable metric space that contains an uncountable nowhere dense set, then 2ℵ0 = 2ℵ1.
In particular, this equality follows from (⋆)R.
Proof. Let Z be an uncountable nowhere dense subset of X. We may assume that Z is of size ℵ1. By (⋆)X, there is a nowhere meager subset Y of X of cardinality ℵ1.
After subtracting Z from Y we may assume that Y and Z are disjoint. For every W ⊆Z let YW = Y ∪W. This gives a collection of 2ℵ1 nowhere meager subsets of X, each of cardinality ℵ1.
By (⋆)X, for each W ⊆Z we may fix fW : Y →YW , a bi-Lipschitz home-omorphism.
Each fW extends to an autohomeomorphism f W of the completion X of X. However, being separable, X has only 2ℵ0 autohomeomorphisms. Hence 2ℵ1 = 2ℵ0.
□ We shall prove below the consistency of 2ℵ0 = ℵ2 together with (⋆)X for every metric space X that is isometric to some separable Banach space or to Urysohn’s universal separable space U, and discuss some of the consequences of this consistency in universality theory.
A metric space X is almost-isometry universal for a class of metric spaces if for every space Y in the class and every K > 1 there exists a K-bi-Lipschitz embedding of Y into X, and X is almost-isometry ultrahomogeneous if for every K > 1, every finite K-bi-Lipschitz map from X to X extends to a K-bi-Lipschitz autohomeomorphism of X. A subspace Y ⊆X is conditionally almost-isometry ultrahomogeneous if for every METRIC BAUMGARTNER THEOREMS AND UNIVERSALITY 10005 K > 1 every finite K-bi-Lipschitz map from Y to Y which extends to a K-bi-Lipschitz autohomeomorphism of X extends to a K-bi-Lipschitz autohomeomorphism of Y 3. Oracle forcing We quote some definitions and results about Shelah’s oracle forcing from Shelah’s .
Some readers may prefer the more detailed presentation of oracle forcing in chapter IV of .
Definition 3.1. A sequence S = (Sα)α<ω of countable transitive models of ZFC except the power set axiom is called an ℵ1-oracle if for all A ⊆ω1, the set {α < ω1 : A ∩α ∈Sα} is stationary in ω1.
Note that if ♦ℵ1 holds, then there is an ℵ1-oracle.
Definition 3.2. Let S be an ℵ1-oracle.
A forcing notion P satisfies the S-chain condition, if for all P′ ⊆P of size ≤ℵ1 there is P′′ ⊆P also of size ≤ℵ1, a 1-1 function f : P′′ →ω1 and a set A ⊆ω1 such that P′ ⊆P′′ and for all limit ordinals α < ω1, if A ∩α ∈Sα, S ∈Sα ∩P(α) and f −1[S] is predense in f −1[α], then f −1[S] is predense in P.
The following Lemma says that forcing notions which satisfy an oracle chain con-dition can be iterated.
Lemma 3.3 (Shelah ). Let S be an ℵ1-oracle.
(a) Suppose that P is a forcing notion satisfying the S-chain condition. Then there is a P-name ˙ T for an ℵ1-oracle such that for every P-name ˙ Q for a forcing notion the following holds: If ⊩P “ ˙ Q satisfies the ˙ T-chain condition”, then P ∗˙ Q satisfies the S-chain condition.
(b) If for some ordinal δ, (Pα)α<δ is a continuous increasing chain of forcing notions satisfying the S-chain condition, then the direct limit S α<δ Pα satisfies the S-chain condition.
The main property of the oracle chain condition which we require is that forcing notions satisfying the oracle chain condition preserve the non-meagerness of the set of ground model reals.
Lemma 3.4 (Shelah ). Assume there is an ℵ1-oracle S′. Then there is an ℵ1-oracle S such that for every forcing notion P satisfying the S-chain condition and every P-generic filter G over the ground model V , in V [G] the set R ∩V is non-meager.
4. Forcing bi-Lipschitz bijections We wish to add sufficiently bi-Lipschitz bijections between two large subsets of a separable Banach space by forcing. The following lemma says that we will not run into trouble at a finite stage of the construction.
10006 Stefan Geschke and Menachem Kojman Lemma 4.1. Let B be a Banach space and K0 > 1. Suppose p is a finite partial function from B to B that extends to a K0-bi-Lipschitz bijection f : B →B. Then for every x0 ∈B \ dom(p) and every K1 > K0 there is a nonempty open set O ⊆B such that for all y ∈O, p ∪{(x0, y)} extends to a K1-bi-Lipschitz bijection from B to B.
Proof. Let x0 ∈B \ dom(p). We may assume that dom(p) is non-empty. Let δ = min{| |x0 −x| | : x ∈dom(p)} and let y0 = f(x0). Since K0 < K1 there is c > 0 such that K0 + c < K1 and 1 K0 −c > 1 K1 .
Let O = y ∈B : | |y −y0| | δ < c .
For every x ∈B and y ∈O let g(x) = ( 0, if | |x −x0| | > δ and (y −y0) · (1 −| |x−x0| | δ ), if | |x −x0| | ≤δ.
It is easily checked that g : B →B is Lipschitz of constant c. Clearly, the function f + g : B →B extends p ∪{(x0, y)}.
We have to check that f + g is a K1-bi-Lipschitz bijection. Let x, x′ be distinct elements of B. Then by the choice of c, | |(f + g)(x) −(f + g)(x′)| | | |x −x′| | ≤| |f(x) −f(x′)| | + | |g(x) −g(x′)| | | |x −x′| | < K0 + c < K1 On the other hand, | |(f + g)(x) −(f + g)(x′)| | | |x −x′| | ≥| |f(x) −f(x′)| | −| |g(x) −g(x′)| | | |x −x′| | > 1 K0 −c > 1 K1 , again by the choice of c.
To see that f + g is onto let z0 ∈B and consider the map h : B →B; x 7→f −1(z0 −g(x)).
By the choice of c, 1 K0 −c > 0 and therefore K0 · c < 1. Since f −1 is Lipschitz of constant K0 and g is Lipschitz of constant c, h is a contraction. Therefore, h has a fixed point x. Now (f + g)(x) = f(h(x)) + g(x) = f(f −1(z0 −g(x))) + g(x) = z0.
In other words, z0 is in the range of f + g.
□ Lemma 4.2. Let S be an ℵ1-oracle and K > 1. Suppose X and Y are nowhere meager subsets of size ℵ1 of a separable Banach space B. Then there is a forcing notion P of size ℵ1 satisfying the S-chain condition such that ⊩P “There is a K-bi-Lipschitz bijection between X and Y ”.
METRIC BAUMGARTNER THEOREMS AND UNIVERSALITY 10007 Proof. We will force by finite approximations of K-bi-Lipschitz bijections from B to B. A finite partial function p from B to B is a potential condition if for some K′ ∈(1, K), p extends to a K′-bi-Lipschitz bijection from B to B.
By recursion on α < ω1 we will construct sequences (Xα)α<ω and (Yα)α<ω1 of subsets of X, respectively Y , a sequence (Pα)α≤ω1 of forcing notions and a sequence (fα)α<ω1 of functions such that the following conditions are satisfied: (1) Every Xα is a countable dense subset of X and every Yα is a countable dense subset of Y .
(2) The sets Xα are pairwise disjoint and so are the sets Yα. Moreover, X = S α<ω1 Xα and Y = S α<ω1 Yα.
(3) For sequences (A0 γ)γ<δ and (A1 γ)γ<δ of subsets of B let P((A0 γ)γ<δ, (A1 γ)γ<δ) be the set of finite partial functions p from S γ<δ A0 γ to S γ<δ A1 γ that are potential conditions such that for all x ∈dom(p) and all γ < α, x ∈A0 γ ⇔ p(x) ∈A1 γ.
P((A0 γ)γ<δ, (A1 γ)γ<δ) is ordered by reverse inclusion.
Then Pα = P((Xγ)γ<α, (Yγ)γ<α).
(4) For every α < ω1 with α > 0, fα is a bijection from Pα onto ω · α.
If α < β < ω1, then fβ is an extension of fα.
(5) For all α < ω1, if for some β ≤α and some S ∈Sβ, f −1 β [S] is predense in Pβ, then every condition in Pα+1 is compatible (in Pα+1) with an element of f −1 β [S], i.e., f −1 β [S] is predense in Pα+1.
Suppose the recursion can be carried out. We claim that P = Pω1 works for the lemma.
Let G be P-generic over the ground model.
Then clearly, b = S G is a K-bi-Lipschitz mapping from a subset of X to a subset of Y .
We have to check that dom(b) = X and ran(b) = Y . We only give the argument for dom(b) = X. The proof of ran(Y ) = Y is symmetric.
Let x be a point in X and let p be a condition in P. Suppose that x ̸∈dom(p).
Let α < ω1 be such that x ∈Xα. By Lemma 4.1, there is an open set O ⊆R such that for all y ∈O, p ∪{(x, y)} is a potential condition.
Since Yα is dense in Y and Y is dense in B, there is y ∈O∩Yα. Now p∪{(x, y)} ∈P extends p and is again a condition in P. It follows that the set of conditions that have x in their domain is dense in P. So by genericity, x ∈dom(b).
We now show that f = S α<ω1 fα witnesses the S-chain condition of P.
Let α be a limit ordinal below ω1. Then for some β < ω1, α = ω · β and hence, by (4), f −1[α] = Pβ. Suppose that for some S ∈Sα, f −1[S] is predense in f −1[α] = Pβ.
Then, by (5), f −1[S] is predense in Pγ+1 for every γ < ω1 with γ ≥β. Hence f −1[S] is predense in P.
Let us turn to the recursive construction. For Z ∈{X, Y } let ≺Z be a wellordering on Z of order type ω1. Let X0 and Y0 be countable dense subsets of X, respectively Y . Suppose for some α < ω1 we have constructed (Xγ)γ<α, (Yγ)γ<α and fα.
By recursion on k < ω we will define sequences (xk)k<ω and (yk)k<ω and then put Xα = {xk : k < ω} and Yα = {yk : k < ω}. Extending fα to fα+1 is straight forward, so we will not mention it anymore.
10008 Stefan Geschke and Menachem Kojman Fix an enumeration (On)n∈ω of all nonempty members of some countable basis for the topology on B. To make sure that Xα and Yα are dense, for all k > 0 we will choose xk ∈X ∩Ok and yk ∈Y ∩Ok.
Suppose α is an even ordinal. Then let x0 be the ≺X-minimal element of X \ S γ<α Xγ.
If α is odd, let y0 be the ≺Y -minimal element of Y \ S γ<α Yγ.
This guarantees that in the end we have X = S γ<ω1 Xγ and Y = S γ<ω1 Yγ.
If α is even, then we already know x0 and choose y0 next, then x1, then y1 and so on. If α is odd, then we already know y0 and choose x0 next, then y1, then x1 and so on.
To make sure that (5) is satisfied for α, we will choose (xk)k<ω and (yk)k<ω so that for even α it holds that (even) for all k < ω, (5) is true for P((Xγ)γ<α⌢{xl : l < k}, (Yγ)γ<α⌢{yl : l < k}) and P((Xγ)γ<α⌢{xl : l ≤k}, (Yγ)γ<α⌢{yl : l < k}) instead of Pα+1 and so that for odd α it holds that (odd) for all k < ω, (5) is true for P((Xγ)γ<α⌢{xl : l < k}, (Yγ)γ<α⌢{yl : l < k}) and P((Xγ)γ<α⌢{xl : l < k}, (Yγ)γ<α⌢{yl : l ≤k}) instead of Pα+1.
We concentrate on one case, namely that α is odd and we have already defined (xl)l<k and (yl)l 0, and we are now looking for yk. The other three cases are symmetric.
In order to satisfy (odd), yk has to meet countably many constraints. Namely, whenever β ≤α is a limit ordinal, S ∈Sβ, f −1 β [S] is predense in Pβ and p ∈ P((Xγ)γ<α⌢{xl : l < k}, (Yγ)γ<α⌢{yl : l ≤k}) uses yk, then p is compatible with some element of f −1 β [S].
We will show that for each constraint, the set of y ∈Y satisfying it is open and dense.
Since Y is nowhere meager, there is y ∈ Ok−1 ∩Y \ (S γ<α Yγ ∪{yl : l < k}) satisfying all these constraints. Putting yk = y ensures (odd).
We finish the proof of the lemma by showing that the set of y meeting a single constraint is open and dense.
Let p ∈P((Xγ)γ<α⌢{xl : l < k}, (Yγ)γ<α⌢{yl : l < k}) and β ≤α. Suppose that x ∈{xl : l < k} is not in dom(p). Suppose further that S ∈Sβ is such that f −1 β [S] is predense in Pβ. Finally, let O be a non-empty open subset of B. We show that there is a non-empty open set U ⊆O such that for all y ∈U, if p ∪{(x, y)} ∈P((Xγ)γ<α ⌢{xl : l < k}, (Yγ)γ<α ⌢({yl : l < k} ∪{y})), then in that partial order, p ∪{(x, y)} is compatible with some element of f −1 β [S].
If there is no y ∈O such that p ∪{(x, y)} is a potential condition, then we can choose U = O. Now suppose that for some y ∈O, p∪{(x, y)} is a potential condition.
By Lemma 4.1, there is c > 0 such that for all y′ ∈R with |y −y′| < c, y′ ∈O and p ∪{(x, y′)} is a potential condition.
By the density of X0 and Y0 and by Lemma 4.1, there are a ∈X0 and b ∈Y0 such that | |a −x| | < c 2K and p ∪{(a, b), (x, y)} is a potential condition. Now p ∪{(a, b)} ∈P((Xγ)γ<α ⌢{xl : l < k}, (Yγ)γ<α ⌢({yl : l < k})) and for all y′ such that p ∪{(a, b), (x, y′)} is a potential condition, | |y′ −b| | < c 2 and by | |y −b| | < c 2, | |y′ −y| | < c. In particular, whenever p ∪{(a, b), (x, y′)} is a potential condition, then y′ ∈O. This argument shows that by passing to a stronger condition METRIC BAUMGARTNER THEOREMS AND UNIVERSALITY 10009 if necessary, we may assume that all y′ such that p ∪{(x, y′)} is a potential condition are elements of O.
By (5), p is compatible with some condition q ∈f −1 β [S]. In particular, p ∪q is a potential condition. By Lemma 4.1, there is an open set U such that for all y′ ∈U, p ∪q ∪{(x, y′)} is still a potential condition. By our assumption on p, U ⊆O.
□ Lemma 4.2 remains true if the separable Banach space B is replaced by the Urysohn space U.
Lemma 4.3. Let S be an ℵ1-oracle and K > 1. Suppose X and Y are nowhere meager subsets of size ℵ1 of the Urysohn space U. Then there is a forcing notion P of size ℵ1 satisfying the S-chain condition such that ⊩P “There is an K-bi-Lipschitz bijection between X and Y ”.
Proof. By Claim 16 in , U is almost-isometry ultrahomogeneous. This implies the analog of Lemma 4.1 for the Urysohn space easily. Now the proof of Lemma 4.3 is the same as the proof of Lemma 4.2 □ Lemma 4.4. If (⋆)U holds, then every nowhere meager subset A ⊆U of size ℵ1 is almost-isometry ultrahomogeneous.
If X is a separable Banach space and (⋆)X holds, then every nowhere meager subspace A ⊆X of cardinality ℵ1 is conditionally almost-isometry ultrahomogeneous, namely, for every K > 1, every finite K-bi-Lipschitz map from A to A extends to a K-bi-Lipschitz autohomeomorphism of A — provided it extends to a K-bi-Lipschitz autohomeomorphism of X.
Proof. Since almost-isometry ultrahomogeneity is preserved under almost-isometry, it suffices to prove that there exists at least one almost-isometry ultrahomogeneous nowhere meager subset of U whose cardinality is ℵ1. Since U is almost-isometry ul-trahomogeneous, any subspace A ⊆U can be enlarged to an almost-isometry ultraho-mogeneous subspace of the same cardinality by a standard closure under bi-Lipschitz autohomeomorphisms argument (using only rational K > 1 as constants).
The proof of the second clause in the Lemma is similar.
□ 5. The results Theorem 5.1. If ZFC is consistent, then so is ZFC + 2ℵ0 = ℵ2 + “for every metric space X, if X is isometric to the Urysohn space or to some separable Banach space, then (⋆)X holds”, where (⋆)X is the statement X has a nowhere meager subset of size ℵ1 and any two nowhere mea-ger subsets of X of size ℵ1 are almost isometric to each other.
Proof. We start from a model of ♦ℵ1 + 2ℵ1 = ℵ2. In this model there is an ℵ1-oracle S as in Lemma 3.4. We then perform a finite support iteration ((Pα)α≤ω2, ( ˙ Qα)α<ω2) such that for each α < ω2, (1) Pα+1 = Pα ∗˙ Qα and ⊩Pα | ˙ Qα| = ℵ1. (This guarantees that each Pβ has a dense subset of size ℵ1.) 10010 Stefan Geschke and Menachem Kojman (2) For some Pα-name ˙ T for an ℵ1-oracle as in Lemma 3.3 a), ⊩Pα “ ˙ Qα satisfies the ˙ T-chain condition”, so that Pα+1 satisfies the S-chain condition. (This guarantees that all Pβ satisfy the S-chain condition.) The forcing notion Pω2 then satisfies the S-chain condition by Lemma 3.3 b) and hence, by Lemma 3.4, in every Pω2-generic extension the ground model reals will form a non-meager, in fact a nowhere meager, subset of R of size ℵ1.
By a theorem of Kuratowski , for any two perfect Polish spaces X and Y there are meager Borel sets A ⊆X and B ⊆Y so that X \A and Y \B are homeomorphic.
Thus, if R has a a nowhere meager set of size ℵ1 then every perfect Polish space has a nowhere meager set of size ℵ1.
If in any Pω2-generic extension of the ground model, if A and B are nowhere meager subsets of size ℵ1 of either the Urysohn space or of a given separable Banach space X, then there is α < ω2 such that A and B are already elements of the corresponding Pα-generic extension and they will be nowhere meager also in that intermediate model.
Note that if A and B are nowhere meager subsets of a Banach space X, then they are dense in X and therefore X as a metric space can be reconstructed from either A or B.
By using some suitable book-keeping we can make sure that for every K > 1 and A, B as above, there is some β < ω2 with α ≤β such that the forcing notion with the name ˙ Qβ adds an K-bi-Lipschitz bijection between A and B. The existence of such a forcing notion is guaranteed by Lemma 4.2 and by Lemma 4.3. For the book-keeping we use the facts that (i) at every initial stage of the iteration we have 2ℵ1 = ℵ2 and (ii) there are only (2ℵ0)ℵ1 = 2ℵ1 separable metric spaces of size ℵ1 (up to isome-try).
It follows that in the Pω2-generic extension, any two sets A and B as above are almost isometric to each other.
□ Remark 5.2. The case X = R in Theorem 5.1 follows also from Burke’s Theorem 1.7 in : it is consistent that there is a nowhere meager subset of R of size ℵ1 and any two such sets are order isomorphic via a restriction to R of an entire function with a first derivative arbitrarily close to 1.
Theorem 5.3. If ZFC is consistent, then so is ZFC + 2ℵ0 = ℵ2 + (a) +(b) + (c), where: (a) There is an almost-isometry ultrahomogeneous and almost-isometry universal element in the class of all separable metric spaces of size ℵ1.
(b) For every separable Banach space B there is a conditionally almost-isometry ultrahomogeneous and almost-isometry universal element in the class of sub-spaces of B of size ℵ1.
(c) For every finite-dimensional Banach space B, the property of being almost-isometry universal in the class of subspaces of B of cardinality ℵ1 determines a space up to almost-isometry.
METRIC BAUMGARTNER THEOREMS AND UNIVERSALITY 10011 Remark 5.4. Condition (c) easily fails for U.
Suppose X ⊆U is universal and dense in U, and let X ∪{x} for a new point x be so that x has distance ≥1 to any point in X. So both X and X ∪{x} are separable and almost-isometry universal, but they are not almost isometric to each other (and X ∪{x} is not almost-isometry ultrahomogeneous). Similar examples exists in every separable Banach space which is isomorphic to one of its hyperplanes, e.g., the infinite dimensional Hilbert space.
Proof. We work in a model of set theory as in Theorem 5.1.
To prove (a) fix a nowhere meager subset Y ⊆U of cardinality ℵ1. By Lemma 4.4, Y is almost-isometry ultrahomogeneous. To prove that Y is almost-isometry universal suppose that X is a separable metric space of size ℵ1.
Since X is isometric to a subspace of U, we assume that X is a subspace of U. Now X ∪Y is nowhere meager in U and of size ℵ1, hence Y and X ∪Y are almost isometric. This implies that X is almost isometrically embedded into Y .
The proof of (b) is similar to that of (a).
To prove (c) suppose X is a finite-dimensional Banach space and that A ⊆X is of size ℵ1 and is almost-isometry universal in the class of subspaces of X of cardinality ℵ1. We claim that A is nowhere meager in X; the almost-isometry uniqueness of A follows from this claim and (⋆)X.
Fix some set B ⊆X which is nowhere meager in X and of cardinality ℵ1 and fix, by the almost-isometry universality of A, some K-bi-Lipschitz embedding f : B →A for some K > 1. Denote by ˆ f the continuous extension of f to X. Also ˆ f is bi-Lipschitz.
Since X is homeomorphic to Rn for some n, Brouwer’s preservation of domain theorem applies to X, and therefore ˆ f is open; since ˆ f is bi-Lipschitz, ˆ f is also closed.
Therefore ran ˆ f = Rn by connectedness of X, and ˆ f is thus an autohomeomorphism of X. It follows that ˆ f[B] is a nowhere meager subset of X. Since ˆ f[B] ⊆A, A is nowhere meager in X.
□ 6. Open problems Let us state the central open problem first: Problem 6.1. Is it consistent to have an almost-isometry universal metric space in cardinality ℵ1 < 2ℵ0?
We remark, again, that with any uncountable regular κ > ℵ1 substituted for ℵ1 in this problem, the answer is negative. The analogous problem for linearly ordered sets and graphs and other relational structures [13, 10] have positive answers.
Problem 6.2. Is it consistent that there is a nowhere meager subset of U of cardi-nality ℵ2 and that any two nowhere meager subsets of U of cardinality ℵ2 are almost isometric?
A positive solution will solve positively also the next problem: Problem 6.3. Is it consistent that 2ℵ0 > ℵ2 and that there is an almost-isometry universal separable metric space of size ℵ2?
Finally: Problem 6.4. Is (⋆)U consistent with 2ℵ0 > ℵ2?
10012 Stefan Geschke and Menachem Kojman Acknowledgements The research was supported by a German-Israeli Foundation grant number I-802-195.6/2003. This note was written during a Research-in-Pairs stay at Mathematisches Forschungsinstitut Oberwolfach. We thank the referee for pointing out that Fact 2.2 holds for separable metric spaces, not just for perfect Polish spaces.
References U. Abraham, M. Rubin and S. Shelah, On the consistency of some partition theorems for con-tinuous colorings, and the structure of ℵ1-dense real order types. Ann. Pure Appl. Logic 29 (1985), no. 2, 123206.
U. Avraham, S. Shelah, Martin’s axiom does not imply that every two ℵ1-dense sets of reals are isomorphic, Israel J. Math. 38 (1981), no. 1-2, 161–176.
J. E. Baumgartner. All ℵ1-dense sets of reals can be isomorphic, Fund. Math. 79 (1973), 101–106.
J. E. Baumgartner. Applications of the proper forcing axiom. in Handbook of set-theoretic topology. K. Kunen and J. Vaughan eds., North-Holland, Amsterdam, 1984, 913–959.
Maxim R. Burke. Entire functions mapping uncountable dense sets of reals onto each other monotonically. preprint.
M. Dˇ zamonja. Club Guessing and the Universal Models. Notre Dame J. Formal Logic 46, no. 3 (2005), 283–300 M. Kojman and S. Shelah. Non existence of universal linear orders in many cardinalities. The Journal of Symbolic Logic, 57(3) 1992 pp. 857–891.
M. Kojman, S. Shelah, Almost isometric embeddings between metric spaces, Israel J. Math. in press.
K. Kuratowski. Topology, volume I, London, Panstwowe Wydawnictwo Naukowe, Warsaw, 1966.
A. H. Mekler, Universal structures in power ℵ1. J. Symbolic Logic 55 (1990), no. 2, 466–477.
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S. Shelah, Proper and Improper Forcing, Perspectives in Mathematical Logic, Springer-Verlag Berlin Heidelberg (1998) W. Sierpinski, Sur les types d’ordre des ensembles linaires, Fundam. Math. 37 (1950), 253–264.
B. Velickovic, Forcing axioms and stationary sets, Adv. Math. 94, No.2, 256–284 (1992).
S. Todorcevic, Comparing the continuum with the first two uncountable cardinals, Dalla Chiara, Maria Luisa (ed.) et al., Logic and scientific methods, Volume one of the proceedings of the tenth international congress of logic, methodology and philosophy of science, Florence, Italy, August 19–25, 1995, Kluwer Academic Publishers. Synth. Libr. 259, 145–155 (1997).
II. Mathematisches Institut, Freie Universit¨ at Berlin, Arnimallee 3, 14195 Berlin, Ger-many and Boise State University, 1910 University Drive, Boise, ID 83725-1555, USA E-mail address: geschke@math.fu-berlin.de Department of Mathematics, Ben-Gurion University of the Negev, Beer Sheva, Israel E-mail address: kojman@math.bgu.ac.il |
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3rd grade math
Course: 3rd grade math > Unit 11
Lesson 1: Perimeter
Perimeter: introduction
Perimeter of a shape
Find perimeter by counting unit squares
Find perimeter by counting units
Find perimeter when given side lengths
Perimeter review
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Perimeter
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Perimeter of a shape
CCSS.Math: 3.MD.D.8
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To find the perimeter of a shape, you add up the lengths of all the sides. For a square or pentagon, it's even easier - you can multiply the length of one side by the number of sides to get the total perimeter. Created by Sal Khan.
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Ro_Demon
5 years ago Posted 5 years ago. Direct link to Ro_Demon's post “Where does the word Perim...”
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Where does the word Perimeter come from?
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Sadie D
a year ago Posted a year ago. Direct link to Sadie D's post “the word is pronounced pe...”
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the word is pronounced perimeter because peri-means around and the word meter-is referring to measurement so they put it all together and it makes perimeter.
have a blessed day today
have a blessed day
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240480
4 years ago Posted 4 years ago. Direct link to 240480's post “why is it called a perime...”
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why is it called a perimeter?
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Johanna
4 years ago Posted 4 years ago. Direct link to Johanna's post ““Peri-“ means “around”, a...”
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“Peri-“ means “around”, and “meter” refers to measurement. That makes sense because the perimeter is the measure around a shape.
I hope this helps!
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Ylara429
9 years ago Posted 9 years ago. Direct link to Ylara429's post “So perimeter is basically...”
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So perimeter is basically adding up al the sides of the shape?
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mijuraut
9 years ago Posted 9 years ago. Direct link to mijuraut's post “Yes, perimeter is the sum...”
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Yes, perimeter is the sum of the side lengths.
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feven
5 years ago Posted 5 years ago. Direct link to feven's post “but what if the permiter ...”
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but what if the permiter and the area
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Zoe #animalrightsmatter
4 years ago Posted 4 years ago. Direct link to Zoe #animalrightsmatter's post “I don't really know what ...”
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I don't really know what you mean by "but what if the perimeter and the area" but I will try to answer this the best I can.A perimeter is the outside of an area. It is the border surrounding something. The area is what's inside that perimeter, what's inside that border. I hope this helps. ✌🏻♥️🦊 Zo
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ariana.rahman
2 months ago Posted 2 months ago. Direct link to ariana.rahman's post “Thank u so much this is r...”
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Thank u so much this is really helpful
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Nitika Nagpal
a month ago Posted a month ago. Direct link to Nitika Nagpal's post “youre really welcome! Gla...”
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youre really welcome! Glad we could help
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Henry Xiong
3 months ago Posted 3 months ago. Direct link to Henry Xiong's post “There is a video in a vid...”
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There is a video in a video.
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elbrown1
3 months ago Posted 3 months ago. Direct link to elbrown1's post “If there are a game insi...”
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If there are a game inside a game sense its 1 game and 1 game its going to be 2x2=4
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Silke
11 years ago Posted 11 years ago. Direct link to Silke's post “Is there a quicker way of...”
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Is there a quicker way of counting the perimeter?
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Sharon
11 years ago Posted 11 years ago. Direct link to Sharon's post “Well, it depends on how y...”
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Well, it depends on how you count your perimeter now.
Ex. Rectangles length is 13 and width is 5
You could do it like this: 13+13+5+5
Or
You could do it like 132+52 (Which is how I personally solve it)
I think the second one is easier. There are probably more methods but that is all I know of now.
I hope this helped :)
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JosephS
2 months ago Posted 2 months ago. Direct link to JosephS's post “Whoever doesnt know what ...”
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Whoever doesnt know what perimeter means heres a explanation perimeter is measuring the outside not inside it is not like area perimeter your adding 4 numbers and in area you multiply 2 numbers and and here are perimeter key words fence outside hope this helps out!
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Sid
9 days ago Posted 9 days ago. Direct link to Sid's post “perimeter is just adding ...”
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perimeter is just adding while area is multiplying
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suhan.nishad
7 years ago Posted 7 years ago. Direct link to suhan.nishad's post “I know what Perimeter is ...”
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I know what Perimeter is it is side+side+side+side and you add them and you get an answer for the shape that you wanted to find out the perimeter and area is length x Width so like length times Width, it is like doing addition and multiplication but different ways.
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Video transcript
What is the perimeter of the shape? Each square in the grid is a 1 by 1 centimeter square. So all we have to do is add up the lengths of these blue segments right over here. And they put it on this nice grid. And each box here is 1 by 1. So let's say we start up here. We want to make sure that we only go to where we started and we don't double count. So this perimeter is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24 centimeters long. So it is 24 centimeters. The other way we could have thought about this is you could have looked at the length of each of these sides. So this is 1 plus 2 plus 3 plus 1 plus 3 plus 1 plus-- what is this-- 5 plus 2 plus 4 plus 2. And you would have also gotten 24 centimeters. Let's do a couple more of these. What is the perimeter of the square? So once again, it's the length of all of the segments that define the outside boundary of the square. And by definition, a square-- all of its sides are equal. So you have 4 sides that are all 7 meters long. So you could say it's 7 meters plus 7 meters plus 7 meters plus 7 meters, or it's 4 times 7 meters or 28 meters. Let's do one more. What is the perimeter of the regular pentagon? So it's a regular pentagon, which means all of its sides have the same length. And they give us a side of one. The length of one is 2. So all of the sides have length 2. So it's going to be 2 plus 2 plus 2 plus 2 plus 2, or essentially five 2's. Or another way of thinking about it is it's 5 times 2, which is going to be 10. You have five sides of this pentagon. Each of those sides are 2 units long. So 2 units long times 5 sides is going to be 10 units.
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189946 | https://scienceinschool.org/article/2021/virtual-labs-real-science/ | Virtual labs, real science – scienceinschool.org
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Issue 52
Virtual labs, real science Inspire article
March 31, 2021
Issue 52
Ages: 11-14, 14-16, 16-19
Topics: Chemistry, Physics, Resources
Keywords: Forces, Informatics, States of matter, Teaching resources
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Author(s):Subhadip Senapati
Not just for remote teaching: virtual labs really came into their own during the COVID-19 pandemic, but they can generally be a useful addition to the STEM teacher’s toolbox.
Introduction
Textbook content and practical experiments are both critical for science education. Experiments and demonstrations help to engage students’ attention and can additionally help them to understand the concepts learned.
With many teachers having to teach remotely during the COVID-19 pandemic, practical experiments have become difficult. One option to introduce a more practical element into online lessons is the use of virtual labs. As the name suggests, a virtual laboratory is a computer-based activity where students interact with simulated computer interfaces to perform experiments using virtual tools, apparatus, or materials.[1,2]
However, many teachers are hesitant to try virtual labs due to concerns that they might be expensive or complicated, or that it isn’t worth setting everything up to use them when they won’t be needed after the pandemic. In fact, many of these tools are easy to use and some, like PhET Interactive Simulations (PhET), are available for free. Furthermore, while virtual labs can’t replace the experience of doing physical experiments, they have features that are complementary to practical classes and be a useful supplement even during normal teaching. For example, the virtual models in these programs allow students to visualize processes, like molecular motion, that would normally be invisible.
This article introduces PhET – its basic features and some of the simulations it offers – in the hope of inspiring teachers to give these systems a try. A comparison with some other simulation platforms is provided at the end.
PhET Interactive Simulations
PhET is one of the most popular simulation platforms, is available for free, and provides interactive simulations for a range of scientific topics.[4,5] It was launched in 2002 by a group of physicists at the University of Colorado Boulder (USA), led by Nobel laureate Dr Carl Wieman.
PhET is not technically difficult to use. Teachers can either download the simulations or use them online, and instructions are provided under ‘Teacher Tips’ on the homepage for each simulation. All simulations are accompanied by explanations of the basic theory, handouts (including tips for teachers), activity worksheets, and discussion questions. Teachers can add their own set of questions or update the existing set to check the conceptual understanding of the students.
PhET stands for Physics Education Technology and initially offered only physics-based simulations. Since then, it has been expanded to simulations involving chemistry, biology, mathematics, and earth science (figure 1). PhET simulations are designed to engage students through an intuitive, story-telling or game-like environment, where students learn through exploration. Currently, PhET has a total of 159 interactive simulations for primary and secondary school students across five disciplines.
Figure 1: Example images from PhET simulations offered for physics, biology, chemistry, and mathematics
Image: PhET Interactive Simulations/ University of Colorado Boulder
Examples of PhET simulations
States of matter is a fundamental topic in chemistry and is critical for students’ conceptual understanding before moving on to more complex concepts. Unfortunately, students often have difficulty visualizing the fundamental particles (atoms and molecules) and their motion. PhET provides an opportunity not only to visualize the organization of the atoms and molecules in a particular state, but also to see the changes that take place during changes of state (figure 2).
Figure 2: The State of Matter simulation. A plot of interaction potential and a phase diagram can optionally be shown in addition to the arrangement of particles.
Image: PhET Interactive Simulations/ University of Colorado Boulder
From the simulation screen, users can select different molecules from a list (neon, argon, oxygen, and water). They can then change the temperature and pressure of the system and see how the arrangement and behaviour of the particles change (figure 3).
Figure 3: Arrangement of oxygen molecules in three different states (solid, liquid, gas), obtained by changing the temperature
Image: PhET Interactive Simulations/ University of Colorado Boulder
This is particularly useful for explaining why ice floats on water, which often confuses students based on their prior knowledge that solids are denser than the corresponding liquid. With this simulation, students can easily visualize how water molecules assemble into a cage-like structure when the temperature is lowered, thus making ice less dense than liquid water (figure 4).
Figure 4: Arrangement of water molecules in three different states (solid, liquid, gas). The cage-like structure of ice is different from that of solid oxygen shown in figure 2.
Image: PhET Interactive Simulations/ University of Colorado Boulder
There is also a range of simulations relating to electricity, covering topics such as electrical charge and conductivity, electrostatics, resistance and conductance, and circuit construction. An example is ‘Electric Field Hockey’, in which students can learn about electrical charges and their interaction by trying to score goals in a simulated ice-hockey game, and there are several other fun activities presented in a game format that students can explore themselves (figure 5).
Figure 5: A game-like simulation from PhET to introduce the concept of force
Image: PhET Interactive Simulations/ University of Colorado Boulder
PhET can also be useful to demonstrate the phenomena that influence climate change and global warming. The greenhouse effect simulation, for example, demonstrates how the earth absorbs sunlight and re-emits it as infrared radiation, and how clouds and greenhouse gases prevent this radiation from escaping into space (figure 6). In the photon absorption tab, students can choose different atmospheric molecules and see how they interact with photons of sunlight or infrared radiation.
Figure 6: a) The earth absorbs photons of sunlight (yellow) and re-emits them as photons of infrared radiation (red), which cannot escape into space in the presence of clouds and greenhouse gases. b) Infrared photons can interact with molecules like methane in the atmosphere and may be absorbed or scattered.
Image: PhET Interactive Simulations/ University of Colorado Boulder
Other virtual labs
PhET is a pioneer in virtual simulations and remains relevant and heavily accessed even after almost two decades. Unfortunately, PhET currently offers relatively few simulations for biology, mathematics, and earth science, although new simulations are continually being added. Furthermore, the interface may appear slightly outdated compared to some of the commercially available virtual labs, and some of the simulations may be difficult for the students to complete without guidance from their teachers. It would also be useful if PhET were to offer a proper assessment option.
There are several other virtual labs, either commercially available or open-source, that offer an alternative. Table 1 provides a comparison of four of these (PhET, OLabs, Beyond Labz, and Labster), chosen based on their features and applicability for a wide range of subjects for school education, integration with different institutions, and a large number of users [4, 7–9]. There are several other virtual labs that offer mostly subject-specific simulations, and a comprehensive list can be found in Ref. .
Table 1. Comparison of the features of four virtual labs: PhET, OLabs, Beyond Labz, Labster.| | Phet | OLabs | Beyond Labz | Labster |
--- ---
| Topics | physics, chemistry, biology, mathematics, earth science | physics, chemistry, biology, maths | general chemistry, organic chemistry, biology, physics, physical science | physics, chemistry, biology, medicine, engineering |
| Target audience | Primary school to secondary school | middle school to secondary school | middle school upwards | high school upwards |
| Platform | website, app | website | website | website, app (beta version available only in selected countries) |
| Connectivity | online/offline | online | online/ offline | online |
| Access | open-source/free | open-source/free | free trial period; then paid access | free trial period, free lab safety simulation; then paid access |
More realistic and sophisticated virtual labs are usually behind a paywall, which can be a hurdle for access unless the educational institution can pay for it.
Summary
Virtual labs have been slowly gaining momentum for the past few years, and many research groups, start-ups, and big companies have been working on improving existing simulations or developing new ideas. However, with widespread remote teaching during the COVID-19 pandemic, these tools really came into their own, and they have been extensively used to support online STEM teaching. PhET, for examples, saw a significant increase in its usage worldwide, and up to a 500% increase in countries like France and Italy. These tools are useful beyond remote teaching though, and can be a valuable supplement to practical classroom experiments.
References
Woodfield et al. (2004)The Virtual ChemLab Project: A Realistic and Sophisticated Simulation of Inorganic Qualitative Analysis.J. Chem. Educ.81:1672–1678. doi:10.1021/ed081p1672
Rutten N,van Joolingen WR, van der Veena JT (2012)The learning effects of computer simulations in science education.Computers & Education58:136–153. doi:10.1016/j.compedu.2011.07.017
Ali N, Ullah SReview to Analyze and Compare Virtual Chemistry Laboratories for Their Use in Education.J. Chem. Educ.97:3563–3574. doi:10.1021/acs.jchemed.0c00185
PhET:
Moore EB et al. (2014)PhET Interactive Simulations: Transformative Tools for Teaching Chemistry.Journal of Chemical Education91:1191–1197. doi:10.1021/ed4005084
Finkelstein N et al. (2006)High tech tools for teaching physics: The physics education technology project. MERLOT Journal of Online Learning and Teaching2:110–121.
OLabs:
Beyond Labz:
Labster:
Open Educational Resources: Simulations and Virtual Labs,
Leytham-Powell C (2020)PhET simulations keep students engaged while learning science remotely.Colorado Arts and Science Magazine
Resources
PhET interactive simulations:
A collection of Science in School articles ideal for remote teaching: Watt S (2020)Science at home: ideas for remote teaching. Science in School50
Some remote learning ideas from EMBL:Dobreva T, Haas E (2020)Science at home: distance learning with EMBL. Science in School50
Other online resources for students: Prior E (2020)Keeping science engaging: Online resources for students during COVID-19.Science in School50
Author(s)
Dr Subhadip Senapati obtained his PhD degree in Chemistry from Arizona State University (USA). At present, he is a Senior Researcher at Prayoga Institute of Education Research, Bangalore (India) and working on different projects in the field of science education research and small-scale domain research.
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189947 | https://courses.lumenlearning.com/ccbcmd-math/chapter/draw-angles-in-standard-position/ | Draw angles in standard position | Applied Algebra and Trigonometry
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Applied Algebra and Trigonometry
Chapter 4.1: Angles
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Draw angles in standard position
Properly defining an angle first requires that we define a ray. A ray consists of one point on a line and all points extending in one direction from that point. The first point is called the endpoint of the ray. We can refer to a specific ray by stating its endpoint and any other point on it. The ray in Figure 1 can be named as ray EF, or in symbol form →E F E F→.
Figure 1
An angle is the union of two rays having a common endpoint. The endpoint is called the vertex of the angle, and the two rays are the sides of the angle. The angle in Figure 2 is formed from →E D E D→ and →E F E F→. Angles can be named using a point on each ray and the vertex, such as angle DEF, or in symbol form \hspace{0.17em}\angle D E F\hspace{0.17em}\angle D E F.
Figure 2
Greek letters are often used as variables for the measure of an angle. The table below is a list of Greek letters commonly used to represent angles, and a sample angle is shown in Figure 2.
θ θ ϕ or φ ϕ or φ α α β β γ γ
theta phi alpha beta gamma
Figure 3.Angle theta, shown as ∠θ∠θ
Figure 4
Angle creation is a dynamic process. We start with two rays lying on top of one another. We leave one fixed in place, and rotate the other. The fixed ray is the initial side, and the rotated ray is the terminal side. In order to identify the different sides, we indicate the rotation with a small arc and arrow close to the vertex as in Figure 4.
The following video provides an illustration of angles in standard position.
As we discussed at the beginning of the section, there are many applications for angles, but in order to use them correctly, we must be able to measure them. The measure of an angle is the amount of rotation from the initial side to the terminal side. Probably the most familiar unit of angle measurement is the degree. One degree is 1 360 1 360 of a circular rotation, so a complete circular rotation contains 360 degrees. An angle measured in degrees should always include the unit “degrees” after the number, or include the degree symbol °. For example, 90 degrees = 90°.
Figure 5
To formalize our work, we will begin by drawing angles on an x–y coordinate plane. Angles can occur in any position on the coordinate plane, but for the purpose of comparison, the convention is to illustrate them in the same position whenever possible. An angle is in standard position if its vertex is located at the origin, and its initial side extends along the positive x-axis.
If the angle is measured in a counterclockwise direction from the initial side to the terminal side, the angle is said to be a positive angle. If the angle is measured in a clockwise direction, the angle is said to be a negative angle.
Figure 6
Drawing an angle in standard position always starts the same way—draw the initial side along the positive x-axis. To place the terminal side of the angle, we must calculate the fraction of a full rotation the angle represents. We do that by dividing the angle measure in degrees by 360°. For example, to draw a 90° angle, we calculate that 90∘360∘=1 4 90∘360∘=1 4. So, the terminal side will be one-fourth of the way around the circle, moving counterclockwise from the positive x-axis. To draw a 360° angle, we calculate that 360∘360∘=1 360∘360∘=1. So the terminal side will be 1 complete rotation around the circle, moving counterclockwise from the positive x-axis. In this case, the initial side and the terminal side overlap.
Since we define an angle in standard position by its terminal side, we have a special type of angle whose terminal side lies on an axis, a quadrantal angle. This type of angle can have a measure of 0°, 90°, 180°, 270° or 360°.
Figure 7. Quadrantal angles have a terminal side that lies along an axis. Examples are shown.
A General Note: Quadrantal Angles
Quadrantal angles are angles whose terminal side lies on an axis, including 0°, 90°, 180°, 270°, or 360°.
How To: Given an angle measure in degrees, draw the angle in standard position.
Express the angle measure as a fraction of 360°.
Reduce the fraction to simplest form.
Draw an angle that contains that same fraction of the circle, beginning on the positive x-axis and moving counterclockwise for positive angles and clockwise for negative angles.
Example 1: Drawing an Angle in Standard Position Measured in Degrees
Sketch an angle of 30° in standard position.
Sketch an angle of −135° in standard position.
Solution
Divide the angle measure by 360°. 30∘360∘=1 12 30∘360∘=1 12
To rewrite the fraction in a more familiar fraction, we can recognize that
1 12=1 3(1 4)1 12=1 3(1 4)
One-twelfth equals one-third of a quarter, so by dividing a quarter rotation into thirds, we can sketch a line at 30° as in Figure 8.
Figure 8
2. Divide the angle measure by 360°. −135∘360∘=−3 8−135∘360∘=−3 8
In this case, we can recognize that
−3 8=−3 2(1 4)−3 8=−3 2(1 4)
3. Negative three-eighths is one and one-half times a quarter, so we place a line by moving clockwise one full quarter and one-half of another quarter, as in Figure 9.
Figure 9
Try It 1
Show an angle of 240° on a circle in standard position.
Solution
Watch this video for more examples of determining angles of rotation.
Candela Citations
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Precalculus. Authored by: OpenStax College. Provided by: OpenStax. Located at: License: CC BY: Attribution
All rights reserved content
Animation: Angles in Standard Position. Authored by: Mathispower4u. Located at: License: All Rights Reserved. License Terms: Standard YouTube License
Examples: Determine Angles of Rotation. Authored by: Mathispower4u. Located at: License: All Rights Reserved. License Terms: Standard YouTube License
Licenses and Attributions
CC licensed content, Specific attribution
Precalculus. Authored by: OpenStax College. Provided by: OpenStax. Located at: License: CC BY: Attribution
All rights reserved content
Animation: Angles in Standard Position. Authored by: Mathispower4u. Located at: License: All Rights Reserved. License Terms: Standard YouTube License
Examples: Determine Angles of Rotation. Authored by: Mathispower4u. Located at: License: All Rights Reserved. License Terms: Standard YouTube License
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189948 | https://fiveable.me/thinking-like-a-mathematician/unit-7/permutations/study-guide/lROFQGmk7wvlD33k | printables
🧠Thinking Like a Mathematician Unit 7 Review
7.1 Permutations
🧠Thinking Like a Mathematician Unit 7 Review
7.1 Permutations
Written by the Fiveable Content Team • Last updated September 2025
Written by the Fiveable Content Team • Last updated September 2025
APA
🧠Thinking Like a Mathematician
Unit & Topic Study Guides
7.1 Permutations
7.2 Combinations
7.3 Binomial theorem
7.4 Recurrence relations
7.5 Graph representations
7.6 Graph traversals
7.7 Trees
7.8 Network flows
Permutations are a fundamental concept in combinatorics, crucial for understanding probability and statistics. They involve arranging objects in specific orders, applying to real-world scenarios like seating arrangements and password generation. The fundamental counting principle forms the basis for calculating permutations.
Various types of permutations exist, including linear, circular, and those with repetition. Each type addresses different scenarios and constraints in mathematical problem-solving. Understanding these variations enhances analytical thinking and problem-solving skills, preparing students for more complex mathematical concepts.
Definition of permutations
Fundamental concept in combinatorics involves arranging objects in a specific order
Crucial for understanding probability, statistics, and abstract algebra in mathematical thinking
Applies to various real-world scenarios, from seating arrangements to password generation
Fundamental counting principle
States that if one event can occur in m ways, and another independent event can occur in n ways, then the two events can occur together in m × n ways
Provides the foundation for calculating more complex permutations
Applies to scenarios with multiple independent choices (selecting shirt, pants, and shoes)
Extends to any number of independent events, multiplying the number of possibilities for each
Factorial notation
Represents the product of all positive integers less than or equal to a given number n
Denoted by n! and calculated as n × (n-1) × (n-2) × ... × 3 × 2 × 1
Simplifies calculations involving large numbers of permutations
Plays a crucial role in many permutation formulas
Special case: 0! is defined as 1, which is important in certain permutation calculations
Types of permutations
Diverse arrangements of objects or elements in specific orders
Each type addresses different scenarios and constraints in mathematical problem-solving
Understanding various permutation types enhances analytical thinking and problem-solving skills
Linear permutations
Arrangements of n distinct objects in a line or sequence
Total number of linear permutations of n objects is n!
Applies to situations like arranging books on a shelf or people in a queue
Order matters in linear permutations (ABC ≠ CBA)
Can involve partial permutations where only r objects out of n are arranged
Circular permutations
Arrangements of n objects around a circle or closed loop
Total number of circular permutations of n objects is (n-1)!
Rotations of the same arrangement are considered identical (clockwise rotations)
Applies to scenarios like seating arrangements at a round table
Requires different counting techniques compared to linear permutations
Permutations with repetition
Arrangements where objects can be repeated or used multiple times
Total number of permutations with repetition is n^r, where n is the number of types of objects and r is the length of the arrangement
Applies to situations like creating passwords with repeated characters
Allows for more possibilities than permutations without repetition
Includes scenarios where some objects are identical (permutations with indistinguishable objects)
Calculating permutations
Involves applying specific formulas and techniques to determine the number of possible arrangements
Requires careful consideration of problem constraints and conditions
Develops logical reasoning and systematic approach to problem-solving in mathematics
Formula for permutations
General formula for permutations of n objects taken r at a time: P(n,r) = n! / (n-r)!
Applies when selecting and arranging r objects from a set of n distinct objects
Simplifies to n! when r = n (all objects are used in the arrangement)
Derivation based on the multiplication principle and factorial notation
Variations exist for specific types of permutations (circular, with repetition)
Permutations vs combinations
Permutations consider the order of selection, while combinations do not
Permutations typically yield a higher count than combinations for the same n and r
Formula for combinations: C(n,r) = n! / (r! × (n-r)!)
Choosing permutations or combinations depends on whether order matters in the problem
Understanding the distinction helps in correctly modeling real-world scenarios
Permutations with restrictions
Involve additional constraints or conditions on the arrangements
Include scenarios like arranging objects with some fixed positions
May require breaking down the problem into smaller sub-problems
Often solved using the complementary counting method or the multiplication principle
Examples include derangements (no element in its original position) or permutations with adjacency restrictions
Applications of permutations
Demonstrate the practical relevance of permutation theory in various fields
Highlight the importance of combinatorial thinking in solving real-world problems
Encourage students to connect abstract mathematical concepts with tangible applications
Probability problems
Calculate the likelihood of specific arrangements or outcomes
Used in analyzing game strategies and gambling odds (card games, lottery)
Apply to problems involving random selection and ordering
Help in understanding and predicting rare events
Crucial in risk assessment and decision-making under uncertainty
Cryptography and codes
Permutations form the basis of many encryption algorithms
Used in creating and breaking substitution ciphers
Contribute to the security of modern cryptographic systems
Apply to the generation of secure passwords and keys
Involve concepts like permutation groups and cycle notation
Genetic sequencing
Analyze possible arrangements of DNA nucleotides
Used in studying gene mutations and evolutionary relationships
Apply to protein folding problems in molecular biology
Help in understanding genetic diversity within populations
Contribute to advancements in personalized medicine and genetic engineering
Permutations in algebra
Extend the concept of permutations to abstract algebraic structures
Provide a foundation for understanding symmetry in mathematics
Develop skills in abstract thinking and pattern recognition
Connect combinatorial ideas with group theory and abstract algebra
Permutation groups
Algebraic structures formed by the set of all permutations on a given set
Obey group axioms (closure, associativity, identity, inverse)
Symmetric group Sn contains all permutations of n elements
Subgroups of permutation groups have important applications in Galois theory
Study of permutation groups leads to insights in abstract algebra and number theory
Cycle notation
Compact way to represent permutations as products of disjoint cycles
Simplifies composition and analysis of complex permutations
Cycle structure determines important properties of permutations
Used to calculate the order of a permutation in a group
Helps in understanding orbits and fixed points of permutations
Parity of permutations
Classifies permutations as even or odd based on the number of transpositions
Even permutations form a subgroup called the alternating group An
Parity is preserved under composition of permutations
Plays a crucial role in the solvability of polynomial equations
Applications in physics (fermion and boson statistics) and computer science (sorting algorithms)
Advanced permutation concepts
Explore more sophisticated aspects of permutation theory
Build upon basic permutation ideas to solve complex problems
Develop advanced problem-solving skills and mathematical intuition
Connect permutations with other areas of mathematics and computer science
Derangements
Permutations where no element appears in its original position
Number of derangements denoted by !n (subfactorial n)
Calculated using the principle of inclusion-exclusion
Formula: !n = n! × (1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n/n!)
Applications in probability (hat-check problem) and combinatorial optimization
Inversions in permutations
Pairs of elements that are out of their natural order in a permutation
Number of inversions measures how "scrambled" a permutation is
Related to the concept of parity in permutations
Used in analyzing sorting algorithms and in some statistical tests
Can be efficiently counted using techniques like merge sort or binary indexed trees
Lehmer code
Unique encoding of permutations as sequences of non-negative integers
Provides a bijection between permutations and a set of integer sequences
Useful in generating permutations and in certain combinatorial algorithms
Relates to the factorial number system and combinatorial number systems
Applications in computer algebra systems and permutation-based encryption
Permutations in computer science
Illustrate the importance of permutation concepts in algorithmic design and analysis
Demonstrate how mathematical thinking applies to computational problems
Encourage students to explore connections between mathematics and computer science
Generating permutations algorithmically
Techniques for systematically listing all permutations of a set
Include algorithms like Heap's algorithm and the lexicographic algorithm
Efficiency and memory usage are important considerations
Used in combinatorial optimization and exhaustive search problems
Applications in generating test cases and in cryptographic protocols
Permutation-based sorting algorithms
Sorting methods that work by repeatedly permuting elements
Include algorithms like bubble sort, insertion sort, and selection sort
Analysis involves studying the average and worst-case number of permutations
Relate to the concept of inversions in permutations
Provide insights into algorithm efficiency and computational complexity
Permutation matrices
Square matrices representing permutations in linear algebra
Properties include orthogonality and determinants of ±1
Used in solving systems of linear equations and matrix factorizations
Applications in computer graphics (transformations) and network theory
Connect permutation theory with linear algebra and matrix theory
Historical significance
Trace the development of permutation theory through time
Highlight key contributions that shaped our understanding of permutations
Provide context for the evolution of mathematical thinking in combinatorics
Origins of permutation theory
Roots in ancient civilizations' study of combinations and arrangements
Early applications in divination and religious practices (I Ching)
Formalized study began in the Renaissance with work on algebraic equations
Gradual recognition of permutations as fundamental mathematical objects
Evolution from practical problem-solving to abstract mathematical theory
Contributions of key mathematicians
Levi ben Gerson (13th-14th century) early work on combinatorics
Blaise Pascal (17th century) developments in combinatorial probability
Joseph Louis Lagrange (18th century) applications to algebraic equations
Augustin Louis Cauchy (19th century) group theory and permutations
Arthur Cayley (19th century) abstract group theory and permutation groups
Évariste Galois (19th century) revolutionary work on permutation groups and solvability of equations
Problem-solving strategies
Develop systematic approaches to tackling permutation problems
Enhance critical thinking and analytical skills in mathematical contexts
Prepare students for applying permutation concepts in various scenarios
Identifying permutation problems
Look for keywords like "arrange," "order," or "sequence" in problem statements
Determine if the order of selection matters in the given scenario
Consider whether repetition is allowed or if all elements are distinct
Assess if there are any restrictions or conditions on the arrangements
Recognize scenarios where circular arrangements or partial permutations apply
Common pitfalls and misconceptions
Confusing permutations with combinations when order matters
Overlooking restrictions or special conditions in the problem statement
Misapplying the fundamental counting principle to dependent events
Failing to account for repeated elements or circular arrangements
Misinterpreting factorial notation or incorrectly applying permutation formulas
Step-by-step approach
Carefully read and analyze the problem statement
Identify the total number of elements and how many are being arranged
Determine if there are any restrictions or special conditions
Choose the appropriate permutation formula or technique
Break down complex problems into simpler sub-problems if necessary
Apply the chosen method and perform calculations
Verify the result by considering extreme cases or using alternative methods
Interpret the solution in the context of the original problem |
189949 | https://www.rcsdk12.org/cms/lib/NY01001156/Centricity/Domain/5280/Algebra1Block%20-%20Notes%20-%20Final%20Exam%20Review%20Day%205.pdf | Name Per Date Notes: Final Exam Review Day #5 Algebra 1 Block Exponential Growth & Decay Exponential growth occurs when an amount exponentially. Exponential decay occurs when an amount exponentially. If the growth/decay rate is greater than 1, it is a function. If the growth/decay rate is less than 1, it is a function. Exponential Growth: F = P(1 + r)t F = P = Exponential Decay: F = P(1 – r)t r = t = YOUR RATE MUST BE WRITTEN AS A DECIMAL! #1 A new SUV purchased for $34,500 depreciates 13% each year. What is the SUV worth after 6 years? Round your answer to the nearest dollar. Growth or decay: Formula: Starting Value: Growth/Decay rate: Answer: #2 A $500 investment increases 6% each year. How much is the investment worth after 17 years? Round your answer to the nearest cent. Growth or decay: Formula: Starting Value: Growth/Decay rate: Answer: #3 A stamp collection is worth $4,200 and appreciates 9% in value each year. How much is the collection worth after 20 years? Round your answer to the nearest dollar. Growth or decay: Formula: Starting Value: Growth/Decay rate: Answer: #4 A boat purchased for $45,000 loses 7.5% of its value each year. What will the boat be worth in 20 years? Round your answer to the nearest dollar. Growth or decay: Formula: Starting Value: Growth/Decay rate: Answer: Linear vs. Exponential Linear Relationships Linear functions use the general form The graph of a linear function looks like a A table that increases or decreases by the same is linear Since arithmetic sequences deal with patterns involving or , they are considered to be linear Exponential Relationships Exponential functions use the general form The graph of an exponential function looks like a A table that increases or decreases at the same is exponential Since geometric sequences deal with patterns involving or , they are considered to be exponential Determine if each sequence could be modeled by a linear or exponential function. #5 10, 13, 16, 19,… #6 16, 12, 8, 4,… #7 2, 4, 8, 16,… #8 90, 80, 70, 60,… Determine if each situation should be modeled by a linear or exponential function. #9 Phillip has $500 in his savings #10 Every time Robert gets an A on account and plans on depositing his math test, his parents $40 every month. give him $10. #11 A savings account that contains $1200 #12 A deserted island has a population earns 4.7% interest every year, of 200 rabbits which doubles compounded annually. every year. #13 A rare painting worth $85,000 #14 Tony is buying candy bars for appreciates in value 4.5% every year. his friends that cost $0.75 each. Determine if the following tables could be represented by a linear or exponential function. #15 #16 #17 #18 x 1 2 3 4 5 f(x) 5 10 15 20 25 x 0 1 2 3 4 f(x) 800 400 200 100 50 x 1 2 3 4 5 f(x) 0.25 0.50 1 2 4 x -2 -1 0 1 2 f(x) 1 4 7 10 13 Functions, Domain, & Range A relation is a function is none of the values repeat. You can test to see if a graph is a function by using the _ ______ The input values are also called the The output values are also called the #19 Determine if the relation is a function. Explain your reasoning. #20 Determine if the relation is a function. Explain your reasoning. #21 Which representations are functions? #22 Which table represents a function? 1) 2) 3) 4) 1) I and II 2) II and IV 3) III, only 4) IV, only |
189950 | https://www.cantorsparadise.com/why-does-foil-even-work-bbddb9548fda | Why does the FOIL method even work? | Alex Lin | Cantor’s Paradise
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Why does the FOIL method even work?
A breakdown of the math behind the binomial multiplication trick.
Alex Lin
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Jul 25, 2021
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“First… Inside… Outside… Last…”
H idden behind this simple acronym lies a step-by-step guide to solving a seemly difficult mathematical problem.
Coined by William Betz in his 1929 textbook, Algebra for Today, the FOIL technique of multiplying two binomials is widely known by children and adults all around the world.
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FOIL’s general form equation by Calcworkshop.com
With this versatile tool secured in our mental toolkit, we tend to never think twice before employing it upon our binomial adversaries; but, have you ever wondered why it works?
Imagine I present you (x+y)(u+v) to expand — however, you aren’t allowed to use the FOIL method. Any ideas on where to start?
I didn’t have a clue. For much of my mathematics education, the FOIL method had become the way to multiply binomials. Having taken this trick for granted, I knew very little about why FOIL even worked? Here’s what I learned.
A proof-based breakdown
Starting at the top, we are given the standalone equation of
Expansion of binomials equation
Our goal is to show that the left side of the equation can be manipulated to become the right side. One of the first tricks of this proof is to realize that the above goal can be restated. It is equivalent to show that the right side of the equation can become the left side… Hence, the equal in the equation!
Flipped expansion of binomials equation
Now, given the left side of the equation, we can probably come up with a couple more ideas on how to manipulate the variables.
Our algebraic context
Before we dive right into the algebraic tricks, let’s take a small detour to further explore the assumptions and properties of this proof. For now, we’ll assume our binomial variables live on the real numbers.
It might be clear we plan on using some additive or multiplicative properties, but why is this allowed?
This question points out another assumption many of us love to make in mathematics: our set of numbers permits us to use the additive or multiplicative axioms.
Without diving too deeply into the weeds of algebraic structures, we will want to understand fields. A field is a set that abides by the following axioms:
Associativity of addition: a + (b + c) = (a + b) + c
Associativity of multiplication: a ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c
Commutativity of addition: a + b = b + a
Commutativity of multiplication: a ⋅ b = b ⋅ a
Additive identity: there exist an element called 0 in the field such that for any a in the field, a + 0 = a
Multiplication identity: there exist an element called 1 in the field such that for any a in the field, a⋅ 1 = a
Additive inverses: for every a in the field, there exists an element in the field, denoted −a, called the additive inverse of a, such that a + (−a) = 0.
Multiplicative inverses: for every a ≠ 0 in the field, there exists an element in the field, denoted 1/a, called the multiplicative inverse of a, such that a ⋅ 1/a= 1.
Distributivity of multiplication over addition: a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c).
These axioms all seem very obvious to us — we have learned these properties from a young age. However, therein lies the assumption that we are always working with a field. This is not always the case.
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Now, how does this apply to our problem? Well, we assumed that our binomial variables exist on the real numbers. Real numbers are proven to be a field, thus allowing us to take advantage of the above axioms in our algebraic manipulations.
As we will see in the next section, the distributivity of multiplication and commutativity of addition axioms play a large role in the proof. In fact, as long as the binomial variables come from a set that satisfies these two pre-requisite axioms, then the FOIL equation will hold.
Algebraic Manipulations
With our algebraic context now set, we are armed with the tools to finally prove FOIL’s equation!
Starting with ac + ad + bc + bd, we can first rearrange the terms by using the commutativity of the addition axiom. The reason for this rearrangement will become clear in the following steps.
Application of commutativity of addition axiom
Now, we notice c appears in the first two terms, while d appears in the last two terms. Let’s apply the distributivity of the multiplication axiom to pull out these two variables.
Application of commutativity of multiplication axiom
At this point, we notice both terms have (a+b). Let’s go ahead and apply the distributivity of multiplication axiom again to pull this term out.
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We have now reached our goal form! Hence, we have now shown that:
Proof finished
Proof conclusion and relation to the FOIL method
Pheww! This proof may have been just a tad bit pedantic, but look at the bright side — we have now shown that the above equation holds. From this, we can see where the origins of the FOIL method sprouted from.
Since we know the equation will hold for all real numbers, Betz came up with a neat acronym to help his students quickly multiply out binomials. Since then, this trick has perverse the math world, helping countless numbers of students in their daily mathematical endeavors!
Neat additional insights
As mentioned in the algebraic context, the FOIL equation holds as long as its variables live on a set that satisfies our two prerequisite axioms.
As a field always satisfies these two axioms, we can claim that the FOIL equation applies to all fields! This is why FOIL also works on complex numbers.
Let’s now see an example of the FOIL method working on a non-field set.
Consider a set of polynomials. Let’s say that p(x), q(x), s(x), and t(x) all live on a set where they have no multiplicative inverses. Our set is surely not a field as it does not satisfy the multiplicative inverse axiom.
However, the application of FOIL to these polynomials still holds as the distributivity of multiplication and commutativity of addition axioms are satisfied.
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FOIL equation applied to our polynomial set
Another fun, yet common, the occurrence of this can be explored in how we manipulate matrices.
Like many of my professors say, “Once we prove the theorem in class, you may use it freely.” So, to my fellow reader, you may now take for granted — the FOIL method.
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What is corrosive or aggressive water?When should I test the pH of my well water?How do I adjust the pH of my well water?Where can I get local help and information about pH in my well water?
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What pH should my well water be?
What pH should my well water be?
Updated over 3 months ago
Table of contents
What is corrosive or aggressive water?When should I test the pH of my well water?How do I adjust the pH of my well water?Where can I get local help and information about pH in my well water?
pH test/©Adobe
The measurement of acidity, or the amount of hydrogen ions, is called pH. It ranges from 0 (acidic) to 14 (basic), with 7 being neutral. A pH level between 6.5 and 8.5 is generally considered acceptable for drinking water. If the pH level of your well water is outside of this range, it may indicate water quality issues including the presence of contaminants.
The USEPA lists pH as a secondary standard as it is considered an aesthetic quality, not a health risk. Signs of low pH include a bitter metallic tase and corrosion, while high pH signs are a slippery feel, soda taste, or deposits.
What is corrosive or aggressive water?
Corrosive or aggressive water refers to water that has the ability to dissolve or erode metals, such as lead, copper, or galvanized pipes. It occurs when the water lacks certain minerals or has certain components that cause the metal to oxidize and dissolve into the water. Factors that contribute to corrosive or aggressive water include low pH, high levels of dissolved solids, chlorides, and sulfates. The solubility of the metals in the water determines whether it is corrosive or not. If the water can solubilize the metals, it is considered aggressive and corrosive. To determine if your water is corrosive or aggressive, it is recommended to have a water sample tested.
When should I test the pH of my well water?
A water test for pH will provide information on the level in your water. To ensure you are protecting your family’s health, we have developed a set of recommendations for testing that are provided in our article,"What do I need to know about sampling my well water?", that includes pH. When testing your well water you will very rarely be testing for only one thing, like pH, and instead testing for a suite of constituents that are important to identify if in your drinking water.
How do I adjust the pH of my well water?
Once you have tested and determined what the current pH level of your well water is, an acid or alkaline substances can be add to adjust the pH. For example, if the pH is too high (alkaline), you can add substances like vinegar or citric acid to lower the pH. On the other hand, if the pH is too low (acidic), you can add substances like baking soda or lime to raise the pH.
If you are unsure about adjusting the pH of your well water or if you encounter any difficulties, it is recommended to seek assistance from a local water systems expert or a professional contractor who specializes in well water.
Where can I get local help and information about pH in my well water?
Contact your local health department – They can help you interpret your sample results and may have information on the best treatment options for your situation. If they personally aren’t able to answer your questions, they will know who to contact who will be able to help you, likely at your state health department.
Contact a water treatment professional - we would recommend contacting a water treatment professional who has been certified through the Water Quality Association’s Professional Certification Program.
State or Federal Geologic or Water Resource Agency – they may have completed field studies of your area or collected samples from water wells to map and understand the water quality in your state. In addition, they may house water well logs, including yours, or be able to provide additional information about the aquifer your well is getting water from.
Consult your well driller or drilling contractor – the contractor that drilled your well, and other local contractors, may be aware of the water quality issues nearby including high iron, and might have worked with other well owners in your area to solve this issue .
Research online – you should be inquisitive; it will help you significantly because there may be great information available online. Use search terms like “my state (CA, ME, IL, etc.) water well logs”, “my state groundwater quality”, or “my state well water quality”.
Contact us directly – if these other local sources don’t work out for you, we may be able to help. We may be able to provide direct assistance, and/or direct you to a technical assistance provider in your area who can help answer your questions (free service funded by USEPA).
Want to learn more about your private well and how to care for it?
Sign up for the free 10-week email course from PrivateWellClass.org. The class is a project of RCAP and the University of Illinois, with funding from USEPA.
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Mentally calculating logarithms
Posted on 22 March 2021 by John
The previous post looked at approximations for trig functions that are simple enough to compute without a calculator. I wondered whether I could come up with something similar for logarithms. I start with log base 10. Later in the post I show how to get to find logs in other bases from logs base 10.
Let
x = m × 10 p.
where 1 ≤ m ≤ 10. Then
log 10(x) = log 10(m) + p
and so without loss of generality we can assume 1 ≤ x ≤ 10.
But we can narrow our range a little more. If x>3, compute the log of x‘=x/10, and if x<0.3 then compute the log of 10 x. So we will assume
0.3 ≤ x ≤ 3.
For x in this range
log 10(x) ≈ (x − 1)/(x + 1)
is a remarkably good approximation. The absolute error is less than 0.0327 on the interval [0.3, 3].
Examples
log 10 0.6 ≈ (0.6 − 1)/(0.6 + 1) = −1/4 = −0.25. Exact: −0.2218
log 10 1776 = 3 + log 10 1.776 ≈ 3 + 0.776/2.776 = 3.2795. Exact: 3.2494
log 10 9000 = 4 + log 10 0.9 ≈ 4 − 0.1/1.9 = 3.9473. Exact: 3.9542
Other bases
Logs to all bases are proportional, so you can convert between log base 10 and log to any other base by multiplying by a proportionality constant.
log b(x) = log 10(x) / log 10(b).
So suppose, for example, you want to compute log 2 48. Since 48 = 32 × 1.5 we have
log 2 48 = log 2 32 + log 2 1.5 = 5 + log 10 1.5 / log 10 2
We can approximate log 10 1.5 as 1/5 and log 10 2 as 1/3 to get
log 2 48 = 5 + 3/5 = 5.6.
The exact value is 5.585.
If you want to use this for natural logs, you might want to memorize
1/log 10 e = log e 10 = 2.3.
Update: There’s a better way to work with other bases. See this post. The same post explains why the approximation is particularly simple for logs base 10.
More accuracy
Abramowitz and Stegun refines the approximation t = (x − 1)/(x + 1). They use the interval [1/√10, √10] rather than [0.3, 3]. This slightly different interval is symmetric about 0 when you convert to t. Equation 4.1.41 runs t through a cubic polynomial and lowers the absolute error to less than 6 × 10−4. Equation 4.1.42 uses a 9th degree polynomial in t to lower the absolute error below 10−9.
Next
My next post will show how to compute 10 x analogously.
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4 thoughts on “Mentally calculating logarithms”
Tom 22 March 2021 at 12:26 Wow, nice.
buba 22 March 2021 at 23:07 There is this easy formula as well. Not as convenient for mental calculation though:
Rick Wicklin 23 March 2021 at 10:43 You’ll be happy to know that Abramowitz and Stegun and now available online as the NIST Digital Library of Mathematical Functions. You can link directly to the page that has the equations that you reference in the “More Accuracy” section.
Tom 23 March 2021 at 13:03 Rick, that’s awesome-
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189953 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10216919/ | Pediatric Gaucher Disease Presenting with Massive Splenomegaly and Hepatic Gaucheroma - PMC
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Children (Basel)
. 2023 May 12;10(5):869. doi: 10.3390/children10050869
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Pediatric Gaucher Disease Presenting with Massive Splenomegaly and Hepatic Gaucheroma
Gianluca Bossù
Gianluca Bossù
1 Pediatric Clinic, Department of Medicine and Surgery, University Hospital of Parma, 43126 Parma, Italy; gianluca.bossu@unipr.it (G.B.); laurapedre@hotmail.it (L.P.)
Find articles by Gianluca Bossù
1, Laura Pedretti
Laura Pedretti
1 Pediatric Clinic, Department of Medicine and Surgery, University Hospital of Parma, 43126 Parma, Italy; gianluca.bossu@unipr.it (G.B.); laurapedre@hotmail.it (L.P.)
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1, Lorenzo Bertolini
Lorenzo Bertolini
2 Unit of Radiologic Sciences, Department of Medicine and Surgery, University Hospital of Parma, 43126 Parma, Italy; lobertolini@ao.pr.it
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2, Susanna Esposito
Susanna Esposito
1 Pediatric Clinic, Department of Medicine and Surgery, University Hospital of Parma, 43126 Parma, Italy; gianluca.bossu@unipr.it (G.B.); laurapedre@hotmail.it (L.P.)
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Editor: Luisa De Sanctis
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1 Pediatric Clinic, Department of Medicine and Surgery, University Hospital of Parma, 43126 Parma, Italy; gianluca.bossu@unipr.it (G.B.); laurapedre@hotmail.it (L.P.)
2 Unit of Radiologic Sciences, Department of Medicine and Surgery, University Hospital of Parma, 43126 Parma, Italy; lobertolini@ao.pr.it
Correspondence: susannamariaroberta.esposito@unipr.it; Tel.: +39-0521-704790
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Received 2023 Mar 31; Revised 2023 May 5; Accepted 2023 May 8; Collection date 2023 May.
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Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
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PMCID: PMC10216919 PMID: 37238417
Abstract
Gaucher Disease (GD) is a condition resulting from an autosomal recessive inheritance pattern, characterized by a deficiency of the lysosomal enzyme beta-glucocerebrosidase. This leads to the accumulation of glucocerebroside and other glycolipids in multiple tissues, causing damage to various organ systems. The diagnosis of GD can be challenging due to its heterogeneity, non-specific symptoms, and variability across different geographic regions and age groups. Although GD is suspected based on symptoms or signs, the diagnosis is confirmed through the measurement of deficient b-glucocerebrosidase activity and the identification of biallelic pathogenic variants in the GBA gene. Intravenous enzyme replacement therapy (ERT) is recommended for GD patients. In this paper, we report a case of a 2-year and 8-month-old girl with massive splenomegaly and radiological finding of hepatic gaucheroma, in whom a genetic study showed homozygous mutation on the GBA gene at c.1448T>C (p.Leu483Pro) and certified the diagnosis of GD. This patient represents the youngest child reported to have gaucheroma and also the first one presenting with gaucheroma at the diagnosis and not during the follow up, highlighting that GD should be routinely included in the differential diagnosis of children presenting with splenomegaly and hepatomegaly, taking into account that the early start of ERT can change the natural history of the disease-preventing serious complications.
Keywords: Gaucher disease, gaucheroma, enzyme replacement therapy, hepatomegaly, splenomegaly
1. Background
Gaucher Disease (GD) is a condition resulting from an autosomal recessive inheritance pattern, characterized by a deficiency of the lysosomal enzyme beta-glucocerebrosidase. This leads to the accumulation of glucocerebroside and other glycolipids in multiple tissues, causing damage to various organ systems [1,2,3]. The genetic basis of GD involves biallelic pathogenic variants in the GBA gene, with more than 400 variants identified to date. The incidence of GD at birth varies in different regions of the world, ranging from approximately 0.39 to 5.80 per 100,000 individuals .
The diagnosis of GD can be challenging due to its heterogeneity, non-specific symptoms, and variability across different geographic regions and age groups . GD is divided into three main phenotypes: type 1 (chronic, non-neuronopathic), type 2 (acute neuronopathic), and type 3 (chronic neuronopathic) . Children with GD1 commonly exhibit symptoms such as splenomegaly, hepatomegaly, anemia, thrombocytopenia, epistaxis, bruising, delayed growth, and delayed puberty, as well as acute and chronic pain associated with bone disorders. Although GD1 is usually presented in childhood, diagnosis can be delayed until adulthood. Type 2 GD is characterized by early neurological involvement leading to fatality, while type 3 GD has slower progressive neurological involvement. GD3 exhibits the visceral manifestations described in GD1, although they tend to be more severe. Over time, GD3 patients develop neurological symptoms, such as cognitive impairment, myoclonic seizures, ataxia, spasticity, difficulty initiating horizontal eye movements, incomplete vertical eye movements, abnormally slow tracking of objects, and convergence problems with eye squinting and muscle weakness [1,2].
In the majority of cases, oculomotor neurological implication, usually associated with visceral manifestations, occurs before the age of 20. As GD1, GD3 phenotypes are heterogeneous, especially concerning neurological implications. Furthermore, neurological signs may appear several years after the visceral symptoms, even in patients originally diagnosed and treated as affected by GD1 .
Patients affected by GD1 have a life expectancy that is usually around ten years shorter than that of the general population, even though advanced therapies could affect the survival of these patients. On the other hand, patients with GD2 usually do not survive beyond two years of age. Children affected by GD3 experience a slower progression of the disease and may live into adulthood. Often GD is not diagnosed or is diagnosed late; this leads to complications and the persistence of symptoms . Although GD is suspected based on symptoms or signs, the diagnosis is confirmed through the measurement of deficient b-glucocerebrosidase activity and the identification of biallelic pathogenic variants in the GBA gene .
In pediatric patients with symptomatic GD1 and GD3, intravenous enzyme replacement therapy (ERT) is recommended. Although ERT does not slow down or stop the progression of a neurological condition, it has been used off-label for palliative care in some cases of GD2 [4,5]. An oral alternative, substrate reduction therapy (SRT), is available for adult patients with GD1. Additionally, symptomatic therapy may also be provided.
Finally, next-generation SRT, gene therapy, pharmacological chaperone therapy, and histone deacetylase inhibitors represent potential new treatments in development [4,5].
In this paper, we report a case of GD diagnosed in a child with massive splenomegaly and a radiological finding of hepatic gaucheroma, highlighting that GD should be routinely included in the differential diagnosis of children presenting with splenomegaly and hepatomegaly because the early start of ERT can change the natural history of the disease-preventing serious complications.
2. Case Report
A 2-year and 8-month-old girl was brought to our attention by her primary care pediatrician. The parents reported persistent abdominal distension during the last 18 months; a series of blood exams performed during the previous year showed anemia, thrombocytopenia, adequate hepato-renal function, and normal hemoglobin chain analysis. At physical examination, significant abdominal distension underlying massive splenomegaly (the splenic pole reached the low hypogastric region) and hepatomegaly was noted without any other significant symptom: general clinical conditions and neurological examination were in order. Development and growth also appeared adequate. Subsequently, the patient was admitted to our Pediatric Clinic Unit for further examination.
Blood exams confirmed anemia and thrombocytopenia (Hb 9.4 g/dL, PLT 62,000/µL) with normal white cell count; blood smear showed anysopoikilocytosis without immature forms. Immunoglobulins and immunophenotype on peripheral blood were in order. Infective diseases (i.e., hepatotropic viruses, leishmania, salmonellae, brucellae, and borrelia) were ruled out; an auto-immune panel comprehensive of ANA, LKM1-2, ASMA, and AMA was carried out and resulted normal for age.
Radiological assessment of the visceromegaly was performed: computed tomography (CT) and magnetic resonance imaging (MRI) described the massive splenomegaly and highlighted a hepatic focal lesion (Figure 1).
Figure 1.
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Magnetic resonance imaging shows massive splenomegaly and hepatomegaly (A), with a focal hepatic lesion characterized as gaucheroma (B).
At this point, a dried blood spot (DBS) test for GD and Nieman Pick A/B disease was carried out, revealing decreased activity of β-glucocerebrosidase A; genetic study was then conducted, which confirmed a homozygous mutation in the GBA gene at c.1448T>C (p.Leu483Pro), and quantitative measurement of lyso-Gb1 was performed, which showed an elevated concentration (732.4 ng/mL). Therefore, the diagnosis of GD was made.
The case was discussed with pediatric radiologists and a pediatric surgeon, and since alfa-fetoprotein and βHCG came out negative, liver biopsy was excluded, and the hepatic lesion was characterized as hepatic gaucheroma.
A neurological examination, including an EEG, was performed, which did not reveal any neurological changes, and no gaze palsy was observed.
The patient was admitted to the referral center for GD where she began treatment and continued monitoring of the disease. Before starting the treatment, the enzyme glucocerebrosidase was measured again, and it remained deficient (0.3 microMol/L/h). Additionally, the value of lysoGb1 was confirmed to be elevated (782 ng/mL). At the age of 2 years and 10 months, she started ERT with intravenous imiglucerase (Cerezyme) at a dosage of 60 U/kg/dose every 14 days without showing any local or systemic reaction. A lower limb MRI was carried out, which showed marked signs of spinal cord impairment according to the underlying pathology. An echocardiogram and ECG were performed, with normal results. The ophthalmologic examination did not show any pathological changes.
Regarding laboratory tests, 3 months after the start of treatment hemoglobin values recovered (11.2 g/dL), while platelets remained stable (60 × 10 3/µL). The CYP2D6 genotype study showed normal activity with the following CYP2D6 genotype 2A/2A, making her a normal metabolizer of CYP2D6 substrates, allowing her to be switched to oral therapy with a substrate inhibitor (Eliglustat-Cerdelga) at the age of 18.
According to the pathology follow-up protocol, blood markers of pathology should be monitored every 6 months . In a few months’ time, a check-up of chitotriosidase, lysoGb1, ferritin, acid phosphatase (first value 46.1 U/L; normal value < 4.7), ACE (first value 155 U/L; normal value 35–114), as well as a metabolic–nutritional routine and a complete abdominal ultrasound will be scheduled.
3. Discussion
A child presenting with splenomegaly and hepatomegaly like the one described in this case report represents a challenge for pediatricians. Different pathologies in children can cause an enlarged spleen and liver: infectious agents, hematologic disorders, infiltrative diseases, and auto-immune diseases are the most common . In the differential diagnosis, it is important to include GD because, owing to its clinical heterogeneity, there is often a delay in the diagnosis that can lead to severe complications. This child came to our attention several months after the onset of the first symptoms (anemia, thrombocytopenia, and splenomegaly).
To diagnose GD in our patient, DBS was performed to evaluate the enzymatic activity of β-glucocerebrosidase (GCase), followed by an analysis of the GBA gene. Supporting the diagnosis, an increase in the lyso-Gb1 biomarker was identified. For a long time, the standard method for diagnosing GD has been to determine the reduced activity of β-glucocerebrosidase (GCase) in peripheral blood cells and to analyze GBA1 mutations. DBS samples have several advantages, including ease of collection, minimal blood requirement, and straightforward transportation. Nevertheless, DBS has limitations when it comes to measuring GCase activity. Bone-marrow aspiration and biopsies to identify Gaucher cells are no longer considered diagnostic tools for GD and should only be done when assessing another hematologic comorbidity . Additionally, distinguishing Gaucher cells from similar cells observed in hematological diseases or infectious diseases (e.g., chronic myeloid leukemia, atypical mycobacteria, chronic myeloid leukemia, or myelodysplasia, etc.) may be difficult . Given this, we did not perform bone marrow aspiration or biopsies because we were confident in the diagnosis of GD, as there were no warning signs suggesting malignant or hematologic comorbidity.
Regarding plasma biomarkers, they have been utilized for a significant period of time to diagnose and monitor patients with GD. However, traditional biomarkers, such as ACE, ferritin, alkaline phosphatase, and high-density lipoprotein, are not unique to GD, and more specific biomarkers, such as chitotriosidase and CCL18, have limited usefulness. On the other hand, Glucosylsphingosine (lyso-Gb1), the deacylated form of glucocerebroside, has been identified as a potential biomarker with high sensitivity and specificity for the diagnosis of GD .
Pathological macrophages in GD release chitotriosidase. While higher plasma chitotriosidase activity is typically observed in type 1 GD patients compared to those with types 2 and 3, it is important to note that increased enzyme activity is not exclusive to GD. In fact, modest elevations in chitotriosidase activity can also be found in various lysosomal and non-lysosomal diseases (i.e., Niemann–Pick disease type C, sarcoidosis, arthritis, multiple sclerosis, thalassemia, and malaria ext.). CCL18 is a member of the C-C chemokine family, and like chitotriosidase, it accumulates in the alternatively activated macrophages present in Gaucher cells in GD. An increase in CCL18 can also be found in various lysosomal and non-lysosomal diseases, similar to chitotriosidase .
Lyso-Gb1 is involved in GD-associated bone pathology and chronic inflammation, making it a relevant biomarker. During GD treatment, lyso-Gb1 levels decline rapidly, like other biomarkers. Plasma levels of lyso-Gb1 are much higher in untreated GD patients than in healthy controls and correlate with hepatomegaly, splenomegaly, and platelet counts. Patients with neuronopathic GD have notably elevated levels of plasma lyso-Gb1 compared to those with non-neuronopathic GD . The reliability of diagnosing GD by measuring the lysoGb1 enzyme in DBS and subsequently performing genetic analysis has been demonstrated in a cross-sectional study. Combining lyso-Gb1 measurement with whole-gene sequencing allowed for a 100% accurate diagnosis of GD. This implies that it could potentially become the new standard for screening patients suspected of having GD and it may also serve as a useful tool for patient evaluation and monitoring .
We observed hepatomegaly in addition to splenomegaly, and MRI showed a hepatic focal lesion radiologically defined as gaucheroma. Gaucheroma is an uncommon disorder characterized by the formation of a “pseudotumor” consisting of a group of Gaucher cells, typically found in the liver, spleen, bone, and lymph nodes .
The prevalence of splenic and hepatic lesions in GD has been reported to range from 19% to 33% and 6% to 20%, respectively, across all age groups . Gaucheromas grow slowly and can appear in both adults and children. Hepatic gaucheromas should be distinguished from hepatocellular carcinoma or lymphoma. Hypodense lesions of the liver and spleen are the distinctive feature of gaucheroma on imaging . In GD1, it has been reported that splenectomy considerably increases the risk of liver gaucheroma. Although the risk of gaucheroma may be higher in cases where GD diagnosis and treatment are delayed, there is no clear association between the risk of gaucheroma and the severity of GD or the presence of lymphadenopathy or malignant alterations . In patients with GD, routine imaging should be performed to identify gaucheromas. If diagnosed with liver Gaucheroma, patients should be referred to a hepatology specialist and monitored regularly using MRI or CT for adults and ultrasound for pediatric patients. For spleen gaucheroma, patients should be re-evaluated after one year and then monitored every 2–3 years using ultrasound or MRI .
According to our knowledge and the literature available [10,11,12,13], this case represents the youngest patient reported as having gaucheroma and also the first one presenting with gaucheroma at the diagnosis and not during the follow-up. Since a biopsy of gaucheromas may pose a risk of further seeding and could be dangerous in this patient for the presence of massive splenomegaly, after consultation with a pediatric surgeon, we felt confident that liver biopsy could be avoided. Tumoral markers were negative, according to the radiologist, the lesion could be well characterized [11,14], and a genetic study confirmed GD diagnosis.
As previously mentioned, over 400 mutations have been identified in the GBA1 gene: c.1226A>G (N370S), c.1448T>C (L444P), c.84dup, c.115+1G>A (IVS2+1G>A) and RecNciI are the most common . The association between genotype and phenotype can assist in determining the risk level and clinical approach for individuals with inherited diseases. GBA genotype may be considered a significant contributor to GD phenotype. On the other hand, determining GBA gene variants may have a limited value in predicting organ involvement and disease severity. Patients with the same GBA genotype, and even siblings, may have different clinical manifestations. In addition, the GBA genotype does not seem to determine the response to ERT.
Despite these limitations, the GBA genotype can be useful to distinguish between the classic neuronopathic and non-neuronopathic forms . If the c.1226A>G (N370S) mutation is identified in homozygosity or heterozygosity, the risk of neurological involvement (GD2 or GD3) can be excluded. Patients with the N370S mutation may remain asymptomatic for a long time, while in the case of homozygosity for the L444P mutation, there is an increased risk for the development of neurological alterations (GD2 or GD3) . Patients homozygous for the rare c.1342G>C (D409H) mutation are at a higher risk of developing damage to their cardiac valves. Heterozygous or homozygous mutation in the GBA1 gene (i.e., c.1226A>G (N370S), c.1448T>C (L444P) ext.), gives an increased risk of developing Parkinson’s disease .
Homozygosity for c.1448T>C (p.Leu483Pro) found in our patient has been usually associated with GD3 [1,12,13], which has usually a slower neurological progression than GD2. Despite characteristic manifestations, there is currently no consensus for the diagnosis of this form other than neurological involvement in a patient with proven GD not explained by other causes [13,14]. Neurological signs may appear years after the onset of visceral manifestations. Indeed, even patients initially diagnosed with GD1 may be affected by GD3 .
Since our patient had no neurologic signs, both clinical and instrumental, it was not possible to define if she was affected by GD1 or GD3, and this will presumably become more evident during the monitoring of the disease.
In the diagnostic algorithm of GD, an eventual GD3 definition is of significative importance because ERT does not cross the blood–brain barrier at therapeutic levels, thus potentially having no impact on neurological symptoms and deterioration [15,16,17,18]. Different pharmacological trials regarding novel therapies for GD3 are ongoing, including gene therapy and molecular chaperones ; moreover, a trial on the safety and pharmacokinetics of eliglustat (SRT) in pediatric patients with GD1 and GD3 is underway and will hopefully provide newer and more feasible tools to treat these patients . Nevertheless, in pediatric patients with GD1 or GD3 presenting with symptoms, an immediate start of treatment is recommended [4,5]. ERT therapy for moderately affected patients starts at 30 U/kg per 2 weeks; the dosage can be increased if therapeutic goals are not met, and usually, patients with severe symptoms start at 60 U/kg . It is proven that ERT has a rapid positive effect on the most frequent manifestations, such as hepatosplenomegaly, anemia, and thrombocytopenia, thus preventing major complications in the short-term period . In the youngest patients, ERT also favors an adequate development of the bone, improving bone density and reducing osteonecrosis [24,25]. In GD, skeletal damages such as vertebral collapse and fractures are serious complications that can heavily impair children’s development, causing long-term disability.
Although individuals with the p.Leu483Pr genotype often show neurologic manifestations, cases of adults treated with ERT showing no neurological worsening up to the age of 50 years are reported in the literature . These adults were treated with ERT, and it is suspected that ERT reduces the general proinflammatory state seen in GD with an indirect effect on nervous system integrity .
Regarding gaucheroma, at the moment there is no clear indication in the literature on treatment regimen . Similar cases to the one reported in our patient have been described in older children and appeared during treatment with ERT [10,11]. Recently, the case of an adult with GD3 presenting a large mesenteric gaucheroma has been described : the patient was already receiving ERT, so SRT was implemented. Follow-up imaging studies after the beginning of combination therapy of SRT and ERT performed after 31 months showed a significant reduction of the mesenteric mass, and the patient reported an improvement in abdominal discomfort. Early start of SRT therapy in childhood might prevent the development of gaucheroma, and combination therapy could be the key to treating this pseudotumor, although data on SRT pediatric pharmacokinetics are awaited to establish appropriate SRT dosage for children.
4. Conclusions
This case shows that GD should be routinely included in the differential diagnosis of children presenting with splenomegaly and hepatomegaly, considering that the early start of ERT can change the disease progression, preventing serious complications. There is no clear evidence regarding the efficacy of ERT for gaucheroma, but hopefully, new effective and safe treatments will be available in the next few years.
Author Contributions
G.B. and L.P. wrote the first draft of the manuscript, participated in patient management, and performed literature review; L.B. performed the radiologic examinations; S.E. was in charge of patient management, gave a clinical and scientific contribution as well as critically revised the manuscript. All authors have read and agreed to the published version of the manuscript.
Institutional Review Board Statement
These case reports were approved by the Ethics Committee of AVEN on 28 February 2023 (PED-2023-01), and parents provided written informed consent for the performed evaluation.
Informed Consent Statement
The Ethics Committee of AVEN approved the publication of this case report (PED-2023-01), and parents provided written informed consent for the publication of this manuscript.
Data Availability Statement
All the data are included in the manuscript.
Conflicts of Interest
The authors declare no conflict of interest.
Funding Statement
This research received no external funding.
Footnotes
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Chapter 2 - Introduction to Electrodynamics | David Griffiths 4th Edition
by Physics Solutions PH
Playlist•29 videos•68,209 views
Some Solved Problems from Chapter 2 (Electrostatics) of Introduction to Electrodynamics by Griffiths 4th EditionSome Solved Problems from Chapter 2 (Electrostatics) of Introduction to Electrodynamics by Griffiths 4th Edition
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189955 | https://www.files.ethz.ch/isn/20287/01.2003.pdf | Cluster of Competence The rehabilitation of war-torn societies A Project co-ordinated by the Centre for Applied Studies in International Negotiations (CASIN) ADMINISTRATION AND GOVERNANCE IN KOSOVO: LESSONS LEARNED AND LESSONS TO BE LEARNED Robert Muharremi, Lulzim Peci, Leon Malazogu Verena Knaus and Teuta Murati; Editor: Isa Blumi Pristina/Geneva, January 2003 This study was prepared by Robert Muharremi, Lulzim Peci, Leon Malazogu, Verena Knaus and Teuta Murati, and edited by Isa Blumi under the auspices of the Kosovar Institute for Policy Research and Development (KIPRED). KIPRED aims to support and promote democratic values in Kosova by offering trainings, conducting research and independent analysis, in order to help policymakers develop professional public policy. The Cluster of competence Rehabilitation of war-torn societies is a project of the Swiss Interdepartmental Co-ordination Committee for Partnership for Peace which is part of the activities of Switzerland in the Partnership for Peace. This Cluster is co-ordinated by Jean F. Freymond, Director of the Centre for Applied Studies in International Negotiations (CASIN). The opinions expressed in this paper only reflect those of the authors and not of the institutions to which they are or were affiliated. CASIN Mission Statement The Centre for Applied Studies in International Negotiations (CASIN), established in 1979 as a Swiss non-profit independent foundation, aims at • Training leaders from governments, business and civil society in governance, diplomacy, negotiation and conflict management, as well as in development and trade issues; • Problem Solving and Facilitation through dialogues in areas such as national and international governance, sustainable development, international trade and globalisation along with good offices in case of conflict; • Research and Coaching to assist policy-makers, negotiators and senior managers in their search for policy options in relation to the smoother functioning of society and the international system. CASIN strives to develop new approaches to governance such as a capacity to anticipate and identify long-term implications, to grasp complex and widening contexts, to perceive and understand the points of view of other parties, and to develop innovative policies and strategies for mutual gain. In other words, CASIN seeks to sharpen negotiators’ and decision-makers’ analytical, interactive, and joint problem solving skills, as well as to broaden their comprehension of the global environment in which they operate and their understanding of the increasingly complex and inter-related issues on the international agenda. In pursuit of these aims CASIN organises: professional training programmes, issue and policy dialogues, informal negotiation sessions, conflict management related activities and carries on studies and applied research under four thematic programmes: Programme on Governance Programme on the Management of Interdependence Programme on the Management of the Global Commons Programme on NGOs and Civil Society For further information on CASIN Centre for Applied Studies in International Negotiations (CASIN), Avenue de la Paix 7 bis, Boite postale 1340, 1211 Geneva 1, Switzerland, Telephone: +41 (0) 22 730 86 60 - Telefax: + 41 (0) 22 730 86 90 - e.mail: casinfo@casin.ch - Abstract This paper aims to study the dynamics of the post-war administration of Kosovo by the international community. Such a study will use the experiences of the last three years in order to formulate new research questions that may enhance the ability of policy-makers to make knowledgeable decisions. By assessing the UN Mission in Kosovo and its implementation of both civilian and military components of its mandate, this paper identifies key successes and failures related to the administrative dynamics post-war Kosovo. The paper is organized in four different sections, each one addressing a particular issue related to UNMIK’s performance and drawing the reader’s attention to possible lessons that may be learned from the experience. In the first section, we concentrate on key aspects of Kosovo’s recent history and its place in the regional and international context. In this framework, we explore the legal foundations of the current administration of Kosovo and the performance of the international community (IC) in transferring competencies/responsibilities to newly established local structures. This is followed by a section that examines the performance of three security agencies -- the international military force led by NATO, the international civilian police and the local police service. We then analyze interethnic relations after the war and evaluate the efforts of various international and local actors in promoting reconciliation, dealing with property issues that affect community relations and facilitating the return of displaced populations. Finally, the paper looks at the divided city of Mitrovica as a case study, testing the saliency of issues of international engagement in local levels of administration in post-war Kosovo. The progress and achievements in post-war Kosovo could not be imagined without an international presence. Despite policy efforts to hand over authority to local structures in some sectors, international presence remains a necessity. All the salient features of post-war and transitional countries in the Balkans have been attributed to Kosovo. As many in the IC claim, Kosovo is being rebuilt from scratch, while undergoing a transition to a market economy. It is administered by an international mission, which at times is confused about its role and vision. Adding to the complication is the external status of Kosovo, itself a taboo topic of discussion. Public life is being built from the bottom-up, starting from municipal structures to central structures. The UNSC Resolution 1244 that mandated the UN Mission in Kosovo was as good as the political compromise that gave rise to it. Unlike many resolutions, 1244 has had quite an impact on the situation on the ground. Ambiguous aspects in the resolution led to conflicting interpretations by different stakeholders, such as Kosovo’s legitimate political representatives, UNMIK, and Serb political entities. It is in these conflicting interpretations and their expectations that we ultimately conclude that 1244 was not necessarily the best mechanism to enable rebuilding Kosovo. The expectations of the international community may at times be too high. The rehabilitation process is a long one. It needs greater synergy among international and local agencies in order to consider past and recent history, culture and traditions and build viable, sustainable structures of democratic decision and policymaking. Thus, peace-building processes will be better assessed and will lead to the higher quality of administration and governance in conflict-ridden areas. Table of Content Executive summary 2 Introduction 4 Short historical background of Kosovo and UNMIK 4 AN OVERALL ASSESSMENT 8 LEGAL FRAMEWORK 8 Applicable Law: Theory and Praxis 8 UNMIK legislation between civil law and common law 10 Legislation of the Provisional Institutions of Self-Government 11 a. Reserved powers and transferred responsibilities 11 b. Harmonization of Kosovar Legislation with the Acquis Communautaire 12 c. Some Aspects of the Current Legislative Practices 13 Policy Recommendations 14 INTERNAL SECURITY MANAGEMENT IN KOSOVO 14 Introduction 14 Deployment and Structuring of International Military Force and Civil Police 15 Enforcement of Law and Order 17 Building sustainability: Kosovo Police Service (KPS) 20 Policy Recommendations 22 INTERETHNIC RELATIONS IN KOSOVO: LOOKING AHEAD 22 Introduction 22 Relations with non-Serb communities 23 Sustainable Returns 24 Kosovar institutions and the overall climate 26 Parallel Institutions 27 Reconstruction and Economic Opportunities 28 Property Management 29 Group right and minority overrepresentation 30 Good practices 32 Lessons Learned for future missions 32 Policy Recommendations 33 THE EUROPEAN UNION’S FIRST MISSION 34 Reviving the ghosts 35 The unknown private sector 37 Property is key 38 Property creation is peacekeeping 40 From reconstruction to development 41 Managing ignorance 41 REHABILITATION OF MITROVICA 42 Security 43 Administration 46 Local Serbs and Belgrade 47 The Kosovo Albanians 49 Recommendations 50 CONCLUSION 50 Appendix: Abbreviations and Language Clarifications 52 About KIPRED 53 Endnotes 54 Introduction This paper identifies several key issues in administering and governing post-war Kosovo. It explains how various stakeholders have been dealing with these issues, drawing lessons from the interaction between locals and the IC and thus inform our policy recommendations for the mission in Kosovo. Through this process, we hope our readers will draw lessons for similar missions that are bound to emerge in the near future. After an overall assessment of the general issues, the paper looks at several areas of governance and administration by studying the performance of the international administration as well as at the transfer of competencies in this regard to local actors. The first section delves into the question of the legal framework of Kosovo. The second section looks at the salient issue of security while analyzing the performances of the international military presence led by NATO, the international civilian police and the local police service. An analysis of interethnic relations after the war and institutional behavior in this regard follows in the third section. The fourth section looks at the postwar economic development and reconstruction. Finally, the paper looks at the specific case of the divided city of Mitrovica, the most difficult issue in the IC’s administration of post-war Kosovo. The main stakeholders to whom the paper is directed are: the international community present in Kosovo (UNMIK, OSCE, UNHCR, EUMIK and foreign diplomatic and liaison offices), the Kosovar government structures at all levels and political parties. However, the lessons drawn for future missions are in turn directed to the academic and peacekeeping communities at large whether involved with Kosovo or not. The paper makes an attempt to analytically describe the situation while leaving it up to the reader to what degree the situation constitutes a success or not. Short historical background of Kosovo and UNMIK Kosovo was a province of former Yugoslavia with its status upgraded to an autonomous federal entity since 1974. In the late eighties, Kosovo’s autonomy was revoked by the Republic of Serbia. Kosovar Albanians (estimated to make up 90% of around the two million inhabitants of the prewar population) declared independence and organized their daily lives and parallel institutions around a peaceful resistance movement led by the Democratic League of Kosovo. Kosovo remained under a Serbian repressive security regime and unrest gradually increased, culminating in armed warfare in 1998 and with the forceful intervention of NATO in 1999. After the war, Kosovo has been administrated by a civilian mission, headed by the United Nations with the military component of the NATO-led Kosovo Force. This mission aimed to administer Kosovo without prejudging its external status, is defined by the UN Security Council Resolution 1244,1 and described in greater detail in the first section. For the first time in history, the UN was given an unprecedented mandate, both in scope and structural complexity, to replace the role of the state. Resolution 1244 gave rise to UNMIK, and called upon it to: perform basic civilian administrative functions, promote the establishment of substantial autonomy and self-government in Kosovo, facilitate a political process to determine Kosovo’s future status, coordinate humanitarian and disaster relief of all international agencies, support the reconstruction of key infrastructure, maintain civil law and order, promote human rights and assure the safe and unimpeded return of all refugees and displaced persons to their home in Kosovo. The operational framework of UNMIK has been divided into four pillars led by various international agencies that for the first time act as part of a government and enjoy a high degree of autonomy in creating and implementing policy. The humanitarian pillar led by the 8 Administration and Governance in Kosovo ___________ UNHCR, was phased out in June 2000. In the fall of 2002, the pillars were: (i) Police and Justice, under the direct leadership of the UN; (ii) Civil Administration (UN); (iii) Democratization and Institution Building (OSCE); (iv) Reconstruction and Economic Development (EU). AN OVERALL ASSESSMENT Although the international administration of territories has historical precedence (e.g. Namibia, Cambodia, Eastern Slavonia, West Irian), the case of Kosovo along with a number of other cases in the nineties (such as Bosnia and Herzegovina and East Timor) presents a new typology for scholars in this field. The literature so far has been constrained to internal productions by the UN, and it has been only recently that relevant independent research has been made in this direction. Richard Caplan produced a concise analysis in this regard, highlighting a number of issues that are related to such international missions.2 This paper treats several salient issues regarding Kosovo specifically. Legitimacy is one of the key factors that an international mission needs to assert before moving into an area. UNMIK had legitimacy, although far from the level that was enjoyed by the NATO-led Kosovo force. In general, Kosovars shared the sentiment that Kosovo needed the security of the peacekeepers and not an omnipresent tutorship at all levels and spheres of public life. Nevertheless, all Kosovars and the two main leaders, Rugova and Thaçi, who publicly renounced the governing structures that they had created, either during the resistance or in the immediate aftermath of the war, accepted the mission. This was not the case with the Serb leadership in northern Mitrovica who have not renounced Belgrade leadership. Despite the challenges, the UN was the only viable organization that claimed its neutrality and multilateral enough to be acceptable to all concerned parties. Regarding the role of third parties in general, Stephen Stedman highlights the role of the “custodian” and the importance of gaining the acceptance from all parties concerned. “An essential prerequisite for successful peace processes is acceptable ‘custodians,’” described by Stephen Stedman as “international actors whose task is to oversee the implementation of the peace processes.”3 One of the duties of UNMIK, and arguably the most important one in the long run, has been to bring the territory to a level where the questions surrounding its external political status can be addressed. Whether it is an issue of the intractability of the conflict or that UNMIK did not offer the good offices that it was supposed to, it can be argued that while UNMIK has been well accepted, three and a half years after the conflict, it never managed to assume the role of a proper “custodian.” One of the major conditions for a successful mission is a clear vision of the administered territory’s final political and economic status. “No international administration can function without having a political vision, implied or stated, for the society it is administering.”4 In the case of East Timor, Eastern Slavonia and Bosnia, the final goal was more or less clear. The unresolved status of Kosovo, on the other hand, continuously politicizes the peace process and fuels sentiment for radical movements. The taboo of this topic was broken in late 2002 with a regional roundtable organized by an independent American NGO quickly followed up by various initiatives. This issue continues to divide communities, breed fears and make peace-building efforts difficult. Perceiving a crisis, communities continue to withdraw within themselves, branding the UN as biased, furthering social cohesion and homogenization of each group with a psychological sense that some of the dynamics of the 1999 conflict still remain. An important issue related to the completion of the mission is that of the level of involvement of various actors. This involvement and the nature of its implementation is predicated on (a) the organizational structure of the UN, (b) the nature of the mission and (c) staffing possibilities. Using its mandate, the UN has deeply permeated Kosovar society. The nature of this interaction, however, is not immune from criticism. Indeed, one of the first complaints about the mission is the lack of knowledge about Kosovo among those charged with 10 Administration and Governance in Kosovo __________ administrating the region. Unlike the British colonial strategy of trying to understand the community that it governed, the UN mission did little to elicit input from the community or from the local leadership, while at the same time enjoying absolute immunity from legal and political accountability. A lack of devolution of authority to the local structures has rendered all levels of government dependent on international structures. This has created increasing dependency on the UN. As Caplan rightly observed: “Early devolution also helps to prevent the administrative equivalent of aid dependency, in which the local population becomes accustomed to international representatives making decisions for them, including some of the harder decisions that they can thus choose to ignore.”5 This dependency has encouraged a culture of passivity. Local authorities were stripped of any decision-making responsibility or accountability. There is an argument to be made here that local authorities should have been allowed to make blunders even at the expense of their own community as freely elected representatives. The UN’s position was that they should be tutored for a longer period of time under the argument that they are not mature or tolerant enough. The argument regarding the level of involvement also extends to almost dictatorial authorities of the Special Representative of the Secretary General (SRSG). Although the use of an all-powerful representative has been used far less than in Bosnia, where radicals have been winning elections continuously, heavy criticism has been received even from internationals such as the Ombudsperson of Kosovo. In violation to the European Charter on Human Rights, applicable in Kosovo, numerous people have been kept in detention with a special order by the SRSG for months and released later for the lack of evidence. Instead of serving as a model for democracy in a place where the latter never existed, the powers of the SRSG have greatly damaged the integrity and the credibility of the UN administration. The extra judicial powers of the international community have been often perceived by the Kosovar leadership and community as political in nature. The perception among Kosovars is that the international administration was often politically oriented and one-sided regarding its influence. While the Serb community has often accused UNMIK of being pro-Albanian and doing little for the return of displaced Serbs, Albanians have often accused it of aiming to marginalize selected political forces and pushing Kosovo towards Serbia and blocking the functioning of its institutions under the disguise of political sensitivity. Although the role of international administration in Kosovo has been given the duty to prepare the region for resolving the final status, it is not clear whether there is any implicit role in its influencing the overall political outcome. Preparing the region for the final status and not giving it a political direction only leaves one path open for the near future: help the establishment of local governing institutions at all levels and build up the economy. A faster transfer of competencies is the key and is a natural development of such an interim mission. There were different opinions among the international community regarding their attitude towards the existing local structures as illegitimate, inexistent, inadequate, or undemocratic. Despite the lack of a vision and the political will to resolve it, the primary objective “of a transitional regime must be to empower the local population to manage its own affairs.”6 Due to the political vision of the administration, this was considered problematic by many as the mere build-up of institutions was considered that it would lead to a finished act where Kosovars would be capable of running their own affairs and would thus no longer be as vulnerable as before to international pressure. The lack of local involvement has stripped the Kosovars of formal and informal leadership responsibility. The international administration reports to the Security Council and to Administration and Governance in Kosovo 11 ___________ member states only, giving no authority to local leaders, reproducing the lack of participation and accountability that Kosovars experienced for many years under Yugoslav rule. This was manifested in the latest elections, a useful barometer of the expectations of the Kosovar society regarding its own power for change. Despite the widespread euphoria in Kosovo after the war, the electoral turnout has fallen three years after the war to less than 50%. Such a lack of voter participation is mainly blamed on Kosovar perceptions that the local structures have no responsibility or competencies. As Darby claims, “…in a divided society, internal responsibility for security instantly brands it as partial, whereas external, multinational bodies can serve as convenient scapegoats for unpopular policies.”7 Kosovar politicians often find it easy to blame failures on UNMIK or to let them take the politically sensitive decisions. The record low electoral turnout in the Fall of 2002 is partly blamed on the low level of authority and competence by the local political elite. This accountability has also been partially reduced at the other end, by a great division between the political appointees and of the well-cushioned administration. Kosovar citizens, who are not even used to asking for transparency and accountability often express delusion at what they see as the first model of “democracy” up close. Not only has this rendered local politicians unaccountable (though powerless too), but it has also strained the relations with representatives of the local communities, who were often given free hands to devise policies at local levels as they wished, resulting, for example, in different ways of managing municipalities depending on the skill, the experience and the tradition from which the international municipal administrator came. Overall, Kosovars accepted UNMIK but never really felt it theirs. They saw it as political, secretive and unaccountable to the local community in any way. “The idea of international rule over a foreign territory can be legitimate only if that rule is exercised on behalf of, and for the benefit of, the foreign population.“8 If one goes by this definition, UNMIK’s legitimacy is in serious doubt, partially due to its imposed mandate from the top, but also due to the way it has been implemented. A number of Kosovars have even started to brand some features of UNMIK as neo-colonial. Operationally, one of the issues mentioned in the sections below is the staffing politics of international agencies who are often unable to recruit the required number and quality of people from member states, who either to do not have excess judges, policemen to send to a country or that their internal mechanisms make it difficult to do so. One of the main criticisms to the civilian component of peacekeeping was the lack of planning. Unlike the military component that had a major proponent behind, NATO in general, and Britain and the US in particular, the UN found it politically sensitive to make any plans until very late in the process. Not only had UNMIK failed to plan before it began its mission, “the first UNMIK strategic planning document was not produced until 5 Dec 1999, six months after the start of the Mission.”9 Caplan also concludes that the “organization is seldom prepared for the worst, missions are planned hurriedly, and support for them is insufficient long after their initial deployment.”10 As the international system is composed of states as basic units, there is a natural tendency to move against this trend as a threat to the state system. Hence, there are current impediments that stem from: (a) political circumstances, (b) the organizational structure and the lack of interdepartmental planning and (c) the inability to coordinate with local structures on the ground. Perhaps one of the most important recommendations that this paper makes is that the UN ought to plan well ahead even for crises that might never erupt. Standards for international administration of disputed war-torn territories such as Kosovo, and of failed states should be 12 Administration and Governance in Kosovo ___________ developed just like for military and humanitarian operations. If early planning is not possible, constant research and intelligence gathering on areas of conflict remains an option. The DPA and the DPKO should establish a joint office that does research on economy, policing, law and a host of other issues about a number of crisis situations in the world. LEGAL FRAMEWORK OF KOSOVO UNDER THE UNITED NATIONS’ INTERIM ADMINISTRATION Applicable Law: Theory and Praxis By virtue of UNSC Resolution 1244 (1999) the Security Council of the United Nations authorized the Secretary-General to establish an international civil presence known as the United Nations Interim Administration Mission in Kosovo (UNMIK) with the mandate “to provide an interim administration for Kosovo under which the people of Kosovo can enjoy substantial autonomy within the Federal Republic of Yugoslavia and to provide transitional administration while establishing and overseeing the development of provisional democratic self-governing institutions to ensure conditions for a peaceful and normal life for all inhabitants of Kosovo.”11 The resolution is the result of diplomatic efforts before and during NATO’s air campaign against Yugoslavia (24 March to 10 June 1999). Despite the failure to reach a peaceful settlement at Rambouillet, the military offensive of the Yugoslav army against the Kosovo Liberation Army (which led to the displacement of approximately half of the population of Kosovo), and the ensuing NATO air campaign, diplomatic efforts resulted in a political solution to the conflict that avoided a hostile intervention of NATO ground troops in Kosovo.12 In early June 1999, the envoy of the European Union, Mr. Martti Ahtisaari, and the envoy of the President of the Russian Federation, Mr. Victor Chernomyrdin, submitted a plan to the Yugoslav Government for the peaceful settlement of the conflict.13 The peace-plan was prepared by the G-8 Foreign Ministers on 6 May 1999 and envisaged the establishment of an interim administration for Kosovo under the auspices of the United Nations. On 2 June 1999, the Yugoslav Government accepted the peace-plan and also agreed to the establishment of the international civil and security presences of the United Nations in Kosovo. The resolution incorporated the peace-plan and established UNMIK as a peacekeeping mission14 to provide an interim administration for Kosovo, pending a political process and settlement in regards to the final status of Kosovo. Although not expressly stated in the resolution, the mandate to provide an interim administration for the people of Kosovo implies the authority to legislate. With reference to the resolution, the Special Representative of the Secretary-General (SRSG) in Kosovo stipulated in the very first UNMIK legislative act, namely UNMIK Regulation No. 1999/1 of 25 July 1999, that all legislative and executive authority with respect to Kosovo, including the administration of the judiciary, is vested in UNMIK and is exercised by the SRSG.15 In the performance of the duties vested in UNMIK, the SRSG would issue legislative acts in the form of regulations.16 However, when the United Nations entered Kosovo, Kosovo was not a “law-free-area.” The Security Council preserved the sovereignty of the Federal Republic of Yugoslavia over Kosovo and with respect to that sovereignty, UNMIK provided that the Yugoslav laws applicable in the territory of Kosovo prior to 24 March 1999, shall continue to apply insofar as they do not conflict with internationally recognized human rights standards, the fulfilment of the mandate given to UNMIK or any regulation issued by UNMIK.17 The local legislation includes legislation of the Yugoslav Federation, of the Republic of Serbia and of the Province of Kosovo. 14 Administration and Governance in Kosovo __________ A precise rule on conflicts between local legislation and UNMIK legislation was introduced a few months later through UNMIK Regulation No. 1999/24 of 12 December 1999 on the Law Applicable in Kosovo. Thereby, the SRSG determined that in case of a conflict, the regulations and subsidiary instruments issued by UNMIK shall take precedence with respect to the local legislation in force.18 The real need for the issuance of said regulation was, however, caused by the determined refusal of the local judges and public prosecutors to apply the local legislation, which was in force on 24 March 1999. They considered this legislation as discriminatory against the Kosovar Albanian population of Kosovo since it was issued after the elimination of the autonomy status of Kosovo and thus without the participation of Kosovo in the respective legislative procedures.19 The local legal community started applying the legislation, which was in force before the suspension of the autonomy status of Kosovo, in particular in criminal matters. In order to avoid a discrepancy between the “law in books” and “the law in action” the SRSG decided to codify the described legal praxis determining that besides UNMIK legal acts the applicable legislation in Kosovo should also include the law in force on 22 March 1989.20 Only if a situation was not covered by said legislation, laws in force after 22 March 1989 could be applied insofar as they were not discriminatory or in violation of internationally recognized human rights standards.21 This amendment as regards the applicable local legislation demonstrates the insufficient inclusion of the local legal community when UNMIK decided on the applicable legislation. From a theoretical point of view the definition of the applicable legislation seemed to be clear. But from a practical point of view it was not clear at all. After the rapid deployment of the international civil presence in Kosovo, international staff members from various countries without prior training or adequate experience in providing interim administrative services under the auspices of the United Nations were confronted with the necessity to restore public order and safety, including the combating of criminal structures. In that context many persons were arrested and held in pre-trial detention. However, at this stage there was absolutely no knowledge about the applicable legislation in Kosovo on the basis of which a detention could be ordered and executed. Police officers from various countries with different legal background applied the legal principles and provisions that were used in their countries. This situation did not improve much in the subsequent months of the Mission. International lawyers, mainly consultants brought to Kosovo to prepare a draft piece of legislation within a relatively short period of time, did not pay sufficient attention to the local legislation and thus prepared legislation that would hardly fit into the existing legal system. The major reason for this behaviour lay in the insufficient information on the applicable local legislation due to the difficult access of official sources on such legislation. Official Gazettes, which contained the relevant laws, were either lost or destroyed during the armed conflict. In addition to this, the lack of proficiency in the Albanian or Serbian language was a major obstacle for international lawyers to gain access to the local legislation. Even today, three years after the deployment of the Mission, there is still no database on the local legislation, although institutions such as the Kosovo Law Center, the former Administrative Department of Public Services and the OSCE improved the level of information through collections of laws and other publications.22 UNMIK legislation between common law and civil law Between June 1999 and September 2002 UNMIK passed 144 regulations and 86 administrative directions covering all areas of law, such as criminal law, civil/commercial law and public/administrative law.23 This legislative activity has had an enormous impact on the legal system of Kosovo, in particular with respect to the civil law system. Despite all Administration and Governance in Kosovo 15 ___________ socialist elements, the legal system of Kosovo as part of the overall Yugoslav legal system is to be classified as belonging to the continental civil law system based on Roman Law. The civil legislation included among other specific legislation a Law on Obligations,24 a Law on Property Relations,25 a Law on Civil Procedure,26 thus reflecting a classical civil law system. With the involvement of a number of international lawyers with a common law background in the legislative activities of UNMIK, and taking also into consideration the aforementioned lack of information on the legal system in Kosovo, UNMIK enacted legislation in many areas, which was evidently based on common law principles, such as the regulation on business organizations or the regulation on pledges. In the field of contract law, UNMIK simply transformed the UN Convention on the International Sale of Goods into domestic contract law without realizing the need for harmonizing this piece of legislation with the remaining legal system.27 This approach did not only introduce common law terminology unknown to local lawyers trained in continental civil law (e.g. the term “collateral” in the context of pledges and mortgages) but also caused legal incoherence and contradictions, which in the end led to legal uncertainty. This result was also supported by the legislative policy pursued by UNMIK, and which consisted in the singular regulation of specific matters (such as pledges) and avoiding the general and final regulation of a matter (such as regulating pledges in the context of a general Property Law).28 This legislative policy is understandable when taking into account that UNMIK considered itself to be only a transitional administration and that it therefore did not want to undertake a long-term and complex legislative operation, such as drafting a new Civil Code for Kosovo. Legislation of the Provisional Institutions of Self-Government a. Reserved powers and transferred responsibilities On 15 May 2001, UNMIK promulgated the Constitutional Framework for Provisional Self-Government (PISG) in Kosovo drafted by international and local experts thereby setting up provisional governmental institutions, including the Assembly of Kosovo, the Government and the Office of the President of Kosovo (UNMIK Regulation No. 2001/9). However, the authority of the SRSG over the PISG is expressed in chapter 8.1 of the Constitutional Framework, which authorizes the SRSG to dissolve the Assembly of Kosovo and to call for new elections in circumstances where the PISG are deemed to act in a manner, which is not compatible with UNSC Resolution 1244 (1999).29 In addition, the Constitutional Framework expressly states that the exercise of the responsibilities of the PISG under the Constitutional Framework does not diminish or affect the authority of the SRSG to ensure full implementation of UNSC Resolution 1244 (1999), including overseeing the PISG, its officials and its agencies.30 With the Constitutional Framework, the SRSG vested certain responsibilities in the PISG with respect to preparing laws, for example in the field of education, youth, culture, health, labor and social welfare, transport and telecommunications, public administration, agriculture, spatial planning and environmental protection, domestic and foreign trade and industry, statistics, tourism as well as good governance and human rights.31 In these fields both the Government as well as the Assembly have the right to initiate and to adopt laws and resolutions, the latter being non-binding declarations.32 Once the legislation has been adopted by the Assembly as the highest legislative body of the PISG it only enters into force if signed and thus promulgated by the SRSG.33 A high UNMIK official describes this sort of legislative competence of the PISG as the exercise of legislative responsibility for and under 16 Administration and Governance in Kosovo ___________ the authority of the SRSG.34 The same principle applies to international agreements, which may be prepared by the PISG as an exercise if the subject matter falls under its responsibilities as referred to in chapter 5 of the Constitutional Framework.35 On the other side, the SRSG retains some so-called reserved powers, where the SRSG has the sole and exclusive legislative power. These fields cover mostly, among others, sovereignty-related matters, such as external relations, the administration of public, state and socially owned property and enterprises, railways and civil aviation, control over the civil registry database, as well as matters relating to the appointment and removal of judges and public prosecutors, the protection of the rights of communities and co-operation with KFOR, the enforcement of public safety and order, defense, civil emergency and security preparedness.36 This differentiation between legislative responsibilities transferred to the PISG and reserved powers of the SRSG, as well as the vague and extremely “open texture” of the relevant provisions in the Constitutional Framework concerning which area is transferred and which is not, proved to become one of the major sources of conflict between the PISG and UNMIK regarding the implementation of the Constitutional Framework. The example of the 23 February 2001 bilateral agreement on the determination of the border between the Federal Republic of Yugoslavia (FRY) and the Former Yugoslav Republic of Macedonia (FYROM) may illustrate the dimensions of the problem In order to avoid future disputes of this nature, UNMIK and the PISG have established - at least for economic, fiscal and commercial legislation - under the umbrella of the Economic and Fiscal Council the first rudimentary principles and procedures for clarifying which subject-matters fall under reserved/transferred responsibilities, and for co-operation in legislative issues. Although this matter is the crucial issue in the future implementation of the Constitutional Framework and the further transfer of responsibilities to the PISG, legal clarity and effective co-ordination are, as of yet, far from being reached. Case Study: Borderline The borderline in questions was, however, also the borderline between Kosovo and FYROM. The result of the agreement was that approximately 2400 acres of land belonging to Kosovo were transferred to FYROM. The PISG considered this agreement to be in violation of the territorial integrity of Kosovo, which was protected by Resolution 1244 and chapter 1.2 of the Constitutional Framework.1 Since the Security Council accepted the border agreement between FRY and FYROM, the Assembly of Kosovo adopted, on 23 May 2002, a resolution by which it did not recognize the border agreement and declared its effects to be null and void with respect to Kosovo (Assembly of Kosovo 2002). Its main argument was that the people of Kosovo were not consulted on this issue and that the borders of Kosovo could not be changed without the consent of the people of Kosovo, arguments deriving from the internationally recognized principle of internal self-determination of peoples. In an immediate reaction to the adoption of the resolution by the Assembly, the SRSG declared the resolution to be null and void. The SRSG’s determination was based on the argument that the PISG is only entitled to pass a resolution if the subject matter in question falls under chapter 5 of the Constitutional Framework. Since “territorial integrity” was not enlisted in chapter 5 but was referred to in chapter 2, the PISG formally did not have the authority to issue any resolution on this matter. The outcome of this dispute was legalistic formality versus internal self-determination. The situation has remained as it was: the border agreement has not been implemented, yet, the PISG sticks to the resolution of the Assembly and UNMIK considers the resolution of the Assembly as nonexistent. Administration and Governance in Kosovo 17 ___________ b. Harmonization of Kosovar Legislation with the Acquis Communautaire Another aspect, which characterizes the Constitutional Framework with respect to legislation, is the commitment of the PISG to closer European integration. Pursuant to chapter 5.7 of the Constitutional Framework, the PISG are responsible for aligning their legislation and practices in all areas of responsibility with relevant European and international standards and norms. This provision contains a clear quasi-constitutional obligation to harmonize the Kosovar legislation with the acquis communautaire of the European Union. With this provision, the Constitutional Framework of Kosovo has become the most developed constitutional text in the Balkan Peninsula as it regards commitment to European integration. In practice, however, the PISG lacks the necessary capacity and resources to indeed be able to undertake the required harmonization. In that context, the European Union has offered its support by providing legislative and administrative assistance through various institutions such as the European Agency for Reconstruction, UNMIK EU Pillar IV and the newly established EU Office in Kosovo, although a long-term, streamlined and coordinated approach is still missing. The major problem that is likely to arise in the context of harmonization with EU standards, is the underdeveloped capacity and the lack of sufficient resources on the Kosovar side to undertake legislative harmonization on their own, without the need to refer to sources outside of Kosovo. Capacity building in this field is of utmost importance for Kosovo in order to support a self-sustaining process for integration into EU structures. c. Some Aspects of the Current Legislative Practices Since the PISG are recently established institutions it is difficult to identify legislative practices and standards regarding the drafting and processing of legislation. However, in the first months after their establishment it can be observed that the legislative initiative is with the Government rather than with the Assembly. In fact, the Assembly takes a very passive role in this respect, criticizing the Government for not producing fast enough draft legislation but lacking in making any attempt to exercise its own right of legislative initiative. Another aspect worth being illuminated is that of international legal advisers funded by various foreign agencies who are involved in the drafting and processing procedure within both the Government and the Assembly and who have their main focus on commercial and economic legislation. The legislation in this area is drafted mainly by those international advisers who offered the final product with little or even no participation of the local lawyers, employed in the PISG and without providing any capacity building for local lawyers, in order to become capable to draft such legislation on their own. A third characteristic to be considered is that of a large number of international non-governmental organizations (NGO’s) and agencies engaged in the administrative structures of the Assembly through “capacity-building-programs” for the civil service of the Assembly and the members of the Assembly. Lectures provided by international experts and consultants, study-trips to foreign legislative institutions, as well as advisory services are the major elements of such programs. On one side, these programs help both the civil servants and the members of the Assembly to become familiar with the functioning of an Assembly. On the other side, the programs often are not tailored to cope with the specific circumstances and needs of the Kosovo Assembly as such. Up until now, there also is no sustainable capacity-building in the field of human resources, administrative management and technical infrastructure. The most dangerous side-effect of such an intense engagement of foreign NGO’s and agencies in the day-to-day business of the Assembly (some NGO’s may even participate in the proceedings of Assembly committees), 18 Administration and Governance in Kosovo __________ however, is that of their uncontrolled and direct possibility of influencing the members of the Assembly during the political decision-making process. This situation bears in itself the risk that it might not be the electorate that the members of the Assembly are listening to but rather financially potent international NGO’s and agencies, a situation, which is questionable from a democracy point of view. 1.4. Policy Recommendations The establishment of the United Nations Mission in East Timor a few months after the establishment of UNMIK shows that both Missions have an almost identical organizational structure, with the same extensive legislative powers, policies and techniques and which could serve as a model for future interim administration missions of the United Nations.37 The legislative experience gained in Kosovo could very well serve as a “lessons-learned module” for future Missions. (a) The major problem that has been identified in this short analysis is the insufficient knowledge about the local legal system. It can hardly be imagined that the United Nations would have the administrative capacity to establish databases of laws for each and every country and to maintain units of trained and experienced lawyers to be “parachuted” into a country where a UN interim administration is conducted. The preferable solution would be a United Nations Criminal Code and a United Nations Administrative Code, which could be applied whenever the United Nations is supposed to conduct the administration of a territory. (b) The conflict-of-laws rule as developed by UNMIK in relation to the local legislation could very well apply as a general rule for determining the relationship between the UN Codes and the respective local laws. Bearing in mind that such UN Codes might face the risk of not being fully compatible with the domestic legislation, they would certainly avoid legal uncertainty, which inevitably exists in the context of any UN interim administration. Peacekeepers could be trained in these codes and by applying them properly, they would ensure an internationally accepted standard of rule of law. (c) Whenever the United Nations have the mandate to legislate in a country, the local legal traditions and systems should be respected. A confusing mixture between common and civil law elements, as exists in Kosovo, should not be repeated. In a country where civil law traditions dominates, the UN should deploy lawyers trained in civil law and in common law countries lawyers with a common law background. (d) In general, the United Nations have to reconsider their peacekeeping approach and shift from an originally military undertaking to a more administrative approach. The future peacekeepers will need to have the profile of public administrators, accountants and lawyers rather than that of soldiers. It is now, that the United Nations needs to start making serious changes in their approach and organization in order to be able to cope with the future challenges of peacekeeping. INTERNAL SECURITY MANAGEMENT IN KOSOVO Introduction This section aims to analyze the international community’s management of Kosovo’s internal security. Its primary strategy towards stabilizing Kosovo and ensuring long-term stability was to build and develop the capacities of local authorities and ultimately transfer its security structures to them. The analysis will focus on three main stakeholders in this process, which ostensibly is the “transfer of competencies” to local Kosovar institutions. The first and perhaps most important security component to the current state of Kosovo is the NATO led Kosovo Force (KFOR). A second key component to the IC effort to enforce stability in Kosovo is the police force administered by the United Nations mission in Kosovo. The last component is the Kosovo Police Service (KPS), a local police force trained by OSCE. Within this triad remain a number of issues in how the considerable resources available to the international community may have affected the performance of these security institutions. The legal framework to the IC’s management of Kosovo’s security has been outlined in UNSC Resolution 1244 (1999), citing Chapter VII of the UN Charter. The resolution identifies the international community as the sole legitimate entity with a mandate to law enforcement and the use of power in the region. The enforcement of peace, as stipulated by the resolution was granted solely to KFOR, whereas KFOR and UNMIK Police (CIVPOL) were given a mandate to exercise law enforcement jointly. A second mandate granted to UNMIK Police is developing the Kosovo Police Service (KPS). What the resolution fails to specify is where and how accountability and responsibility in managing the internal security in Kosovo are to be drawn between the international military and civil presence. In its command structure, there are two quite distinct and often conflicting bodies at play in Kosovo. KFOR forces are responsible to NATO Headquarters in Brussels and are not answerable to UNMIK, whereas UNMIK Police and KPS are directly responsible to the SRSG. This bifurcation of command and control has led to some debilitating conflicts in policy as the case in Mitrovica, provided later, amply shows. More importantly to the stated ambition of the international mission to “transfer competences” to local institutions, there are significant contradictions in the Constitutional Framework of Kosovo (CFK), promulgated by former SRSG, Hans Haekerup, on 15 May 2001. In the CFK, police and justice remain under UNMIK and are non-transferable competencies thereby denying the Kosovar Government any formal responsibility in the enforcement of law and order. The CFK does state that in the last phase of its mission, UNMIK police will transfer the primary responsibility to KPS with UNMIK Police assuming a supportive and monitoring role. The problem lies in the lack of any real timeframe in which the UN mission may be deemed ready to actually handover those responsibilities, creating institutional tensions between the Kosovar Government and UNMIK.38 Deployment and Structuring of International Military Force and Civil Police The main challenge to the UN Mission in Kosovo in regards to security is that it is the first-ever United Nations Mission that had a law enforcement mandate. Without any operational model to rely on, the attempts to, among other things, address the (in)security vacuum that emerged in Kosovo after the withdrawal of Yugoslav/Serbian armed forces in June 1999 was not coherent and often at odds with other tasks facing the mission.39 Other challenges, such as supervising the demilitarization and transformation of the KLA (UÇK) into a civilian force 20 Administration and Governance in Kosovo ___________ were compromised by the uncertainties of the mission.40 Perhaps most interesting to point out in the often overlapping and contradictory responsibilities for security in Kosovo, was KFOR’s role in law enforcement in the initial months of the mission. NATO troops, trained in combat were either not prepared or unwilling to fulfill the tasks of a civilian police force. This caused an initial period of chaos and cost Kosovo many months in the rebuilding process. The confusion over the mission begins prior to NATO direct intervention in the Kosovo war. KFOR’s original function was to contain the Kosovo conflict within its borders. Under the name of Operation Joint Guardian, the initial allocation of forces and funds took place in the summer of 1998 and was halted from October 1998 to February 1999.41 The planning phase that restarted in February 1999 for the inevitable intervention of NATO troops into Kosovo initiated a deployment of troops to Macedonia and Albania, eventually reaching a number of 17,500 troops by the June 9th, 1999 Military Technical Agreement signed between NATO and Yugoslav/Serbian authorities in Macedonia. As agreed upon in Kumanovo/ë, NATO troops, numbering 16,100 entered Kosovo during the last day of withdrawal of Yugoslav/Serbian Forces on 18 June 1999. An additional 11,000 soldiers were made available in bases in FYROM.42 Within a few days, KFOR Headquarters was relocated from FYROM to Prishtina and five KFOR Regional headquarters at the level of brigades were established, corresponding with the nations (USA, Great Britain, France, Germany and Italy) in charge of regional commands. By the end of 1999, with the additional participation of more than 30 partner countries, KFOR’s total troop deployment reached 45,000.43 The rapid deployment of KFOR in Kosovo proved to be crucial to creating a secure environment in the region. This deployment made the return of 1.3 million Kosovar Albanian refugees possible.44 The rate of return, however, far exceeded the expectations of the intervening institutions charged with providing humanitarian assistance and of course, security.45 The end result was that the United Nations was not yet prepared to take over policing duties with its anticipated UNMIK Police force. KFOR, left with the responsibility of providing security and enforcing law and order, failed by all accounts. The murder rate in Kosovo by the second part of June 1999 reached the level of 50 murders per week.46 This constituted a significant failure, compounded by the fact that Serb minorities were seen as targets for revenge on a large scale and that KFOR failed to prevent Belgrade from realizing the partition of Mitrovica and creation of ethnically-pure enclaves. To its credit, KFOR vastly improved security over the next few months. Hundreds of suspected criminals were arrested, weapons and ammunition confiscated and to most observers, security and stability had been restored.47 By the beginning of 2000, the murder rate declined to 5 per week.48 In the context of responding to what appeared to be KFOR’s mishandling of the protection of Serbs, since early 2000, more than half of KFOR’s total troop engagement has been involved in protecting Serbs and other minorities.49 As far as NATO was concerned, KFOR would have been more effective in performing its primary tasks if Kosovo had a stronger international police presence as well as a properly functioning judicial system.50 In contrast to KFOR’s activities, the international civilian police force has faced serious problems in that significant delay in the deployment of officers and the building of an operating structure forced the project to implement long-term security plans in Kosovo that dragged on well into 2000.51 The formal guidance for the functions of UNMIK Police, that are the bases for its operational planning over the long term, were initiated by the report of the UN Secretary General Kofi Annan on 12 July 1999, already a month after the KFOR Administration and Governance in Kosovo 21 ___________ deployment.52 The end result of the United Nation’s lack of preparation was that only a small initial force of 200 unarmed civilian personnel serving in the UN Mission in Bosnia and Herzegovina arrived in Kosovo in the end of June 1999.53 The initial plan was that UNMIK Police would consist of 3110 police officers. These plans were quickly modified after receiving recommendations from then SRSG Bernard Kouchner, increasing the number to 4718.54 Reinforcements started to arrive at the end of August 1999, first to Prishtina and finally arriving in Peja by June 2000. The current structure (regions and stations) was fully established by mid-2000 with the exception of the region of Mitrovica where UNMIK Police has full investigative authority but KFOR kept its technical primacy.55 The problem with UNMIK Police ultimately lies in their competencies. More than half of the total number of international police officers deployed in Kosovo are from autocratic or highly corrupted countries. By importing police officers who have no experience in enforcing laws in democratic societies and operating in a courteous and helpful manner, UNMIK has been making matters worse. With such a profile of “international police” officers, it was clear they could not significantly contribute to the development of local policing and antagonized relations with the local population.56 Moreover, the cost of engaging international police is very high and not sustainable for such international projects. The UN budget allocates approximately $115,000,000 per year only for the UNMIK Police Officer’s Mission Subsistence Allowance (MSA) ($71 per day for one UNMIK Police Officer).57 Due to the fact that a large number of UNMIK Police in Kosovo are insufficiently “police officers,” the cost benefit of this undertaking needs to be seriously reassessed by United Nations. The above analysis of the deployment and performance of KFOR and UNMIK Police in Kosovo highlights the following key lessons that might be used by the international community for possible future missions: (a) If the peace keeping/enforcement military component is compelled to perform the tasks of a civilian police force, the forces involved need additional training in executing tasks of public order and law enforcement. In this regard, military forces with specific civilian policing experience (like the French Gendarmerie, Italian Carabinieri or Spanish Guardia Civil) may play a crucial role in planning and performing civilian police duties in future missions. (b) There is a clear need for an advance police force with operational planning that should be accompanied by pre-deployment coordination with military forces. These efforts can be co-coordinated with national governments of the countries that contribute to the military component in order to avoid the lack of trained personnel flooding the field, as had happened in Kosovo. (c) The experience of Kosovo demonstrates that there is a need for re-approaching peace-keeping/enforcement operations by putting together, in a common package, military and police planning for future cases. Enforcement of Law and Order The international military and civilian mission in Kosovo was, in the beginning, operating in an atmosphere of legal uncertainty in respect to which laws to apply. This shortcoming was compounded by the fact that there was no supporting legal infrastructure in place at the time. This situation forced KFOR troops to process cases according to their national laws, because what could international police officers do in such a complex situation was to act to the best of their knowledge. In the case of criminal procedures, definitions of crimes and the concept of due process was not universally recognized by those enforcing the peace. As a 22 Administration and Governance in Kosovo __________ consequence of this situation, Kosovo citizens were living in legal chaos and their rights were often abused. Efforts were made to instruct law enforcement personnel of the applicable law in Kosovo at the time (UNMIK declared that laws in Kosovo in force prior to 23 March 1989 would serve this purpose). This failed miserably as copies of the law of Kosovo was only made available in English to international mission ten months after the mission started.58 These circumstances resulted in UNMIK Police making many mistakes. In this period of legal chaos in executing policing functions, UNMIK Police had put forward a proposal calling for the implementation of a “generic criminal code” in which mission personnel could receive pre-training and apply it in circumstances of legal uncertainty. Unfortunately, this proposal had been ruled out by UNMIK.59 Even more damaging, until January 2001, the UNMIK Police was not allowed to undertake undercover surveillance activities leading to serious inefficiencies in intelligence.60 Despite these administrative barriers, many of these activities had been carried out with the self-initiative of a number of UNMIK Police officers.61 On the other side, the judicial system of Kosovo suffered a general lack of professionalism due to the disconnection of local Albanian prosecutors and judges from the system for a decade. The situation in the judicial system is being exasperated by the fact that it had become more complicated due to the cases of intimidation and coercion that were facing judicial workers.62 Moreover, the lack of a witnesses protection program has significantly hampered the implementation of the rule of law in Kosovo that is resulting in little co-operation from citizens with the police as far as serious crimes are concerned.63 Initially there was tension between the judiciary and international police due to the fact that many police officers came from the countries where there is no tradition of investigative judges as is the case in Kosovo. In Kosovo, the investigative judge has the leading role in a criminal investigation. This investigative role has been hampered, however, by the fact that the Kosovo judiciary has limited capacities to properly do their work and UNMIK Police often circumvent the traditional procedures.64 The other complicating issue in fighting crime in Kosovo is the lack of jail capacity; Kosovo’s jail capacity has room for only 600 detainees, whereas the daily arrest rate is between 50 and 60 persons.65 This situation forces UNMIK Police to release many individuals who have committed crimes. This issue has still not been addressed properly by UNMIK. In regards to the enforcement of the overall rule of law in Kosovo, KFOR is facing serious obstacles. Despite the fact that NATO decided early on to give COMKFOR command and control over all forces in Kosovo, there is still little co-ordination among the national brigades with the exception of British forces that are available to COMKFOR for deployment outside of their brigade sector. Every other national force goes its own way, whatever the situation might require, resulting in often conflicting methods of engagement and contradictory agendas.66 Contrary to KFOR, UNMIK Police forces have a single chain of command and they are solely responsible to the UNMIK Police Commissioner, who is responsible to SRSG. UNMIK police has in place special anti-riot forces that might be located at any time and placed within Kosovo. These distinct communication and coordination issues have served as points of tension between the two entities over the last three years. The limits of NATO control over forces in Kosovo have affected the enforcement of order and the rule of law in Northern Mitrovica, highlighted during periods of crisis. This is the most glaring problem in Kosovo today, as it is in this part of Kosovo where the UNMIK Administration and Governance in Kosovo 23 ____________ police have no power to act. First, there is a lack of cooperation among the Serbian population; secondly, UNMIK officials have conceded that police operatives from Serbia are patrolling Mitrovica’s streets in civilian clothes. Not only is this a breach of the Military Technical Agreement of 1999, it also makes it impossible for the local population to feel safe enough to cooperate with UNMIK Police if they wanted.67 In response to criticisms, UNMIK Police claims that only KFOR can act and conduct military operations with the use of force. This bickering becomes more apparent when KFOR denies that it has any knowledge about Serb police officials operating in Mitrovica.68 This denial has been impossible to sustain, however, with UNMIK Police discovering a list of 70 police officers from Serbia, who are under command of the General of Serbian Interior Ministry, Svetislav Gjorgjeviq.69 It is only in late November 2002 that UNMIK started to extend its administrative duties to Northern Mitrovica. What ultimately explains this sudden turn around in UNMIK’s success speaks more of UNMIK’s political efforts than law enforcement efforts by KFOR and UNMIK Police. In analyzing the experience of maintaining law and order in Kosovo, the following key lessons may be learned: (a) The initial lack of basic laws for conducting a criminal investigation and the development of special police agencies for fighting crime is proving to be one of the main obstacles to ensuring public order and safety in internationally administered territories. The UN needs, therefore, to formalize a package of laws that can be used by both peacekeeping troops and international police in order to avoid the vacuum created by a legal framework that actually hinders law enforcement by inviting procedural chaos. (b) Training officers in the international military and police forces in local criminal law is essential for ensuring inter-institutional effectiveness. (c) The weak chain of command with COMKFOR in relation with national chains of command makes it almost impossible to use NATO troops at the inter-regional level. This circumstance is affecting overall law enforcement in Kosovo, including the effective co-operation between KFOR and UNMIK Police in riot situations. A solution might be found in establishing a contingent of NATO multi-national troops under the direct command of COMKFOR, a force that can be used in situations when a single brigade is not sufficient. (d) Since both KFOR and UNMIK Police have the same law and order enforcement mandate, their co-operation and joint action in disabling the illegal operation of Serbian police forces in Kosovo is vital for securing the normal functioning of UNMIK and the Provisional Institutions of Self-Government administration in the entire territory. (e) The lack of detention centers is hindering efforts to fight crime in Kosovo. This issue needs to be urgently addressed by UNMIK and the international donor community. Building Sustainability: Kosovo Police Service The international community has shown a reluctance to use former KLA fighters and former police officers fired in 1989 to form the basis of Kosovo’s civilian police force. According to UNMIK police, the rationale behind not giving back jobs to former police officers was due to the consequences of apartheid over the last decade in which few KPS officers had policing experience and many faced a decade of unemployment, requiring total retraining.70 Under these conditions, the building of the Kosovo Police Service as a local police service started from scratch in the second half of 1999. The efforts of UNMIK in developing KPS are being 24 Administration and Governance in Kosovo __________ done through Pillar I of Police and Justice (Field training, supervision and monitoring) and Pillar III of Institutional Building, respectively through the OSCE Police School. Candidates selected for training and subsequently recruited into the Kosovo Service must be residents of Kosovo between the ages of 21 to 55 and have finished secondary school. Apart from these general conditions, an internal investigation is being conducted by UNMIK police regarding the verification of the candidate’s past. Thereafter, candidates undergo 12 weeks of classroom basic instruction at the Kosovo Police Service School (KPSS) and 15 weeks of structured training with UNMIK Police71 and starting a few months ago, with the KPS as well.72 In the building process of KPS there have been initial obstacles mainly related to the training infrastructure (place and cadres), as well as to the type and style of training that should be given. Eventually, KPS adopted US Policing techniques and equipment due to the fact that the US government covers the main burden of financing the program. According to US standards, four to five years of experience are needed for police officers to be considered for special duties while ten are needed for supervisory ones. This model cannot, however, be fully applied towards the KPS due to the urgency of developing its structure. Nevertheless, KPS personnel have rapidly developed. By September 2002, around 4500 officers from all of Kosovo’s communities are members of the KPS.73 At this stage, KPS has been implanted in all corners of the territory of Kosovo, including Northern Mitrovica.74 Apart from these training programs conducted by OSCE, a re-certification/re-qualification program has been developed and delivered for verifying the proficiency of KPS Officers in specific areas. There have also been specialized trainings organized in necessary police disciplines used in law enforcement as well as other advanced training programs delivered by foreign experts, mainly Americans.75 In addition, KPSS is organizing courses for supervisors, mid-management and one Senior Management Course. Graduates however, while obtaining ranks do not have any commanding function. The Colonel of KPS serves as an adviser to the UNMIK Police Commissioner, whereas Lieutenant Colonels serve are advisers to Regional UNMIK Commanders. KPS Police Officers are only allowed to occupy the first line supervisory/commanding responsibilities, once they are themselves fully “integrated” within UNMIK Police, including the Intelligence Unit on Organized Crime.76 Graduates from the academy operate under the supervision from UN international police “field training officers.” Most of these officers have little training in mentoring and lack familiarity with the region.77 Moreover, a number of UNMIK Police Officers have little professional expertise. In many cases, KPS Police Officers are more professional than their UNMIK Police counterparts.78 Due to the fact that KPS police officers are currently under UNMIK Police supervision, low professionalism of a number UNMIK Police Officers is causing tensions with the KPS Police force.79 A promising sign that this is being addressed is that international police officers with a high level of expertise have been granted senior positions within UNMIK Police. This should prove promising for KPS as it increases its professionalism. Despite the fact that UNMIK Police recognizes that KPS is more professional than some police forces in the region, UNMIK police do not yet consider it a stable civil servant force.80 On the other hand, by KPS the key components for achieving stability are professionalism and the impartiality of KPS police officers as well as support from the community and respect for the internal hierarchy.81 The last component is very important for ensuring the Administration and Governance in Kosovo 25 ___________ independence of the KPS. There have been cases in which individual KPS Police Officers have been promoted by UNMIK to the rank of Lieutenant Colonel without passing through the full cycle of promotion, which begins with the rank of sergeant. This accelerated career trajectory for some is causing dissatisfaction within KPS. The full staffing of KPS will have a final authorized strength of 6283 complemented by over 2000 civilian support staff. It is projected that a complete transfer of policing services for Kosovo will be completed by 2005, thus taking a period of six years for completing the build up and transfer of power by UNMIK Police.82 As of 2005, UNMIK Police will assume monitoring and training duties. There are inconsistencies to this transition process however. According to the Constitutional Framework of Kosovo, Police and Justice remain under UNMIK’s control as one of the non-transferable competencies it has secured for itself. During this period, therefore, the transfer of competences from UNMIK Police to KPS is not accompanied with the building of Kosovo officials’ capacity to administer the Department of Police. The current situation can affect in the long run the sustainability of KPS in relation to its administration by Kosovar Government. UNMIK needs to address this issue by developing a clear transfer of competencies to Kosovar authorities, including the possibility of a joint administration of the Department of Police at an early stage. The full transfer of powers in policing to Kosovar authorities might be completed with the final evolution of Department of Police into Kosovo Ministry of Interior Affairs. The above review of the last three years highlights the following key lessons: (a) Building a local police service up from scratch takes a long period of time. Although Kosovo has a relatively small population, it still will take six years to implement a full transition of powers. In this case, a new method of ensuring the speedy transfer of competencies is needed. (b) Building a credible and respected carrier as well as establishing standards to promote officers is a crucial issue for ensuring long-term institutional stability and sustainability for the local police force. In addition, there needs to be stronger professional criteria for monitoring by the international police force. (c) In order to ensure the sustainability of the administration of internal security in Kosovo, the transfer of competencies from UNMIK Police Department to Kosovar Government should be completed in parallel with the transferring of competencies from UNMIK Police to KPS. Policy Recommendations: (a) The experience of Kosovo demonstrates that there is a need for re-approaching peace-keeping/enforcement operations by putting in a common package of military and police planning for cases when both are mandated to maintain and ensure law enforcement in internationally administered territories. (b) There is a need for early planning and a pre-deployment of international police officers who are on good communicative terms with military forces. As far as expertise is concerned, early joint planning of policing between international police and military forces can be enhanced with the co-operation of international police with military forces that have specific civilian policing experience. 26 Administration and Governance in Kosovo ___________ (c) The UN needs to formalize a package of laws for criminal procedures and investigation that can be used by both peacekeeping troops and international police in order to avoid a vacuum in the legal framework of the administered territory. (d) The transfer of competencies from UNMIK Police Department to Kosovar Government should be completed parallel with the transference of competencies from UNMIK Police to KPS. INTERETHNIC RELATIONS IN KOSOVO: LOOKING AHEAD Introduction This section analyzes, assesses and provides recommendations to the international, regional and local institutions as well as community leaders that deal with interethnic relations in one way or another. The section makes an initial analysis of the benchmarks and the thinking of the international administration at the outset of the mission and the latter developments in the field with the aim of learning lessons of good practices regarding interethnic relations. Interethnic relations in Kosovo today inherit a ten year legacy of conflict between the state and the Albanian majority in Kosovo that gradually escalated from persecution and massive human rights violations into armed warfare that resulted with over ten thousand persons killed or missing and 600 villages razed to the ground. After the end of the war, the previously privileged ethnic group found itself at the receiving end of a seemingly dis-organized campaign of revenge and take-over of apartments, left over by the fleeing ones who often did so under pressure. At the outset of the mission, the primary priority of UNMIK regarding ethnic relations was “to promote the right to return through a return planning strategy which emphasizes return to multiple geographic areas in an incremental, low-profile and orderly fashion.”83 It stressed the need to be transparent, engage in maximum consultations and create confidence-building measures whenever possible. The fields deemed important were: Security, Freedom of movement, Property, Housing, Infrastructure, Public utilities, Health, Social services, Income generation, and Humanitarian assistance. As the scope of this paper does not allow for a comprehensive analysis of all the related issues, it focuses on the most salient ones: sustainable returns, the role of Kosovar institutions in shaping an overall climate, property disputes, the inability to deal with parallel institutions, security strategy in ensuring safety and freedom of movement and the exclusion of the majority leadership. Institutionally, even three years after the war, minority issues remain the exclusive competence of the SRSG. While there has been pressure on local Kosovar institutions, the major area where Kosovar institutions can have any influence is in shaping the public opinion. Kosovar politicians often complain that while they bear the brunt of the criticism and pressure when ethnic relations suffer, their authority in improving the situation on the ground is very limited, as they do not control the police nor can they influence the judiciary or other relevant institutions. Even local grievances within each municipality have a mechanism to sort their problems to the attention of the international administration at the central level. Regardless of specific minority concerns, post-war democratization taking place in Kosovo continues to be fragile at best and flawed in many regards: (a) extremely centralized power structures by UNMIK amounting to almost dictatorial powers of the SRSG, (b) fledgling and relatively incompetent Kosovar governing structures, (c) slow rate of transfer of competencies. Although belated, the establishment of Kosovar institutions is expected to increase accountability. However, the numerous reserved powers that remain with the top UN administration have been subject to discontent among Albanians and Serbs alike. Transfer of competencies does not only refer to political and security issues, but even to daily management of Kosovar municipalities, regardless of the ethnic composition of the 28 Administration and Governance in Kosovo __________ municipality. Even a document issued by the Serbian Coordination Center among others complained of low competencies to municipalities, this being in their interest to increase the control of Serb dominated northern municipalities. As a result, the Coordination Center (in conjunction with the Serb coalition Povratak), produced a decentralization plan for Kosovo that amounts to doubling the already swollen governing and administrative institutions in Kosovo, suggesting bicameral local, regional and national legislative bodies. Overall, the present legal framework provides for a fairly decentralized local government in Kosovo. Despite the small size of Kosovo (one third of Wales), 30 municipalities retain control over most competencies on all public services, maintenance of historical monuments and sights, primary and secondary education, youth work, integration of minorities, economic framework, municipal fees, employment, implementation of central legal acts and procedures, relations to UNMIK, primary health provision, infrastructure, spatial and urban planning. Despite global trends of decentralization, due to low capacities in terms of human resources and finances, one might argue that what Kosovo needs is the opposite, the strengthening of central institutions. Relations with non-Serb communities Due to the salience of the relations of the Albanian majority with the Serb community, this is the main focus of this paper. It is however important to briefly note the situation of other minorities. Aside of Serbs, and to some degree of the Roma, the problems of other minorities are mainly of social nature and most are common with those of the mainstream Albanian population. For example, the Ashkali and the Egyptians are stigmatized and kept away from well-paid jobs due to their historical discrimination and low qualifications as a result and due to prejudices common throughout South-Eastern Europe. The Ashkali and the Egyptians were previously categorized as Roma and today there is a standard practice to treat them under one heading as RAE communities (Roma-Ashkali-Egyptian) due to the commonality of their problems. The major difference is that the Ashkali and the Egyptians speak Albanian as their native language and are fairly well integrated within the Albanian community. They have also traditionally lived in Albanian neighborhoods or in close proximity. An additional factor for this differentiation was the after-war resentment of the Albanian population towards the Roma as members of the Roma community were accused of having collaborated with the Milosevic’s regime. This served as extra stimuli to the Ashkali and the Egyptian communities to distinguish themselves from the other groups. Turks in turn are very well integrated and one might even claim that their economic situation is relatively well off even compared to the Albanian majority. They speak Albanian, belong to the same religious affiliation and have traditionally been close to the Albanian community. Despite initial accusations by members of the majority that the local Turkish leadership had sided with Milosevic during the ten year rule and their non participation in the first local elections, they faced no major obstacles in their integration and no violence was exercised against members of this community. The Bosnjak community is also in very good relations with the Albanian community mainly due to a common sense of victimhood during the war and due to shared religious affiliation. However, divisions remain, and as a result, this community faces indirect discrimination and disadvantages especially in higher education. Another local community, the Goranis that live almost exclusively in the most southern (and one of the most backward) municipalities finds Administration and Governance in Kosovo 29 ___________ itself in a similar situation as the Bosnjak community, but due to their geographic isolation, they face no security issues. The rest of the paper will focus exclusively on relations with Serbs with an emphasis on the most important issue that hinges on overall relations, that of returns. Sustainable Returns Perhaps the most salient aspect of interethnic relations is the much politicized issue of the returns of internally displaced persons (IDPs). Ever since 2000, the United Nations Interim Mission in Kosovo has been very engaged in planning and implementing a policy on returns. Lead by the main humanitarian pillar of UNMIK, the United Nations Higher Commission on Refugees (UNHCR), the returns of displaced persons was promoted as a human right to a sustainable return. The right to private property was in turn seen as an inviolable right of any person, regardless of any change of borders or international status of the territory. Submitting to the will of each person, UNMIK sought to ensure sustainable individual returns to previous habitual residence. In general, “[t]he role of UNMIK or any government authority is not to determine the location of the returns or to dictate IDPs and refugees how and when they can return, but to help improve the condition in a way that IDPs and refugees have the ability to exercise their individual decision to return.”84 This policy has been counteracted by Belgrade and especially by the Coordination Center of FRY and of the Republic of Serbia, who from the beginning, sought to earn political points by promoting pompous organized returns. During the first year of UNMIK administration, this issue was portrayed as a failure by Milosevic as a way to discredit UNMIK, by paying salaries and giving out apartments to IDP and by showing Kosovo in a horrific light as a way to keep his rule and preserve his traditional support base, the Kosovo Serbs. The mayor of Leposavic, Nenad Radosavljevic, who was later appointed as advisor to the SRSG, claimed that these policies later hurt the efforts to return IDPs as they had made a good living in Serbia itself.85 After the fall of Milosevic, this policy subsided but was kept on track by the ensuing Yugoslav president, Vojislav Kostunica and his Kosovo envoy, the head of the Coordination Center, Nebojsa Covic. Belgrade would always remain the major partner, both as a host of the IDPs and as their mother nation to which ethnic Serbs look towards, rather than Prishtina. After the fall of Milosevic, following an all-European exuberance, UNMIK tried to appease Belgrade with a host of agreements that in turn had a tremendous cost in the relations of UNMIK with its own Kosovar population. Belgrade was often accused of having discouraged local Serbs’ participation in the newly created institutions; to have acted obstructively in a host of bilateral issues, such as police cooperation or returning civil registry books, and that they have used extremist rhetoric created for internal public consumption as a way of showing that they are doing something to bring the “cradle of Serbia” back. The international community managed to get some concessions in return, most notably, the release of Albanian political prisoners, however, the issue of returns still remains the most lively debated even while this paper is being written. In practical terms and in terms of location, the returns were planned in three phases: (i) to uninhabited areas or to present enclaves (if originally inhabited there); (ii) to present enclaves or close, (iii) to urban areas such as Prishtina/Albanians in the north. The strategy that UNMIK followed in order to conduct returns in a sustainable manner was the creation of Regional and Local Working Groups on Returns, known as RWGs and LWGs respectively in 30 Administration and Governance in Kosovo ___________ the UNMIK jargon. Headed by the respective UNMIK Municipal Administrators, these bodies include representatives from the UNHCR, KFOR, specialized international agencies and NGOs involved in return activities and the local communities themselves (both majority and minority).86 This has been effective only in some areas. Despite public statements by UNMIK that the return process must be conducted at a grass-root level by meeting with neighbors in order to create a better zone of comfort, this was implemented in a top-down manner. Despite the involvement of various NGOs, the wide community and the municipal government were initially kept at bay. While the role of Kosovar participants was stressed as crucial, they were invited only at a later stage. This caused resentment of local politicians who felt that they were being called upon a done deal only to legitimize a manipulated process that they had no control over. Suspicions that a number of enclaves in Kosovo are being connected through back roads and that one soon might see their status upgraded to a canton, did not help either and made any Kosovar Albanian participant nervous. Despite a bad strategy, under informal pressure, several municipalities have been cooperating with the LWGs, welcoming both organized and spontaneous returnees.87 “The Framework outlines the major actors who will be involved in return planning and/or implementation, and their respective roles and responsibilities. The actors include not only those members actively participating in the JCR and RWGs (UNHCR, UNMIK, KFOR, OSCE and Serb representatives) but also important players such as NGOs and the donor community.”88 The thinking among the international administration does recognize the importance of creating “progress within and between communities to enhance tolerance and willingness to co-exist.”89 They also recognize “the active and responsible participation of the Kosovo Albanian community, through their leaders” as a key element of successful return planning.90 However, the role of Kosovar Albanian leaders remains tentative where the presence of “…Albanian leaders at all levels in the return planning and implementation process is also emphasized.”91 In principle, however, stripping Kosovars of responsibility and ownership over the process is retrospectively evaluated as a mistake. Even at the municipal level, mechanisms were created to avoid over-voting by the Albanian majority. While local minorities can regularly be outvoted, the UNMIK Regulation 2000/45 on Local Self-Government provides for an international Local Community Officer in each municipality and the creation of two municipal mechanisms, the Mediation Committee and the Communities Committee, which have the power to bring an issue to the attention of the SRSG. The SRSG can then create a mixed three-member panel (international-Albanian-Serb) and if no resolution is decided, he then has the authority to decide. The role of local Kosovar leadership from the receiving end is crucial if one is to ensure integration and long-lasting returns. There have been doubts from both ends on the readiness of the communities to accept returns and the willingness of the returnees to accept their own integration and accept the new reality created after the war. The host community on one hand complains that returnees might be war criminals, also fearing that more returns will make some type of Yugoslav/Serbian rule more probable. UNMIK’s attempts to alleviate fear of war criminals did little to pacify the people since the courts that must deal with this issue have been dysfunctional at best. Administration and Governance in Kosovo 31 ___________ Kosovar institutions and the overall climate Since all the key competencies regarding interethnic relations lie with the SRSG, the only responsibilities that remain with the Kosovar institutions is to influence the overall climate for improving interethnic relations. Even two years following the war, the activities of Kosovar institutions did not go beyond conciliatory remarks aiming to please the international community. However, the findings of a poll in mid 2002 point to overall acceptance of the return of the Serb community by the vast majority. As delineated above, the role of municipal politicians and the desire of central governments to show a constructive policy have had an impact on the people. However, the opinion making needs to be matched with real policies when it comes to dealing with disputes quickly and effectively, otherwise politicians often find themselves advocating against the general will. While on one hand international diplomats convince politicians that it is in their interest to advocate tolerance, politicians in turn are not able to explain and justify to the voters why parallel Serb institutions remain in place or why can’t Albanians go back to northern Mitrovica. Some politicians have remembered that fiery rhetoric remarks against other communities are the easiest way to garner support. While the number of incidents has gone to an all-time low, the trust is not being regained easily due to the physical separation of the two communities and the lack of contacts as a result. Despite efforts to calm down the scene, overall political constellations perceived to be pushing Kosovo towards Serbia (primarily driven by the EU to create what many sarcastically refer to as “Solania”), the insoluble problems of Mitrovica, the recent border dispute with Macedonia, lobbying by Belgrade to march IDPs back to Kosovo massively, attempts to disguise cantonization under the heading of decentralization make it more difficult for politicians who advocate reconciliation and slow down any healing process of local communities, NGOs and the international community. Sustainability hinges on many other issues. Knowing the Albanian language in the Albanian dominated Kosovo is a necessity for this community to integrate and to be able to make a living. Previously privileged, seemingly few Serbs speak Albanian, although most understand it. Also, previous privileges also make it very hard for Serbs to accept a numerically subordinated role that ensures equal treatment at best. Many Serb intellectuals will privately admit that Serbs continue to hold racist attitudes towards Albanians, which will make them more likely to flee to Serbia if they cease to be the masters in the area. Some claim that without defining the status of Kosovo, conducting the returns will be a difficult issue. According to an influential journalist, “as long as Albanians fear that any returns of Serbs can bring about the return of the Serbian regime” it will be very difficult to conduct the returns.92 Others put the returns under a slightly different context, by claiming that the returns of Serbs to Kosovo can only succeed if they accept the new reality in Kosovo. Other communities have in one way or another agreed to this and made this public except the Serb community.93 This declaration paved the way for a “Platform for Joint Action” by Albanian and RAE leaders. This platform among others rejected the notion of collective guilt, declared steps for better understanding of the needs of Kosovar RAE, committed to a process of returns, took steps to change the perception through the media, committed to develop a special program for RAE children to have access to education, and emancipation through participation. Serb returnees on the other hand often preferred to remain loyal to Belgrade and to the leadership of northern Kosovo that aspired partition, as opposed to Serbs in the rest of 32 Administration and Governance in Kosovo __________ Kosovo who had more interest in improving relations with the international community and with the Albanians without alienating Belgrade. It is often said that an agreement always needs two willing parties. According to UNMIK, “Inter-ethnic dialogue requires not only the positive engagement of the Kosovo Albanian community, but also a willingness on the part of Kosovo Serb remainees and returnees to engage in positive communication and confidence building measures with Kosovo Albanians through municipal leadership structures and through community counterparts.”94 As the willingness to embrace the newly created structures by local Serbs was very low, UNMIK implicitly accepted this dual pacification of both communities, justifying a separate parallel system for the Serb community. Parallel Institutions While the parallel system might have been justified at the very early stage as means of preservation, it quickly became clear that in the long run this was going to cause a lot of havoc. Over three years later, although officially illegal for the UNMIK administration, parallel Serbian institutions operate almost in every Serbian enclave within Kosovo, while the northern part has even been illegally attached to the Yugoslav Telecom and to the Serbian energetic system. As a result, communities remain physically sealed against each other with very limited levels of communication. This short-term “guarantee” implies a perpetual isolation of minorities and a long-term failure. A particular issue that is addressed, within the context of an emerging self-government in Kosovo, is that of continued existence and entrenchment of parallel structures, which are becoming increasingly detrimental to ensuring access to essential services for minorities.95 Three years later, combating these institutions would in the long run improve interethnic relations, stimulate integration and create a local Serb elite that would safeguard its own interests better than Belgrade does. While one might argue that in some cases some parallel structures continue to be inevitable as an interim measure due to insecurity and restrictions of freedom of movement, these structures ultimately provide an unsustainable second-class service for minorities and inhibit important forms of inter-ethnic interaction. It has been found that little progress exists in this issue. As a matter of fact, some aspects in the divided city of northern Mitrovica have been getting worse gradually. Despite commitment and commendable efforts on the part of many dedicated individuals within the relevant JIAS department, the municipal structures and numerous NGOs, the provision of health care services is still largely marked by separate service provision for different ethnic groups.96 So far, parallelisms have been most salient in health and education (approved by UNMIK and paid both by UNMIK and often paid double by Belgrade structures) and courts (not approved by UNMIK, but tacitly allowed to operate). As an example of UNMIK trying to deal with already swollen parallel structures and of greater entrenchment is the recent case when UNMIK appointed a new director of the hospital in Gracanica, who was not accepted by the majority of the workers who would receive lower salaries from the consolidated Kosovar budget.97 This case also reflected disagreements among various political forces. Reconstruction and Economic Opportunities Another major precondition to returns was the reconstruction of destroyed homes and infrastructure, in which many donors have given millions of dollars for. The US, Italy, Germany, Switzerland, European Agency for Reconstruction, Serbia, UNHCR and other agencies have given close to twenty million dollars and more have been pledged. However, Administration and Governance in Kosovo 33 ____________ while building houses is necessary, if this approach is to guarantee sustainability, it needs to be coupled with a political process that is nowhere in sight. As it stood in late 2002, most returnees are more likely to sell their property and leave for Serbia once again. The sale of property presents another issue, the economic viability of returns. Regarding returns, UNMIK has mainly focused on operational issues. While the economic side of returns has not been overlooked, this issue is to some degree beyond their control. While it is understandable that certain communities may profit disproportionately from the economy, the economic viability of the Serb minority is certainly a function of overall Kosovar economy, which has in turn been a failure. Inadequate approaches to property issues, legal framework, inadequate customs policy, lack of a commercial law, inadequate educational system that does not match market demands, an increasing illiteracy rate, organized crime, corruption and nepotism, inadequate approach to socially owned enterprises98 have all contributed to overall low economic prospects for Kosovo. Naturally, even the initial boosts of large amounts of money by the tremendous international presence mainly went to the benefit of the majority in the capital and seems not to have been invested in long-term self-generating business, but rather in unsustainable and badly planned investment that went down similarly to the high technology bubble in the United States. One can also identify a natural flow of population of Serbs from Kosovo since the 1980s. Even before the conflict and despite the fact that a disproportionate share of high paid jobs went to the Serb population, Serbs and especially the youth have had a propensity to migrate to Belgrade from a relatively rural Kosovo. The interior Serbia continues to face a similar trend today. Furthermore, when compared to Serbia, Kosovo offers limited economic prospects even to Albanians. These arguments do not aim to present a bleak picture of what might be expected in this regard, but to present a realistic argument of what the UN administration and an increasingly receptive local government face. Even Serb leaders wonder about the number of Serbs who really want to live in Kosovo, a number, which remains totally unknown (Radosavljevic 2000/28/10). A famous human rights advocate, Adem Demaçi, cites how Bosnjaks who have been very well accepted are increasingly leaving after trying their luck for several years after the war. “We are attempting to return some, while others are leaving after having no economic prospects for three years,” Demaçi claims.99 Avoiding this question (how many really want to return?) also brings to the issue of the immeasurability of the success in this regard. Regardless of the number who returned, this issue can be praised as a success or it can be criticized, depending on regional and global friendship constellations. Property Management Kosovo has not had a satisfactory property system for as long as anyone can remember. Despite a legacy of problems throughout the century, the situation in Kosovo since 1989 (when the autonomy was stripped), has created a lot of problems. “The property system began to collapse as of 1989, when the Belgrade regime instituted increasingly discriminatory property laws on the majority Albanian population. As a result, a significant numbers of Albanians lost their occupancy rights to socially owned properties, which were reallocated to Kosovo Serbs or Croatian Serb refugees. As a consequence, most property transactions among the Albanian community during this time were carried out informally (i.e. without legal records) which led to the property and cadastral records, the key to any functioning property system, losing most of their value as accurate documentation.”100 34 Administration and Governance in Kosovo __________ After the war, deciding on the applicable law regarding property was extremely confusing, and many made the question whether all property transactions after the repeal of the autonomy were to be voided. Moreover, property was sold massively after the war, some of it under duress. Many people who had lost it during the ten years also tried to reclaim what they saw as unjustly taken over by the regime. Also, after the war, there were no institutions to provide for the legality of all property transactions and much property ended up being sold to more than one new owner, which resulted in many disputes. “Many properties have multiple claimants to ownership, and proving ownership of a property is exceptionally difficult. Given the institutional vacuum and the lack of respect of the rule of law, it is hardly surprising that illegal occupations and constructions are widespread. The problems are further compounded by the wave of voluntary and involuntary returns of Kosovars to the province, since the arrival of UNMIK and KFOR.”101 When UNMIK realized the importance of property management and of good records by the cadastre, it set up the Housing and Property Directorate (HPD) managed by Habitat, often considered one of the major failures of the international mission. Even three years after the war, this issue remains a major obstacle not only to returns of displaced populations, but also to the development of the economy, to an end to many disputes, to foreign investment, and to privatization. An important problem also became what the international community called “strategic sales” of minority-owned property, which can be defined as the practice whereby the minority owners of property, located in strategically important locations within minority areas, are induced to sell their property, as part of what appears to be an organized campaign.102 One attempt was made to curb this by requiring permission when conducting interethnic transactions of real estate. It was quickly put down as an idea. Even for the IDPs who had planned to return, the more time passes by, the better they get stabilized in their new settlements, and the less likely they are to return. Skyrocket prices of real estate in downtown Prishtina, boosted by the very high number of foreigners and by a rapidly increasing Prishtina, whose population doubled after the war, served as an extra incentive to sell property. Much of the minority owned property in downtown Prishtina has been sold, most of it over the market price due to the extremely high demand for housing after the war. An apartment in Prishtina was normally sold double the price of a similar one in interior Serbia and usually higher than in Belgrade. Behind housing purchases, one should also look at basic market laws and the extremely limited supply for housing after the war. The trend is also that it is mostly in Albanian neighborhoods or towns that apartments are sold whereas those in the enclaves are not. The reason is that Albanians will not buy property in which they cannot move in right away. There is not much that UNMIK could do in these cases except privately conclude that Serbs with no property will probably not want to return to Kosovo. Group rights and minority overrepresentation The international community has also made advances in working along side the local leadership, but often railroading everyone, in creating special rights for minority communities. Most notable was the creation of a strict proportional electoral system and twenty set-aside seats on top of that, ten for Serbs and ten for other minorities. We could think of no minorities in Europe that are over represented to this degree. Out of 120 seats of the Kosovar Administration and Governance in Kosovo 35 ___________ Assembly, the overall Serb representation amounts to be in a strong third political block with 21 seats (11 won, 10 set aside), just five less than the second strongest Albanian block. Municipal assemblies are also elected in a proportional system, though without any set aside seats. It is expected that Serbs will take over at least three municipalities (all three in the North) and are potentially able to take over two more (Novo Bërdë/Novo Brdo/Artanë and Shtërpc/Strpce). It is worth mentioning that three out of these five were upgraded into municipalities during the Milosevic regime (Zubin Potok, Novo Bërdë and Strpce) and one after the arrival of the international administration (Zvecan). Although fairly skewed towards the Serb minority, the rules of the game have in general been accepted by all, though the Serb community made two major conditions/demands that have been rejected by UNMIK and by the local administration so far: the creation of a second municipality in Mitrovica and the implementation of a decentralization plan. In response, UNMIK has created a joint UNMIK-Kosovar government Joint group on Decentralization. It has also been decided that no new municipalities were to be formed, although the Constitutional Framework does not specify the overall number of municipalities. In terms of ethnic political accommodation, the situation is not gloomy at all. Serbs are effectively in the government, not as a coalition partner, but through an agreement that grants one ministerial post to the Serbs and one to other minorities. Effectively, Serbs and other minorities (who take this post by rotation) are represented in the government. Minorities have also had high posts in municipal governments: an Egyptian was vice-president of the Municipal Assembly of Gjakova, a Serb was vice-president of Gjilan, and one Serb of Viti. Minorities in Kosovo have also seen their concerns and priorities change somewhat over time. While security was never an issue with the Turkish community, it has decreased drastically for the Bosnjak, the Gorani, the Ashkali and the Egyptian communities, but it still remains somewhat of an issue for the Serbs and the Roma. Overall, areas of minority residence are safe, however, freedom of movement remains an issue for some local minorities throughout Kosovo. Although many Serbs drive their own vehicles to Serbia through Prishtina and the rest of Kosovo, there is a doze of fear involved nevertheless, due to objective and perceived risks. Albanians in turn prefer not to travel to enclaves, while traveling to northern Kosovo is out of the question. Local minorities in this context also include Albanians who live in the Serb dominated north of Kosovo, an issue that generally goes underreported. UNHCR and OSCE report of gradual decrease in crime and violence against all minorities. While the level of interethnic crime grew sharply several months after the war, it has been falling ever since. Overall, there has been a decrease in serious security incidents. What goes unreported in these reports is a general increase in the overall rate of crime. The issue of security has been discussed in greater detail in the previous section of this paper. One major positive development in this regard was a change of strategy by the NATO’s Kosovo Force in guarding enclaves. Changing from an obvious physical military presence to a less intrusive presence with frequent mobile patrolling has proved very productive. With the gradual strengthening of law enforcement authorities (although unsatisfactory at best) the lack of watchtowers and tanks at the entrance of minority villages produces a positive psychological effect both among the external majority population as well as reduces the sense of isolation among the minorities. Aside of several minor incidents, the incidence rate is the same. Mobile patrolling also improves freedom of movement in major transport arteries and assures safer mobility of minorities. 36 Administration and Governance in Kosovo ___________ Perhaps one of the most unhelpful decisions of the international community, which by and large was enforced without the previous blessing of the local political elite, was to take the date of 1 January 1998 as decisive for enfranchising residents of Kosovo. Much to the discontent of the Albanian population, this decision enfranchised Serb refugees from Croatia and Bosnia who were temporarily placed in Kosovo by the Milosevic regime, while it disenfranchised hundreds of thousands of Kosovar diasporas of all ethnic backgrounds who had left during the ten year long repression. According to the population registration in 1991, there were 194,190 Serbs in Kosovo.103 On the other hand, UNHCR/OSCE put the number of minorities displaced from Kosovo (in Serbia, Macedonia and Montenegro) at 235,000 (200,000 of them Serbs) while 20,000 are displaced internally within Kosovo.104 The number or registered voters alone was put at 170,000. There are plenty of controversies related to these figures, which have raised suspicions on possible hidden motivations for various statistics games. Good practices There are success stories also in the aftermath of the war in Kosovo. Aside of some cases of returns, we have identified several issues that might shed light on what has worked in the past. The city of Gjilan is an example where Serbs have walked freely throughout the town with full freedom of speech. This is often attributed to (i) a young and energetic mayor and (ii) to a Serb leader who was ready to grab the hand of friendship that was offered to him. The skiing resort managed by local Serbs in Strpce started to be used by Albanians two and half years after the conflict. This is a sign how the motivation of money and the engagement of the international community made this happen. The ownership of this resort is disputable though. Albanians, Serbs and others reportedly have excellent relations in downtown Prishtina at the home for pensioners and the disabled.105 Most important of all is a change of attitude among the media who now cover good interethnic stories extensively. Some media have contributed to the reconciliation efforts by seeking out good stories of interethnic coexistence and publicizing this widely in positive light. The most successful move was the establishment of “a common license plate”, copied after its success in Bosnia. Despite a moderate success in Kosovo, it never reached the level of Bosnia, as Serbs felt a dose of estrangement towards the newly Kosovar plates. Serbs today use Kosovar plates to travel within Kosovo and replace them with the old plates issued by the Serbian authorities. Lessons Learned for future missions: (a) For any long-lasting solutions to stand the test of time, the international community needs to involve local communities in the decision-making processes; transfer administrative and political competencies early on (increases local ownership and accountability with its constituency); (b) Establish transparent structures and better accountability. Staff from hundreds of countries have different interests and ways of working and the absence of a clear hierarchy and accountability gives rise to corruption, political unaccountability, inefficiencies; (c) To be mindful of ethnic balances and previous power balances of local ethnic groups and political forces (favoring a minority that used to exercise an apartheid-like regime may remind the majority of previous injustice); (d) An administrative structure should always try to stay away from political decisions, despite the inevitability to do so; it should early on establish itself as a legitimate local Administration and Governance in Kosovo 37 ___________ body. Maintaining closer contacts with the community will help the international mission understand events better and it will build a healthier relationship with the local constituency; (e) Establish a property dispute resolution system and a well functioning cadastre early on. Property is key to settling interethnic disputes. Policy Recommendations (a) Returns: Promote the involvement of Kosovar institutions for a sustainable process of returns by giving true authority and responsibility; Depoliticize the returns process; Controversies in numbers enable the politicization of the issue and allow room for egregious declarations from all sides by parties who aim at aggravating the conflict further. As such, the Kosovar public and the government are more likely to feel a sense of ownership towards the process and commit funding from the Kosovar budget designated for returns; Assure conditions for a voluntary and safe returns that brings back the dignity to the returnees with adequate public services and support by the host community; (b) Regional politics: Pressure Belgrade through the international community not to support parallel institutions; Abolish the parallel system of justice immediately. Gradually work with health and education. The practice where minority judge minority cases should be replaced with close monitoring by the OSCE.106 (c) Interethnic interaction: Public support for political parties and non-governmental organizations that conduct work in improving interethnic relations; Search for practical steps to increase overall interethnic communication, such as joint marketplaces, or similar; Mixed police patrols in all minority areas. (d) Establish rule of law across Kosovo: Ordinary citizens in northern Kosovo support the establishment of the rule of law and dare not to confront local organized crime with strong ties to the political elite. (e) Start to approach the status issue in informal forums. This is a sensitive issue, however, many other components are held hostage to this question. (f) The need for an exit strategy of the international civilian mission. The international community should prepare the local communities to co-exist and govern in case of a sharp cut down of budget, staff and competencies of the international mission. This opens a lot of various interpretations ranging from: threat to Albanian institutions that Serbian ones might return, threat to the Serb population, pressure to solve the status. As funds brought by the international mission decrease, the less necessary they are viewed by the locals and the more they will insist on competencies and authority. (g) In order to improve the trust among the communities, all stakeholders should consider forming a Truth and Reconciliation Committee with representatives of all communities and of the international community; Trying war criminals will also bring a sense of justice among the victims and help them direct their anger towards specific individuals tried for the crimes instead of blaming the community collectively; Pressure Belgrade for a public apology/acknowledgment of suffering. (h) Improve overall economic development: Invest in rural infrastructure, as this was one of the main legacies of the communist system. This is a precondition to rural development; Create a “solidarity principle” (a fund for underdeveloped regions within 38 Administration and Governance in Kosovo __________ Kosovo) to ensure that any region does not lag behind others drastically, which would benefit to minorities and rural populations alike. (i) Support non-territorial autonomy for all communities (high degree of group rights) instead of renewed ideason how to carve-up territories within municipalities; Promote special culture ties to mother communities (Serbs with Belgrade, Turks with Ankara as a way to promote language instruction and study their own cultures); Approach the decentralization proposal by the Coordination Center (drafted by Povratak) carefully as it might breed cantonization, or what Albanians fear “a separate Serb geographic entity within Kosovo. Ideas for further decentralization should be supported, though with caution due to limited administrative resources and human capacity. Aside of political implications, Kosovo should seek ways to streamline and cut down in administration and local government and not bureaucratize them further to the point becoming unsustainable; (j) Help the Assembly to pass anti-discriminatory laws in general; Serbs and other outstanding groups will benefit with an overall positive result in the level of tolerance of the Kosovar society. THE EUROPEAN UNION’S FIRST MISSION While the last rounds of pre-accession negotiations are continuing with ten prospective new member states seeking to enter the European Union, Kosovo remains a unique case. Kosovo today is already using the Euro as its official currency, Prishtina is covered in blue and yellow flags to celebrate Europe Day on May 9th and Kosovo’s first minister of Finance was a European Commission official. The promise of Kosovo’s ‘Europeanisation’ and integration into the European family is continuously raised as a carrot to push for further reforms at all levels. Omnipresent Europe and its institutions are in Kosovo today, the promise of Europe remains uncomfortably vague however. Unless this ambiguity will change, the Balkans will remain an island of instability in the heart of Europe, exporting migrants and importing peacekeepers.107 Peacekeepers, consultants and experts of all sorts are an abundant species in Kosovo. On June 10th 1999, the work of the European Union and other international organizations were welcomed by locals to develop a comprehensive approach to the economic development and stabilization of the region affected by the Kosovo crisis.108 This was the only direct reference in the Security Council Resolution 1244 to the European Union, and it was to become the basis for a serious commitment by the European Union and its member states to rebuild, reconstruct and develop Kosovo and its economy. There were no internal procedures or precedents to fall back upon - Europe had never been ‘on a mission’ itself - the European Union Pillar (or Pillar IV) as part of the United Nations Mission in Kosovo (UNMIK) had to be invented from scratch. In August 1999, the EU Pillar was a mere handful of seven commission experts and support staff headed by Jolly Dixon.109 During the first year of its existence, the EU Pillar was not European in the sense that it was made up of staff from the European Union. Indeed, arguably one of its earliest institution building success stories, the Kosovo Central Fiscal Authority was implemented largely by Americans and Australians paid by USAID. Contractual arrangements varied greatly; staff was paid out of different budgets, staff even in the same department worked according to, at times, different benchmarks. For a few months in 2000, an Australian consultant headed the EU Pillar. Over the past three years, the EU Pillar has been in constant flux. In order to understand EU Pillar policies, one needs to consider its unique set up and development. Its mandate and responsibilities have changed, resources have changed and its internal structures have also changed on several occasions. In addition, the EU Pillar has seen a rapid turnover of staff. The position of Spokesperson has changed hands four times. In the end, no other international institution in Kosovo has experienced so many dramatic changes between the summer of 1999 and today. Today, the EU Pillar and the institutions in which its staff plays a leading role are responsible for an astonishing array of assets, people and tasks. The Kosovo Trust Agency110 is responsible for the privatization or liquidation of more than 400 socially owned and publicly owned enterprises and their assets all over Kosovo. The Banking and Payments Authority, Kosovo’s revenue authority, customs, all public utilities, the airport and major concerns like the Trepca company are all managed, administered or supervised by the EU Pillar. The EU Pillar has always been the smallest of the four UNMIK pillars, but the only one that has grown continuously in terms of personnel and responsibilities. In 2002, the EU Pillar controls large resources, takes key economic decisions and directly manages key institutions. EU Pillar decisions affect all socially owned enterprises, over 50,000 registered private businesses and almost every single household in Kosovo. 40 Administration and Governance in Kosovo ___________ To manage such a wide array of assets and responsibilities effectively requires a strong presence in the field, a good understanding of the Kosovo specific institutional and legal context, familiarity with the legacy of Yugoslav socialism and public institutions, and a system to widely share and manage Kosovo’s scarcest resource: reliable information. The EU Pillar has always had the smallest part of its staff outside Prishtina, having built up a modest field presence only since early 2000.111 The EU Pillar’s first regional office in Mitrovica was opened in April 2000, with one international and one local staff. The UN regional structure in Mitrovica at that time, without the municipal administrations in the region, employed about thirty internationals. In the summer of 2001, as few as twenty-one staff working for the Department of Trade and Industry were responsible for more than 400 socially owned enterprises in Kosovo. Few of these international experts were familiar with Yugoslav socialism and its peculiarities like workers self-management and socially owned property. The legal and practical meaning of specifically Yugoslav concepts like 'socially owned property' continues to confuse international experts today. Reviving the ghosts The evolution of the EU Pillar and its limitations are best captured by the example of its growing authority over the fruits of Kosovo’s socialist development - Kosovo’s four hundred or more socially owned enterprises (SOEs). Constrained by a severe lack of reliable inform-ation about the reality on the ground, hamstrung by its own limited administrative resources and misguided by uninformed staff, the EU Pillar accidentally revived the ghosts of socialist Yugoslavia. For the first year and a half there was no EU Pillar presence in the field, and the authority to act as administrator or trustee with respect to all enterprises that are industrial or commercial in nature112 was only transferred to the Department of Trade and Industry (DTI) in December 2000.113 From late 1999 to the end of 2000, UN administrators at the municipal level tried their best to direct the activities of socially owned enterprises on their territory by reinstating socialist-era reporting requirements and institutional controls. Following the first Kosovo-wide municipal elections, several disputes broke out over the control of these SOEs,114 especially in those enterprises that produced a steady stream of cash flow from renting out parts of their premises or selling off stock. According to the applicable law, it was illegal to rent out socially owned assets and to pocket or re-distribute rent proceeds among the workforce instead of reinvesting them to enhance the company’s value. Following a crisis in the metal production company Zahir Pajaziti (previously known as FAN or FAGARA) in Podujevo in February 2001, the EU Pillar tried to resolve conflicts between self-appointed directors, newly elected municipal assemblies and the workforce by reviving the institution of the workers’ council as chief executive body of an enterprise, elected by the workers’ collective as a whole. The workers in these enterprises were again able to elect and dismiss their own managers and to reallocate profits among themselves. The system of workers’ self-management was at the heart of Yugoslav socialism. It institutionalized workplace ‘democracy’ within the constraints of a one-party system committed to building a socialist society. At the same time, workers’ collectives were made subject to a complex web of legal and institutional constraints. The result was an economic system marked by constant conflict, in which interminable negotiations rivalled production as the principal activity. Within the system enterprises were paralyzed despite sharply deteriorating economic outcomes. In the absence of any real market discipline the communist party was central to resolving conflicts over limited resources. Administration and Governance in Kosovo 41 ___________ DTI’s decision to re-introduce workers’ self-management was based on an ad hoc decision by a few uninformed individuals and it turned out to be a fatal mistake. Enterprises managed by workers councils are per definition opposed to restructuring and downsizing, and naturally resist new investors and new ideas. Holding workers’ council elections in about 100 enterprises did not help to assert DTI’s115 power as administrator of all socially owned enterprises. In fact, control over some of Kosovo’s most valuable assets continues to be decided outside the legal framework in various local disputes. The rule of the local strongman and not the rule of law prevails. In some enterprises workers council elections have been held, in others directors appointed by the post-war UCK government held on to power, in some instances pre-1989 directors returned to their posts; and in six cases DTI assumed direct administration. The Department of Trade and Industry lacked the most basic information that was needed to formulate a sound policy. Until today, the actual number of socially owned enterprises in Kosovo is unknown, so is the number of employees working in these companies and nobody can tell you how many hectares of land are held by socially owned enterprises. The re-introduction of workers’ self-management is a perfect example of how inadequate inform-ation eventually leads to inadequate policies. Instead of resolving the problem of corporate governance and management control in socially owned enterprises, it created additional impediments to economic development. The short-term policy to resolve disputes in socially owned enterprises by electing workers’ councils based on the 1988 Law on Enterprise had been formulated without understanding Kosovo’s complex institutional environment, without attempting to understand the legacy of Yugoslav socialism and without taking into account the limitations of DTI’s own authority. As the World Bank repeatedly stressed in its Development Report 2002, much of the important work in building institutions lies in modifying those that already exist in order to complement better other institutions and to recognize what not to build in a particular context, as much as what to build.116 Even Yugoslav communist reformers had recognized the institution of workers’ self-management as the Achilles’ heel of the Yugoslav economic system. As John Allcock noted, “taken together, the constitution of 1974 and the Law on Associated Labor contributed as much as any other feature of Yugoslavia’s history to its eventual collapse.”117 By accident, DTI turned Kosovo into Europe’s socialist museum. The unknown private sector Limited information also affected other policy areas that fell under the EU Pillar’s mandate. The nature and structure of the private sector in Kosovo has been as much a mystery to most international observers, as has the Yugoslav system of workers’ self management. In July 1999, Jolly Dixon proclaimed that “the first key challenge is to enable the private sector to become the major engine of growth.”118 Many foreign observers were struck by the speed with which a new private sector emerged in Kosovo after the conflict of 1999. But in many respects it was a return to a private sector of the past, which had roots in pre-socialist times, persisted throughout the Yugoslav era as mala privreda (small business) and expanded substantially in the early 1990s, only to stagnate from 1995 onwards.119 Kosovo’s private sector is predominantly small-scale and not capital intensive. 58.3% of registered private businesses are engaged in trade and construction activities and 49.9% of registered businesses in September 2002 are one-person enterprises.120 In many respects Kosovo’s private sector in 2002 is similar to the US economy of the 1830s, with most private businesses centering on the family and small mini-markets selling a limited range of imported 42 Administration and Governance in Kosovo __________ goods in kiosks and corner stores. Most traders are simultaneously their own bankers, transporters, managers and insurers. Despite the fact that private sector led growth was always considered a top priority, there was no common EU Pillar strategy on how to assist the local business community. The private sector never received the same amount of attention as did the old socialist industries and even fewer resources were allocated to gather information about its nature. In many respects Kosovo’s private sector might appear “black” or “grey,” but only because little effort was spent by international and local institutions on trying to understand this new private economy. In late 2000, DTI set up a Private Sector Development Team in each of the five regions and employed seven internationals and four local staff. These teams spent most of their time collecting information about other NGOs or micro credit institutions, but no time was spent on exploring the real economy. The concerns of private traders remained unheard and obstacles to private sector led growth remained unknown to policy makers in Prishtina. The policies that really affect the lives of Kosovo traders and producers were formulated in the offices of the Kosovo Customs Mission, discussed among tax administration officials and modified by new legislation developed by the Ministry of Labor and Social Welfare. Tax policies, the recognition of Kosovo license plates by other countries, free trade agreements and employment policies taken together created the framework within which private businesses can operate. In Prishtina town, 60% of micro enterprises are currently at the survival level.121 Any small market fluctuations or the introduction of a new tax policy might force them to end their business activities. The development of policies aimed at strengthening the private sector is hindered by the limited and uncoordinated efforts to gather basic information about Kosovo’s private economy. The whole system of business registrations and the official number of private businesses that applied for a provisional business registration is highly unreliable. According to tax administration officials about 10,000 of the registered 51,000 private businesses have disappeared again or never started any business activity.122 Estimates about the level of unemployment in Kosovo range from 30 to 60 percent. The number of people regularly employed by the private sector is equally unknown. As foreign assistance and international donor money is expected to decline sharply in 2003, and as international organizations continue to cut their staff and budgets, the pressure on the private sector to absorb the unemployed increases. As official employment in Kosovo’s socially owned enterprises is also expected to drop as a result of the KTA’s privatization and liquidation policy - it is only in the private sector that new jobs will be created and new tax revenues for Kosovo’s budget will be generated. If there is no future policy to address Kosovo’s high levels of unemployment and the poor state of infrastructure, the European Union’s first mission in Kosovo will fail. As Kosovo would drift further away from Europe, the promise of European integration will gradually lose its appeal and the carrot that now pushes for reform would wither away. The success of the EU Pillar depends on Kosovo’s private sector; in other words, the success of the mission will depend on UNMIK’s ability to establish clear property rights. Property is key Field research in Kosovo clearly shows that what the private sector is lacking is access to serviced land, clear property titles and major improvements in basic infrastructure. In Kosovo and all over the region, new businesses have been forced to ‘piggy-back’ on the centers of socialist development in order to gain access to infrastructure, while most of the municipality Administration and Governance in Kosovo 43 ____________ remains closed to development. There is a marked shortage of suitable premises available for private business, and small shops and businesses run out of space.123 In former Yugoslavia, the central planning of infrastructure development was conducted purely to favor the SOE sector. There was no tradition of investing in infrastructure to develop rural areas, and certainly not to encourage private-sector growth. The poor state of infrastructure is a serious constraint on private-sector development throughout most of the region.124 As a result, we can see wild urbanization along Kosovo’s main roads - causing uneconomic use of space, raising infrastructure costs and damaging both the ecological and agricultural profile of rural areas. Each day, more agricultural land is converted into construction sites or transformed into another wild industrial zone. The Kosovo Ministry of Agriculture claims that between 15.000-20.000 hectares of agricultural land have disappeared since 1989.125 Clarity of property titles would greatly reduce the transaction costs on the land market and free private capital. The prevailing confusion about the future of socially owned and publicly owned property results in real estate prices far exceeding the budgets of average Kosovar businessmen and absorbs valuable investment capital that could otherwise be spent on productive activities. Most of the old socialist laws regulating property relations are still applicable in Kosovo today. Half of Kosovo’s total landmass, or 466,000 hectares of land are publicly owned. This includes land underneath private homes in every town, huge parcels of land in various industrial zones as well as roads and rivers. All land covered by an urban plan within the boundaries of a city is by definition publicly owned. There is no private ownership on publicly owned land; owners of private homes only hold a user right - the right to use the property as long as the house is intact. Until today, all publicly owned land is still subject to the Law on Expropriation, and a competent authority can at any time expropriate the current user. The law foresees only limited compensation for the private “user.” Only land and properties outside an urban plan and thus out of reach of the Law on Land for Construction or the Law on Expropriation, has a clear and full ownership title.126 Only land that has not been nationalized, confiscated or allocated to agro-cooperatives remains fully privately owned, encumberable and tradable. Property is key to economic development. Clear property rights are the basis for social and political stability by making everybody a stakeholder. Clear property rights are closely linked to the rule of law and confusion over property rights is harmful to economic development. Hernando De Soto convincingly argued that unless property rights are clearly established, property is nothing but a dead asset. In Kosovo, like in other developing countries, “resources are held in defective forms: houses built on land whose ownership rights are not adequately recorded, unincorporated businesses with undefined liabilities, industries located where financiers and investors cannot see them. Because the rights to these possessions are not adequately documented, these assets cannot readily be turned into capital, cannot be traded outside of narrow local circles where people know and trust each other and cannot be used as collateral for a loan or a share against an investment.”127 Kosovo’s housing stock has an estimated value of around € 2 billion, but most property in Kosovo is a dead asset: It cannot be used to raise capital and it cannot be mortgaged.128 Vast areas of prime commercial real estate are locked in dysfunctional socially owned enterprises; 44 Administration and Governance in Kosovo __________ vast areas of publicly owned land in Kosovo’s towns are not freely tradable and encumberable. A new law on mortgages exists only in draft form. The private sector is starved of space and investment capital. The Special Representative of the Secretary General Michael Steiner even raised the issue in his inauguration speech delivered on 19 February 2002: “Property reform and privatization will have to be tackled. Every house, every shop and every inch of land should have a clear owner. Valuable assets should be developed. Otherwise, they just remain dead capital.”129 Property creation is peacekeeping It is UNMIK’s basic goal to create a viable economy under the rule of law. This entails replacing the old system with a functioning property regime. The failure to do so carries significant economic and social costs. Without clear property titles there will be no investment in Kosovo. Without tradable and encumberable property titles, the private sector will be starved off fresh capital. Without a government capable of enforcing property rights, shopkeepers and small businessmen across Kosovo have to resort to private security companies to protect their interests. The absence of clear, enforceable property titles creates a market for private protection agencies, substituting for the weak state. This undermines all international efforts to establish the rule of law and a safe and secure environment. Property creation is a political problem par excellence. Resolving the status of social ownership in Kosovo is not just a preserve for lawyers and economists. No genuine social change can be brought about without identifying and working with constituencies and local interests with a stake in such reforms. Who are these local constituencies in Kosovo? A study on the Bosnian municipality of Kalesija showed clearly that given the right conditions, the new private sector can be the main engine of growth. In Kalesija, the private sector expanded from 5 to 42 percent of total official employment within a few years.130 The private sectors is the key beneficiary of any property reform and private businesses will profit most from successful privatizations and liquidations that would free the assets of socially owned enterprises for private investment. Besides the private sector, municipalities would also benefit directly from clarity regarding municipal, socially owned and publicly owned property. Whereas before municipalities were able to allocate publicly owned construction land for commercial zones or social housing, today the newly established Kosovo Trust Agency131 as trustee of all publicly-owned and socially-owned enterprises and related assets, is in control of large parts of municipal property. Many of the prime real estate assets are now in the hands of the KTA and the KTA is burdened with a huge social and political responsibility to develop, sell, allocate and designate such parcels of land. Today, municipalities are left with the legacy of bad socialist planning and social ownership, but they are stripped of the means to tackle the problems. Most municipalities are still operating on the basis of urban and zoning plans that have been developed for the needs of socialist planning dating back to the mid-80s. Socialist Yugoslav town planning focused on large industries and not on the needs of the private sector. Kosovo’s private sector urgently needs a functioning property regime in order to develop and to provide new jobs for future generations. What is needed is a policy that addresses simultaneously two issues: create secure and clear property title and address the problem of Kosovo’s poor infrastructure. Administration and Governance in Kosovo 45 ___________ From reconstruction to development Kosovo has experienced an impressive and unprecedented flow of international reconstruction aid. More than 500 kilometers of roads have been repaired, water systems have been rehabilitated and more than twenty thousand houses have been rebuilt. However, not a single new kilometer of road has been paved and sixty percent of Kosovo’s population still lives in villages without any paved roads. Many of these village dirt roads become inaccessible on rainy days. In Kline municipality, only 19% of inhabitants are connected to the local water system. In Viti/Vitina municipality garbage collection is limited to Vitina town; there is no waste disposal system for the remaining 42 villages. In the absence of the state most local infrastructure in villages around Kosovo has been privately financed. Villagers themselves organize the collection of funds within the village and among the Diaspora community, call for tender offers and contribute their own labor to pave a road, install a sewerage system or build a youth center. In Stublle, the village committee has collected funds for 3 kilometers of asphalt, a youth center and a local ambulance. In Sllatina e Epërme, € 130.000 have been raised by the village religious leader for the construction of a sewerage system. Rural areas in Kosovo traditionally missed out on the benefits of Yugoslav development. Socialist planners deliberately ignored the needs of the private sector and forgot about the countryside. The economic incentives to pursue more than subsistence agriculture are so low that a high proportion of agricultural land in Kosovo has been abandoned altogether or sold dearly to the new private sector. The agricultural sector cannot absorb the unemployed, and unless the patterns of development become more even between towns, industrial zones and rural areas, Kosovo will continue to experience labor migration to towns and export its labor force to Europe.132 Managing ignorance What distinguishes effective development strategies from pure reconstruction is that they focus on developing local institutions and capacities. Only by understanding the strengths and weaknesses of existing structures is it possible to accomplish the lasting institutional change, which is a precondition for sustained economic development.133 Identifying your own information deficits has to be priority number one for every international organization aiming to reform an inherited institution or economic system. A sound and realistic reform agenda has to be based on accurate information about the reality on the ground. An economic strategy has to take stock of the status quo in the country, region or sector concerned. International organizations generally invest a lot in their human capital, their IT equipment and public relations. Too little time and resources are however spent on gathering basic information in the field, analyzing the available data, sharing these data with other agencies, seeking out local expertise and retaining institutional memory. The success of the European Union’s first mission will depend on its ability to learn how to collect information and manage ignorance better. A better-informed policy debate could have prevented the misguided policy of reconstituting workers’ self-management in socially owned enterprises. Donor resources could have been allocated more efficiently to improve rural infrastructure and access to serviced land based on more accurate information about the needs of the private sector. Property reform should have been on the agenda for the last three years. 46 Administration and Governance in Kosovo ___________ The real challenge of international institutions working in an unfamiliar and ever-changing environment like Kosovo, is to acknowledge ignorance and to manage it in the most effective way. The ad hoc nature and temptation of international missions to engage in continuous short-term fire fighting, clearly works against time-consuming efforts to gather and collect the necessary information for an informed policy debate. However, the long-term costs of a policy that failed as a result of ignorance outweigh the short-term benefits of quick solutions. The future of Kosovo depends on a strong private sector. None of the new provisional institutions of government, the education system or the healthcare system can function without a strong private sector. It is only in the private economy, where the tax base of this country is generated. In 2001, the private sector alone generated € 24 million in taxes. For the EU Pillar to successfully contribute to the economic development and stabilization of the region, as foreseen in Security Council Resolution 1244, more time and resources need to be allocated to the triple challenge of gathering information, consulting with key stakeholders and understanding the reality on the ground.134 The main lesson from the EU Pillar’s short-lived attempt to develop Kosovo’s private sector has to be that, in future missions, international experts have to switch off their computers and go out into the streets to discover the real economy. Private sector development is about identifying the real constraints on private-sector development. The private sector of any country is not something esoteric or abstract, it is about traders, producers, consumers, electricity providers, tax officials and infrastructure planners. REHABILITATION IN MITROVICA Mitrovica, and its future status remains one of the most intractable problems in Europe, complementing the tremendous obstacles facing Kosovo as a whole. The situation, therefore, deserves individual attention due to the severity of its obstacles, particularly in the areas of security and the rule of law, administration and governance, as well as interethnic relations. This section addresses how approaches taken in other parts of Kosovo were not applied to Mitrovica and analyze what this neglect means for the stability of the area. Opting for such stability requires the application of sustainable solutions to deep-rooted problems; these solutions must be sought out through research and then implemented by all stakeholders involved, including UNMIK, KFOR, the Kosovo government, the local administration and Belgrade. Mitrovica is located in the north of Kosovo, having an estimated pre-war population of about 300,000 of which about 43,000 were Serbs.135 The Ibar River runs across the high plains of Kosovo and passes under the five bridges of Mitrovica which in better times were used by Kosovo’s population to cross the river. The traditional function of the bridges was significantly different after the war, serving as the center piece for a peculiar surrounding consisting of international peacekeepers standing guard at armed checkpoints with tanks and stumbles of razor wire deployed to prevent people from using the bridge as they once did. During the conflict, the Kosovo Albanian majority residences suffered heavy damage, with an estimated 65% of homes looted and destroyed, while the predominately Serb neighborhoods were barely touched.136 Cynicism towards reaching a political solution that is satisfying to both communities, in an area where the “outsider has been reduced to hatred and revenge and a river with a divide both narrow and vast,” prevails. 137 Even with the presence of foreign troops, there have been strict rules limiting movement in between zones.138 The city has subsequently accommodated three administrative structures, -- Kosovo Albanian, Kosovo Serb and the UN civil administration. The northern area of the town now contains about 12,000 Kosovo Serbs, 3,000 Kosovo Albanians, 600 Turks, 500 Roma, as well as some 5,000 Internally Displaced Persons (IDP’s) from the north.139 On the other hand, the South contains close to 50,000 Kosovo Albanians, with 300 Serb families who had been residing there before the war.140 In recent weeks, the UN’s SRSG, Michael Steiner, has declared UNMIK authority in northern Mitrovica, reporting on the official closing of all Serbian Belgrade-funded parallel institutions that took root subsequent to the war. On 2 October 2002, Steiner announced his seven-point plan for Mitrovica, emphasizing that the people of Kosovo must decide if they long for “stagnation or change.” He argued that leaving things as they are would contribute to “lawlessness, insecurity, fear and political marginalisation.”141 The Albanians and Serbs immediately responded to the plan, mostly with skepticism; Mitrovica Serbs were against “the manner” in which the plan was formulated, while Albanian inhabitants were pessimistic, not because of the content of the plan, but because of “the failure of previous plans” announced by UNMIK.142 In other words, Steiner’s plan incited concern within the Albanian community, mainly because the international community “hasn’t been efficient thus far, in penetrating northern Mitrovica with its administration.”143 On the other hand, the Serbs see the plan as more of an ultimatum as the Serbian representatives argue that they were not consulted with the details of the plan prior to the declaration.144 Notwithstanding the fact that the UN has finally claimed its mandated authority over the city, Kosovar inhabitants question the determination and willingness of the international community in moving towards the achievement of expressed goals. This requires a 48 Administration and Governance in Kosovo __________ determined approach on UNMIK’s part, in putting formal ideas into meaningful action; this will be a gradual process but it must be a visibly successful one. This process started on November 25, 2002, as UNMIK declared its authority on Mitrovica in order to bring “normalcy to the city.”145 Security One of the areas that requires improvement is security. Authority needs to be reintroduced in the North of Mitrovica in order to create a safe and secure environment that will stimulate the return of refugees and the freedom of movement. The reasons for the present situation stems from organized efforts to maintain ethnic separation after the war by forces loyal to the then Serbian leader, Milosevic. Serb agents set up illegal checkpoints, restricting movement between the two sides. This group of Kosovo Serbs -- the “Group from the Bridge – was “able to mobilize hundreds of people within minutes to prevent any attempts by Kosovo Albanians from moving into the North of the town.”146 The Group’s main preoccupation continues to be its opposition to the return of Kosovo Albanians to the North; the group has argued that a return policy would result in an expulsion of all Serbs from the North and eventually the whole region.147 ICG argues that “if the Serb hardliners can demonstrate that KFOR and UNMIK are incapable of creating a secure environment for non-Albanians and functioning institutions in strife-ridden areas, then Belgrade can push its case more forcefully.”148 The group has engaged in several criminal acts north of the city, including the intimidation of members of the local Serb population who have cooperated with UNMIK.149 Compounding the problem is the fact that there has been little effort on the part of KFOR and UNMIK to “crack down” on the group.150 This dilemma requires the arrest of radical criminals who are pushing for the partitioning of Mitrovica and who may stand in the way of functioning multi-ethnic institutions. Such measures are necessary in order to build a sense of security in people who want to return home. A similar dilemma existed in East Timor, where angry militias were not ready to admit defeat, causing troops to spread to the border region to block their attempts to sabotage the UN efforts there. Jonathon Steel deemed the efforts to counter Indonesia efforts to disrupt UN efforts as “vital in capturing former militia members who attempted to go back to their villages as returning refugees.” 151 Steiner has argued that UNMIK and KFOR have taken a more robust and active approach towards the so-called “bridgewatchers” that has helped to decrease the level of violence in the North.152 However, as stated in the security section of this assessment, UNMIK claims that only KFOR can act and conduct military operations by force. At the same time, UNMIK police has no power to act in the North due to the non-consent of the Serbian population, another challenge in building a multi-ethnic police force.153 An incident that occurred on 8 April 2002 reinforces the obstacles that remain in the area of security and safety, regardless of the fact that violence has abated since 1999.154 Violence erupted after UNMIK attempted to arrest a known “bridgewatcher” for inciting riots in February 2002; the individual was also accused of other crimes committed prior to the incident.155 Consequently, he resisted arrest, which provoked a riot and ignited the crowd to throw grenades at the police, injuring a total of 26 UNMIK officers and some Serb civilians. UNMIK, KFOR and the Serbs responded to the incident in different ways; however, little was done to address the severity of the incident.156 SRSG Michael Steiner’s response consisted of a mere condemnation, with no official statement; KFOR command claimed that they were “not informed of the action, which meant Administration and Governance in Kosovo 49 ___________ there was no KFOR intervention, although they “did safeguard their existing checkpoints and called for reinforcements.”157 UNMIK police, on the other hand, discerned the incident as “routine” and described their effort to enforce the rule of law as consistent. Notwithstanding these claims, “UNMIK administration and KFOR officials privately claim that the attack was attributed partly to a poorly led police action and a breakdown of coordination between the police and KFOR.”158 The roots of failure can be traced to the beginning of the NATO mission with insufficient measures taken by KFOR in providing security in the city; failing to insist on an undivided city. A similar strategy was adopted in Bosnia in a “belief that this was the best way to maintain security.” 159 Perhaps for the short term, but reversing the strategy has proven more difficult. These security issues tie directly into the concerns of inhabitants who desire to return to their former homes, to obtain access to jobs and better services. If UNMIK continues with its administration without addressing these issues, multi-ethnic cooperation will not be possible, especially if more determined security measures are not undertaken in order to prevent groups from harming one another. As UN administrator for Mitrovica, John Rodgers pointed out, things such as freedom of movement will not come over night but the important thing is putting together the basic structures that are needed. This includes adequate security to be provided by UNMIK and KPS with the assistance of KFOR when necessary. Without security and the rule of law, taking advantage of freedom of movement and access to institutions will not be feasible, even in the long future. The first two points of Steiner’s plan for Mitrovica tie directly into these security issues; the first point states that UNMIK police and KFOR should keep close watch of the bridge, while the second point states that the Serbs will move forward in being integrated into KPS as full time police officers.160 The former has been the case until now, but criminals still reside at the bridge and continue to interfere with law and order with little measures taken to arrest them; therefore, more aggressive action needs to be taken in this respect. Similarly, for a multi-ethnic force to function in harmony, the force must fully accept that it has been integrated in the Kosovo Police Force, breaking ties with Belgrade and working together with other ethnic groups. They must also push for the arrest of known criminals in the north as they do for any other criminal in Kosovo. In East Timor, the UN peacekeeping mission has been successful in its primary goals of the mission, which included protecting East Timorize from reprisals by the militias and “providing an environment for most refugees, one-third of the population, to come home.”161 This happened because peacekeepers continued to guard East Timor’s borders from infiltrating trouble-makers. Such an approach is essential for Kosovo over the next couple of years. Along these lines, it is important that peacekeeping missions are generously and quickly funded from the start. As seen in Srebrenica, lack of adequate troop strength to provide a “safe haven” intensified the problem. If security is not exercised adequately and properly, the goal of refugee return will not be possible. At the same time, judicial institutions and tribunals are central to the future success of a democratic Kosovo. UNMIK must demonstrate its determination in asserting its authority in North Mitrovica in order to stimulate the improvement of rule of law and functioning judicial systems. Secretary General Annan declared that sustained “support for the missions’ fight against crimes, through criminal investigation leading to arrests and capacity building of the local police and the judiciary, will lead Kosovo towards normalization.”162 These gestures must be backed with action. 50 Administration and Governance in Kosovo ___________ Sustainability is especially vital in the area of security in Mitrovica. The significance of sustainability is also illustrated in the case of Cambodia where the United Nations attempted to establish a strong state and to protect human rights in the country.163 Democracy was to be the basis of these efforts. This process failed because the efforts were not sustained. While constructive intervention was to be part of peacebuilding, in the view of local people and elites, such intervention was not “moving in the rights direction,” mainly because some sectors felt threatened by the process. Mitrovica remains a challenge to the implementation of the Security Council mandates on Kosovo, with the mission seeking to stabilize the situation on the ground by continuing dialogue with the Federal Republic of Yugoslavia. The Albanian relationship with the French peacekeepers remains ambivalent with many Albanians claiming the troops lack objectivity in their mission and do little to tackle different criminal activity by violent and aggressive groups.164 At the same time, the Serbs will not let anyone cross the bridge without interrogation, “deciding who may enter and threatening others.” The results of UNMIK’s recent action remain to be seen, but what is clear is that there will be diminutive if meaningful enforcement of its mandate in the North is not taken. Administration UN Administrator for Mitrovica, John Rodgers, explained that in terms of administration, the most important factor is making sure proper services are provided to all persons throughout the municipality. However, he argues that what remains to be managed is the “right to accept and receive these services.”165 Rodgers is optimistic in Steiner’s plan for Mitrovica in bringing order to the city, although he emphasized that such a process requires time and continuous cooperation; he is not alone in this optimism. Among other issues is the issue of employment within UNMIK institutions with seventy job vacancies recently being announced. The UNMIK administration has explained the procedure for hiring new employees as being based on a potential candidates credentials; they must be “skilled and competent people, making sure that jobs that are done, are done properly.”166 He stressed the importance of hiring people who can complete jobs efficiently and effectively, and in a professional manner. However one concern is related to the likelihood of creating a multi-ethnic work force. If this issue is not addressed, the working environment may consist of UNMIK working with the Serbs alone in the north, stimulating the development of a separate municipality, and embarking in such a set-up that will stimulate a dangerous precedent-partitioning of the city. UNMIK was unable to establish authority in the North, which meant they were unable to maintain a multiethnic hospital, court structure, and other public services; institutions such as education and health report to respective ministries in Belgrade.167 This has driven frustration in the Albanians who have been denied access to these institutions; thus, lack of emloyment in the South further aggravates the feeling of alientation of the ethnic Albanians. The main concerns of the Albanians in the South is returning to their property, and making use of facilities, such as health and education insitutions; but more importantly, they are seeking improvement in their economic well-being. Most see UNMIK’s plan as neglecting to address this issue of employment, moving backward in improving the economic standings of many unemployed Albanians now residing in the South.168 While building common institutions is important, the international community must not neglect the issue of economic development. In East Timor, many of the problems arose from inadequcies in the UNTAET’s Administration and Governance in Kosovo 51 ___________ internal management, insufficent training to Timorese and a tendency to highlight institution building too far above social and economic development. These operations and the insensitive behavior of many of its staff prompted justifiable complaints from local people. Points five and six of Steiner’s plan for Mitrovica include improving the economic standings of Mitrovica, as well as providing investment. At the same time, Steiner argues that without legitimate institutions, the implementation of such investment will not be feasible; however, the international community should more strongly promote such an issue since this investment could directly contribute to the making of stable institutions and their sustainability, as well as people’s economic standings. UNMIK must take into consideration all aspects of the problem if lasting change is to eventuate; it is clear that if economic support is not sustained in the area, the situation will not progress. This consideration is especially necessary, since the Serbs lack the economic capacity to support Serb refugee return while “the international community lacks political will and resources, and facing a draw down in funding, will probably not be able to invest substantially either in refugee returns or economic growth.”169 Ramadan Kelmendi, LDK representative in Mitrovica claims that UNMIK has illusions in regards to the problems involved, explaining UNMIK’s actions as steps that should have been attempted three years ago. He claimed that there are 23,000 apartments and houses that need to be returned to, including their renovations, access to health and faculties, employment, authority and competence.170 Prime Minister, Bajram Rexhepi explains that things will happen gradually in moving from the zero level of activity, but that stakeholders need to be realistic, especially in respect to the disposal of parallel institutions. At the same time, the return of refugees will be a slow process that must be countered not only by the international community and governments but also by the citizens who need to be ready to do their part.171 Institutions need to be mixed in their ethnicity with expertise, not party affiliation being the basis for the hiring and selection within these institutions. Point three of Steiner’s plan focuses on decisions that are jointly to be made at the municipal level under UNMIK auspices, guarding against outvoting the majority coalition agreement between the Serb and Albanian political parties in the Municipal Assembly. The agreement would assume that, irrespective of their size, the communities have a say at the municipal level. However, Steiner concludes that in order to move forward, political participation in the municipality is necessary, and that without legitimate institutions, there can be no descent and no investment. The Local Serbs and Belgrade On 14 November 1999 The New York Times reported that Slobodan Milosevic was assisting the organization of security and ordinary life in the northern part of the Mitrovica and in northern Kosovo itself.172 Most Westerners argued that Belgrade was using the North, which it supplies with money and goods, to try to destabilize Kosovo or to prevent its “normalization” in the aftermath of Milosevic's bloody campaign against Albanians. It seems that such a strategy is still in place, three years later. It was Milosevic’s aim to attempt to at least keep part of Kosovo within Serbia and under the influence of Belgrade, which feels that Western governments have flouted Serbian sovereignty over the region with the fear that Kosovo will eventually desire real independence. Much of the culpability for the difficulties in Mitrovica is attributed to hard-line elements among the Serbs in both Belgrade and Mitrovica who have been “unwilling to submit to UNMIK rule, or to accept integration of the northern part of the city into Kosovo society and 52 Administration and Governance in Kosovo __________ political institutions.”173 At the same time, Belgrade is viewed as “a crucial factor in determining the political behavior of Kosovo’s remaining Serbs.”174 ICG describes the reasons for this as “partly political, partly ethnic and partly criminal” and emphasizes that UNMIK and KFOR must understand these “interlocked motivations” to deal successfully with Mitrovica. At the same time, Belgrade has been using scarce resources to support parallel institutions with lack of transparency in terms of the budget that has concealed the distribution of these funds. Belgrade has funded Kosovo with at least 50 million during the first months of 2001 and as many as 29,800 people in Kosovo were on the payroll of Serbia in 2001. While UNMIK’s resolve in the area has been weak, they did open a community office in northern Mitrovica in early 2002. However, even though such an office exists, “Belgrade uses every available means to maintain its grip on the North with the intention of partitioning the province,” as opposed to promoting strategies to unite Mitrovica.175 Local Serbs claim that the international community has not provided sufficient protection against violent attacks, leading to distrust; therefore, they turned to Belgrade for civil authority. In the past, Serbia has paid the salaries of northern Mitrovica civil servants even though elsewhere the UN has been in charge. Nebojsa Covic, Serbia’s Vice Premier and the Minister responsible for Kosovo, “salutes” Steiner’s plan, but he also challenges it, since Serbian representatives were not involved in its formulation. The Kosovo government expresses Belgrade’s intention to once again sabotage the international community and the SRSG’s efforts for normalizing the situation and unifying Mitrovica.176 Although it is too soon to analyze the progress that has been made since UNMIK declared its authority in northern Mitrovica, some have claimed that the documentation of 120 workers of parallel institutions have been moved to residential buildings near the hospital, with UNMIK being aware that these illegal structures exist. UNMIK denied these accusations, claiming that parallel institutions will no longer exist, and that the documentation will remain enact due to the benefit of the Serbs in this respect.177 However, they stated that they would deal with the location of such documentation after buildings are renovated; UNMIK should not delay addressing this issue. UNMIK must consider the distrust that has been expressed towards them and the support that the Serbs have received from Belgrade, leaving the impression that “the Serbs lack an equivalent incentive to cooperate in its arrangement,” even if UNMIK has now made clear its priority in extending the rule of law throughout Kosovo.178 Albanians and Serbs have little confidence in the Albanian-controlled municipality treating them with equitably or providing them with services. The Serbs in northern Mitrovica are far from satisfied to losing the whole of Kosovo to the Albanians, claiming that if Kosovo became fully independent from Yugoslavia, they would seek to secede from Kosovo. In this regards, some Western officials make a case for the partitioning of Kosovo, giving the Serbs the north of the region and the Albanians the remaining parts. This is a dangerous scenario, which is not democratically feasible; hence constructing such a division would set a dangerous precedent. The first obstacle to any easy partition plan around Mitrovica is ringed with mines, providing gold, zinc, and other precious metals. These mines conquered by Hitler in WWII and vital to the Yugoslav economy during the Cold War, have taken on extreme mystical importance to the locals.179 From the beginning, Steiner stressed that “business and investment would not flow into the gray zone of illegal parallel structures.”180 Overall, Kosovo’s economy as a whole is still far from being self-sustainable with a framework of privatization being created. Administration and Governance in Kosovo 53 ____________ At the same time, privatization must be carried out in a transparent manner to enable the international community to follow development. The Kosovo Albanians The economic challenges are more visible to decision makers away from Kosovo, while cultural and political issues are more clear to decision makers within Kosovo with rapid communication not solving the challenge of developing a common perspective from a distance.181 For the Albanians, “the need to construct some form of self-sufficient economy and alleviate unemployment made ownership and control of Trepca an economic necessity.”182 This includes the existence of foreign investment. In this respect, the Serbs have attempted to “separate some of the processing capacity from less profitable raw material production and “strengthen ties to the rest of Serbia, reopening mutually beneficial trade links and providing some economic security.”183 The fate of Trepca along with its resources divided between majority Albanian and majority Serb areas will continue to be an important issue for Kosovo. The Kosovo government calls for freedom of movement, a disbanding of the “bridgewatchers,” return to institutions such as health and education, and return to property. More pessimism on the Kosovo Albanian part deals with the spreading of rumors, regarding the signing of an agreement between Covic and UNMIK. This has caused fear within the Albanians that UNMIK has agreed to the creation of a new municipality in northern Mitrovica. This is something that would be unacceptable to the ethnic Albanians; if this occurred, violence would result. The lessons learned by the following analysis include: (a) UNMIK has not been efficient in effecting northern Mitrovica with its administration leading to diminutive progress in the area. Action to eliminate parallel institutions and to formulate and administer lawful structures, generally recognized throughout the territory of the municipality, has only recently been declared. (b) Although violence has decreased, obstacles remain in the area of security and the rule of law, especially in arresting criminals responsible for violence; lack of coordination between UNMIK and KFOR has contributed to the problem. (c) Belgrade has continually supported the parallel structures in northern Mitrovica, including radical elements, unwilling to submit to UNMIK rule or to accept the integration of northern Mitrovica. (d) The rule of law has not been extended to northern Mitrovica, denying freedom of movement, return of refugees and access to health and education institutions and employment. (e) UNMIK has not been transparent in the actions it has taken to establish authority in northern Mitrovica. The following areas must be sustained if meaningful change is to transpire, including refugee return, access to institutions and investment. Cooperation between all actors must progress in implementing an effective and multi-ethnic administration. Efficient and democratic leadership is a significant factor in striving for change. 54 Administration and Governance in Kosovo __________ Recommendations: (a) To UNMIK: Stand firm in establishing authority in Northern Mitrovica by putting formal expressions into meaningful action. This includes practicing sustainability in formulating and administering structures generally recognized throughout the territory of the municipality; confirming the elimination of parallel structures; and opting for a multi-ethnic working environment by treating both ends of Mitrovica as one municipality (b) Practice transparency in informing the public on the steps that are taken in attempting to establish efficient and democratic governance in Northern Mitrovica. (c) UNMIK police: Take more aggressive measures in arresting criminals, where there is sufficient evidence at hand, as well as cracking down on parallel institutions that continue to exist and promoting the rule of law. This entails continual cooperation with KFOR, who need to support the process and provide security and adequate border control. (d) To Belgrade: commit to the authority of UNMIK and cease funding of parallel institutions. (e) Kosovo Albanians and Serbs: accept the integrity of the Mitrovica municipality and the return of all refugees. CONCLUSION As the international community is faced with potentially another disastrous post war situation in Iraq, lessons learned from what has transpired in Kosovo over the last three years becomes of paramount importance. The international community’s allergic behavior towards criticism of its activities in Kosovo reflect a general reactive sensibility that is not conducive to improving the operational capacities of the many agencies bound to be involved in post conflict Iraq. As the above has suggested in detail, there are a number of factors that need to be addressed when taking on the responsibility of rebuilding a society, which has faced decades of social and economic marginality as well as political and cultural repression. In failing to adapt a more flexible and at the same time, ethnically consistent approach to the administration of Kosovo, the noted failures have intensified the factors that initiated the conflicts of the late twentieth century in the Balkans, rather than dilute them. Failing to adopt an appreciation for the fundamental dynamism of human communities along with the every-changing needs of these communities has left Kosovo’s inhabitants often at a loss as to where their future will take them. Tragically, the momentary sense of security they may have felt in the summer of 1999 has vanished. The subsequent intensification of collective fear and a growing sense that what has transpired over the last three years does not warrant the explicit support of Kosovo’s population should serve as a warning when practitioners of future projects enter into potentially far more hostile environments. Iraq, much like Kosovo, is a complex and ethnically diverse entity that will not be easily “reformed;” assumed sensibilities will always turn out more complicated than initially assumed. The appropriation of state and society building roles by the international community has exhibited the most fundamental flaws of the developed world in respect to its relationship to the other societies. The events on September 11, 2001 should be a warning that institutional neglect and the arrogance and patronizing attitudes of the post industrial world will not translate into effective governance. Unless future stake-holders in the resuscitation of Iraq or other future missions indulge in long-term and thorough research into the dynamics and collective as well as individual needs of the subject population, Kosovo’s lessons will not have been learned and the stability of other regions of the world will be permanently under threat as is currently the case in the Balkans. The above contribution has illuminated a parochialism in the administration of post conflict societies that if allowed to persist, will enable anti-democratic forces around the world to continuously threaten the humanitarian interests of the people most in need. 56 Administration and Governance in Kosovo __________ Abbreviations and Language clarifications CFK Constitutional Framework of Kosovo CIVPOL UNMIK Police COMKFOR Commander KFOR DTI Department of Trade and Industry EU European Union EUMIK European Union Mission in Kosovo FRY Federal Republic of Yugoslavia FYROM Former Yugoslav Republic of Macedonia ICG International Crisis Group IDP internally displaced persons JCR Joint Committee on Returns JIAS Joint Interim Administrative Structures (joint UNMIK/Kosovar structures serving as interim ministerial functions) KFOR Kosovo Force (NATO) KLA Ushtria Çlirimtare e Kosovës (Kosovo Liberation Army) KTA Kosovo Trust Agency OSCE Organization for Security and Cooperation in Europe OSRSG Office of the Special Representative of the Secretary-General HPD Housing and Property Directorate HPCC Housing and Property Claims Commission IC International Community LCO Local Community Officer LWG Local Working Groups on Return KPS Kosovo Police Service KPSS Kosovo Police Service School LDK Mr. Rugova’s Party Minority is used to describe a numerically inferior community and does not necessarily bear political and legal implications. It is often used for local minorities, members of the majority situation within a minority dominated geographic area. MSA Mission Subsistence Allowance PISG Provisional Institutions of Self-Government “RAE communities” is used to denote Roma-Ashkali-Egyptian communities under one heading as they share most problems. RWG Regional Working Groups on Return SC/JCR Steering Committee of the Joint Committee on Returns SRSG Special Representative of the Secretary-General of the UN UCK (KLA) Ushtria Çlirimtare e Kosovës (Kosovo Liberation Army) UNHCR United Nations High Commissioner for Refugees UNMIK United Nations Interim Administration Mission in Kosovo UNSC United Nations Security Council Names of towns/villages are quoted according to OSCE’s custom. New names, after the war, are occasionally used if the Editorial Board deems that this version has entered into widespread use. About KIPRED The Kosovar Institute for Policy Research and Development aims to support and promote democratic values in Kosova by offering trainings, conducting research and independent analysis, in order to help policymakers develop professional public policy. A Professional Council ensures overall direction and criteria for KIPRED and serves as an Editorial Board, members of which also engage in team research. KIPRED focuses, and does not restrict itself to: development of political parties, public administration and local government, public policy, interethnic relations, regional cooperation and political economy. KIPRED publishes three types of analyses: policy briefs, policy reports/analyses and academic papers. KIPRED also translates analysis from abroad for Albanian readers. For students, policymakers and independent experts, KIPRED offers an online catalogue of resources through its web site. The training pillar is dedicated to professional development of politicians and political activists. One of the main projects of the Institute is the Internet Academy for Democracy, developed in cooperation with the main donor or KIPRED, the Olof Palme International Center. Created by Swedish academics and political practitioners the Academy offers interactive modules of teaching in the following fields: democracy and ideologies, structures and political party organization, electoral campaigns and ethics in politics. Contact Information: KIPRED/KCSF Kodra e Diellit, Rruga 3, Lam. 36, Prishtina, Kosova/o Tel/Fax: +381 (0)38 555 887; kipred@hotmail.com; www.kipred.org Endnotes: 1 Henceforth “the resolution”. 2 Caplan, Robert, “A New Trusteeship? The International Administration of War-torn Territories.” Adelphi Paper 341, February 2002. 3 Stedman, Stephen “Spoiler Problems in Peace Processes,” International Security 22, no. 2 (Fall 1997): 5-53. 4 Caplan 76. 5 Caplan 51-52 6 Caplan 51. 7 Darby, John “The Effects of Violence on Peace Processes,” Washington D.C.: United States Institute of Peace 2001, p: 8. 8 Caplan 57-58. 9 Caplan 49. 10 Caplan 66. 11 UNSC Resolution S/RES/1244 (1999) of 10 June 1999, in: M. Bothe/ Th. Dorschel: UN Peacekeeping – A Documentary Introduction, The Hague/ London/ Boston, p. 249. 12 The Independent International Commission on Kosovo (2002 August 15) The Kosovo Report, Oxford: 2000, 67. 13 S/1999/649 of 7 June 1999 [The so-called “Chernomyrdin-Ahtisaari-Document”] to be found at 14 It is a disputed matter whether UNMIK is based directly on Chapter VII of the UN-Charter or has its legal basis in the agreement of the Yugoslav Government. See Perrit H.H. and J.M. Scheib (2000, May 8). “Rebuilding Kosovo - UNMIK as a Trustee Occupant, Paper delivered at the Southeastern European Business Conference” in Matheson M.J. “United Nations Governance of Post-Conflict Societies” The American Journal of International Law 93, 857. 15 UNMIK Regulation No. 1999/1 of 25 July 1999, 1.1. 16 UNMIK REG1999/1, 4. 17 UNMIK REG 1999/1, 3. 18 UNMIK Regulation No. 1999/24 of 12 December 1999, 1.1. 19 Report of the Secretary-General 1999, S/1999/1250, Par. 55. 20 UNMIK REG 1999/24, 1.1. 21 UNMIK REG 1999/24, 1.2. 22 In order to involve the local legal community in legislative affairs, UNMIK established the Joint Advisory Committee on Legal Matters as a consultative body for the SRSG, and which consisted of local lawyers mainly with academic and less with practical background. 23 For a full and detailed overview on this legislative activity see the official UNMIK webpage at 24 FRY Official Gazette 78/29. 25 FRY Official Gazette 80/6. 26 FRY Official Gazette 77/4. 27 See UNMIK Regulation No. 2000/68 on Contracts for the Sale of Goods. 28 A general Criminal Code was prepared by UNMIK with the support of the European Council. The Criminal Code is still a draft and is despite persistent demands of the Kosovar legal community still not promulgated. 29 UNMIK Regulation No. 2001/9 of 15 May 2001, chapter 8.1. 30 UNMIK REG 2001/9, 12. 31 UNMIK REG 2001/9, 5.1. 32 UNMIK REG 2001/9, 9.1.26. 33 UNMIK REG2001/9, 9.1.45. 34 Hoefer-Wissing, Neithart. “UNMIK holds on to a share of power,” Focus Kosovo, 2002 February, 11. 35 UNMIK REG2001/9, 9.1.26f. 36 UNMIK REG2001/9, 8.1 and 8.2. 37 Stahn, C. “The United Nations Transitional Administration in Kosovo and East Timor – A First Analysis,” in: Max Planck Yearbook of United Nations Law, (2001) Vol. 5, 105 (109). 38 As is well known, there is ambiguity in Resolution 1244 (1999) in relation to the future status of Kosovo and its aim to establish a “substantial autonomy” for the region. Bernard Kouchner is said to have read the text of the UN Security Council Resolution 1244 twice every morning to understand what was meant by “substantial autonomy.” (Richard Caplan 77). 39 Conducted as a result of the Military Technical Agreement signed between the International Security Force (KFOR) and the Governments of the Federal Republic of Yugoslavia and the Republic of Serbia, signed on June 9th, 1999. Source: Administration and Governance in Kosovo 59 ____________ 40 A demilitarization agreement was signed between the KLA leadership and the IC on June 20th, 1999. Source: 41 Gordon IV J., McGinn J., Nardulli B., Perry W., Pirme B., Disjointed War: Military Operation in Kosovo, 1999, Chapter Five: Enforcing the Peace, RAND 2002, 42 NATO, Morning Briefing, Jamie Shea, June 9th, 1999, Source: 43 KFOR, Structure: 44 Lord Robertson of Port Ellen, Secretary General of NATO, Kosovo One Year On: Achievement and Challenge, pp. 16, March 21st, 2001 Source: 45 Ibid, p. 11. 46 Ibid, p. 16. 47 KFOR , Objectives and Mission, 48 Lord Robertson of Port Ellen, Secretary General of NATO, Kosovo One Year On: Achievement and Challenge, pp. 16, March 21st, 2001 Source: 49 Source: 50 Lord Robertson of Port Ellen, Secretary General of NATO, Kosovo One Year On: Achievement and Challenge, pp. 16, March 21st, 2001 Source: 51 Caplan, op cit., 77. 52 Source: 53 UNMIK Police Annual Report 2000, p.: 11, Source: www.unmikonline/civpol/reports/report2000.pdf 54 Source: (Police Personnel) 55 UNMIK Police Annual Report 2000, p.: 11, Source: www.unmikonline/civpol/reports/report2000.pdf 56 Source: www.civpol.org/unmik/balance.html 57 Source: 58 Chappel, Derek and Barry Fletcher. Personal interview, UNMIK Police Spokespersons, September 2002. 59 Ibid. 60 The Organized Crime Intelligence Unit was established on January 15th, 2001, UNMIK Police Press Release, January 15th, 2002. 61 Chappell and Fletcher. 62 International Crisis Group “Finding the Balance: The Scales of Justice System in Kosovo” p. 5. 12 September 2002, ICG Balkans Report No. 34, Pristina/Brussels. 63 Chappell and Fletcher. 64 ICG, “Finding the Balance: The Scales of Justice System in Kosovo”. 65 Chappell and Fletcher. 66 Gordon et al. 67 Chappell and Fletcher. 68 Lynch, US General, Personal Interview, Prishtina, October 2002. 69 Koha Ditore, p. 1. November 16th, 2002, Prishtina. 70 Chappell and Fletcher. 71 For details see: UNMIK Police Annual Report 2000, www.unmikonline/civpol/reports/report2000.pdf 72 Chappell and Fletcher. 73 KPS Human Resources Statistics Weekly Report, September 14th, 2002. 74 UNMIK Police Press Release, September 18th 2002. 75 Chappell and Fletcher. 76 Morina, Shefki, Kosovo Police Service Spokesperson, Personal Interview, September 2002. 77 Gordon et al. 78 Coster, Dennis, Personal Interview, Chief Deputy Commander of US Police Contingent, UNMIK Police. 79 Morina. 80 Chappell and Fletcher. 81 Morina. 82 UNMIK Police, “Enhanced Comprehensive Development Framework Matrix for Kosovo Police Service” 2002. 83 UNHCR/UNMIK (2001 January 13). Joint Committee on the Return of Kosovo Serbs: Framework for Return 2001. Prishtina. 84 United Nations Interim Administration Mission in Kosovo (2002 May 17). “E drejta për kthim të qëndrueshëm: Fletëkoncept [The rigth to sustainable return: A concept paper,” Zyra e kthimit dhe e komuniteteve [The office of the returns of communities, The Office of the SRSG], Prishtina. 85 Radosavljevic, Nenad (2000 October 28). Platforma za dugorocno resavanje pitanja Kosova. [Public letter]. 86 United Nations Interim Administration Mission in Kosovo (2002 August). “Returns to Kosovo: a new approach (as of August 2002),” Focus Kosovo, Pristina. 60 Administration and Governance in Kosovo __________ 87 Manuel, Susan (2002 August). Pioneers pave the way. Focus Kosovo, pp.: 13-14. 88 UNHCR/UNMIK 2001. op cit. 89 ibid. 2. 90 Ibid. p.: 2. 91 Ibid. 92 Palokaj, Augustin (2002 June 17). “Projektet e kthimit të serbëve në Kosovë do të dështojnë në qoftë se nuk pranohet realiteti (Return projects of Serbs to Kosovo will fail if the new reality is not accepted),” Koha Ditore, No. 1790, 3. 93 For a declaration regarding RAE communities see UNHCR/UNMIK (2000 April 12). Declaration from Humanitarian Round Table (also “Platform for Joint Action regarding Kosovar Roma, Ashkalija and Egyptian Communities” and “Return of Roma, Ashkaelia and Egyptians Statement of Principles”), Prishtina. 94 UNMIK 2001, op cit. 2-3. 95 UNHCR/OSCEa. Ninth Assessment of the Situation of Ethnic Minorities in Kosovo: Period covering September 2001 to April 2002. 96 UNHCR/OSCE (26 March 2001). Assessment of the Situation of Ethnic Minorities In Kosovo: period (covering October 2000 through February 2001), 2. 97 Bytyçi, Syzana (2002 August 30). “Beogradi vendos për drejtorin në Graçanicë,” Koha Ditore, No. 1864, p: 1. 98 Lessons Learned Analysis (LLA) Unit of the European Union Mission in Kosovo (2002 8 August 2002). The Ottoman Dilemma: Power and Property Relations Under the United Nations Mission in Kosovo.” 99 Mustafa, Artan (2002 September 6). “Demaçi: Pa definimin e statusit politik të Kosovës, kthimi i serbëve do të jetë i vështirë (The return of Serbs will be difficult without defining the political status),” Epoka e Re, No. 729, 3. 100 UNMIK 2001, Op cit. Annex 2. 101 Ibid. 102 What constitutes a strategic location is not always easily defined as it has a lot to do with the perception of the communities themselves. It should be understood to include properties which link or bridge clusters of minority houses to each other or to services such as shops or clinics. The fact that the sale of such a property would have the result of reducing further still the general area inhabited by the minority group obliging them to transit a majority area is a common basis upon which communities classify certain properties as strategic. UNHCR/OSCE 2001 op cit. 16. 103 Savezni Zavod za Statistiku (1995). Popis stanovnistva i domacinstva u SR Jugoslavije prema popisu 1991, Stanovnistvo 47, Centar za demografska istrazivanja Instituta drustvenih nauka, Beograd, 213. 104 UNHCR/OSCE (1999 September). Assessment of the Situation of Ethnic Minorities in Kosovo. 105 Uka, Afërdita and Besa Bytyqi (2002 August 15). “Një kulm i përbashkët për shumë nacionalitete.” Zëri, 6. 106 UNHCR/OSCE. Assessment of the Situation of Ethnic Minorities In Kosovo: period covering March 2001 through August 2001. 107 ESI/LLA, “Western Balkans 2004: Assistance, Cohesion and the new boundaries of Europe. A call for policy reform, 3 November 2002, www.esiweb.org 108 United Nations Security Council Resolution S/R/1244 (1999), www.un.org 109 Please note the difference between the European Agency for Reconstruction (EAR,www.ear.int), the donor arm of the European Commission, and the EU Pillar, responsible for the development of economic policies and an integral part of UNMIK (www.euinkosovo.org). 110 The Kosovo Trust Agency (KTA) was created per UNMIK Regulation 2002/12 on 13 June 2002. 111 The EU Pillar’s regional offices are to be closed by the end of 2002. 112 UNMIK Regulation 2000/63 promulgated on 7 December 2000. 113 Established per UN Security Council Resolution 2000/63 in December 2000. 114 Socially owned enterprises (SOEs). 115 Department of Trade and Industry. 116 World Bank Development Report 2002. 117 Allcock, John B, Explaining Yugoslavia, Hurst & Company, 2000. 118 Reconstruction and Investment 2001-2003. 119 ESI/LLA, op cit. 120 Statistical Overview of Registered Businesses in Kosovo, September 2001. 121 Prishtina. The Kosovo Big Small-Business Market, Small Scale Enterprise, Situation and Perspective, published by the Administrative Department of Labor and Employment, August 2002. 122 Draft Statistical Overview of Registered Businesses in Kosovo, September 2002. 123 Governance and Development. A real life story of private sector growth in Bosnia and Herzegovina, LLA, 13 February 2002, 124 Ibid. Administration and Governance in Kosovo 61 _____________ 125 Land Administration Seminar, 14 November 2002. 126 Law on Expropriation, Official Gazette of SAP Kosova, No 21/78 and Law on Land for Construction, Official Gazette of SAP Kosova, No. 14/80. 127 De Soto, Hernando. The Mystery of Capital, Basic Books, 2000. 128 Macroeconomic Unit of the Central Fiscal Authority, Kosovo, October 2002. 129 Speeches by the SRSG on www.unmikonline.org 130 Governance and Development. A real life story of private sector growth in Bosnia and Herzegovina, ESI/LLA, 13 February 2002, www.esiweb.org 131 UNMIK Regulation 2002/13 promulgated on 13 June 2002. 132 ESI/LLA Western Balkans 2004, op cit. 133 ESI/LLA Kalesija Report. www.esiweb.org 134 UN SCR 1244, 10 June 1999. 135 OSCE, As Seen As Told: An Analysis of Human Rights Violations in Kosovo (Warsaw: Office of Democratic Institutions and Human Rights, 1999), p. 89. 136 Ibid. p. 89. 137 Steven Erlanger, “Fears Grow Over the De Facto Partition of Kosovo,” New York Times, 14 November 1999. 138 OSCE op cit., p. 89. 139 OSCE population estimate cited in International Crisis Group (ICG), UNMIK’s Kosovo Albatross; Tackling Division in Mitrovica (Pristina: International Crisis Group, 3 June 2002), p. 3. 140 Ibid. p. 3. 141 Micheal Steiner, Special Representative for the Secretary General, A Choice for Mitrovica 1 October, 2002. 142 UNMIK Division of Public Information-Media Monitoring, “Serbs Consider Seven Point Plan an Ultimatum, Albanians are Skeptical,” Koha Ditore, 6 October 2002. 143 Ibid. 144 Ibid. 145 UNMIK, New Coverage, UN Assumes Administrative Control in Mitrovica, November 26, 2002. 146 OSCE, As Seen As Told: An Analysis of Human Rights Violations in Kosovo (Warsaw: Office of Democratic Institutions and Human Rights, 1999), p. 90. 147 Ibid. 148 ICG, UNMIK’s Kosovo Albatross; Tackling Division in Mitrovica (Pristina: International Crisis Group, 3 June 2002) (note 5), p. 7. 149 Ibid. p. 3. ICG describes the “Bridgewatchers” as being paid by the Serbian Ministry of the Interior (MUP) as members of State Security (DB); consequently, directly violating UNSCR 1244. They force shops to pay “protection” fees and distribute occupied apartments for rent. See ICG, Interview with KFOR and UNMIK (note 4), p. 3. ICG writes that the “Bridgewatchers” supplement their funds through organized crime including smuggling and prostitution. They also prevent many Albanians and other minorities from returning to their homes and frustrate the efforts of the international community to establish a presence in the North. 150 Andrew Pervis, A Legacy of Hate. Time Europe, 18 February 2002 (note 4) states that Covic attempted to persuade one of the first leaders of the “Bridgewatcher”, Ivanovic, to support the establishment of a multi-ethnic police force in which Ivanovic agreed to only if they has their own budget and uniforms. This is unacceptable in the drive for a multi-ethnic police force. It is also unacceptable in terms of its separation with northern Mitrovica moving, more or less, into its own municipal structure. 151 Jonathon Steel, “Nation Building in East Timor,” World Policy Journal, Summer 2002, Vol 11, Issue 2, p. 56. 152 United Nations. Security Council Press Release, “Council Members Regret Stalemate in Installation of Kosovo’s Newly Elected Institutions,” 21 January 2002, SC/7278, p. 13. 153 See Section 2 on Internal and External Security. 154ICG (note 4), p. 3. In February 2000, a rocket-propelled grenade attack on a UNHCR bus carrying Serbs from Mitrovica killed three and wounded several others. This set off revenge attacks and a cycle of violence. 155 OSCE, As Seen As Told, p. 90. 156 ICG, UNMIK’s Kosovo Albatross, p. 4 157 Ibid. 158 Ibid, p. 5. 159 Independent International Commission on Kosovo. Kosovo Report. (New York: Oxford, 2000), p.109. 160 Michael Steiner, Special Representative to the UN Secretary General, The Seven Point Plan: A Choice for Mitrovica, 1 October 2002. 161 Steel, op cit. 162 United Nations. “Kosovo Making Progress in Security, Inter-ethnic Relations, Annan Reports.” 26 July 2002. 62 Administration and Governance in Kosovo _______________ 163 Lizee, Pierie. “Conflict Resolution as a Construction of Security: Peace-Building, Constructive Intervention, and Human Security in Southeast Asia.” Development and Security in Southeast Asia Symposium (Toronto: York University 1999) 25 October. 164 Guy Lawson, “The View from the Bridge,” New York Times, 20 August 2000. 165 Radio Mitrovica, Op cit. 166 Ibid. 167ICG, p. 3. Serbian Interior Ministry forces operate in the North, and suspects arrested by them are brought to trial at courts in Serbia. 168 The Economist, First of All, be Nicer to Each Other, 5 October 2002, Vol. 365, Issue 8293. 169 The Economist, Op cit. 170 Radio Mitrovica, Op cit. 171 Interview with Bajram Rexhepi . KTV Direct, December 6, 2002. 172 Steven Erlanger, “Fears Grow over the De Facto Partition of Kosovo.” New York Times. 14 November 1999 173 ICG, UNMIK’s Kosovo Albatross, Op cit. p. 6. 174Ibid. 175 Andrew Pervis, A Legacy of Hate. Time Europe. February 18, 2002, Vol.59, No. 7. 176Musa Mustafa and Violeta Hyseni, “Serbian Parallel Institutions Continue in New Location.” Koha Ditore, 3 December 2002. 177 UNMIK Division of Public Information, Media Monitors, Bota Sot, 6 October 2002. 178 ICG op cit. p.: 6. 179 Carlotta Gall, “NATO-Led Troops Caught in Battle over Kosovo Town.” New York Times, 14 February 2000. 180 United Nations, Security Council Press Release, “Council Reviews Persistent Challenges Facing the Province.” 30 July 2002, no SC/7472. 181 Dana Eyre. “Partnership for Peace: Revitalizing Trepca and the Building of Peace in Kosovo.” Peacekeeping and International Relations, vol. 30, no. 4, pp19-22 182 OSCE. As Seen as Told: An Analysis of the Human Rights Findings of the OSCE Verification Mission. October 1998 to June 1999, p. 93. 183 Ibid. |
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5.3 Integer Exponents and Scientific Notation
Learning Objectives
By the end of this section, you will be able to:
Use the definition of a negative exponent
Simplify expressions with integer exponents
Convert from decimal notation to scientific notation
Convert scientific notation to decimal form
Multiply and divide using scientific notation
Use the Definition of a Negative Exponent
We saw that the Quotient Property for Exponents introduced earlier in this chapter, has two forms depending on whether the exponent is larger in the numerator or the denominator.
Quotient Property for Exponents
If is a real number, , and are whole numbers, then
What if we just subtract exponents regardless of which is larger?
Let’s consider .
We subtract the exponent in the denominator from the exponent in the numerator.
We can also simplify by dividing out common factors:
This implies that and it leads us to the definition of a negative exponent.
Negative Exponent
If is an integer and , then .
The negative exponent tells us we can re-write the expression by taking the reciprocal of the base and then changing the sign of the exponent.
Any expression that has negative exponents is not considered to be in simplest form. We will use the definition of a negative exponent and other properties of exponents to write the expression with only positive exponents.
For example, if after simplifying an expression we end up with the expression , we will take one more step and write . The answer is considered to be in simplest form when it has only positive exponents.
EXAMPLE 1
Simplify: a) b) .
Solution
| |
| a) |
| Use the definition of a negative exponent, . |
| Simplify. |
| b) |
| Use the definition of a negative exponent, . |
| Simplify. |
TRY IT 1.1
Simplify: a) b) .
Show answer
a) b)
TRY IT 1.2
Simplify: a) b) .
Show answer
a) b)
In (Example 1) we raised an integer to a negative exponent. What happens when we raise a fraction to a negative exponent? We’ll start by looking at what happens to a fraction whose numerator is one and whose denominator is an integer raised to a negative exponent.
|
| |
| Use the definition of a negative exponent, . |
| Simplify the complex fraction. |
| Multiply. |
This leads to the Property of Negative Exponents.
Property of Negative Exponents
If is an integer and , then .
EXAMPLE 2
Simplify: a) b) .
Solution
| |
| a) |
| Use the property of a negative exponent, . |
| b) |
| Use the property of a negative exponent, . |
| Simplify. |
TRY IT 2.1
Simplify: a) b) .
Show answer
a) b)
TRY IT 2.2
Simplify: a) b) .
Show answer
a) b)
Suppose now we have a fraction raised to a negative exponent. Let’s use our definition of negative exponents to lead us to a new property.
|
| |
| Use the definition of a negative exponent, . |
| Simplify the denominator. |
| Simplify the complex fraction. |
| But we know that is . |
| This tells us that: |
To get from the original fraction raised to a negative exponent to the final result, we took the reciprocal of the base—the fraction—and changed the sign of the exponent.
This leads us to the Quotient to a Negative Power Property.
Quotient to a Negative Exponent Property
If are real numbers, , and is an integer, then .
EXAMPLE 3
Simplify: a) b) .
Solution
| |
| a) |
| Use the Quotient to a Negative Exponent Property, . |
| Take the reciprocal of the fraction and change the sign of the exponent. |
| Simplify. |
| b) |
| Use the Quotient to a Negative Exponent Property, . |
| Take the reciprocal of the fraction and change the sign of the exponent. |
| Simplify. |
TRY IT 3.1
Simplify: a) b) .
Show answer
a) b)
TRY IT 3.2
Simplify: a) b) .
Show answer
a) b)
When simplifying an expression with exponents, we must be careful to correctly identify the base.
EXAMPLE 4
Simplify: a) b) c) d) .
Solution
| |
| a) Here the exponent applies to the base . |
| Take the reciprocal of the base and change the sign of the exponent. |
| Simplify. |
| b) The expression means “find the opposite of .” Here the exponent applies to the base . |
| Rewrite as a product with . |
| Take the reciprocal of the base and change the sign of the exponent. |
| Simplify. |
| c) Here the exponent applies to the base . |
| Take the reciprocal of the base and change the sign of the exponent. |
| Simplify. |
| d) The expression means “find the opposite of .” Here the exponent applies to the base . |
| Rewrite as a product with . |
| Take the reciprocal of the base and change the sign of the exponent. |
| Simplify. |
TRY IT 4.1
Simplify: a) b) c) d) .
Show answer
a) b) c) 25 d)
TRY IT 4.2
Simplify: a) b) , c) d) .
Show answer
a) b) c) 49 d)
We must be careful to follow the Order of Operations. In the next example, parts (a) and (b) look similar, but the results are different.
EXAMPLE 5
Simplify: a) b) .
Solution
| |
| a) Do exponents before multiplication. |
| Use . |
| Simplify. |
| b) |
| Simplify inside the parentheses first. |
| Use . |
| Simplify. |
TRY IT 5.1
Simplify: a) b) .
Show answer
a) b)
TRY IT 5.2
Simplify: a) b) .
Show answer
a) 2 b)
When a variable is raised to a negative exponent, we apply the definition the same way we did with numbers. We will assume all variables are non-zero.
EXAMPLE 6
Simplify: a) b) .
Solution
| |
| a) |
| Use the definition of a negative exponent |
| b) |
| Use the definition of a negative exponent . |
| Simplify. |
TRY IT 6.1
Simplify: a) b)
Show answer
a) b)
TRY IT 6.2
Simplify: a) b) .
Show answer
a) b)
When there is a product and an exponent we have to be careful to apply the exponent to the correct quantity. According to the Order of Operations, we simplify expressions in parentheses before applying exponents. We’ll see how this works in the next example.
EXAMPLE 7
Simplify: a) b) c) .
Solution
| |
| a) Notice the exponent applies to just the base. |
| Take the reciprocal of and change the sign of the exponent. |
| Simplify. |
| b) Here the parentheses make the exponent apply to the base. |
| Take the reciprocal of and change the sign of the exponent. |
| Simplify. |
| c) The base here is . |
| Take the reciprocal of and change the sign of the exponent. |
| Simplify. |
| Use |
TRY IT 7.1
Simplify: a) b) c) .
Show answer
a) b) c)
TRY IT 7.2
Simplify: a) b) c) .
Show answer
a) b) c)
With negative exponents, the Quotient Rule needs only one form , for . When the exponent in the denominator is larger than the exponent in the numerator, the exponent of the quotient will be negative.
Simplify Expressions with Integer Exponents
All of the exponent properties we developed earlier in the chapter with whole number exponents apply to integer exponents, too. We restate them here for reference.
Summary of Exponent Properties
If are real numbers, and are integers, then
EXAMPLE 8
Simplify: a) b) c) .
Solution
|
| |
| Use the Product Property, . |
| Simplify |
|
| |
| Notice the same bases, so add the exponents. |
| Simplify. |
| Use the definition of a negative exponent, . |
|
| |
| Add the exponents, since the bases are the same. |
| Simplify. |
| Take the reciprocal and change the sign of the exponent, using the definition of a negative exponent. |
TRY IT 8.1
Simplify: a) b) c) .
Show answer
a) b) c)
TRY IT 8.2
Simplify: a) b) c) .
Show answer
a) b) c)
In the next two examples, we’ll start by using the Commutative Property to group the same variables together. This makes it easier to identify the like bases before using the Product Property.
EXAMPLE 9
Simplify: .
Solution
|
| |
| Use the Commutative Property to get like bases together. |
| Add the exponents for each base. |
| Take the reciprocals and change the signs of the exponents. |
| Simplify. |
TRY IT 9.1
Simplify: .
Show answer
TRY IT 9.2
Simplify: .
Show answer
If the monomials have numerical coefficients, we multiply the coefficients, just like we did earlier.
EXAMPLE 10
Simplify: .
Solution
|
| |
| Rewrite with the like bases together. |
| Multiply the coefficients and add the exponents of each variable. |
| Use the definition of a negative exponent, . |
| Simplify. |
TRY IT 10.1
Simplify: .
Show answer
TRY IT 10.2
Simplify: .
Show answer
In the next two examples, we’ll use the Power Property and the Product to a Power Property.
EXAMPLE 11
Simplify: .
Solution
|
| |
| Use the product to a Power Property, . |
| Use the Power Property, . |
| Use the Definition of a Negative Exponent, . |
| Simplify. |
TRY IT 11.1
Simplify: .
Show answer
TRY IT 11.2
Simplify: .
Show answer
EXAMPLE 12
Simplify: .
Solution
|
| |
| Use the Product to a Power Property, . |
| Simplify and multiply the exponents of using the Power Property, . |
| Rewrite by using the Definition of a Negative Exponent, . |
| Simplify. |
TRY IT 12.1
Simplify: .
Show answer
TRY IT 12.2
Simplify: .
Show answer
To simplify a fraction, we use the Quotient Property and subtract the exponents.
EXAMPLE 13
Simplify: .
Solution
|
| |
| Use the Quotient Property, . |
| Simplify. |
TRY IT 13.1
Simplify: .
Show answer
TRY IT 13.2
Simplify: .
Show answer
Convert from Decimal Notation to Scientific Notation
Remember working with place value for whole numbers and decimals? Our number system is based on powers of 10. We use tens, hundreds, thousands, and so on. Our decimal numbers are also based on powers of tens—tenths, hundredths, thousandths, and so on. Consider the numbers 4,000 and . We know that 4,000 means and 0.004 means .
If we write the 1000 as a power of ten in exponential form, we can rewrite these numbers in this way:
In scientific notation, a number is expressed as the product of a coefficient and an exponential expression with a base of 10 and an integer power. The coefficient is a decimal number greater than or equal to 1 but less than 10, and the power of 10 is always an integer.
Scientific Notation
A number is expressed in scientific notation when it is of the form
It is customary in scientific notation to use as the multiplication sign, even though we avoid using this sign elsewhere in algebra.
If we look at what happened to the decimal point, we can see a method to easily convert from decimal notation to scientific notation.
In both cases, the decimal was moved 3 places to get the coefficient between 1 and 10
EXAMPLE 14
How to Convert from Decimal Notation to Scientific Notation
Write in scientific notation: 37,000.
Solution
| |
| Step 1. Identify the decimal point’s current position in the number: In this case, the decimal point is after the last digit, so it’s located at the end of the number. |
| Step2. Count the number of places you need to move the decimal point to make the number between 1 and 10: In this case, you need to move the decimal point 4 places to the left. |
| Step 3. Write the number as a decimal between 1 and 10, followed by the multiplication symbol and 10 raised to the power of the number of places the decimal point moved: + Decimal between 1 and 10: 3.7 (move the decimal point from 37000 four places to the left) + Power of 10: |
| Step 4. Check: Check to see if your answer makes sense. | is 10,000 and will be . |
TRY IT 14.1
Write in scientific notation: .
Show answer
TRY IT 14.2
Write in scientific notation: .
Show answer
HOW TO: Convert from decimal notation to scientific notation
Move the decimal point so that the first number is greater than or equal to 1 but less than 10.
Count the number of decimal places, n, that the decimal point was moved.
Write the number as a product with a power of 10.If the original number is:
greater than 1, the power of 10 will be 10n.
between 0 and 1, the power of 10 will be 10−n.
Check.
EXAMPLE 15
Write in scientific notation: .
Solution
The original number, , is between 0 and 1 so we will have a negative power of 10
|
| |
| Move the decimal point to get 5.2, a number between 1 and 10. |
| Count the number of decimal places the point was moved. | 3 places |
| Write as a product with a power of 10. |
| Check. |
TRY IT 15.1
Write in scientific notation: .
Show answer
TRY IT 15.2
Write in scientific notation: .
Show answer
Convert Scientific Notation to Decimal Form
How can we convert from scientific notation to decimal form? Let’s look at two numbers written in scientific notation and see.
If we look at the location of the decimal point, we can see an easy method to convert a number from scientific notation to decimal form.
In both cases the decimal point moved 4 places. When the exponent was positive, the decimal moved to the right. When the exponent was negative, the decimal point moved to the left.
EXAMPLE 16
How to Convert Scientific Notation to Decimal Form
Convert to decimal form: .
Solution
| | |
--- |
| Step 1. Determine the exponent, on the factor . | The exponent is . |
| Step 2. Move the decimal places, adding zeros if needed. If the exponent is positive, move the decimal point places to the right. If the exponent is negative, move the decimal point places to the left. | The exponent is positive, so move the decimal point 3 places to the right. We need to add 2 zeros as placeholders. | |
| Step 3. Check to see if your answer makes sense. | is and times will be . |
TRY IT 16.1
Convert to decimal form: .
Show answer
1,300
TRY IT 16.2
Convert to decimal form: .
Show answer
92,500
The steps are summarized below.
HOW TO: Convert scientific notation to decimal form.
To convert scientific notation to decimal form:
Identify the exponent, , that represents the power of 10 in the scientific notation.
Move the decimal places, adding zeros if needed.
If the exponent is positive, move the decimal point places to the right.
If the exponent is negative, move the decimal point places to the left.
Check.
EXAMPLE 17
Convert to decimal form: .
Solution
|
| |
| Identify the exponent, , that represents the power of 10 in the scientific notation. | The exponent is |
| Since the exponent is negative, move the decimal point 2 places to the left. |
| Add zeros as needed for placeholders. |
TRY IT 17.1
Convert to decimal form: .
Show answer
0.00012
TRY IT 17.2
Convert to decimal form: .
Show answer
0.075
Multiply and Divide Using Scientific Notation
Astronomers use very large numbers to describe distances in the universe and ages of stars and planets. Chemists use very small numbers to describe the size of an atom or the charge on an electron. When scientists perform calculations with very large or very small numbers, they use scientific notation. Scientific notation provides a way for the calculations to be done without writing a lot of zeros. We will see how the Properties of Exponents are used to multiply and divide numbers in scientific notation.
EXAMPLE 18
Multiply. Write answers in decimal form: .
Solution
|
| |
| Use the Commutative Property to rearrange the factors. |
| Multiply. |
| Change to decimal form by moving the decimal two places left. |
TRY IT 18.1
Multiply . Write answers in decimal form.
Show answer
0.06
TRY IT 18.2
Multiply . Write answers in decimal form.
Show answer
0.009
EXAMPLE 19
Divide. Write answers in decimal form: .
Solution
|
| |
| Rewrite as the product of two fractions. |
| Divide. |
| Change to decimal form by moving the decimal five places right. |
TRY IT 19.1
Divide . Write answers in decimal form.
Show answer
400,000
TRY IT 19.2
Divide . Write answers in decimal form.
Show answer
Access these online resources for additional instruction and practice with integer exponents and scientific notation:
Negative Exponents
Scientific Notation
Scientific Notation 2
Key Concepts
Property of Negative Exponents
If is a positive integer and , then
Quotient to a Negative Exponent
If are real numbers, and is an integer , then
To convert a decimal to scientific notation:
Move the decimal point so that the coefficient is greater than or equal to 1 but less than 10.
Count the number of decimal places, , that the decimal point was moved.
Write the number as a product with a power of 10. If the original number is:
greater than 1, the power of 10 will be
between 0 and 1, the power of 10 will be
Check.
To convert scientific notation to decimal form:
Identify the exponent, denoted as , that represents the power of 10 in the scientific notation.
Move the decimal places, adding zeros if needed.
If the exponent is positive, move the decimal point places to the right.
If the exponent is negative, move the decimal point places to the left.
Check
Practice Makes Perfect
Use the Definition of a Negative Exponent
In the following exercises, simplify.
| | |
--- |
| 1. a) b) | 2. a) b) |
| 3. a) b) | 4. a) b) |
| 5. a) b) | 6. a) b) |
| 7. a) b) | 8. a) b) |
| 9. a) b) | 10. a) b) |
| 11. a)b) | 12. a) b) |
| a) b) c) d) | a) b) c) d) |
| a) b) c) d) | a) b) c) d) |
| 17. a) b) | 18. a) b) |
| 19. a) b) | 20. a) b) |
| 21. a) b) | 22. a) b) |
| 23. a) b) | 24. a) b) |
| a) b) c) | a) b) c) |
| a) b) c) | a) b) c) |
Simplify Expressions with Integer Exponents
In the following exercises, simplify.
| | |
--- |
| a) b) c) | 30. a) b) c) |
| a) b) c) | a) b) c) |
| 33. | 34. |
| 35. | 36. |
| 37. | 38. |
| 39. | 40. |
| 41. | 42. |
| 43. | 44. |
| 45. | 46. |
| 47. | 48. |
| 49. |
Convert from Decimal Notation to Scientific Notation
In the following exercises, write each number in scientific notation.
| | |
--- |
| 50. 57,000 | 51. 340,000 |
| 52. 8,750,000 | 53. 1,290,000 |
| 54. 0.026 | 55. 0.041 |
| 56. 0.00000871 | 57. 0.00000103 |
Convert Scientific Notation to Decimal Form
In the following exercises, convert each number to decimal form.
| | |
--- |
| 58. | 59. |
| 60. | 61. |
| 62. | 63. |
| 64. | 65. |
Multiply and Divide Using Scientific Notation
In the following exercises, multiply. Write your answer in decimal form.
| | |
--- |
| 66. | 67. |
| 68. | 69. |
In the following exercises, divide. Write your answer in decimal form.
| | |
--- |
| 70. | 71. |
| 72. | 73. |
Everyday Math
| | |
--- |
| 74. The population of the United States on July 1, 2010 was about 34,000,000. Write the number in scientific notation. | 75. The population of the world on July 1, 2010 was more than 6,850,000,000. Write the number in scientific notation |
| 76. The average width of a human hair is 0.0018 centimetres. Write the number in scientific notation. | 77. The probability of winning the 2010 Megamillions lottery was about 0.0000000057. Write the number in scientific notation. |
| 78. In 2010, the number of Facebook users each day who changed their status to ‘engaged’ was . Convert this number to decimal form. | 79. At the start of 2012, the US federal budget had a deficit of more than . Convert this number to decimal form. |
| 80. The concentration of carbon dioxide in the atmosphere is . Convert this number to decimal form. | 81. The width of a proton is of the width of an atom. Convert this number to decimal form. |
| 82. Health care costs The Centers for Medicare and Medicaid projects that American consumers will spend more than $4 trillion on health care by 2017 1. Write 4 trillion in decimal notation. 2. Write 4 trillion in scientific notation. | 83. Coin production In 1942, the U.S. Mint produced 154,500,000 nickels. Write 154,500,000 in scientific notation. |
| 84. Distance The distance between Earth and one of the brightest stars in the night star is 33.7 light years. One light year is about 6,000,000,000,000 (6 trillion), miles. a) Write the number of miles in one light year in scientific notation. b)Use scientific notation to find the distance between Earth and the star in miles. Write the answer in scientific notation. | 85. Debt At the end of fiscal year 2019 the gross Canadian federal government debt was estimated to be approximately $685,450,000,000 ($685.45 billion), according to the Federal Budget. The population of Canada was approximately 37,590,000 people at the end of fiscal year 2019 a) Write the debt in scientific notation. b) Write the population in scientific notation. c) Find the amount of debt per person by using scientific notation to divide the debt by the population. Write the answer in scientific notation. |
Writing Exercises.
| | |
--- |
| 86. a) Explain the meaning of the exponent in the expression . b) Explain the meaning of the exponent in the expression . | 87. When you convert a number from decimal notation to scientific notation, how do you know if the exponent will be positive or negative? |
Answers
| | | |
---
| 1. a) b) | 3. a) b) | 5. a) b) 25 |
| 7. a) b) 10000 | 9. a) b) | 11. a) b) |
| 13. a) b) c) 49 d) | 15. a) b) c) d) | 17. a) b) |
| 19. a) b) | 21. a) b) | 23. a) b) |
| 25. a) b) c) | 27. a) b) c) | 29. a) b) c) |
| 31. a) 1 b) c) | 33. | 35. |
| 37. | 39. | 41. |
| 43. | 45. | 47. |
| 49. | 51. | 53. |
| 55. | 57. | 59. 830 |
| 61. 16,000,000,000 | 63. 0.038 | 65. 0.0000193 |
| 67. 0.02 | 69. | 71. 500,000,000 |
| 73. 20,000,000 | 75. . | 77. |
| 79. 15,000,000,000,000 | 81. 0.00001 | 83. |
| 85. | 87. Answers will vary |
Attributions
This chapter has been adapted from “Integer Exponents and Scientific Notation” in Elementary Algebra (OpenStax) by Lynn Marecek and MaryAnne Anthony-Smith, which is under a CC BY 4.0 Licence. Adapted by Izabela Mazur. See the Copyright page for more information.
License
Intermediate Algebra II Copyright © 2021 by Pooja Gupta is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.
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189957 | https://www.pw.live/school-prep/exams/half-angle-formulas | Half Angle Formulas, Sin, Cos and Tan
The half angle formulas are used to find the sine and cosine of half of an angle A, making it easier to work with trigonometric functions
Anchal Singh15 Oct, 2023
The
half angle formulas
are trigonometric identities that express the trigonometric functions of half an angle in terms of the trigonometric functions of the original angle. These formulas are particularly useful in trigonometry and calculus when dealing with angles that are smaller or more manageable than the original angle. There are different half-angle formulas for various trigonometric functions. Here are the half-angle formulas for sine, cosine, and tangent:
These half angle formulas are helpful for simplifying trigonometric expressions and for solving problems involving trigonometric identities. They are often used in calculus, geometry, and engineering to manipulate and solve equations involving trigonometric functions.
Half Angle Identities
Here are the popular half-angle identities for the trigonometric functions sine, cosine, and tangent:
Half angle formula of sin:
sin A/2 = ±√[(1 - cos A) / 2]
Half angle formula of cos:
cos A/2 = ±√[(1 + cos A) / 2]
Half angle formula of tan:
tan A/2 = ±√[1 - cos A] / [1 + cos A] (or) sin A / (1 + cos A) (or) (1 - cos A) / sin A
Half Angle Formulas Derivation Using Double Angle Formulas
To derive the half angle formulas, we start by using the double angle formulas, which express trigonometric functions in terms of double angles like 2θ, 2A, 2x, and so on. These double angle formulas are well-known:
For sine (sin):
sin(2x) = 2sin(x)cos(x)
For cosine (cos):
cos(2x) = cos^2(x) - sin^2(x)
or
cos(2x) = 1 - 2sin^2(x)
or
cos(2x) = 2cos^2(x) - 1
For tangent (tan):
tan(2x) = 2tan(x) / (1 - tan^2(x))
Now, if we replace x with A/2 in each of these double angle formulas, we obtain the half-angle identities because 2x is equivalent to 2(A/2), which simplifies to A:
For sine (sin):
sin(A) = 2sin(A/2)cos(A/2)
For cosine (cos):
cos(A) = cos^2(A/2) - sin^2(A/2)
or
cos(A) = 1 - 2sin^2(A/2)
or
cos(A) = 2cos^2(A/2) - 1
For tangent (tan):
tan(A) = 2tan(A/2) / (1 - tan^2(A/2))
These half angle identities are helpful in trigonometry for simplifying expressions and solving problems involving trigonometric functions when working with angles that are half of the original angles (A/2 in this case).
Half Angle Formulas of Sin Proof
To derive the half-angle formula for the sine function. Here's the step-by-step derivation:
Starting with:
cos(A) = 1 - 2sin^2(A/2)
You can isolate sin^2(A/2) by rearranging the terms:
2sin^2(A/2) = 1 - cos(A)
Then, divide both sides by 2 to find sin^2(A/2):
sin^2(A/2) = (1 - cos(A)) / 2
Finally, to find sin(A/2), take the square root of both sides, noting that sin(A/2) could be either positive or negative depending on the quadrant in which A/2 lies:
sin(A/2) = ±√[(1 - cos(A)) / 2]
This result gives you the half-angle formula for the sine function. The ± sign accounts for the possibility of sine being positive or negative depending on the quadrant, as you correctly mentioned.
Half Angle Formulas of Cos Derivation
the half-angle formula for the cosine function. Here's the step-by-step derivation:
Starting with:
cos(A) = 2cos^2(A/2) - 1
You can isolate cos^2(A/2) by rearranging the terms:
2cos^2(A/2) = 1 + cos(A)
Then, divide both sides by 2 to find cos^2(A/2):
cos^2(A/2) = (1 + cos(A)) / 2
Finally, to find cos(A/2), take the square root of both sides, noting that cos(A/2) could be either positive or negative depending on the quadrant in which A/2 lies:
cos(A/2) = ±√[(1 + cos(A)) / 2]
This result gives you the half angle formulas for the cosine function. The ± sign accounts for the possibility of cosine being positive or negative depending on the quadrant, as you correctly mentioned.
Half Angle Formulas of Tan Derivation
We have tan (A/2) = [sin (A/2)] / [cos (A/2)]
Using the half angle formulas of sin and cos,
tan (A/2) = [±√(1 - cos A)/2] / [±√(1 + cos A)/2]
=
±√[(1 - cos A) / (1 + cos A)]
This is one of the formulas of tan (A/2). Let us derive the other two formulas by rationalizing the denominator tan (A/2) = ±√[(1 - cos A) / (1 + cos A)] × √[(1 - cos A) / (1 - cos A)]
= √[(1 - cos A)
2
/ (1 - cos
2
A)]
= √[(1 - cos A)
2
/ sin
2
A]
=
(1 - cos A) / sin A
This is the second formula of tan (A/2). To derive another formula, let us multiply and divide the above formula by (1 + cos A). Then we get
tan (A/2) = [(1 - cos A) / sin A] × [(1 + cos A) / (1 + cos A)]
= (1 - cos
2
A) / [sin A (1 + cos A)]
= sin
2
A / [sin A (1 + cos A)]
=
sin A / (1 + cos A)
Thus, tan (A/2) = ±√[(1 - cos A) / (1 + cos A)] = (1 - cos A) / sin A = sin A / (1 + cos A).
Half Angle Formulas Using Semi perimeter
The half-angle formulas for cosine, sine, and tangent functions using the semi-perimeter of a triangle. Here's a summary of the formulas you've derived:
Cosine Half-Angle Formula:
cos(A/2) = √[s(s - a) / bc]
Sine Half-Angle Formula:
sin(A/2) = √[(s - b)(s - c) / bc]
Tangent Half-Angle Formula:
tan(A/2) = sin(A/2) / cos(A/2)
These formulas express the trigonometric functions of half an angle (A/2) in terms of the sides of a triangle (a, b, c) and the semi-perimeter (s). These identities are particularly useful in trigonometry and geometry when you're working with triangles and angles related to them.
Half Angle Formulas Applications
Trigonometric Simplification:
Half-angle formulas are used to simplify trigonometric expressions, making them more manageable and easier to work with. They can simplify complex expressions involving trigonometric functions, which is especially valuable in calculus and other advanced mathematics.
Geometric Problems:
In geometry, half-angle formulas are applied to solve problems involving angles and shapes. They can be used to find missing angles, determine side lengths in triangles, and solve geometric constructions.
Navigation:
Half-angle formulas are essential in navigation, such as in aviation and marine navigation. They help in calculating angles and distances, aiding pilots and navigators in determining their position and course.
Engineering:
Engineers use half-angle formulas to analyze and design various structures and systems. For instance, in mechanical engineering, these formulas can be applied to determine angles in linkages, cam profiles, and gear mechanisms.
Physics:
Half-angle formulas are employed in physics to solve problems related to wave propagation, interference, and diffraction. They are also useful in analyzing the behavior of light and electromagnetic waves.
Trigonometric Identities:
Half-angle formulas are part of a larger set of trigonometric identities. These identities are widely used in mathematics, particularly in calculus, differential equations, and complex analysis.
Signal Processing:
In signal processing and electrical engineering, half-angle formulas are used to analyze and process signals, such as determining phase shifts or analyzing oscillations.
Statistics:
In statistical analysis, half-angle formulas can be applied to calculate measures of central tendency and dispersion, particularly in contexts where periodic or cyclical data is involved.
| | |
--- |
| Related Links | |
| Fibonacci Sequence Formula | Eulers Formula |
| Exponential Formula | Factorial Formula |
Half Angle Formulas
What are half-angle formulas?
Half-angle formulas are trigonometric identities that express the trigonometric functions of half of an angle in terms of the trigonometric functions of the original angle.
What are the primary half-angle formulas?
The primary half-angle formulas include those for sine, cosine, and tangent of half an angle, and they are derived from the double-angle formulas.
When are half-angle formulas used?
Half-angle formulas are used in trigonometry to simplify trigonometric expressions and solve problems involving angles that are half of the original angles. They are particularly valuable in calculus, geometry, and engineering.
Why are half-angle formulas useful?
Half-angle formulas make it easier to work with trigonometric functions of smaller angles. They are helpful for reducing the complexity of mathematical expressions and for solving problems that involve angles that are smaller and more manageable.
How do I determine which sign (±) to use in half-angle formulas?
The choice of sign in the formulas depends on the quadrant in which the half-angle lies. You choose the sign to ensure the appropriate positive or negative value for the trigonometric function in that quadrant.
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189958 | https://www.bbc.co.uk/bitesize/guides/zcqbdxs/revision/1 | GCSE
AQA
Evolution - AQAPrinciples of evolution by natural selection
Evolution is the change of inherited characteristics within a population over time through natural selection, which may result in the formation of a new species.
Part of Biology (Single Science)Inheritance, variation and evolution
Save to My Bitesize
Principles of evolution by natural selection
The idea behind the theory of evolutionThe process of change in the inherited traits of a population of organisms from one generation to the next. through the process of natural selection is that all speciesA type of organism that is the basic unit of classification. Individuals of different species are not able to interbreed successfully. of living things have evolved from simple life forms over a period of time. The Earth is about 4.5 billion years old and there is scientific evidence to suggest that life on Earth began more than three billion years ago.
This slideshow shows key events in evolution, from the first bacteriaSingle-celled microorganisms, some of which are pathogenic in humans, animals and plants. Singular is bacterium. to humans.
Image gallery
Skip image gallery
1 of 18
Slide 1 of 18, Archean eon. About 3,500,000,000 years ago: the first bacteria appeared (prokaryotes)., An evolution timeline Archean eon
End of image gallery
Natural selection
The accepted theory of evolution explains that it happens by natural selectionThe natural process whereby the best-adapted individuals survive longer, have more offspring and thereby spread their characteristics. Sometimes referred to as 'survival of the fittest'.. The key points are:
Individuals in a species show a wide range of variationDifference between individuals, distance from the norm. and this variation is because of differences in their geneThe basic unit of genetic material inherited from our parents. A gene is a section of DNA which controls part of a cell's chemistry - particularly protein production..
Individuals with characteristics most suited to their environment are more likely to survive and reproduce. This is commonly known as 'survival of the fittest'. The genes that allow these individuals to be successful within their environment are passed on to their offspring, which results in these specific genes becoming more common.
Those that are poorly adapted to their environment are less likely to survive and reproduce. Their genes are less likely to be passed on to the next generation.
Over a period of time, a species will gradually evolve.
Both genes and the environment can cause variation, but only genetic variation can be passed on to the next generation.
If two populations of one species become increasingly different in phenotypeThe visible characteristics of an organism which occur as a result of its genes. that they can no longer interbreed to form fertile offspring, this can result in the formation of two species.
A simple example can be seen in peacocks:
females choose a mate based on their colourful tail feathers
the more colourful the tail of a peacock, the more likely they are to mate and pass on these genes
over time, the tails of peacocks have become more colourful
Next page
The work of Lamarck
More guides on this topic
Reproduction, the genome and gene expression - AQA
Genetic inheritance - AQA
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189959 | https://www.mathsisfun.com/geometry/area-moments.html | Moments of Area
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Moments of Area
First and Second Moment of Area
Moment in Physics
In Physics Moment (or Torque) is force times distance:
But there are other Moments, read on!
First Moment of Area
First Moment of Area is area times distance (to some reference line):
First Moment of Area = A x d
For this simple case we can multiply the whole area by the distance (from its middle to the reference line).
Example:
Area is 20 mm x 10 mm = 200 mm 2
First Moment of Area (relative to the bottom line) = 200 mm 2 x 25 mm = 5000 mm 3
(Note the unit is mm 3, but is not a volume!)
One of it's great uses is to find the centroid , which is the average position of all the points of an object:
A plane shape cut from a piece of card will balance perfectly on its centroid.
To find the distance to the centroid from any axis, we divide the First Moment of Area by the Total Area:
Distance to Centroid = First Moment of AreaTotal Area
When we do that for both the x-axis and y-axis we get the centroid.
We can estimate where the centroid is using squares:
images/area-estim.js?mode=mom1
Second Moment of Area
For the Second Moment of Area we multiply the area by the distance squared:
(need infinitely many tiny squares)
But be careful! We need to multiply every tiny bit of area by its distance squared, because area further away has a bigger effect (due to the distance being squared).
It is called "Second" moment because we square the distance "x 2"
It is also called the area moment of inertia.
We can estimate the second moment using squares, but it is very inaccurate:
images/area-estim.js?mode=mom2
We can use the x-axis or y-axis as the reference line, or we can use the centroid for the reference line. You can try that option above.
Notation: The symbol is an "I" followed by a little "x" or "y" for the reference axis.
I x is in relation to the x axis (and we use y distances times area)
I y is in relation to the y axis (and we use x distances times area)
The letter I refers to Inertia in "area moment of inertia".
Where possible use an accurate formula such as:
I x = bh 33
I y = b 3 h3
I x = bh 312
I y = b 3 h12
I x = bh 312
I y = b 3 h+b 2 ha+bha 212
I x = π r 44
I y = π r 44
Many more!
Engineers use the second moment of area to work out how rigid (hard to bend) a beam is.
Example: A beam that is 100 mm by 24 mm
Lying flat it looks like this:
I x = bh 312= 100 × 24 312=115,200 mm 4
But sitting upright it is:
I x = bh 312= 24 × 100 312=2,000,000 mm 4
It is nearly 20 times as rigid sitting upright!
And that is why beams sit up like this:
Try bending a ruler about each axis to experience it for yourself:
Engineers Love I-Beams
Here we have two equal-sized beams, but one is solid, the other shaped like an "I"
The solid beam is a bit stiffer against bending (I x = 333 vs 205) but very much heavier (40kg vs 16kg).
In practice we could have a slightly bigger I-Beam and still save a lot of money in steel, transport and handling.
To calculate the second moment of area for an I-Beam we can break it down into rectangles and get these formulas:
I x = wh 3 − (w−b)(h−2t)312
I y = w 3 h − (w−b)3(h−2t)12
But in practice it is best to use the manufacturer's tables of beam properties as they account for any odd shapes.
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189960 | https://www.youtube.com/shorts/ALFJyzeMxPk | Chord of Contact of a Parabola #jeedailyconcepts #parabola - YouTube
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Chord of Contact of a Parabola #jeedailyconcepts#parabola
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Chord of Contact of a Parabola #jeedailyconcepts#parabola
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2023 Mar 4
From any point outside a parabola, you can draw 2 tangents to it. And the chord of the parabola joining he points of intersection of the tangents and the parabola is known as the chord of contact of the parabola. For more such videos, subscribe to the channel:
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189961 | https://samjshah.com/2009/05/04/my-exponential-function-unit/ | Continuous Everywhere but Differentiable Nowhere
I have no idea why I picked this blog name, but there's no turning back now
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My Exponential Function Unit
My Exponential Function Unit for Algebra II
Basic Context: This unit is coming right on the heels of function transformations. Students are familiar with translating functions up, down, left, and right; reflecting functions over the x- and y-axes; and vertically and horizontally stretching and shrinking functions.
Structure: The work on exponential functions is broken into four parts.
Part 0: Preliminary Diversion into Inverse Functions
Part I: Graphing exponential functions
Part II: Solving basic exponential function equations
Part III: Applications of exponential functions (carbon dating and compound interest)
Time: This took a total of 13 days — including an introductory activity day, a review day, a day where we did an exponential decay simulation as an entre to carbon dating, and two assessment days.
Nature of Class: I teach 15 students in a non-accelerated Algebra II class. The ability level of the students range the gamut. Many have a hard time thinking abstractly. All have graphing calculators and know how to use them at the basic to intermediate level. We meet 4 days a week for 50 minutes each day.
Broad Goal: The goal for this unit was to really drive home the concept of exponential functions.
Major Failures: I see two major failures. One is not seriously talking about how fast exponential functions grow. This would have been a really fun day, working on a problem like: “Would you rather have (A) $1,000,000 a day for the month of May, or (B) $1 on the first day, $2 on the second day, $4 on the third day, $8 on the fourth day, etc.” The second is just not having a lot of fun with this. The exponential decay simulation we did could have been so much more powerful, and changed in so many fun and really great ways. We could also have done an activity for exponential growth, using real data — population growth, Moore’s Law, or something to do with the Supreme Court. It would have been nice to finish off with a nice 2 day research activity. If for nothing else, to let my students produce something they could be proud of.
Major Strengths: In terms of getting students to understand exponential functions conceptually, I think I’ve done a pretty good job. My students can relate tables, graphs, and equations. They understand why the functions look the way they look. By the time we finished the exponential application days, students were coming up with the formula for the depreciated value of an object without any help.
Materials [NOTE: If you are opening these docs on a Mac, “Select All” and change the font to “Gill Sans.”]
Part 0: Preliminary Excursion into Inverse FunctionsPDFs of My Smartboards before class: 1, 2, 3.
Part I: Graphing Exponential Functions
1. Introductory exercise introducing students to exponential growth and decay (.doc)
2. Introduction to exponential functions, and graphing basic exponential functions (.doc); HW (.doc)
Part II: Solving Basic Exponential Equations1. PDF of My Smartboard before class: 1
Review Sheet on Part 0, Part I, and Part II to prepare students for the assessment (.doc)
Part III: Applications of Exponential Functions
1. Coin Drop Simulation for Exponential Decay (.doc); HW (.doc)
2. Carbon Dating (.doc); HW (.doc)
3. Compound Interest (.doc)
What I Want You To Know: Looking at just the stark documents, this whole unit seems like it might be a bit formulaic. However, particular moments of the guided notes, or the SmartBoards, or during the activities, were actually designed to be places where we have classroom discussions. For example, when one of the worksheets reads:
we actually had a great 5-7 minutes talking about the answers! So I’m afraid these resources make it seem like we might not have really interrogated exponential functions. But we did.
You can really see what I mean because… during this unit, my friend came to observe my class. (It was an assignment for a class she was taking for her Masters.) It happened to be the class where we first talked about exponential decay. While I was teaching, she decided to make a (partial) transcript of the entire class. The transcript is very rough and partial, and you can’t really tell what’s going on exactly, but you can get a sense of what the class was like:
Transcription (with student names redacted) after the Jump
today is the day that it’s all supposed to come together
let’s see if the click happens
I am collecting the homework
I am not grading it on correctness, so don’t freak out
we’ll talk about it a little bit, and then have the rest of the lesson to try and have it make sense
(some are late, he looks at the watch)
anyone else?
based on class yesterday: what made sense? what didn’t make sense?
on the HW
document camera set up, whiteboard ready
question 2, 1b, 1c
i didn’t think about it this way
(is it wrong?)
no! i didn’t think about it this way! this is the best way! i am getting giddy!
(is there another way?)
yes, there is another way
use the formula, [a student] explains
someone else plots it on the calculator
so excited that
what would be a resonable other answer? 87? 72?
what would not make sense? 4?
could I have started out with 5 coins (i coulc have in theory, but the chance of that happening would be achhhhhh!)
let’s start on 1b before we start on today’s lesson
calling on [student],
I know you don’t have your paper, but we can still do this?
she is unsure, having a tough time
how would this equation change if we started out with 100 experiments vs, 537
come up with the equation: labels the parts of the equation
then go to part c
questions? no?
OK, if you have feedback on this I was definitely serious about giving it to me whether you liked it or not (an email, or a note on my desk)
but not cut out letters: that will scare me?
Today we are going to learn about carbon dating?
Practical example of this type of experiment
What do we know about carbon dating?
(any history channel buffs?)
OK…there’s a thing called carbon 14, and there’s a half-life (whatever that is)
makes them feel comfortable if they don’t know about it
[student]: every organism goes through a life cycle
plants go through the chlorophyl thing…what is it called? photosynthesis?
there’s a krebs cycle, i don’t even know that that is…
the amount of C14 you have in your body is constant, as soon as you die, the C14 starts decaying.
i think collectively with all of our knowledge together: that’s essentially carbon 14
if you don’t get it, it’s OK; I looked it up on line, it’s very complicated
1) it’s radioactive: it has a half life
2) when you die there’s 100% of the carbon in your body
get in groups of 3, only restriction is not to be in the same group as yesterday
float around the room
checking the groups
what i want you to do is fill out this chart
let me explain to you what this chart is saying
what would be a good number to fill in in the section that says “years since organic matter was alive”
we were thinking about using the half life?
why would that be a good thing to use?
DO you know any other information for the intervals?
checking in with [student]; asking this question
could you fill in the rest of this table using this number? try it and see if it turns out to be good
we’re stuck!
would 1 year make sense?
could you fill in the other stuff?
no, what do you need to choose
some groups checking in with other groups!
what did you guys do…we are doubling the half life
looks good to me, but then my question is what would be useful here?
I am going to come back to you
they are working but look confused
this looks good! but you need to fill in the middle column
I am liking these numbers…can you fill in what these would be?
why don’t you erase the second number
(gives them the example, after 5750)
I can’t go any further
discussion of what they use carbon dating for
Oh, it takes this amount of years
this is making sense to me, now explain it to this one
wait…what are you doing?
i thought we were supposed to go to zero!
will it ever go completely to zero?!
no, what do we know about this?
but the practicality of
what about this middle column?
[Student Name]!
he’s distracted, apologizes and focuses
Oh really, it’s so fascinating!
their tongues?!
out of context, i can see that it would be a strange comment
i just care that there are no numbers! no 10 tongues!
finish this on task and then do your graph
are you finished?!
Fill out the graph!
you want to finish and then you want to graph!
chop chop!
[Student] still needs the example
you may just want to make one graph between the two of you?
What’s that old?
Dinosaurs!
anything with any organic material
Oh really, i am so interested in UNC, but I am more interested in this…
I came back
why would you double it instead of halve it?
[Student], you need to understand the concept
you need to wait 5750 years each time, using his hands to explain
Now the question is do you connect the dots?
in the other one we didn’t? why?
does it make sense to connect the dots?
does it make sense to have a 1.25 experiments?
you guys have been dawdling, you need to have this graph done
make sure you fill in the years
instead of making them nice numbers, what would a really nice and easy thing to do with the x axis
no it’s not, I am horrified by this? What are these numbers?
[Student] gets it! Explain to [Student] and [Student]
keep on adding that number
Oh really? there’s over 100% of carbon 14? how is that possible?
how’s your graph?
then if your graph is done see if you can’t answer just by looking at the graph: part A and B!
15 min left in class
just an estimation. you don’t have to be super accurate.
how did you get that? Oh, that makes sense. I totally buy that!
Great handout!
Once you finish A and B call me over!
[Student] clarifying that the number/percentage is what is gone or what remains
you are getting it!
How do you figure this out?
I’m not sure, this graph is a little off
You do need to label it!
why are we drawing a line? In the last one we didn’t!
OK? We’ll talk about it
there is a better reason!
I think you should plot the point 0,100
Since you started it, you might as well finish it
what’s going on over here?
I’m not sure I get that!
you can connect it with a line.
Why didn’t we connect the last one?
no! (cheerful, but dismissive)
hey hey language!
be friends!
back to [student]’s group
OK, last time: why are you getting that?
Let me ask you: what does this column represent
[Student], you want to listen and focus, I am helping your group here
you want to think about part c a little more: I’m not sure I like your answer
what happens when t is zero…
how are you going to modify your equation so that you have something that makes sense for t to = zero
read question B very carefully to answer this
it’s not that 22.3% remains, it’s that 22.3% is lost
If you start with 100% how much is left after 22.3% is lost
sounds like they are hearing what is going on with the other groups, and
still working on their equation
what if t were 1?
you are very close! i like the division!
OK, I am going to bring us all together so we can go over parts A and B!
I need someone to volunteer their graph!
Why did we connect them, when in the last one we didn’t ([Student])
so everyone gets the answers that many answered in their groups
guesstimating: close to 5750: 4500, 4000?
part b: [student] help me out! If you start out with a dollar and you lose 22.3% of it, how much is left?
it’s about halfway in…what akeim did was take about half of 5750
I think it’s like the other one we did
Gives an equation: why the 5750 here? Oh, it doesn’t make sense
we know that we’re wrong, but I like where we are going, this is a great start!
so watch this! fills in the division in the exponent!
(oh snap!)
if you didn’t see it this time, that’s OK. I didn’t mean for you all to see it the first time, but hopefully you will be able to see it and understand it for next time
what I need you to do for HW is part D and E
and then I am changing the HW to only parts 1 and 2, not 3 and 4, I will post it on the conference
(I am grading it for a serious attempt)
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10 comments
little “labs” you could do for half-life data:
PENNIES
1) start with 20 pennies in a cup, shake and dump out on table. (n = 20)
2) remove any pennies that are tails. count # of heads pennies. (n = 10 or so)
3) put the n “heads pennies back in cup, shake and dump out on table.
4) remove any pennies that are tails. count # of heads pennies. (n = 5 or so)
5) repeat steps 3-4 until n = 0
6) graph data
(also try using with die, where you remove only when face value is 5 or 6… this gives a longer half-life)
BEER
supposedly, the foam head on beer “decays” more or less exponentially. i haven’t tried this yet, but a friend told me that you can have kids take data on the foam of non-alcoholic(!) beer poured into a graduated cylinder. the mL markings offer a metric for amt of foam. use a stopwatch for time. measure every x seconds. graph… calculate half-life of beer foam head.
@mjs: In fact, I did that penny drop exercise ( — but instead of getting the pennies, I had students run a calculator simulation. I thought we could get the same benefit, but without all the time spent counting. Plus, I wanted students to drop 100 pennies. As for the die, I actually had that as a homework problem this year!
The beer thing is new to me… looking online, there’s an IgNobel that went to it, but for some reason I can’t access the paper…
I know why everything on your blog went italics. You put an em tag (for the italics) in this post right before the fold, but when you page normally loads, it doesn’t load the below the fold part where the tag is closed. That’s my read. Basically, close the em tag before you add the option to expand the post.
@Nick: You fixed it! Wow, good troubleshooting. Thanks a zillion. It was annoying me to no end!
Enjoyed reading this – it gave me some good ideas for next year’s exponential functions unit in Algebra 2.
One question, though – why do you just meet 4 days/week? Just curious. :)
Kristen
@Kristen: we have a rotating schedule at my school. And the way it rotates, students meet for classes 4 days a week, instead of 5. And the classes meet at different times of the day, each day.
Note: The exponential equation y = a x, where a > 0 and a is a one-to-one function and has an inverse which is defined implicitly by the equation x = a y.
The log of a number is the exponent when written in exponential form.
If the base of a logarithmic function is the irrational number e, then we have the natural
logarithm function. This function is given a special symbol. That is .
are inverse functions.
Logarithms to the base 10 are called common logarithms. This logarithm is abbreviated as
y = log x.
Change of base formula:
. Equations that contain logarithms are called logarithmic equations. In solving this type of equation be sure to check each apparent solution in the original equation and discard any that are extraneous. One of the most powerful notions in mathematics is the idea of approximating a function with other functions. Students’ first exposure to this concept typically is Taylor approximations at the end of second semester calculus where a function f(x) is approximated by a polynomial, which can be thought of as a linear combination of power functions with non-negative integer exponents. Thus, these power functions can be thought of as a basis for the vector space of Taylor polynomial approximations.
The next exposure to this concept for those students majoring in mathematics and some related fields is the notion of Fourier series in differential equations or a more advanced course. Here, a function f(x), usually a periodic function, is approximated by a linear combination of sinusoidal functions of the form sin (nx) and cos (nx). In this case, the sinusoidal functions can be thought of as a basis for a vector space. However, by the time students Finally, when we speak of the agreement between a function f and an approximation En of order n, we will use the interpretation that f and En agree in value at the indicated point and that all derivatives up to order n also agree at that point. Thus, at x = 0, say, we require that
f ’(0) = En’(0), f “(0) = En”(0), …, f (n)(0) = En(n) (0).
and prof dr mircea orasanu
shows that
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189962 | https://www.engineersedge.com/fluid_flow/pressure_drop/pressure_drop.htm | Engineers Edge utilizes cookies to enable essential site functionality, and targeted advertising. To learn more, see our Privacy Policy.
Pressure Drop Along Pipe Length - Fluid Flow Hydraulic and Pneumatic
Fluid Pressure Drop Along Pipe Length of Uniform Diameter
Fluid Flow Table of Contents
Hydraulic and Pneumatic Knowledge
Pressure drop in pipes is caused by:
Friction
Vertical pipe difference or elevation
Changes of kinetic energy
Calculation of pressure drop caused by friction in circular pipes
To determine the fluid (liquid or gas) pressure drop along a pipe or pipe component, the following calculations, in the following order.
Equation Reynolds Number:
| | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- |
| Re = ω D / v Re = ρ v l / µ Re = ω l / v Re = Reynolds Number (unitless) ω = Velocity of fluid flow (m/sec) D = Diameter of Pipe (m) v = Kinematic Viscosity (m2/s) ρ = Density of fluid (kg/m3) l = Characteristic length, thoe chord of an airfoil Kinematic Viscosity Example kinematic viscosity values for air and water at 1 atm and various temperatures. Air Kinematic Viscosity m2/a | | | | --- | 1.2462E-5 | -10 | 14 | | 1.3324E-5 | 0 | 32 | | 1.4207E-5 | 10 | 50 | | 1.5111E-5 | 20 | 68 | Water Kinematic Viscosity m2/ a | | | | --- | 1.6438E-6 | 1 | 33.8 | | 1.267E-6 | 10 | 50 | | 9.7937E-7 | 20 | 6 | Kinematic Viscosity Table Chart of Liquids |
If the Reynolds number < 2320, than you have laminar flow.
Laminar flow is characterized by the gliding of concentric cylindrical layers past one another in orderly fashion. The velocity of the fluid is at its maximum at the pipe axis and decreases sharply to zero at the wall. The pressure drop caused by friction of laminar flow does not depend of the roughness of pipe.
If the Reynolds number > 2320, you have turbulent flow.
There is an irregular motion of fluid particles in directions transverse to the direction of the main flow. The velocity distribution of turbulent flow is more uniform across the pipe diameter than in laminar flow. The pressure drop caused by friction of turbulent flow depends on the roughness of pipe.
Select pipe friction Coefficient:
The pipe friction coefficient is a dimensionless number. The friction factor for laminar flow condition is a function of Reynolds number only, for turbulent flow it is also a function of the characteristics of the pipe wall.
Determine Pipe friction coefficient at laminar flow:
λ = 64 / Re
Where:
λ = Pipe Friction Coefficient
Re = Reynolds number
Note: Perfectly smooth pipes will have a roughness of zero.
Determine Pipe friction coefficient at turbulent flow (in the most cases) Colbrook Equation:
or
Where:
λ = Pipe Friction Coefficient
g = Acceleration of Gravity (9.8 m/s2)
Re = Reynolds Number (unitless)
k = Absolute Roughness (m)
D = Diameter of Pipe (m)
lg = Short for Log
The solutions to this calculation is plotted vs. the Reynolds number to create a Moody Chart.
Following table gives typical roughness values in millimeters for commonly used piping materials.
| | |
--- |
| Surface Material | Absolute Roughness Coefficient - k (mm) |
| Aluminum, Lead | 0.001 - 0.002 |
| Drawn Brass, Drawn Copper | 0.0015 |
| Aluminum, Lead | 0.001 - 0.002 |
| PVC, Plastic Pipes | 0.0015 |
| Fiberglass | 0.005 |
| Stainless steel | 0.015 |
| Steel commercial pipe | 0.045 - 0.09 |
| Stretched steel | 0.015 |
| Weld steel | 0.045 |
| Galvanized steel | 0.15 |
| Rusted steel | 0.15 - 4 |
| Riveted steel | 0.9 - 9 |
| New cast iron | 0.25 - 0.8 |
| Worn cast iron | 0.8 - 1.5 |
| Corroding cast iron | 1.5 - 2.5 |
| Asphalted cast iron | 0.012 |
| Galvanized iron | 0.015 |
| Smoothed cement | 0.3 |
| Ordinary concrete | 0.3 - 3 |
| Well planed wood | 0.18 - 0.9 |
| Ordinary wood | 5 |
Determine Pressure drop in circular pipes:
Δp = λ · L / D · ρ / 2 · ω2
Where:
Δp = Pressure Drop (Pa or kg/ms 2)
λ = Pipe Friction Coefficient
L = Length of Pipe (m)
D = Pipe Diameter (m)
p = Density (kg/m3)
ω = Flow Velocity (m/s)
If you have valves, elbows and other elements along your pipe then you calculate the pressure drop with resistance coefficients specifically for the element. The resistance coefficients are in most cases found through practical tests and through vendor specification documents. If the resistance coefficient is known, than we can calculate the pressure drop for the element.
Δp = Γ · ρ / 2 · ω2
Where:
= Pressure Drop (kg/m2)
Γ = Resistance Coefficient (determined by test or vendor specification)
p = Density (kg/m3)
ω = Flow Velocity
Pressure drop by gravity or vertical elevation
Δp = ρ · g · ΔH
Where:
Δp = Pressure Drop
(kg/m2)
p = Density (kg/m3)
g = Acceleration of Gravity (9.8 m/s/s)
ΔH = Vertical Elevation or Drop
(m)
Pressure drop of gasses and vapor
Compressible fluids expands caused by pressure drops (friction) and the velocity will increase. Therefore is the pressure drop along the pipe not constant.
Where:
p1 = Pressure incoming (kg/m2)
T1 = Temperature incoming (°C)
p2 = Pressure leaving (kg/m2)
T2 = Temperature leaving (°C)
We set the pipe friction number as a constant and calculate it with the input-data. The temperature, which is used in the equation, is the average of entrance and exit of pipe.
Note: You can calculate gases as liquids, if the relative change of density is low (change of density/density = 0.02).
Head Loss Head loss is the reduction in the total head or pressure of the fluid as it moves through a fluid system.
Friction Factor Friction factor has been determined to depend on the Reynolds number for the flow and the degree of roughness of the pipes
Darcy's Equation Darcy's Equation Fluids Flow Equation - also called Darcy Weisbach equation.
Minor Losses Pressure losses that occur in pipelines due to bends, elbows, joints, valves, etc. are sometimes called minor losses.
Equivalent Piping Length Equivalent Piping Length Head Loss Equation Fluids
Loss of Air Pressure Due To Pipe Friction Table 1 Loss of Air Pressure Due To Pipe Friction Table
Loss of Air Pressure Due To Pipe Friction Table 2 Loss of Air Pressure Due To Pipe Friction Table
Loss of Pressure Through Screw Pipe Fittings Loss of Pressure Through Screw Pipe Fitting. Each pipeline fitting within the system
Hazen-Williams Method for Pipe Friction and Pressure Drop Hazen-Williams formula:
Estimated Pressure Drop for Thermoplastic Lined Fittings and Valves Estimated Pressure Drop for Thermoplastic Lined Fittings and Valves.
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189963 | https://fspieksma.win.tue.nl/papers/SIDMA2006paper.pdf | SIAM J. DISCRETE MATH.
c ⃝2006 Society for Industrial and Applied Mathematics Vol. 20, No. 3, pp. 748–768 APPROXIMATION ALGORITHMS FOR RECTANGLE STABBING AND INTERVAL STABBING PROBLEMS∗ SOFIA KOVALEVA† AND FRITS C. R. SPIEKSMA‡ Abstract. In the weighted rectangle stabbing problem we are given a grid in R2 consisting of columns and rows each having a positive integral weight, and a set of closed axis-parallel rectangles each having a positive integral demand. The rectangles are placed arbitrarily in the grid with the only assumption being that each rectangle is intersected by at least one column or row. The objective is to find a minimum-weight (multi)set of columns and rows of the grid so that for each rectangle the total multiplicity of selected columns and rows stabbing it is at least its demand. A special case of this problem, called the interval stabbing problem, arises when each rectangle is intersected by exactly one row. We describe an algorithm called STAB, which is shown to be a constant-factor approximation algorithm for different variants of this stabbing problem.
Key words. rectangle stabbing, approximation algorithms, combinatorial optimization AMS subject classifications. 68W25, 68R05, 90C27 DOI. 10.1137/S089548010444273X 1. Introduction. The weighted rectangle stabbing problem (WRSP) can be de-scribed as follows: given are a grid in R2 consisting of columns and rows each having a positive integral weight, and a set of closed axis-parallel rectangles each having a positive integral demand. The rectangles are placed arbitrarily in the grid with the only assumption being that each rectangle is intersected by at least one column or row. The objective is to find a minimum-weight (multi)set of columns and rows of the grid so that for each rectangle the total multiplicity of selected columns and rows stabbing this rectangle equals at least its demand. (A column or row is said to stab a rectangle if it intersects it.) A special case of the WRSP is the case where each rectangle is intersected by exactly one row; we will refer to the resulting problem as the weighted interval stabbing problem (WISP), or ISP in the case of unit weights (see Figure 1 for an example of an instance of the ISP).
Fig. 1. An instance of ISP with unit demands. The rectangles (or intervals in this case) are in grey; the columns and row in black constitute a feasible solution.
∗Received by the editors March 31, 2004; accepted for publication (in revised form) August 16, 2005; published electronically October 12, 2006. This work grew out of the Ph.D. thesis ; a pre-liminary version of this paper appeared in the Proceedings of the 12th Annual European Symposium on Algorithms . This research was supported by EU-grant APPOL, IST 2001-30027.
†Corresponding author. Department of Quantitative Economics, Maastricht University, P.O. Box 616, NL-6200 MD Maastricht, The Netherlands (sonja.kovaleva@mail.com).
‡Department of Applied Economics, Katholieke Universiteit Leuven, Naamsestraat 69, B-3000, Leuven, Belgium (frits.spieksma@econ.kuleuven.be).
748 APPROXIMATION ALGORITHMS FOR STABBING PROBLEMS 749 Motivation. Although at first sight the WRSP may seem rather specific, it is not difficult to see that the following two problems can be reduced to WRSP.
• Solving special integer programming problems.
The following type of inte-ger linear programming problem can be reformulated as instances of WRSP: minimize{wx| (B|C)x ≥b, x ∈Zl}, where B and C are both 0,1-matrices with consecutive 1’s in the rows (so-called interval matrices; see, e.g., Schri-jver ), b ∈Zn +, w ∈Zl +. Indeed, construct a grid which has a column for each column in B and a row for each column in C. For each row i of matrix B|C, draw a rectangle i such that it intersects only the columns and rows of the grid corresponding to the positions of 1’s in row i. Observe that this construction is possible since B and C have consecutive 1’s in the rows. To complete the construction, assign demand bi to each rectangle i and a cor-responding weight wj to each column and row of the grid. Let the decision variables x describe the multiplicities of the columns and rows of the grid.
In this way we have obtained an instance of WRSP. In other words, integer programming problems where the columns of the constraint matrix A can be permuted such that A = (B|C), with B and C each being an interval matrix, are special cases of WRSP.
• Stabbing geometric figures in the plane. Given a set of arbitrary connected closed geometric sets in the plane, use a minimum number of straight lines of two given directions to stab each of these sets at least once.
Indeed, by introducing a new coordinate system specified by the two directions and by replacing each closed connected set by a closed rectangle defined by the projections of the set to the new coordinate axes, we obtain an instance of the problem of stabbing rectangles using a minimum number of axis-parallel lines.
More specifically, we define a grid whose rows and columns are axis-parallel lines containing the rectangles’ edges. We can restrict attention to those lines since any axis-parallel line stabbing some set of rectangles can be replaced by a line stabbing this set and containing a rectangle’s edge.
Therefore, the problem of stabbing the rectangles with axis-parallel lines reduces to the problem of stabbing them with the rows and columns of the grid.
Literature. The WRSP and its special case WISP have already received at-tention in the literature. Motivated by an application in parallel processing, Gaur, Ibaraki, and Krishnamurti present a 2-approximation algorithm for the WRSP with unit weights and demands, which admits an easy generalization to arbitrary weights and demands. Furthermore, Hassin and Megiddo (mentioning military and medical applications) study a number of special cases of the problem of stabbing geometric figures in R2 by a minimum number of straight lines. In particular, they present a 2-approximation algorithm for the task of stabbing connected figures of the same shape and size with horizontal and vertical lines. Moreover, they study the case of stabbing horizontal line segments of length K, whose endpoints have integral x-coordinates, with a minimum number of horizontal and vertical lines, and give a 2 −1 K -approximation algorithm for this problem. In our setting this corresponds to the ISP with unit demands, where each rectangle in the input is intersected by exactly K columns. Finally, Cˇ alinescu et al. , mentioning applications in embedded sensor networks, show that the problem of separating n points in the plane with a minimum number of axis-parallel lines is a special case of the unweighted rectangle stabbing problem.
Concerning computational complexity, a special case of ISP where each rectangle is stabbed by at most two columns is shown to be APX-hard in .
750 SOFIA KOVALEVA AND FRITS C. R. SPIEKSMA Our results. We present here an approximation algorithm called STAB for dif-ferent variants of WISP (see, e.g., Vazirani for an overview on approximation algorithms). First, we show that STAB is a 1 (1−(1−1/k)k)-approximation algorithm for ISPk, the variant of ISP where each row intersects at most k rectangles (e.g., the instance depicted in Figure 1 is an instance of ISP3). Observe that STAB is a 4 3-approximation algorithm for the case k = 2, and that STAB is an e e−1-approximation algorithm for the case where the number of rectangles sharing a row is unlimited (k = ∞). Thus, STAB improves upon the results described in Hassin and Megiddo (for K ≥3) and does not impose any restrictions on the number of columns inter-secting rectangles. Second, we show that STAB is an e e−1-approximation algorithm for the weighted case of ISP∞, i.e., the case where the columns and the rows of the grid have arbitrary positive integral weights.
Third, we state here that the algo-rithm described by Gaur, Ibaraki, and Krishnamurti can be generalized to yield a q+1 q -approximation algorithm for WRSP where the demand of each rectangle is bounded from below by an integer q. Observe that this provides a 2-approximation algorithm for the WRSP described in the introduction, where q = 1. Thus, this is an improvement upon the approximation ratio of the algorithm of Gaur, Ibaraki, and Krishnamurti for instances with a lower bound on the rectangles’ demands that is larger than 1. For the proof of this result, we refer to Kovaleva .
Our algorithms are based on rounding the linear programming relaxation of an integer programming formulation in an interesting way. We use the following property present in our formulation: The variables can be partitioned into two sets such that when the values of one set are fixed, one can compute the optimal values of the other variables in polynomial time, and vice versa. Next, we consider different ways of rounding one set of variables and compute each time the optimal values of the remaining variables, while keeping the best solution.
We also show that there exist instances of ISP2 and ISP∞(see section 3) and WRSP (see ) for which the ratio between the values of a natural integer linear pro-gramming (ILP) formulation and its linear programming relaxation (LP-relaxation) is equal (or arbitrarily close) to the obtained approximation ratios. This suggests that these approximation ratios are unlikely to be improved by an LP-rounding algorithm based on the natural ILP formulation.
2. Preliminaries. Let us formalize the definition of WRSP. Let the grid in the input consist of t columns and m rows, numbered consecutively from left to right and from bottom to top, with positive weight wc (vr) attached to each column c (row r).
Further, we are given n rectangles such that rectangle i has demand di ∈Z+ and is specified by leftmost column li, rightmost column ri, top row ti, and bottom row bi.
Let us give a natural ILP formulation of WRSP. In this paper we use notation [a : b] for the set of integers {a, a + 1, . . . , b}. The decision variables yc, zr ∈Z+, c ∈[1 : t], r ∈[1 : m], denote the multiplicities of column c and row r, respectively.
Minimize m r=1 vrzr + t c=1 wcyc (1) subject to r∈[bi:ti] zr + c∈[li:ri] yc ≥di ∀i ∈[1 : n], (2) zr, yc ∈Z1 + ∀r, c.
(3) APPROXIMATION ALGORITHMS FOR STABBING PROBLEMS 751 In a vector notation this can be represented as Minimize vz + wy (4) subject to Bz + Cy ≥d, (5) z ∈Zm +, y ∈Zt +, (6) where B ∈{0, 1}n×m and C ∈{0, 1}n×t are the constraint matrices of inequalities (2).
The linear programming relaxation is obtained by replacing the integrality constraints (6) by the nonnegativity constraints z ∈Rm +, y ∈Rt +.
For an instance I of WRSP and a vector a ∈Zn, we introduce two auxiliary ILP problems: IPz(I, a): (7) Minimize vz subject to Bz ≥a, z ∈Zm +.
IPy(I, a): (8) Minimize wy subject to Cy ≥a, y ∈Zt +.
Lemma 2.1. For any a ∈Zn, the LP-relaxation of each of the problems IPz(I, a) and IPy(I, a) is integral.
Proof.
As was previously observed in , matrices B and C have a so-called consecutive 1’s property.
This implies that these matrices are totally unimodular (see, e.g., Schrijver ), which implies the lemma.
Corollary 2.2. The optimum value of IPz(I, a) (IPy(I, a)) is smaller than or equal to the value of any feasible solution to its LP-relaxation.
Corollary 2.3. The problem IPz(I, a) (IPy(I, a)) can be solved in polynomial time. Its optimal solution coincides with that of its LP-relaxation.
In fact, the special structure of IPz(I, a) and IPy(I, a) allows us to solve it via a minimum cost flow algorithm: Let MCF(p, q) denote the time needed to solve the minimum cost flow problem on a network with p nodes and q arcs. A proof of the following lemma can also be found in Veinott and Wagner .
Lemma 2.4. The problem IPz(I, a) (IPy(I, a)) can be solved in time O(MCF(t, n+ t)) (O(MCF(m, n + m))).
Proof. Consider the LP-relaxation of formulation IPy(I, a) and substitute the current variables by new variables u0, . . . , ut as yc = uc −uc−1 ∀c ∈[1 : t]. Then it transforms into (9) Minimize −w1u0 + (w1 −w2)u2 + · · · + (wt−1 −wt)ut−1 + wtut subject to uri −uli−1 ≥ai ∀i ∈[1 : n], uc −uc−1 ≥0 ∀c ∈[1 : t].
Let us denote the vector of objective coefficients, the vector of right-hand sides, and the constraint matrix by w, a, and C, respectively, and the vector of variables by u. Then (8) can be represented as {minimize wu| Cu ≥a}. Its dual is {maximize ax| CT x = w, x ≥0}. Observe that this is a minimum cost flow formulation with flow conservation constraints CT x = w, since CT has exactly one 1 and one -1 in each column. Given an optimal solution to the minimum cost flow problem, one can obtain the optimal dual solution u0, . . . , ut via a shortest path computation (see Ahuja, Magnanti, and Orlin ), and thus optimal y1, . . . , yt values as well.
752 SOFIA KOVALEVA AND FRITS C. R. SPIEKSMA 3. Algorithm STAB. Recall that the interval stabbing problem WISP refers to the restriction of WRSP, where each rectangle in the input is intersected by exactly one row. We also refer by WISPk to WISP, where each row intersects at most k rectangles. We assume in this section that all demands are unit (di = 1, i ∈[1 : n]), thus resulting in the following formulation: Minimize m r=1 vrzr + t c=1 wcyc (10) subject to zρi + c∈[li:ri] yc ≥1 ∀i ∈[1 : n], (11) zr, yc ∈Z1 + ∀r, c.
(12) Here we denote by ρi the index of the row intersecting rectangle i.
First we describe algorithm STAB for WISP. In subsection 3.1 we show that it achieves a ratio of 1 1−(1−1/k)k for the unweighted version of WISPk: ISPk.
In subsection 3.2 we prove that STAB achieves a ratio of e e−1 for WISP. Subsection 3.3 shows that the integrality gap between the values of a natural integer programming formulation of ISPk and its LP-relaxation for k = 2 and k = ∞coincides with the approximation ratio of the algorithm. An alternative algorithm for the case k = 2 yielding the same worst-case ratio (i.e., 4 3) is described in Kovaleva and Spieksma .
Informally, algorithm STAB can be described as follows: Solve the LP-relaxation of (10)–(12), and denote the solution found by (ylp, zlp).
Assume, without loss of generality, that the rows are sorted as z lp 1 ≥z lp 2 ≥· · · ≥zlp m.
At each iteration j (j = 0, . . . , m) we solve the problem (10)–(12) with a fixed vector z, the first j elements of which are set to 1, and the others to 0. As shown in Lemma 2.4, this can be done in polynomial time using a minimum cost flow algorithm. Finally, we take the best of the resulting m + 1 solutions. A formal description of STAB is shown in Figure 2.
We use notation value(y, z) ≡t c=1 yc + m r=1 zr, value(y) ≡t c=1 yc, and value(z) ≡m r=1 zr.
1. solve the LP-relaxation of (10)–(12), and obtain its optimal solution (ylp, zlp); 2. reindex the rows of the grid so that z lp 1 ≥z lp 2 ≥· · · ≥zlp m; 3. V ←∞; 4. for j = 0 to m for i = 1 to j ¯ zi ←1, for i = j + 1 to m ¯ zi ←0.
solve IPy(I, b), where bi = 1 −¯ zρi, ∀i ∈[1 : n], and obtain ¯ y; if value(¯ y, ¯ z) < V , then V ←value(¯ y, ¯ z), y∗←¯ y, z∗←¯ z; 5. return (y∗, z∗).
Fig. 2. Algorithm STAB.
3.1. The approximation result for ISPk. In this subsection we show that algorithm STAB is a 1 1−(1−1/k)k -approximation algorithm for ISPk. Let us first adapt APPROXIMATION ALGORITHMS FOR STABBING PROBLEMS 753 the ILP formulation (10)–(12) to ISPk with unit demands: Minimize t c=1 yc + m r=1 zr (13) subject to zρi + c∈[li:ri] yc ≥1 ∀i ∈[1 : n], (14) zr, yc ∈Z+ ∀r, c.
(15) Theorem 3.1.
Algorithm STAB is a 1 1−(1−1/k)k -approximation algorithm for ISPk.
Proof. Consider an instance I of ISPk, and let (ylp, zlp) and (y∗, z∗) be, respec-tively, an optimal LP solution and the solution returned by the algorithm for I. We prove the theorem by establishing that (16) value(y∗, z∗) ≤ 1 1 −(1 −1/k)k value(y lp, z lp).
It is enough to prove the result for instances satisfying the following assumption: We assume that the optimal LP solution satisfies constraints (14) at equality; i.e., (17) z lp ρi + c∈(li:ri) y lp c = 1 ∀i ∈[1 : n].
We now sketch why we can assume that (17) holds. Indeed, suppose that (17) does not hold for some intervals i of some instance I. Then we modify I by shortening those intervals for which (17) does not hold. More precisely, by splitting the columns with ylp-values we shorten the appropriate intervals so that the assumption becomes true (see Figure 3 for an example). Thus, given I and (ylp, zlp), we create an instance I′ for which (17) holds. It is now easy to check that an optimal LP solution for I (with the split columns) is also an optimal LP solution for I′. Since in I′ the intervals have become shorter, algorithm STAB applied to I′ returns a solution with a value equal to or larger than the value of the solution returned for I. Then inequality (16) proven for I′ implies this inequality for I as well.
Fig. 3. Example of an initial instance (left) and a new instance satisfying the assumption (right).
We order the rows of the grid in order of nonincreasing zlp-values, and we denote by l (l ≥0) the number of zlp-values equal to 1. Then z lp 1 = · · · = z lp l = 1, 1 > z lp l+1 ≥ · · · ≥zlp m ≥0. We assume that value(ylp) is positive (otherwise all the zlp-values have to be equal to 1 and the theorem obviously holds).
By construction, (18) value(y∗, z∗) = min j∈[0:m] value(yj, zj) ≤min j∈[l:m] value(yj, zj), 754 SOFIA KOVALEVA AND FRITS C. R. SPIEKSMA where (yj, zj) is the jth solution generated in step 4 of STAB.
Let us proceed by defining a number qj = qj(Δ, β) ∈R for each j ∈[0 : m] that depends on a given Δ ∈[0, 1]m and β > 0 as follows: (19) ⌊qj⌋ k=1 (1 −Δj+k) + (qj −⌊qj⌋)(1 −Δj+⌈qj⌉) = β, where we put Δj = 0 if j > m. Since the left-hand side is 0 at qj = 0 and continuously increases to infinity as qj grows, there always exists a unique point qj satisfying the equality.
We will prove the following lemma.
Lemma 3.2.
value(yj, zj) ≤j + k · qj z lp, value(ylp) k ∀j ∈[l : m].
Then, assuming that Lemma 3.2 holds, it follows from (18) that (20) value(y∗, z∗) ≤min j∈[l:m] j + k · qj z lp, value(ylp) k .
Theorem 3.1 follows now from the following lemma, the proof of which can be found in the appendix.
Lemma 3.3. Given are real numbers 1 ≥Δ1 ≥Δ2 ≥· · · ≥Δm ≥0, a positive real number Y , an integer p ≥2, and an integer l ≥0. Then the following holds: (21) min i∈[l:m] (i + p · qi(Δ, Y/p)) ≤ 1 1 −(1 −1/p)p Y + m r=l+1 Δr + l.
By applying this lemma with p = k, Δ = zlp, and Y = value(ylp), the right-hand side of (20) can be bounded by 1 1−(1−1/k)k value(y lp) + m r=l+1 z lp r +l ≤ 1 1−(1−1/k)k value(y lp) + m r=l+1 z lp r + l , and since z lp 1 = · · · = z lp l = 1, the right-hand side of this last expression is equal to 1 1 −(1 −1/k)k value(y lp, z lp).
The theorem is then proved.
Proof of Lemma 3.2. Consider (yj, zj); for some j ∈[l : m], let us find an upper bound for value(yj, zj). By construction, −zj r = 1 ∀r ≤j, −zj r = 0 ∀r ≥j + 1, −yj is an optimal solution to IPy(I, b), where bi = 1 −zj ρi ∀i ∈[1 : n].
Obviously, value(zj) = j. In order to bound value(yj) we introduce a solution y′j, which is feasible to the LP-relaxation of IPy(I, b). Then, Corollary 2.2 implies that value(yj) ≤value(y′j).
APPROXIMATION ALGORITHMS FOR STABBING PROBLEMS 755 First, let us define subsets S1, S2, . . . , Sm, where Sr ⊂[1 : t] ∀r = 1, . . . , m (i.e., each subset consists of a set of columns of the grid), in the following way: Sr = i:ρi=r [li : ri].
Thus, Sr is the set of columns stabbing intervals in row r.
Fix now some j ∈[l : m], and construct vector y′j as follows: For each column c ∈[1 : t], – if c ∈Sj+1 ∪· · · ∪Sm, then denote by t the minimum index such that c ∈St and let y′j c = 1 (1−zlp t )ylp c (recall that zlp r < 1 ∀r ∈[l + 1 : m]); – otherwise, let y′j c = ylp c .
Let us now establish feasibility of y′j with respect to the LP-relaxation of IPy(I, b).
For any interval i we show that the following inequality holds: (22) c∈[li:ri] y′j c ≥1 −zj ρi.
If ρi < j + 1, where ρi is the row number of interval i, then zj ρi = 1, and the inequality holds automatically. Consider the case ρi ≥j + 1. For any c ∈Sρi, y′j c is defined as ylp c /(1 −z lp t ), where t ≤ρi. Since z lp t are nonincreasing with t, we have y′j c ≥ylp c /(1 −zlp ρi). Then, since [li : ri] ⊆Sρi, we have y′j c ≥ylp c /(1 −zlp ρi) for any c ∈[li : ri]. Using this, and remembering that (ylp, zlp) satisfies zlp ρi + c∈[li:ri] ylp c ≥1, we have c∈[li:ri] y′j c ≥ 1 (1 −z lp ρi) c∈[li:ri] y lp c ≥ 1 −zlp ρi 1 −z lp ρi = 1.
Thus, we have shown that inequality (22) holds for any i ∈[1 : n], and therefore y′j is feasible to the LP-relaxation of IPy(I, b). Now Corollary 2.2 implies that (23) value(yj) ≤value(y′j).
In what follows we show that value(y′j) ≤k · qj(zlp, value(ylp) k ) ∀j ∈[l : m]. By construction of y′j, using notation Y (S) = c∈S ylp c , (24) value(y′j) = 1 1−zlp j+1 Y (Sj+1) + 1 1−zlp j+2 Y (Sj+2\Sj+1) + · · · + 1 1−zlp m Y (Sm(Sj+1∪Sj+2∪· · ·∪Sm−1)) + Y (1 : t).
Observe that for the Y (·)-terms the following equality holds: (25) Y (Sj+1) + Y (Sj+2\Sj+1) + · · · + Y (Sm(Sj+1 ∪Sj+2 ∪· · · ∪Sm−1)) + Y (1 : t) = t c=1 y lp c = value(y lp).
Moreover, using the definition of Sr, our assumption (17), and the fact that there are 756 SOFIA KOVALEVA AND FRITS C. R. SPIEKSMA at most k intervals per row, we have for each r = j + 1, . . . , m (26) Y (Sr(Sj+1 ∪Sj+2 ∪· · · ∪Sr−1)) ≤Y (Sr) = c∈Sr y lp c ≤ i:ρi=r c∈[li:ri] y lp c = i:ρi=r (1 −z lp ρi) ≤k(1 −z lp r ).
Now consider the following optimization problem: max Yj+1,Yj+2,...
1 1 −z lp j+1 Yj+1 + 1 1 −z lp j+2 Yj+2 + · · · + 1 1 −z lp m Ym + ∞ r=m+1 Yr subject to Yj+1 + · · · + Ym + ∞ r=m+1 Yr ≤value(y lp), (27) 0 ≤Yr ≤k(1 −z lp r ) ∀r = j + 1, . . . , m, (28) 0 ≤Yr ≤k ∀r = m + 1, . . . , ∞.
(29) Due to (25) and (26) the following solution is feasible to this optimization problem: Yr = Y (Sr(Sj+1 ∪Sj+2 ∪· · · ∪Sr−1)) for each r = j + 1, . . . , m, and ∞ r=m+1 Yr = Y ((1 : t)(Sj+1 ∪Sj+2 ∪· · · ∪Sm)) (distributed arbitrarily among the components of the sum while satisfying (29)). Therefore the optimum value of this optimization problem is an upper bound on the right-hand side of (24).
What does the optimum solution to this optimization problem look like? Notice that the constraint matrix of (27)–(29) is a so-called greedy matrix (see Hoffman, Kolen, and Sakarovitch ). Together with the fact that the objective coefficients are nonincreasing, a result from implies that successive maximization of the variables Yj+1, Yj+2, . . . in this order produces an optimum solution.
Thus, we obtain the following optimal solution: Yj+1 = k(1 −z lp j+1), Yj+2 = k(1 −z lp j+2), . . . , Yj+⌊q⌋= k(1 −z lp j+⌊q⌋), Yj+⌊q⌋+1 = (q −⌊q⌋)k(1 −z lp j+⌊q⌋+1) for some number q ∈R+, which due to (27) has to satisfy k(1 −z lp j+1) + k(1 −z lp j+2) + · · · + k(1 −z lp j+⌊q⌋) + k(q −⌊q⌋)(1 −z lp j+⌊q⌋+1) = value(y lp), where we put zlp r = 0 for any r > m. Notice that q ≡qj(zlp, value(ylp) k ) (see (19)), and the optimum value of the problem (27)–(29), which bounds the right-hand side of (24) from above, is k · qj(zlp, value(ylp) k ). This proves Lemma 3.2.
3.2. The approximation result for WISP. In this section we consider the weighted version of ISP, without any limitation on the number of rectangles sharing a row, and prove the following result.
Theorem 3.4. Algorithm STAB is an e/(e−1) ≈1.582-approximation algorithm for WISP.
Proof. Consider an instance I of WISP, and let (ylp, zlp) and (y∗, z∗) be, respec-tively, an optimal solution to the LP-relaxation of (10)–(12) and the solution returned by the algorithm for I. We show that their values are related as follows: (30) value(y∗, z∗) ≤ e e −1value(y lp, z lp).
APPROXIMATION ALGORITHMS FOR STABBING PROBLEMS 757 Since value(ylp, zlp) is a lower bound for the optimal value of WIS, the theorem follows.
Assume, without loss of generality, that the rows of the grid are sorted so that z lp 1 ≥z lp 2 ≥· · · ≥zlp m. Further, suppose there are l zlp-values equal to 1, i.e., z lp 1 = . . . z lp l = 1, and 1 > z lp l+1 ≥z lp p+2 ≥· · · ≥zlp m ≥0.
Let (yj, zj) be candidate solution number j constructed by STAB for I ∀j ∈[0 : m]. From the design of STAB we know that (31) value(y∗, z∗) = min j∈[0:m] value(yj, zj) ≤min j∈[l:m] value(yj, zj).
Claim 1.
value(yj, zj) ≡wyj + vzj ≤ j r=1 vr + wylp 1 −z lp j+1 for any j ∈[l : m].
Let us prove it. Consider (yj, zj) for some j ∈[l : m]. By construction, – zj r = 1 ∀r ≤j, – zj r = 0 ∀r ≥j + 1, – yj is an optimal solution to IPy(I, b) with bi = 1 −zρi ∀i ∈[1 : n].
Clearly, vzj ≡m r=1 vrzj r = j r=1 vr. Let us show that (32) wyj ≤ wylp 1 −z lp j+1 .
To prove this, we establish that the fractional solution (33) 1 1 −z lp j+1 y lp, where we set z lp m+1 = 0, is feasible to the LP-relaxation of IPy(I, b).
Since yj is optimal to IPy(I, b), Corollary 2.2 implies (32). So, let us prove the following claim.
Claim 1.1. Solution (33) is feasible to the LP-relaxation of IPy(I, b) with bi = 1 −zρi ∀i ∈[1 : n]. We show that constraint (8) is satisfied: (34) 1 1 −z lp j+1 c∈[li,ri] y lp c ≥1 −zj ρi for any i ∈[1 : n].
Indeed, in case zj ρi = 1, the inequality trivially holds. Otherwise, if zj ρi = 0, it follows from the construction of zj that ρi ≥j +1. The ordering of the zlp-values implies that zlp ρi ≤z lp j+1. Then, using this and the fact that solution (ylp, zlp) satisfies constraint (14), we have 1 1 −z lp j+1 c∈[li,ri] y lp c ≥ 1 1 −z lp j+1 (1 −z lp ρi) ≥ 1 1 −z lp j+1 (1 −z lp j+1) = 1.
This proves (34) and, subsequently, Claims 1.1 and 1.
From (31) and Claim 1, value(y∗, z∗) ≤min j∈[l:m] j r=1 vr + wylp 1 −z lp j+1 758 SOFIA KOVALEVA AND FRITS C. R. SPIEKSMA = l r=1 vr + min j∈[l:m] j r=l+1 vr + wylp 1 −z lp j+1 .
Lemma 3.5 given below implies now that the last expression can be upper bounded by l r=1 vr + e e −1 m r=l+1 vrz lp r + wy lp ≤ e e −1 l r=1 vr + m r=l+1 vrz lp r + wy lp .
Since z lp 1 = · · · = z lp l = 1, the last expression can be rewritten as e e −1 m r=1 vrz lp r + wy lp = e e −1(vz lp + wy lp), which establishes inequality (30) and proves the theorem.
Lemma 3.5.
Suppose we are given numbers 1 > Δ1 ≥Δ2 ≥· · · ≥Δm ≥0 ∀i = 1, . . . , m, and Δm+1 = 0. Further, given are positive numbers a1, a2, . . . , am and Y . Then we have (35) min j=0,...,m j r=1 ar + 1 1 −Δj+1 Y ≤ e e −1 m r=1 arΔr + Y .
We give the proof of this lemma in the appendix.
3.3. Tightness. In this subsection we demonstrate that the ratio between the optimum values of ISPk and the LP-relaxation of its ILP formulation (13)–(15) can be arbitrarily close to the bounds achieved by STAB in case k = 2 and k = ∞(which are, respectively, 4/3 and e/(e −1)).
For the case k = 2 this is shown by the instance of ISP2 depicted in Figure 4 (recall that all the column and row demands and rectangle weights are unit). Here the optimal value of the problem is 2, since at least two elements (columns or rows) are needed to stab the three rectangles, whereas the optimal fractional solution has the value of 3/2.
Fig. 4. An instance of ISP2 and an optimal fractional solution.
In the remainder of the section we consider the problem ISP∞, or simply ISP, without any limitation on the number of rectangles sharing a row.
Theorem 3.6. The integrality gap of (13)–(15) is arbitrarily close to e e−1.
Proof. For each m ∈N we will construct an instance Im of ISP and show that the value of some feasible solution to its LP-relaxation tends to be e e−1 times its optimal value as m increases.
APPROXIMATION ALGORITHMS FOR STABBING PROBLEMS 759 Let us construct Im as follows. Let the grid have m rows and t = m! columns. Let the rows be numbered consecutively and let each row j intersect exactly j rectangles of the instance. Let rectangles intersected by row j be numbered j1, . . . , jj. All these rectangles are disjoint and each intersects exactly m!
j columns (see Figure 5). So, for a rectangle ji we have that its row number ρji is r, and its leftmost and rightmost columns are lji = m!
j (i −1) + 1 and rji = m!
j i. The total number of rectangles in the instance is then n = 1 + 2 + · · · + m.
Fig. 5. Instance I4.
We claim that the following solution (y, z) is feasible to the LP-relaxation of (13)–(15) for Im: (36) zj = 0 ∀j = 1, . . . , P, 1 −P/j ∀j = P + 1, . . . , m, yc = P m!
∀c = 1, . . . , m!, where P = P(m) is the number satisfying 1 m + 1 m −1 + · · · + 1 P + 1 ≤1 and 1 m + 1 m −1 + · · · + 1 P + 1 + 1 P ≥1.
Denote the value of this solution by LP(Im), and observe that LP(Im) = t c=1 yc + m r=1 zr = m −P 1 P + 1 + 1 P + 2 + · · · + 1 m .
Let us show feasibility of (y, z). Take any rectangle ji and show that the constraint zρji + c∈[lji,rji] yc ≥1 is satisfied. Notice that the z-values of our solution also can be expressed as zj = max (1 −P j , 0) ∀j = 1, . . . ., m. Substituting these values, and rewriting the left-hand side of constraints (14) gives max 1 −P ji , 0 + c∈[lji,rji] P m! = max 1 −P ji , 0 + m!
ji P m!
= max 1 −P ji , 0 + P ji .
Clearly, the last expression is at least equal to 1, which proves feasibility of solution (y, z) to the LP-relaxation of (13)–(15) for Im.
760 SOFIA KOVALEVA AND FRITS C. R. SPIEKSMA Now denote by OPT(I) the optimum value to ISP for I, and show that OPT(Im) = m. Consider any optimal integral solution, and denote by k the maximum row num-ber, whose corresponding z-value is 0. First, this means that there are at least m −k rows whose z-values are 1. Second, observe that, since there are k disjoint rectangles on row k and this row is not selected, there are at least k columns needed to stab these rectangles. Therefore, this solutions has to select at least m −k rows and k columns, meaning OPT(Im) ≥m. Since there exists a feasible solution of value m (select all the rows, for instance), we obtain that OPT(Im) = m.
We use Lemma 5.3 given in the appendix to prove that the ratio OPT(Im) LP(Im) = m m −P( 1 m + 1 m−1 + · · · + 1 P +1) approaches e e−1 when m increases. This establishes our tightness result.
As mentioned in the introduction, Theorems 3.1 and 3.6 imply that it is unlikely that a better ratio for ISP∞can be achieved using formulation (13)–(15).
Approximation algorithms with a ratio of e e−1 are not uncommon in the literature; integrality gaps with this ratio seem to appear less frequently.
Another example of a (different) formulation with an integrality gap that equals e e−1 is described in Hoogeveen, Skutella, and Woeginger .
4. Conclusion. We presented an approximation algorithm called STAB for two variants of the weighted rectangle stabbing problem.
STAB achieves a ratio of 1 1−(1−1/k)k for ISPk, the special case where each rectangle is stabbed by a single row and by at most k columns, and where all stabbing lines have unit weight. STAB achieves a ratio of e e−1 for WISP, the special case where each rectangle is stabbed by a single row. STAB considers different ways of rounding the LP-relaxation and outputs the best solution found in this way; it is also shown that the ratio proved equals the integrality gap when k = 2 and when k = ∞.
5. Appendix. In this appendix we give proofs of lemmas which we used in this paper.
Lemma 3.3. Given are real numbers 1 ≥Δ1 ≥Δ2 ≥· · · ≥Δm ≥0, a positive real number Y , an integer p ≥2, and an integer 0 ≤l < m. The following holds: (37) min i=l,...,m (i + p · qi(Δ, Y/p)) ≤ 1 1 −(1 −1/p)p Y + m r=l+1 Δr + l, where qi = qi(Δ, Y/p) for each i ∈[0 : m] is uniquely defined by the equality (38) ⌊qi⌋ k=1 (1 −Δi+k) + (qi −⌊qi⌋)(1 −Δi+⌈qi⌉) = Y/p, where we put Δi = 0 if i > m.
Proof. It is enough to prove this lemma for l = 0. The case of other l < m can be reduced to the case of l = 0 by changing the index to j = i −l and observing that qj+l(Δ, Y/p) = qi(Δ−l, Y/p), where vector Δ−l is obtained by deleting the first l elements from vector Δ. So we will prove that min i=0,...,m (i + p · qi(Δ, Y/p)) ≤ 1 1 −(1 −1/p)p Y + m r=1 Δr .
APPROXIMATION ALGORITHMS FOR STABBING PROBLEMS 761 The proof consists of two lemmas. In Lemma 5.1 we show that the left-hand side of (37) is upper bounded by the following supremum: (39) sup f(·) ∈H G(f(·)), where (40) G(f(·)) = min x ∈R+ (f(x) + p · (f(x + Y/p) −f(x))) , and the class of functions H is defined as (41) H = f(·) : R+ →R+ f(·) is continuous, increasing, concave, f(0) = 0, f(x) ≤x + m r=1 Δr .
In Lemma 5.2 we show that this supremum is upper bounded by the right-hand side of (37), which proves the lemma.
Lemma 5.1.
min i=0,...,m (i + p · qi(Δ, Y/p)) ≤ sup f(·) ∈H G(f(·)), where G(f(·)) and H are defined in (40) and (41).
Proof. To establish this, it is sufficient to exhibit a particular function ˆ f(·) ∈H, such that (42) G( ˆ f(·)) = min i=0,...,m (i + p · qi(Δ, Y/p)) .
Then, the supremum of G(f(·)) over all the possible f(·) ∈H is clearly larger than or equal to G( ˆ f(·)).
Before we describe the function ˆ f(·), let us define an auxiliary function F(·) : R+ →R+ as follows: (43) F(q) ≡ ⌊q⌋ r=1 (1 −Δr) + (q −⌊q⌋) (1 −Δ⌈q⌉), where we set Δr = 0 ∀r ≥m + 1.
Observe that F(·) is – continuous; – increasing, since Δr < 1, and therefore (1 −Δr) > 0 ∀r = 1, . . . , ∞; – convex, since the coefficients Δr are nonincreasing with increasing r, and there-fore the coefficients (1 −Δr) are nondecreasing with increasing r.
Furthermore, – F(0) = 0; – F(q) ≥(q −m r=1 Δr) ∀q ∈R+, since F(q) can be also represented as F(q) = q − ⎛ ⎝ ⌊q⌋ r=1 Δr + (q −⌊q⌋)Δ⌈q⌉ ⎞ ⎠, and obviously (⌊q⌋ r=1 Δr + (q −⌊q⌋)Δ⌈q⌉) ≤m r=1 Δr ∀q ∈R+; 762 SOFIA KOVALEVA AND FRITS C. R. SPIEKSMA – F(q) is linear on each of the intervals [i, i+1], i = 0, . . . , m−1, and on [m, +∞).
We are now ready to present ˆ f(·) : R+ →R+. We define ˆ f(·) ≡F −1(·) (since F(·) is increasing, F −1(·) exists).
We claim that ˆ f(·) ∈H. Indeed, ˆ f(·) has the following properties: – ˆ f(·) : R+ →R+ since F(·) : R+ →R+; – ˆ f(·) is continuous, increasing, and concave, since F(·) is continuous, increasing, and convex; – ˆ f(0) = 0, since F(0) = 0; – ˆ f(x) ≤x + m r=1 Δr ∀x ∈R+.
This can be obtained from F(q) ≥(q − m r=1 Δr) ∀q ∈R+, using F(q) = x, q = ˆ f(x).
This proves that ˆ f(·) ∈H.
To prove the lemma it remains to show that G( ˆ f(·)) = min i=0,...,m (i + p · qi(Δ, Y/p)) .
Comparing the definition of qi(Δ, Y/p) (see (38)) and F(·) (see (43)), observe that for each i ∈[0 : m] qi satisfies (44) F(i + qi) −F(i) = Y/p.
Thus, qi = F −1(F(i) + Y/p) −i.
Setting xi ≡F(i) ∀i = 0, . . . , m, we find that i = F −1(xi) and qi = F −1(xi +Y/p))−F −1(xi). Replacing F −1(·) by ˆ f(·), we obtain qi = ˆ f(xi + Y/p) −ˆ f(xi) ∀i = 0, . . . , m.
Using this together with i = F −1(xi) = ˆ f(xi), we can rewrite (45) min i=0,...,m (i + p · qi(Δ, Y/p)) = min i = 0, . . . , m xi = ˆ f −1(i) ˆ f(xi) + p( ˆ f(xi + Y/p) −ˆ f(xi)) .
Now we need to show that the latter expression is equal to (46) G( ˆ f(·)) ≡ min x ∈R+ ˆ f(x) + p( ˆ f(x + Y/p) −ˆ f(x)) .
We do this by showing that the function ˆ f(x) + p( ˆ f(x + Y/p) −ˆ f(x)) is continuous and concave in each of the intervals [xi, xi+1] ∀i = 0, . . . , m −1, and is increasing in [xm, +∞). Therefore the minimum can be achieved only at one of the endpoints x0, x1, . . . , xm.
Indeed, consider function ˆ f(x)+p( ˆ f(x+Y/p)−ˆ f(x)) in [xi, xi+1] for some i ∈[0 : m−1]. It can also be written as p ˆ f(x + Y/p) −(p −1) ˆ f(x). We know that ˆ f(x + Y/p) is concave on [xi, xi+1], since it is concave everywhere in R+. Furthermore, ˆ f(x) is linear on each [xi, xi+1], i ∈[0 : m−1], since F(·) is linear on [i, i + 1], i ∈[0 : m−1].
Obviously, a concave function minus a linear function is again concave.
Now we show that p ˆ f(x + Y/p) −(p −1) ˆ f(x) is increasing in [xm, +∞). Since ˆ f(x) = F −1(·) is increasing and linear in [xm, +∞), the growth rate of ˆ f(x) is the APPROXIMATION ALGORITHMS FOR STABBING PROBLEMS 763 same as the growth rate of ˆ f(x + Y/p) in [xm, +∞), and thus the growth rate of p ˆ f(x + Y/p) −(p −1) ˆ f(x) is positive. We have proved that the minimum in (46) is always achieved at one of the points x0, x1, . . . , xm, and therefore (46) is equal to (45). This completes the proof of Lemma 5.1.
Lemma 5.2.
sup f(·)∈H G(f(·)) ≤ 1 1 −(1 −1/p)p C, where C = Y + m r=1 Δr, G(f(·)) = min x∈R+ (f(x) + p(f(x+Y/p)−f(x))) , and the set of functions H (via notation C) is H = f(·) : R+ →R+ f(·) is continuous, increasing, concave, f(0) = 0, f(x) ≤x + C −Y .
Proof. We will prove several claims and subclaims.
Claim 1.
sup f(·)∈H G(f(·)) = sup g : f g(·)∈H g, where for each g ∈R+ function f g(·) is defined as follows: – f g(j · Y/p) = g(1 −(1 −1/p)j) ∀j ∈0 ∪N; – f g(x) is continuous in [0, +∞) and linear in each [(j −1) · Y/p, j · Y/p], j ∈N.
Notice that f g(·) is completely defined by the above characterization.
To prove this claim it is enough to show that for any f(·) ∈H there exists a function f ˆ g(·) ∈H, with ˆ g ≥0, such that G(f(·)) = G(f ˆ g(·)) = ˆ g.
To show that, we prove two subsidiary claims.
Claim 1.1. For any g ≥0, G(f g(·)) ≡ min x∈R+ (f g(x) + p(f g(x+Y/p)−f g(x))) = g.
Indeed, by construction f g(x) is linear in each of the intervals [(j −1) · Y/p, j · Y/p], j ∈N. This implies that function (f g(x) + p(f g(x+Y/p)−f g(x))) is linear in each of these intervals as well. Therefore the minimum over all x ≥0 is achieved in one of the endpoints 0, Y/p, 2Y/p, . . . . Consider (f g(x) + p · (f g(x+Y/p)−f g(x))) at the point x = j · Y/p for some j ∈N ∪0: f g(j · Y/p) + p · (f g((j + 1) · Y/p)−f g(j · Y/p)).
764 SOFIA KOVALEVA AND FRITS C. R. SPIEKSMA Using the definition of f g(·) we can rewrite it as follows: g · (1 −(1 −1/p)j) + p · (g(1 −(1 −1/p)j+1) −g · (1 −(1 −1/p)j)).
With simple computations one can verify that the last expression is equal to g. This proves Claim 1.1.
Claim 1.2. For any f(·) ∈H it holds that f ˆ g(·) ∈H, where ˆ g = G(f(·)).
Clearly, f ˆ g(x) is concave. To prove that f ˆ g(x) ≤x + C −Y ∀x ∈R+, it is sufficient to show that f ˆ g(x) ≤f(x), since f(·) ∈H means, e.g., f(x) ≤x + C −Y ∀x ∈R+.
So, let us establish that f ˆ g(x) ≤f(x) ∀x ∈R+. Recall that f ˆ g(x) is linear in each of the intervals [(j −1) · Y/p, j · Y/p], j ∈N, and f(x) is concave in R+. Then it is sufficient to show that f ˆ g(x) ≤f(x) ∀x = j · Y/p, j ∈0 ∪N.
We use mathematical induction on j. For j = 0, f ˆ g(0) = f(0) = 0 and the inequality trivially holds. Suppose, for j−1 we have proved that f ˆ g((j−1)·Y/p) ≤f((j−1)·Y/p), and let us show that f ˆ g(j · Y/p) ≤f(j · Y/p).
Observe that f ˆ g(·) can be represented in a recursive way as follows: (47) f ˆ g(j · Y/p) = ˆ g/p + f ˆ g((j−1) · Y/p) (1 −1/p).
Since ˆ g = G(f(·)) we know that ˆ g ≤f((j−1) · Y/p) + p · (f(j · Y/p)−f((j−1) · Y/p)).
Rearranging the expression, we obtain f(j · Y/p) ≥ˆ g/p + f((j−1) · Y/p) (1 −1/p).
By the induction hypothesis and (47) we can bound the right-hand side by ˆ g/p + f((j−1) · Y/p) (1 −1/p) ≥ˆ g/p + f ˆ g((j−1) · Y/p) (1 −1/p) = f ˆ g(j · Y/p).
This proves Claim 1.2.
These two claims imply that for any f(·) ∈H, there exists f ˆ g(·) ∈H, with ˆ g ≥0, such that G(f(·)) = G(f ˆ g(·)) = ˆ g.
This implies Claim 1.
Claim 2.
sup g : f g(·)∈H g ≤ 1 1 −(1 −1/p)p C.
Indeed, f g(·)∈H implies f g(x) ≤x + C −Y ∀x ∈R+ and, in particular, for x = Y.
From this, using the definition of f g(·), we obtain f g(Y ) ≡f g(p · Y/p) ≡g(1 −(1 −1/p)p) ≤Y + C −Y = C, and from the last inequality, we obtain g ≤ 1 (1 −(1 −1/p)p) C, APPROXIMATION ALGORITHMS FOR STABBING PROBLEMS 765 which proves Claim 2 and establishes Lemma 5.2.
Now we give a proof of Lemma 3.5. This version of the proof is due to Sgall (see the acknowledgments).
Lemma 3.5. Suppose we are given numbers 1 > Δ1 ≥Δ2 ≥· · · ≥Δm ≥0 and Δm+1 = 0. Further, given are positive numbers a1, a2, . . . , am and Y . Then we have (48) min j=0,...,m j r=1 ar + Y 1 −Δj+1 ≤ e e −1 m r=1 arΔr + Y .
Proof. We use mathematical induction on the size of inequality m. For m = 0, the statement trivially holds. Suppose that the lemma was proved for any inequality of size smaller than m. First, consider the case Δ1 ≥e−1 e . We can write min j=0,...,m j r=1 ar + Y 1 −Δj+1 ≤ min j=1,...,m j r=1 ar + Y 1 −Δj+1 = a1 + min j=1,...,m j r=2 ar + Y 1 −Δj+1 .
The latter minimum is the left-hand side of (48) for a smaller sequence: Δ2, . . . , Δm and a2, . . . , am. Applying the induction hypothesis, we can bound the last expression from above as follows (we also use our bound on Δ1): a1 + e e −1 m r=2 arΔr + Y ≤a1 · Δ1 e e −1 + e e −1 m r=2 arΔr + Y = e e −1 m r=1 arΔr + Y .
Thus, we have shown an induction step for the case Δ1 ≥e−1 e . For the remaining case, Δ1 < e−1 e , we give a direct proof below.
Suppose Δ1 < e−1 e . Denote the left-hand side of (48) by X, and notice that (49) j r=1 ar ≥X − 1 1 −Δj+1 Y for 0 ≤j ≤m.
The following steps are justified below: m r=1 arΔr + Y = m j=1 (Δj −Δj+1) j r=1 ar + Y ≥(1) m j=1 (Δj −Δj+1)X − ⎛ ⎝ m j=1 Δj −Δj+1 1 −Δj+1 ⎞ ⎠Y + Y = Δ1X − ⎛ ⎝ m j=1 Δj −1 1 −Δj+1 + 1 ⎞ ⎠Y + Y 766 SOFIA KOVALEVA AND FRITS C. R. SPIEKSMA = Δ1X + ⎛ ⎝1 −m + m j=1 1 −Δj 1 −Δj+1 ⎞ ⎠Y ≥(2) Δ1X + 1 −m + m(1 −Δ1) 1 m Y ≥(3) Δ1X + 1 −m + m(1 −Δ1) 1 m (1 −Δ1)X = 1 + m(−1 + (1 −Δ1) 1 m )(1 −Δ1) X ≥(4) 1 −1 e X.
(1) Here we use the ordering of the deltas and inequality (49).
(2) This inequality follows from the arithmetic-geometric mean inequality m j=1 xj ≥ m(m j=1 xj)1/m used for positive numbers xj = 1−Δj 1−Δj+1 .
(3) Here we use inequality Y ≥(1 −Δ1)X, which is implied by (49) for j = 0 and the fact that the coefficient of Y is nonnegative, which follows from 1 −Δ1 ≥1 e ≥ (1 −1 m)m.
(4) This inequality is elementary calculus: The minimum of the left-hand side over all Δ1 is achieved for 1 −Δ1 = ( m m+1)m, and, after substituting this value, it reduces to 1 −( m m+1)m+1 ≥1 −1 e.
Lemma 5.3. Let P(m) ∈N be defined as follows: (50) 1 m + 1 m −1 + · · · + 1 P(m) + 1 ≤1 and (51) 1 m + 1 m −1 + · · · + 1 P(m) + 1 + 1 P(m) ≥1.
Then, lim m→∞ m m −P(m)( 1 m + 1 m−1 + · · · + 1 P (m)+1) = e e −1.
Proof. Let us first find limm→∞P(m)/m. Observe that the following inequalities hold: 1 m + 1 m −1 + · · · + 1 P(m) + 1 ≥ m+1 P (m)+1 1 x dx = ln m + 1 P(m) + 1, 1 m + 1 m −1 + · · · + 1 P(m) ≤ m P (m)−1 1 x dx = ln m P(m) −1 (the equalities follow from b a 1/x dx = ln b/a). Then (50) and (51) imply 1 ≥ln m + 1 P(m) + 1, 1 ≤ln m P(m) −1.
APPROXIMATION ALGORITHMS FOR STABBING PROBLEMS 767 From this we have m + 1 e −1 ≤P(m) ≤m e + 1.
Dividing by m, 1 + 1/m e −1/m ≤P(m) m ≤1 e + 1/m.
Now we see that limm→∞P(m)/m = 1/e.
Let us now find limm→∞( 1 m + 1 m−1 + · · · + 1 P (m)+1). From (50) and (51) we have 1 − 1 P(m) ≤1 m + 1 m −1 + · · · + 1 P(m) + 1 ≤1.
Since we already know that limm→∞P(m) = ∞, we have lim m→∞ 1 m + 1 m −1 + · · · + 1 P(m) + 1 = 1.
Now consider m m −P(m)( 1 m + 1 m−1 + · · · + 1 P (m)+1) = 1 1 −P (m) m ( 1 m + 1 m−1 + · · · + 1 P (m)+1) .
Using limm→∞ P (m) m = 1/e and limm→∞( 1 m + 1 m−1 + · · · + 1 P (m)+1) = 1 we have lim m→∞ 1 1 −P (m) m ( 1 m + 1 m−1 + · · · + 1 P (m)+1) = 1 1 −1/e = e e −1, which establishes the lemma.
Acknowledgments. We are very grateful to professor Jiˇ r´ ı Sgall from the Math-ematical Institute of the Academy of Sciences of the Czech Republic, for allowing us to include his proof of Lemma 3.5. We also thank an anonymous referee whose comments improved the paper.
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The Physics Classroom » Physics Tutorial » Sound Waves and Music » Resonance
Sound Waves and Music - Lesson 5 - Physics of Musical Instruments
Resonance
Resonance
Guitar Strings
Open-End Air Columns
Closed-End Air Columns
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The goal of Unit 11 of The Physics Classroom Tutorial is to develop an understanding of the nature, properties, behavior, and mathematics of sound and to apply this understanding to the analysis of music and musical instruments. Thus far in this unit, applications of sound wave principles have been made towards a discussion of beats, musical intervals, concert hall acoustics, the distinctions between noise and music, and sound production by musical instruments. In Lesson 5, the focus will be upon the application of mathematical relationships and standing wave concepts to musical instruments. Three general categories of instruments will be investigated: instruments with vibrating strings (which would include guitar strings, violin strings, and piano strings), open-end air column instruments (which would include the brass instruments such as the trombone and woodwinds such as the flute and the recorder), and closed-end air column instruments (which would include some organ pipe and the bottles of a pop bottle orchestra). A fourth category - vibrating mechanical systems (which includes all the percussion instruments) - will not be discussed. These instrument categories may be unusual to some; they are based upon the commonalities among their standing wave patterns and the mathematical relationships between the frequencies that the instruments produce.
Resonance
As was mentioned in Lesson 4, musical instruments are set into vibrational motion at their natural frequency when a person hits, strikes, strums, plucks or somehow disturbs the object. Each natural frequency of the object is associated with one of the many standing wave patterns by which that object could vibrate. The natural frequencies of a musical instrument are sometimes referred to as the harmonics of the instrument. An instrument can be forced into vibrating at one of its harmonics (with one of its standing wave patterns) if another interconnected object pushes it with one of those frequencies. This is known as resonance - when one object vibrating at the same natural frequency of a second object forces that second object into vibrational motion.
The word resonance comes from Latin and means to "resound" - to sound out together with a loud sound. Resonance is a common cause of sound production in musical instruments. One of our best models of resonance in a musical instrument is a resonance tube (a hollow cylindrical tube) partially filled with water and forced into vibration by a tuning fork. The tuning fork is the object that forced the air inside of the resonance tube into resonance. As the tines of the tuning fork vibrate at their own natural frequency, they created sound waves that impinge upon the opening of the resonance tube. These impinging sound waves produced by the tuning fork force air inside of the resonance tube to vibrate at the same frequency. Yet, in the absence of resonance, the sound of these vibrations is not loud enough to discern. Resonance only occurs when the first object is vibrating at the natural frequency of the second object. So if the frequency at which the tuning fork vibrates is not identical to one of the natural frequencies of the air column inside the resonance tube, resonance will not occur and the two objects will not sound out together with a loud sound. But the location of the water level can be altered by raising and lowering a reservoir of water, thus decreasing or increasing the length of the air column. As we have learned earlier, an increase in the length of a vibrational system (here, the air in the tube) increases the wavelength and decreases the natural frequency of that system. Conversely, a decrease in the length of a vibrational system decreases the wavelength and increases the natural frequency. So by raising and lowering the water level, the natural frequency of the air in the tube could be matched to the frequency at which the tuning fork vibrates. When the match is achieved, the tuning fork forces the air column inside of the resonance tube to vibrate at its own natural frequency and resonance is achieved. The result of resonance is always a big vibration - that is, a loud sound.
Another common physics demonstration that serves as an excellent model of resonance is the famous "singing rod" demonstration. A long hollow aluminum rod is held at its center. Being a trained musician, teacher reaches in a rosin bag to prepare for the event. Then with great enthusiasm, he/she slowly slides her hand across the length of the aluminum rod, causing it to sound out with a loud sound. This is an example of resonance. As the hand slides across the surface of the aluminum rod, slip-stick friction between the hand and the rod produces vibrations of the aluminum. The vibrations of the aluminum force the air column inside of the rod to vibrate at its natural frequency. The match between the vibrations of the air column and one of the natural frequencies of the singing rod causes resonance. The result of resonance is always a big vibration - that is, a loud sound.
The familiar sound of the sea that is heard when a seashell is placed up to your ear is also explained by resonance. Even in an apparently quiet room, there are sound waves with a range of frequencies. These sounds are mostly inaudible due to their low intensity. This so-called background noise fills the seashell, causing vibrations within the seashell. But the seashell has a set of natural frequencies at which it will vibrate. If one of the frequencies in the room forces air within the seashell to vibrate at its natural frequency, a resonance situation is created. And always, the result of resonance is a big vibration - that is, a loud sound. In fact, the sound is loud enough to hear. So the next time you hear the sound of the sea in a seashell, remember that all that you are hearing is the amplification of one of the many background frequencies in the room.
Resonance and Musical Instruments
Musical instruments produce their selected sounds in the same manner. Brass instruments typically consist of a mouthpiece attached to a long tube filled with air. The tube is often curled in order to reduce the size of the instrument. The metal tube merely serves as a container for a column of air. It is the vibrations of this column that produces the sounds that we hear. The length of the vibrating air column inside the tube can be adjusted either by sliding the tube to increase and decrease its length or by opening and closing holes located along the tube in order to control where the air enters and exits the tube. Brass instruments involve the blowing of air into a mouthpiece. The vibrations of the lips against the mouthpiece produce a range of frequencies. One of the frequencies in the range of frequencies matches one of the natural frequencies of the air column inside of the brass instrument. This forces the air inside of the column into resonance vibrations. The result of resonance is always a big vibration - that is, a loud sound.
Woodwind instruments operate in a similar manner. Only, the source of vibrations is not the lips of the musician against a mouthpiece, but rather the vibration of a reed or wooden strip. The operation of a woodwind instrument is often modeled in a Physics class using a plastic straw. The ends of the straw are cut with a scissors, forming a tapered reed. When air is blown through the reed, the reed vibrates producing turbulence with a range of vibrational frequencies. When the frequency of vibration of the reed matches the frequency of vibration of the air column in the straw, resonance occurs. And once more, the result of resonance is a big vibration - the reed and air column sound out together to produce a loud sound. As if this weren't silly enough, the length of the straw is typically shortened by cutting small pieces off its opposite end. As the straw (and the air column that it contained) is shortened, the wavelength decreases and the frequency was increases. Higher and higher pitches are observed as the straw is shortened. Woodwind instruments produce their sounds in a manner similar to the straw demonstration. A vibrating reed forces an air column to vibrate at one of its natural frequencies. Only for wind instruments, the length of the air column is controlled by opening and closing holes within the metal tube (since the tubes are a little difficult to cut and a too expensive to replace every time they are cut).
Resonance is the cause of sound production in musical instruments. In the remainder of Lesson 5, the mathematics of standing waves will be applied to understanding how resonating strings and air columns produce their specific frequencies.
Next Section:
Guitar Strings
Open-End Air Columns
Closed-End Air Columns
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Computing the Intersection of Two Lines
Posted on March 3, 2021 by Clinton Freeman
Two lines and intersect when . We can solve for and in terms of the coefficients as follows. First, solve for :
(1)
Second, plug the solution into the second line equation:
(2)
Repeat this process, solving for first, to obtain an expression for :
(3)
Plug the solution into the second line equation:
(4)
Note that the denominators of both coordinates are the same modulo sign, so we can multiply either by to normalize the denominator and make its calculation reusable (below we apply this to ). This leaves us with the final expression:
(5)
The following demo allows you to specify two lines via two points each and observe the intermediate intersection point calculations.
x1: y1:
x2: y2:
x3: y3:
x4: y4:
intX numerator:
intX denominator:
intY numerator:
intY denominator:
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189967 | https://libjournal.uncg.edu/ncjms/article/download/2173/1538 | North Carolina Journal of Mathematics and Statistics Volume 7, Pages 13–23 (Accepted July 5, 2021, published July 12, 2021) ISSN 2380-7539
Geometry of a Family of Quartic Polynomials
Christopher Frayer and Lukas Smith ABSTRACT . For a fixed A ∈ C with |A| = 1 , let P denote the family of complex-valued polyno-mials of the form p(z) = ( z − 1)( z − A )( z − r1)( z − r2) with |r1| = |r2| = 1 . By the Gauss-Lucas Theorem, the critical points of a polynomial in P lie in the unit disk. This paper characterizes the location and structure of these critical points. We show that the unit disk contains ‘desert’ regions in which critical points of polynomials in P do not occur. In fact, depending on the location of A, the unit disk contains one or two desert regions bounded by the curve implicitly defined by
|2z − (A + 1) | = |4z2 − 3( A + 1) z + 2 A| .
In addition to determining where critical points of polynomials in P are located, we also show that almost every c inside the unit disk and outside the desert region(s) is the critical point of a unique polynomial in P.
Introduction
The Gauss-Lucas Theorem (Marden, 1966) guarantees that the critical points of a complex-valued polynomial lie in the convex hull of the roots of that polynomial. Moreover, a critical point lies on the boundary of the convex hull if and only if the critical point is a multiple root of the polynomial. Refinements of the Gauss-Lucas Theorem (see Steinerberger (2020) and the numerous references within) seek to characterize regions which contain all, some, or none of the critical points of a polynomial. One such refinement (Steinerberger, 2020) shows that if a polynomial p has m + n roots with n
roots inside the unit disk and m roots outside the unit disk, then there exists a constant d0 > 1 such that n − 1 critical points of p lie inside the unit disk and the other m critical points have modulus larger than d0. That is, there exists an annular region {z : 1 < |z| ≤ d0} containing no critical points of p. Another refinement (R¨ udinger, 2014) investigates polynomials with a zero inside the convex hull of its roots. Of particular interest is a degree four polynomial with four distinct roots forming a concave quadrilateral. If p is such a polynomial, then one of the three triangles formed by the roots of p contains no critical points of p.Investigating polynomials whose roots lie on a circle leads to similar refinements of the Gauss-Lucas Theorem. By changing coordinates, such a polynomial can be normalized to have roots on the unit circle with one root located at z = 1 . By the Gauss-Lucas Theorem, the critical points of such a polynomial must lie in the unit disk. Families of polynomials of this form are investigated in (Frayer et al., 2014; Frayer, 2017; Frayer and Gauthier, 2018), and in each case, the unit disk contains ‘desert’ regions where critical points do not occur. For a fixed a ∈ [−1, 1] , Frayer and
Received by the editors June 18, 2021. 2010 Mathematics Subject Classification. 30C15.
Key words and phrases. geometry of polynomials; critical points; Gauss-Lucas Theorem.
©2021 The Author(s). Published by University Libraries, UNCG. This is an OpenAccess 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.
13 14 C. Frayer and L. Smith.
Thomson (2020) extends these results to the family of complex-valued polynomials
Ωa = {p : C → C | p(z) = ( z − 1)( z − a)( z − r1)( z − r2), |r1| = |r2| = 1 } .
For fixed z1 and z2 on the unit circle, it follows from (R¨ udinger, 2014) and/or (Steinerberger, 2020), depending on the position of a relative to z1 and z2, that a region inside the unit disk contains no critical points of (z − 1)( z − a)( z − z1)( z − z2) ∈ Ωa. Surprisingly, as r1 and r2 vary around the unit circle, the unit disk contains desert regions which contain no critical points of polynomials in
Ωa. Furthermore, almost every c inside the unit disk and outside the desert regions is the critical point of a unique polynomial in Ωa.For a fixed A ∈ C with |A| = 1 , this paper extends the ideas of Frayer and Thomson (2020) to the family of quartic polynomials
P = {p : C → C | p(z) = ( z − 1)( z − A )( z − r1)( z − r2), |r1| = |r2| = 1 } .
Once again, the unit disk contains desert regions where critical points of polynomials in P do not occur. In fact, depending upon the location of A, the unit disk contains one or two desert regions, and almost every c inside the unit disk and outside the desert region(s) is the critical point of a unique polynomial in P (see our Theorem 6).
Critical Points
Suppose A ∈ C with |A| = 1 and |r1| = |r2| = 1 . Then, by the Gauss-Lucas Theorem
p(z) = ( z − 1)( z − A )( z − r1)( z − r2) ∈ P
has three critical points, the zeros of p′, that lie in the unit disk.
2.1. Preliminary Information
We begin our analysis of these critical points by introducing a result from Frayer et al. (2014). Given α > 0, we let Tα denote the circle of diameter α that passes through 1 and 1 − α in the complex plane.
Theorem 1. (Frayer et al., 2014). Let f (z) = ( z − 1)( z − z1) · · · (z − zn), where zk = eiθ k for each k. Let c1, . . . , c n denote the critical points of f (z), and suppose that 1 6 = ck ∈ Tαk for each
k. Then n∑
k=1
1
αk
= n.
Applying Theorem 1 to polynomials in P gives a disk, |z − 34 | < 14 , where critical points cannot occur.
Theorem 2. No polynomial in P has a critical point strictly inside T1/2.Proof. Let c1, c2, and c3 be critical points of p(z) = ( z − 1)( z − A )( z − r1)( z − r2) ∈ P with
c1 ∈ Tα1 , c2 ∈ Tα2 , and c3 ∈ Tα3 . Theorem 1 implies
1
α1
1
α2
1
α3
= 3 .
Suppose to the contrary that α3 < 12 . Then,
1
α1
1
α2
= 3 − 1
α3
= β < 1Geometry of a Family of Quartic Polynomials 15 implies α1 = α2 = 2
β
, or α1 < 2
β
and α2 > 2
β
. Either possibility is a contradiction as 2
β
2 and all three critical points must lie in the unit disk.
Similarly, as a consequence of Theorem 2, no polynomial in P has a critical point inside the disk ∣∣z − 34 A∣∣ < 14 .For p(z) = ( z − 1)( z − A )( z − r1)( z − r2) ∈ P , rotation about the origin by 12 Arg (A) radians in the clockwise direction forces the roots located at A and 1 to become complex conjugates. This symmetry is conducive to further analysis. For a fixed A ∈ C with |A| = 1 we let PA denote the family of polynomials
PA = {p : C → C | p(z) = ( z − A)( z − A)( z − r1)( z − r2), |r1| = |r2| = 1 } .
This paper will characterize the critical points of polynomials in PA (see Theorem 5), and then apply those results to the original family of polynomials P (see Theorem 6). As an initial observation, Theorem 2 implies that the unit disk contains two disks in which critical points of polynomials in PA cannot occur.
Corollary 1. No polynomial in PA has a critical point in the open disk ∣∣z − 34 A∣∣ < 14 or ∣∣z − 34 A∣∣ <
14
.
We now investigate several examples. For convenience, let U denote the unit circle.
Example 1. A polynomial p ∈ PA has a critical point at A whenever A is a repeated root of p. So, for each r ∈ U ,
(z − A)2(z − A)( z − r) ∈ PA
has a critical point at A. Similarly, for each r ∈ U ,
(z − A)( z − A)2(z − r) ∈ PA
has a critical point at A.
As Example 1 describes the only polynomials in PA with a critical point at A or A, we will assume that c / ∈ { A, A } as necessary through the remainder of the paper. Before exploring another example, we define and analyze an important family of curves.
Definition 1. Let A ∈ U with Re (A) = α and define
DA = {z ∈ C : ∣∣2z2 − 3αz + 1 ∣∣ = |z − α|} .
The set DA depends upon A ∈ U . To visualize DA, we explain how DA changes as A moves around the unit circle. If A starts at z = 1 and moves around the unit circle in the counterclockwise direction, DA is a simple closed curve tangent to U at A and A. When Re (A) = 79 , the curve bifur-cates into two disconnected simple closed curves tangent to U at A and A. See Figure 2.1. By sym-metry, as A moves past z = i the two curves come back together when Re (A) = −79 . For future use we note that both 34 A and 34 A are contained inside DA. A GeoGebra animation illustrating DA for varying values of A can be found at ∼frayerc/DA.html .Example 1 shows that infinitely many polynomials in PA have a critical point at c ∈ { A, A }.Similarly, one might wonder how many polynomials in PA will have a critical point at r ∈ U {A, A }.
Example 2. By the Gauss-Lucas Theorem,
pr(z) = ( z − A)( z − A)( z − r)2 ∈ PA16 C. Frayer and L. Smith. -1 -1 11
-1 -1
11
00
A-1 -1 11
-1 -1
11
00
A
FIGURE 2.1. The curve DA when 0 < Re (A) < 79 on the left, and when 79 <
Re (A) < 1 on the right.
is the only polynomial in PA with a critical point at r ∈ U { A, A }. Differentiating and simplifying gives
p′
r
(z) = 2( z − r)(2 z2 − (r + 3 α)z + rα + 1) .
Thus, pr has a critical point at z = r, as expected, and two other critical points that satisfy
2c2 − (r + 3 α)c + rα + 1 = 0 . Rewriting as 2c2 − 3αc + 1 = r(c − α), taking the modulus of both sides, and noting that r ∈ U yields ∣∣2c2 − 3αc + 1 ∣∣ = |c − α| .
To summarize, pr(z) = ( z − A)( z − A)( z − r)2 is the unique polynomial in PA with a critical point at r ∈ U \ { A, A }. Furthermore, the other two critical points of pr lie on the curve DA.
2.2. The General Case
A polynomial of the form p(z) = ( z − A)( z − A)( z − r1)( z − r2) ∈ PA has three critical points in the unit disk. To further understand these critical points, we investigate how a critical point of p
is related to r1 and r2. For α = Re (A),
p(z) = ( z2 − 2αz + 1)( z − r1)( z − r2)
and differentiation gives
p′(z) = 4 z3 − (3 r1 + 3 r2 + 6 α)z2 + (4 αr 1 + 4 αr 2 + 2 r1r2 + 2) z − r1 − r2 − 2αr 1r2.
If c is a critical point of p(z), then
0 = p′(c) = 4 c3 − (3 r1 + 3 r2 + 6 α)c2 + (4 αr 1 + 4 αr 2 + 2 r1r2 + 2) c − r1 − r2 − 2αr 1r2.
Solving for r1 gives
r1 = (3 c2 − 4αc + 1) r2 − (4 c3 − 6αc 2 + 2 c)(2 c − 2α)r2 − (3 c2 − 4αc + 1) .
Definition 2. Given c ∈ C, we define
fc(z) = (3 c2 − 4αc + 1) z − (4 c3 − 6αc 2 + 2 c)(2 c − 2α)z − (3 c2 − 4αc + 1)
and let Sc = fc(U ).Geometry of a Family of Quartic Polynomials 17 Observe that fc is a Mobius transformation with fc(r2) = r1. Furthermore, for c ∈ C \ { A, A },
(fc)−1 = fc, and it follows that fc(r1) = r2. We have established the following result.
Theorem 3. A polynomial p(z) = ( z − A)( z − A)( z − r1)( z − r2) ∈ PA has a critical point at c
if and only if fc(r2) = r1.
Since r1, r 2 ∈ U , fc(r1) = r2 ∈ Sc and fc(r2) = r1 ∈ Sc implies {r1, r 2} ⊆ Sc ∩ U . Lemma 1 investigates |Sc ∩ U | and is a direct extension of a result from (Frayer and Thomson, 2020).
Lemma 1. Suppose c ∈ C.(1) If Sc ∩ U = ∅, then no polynomial in PA has a critical point at c.(2) If Sc = U , then infinitely many polynomials in PA have a critical point at c.(3) If |Sc ∩ U | ∈ { 1, 2}, then c is the critical point of a unique polynomial in PA.
As Sc = fc(U ), characterizing the critical points of polynomials in PA requires a better under-standing of the Mobius transformation fc. Direct computations show that the Mobius transforma-tion fc has two fixed points, c and q = 2c2 − 3αc + 1
c − α , and pole z∞ = 3c2 − 4αc + 1 2c − 2α (also the pole of inversion) which is the midpoint of the line segment connecting c and q. The normal form of the Mobius transformation is given by
fc(z) − cfc(z) − q = re iθ z − cz − q .
Observing that fc(z∞) = ∞ and manipulating algebraically gives re iθ = −1. Therefore, fc is an elliptic Mobius transformation. See (Hitchman, 2018). Elliptic Mobius transformations can be expressed as a composition of two inversions about clines (circles or lines). In this case, as z∞ is on the line L passing through c and q, for Ω the circle centered at z∞ passing through c and q,
fc(z) = iΩ(rL(z))
where iΩ is inversion about the circle Ω and rL is reflection about the line L. This allows us to visualize Sc = fc(U ) geometrically and will be useful for future observations. See Figure 2.2.
Properties of Sc
Suppose c / ∈ { A, A }. Since fc is a Mobius transformation and U is a circle, Sc = fc(U ) is a circle or a line. To further understand Sc, we begin with a special case. To determine the values of
c for which Sc = U , we make use of the following result.
Theorem 4. (see Frayer, 2017, Theorem 2) A Mobius transformation T sends the unit circle to the unit circle if and only if
T (z) = αz − ββz − α
for some α, β ∈ C with |αβ | 6 = 1 .
Applying Theorem 4 to
fc(z) = (3 c2 − 4αc + 1) z − (4 c3 − 6αc 2 + 2 c)(2 c − 2α)z − (3 c2 − 4αc + 1) 18 C. Frayer and L. Smith. -1 -1 11
-1 -1
11
00
FIGURE 2.2. The Mobius transformation fc can be visualized as the composition of two inversions: reflection about the line L and inversion about the circle Ω.implies that Sc = U whenever c satisfies
3c2 − 4αc + 1 = 3 c2 − 4α + 1 and 2c − 2α = 4 c3 − 6αc 2 + 2 c. (3.1) Manipulating the left equation in (3.1) and setting c = x + iy gives
3c2 − 4αc + 1 = 3 c2 − 4αc + 1 (c − c)(6 x − 4α) = 0 (−2yi )(6 x − 4α) = 0 .
Therefore c = x + iy satisfies the left equation in (3.1) whenever y = 0 or x = 23 α. Substituting
c = 23 α + iy into the right equation in (3.1) and equating real and imaginary parts eventually gives Re : 2y2 + 40 27 α2 − 2 = 0
Im : 4y3 + 24 9 α2y − 4y = 0 .
Since this system of equations has no solution, c = 23 α + iy does not satisfy (3.1), and the only remaining possibility is y = 0 . Substituting c = x into the right equation in (3.1) gives
4x3 − 6αx 2 + 2 α = 0 . (3.2) For α ∈ (−1, 1) , (3.2) has a unique solution, call it c∞, with c∞ ∈ (−1, 1) . Furthermore, when
α = 1 we have A = A = 1 and (3.2) has two solutions: −12 and 1. As Example 1 characterized the polynomials in PA with a critical point at A (or A) we only have one solution of interest, c∞ = −12 .Similarly, when α = −1 we have A = A = −1 and c∞ = 12 . We have established the following result.
Lemma 2. Suppose A ∈ U with Re (A) = α and c∞ the unique value in (−1, 1) with 4( c∞)3 −
6α(c∞)2 + 2 α = 0 . Then, Sc = U if and only if c = c∞.Geometry of a Family of Quartic Polynomials 19 As another special case, note that Sc is a line whenever there exists a z0 ∈ U with
(2 c − 2α)z0 − (3 c2 − 4αc + 1) = 0 .
Adding the right term to both sides, taking the modulus, and noting that |z0| = 1 implies Sc is a line if and only if
|2c − 2α| = |3c2 − 4αc + 1 |. (3.3) For |2c − 2α| 6 = |3c2 − 4αc + 1 |, Sc is a circle. By the definition of Sc, z ∈ Sc if and only if there exists some w ∈ U with fc(w) = z. Equivalently, f −1
c
(z) = fc(z) = w implies |fc(z)| = |w| = 1 ,and so ∣∣∣∣
(3 c2 − 4αc + 1) z − (4 c3 − 6αc 2 + 2 c)(2 c − 2α)z − (3 c2 − 4αc + 1)
∣∣∣∣ = 1 .
Therefore, z ∈ Sc if and only if
∣∣∣∣z − 3c2 − 4αc + 1 2c − 2α
∣∣∣∣ =
∣∣∣∣
3c2 − 4αc + 1 2c − 2α
∣∣∣∣∣∣∣∣z − 4c3 − 6αc 2 + 2 c
3c2 − 4αc + 1
∣∣∣∣ . (3.4) When d 6 = 1 , the solution set of
|z − u| = d|z − v|
is a circle of Appollonius (see Partenskii (2008)) and has center C and radius R satisfying
C = d2v − ud2 − 1 and R = |v − u|
∣∣∣∣
dd2 − 1
∣∣∣∣ . (3.5) When d =
∣∣∣3c2−4αc +1 2c−2α
∣∣∣ = 1 in (3.4), Sc is a line (verifying our previous observation in equation (3.3)). When d 6 = 1 , Sc is a circle with center
C =
∣∣∣3c2−4αc +1 2c−2α
∣∣∣2 4c3−6αc 2+2 c
3c2−4αc +1
− 3c2−4αc +1 2c−2α
∣∣3c2−4αc +1 2c−2α
∣∣2 − 1 . (3.6) As one last case of interest, we determine when Sc is tangent to U . According to Lemma 1 and Example 2, if Sc is tangent to U at r / ∈ { A, A }, then c is a critical point of the unique polynomial
pr(z) = ( z − A)( z − A)( z − r)2 ∈ PA.
Furthermore, as seen in Example 2, the three critical points of pr satisfy c1 = r and c2,3 ∈ DA.Therefore, if Sc is tangent to U , then c ∈ U ∪ DA. Let’s explore Sc when c ∈ U ∪ DA.
Lemma 3. If c ∈ DA \ { A, A }, then Sc is internally tangent to U at 2c2 − 3αc + 1
c − α .
Proof. Let c ∈ DA. Then |2c2 − 3αc + 1 | = |c − α| and it follows that
∣∣∣∣
2c2 − 3αc + 1
c − α
∣∣∣∣ = 1 .20 C. Frayer and L. Smith.
Letting eiφ = 2c2 − 3αc + 1
c − α , direct calculations show that fc(eiφ ) = eiφ ∈ Sc. Furthermore, using standard manipulations and the definition of DA, the center of Sc becomes
C =
3c2−4αc +1 2c−2α
4c3−6αc 2+2 c
2c−2α
− 3c2−4αc +1 2c2−3αc +1 2c2−3αc +1 2c−2α
2c2−3αc +1 2c2−3αc +1
∣∣3c2−4αc +1 2c−2α
∣∣2 − 1= (3 c2 − 4αc + 1)(2 c − 2α)ce iφ − (3 c2 − 4αc + 1)(2 c2 − 3αc + 1)2 eiφ
|3c2 − 4αc + 1 |2 − | 2c − 2α|2
= CN
CD
eiφ .
Observing that CD ∈ R and further simplifying CN eventually gives C = Ke iφ with K ∈ R.Then, as eiφ ∈ SC ∩ U is on the line segment connecting the centers of Sc and U , Sc is tangent to
U at eiφ . It remains to show that Sc is internally tangent to U .To show that Sc is internally tangent to U , we verify that there exists a z0 ∈ U with |fc(z0)| < 1.We do so by recalling the geometric interpretation of fc. Letting L represent the line through
c and eiφ (the two fixed points of fc) and Ω the circle with diameter passing through c and eiφ ,
fc(z) = iΩ(rL(z)) where iΩ is inversion about the circle Ω and rL is reflection about the line L.See Figure 3.1. If E represents the second point of intersection between L and U , then fc(E) =
iΩ(rL(E)) = iΩ(E) ∈ Sc. Furthermore, as E is outside of Ω, fc(E) = iΩ(E) is inside Ω with
|fc(E)| < 1. Therefore Sc is internally tangent to U at eiφ .
Similar, but less involved computations show that when c ∈ U , Sc is externally tangent to U at
c with C = 43 c. We have established the following result.
Lemma 4. Suppose c ∈ C \ { A, A }.
(1) Sc is internally tangent to U if and only if c ∈ DA.(2) Sc is externally tangent to U if and only if c ∈ U .-1 -1 11
-1 -1
11
00
FIGURE 3.1. When c ∈ DA, eiφ ∈ U and E ∈ L ∩ U is outside of Ω. Therefore,
fc(E) = iΩ(rL(E)) = iΩ(E) ∈ Sc with |fc(E)| < 1.Geometry of a Family of Quartic Polynomials 21
Main Results
We let OA represent the open region inside the unit disk and outside the region(s) bounded by DA. Visually, in Figure 2.1, OA is the region inside, but not on, the unit circle and enclosed between, but not on, the DA curves. Denote the closure of OA by OA.When c ∈ DA ∪ U , Sc is tangent to U . Lemmas 5 and 6 determine |Sc ∩ U | when c / ∈ DA ∪ U .
Lemma 5. If c is contained inside DA, then Sc ∩ U = ∅.Proof. Let c be contained inside DA and recall that 34 A and 34 A are also contained inside DA. As
c / ∈ DA ∪ U , Sc is not tangent to U . Suppose to the contrary that |Sc ∩ U | = 2 . As we drag c to 3A
4
or 3A
4
(which ever is closer) along a line segment contained in DA, Sc is continuously transformed into a circle not intersecting U . By the Intermediate Value Theorem, there must exist a c0 on the line segment with Sc0 tangent to U . However, as the line segment does not intersect DA ∪ U , this contradicts Lemma 4 and it follows that Sc ∩ U = ∅.
A similar argument can be used to prove Lemma 6.
Lemma 6. If c ∈ O A \ { c∞}, then |Sc ∩ U | = 2 .
We are now ready to characterize the critical points of polynomials in PA.
Theorem 5. Let c ∈ C.(1) If c / ∈ O A, then no polynomial in PA has a critical point at c.(2) If c ∈ { c∞, A, A }, then infinitely many polynomials in PA have a critical point at c.(3) If c ∈ O A \ { c∞, A, A }, then a unique polynomial in PA has a critical point at c.Proof. Let c ∈ C.(1) If c is inside DA, Lemmas 5 and 1 imply that no polynomial in PA has a critical point at
c. Furthermore, by the Gauss-Lucas Theorem, no polynomial in PA has a critical point outside the unit disk. (2) If c = c∞, then Lemma 2 implies Sc = U . Therefore, by Lemma 1 and Theorem 3,
(z − A)( z − A)( z − r)( z − fc(r)) ∈ PA
has a critical point at c for each r ∈ U . By Example 1, there are infinitely many polynomi-als in PA with a critical point at c ∈ { A, A }.(3) If c ∈ O A \ { c∞, A, A }, c ∈ O A \ { c∞} or c ∈ { U ∪ DA} \ { A, A }. When c ∈ O A \ { c∞}
Lemma 6 implies |Sc ∩ U | = 2 . When c ∈ { U ∪ DA} \ { A, A } Lemma 4 implies that
|Sc ∩ U | = 1 . Therefore, by Lemma 1, there is a unique polynomial in PA with a critical point at c
We finish our discussion by revisiting the family of polynomials P. For a fixed A ∈ U and some
r1, r 2 ∈ U ,
(z − A)( z − A)( z − r1)( z − r2) ∈ PA.
Counterclockwise rotation by Arg (A) gives
(z − A2)( z − 1)( z − ˜r1)( z − ˜r2) ∈ P 22 C. Frayer and L. Smith.
and demonstrates a one-to-one correspondence between PA and P. In fact, by redefining DA, OA
and c∞, Theorem 5 can easily be restated for polynomials in P.For A ∈ U , we let D denote the set of complex numbers satisfying
|2z − (A + 1) | = |4z2 − 3( A + 1) z + 2 A|
and O represent the open region enclosed inside the unit disk and outside the region(s) bounded by
D. To visualize D and O, rotate the images in Figure 2.1 so that A becomes z = 1 . Furthermore, for α = Re (A) and k the unique real solution of 8( α + 1) k3 − 6( α + 1) k2 + 1 = 0 , we define
˜c∞ = k(A + 1) . We are now able to restate Theorem 5 for polynomials in P.
Theorem 6. Let c ∈ C.(1) If c / ∈ O , then no p ∈ P has a critical point at c.(2) If c ∈ { ˜c∞, A, 1}, then infinitely many polynomials in P have a critical point at c.(3) If c ∈ O \ { ˜c∞, A, 1}, then a unique polynomial in P has a critical point at c.
This completes our analysis of critical points of polynomials in P, and as usual, many interesting questions remain. It would be nice to characterize critical points of families of polynomials similar to P and Ωa. For example, when A ∈ C \ R with |A| < 1, what can be said about critical points of polynomials of the form
p(z) = ( z − 1)( z − A)( z − r1)( z − r2)
with |r1| = |r2| = 1 ? Such a polynomial lies ‘inbetween’ P and Ωa. Preliminary analysis suggests that our methods apply nicely to this scenario. For fixed z1 and z2 on the unit circle, it follows from (R¨ udinger, 2014) and/or (Steinerberger, 2020), depending on the position of A relative to z1 and
z2, that a region inside the unit disk contains no critical points of (z − 1)( z − A)( z − z1)( z − z2).But what more can we say about the existence of a desert region(s) as r1 and r2 vary around the unit circle? Furthermore, for a specified c ∈ C, how many, if any, such polynomials will have a critical point at c? Much more is waiting to be investigated.
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C. Frayer. Geometry of polynomials with three roots. Missouri Journal of Mathematical Sciences ,29(2):161–175, 2017. C. Frayer and L. Gauthier. A tale of two circles: Geometry of a class of quartic polynomials.
Involve , 11(3):489–500, 2018. C. Frayer and P. Thomson. Geometry of a family of quartic polynomials. Pi Mu Epsilon Math Journal , Fall, 2020. C. Frayer, M. Kwon, C. Shafhauser, and J. Swenson. The geometry of cubic polynomials. Math. Magazine , 87(2):113–124, 2014. M. Hitchman. Geometry with an Introduction to Cosmic Topology . This work is licensed under the Creative Commons Attribution-ShareAlike 4.0 International License, 2018. M. Marden. Geometry of polynomials . Second edition. Mathematical Surveys, No. 3. American Mathematical Society, 1966. M. Partenskii. The circle of apollonius and its applications in introductory physics. The Physics Teacher , 46:104–108, 2008. A. R¨ udinger. Strengthening the gauss-lucas theorem for polynomials with zeros in the interior of the convex hull. arXiv preprint arXiv:1405.0689 , 2014. Geometry of a Family of Quartic Polynomials 23 S. Steinerberger. A stability version of the gauss–lucas theorem and applications. Journal of the Australian Mathematical Society , 109(2):262–269, 2020.
(C. Frayer) D EPARTMENT OF MATHEMATICS , U NIVERSITY OF WISCONSIN -P LATTEVILLE , P LATTEVILLE , WI 53818, USA
Email address , Corresponding author: frayerc@uwplatt.edu
(L. Smith) D EPARTMENT OF MATHEMATICS , U NIVERSITY OF WISCONSIN -P LATTEVILLE , P LATTEVILLE , WI 53818, USA
Email address : smithluk182@gmail.com |
189968 | https://www.youtube.com/playlist?list=PL3QGWUBVReqRsGzlkOn6hnJPfSHdCs85c | Ordering Setups | Analytical Reasoning | LSAT - YouTube
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189969 | https://www.math.ucla.edu/~mt/131a.1.02s/131A-HW-Sol.pdf | SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 5. Problem 8. Prove that if x and y are real numbers, then 2xy ≤x2 + y2.
Proof. First we prove that if x is a real number, then x2 ≥0. The product of two positive numbers is always positive, i.e., if x ≥0 and y ≥0, then xy ≥0. In particular if x ≥0 then x2 = x · x ≥0. If x is negative, then −x is positive, hence (−x)2 ≥0. But we can conduct the following computation by the associativity and the commutativity of the product of real numbers: 0 ≥(−x)2 = (−x)(−x) = ((−1)x)((−1)x) = (((−1)x))(−1))x = (((−1)(x(−1)))x = (((−1)(−1))x)x = (1x)x = xx = x2.
The above change in bracketting can be done in many ways. At any rate, this shows that the square of any real number is non-negaitive. Now if x and y are real numbers, then so is the difference, x −y which is defined to be x + (−y). Therefore we conclude that 0 ≤(x + (−y))2 and compute: 0 ≤(x + (−y))2 = (x + (−y))(x + (−y)) = x(x + (−y)) + (−y)(x + (−y)) = x2 + x(−y) + (−y)x + (−y)2 = x2 + y2 + (−xy) + (−xy) = x2 + y2 + 2(−xy); adding 2xy to the both sides, 2xy = 0 + 2xy ≤(x2 + y2 + 2(−xy)) + 2xy = (x2 + y2) + (2(−xy) + 2xy) = (x2 + y2) + 0 = x2 + y2.
Therefore, we conclude the inequality: 2xy ≤x2 + y2 for every pair of real numbers x and y.
♥ 1 2 SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 5. Problem 11. If a and b are real numbers with a < b, then there exists a pair of integers m and n such that a < m n < b, n ̸= 0.
Proof. The assumption a < b is equivalent to the inequality 0 < b −a. By the Archimedian property of the real number field, R, there exists a positive integer n such that n(b −a) > 1.
Of course, n ̸= 0. Observe that this n can be 1 if b −a happen to be large enough, i.e., if b−a > 1. The inequality n(b−a) > 1 means that nb−na > 1, i.e., we can conclude that na + 1 < nb.
Let m be the smallest integer such that na < m. Does there exists such an integer? To answer to the question, we consider the set A = {k ∈Z : k > na} of integers. First A ̸= ∅. Because if na ≥0 then 1 ∈A and if na > 0 then by the Archimedian property of R, there exists k ∈Z such that k = k · 1 > na.
Hence A ̸= ∅. Choose ℓ∈A and consider the following chain: ℓ> ℓ−1 > ℓ−2 > · · · > ℓ−k, k ∈N.
This sequence eventually goes down beyond na. So let k be the first natural number such that ℓ−k ≤na, i.e., the natural number k such that ℓ−k ≤ na < ℓ−k + 1. Set m = ℓ−k + 1 and observe that na < m = ℓ−k ≤+1 ≤na + 1 < nb.
Therefore, we come to the inequality na < m < nb. Since n is a positive integer, we devide the inequlity by n withoug changing the direction of the inequality: a = na n < m n < nb n = b.
♥ Page 14, Problem 6. Generate the graph of the following functions on R and use it to determine the range of the function and whether it is onto and one-to-one: a) f(x) = x3.
b) f(x) = sin x.
c) f(x) = ex.
d) f(x) = 1 1+x4 .
SOLUTION SET FOR THE HOMEWORK PROBLEMS 3 Solution. a) The function f is bi-jection since f(x) < f(y) for any pair x, y ∈R with the relation x < y and for every real number y ∈R there exists a real numbe x ∈ R such that y = f(x).
b) The function f is neither in-jective nor surjective since f(x + 2π) = f(x) x + π ̸= x, x ∈R, and if y > 1 then there is no x ∈R such that y = f(x).
c) The function f is injective because f(x) < f(y) if x < y, x, y ∈R, but not surjec-tive as a map from R to R, be-cause there exists no x ∈R such that f(x) = −1.
4 SOLUTION SET FOR THE HOMEWORK PROBLEMS d) The function f is not injective as f(x) = f(−x) and x ̸= −x for x ̸= 0, nor surjective as there is no x ∈R such that f(x) = −1.
♥ Page 14, Problem 8. Let P be the set of polynomials of one real variable. If p(x) is such a polynomial, define I(p) to be the function whose value at x is I(p)(x) ≡ Z x 0 p(t)dt.
Explain why I is a function from P to P and determine whether it is one-to-one and onto.
Solution. Every element p ∈P is of the form: p(x) = a0 + a1x + a2x2 + · · · + an−1xn−1, x ∈R, with a0, a1, · · · , an−1 real numbers. Then we have I(p)(x) = Z x 0 (a0 + a1t + a2t2 + · · · + an−1tn−1)dt = a0x + a1 2 x2 + a2 3 x3 + · · · + an−1 n xn.
Thus I(p) is another polynomial, i.e., an element of P. Thus I is a function from P to P.
We claim that I is injective: If p(x) = a0 + a1x + a2x2 + · · · + am−1xm−1; q(x) = b0 + b1x + b2x2 + · · · + bn−1xn−1 have I(p)(x) = I(q)(x), x ∈R,i.e., a0x + a1 2 x2 + a2 3 x3 + · · · + am−1 m xm = b0x + b1 2 x2 + b2 3 x3 + · · · + bn−1 n xn.
Let P(x) = I(p)(x) and Q(x) = I(q)(x). Then the above equality for all x ∈R allows us to differentiate the both sides to obtain P ′(x) = Q′(x) for every x ∈R, SOLUTION SET FOR THE HOMEWORK PROBLEMS 5 in particular a0 = P ′(0) = Q′(0) = b0. The second differentiation gives P ′′(x) = Q′′(x) for every x ∈R, in particular a1 = P ′′(0) = Q′′(0) = b1.
Suppose that with k ∈N we have P (k)(x) = Q(k)(x) for every x ∈R. Then the differen-tiation of the both sides gives P (k+1)(x) = Q(k+1)(x) forevery x ∈R, in particular ak+1 = P (k+1)(0) = Q(k+1)(0) = bk+1. Therefore the mathematical induction gives a0 = b0, a1 = b1, · · · , am−1 = bm−1 and m = n, i.e., p = q. Hence the function I is injective.
We claim that I is not surjective: As I(p)(0) = 0, the constant polynomial q(x) = 1 cannot be of the form q(x) = I(p)(x) for any p ∈P, i.e., there is no p ∈P such that I(p)(x) = 1. Hence the constant polynimial q is not in the image I(P).
♥ Page 19, Problem 3. Prove that: a) The union of two finite sets is finite.
b) The union of a finite sent and a countable set is countable.
c) The union of two contable sets is countable.
Proof. a) Let A and B be two finite sets. Set C = A ∩B and D = A ∪B. First, let a, b and c be the total number of elements of A, B and C respectively. As C ⊂A and C ⊂B, we know that c ≤a and c ≤b. We then see that the union: D = C ∪(A\C) ∪(B\C) is a disjoint union, i.e., the sets C, A\C and B\C are mutually disjoint. Thus the total number d of elements of D is precisely c + (a −c) + (b −c) = a + b −c which is a finite number, i.e., D is a finite set with the total number d of elements.
b) Let A be a finite set and B a countable set. Set C = A ∩B and D = A ∪B. Since C is a subset of the finite set A, C is finite. Let m be the total number of elements of C and {c1, c2, · · · , cm} be the list of elemtns of C. Let n be the total number of elements of A and let {a1, a2, · · · , an−m} be the leballing of the set A\C. Arrange an enumeration of the elements of B in the following fashion: B = {c1, c2, · · · , cm, bm+1, bm+2, · · · }.
Arranging the set A in the following way: A = {a1, a2, · · · , an−m, c1, c2, · · · , cm}, 6 SOLUTION SET FOR THE HOMEWORK PROBLEMS we enumerate the elements of D = A ∪B in the following way: di = ai for 1 ≤i ≤n −m; ci−n−m for n −m < i ≤n; bi−n+m for i > n.
This gives an enumeration of the set D. Hence D is countable.
c) Let A and B be two countable sets. Let A = {an : n ∈N} and B = {bn : n ∈N} be enumerations of A and B respectively. Define a map f from the set N of natural numbers in the following way: f(2n −1) = an, n ∈N; f(2n) = bn, n ∈N.
Then f maps N onto D = A ∪B. The surjectivity of the map f guarantees that f −1(d) ̸= ∅ for every d ∈D.
For each d ∈N, let g(d) ∈N be the first element of f−1(d).
Since f −1(d) ∩f−1(d′) = ∅for every distinct pair d, d′ ∈D, g(d) ̸= g(d′) for every distinct pair d, d′ ∈D. Hence the map g is injective. Now we enumerate the set D by making use of g. Let d1 ∈D be the element of D such that g(d1) is the least element of g(D). After {d1, d2, · · · , dn} were chosen, we choose dn+1 ∈D as the element such that g(dn+1) is the least element of g(D{d1, d2, · · · , dn}). By induction, we choose a sequence {dn} of elements of D. Observe that 1 ≤g(d1) < g(d2) < · · · < g(dn) < · · · in N. Hence we have n ≤g(dn).
This means that every d ∈N appears in the list {d1, d2, · · · }. Hence D is countable.
♥ Page 24, Problem 1. Give five examples which show that P implies Q does not necessarily mean that Q implies P.
Examples. 1) P is the statement that x = 1 and Q is the statement that x2 = 1.
2) P is the statement that x ≤1 and Q is the statement that x ≤2.
3) Let A be a subset of a set B with A ̸= B. P is the statement that x is an element of A and Q is the statement that x is an element of B.
4) P is the statement that x is a positive real number and Q is the statement that x2 is a positive real number.
5) P is the statement that x = 0 or x = 1 and Q is the statement that x(x−1)(x−2) = 0.
♥ Page 24, Problem 3. Suppose that a, b, c, and d are positive real numbers such that a/b < c/d. Prove that a b < a + c b + d < c d.
Proof. The inequality a/b < c/d is equivalent to the inequality bc −ad > 0. We compare two numbers by subtracting one from the other. So we compare the first two of the above SOLUTION SET FOR THE HOMEWORK PROBLEMS 7 three fractions first and then the second pair of the fractions: a + c b + d −a b = b(a + c) −a(b + d) b(b + d) = bc −ad b(b + d) > 0; c d −a + c b + d = c(b + d) −(a + c)d c(b + d) = bc −ad c(b + d) > 0.
Therefore the desired inequalities follows.
♥ Page 24, Problem 4. Suppose that 0 < a < b. Prove that a) a < √ ab < b.
b) √ ab < a+b 2 .
Proof. a) We compute, based on the fact that the inequality a < b implies the inequality √a < √ b, b = √ b √ b > √a √ b = √ ab > √a√a = a.
b) We simply compute: a + b 2 − √ ab = (√a)2 + ( √ b)2 −2√a √ b 2 = (√a − √ b)2 2 > 0, where we used the fact that (x + y)2 = x2 −2xy + y2 which follows from the distributive law and the commutativity law in the field of real numbers as seen below: (x −y)2 = (x −y)x −(x −y)y by the distributive law = x2 −yx −xy + y2 = x2 −2xy + y2 by the commutativity law.
♥ Remark. In the last computation, is (x −y)2 = x2 −2xy + y2 obvious?
If so, take x = µ 0 1 0 0 ¶ and y = µ 0 0 −1 0 ¶ 8 SOLUTION SET FOR THE HOMEWORK PROBLEMS and compute x −y = µ 0 1 1 0 ¶ and (x −y)2 = µ 1 0 0 1 ¶ ; x2 = µ 0 0 0 0 ¶ , y2 = µ 0 0 0 0 ¶ , xy = µ −1 0 0 0 ¶ , yx = µ 0 0 0 −1 ¶ ; x2 −2xy + y2 = µ 2 0 0 0 ¶ ̸= µ 1 0 0 1 ¶ = (x −y)2.
Therefore, the formula (x −y)2 = x2 −2xy + y2 is not universally true.
This is a consequence of the distributive law and the commutative law which governs the field R of real numbers as discussed in the very early class.
Page 24, Problem 5. Suppose that x and y satisfy x 2 + y 3 = 1. Prove that x2 + y2 > 1.
SOLUTION SET FOR THE HOMEWORK PROBLEMS 9 Proof. The point (x, y) lies on the line: x 2 + y 3 = 1 (L) which cuts through x-axis at (2, 0) and y-axis at (0, 3). The line L is also discribed by parameter (2t, 3(1 −t)), t ∈R. So we compute x2 + y2 = (2t)2 + (3(1 −t))2 = 4t2 + 9(1 −t)2 = 4t2 + 9t2 −18t + 9 = 13t2 −18t + 9 = 13 µ t2 −18 13t + 9 13 ¶ = 13 (µ t −9 13 ¶2 + 9 13 − µ 9 13 ¶2) > 0 o for all t ∈R as 9 13 < 1.
♥ (L) Page 25, Problem 8. Prove that for all positive integers n, 13 + 23 + · · · + n3 = (1 + 2 + · · · + n)2.
Proof. Suppose n = 1. Then the both sides of the above identity is one. So the formula hold for n = 1.
Suppose that the formula hold for n, i.e., 13 + 23 + · · · + n3 = (1 + 2 + · · · + n)2.
Adding (n + 1)3 to the both sides, we get 1 + 2 + · · · + n3 + (n + 1)3 = (1 + 2 + · · · + n)2 + (n + 1)3 = µn(n + 1) 2 ¶2 + (n + 1)3 by Proposition 1.4.3 = (n + 1)2 µn2 + 4n + 4 4 ¶ = (n + 1)2(n + 2)2 4 = (1 + 2 + · · · + n + n + 1)2 by Proposition 1.4.3.
Thus the formula holds for n+1. Therefore mathematical induction assures that the formula holds for every n ∈N.
♥ 10 SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 25, Problem 9. Let x > −1 and n be a positive integer. Prove Bernoulli’s inequality: (1 + x)n ≥1 + nx.
Proof. If x ≥0, then the binary expansion theorem for real numbers gives (1 + x)n = 1 + nx + µ n 2 ¶ x2 + µ n 3 ¶ x3 + · · · + µ n n −1 ¶ xn−1 + xn ≥1 + nx as x2, x3, · · · , xn−1 and xn are all non-negative. If −1 < x < 0, then the inequality is more delicate. But we can proceed in the following way: (1 + x)n −1 = x µ (1 + x)n−1 + (1 + x)n−1 + · · · + (1 + x) + 1 ¶ ; n ≥(1 + x)n−1 + (1 + x)n−1 + · · · + (1 + x) + 1 as 1 + x < 1.
As x < 0, we get nx ≤x µ (1 + x)n−1 + (1 + x)n−1 + · · · + (1 + x) + 1 ¶ , consequently the desired inequality: (1 + x)n −1 = x µ (1 + x)n−1 + (1 + x)n−1 + · · · + (1 + x) + 1 ¶ ≥nx.
♥ Page 25, Problem 11. Suppose that c < d.
a) Prove that there is a q ∈Q so that |q − √ 2| < d −c.
b) Prove that q − √ 2 is irrational.
c) Prove that there is an irrational number between c and d.
Proof. a) Choose a1 = 1 and b1 = 2 and observe that a2 1 = 1 < 2 < 4 = b2 1 consequently a1 < √ 2 < b1.
Consider (a1 + b1)/2 and square it to get µa1 + b1 2 ¶2 = 9 4 > 2 SOLUTION SET FOR THE HOMEWORK PROBLEMS 11 and put a2 = a1 and b2 = (a1 + b1)/2. Suppose that a1, a2, · · · , an and b1, b2, · · · , bn were chosen in such a way that i) if µak−1 + bk−1 2 ¶2 < 2, then ak = (ak−1 + bk−1)/2 and bk = bk−1; ii) if µak−1 + bk−1 2 ¶2 > 2, then ak = ak−1 and bk = (ak−1 + bk−1)/2.
Thus we obtain sequences {an} and {bk} of rational numbers such that a1 ≤a2 ≤a3 ≤· · · an ≤· · · · · · bn ≤bn−1 ≤· · · b2 ≤b1 and bk −ak = bk−1 −ak−1 2 ; a2 n < 2 < b2 n, n ∈N, hence an < √ 2 < bn.
As b1 −a1 = 1, we have bn −an = 1/2n. For a large enough n ∈N we have 1/2n < d −c.
Now we conclude 0 < √ 2 −an ≤bn −an = 1 2n < d −c and an ∈Q.
Thus q = an has the required property.
b) Set p = √ 2 −q. If p ∈Q, then √ 2 = p + q ∈Q which is impossible. Therefore, p cannot be rational.
c) From (b), p = √ 2 −q is an irrational number and 0 < p < d −c from (a). Thus we get c < c + p < d. As seen in (b), c + p cannot be rational. Because if c + p = a is rational, then p = a −c has to be rational which was just proven not to be the case.
♥ 12 SOLUTION SET FOR THE HOMEWORK PROBLEMS Problems.
1) Negate the following statement on a function f on an interval [a, b], a < b. The function f has the property: f(x) ≥0 for every x ∈[a, b].
2) Let f(x) = x2 + bx + c, x ∈R.
What can you say about the relation on the constants b and c in each of the following cases?
a) f(x) ≥0 for every x ∈R.
b) f(x) ≥0 for som x ∈R.
3) Let f(x) = x2 + 4x + 3, x ∈R.
Which of the following statements on f is true? State the proof of your conclusion.
a) f(x) < 0 for some x ∈R; b) f(x) > 0 for some x ∈R; c) f(x) ≥0 for every x ∈R.
d) f(x) < 0 for every x ∈R.
Solution. 1) There exists an x0 ∈[a, b] such that f(x0) < 0.
2-a) First, we look at the function f closely: f(x) = x2 + bx + c = x2 + bx + b2 4 + c −b2 4 = µ x + b 2 ¶2 + 4c −b2 4 ≥4c −b2 4 for every x ∈R.
SOLUTION SET FOR THE HOMEWORK PROBLEMS 13 The function f assumes its smallest value f µ −b 2 ¶ = 4c −b2 4 at x = −b 2. Thus f(x) ≥0 for every x ∈R if and only if f µ −b 2 ¶ = 4c −b2 4 ≥0 if and only if 4c ≥b2.
2-b) If 4c ≥b2, then f(x) ≥0 for every x ∈R, in particular f(0) ≥0. If 4c < b2, then we have f à −b + √ b2 −4c 2 !
= 0.
Therefore the condition that f(x) ≥0 for some x ∈R holds regardless of the values of b and c. So we have no relation between b and c.
3) First, we factor the polynomial f and draw the graph: f(x) = x2 + 4x + 3 = (x + 3)(x + 1).
We conclude that f(x) ≤0 is equivalent to the condition that −3 ≤x ≤−1. Therefore we conclude that (a) and (b) are both true and that (c) and (d) are both false.
♥ Page 25, Problem 12. Prove that the constant e = ∞ X k=0 1 k!
is an irrational number.
Proof. Set sk = k X j=0 1 j!, k ∈N.
14 SOLUTION SET FOR THE HOMEWORK PROBLEMS Clearly we have 2 = s1 ≤s2 ≤s3 ≤· · · ≤sk ≤· · · , i.e., the sequence {sk} is an increasing sequence. If k ≥3, we have sk = 1 + 1 + 1 1 · 2 + 1 1 · 2 · 3 + · · · + 1 1 · 2 · 3 · · · · k ≤1 + 1 + 1 2 + 1 2 · 2 + · · · + 1 2k−1 = 1 + 1 − 1 2k 1 −1 2 < 1 + 2 = 3.
Thus the sequence {sk} is bounded. The Bounded Monotonce Convergence Axiom (MC) guarantees the convergence of {sk}. Thus the limit e of {sk} exists and e ≤3.
a) If n > k, then we have sn −sk = 1 + 1 + 1 2! + 1 3! + · · · + 1 k! + 1 (k + 1)! + · · · + 1 n!
− µ 1 + 1 + 1 2! + 1 3! + · · · + 1 k!
¶ = 1 (k + 1)! + 1 (k + 2)! + · · · + 1 n!
= 1 (k + 1)!
× µ 1 + 1 k + 2 + 1 (k + 2)(k + 3) + 1 (k + 2)(k + 3)(k + 4) + · · · + 1 (k + 1) · · · n ¶ .
Since k + 1 < k + 2 < k + 3 < · · · < n, and 1 k + 1 > 1 k + 2 1 k + 3 > · · · > 1 n, we have 1 (k + 1)2 > 1 (k + 1)(k + 2), 1 (k + 1)3 > 1 (k + 1)(k + 2)(k + 3), · · · 1 (k + 1)n−k−1 > 1 (k + 1)(k + 2) · · · n and sn −sk < 1 (k + 1)!
n−k−1 X ℓ=0 1 (k + 1)ℓ= 1 (k + 1)!
1 − 1 (k+1)n−k 1 − 1 (k+1) < 1 k(k + 1)!.
SOLUTION SET FOR THE HOMEWORK PROBLEMS 15 Taking the limit of the left hand side as n →∞, we get e −sk ≤ 1 k(k + 1)! < 1 k · k!.
b) Now suppose that e is rational, i.e., e = p/q for some p, q ∈N. Then we must have p q −sq ≤ 1 q · q!
and 0 < q!
µp q −sq ¶ < 1 q .
But now q!(p/q) is an integer and p!sq is also because 2!, 3!, · · · and q!, appearing in the denominators of the summation of sq, all divide q!, i.e., q!sq = q!
µ 1 + 1 + 1 2! + 1 3! + · · · + 1 q!
¶ = q! + q! + (3 · 4 · · · q) + (4 · 5 · · · q) + · · · + (k · (k + 1) · · · q) + · · · + q + 1.
Thus the number q!(p/q −sq) is a positive integer smaller than 1/q, which is not possible.
This contradiction comes from the assumption that e = p/q, p, q ∈N. Therefore we conclude that there is no pair of natural numbers p, q such that e = p/q, i.e., e is an irrational number.
♥ Page 33, Problem 1. Compute enough terms of the following sequences to guess what their limits are: a) an = n sin 1 n.
b) an = µ 1 + 1 n ¶n .
c) an+1 = 1 2an + 2, a1 = 1 2.
d) an+1 = 5 2an(1 −an), a1 = 0.3.
16 SOLUTION SET FOR THE HOMEWORK PROBLEMS Answer. a) It is not easy to compute sin 1/2, sin 1/3 and so on.
So let us take a closer look at the function sin x near x = 0: Consider the circle of radius 1 with center 0, i.e., 0A = 1, and draw a line 0B with angle ∠A0B = x.
Let C be the intersection of the line 0B and the circle. Draw a line CD through C and perpen-dicular to the line 0A with D the intersection of 0 A B C D x the new line and the line 0A. So obtain the figure on the right. Now the length CD = sin x.
We have the arc length ⌢ AC = x and the line length AB = tan x. To compare the sizes of x, sin x and tan x, we consider the areas of the triangle △0AC, the piza pie cut shape ∢0AC and △0AB which are respectively (sin x)/2, x/2 and (tan x)/2. As these three figures are in the inclusion relations: △0AC ⊂∢0AC ⊂△0AB, we have sin x 2 ≤x 2 ≤tan x 2 = 1 2 sin x cos x Consequently we conclude that cos x ≤sin x x ≤1.
Therefore we have cos µ 1 n ¶ ≤n sin µ 1 n ¶ ≤1 and lim n→∞n sin µ 1 n ¶ = 1.
b) We simply compute a few terms: a1 = µ 1 + 1 1 ¶1 = 2, a2 = µ 1 + 1 2 ¶2 = 1 + 1 + 1 4 = 2.25.
a3 = µ 1 + 1 3 ¶3 = 1 + 1 + 3 9 + 1 27 = 210 27, a4 = µ 1 + 1 4 ¶4 = 1 + 1 + 6 · 1 16 + 4 · 1 64 + 1 256 = 26 · 16 + 4 · 4 + 1 256 = 296 + 16 + 1 256 = 2103 256, a5 = µ 1 + 1 5 ¶5 = 1 + 1 + µ 5 2 ¶ 1 25 + µ 5 3 ¶ 1 125 + µ 5 4 ¶ 1 625 + 1 3125 = 210 · 125 + 10 · 25 + 5 · 5 + 1 3125 = 21256 3125.
SOLUTION SET FOR THE HOMEWORK PROBLEMS 17 It is still hard to make the guess of the limit of (1 + 1/n)n. So let us try something else.
an = µ 1 + 1 n ¶n = 1 + 1 + µ n 2 ¶ 1 n2 + µ n 3 ¶ 1 n3 + · · · + µ n k ¶ 1 nk + · · · + 1 nn = 2 + 1 2!
n(n −1) n2 + 1 3!
n(n −1)(n −2) n3 + · · · + 1 k!
n(n −1)(n −2) · · · (n −k + 1) nk + · · · + 1 n!
n(n −1)(n −2) · · · 2 · 1 nn = 2 + 1 2!
µ 1 −1 n ¶ + 1 3!
µ 1 −1 n ¶ µ 1 −2 n ¶ + 1 k!
µ 1 −1 n ¶ µ 1 −2 n ¶ · · · µ 1 −k n ¶ + · · · + 1 n!
µ 1 −1 n ¶ µ 1 −2 n ¶ · · · µ 1 −n −1 n ¶ < n X k=0 1 k! = sn.
an+1 = µ 1 + 1 n + 1 ¶n+1 = 2 + 1 2!
µ 1 − 1 n + 1 ¶ + 1 3!
µ 1 − 1 n + 1 ¶ µ 1 − 2 n + 1 ¶ + + 1 k!
µ 1 − 1 n + 1 ¶ µ 1 − 2 n + 1 ¶ · · · µ 1 − k n + 1 ¶ + · · · + 1 (n + 1)!
µ 1 − 1 n + 1 ¶ µ 1 −2 n+ ¶ · · · µ 1 − n n + 1 ¶ As each term of an+1 is greater than the corresponding term of an, we have an ≤an+1, n ∈N, i.e., the sequence {an} is increasing and bounded by e as an ≤sn ≤e. Therefore we conclude that the sequence {an} converges and the limit is less than or equal to e.1 c) Skip.
d) Let us check a few terms: a1 = 0.3, a2 = 5 2a1(1 −a1) = 0.525 a3 = 5 2a2(1 −a2) = 0.6234375 a4 =0.58690795898, a5 = 0.60611751666, a6 = 0.59684768164 a7 = 0.6015513164, a8 = 0.59921832534, a9 = 0.60038930979, a10 = 0.5998049662 With f(x) = 5 2x(1 −x) = 5 2(x −x2) = 5 2 Ã 1 4 − µ1 2 −x ¶2!
≤5 8 < 1 1In fact the limit of {an} is the natural logarithm number e, which will be shown later.
18 SOLUTION SET FOR THE HOMEWORK PROBLEMS we have 0 ≤f(x) ≤5 8 = 0.625 for all x ∈[0, 1], and consequently 0 ≤an+1 = f(an) ≤5 8 = 0.625, n ≥3.
To compare an and an+1 = f(an), we consider x −f(x) = x −5 2x(1 −x) = 2x −5(x −x2) 2 = 5x2 −3x 2 = x(5x −3) 2 ½ ≤0 for all x ∈[0, 0.6] ≥0 for all x / ∈[0, 0.6] .
This means that an+1 ≤an if 0 ≤an ≤0.6 and an+1 ≥an if an < 0 or an > 0.6. But the case an < 0 has been excluded by the above arguments. From the computation of the first three terms we observe that the sequence {an} seems to oscillate. At any rate, if the sequence {an} converges, then we must have a = limn→∞an = limn→∞an+1, i.e., we must have a = f(a), which narrows the candidate of the limit down to either 0 or 3/5 = 0.6. Let us examine the candidate 3/5 first. So we compute the error ¯ ¯ ¯ ¯ 3 5 −f(x) ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ 3 5 −5 2x(1 −x) ¯ ¯ ¯ ¯ = |6 −25x(1 −x)| 10 = |25x2 −25x + 6| 10 = |(5x −2)(5x −3)| 10 = 1 2|5x −2| ¯ ¯ ¯ ¯x −3 5 ¯ ¯ ¯ ¯ = 1 2 ¯ ¯ ¯ ¯5 µ x −3 5 ¶ + 1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯x −3 5 ¯ ¯ ¯ ¯ Therefore, if |x −3/5| = δ, then ¯ ¯ ¯ ¯ 3 5 −f(x) ¯ ¯ ¯ ¯ ≤1 2(5δ + 1)δ.
Thus if δ < 1/5, then with r = (5δ + 1)/2 < 1 we have ¯ ¯ ¯ ¯ 3 5 −f(x) ¯ ¯ ¯ ¯ ≤rδ, SOLUTION SET FOR THE HOMEWORK PROBLEMS 19 in other words rk ¯ ¯ ¯ ¯ 3 5 −an ¯ ¯ ¯ ¯ ≥rk−1 ¯ ¯ ¯ ¯ 3 5 −an+1 ¯ ¯ ¯ ¯ ≥rk−2 ¯ ¯ ¯ ¯ 3 5 −an+2 ¯ ¯ ¯ ¯ ≥rk−3 ¯ ¯ ¯ ¯ 3 5 −an+3 ¯ ¯ ¯ ¯ ≥· · · ≥ ¯ ¯ ¯ ¯ 3 5 −an+k ¯ ¯ ¯ ¯ .
Therefore, if we get |3/5 −an| < 1/5 for some n ∈N, then we have lim k→∞ak = 3 5.
But we know ¯ ¯ ¯ ¯ 3 5 −a3 ¯ ¯ ¯ ¯ = 0.6234375 −0.6 = 0.0234375 < 0.2 = 1 5.
Therefore the limit of the sequence {an} is 0.6 as seen in the first computation.
♥ Page 33, Problem 2. Prove directly that each of the following sequences converges by letting ε > 0 be given and finding N(ε) so that |a −an| < ε for every n ≥N(ε).
(1) a) an = 1 + 10 2 √n.
b) an = 1 + 1 3 √n.
c) an = 3 + 2−n.
d) an = r n n + 1.
Solution. a) Obviously our guess on the limit a is a = 1. So let us try with a = 1 to find N(ε) which satisfy the condition (1): |1 −an| = 10 2 √n < ε for every n ≥N(ε), 20 SOLUTION SET FOR THE HOMEWORK PROBLEMS which is equivalent to the inequality: √n > 10 ε ⇔n > 100 ε2 for every n ≥N(ε). Thus if we choose N(ε) to be N(ε) = ·100 ε2 ¸ + 1, where [x], x ∈R, means the largest integer which is less than or equal to x, i.e., the integer m such that m ≤x < m + 1, then for every n ≥N(ε), we have 100 ε2 < N(ε) ≤n, hence ε2 > 100 n and ε > 10 2 √n = |1 −an|.
This shows that lim n→∞ µ 1 + 10 2 √n ¶ = 1.
b) It is also easy to guess that the limit a of {an} is 1. So let ε > 0 and try to find N(ε) which satisfy the condition (1) above which is: 1 3 √n = |1 −an| < ε for every n ≥N(ε).
So we look for the smallest integer N which satisfy 1 3 √n < ε equivalently 1 ε < 3 √n, which is also equivalent to n > 1 ε3 .
So with N(ε) = £ 1/ε3¤ + 1, if n ≥N(ε), then 1 ε3 < N(ε) ≤n consequently 1 3 √n < ε.
d) First, we make a small change in the form of an: an = r n n + 1 = s 1 1 + 1 n , SOLUTION SET FOR THE HOMEWORK PROBLEMS 21 and guess that the limit a of {an} would be 1. So we compute: ¯ ¯ ¯ ¯1 − r n n + 1 ¯ ¯ ¯ ¯ = √n + 1 −√n √n + 1 = (√n + 1 −√n)(√n + 1 + √n) √n + 1(√n + 1 + √n) = n + 1 −n n + 1 + p (n + 1)n ≤1 n.
Hence if n ≥[1/ε] + 1, then 1/n ≤1/([1/ε] + 1) < 1/(1/ε) = ε, i.e., ¯ ¯ ¯ ¯1 − r n n + 1 ¯ ¯ ¯ ¯ < ε for every n ≥N(ε).
♥ Page 33, Problem 3. Prove directly that each of the following sequences converges by letting ε > 0 be given and finding N(ε) so that |a −an| < ε for every n ≥N(ε).
(1) a) an = 5 − 2 ln n for n ≥2.
b) an = 3n + 1 n + 2 .
c) an = n2 + 6 2n2 −2 for n ≥2.
d) an = 2n n! .
Solution. a) From the form of the sequence, we guess that the limit a would be 5. So we try 5 as a: |5 −an| = ¯ ¯ ¯ ¯5 − µ 5 + 2 ln n ¶¯ ¯ ¯ ¯ = 2 ln n, which we want to make smaller than a given ε > 0. So we want find how large n ought to be in order to satisfy the inequlity: ε > 2 ln 2.
22 SOLUTION SET FOR THE HOMEWORK PROBLEMS This inequality is equivalent to ln n > 2/ε. Taking the exponential of the both sides, we must have n > exp(2/ε). So if we take N(ε) = · exp µ2 ε ¶¸ + 1, then for every n ≥N(ε) the inequality (1) holds.
b) First we change the form of each term slightly: an = 3n + 1 n + 2 = 3 + 1 n 1 + 2 n , to make a guess on a. This indicates that the limit a would be 3. So we try to fulfil the requirement of (1) with a = 3: ¯ ¯ ¯ ¯3 −3n + 1 n + 2 ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ 3(n + 2) −(3n + 1) n + 2 ¯ ¯ ¯ ¯ = 5 n + 2 < 5 n.
So if the inequality 5/n < ε holds, then |3−an| < ε holds. Thus N(ε) = [5/ε]+ 1 gives that |5 −an| < ε for every n ≥N(ε).
c) We alter the form of the sequence slightly: an = n2 + 6 2n2 −2 = 1 + 6 n2 2 − 2 n2 , in order to make a good guess on the limit a, which looks like 1/2. Let us try with this a: ¯ ¯ ¯ ¯ 1 2 −n2 + 6 2n2 −2 ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ (n2 −1) −(n2 + 6) 2n2 −2 ¯ ¯ ¯ ¯ = 7 2n2 −2, for n ≥2.
If n ≥2, then (2n2 −2) −2(n −1)2 = 4n > 0, so that 2(n −1)2 < 2n2 −2 and therefore ¯ ¯ ¯ ¯ 1 2 −n2 + 6 2n2 −2 ¯ ¯ ¯ ¯ < 7 2(n −1)2 .
Thus if N(ε) = ·p 7/(2ε) ¸ + 2, then for every n ≥N(ε) we have ¯ ¯ ¯ ¯ 1 2 −n2 + 6 2n2 −2 ¯ ¯ ¯ ¯ < 7 2(n −1)2 ≤ 7 2(N(ε) −1)2 = 7 2 ³hq 7 2ε i + 1 ´2 < 7 2 · 7 2ε = ε.
SOLUTION SET FOR THE HOMEWORK PROBLEMS 23 Therefore lim n→∞an = 1 2.
d) With an = 2n/n!
we look at the ratio an/an+1: an an+1 = 2n n! · (n + 1)!
2n+1 = n + 1 2 ≥2 for n ≥3.
Therefore, we have for every k ≥2 a3 ≥2ka3+k equivalently ak+3 ≤a3 2k = 8 6 · 2k = 1 3 · 2k−2 < 1 2k−1 < 1 k −1.
So for any ε > 0 if n ≥N(ε) = £ 1 ε ¤ + 5, then we have 0 < an < ε.
♥ Page 24, Problem 6. Suppose that an →a and let b be any number strictly less than a.
Prove that an > b for all but finitely many n.
Proof. The assumption a > b yields b −a > 0 so that there exists N ∈N such that |a −an| < b −a for every n ≥N, equivalently b −a < a −an < a −b for every n ≥N, hence b < an for every n ≥N.
Thus the total number of n with b ≥an is at most N −1 which is of course finite. Thus an > b for all but finitely many n.
♥ Page 34, Problem 9. a) Find a sequence {an} and a real number a so that |an+1 −a| < |an −a| for each n, but {an} does not converge to a.
b) Find a sequence {an} and a real number a so that an →a but so that the above inequality is violated for infinitely many n.
Answer. a) Take an = 1/n and a = −1. Then |an+1 −a| = 1 n + 1 + 1 < 1 n + 1 = |an −a| but an ↛ a.
b) Set an = 1 n µ 1 + (−1)n 2 ¶ and a = 0.
24 SOLUTION SET FOR THE HOMEWORK PROBLEMS Then we have an = ½ 1 2n for odd n; 3 2n for even n and an →0.
If n is odd, then an+1 = 3 2(n + 1) > 1 2n = an.
This occurs infinitely many times, i.e., at every odd n.
♥ Page 39, Problem 1. Prove that each of the following limits exists: a) an = 5 µ 1 + 1 3 √n ¶2 .
b) an = 3n + 1 n + 2 .
c) an = n2 + 6 3n2 −2.
d) an = 5 + ¡ 2 3n ¢2 2 + 2n+5 3n−2 Proof. a) First limn→∞1/ 3 √n = 0 because for any given ε > 0 if n ≥N(ε) = [1/ε3]+1 then 1 3 √n ≤ 1 3 p N(ε) = 1 3 q£ 1 ε3 ¤ + 1 < 1 3 q 1 ε3 = 1 ¡ 1 ε ¢ = ε.
Hence we get lim n→∞5 µ 1 + 1 3 √n ¶2 = 5 by the combination of Theorem 2.2.3, Theorem 2.2.4 and Theorem 2.2.5 as seen below: 1 + 1 3 √n →1 by Theorem 2.2.3 ⇒ µ 1 + 1 3 √n ¶2 →1 by Theorem 2.2.5 ⇓ 5 µ 1 + 1 3 √n ¶2 →5 by Theorem 2.2.4.
SOLUTION SET FOR THE HOMEWORK PROBLEMS 25 b) We change the form of each term an slightly: an = an = 3n + 1 n + 2 = 3 + 1 n 2 + 2 n .
We know that 1/n →0 and 2/n →0 as n →∞.
Thus we get the following chain of deduction: 3 + 1 n →3 and 2 + 2 n →2 by Theorem 2.2.3 ⇓ 3 + 1 n 2 + 2 n →3 2 by Theorem 2.26.
c) We change the form of each term an in the following way: an = n2 + 6 3n2 −2 = 1 + 6 n2 3 − 2 n2 .
As 1/n →0, Theorem 2.2.5 yields that 1/n2 →0 and therefore 1 + 6 n2 3 − 2 n2 →1 + 6 · 0 3 −2 · 0 = 1 3 by Theorem 2.2.4 and Theorem 2.2.6.
d) As seen before, we have 1 3n ≤1 n →0 and 2n + 5 3n −2 = 2 + 5 n 3 −2 n →2 3.
Thus we get an = 5 + ¡ 2 3n ¢2 2 + 2n+5 3n−2 →5 + 4 · 0 · 0 2 + 2 3 = 15 8 by a combination of Theorem 2.2.3, Theorem 2.2.4 and Theorem 2.2.6.
♥ Page 39, Problem 6. Let p(x) be any polynomial and suppose that an →a. Prove that lim n→∞p(an) = p(a).
Proof. Suppose that the polynomial p(x) has the form: p(x) = pkxk + pk−1xk−1 + · · · + p1x + p0.
We claim that aℓ n →aℓfor each ℓ∈N. If ℓ= 1, then certainly we have the convergence: a1 n = an →a = a1. Suppose aℓ−1 n →aℓ−1. Then by Theorem 2.2.5 we have aℓ n = aℓ−1 n an → aℓ−1a = aℓ. By mathematical induction we have aℓ n →aℓfor each ℓ∈N. Therefore, each term pℓaℓ n converges to pℓaℓfor ℓ= 1, 2, · · · , k. A repeated use of Theorem 2.2.3 yields that p(an) = pkak n + pk−1ak−1 n + · · · + p1an + p0 →pkak + pk−1ak−1 + · · · + p1a + p0 = p(a).
♥ 26 SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 39, Problem 7. Let {an} and {bn} be sequences and suppose that an ≤bn for all n and that an →∞. Prove that bn →∞.
Proof. The divergence an →∞means that for every M there exists N ∈N such that an > M for every n ≥N. The assumption that an ≤bn gives M < an ≤bn for every n ≥N. Hence bn →∞.
♥ Page 39, Problem 9. a) Let {an} be the sequence given by an+1 = 1 2an + 2, a1 = 0.5 Prove that an →4.
b) Consider the sequence defined by an+1 = αan + 2.
Show that if |α| < 1, then the sequence has a limit independent of a1.
Proof. a) Based on the hint, we compute an+1 −4 = 1 2an + 2 −4 = 1 2an −2 = 1 2(an −4).
Hence we get |an −4| = 1 2|an−1 −4| = 1 22 |an−2 −4| = · · · = 1 2n−1 |a1 −4| = 3.5 2n−1 →0.
b) We just compute an+1 − 2 1 −α = αan + 2 − 2 1 −α = αan + 2(1 −α) −2 1 −α = αan − 2α 1 −α = α µ an − 2 1 −α ¶ ; an − 2 1 −α = α µ an−1 − 2 1 −α ¶ = α2 µ an−2 − 2 1 −α ¶ = · · · = αn−1 µ a1 − 2 1 −α ¶ − →0 as |α| < 1.
Hence {an} converges and lim n→∞an = 2 1 −α which is independent of a1.
♥ SOLUTION SET FOR THE HOMEWORK PROBLEMS 27 Page 40, Problem 10. For a pair (x, y) of real numbers, define ∥(x, y)∥= p x2 + y2.
a) Let x1, x2, y1, y2 be real numbers. Prove that |x1x2 + y1y2| ≤ q x2 1 + x2 2 q y2 1 + y2 2.
b) Prove that for any two dimensional vectors (x1, y1), (x2, y2) ∈R2 ∥(x1, y1) + (x2, y2)∥≤∥(x1, y1)∥+ ∥(x2, y2)∥.
c) Let pn = (xn, yn) be a sequence of points in the plane R2 and let p = (x, y). We say that pn →p if ∥pn −p∥→0. Prove that pn →p if and only if xn →x and yn →y.
Proof. a) Let us compute: (x2 1 + x2 2)(y2 1 + y2 2) −(x1y1 + x2y2)2 = x2 1y2 1 + x2 1y2 2 + x2 2y2 1 + x2 2y2 2 −(x2 1y2 1 + 2x1y1x2y2 + x2 2y2 2) = x2 1y2 2 + x2 2y2 1 −2x1y1x2y2 = (x1y2 −x2y1)2 ≥0.
b) We also compute directly: ∥(x1, y1) + (x2, y2)∥2 = ∥(x1 + x2, y1 + y2)∥2 = (x1 + x2)2 + (y1 + y2)2 = x2 1 + 2x1x2 + x2 2 + y2 1 + 2y1y2 + y2 2 = x2 1 + x2 2 + 2(x1x2 + y1y2) + y2 1 + y2 2 ≤x2 1 + y2 1 + 2 q x2 1 + y2 1 q x2 2 + y2 2 + x2 2 + y2 2 = µq x2 1 + x2 2 + q y2 1 + y2 2 ¶2 = µ ∥(x1, y1)∥+ ∥(x2, y2)∥ ¶2 .
This shows the inequality: ∥(x1, y1) + (x2, y2)∥≤∥(x1, y1)∥+ ∥(x2, y2)∥.
c) Since we have the inequalities: max{|xn −x|, |yn −y|} ≤ p (xn −x)2 + (yn −y)2 ≤2 max{|xn −x|, |yn −y|}, show that ∥pn −p∥→0 if and only if max{|xn −x|, |yn −y|} →0 if and only if |xn −x| →0 and |yn −y| →0.
♥ 28 SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 50, Problem 1. Prove directly that an = 1 + 1 √n is a Cauchy sequence.
Proof. We just compute for m < n: |am −an| = ¯ ¯ ¯ ¯ µ 1 + 1 √m ¶ − µ 1 + 1 √n ¶¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ 1 √m − 1 √n ¯ ¯ ¯ ¯ ≤ 1 √m + 1 √n ≤ 2 √m since m < n.
So if ε > 0 is given, then we take N to be [(2/ε)2] + 1 so that for every n > m ≥N we have |am −an| < 2 √m ≤ 2 rh¡ 2 ε ¢2i + 1 < 2 q¡ 2 ε ¢2 = ε.
♥ Page 50, Problem 2. Prove that the rational numbers are dense in the real numbers.
Proof. We have to prove that for every ε > 0 (a −ε, a + ε) ∩Q ̸= ∅. Choose m = [1/ε] + 1 so that 1/m < ε. If a −ε > 0, then the Archimedian property of R yields the existence of k ∈N such that k m = k · 1 m > a −ε.
Let n be the first such a number. Then we have n −1 m ≤a −ε < n m = n −1 m + 1 m ≤a −ε + 1 m < a −ε + ε = a.
Therefore, we conclude that a −ε < n m < a therefore n m ∈(a −ε, a + ε) ∩Q.
If a −ε < 0, then we apply the Archimedian property of R to the pair 1/m and ε −a > 0 to find a natural number k ∈N such that ε −a < k · 1 m = k m.
Let n ∈N be the smallest natural number such that n/m ≥ε −a, so that n −1 m < ε −a ≤n m, equivalently −n m ≤a −ε < 1 −n m = −n m + 1 m.
As we have chosen m ∈N so large that 1/m < ε, the above inequality yields 1 −n m = 1 m −n m < ε −n m ≤ε + (a −ε) = a, i.e., a −ε < 1 −n m < a.
Therefore we have 1 −n m ∈(a −ε, a + ε) ∩Q consequently (a −ε, a + ε) ∩Q ̸= ∅ for arbitrary ε > 0. Hence a is a limit point of Q.
SOLUTION SET FOR THE HOMEWORK PROBLEMS 29 Page 59, Problem 3. Suppose that the sequence {an} converges to a and d is a limit point of the sequence {bn}. Prove that ad is a limit point of the sequence {anbn}.
Proof. By the assumption on the sequence {bn}, there exists a subsequence {bnk} of the sequence {bn} such that lim k→∞bnk = d.
The subsequence {ankbnk} of the sequence {anbn} converges to ad because the subsequence {ank} of {an} converges to the same limit a. Hence ad is a limit point of {anbn}.
♥ Page 59, Problem 6. Consider the following sequence: a1 = 1 2; the next three terms are 1 4, 1 2, 3 4; the next seven terms are 1 8, 1 4, 3 8, 1 2, 5 8, 3 4, 7 8; · · · and so forth. What are the limit points.
Answer. The sequence {an} consists of the numbers {k/2n : k = 1, 2, · · · , 2n −1, n ∈N}.
Fix x ∈[0, 1].
We are going to construct a subsequence {bn} of the sequence {an} by induction. For n = 1, choose k1 = ½ 0 if x ≤1 2; 1 if 1 2 < x ≤1.
For each n > 1, let kn be the natural number such that kn 2n ≤x < kn + 1 2n .
Then the ratio bn = kn/2n is in the sequence {an} and |bn −x| < 1 2n →0 as n →∞.
Hence limn→∞bn = x. Therefore, every x ∈[0, 1] is a limit point of {an}. Thus the sequence {an} is dense in the closed unit interval [0, 1].
♥ Page 59, Problem 8. Let {Ik : k ∈N} be a nested family of closed, finite intervals; that is, I1 ⊃I2 ⊃· · · . Prove that there is a point p contained in all the intervals, that is p ∈∩∞ k=1Ik.
Proof. The assumption means that if Ik = [ak, bk], k ∈N, then a1 ≤a2 ≤· · · ≤ak ≤· · · · · · ≤bk ≤bk−1 ≤· · · ≤b2 ≤b1.
The sequence {ak} is increasing and bounded by any of {bℓ}. Fix k ∈N. Then we have a = lim n→∞an ≤bk for k ∈N, 30 SOLUTION SET FOR THE HOMEWORK PROBLEMS where the convergence of {an} is guaranteed by the boundedness of the sequence. Now look at the sequence {bk} whis is decreasing and bounded below by a. Hence it converges to b ∈R and a ≤b. Thus the situation is like the following: a1 ≤a2 ≤· · · ≤ak ≤· · · a ≤b ≤· · · ≤bk ≤bk−1 ≤· · · ≤b2 ≤b1.
Hence the interval [a, b] is contained in the intersection ∩∞ k=1Ik. Any point p in the interval [a, b] is a point of ∩∞ k=1Ik; in fact [a, b] = ∩∞ k=1Ik.
♥ Page 59, Problem 9. Suppose that {xn} is a monotone increasing sequence of points in R and suppose that a subsequence of {xn} converges to a finite limit. Prove that {xn} converges to a finite limit.
Proof. Let {xnk} be the subsequence converging to the finite limit x0. As n1 < n2 < · · · < nk < · · · , we have k ≤nk for every k ∈N. If ε > 0 is given, then choose K so large that |xnk −x0| < ε for every k ≥K, i.e., x0 −ε < xnk ≤x0 for every k ≥K. Set N = nK. Then if m ≥N, then we have x0 −ε < xnK = xN ≤xm ≤xnm ≤x0.
Hence we have 0 ≤x0 −xm < ε for every m ≥N. Hence {xn} converges to the same limit x0.
♥ Page 79, Problem 3. Let f(x) be a continuous function. Prove that |f(x)| is a continuous function.
Proof. Let x ∈[a, b] be a point in the domain [a, b] of the function f. If ε > 0 is given, then choose a δ > 0 so small that |f(x) −f(y)| < ε whenever |x −y| < δ. If |x −y| < δ, then ¯ ¯|f(x)| −|f(y)| ¯ ¯ ≤|f(x) −f(y)| < ε.
Hence |f| is continuous at x.
♥ Page 79, Problem 5. Suppose that f is a continuous function on R such that f(q) = 0 for every q ∈Q. Prove that f(x) = 0 for every x ∈R.
Proof. Choose x ∈R and ε > 0. Then there exists δ > 0 such that |f(x) −f(y)| < ε whenever |x −y| < δ. Take q ∈Q ∩(x −δ, x + δ), then |f(x)| = |f(x) −f(q)| < ε.
Thus |f(x)| is less than any ε > 0 which is possible only when |f(x)| = 0.
♥ SOLUTION SET FOR THE HOMEWORK PROBLEMS 31 Page 79, Problem 7. Let f(x) = 3x −1 and let ε > 0 be given. How small δ be chosen so that |x −1| ≤ε implies |f(x) −2| < ε?
Answer. To determine the magnitude of δ, assume that |x −1| < δ and see how the error becomes: |f(x) −2| = |3x −1 −2| = |3x −3| = 3|x −1| < 3δ.
Thus if 3δ ≤ε, i.e., if δ ≤ε/3, then |x −1| < δ implies |f(x) −2| < ε.
♥ Page 79, Problem 8. Let f(x) = x2 and let ε > 0 be given.
a) Find a δ so that |x −1| ≤δ implies |f(x) −1| ≤ε.
b) Find a δ so that |x −2| ≤δ implies |f(x) −2| ≤ε.
c) If n > 2 and you had to find a δ so that |x −n| ≤δ implies |f(x) −n2| ≤ε, would the δ be larger or smaller than the δ for parts (a) and (b)? Why?
Answer. a) Choose δ > 0 and see how the error grows from |x −1| ≤δ: |f(x) −1| = |x2 −1| = |(x + 1)(x −1)| = |x + 1||x −1| ≤|x + 1|δ = |x −1 + 1 + 1|δ ≤(|x −1| + 2)δ ≤(δ + 2)δ.
So we want to make (δ + 2)δ ≤ε. Let us solve this inequality: 0 ≥δ2 + 2δ −ε = (δ + 1)2 −ε −1 ⇔ ε + 1 ≥(δ + 1)2 ⇔ − √ ε + 1 −1 ≤δ ≤ √ ε + 1 −1.
But we know that δ must be positive. Hence 0 < δ ≤√ε + 1 −1 = ε/(√ε + 1 + 1). If δ is chosen in the interval (0, √ε + 1 −1), then the above calculation shows that |x −1| ≤δ ⇒ |f(x) −1| ≤ε.
b) Now we continue to examine the case |x −2| ≤δ: |f(x) −4| = |x2 −4| = |(x + 2)(x −2)| = |x + 2||x −2| ≤δ|x + 2| = δ|x −2 + 4| ≤δ(|x −2| + 4) ≤δ(δ + 4).
So we want to make (δ + 4)δ ≤ε, equivalently: ε ≥δ(δ + 4) = δ2 + 4δ ⇔ δ2 + 4δ −ε ≤0 ⇔ − √ ε + 4 −2 ≤δ ≤ √ ε + 4 −2 = (√ε + 4 −2)(√ε + 4 + 2) √ε + 4 + 2 = ε √ε + 4 + 2.
32 SOLUTION SET FOR THE HOMEWORK PROBLEMS Hence if we take 0 < δ ≤ε/(√ε + 4 + 2), then |x −2| < δ ⇒ |f(x) −4| < ε.
c) Similarly, we examine the case |x −n| ≤δ: |f(x) −n2| = |x2 −n2| = |x + n||x −n| ≤δ|x + n| ≤δ|x −n + 2n| ≤δ(|x −n| + 2n) ≤δ(δ + 2n).
So we want to make (δ + n)δ ≤ε, equivalently: ε ≥δ(δ + 4) = δ2 + 2nδ ⇔ δ2 + 2nδ −ε ≤0 ⇔ − p ε + n2 −n ≤δ ≤ p ε + n2 −n = ( √ ε + n2 −n)( √ ε + n2 + n) √ ε + n2 + n = ε √ ε + n2 + n.
Hence if we take δ > 0 so small that 0 < δ ≤ε/( √ ε + n2 + n), then |x −n| ≤δ ⇒ |f(x) −n| ≤ε.
The largest possible δ = ε/( √ ε + n2 + n) is squeezed to zero when n glows indefinitely.
Page 79, Problem 11. Let f(x) = √x with domain {x : x ≥0}.
a) Let ε > 0 be given. For each c > 0, show how to choose δ > 0 so that |x −c| ≤δ implies |√x −√c| ≤ε.
b) Give a separate argument to show that f is continuous at zero.
Solution. a) Once again we examine the growth of error by letting |x −c| ≤δ and compute: |√x −√c| = | √ x −c + c −√c = |x −c| √x −c + c + √c ≤ δ √ c −δ + √c ≤ δ p c 2 + √c ³ under the assumption δ ≤c 2 ´ ≤2δ √c.
Thus if 0 < δ ≤min{c/2, ε√c/2} , then |x −c| ≤δ ⇒ |√x −√c| ≤ε.
Hence f is continuous at c > 0.
b) If 0 ≤x ≤δ, then |√x − √ 0| = √x ≤ √ δ.
Hence if 0 < δ ≤ε2, then 0 ≤x ≤δ ⇒0 ≤√x ≤ε.
Therefore f is continuous at 0.
♥ |
189970 | https://jillianstarrteaching.com/addition-facts/ | Welcome, Fellow Math Enthusiast! I’m so happy you’re here!
Teaching with Jillian Starr
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Grab Your FREE Gift
Addition Centers Freebie
Looking for fun and engaging centers to help your students practice their addition fact fluency? Be sure to grab these freebies!
Addition Centers Freebie
Looking for fun and engaging centers to help your students practice their addition fact fluency? Be sure to grab these freebies!
7 Fun Center Activities to Build Fluency with Addition Facts
Building fluency with addition facts is a critical building block in our students’ mathematical understanding. We know that the more time we give them to become truly fluent, the stronger the mathematician they will become.
But what does it mean to be fluent? Fluency involves three components: accuracy, efficiency, and flexibility. Notice that I never mentioned speed or rote memorization (even though students will come to answers more quickly as they build flexible and efficient strategies).
So how do we help build fluency? Well, like I said, we have to give students lots of opportunities to experience addition and subtraction so they can begin to develop their own strategies, and then give them time to practice! I’ve got a list of fun center activities that will do just that. Ready? Let’s dive in.
Addition within Ten
In addition to making ten, students must quickly recall addition facts within ten. Enter: Fact family activities.
Domino Addition
When I taught first grade, fact families were a big part of our math scope and sequence. We would create fact family books. And, I used a range of fact family activities during math centers. It is hard to pick a favorite. But here’s one tried and true: Domino Addition.
Domino Addition is a simple activity that combines hands-on manipulatives (dominos) and practice writing number sentences within 10.
The directions are simple. Students choose a domino from a pile. Then, they draw the domino into the empty domino template. Finally, they write the number sentence to match the domino.
Addition Facts Logic Puzzles within 10
Are your students ready for a challenge? It is time for addiction facts logic puzzles.
Logic puzzles or logic tiles are perfect for partner or independent center work. Logic puzzles and logic tiles (I used both) include 10 number sentences with ten missing digits (0-9). There is only one correct way to solve a logic puzzle. And, in order to solve it, students will need to recall their addition facts with ease.
These are a great way to level up fast finishers. Students practice recall of their addition facts while also strengthening their problem-solving skills.
Using Ten As A Benchmark
When we think forward in our students’ learning trajectory, we can see that addition, subtraction, place value, and missing addends are ALL supported by fluency within ten. If students do not have a solid fluency foundation within ten, they will hit a wall as they progress into multi-digit numbers and more complex mental math.
But, then what? It’s time to build on making ten by using those addition facts in numbers that go beyond 10.
Make a Ten To Add
Let’s take this awesome activity, Bridges to Ten, for a spin. Literally.
Bridges to Ten is a printable activity that takes seconds to prep and provides templates to help students organize their process of adding two single-digit numbers to form a sum over ten.
In Bridges to Ten, students will use two different spinners to find their addends and build those numbers into the corresponding tens frames.
5 In A Row – Make a Ten to Add
5 in a Row is another great activity to practice using ten as a benchmark when adding. 5 in a Row is also a printable activity with spinners—this time with a gameboard. Here’s how to play!
Bonds to Make Bridges
Looking for one more step in this progression? Bonds to Make Bridges is a Bridges to Ten activity with a twist.
Bonds to Make Bridges combines Bridges to Ten with number bonds. Students follow similar instructions as Bridges to Ten. However, there is one added step. Students break the first added into parts. One of those parts needs to help make ten on their ten-frame.
2-Digit Addition Centers
One of the most challenging transitions is moving from 1-digit to 2-digit addition with regrouping. Strong addition facts support this transition. However, sometimes, students need activities that make the connection more concrete. While there are many 2-digit addition with regrouping activities that I’d love to share, here are two quick and easy centers.
2-Digit Addition Puzzles
Visuals are a great way to make addition facts (as building blocks) for 2-digit addition with regrouping more obvious to students. Addition puzzles showcase a single number sentence in four different visuals. Each piece of the puzzle shows one strategy for finding the sum.
Construction Zone Addition
Construction Zone Addition is one of my favorite, more advanced addition activities.
Here’s how to play:
Fun warning: Students LOVE pretending to be builders. And, in addition to all that fun, students are guided on a hands-on journey to finding the sum of multi-digit numbers. Even better, students can play as they progress. There is a version for:
It is not always easy to find quality addition center activities that support students’ addiction facts fluency while building toward more complex number problems. I hope this taste of some of them is helpful! And I hope you can now see how supporting the development of addition facts doesn’t have to be mundane or difficult!
Do you have any favorite activities to practice addition facts? Let me know in the comments! I’d love to hear them!
Addition & Subtraction Resources
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189971 | https://openstax.org/books/college-physics-ap-courses/pages/2-section-summary | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
College Physics for AP® Courses
Section Summary
College Physics for AP® CoursesSection Summary
Search for key terms or text.
## 2.1 Displacement
Kinematics is the study of motion without considering its causes. In this chapter, it is limited to motion along a straight line, called one-dimensional motion.
Displacement is the change in position of an object.
In symbols, displacement is defined to be
where is the initial position and is the final position. In this text, the Greek letter (delta) always means “change in” whatever quantity follows it. The SI unit for displacement is the meter (m). Displacement has a direction as well as a magnitude.
When you start a problem, assign which direction will be positive.
Distance is the magnitude of displacement between two positions.
Distance traveled is the total length of the path traveled between two positions.
## 2.2 Vectors, Scalars, and Coordinate Systems
A vector is any quantity that has magnitude and direction.
A scalar is any quantity that has magnitude but no direction.
Displacement and velocity are vectors, whereas distance and speed are scalars.
In one-dimensional motion, direction is specified by a plus or minus sign to signify left or right, up or down, and the like.
## 2.3 Time, Velocity, and Speed
Time is measured in terms of change, and its SI unit is the second (s). Elapsed time for an event is
where is the final time and is the initial time. The initial time is often taken to be zero, as if measured with a stopwatch; the elapsed time is then just .
Average velocity is defined as displacement divided by the travel time. In symbols, average velocity is
The SI unit for velocity is m/s.
Velocity is a vector and thus has a direction.
Instantaneous velocity is the velocity at a specific instant or the average velocity for an infinitesimal interval.
Instantaneous speed is the magnitude of the instantaneous velocity.
Instantaneous speed is a scalar quantity, as it has no direction specified.
Average speed is the total distance traveled divided by the elapsed time. (Average speed is not the magnitude of the average velocity.) Speed is a scalar quantity; it has no direction associated with it.
## 2.4 Acceleration
Acceleration is the rate at which velocity changes. In symbols, average acceleration is
The SI unit for acceleration is .
Acceleration is a vector, and thus has a both a magnitude and direction.
Acceleration can be caused by either a change in the magnitude or the direction of the velocity.
Instantaneous acceleration is the acceleration at a specific instant in time.
Deceleration is an acceleration with a direction opposite to that of the velocity.
## 2.5 Motion Equations for Constant Acceleration in One Dimension
To simplify calculations we take acceleration to be constant, so that at all times.
We also take initial time to be zero.
Initial position and velocity are given a subscript 0; final values have no subscript. Thus,
The following kinematic equations for motion with constant are useful:
In vertical motion, is substituted for .
## 2.6 Problem-Solving Basics for One Dimensional Kinematics
The six basic problem solving steps for physics are:
Step 1. Examine the situation to determine which physical principles are involved.
Step 2. Make a list of what is given or can be inferred from the problem as stated (identify the knowns).
Step 3. Identify exactly what needs to be determined in the problem (identify the unknowns).
Step 4. Find an equation or set of equations that can help you solve the problem.
Step 5. Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units.
Step 6. Check the answer to see if it is reasonable: Does it make sense?
## 2.7 Falling Objects
An object in free-fall experiences constant acceleration if air resistance is negligible.
On Earth, all free-falling objects have an acceleration due to gravity , which averages
Whether the acceleration a should be taken as or is determined by your choice of coordinate system. If you choose the upward direction as positive, is negative. In the opposite case, is positive. Since acceleration is constant, the kinematic equations above can be applied with the appropriate or substituted for .
For objects in free-fall, up is normally taken as positive for displacement, velocity, and acceleration.
## 2.8 Graphical Analysis of One Dimensional Motion
Graphs of motion can be used to analyze motion.
Graphical solutions yield identical solutions to mathematical methods for deriving motion equations.
The slope of a graph of position vs. time is velocity .
The slope of a graph of velocity vs. time graph is acceleration .
Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs.
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Authors: Gregg Wolfe, Erika Gasper, John Stoke, Julie Kretchman, David Anderson, Nathan Czuba, Sudhi Oberoi, Liza Pujji, Irina Lyublinskaya, Douglas Ingram
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Book title: College Physics for AP® Courses
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Ultimate Guide on How to Calculate Peak-to-Peak Voltage
If you want to know how to calculate peak-to-peak voltage, here is the formula you'll need to use:
VP-P = VP × 2
Where,
VP is the peak voltage
VP-P is the peak-to-peak voltage
Let's say you wish to find the peak-to-peak voltage when the peak voltage is 240V. Upon substituting the values in the above equation, we get:
VP-P = 240V × 2 = 480V
It's worth noting that the peak-to-peak voltage calculation is relevant for AC waveforms. Hence, it may not apply directly to complex or non-sinusoidal waveforms.
Jackery Portable Power Stations are battery-powered inverter generators that can supply reliable electricity to appliances, including refrigerators, water heaters, ACs, sump pumps, LED lights, and coolers. They have fully upgraded BMS technology that prevents equipment damage from short circuits and voltage fluctuations.
What Is Peak to Peak Voltage?
Voltage is the electric potential within an electric circuit and provides the potential for current flow. Peak-to-peak voltage is the distance from the trough or lowest negative amplitude to the highest positive amplitude (or crest) of the AC waveform.
Peak-to-peak voltage is equal to the full height of the waveform and can be found using the peak or RMS voltage. In other words, the peak-to-peak value of an AC (alternating current) is twice the peak value or 2.828 times the root-mean-square value. It is typically indicated by VP-P.
Here are some of the terms you must keep in mind when using the formula:
Voltage Peak: The maximum instantaneous value of any function measured from the zero volt level. If the average value of the function is zero volts, the peak amplitude and peak value are the same. It is denoted by VP.
Voltage Peak to Peak: It is the total voltage between positive and negative waveform peaks. In other words, it is the sum of the magnitude of both positive and negative peaks. It is denoted by VP-P.
Voltage RMS: It is the practical value of the root mean square of the waveform. In other words, it is the direct current equivalent voltage of alternating current power. It is denoted by VRMS.
Voltage Average: The waveform level defined by the condition that the curve enclosing the area above the level is precisely equal to the area below the level. It is typically indicated by Vavg.
Here's the chart revealing the relationship between the peak-to-peak voltage, peak voltage, and RMS voltage on an AC waveform.
How to Calculate Peak-to-Peak Voltage?
There are two methods to calculate peak-to-peak voltage: either from peak voltage or from RMS voltage. If you know either the peak voltage or RMS voltage, you can use the formulas to determine the peak-to-peak voltage. Here's how to calculate peak-to-peak voltage in two different ways:
Peak Voltage to Peak-to-Peak Voltage Formula
If you know the peak voltage of the AC waveform, you can use the following formula:
VP-P = VP × 2
Where,
VP is the peak voltage
VP-P is the peak-to-peak voltage
In other words, the peak-to-peak voltage is equal to twice the peak voltage.
For example, if the peak voltage is 120V, here's how do you calculate peak-to-peak voltage.
VP-P = 120V × 2 = 240V
RMS Voltage to Peak-to-Peak Voltage Formula
Root Mean Square or RMS is the alternating voltage or square root of the mean square of instantaneous values of the voltage signal. In an AC waveform, it is the amount of AC power that produces the same heating effect as DC Power.
If you know the value of VRMS, you can easily calculate the peak-to-peak voltage using the below formula:
VP-P = 2 × √2 × VRMS
Where,
VRMS is the RMS voltage
VP-P is the peak-to-peak voltage
In other words, the peak-to-peak voltage equals twice the square root of two times the RMS voltage.
For instance, if the RMS voltage is 120V, the peak-to-peak voltage can be calculated as follows:
VP-P = 2 × √2 × 120V = 340V
What Are The Differences Between Peak-to-Peak, Peak, and RMS Voltages?
The voltage typically goes up and down in a smooth sine wave in alternating current. For example, it goes up to meet the maximum positive voltage, back to zero, then to the maximum negative value, and finally back to zero to start all over again.
As you can see from the battery voltage graph, there is a voltage change over time. For this reason, there are different ways to express the voltage change over time, depending on what you are trying to emphasize. Let's explain how these terms differ and the related formulas:
Peak Voltage (VP): The peak voltage has the highest value compared to zero volts. The amplitude is similar to the peak voltage value as it is the highest value of the waveform. The peak and amplitude values are used in waveform analysis but not in the AC electrical work.
Peak-to-Peak Voltage (VP-P): The peak-to-peak voltage is the difference between the lowest and highest values in an AC waveform. Generally speaking, the peak-to-peak value will be twice the peak value. In mathematical terms, VP-P = VP × 2.
RMS Voltage: Root Mean Square (RMS) is the direct current equivalent to the voltage of AC power. It gives the effective voltage of AC power for watts and other calculations. It's worth noting that RMS is used while calculating Ohm's law. In mathematical terms, the peak-to-peak voltage is equal to twice the square root of two times VRMS.
If you want to understand the differences between peak, peak-to-peak, and RMS voltage, let's take an example:
120 VRMS equals 170 volts peak (VP), and 170 volts peak (VP) equals 340 volts peak to peak (VP-P), which is the difference between +170V and -170V.
Mathematically, the peak-to-peak voltage can be expressed using the following formula:
VP-P = 2 × VP = 2 × √2 × VRMS or VP = √2 × VRMS
If the VRMS is 120V, the peak voltage will be equal to √2 × 120V = 170V, and the peak-to-peak voltage will be equal to 2 × 170V = 340V.
What Is The Battery Voltage?
Simply put, battery voltage is the electric potential difference between both (positive and negative) battery terminals. The difference is generally created by the lack or excess of electrons flowing in the circuits. You can think of voltage like a spring in the battery. The more we push the spring down, the greater its potential.
Understanding the basic battery voltage helps you determine how much power it can supply. One important thing to remember is that different electrical systems have varying voltages. For example, it can range from 12V - 48V direct current systems to 110V, 220V, and higher in AC applications.
Battery voltage typically depends on the chemistry and cell count. For example, lithium-ion batteries have a voltage range of around 13.6V when charged compared to 12.7V of lead-acid batteries. You can check the battery voltage charts to understand the voltages of different batteries. Understanding the voltages of various types of batteries is essential if you want to choose the best battery for a specific device.
Let's explain the two types of batteries briefly:
Lithium Batteries
A lithium battery is a popular technology with high energy density and higher capacity. When charged, a 12V battery boasts a normal maximum voltage of 13.6V in the lithium-ion battery voltage chart. It can deliver constant voltage when there is battery discharge, ensuring safety and longevity.
The LiFePO4 batteries are the safest among all and come with superior BMS technology to provide optimum charging and discharging. You can check the LiFePO4 battery voltage chart to determine how to enhance the battery lifespan and safely charge the appliances.
Lithium batteries are safe and reliable, so many power stations or battery backups are manufactured using them. Compared to their counterparts, lithium batteries can safely charge electrical appliances and keep devices running for long hours.
Lead-Acid Batteries
Lead-acid batteries are mainly used as starter batteries in automotive and marine applications. When charged, they typically have a voltage of 12.7V. Compared to lithium batteries, they do not offer the same energy density.
Here's a table revealing the voltage, energy density, and unique characteristics of different types of batteries.
| | | | | |
--- ---
| Battery | Anode | Cathode | Voltage | Density |
| Alkaline | Zinc | Manganese dioxide | 1.5V | 0.5 MJ/Kg |
| Zinc-carbon | Zinc | Manganese dioxide | 1.5V | 0.13 MJ/Kg |
| Lithium (BR) | Lithium | Carbon monofluoride | 3V | 1.3 MJ/Kg |
| Lithium (CR) | Lithium | Manganese dioxide | 3V | 1 MJ/Kg |
| Lithium-thionyl chloride | Lithium | Sulfur-oxygen chlorine | 3.6V | 1.04 MJ/Kg |
| Zinc-air | Zinc | Oxygen | 1.4V | 1.69 MJ/Kg |
Battery
Anode
Cathode
Voltage
Density
Alkaline
Zinc
Manganese dioxide
1.5V
0.5 MJ/Kg
Zinc-carbon
Zinc
Manganese dioxide
1.5V
0.13 MJ/Kg
Lithium (BR)
Lithium
Carbon monofluoride
3V
1.3 MJ/Kg
Lithium (CR)
Lithium
Manganese dioxide
3V
1 MJ/Kg
Lithium-thionyl chloride
Lithium
Sulfur-oxygen chlorine
3.6V
1.04 MJ/Kg
Zinc-air
Zinc
Oxygen
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1.69 MJ/Kg
Jackery Portable Power Stations feature reliable and robust NMC or LiFePO4 batteries. They have fully upgraded BMS (Battery Management System) technology to offer a stable power supply to the appliances. There are built-in temperature sensors to ensure the portable power station works in varying temperature conditions.
How to Measure the Battery Voltage
Regardless of what applications you use batteries for, it is essential to understand how to measure the voltage throughout their use. Here are two ways to measure and monitor battery voltage:
Use Multimeter
A straightforward way to measure battery voltage is to use a multimeter. To do this, set the battery to DC voltage and connect the red lead directly to the positive battery terminal, followed by attaching the black lead to the negative. You'll then see the battery voltage displayed on the multimeter. This also helps you determine the battery's state of charge and maintain its health.
Battery Monitor and Sensors
These devices measure and report on battery status, including temperature, voltage, and current. They provide real-time data for assessment and monitoring to anticipate battery health and performance. If the BMS uses battery monitors and sensors, this means the devices remain functional and powered.
Jackery Portable Power Stations Explained
Jackery is a global leader in manufacturing safe solar generators, portable power stations, and solar panels. The Jackery Portable Power Stations are compact and powerful battery-powered inverter generators with multiple input and output ports. The Jackery SolarSaga Solar Panels absorb and convert the sun's energy into DC electricity, which is converted into AC and stored in the Jackery Portable Power Stations to charge appliances.
Equipped with NMC or LiFePO4 batteries, these portable power stations feature BMS technology, 12 layers of protection, nine temperature sensors, and top shock-resistant level 9 to take safety to the next level. The Jackery Solar Generators have an ergonomic design and foldable handles and are designed with exploration in mind. You can carry these solar generators on your camping trips or use them as an emergency backup during extended power outages.
Jackery Explorer 3000 Pro Portable Power Station
The Jackery Explorer 3000 Pro Portable Power Station is equipped with an NMC battery and can charge up to 99% of household or outdoor appliances. It has a unique quiet canyon cooling system to ensure the emergency battery backup power source works by emitting less than 30dB of noise. The fully upgraded battery management system offers protection against overvoltage and short circuits. The built-in temperature sensors determine the peak temperature position depending on the simulation experiments.
Appliances running time:
Customer Review
"We got the 3000 Pro with four 200-watt panels. Since then, I've tried out my washing machine, gas dryer, and electric lawn mower and charged my phone. I've even charged the generator while powering devices with it. My idea of what solar and batteries can do has been greatly expanded. This unit delivers real power." — P.V.
Jackery Explorer 2000 Pro Portable Power Station
The powerful and compact Jackery Explorer 2000 Pro Portable Power Station is one of the best charging solutions for power outages and outdoor adventures. Its industry-leading BMS technology and built-in temperature sensors take safety to the next level. It enables efficient voltage control, temperature control, short circuit protection, and more advanced safety operations. With its multiple output ports, you can charge eight appliances at once. It houses two dual protection chips for safe and reliable charging.
Appliances running time:
Customer Review
"I purchased this to give my husband and me peace of mind during what has become more frequent power outages in our home. My husband uses an oxygen concentrator at night. The unit quickly charged up, and testing has shown it will power the concentrator for 7.1 hours, which will get him through the night in an emergency." — Judy Browne.
Jackery Explorer 2000 Plus Portable Power Station
If you're looking for a versatile and safe charging solution, you might consider the Jackery Explorer 2000 Plus Portable Power Station your best bet. Its outstanding LiFePO4 battery has a lifespan of 10 years. It is ideal for off-grid living and can also supply power during extended blackouts. The innovative ChargeShield technology provides 12 protective algorithms, 62 protective mechanisms, and four types of physical safety protection. The unique stepped variable speed charging algorithm improves safety and boosts the battery lifespan by 50%.
Customer Review
"Peace of mind knowing we have a backup in the event of a power failure. A very nice, powerful unit that charges quickly. Have run several appliances in the house. I have purchased a second Explorer 2000 Pro to alternate when one needs recharging with the solar panels. We are now prepared for an emergency. Highly recommended." — Richard W.
Is Battery Voltage Dangerous?
Battery voltage is not dangerous until it exceeds 50V. The human body can usually handle up to 50V of shock with unlimited current capacity. The theory is based on the power distribution of the human body, which indicates that the arms and legs are at least 500 ohms.
In this case, if the volts are 50V, no lethal current would be passed into the heart. On the other hand, if the voltage exceeds 50V, the human body acts as a conductor. Some dangers associated with high voltage include hearing loss, broken bones, cardiac arrest, eye injuries, and burns.
So, what is the normal voltage of the battery? The correct answer depends on the type of battery you have. For example, a household AAA battery will have a different voltage compared to a car battery. The reason behind the difference is the chemical reaction that takes place in the cell. A fully charged car battery will measure around 12.6V, whereas an AAA battery will have 1.5 volts.
How to Calculate Peak-to-Peak Voltage FAQs
What is the formula for peak voltage?
The formula for peak voltage is Peak Voltage = Peak-to-Peak Voltage ÷ 2.
Or, VPeak = VP-P ÷ 2
If the peak-to-peak voltage is 120V, the peak voltage will be 120V ÷ 2 = 60V.
If you want to calculate peak voltage using RMS voltage, you can use the below formula:
VPeak = 1.414 x VRMS
The peak voltage for 120V RMS will be 1.414 × 120V = 170V.
What is the peak voltage of 480V?
If you're calculating peak voltage for the three-phase system, the formula will be VPeak(L-L) = √2 × VL-L. When the utility voltage is 480 VAC, the peak voltage will be about 678.7V, and the peak-to-peak voltage is nearly twice that, at about 1357.4V.
What is the peak-to-peak voltage of 220V?
The peak-to-peak voltage of a 220VAC RMS sine wave at 50-60 Hz is 620V. The peak voltage of a 220V AC source is 311V, which is calculated by multiplying the RMS voltage (220V) by 1.414 (the square root of 2).
What is the maximum peak-to-peak voltage?
The maximum peak-to-peak voltage of an AM wave is equal to 16 millivolts (mV), and the minimum is 4 millivolts (mV).
How do you calculate 3-phase peak voltage?
You can quickly calculate the peak voltage using the line-to-line and line-to-neutral voltage for three-phase systems. For example, if the line-to-line voltage is 480V AC, the peak voltage will be:
VP = √2 × VL-L = 678.7V.
On the other hand, if neutral wiring is present in the three-phase system, you can use the following formula to calculate line-to-neutral and peak voltage.
VL-N = VL-L ÷ √3 = 480V AC ÷ √3 = 277V AC
The peak voltage will be:
VP = √2 × VL-N = 392V.
Final Thoughts
Peak-to-peak voltage helps you understand the maximum voltage variation of the AC supply. It plays a crucial role in both designing and testing electronic circuits. It also ensures all the components in the circuits are rated to handle the maximum voltages. Understanding how to calculate the peak-to-peak voltage can also help In waveform analysis and amplifier design.
If you are looking for a portable power station that can offer safe charging without worrying about voltage fluctuations, you may consider investing in the Jackery Portable Power Stations. They are built with upgraded BMS and state-of-the-art temperature sensors to supply charge without equipment damage. Additionally, they have a large battery capacity, helping you charge 99% of household or outdoor appliances.
Disclaimer:
The runtime mentioned for appliances powered by Jackery is for reference only. Actual runtime may vary under different conditions. Please refer to real-world performance for accurate results.
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4 декабря 2024, 18:38 , Обновлено 5 декабря 09:08
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Нефтегазовые доходы бюджета РФ в ноябре 2024 г. снизились в 1,5 раза к октябрю. Это нормально
В январе-ноябре 2024 г. нефтегазовые доходы бюджета составили 10,3409 трлн руб. при том, что в действующем законе о бюджете нефтегазовые доходы запланированы в объеме 10,985 трлн руб.
Источник: Минфин РФ
Москва, 4 дек - ИА Neftegaz.RU. Нефтегазовые доходы бюджета РФ в ноябре 2024 г. снизились на 16,6% в годовом сравнении, а по итогам 11 месяцев рост составил 26,5%.
Об этом Минфин РФ сообщил 4 декабря 2024 г.
Отчет за октябрь 2024 г.
Статистика нефтегазовых доходов за месяц
Нефтегазовые доходы (НГД) бюджета в ноябре 2024 г. составили 801,7 млрд руб., что на 16,6% меньше по сравнению с ноябрем 2023 г. и в 1,5 раза меньше по сравнению с октябрем 2024 г.
Значительное снижение НГД в месячном сравнении обусловлено эффектом поступлений по налогу на дополнительный доход (НДД), который выплачивается по итогам квартала 4 раза в год (в т.ч. в октябре).
В ноябре 2024 г.:
поступления по налогу на добычу полезных ископаемых (НДПИ) снизились на 13,7% в годовом сравнении, но выросли на 11,7% в месячном, составив 1,0143 трлн руб., в т.ч.:
на нефть - 823 млрд руб. (снижение на 17,4% в годовом сравнении и рост на 10,9% в месячном),
на газ - 129,7 млрд руб. (рост на 6,2% в годовом сравнении и на 16,2% в месячном),
на газовый конденсат - 61,6 млрд руб. (рост на 10,2% и на 13,2% в годовом и месячном сравнении);
поступления от экспортной пошлины - 69,1 млрд руб., что в 1,8 раза меньше в годовом сравнении и на 28,7% больше - в месячном:
в связи с завершением финального витка большого налогового маневра в нефтяной отрасли промышленности, поступления по этой статье обеспечивает газ,
экспортная пошлина на газ составила 66,9 млрд руб., сократившись на 6,6% в годовом сравнении и увеличившись на 28,7% в месячном.
Базовый месячный объем НГД в ноябре - 727,4 млрд руб (1,0431 трлн руб. - октябрь), дополнительные НГД - 74,3 млрд руб (168,7 млрд руб. - октябрь).
В ноябре 2023 г. В. Путин подписал закон о возврате к расчету бюджетного правила исходя из базовой цены на нефть в 60 долл. США/барр. с индексацией на 2%/год начиная с 2027 г. (до этого Бюджетный кодекс устанавливал номинальный объем базовых НГД на уровне 8 трлн руб./год с индексацией на 4%/год начиная с 2026 г.).
Нарастающим итогом
В январе-ноябре 2024 г. нефтегазовые доходы бюджета составили 10,3409 трлн руб., что на 26,5% больше, чем за 11 месяцев 2023 г.:
поступления от НДПИ - рост на 32,2%, до 11,2342 трлн руб., в т.ч.:
нефть - 9,3095 трлн руб., рост на 33,2%,
газ - 1,3104 млрд руб., рост на 19,0%,
газовый конденсат - 614,4 млрд руб., рост в 1,5 раза;
поступления от экспортной пошлины снизились в 2,1 раза, до 402,9 млрд руб., в т.ч.:
по газу - 409,1 млрд руб., снижение на 17,4%;
поступления по НДД выросли в 1,6 раза, до 2,0463 трлн руб.
Базовые НГД составили 9,1001 трлн руб., дополнительные - 1,2408 трлн руб.
Напомним, что в действующем законе о бюджете нефтегазовые доходы запланированы в объеме 10,985 трлн руб., в т.ч. 9,932 трлн руб. базовые НГД.
Этот уровень будет достигнут, даже если показатель декабря 2024 г. окажется на уровне традиционно одного из наиболее слабых месяцев года - января (в 2024 г. в январе НГД составили 675,2 млрд руб.).
В пояснительной записке к недавно утвержденному федеральному бюджету на 2025 г. и плановый период 2026-2027 гг, оценка НГД на 2024 г. была повышена на 3%, до 11,3095 трлн руб., в т.ч. базовые - 9,7508 трлн руб., дополнительные - 1,5587 трлн руб.
А вот этого уровня достичь вряд ли удастся.
Выплаты по демпферу
Выплаты нефтяным компаниям из бюджета в рамках демпфирующего механизма составили:
за ноябрь 2024 г. - 132,2 млрд руб., что на 31,4% меньше в годовом сравнении и на 24,0% больше в месячном сравнении (106,6 млрд руб. - в октябре 2024 г.),
за 11 месяцев 2024 г. - 1,6762 трлн руб., рост на 26%.
Операции с валютой без зеркала
Ожидаемый объем дополнительных НГД федерального бюджета в декабре 2024 г. прогнозируется в размере 95,9 млрд руб.
Суммарное отклонение фактически полученных НГД от ожидаемого месячного объема НГД и оценки базового месячного объема НГД от базового месячного объема НГД по итогам ноября 2024 г. составило 18,5 млрд руб.
В связи с этим:
совокупный объем средств, направляемых на покупку иностранной валюты и золота, составляет 114,4 млрд руб.,
операции будут проводиться в период с 6 декабря по 14 января 2025 г.,
объем покупки иностранной валюты и золота составит в эквиваленте 5,4 млрд руб./сутки.
В целях снижения волатильности финансовых рынков Банком России приостановлена трансляция покупок иностранной валюты на внутренний валютный рынок с 28 ноября и до конца 2024 г.
Решение о возобновлении Банком России операций на внутреннем валютном рынке в рамках зеркалирования регулярных операций, связанных с реализацией бюджетного правила, будет приниматься с учетом складывающейся ситуации на финансовых рынках.
Отложенные покупки будут осуществлены в течение 2025 г.
Авторы:
Е. Алифирова
Специальный корреспондент Neftegaz.RU
#минфин рф#бюджет#доходы#нефтегазовые доходы#ндпи#демпфер#покупка валюты#ноябрь 2024 г.
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189974 | https://online.stat.psu.edu/stat415/lesson/2/2.5 | 2.5 - A t-Interval for a Mean | STAT 415
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2
2.5
2.5 - A t-Interval for a Mean
Our work so far Section
So far, we have shown that the formula:
x¯±z α/2(σ n)
is appropriate for finding a confidence interval for a population mean if two conditions are met:
The population standard deviation σ is known, and
X 1,X 2,…,X n are normally distributed. (The truth is that X 1,X 2,…,X n need not be normally distributed as long as the sample size n is large enough for the Central Limit Theorem to apply. In this case, the confidence interval is an approximate confidence interval.)
Now, as suggested earlier in this lesson, it is unrealistic to think that we'd ever be in a situation where the first condition would be met. That is, when would we ever know the population standard deviation σ, but not the population mean μ? Let's entertain, then, the realistic situation in which not only the population mean μ is unknown, but also the population standard deviation σ is unknown.
What if σ is unknown? Section
Try It!
What would be a reasonable thing to do if the population standard deviation σ is unknown?
Answer
Estimate it with the sample standard deviation, S!
Yes, the reasonable thing to do is to estimate the population standard deviation σ with the sample standard deviation:
S=1 n−1∑i=1 n(X i−X¯)2
Then, in deriving the confidence interval, we'd start out with:
X¯−μ S/n
instead of:
X¯−μ σ/n∼N(0,1)
Then, to derive the confidence interval, in this case, we just need to know how:
T=X¯−μ S/n
is distributed!
How is T=X¯−μ S/n distributed? Section
Given that the ratio is typically denoted by the capital letter T, we probably shouldn't be surprised that the ratio follows a T distribution!
Theorem
If X 1,X 2,…,X n are normally distributed with mean μ and variance σ 2, then:
T=X¯−μ S/n
follows a T distribution with n−1 degrees of freedom.
Proof
The proof is as simple as recalling a few distributional results from our work in Stat 414. Recall the definition of a T random variable, namely if Z∼N(0,1) and U∼χ(r)2 are independent, then:
T=Z U/r
follows the T distribution with r degrees of freedom. Furthermore, recall that if X 1,X 2,…,X n are normally distributed with mean μ and variance σ 2, then:
Z=X¯−μ σ/n∼N(0,1)
(n−1)S 2 σ 2∼χ n−1 2
X¯ and S 2 are independent
Now, we just have to put all that we've remembered together:
T=x¯−μ σ/n(n−1)s 2 σ 2 n−1=x¯−μ σ/n(σ s)=x¯−μ s/n∼t n−1
The first equality simply defines a T random variable using the first, second, and third bullet point above. The second equality comes from canceling out the n−1 terms in the denominator. The third equality comes from canceling out the σ terms, leaving us with:
T=X¯−μ S/n
following a T distribution with n−1 degrees of freedom, as was to be proved!
Now that we have the distribution of T=X¯−μ S/n behind us, we can derive the confidence interval for a population mean in the realistic situation that σ is unknown.
Theorem
If X 1,X 2,…,X n are normally distributed random variables with mean μ and variance σ 2, then a (1−α)100% confidence interval for the population mean μ is:
x¯±t α/2,n−1(s n)
This interval is often referred to as the "t-interval for the mean."
Proof
The proof is very similar to that for the Z-interval for the mean. We start by drawing a picture of a T-distribution with n−1 degrees of freedom:
From the diagram, we can see that the following probability statement is true:
P[−t α/2,n−1≤T≤t α/2,n−1]=1−α
Then, simply replacing T, we get:
P[−t α/2,n−1≤X¯−μ s/n≤t α/2,n−1]=1−α
Let's again focus only on the inequality inside the brackets for a bit. Because we manipulate each of the three sides of the inequality equally, each of the following statements are equivalent:
−t α/2,n−1≤X¯−μ s/n≤t α/2,n−1−t α/2,n−1(s n)≤X¯−μ≤+t α/2,n−1(s n)−X¯−t α/2,n−1(s n)≤−μ≤−X¯+t α/2,n−1(s n)X¯−t α/2,n−1(s n)≤μ≤X¯+t α/2,n−1(s n)
That is, we have shown that a (1−α)100% confidence interval for the mean μ is:
[X¯−t α/2,n−1(s n),X¯+t α/2,n−1(s n)]
as was to be proved.
Just one more thing. Before we go off and work through an example, let's clarify a bit of confidence interval terminology.
t-interval
With the formula for the t-interval:
x¯±t α/2,n−1(s n)
in mind, we say that:
x¯ is a "point estimate" of μ
x¯±t α/2,n−1(s n) is an "interval estimate" of μ
s n is the "standard error of the mean"
t α/2,n−1(s n) is the "margin of error"
Now, let's take a look at an example!
Example 2-2 Section
A random sample of 16 Americans yielded the following data on the number of pounds of beef consumed per year:
118 115 125 110 112 130 117 112 115 120 113 118 119 122 123 126
What is the average number of pounds of beef consumed each year per person in the United States?
Answer
To help answer the question, we'll calculate a 95% confidence interval for the mean. As the above theorem states, in order for the t-interval for the mean to be appropriate, the data must follow a normal distribution. We can use a normal probability plot to provide evidence that the data are (sufficiently) normally distributed:
That is, because the data points fall at least approximately on a straight line, there's no reason to conclude that the data are not normally distributed. That's convoluted statistician talk for "we're good to go." Now, punching the n=16 data points into a calculator (or statistical software), we can easily determine that the sample mean is 118.44 and the sample standard deviation is 5.66. For a 95% confidence interval with n=16 data points, we need:
t 0.025,15=2.1314
Now, we have all of the necessary elements to calculate the 95% confidence interval for the mean. It is:
x¯±t 0.025,15(s n)=118.44±2.1314(5.66 16)
Simplifying, we get:
118.44±3.016
or:
(115.42,121.46)
That is, we can 95% confident that the average amount of beef consumed each year per person in the United States is between 115.42 and 121.46 pounds. Wow, that's a lot of beef!
Minitab®
Using Minitab Section
Again, statistical software, such as Minitab, can make calculating confidence intervals easier. To ask Minitab to calculate a t-interval for a mean μ, you need to do this:
Enter the data in one of the columns. Here's the data from the above example entered in the C1 column:
Convince yourself that the data come from a normal distribution... either from your previous experience or by creating a normal probability plot. To ask Minitab to generate a normal probability plot, under the Stat menu, select Basic Statistics, and then select Normality Test...:
In the pop-up window that appears, select the data (column) to be plotted so that it appears in the box labeled Variable:
Select OK. When you do so, a new graphics window should appear containing the normal probability plot:
(The plot appearing in the example above was generated in Minitab using different commands. That's why it looks different from this one.)
Then, after convincing yourself that the normality assumption is appropriate, under the Stat menu, select Basic Statistics, and then select 1-Sample t...:
In the pop-up window that appears, select the column (data) to be analyzed so that it appears in the box labeled Samples in columns:
Select OK. The confidence interval output will appear in the Session window. Here is what the Minitab output looks like for the beef example:
One-Sample T: beef
| Variable | N | Mean | StDev | SE Mean | 95% CI |
--- --- --- |
| beef | 16 | 118.44 | 5.66 | 1.41 | (115.42, 121.45) |
«Previous 2.4 - An Interval's Length
Next 2.6 - Non-normal Data»
Lesson
Section 1: Estimation
Lesson 1: Point Estimation
1.1 - Definitions
1.2 - Maximum Likelihood Estimation
1.3 - Unbiased Estimation
1.4 - Method of Moments
Lesson 2: Confidence Intervals for One Mean
2.1 - The Situation
2.2 - A Z-Interval for a Mean
2.3 - Interpretation
2.4 - An Interval's Length
2.5 - A t-Interval for a Mean
2.6 - Non-normal Data
Lesson 3: Confidence Intervals for Two Means
3.1 - Two-Sample Pooled t-Interval
3.2 - Welch's t-Interval
3.3 - Paired t-Interval
Lesson 4: Confidence Intervals for Variances
4.1 - One Variance
4.2 - The F-Distribution
4.3 - Two Variances
Lesson 5: Confidence Intervals for Proportions
5.1 - One Proportion
5.2 - Two Proportions
Lesson 6: Sample Size
6.1 - Estimating a Mean
6.2 - Estimating a Proportion for a Large Population
6.3 - Estimating a Proportion for a Small, Finite Population
Lesson 7: Simple Linear Regression
7.1 - Types of Relationships
7.2 - Least Squares: The Idea
7.3 - Least Squares: The Theory
7.4 - The Model
7.5 - Confidence Intervals for Regression Parameters
7.6 - Using Minitab to Lighten the Workload
Lesson 8: More Regression
8.1 - A Confidence Interval for the Mean of Y
8.2 - A Prediction Interval for a New Y
8.3 - Using Minitab to Lighten the Workload
Section 2: Hypothesis Testing
Lesson 9: Tests About Proportions
9.1 - The Basic Idea
9.2 - More Examples
9.3 - The P-Value Approach
9.4 - Comparing Two Proportions
9.5 - Using Minitab
Lesson 10: Tests About One Mean
10.1 - Z-Test: When Population Variance is Known
10.2 - T-Test: When Population Variance is Unknown
10.3 - Paired T-Test
10.4 - Using Minitab
Lesson 11: Tests of the Equality of Two Means
11.1 - When Population Variances Are Equal
11.2 - When Population Variances Are Not Equal
11.3 - Using Minitab
Lesson 12: Tests for Variances
12.1 - One Variance
12.2 - Two Variances
12.3 - Using Minitab
Lesson 13: One-Factor Analysis of Variance
13.1 - The Basic Idea
13.2 - The ANOVA Table
13.3 - Theoretical Results
13.4 - Another Example
Lesson 14: Two-Factor Analysis of Variance
14.1 - An Example
Lesson 15: Tests Concerning Regression and Correlation
15.1 - A Test for the Slope
15.2 - Three Tests for Rho
15.3 - An Approximate Confidence Interval for Rho
Section 3: Nonparametric Methods
Lesson 16: Chi-Square Goodness-of-Fit Tests
16.1 - The General Approach
16.2 - Extension to K Categories
16.3 - Unspecified Probabilities
16.4 - Continuous Random Variables
16.5 - Using Minitab to Lighten the Workload
Lesson 17: Contingency Tables
17.1 - Test For Homogeneity
17.2 - Test for Independence
Lesson 18: Order Statistics
18.1 - The Basics
18.2 - The Probability Density Functions
18.3 - Sample Percentiles
Lesson 19: Distribution-Free Confidence Intervals for Percentiles
19.1 - For A Median
19.2 - For Any Percentile
Lesson 20: The Wilcoxon Tests
20.1 - The Sign Test for a Median
20.2 - The Wilcoxon Signed Rank Test for a Median
20.3 - Tied Observations
Lesson 21: Run Test and Test for Randomness
21.1 - The Run Test
21.2 - Test for Randomness
Lesson 22: Kolmogorov-Smirnov Goodness-of-Fit Test
22.1 - The Test
22.2 - Two Examples
22.3 - A Confidence Band
Section 4: Bayesian Methods
Lesson 23: Probability, Estimation, and Concepts
23.1 - Subjective Probability
23.2 - Bayesian Estimation
Section 5: More Theory & Practice
Lesson 24: Sufficient Statistics
24.1 - Definition of Sufficiency
24.2 - Factorization Theorem
24.3 - Exponential Form
24.4 - Two or More Parameters
Lesson 25: Power of a Statistical Test
25.1 - Definition of Power
25.2 - Power Functions
25.3 - Calculating Sample Size
Lesson 26: Best Critical Regions
26.1 - Neyman-Pearson Lemma
26.2 - Uniformly Most Powerful Tests
Lesson 27: Likelihood Ratio Tests
27.1 - A Definition and Simple Example
27.2 - The T-Test For One Mean
Lesson 28: Choosing Appropriate Statistical Methods
28.1 - One Categorical Response
28.2 - One Continuous Response
28.3 - Two Continuous Measurements
28.4 - Practice
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189975 | https://papers.ssrn.com/sol3/Delivery.cfm/SSRN_ID2543807_code2292577.pdf?abstractid=2543807&mirid=1&type=2 | The Rise and Fall of the Ultra Vires Doctrine in United States, United Kingdom, and Commonwealth Caribbean Corporate Common Law: A Triumph of Experience Over Logic by Stephen Leacock :: SSRN
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The Rise and Fall of the Ultra Vires Doctrine in United States, United Kingdom, and Commonwealth Caribbean Corporate Common Law: A Triumph of Experience Over Logic
DEPAUL BUSINESS & COMMERCIAL LAW JOURNAL Vol. 5, No. 1, 2006
38 Pages Posted: 14 Jan 2015
See all articles by Stephen Leacock
Stephen Leacock
Barry University - Dwayne O. Andreas School of Law
Date Written: 2006
Abstract
In free market economies, corporate laws change over time. Modernization of corporate law is also crucial to commercial growth and to free market economies. The author argues that concepts that have outlived their utility, such as the ultra vires doctrine, should be promptly eliminated where feasible. He proposes a thorough and complete reformation of the ultra vires doctrine in England, where change is only partially complete at this point, to mirror the United States reformation.
After the introduction, the author generally discusses the formation of companies in common law jurisdictions, identifies similarities in this process globally, and highlights differences with regard to individual jurisdictions. He then analyzes the ultra vires doctrine and discusses its development in England and in the United States, identifies pertinent factors that have led to its progressive reform in England, the United States and other Common Law Jurisdictions globally, and ends by articulating the doctrine’s current status in modern corporate law.
Keywords: corporate law, ultra vires doctrine, international law, common law jurisdictions
JEL Classification: K10, K22, K33
Suggested Citation:Suggested Citation
Leacock, Stephen, The Rise and Fall of the Ultra Vires Doctrine in United States, United Kingdom, and Commonwealth Caribbean Corporate Common Law: A Triumph of Experience Over Logic (2006). DEPAUL BUSINESS & COMMERCIAL LAW JOURNAL Vol. 5, No. 1, 2006, Available at SSRN:
Stephen Leacock (Contact Author)
Barry University - Dwayne O. Andreas School of Law ( email )
6441 East Colonial Drive
Orlando, FL 32807
United States
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25 References
MODEL Bus. CORP. AcT §, volume 3 Posted: 1969 Crossref
Except as provided in subsection (b), the validity of corporate action may not be challenged on the ground that the corporation lacks or lacked power to act. (b) A corporation's power to act may be challenged
) in a proceeding by the corporation, directly, derivatively, or through a receiver, trustee, or other legal representative, against an incumbent or former director, officer, employee, or agent of the corporation; or (3) in a proceeding by the Attorney General
shareholder's proceeding under subsection (b)(1) to enjoin an unauthorized corporate act, the court may enjoin or set aside the act, if equitable and if all affected persons are parties to the proceeding, and may award damages for loss (other than anticipated profits) suffered by the corporation or another party because of enjoining the unauthorized act
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Paúl Noboa-Velasco La Doctrina Ultra Vires en el Contexto Societario Ecuatoriano (The Ultra Vires Doctrine and its Regulation in Ecuador) 4 Pages ·Posted: 26 May 2020 ·Downloads: 274 Download PDF Add Paper to My Library
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189976 | https://www.codecademy.com/resources/docs/numpy/built-in-functions/correlation-coefficient | Skip to Content
Python:NumPy .corrcoef()
Preview: @Sriparno08 242 total contributions
@Sriparno08
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Published Jul 24, 2024
Contribute to Docs
In NumPy, the .corrcoef() method computes the Pearson correlation coefficient of two specified arrays and returns an array as the result.
Career path
### Data Scientist: Machine Learning Specialist
Machine Learning Data Scientists solve problems at scale, make predictions, find patterns, and more! They use Python, SQL, and algorithms.
+ Includes 27 CoursesIncludes 27 Courses
+ With Professional CertificationWith Professional Certification
+ Beginner Friendly.Beginner Friendly
95 hours95 hours
Course
### Learn Python 3
Learn the basics of Python 3.12, one of the most powerful, versatile, and in-demand programming languages today.
+ With CertificateWith Certificate
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Syntax
numpy.corrcoef(x, y=None, rowvar=True, dtype=None)
x: The first array to be used for computing the Pearson correlation coefficient.
y (Optional): The second array to be used for computing the Pearson correlation coefficient.
rowvar (Optional): If True (default), then each row represents a variable and each column contains an observation. If False, then the roles are reversed.
dtype (Optional): Specifies the return data type.
Example
The following example demonstrates the usage of the .corrcoef() method:
```
```
Importing the NumPy library
import numpy as np
Defining two arrays
arr1 = np.array([6, 21, 37])
arr2 = np.array([1, 25, 51])
Using the .corrcoef() method
res = np.corrcoef(arr1, arr2)
Printing the result
print(res)
```
Copy to clipboard
Copy to clipboard
```
The above code produces the following output:
```
```
[[1. 0.99999002]
[0.99999002 1. ]]
```
Copy to clipboard
Copy to clipboard
```
Codebyte Example
Run the following codebyte example to understand the use of the .corrcoef() method:
```
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```
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Career path
### Data Scientist: Machine Learning Specialist
Machine Learning Data Scientists solve problems at scale, make predictions, find patterns, and more! They use Python, SQL, and algorithms.
+ Includes 27 CoursesIncludes 27 Courses
+ With Professional CertificationWith Professional Certification
+ Beginner Friendly.Beginner Friendly
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Course
### Learn Python 3
Learn the basics of Python 3.12, one of the most powerful, versatile, and in-demand programming languages today.
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24 hours24 hours |
189977 | https://www.arxiv.org/pdf/2410.18543 | Suppressing chaos with mixed superconducting qubit devices
Ben Blain, 1, 2 Giampiero Marchegiani, 1, ∗ Luigi Amico, 1, 3, 4 and Gianluigi Catelani 5, 1
1
Quantum Research Center, Technology Innovation Institute, Abu Dhabi 9639, UAE
2
School of Physics and Astronomy, University of Kent, Canterbury CT2 7NH, United Kingdom
3
Dipartimento di Fisica e Astronomia, Via S. Sofia 64, 95123 Catania, Italy
4
INFN-Sezione di Catania, Via S. Sofia 64, 95127 Catania, Italy
5
JARA Institute for Quantum Information (PGI-11), Forschungszentrum J¨ ulich, 52425 J¨ ulich, Germany
In quantum information processing, a tension between two different tasks occurs: while qubits’ states can be preserved by isolating them, quantum gates can be realized only through qubit-qubit interactions. In arrays of qubits, weak coupling leads to states being spatially localized and strong coupling to delocalized states. Here, we study the average energy level spacing and the relative entropy of the distribution of the level spacings (Kullback-Leibler divergence from Poisson and Gaussian Orthogonal Ensemble) to analyze the crossover between localized and delocalized (chaotic) regimes in linear arrays of superconducting qubits. We consider both transmons as well as capacitively shunted flux qubits, which enables us to tune the qubit anharmonicity. Arrays with uniform anharmonicity, comprising only transmons or flux qubits, display remarkably similar dependencies of level statistics on the coupling strength. In systems with alternating anharmonicity, for typical disorder in the qubit frequencies the localized regime is found to be more resilient to the increase in qubit-qubit coupling strength in comparison to arrays with a single qubit type. Our results, which we also confirm using generalized Bose-Hubbard models, support designing devices that incorporate different qubit types to achieve higher performances.
I. INTRODUCTION
Superconducting qubits provide a feasible platform for quantum computing and simulation purposes . For computation, fast quantum gates can be attained if the qubits are coupled to each other with sufficient strength [2–4]. On the other hand, qubits need to be sufficiently isolated to minimize the effects of residual in-teractions that negatively affect information processing protocols. As such, studying this type of trade-off is a problem of central importance which is attracting consid-erable attention in the field [5–7]. It has indeed been ar-gued that qubit arrays for quantum computation should be in a localized regime that can be achieved through a certain amount of spread (disorder) in qubit transi-tion frequencies . Subsequent works both showed that quasiperiodic parameter modulation can be more effec-tive than random disorder in keeping the system in the localized regimes and considered the connection to classical chaos in coupled nonlinear oscillators . Fixed qubit-qubit couplings lead to so-called residual ZZ inter-actions, which can impact gate fidelities in multi-qubit systems [10–12]. Interestingly enough, such interaction can be suppressed when coupling qubits with opposite anharmonicity [13, 14]. The interplay between localized and extended corre-lations of qubits also provides a valuable view point for quantum simulation . Specifically, linear ar-rays of capacitively-coupled transmons have been demonstrated to provide a quantum simulator for driven-dissipative bosonic systems with attractive interac-
∗giampiero.marchegiani@tii.ae
tions [17, 18]. In this 1D platform, the many-body lo-calized to ergodic transition has been studied both ex-perimentally and theoretically [19–22], and recently the experimental capabilities have been extended to lad-ders and 2D arrays [24–29]. In addition, theoreti-cal works have analyzed low-energy states of the system characterized by localized excitations of (bright) solitonic type [30–32]. In this work we study the localization properties in qubit arrays through the statistical distribution of the energy level spacings of transmons and/or capacitively shunted flux qubits (CSFQs) [33–35]. We employ two different approaches: the first one is based on the aver-age level-spacing ratio, see also Ref. ; for the second one we consider a relative entropy of the level spacing distribution known as Kullback–Leibler divergence . Both approaches show that the repulsive (CSFQ) and at-tractive (transmon) cases are characterized by equivalent spectral statistics, a result that we explain by resorting to an appropriate Bose-Hubbard model. The suppression of unwanted ZZ interactions motivates us to consider ar-rays with alternating values of anharmonicity, that is, mixed transmon-CSFQ devices. Interestingly, we find that the localized phase can persist up to stronger cou-pling strength than in arrays with uniform anharmonicity but otherwise identical parameters; the increase can be significant (over 40% in linear arrays) for typical disor-der, but becomes negligible when the disorder is stronger than the anharmonicity. We also show that increasing the mismatch between the anharmonicities beyond the value needed for optimal ZZ suppression further stabi-lizes the localized phase. The latter result is obtained by considering a generalized Bose-Hubbard model; hence, the investigation of such models in superconducting qubit
arXiv:2410.18543v3 [quant-ph] 29 Aug 2025
2platforms can complement their study in the context of bosonic atoms in optical lattices . The article is organized as follows: in Sec. II, we in-troduce the models investigated in this work. Then, in Sec. III, we describe the diagnostic tools we use for analyzing the level statistics. Our results on the comparison between the level statistics of uniform and alternating-sign one-dimensional qubit arrays are pre-sented in Sec. IV. The extension of our findings to higher-connectivity arrays is briefly explored in Sec. V. Finally, we summarize our results and discuss possible future di-rections in Sec. VI.
II. MODELS
An array of M capacitively-coupled qubits is described by the Hamiltonian ˆH =
M
X
i=1
ˆHQ
i
K
M−1
X
i=1
ˆNi ˆNi+1 . (1) where ˆHQ
i
is the Hamiltonian of the qubit at site i, either a transmon (Q = T) or a CSFQ (Q = F), K character-izes the coupling strength between neighboring devices, and ˆNi counts the number of excess Cooper pairs on the island of qubit i. Figure 1(a) schematically shows (top to bottom) transmon and CSFQ arrays, and an array with CSFQs/transmons at odd/even sites. The Hamiltonian for the transmon in Fig. 1(b) is ˆHT
i
= 4 EC ˆN 2
i
− EJi cos ˆ φi, (2) where EC is the charging energy, assumed to be the same for all transmons in an array, EJi is the Josephson en-ergy, and ˆ φi, conjugated to ˆNi, is the phase difference across the Josephson junction at site i. We consider the transmon regime EJi ≫ EC , in which the spectrum is weakly anharmonic. The site-index dependence of the Josephson energy can account for variations due to fab-rication and/or design choices. We note that while EC
depends on the dimensions of capacitors (typically tens to hundreds of microns) that can be fabricated accurately, the nanometer-thick tunnel barrier of the junction, deter-mining EJ , is more easily subject to random fluctuations. For the CSFQ we use the Hamiltonian ˆHF
i
= 4 ECF ˆN 2
i
EJF i {− 2 cos( ˆ φi) + α cos(2 ˆ φi)} , (3) where ECF is the charging energy of the CSFQ, EJF i is the Josephson energy of the two identical large junctions of the ith CSFQ, and α is the ratio between the Joseph-son energy of the third (smaller) Josephson junction and
EJF i , see Fig. 1(c). Both ECF and α are taken as inde-pendent of i and we have assumed to apply an external magnetic field such that half a flux quantum threads the loop formed by the three junctions. Here ˆ φi is a collective variable, the average of the phase differences across the
Transmon CSFQ
AiEJi/(8 EC)(1 −2α)EJF i/(4 ECF )
ℏω01 i
p8EJiEC−EC
p16(1 −2α)EJF iECF −ECF 1−8α
1−2α
Ui−ECECF (8 α−1) /(1 −2α)
Ji,i +1 K
2
4
√AiAi+1
TABLE I. Relations between the parameters of the qubit ar-ray Hamiltonian, Eq. (1), and the Bose-Hubbard one, Eq. (4).
two large junctions, and we ignore the mode associated with the difference of the phases as it has much higher energy . Equation (1) can be approximately cast in the form of a generalized Bose-Hubbard model. Specifically, we intro-duce bosonic annihilation, ˆbi, and creation, ˆb†
i
, operators at each site via the relations ˆ φi = (4 Ai)−1/4(ˆb†
i
ˆbi) and ˆNi = ı(Ai/4) 1/4(ˆb†
i
−ˆbi) , where Ai for each qubit type is given in Table I. We then expand the cosines up to the fourth power of ˆ φi and for consistency we keep only terms that commute with the number operator ˆ ni = ˆb†
i
ˆbi
(that is, we consider terms up to the next-to-leading or-der in p1/A i ≪ 1). Making a similar approximation in the term coupling neighboring qubits, where we keep the contributions that commute with the total number operator ˆ n = P
i
ˆni, we obtain ˆHBH =
M
X
i=1
ℏω01 i ˆni +
M
X
i=1
Ui
2 ˆni(ˆ ni − 1) +
M−1
X
i=1
Ji,i +1
ˆb†
i+1
ˆbi + H.c.
, (4) where ℏ = h/ 2π is the reduced Planck constant. The pa-rameters of the Bose-Hubbard Hamiltonian are related to those of the qubits as in Table I. Note that with our approximations the frequencies ω01 i and hopping coef-ficients Ji,i +1 depend on the site, while the interaction strengths Ui = U Q depend only on the qubit type, U T
being always negative for transmons and U F for CSFQs being positive when 1 /8 < α < 1/2. Note that since [ ˆHBH , ˆn] = 0, we can subtract a term ¯ ω01 ˆn from the right hand side, where ¯ ω01 is the average of the frequency over the sites; this shows that the relevant energy scales are the typical fluctuation in the frequency, the interaction energies, and the (average) hopping amplitude.
III. DIAGNOSTIC TOOLS FOR LEVEL STATISTICS
Studying the statistics of the difference between eigenenergies is a well-established tool to determine whether a system is localized or chaotic. Poisson statis-tics characterizes localized systems and Wigner-Dyson statistics chaotic ones [37, 38]. Here we consider the dis-3(a) Arrays:
Uniform
Alternating
(b) (c) Transmon
≡ JJ
CSFQ
≡ JJ
α
FIG. 1. (a) The systems considered in this work: uniform arrays of transmons (top) and CSFQs (middle), and an array comprising both devices (bottom) with alternating sign anharmonicity. (b) Circuit diagram of a single transmon qubit. (c)
Circuit diagram of a single CSFQ. The parameter α represents the ratio of the Josephson energy of the smaller Josephson junction (marked JJ) to the two identical larger Josephson junctions.
tribution P (r) of the consecutive level spacing ratio
rn = min
En+1 − En
En − En−1
, En − En−1
En+1 − En
, (5) where index n = 2 , . . . , D − 1 counts the eigenstates in ascending order of energy and D is the dimension of the considered sector of the Hilbert space with a fixed num-ber N of excitations. By definition, 0 ≤ rn ≤ 1, and, more importantly for our purposes, the distribution of rn
is independent of the local density of states [39, 40]. The Poisson statistics for energy levels results in the probabil-ity distribution for rn being P0(r) = 2 /(1 + r)2. No an-alytical formulas are in general available for the Wigner-Dyson statistics. Here, since the Hamiltonians have only real entries, we are interested in the distribution for the Gaussian Orthogonal Ensemble (GOE) [41, 42], for which we use the expression
P1(r) = 2 C1
r + r2
[(1 + r)2 − 0.875 r]5/2 (6) where C1 ≃ 3.662 is the normalization constant. The subscript β = 0 , 1 in Pβ is the so-called Dyson index, denoting the absence ( β = 0) or presence ( β > 0) of level repulsion. The choice of the form of P1 and its relation to the “Wigner surmise” are discussed in Appendix A (cf. Ref. ). To quantitatively characterize the crossing point be-tween localization and chaos as we change parameters in the Hamiltonians, we consider two diagnostic tools: 1. the average level spacing ratio ¯ r = ⟨P
n
rn/(D − 2) ⟩,with ⟨. . . ⟩ denoting averaging over a Gaussian distribu-tion of Josephson energies. The values for the Pois-son and GOE distributions are ¯r0 = R 10 dr P 0(r)r =2 log 2 − 1 ≃ 0.3863 and ¯ r1 ≃ 0.5308, and we define the crossing point by requiring that ¯ r = ( ¯ r0 + ¯ r1)/2; 2. the Kullback-Leibler (KL) divergence DKL ,
DKL (P || Q) = X
k
pk log
pk
qk
, (7) which gives the entropy of distribution P relative to dis-tribution Q. Here P is the numerically calculated dis-tribution P (r), with index k in pk denoting the kth bin 00.20.40.60.811.21.41.61.820 0.2 0.4 0.6 0.8 1
Probability density
rn
510 15 20 25 30
K/h [MHz]
10 20 30 40 50 60 70
J/h [MHz]
FIG. 2. The distribution of the level spacing ratio rn [Eq. (5)] for a M = 10 CFSQ array (solid lines) with N = 5 exci-tations for different coupling strengths K in Eq. (1) (color scale). The qubit parameters are α = 0 .35, ECF /h = 54 MHz, and EJF is randomly selected from a Gaussian distribu-tion with mean EJF /h = 301 GHz and standard deviation
δE JF /h = 8 .51 GHz; these parameters match those of trans-mon devices . The data is averaged over 5000 disorder re-alizations. The dashed lines show the distributions P0 (black) and P1 (gray). The dot-dashed line shows the level statistics for the Bose-Hubbard model [Eq. (4)] with average hopping amplitude J equivalent to K/h = 5 MHz. The average values of J (left axis in the color bar) are computed from K accord-ing to Table I.
to which the rn are assigned (see Appendix B for more details on the numerical approach), and Q is either P0
or P1; the crossing point is identified by requiring that
DKL (P || P0) = DKL (P || P1).
IV. LEVEL STATISTICS IN ONE-DIMENSIONAL ARRAYS
Having introduced the relevant models and the diag-nostic tools in Secs. II and III respectively, we move to the discussion of the level statistics for some concrete cases. To minimize finite-size effects, we focus on one-dimensional arrays in this section, while extensions to two-dimensional arrays are considered below in Sec. V. 40.40.44 0.48 0.52 00.10.20.30 50 100 150 200 250 300
Transmon CSFQ Alternating
¯r
(a) DKL (P || Q)
J/h [MHz]
(b)
0.04 0.06 0.08 60 80 100
FIG. 3. (a) Average level spacing ratio and (b) KL divergence for transmon (red), CSFQ (blue), and alternating transmon– CSFQ (green) arrays as functions of average hopping ampli-tude J.The dotted horizontal lines in panel (a) correspond to (top to bottom) ¯ r1, (¯ r0+ ¯ r1)/2, and ¯ r0.In panel (b), the solid curves are for DKL (P|| P0) and the dashed ones for
DKL (P|| P1). Parameters for CSFQs are as in Fig. 2, the cor-responding parameters for transmons are EC/h = 250 MHz,
EJ/h = 44 GHz, δE J/h = 1 .17 GHz . The inset zooms into the crossing points, highlighting the positive shift in cou-pling strength for the alternating array when compared to the transmon- and CSFQ-only systems.
Figure 2 shows the disorder-averaged distribution of
rn for a CSFQ array [Eq. (1) with Q = F ] for a range of coupling strengths. We consider the case of half fill-ing (the total number of excitations being half the length of the array), as states with maximum local occupancy 1 within this subspace are expected to be representa-tive of the computational subspace . The distribution nearly follows P0 (localized) for the lowest coupling con-sidered, K/h = 1 MHz, and P1 (chaotic) for the largest,
K/h = 30 MHz. With increasing coupling strength, the distributions evolve from P0 to P1. This evolution can be investigated using so-called intermediate statistics ; here we only point out that, similarly to the case of trans-mon arrays , we find that the level statistics of the Bose-Hubbard model [Eq. (4)] agrees with that of the CSFQ array, as we show explicitly for one intermediate coupling value. Now turning to our diagnostic tools, we plot in Fig. 3 the average level spacing and KL divergences for trans-mon [red curve , Q = T in Eq. (1)], CSFQ (blue curve,
Q = F ), and alternating transmon–CSFQ (green curve,
Q = T /F for even/odd site) arrays as a function of the qubit-qubit coupling strength expressed in terms of the (average) hopping amplitude J. The qubit param-eters are chosen such that the average frequency ¯ ω01 as well as its standard deviation δω 01 are the same for all three array types, and the anharmonicities are opposite for transmons and CSFQs, U F = −U T ≡ U . Both ¯ r
and DKL change smoothly with coupling strength (up to fluctuations whose amplitude decreases by increas-ing the number of disorder realizations). Clearly, at low coupling all of the distributions are in the localized regime, since ¯ r ≃ r0 and DKL (P || P0) ≪ DKL (P || P1), and at large coupling in the chaotic one, ¯ r ≃ r1 and
DKL (P || P0) ≫ DKL (P || P1). The transmon and CSFQ arrays display remarkably similar dependencies of ¯ r and DKL on J (the transmons’
DKL curves would agree with those in Ref. if plot-ted with the same normalization); both quantities are found to be close to the values obtained for the Bose-Hubbard model. This feature reflects a remarkable sym-metry that holds for the level statistics of Bose-Hubbard model under the exchange U ↔ − U , if the same disorder distribution symmetric around ¯ ω01 is used for the two signs of U (the derivation is given in Appendix C). For all three qubit array types, the crossing points in panel (a) of Fig. 3 (from ¯ r) are comparable to those in panel (b) (from DKL ). This finding cross-validates the two ap-proaches. For concreteness, in what follows we define the crossing point hopping amplitude JQC as the value of J
for which the divergences are the same for arrays of type
Q = T, F, A (with A denoting arrays with alternating transmon/CSFQ qubits). The inset in Fig. 3(a) highlights one of our main re-sults, namely that, for disorder strength as in typical devices, the crossing point hopping amplitude is larger for alternating arrays, JAC > JTC ≃ JFC . We quan-tify this enhancement by introducing ∆ JC = JAC − JFC ;in Fig. 4(a) we plot ∆ JC /J FC [calculated for the Bose-Hubbard model, Eq. (4)] as function of δω 01 /U . The rel-ative enhancement of JC is largest for weak disorder and decreases monotonically with disorder strength. Such a result can be qualitatively understood by considering the level repulsion between next-nearest-neighbor sites: at small disorder, the fact that the multi-excitation levels of the site placed between the next-nearest-neighbors are detuned due to the opposite sign of U favors localization, but this effect becomes less important with increasing disorder (see Appendix D). In fact, for strong disorder,
δω 01 /U ≳ 1.7, we find that JAC < J FC , even though for all array types the crossing point hopping increases with disorder, see the inset in Fig. 4(a). So far, we have set U T = −U F for the alternating arrays, a choice motivated by the cancellation of ZZ in-teractions between neighboring qubits [13, 14]. However, from the point of view of localization, we find this is not necessarily the optimal choice. This remarkable feature is highlighted in Fig. 4(b). Indeed, JC in an alternating array can be increased or decreased by changing the ra-tio η = −U F /U T . In other words, our results show that the localization enhancement in arrays with alternating sign anharmonicity is unrelated to the suppression of the residual ZZ interaction in such systems. 5−0.200.20.40.60.811.20 0.5 1 1.5 2 2.5 3
0200 400
0 1 2 360 90 120 0 0.2 0.4 0.6 0.8 1 1.2
∆JC /J FC
Disorder strength δω 01 /U
(a)
JC /h [MHz]
JAC /h [MHz]
η
(b)
FIG. 4. (a) Relative shift in crossing point ∆ JC/J FCas a func-tion of the disorder strength δω 01 /U for the Bose-Hubbard model. The dashed vertical line shows the disorder strength used in all other figures (cf. Ref. ). The inset shows the numerically calculated values of JFC(blue symbols) and JAC
(green symbols), fitted by quadratic functions (solid curves); the relative difference between those two curves gives the solid line in the main panel. (b) Crossing point hopping amplitude
JACfor an array with alternating transmons and CSFQs vs. ratio of anharmonicites η=−UF/U T(we change UFwhile keeping fixed UT=−250 MHz h, other parameters as in Fig. 3, mapped to the Bose-Hubbard model according to Ta-ble I). The linear best-fit line is shown as a guide to the eye. The points in all panels are obtained by averaging over 1000 disorder realizations.
V. LEVEL STATISTICS OF TWO-DIMENSIONAL ARRAYS
In the previous section we showed that one-dimensional alternating-sign anharmonicity qubit archi-tectures are more resilient to chaos; the arrays used in actual devices for quantum computing purposes are usu-ally two-dimensional , so here we explore to what de-gree higher connectivity affects our results. For the com-putation of the full spectrum we use the exact diago-nalization technique, which is typically limited to small size and few particles, M ∼ 10 − 12 at half-filling [9, 21] since DMM/ 2 ∝ 2.6M /√M for large M , and so not par-ticularly suitable for higher-dimensionality arrays (here
DM
N
= N +M −1
N
is the dimension of the N -particle sec-tor). However, we can explore two minimal yet rele-vant case studies, in analogy with the discussion given in , i) the surface-7 chip architecture , and ii) a 3 × 3 array. The qubit connections for the two cases are shown on the left of Fig. 5(a) and (b), respectively; for instance, the surface-7 chip is obtained from a lin-ear chain of seven qubits by connecting the middle one (qubit 4) to the first and the last ones (qubits 1 and 7), and so the model Hamiltonian is obtained by adding the terms J4,1ˆb†
4
b1 + J4,7ˆb†
4
b7 + H .c. to the Hamiltonian in Eq. (4). Our goal is to compare the level statistics in uniform and alternating-sign anharmonicity arrays, the latter case having qubits with opposite anharmonicity (η = 1) as shown in Fig. 5. The solid curves in the main panels of Fig. 5 show the average level spacing ratio ¯ r as function of the hopping J
(in units of U ) for the surface-7 chip with N = 4 (panel a) and a 3 × 3 array with N = 5 (panel b), both for a trans-mon array (purple) and an alternating transmon–CSFQ array (green). For comparison, we also display the aver-age level spacing ratio for linear chains with the same N
and M (dashed curves in Fig. 5). The crossing points for the surface-7 chip and the 3 × 3 array – the intercepts be-tween the curves and the middle horizontal line, see main text – are sizably reduced compared to the linear case, since the higher connectivity suppresses the localization effects (analog results have been reported in Ref. for transmon-only arrays). Nonetheless, the use of alter-nating transmon-CSFQ arrays still leads to stronger re-silience to chaos, although in a reduced fashion: for the surface-7 chip the relative increase of the crossing point, (JAC − JFC )/J FC , decreases from 59% to 34%, and for the 3 × 3 array from 52% to 36%. These results confirm that the use of alternating-sign anharmonicity qubits is bene-ficial also for two-dimensional arrays.
VI. CONCLUSIONS
We have investigated the crossover from localized to chaotic regimes in arrays of superconducting qubits comprising C-shunted flux qubits (CSFQ) and trans-mon qubits displaying positive and negative anharmonic-ity, respectively, as described by the Hamiltonians in Eqs. (1)-(3). Once the parameters of the models are appropriately matched, the statistics of the level spac-ing ratio can also be reproduced by using the disordered Bose-Hubbard model of Eq. (4). We note that the level statistics of the Bose-Hubbard model is found to be in-dependent of the sign of the onsite interaction U . Such finding is in contrast with the ground-state properties of the model being clearly different in the attractive or repulsive cases (see Ref. and references therein). Our results indicate that for small disorder strength relative to the anharmonicity, arrays with alternating transmons and CSFQs remain localized up to a higher coupling strength compared with chains consisting of one qubit type only, see Figs. 3 and 4. Interestingly, the on-set of the chaotic behavior arises at even higher coupling when increasing the difference between nearest-neighbor anharmonicities. For large disorder strength compared to the anharmonicity, arrays with a single qubit type or alternating qubits display similar localization properties. 60.40.44 0.48 0.52 0.40.44 0.48 0.52 0 0.2 0.4 0.6 0.8 1
XLinear Uniform-U
Alternating-U
¯r¯rJ/U (a)
X = Surface-7
(b) X = 3 ×3
≡Uniform
Alternating
FIG. 5. Level statistics in two-dimensional arrays. The solid curves show the average level spacing ratio ¯ ras function of the hopping amplitude Jfor (a) a surface-7 chip with N= 4 particles and (b) a 3 ×3 lattice with N= 5 particles for uni-form (purple) and alternating-sign interaction (green). For comparison, the average level spacing ratio for linear chains with the same particle and site numbers are displayed with dashed lines. The dotted horizontal lines in both panels cor-respond to (top to bottom) ¯ r1, (¯ r0+ ¯ r1)/2, and ¯ r0. For each value of J(assumed homogeneous throughout the array), we average over 1000 disorder realizations taken from a Gaus-sian distribution with standard deviation δω/ |U|= 0 .47 for the Hamiltonian in Eq. (4), where additional hopping terms are included according to the connectivity schematized on the left of the panels.
For quantum computing applications, future research could extend our investigation of the localized to chaotic crossover to additional diagnostic tools, such as the Walsh transform . Larger systems, especially in higher dimensions, could be studied via a classical approach . On the quantum simulation side, the possibility of engi-neering the strength and nature (attractive/repulsive) of the on-site interaction provides a new platform for the investigation of many-body localization. More broadly, we highlight that superconducting qubits enable tailoring the interaction properties of the system at a local level, beyond what has been achieved so far in other platforms, for example the cold atoms one [46, 47].
ACKNOWLEDGMENTS
Specialist and High Performance Computing systems provided by Information Services at the University of Kent.
Appendix A: Distributions for the consecutive level spacing ratio
The study of the distribution of level spacings in quan-tum systems has a long history, in particular within Ran-dom Matrix Theory [41, 42]. The well-known generaliza-tion of the Wigner surmise for the level spacing distribu-tion PW (s) of random matrices, corresponding to exact results for 2 × 2 matrices, is
PW (s) = aβ sβ e−bβ s2
, (A1) where sn = En+1 − En, and aβ and bβ are known con-stants . The Dyson index β depends on the symme-tries of the considered random matrices, with β = 1 for the Gaussian Orthogonal Ensemble (GOE) of real sym-metric matrices, β = 2 for the Gaussian Unitary Ensem-ble (Hermitian matrices), and β = 4 for the Gaussian Symplectic Ensemble (Hermitian quaternionic matrices). In analogy with the Wigner surmise, an approximate distribution PW (r) for the level spacing ratios [Eq. (5)], which corresponds to the exact distribution for 3 × 3 ran-dom matrices, was derived in Ref. ,
PW (r) = 2
Zβ
(r + r2)β
(1 + r + r2)1+3 β/ 2 , (A2) where Z1 = 8 /27 for the GOE. To be able to both cap-ture the distribution for large matrices and interpolate between localized and chaotic regimes, a generalization of Eq. (A2) was proposed in Ref. ,
Pβ (r; γ) = 2 Cβγ
(r + r2)β
[(1 + r)2 − γr ]1+3 β/ 2 , (A3) where Cβγ is a normalization coefficient such that
R 10 dr P β (r; γ) = 1. Equation (A3) with β = 0 , γ = 0 reduces to the distribution P0 of the main text, and with
γ = 1 to PW (r) in Eq. (A2). Since the models which we study in this work only have real entries in the Hamil-tonian, we are interested in the GOE, β = 1; to find corresponding value for γ, we fit P1(r; γ) to estimates for the distribution obtained from numerical data for large matrices in Refs. and , finding γ ≃ 0.875 (cf. Figs. 6 and 8), see Eq. (6) of the main text. As mentioned above, Eq. (A3) can interpolate between the localized and chaotic regimes ; here we use this feature to capture the localized–chaotic crossing point by assuming β ∈ [0 . . . 1]. We perform a two-parameter fit of Eq. (A3) to the binned frequency distributions at each value of hopping amplitude J, such as those shown in Fig. 2, to find β and γ. Figure 6 shows the value of the fitted β as a function of J for the Bose-Hubbard (BH) model in Eq. (4) with uniform interaction. The index β
monotonically increases with J from 0 (localized regime) to 1 (chaotic); we define the crossing point hopping JC
as the value of J for which β = 0 .5, identified by the purple dashed vertical line. The other vertical dashed 700.20.40.60.810 50 100 150 200 250 300
β
J/h [MHz]
-0.3 00.3 0.6 0.9 1.2 050 100 150 200 250 300
γ
FIG. 6. The values of β (main panel) and γ (inset) found by fitting Eq. (A3) to numerical level spacing ratio distributions for the uniform-interaction Bose-Hubbard as a function of coupling strength J. The dashed horizontal line is at β = 0 .5, and the intersection with the data gives the value of the cross-ing point between localized and chaotic regimes JC (dashed purple vertical line). For comparison, we also show the cross-ing points determined with the methods of the main text, the
DKL crossing (dashed green line) and ¯ r = 0 .4585 (dashed blue line). The horizontal line in the inset is at γ = 0 .875. Param-eters are as in Fig. 2, mapped to the Bose-Hubbard model according to Table I, with 5000 disorder realizations used.
lines in Fig. 6 show the crossing points estimated using the methods used in the main text, the DKL crossing (green) and ¯ r (blue): the value of JC from β = 0 .5 is approximately 13 MHz higher than that from the DKL
crossing, which in turn is about 3 MHz higher than that from the ¯ r method. Despite the quantitative differences in the estimates for JC , all three methods give a consistent picture of the evolution of the crossing point with disorder, as shown in Fig. 7: for both uniform and alternating interaction in the BH model, JC increases monotonically with disorder strength δω 01 . Moreover, all three methods agree on the value of δω 01 /U ∼ 1.7 above which JC for the uniform case become larger than for the alternating one. In closing this section, we briefly discuss a potential limitation in using Eq. (A3) to study the localized to chaotic crossover. In proposing Eq. (A3), the authors of Ref. already noted that the relation between γ and
β is not universal, since it depends on the model being studied: if there were a unique relationship determining γ
as function of β, then the crossover could be analyzed in terms of β only. Our results in Fig. 8 show not only that for the BH model the relation differs from that of other models investigated in Ref. , but also that it displays a marked dependence on the number of excitations N
but not on system size M .050 100 150 200 250 300 350 400 450 500 0 0.5 1 1.5 2 2.5 3
DKL Crossing
β= 0 .5¯r= 0 .4585
Uniform Alternating Crossing point JC/h [MHz] Disorder strength δω 01 /U
FIG. 7. The effect of disorder strength on the crossing point
JC , extracted from the DKL crossing (squares), β = 0 .5 (cir-cles), and ¯ r = 0 .4585 (triangles), for the uniform-interaction (purple) and alternating-interaction (green) Bose-Hubbard model. The solid lines are quadratic polynomial fits to the
DKL crossing values. The dashed vertical line represents the disorder strength used throughout this work (cf. Fig. 4). Pa-rameters are as in Fig. 2 in the main text. Data is for 1000 disorder realizations. −0.400.40.81.20 0.2 0.4 0.6 0.8 1
N = 3 4 5 6 7
M = 6 8 10 12
γ
β
FIG. 8. Plot of γ as a function of β, both found by two-parameter fits of Eq. (A3) to numerical level spacing ratio distributions for the uniform-interaction Bose-Hubbard model [Eq. (4)], for various numbers of excitations N and sites M .Each point corresponds to one value of J, with 1000 disor-der realizations used. Other parameters are as in Fig. 2 and mapped to the Bose-Hubbard model according to Table I. The dashed horizontal line shows γ = 0 .875.
Appendix B: Numerical methods
For the numerical studies of arrays with uniform and mixed qubit types, we consider a fixed number of exci-tations N . We first diagonalize the Hamiltonian ˆHQ
i
of 8each single qubit at site i, where the N lowest eigenstates
|u⟩i, u = 0 , . . . , N , with energy EQi u are computed in the charge basis (eigenstates of operator ˆNi) restricted to
Ni ∈ {− 50 . . . 50 }. We then consider an array of qubits, Eq. (1), for which the Hilbert space is the tensor prod-uct of the individual qubits’ Hilbert spaces, and keep the leading-order particle-number-conserving hopping terms in the expansion of ˆNi ˆNi+1 , resulting in the Hamiltonian ˆHC =
M
X
i=1
N
X
u=0
EQi u |u⟩i⟨u|i+
K
M−1
X
i=1
N − 1
X
u=0
N
X
v=1
giuv |u + 1 ⟩i⟨u|i⊗|v − 1⟩i+1 ⟨v|i+1 + H.c. (B1) with giuv = ⟨u + 1 |i ˆNi|u⟩i⟨v − 1|i+1 ˆNi+1 |v⟩i+1 .The values of rn are computed from the eigenvalues found by exact diagonalization of Eq. (B1) and are av-eraged over many realizations of disorder in Josephson energy, each randomly selected from a Gaussian distri-bution with means EJ or EJF and a standard deviation computed such that the distributions of ω01 for the two devices match (see below). To calculate the KL diver-gence, Eq. (7), we bin the resulting data into 50 equally wide bins, and the values of pk are the frequencies in each bin, while qk = Pi(rk), i = 0 , 1, is computed with
rk being the midpoint of the kth bin. As indicated above, to make a suitable comparison between transmon and CSFQ arrays, or to attain the cancellation of residual ZZ interaction, some care must be taken in selecting the parameters of the qubits. For transmons, we simply used the same (experimentally mo-tivated) parameters as in Ref. . For CSFQs, we need to determine values of EJF , its standard deviation δE JF ,
ECF , and α such that the average frequency ¯ ω01 , its stan-dard deviation δω 01 , and (the absolute value of) the ana-harmonicity A match those of the transmons; the anhar-monicity of a qubit is given by A = ℏ(ω12 − ω01 ), where
ℏωij = EQj − EQi .To numerically find suitable parameter values, let us consider the following approximate expressions for ω01
and A:
ℏω01 ≈ p16 EJF (1 − 2α)ECF − ECF
1 − 8α
1 − 2α + ECF
s
4ECF
EJF (1 − 2α)
"
181 − 32 α
1 − 2α − 14
1 − 8α
1 − 2α
2#
, (B2)
A ≈ − ECF
1 − 8α
1 − 2α + ECF
s
4ECF
EJF (1 − 2α)
"
141 − 32 α
1 − 2α − 17 32
1 − 8α
1 − 2α
2#
, (B3) valid for EJF (1 − 2α) ≫ ECF . Equations (B2) and (B3) can be also used for the transmon with the replacements:
α → 0, ECF → EC and EJF → EJ /2 [compare Eqs. (2) and (3)]. To determine starting points for our numerical determination of the CSFQ’s parameters, we drop the last term in each of the two equations and invert them to find
EJF ≈ 8α − 11 − 2α
√8EC EJ − 2EC
2
16 EC (1 − 2α) ; (B4)
ECF ≈ EC
1 − 2α
8α − 1 . (B5) Note that Eq. (B5) can be satisfied only for 1 /8 < α <
1/2. Similarly, for the standard deviation δE JF using Eqs. (B4) and (B5) we obtain the following expression for its starting value:
δE JF
EJF
≈ 2 ℏδω 01
ℏω01 − EC
. (B6) Setting α = 0 .35 roughly in the middle of the allowed range, we find experimentally realistic values for EJF ,cf. Ref. [35, 48, 49].
Appendix C: Level statistics of the Bose-Hubbard model
The Hamiltonian for the disordered Bose-Hubbard model [cf. Eq. (4)] can be written in the form ˆHBH =
M
X
i=1
ℏδω 01 i ˆni + U
2
M
X
i=1
ˆni(ˆ ni − 1) +
M−1
X
i=1
Ji,i +1
ˆb†
i+1
ˆbi + H.c.
. (C1) where δω 01 i is the deviation of the frequency at site
i from the average frequency of the array (subtracting the average frequency, ˆHBH → ˆHBH − ¯ω01 ˆn, does not change the level spacings when considering a sector of the Hilbert space with a given number of excitations). Clearly, ˆH′
BH
= − ˆHBH has the same level spacings as ˆHBH , the opposite frequencies, and the opposite sign for the interaction term; the sign of Ji,i +1 is unimportant and can be changed by redefining ˆbi → (−1) iˆbi (note that this transformation preserves both ˆ ni and the bosonic com-mutation relations). Thus, differences in the level spacing 90.40.44 0.48 0.52 00.10.20.30 50 100 150 200 250 300
Transmon CSFQ Alt. Unif.-U BH Alt.-U BH
¯r
(a) DKL (P || Q)
J/h [MHz]
(b)
FIG. 9. Crossover in qubit arrays vs Bose-Hubbard model.
(a) Average level spacing ratio and (b) KL divergence for transmon (solid red), CSFQ (solid blue), alternating transmon–CSFQ (solid green) arrays, uniform interaction (dashed purple) and alternating-sign interaction (dashed green) BH model as functions of (average) hopping ampli-tude J.The dotted horizontal lines in panel (a) correspond to (top to bottom) ¯ r1, (¯ r0+ ¯ r1)/2, and ¯ r0. In panel (b), the increasing curves are for DKL (P|| P0) and the decreasing ones for DKL (P|| P1). The parameters used are as in Table I and mapped for the BH model as in Fig. 3, with 5000 disorder realizations used.
ratios under the exchange U ↔ − U are only due to the frequencies. For a distribution of frequencies symmet-ric around the mean, for each set of δω 01 i there exists a corresponding set with opposite values of δω 01 i; in aver-aging over disorder (that is, over frequencies), both sets contribute to the level spacing statistics, and hence the cases with opposite values of U have the same statistics, as stated in the main text. In Fig. 9, we show the same data as in Fig. 3 (solid lines) and, for comparison, the corresponding results for the BH model (dashed), both with uniform and alternat-ing interaction. We recall that the BH model is expected to be a good approximation to transmon and CSFQ ar-rays in the limit EJ ≫ EC , (for the CSFQ, replace
EJ → 2(1 − 2α)EJF , E C → ECF , cf. Table I); indeed, we find that all quantities are within a few percent of each other. In fact, the uniform-interaction BH model curves are hardly distinguishable from those for the CSFQ ar-ray, whose EJ /E C ratio is larger than for the transmon array.
Counter-rotating terms
The results presented in this work are obtained us-ing the rotating-wave approximation ; in other words, terms which do not preserve the total number of exci-tations are neglected [cf. Eq. (4)]. This approximation holds for small coupling compared to the average qubit frequency J/ ¯ω01 ≪ 1, with relative corrections to the energy levels being ∼ (J/ ¯ω01 )2 at the leading order in perturbation theory. Here we show that the counter-rotating (CR) terms give a negligible contribution to the level statistics of the investigated models. For concrete-ness, we focus on the disordered Bose-Hubbard Hamilto-nian, and include the CR terms as follows: ˆHBH-CR = ˆHBH +
M−1
X
i=1
Ji,i +1
ˆb†
i+1
ˆb†
i
ˆbi+1 ˆbi
, (C2) where the first (second) term in the round parenthe-ses raises (lowers) excitation numbers of two neighbor-ing sites by one excitation each. Thus, last term of the Hamiltonian in Eq. (C2) does not conserve the total num-ber of excitations, [ ˆHBH −CR , ˆn]̸ = 0, and connects all the sectors of the Hilbert space with the same parity, yield-ing an infinite-dimensional Hilbert space which needs to be truncated in numerical calculations. We account for the effect of the CR terms on the spectrum of the N -particle sector [with dimension DM
N
= N +M −1
N
] at the second-order in perturbation theory. Concretely, we con-sider the matrix representation of ˆHBH −CR on a basis set including the N -particle sector Fock states |ψk⟩ with
k = 1 , . . . , D M
N
, and all the states |ϕl⟩ with l = 1 , . . . , ˜DM
N
for which ⟨ϕk| PM −1
i=1
ˆb†
i+1
ˆb†
i
ˆbi+1 ˆbi
|ψl⟩̸ = 0 for at least one k. Clearly, these states belong to the N − 2or the N + 2-particle sectors; their number ( ˜DM
N
) can be evaluated in a closed form, reading (for M > 2 and
N ≥ 2) ˜DM
N
= ( M − 1) DM
N
(N + M − 4)( N + M − 3)
M − 2 DM −2
N − 2
(C3) and ˜DM =2
N
= 2 N for M = 2. The first (second) term on the right-hand side of Eq. (C3) accounts for the states reached by acting on the N -particle sector with the two creation (annhiliation) operators in the CR term. This formula shows that the dimension of the Hilbert space we need to consider to include perturbatively the CR term is at least M times that of the original one, and so the computation of the spectrum is more costly. Taking for example the case N = 4, M = 10, we have DM
N
= 715 and ˜DM
N
= 6930, and the dimension of the Hilbert space increases by approximately one order of magnitude. Af-ter computing the DM
N
˜DM
N
eigenvalues for each disorder realization, the level statistics is computed by selecting the DM
N
eigenvalues for which the expectation value of the total particle number operator ˆ n on the correspond-ing eigenstates rounds to N ; this procedure is robust be-10 0.40.44 0.48 0.52 00.10.20.30 50 100 150 200
Uniform-U
Alternating-U
ˆHBH-CR
ˆHBH
¯r
(a) DKL (P || Q)
J/h [MHz]
(b)
FIG. 10. Influence of the counter-rotating terms on the level statistics. (a) Average level spacing ratio and (b) KL diver-gence for uniform (purple), and alternating (green) BH arrays as functions of the hopping amplitude J, obtained includ-ing (solid) or neglecting (dashed) the counter-rotating (CR) terms. The dotted horizontal lines in panel (a) correspond to (top to bottom) ¯ r1, (¯ r0+ ¯ r1)/2, and ¯ r0.We consider 5000 disorder realizations for the dashed curves, where CR terms are neglected. For the solid curves, due to the larger Hilbert space (see text) we consider 1000 disorder realizations for 50 < J/h < 150 MHz, and 500 realizations for the remain-ing values of J. For all the data, we consider N= 4 particles and M= 10 sites. The parameters used are as in Table I and mapped to the BH model as in Fig. 3.
cause the relative deviation ( ⟨ˆn⟩ − N ) /N over the these states is of order J2/¯ω201 .Figure 10 shows the average level spacing ratio and the KL divergence for the spectrum of a chain with N = 4,
M = 10 and averaged over disorder realizations (see caption), either including [solid, Eq. (C2)] or neglecting [dashed, Eq. (4)] the CR terms. The differences between the two cases for both the uniform (purple) and the alter-nating (green) systems are comparable to the statistical uncertainty in our dataset, showing that CR terms do not significantly affect the level statistics; this result also validates our perturbative approach, since the influence of higher order terms is expected to be even smaller.
Appendix D: Level repulsion in a 3-site array
The enhanced resilience to chaos in alternating transmon-CSFQ chains compared to uniform chains for small disorder strengths ℏδω 01 ≪ U can be captured in a minimal setting, consisting of three sites ( M = 3) and two excitations ( N = 2). For convenience, we set
ℏ = 1 hereafter. With no loss of generality, we consider All transmons Alternating chain
(a)
|2, 0, 0⟩
|1, 1, 0⟩ |1, 0, 1⟩
|0, 2, 0⟩
|0, 0, 2⟩
|0, 1, 1⟩ δω
0
−U + 2 δω
−U
Energy
√2J √2J
J
√2J
J
√2J
(b)
|2, 0, 0⟩
|1, 1, 0⟩
|0, 2, 0⟩
|1, 0, 1⟩
|0, 0, 2⟩
|0, 1, 1⟩
UC + 2 δω
δω
0
−U
Energy
√2J √2J
J
√2J
J
√2J
FIG. 11. Level diagrams in a three-site ( M= 3) and two par-ticle ( N=2) array with resonant frequency end-sites for (a) transmon and (b) alternating chains. The level repulsion be-tween the states |2,0,0⟩and |0,0,2⟩depends on the energy of the state |0,2,0⟩, and so on the anharmonicity of the central qubit. For small disorder, δω ≪U, arrays with alternating sign anharmonicity have a smaller level repulsion, thus favor-ing localization (see text).
arrays with transmons at the lateral sites (Q=T for
i = {1, 3}), and a generic qubit type in the central site
C. To compare localization in systems with uniform and alternating-sign anharmonicity (interaction), we investigate the level repulsion between the second excited states in next-to-nearest sites, i.e., the states |2, 0, 0⟩
and |0, 0, 2⟩ in our minimal setting, where a suppressed level repulsion favors localization. More specifically, in a worst-case scenario for localization, we take the trans-mons to have the same frequency ( ω) and anharmonicity (−U ), so that |2, 0, 0⟩ and |0, 0, 2⟩ are degenerate. Due to disorder, the qubit frequency in the central site is detuned by δω , and hence the state |0, 2, 0⟩ is detuned from |2, 0, 0⟩ and |0, 0, 2⟩; specifically, the detuning amounts to 2 δω and U + UC + 2 δω for all-transmon and alternating chains, respectively (see level diagrams in Fig. 11). In fact, this difference determines the enhanced localization for alternating chains, as we detail here. The Hamiltonian of the disordered Bose-Hubbard model [Eq. (4)] for N = 2 and M = 3 can be written in the ba-sis {| 2, 0, 0⟩ , |1, 1, 0⟩ , |1, 0, 1⟩ , |0, 2, 0⟩ , |0, 1, 1⟩ , |0, 0, 2⟩} 11 as the 6 × 6 matrix, ˆHBH =
−U √2J 0 0 0 0
√2J δω J √2J 0 00 J 0 0 J 00 √2J 0 UC + 2 δω √2J 00 0 J √2J δω √2J
0 0 0 0 √2J −U
, (D1) where we subtracted the irrelevant constant 2 ω. The eigenvalues of the Hamiltonian in Eq. (D1) can be com-puted explicitly for arbitrary values of the parameters U ,
UC , δω , and J since the characteristic polynomial of the matrix factorizes in a quadratic and a quartic equation. For our goals, it is sufficient to estimate the energy level splitting (∆ E) of states |2, 0, 0⟩ and |0, 0, 2⟩ in the limit
J ≪ U, |U + δω |, |U + UC + 2 δω |. Using a standard perturbative approach for degenerate levels, the energy splitting at the leading order can be expressed as ∆E(η) ≃ 4J4
U 3(1 + δω/U )2 1 + 21 + η + 2 δω/U , (D2) where η = UC /U . Clearly, ∆ E ∝ J4 as four hop-ping processes are necessary to connect the states |2, 0, 0⟩
and |0, 0, 2⟩ (see Fig. 11). The level splitting is ob-tained by summing over the two possible paths; both paths involve the states |1, 1, 0⟩ and |0, 1, 1⟩ and either the state |1, 0, 1⟩ and |0, 2, 0⟩. Each hopping between the states j and j′ contributes a factor Vjj ′ /(Ej′ + U ), where Vjj ′ = ⟨j| ˆHBH |j′⟩, Ej′ = ⟨j′| ˆHBH |j′⟩, and j′ runs only over the intermediate states (that is, the denomi-nator for the last hop is unity); clearly, the splitting for the path involving |0, 2, 0⟩ depends on the central qubit type. Equation (D2) captures the enhanced localization of an alternating chain (with η > 0) over a uniform chain (η = −1); in fact at low disorder, δω/U ≪ 1, the level splitting for the alternating chain in comparison to the uniform one is suppressed as ∆E(η)∆E(−1) ≃ |δω |
U
3 + η
1 + η , (D3) and decreases monotonically with η for a fixed disorder strength δω , consistently with the increased value of JC
with η [cf. Fig. 4(b)]. In contrast, the relative split-ting increases with |δω |/U , qualitatively in agreement with the reported decrease in ∆ JC /J C with disorder, see Fig. 4(a). The result in Eq. (D2) enables us to consider the effect of disorder in the anharmonicity on localization in uni-form arrays. Let us take η = −1 + δU/U with δU ≪ U ;in this case we find (still assuming δω ≪ U )∆E(η)∆E(−1) ≃ 1 + δU
2δω
−1
(D4) Therefore, for deviation δU from uniform anharmonicity large compared to the frequency disorder δω , we expect localization to persist to stronger hopping J than in the case of nearly-uniform anharmonicity δU ≪ δω .
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189978 | https://www.sec.gov/Archives/edgar/data/51434/000005143416000042/ip10-k123115.htm | 10-K 1 ip10-k123115.htm 10-K 10-K
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UNITED STATES
SECURITIES AND EXCHANGE COMMISSION
WASHINGTON, D.C. 20549
FORM 10-K
(Mark One)
ý ANNUAL REPORT PURSUANT TO SECTION 13 OR 15(d) OF THE SECURITIES EXCHANGE ACT OF 1934 for the fiscal year ended December 31, 2015
or
¨TRANSITION REPORT PURSUANT TO SECTION 13 OR 15(d) OF THE SECURITIES EXCHANGE ACT OF 1934 For the transition period from to
Commission File No.1-3157
INTERNATIONAL PAPER COMPANY
(Exact name of registrant as specified in its charter)
New York 13-0872805
(State or other jurisdiction of incorporation or organization)(I.R.S. Employer Identification No.)
6400 Poplar Avenue
Memphis, Tennessee
(Address of principal executive offices)
38197
(Zip Code)
Registrant’s telephone number, including area code: (901)419-9000
Securities registered pursuant to Section 12(b) of the Act:
Title of each class Name of each exchange on which registered
Common Stock, $1 per share par value New York Stock Exchange
Securities Registered Pursuant to Section 12(g) of the Act: None
Indicate by check mark if the registrant is a well-known seasoned issuer, as defined in Rule 405 of the Securities Act.
Yes ý No ¨
Indicate by check mark if the registrant is not required to file reports pursuant to Section 13 or Section 15(d) of the Act.
Yes ¨No ý
Indicate by check mark whether the registrant (1)has filed all reports required to be filed by Section 13 or 15(d) of the Securities Exchange Act of 1934 during the preceding 12 months (or for such shorter period that the registrant was required to file such reports), and (2)has been subject to such filing requirements for the past 90 days. Yes ý No ¨
Indicate by check mark whether the registrant has submitted electronically and posted on its corporate Website, if any, every Interactive Data File required to be submitted and posted pursuant to Rule 405 of Regulation S-T (section 232.405 of this chapter) during the preceding 12 months (or for such shorter period that the registrant was required to submit and post such files). Yes ý No ¨
Indicate by check mark if disclosure of delinquent filers pursuant to Item 405 of Regulation S-K (section 229.405) is not contained herein, and will not be contained, to the best of registrant’s knowledge, in definitive proxy or information statements incorporated by reference in Part III of this Form 10-K or any amendment to this Form 10-K. ý
Indicate by check mark whether the registrant is a large accelerated filer, an accelerated filer, a non-accelerated filer or a smaller reporting company. See definitions of “large accelerated filer,” “accelerated filer” and “smaller reporting company” in Rule 12b-2 of the Exchange Act. (Check one):
Large accelerated filer x Accelerated filer Non-accelerated filer Smaller reporting company
(Do not check if a smaller reporting company)
Indicate by check mark whether the registrant is a shell company (as defined in Rule 12b-2 of the Act). Yes ¨No ý
The aggregate market value of the Company’s outstanding common stock held by non-affiliates of the registrant, computed by reference to the closing price as reported on the New York Stock Exchange, as of the last business day of the registrant’s most recently completed second fiscal quarter (June 30, 2015) was approximately $21,026,985,885.
The number of shares outstanding of the Company’s common stock as of February 19, 2016 was 411,157,696.
Documents incorporated by reference:
Portions of the registrant’s proxy statement filed within 120 days of the close of the registrant’s fiscal year in connection with registrant’s 2016 annual meeting of shareholders are incorporated by reference into Part III of this Form 10-K.
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INTERNATIONAL PAPER COMPANY
INDEX TO ANNUAL REPORT ON FORM 10-K
FOR THE YEAR ENDED DECEMBER 31, 2015
PART I.1
ITEM 1.BUSINESS.1
General1
Financial Information Concerning Industry Segments1
Financial Information About International and U.S. Operations1
Competition and Costs1
Marketing and Distribution2
Description of Principal Products2
Sales Volumes by Product2
Research and Development3
Environmental Protection3
Climate Change3
Employees5
Executive Officers of the Registrant5
Raw Materials6
Forward-looking Statements6
ITEM 1A.RISK FACTORS.7
ITEM 1B.UNRESOLVED STAFF COMMENTS.11
ITEM 2.PROPERTIES.11
Forestlands11
Mills and Plants11
Capital Investments and Dispositions11
ITEM 3.LEGAL PROCEEDINGS.11
ITEM 4.MINE SAFETY DISCLOSURES.11
PART II.12
ITEM 5.MARKET FOR REGISTRANT’S COMMON EQUITY, RELATED STOCKHOLDER MATTERS AND ISSUER PURCHASES OF EQUITY SECURITIES.12
ITEM 6.SELECTED FINANCIAL DATA.14
ITEM 7.MANAGEMENT’S DISCUSSION AND ANALYSIS OF FINANCIAL CONDITION AND RESULTS OF OPERATIONS.17
Executive Summary17
Corporate Overview20
Results of Operations21
Description of Industry Segments24
Industry Segment Results25
Liquidity and Capital Resources30
Critical Accounting Policies and Significant Accounting Estimates35
Recent Accounting Developments38
Legal Proceedings38
Effect of Inflation38
Foreign Currency Effects38
Market Risk39
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INTERNATIONAL PAPER COMPANY
INDEX TO ANNUAL REPORT ON FORM 10-K
FOR THE YEAR ENDED DECEMBER 31, 2015
ITEM 7A.QUANTITATIVE AND QUALITATIVE DISCLOSURES ABOUT MARKET RISK.39
ITEM 8.FINANCIAL STATEMENTS AND SUPPLEMENTARY DATA.40
Report of Management on Financial Statements, Internal Control over Financial Reporting and Internal Control Environment and Board of Directors Oversight40
Reports of Deloitte & Touche LLP, Independent Registered Public Accounting Firm42
Consolidated Statement of Operations44
Consolidated Statement of Comprehensive Income45
Consolidated Balance Sheet46
Consolidated Statement of Cash Flows47
Consolidated Statement of Changes in Equity48
Notes to Consolidated Financial Statements49
Interim Financial Results (Unaudited)83
ITEM 9.CHANGES IN AND DISAGREEMENTS WITH ACCOUNTANTS ON ACCOUNTING AND FINANCIAL DISCLOSURE.83
ITEM 9A.CONTROLS AND PROCEDURES.85
ITEM 9B.OTHER INFORMATION.86
PART III.86
ITEM 10.DIRECTORS, EXECUTIVE OFFICERS AND CORPORATE GOVERNANCE.86
ITEM 11.EXECUTIVE COMPENSATION.86
ITEM 12.SECURITY OWNERSHIP OF CERTAIN BENEFICIAL OWNERS AND MANAGEMENT AND RELATED STOCKHOLDER MATTERS.86
ITEM 13.CERTAIN RELATIONSHIPS AND RELATED TRANSACTIONS, AND DIRECTOR INDEPENDENCE.86
ITEM 14.PRINCIPAL ACCOUNTANT FEES AND SERVICES.87
PART IV.87
ITEM 15.EXHIBITS AND FINANCIAL STATEMENT SCHEDULES.87
Additional Financial Data87
Schedule II – Valuation and Qualifying Accounts90
SIGNATURES91
APPENDIX I2015 LISTING OF FACILITIESA-1
APPENDIX II2015 CAPACITY INFORMATIONA-4
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PART I.
ITEM 1. BUSINESS
GENERAL
International Paper Company (the “Company” or “International Paper,” which may also be referred to as “we” or “us”) is a global paper and packaging company with primary markets and manufacturing operations in North America, Europe, Latin America, Russia, Asia, Africa and the Middle East. We are a New York corporation, incorporated in 1941 as the successor to the New York corporation of the same name organized in 1898. Our home page on the Internet is www.internationalpaper.com. You can learn more about us by visiting that site.
In the United States, at December 31, 2015, the Company operated 24 pulp, paper and packaging mills, 169 converting and packaging plants, 16 recycling plants and three bag facilities. Production facilities at December 31, 2015 in Europe, Asia, Africa, India, Latin America and South America included 16 pulp, paper and packaging mills, 67 converting and packaging plants, and two recycling plants. We operate a printing and packaging products distribution business principally through 12 branches in Asia. At December 31, 2015, we owned or managed approximately 335,000 acres of forestland in Brazil and had, through licenses and forest management agreements, harvesting rights on government-owned forestlands in Russia. Substantially all of our businesses have experienced, and are likely to continue to experience, cycles relating to industry capacity and general economic conditions.
For management and financial reporting purposes, our businesses are separated into three segments: Industrial Packaging; Printing Papers; and Consumer Packaging. A description of these business segments can be found on pages 24 and 25 of Item 7. Management’s Discussion and Analysis of Financial Condition and Results of Operations. The Company’s 50% equity interest in Ilim Holding S.A. is also a separate reportable industry segment.
From 2011 through 2015, International Paper’s capital expenditures approximated $6.6 billion, excluding mergers and acquisitions. These expenditures reflect our continuing efforts to improve product quality and environmental performance, as well as lower costs and maintain reliability of operations. Capital spending in 2015 was approximately $1.5 billion and is expected to be approximately $1.3 billion in 2016. You can find more information about capital expenditures on page 31 of Item 7. Management’s Discussion and Analysis of Financial Condition and Results of Operations.
Discussions of acquisitions can be found on pages 31 and 32 of Item 7. Management’s Discussion and
Analysis of Financial Condition and Results of Operations.
You can find discussions of restructuring charges and other special items on pages 22 through 24 of Item 7. Management’s Discussion and Analysis of Financial Condition and Results of Operations.
Throughout this Annual Report on Form 10-K, we “incorporate by reference” certain information in parts of other documents filed with the Securities and Exchange Commission (SEC). The SEC permits us to disclose important information by referring to it in that manner. Please refer to such information. Our annual reports on Form 10-K, quarterly reports on Form 10-Q and current reports on Form 8-K, along with all other reports and any amendments thereto filed with or furnished to the SEC, are publicly available free of charge on the Investor Relations section of our Internet Web site at www.internationalpaper.com as soon as reasonably practicable after we electronically file such material with, or furnish it to, the SEC. The information contained on or connected to our Web site is not incorporated by reference into this Form 10-K and should not be considered part of this or any other report that we filed with or furnished to the SEC.
FINANCIAL INFORMATION CONCERNING INDUSTRY SEGMENTS
The financial information concerning segments is set forth in Note 19 Financial Information by Industry Segment and Geographic Area on pages 81 and 82 of Item 8. Financial Statements and Supplementary Data.
FINANCIAL INFORMATION ABOUT INTERNATIONAL AND U.S. OPERATIONS
The financial information concerning international and U.S. operations and export sales is set forth in Note 19 Financial Information by Industry Segment and Geographic Area on page 82 of Item 8. Financial Statements and Supplementary Data.
COMPETITION AND COSTS
The markets in the pulp, paper and packaging product lines are large and fragmented. The major markets, both U.S. and non-U.S., in which the Company sells its principal products are very competitive. Our products compete with similar products produced by other forest products companies. We also compete, in some instances, with companies in other industries and against substitutes for wood-fiber products.
Many factors influence the Company’s competitive position, including price, cost, product quality and services. You can find more information about the impact of these factors on operating profits on pages 17 through 30 of Item 7. Management’s Discussion and Analysis of Financial Condition and Results of
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Operations. You can find information about the Company’s manufacturing capacities on page A-4 of Appendix II.
MARKETING AND DISTRIBUTION
The Company sells packaging products, paper products and other products directly to end users and converters, as well as through agents, resellers and paper distributors.
DESCRIPTION OF PRINCIPAL PRODUCTS
The Company’s principal products are described on pages 24 and 25 of Item 7. Management’s Discussion and Analysis of Financial Condition and Results of Operations.
SALES VOLUMES BY PRODUCT
Sales volumes of major products for 2015, 2014 and 2013 were as follows:
Sales Volumes by Product (1)
In thousands of short tons 2015 2014 2013
Industrial Packaging
North American Corrugated Packaging 10,284 10,355 10,393
North American Containerboard 3,110 3,035 3,273
North American Recycling 2,379 2,459 2,379
North American Saturated Kraft 156 186 176
North American Gypsum/Release Kraft 171 168 157
North American Bleached Kraft 23 26 132
EMEA Industrial Packaging 1,417 1,379 1,342
Asian Box 359 408 416
Brazilian Packaging 305 318 297
Industrial Packaging 18,204 18,334 18,565
Printing Papers
U.S. Uncoated Papers 1,879 1,968 2,508
European and Russian Uncoated Papers 1,493 1,531 1,413
Brazilian Uncoated Papers 1,125 1,141 1,150
Indian Uncoated Papers 241 231 232
Uncoated Papers 4,738 4,871 5,303
Market Pulp (2)1,736 1,776 1,711
Consumer Packaging
North American Consumer Packaging 1,425 1,486 1,556
European and Russian Coated Paperboard 381 354 355
Asian Coated Paperboard 958 1,358 1,430
Consumer Packaging 2,764 3,198 3,341
(1)Includes third-party and inter-segment sales and excludes sales of equity investees.
(2)Includes North American, European and Brazilian volumes and internal sales to mills.
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RESEARCH AND DEVELOPMENT
The Company operates its primary research and development center in Loveland, Ohio, as well as several product laboratories. Additionally, the Company has an interest in ArborGen, Inc., a joint venture with certain other forest products companies.
We direct research and development activities to short-term, long-term and technical assistance needs of customers and operating divisions, and to process, equipment and product innovations. Activities include product development within the operating divisions; studies on innovation and improvement of pulping, bleaching, chemical recovery, papermaking, converting and coating processes; packaging design and materials development; mechanical packaging systems, environmentally sensitive printing inks and reduction of environmental discharges; re-use of raw materials in manufacturing processes; recycling of consumer and packaging paper products; energy conservation; applications of computer controls to manufacturing operations; innovations and improvement of products; and development of various new products. Our development efforts specifically address product safety as well as the minimization of solid waste. The cost to the Company of its research and development operations was $27 million in 2015, $16 million in 2014 and $18 million in 2013.
We own numerous patents, copyrights, trademarks, trade secrets and other intellectual property rights relating to our products and to the processes for their production. We also license intellectual property rights to and from others where advantageous or necessary. Many of the manufacturing processes are among our trade secrets. Some of our products are covered by U.S. and non-U.S. patents and are sold under well known trademarks. We derive a competitive advantage by protecting our trade secrets, patents, trademarks and other intellectual property rights, and by using them as required to support our businesses.
ENVIRONMENTAL PROTECTION
International Paper is subject to extensive federal and state environmental regulation as well as similar regulations internationally. Our continuing objectives include: (1)controlling emissions and discharges from our facilities into the air, water and groundwater to avoid adverse impacts on the environment, and (2)maintaining compliance with applicable laws and regulations. The Company spent $93 million in 2015 for capital projects to control environmental releases into the air and water, and to assure environmentally sound management and disposal of waste. The 2015 spend included costs associated with the U.S. Environmental Protection Agency’s (EPA) Boiler MACT (maximum achievable control technology)regulations. We expect
to spend $118 million in 2016 for similar capital projects. Capital expenditures for 2017 environmental projects are anticipated to be approximately $114 million. Capital expenditures for 2018 environmental projects are estimated to be $83 million. On January 31, 2013, EPA issued the final suite of Boiler MACT regulations. These regulations require owners of specified boilers to meet revised air emissions standards for certain substances. Several lawsuits have been filed to challenge all or portions of the Boiler MACT regulations. On December 3, 2015, the U.S. Court of Appeals for the D.C. Circuit heard oral arguments of the petitioners challenging these regulations. As such, the projected capital expenditures for environmental projects represent our current best estimate of future expenditures with the recognition that the Boiler MACT regulations could change as a result of the pending court decision.
In the U.S., revisions to National Ambient Air Quality Standards (NAAQS) for sulfur dioxide (SO2), nitrogen dioxide (NO2), and fine particulate (PM2.5) finalized between 2010 and 2012, and a promulgated revision to the NAAQS for ozone on October 1, 2015, have not had a material impact on the Company. Similarly, regulations addressing specific implementation issues related to the SO2 NAAQS were released in 2015 by the EPA and are being implemented during the next two to four years. Potentially material capital investment may be required in response to these emerging requirements, but evaluations are ongoing.
CLIMATE CHANGE
Climate change refers to any significant change in the measure of the earth’s climatic conditions such as temperature, precipitation, or winds that persist for decades or longer. Climate change can be caused by natural factors, such as changes in the sun’s intensity and ocean circulation, and human activities can also affect the composition of the earth’s atmosphere, such as from the burning of fossil fuels. In an effort to mitigate the potential of climate change impacts from human activities, various international, national and sub-national (regional, state and local) governmental actions have been undertaken. Presently, these efforts have not materially impacted International Paper, but such efforts may have a material impact on the Company in the future.
International Efforts
The 1997 Kyoto Protocol established emission reduction obligations for certain countries where the Company had and continues to have operations. Though the Kyoto Protocol expired in 2012, several countries, and most notably the European Union (EU), extended their emissions commitments until 2020. A successor program to the Kyoto Protocol is the subject
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of on-going international negotiations including a Conference of the Parties (COP21) to the Kyoto Protocol. COP21 took place in December 2015 and although well short of reaching another international agreement, many countries, including the U.S. and EU member states, did establish non-binding emissions reduction targets. The U.S non-binding commitment is for greenhouse gas (GHG) emissions to be 26% to 28% below 2005 GHG emissions levels by 2025. Other countries in which we do business made similar non-binding commitments. The Company’s voluntary GHG reductions, which are set out in the Company’s annual Sustainability Report, are roughly in line with the percentages of the U.S. non-binding commitment. It is not clear at this time what, if any, further reductions by the Company might be required by the countries in which we operate. Due to this uncertainty, it is not possible at this time to estimate the potential impacts of future international agreements on the Company.
To assist member countries in meeting obligations under the Kyoto Protocol, the EU established and continues to operate an Emissions Trading System (EU ETS). Currently, we have two sites directly subject to regulation under Phase III of the EU ETS, one in Poland and one in France. Other sites that we operate in the EU experience indirect impacts of the EU ETS through purchased power pricing. Neither the direct nor indirect impacts of the EU ETS have been material to the Company, but they could be material to the Company in the future depending on how the 2015 non-binding commitments or allocation of and market prices for GHG credits under existing rules evolve over the coming years.
National Efforts
In the U.S., the Kyoto Protocol was not ratified and Congress has not passed GHG legislation. EPA has enacted (i) regulations to control GHGs from mobile sources (through transportation fuel efficiency standards), (ii) New Source Performance Standards (NSPS) for new Electrica Generating Units (EGUs), (iii) regulations requiring reporting of GHGs from sources of GHGs greater than 25,000 tons per year, and (iv) in 2015, requirements for states to develop plans to reduce GHGs from utility electric generating units (EGUs). In 2015, the Company reported to EPA the GHG emissions from 21 of our U.S. manufacturing sites and 9 landfills.
On November 19, 2014, EPA issued a revised draft carbon accounting framework addressing the circumstances under which biomass combustion can be considered carbon neutral. EPA has stated it intends to issue future rulemakings to address how states may use the revised framework in implementing state permit rules and in developing plans for regulating GHGs from utility electric generators. Given the uncertainties
regarding the framework and scope of future GHG rulemaking, it is unclear what impacts, if any, EPA’s actions in this area will have on the Company’s operations. To date there have been only minor permitting considerations and no substantive impacts.
In 2013, EPA issued final regulations establishing NSPS for new (EGUs). This regulation is the first of several expected NSPSs that EPA will implement over the coming years. The EPA has not yet identified the pulp and paper industry in the first phase of sectors to be covered by the new standards. However, we anticipate that at some future time pulp and paper sources may be subject to new GHG NSPS rules. It is unclear what impacts, if any, future GHG NSPS rules will have on the Company’s operations.
On August 3, 2015, EPA promulgated the Clean Power Plan (CPP) rule to address climate change by reducing carbon dioxide (CO2) and other designated green house gas pollutant emissions from utility EGUs. In response, states are to develop EGU pollutant reduction plans over the next 1 to 3 years to reduce emissions over the 2022 to 2033 timeframe by about 32 percent from 2005 levels. These plans, or the federal plan that would take effect if the states do not act, pose potential cost increases for electricity purchased by the Company. EPA estimated that the proposed rule would increase purchased electricity prices by less than seven percent, but some utilities are estimating significantly higher price increases from the final rule (11 to 14%, or more). The magnitude of the cost increase to the Company will not be possible to estimate reliably until the plans and the utility industries’ responses are better defined over the next few years. Adding to the uncertainty, states and some industry parties have filed lawsuits challenging the rule, the result of which could materially affect the scope and stringency of the regulations. On February 9, 2016, the U.S. Supreme Court granted a stay of the Clean Power Plan. The stay will remain in effect until final disposition of the case, and as such, the rule’s potential impact on the Company remains unclear.
State, Regional and Local Measures
A few U.S. states have enacted or are considering legal measures to require the reduction of emissions of GHGs by companies and public utilities, primarily through the development of GHG emission inventories or regional GHG cap-and-trade programs. One such state is California. The Company does not have any sites currently subject to California's GHG regulatory plan. There may be indirect impacts from changing input costs (such as electricity) at some of our California converting operations but these have yet to manifest themselves in material impacts. Although we are monitoring proposed programs in other states, it is unclear what impacts, if any, state-level GHG rules will
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have on the Company’s operations. Further state measures are under substantive review as they respond to EPA’s 2015 Clean Power Plan and develop an implementation plan over the next 1 to 3 years. The CPP allows significant flexibility in how states develop their plans, so the uncertainty regarding potential impacts will remain high until more specificity is reached and individual power companies develop their compliance strategies.
Summary
Regulation of GHGs continues to evolve in various countries in which we do business. While it is likely that there will be increased governmental action regarding GHGs and climate change, any material impact to the company is not likely to occur before 2020 and at this time it is not reasonably possible to estimate Company costs of compliance with rules that have not yet been adopted or implemented and may not be adopted or implemented in the future. In addition to possible direct impacts, future legislation and regulation could have indirect impacts on International Paper, such as higher prices for transportation, energy and other inputs, as well as more protracted air permitting processes, causing delays and higher costs to implement capital projects. International Paper has controls and procedures in place to stay informed about developments concerning possible climate change legislation and regulation in the U.S. and in other countries where we operate. We regularly assess whether such legislation or regulation may have a material effect on the Company, its operations or financial condition, and whether we have any related disclosure obligations.
Additional information regarding climate change and International Paper is available in our Sustainability Report found at EN/Company/Sustainability/SustainabilityReport.html, though this information is not incorporated by reference into this Form 10-K and should not be considered part of this or any other report that we file with or furnish to the SEC.
EMPLOYEES
As of December 31, 2015, we have approximately 56,000 employees, nearly 34,000 of whom are located in the United States. Of the U.S. employees, approximately 26,000 are hourly, with unions representing approximately 14,000 employees. Approximately 11,000 of this number are represented by the United Steelworkers union (USW).
International Paper, the USW, and several other unions have entered into two master agreements covering various mills and converting facilities. These master agreements cover several specific items, including
wages, select benefit programs, successorship, employment security, and health and safety. Individual facilities continue to have local agreements for other subjects not covered by the master agreements. If local facility agreements are not successfully negotiated at the time of expiration, under the terms of the master agreements the local contracts will automatically renew with the same terms in effect. The mill master agreement covers 19 of our U.S. pulp, paper, and packaging mills; the converting agreement includes 61 of our converting facilities. In addition, International Paper is party to a master agreement with District Council 2, International Brotherhood of Teamsters, covering 13 additional converting facilities.
During 2015, local labor agreements were negotiated at five mills and 13 converting facilities. In 2016, local labor agreements are scheduled to be negotiated at 29 facilities, including four mills and 25 converting facilities. 26 of these agreements will automatically renew under the terms of the applicable master agreement if new agreements are not reached.
EXECUTIVE OFFICERS OF THE REGISTRANT
Mark S. Sutton, 54, chairman (since January 1, 2015) & chief executive officer (since November 1, 2014). Mr. Sutton previously served as president & chief operating officer from June 1, 2014 to October 31, 2014, senior vice president - industrial packaging from November 2011 to May 31, 2014, senior vice president - printing and communications papers of the Americas from 2010 until 2011, senior vice president - supply chain from 2008 to 2009, vice president - supply chain from 2007 until 2008, and vice president - strategic planning from 2005 until 2007. Mr. Sutton joined International Paper in 1984.
W. Michael Amick, Jr., 52, senior vice president - North American papers, pulp & consumer packaging since November 1, 2014. Mr. Amick previously served as vice president - president, IP India, from August 2012 to October 31, 2014, and vice president and general manager for the coated paperboard business from 2010 to 2012. Mr. Amick joined International Paper in 1990.
C. Cato Ealy, 59, senior vice president - corporate development since 2003. Mr. Ealy is a director of Ilim Holding S.A., a Swiss holding company in which International Paper holds a 50% interest, and of its subsidiary, Ilim Group. Mr. Ealy joined International Paper in 1992.
William P. Hoel, 59, senior vice president, Container The Americas, since February 2012. Mr. Hoel previously served as vice president, Container The Americas, from 2005 until 2012, senior vice president, corporate sales and marketing, from 2004 until 2005,
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and vice president, Wood Products, from 2000 until 2004. Mr. Hoel joined International Paper in 1983.
Tommy S. Joseph, 56, senior vice president - manufacturing, technology, EH&S and global sourcing since January 2010. Mr. Joseph previously served as senior vice president - manufacturing, technology, EH&S from February 2009 until December 2009, and vice president - technology from 2005 until February 2009. Mr. Joseph is a director of Ilim Holding S.A., a Swiss Holding Company in which International Paper holds a 50% interest, and of its subsidiary, Ilim Group. Mr. Joseph joined International Paper in 1983.
Thomas G. Kadien, 59, senior vice president - human resources, government relations & global citizenship, since November 1, 2014. Mr. Kadien previously served as senior vice president - consumer packaging and IP Asia from January 2010 to October 31, 2014, and senior vice president and president - xpedx from 2005 until 2009. Mr. Kadien serves on the board of directors of The Sherwin-Williams Company. Mr. Kadien joined International Paper in 1978.
Glenn R. Landau, 47, senior vice president - president, IP Latin America since November 1, 2014. Mr. Landau previously served as vice president - president IP Latin America from 2013 to October 31, 2014, vice president - investor relations from 2011 to 2013, and vice president and general manager, containerboard and recycling from 2007 to 2011. Mr. Landau joined International Paper in 1991.
Timothy S. Nicholls, 54, senior vice president - industrial packaging since November 1, 2014. Mr. Nicholls previously served as senior vice president - printing and communications papers of the Americas from November 2011 to October 31, 2014, senior vice president and chief financial officer from 2007 until 2011, vice president and executive project leader of IP Europe during 2007, and vice president and chief financial officer - IP Europe from 2005 until 2007. Mr. Nicholls joined International Paper in 1991.
Jean-Michel Ribieras, 53, senior vice president - president, IP Europe, Middle East, Africa & Russia since June 2013. Mr. Ribieras previously served as president - IP Latin America from 2009 until 2013. Mr. Ribieras is a director of Ilim Holding S.A., a Swiss holding company in which International Paper holds a 50% interest, and of its subsidiary, Ilim Group. Mr. Ribieras joined International Paper in 1993.
Carol L. Roberts, 56, senior vice president & chief financial officer since November 2011. Ms. Roberts previously served as senior vice president - industrial packaging from 2008 until 2011 and senior vice president - IP packaging solutions from 2005 until 2008. Ms. Roberts serves on the board of directors of
Alcoa Inc. Ms. Roberts joined International Paper in 1981.
Sharon R. Ryan, 56, senior vice president, general counsel & corporate secretary since November 2011. Ms. Ryan previously served as vice president, acting general counsel & corporate secretary from May 2011 until November 2011, vice president from March 2011 until May 2011, associate general counsel, chief ethics and compliance officer from 2009 until 2011, and associate general counsel from 2006 until 2009. Ms. Ryan joined International Paper in 1988.
RAW MATERIALS
Raw materials essential to our businesses include wood fiber, purchased in the form of pulpwood, wood chips and old corrugated containers (OCC), and certain chemicals, including caustic soda and starch. Information concerning fiber supply purchase agreements that were entered into in connection with the Company’s 2006 Transformation Plan and the CBPR acquisition in 2008 is presented in Note 11 Commitments and Contingent Liabilities on page 61 of Item 8. Financial Statements and Supplementary Data.
FORWARD-LOOKING STATEMENTS
Certain statements in this Annual Report on Form 10-K that are not historical in nature may be considered “forward-looking” statements within the meaning of the Private Securities Litigation Reform Act of 1995. These statements are often identified by the words, “will,” “may,” “should,” “continue,” “anticipate,” “believe,” “expect,” “plan,” “appear,” “project,” “estimate,” “intend,” and words of a similar nature. These statements are not guarantees of future performance and reflect management’s current views with respect to future events, which are subject to risks and uncertainties that could cause actual results to differ materially from those expressed or implied in these statements. Factors which could cause actual results to differ include but are not limited to: (i) the level of our indebtedness and changes in interest rates; (ii) industry conditions, including but not limited to changes in the cost or availability of raw materials, energy and transportation costs, competition we face, cyclicality and changes in consumer preferences, demand and pricing for our products; (iii) global economic conditions and political changes, including but not limited to the impairment of financial institutions, changes in currency exchange rates, credit ratings issued by recognized credit rating organizations, the amount of our future pension funding obligation, changes in tax laws and pension and health care costs; (iv) unanticipated expenditures related to the cost of compliance with existing and new environmental and other governmental regulations and to actual or potential litigation; (v) whether we experience a material disruption at one of our
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manufacturing facilities; (vi) risks inherent in conducting business through a joint venture; (vii) the execution of a definitive agreement to sell our corrugated box business in China and Southeast Asia, and the successful closing of the transaction within the estimated timeframe; and (viii) our ability to achieve the benefits we expect from strategic acquisitions, divestitures and restructurings. These and other factors that could cause or contribute to actual results differing materially from such forward looking statements are discussed in greater detail below in “Item 1A. Risk Factors.” We undertake no obligation to publicly update any forward-looking statements, whether as a result of new information, future events or otherwise.
All financial information and statistical measures regarding our 50/50 Ilim joint venture in Russia (“Ilim”), other than historical International Paper Equity Earnings and dividends received by International Paper, have been prepared by the management of Ilim. In providing this information in this filing, we are relying on the effectiveness of Ilim's internal control environment. Any projected financial information and statistical measures reflect the current views of Ilim management and are subject to the risks and uncertainties that could cause actual results to differ materially from those expressed or implied by such projections.
ITEM 1A. RISK FACTORS
In addition to the risks and uncertainties discussed elsewhere in this Annual Report on Form 10-K (particularly in Item 7. Management’s Discussion and Analysis of Financial Condition and Results of Operations), or in the Company’s other filings with the Securities and Exchange Commission, the following are some important factors that could cause the Company’s actual results to differ materially from those projected in any forward-looking statement.
RISKS RELATING TO INDUSTRY CONDITIONS
CHANGES IN THE COST OR AVAILABILITY OF RAW MATERIALS, ENERGY AND TRANSPORTATION COULD AFFECT OUR PROFITABILITY. We rely heavily on the use of certain raw materials (principally virgin wood fiber, recycled fiber, caustic soda and starch), energy sources (principally natural gas, coal and fuel oil) and third-party companies that transport our goods. The market price of virgin wood fiber varies based upon availability and source. In addition, the increase in demand of products manufactured, in whole or in part, from recycled fiber, on a global basis, may cause an occasional tightening in the supply of recycled fiber. Energy prices, in particular prices for oil and natural gas, have fluctuated dramatically in the past and may continue to fluctuate in the future. Our profitability has been, and will continue to be, affected by changes
in the costs and availability of such raw materials, energy sources and transportation sources.
THE INDUSTRIES IN WHICH WE OPERATE EXPERIENCE BOTH ECONOMIC CYCLICALITY AND CHANGES IN CONSUMER PREFERENCES. FLUCTUATIONS IN THE PRICES OF, AND THE DEMAND FOR, OUR PRODUCTS COULD MATERIALLY AFFECT OUR FINANCIAL CONDITION, RESULTS OF OPERATIONS AND CASH FLOWS. Substantially all of our businesses have experienced, and are likely to continue to experience, cycles relating to industry capacity and general economic conditions. The length and magnitude of these cycles have varied over time and by product. In addition, changes in consumer preferences may increase or decrease the demand for our fiber-based products and non-fiber substitutes. These consumer preferences affect the prices of our products. Consequently, our operating cash flow is sensitive to changes in the pricing and demand for our products.
COMPETITION IN THE UNITED STATES AND INTERNATIONALLY COULD NEGATIVELY IMPACT OUR FINANCIAL RESULTS. We operate in a competitive environment, both in the United States and internationally, in all of our operating segments. Product innovations, manufacturing and operating efficiencies, and marketing, distribution and pricing strategies pursued or achieved by competitors could negatively impact our financial results.
RISKS RELATING TO MARKET AND ECONOMIC FACTORS
ADVERSE DEVELOPMENTS IN GENERAL BUSINESS AND ECONOMIC CONDITIONS COULD HAVE AN ADVERSE EFFECT ON THE DEMAND FOR OUR PRODUCTS AND OUR FINANCIAL CONDITION AND RESULTS OF OPERATIONS. General economic conditions may adversely affect industrial non-durable goods production, consumer spending, commercial printing and advertising activity, white-collar employment levels and consumer confidence, all of which impact demand for our products. In addition, volatility in the capital and credit markets, which impacts interest rates, currency exchange rates and the availability of credit, could have a material adverse effect on our business, financial condition and our results of operations.
THE LEVEL OF OUR INDEBTEDNESS COULD ADVERSELY AFFECT OUR FINANCIAL CONDITION AND IMPAIR OUR ABILITY TO OPERATE OUR BUSINESS. As of December 31, 2015, International Paper had approximately $9.3 billion of outstanding indebtedness, including $9.3 billion of indebtedness outstanding under our floating and fixed rate notes. There was no indebtedness outstanding under our
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credit facilities as of December 31, 2015. The level of our indebtedness could have important consequences to our financial condition, operating results and business, including the following:
•it may limit our ability to obtain additional debt or equity financing for working capital, capital expenditures, product development, dividends, share repurchases, debt service requirements, acquisitions and general corporate or other purposes;
•a portion of our cash flows from operations will be dedicated to payments on indebtedness and will not be available for other purposes, including operations, capital expenditures and future business opportunities;
•the debt service requirements of our indebtedness could make it more difficult for us to satisfy other obligations;
•our indebtedness that is subject to variable rates of interest exposes us to increased debt service obligations in the event of increased interest rates;
•it may limit our ability to adjust to changing market conditions and place us at a competitive disadvantage compared to our competitors that have less debt; and
•it may increase our vulnerability to a downturn in general economic conditions or in our business, and may make us unable to carry out capital spending that is important to our growth.
In addition, we are subject to agreements that require meeting and maintaining certain financial ratios and covenants. A significant or prolonged downturn in general business and economic conditions may affect our ability to comply with these covenants or meet those financial ratios and tests and could require us to take action to reduce our debt or to act in a manner contrary to our current business objectives.
CHANGES IN CREDIT RATINGS ISSUED BY NATIONALLY RECOGNIZED STATISTICAL RATING ORGANIZATIONS COULD ADVERSELY AFFECT OUR COST OF FINANCING AND HAVE AN ADVERSE EFFECT ON THE MARKET PRICE OF OUR SECURITIES. Maintaining an investment-grade credit rating is an important element of our financial strategy, and a downgrade of the Company’s ratings below investment grade may limit our access to the capital markets, have an adverse effect on the market price of our securities, increase our cost of borrowing and require us to post collateral for derivatives in a net liability position. The Company’s desire to maintain its investment grade rating may cause the Company to take certain actions designed to improve its cash flow, including sale of assets, suspension or reduction of our
dividend and reductions in capital expenditures and working capital.
Under the terms of the agreements governing approximately $2.5 billion of our debt as of December 31, 2015, the applicable interest rate on such debt may increase upon each downgrade in our credit rating. As a result, a downgrade in our credit rating may lead to an increase in our interest expense. There can be no assurance that such credit ratings will remain in effect for any given period of time or that such ratings will not be lowered, suspended or withdrawn entirely by the rating agencies, if, in each rating agency’s judgment, circumstances so warrant. Any such downgrade of our credit ratings could adversely affect our cost of borrowing, limit our access to the capital markets or result in more restrictive covenants in agreements governing the terms of any future indebtedness that we may incur.
DOWNGRADES IN THE CREDIT RATINGS OF BANKS ISSUING CERTAIN LETTERS OF CREDIT WILL INCREASE OUR COST OF MAINTAINING CERTAIN INDEBTEDNESS AND MAY RESULT IN THE ACCELERATION OF DEFERRED TAXES. We are subject to the risk that a bank with currently issued irrevocable letters of credit supporting installment notes delivered to Temple-Inland in connection with Temple-Inland's 2007 sales of forestlands may be downgraded below a required rating. Since 2007, certain banks have fallen below the required ratings threshold and were successfully replaced, or waivers were obtained regarding their replacement. As a result of continuing uncertainty in the banking environment, a number of the letter-of-credit banks currently in place remain subject to risk of downgrade and the number of qualified replacement banks remains limited. The downgrade of one or more of these banks may subject the Company to additional costs of securing a replacement letter-of-credit bank or could result in an acceleration of payments of up to $840 million in deferred income taxes if replacement banks cannot be obtained. The deferred taxes are currently recorded in the Company's consolidated financial statements. See Note 12, Variable Interest Entities, on pages 64 through 66, and Note 10, Income Taxes, on pages 59 through 61, in Item 8. Financial Statements and Supplementary Data for further information.
OUR PENSION AND HEALTH CARE COSTS ARE SUBJECT TO NUMEROUS FACTORS WHICH COULD CAUSE THESE COSTS TO CHANGE. We have defined benefit pension plans covering substantially all U.S. salaried employees hired prior to July 1, 2004 and substantially all hourly and union employees regardless of hire date. We provide retiree health care benefits to certain of our U.S. salaried and certain hourly employees. Our pension costs are dependent upon numerous factors resulting from actual plan experience and assumptions of future experience.
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Pension plan assets are primarily made up of equity and fixed income investments. Fluctuations in actual equity market returns, changes in general interest rates and changes in the number of retirees may result in increased pension costs in future periods. Likewise, changes in assumptions regarding current discount rates and expected rates of return on plan assets could increase pension costs. Health care reform under the Patient Protection and Affordable Care Act of 2010 could also increase costs with respect to medical coverage of the Company’s full-time employees. Significant changes in any of these factors may adversely impact our cash flows, financial condition and results of operations.
OUR PENSION PLANS ARE CURRENTLY UNDERFUNDED, AND OVER TIME WE MAY BE REQUIRED TO MAKE CASH PAYMENTS TO THE PLANS, REDUCING THE CASH AVAILABLE FOR OUR BUSINESS. We record a liability associated with our pension plans equal to the excess of the benefit obligation over the fair value of plan assets. The benefit liability recorded under the provisions of Accounting Standards Codification (ASC) 715, “Compensation – Retirement Benefits,” at December 31, 2015 was $3.6 billion. The amount and timing of future contributions will depend upon a number of factors, including the actual earnings and changes in values of plan assets and changes in interest rates. As described elsewhere in this Annual Report on Form 10-K, during the first half of 2016, former employees who are participants in our pension plan will be able to request early payment of their entire plan benefit in the form of a single lump sum payment. While all payments will be made from the plan's trust assets, the target population has a total liability of $3.0 billion. For further information, see Item 7. Management's Discussion and Analysis of Financial Condition and Results of Operations - Liquidity and Capital Resources on page 34.
CHANGES IN INTERNATIONAL CONDITIONS COULD ADVERSELY AFFECT OUR BUSINESS AND RESULTS OF OPERATIONS. Our operating results and business prospects could be substantially affected by risks related to the countries outside the United States in which we have manufacturing facilities or sell our products. Specifically, Russia, Brazil, Poland, India, and Turkey, where we have substantial manufacturing facilities, are countries that are exposed to economic and political instability in their respective regions of the world. Fluctuations in the value of local currency versus the U.S. dollar, downturns in economic activity, adverse tax consequences, nationalization or any change in social, political or labor conditions in any of these countries or regions could negatively affect our financial results. Trade protection measures in favor of local producers of competing products, including governmental subsidies, tax benefits and other measures giving local producers a competitive
advantage over International Paper, may also adversely impact our operating results and business prospects in these countries. In addition, our international operations are subject to regulation under U.S. law and other laws related to operations in foreign jurisdictions. For example, the Foreign Corrupt Practices Act prohibits U.S. companies and their representatives from offering, promising, authorizing or making payments to foreign officials for the purpose of obtaining or retaining business abroad. Failure to comply with domestic or foreign laws could result in various adverse consequences, including the imposition of civil or criminal sanctions and the prosecution of executives overseeing our international operations.
RISKS RELATING TO LEGAL PROCEEDINGS AND COMPLIANCE COSTS
WE ARE SUBJECT TO A WIDE VARIETY OF LAWS, REGULATIONS AND OTHER GOVERNMENT REQUIREMENTS THAT MAY CHANGE IN SIGNIFICANT WAYS, AND THE COST OF COMPLIANCE WITH SUCH REQUIREMENTS COULD IMPACT OUR BUSINESS AND RESULTS OF OPERATIONS. Our operations are subject to regulation under a wide variety of U.S. federal and state and non-U.S. laws, regulations and other government requirements -- including, among others, those relating to the environment, health and safety, labor and employment and health care. There can be no assurance that laws, regulations and government requirements will not be changed, applied or interpreted in ways that will require us to modify our operations and objectives or affect our returns on investments by restricting existing activities and products, subjecting them to escalating costs. For example, we have incurred, and expect that we will continue to incur, significant capital, operating and other expenditures complying with applicable environmental laws and regulations. There can be no assurance that future remediation requirements and compliance with existing and new laws and requirements, including with global climate change laws and regulations, Boiler MACT and NAAQSs, will not require significant expenditures, or that existing reserves for specific matters will be adequate to cover future costs. We could also incur substantial fines or sanctions, enforcement actions (including orders limiting our operations or requiring corrective measures), natural resource damages claims, cleanup and closure costs, and third-party claims for property damage and personal injury as a result of violations of, or liabilities under, environmental laws, regulations, codes and common law. The amount and timing of environmental expenditures is difficult to predict, and, in some cases, liability may be imposed without regard to contribution or to whether we knew of, or caused, the release of hazardous substances. As another example, we are subject to a number of labor
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and employment laws and regulations that could significantly increase our operating costs and reduce our operational flexibility.
RESULTS OF LEGAL PROCEEDINGS COULD HAVE A MATERIAL EFFECT ON OUR CONSOLIDATED FINANCIAL STATEMENTS. The costs and other effects of pending litigation against us cannot be determined with certainty. Although we do not believe that the outcome of any pending or threatened lawsuits or claims will have a material effect on our business or consolidated financial statements, there can be no assurance that the outcome of any lawsuit or claim will be as expected.
RISKS RELATING TO OUR OPERATIONS
MATERIAL DISRUPTIONS AT ONE OF OUR MANUFACTURING FACILITIES COULD NEGATIVELY IMPACT OUR FINANCIAL RESULTS. We operate our facilities in compliance with applicable rules and regulations and take measures to minimize the risks of disruption at our facilities. A material disruption at our corporate headquarters or one of our manufacturing facilities could prevent us from meeting customer demand, reduce our sales and/or negatively impact our financial condition. Any of our manufacturing facilities, or any of our machines within an otherwise operational facility, could cease operations unexpectedly due to a number of events, including:
•fires, floods, earthquakes, hurricanes or other catastrophes;
•the effect of a drought or reduced rainfall on its water supply;
•the effect of other severe weather conditions on equipment and facilities;
•terrorism or threats of terrorism;
•domestic and international laws and regulations applicable to our Company and our business partners, including joint venture partners, around the world;
•unscheduled maintenance outages;
•prolonged power failures;
•an equipment failure;
•a chemical spill or release;
•explosion of a boiler;
•damage or disruptions caused by third parties operating on or adjacent to one of our manufacturing facilities;
•disruptions in the transportation infrastructure, including roads, bridges, railroad tracks and tunnels;
•widespread outbreak of an illness or any other communicable disease, or any other public health crisis;
•labor difficulties; and
•other operational problems.
Any such downtime or facility damage could prevent us from meeting customer demand for our products and/or require us to make unplanned expenditures. If one of these machines or facilities were to incur significant downtime, our ability to meet our production targets and satisfy customer requirements could be impaired, resulting in lower sales and having a negative effect on our business and financial results.
WE ARE SUBJECT TO INFORMATION TECHNOLOGY RISKS RELATED TO BREACHES OF SECURITY PERTAINING TO SENSITIVE COMPANY, CUSTOMER, EMPLOYEE AND VENDOR INFORMATION AS WELL AS BREACHES IN THE TECHNOLOGY USED TO MANAGE OPERATIONS AND OTHER BUSINESS PROCESSES. Our business operations rely upon secure information technology systems for data capture, processing, storage and reporting. Despite careful security and controls design, implementation, updating and independent third party verification, our information technology systems, and those of our third party providers, could become subject to employee error or malfeasance, cyber attacks, or natural disasters. Network, system, application and data breaches could result in operational disruptions or information misappropriation including, but not limited to, interruption to systems availability, denial of access to and misuse of applications required by our customers to conduct business with International Paper. Access to internal applications required to plan our operations, source materials, manufacture and ship finished goods and account for orders could be denied or misused. Theft of intellectual property or trade secrets, and inappropriate disclosure of confidential company, employee, customer or vendor information, could stem from such incidents. Any of these operational disruptions and/or misappropriation of information could result in lost sales, business delays, negative publicity and could have a material effect on our business.
CERTAIN OPERATIONS ARE CONDUCTED BY JOINT VENTURES THAT WE CANNOT OPERATE SOLELY FOR OUR BENEFIT. Certain operations in Russia are carried on by a joint venture, Ilim. In joint ventures, we share ownership and management of a company with one or more parties who may or may not have the same goals, strategies, priorities or resources
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as we do. In general, joint ventures are intended to be operated for the benefit of all co-owners, rather than for our exclusive benefit. Operating a business as a joint venture often requires additional organizational formalities as well as time-consuming procedures for sharing information and making decisions. In joint ventures, we are required to pay more attention to our relationship with our co-owners as well as with the joint venture, and if a co-owner changes, our relationship may be adversely affected. In addition, the benefits from a successful joint venture are shared among the co-owners, so that we do not receive all the benefits from our successful joint ventures.
WE MAY NOT ACHIEVE THE EXPECTED BENEFITS FROM STRATEGIC ACQUISITIONS, JOINT VENTURES, DIVESTITURES AND OTHER CORPORATE TRANSACTIONS. Our strategy for long-term growth, productivity and profitability depends, in part, on our ability to accomplish prudent strategic acquisitions, joint ventures, divestitures and other corporate transactions and to realize the benefits we expect from such transactions, and we are subject to the risk that we may not achieve the expected benefits. Among the benefits we expect from potential as well as completed acquisitions and joint ventures are synergies, cost savings, growth opportunities or access to new markets (or a combination thereof), and in the case of divestitures, the realization of proceeds from the sale of businesses and assets to purchasers placing higher strategic value on such businesses and assets than does International Paper.
ITEM 1B. UNRESOLVED STAFF COMMENTS
None.
ITEM 2. PROPERTIES
FORESTLANDS
As of December 31, 2015, the Company owned or managed approximately 335,000 acres of forestlands in Brazil, and had, through licenses and forest management agreements, harvesting rights on government-owned forestlands in Russia. All owned lands in Brazil are independently third-party certified for sustainable forestry under the Brazilian National Forest Certification Program (CERFLOR) and the Forest Stewardship Council (FSC).
MILLS AND PLANTS
A listing of our production facilities by segment, the vast majority of which we own, can be found in Appendix I hereto, which is incorporated herein by reference.
The Company’s facilities are in good operating condition and are suited for the purposes for which they are presently being used. We continue to study the economics of modernization or adopting other alternatives for higher cost facilities.
CAPITAL INVESTMENTS AND DISPOSITIONS
Given the size, scope and complexity of our business interests, we continually examine and evaluate a wide variety of business opportunities and planning alternatives, including possible acquisitions and sales or other dispositions of properties. You can find a discussion about the level of planned capital investments for 2016 on page 33 and 34, and dispositions and restructuring activities as of December 31, 2015, on pages 20 through 24 of Item 7. Management’s Discussion and Analysis of Financial Condition and Results of Operations, and on page 54 and pages 56 and 57 of Item 8. Financial Statements and Supplementary Data.
ITEM 3. LEGAL PROCEEDINGS
Information concerning the Company’s legal proceedings is set forth in Note 11 Commitments and Contingencies on pages 61 through 64 of Item 8. Financial Statements and Supplementary Data.
ITEM 4. MINE SAFETY DISCLOSURES
Not applicable.
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PART II.
ITEM 5. MARKET FOR REGISTRANT’S COMMON EQUITY, RELATED STOCKHOLDER MATTERS AND ISSUER PURCHASES OF EQUITY SECURITIES
Dividend per share data on the Company’s common stock and the high and low sales prices for the Company’s common stock for each of the four quarters in 2015 and 2014 are set forth on page 83 of Item 8. Financial Statements and Supplementary Data. As of
the filing of this Annual Report on Form 10-K, the Company’s common shares are traded on the New York Stock Exchange. As of February 19, 2016, there were approximately 12,705 record holders of common stock of the Company.
The table below presents information regarding the Company’s purchase of its equity securities for the time periods presented.
PURCHASES OF EQUITY SECURITIES BY THE ISSUER AND AFFILIATED PURCHASERS.
Period Total Number of Shares Purchased (a)Average Price Paid per Share Total Number of Shares (or Units) Purchased as Part of Publicly Announced Programs Maximum Number (or Approximate Dollar Value) of Shares that May Yet Be Purchased Under the Plans or Programs (in billions)
October 1, 2015 - October 31, 2015—$——$1.13
November 1, 2015 - November 30, 2015 2,028,004 41.05 2,027,636 1.05
December 1, 2015 - December 31, 2015 404,562 41.80 402,163 1.03
Total 2,432,566
(a)2,767 shares were acquired from employees from share withholdings to pay income taxes under the Company’s restricted stock programs. The remainder were purchased under a share repurchase program that was approved by our Board of Directors and announced on July 8, 2014. Through this program, which does not have an expiration date, we were authorized to purchase, in open market transactions (including block trades), privately negotiated transactions or otherwise, up to $1.5 billion of shares of our common stock. As of February 19, 2016, approximately $933 million of shares of our common stock remained authorized for purchase under our share repurchase programs.
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PERFORMANCE GRAPH
The performance graph shall not be deemed to be “soliciting material” or to be “filed” with the Commission or subject to Regulation 14A or 14C, or to the liabilities of Section 18 of the Exchange Act of 1934, as amended.
The following graph compares a $100 investment in Company stock on December 31, 2010 with a $100
investment in our Return on Invested Capital (ROIC) Peer Group and the S&P 500 also made at market close on December 31, 2010. The graph portrays total return, 2010–2015, assuming reinvestment of dividends.
Note 1: The companies included in the ROIC Peer Group are Domtar Inc., Fibria Celulose S.A., Klabin S.A., Metsa Board Corporation, Mondi Group, Packaging Corporation of America, Smurfit Kappa Group, Stora Enso Group, and UPM-Kymmene Corp. MeadWestvaco Corp. and Rock-Tenn Company are included in the ROIC Peer Group results through 2014.
Note 2: Returns are calculated in $USD.
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ITEM 6. SELECTED FINANCIAL DATA
FIVE-YEAR FINANCIAL SUMMARY (a)
Dollar amounts in millions, except per share amounts and stock prices 2015 2014 2013 2012 2011
RESULTS OF OPERATIONS
Net sales$22,365$23,617$23,483$21,852$19,464
Costs and expenses, excluding interest 20,544 22,138 21,643 20,214 17,528
Earnings (loss) from continuing operations before income taxes and equity earnings 1,266(b)872(d)1,228(g)967(j)1,395(m)
Equity earnings (loss), net of taxes 117(200)(39)61 140
Discontinued operations, net of taxes—(13)(e)(309)(h)77(k)82(n)
Net earnings (loss)917(b-c)536(d-f)1,378(g-i)799(j-l)1,336(m-o)
Noncontrolling interests, net of taxes(21)(19)(17)5 14
Net earnings (loss) attributable to International Paper Company 938(b-c)555(d-f)1,395(g-i)794(j-l)1,322(m-o)
FINANCIAL POSITION
Current assets less current liabilities$2,553$3,050$3,898$3,907$5,718
Plants, properties and equipment, net 11,980 12,728 13,672 13,949 11,817
Forestlands 366 507 557 622 660
Total assets 30,587 28,684 31,528 32,153 27,018
Notes payable and current maturities of long-term debt 426 742 661 444 719
Long-term debt 8,900 8,631 8,827 9,696 9,189
Total shareholders’ equity 3,884 5,115 8,105 6,304 6,645
BASIC EARNINGS PER SHARE ATTRIBUTABLE TO INTERNATIONAL PAPER COMPANY COMMON SHAREHOLDERS
Earnings (loss) from continuing operations$2.25$1.33$3.85$1.65$2.87
Discontinued operations—(0.03)(0.70)0.17 0.19
Net earnings (loss)2.25 1.30 3.15 1.82 3.06
DILUTED EARNINGS PER SHARE ATTRIBUTABLE TO INTERNATIONAL PAPER COMPANY COMMON SHAREHOLDERS
Earnings (loss) from continuing operations$2.23$1.31$3.80$1.63$2.84
Discontinued operations—(0.02)(0.69)0.17 0.19
Net earnings (loss)2.23 1.29 3.11 1.80 3.03
Cash dividends 1.640 1.450 1.250 1.088 0.975
Total shareholders’ equity 9.43 12.18 18.57 14.33 15.21
COMMON STOCK PRICES
High$57.90$55.73$50.33$39.88$33.01
Low 36.76 44.24 39.47 27.29 21.55
Year-end 37.70 53.58 49.03 39.84 29.60
FINANCIAL RATIOS
Current ratio 1.7 1.6 1.8 1.8 2.2
Total debt to capital ratio 0.71 0.65 0.54 0.62 0.60
Return on shareholders’ equity 20.0%(b-c)7.7%(d-f)20.2%(g-i)11.6%(j-l)17.9%(m-o)
CAPITAL EXPENDITURES$1,487$1,366$1,198$1,383$1,159
NUMBER OF EMPLOYEES 56,000 58,000 64,000 65,000 56,000
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FINANCIAL GLOSSARY
Current ratio—
current assets divided by current liabilities.
Total debt to capital ratio—
long-term debt plus notes payable and current maturities of long-term debt divided by long-term debt, notes payable and current maturities of long-term debt and total shareholders’ equity.
Return on shareholders’ equity—
net earnings attributable to International Paper Company divided by average shareholders’ equity (computed monthly).
FOOTNOTES TO FIVE-YEAR FINANCIAL SUMMARY
(a)All periods presented have been restated to reflect the xpedx business and the Temple-Inland Building Products business as discontinued operations, if applicable.
2015:
(b) Includes the following pre-tax charges (gains):
In millions 2015
Riegelwood mill conversion costs, net of proceeds from sale of Carolina Coated Bristols brand$8
Timber monetization restructuring 16
Early debt extinguishment costs 207
IP-Sun JV impairment 174
Brazil Packaging impairment 137
Legal liability reserve adjustment 15
Refund of state tax credits(4)
Other items 6
Total$559
(c) Includes the following tax expenses (benefits):
In millions 2015
IP-Sun JV impairment$(67)
Cash pension contribution 23
Other items 7
Total$(37)
2014:
(d) Includes the following pre-tax charges (gains):
In millions 2014
Temple-Inland integration$16
Courtland mill shutdown 554
Early debt extinguishment costs 276
India legal contingency resolution(20)
Multi-employer pension plan withdrawal liability 35
Foreign tax amnesty program 32
Asia Industrial Packaging goodwill impairment 100
Loss on sale by investee and impairment of investment 47
Other items 12
Total$1,052
(e) Includes the after-tax operating earnings of the xpedx business prior to the spin-off and the following after-tax charges (gains):
In millions 2014
xpedx spinoff$16
Building Products divestiture 9
xpedx restructuring(1)
Total$24
(f) Includes the following tax expenses (benefits):
In millions 2014
State legislative tax change$10
Internal restructuring(90)
Other items(1)
Total$(81)
2013:
(g) Includes the following pre-tax charges (gains):
In millions 2013
Temple-Inland integration$62
Courtland mill shutdown 118
Early debt extinguishment costs 25
Insurance reimbursement related to legal settlement(30)
Shut down of paper machine at Augusta mill 45
India Papers tradename and goodwill impairment 127
Fair value adjustment of company airplanes 9
Cass Lake environmental reserve 6
Bargain purchase adjustment - Turkey(13)
Other items(5)
Total$344
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(h) Includes the after-tax operating earnings of the xpedx business for the full year and the Temple-Inland Building Products business through the date of sale in July 2013. Also includes the following after-tax charges (gains):
In millions 2013
xpedx spinoff$14
xpedx goodwill impairment 366
Building Products divestiture 19
xpedx restructuring 19
Total$418
(i) Includes the following tax expenses (benefits):
In millions 2013
Settlement of U.S. federal tax audits$(744)
Income tax reserve release(31)
Other items 1
Total$(774)
2012:
(j) Includes the following pre-tax charges (gains):
In millions 2012
Temple-Inland integration$164
Early debt extinguishment costs 48
EMEA packaging business restructuring 17
Temple-Inland inventory fair value adjustment 20
Hueneme mill long-lived asset fair value adjustment 62
Containerboard mill divestitures 29
Total$340
(k) Includes the after-tax operating earnings of the xpedx business and the Temple-Inland Building Products business for the full year. Also includes the following after-tax charges (gains):
In millions 2012
Building Products divestiture$9
xpedx restructuring 28
Total$37
(l) Includes the following tax expenses (benefits):
In millions 2012
Internal restructuring$14
Deferred tax asset adjustment related to Medicare Part D reimbursement 5
Total$19
2011:
(m) Includes the following pre-tax charges (gains):
In millions 2011
Temple-Inland acquisition costs$20
Early debt extinguishment costs 32
APPM acquisition costs 18
Reversal of environmental and other reserves related to repurposing at Franklin mill(24)
Cass Lake environmental reserve 27
North American Shorewood business fixed asset impairment 129
Shorewood business impairment 78
Inverurie, Scotland mill asset impairment 11
Total$291
(n) Includes the following after-tax charges (gains):
In millions 2011
Gain for earnout provision - sale of Kraft Papers business$(30)
Tax benefit - Brazilian Coated Papers business sale(15)
Interest income on tax benefit - Brazilian Coated Papers business sale(4)
xpedx restructuring 34
Total$(15)
(o) Includes the following tax expenses (benefits):
In millions 2011
Internal restructuring$24
Tax benefit related to reduction of the carrying value of the Shorewood business and write-off of the associated deferred tax liability(222)
Tax expense for APPM acquisitions costs 9
Release of deferred tax asset valuation allowance 13
Other items 2
Total$(174)
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ITEM 7. MANAGEMENT’S DISCUSSION AND ANALYSIS OF FINANCIAL CONDITION AND RESULTS OF OPERATIONS
EXECUTIVE SUMMARY
Operating Earnings (a non-GAAP measure) is defined as net earnings from continuing operations (a GAAP measure) excluding special items and non-operating pension expense. International Paper generated Operating Earnings per diluted share attributable to common shareholders of $3.65 in 2015, compared with $3.00 in 2014, and $3.06 in 2013. Diluted earnings (loss) per share attributable to common shareholders were $2.23 in 2015, compared with $1.29 in 2014 and $3.11 in 2013.
International Paper delivered solid results during 2015 driven by strong margins and earnings in our North American Industrial Packaging business and record performance from the Ilim joint venture. We generated $1.8 billion of free cash flow which enabled the Company to return cash to our shareholders in the form of approximately $500 million in share repurchases and a 10% increase in the quarterly dividend beginning with the 2015 fourth quarter dividend payment. During 2015, we successfully completed the restructuring of the 2006 timber monetization to achieve our objectives of reducing risk and preserving financial flexibility, while maintaining the deferral of $1.4 billion of deferred income taxes. Finally, with respect to our balanced use of cash, we completed a $2 billion bond issue and related tender offer along with making a $750 million voluntary pension contribution.
Our 2015 results reflect the benefits of favorable input costs offset by price and mix declines across our North American businesses. Volumes were generally flat compared to 2014 except for lower volumes in our North American Industrial Packaging business due to lower containerboard export tons. Input costs decreased versus 2014 largely due to lower energy, chemicals and freight costs. Price declined relative to 2014 driven mainly by lower pricing in our North American Industrial Packaging and Printing Papers and Pulp businesses. Our Ilim joint venture generated record results in 2015 driven by improved operations and increased margins. The positive results were partially offset by the unfavorable impact of non-cash foreign currency movements associated with Ilim’s US dollar denominated debt. Finally, during 2015 we completed the divestiture of our interest in the IP-Sun joint venture, generating $23 million in cash proceeds and removing approximately $400 million of debt from our balance sheet upon completion of the deal.
Overall, 2015 reflects solid performance in what continues to be a challenging economic environment. We once again generated returns in excess of our cost of capital while returning cash to our shareholders in the
form of increased dividends and share repurchases. Our focus on maximizing free cash flow generation and deploying capital in a way that creates additional value for our shareholders has positioned us for another successful year in 2016.
Looking ahead to the 2016 first quarter, we expect seasonally lower volumes in our North American Industrial Packaging business, with some offset from higher export volume which carries a lower margin. Additionally, we expect seasonally lower volumes in our Brazilian Printing Papers business as the fourth quarter historically represents the strongest volume quarter for this business. Pricing is expected to be lower for our North American Printing Papers and Pulp business, primarily driven by lower pulp prices. Additionally, pricing is expected to be lower in our North American Industrial Packaging business due to lower export pricing and price index changes. We expect price improvements in our EMEA Printing Papers business, including Russia, and Brazilian Printing Papers business following announced price increases although these will be largely offset by inflationary cost pressures. We expect operating performance to be in line with the 2015 fourth quarter with some modest improvement in our North American Industrial Packaging business. Planned maintenance downtime costs should increase, primarily driven by outages in our North American Industrial Packaging and Printing Papers businesses, including costs associated with the Riegelwood mill conversion. Equity earnings from our Ilim joint venture are expected to benefit from strong operations offset by softwood pulp price pressure and normal seasonality. Additionally, we expect Ilim’s earnings to be impacted by the absence of the positive impact from foreign currency movements driven by Ilim’s U.S. dollar denominated debt as we assume no change in foreign currency rates in our outlook.
For the 2016 full year, we continue to face an uncertain macroeconomic environment but believe we are well positioned to deal with whatever the market brings. We will continue to improve our North American Industrial Packaging business by further realizing optimization opportunities during 2016. We expect to complete the Riegelwood mill conversion during first half of 2016 and be fully ramped by the 2016 fourth quarter, initially producing softwood market pulp. Additionally, we will continue executing against our plan to drive profitable growth following the recent expansion within the Foodservice business as well as optimizing commercial opportunities and mix within the North American Printing Papers portfolio. Finally, we will remain focused on maximizing free cash flow generation and deploying that capital in a way that creates additional value for our shareholders.
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Free cash flow (a non-GAAP measure) of $1.8 billion generated in 2015 was lower than the $2.1 billion generated in 2014 and even with the $1.8 billion generated in 2013 (see reconciliation on page 30).
Operating Earnings per share attributable to common shareholders of $0.87 in the 2015 fourth quarter were lower than the $0.97 in the 2015 third quarter, but higher than the $0.53 in the 2014 fourth quarter. Diluted earnings (loss) per share attributable to common shareholders were $0.43 in the 2015 fourth quarter, compared with $0.53 in the 2015 third quarter and $0.32 in the 2014 fourth quarter.
Free cash flow of $501 million generated in the 2015 fourth quarter was lower than the $512 million generated in the 2015 third quarter and the $739 million generated in the 2014 fourth quarter (see reconciliation on page 30).
Operating Earnings and Operating Earnings Per Share are non-GAAP measures. Diluted earnings (loss) per share attributable to International Paper Company common shareholders is the most directly comparable GAAP measure. The Company calculates Operating Earnings by excluding the after-tax effect of items considered by management to be unusual from the earnings reported under GAAP, non-operating pension expense and discontinued operations. Management uses this measure to focus on on-going operations, and believes that it is useful to investors because it enables them to perform meaningful comparisons of past and present operating results. The Company believes that using this information, along with the most directly comparable GAAP measure, provides for a more complete analysis of the results of operations. The following are reconciliations of Operating Earnings per share attributable to International Paper Company common shareholders to diluted earnings (loss) per share attributable to International Paper Company common shareholders.
2015 2014 2013
Operating Earnings (Loss) Per Share Attributable to Shareholders$3.65$3.00$3.06
Non-operating pension expense(0.38)(0.30)(0.44)
Special items(1.04)(1.39)1.18
Diluted Earnings (Loss) Per Share from Continuing Operations 2.23 1.31 3.80
Discontinued operations—(0.02)(0.69)
Diluted Earnings (Loss) Per Share Attributable to Shareholders$2.23$1.29$3.11
Three Months Ended December 31, 2015 Three Months Ended September 30, 2015 Three Months Ended December 31, 2014
Operating Earnings (Loss) Per Share Attributable to Shareholders$0.87$0.97$0.53
Non-operating pension expense(0.09)(0.11)(0.07)
Special items(0.35)(0.33)(0.12)
Diluted Earnings (Loss) Per Share from Continuing Operations 0.43 0.53 0.34
Discontinued operations——(0.02)
Diluted Earnings (Loss) Per Share Attributable to Shareholders$0.43$0.53$0.32
Results of Operations
Industry segment operating profits are used by International Paper’s management to measure the earnings performance of its businesses. Management believes that this measure allows a better understanding of trends in costs, operating efficiencies, prices and volumes. Industry segment operating profits are defined as earnings before taxes, equity earnings, noncontrolling interests, interest expense, corporate items and corporate special items. Industry segment operating profits are defined by the Securities and Exchange Commission as a non-GAAP financial measure, and are not GAAP alternatives to net income or any other operating measure prescribed by accounting principles generally accepted in the United States.
International Paper operates in three segments: Industrial Packaging, Printing Papers and Consumer Packaging.
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The following table presents a reconciliation of net earnings (loss) attributable to International Paper Company to its total industry segment operating profit:
In millions 2015 2014 2013
Net Earnings (Loss) Attributable to International Paper Company$938$555$1,395
Deduct – Discontinued operations:
(Earnings) from operations—(11)(109)
Special items (gain) loss—24 418
Earnings (Loss) From Continuing Operations Attributable to International Paper Company 938 568 1,704
Add back (deduct):
Income tax provision 466 123(498)
Equity (earnings) loss, net of taxes(117)200 39
Net earnings (loss) attributable to noncontrolling interests(21)(19)(17)
Earnings (Loss) From Continuing Operations Before Income Taxes and Equity Earnings 1,266 872 1,228
Interest expense, net 555 601 612
Noncontrolling interests / equity earnings included in operations 8 2(1)
Corporate items 36 51 61
Special items:
Restructuring and other charges 238 282 10
Net losses (gains) on sales and impairments of businesses—38—
Non-Operating Pension Expense 258 212 323
$2,361$2,058$2,233
Industry Segment Operating Profit
Industrial Packaging$1,853$1,896$1,801
Printing Papers 533(16)271
Consumer Packaging(25)178 161
Total Industry Segment Operating Profit$2,361$2,058$2,233
Industry segment operating profits in 2015 included a net loss from special items of $321 million compared with $732 million in 2014 and $336 million in 2013.Operationally, compared with 2014, the benefit from lower input costs ($232 million) was offset by lower average sales price realizations and mix ($226 million), lower sales volumes ($38 million), higher operating costs ($16 million), higher maintenance outage costs ($37 million) and higher other costs ($23 million).
The principal changes in operating profit by segment were as follows:
•Industrial Packaging’s profits of $1.9 billion were $43 million lower than in 2014 as the benefit of lower input costs was offset by lower average sales price realizations and mix, lower sales volumes, higher operating costs and higher maintenance outage costs. In addition, 2015 operating profits included a goodwill and trade name impairment charge of $137 million related to our Brazil Packaging business. Operating profits in 2014 included $16 million of costs associated with the integration of Temple-Inland, a goodwill impairment charge of $100 million related to our Asia Industrial Packaging business, a charge of $35 million for costs associated with a multi-employer pension plan withdrawal liability and a net charge of $7 million for other items.
•Printing Papers’ profits of $533 million represented a $549 million increase in operating profits from 2014. The benefits from lower input costs, lower costs associated with the closure of our Courtland, Alabama mill and lower foreign exchange impact were offset by lower average sales price realizations and mix, lower sales volumes, higher operating costs and higher maintenance outage costs. The 2014 operating loss included a special items charge of $554 million for costs associated with the shutdown of our Courtland, Alabama mill, a gain of $20 million for the resolution of a legal contingency in India and a charge of $32 million for costs associated with a foreign tax amnesty program.
•Consumer Packaging’s operating loss of $25 million represented a $203 million reduction in operating profits from 2014. The benefits from higher sales volumes, lower planned maintenance downtime costs and lower input costs were offset by lower average sales price realizations and mix, higher operating costs, and higher foreign exchange and other expenses. In addition, 2015 operating profits included an asset impairment charge of $174 million
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related to the sale of our 55% equity share of the IP-Sun JV in Asia, a net cost of $8 million related to costs to convert our Riegelwood mill to 100% pulp production, net of proceeds from the sale of the Carolina Coated Bristols brand, and $2 million of sheet plant closure costs. Operating profits in 2014 included $8 million of sheet plant closure costs.
Corporate items, net, of $36 million of expense in 2015 were lower than the $51 million of expense in 2014 due to the absence of a one-time non-cash foreign exchange charge related to the administrative restructuring of some international entities in 2014. The decrease in 2014 from the expense of $61 million in 2013 is due to lower pension costs partially offset by the one-time non-cash foreign exchange charge.
Corporate special items, including restructuring and other items and net losses on sales and impairments of businesses were a loss of $238 million in 2015 compared with a loss of $320 million in 2014 and a loss of $4 million in 2013. The loss in 2015 is due to debt premium costs, costs associated with the restructure of our timber monetization and a legal liability reserve adjustment. The loss in 2014 is primarily due to debt extinguishment costs and a loss on the sale of a business by ASG, which was formerly referred to as AGI-Shorewood and in which we hold an investment, and the subsequent partial impairment of our ASG investment.
Interest expense, net, was $555 million in 2015 compared with $607 million ($601 million excluding special items net interest expense reported in the Printing Papers business segment) in 2014 and $612 million in 2013. The decrease in 2015 compared with 2014 is due to lower average interest rates. The decrease in 2014 compared with 2013 also reflects lower average interest rates.
A net income tax provision of $466 million was recorded for 2015, including a tax benefit of $62 million related to internal restructurings, an expense of $23 million for the tax impact of the 2015 cash pension contribution of $750 million and a tax expense of $2 million for other items. The 2014 income tax provision of $123 million includes a tax benefit of $90 million related to internal restructurings and a net tax expense of $9 million for other items. The 2013 income tax benefit of $498 million includes a tax benefit of $770 million associated with the settlement of tax audits and a net tax benefit of $4 million for other items.
Discontinued Operations
2014: On July 1, 2014, International Paper completed the spinoff of its distribution business, xpedx, which subsequently merged with Unisource Worldwide, Inc., with the combined companies now operating as Veritiv Corporation (Veritiv). The xpedx business had historically represented the Company's Distribution reportable segment.
The spinoff was accomplished by the contribution of the xpedx business to Veritiv and the distribution of 8,160,000 shares of Veritiv common stock on a pro-rata basis to International Paper shareholders. International Paper received payments of approximately $411 million, financed with new debt in Veritiv's capital structure.
2013: On April 1, 2013, the Company finalized the sale of Temple-Inland's 50% interest in Del-Tin Fiber L.L.C. to joint venture partner Deltic Timber Corporation for $20 million in assumed liabilities and cash.
On July 19, 2013 the Company finalized the sale of its Temple-Inland Building Products division to Georgia-Pacific Building Products, LLC for approximately $726 million in cash.
Liquidity and Capital Resources
For the year ended December 31, 2015, International Paper generated $2.6 billion of cash flow from operations compared with $3.1 billion in 2014 and $3.0 billion in 2013. Cash flow from operations included $750 million, $353 and $31 million of cash pension contributions in 2015, 2014 and 2013, respectively. Capital spending for 2015 totaled $1.5 billion, or 115% of depreciation and amortization expense. Net decreases in debt totaled $74 million. Our liquidity position remains strong, supported by approximately $2.1 billion of credit facilities that we believe are adequate to meet future liquidity requirements. Maintaining an investment-grade credit rating for our long-term debt continues to be an important element in our overall financial strategy.
We expect to generate strong free cash flow again in 2016 and will continue our balanced use of cash through investments in capital projects, the reduction of total debt, including the Company’s unfunded pension obligation, returning value to shareholders and strengthening our businesses through strategic acquisitions, as appropriate.
Capital spending for 2016 is targeted at $1.3 billion, or about 100% of depreciation and amortization.
Legal
See Note 11 Commitments and Contingent Liabilities on pages 61 through 64 of Item 8. Financial Statements and Supplementary Data for a discussion of legal matters.
CORPORATE OVERVIEW
While the operating results for International Paper’s various business segments are driven by a number of business-specific factors, changes in International Paper’s operating results are closely tied to changes in general economic conditions in North America, Europe, Russia, Latin America, Asia, Africa and the Middle East. Factors that impact the demand for our products include
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industrial non-durable goods production, consumer spending, commercial printing and advertising activity, white-collar employment levels, and movements in currency exchange rates.
Product prices are affected by general economic trends, inventory levels, currency exchange rate movements and worldwide capacity utilization. In addition to these revenue-related factors, net earnings are impacted by various cost drivers, the more significant of which include changes in raw material costs, principally wood, recycled fiber and chemical costs; energy costs; freight costs; salary and benefits costs, including pensions; and manufacturing conversion costs.
The following is a discussion of International Paper’s results of operations for the year ended December 31, 2015, and the major factors affecting these results compared to 2014 and 2013.
RESULTS OF OPERATIONS
For the year ended December 31, 2015, International Paper reported net sales of $22.4 billion, compared with $23.6 billion in 2014 and $23.5 billion in 2013.International net sales (including U.S. exports) totaled $7.8 billion or 35% of total sales in 2015. This compares with international net sales of $9.3 billion in 2014 and $9.5 billion in 2013.
Full year 2015 net earnings attributable to International Paper Company totaled $938 million ($2.23 per share), compared with net earnings of $555 million ($1.29 per share) in 2014 and $1.4 billion ($3.11 per share) in 2013. Amounts in all periods include the results of discontinued operations.
Earnings from continuing operations attributable to International Paper Company after taxes in 2015 were $938 million, including $439 million of net special items charges and $157 million of non-operating pension expense compared with $568 million, including $599 million of net special items charges and $129 million of non-operating pension expense in 2014, and $1.7 billion, including $528 million of net special items gains and $197 million of non-operating pension expense in 2013. Compared with 2014, the benefits from lower input costs, lower corporate and other costs and lower interest expense were offset by lower average sales price realizations and mix, lower sales volumes, higher operating costs, higher maintenance outage costs, and higher tax expense. In addition, 2015 results included higher equity earnings, net of taxes, relating to the Company’s investment in Ilim Holdings, SA.
See Industry Segment Results on pages 25 through 30 for a discussion of the impact of these factors by segment.
Discontinued Operations
2014:
In 2014, $24 million of net income adjustments were recorded relating to discontinued businesses, including $16 million of costs associated with the spin-off of the xpedx business and $9 million of costs associated with the divestiture of the Temple-Inland Building Products business. Also included are the operating earnings of the xpedx business prior to the spin-off on July 1, 2014.
2013:
In 2013, $418 million of net income adjustments were recorded relating to discontinued businesses, including goodwill impairment charges of $366 million associated with the xpedx business, $19 million for costs associated with the restructuring of the xpedx business, $14 million for costs associated with the spin-off of the xpedx business and $19 million for costs associated with the sale of the Temple-Inland Building Products business. Also included are the operating profits for the xpedx business for the full year and for the Temple-Inland Building Products business through the date of sale of July 19, 2013.
Income Taxes
A net income tax provision of $466 million was recorded for 2015, including a tax benefit of $62 million related to internal restructurings, a tax expense of $23 million for the tax impact of the 2015 cash pension contribution of $750 million and a $2 million tax expense for other items. Excluding these items, an $83 million net tax benefit for other special items and a $101 million tax benefit related to non-operating pension expense, the tax provision was $687 million, or 33% of pre-tax earnings before equity earnings.
A net income tax provision of $123 million was recorded for 2014 including a tax benefit of $90 million related to internal restructurings and a net $9 million tax expense for other items. Excluding these items, a $372 million
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net tax benefit for other special items and a $83 million tax benefit related to non-operating pension expense, the tax provision was $659 million, or 31% of pre-tax earnings before equity earnings.
A net income tax benefit of $498 million was recorded for 2013, including a tax benefit of $770 million related to the settlement of tax audits and a net benefit of $4 million for other items. Excluding these items, a $95 million net tax benefit for other special items and a $126 million tax benefit related to non-operating pension expense, the tax provision was $497 million, or 26% of pre-tax earnings before equity earnings.
Equity Earnings, Net of Taxes
Equity earnings, net of taxes in 2015, 2014 and 2013 consisted principally of the Company’s share of earnings from its 50% investment in Ilim Holding S.A. in Russia (see page 30).
Corporate Items and Interest Expense
Corporate items totaled $36 million of expense for the year ended December 31, 2015 compared with $51 million in 2014 and $61 million in 2013. The decrease in 2015 from 2014 reflects the absence of a one-time non-cash foreign exchange charge related to the administrative restructuring of some international entities that occurred in 2014. The decrease in 2014 from 2013 reflects lower pension expenses partially offset by a one-time non-cash foreign exchange charge related to the administrative restructuring of some international entities.
Net corporate interest expense totaled $555 million in 2015, $601 million in 2014 and $612 million in 2013.The decrease in 2015 compared with 2014 reflects lower average interest rates. The decrease in 2014 compared with 2013 also reflects lower average interest rates.
Net earnings attributable to noncontrolling interests totaled a loss of $21 million in 2015 compared with a loss of $19 million in 2014 and a loss of $17 million in 2013. The decrease in 2015 reflects the sale of our equity share of the IP-Sun JV and lower earnings for the Shandong IP & Sun Food Packaging Co., Ltd joint venture in China prior to its divestiture. The decrease in 2014 compared with 2013 reflects the impact of the acquisition of the remaining 25% share of Orsa IP from the joint venture partner.
Special Items
Restructuring and Other Charges
International Paper continually evaluates its operations for improvement opportunities targeted to (a)focus our portfolio on our core businesses, (b)rationalize and realign capacity to operate fewer facilities with the same
revenue capability and close high cost facilities, and (c)reduce costs. Annually, strategic operating plans are developed by each of our businesses. If it subsequently becomes apparent that a facility’s plan will not be achieved, a decision is then made to (a)invest additional capital to upgrade the facility, (b)shut down the facility and record the corresponding charge, or (c)evaluate the expected recovery of the carrying value of the facility to determine if an impairment of the assets have occurred. In recent years, this policy has led to the shutdown of a number of facilities and the recording of significant asset impairment charges and severance costs. It is possible that additional charges and costs will be incurred in future periods in our core businesses should such triggering events occur.
2015: During 2015, corporate restructuring and other charges totaling $242 million before taxes ($155 million after taxes) were recorded. These charges included:
•a $207 million charge before taxes ($133 million after taxes) for premiums paid on a cash tender offer on outstanding debt (see Note 13 Debt and Lines of Credit on pages 66 and 67 of Item 8. Financial Statements and Supplementary Data),
•a $16 million charge before taxes ($10 million after taxes) for costs related to the restructuring of our 2006 timber monetization,
•a $15 million charge before taxes ($9 million after taxes) for legal reserve adjustments, and
•a $4 million charge before taxes ($3 million after taxes) for other items.
In addition, restructuring and other charges totaling $10 million before taxes ($6 million after taxes) were recorded in the Consumer Packaging industry segment including:
•an $8 million net charge before taxes ($4 million after taxes) related to costs associated with the conversion of the Riegelwood, North Carolina facility to 100% pulp production, net of proceeds from the sale of the Carolina Coated Bristols brand, and
•a $2 million charge (before and after taxes) for other items.
2014:During 2014, corporate restructuring and other charges totaling $277 million before taxes ($169 million after taxes) were recorded. These charges included:
•a $276 million charge before taxes ($169 million after taxes) for costs related to the early extinguishment of debt (see Note 13 Debt and Lines of Credit on pages 66 and 67 of Item 8. Financial Statements and Supplementary Data)
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In addition, restructuring and other charges totaling $569 million before taxes ($349 million after taxes) were recorded in the Industrial Packaging, Printing Papers and Consumer Packaging industry segments including:
•a $554 million charge before taxes ($338 million after taxes) for costs related to the shutdown of the Courtland, Alabama mill, and
•a $15 million charge before taxes ($11 million after taxes) for other items.
2013:During 2013, corporate restructuring and other charges totaling a gain of $5 million before taxes ($3 million after taxes) were recorded. These charges included:
•a $25 million charge before taxes ($16 million after taxes) for costs related to the early extinguishment of debt (see Note 13 Debt and Lines of Credit on pages 66 and 67 of Item 8. Financial Statements and Supplementary Data), and
•a $30 million gain before taxes ($19 million after taxes) for insurance reimbursements related to the Guaranty Bank legal settlement.
In addition, restructuring and other charges totaling $161 million before taxes ($101 million after taxes) were recorded in the Industrial Packaging, Printing Papers and Consumer Packaging industry segments including:
•a $118 million charge before taxes ($72 million after taxes) for costs related to the shutdown of the Courtland, Alabama mill,
•a $45 million charge before taxes ($28 million after taxes) for costs related to the shutdown of a paper machine at the Augusta, Georgia mill, and
•a $2 million gain before taxes (loss of $1 million after taxes) for other items.
Impairments of Goodwill
In the fourth quarter of 2015, in conjunction with the annual testing of its reporting units for possible goodwill impairments, the Company calculated the estimated fair value of its Brazil Packaging business using the discounted future cash flows and determined that all of the goodwill in the business, totaling $137 million, should be written off. The decline in the fair value of the Brazil Packaging business and resulting impairment charge was due to the negative impacts on the cash flows of the business caused by the continued decline of the overall Brazilian economy.
In the fourth quarter of 2014, in conjunction with the annual testing of its reporting units for possible goodwill impairments, the Company calculated the estimated
fair value of its Asia Industrial Packaging business using expected discounted future cash flows and determined that due to a change in the strategic outlook, all of the goodwill of this business, totaling $100 million, should be written off. The decline in the fair value of the Asia Industrial Packaging business and resulting impairment charge was due to a change in the strategic outlook for the business.
In the fourth quarter of 2013, in conjunction with the annual testing of its reporting units for possible goodwill impairments, the Company calculated the estimated fair value of its India Papers business using the discounted future cash flows and determined that all of the goodwill of this business, totaling $112 million, should be written off. The decline in the fair value of the India Papers reporting unit and resulting impairment charge was due to a change in the strategic outlook for the India Papers operations.
Also in the fourth quarter of 2013, the Company calculated the estimated fair value of its xpedx business using the discounted future cash flows and wrote off all of the goodwill of its xpedx business, totaling $400 million, which has been included in Discontinued operations in the accompanying consolidated statement of operations. The decline in the fair value of the xpedx reporting unit and resulting impairment charge was due to a significant decline in earnings and a change in the strategic outlook for the xpedx operations.
Also during 2013, the Company recorded a pre-tax charge of $15 million ($7 million after taxes and noncontrolling interest) for the impairment of a trade name intangible asset related to our India Papers business.
Net Losses on Sales and Impairments of Businesses
Net losses on sales and impairments of businesses included in special items totaled a pre-tax loss of $174 million ($113 million after taxes) in 2015, a pre-tax loss of $38 million ($31 million after taxes) in 2014 and a pre-tax loss of $3 million ($1 million after taxes) in 2013. The principal components of these losses were:
2015: On October 13, 2015, the Company finalized the sale of its 55% interest in IP Asia Coated Paperboard (IP-Sun JV) business, within the Company's Consumer Packaging segment, to its Chinese coated board joint venture partner, Shandong Sun Holding Group Co., Ltd. for RMB 149 million (approximately USD $23 million). During the third quarter of 2015, a determination was made that the current book value of the asset group exceeded its estimated fair value of $23 million, which was the agreed upon selling price. The 2015 loss includes the pre-tax impairment charge of $174 million ($113 million after taxes). A pre-tax charge of $186
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million was recorded during the third quarter in the Company's Consumer Packaging segment to write down the long-lived assets of this business to their estimated fair value. In the fourth quarter of 2015, upon the sale and corresponding deconsolidation of IP-Sun JV from the Company's consolidated balance sheet, final adjustments were made resulting in a reduction of the impairment of $12 million. The amount of pre-tax losses related to noncontrolling interest of the IP-Sun JV included in the Company's consolidated statement of operations for the years ended December 31, 2015, 2014 and 2013 were $19 million, $12 million and $8 million, respectively. The amount of pre-tax losses related to the IP-Sun JV included in the Company's consolidated statement of operations for the years ended December 31, 2015, 2014 and 2013 were $226 million, $51 million and $41 million, respectively.
The net 2015 loss totaling $174 million related to the impairment of Sun-JV is included in Net (gains) losses on sales and impairments of businesses in the accompanying consolidated statement of operations.
2014: During 2014, the Company recorded net pre-tax charges of $47 million ($36 million after taxes) for a loss on the sale of a business by ASG (formerly referred to as AGI-Shorewood), in which we hold an investment and the subsequent partial impairment of our ASG investment, and a pre-tax gain of $9 million ($5 million after taxes) related to the sale of an investment.
2013: During 2013, the Company recorded net pre-tax charges of $3 million ($1 million after taxes) for adjustments related to the divestiture of three containerboard mills in 2012 and the sale of the Shorewood business.
Industry Segment Operating Profits
Industry segment operating profits of $2.4 billion in 2015 decreased from $2.1 billion in 2014. The benefit from lower input costs ($232 million) was offset by lower average sales price realizations and mix ($226 million), lower sales volumes ($38 million), higher operating costs ($16 million), higher maintenance outage costs ($37 million) and higher other costs ($23 million).
Special items were a $321 million net loss in 2015 compared with a net loss of $732 million in 2014.
Market-related downtime in 2015 increased to approximately 440,000 tons from approximately 281,000 tons in 2014.
DESCRIPTION OF INDUSTRY SEGMENTS
International Paper’s industry segments discussed below are consistent with the internal structure used to manage these businesses. All segments are differentiated on a common product, common customer basis consistent with the business segmentation generally used in the forest products industry.
Industrial Packaging
International Paper is the largest manufacturer of containerboard in the United States. Our U.S. production capacity is over 13 million tons annually. Our products include linerboard, medium, whitetop, recycled linerboard, recycled medium and saturating kraft. About 80% of our production is converted domestically into corrugated boxes and other packaging by our 165 U.S. container plants. Additionally, we recycle approximately one million tons of OCC and mixed and white paper through our 18 recycling plants. In EMEA, our operations include two recycled fiber containerboard mills in Morocco and Turkey and 26 container plants in France, Italy, Spain, Morocco and Turkey. In Brazil our operations include three containerboard mills and four box plants. In Asia, our operations include 16 container plants in China and additional container plants in Indonesia, Malaysia, Singapore, and Thailand. Our container plants are supported by regional design centers, which offer total packaging solutions and supply chain initiatives.
Printing Papers
International Paper is one of the world’s leading producers of printing and writing papers. Products in this segment include uncoated papers and pulp.
Uncoated Papers:This business produces papers for use in copiers, desktop and laser printers and digital imaging. End use applications include advertising and promotional materials such as brochures, pamphlets, greeting cards, books, annual reports and direct mail. Uncoated papers also produces a variety of grades that are converted by our customers into envelopes, tablets, business forms and file folders. Uncoated papers are sold under private label and International Paper brand names that include Hammermill, Springhill, Williamsburg, Postmark, Accent, Great White, Chamex, Ballet, Rey, Pol, and Svetocopy. The mills producing uncoated papers are located in the United States, France, Poland, Russia, Brazil and India. The mills have uncoated paper production capacity of approximately 4 million tons annually. Brazilian operations function through International Paper do Brasil, Ltda, which owns or manages approximately 335,000 acres of forestlands in Brazil.
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Pulp: Pulp is used in the manufacture of printing, writing and specialty papers, towel and tissue products and filtration products. Pulp is also converted into products such as diapers and sanitary napkins. Pulp products include fluff, and southern softwood pulp, as well as southern and birch hardwood paper pulps. These products are produced in the United States, France, Poland, Russia, and Brazil and are sold around the world. International Paper facilities have annual dried pulp capacity of about 1.8 million tons.
Consumer Packaging
International Paper is one of the world’s largest producers of solid bleached sulfate board with annual U.S. production capacity of about 1.2 million tons (reduced from about 1.6 million tons) after initiating the conversion of the Riegelwood Mill to 100% pulp production in late December of 2015. Our coated paperboard business produces high quality coated paperboard for a variety of packaging and foodservice end uses. Our Everest®, Fortress®, and Starcote® brands are used in packaging applications for everyday products such as food, cosmetics, pharmaceuticals and tobacco products. The Carolina® brand, which was sold to MeadWestvaco Corporation in April 2015, was used in commercial printing end uses. Our U.S. capacity is supplemented by about 379,000 tons of capacity at our mills producing coated board in Poland and Russia and, prior to its sale in October 2015, by our International Paper & Sun Cartonboard Co., Ltd. joint venture in China which had an annual capacity of 1.4 million tons.
Our Foodservice business produces cups, lids, food containers and plates through three domestic plants and four international facilities.
Ilim Holding S.A.
In October 2007, International Paper and Ilim Holding S.A. (Ilim) completed a 50:50 joint venture to operate a pulp and paper business located in Russia. Ilim’s facilities include three paper mills located in Bratsk, Ust-Ilimsk, and Koryazhma, Russia, with combined total pulp and paper capacity of over 3.4 million tons. Ilim has exclusive harvesting rights on timberland and forest areas exceeding 14.8 million acres (6.0 million hectares).
Products and brand designations appearing in italics are trademarks of International Paper or a related company.
INDUSTRY SEGMENT RESULTS
Industrial Packaging
Demand for Industrial Packaging products is closely correlated with non-durable industrial goods production, as well as with demand for processed foods, poultry, meat and agricultural products. In addition to prices and volumes, major factors affecting the profitability of Industrial Packaging are raw material and energy costs, freight costs, manufacturing efficiency and product mix.
Industrial Packaging net sales for 2015 decreased 3% to $14.5 billion compared with $14.9 billion in 2014, and 2% compared with $14.8 billion in 2013. Operating profits were 2%lower in 2015 than in 2014 and 3%higher than in 2013. Excluding costs associated with the acquisition and integration of Temple-Inland, goodwill impairment charges, costs associated with a multi-employer pension liability and other special items, operating profits in 2015 were 3%lower than in 2014 and 8%higher than in 2013. Benefits from lower input costs ($175 million) were offset by lower average sales price realizations and mix ($144 million), lower sales volumes ($36 million), higher operating costs ($43 million) and higher maintenance outage costs ($16 million). Additionally, operating profits in 2015 include a goodwill and trade name impairment charge associated with our Brazil Packaging business ($137 million). Operating profits in 2014 include a goodwill impairment charge of $100 million related to our Asia Industrial Packaging business, costs of $16 million associated with the integration of Temple-Inland, a charge of $35 million associated with a multi-employer pension plan withdrawal liability and a net charge of $7 million for other items. Operating profits in 2013 include costs of $62 million associated with the integration of Temple-Inland, a gain of $13 million related to a bargain purchase adjustment on the acquisition of a majority share of our operations in Turkey, and a net gain of $1 million for other items.
Industrial Packaging
In millions 2015 2014 2013
Sales$14,484$14,944$14,810
Operating Profit 1,853 1,896 1,801
North American Industrial Packaging net sales were $12.5 billion in 2015 compared with $12.7 billion in 2014 and $12.5 billion in 2013. Operating profits in 2015 were $2.0 billion compared with $2.0 billion (both including and excluding costs associated with the integration of Temple-Inland, a multi-employer pension withdrawal liability and other special items) in 2014 and $1.8 billion (both including and excluding costs associated with the integration of Temple-Inland and other special items in 2013.
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Sales volumes decreased in 2015 compared with 2014 reflecting slightly lower box shipments and lower shipments of containerboard to export markets. In 2015, the business took about 814,000 tons of total downtime of which about 363,000 were market-related and 451,000 were maintenance downtime. The business took about 622,000 tons of total downtime in 2014 of which 240,000 were market-related and 382,000 were maintenance downtime. Average sales price realizations were lower mostly for Euro-denominated shipments of containerboard to export markets. Input costs were lower, primarily for energy. Distribution costs were flat as lower freight fuel surcharges offset rate increases. Planned maintenance downtime costs were $15 million higher than in 2014. Manufacturing operating costs decreased, but were more than offset by wage and benefit inflation. Depreciation costs were lower.
Looking ahead to the first quarter of 2016, compared with the fourth quarter of 2015, sales volumes for boxes are expected to be seasonally lower, while shipments of containerboard to export markets should increase. Input costs are expected to be higher for energy and wood, but lower for waste fiber. Planned maintenance downtime spending is expected to be about $21 million higher. Manufacturing operating costs are expected to improve.
EMEA Industrial Packaging net sales were $1.1 billion in 2015 compared with $1.3 billion in 2014 and $1.3 billion in 2013. Operating profits in 2015 were $13 million compared with $25 million ($31 million excluding restructuring costs) in 2014 and $43 million ($32 million excluding a gain on a bargain purchase price adjustment on the acquisition of a majority share of our operations in Turkey and restructuring costs) in 2013.
Sales volumes in 2015 were higher than in 2014 reflecting improved market demand and strong commercial initiatives in the Eurozone throughout the year and growth in Morocco and Turkey in the fourth quarter. Net sales decreased primarily due to the negative impact of foreign exchange rates. Higher board costs also contributed to lower average sales margins. Other input costs, primarily for energy, were lower. Operating earnings in 2015 also included a gain of $4 million related to the change in ownership of our OCC collection operations in Turkey.
Entering the first quarter of 2016, compared with the fourth quarter of 2015 sales volumes are expected to be flat. Average sales margins are expected to be favorably impacted by higher box sales prices, lower board costs in Turkey and a favorable mix. Input costs for energy should be slightly higher.
Brazilian Industrial Packaging net sales were $228 million in 2015 compared with $349 million in 2014 and $335 million in 2013. Operating profits in 2015 were a loss
of $163 million (a loss of $26 million excluding goodwill and trade name impairment charges) compared with a loss of $3 million (a loss of $4 million excluding a net gain related to acquisition and integration costs) in 2014 and a loss of $2 million (a gain of $2 million excluding acquisition and integration costs) in 2013.
Sales volumes in 2015 decreased compared with 2014 due to overall weak economic conditions and lower box consumption in the product segments of some of our key customers. Average sales price realizations for boxes were lower. Input costs were slightly higher. Operating costs also increased. Planned maintenance downtime costs were $1 million lower in 2015 compared with 2014.
Looking ahead to the first quarter of 2016, compared with the fourth quarter of 2015 sales volumes are expected to be seasonally lower. Average sales margins should improve reflecting a previously announced sales price increase for boxes. Input costs are expected to be stable and operating costs should reflect the benefits of cost savings initiatives.
Asian Industrial Packaging net sales were $601 million in 2015 compared with $625 million in 2014 and $685 million in 2013. Operating profits were a loss of $6 million in 2015 compared with a loss of $112 million (a loss of $5 million excluding goodwill impairment charges and restructuring costs) in 2014 and a loss of $2 million (a gain of $2 million excluding restructuring costs) in 2013. Compared with 2014, sales volumes for boxes in 2015 were lower and average sales margins decreased due to competitive price pressures and an unfavorable sales mix. However, operating costs were lower.
Looking ahead to the first quarter of 2016, sales volumes are expected to be seasonally lower. On October 8, 2015, the Company announced that it was pursuing strategic options for its corrugated box business in China and Southeast Asia and had signed a non-binding letter of intent with a prospective buyer.
Printing Papers
Demand for Printing Papers products is closely correlated with changes in commercial printing and advertising activity, direct mail volumes and, for uncoated cut-size products, with changes in white-collar employment levels that affect the usage of copy and laser printer paper. Pulp is further affected by changes in currency rates that can enhance or disadvantage producers in different geographic regions. Principal cost drivers include manufacturing efficiency, raw material and energy costs and freight costs.
Printing Papers net sales for 2015 decreased 12% to $5.0 billion compared with $5.7 billion in 2014 and 19%
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compared with $6.2 billion in 2013. Operating profits in 2015 were significantly higher than in both 2014 and 2013. Excluding facility closure costs, impairment costs and other special items, operating profits in 2015 were 3%lower than in 2014 and 4%higher than in 2013. Benefits from lower input costs ($18 million), lower costs associated with the closure of our Courtland, Alabama mill ($44 million) and favorable foreign exchange ($33 million) were offset by lower average sales price realizations and mix ($52 million), lower sales volumes ($16 million), higher operating costs ($18 million) and higher planned maintenance downtime costs ($26 million). In addition, operating profits in 2014 include special items costs of $554 million associated with the closure of our Courtland, Alabama mill. During 2013, the Company accelerated depreciation for certain Courtland assets, and evaluated certain other assets for possible alternative uses by one of our other businesses. The net book value of these assets at December 31, 2013 was approximately $470 million. In the first quarter of 2014, we completed our evaluation and concluded that there were no alternative uses for these assets. We recognized approximately $464 million of accelerated depreciation related to these assets in 2014. Operating profits in 2014 also include a charge of $32 million associated with a foreign tax amnesty program, and a gain of $20 million for the resolution of a legal contingency in India, while operating profits in 2013 included costs of $118 million associated with the announced closure of our Courtland, Alabama mill and a $123 million impairment charge associated with goodwill and a trade name intangible asset in our India Papers business.
Printing Papers
In millions 2015 2014 2013
Sales$5,031$5,720$6,205
Operating Profit (Loss)533(16)271
North American Printing Papers net sales were $1.9 billion in 2015, $2.1 billion in 2014 and $2.6 billion in 2013. Operating profits in 2015 were $179 million compared with a loss of $398 million (a gain of $156 million excluding costs associated with the shutdown of our Courtland, Alabama mill) in 2014 and a gain of $36 million ($154 million excluding costs associated with the Courtland mill shutdown) in 2013.
Sales volumes in 2015 decreased compared with 2014 primarily due to the closure of our Courtland mill in 2014. Shipments to the domestic market increased, but export shipments declined. Average sales price realizations decreased, primarily in the domestic market. Input costs were lower, mainly for energy. Planned maintenance downtime costs were $12 million higher in 2015. Operating profits in 2014 were negatively impacted by costs associated with the shutdown of our Courtland, Alabama mill.
Entering the first quarter of 2016, sales volumes are expected to be up slightly compared with the fourth quarter of 2015. Average sales margins should be about flat reflecting lower average sales price realizations offset by a more favorable product mix. Input costs are expected to be stable. Planned maintenance downtime costs are expected to be about $14 million lower with an outage scheduled in the 2016 first quarter at our Georgetown mill compared with outages at our Eastover and Riverdale mills in the 2015 fourth quarter.
In January 2015, the United Steelworkers, Domtar Corporation, Packaging Corporation of America, Finch Paper LLC and P. H. Glatfelter Company (the Petitioners) filed an anti-dumping petition before the United States International Trade Commission (ITC) and the United States Department of Commerce (DOC) alleging that paper producers in China, Indonesia, Australia, Brazil, and Portugal are selling uncoated free sheet paper in sheet form (the Products) in violation of international trade rules. The Petitioners also filed a countervailing-duties petition with these agencies regarding imports of the Products from China and Indonesia.In January 2016, the DOC announced its final countervailing duty rates on imports of the Products to the United States from certain producers from China and Indonesia. Also, in January 2016, the DOC announced its final anti-dumping duty rates on imports of the Products to the United States from certain producers from Australia, Brazil, China, Indonesia and Portugal. In February 2016, the ITC concluded its anti-dumping and countervailing duties investigations and made a final determination that the U.S. market had been injured by imports of the Products. Accordingly, the DOC’s previously announced countervailing duty rates and anti-dumping duty rates will be in effect for a minimum of five years. We do not believe the impact of these rates will have a material, adverse effect on our consolidated financial statements.
Brazilian Papers net sales for 2015 were $878 million compared with $1.1 billion in 2014 and $1.1 billion in 2013. Operating profits for 2015 were $186 million compared with $177 million ($209 million excluding costs associated with a tax amnesty program) in 2014 and $210 million in 2013.
Sales volumes in 2015 were lower compared with 2014 reflecting weak economic conditions and the absence of 2014 one-time events. Average sales price realizations improved for domestic uncoated freesheet paper due to the realization of price increases implemented in the second half of 2015. Margins were unfavorably affected by an increased proportion of sales to the lower-margin export markets. Raw material costs increased for energy and wood. Operating costs were higher than in 2014, while planned maintenance downtime costs were $4 million lower.
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Looking ahead to 2016, compared with the fourth quarter of 2015 sales volumes in the first quarter are expected to decrease due to seasonally weaker customer demand for uncoated freesheet paper. Average sales price improvements are expected to reflect the partial realization of announced sales price increases in the Brazilian domestic market for uncoated freesheet paper. Input costs are expected to be slightly higher for chemicals and electricity.
European Papers net sales in 2015 were $1.2 billion compared with $1.5 billion in 2014 and $1.5 billion in 2013. Operating profits in 2015 were $133 million compared with $140 million in 2014 and $167 million in 2013.
Compared with 2014, sales volumes for uncoated freesheet paper in 2015 were slightly lower in both Russia and Europe. Average sales price realizations for uncoated freesheet paper increased in Russia, but remained flat in Europe, reflecting tight demand and supply conditions in the first half of the year. Input costs increased slightly as higher costs for wood, chemicals and energy in Russia were largely offset by lower costs in Europe. Planned maintenance downtime costs were $11 million higher in 2015 than in 2014.
Entering 2016, domestic sales volumes in the first quarter are expected to be seasonally weaker in Russia, and stable in Europe. Average sales price realizations for uncoated freesheet paper are expected to reflect the impact of announced price increases in both Europe and Russia. Input costs should be slightly higher for wood and chemicals. Planned maintenance downtime costs should be $1 million lower than in the fourth quarter of 2015.
Indian Papers net sales were $172 million in 2015, $178 million in 2014 and $185 million ($174 million excluding excise duties which were included in net sales in 2013 and prior periods) in 2013. Operating profits were a loss of $11 million in 2015, compared with a gain of $8 million (a loss of $12 million excluding a gain related to the resolution of a legal contingency) in 2014 and a loss of $145 million (a loss of $22 million excluding goodwill and trade name impairment charges) in 2013.
Average sales price realizations decreased in 2015 compared with 2014 reflecting soft market demand. Sales volumes increased, primarily to export markets. Input costs were lower for wood and chemicals. Operating costs were higher in 2015, but planned maintenance downtime costs were even with 2014. Looking ahead to the first quarter of 2016, sales volumes are expected to be seasonally higher. Average sales price realizations are expected to be stable.
U.S. Pulp net sales were $844 million in 2015 compared with $895 million in 2014 and $815 million in 2013.
Operating profits were $46 million in 2015 compared with $57 million in 2014 and $2 million in 2013.
Sales volumes in 2015 decreased from 2014 with lower softwood pulp volumes being partially offset by higher fluff pulp volumes. Average sales price realizations were lower for both fluff pulp and softwood market pulp. Input costs decreased primarily for energy. Operating costs were higher, but distribution costs were lower. Planned maintenance downtime costs were $4 million lower in 2015 than in 2014.
Compared with the fourth quarter of 2015, sales volumes in the first quarter of 2016 are expected to be stable. Average sales price realizations are expected to be lower for fluff pulp and softwood market pulp. Input costs should be higher for fuels and utilities. Planned maintenance downtime costs should be about $45 million higher than in the fourth quarter of 2015 including outage costs associated with the conversion of our Riegelwood mill to 100% pulp production.
Consumer Packaging
Demand and pricing for Consumer Packaging products correlate closely with consumer spending and general economic activity. In addition to prices and volumes, major factors affecting the profitability of Consumer Packaging are raw material and energy costs, freight costs, manufacturing efficiency and product mix.
Consumer Packaging net sales in 2015 decreased 14% from 2014, and decreased 14% from 2013. Operating profits decreased 114% from 2014 and decreased 116% from 2013. Excluding the cost associated with the conversion of our Riegelwood, North Carolina mill to 100% pulp production, net of the proceeds from the sale of the Carolina Coated Bristols brand, costs associated with the impairment of goodwill and other assets of the IP-Sun JV, costs associated with the permanent shutdown of a paper machine at our Augusta, Georgia mill and other special items, 2015 operating profits were 15%lower than in 2014, and 24%lower than in 2013. Benefits from higher sales volumes ($14 million), lower planned maintenance downtime costs ($5 million) and lower input costs ($39 million) were offset by lower average sales price realizations and mix ($30 million), higher operating costs ($44 million), and higher foreign exchange and other costs ($11 million). In addition, operating profits in 2015 include a charge of $174 million for the impairment of goodwill and other assets for the IP-Sun JV, an $8 million cost related to the conversion of our Riegelwood mill to 100% pulp production, net of the proceeds from the sale of the Carolina Coated Bristols brand, and $2 million of costs associated with sheet plant closures, while operating profits in 2014 include $8 million of costs associated with sheet plant closures. Operating profits in 2013 include costs of $45 million related to the permanent shutdown of a paper machine at our
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Augusta, Georgia mill and $2 million of costs associated with the sale of the Shorewood business.
Consumer Packaging
In millions 2015 2014 2013
Sales$2,940$3,403$3,435
Operating Profit (Loss)(25)178 161
North American Consumer Packaging net sales were $1.9 billion in 2015 compared with $2.0 billion in 2014 and $2.0 billion in 2013. Operating profits were $81 million ($91 million excluding the cost associated with the planned conversion of our Riegelwood mill to 100% pulp production, net of proceeds from the sale of the Carolina Coated Bristols brand, and sheet plant closure costs) in 2015 compared with $92 million ($100 million excluding sheet plant closure costs) in 2014 and $63 million ($110 million excluding paper machine shutdown costs and costs related to the sale of the Shorewood business) in 2013.
Coated Paperboard sales volumes in 2015 were lower than in 2014 reflecting weaker market demand. The business took about 77,000 tons of market-related downtime in 2015 compared with about 41,000 tons in 2014. Average sales price realizations increased modestly year over year as competitive pressures in the current year only partially offset the impact of sales price increases implemented in 2014. Input costs decreased for energy and chemicals, but wood costs increased. Planned maintenance downtime costs were $10 million lower in 2015. Operating costs were higher, mainly due to inflation and overhead costs.
Foodservice sales volumes increased in 2015 compared with 2014 reflecting strong market demand. Average sales margins increased due to lower resin costs and a more favorable mix. Operating costs and distribution costs were both higher.
Looking ahead to the first quarter of 2016, Coated Paperboard sales volumes are expected to be slightly lower than in the fourth quarter of 2015 due to our exit from the coated bristols market. Average sales price realizations are expected to be flat, but margins should benefit from a more favorable product mix. Input costs are expected to be higher for wood, chemicals and energy. Planned maintenance downtime costs should be $4 million higher with a planned maintenance outage scheduled at our Augusta mill in the first quarter. Foodservice sales volumes are expected to be seasonally lower. Average sales margins are expected to improve due to a more favorable mix. Operating costs are expected to decrease.
European Consumer Packaging net sales in 2015 were $319 million compared with $365 million in 2014 and $380 million in 2013. Operating profits in 2015 were $87 million compared with $91 million in 2014 and $100 million in 2013. Sales volumes in 2015 compared with
2014 increased in Europe, but decreased in Russia. Average sales margins improved in Russia due to slightly higher average sales price realizations and a more favorable mix. In Europe average sales margins decreased reflecting lower average sales price realizations and an unfavorable mix. Input costs were lower in Europe, primarily for wood and energy, but were higher in Russia, primarily for wood.
Looking forward to the first quarter of 2016, compared with the fourth quarter of 2015, sales volumes are expected to be stable. Average sales price realizations are expected to be slightly higher in both Russia and Europe. Input costs are expected to be flat, while operating costs are expected to increase.
Asian Consumer Packaging The Company sold its 55% equity share in the IP-Sun JV in October 2015. Net sales and operating profits presented below include results through September 30, 2015. Net sales were $682 million in 2015 compared with $1.0 billion in 2014 and $1.1 billion in 2013. Operating profits in 2015 were a loss of $193 million (a loss of $19 million excluding goodwill and other asset impairment costs) compared with losses of $5 million in 2014 and $2 million in 2013.
Sales volumes and average sales price realizations were lower in 2015 due to over-supplied market conditions and competitive pressures. Average sales margins were also negatively impacted by a less favorable mix. Input costs and freight costs were lower and operating costs also decreased.
On October 13, 2015, the Company finalized the sale of its 55% interest in IP Asia Coated Paperboard (IP-Sun JV) business, within the Company's Consumer Packaging segment, to its Chinese coated board joint venture partner, Shandong Sun Holding Group Co., Ltd. for RMB 149 million (approximately USD $23 million). During the third quarter of 2015, a determination was made that the current book value of the asset group exceeded its estimated fair value of $23 million, which was the agreed upon selling price. The 2015 loss includes the net pre-tax impairment charge of $174 million ($113 million after taxes). A pre-tax charge of $186 million was recorded during the third quarter in the Company's Consumer Packaging segment to write down the long-lived assets of this business to their estimated fair value. In the fourth quarter of 2015, upon the sale and corresponding deconsolidation of IP-Sun JV from the Company's consolidated balance sheet, final adjustments were made resulting in a reduction of the impairment of $12 million. The amount of pre-tax losses related to noncontrolling interest of the IP-Sun JV included in the Company's consolidated statement of operations for the years ended December 31, 2015, 2014 and 2013 were $19 million, $12 million and $8 million, respectively. The amount of pre-tax losses related to the IP-Sun JV included in the Company's
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consolidated statement of operations for the years ended December 31, 2015, 2014 and 2013 were $226 million, $51 million and $41 million, respectively.
Equity Earnings, Net of Taxes – Ilim Holding S.A.
International Paper accounts for its investment in Ilim Holding S.A. (Ilim), a separate reportable industry segment, using the equity method of accounting.
The Company recorded equity earnings, net of taxes, related to Ilim of $131 million in 2015 compared with a loss of $194 million in 2014 and a loss of $46 million in 2013. Operating results recorded in 2015 included an after-tax non-cash foreign exchange loss of $75 million compared with an after-tax foreign exchange loss of $269 million in 2014 and an after-tax foreign exchange loss of $32 million in 2013 primarily on the remeasurement of Ilim's U.S. dollar-denominated net debt.
Sales volumes for the joint venture increased year over year for shipments to China of hardwood pulp and softwood pulp, but decreased for linerboard. Sales volumes in the domestic Russian market increased for hardwood pulp and paper, but decreased for softwood pulp and linerboard. Average sales price realizations were higher in 2015 for sales of hardwood pulp to export markets and linerboard to the domestic market, but were offset by lower average sales price realizations for sales of softwood pulp to export markets. Input costs increased year-over-year for wood, chemicals, fuel and energy. Freight costs also increased. The Company received cash dividends from the joint venture of $35 million in 2015 and $56 million in 2014. No dividends were paid in 2013.
Entering the first quarter of 2016, sales volumes are expected to be seasonally lower than in the fourth quarter of 2015 due to the January holidays in Russia. Average sales price realizations are expected to decrease for exported hardwood pulp, softwood pulp and containerboard, slightly offset by higher average sales price realizations for paper in the domestic market. Input costs for energy, chemicals and wood should be higher and distribution costs are also expected to increase.
LIQUIDITY AND CAPITAL RESOURCES
Overview
A major factor in International Paper’s liquidity and capital resource planning is its generation of operating cash flow, which is highly sensitive to changes in the pricing and demand for our major products. While changes in key cash operating costs, such as energy, raw material and transportation costs, do have an effect on operating cash generation, we believe that our focus
on pricing and cost controls has improved our cash flow generation over an operating cycle.
Cash uses during 2015 were primarily focused on working capital requirements, capital spending, debt reductions and returning cash to shareholders.
Cash Provided by Operating Activities
Cash provided by operations totaled $2.6 billion in 2015 compared with $3.1 billion for 2014 and $3.0 billion for 2013.
The major components of cash provided by operations are earnings from operations adjusted for non-cash income and expense items, cash pension contributions and changes in working capital. Earnings from operations, adjusted for non-cash income and expense items and cash pension contributions decreased by $433 million in 2015 versus 2014 driven mainly by increased cash pension contributions in 2015. Cash used for working capital components, accounts receivable and inventory less accounts payable and accrued liabilities, interest payable and other totaled $222 million in 2015, compared with a cash use of $158 million in 2014 and a cash use of $486 million in 2013.
The Company generated free cash flow of approximately $1.8 billion, $2.1 billion and $1.8 billion in 2015, 2014 and 2013, respectively. Free cash flow is a non-GAAP measure and the most comparable GAAP measure is cash provided by operations. Management uses free cash flow as a liquidity metric because it measures the amount of cash generated that is available to maintain our assets, make investments or acquisitions, pay dividends and reduce debt. The following are reconciliations of free cash flow to cash provided by operations:
In millions 2015 2014 2013
Cash provided by operations$2,580$3,077$3,028
(Less)/Add:
Cash invested in capital projects(1,487)(1,366)(1,198)
Cash contribution to pension plan 750 353 31
Insurance reimbursement for Guaranty Bank settlement——(30)
Free Cash Flow$1,843$2,064$1,831
In millions Three Months Ended December 31, 2015 Three Months Ended September 30, 2015 Three Months Ended December 31, 2014
Cash provided by operations$990$837$1,144
(Less)/Add:
Cash invested in capital projects(489)(325)(405)
Free Cash Flow$501$512$739
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Alternative Fuel Mixture Credit
On July 19, 2011, the Company filed an amended 2009 tax return claiming alternative fuel mixture tax credits as non-taxable income. The amended position has been accepted by the Internal Revenue Service (IRS) in the closing of the IRS tax audit for the years 2006 - 2009. As a result, during 2013, the Company recognized an income tax benefit of $753 million related to the non-taxability of the alternative fuel mixture tax credits.
Investment Activities
Investment activities in 2015 were up from 2014 reflecting an increase in capital spending and the use of $198 million of cash in conjunction with the timber monetization restructuring (see Note 12 Variable Interest Entities and Preferred Securities of Subsidiaries on pages 64 through 66 of Item 8. Financial Statements and Supplementary Data) in 2015. In addition, 2014 investment activity includes the receipt of approximately $400 million in connection with the spin-off of the xpedx distribution business. The Company maintains an average capital spending target around depreciation or amortization levels or modestly above due to strategic plans over the course of an economic cycle. Capital spending was $1.5 billion in 2015, or 115% of depreciation and amortization, compared with $1.4 billion in 2014, or 97% of depreciation and amortization, and $1.2 billion, or 77% of depreciation and amortization in 2013. Across our businesses, capital spending as a percentage of depreciation and amortization ranged from 118% to 100% in 2015.
The following table shows capital spending for operations by business segment for the years ended December 31, 2015, 2014 and 2013.
In millions 2015 2014 2013
Industrial Packaging$858$754$629
Printing Papers 361 318 294
Consumer Packaging 216 233 208
Distribution——9
Subtotal 1,435 1,305 1,140
Corporate and other 52 61 58
Total$1,487$1,366$1,198
Capital expenditures in 2016 are currently expected to be about $1.3 billion, or 100% of depreciation and amortization.
Acquisitions and Joint Ventures
OLMUKSAN
2014: In May 2014, the Company conducted a voluntary tender offer for the remaining outstanding
12.6% public shares of Olmuksan. The Company also purchased outstanding shares of Olmuksan outside of the tender offer. As of December 31, 2014 and 2015, the Company owned 91.7% of Olmuksan's outstanding and issued shares.
2013: On January 3, 2013, International Paper completed the acquisition (effective date of acquisition on January 1, 2013) of the shares of its joint venture partner, Sabanci Holding, in the Turkish corrugated packaging company, Olmuksa International Paper Sabanci Ambalaj Sanayi ve Ticaret A.S., now called Olmuksan International Paper Ambalaj Sanayi ve Ticaret A.S. (Olmuksan), for a purchase price of $56 million. The acquired shares represented 43.7% of Olmuksan's shares. Prior to this acquisition, International Paper held a 43.7% equity interest in Olmuksan.
Because the transaction resulted in International Paper becoming the majority shareholder, owning 87.4% of Olmuksan's outstanding and issued shares, its completion triggered a mandatory call for tender of the remaining public shares which began in March 2013 and ended in April 2013, with no shares tendered. As a result, the 12.6% owned by other parties were considered non-controlling interests. Olmuksan's financial results have been consolidated with the Company's Industrial Packaging segment beginning January 1, 2013, the effective date which International Paper obtained majority control of the entity.
Following the transaction, the Company's previously held 43.7% equity interest in Olmuksan was remeasured to a fair value of $75 million, resulting in a gain of $9 million. In addition, the cumulative translation adjustment balance of $17 million relating to the previously held equity interest was reclassified, as expense, from accumulated other comprehensive income.
The final purchase price allocation indicated that the sum of the cash consideration paid, the fair value of the noncontrolling interest and the fair value of the previously held interest was less than the fair value of the underlying assets by $21 million, resulting in a bargain purchase price gain being recorded on this transaction. The aforementioned remeasurement of equity interest gain, the cumulative translation adjustment to expense, and the bargain purchase gain are included in the Net bargain purchase gain on acquisition of business in the accompanying consolidated statement of operations.
ORSA
2014: On April 8, 2014, the Company acquired the remaining 25% of shares of Orsa International Paper Embalangens S.A. (Orsa IP) from its joint venture
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partner, Jari Celulose, Papel e Embalagens S.A. (Jari), a Grupo Orsa company, for approximately $127 million, of which $105 million was paid in cash with the remaining $22 million held back pending satisfaction of certain indemnification obligations by Jari. International Paper will release the amount held back, or any amount for which we have not notified Jari of a claim, by March 30, 2016. An additional $11 million, which was not included in the purchase price, was placed in an escrow account pending resolution of certain open matters. During 2014, these open matters were successfully resolved, which resulted in $9 million paid out of escrow to Jari and correspondingly added to the final purchase consideration. The remaining $2 million was released back to the Company. As a result of this transaction, the Company reversed the $168 million of Redeemable noncontrolling interest included on the March 31, 2014 consolidated balance sheet. The net difference between the Redeemable noncontrolling interest balance plus $14 million of currency translation adjustment reclassified out of Other comprehensive income less the 25% purchase price was reflected as an increase to Retained earnings on the consolidated balance sheet.
2013: On January 14, 2013, International Paper and Jari formed Orsa IP with International Paper holding a 75% stake. The value of International Paper's investment in Orsa IP was approximately $471 million. Because International Paper acquired a majority control of the joint venture, Orsa IP's financial results have been consolidated with our Industrial Packaging segment from the date of formation on January 14, 2013. The 25% owned by Jari was considered a redeemable noncontrolling interest and met the requirements to be classified outside permanent equity. As such, the Company reported $163 million in Redeemable noncontrolling interest in the December 31, 2013 consolidated balance sheet.
Financing Activities
Amounts related to early debt extinguishment during the years ended December 31, 2015, 2014 and 2013 were as follows:
In millions 2015 2014 2013
Debt reductions (a)$2,151$1,625$574
Pre-tax early debt extinguishment costs (b)207 276 25
(a)Reductions related to notes with interest rates ranging from 2.00% to 9.38% with original maturities from 2014 to 2031 for the years ended December 31, 2015, 2014 and 2013. Includes the $630 million payment for a portion of the Special Purpose Entity Liability (see Note 12 Variable Interest Entities on pages 64 through 66 of Item 8. Financial Statements and Supplementary Data ).
(b)Amounts are included in Restructuring and other charges in the accompanying consolidated statements of operations.
2015: Financing activities during 2015 included debt issuances of $6.9 billion and retirements of $6.9 billion for a net decrease of $74 million.
During 2015, the Company restructured the timber monetization which resulted in the use of $630 million in cash to pay down a portion of the third party bank loans and refinance the loans on nonrecourse terms. (see Note 12 Variable Interest Entities on pages 64 through 66 of Item 8. Financial Statements and Supplementary Data).
International Paper utilizes interest rate swaps to change the mix of fixed and variable rate debt and manage interest expense. At December 31, 2015, International Paper had interest rate swaps with a total notional amount of $17 million and maturities in 2018 (see Note 14 Derivatives and Hedging Activities on pages 67 through 71 of Item 8. Financial Statements and Supplementary Data). During 2015, existing swaps and the amortization of deferred gains on previously terminated swaps decreased the weighted average cost of debt from 5.9% to an effective rate of 5.8%. The inclusion of the offsetting interest income from short-term investments reduced this effective rate to 5.1%.
In 2015, International Paper issued $700 million of 3.80% senior unsecured notes with a maturity date in 2026, $600 million of 5.00% senior unsecured notes with a maturity date in 2035, and $700 million of 5.15% senior unsecured notes with a maturity date in 2046. The proceeds from this borrowing were used to repay approximately $1.0 billion of notes with interest rates ranging from 4.75% to 9.38% and original maturities from 2018 to 2022, along with $211 million of cash premiums associated with the debt repayments. Additionally, the proceeds from this borrowing were used to make a $750 million voluntary cash contribution to the Company's pension plan. Pre-tax early debt retirement costs of $207 million related to the debt repayments, including the $211 million of cash premiums, are included in restructuring and other charges in the accompanying consolidated statement of operations for the twelve months ended December 31, 2015.
Other financing activities during 2015 included the net repurchase of approximately 8.0 million shares of treasury stock, including restricted stock withholding, and the issuance of 62,000 shares of common stock for various plans, including stock options exercises that generated approximately $2.4 million of cash. Repurchases of common stock and payments of restricted stock withholding taxes totaled $604.6 million, including $522.6 million related to shares repurchased under the Company's share repurchase program.
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In October 2015, International Paper announced that the quarterly dividend would be increased from $0.40 per share to $0.44 per share, effective for the 2015 fourth quarter.
2014: Financing activities during 2014 included debt issuances of $2.0 billion and retirements of $2.1 billion, for a net decrease of $113 million.
International Paper utilizes interest rate swaps to change the mix of fixed and variable rate debt and manage interest expense. At December 31, 2014, International Paper had interest rate swaps with a total notional amount of $230 million and maturities in 2018 (see Note 14 Derivatives and Hedging Activities on pages 67 through 71 of Item 8. Financial Statements and Supplementary Data). During 2014, existing swaps and the amortization of deferred gains on previously terminated swaps decreased the weighted average cost of debt from 6.8% to an effective rate of 6.7%. The inclusion of the offsetting interest income from short-term investments reduced this effective rate to 6.3%.
During the second quarter of 2014, International Paper issued $800 million of 3.65% senior unsecured notes with a maturity date in 2024 and $800 million of 4.80% senior unsecured notes with a maturity date in 2044. The proceeds from this borrowing were used to repay approximately $960 million of notes with interest rates ranging from 7.95% to 9.38% and original maturities from 2018 to 2019. Pre-tax early debt retirement costs of $262 million related to these debt repayments, including $258 million of cash premiums, are included in Restructuring and other charges in the accompanying consolidated statement of operations for the twelve months ended December 31, 2014.
Other financing activities during 2014 included the net repurchase of approximately 17.9 million shares of treasury stock, including restricted stock withholding, and the issuance of 1.6 million shares of common stock for various plans, including stock options exercises that generated approximately $66 million of cash. Repurchases of common stock and payments of restricted stock withholding taxes totaled $1.06 billion, including $983 million related to shares repurchased under the Company's share repurchase program.
In September 2014, International Paper announced that the quarterly dividend would be increased from $0.35 per share to $0.40 per share, effective for the 2014 fourth quarter.
2013: Financing activities during 2013 included debt issuances of $241 million and retirements of $845 million, for a net decrease of $604 million.
International Paper utilizes interest rate swaps to change the mix of fixed and variable rate debt and manage interest expense. At December 31, 2013, International Paper had interest rate swaps with a total
notional amount of $175 million and maturities in 2018 (see Note 14 Derivatives and Hedging Activities on pages 67 through 71 of Item 8. Financial Statements and Supplementary Data). During 2013, existing swaps and the amortization of deferred gains on previously terminated swaps decreased the weighted average cost of debt from 6.7% to an effective rate of 6.5%. The inclusion of the offsetting interest income from short-term investments reduced this effective rate to 6.2%.
Other financing activities during 2013 included the net repurchase of approximately 10.9 million shares of treasury stock, including restricted stock withholding, and the issuance of 7.3 million shares of common stock for various plans, including stock options exercises that generated approximately $298 million of cash. Repurchases of common stock and payments of restricted stock withholding taxes totaled $512 million, including $461 million related to shares repurchased under the Company's share repurchase program.
In September 2013, International Paper announced that the quarterly dividend would be increased from $0.30 per share to $0.35 per share, effective for the 2013 fourth quarter.
Variable Interest Entities
Information concerning variable interest entities is set forth in Note 12 Variable Interest Entities on pages 64 through 66 of Item 8. Financial Statements and Supplementary Data for discussion.
Liquidity and Capital Resources Outlook for 2016
Capital Expenditures and Long-Term Debt
International Paper expects to be able to meet projected capital expenditures, service existing debt and meet working capital and dividend requirements during 2016 through current cash balances and cash from operations. Additionally, the Company has existing credit facilities totaling $2.1 billion available at December 31, 2015.
The Company was in compliance with all its debt covenants at December 31, 2015. The Company’s financial covenants require the maintenance of a minimum net worth of $9 billion and a total debt-to-capital ratio of less than 60%. Net worth is defined as the sum of common stock, paid-in capital and retained earnings, less treasury stock plus any cumulative goodwill impairment charges. The calculation also excludes accumulated other comprehensive income/loss and Nonrecourse Financial Liabilities of Special Purpose Entities. The total debt-to-capital ratio is defined as total debt divided by the sum of total debt plus net worth. At December 31, 2015, International Paper’s net worth was $14.1 billion, and the total-debt-to-capital ratio was 39.8%.
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The Company will continue to rely upon debt and capital markets for the majority of any necessary long-term funding not provided by operating cash flows. Funding decisions will be guided by our capital structure planning objectives. The primary goals of the Company’s capital structure planning are to maximize financial flexibility and preserve liquidity while reducing interest expense. The majority of International Paper’s debt is accessed through global public capital markets where we have a wide base of investors.
Maintaining an investment grade credit rating is an important element of International Paper’s financing strategy. At December 31, 2015, the Company held long-term credit ratings of BBB (stable outlook) and Baa2 (stable outlook) by S&P and Moody’s, respectively.
Contractual obligations for future payments under existing debt and lease commitments and purchase obligations at December 31, 2015, were as follows:
In millions 2015 2016 2017 2018 2019 Thereafter
Maturities of long-term debt (a)$426$43$811$427$183$7,436
Lease obligations 118 95 72 55 41 128
Purchase obligations (b)3,001 541 447 371 358 1,579
Total (c)$3,545$679$1,330$853$582$9,143
(a)Total debt includes scheduled principal payments only.
(b)Includes $2.1 billion relating to fiber supply agreements entered into at the time of the 2006 Transformation Plan forestland sales and in conjunction with the 2008 acquisition of Weyerhaeuser Company’s Containerboard, Packaging and Recycling business.
(c)Not included in the above table due to the uncertainty as to the amount and timing of the payment are unrecognized tax benefits of approximately $101 million.
We consider the undistributed earnings of our foreign subsidiaries as of December 31, 2015, to be indefinitely reinvested and, accordingly, no U.S. income taxes have been provided thereon. As of December 31, 2015, the amount of cash associated with indefinitely reinvested foreign earnings was approximately $600 million. We do not anticipate the need to repatriate funds to the United States to satisfy domestic liquidity needs arising in the ordinary course of business, including liquidity needs associated with our domestic debt service requirements.
Pension Obligations and Funding
At December 31, 2015, the projected benefit obligation for the Company’s U.S. defined benefit plans determined under U.S. GAAP was approximately $3.5 billion higher than the fair value of plan assets. Approximately $3.2 billion of this amount relates to plans that are subject to minimum funding requirements. Under current IRS funding rules, the calculation of minimum funding requirements differs from the calculation of the present value of plan benefits
(the projected benefit obligation) for accounting purposes. In December 2008, the Worker, Retiree and Employer Recovery Act of 2008 (WERA) was passed by the U.S. Congress which provided for pension funding relief and technical corrections. Funding contributions depend on the funding method selected by the Company, and the timing of its implementation, as well as on actual demographic data and the targeted funding level. The Company continually reassesses the amount and timing of any discretionary contributions and elected to make contributions totaling $750 million and $353 million for the years ended December 31, 2015 and 2014, respectively. At this time, we do not expect to have any required contributions to our plans in 2016, although the Company may elect to make future voluntary contributions. The timing and amount of future contributions, which could be material, will depend on a number of factors, including the actual earnings and changes in values of plan assets and changes in interest rates. International Paper has announced a voluntary, limited-time opportunity for former employees who are participants in the Retirement Plan of International Paper Company (the Pension Plan) to request early payment of their entire Pension Plan benefit in the form of a single lump sum payment.Eligible participants who wish to receive the lump sum payment must make an election between February 29 and April 29, 2016, and payment is scheduled to be made on or before June 30, 2016.All payments will be made from the Pension Plan trust assets.The target population has a total liability of $3.0 billion. The amount of the total payments will depend on the participation rate of eligible participants, but is expected to be approximately $1.5 billion. Based on the expected level of payments, settlement accounting rules will apply in the period in which the payments are made. This will result in a plan remeasurement and the recognition in earnings of a pro-rata portion of unamortized net actuarial loss.
Ilim Holding S.A. Shareholder’s Agreement
In October 2007, in connection with the formation of the Ilim Holding S.A. joint venture, International Paper entered into a shareholder’s agreement that includes provisions relating to the reconciliation of disputes among the partners. This agreement was amended on May 7, 2014. Pursuant to the amended agreement, beginning on January 1, 2017, either the Company or its partners may commence certain procedures specified under the deadlock provisions. If these or any other deadlock provisions are commenced, the Company may in certain situations, choose to purchase its partners’ 50% interest in Ilim. Any such transaction would be subject to review and approval by Russian and other relevant antitrust authorities. Any such purchase by International Paper would result in the consolidation of Ilim’s financial position and results of operations in all subsequent periods.
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CRITICAL ACCOUNTING POLICIES AND SIGNIFICANT ACCOUNTING ESTIMATES
The preparation of financial statements in conformity with accounting principles generally accepted in the United States requires International Paper to establish accounting policies and to make estimates that affect both the amounts and timing of the recording of assets, liabilities, revenues and expenses. Some of these estimates require judgments about matters that are inherently uncertain.
Accounting policies whose application may have a significant effect on the reported results of operations and financial position of International Paper, and that can require judgments by management that affect their application, include the accounting for contingencies, impairment or disposal of long-lived assets and goodwill, pensions and postretirement benefit obligations, stock options and income taxes. The Company has discussed the selection of critical accounting policies and the effect of significant estimates with the Audit and Finance Committee of the Company’s Board of Directors.
Contingent Liabilities
Accruals for contingent liabilities, including legal and environmental matters, are recorded when it is probable that a liability has been incurred or an asset impaired and the amount of the loss can be reasonably estimated. Liabilities accrued for legal matters require judgments regarding projected outcomes and range of loss based on historical experience and recommendations of legal counsel. Liabilities for environmental matters require evaluations of relevant environmental regulations and estimates of future remediation alternatives and costs.
Impairment of Long-Lived Assets and Goodwill
An impairment of a long-lived asset exists when the asset’s carrying amount exceeds its fair value, and is recorded when the carrying amount is not recoverable through cash flows from future operations. A goodwill impairment exists when the carrying amount of goodwill exceeds its fair value. Assessments of possible impairments of long-lived assets and goodwill are made when events or changes in circumstances indicate that the carrying value of the asset may not be recoverable through future operations. Additionally, testing for possible impairment of goodwill and intangible asset balances is required annually. The amount and timing of any impairment charges based on these assessments require the estimation of future cash flows and the fair market value of the related assets based on management’s best estimates of certain key factors, including future selling prices and volumes, operating, raw material, energy and freight costs, and various other projected operating economic factors. As these
key factors change in future periods, the Company will update its impairment analyses to reflect its latest estimates and projections.
Under the provisions of Accounting Standards Codification (ASC) 350, “Intangibles – Goodwill and Other,” the testing of goodwill for possible impairment is a two-step process. In the first step, the fair value of the Company’s reporting units is compared with their carrying value, including goodwill. If fair value exceeds the carrying value, goodwill is not considered to be impaired. If the fair value of a reporting unit is below the carrying value, then step two is performed to measure the amount of the goodwill impairment loss for the reporting unit. This analysis requires the determination of the fair value of all of the individual assets and liabilities of the reporting unit, including any currently unrecognized intangible assets, as if the reporting unit had been purchased on the analysis date. Once these fair values have been determined, the implied fair value of the unit’s goodwill is calculated as the excess, if any, of the fair value of the reporting unit determined in step one over the fair value of the net assets determined in step two. The carrying value of goodwill is then reduced to this implied value, or to zero if the fair value of the assets exceeds the fair value of the reporting unit, through a goodwill impairment charge.
The impairment analysis requires a number of judgments by management. In calculating the estimated fair value of its reporting units in step one, the Company uses the projected future cash flows to be generated by each unit over the estimated remaining useful operating lives of the unit’s assets, discounted using the estimated cost-of-capital discount rate for each reporting unit. These calculations require many estimates, including discount rates, future growth rates, and cost and pricing trends for each reporting unit. Subsequent changes in economic and operating conditions can affect these assumptions and could result in additional interim testing and goodwill impairment charges in future periods. Upon completion, the resulting estimated fair values are then analyzed for reasonableness by comparing them to earnings multiples for historic industry business transactions, and by comparing the sum of the reporting unit fair values and other corporate assets and liabilities divided by diluted common shares outstanding to the Company’s market price per share on the analysis date.
In the fourth quarter of 2015, in conjunction with the annual testing of its reporting units for possible goodwill impairments, the Company calculated the estimated fair value of its Brazil Packaging business using the discounted future cash flows and determined that all of the goodwill in the business, totaling $137 million, should be written off. The decline in the fair value of the Brazil Packaging business and resulting impairment charge was due to the negative impacts on the cash
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flows of the business caused by the continued decline of the overall Brazilian economy.
In the fourth quarter of 2014, in conjunction with the annual testing of its reporting units for possible goodwill impairments, the Company calculated the estimated fair value of its Asia Industrial Packaging business using the discounted future cash flows and determined that all of the goodwill in this business, totaling $100 million, should be written off. The decline in the fair value of the Asia Industrial Packaging business and resulting impairment charge was due to a change in the strategic outlook for the business.
In the fourth quarter of 2013, in conjunction with the annual testing of its reporting units for possible goodwill impairments, the Company calculated the estimated fair value of its India Papers business using the discounted future cash flows and determined that all of the goodwill of this business, totaling $112 million, should be written off. The decline in the fair value of the India Papers reporting unit and resulting impairment charge was due to a change in the strategic outlook for the India Papers operations.
Also in the fourth quarter of 2013, the Company calculated the estimated fair value of its xpedx business using the discounted future cash flows and wrote off all of the goodwill of its xpedx business, totaling $400 million. The decline in fair value of the xpedx reporting unit and resulting impairment charge was due to a significant decline in earnings and a change in the strategic outlook for the xpedx operations.
As a result, during the fourth quarter of 2013, the Company recorded a total goodwill impairment charge of $512 million ($485 million after taxes and a gain of $3 million related to noncontrolling interest), representing all of the recorded goodwill of the xpedx business and the India Papers business.
Also during 2013, the Company recorded a pre-tax charge of $15 million ($7 million after taxes and noncontrolling interest) for the impairment of a trade name intangible asset related to our India Papers business.
Pension and Postretirement Benefit Obligations
The charges recorded for pension and other postretirement benefit obligations are determined annually in conjunction with International Paper’s consulting actuary, and are dependent upon various assumptions including the expected long-term rate of return on plan assets, discount rates, projected future compensation increases, health care cost trend rates and mortality rates.
The calculations of pension and postretirement benefit obligations and expenses require decisions about a number of key assumptions that can significantly affect liability and expense amounts, including the expected long-term rate of return on plan assets, the discount rate used to calculate plan liabilities, the projected rate of future compensation increases and health care cost trend rates.
Benefit obligations and fair values of plan assets as of December 31, 2015, for International Paper’s pension and postretirement plans were as follows:
In millions Benefit
Obligation Fair Value of
Plan Assets
U.S. qualified pension$14,092$10,923
U.S. nonqualified pension 347—
U.S. postretirement 275—
Non-U.S. pension 204 155
Non-U.S. postretirement 45—
The table below shows assumptions used by International Paper to calculate U.S. pension obligations for the years shown:
2015 2014 2013
Discount rate 4.40%4.10%4.90%
Rate of compensation increase 3.75%3.75%3.75%
Additionally, health care cost trend rates used in the calculation of U.S. postretirement obligations for the years shown were:
2015 2014
Health care cost trend rate assumed for next year 7.00%7.00%
Rate that the cost trend rate gradually declines to 5.00%5.00%
Year that the rate reaches the rate it is assumed to remain 2022 2022
International Paper determines these actuarial assumptions, after consultation with our actuaries, on December 31 of each year to calculate liability information as of that date and pension and postretirement expense for the following year. The expected long-term rate of return on plan assets is based on projected rates of return for current and planned asset classes in the plan’s investment portfolio. The discount rate assumption was determined based on a hypothetical settlement portfolio selected from a universe of high quality corporate bonds.
The expected long-term rate of return on U.S. pension plan assets used to determine net periodic cost for the year ended December 31, 2015 was 7.75%.
Increasing (decreasing) the expected long-term rate of return on U.S. plan assets by an additional 0.25% would decrease (increase) 2016 pension expense by
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approximately $27 million, while a (decrease) increase of 0.25% in the discount rate would (increase) decrease pension expense by approximately $36 million. The effect on net postretirement benefit cost from a 1% increase or decrease in the annual health care cost trend rate would be approximately $1 million.
Actual rates of return earned on U.S. pension plan assets for each of the last 10 years were:
Year Return Year Return
2015 1.3%2010 15.1%
2014 6.4%2009 23.8%
2013 14.1%2008(23.6)%
2012 14.1%2007 9.6%
2011 2.5%2006 14.9%
The 2012, 2013 and 2014 returns above represent weighted averages of International Paper and Temple-Inland asset returns. International Paper and Temple-Inland assets were combined in October 2014. The annualized time-weighted rate of return earned on U.S. pension plan assets was 7.5% and 7.0% for the past five and ten years, respectively. The following graph shows the growth of a $1,000 investment in International Paper’s U.S. Pension Plan Master Trust. The graph portrays the time-weighted rate of return from 2005 – 2015.
ASC 715, “Compensation – Retirement Benefits,” provides for delayed recognition of actuarial gains and losses, including amounts arising from changes in the estimated projected plan benefit obligation due to changes in the assumed discount rate, differences between the actual and expected return on plan assets, and other assumption changes. These net gains and losses are recognized in pension expense prospectively over a period that approximates the average remaining service period of active employees expected to receive benefits under the plans to the extent that they are not offset by gains and losses in subsequent years. The estimated net loss and prior service cost that will be amortized from accumulated other comprehensive income into net periodic pension cost for the U.S. pension plans over the next fiscal year are $374 million and $41 million, respectively.
Net periodic pension and postretirement plan expenses, calculated for all of International Paper’s plans, were as follows:
In millions 2015 2014 2013 2012 2011
Pension expense
U.S. plans (non-cash)$461$387$545$342$195
Non-U.S. plans 6—5 3 1
Postretirement expense
U.S. plans 8 7(1)(4)7
Non-U.S. plans 5 7 7 1 2
Net expense$480$401$556$342$205
The increase in 2015 U.S. pension expense principally reflects a decrease in the discount rate, updated mortality assumptions, higher amortization of unrecognized actuarial losses and a settlement charge in 2015.
Assuming that discount rates, expected long-term returns on plan assets and rates of future compensation increases remain the same as in 2015, projected future net periodic pension and postretirement plan expenses would be as follows:
In millions 2017 (1)2016(1)
Pension expense
U.S. plans (non-cash)$278$364
Non-U.S. plans 4 5
Postretirement expense
U.S. plans 14 14
Non-U.S. plans 8 5
Net expense$304$388
(1)Based on assumptions at December 31, 2015.
The Company estimates that it will record net pension expense of approximately $364 million for its U.S. defined benefit plans in 2016, with the decrease from expense of $461 million in 2015 reflecting an increase in the assumed discount rate to 4.40% in 2016 from 4.10% in 2015, updated demographic assumptions and lower unrecognized losses.
The market value of plan assets for International Paper’s U.S. qualified pension plan at December 31, 2015 totaled approximately $10.9 billion, consisting of approximately 48% equity securities, 33% debt securities, 10% real estate and 9% other assets. Plan assets include an immaterial amount of International Paper common stock.
The Company’s funding policy for its qualified pension plans is to contribute amounts sufficient to meet legal funding requirements, plus any additional amounts that the Company may determine to be appropriate considering the funded status of the plan, tax deductibility, the cash flows generated by the Company,
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and other factors. The Company continually reassesses the amount and timing of any discretionary contributions and could elect to make voluntary contributions in the future. There are no required contributions to the U.S. qualified plan in 2016. The nonqualified defined benefit plans are funded to the extent of benefit payments, which totaled $62 million for the year ended December 31, 2015.
Accounting for Stock-Based Compensation
The Company has a Performance Share Plan, which grants performance-based restricted stock units that are paid out in stock when the awards are earned. Such incentive compensation plans are accounted for under ASC 718, “Compensation - Stock Compensation.” This standard requires that the value of shares to be issued under this plan be recognized as compensation over the period in which the awards are earned based on the fair value of the awards, and requires the use of a number of judgments and assumptions in determining the timing and amount of such charges. Additionally, since a component of these awards is based on the Company’s performance over a specified period compared to other companies, the amount of expense recorded for a given period could require adjustments after the end of the period.
Income Taxes
International Paper records its global tax provision based on the respective tax rules and regulations for the jurisdictions in which it operates. Where the Company believes that a tax position is supportable for income tax purposes, the item is included in its income tax returns. Where treatment of a position is uncertain, liabilities are recorded based upon the Company’s evaluation of the “more likely than not” outcome considering technical merits of the position based on specific tax regulations and facts of each matter. Changes to recorded liabilities are only made when an identifiable event occurs that changes the likely outcome, such as settlement with the relevant tax authority, the expiration of statutes of limitation for the subject tax year, change in tax laws, or a recent court case that addresses the matter.
Valuation allowances are recorded to reduce deferred tax assets when it is more likely than not that a tax benefit will not be realized. Significant judgment is required in evaluating the need for and magnitude of appropriate valuation allowances against deferred tax assets. The realization of these assets is dependent on generating future taxable income, as well as successful implementation of various tax planning strategies.
While International Paper believes that these judgments and estimates are appropriate and reasonable under the circumstances, actual resolution
of these matters may differ from recorded estimated amounts.
The Company’s effective income tax rates, before equity earnings and discontinued operations, were 37%, 14% and (41)% for 2015, 2014 and 2013, respectively. These effective tax rates include the tax effects of certain special items that can significantly affect the effective income tax rate in a given year, but may not recur in subsequent years. Management believes that the effective tax rate computed after excluding these special items may provide a better estimate of the rate that might be expected in future years if no additional special items were to occur in those years. Excluding these special items, the effective income tax rate for 2015 was 33% of pre-tax earnings compared with 31% in 2014 and 26% in 2013. We estimate that the 2016 effective income tax rate will be approximately 34% based on expected earnings and business conditions.
RECENT ACCOUNTING DEVELOPMENTS
There were no new accounting pronouncements issued or effective during the fiscal year which have had or are expected to have a material impact on the Company’s consolidated financial statements. See Note 2 Recent Accounting Developments on pages 51 and 52 of Item 8. Financial Statements and Supplementary Data for a discussion of new accounting pronouncements.
LEGAL PROCEEDINGS
Information concerning the Company’s environmental and legal proceedings is set forth in Note 11 Commitments and Contingencies on pages 61 through 64 of Item 8. Financial Statements and Supplementary Data.
EFFECT OF INFLATION
While inflationary increases in certain input costs, such as energy, wood fiber and chemical costs, have an impact on the Company’s operating results, changes in general inflation have had minimal impact on our operating results in each of the last three years. Sales prices and volumes are more strongly influenced by economic supply and demand factors in specific markets and by exchange rate fluctuations than by inflationary factors.
FOREIGN CURRENCY EFFECTS
International Paper has operations in a number of countries. Its operations in those countries also export to, and compete with, imports from other regions. As such, currency movements can have a number of direct and indirect impacts on the Company’s financial statements. Direct impacts include the translation of international operations’ local currency financial statements into U.S. dollars and the remeasurement
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impact associated with non-functional currency financial assets and liabilities. Indirect impacts include the change in competitiveness of imports into, and exports out of, the United States (and the impact on local currency pricing of products that are traded internationally). In general, a weaker U.S. dollar and stronger local currency is beneficial to International Paper. The currencies that have the most impact are the Euro, the Brazilian real, the Polish zloty and the Russian ruble.
MARKET RISK
We use financial instruments, including fixed and variable rate debt, to finance operations, for capital spending programs and for general corporate purposes. Additionally, financial instruments, including various derivative contracts, are used to hedge exposures to interest rate, commodity and foreign currency risks. We do not use financial instruments for trading purposes. Information related to International Paper’s debt obligations is included in Note 13 Debt and Lines of Credit on pages 66 and 67 of Item 8. Financial Statements and Supplementary Data. A discussion of derivatives and hedging activities is included in Note 14 Derivatives and Hedging Activities on pages 67 through 71 of Item 8. Financial Statements and Supplementary Data.
The fair value of our debt and financial instruments varies due to changes in market interest and foreign currency rates and commodity prices since the inception of the related instruments. We assess this market risk utilizing a sensitivity analysis. The sensitivity analysis measures the potential loss in earnings, fair values and cash flows based on a hypothetical 10% change (increase and decrease) in interest and currency rates and commodity prices.
Interest Rate Risk
Our exposure to market risk for changes in interest rates relates primarily to short- and long-term debt obligations and investments in marketable securities. We invest in investment-grade securities of financial institutions and money market mutual funds with a minimum rating of AAA and limit exposure to any one issuer or fund. Our investments in marketable securities at December 31, 2015 and 2014 are stated at cost, which approximates market due to their short-term nature. Our interest rate risk exposure related to these investments was not material.
We issue fixed and floating rate debt in a proportion consistent with International Paper’s targeted capital structure, while at the same time taking advantage of market opportunities to reduce interest expense as appropriate. Derivative instruments, such as interest rate swaps, may be used to implement this capital structure. At December 31, 2015 and 2014, the net fair
value liability of financial instruments with exposure to interest rate risk was approximately $9.3 billion and $9.8 billion, respectively. The potential loss in fair value resulting from a 10% adverse shift in quoted interest rates would have been approximately $565 million and $410 million at December 31, 2015 and 2014, respectively.
Commodity Price Risk
The objective of our commodity exposure management is to minimize volatility in earnings due to large fluctuations in the price of commodities. Commodity swap or forward purchase contracts may be used to manage risks associated with market fluctuations in energy prices. The net fair value of such outstanding energy hedge contracts at December 31, 2015 and 2014 was approximately a $7 million and a $2 million liability, respectively. The potential loss in fair value resulting from a 10% adverse change in the underlying commodity prices would have been approximately $1 million at December 31, 2015 and 2014, respectively.
Foreign Currency Risk
International Paper transacts business in many currencies and is also subject to currency exchange rate risk through investments and businesses owned and operated in foreign countries. Our objective in managing the associated foreign currency risks is to minimize the effect of adverse exchange rate fluctuations on our after-tax cash flows. We address these risks on a limited basis by financing a portion of our investments in overseas operations with borrowings denominated in the same currency as the operation’s functional currency, or by entering into cross-currency and interest rate swaps, or foreign exchange contracts. At December 31, 2015 and 2014, the net fair value of financial instruments with exposure to foreign currency risk was approximately a $4 million and a $1 million asset, respectively. The potential loss in fair value for such financial instruments from a 10% adverse change in quoted foreign currency exchange rates would have been approximately $30 million and $52 million at December 31, 2015 and 2014, respectively.
ITEM 7A. QUANTITATIVE AND QUALITATIVE DISCLOSURES ABOUT MARKET RISK
See the preceding discussion and Note 14 Derivatives and Hedging Activities on pages 67 through 71 of Item 8. Financial Statements and Supplementary Data.
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ITEM 8. FINANCIAL STATEMENTS AND SUPPLEMENTARY DATA
REPORT OF MANAGEMENT ON:
Financial Statements
The management of International Paper Company is responsible for the preparation of the consolidated financial statements in this annual report and for establishing and maintaining adequate internal controls over financial reporting. The consolidated financial statements have been prepared using accounting principles generally accepted in the United States of America considered appropriate in the circumstances to present fairly the Company’s consolidated financial position, results of operations and cash flows on a consistent basis. Management has also prepared the other information in this annual report and is responsible for its accuracy and consistency with the consolidated financial statements.
As can be expected in a complex and dynamic business environment, some financial statement amounts are based on estimates and judgments. Even though estimates and judgments are used, measures have been taken to provide reasonable assurance of the integrity and reliability of the financial information contained in this annual report. We have formed a Disclosure Committee to oversee this process.
The accompanying consolidated financial statements have been audited by the independent registered public accounting firm, Deloitte& Touche LLP. During its audits, Deloitte& Touche LLP was given unrestricted access to all financial records and related data, including minutes of all meetings of stockholders and the board of directors and all committees of the board. Management believes that all representations made to the independent auditors during their audits were valid and appropriate.
Internal Control Over Financial Reporting
The management of International Paper Company is also responsible for establishing and maintaining adequate internal control over financial reporting. Internal control over financial reporting is the process designed by, or under the supervision of, our principal executive officer and principal financial officer, and effected by our Board of Directors, management and other personnel to provide reasonable assurance regarding the reliability of financial reporting and the preparation of financial statements for external purposes. All internal control systems have inherent limitations, including the possibility of circumvention and overriding of controls, and therefore can provide only reasonable assurance of achieving the designed control objectives. The Company’s internal control
system is supported by written policies and procedures, contains self-monitoring mechanisms, and is audited by the internal audit function. Appropriate actions are taken by management to correct deficiencies as they are identified.
The Company has assessed the effectiveness of its internal control over financial reporting as of December 31, 2015. In making this assessment, it used the criteria described in “Internal Control – Integrated Framework (2013)” issued by the Committee of Sponsoring Organizations of the Treadway Commission (COSO). Based on this assessment, management believes that, as of December 31, 2015, the Company’s internal control over financial reporting was effective.
The Company’s independent registered public accounting firm, Deloitte& Touche LLP, has issued its report on the effectiveness of the Company’s internal control over financial reporting. The report appears on pages 42 and 43.
Internal Control Environment And Board Of Directors Oversight
Our internal control environment includes an enterprise-wide attitude of integrity and control consciousness that establishes a positive “tone at the top.” This is exemplified by our ethics program that includes long-standing principles and policies on ethical business conduct that require employees to maintain the highest ethical and legal standards in the conduct of International Paper business, which have been distributed to all employees; a toll-free telephone helpline whereby any employee may anonymously report suspected violations of law or International Paper’s policy; and an office of ethics and business practice. The internal control system further includes careful selection and training of supervisory and management personnel, appropriate delegation of authority and division of responsibility, dissemination of accounting and business policies throughout International Paper, and an extensive program of internal audits with management follow-up.
The Board of Directors, assisted by the Audit and Finance Committee (Committee), monitors the integrity of the Company’s financial statements and financial reporting procedures, the performance of the Company’s internal audit function and independent auditors, and other matters set forth in its charter. The Committee, which consists of independent directors, meets regularly with representatives of management, and with the independent auditors and the Internal Auditor, with and without management representatives in attendance, to review their activities. The Committee’s Charter takes into account the New York Stock Exchange rules relating to Audit Committees and
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the SEC rules and regulations promulgated as a result of the Sarbanes-Oxley Act of 2002. The Committee has reviewed and discussed the consolidated financial statements for the year ended December 31, 2015, including critical accounting policies and significant management judgments, with management and the independent auditors. The Committee’s report recommending the inclusion of such financial statements in this Annual Report on Form 10-K will be set forth in our Proxy Statement.
MARK S. SUTTON
CHAIRMAN AND CHIEF EXECUTIVE OFFICER
CAROL L. ROBERTS
SENIOR VICE PRESIDENT AND CHIEF FINANCIAL OFFICER
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REPORT OF INDEPENDENT REGISTERED PUBLIC ACCOUNTING FIRM, ON CONSOLIDATED FINANCIAL STATEMENTS
To the Board of Directors and Shareholders of International Paper Company:
We have audited the accompanying consolidated balance sheets of International Paper Company and subsidiaries (the "Company") as of December 31, 2015 and 2014, and the related consolidated statements of operations, comprehensive income, changes in equity, and cash flows for each of the three years in the period ended December 31, 2015. Our audits also included the financial statement schedule listed in the Index at Item 15. These financial statements and financial statement schedule are the responsibility of the Company's management. Our responsibility is to express an opinion on the financial statements and financial statement schedule based on our audits.
We conducted our audits in accordance with the standards of the Public Company Accounting Oversight Board (United States). Those standards require that we plan and perform the audit to obtain reasonable assurance about whether the financial statements are free of material misstatement. An audit includes examining, on a test basis, evidence supporting the amounts and disclosures in the financial statements. An audit also includes assessing the accounting principles used and significant estimates made by management, as well as evaluating the overall financial statement presentation. We believe that our audits provide a reasonable basis for our opinion.
In our opinion, such consolidated financial statements present fairly, in all material respects, the financial position of International Paper Company and subsidiaries as of December 31, 2015 and 2014, and the results of their operations and their cash flows for each of the three years in the period ended December 31, 2015, in conformity with accounting principles generally accepted in the United States of America. Also, in our opinion, such financial statement schedule, when considered in relation to the basic consolidated financial statements taken as a whole, presents fairly, in all material respects, the information set forth therein.
We have also audited, in accordance with the standards of the Public Company Accounting Oversight Board (United States), the Company's internal control over financial reporting as of December 31, 2015, based on the criteria established in Internal Control - Integrated Framework (2013) issued by the Committee of Sponsoring Organizations of the Treadway Commission and our report dated February 25, 2016 expressed an unqualified opinion on the Company's
internal control over financial reporting.
Memphis, Tennessee
February 25, 2016
REPORT OF INDEPENDENT REGISTERED PUBLIC ACCOUNTING FIRM, ON INTERNAL CONTROL OVER FINANCIAL REPORTING
To the Board of Directors and Shareholders of International Paper Company:
We have audited the internal control over financial reporting of International Paper Company and subsidiaries (the "Company") as of December 31, 2015, based on criteria established in Internal Control - Integrated Framework (2013) issued by the Committee of Sponsoring Organizations of the Treadway Commission. The Company's management is responsible for maintaining effective internal control over financial reporting and for its assessment of the effectiveness of internal control over financial reporting, included in the accompanying Report of Management on Internal Control Over Financial Reporting. Our responsibility is to express an opinion on the Company's internal control over financial reporting based on our audit.
We conducted our audit in accordance with the standards of the Public Company Accounting Oversight Board (United States). Those standards require that we plan and perform the audit to obtain reasonable assurance about whether effective internal control over financial reporting was maintained in all material respects. Our audit included obtaining an understanding of internal control over financial reporting, assessing the risk that a material weakness exists, testing and evaluating the design and operating effectiveness of internal control based on the assessed risk, and performing such other procedures as we considered necessary in the circumstances. We believe that our audit provides a reasonable basis for our opinion.
A company's internal control over financial reporting is a process designed by, or under the supervision of, the company's principal executive and principal financial officers, or persons performing similar functions, and effected by the company's board of directors, management, and other personnel to provide reasonable assurance regarding the reliability of financial reporting and the preparation of financial statements for external purposes in accordance with
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generally accepted accounting principles. A company's internal control over financial reporting includes those policies and procedures that (1) pertain to the maintenance of records that, in reasonable detail, accurately and fairly reflect the transactions and dispositions of the assets of the company; (2) provide reasonable assurance that transactions are recorded as necessary to permit preparation of financial statements in accordance with generally accepted accounting principles, and that receipts and expenditures of the company are being made only in accordance with authorizations of management and directors of the company; and (3) provide reasonable assurance regarding prevention or timely detection of unauthorized acquisition, use, or disposition of the company's assets that could have a material effect on the financial statements.
Because of the inherent limitations of internal control over financial reporting, including the possibility of collusion or improper management override of controls, material misstatements due to error or fraud may not be prevented or detected on a timely basis. Also, projections of any evaluation of the effectiveness of the internal control over financial reporting to future periods are subject to the risk that the controls may become inadequate because of changes in conditions, or that the degree of compliance with the policies or procedures may deteriorate.
In our opinion, the Company maintained, in all material respects, effective internal control over financial reporting as of December 31, 2015, based on the criteria established in Internal Control - Integrated Framework (2013) issued by the Committee of Sponsoring Organizations of the Treadway Commission.
We have also audited, in accordance with the standards of the Public Company Accounting Oversight Board (United States), the consolidated financial statements and financial statement schedule as of and for the year ended December 31, 2015 of the Company and our report dated February 25, 2016 expressed an unqualified opinion on those financial statements and financial statement schedule.
Memphis, Tennessee
February 25, 2016
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CONSOLIDATED STATEMENT OF OPERATIONS
In millions, except per share amounts, for the years ended December 31 2015 2014 2013
NET SALES$22,365$23,617$23,483
COSTS AND EXPENSES
Cost of products sold 15,468 16,254 16,282
Selling and administrative expenses 1,645 1,793 1,796
Depreciation, amortization and cost of timber harvested 1,294 1,406 1,531
Distribution expenses 1,406 1,521 1,583
Taxes other than payroll and income taxes 168 180 178
Restructuring and other charges 252 846 156
Impairment of goodwill and other intangibles 137 100 127
Net (gains) losses on sales and impairments of businesses 174 38 3
Net bargain purchase gain on acquisition of business——(13)
Interest expense, net 555 607 612
EARNINGS (LOSS) FROM CONTINUING OPERATIONS BEFORE INCOME TAXES AND EQUITY EARNINGS 1,266 872 1,228
Income tax provision (benefit)466 123(498)
Equity earnings (loss), net of taxes 117(200)(39)
EARNINGS (LOSS) FROM CONTINUING OPERATIONS 917 549 1,687
Discontinued operations, net of taxes—(13)(309)
NET EARNINGS (LOSS)917 536 1,378
Less: Net earnings (loss) attributable to noncontrolling interests(21)(19)(17)
NET EARNINGS (LOSS) ATTRIBUTABLE TO INTERNATIONAL PAPER
COMPANY$938$555$1,395
BASIC EARNINGS (LOSS) PER SHARE ATTRIBUTABLE TO INTERNATIONAL PAPER COMPANY COMMON SHAREHOLDERS
Earnings (loss) from continuing operations$2.25$1.33$3.85
Discontinued operations, net of taxes—(0.03)(0.70)
Net earnings (loss)$2.25$1.30$3.15
DILUTED EARNINGS (LOSS) PER SHARE ATTRIBUTABLE TO INTERNATIONAL PAPER COMPANY COMMON SHAREHOLDERS
Earnings (loss) from continuing operations$2.23$1.31$3.80
Discontinued operations, net of taxes—(0.02)(0.69)
Net earnings (loss)$2.23$1.29$3.11
AMOUNTS ATTRIBUTABLE TO INTERNATIONAL PAPER COMPANY COMMON SHAREHOLDERS
Earnings (loss) from continuing operations$938$568$1,704
Discontinued operations, net of taxes—(13)(309)
Net earnings (loss)$938$555$1,395
The accompanying notes are an integral part of these financial statements.
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CONSOLIDATED STATEMENT OF COMPREHENSIVE INCOME
In millions for the years ended December 31 2015 2014 2013
NET EARNINGS (LOSS)$917$536$1,378
OTHER COMPREHENSIVE INCOME (LOSS), NET OF TAX
Amortization of pension and postretirement prior service costs and net loss:
U.S. plans (less tax of $186, $154 and $195)296 242 307
Pension and postretirement liability adjustments:
U.S. plans (less tax of $206, $798 and $756)(329)(1,253)1,188
Non-U.S. plans (less tax of $0, $5 and $3)(2)(18)(4)
Change in cumulative foreign currency translation adjustment(1,042)(876)(426)
Net gains/losses on cash flow hedging derivatives:
Net gains (losses) arising during the period (less tax of $3, $3 and $2)(3)10—
Reclassification adjustment for (gains) losses included in net earnings (less tax of $8, $1 and $3)12(4)(7)
TOTAL OTHER COMPREHENSIVE INCOME (LOSS), NET OF TAX(1,068)(1,899)1,058
Comprehensive Income (Loss)(151)(1,363)2,436
Net (Earnings) Loss Attributable to Noncontrolling Interests 21 19 17
Other Comprehensive (Income) Loss Attributable to Noncontrolling Interests 6 12 23
COMPREHENSIVE INCOME (LOSS) ATTRIBUTABLE TO INTERNATIONAL PAPER COMPANY$(124)$(1,332)$2,476
The accompanying notes are an integral part of these financial statements.
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CONSOLIDATED BALANCE SHEET
In millions, except per share amounts, at December 31 2015 2014
ASSETS
Current Assets
Cash and temporary investments$1,050$1,881
Accounts and notes receivable, less allowances of $70 in 2015 and $82 in 2014 2,675 3,083
Inventories 2,228 2,424
Deferred income tax assets 312 331
Other current assets 212 240
Total Current Assets 6,477 7,959
Plants, Properties and Equipment, net 11,980 12,728
Forestlands 366 507
Investments 228 248
Financial Assets of Special Purpose Entities (Note 12)7,014 2,145
Goodwill 3,335 3,773
Deferred Charges and Other Assets 1,187 1,324
TOTAL ASSETS$30,587$28,684
LIABILITIES AND EQUITY
Current Liabilities
Notes payable and current maturities of long-term debt$426$742
Accounts payable 2,078 2,664
Accrued payroll and benefits 434 477
Other accrued liabilities 986 1,026
Total Current Liabilities 3,924 4,909
Long-Term Debt 8,900 8,631
Nonrecourse Financial Liabilities of Special Purpose Entities (Note 12)6,277 2,050
Deferred Income Taxes 3,231 3,063
Pension Benefit Obligation 3,548 3,819
Postretirement and Postemployment Benefit Obligation 364 396
Other Liabilities 434 553
Commitments and Contingent Liabilities (Note 11)
Equity
Common stock $1 par value, 2015 & 2014 – 448.9 shares 449 449
Paid-in capital 6,243 6,245
Retained earnings 4,649 4,409
Accumulated other comprehensive loss(5,708)(4,646)
5,633 6,457
Less: Common stock held in treasury, at cost, 2015 – 36.776 shares and 2014 – 28.734 shares 1,749 1,342
Total Shareholders’ Equity 3,884 5,115
Noncontrolling interests 25 148
Total Equity 3,909 5,263
TOTAL LIABILITIES AND EQUITY$30,587$28,684
The accompanying notes are an integral part of these financial statements.
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CONSOLIDATED STATEMENT OF CASH FLOWS
In millions for the years ended December 31 2015 2014 2013
OPERATING ACTIVITIES
Net earnings (loss)$917$536$1,378
Depreciation, amortization, and cost of timber harvested 1,294 1,414 1,547
Deferred income tax provision (benefit), net 281(135)146
Restructuring and other charges 252 881 210
Pension plan contribution(750)(353)(31)
Net bargain purchase gain on acquisition of business——(13)
Periodic pension expense, net 461 387 545
Net (gains) losses on sales and impairments of businesses 174 38 3
Equity (earnings) losses, net of taxes(117)200 39
Release of tax reserves——(775)
Impairment of goodwill and other intangible assets 137 100 527
Other, net 153 167(62)
Changes in current assets and liabilities
Accounts and notes receivable 7(97)(134)
Inventories(131)(103)(114)
Accounts payable and accrued liabilities(89)(18)(110)
Interest payable(17)(18)(57)
Other 8 78(71)
CASH PROVIDED BY (USED FOR) OPERATING ACTIVITIES 2,580 3,077 3,028
INVESTMENT ACTIVITIES
Invested in capital projects(1,487)(1,366)(1,198)
Acquisitions, net of cash acquired——(505)
Proceeds from divestitures 23—726
Proceeds from spinoff—411—
Investment in Special Purpose Entities(198)——
Proceeds from sale of fixed assets 37 61 65
Other(114)34 85
CASH PROVIDED BY (USED FOR) INVESTMENT ACTIVITIES(1,739)(860)(827)
FINANCING ACTIVITIES
Repurchase of common stock and payments of restricted stock tax withholding(605)(1,062)(512)
Issuance of common stock 2 66 298
Issuance of debt 6,873 1,982 241
Reduction of debt(6,947)(2,095)(845)
Change in book overdrafts(14)30(123)
Dividends paid(685)(620)(554)
Acquisition of redeemable noncontrolling interest—(114)—
Debt tender premiums paid(211)(269)—
Redemption of securities——(150)
Other(14)(4)(43)
CASH PROVIDED BY (USED FOR) FINANCING ACTIVITIES(1,601)(2,086)(1,688)
Effect of Exchange Rate Changes on Cash(71)(52)(13)
Change in Cash and Temporary Investments(831)79 500
Cash and Temporary Investments
Beginning of the period 1,881 1,802 1,302
End of the period$1,050$1,881$1,802
The accompanying notes are an integral part of these financial statements.
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CONSOLIDATED STATEMENT OF CHANGES IN EQUITY
In millions Common Stock Issued Paid-in Capital Retained Earnings Accumulated Other Comprehensive Income (Loss)Treasury Stock Total International Paper Shareholders’ Equity Noncontrolling Interests Total Equity
BALANCE, JANUARY 1, 2013$440$6,042$3,662$(3,840)$—$6,304$332$6,636
Issuance of stock for various plans, net 7 421——(20)448—448
Repurchase of stock————512(512)—(512)
Dividends——(567)——(567)—(567)
Dividends paid to noncontrolling interests by subsidiary——————(1)(1)
Noncontrolling interests of acquired entities——(44)——(44)(112)(156)
Comprehensive income (loss)——1,395 1,081—2,476(40)2,436
BALANCE, DECEMBER 31, 2013 447 6,463 4,446(2,759)492 8,105 179 8,284
Issuance of stock for various plans, net 2 69——(212)283—283
Repurchase of stock————1,062(1,062)—(1,062)
xpedx spinoff—(287)———(287)—(287)
Dividends——(633)——(633)—(633)
Acquisition of redeemable noncontrolling interests——46——46—46
Remeasurement of redeemable noncontrolling interest——(5)——(5)—(5)
Comprehensive income (loss)——555(1,887)—(1,332)(31)(1,363)
BALANCE, DECEMBER 31, 2014 449 6,245 4,409(4,646)1,342 5,115 148 5,263
Issuance of stock for various plans, net—35——(198)233—233
Repurchase of stock————605(605)—(605)
Dividends——(698)——(698)—(698)
Transactions of equity method investees—(37)———(37)—(37)
Divestiture of noncontrolling interests——————(96)(96)
Comprehensive income (loss)——938(1,062)—(124)(27)(151)
BALANCE, DECEMBER 31, 2015$449$6,243$4,649$(5,708)$1,749$3,884$25$3,909
The accompanying notes are an integral part of these financial statements.
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NOTES TO CONSOLIDATED FINANCIAL STATEMENTS
NOTE 1 SUMMARY OF BUSINESS AND SIGNIFICANT ACCOUNTING POLICIES
NATURE OF BUSINESS
International Paper (the Company) is a global paper and packaging company with primary markets and manufacturing operations in North America, Europe, Latin America, Russia, Asia, Africa and the Middle East. Substantially all of our businesses have experienced, and are likely to continue to experience, cycles relating to available industry capacity and general economic conditions.
FINANCIAL STATEMENTS
These consolidated financial statements have been prepared in conformity with accounting principles generally accepted in the United States that require the use of management’s estimates. Actual results could differ from management’s estimates.
On July 1, 2014, International Paper completed the spinoff of its distribution business, xpedx, and xpedx's merger with Unisource Worldwide, Inc., with the combined companies now operating as Veritiv Corporation (Veritiv). As a result of the spinoff, all prior year amounts have been adjusted to reflect xpedx as a discontinued operation. See Note 7 for further discussion.
CONSOLIDATION
The consolidated financial statements include the accounts of International Paper and its wholly-owned, controlled majority-owned and financially controlled subsidiaries. All significant intercompany balances and transactions are eliminated.
Investments in affiliated companies where the Company has significant influence over their operations are accounted for by the equity method. International Paper’s share of affiliates’ results of operations totaled earnings (loss) of $117 million, $(200) million and $(39) million in 2015, 2014 and 2013, respectively.
REVENUE RECOGNITION
Revenue is recognized when the customer takes title and assumes the risks and rewards of ownership. Revenue is recorded at the time of shipment for terms designated f.o.b. (free on board) shipping point. For sales transactions designated f.o.b. destination, revenue is recorded when the product is delivered to the customer’s delivery site, when title and risk of loss are transferred. Timber and forestland sales revenue is generally recognized when title and risk of loss pass to the buyer.
SHIPPING AND HANDLING COSTS
Shipping and handling costs, such as freight to our customers’ destinations, are included in distribution expenses in the consolidated statement of operations. When shipping and handling costs are included in the sales price charged for our products, they are recognized in net sales.
ANNUAL MAINTENANCE COSTS
Costs for repair and maintenance activities are expensed in the month that the related activity is performed under the direct expense method of accounting.
TEMPORARY INVESTMENTS
Temporary investments with an original maturity of three months or less are treated as cash equivalents and are stated at cost, which approximates market value.
INVENTORIES
Inventories are valued at the lower of cost or market value and include all costs directly associated with manufacturing products: materials, labor and manufacturing overhead. In the United States, costs of raw materials and finished pulp and paper products, are generally determined using the last-in, first-out method. Other inventories are valued using the first-in, first-out or average cost methods.
PLANTS, PROPERTIES AND EQUIPMENT
Plants, properties and equipment are stated at cost, less accumulated depreciation. Expenditures for betterments are capitalized, whereas normal repairs and maintenance are expensed as incurred. The units-of-production method of depreciation is used for pulp and paper mills, and the straight-line method is used for other plants and equipment. Annual straight-line depreciation rates generally are, for buildings — 2.50% to 5.00%, and for machinery and equipment — 5% to 33%.
FORESTLANDS
At December 31, 2015, International Paper and its subsidiaries owned or managed approximately 335,000 acres of forestlands in Brazil, and through licenses and forest management agreements, had harvesting rights on government-owned forestlands in Russia. Costs attributable to timber are expensed as trees are cut. The rate charged is determined annually based on the relationship of incurred costs to estimated current merchantable volume.
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GOODWILL
Goodwill relating to a single business reporting unit is included as an asset of the applicable segment. For goodwill impairment testing, this goodwill is allocated to reporting units. Annual testing for possible goodwill impairment is performed as of the beginning of the fourth quarter of each year, with additional interim testing performed when management believes that it is more likely than not events or circumstances have occurred that would result in the impairment of a reporting unit’s goodwill.
In performing this testing, the Company estimates the fair value of its reporting units using the projected future cash flows to be generated by each unit over the estimated remaining useful operating lives of the unit’s assets, discounted using the estimated cost of capital for each reporting unit. These estimated fair values are then analyzed for reasonableness by comparing them to historic market transactions for businesses in the industry, and by comparing the sum of the reporting unit fair values and other corporate assets and liabilities divided by diluted common shares outstanding to the Company’s traded stock price on the testing date. For reporting units whose recorded value of net assets plus goodwill is in excess of their estimated fair values, the fair values of the individual assets and liabilities of the respective reporting units are then determined to calculate the amount of any goodwill impairment charge required. See Note 9 for further discussion.
IMPAIRMENT OF LONG-LIVED ASSETS
Long-lived assets are reviewed for impairment upon the occurrence of events or changes in circumstances that indicate that the carrying value of the assets may not be recoverable, measured by comparing their net book value to the undiscounted projected future cash flows generated by their use. Impaired assets are recorded at their estimated fair value.
INCOME TAXES
International Paper uses the asset and liability method of accounting for income taxes whereby deferred income taxes are recorded for the future tax consequences attributable to differences between the financial statement and tax bases of assets and liabilities. Deferred tax assets and liabilities are measured using enacted tax rates expected to apply to taxable income in the years in which those temporary differences are expected to be recovered or settled. Deferred tax assets and liabilities are remeasured to reflect new tax rates in the periods rate changes are enacted.
International Paper records its worldwide tax provision based on the respective tax rules and regulations for the jurisdictions in which it operates. Where the
Company believes that a tax position is supportable for income tax purposes, the item is included in its income tax returns. Where treatment of a position is uncertain, liabilities are recorded based upon the Company’s evaluation of the “more likely than not” outcome considering the technical merits of the position based on specific tax regulations and the facts of each matter. Changes to recorded liabilities are made only when an identifiable event occurs that changes the likely outcome, such as settlement with the relevant tax authority, the expiration of statutes of limitation for the subject tax year, a change in tax laws, or a recent court case that addresses the matter.
While the judgments and estimates made by the Company are based on management’s evaluation of the technical merits of a matter, assisted as necessary by consultation with outside consultants, historical experience and other assumptions that management believes are appropriate and reasonable under current circumstances, actual resolution of these matters may differ from recorded estimated amounts, resulting in charges or credits that could materially affect future financial statements.
STOCK-BASED COMPENSATION
Compensation costs resulting from all stock-based compensation transactions are measured and recorded in the consolidated financial statements based on the grant-date fair value of the equity or liability instruments issued. Compensation cost is recognized over the period that an employee provides service in exchange for the award.
ENVIRONMENTAL REMEDIATION COSTS
Costs associated with environmental remediation obligations are accrued when such costs are probable and reasonably estimable. Such accruals are adjusted as further information develops or circumstances change. Costs of future expenditures for environmental remediation obligations are discounted to their present value when the amount and timing of expected cash payments are reliably determinable.
ASSET RETIREMENT OBLIGATIONS
A liability and an asset are recorded equal to the present value of the estimated costs associated with the retirement of long-lived assets where a legal or contractual obligation exists and the liability can be reasonably estimated. The liability is accreted over time and the asset is depreciated over the life of the related equipment or facility. International Paper’s asset retirement obligations principally relate to closure costs for landfills. Revisions to the liability could occur due to changes in the estimated costs or timing of closures, or possible new federal or state regulations affecting these closures.
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In connection with potential future closures or redesigns of certain production facilities, it is possible that the Company may be required to take steps to remove certain materials from these facilities. Applicable regulations and standards provide that the removal of certain materials would only be required if the facility were to be demolished or underwent major renovations. At this time, any such obligations have an indeterminate settlement date, and the Company believes that adequate information does not exist to apply an expected-present-value technique to estimate any such potential obligations. Accordingly, the Company does not record a liability for such remediation until a decision is made that allows reasonable estimation of the timing of such remediation.
TRANSLATION OF FINANCIAL STATEMENTS
Balance sheets of international operations are translated into U.S. dollars at year-end exchange rates, while statements of operations are translated at average rates. Adjustments resulting from financial statement translations are included as cumulative translation adjustments in Accumulated other comprehensive loss.
NOTE 2 RECENT ACCOUNTING DEVELOPMENTS
Other than as described below, no new accounting pronouncement issued or effective during the fiscal year has had or is expected to have a material impact on the consolidated financial statements.
CLASSIFICATION OF DEFERRED TAXES
In November 2015, the Financial Accounting Standards Board (FASB) issued Accounting Standards Update (ASU) 2015-17, "Balance Classification of Deferred Taxes." This ASU requires entities to offset all deferred tax assets and liabilities (and valuation allowances) for each tax-paying jurisdiction within each tax-paying component. The net deferred tax must be presented as a single noncurrent amount. This ASU is effective for annual reporting periods beginning after December 15, 2016, and interim periods within those years. Early adoption is permitted. The application of the requirements of this guidance is not expected to have a material effect on the consolidated financial statements.
BUSINESS COMBINATIONS
In September 2015, the FASB issued ASU 2015-16, "Business Combinations - Simplifying the Accounting for Measurement Period Adjustments." This ASU provides that an acquirer must recognize adjustments to provisional amounts that are identified during the measurement period in the reporting period in which the adjustment amounts are determined. The ASU also requires acquirers to present separately on the face of the income statement, or disclose in the notes, the
portion of the amount recorded in current-period earnings by line item that would have been recorded in previous reporting periods if the adjustment to the provisional amounts had been recognized at the acquisition date. This ASU is effective for annual reporting periods beginning after December 15, 2016, and interim periods within fiscal years beginning after December 15, 2017. This ASU must be applied prospectively to adjustments to provisional amounts that occur after the effective date. Early adoption is permitted for financial statements that have not been issued. The Company is currently evaluating the provisions of this guidance.
INVENTORY
In July 2015, the FASB issued ASU 2015-11, "Simplifying the Measurement of Inventory." This ASU provides that entities should measure inventory at the lower of cost and net realizable value. Net realizable value is the estimated selling prices in the ordinary course of business less reasonably predictable costs of completion, disposal and transportation. Subsequent measurement is unchanged for inventory measure using LIFO or the retail inventory method. This ASU is effective for annual reporting periods beginning after December 15, 2016, and interim periods within those years. Early adoption is permitted. The Company is currently evaluating the provisions of this guidance.
CLOUD COMPUTING ARRANGEMENTS
In April 2015, the FASB issued ASU 2015-05, "Customer's Accounting for Fees Paid in a Cloud Computing Arrangement." This ASU provides clarification on whether a cloud computing arrangement includes a software license. If a software license is included, the customer should account for the license consistent with its accounting of other software licenses. If a software license is not included, the arrangement should be accounted for as a service contract. This ASU is effective for annual reporting periods beginning after December 15, 2015, and interim periods within those years. Early adoption is permitted. The application of the requirements of this guidance is not expected to have a material effect on the consolidated financial statements.
DEBT ISSUANCE COSTS
In April 2015, the FASB issued ASU 2015-03, "Interest - Imputation of Interest (Subtopic 835-30: Simplifying the Presentation of Debt Issuance Costs)," which simplifies the balance sheet presentation of the costs for issuing debt. This ASU is effective for annual reporting periods beginning after December 15, 2015, and interim periods within those years; however, early adoption is allowed. An entity should apply the new guidance on a retrospective basis, wherein the balance sheet of each individual period presented should be
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adjusted to reflect the period-specific effects of applying the new guidance. The application of the requirements of this guidance is not expected to have a material effect on the consolidated financial statements.
CONSOLIDATION
In February 2015, the FASB issued ASU 2015-02, "Consolidation," which amends the requirements for consolidation and significantly changes the consolidation analysis required. This ASU is effective for annual reporting periods beginning after December 15, 2015, and interim periods within those years. The application of the requirements of this guidance is not expected to have a material effect on the consolidated financial statements.
SHARE-BASED PAYMENT
In June 2014, the FASB issued ASU 2014-12, "Accounting for Share-based Payments When the Terms of an Award Provide That Performance Target Could Be Achieved After the Requisite Service Period." This guidance provides that entities should treat performance targets that can be met after the requisite service period of a share-based payment award as performance conditions that affect vesting. As such, an entity should not record compensation expense related to an award for which transfer to the employee is contingent on the entity's satisfaction of a performance target until it becomes probable that the performance target will be met. This ASU is effective for annual reporting periods beginning after December 15, 2015, and interim periods within those years. The application of the requirements of this guidance is not expected to have a material effect on the consolidated financial statements.
REVENUE RECOGNITION
In May 2014, the FASB issued ASU 2014-09, "Revenue from Contracts with Customers." The guidance replaces most existing revenue recognition guidance and provides that an entity should recognize revenue to depict the transfer of promised goods or services to customers in an amount that reflects the consideration to which the entity expects to be entitled in exchange for those goods and services. This ASU is effective for annual reporting periods beginning after December 15, 2017, and interim periods within those years, and permits the use of either the retrospective or cumulative effect transition method. The Company is currently evaluating the provisions of this guidance.
NOTE 3 EARNINGS PER SHARE ATTRIBUTABLE TO INTERNATIONAL PAPER COMPANY COMMON SHAREHOLDERS
Basic earnings per share is computed by dividing earnings by the weighted average number of common shares outstanding. Diluted earnings per share is computed assuming that all potentially dilutive securities, including “in-the-money” stock options, were converted into common shares.
A reconciliation of the amounts included in the computation of basic earnings (loss) per share from continuing operations, and diluted earnings (loss) per share from continuing operations is as follows:
In millions, except per share amounts 2015 2014 2013
Earnings (loss) from continuing operations$938$568$1,704
Effect of dilutive securities (a)———
Earnings (loss) from continuing operations– assuming dilution$938$568$1,704
Average common shares outstanding 417.4 427.7 443.3
Effect of dilutive securities (a):
Restricted performance share plan 3.2 4.2 4.5
Stock options (b)—0.1 0.3
Average common shares outstanding –assuming dilution 420.6 432.0 448.1
Basic earnings (loss) per share from continuing operations$2.25$1.33$3.85
Diluted earnings (loss) per share from continuing operations$2.23$1.31$3.80
(a)Securities are not included in the table in periods when antidilutive.
(b)Options to purchase shares were not included in the computation of diluted common shares outstanding if their exercise price exceeded the average market price of the Company’s common stock for each respective reporting date.
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NOTE 4 OTHER COMPREHENSIVE INCOME
The following table presents changes in AOCI for the year ended December 31, 2015:
In millions Defined Benefit Pension and Postretirement Items (a)Change in Cumulative Foreign Currency Translation Adjustments (a)Net Gains and Losses on Cash Flow Hedging Derivatives (a)Total (a)
Balance as of December 31, 2014$(3,134)$(1,513)$1$(4,646)
Other comprehensive income (loss) before reclassifications(331)(1,002)(3)(1,336)
Amounts reclassified from accumulated other comprehensive income 296(40)12 268
Net Current Period Other Comprehensive Income(35)(1,042)9(1,068)
Other Comprehensive Income (Loss) Attributable to Noncontrolling Interest—6—6
Balance as of December 31, 2015$(3,169)$(2,549)$10$(5,708)
(a) All amounts are net of tax. Amounts in parentheses indicate debits to AOCI.
The following table presents changes in AOCI for the year ended December 31, 2014:
In millions Defined Benefit Pension and Postretirement Items (a)Change in Cumulative Foreign Currency Translation Adjustments (a)Net Gains and Losses on Cash Flow Hedging Derivatives (a)Total (a)
Balance as of December 31, 2013$(2,105)$(649)$(5)$(2,759)
Other comprehensive income (loss) before reclassifications(1,271)(863)10(2,124)
Amounts reclassified from accumulated other comprehensive income 242(13)(4)225
Net Current Period Other Comprehensive Income(1,029)(876)6(1,899)
Other Comprehensive Income (Loss) Attributable to Noncontrolling Interest—12—12
Balance as of December 31, 2014$(3,134)$(1,513)$1$(4,646)
(a) All amounts are net of tax. Amounts in parentheses indicate debits to AOCI.
The following table presents changes in AOCI for the year ended December 31, 2013:
In millions Defined Benefit Pension and Postretirement Items (a)Change in Cumulative Foreign Currency Translation Adjustments (a)Net Gains and Losses on Cash Flow Hedging Derivatives (a)Total (a)
Balance as of December 31, 2012$(3,596)$(246)$2$(3,840)
Other comprehensive income (loss) before reclassifications 1,184(443)—741
Amounts reclassified from accumulated other comprehensive income 307 17(7)317
Net Current Period Other Comprehensive Income 1,491(426)(7)1,058
Other Comprehensive Income (Loss) Attributable to Noncontrolling Interest—23—23
Balance as of December 31, 2013$(2,105)$(649)$(5)$(2,759)
(a) All amounts are net of tax. Amounts in parentheses indicate debits to AOCI.
The following table presents details of the reclassifications out of AOCI for the three years ended:
Details About Accumulated Other Comprehensive Income Components Amount Reclassified from Accumulated Other Comprehensive Income (a)Location of Amount Reclassified from AOCI
2015 2014 2013
In millions
Defined benefit pension and postretirement items:
Prior-service costs$(33)$(17)$(9)(b)Cost of products sold
Actuarial gains/(losses)(449)(379)(493)(b)Cost of products sold
Total pre-tax amount(482)(396)(502)
Tax (expense)/benefit 186 154 195
Net of tax(296)(242)(307)
Change in cumulative foreign currency translation adjustments:
Business acquisition/divestiture 40 13(17)Net (gains) losses on sales and impairments of businesses or Retained earnings
Tax (expense)/benefit———
Net of tax 40 13(17)
Net gains and losses on cash flow hedging derivatives:
Foreign exchange contracts(20)3 10(c)Cost of products sold
Total pre-tax amount(20)3 10
Tax (expense)/benefit 8 1(3)
Net of tax(12)4 7
Total reclassifications for the period$(268)$(225)$(317)
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(a) Amounts in parentheses indicate debits to earnings/loss.
(b) These accumulated other comprehensive income components are included in the computation of net periodic pension cost (see Note 16 for additional details).
(c) This accumulated other comprehensive income component is included in our derivatives and hedging activities (see Note 14 for additional details).
NOTE 5 RESTRUCTURING CHARGES AND OTHER ITEMS
2015: During 2015, total restructuring and other charges of $252 million before taxes were recorded. These charges included:
In millions 2015
Early debt extinguishment costs (see Note 13)$207
Timber monetization restructuring 16
Legal liability reserve adjustment 15
Riegelwood mill conversion costs net of proceeds from the sale of Carolina Coated Bristols brand (a)8
Other 6
Total$252
(a)Includes $5 million of severance charges, $24 million of accelerated depreciation, sale proceeds of $22 million and $1 million of other charges.
Included in the $252 million of restructuring and other charges is severance expense of $5 million which is related to 69 employees.
2014: During 2014, total restructuring and other charges of $846 million before taxes were recorded. These charges included:
In millions 2014
Early debt extinguishment costs (see Note 13)$276
Courtland mill shutdown (a)554
Other (b)16
Total$846
(a) Includes $464 million of accelerated depreciation, $24 million of inventory impairment charges, $26 million of severance charges and $40 million of other charges which are recorded in the Printing Papers segment.
(b) Includes $15 million of severance charges.
Included in the $846 million of restructuring and other charges is severance expense of $41 million which is related to 957 employees.
2013: During 2013, total restructuring and other charges of $156 million before taxes were recorded. These charges included:
In millions 2013
Early debt extinguishment costs (see Note 13)$25
Courtland mill shutdown (a)118
Box plant closures(13)
Augusta paper machine shutdown (b)45
Insurance reimbursements(30)
Other (c)11
Total$156
(a) Includes $73 million of accelerated depreciation and other non-cash charges, $42 million of severance charges and $3 million of other charges which are recorded in the Printing Papers segment. During 2013, the Company accelerated depreciation for certain Courtland assets, and diligently evaluated certain other assets for possible alternative uses by one of our other businesses. The net book value of these assets at December 31, 2013 was approximately $470 million.
(b) Includes $39 million of accelerated depreciation charges, $2 million of severance charges and $4 million of other charges which are recorded in the Consumer Packaging segment.
(c) Includes $2 million of severance charges.
Included in the $156 million of restructuring and other charges is severance expense of $46 million which is related to 1,384 employees.
ALTERNATIVE FUEL MIXTURE TAX CREDIT
On July 19, 2011 the Company filed an amended 2009 tax return claiming alternative fuel mixture tax credits as non-taxable income. The amended position has been accepted by the Internal Revenue Service (IRS) in the closing of the IRS tax audit for the years 2006 - 2009. As a result, during 2013, the Company recognized an income tax benefit of $753 million related to the non-taxability of the alternative fuel mixture tax credits.
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NOTE 6 ACQUISITIONS AND JOINT VENTURES
OLMUKSAN
2014: In May 2014, the Company conducted a voluntary tender offer for the remaining outstanding 12.6% public shares of Olmuksan. The Company also purchased outstanding shares of Olmuksan outside of the tender offer. As of December 31, 2014 and 2015, the Company owned 91.7% of Olmuksan's outstanding and issued shares.
2013: On January 3, 2013, International Paper completed the acquisition (effective date of acquisition on January 1, 2013) of the shares of its joint venture partner, Sabanci Holding, in the Turkish corrugated packaging company, Olmuksa International Paper Sabanci Ambalaj Sanayi ve Ticaret A.S., now called Olmuksan International Paper Ambalaj Sanayi ve Ticaret A.S. (Olmuksan), for a purchase price of $56 million. The acquired shares represented 43.7% of Olmuksan's shares. Prior to this acquisition, International Paper held a 43.7% equity interest in Olmuksan.
Because the transaction resulted in International Paper becoming the majority shareholder, owning 87.4% of Olmuksan's outstanding and issued shares, its completion triggered a mandatory call for tender of the remaining public shares which began in March 2013 and ended in April 2013, with no shares tendered. As a result, the 12.6% owned by other parties were considered non-controlling interests. Olmuksan's financial results have been consolidated with the Company's Industrial Packaging segment beginning January 1, 2013, the effective date which International Paper obtained majority control of the entity.
Following the transaction, the Company's previously held 43.7% equity interest in Olmuksan was remeasured to a fair value of $75 million, resulting in a gain of $9 million. In addition, the cumulative translation adjustment balance of $17 million relating to the previously held equity interest was reclassified, as expense, from accumulated other comprehensive income.
The final purchase price allocation indicated that the sum of the cash consideration paid, the fair value of the noncontrolling interest and the fair value of the previously held interest was less than the fair value of the underlying assets by $21 million, resulting in a bargain purchase price gain being recorded on this transaction. The aforementioned remeasurement of equity interest gain, the cumulative translation adjustment to expense, and the bargain purchase gain are included in the Net bargain purchase gain on acquisition of business in the accompanying consolidated statement of operations.
The following table summarizes the final allocation of the purchase price to the fair value of assets and liabilities acquired as of January 1, 2013, which was completed in the fourth quarter of 2013.
In millions
Cash and temporary investments$5
Accounts and notes receivable 72
Inventory 31
Other current assets 2
Plants, properties and equipment 106
Investments 11
Total assets acquired 227
Notes payable and current maturities of long-term debt 17
Accounts payable and accrued liabilities 27
Deferred income tax liability 4
Postretirement and postemployment benefit obligation 6
Total liabilities assumed 54
Noncontrolling interest 18
Net assets acquired$155
Pro forma information related to the acquisition of Olmuksan has not been included as it does not have a material effect on the Company's consolidated results of operations.
ORSA
2014: On April 8, 2014, the Company acquired the remaining 25% of shares of Orsa International Paper Embalangens S.A. (Orsa IP) from its joint venture partner, Jari Celulose, Papel e Embalagens S.A. (Jari), a Grupo Orsa company, for approximately $127 million, of which $105 million was paid in cash with the remaining $22 million held back pending satisfaction of certain indemnification obligations by Jari. International Paper will release the amount held back, or any amount for which we have not notified Jari of a claim, by March 30, 2016. An additional $11 million, which was not included in the purchase price, was placed in an escrow account pending resolution of certain open matters. During 2014, these open matters were successfully resolved, which resulted in $9 million paid out of escrow to Jari and correspondingly added to the final purchase consideration. The remaining $2 million was released back to the Company. As a result of this transaction, the Company reversed the $168 million of Redeemable noncontrolling interest included on the March 31, 2014 consolidated balance sheet. The net difference between the Redeemable noncontrolling interest balance plus $14 million of currency translation adjustment reclassified out of Other comprehensive income less the 25% purchase price was reflected as an increase to Retained earnings on the consolidated balance sheet.
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2013: On January 14, 2013, International Paper and Jari formed Orsa IP with International Paper holding a 75% stake. The value of International Paper's investment in Orsa IP was approximately $471 million. Because International Paper acquired a majority control of the joint venture, Orsa IP's financial results have been consolidated with our Industrial Packaging segment from the date of formation on January 14, 2013. The 25% owned by Jari was considered a redeemable noncontrolling interest and met the requirements to be classified outside permanent equity. As such, the Company reported $163 million in Redeemable noncontrolling interest in the December 31, 2013 consolidated balance sheet.
The following table summarizes the final allocation of the purchase price to the fair value of assets and liabilities acquired as of January 14, 2013, which was completed in the fourth quarter of 2013.
In millions
Cash and temporary investments$16
Accounts and notes receivable 5
Inventory 27
Plants, properties and equipment 290
Goodwill 260
Other intangible assets 110
Other long-term assets 2
Total assets acquired 710
Accounts payable and accrued liabilities 68
Deferred income tax liability 37
Total liabilities assumed 105
Noncontrolling interest 134
Net assets acquired$471
The identifiable intangible assets acquired in connection with the Orsa IP acquisition included the following:
In millions Estimated
Fair Value Average
Remaining
Useful Life
Asset Class:(at acquisition
date)
Customer relationships$88 12 years
Trademark 3 6 years
Wood supply agreement 19 25 years
Total$110
Pro forma information related to the acquisition of Orsa IP has not been included as it does not have a material effect on the Company's consolidated results of operations.
Due to the complex organizational structure of Orsa IP's operations, and the extended time required to prepare consolidated financial information in accordance with accounting principles generally accepted in the United
States, the Company reports Orsa IP's operating results on a one-month lag basis.
NOTE 7 DIVESTITURES / SPINOFF
DISCONTINUED OPERATIONS
2014: On July 1, 2014, International Paper completed the spinoff of its distribution business, xpedx, which subsequently merged with Unisource Worldwide, Inc., with the combined companies now operating as Veritiv Corporation (Veritiv). The xpedx business had historically represented the Company's Distribution reportable segment.
The spinoff was accomplished by the contribution of the xpedx business to Veritiv and the distribution of 8,160,000 shares of Veritiv common stock on a pro-rata basis to International Paper shareholders. International Paper received a payment of approximately $411 million, financed with new debt in Veritiv's capital structure.
All current and historical operating results for xpedx are included in Discontinued operations, net of tax, in the accompanying consolidated statement of operations. The following summarizes the major classes of line items comprising Earnings (Loss) Before Income Taxes and Equity Earnings reconciled to Discontinued Operations, net of tax, related to the xpedx spinoff for all periods presented in the consolidated statement of operations:
In millions 2014 2013
Net Sales$2,604$5,597
Costs and Expenses
Cost of products sold 2,309 4,941
Selling and administrative expenses 191 409
Depreciation, amortization and cost of timber harvested 9 16
Distribution expenses 69 149
Restructuring and other charges 25 54
Impairment of goodwill and other intangibles—400
Other, net 3 7
Earnings (Loss) Before Income Taxes and Equity Earnings(2)(379)
Income tax provision (benefit)(1)(25)
Discontinued Operations, Net of Taxes (a)$(1)$(354)
(a) These amounts, along with those disclosed below related to the Temple-Inland Building Products divestitures, are included in Discontinued operations, net of tax, in the consolidated statement of operations.
Total cash provided by operations related to xpedx of $29 million and $81 million for 2014 and 2013, respectively, is included in Cash Provided By (Used For) Operations in the consolidated statement of cash flows. Total cash provided by (used for) investing activities related to xpedx of $3 million and $12 million for 2014
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and 2013, respectively, is included in Cash Provided By (Used For) Investing Activities in the consolidated statement of cash flows.
2013: On April 1, 2013, the Company finalized the sale of Temple-Inland's 50% interest in Del-Tin Fiber L.L.C. to joint venture partner Deltic Timber Corporation for $20 million in assumed liabilities and cash.
On July 19, 2013 the Company finalized the sale of its Temple-Inland Building Products division to Georgia-Pacific Building Products, LLC for approximately $726 million in cash.
Related to these divestitures, the Company recorded income (loss) of $0 million, $(12) million and $45 million for the years ended December 31, 2015, 2014 and 2013, respectively. These amounts are included in Discontinued operations, net of tax in the consolidated statement of operations.
OTHER DIVESTITURES AND IMPAIRMENTS
2015: On October 13, 2015, the Company finalized the sale of its 55% interest in IP Asia Coated Paperboard (IP-Sun JV) business, within the Company's Consumer Packaging segment, to its Chinese coated board joint venture partner, Shandong Sun Holding Group Co., Ltd. for RMB 149 million (approximately USD $23 million). During the third quarter of 2015, a determination was made that the current book value of the asset group exceeded its estimated fair value of $23 million, which was the agreed upon selling price. The 2015 loss includes the net pre-tax impairment charge of $174 million ($113 million after taxes). A pre-tax charge of $186 million was recorded during the third quarter in the Company's Consumer Packaging segment to write down the long-lived assets of this business to their estimated fair value. In the fourth quarter of 2015, upon the sale and corresponding deconsolidation of IP-Sun JV from the Company's consolidated balance sheet, final adjustments were made resulting in a reduction of the impairment of $12 million. The amount of pre-tax losses related to noncontrolling interest of the IP-Sun JV included in the Company's consolidated statement of operations for the years ended December 31, 2015, 2014 and 2013 were $19 million, $12 million and $8 million, respectively. The amount of pre-tax losses related to the IP-Sun JV included in the Company's consolidated statement of operations for the years ended December 31, 2015, 2014 and 2013 were $226 million, $51 million and $41 million, respectively.
The net 2015 loss totaling $174 million related to the impairment of Sun-JV is included in Net (gains) losses on sales and impairments of businesses in the accompanying consolidated statement of operations.
2014: During 2014, the Company recorded a net pre-tax charge of $47 million ($36 million after taxes) for the loss on the sale of a business by our equity method investee, ASG (formerly referred to as AGI-Shorewood), and the subsequent partial impairment of this ASG investment.
The net 2014 loss totaling $38 million, including the ASG impairment discussed above, related to other divestitures and impairments is included in Net (gains) losses on sales and impairments of businesses in the accompanying consolidated statement of operations.
2013: During 2013, the Company recorded net pre-tax charges of $3 million ($1 million after taxes) for adjustments related to the divestiture of three containerboard mills in 2012 and the sale of the Shorewood business. This loss is included in Net (gains) losses on sales and impairments of businesses in the accompanying consolidated statement of operations.
NOTE 8 SUPPLEMENTARY FINANCIAL STATEMENT INFORMATION
TEMPORARY INVESTMENTS
In millions at December 31 2015 2014
Temporary Investments$738$1,480
ACCOUNTS AND NOTES RECEIVABLE
Accounts and notes receivable, net of allowances, by classification were:
In millions at December 31 2015 2014
Accounts and notes receivable:
Trade$2,480$2,860
Other 195 223
Total$2,675$3,083
INVENTORIES
In millions at December 31 2015 2014
Raw materials$339$494
Finished pulp, paper and packaging products 1,248 1,273
Operating supplies 563 562
Other 78 95
Inventories$2,228$2,424
The last-in, first-out inventory method is used to value most of International Paper’s U.S. inventories. Approximately 78% of total raw materials and finished products inventories were valued using this method. If the first-in, first-out method had been used, it would have increased total inventory balances by approximately $345 million and $334 million at December 31, 2015 and 2014, respectively.
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PLANTS, PROPERTIES AND EQUIPMENT
In millions at December 31 2015 2014
Pulp, paper and packaging facilities$31,466$31,805
Other properties and equipment 1,242 1,263
Gross cost 32,708 33,068
Less: Accumulated depreciation 20,728 20,340
Plants, properties and equipment, net$11,980$12,728
In millions 2015 2014 2013
Depreciation expense$1,213$1,308$1,415
INTEREST
Cash payments related to interest were as follows:
In millions 2015 2014 2013
Interest payments$680$718$751
Amounts related to interest were as follows:
In millions 2015 2014 2013
Interest expense (a)$644$677$669
Interest income (a)89 70 57
Capitalized interest costs 25 23 17
(a)Interest expense and interest income exclude approximately $25 million, $38 million and $45 million in 2015, 2014 and 2013, respectively, related to investments in and borrowings from variable interest entities for which the Company has a legal right of offset (see Note 12).
NOTE 9 GOODWILL AND OTHER INTANGIBLES
GOODWILL
The following tables present changes in the goodwill balances as allocated to each business segment for the years ended December 31, 2015 and 2014:
In millions Industrial
Packaging Printing
Papers Consumer
Packaging Total
Balance as of January 1, 2015
Goodwill$3,396$2,234$1,784$7,414
Accumulated impairment losses (a)(100)(1,877)(1,664)(3,641)
3,296 357 120 3,773
Reclassifications and other (b)(70)(95)(3)(168)
Additions/reductions(1)(15)(c)(117)(d)(133)
Impairment loss(137)(e)——(137)
Balance as of December 31, 2015
Goodwill 3,325 2,124 1,664 7,113
Accumulated impairment losses (a)(237)(1,877)(1,664)(3,778)
Total$3,088$247$—$3,335
(a)Represents accumulated goodwill impairment charges since the adoption of ASC 350, “Intangibles – Goodwill and Other” in 2002.
(b)Represents the effects of foreign currency translations and reclassifications.
(c)Reflects a reduction from tax benefits generated by the deduction of goodwill amortization for tax purposes in Brazil.
(d)Reduction due to the sale and de-consolidation of Shandong Sun joint venture in Asia.
(e)Reflects a charge for goodwill impairment related to our Brazil Industrial Packaging business.
In millions Industrial
Packaging Printing
Papers Consumer
Packaging Distribution Total
Balance as of January 1, 2014
Goodwill$3,430$2,311$1,787$400$7,928
Accumulated impairment losses (a)—(1,877)(1,664)(400)(3,941)
3,430 434 123—3,987
Reclassifications and other (b)(34)(57)(3)—(94)
Additions/reductions—(20)(c)——(20)
Impairment loss(100)(d)———(100)
Write off of goodwill———(400)(400)
Write off of accumulated impairment loss———400 400
Balance as of December 31, 2014
Goodwill 3,396 2,234 1,784—7,414
Accumulated impairment losses (a)(100)(1,877)(1,664)—(3,641)
Total$3,296$357$120$—$3,773
(a)Represents accumulated goodwill impairment charges since the adoption of ASC 350, “Intangibles – Goodwill and Other” in 2002.
(b)Represents the effects of foreign currency translations and reclassifications.
(c)Reflects a reduction from tax benefits generated by the deduction of goodwill amortization for tax purposes in Brazil.
(d)Reflects a charge of $100 million for goodwill impairment related to our Asia Industrial Packaging business.
In the fourth quarter of 2015, in conjunction with the annual testing of its reporting units for possible goodwill impairments, the Company calculated the estimated fair value of its Brazil Packaging business using its discounted future cash flows and determined that all of the goodwill in the business, totaling $137 million, should be written off. The decline in the fair value of the Brazil Packaging business and resulting impairment charge was due to the negative impacts on the cash flows of the business caused by the continued decline of the overall Brazilian economy.
In the fourth quarter of 2014, in conjunction with the annual testing of its reporting units for possible goodwill impairments, the Company calculated the estimated fair value of its Asia Industrial Packaging business using the discounted future cash flows and determined that all of the goodwill in this business, totaling $100 million, should be written off. The decline in the fair value of the Asia Industrial Packaging business and resulting impairment charge was due to a change in the strategic outlook for the business.
In the fourth quarter of 2013, in conjunction with the annual testing of its reporting units for possible goodwill impairments, the Company calculated the estimated fair value of its India Papers business using the discounted future cash flows and determined that all of the goodwill of this business, totaling $112 million, should be written off. The decline in the fair value of the India Papers reporting unit and resulting impairment charge was due to a change in the strategic outlook for the India Papers operations.
At December 31, 2013, there was $400 million of goodwill and $400 million of accumulated impairment losses included in the consolidated balance sheet associated with
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the xpedx business (Distribution segment). Effective July 1, 2014, the Company completed the spinoff of its xpedx business which had historically represented the Company's Distribution reportable segment. Following the spinoff of xpedx, the assets and liabilities of this business have been adjusted off of the consolidated balance sheet and are not included in balances as of December 31, 2014.
OTHER INTANGIBLES
Identifiable intangible assets comprised the following:
2015 2014
In millions at
December 31 Gross
Carrying
Amount Accumulated
Amortization Gross
Carrying
Amount Accumulated
Amortization
Customer relationships and lists$495$166$561$157
Non-compete agreements 69 56 74 53
Tradenames, patents and trademarks 61 54 61 44
Land and water rights 33 6 81 9
Software 22 20 23 22
Other 46 29 48 24
Total$726$331$848$309
The Company recognized the following amounts as amortization expense related to intangible assets:
In millions 2015 2014 2013
Amortization expense related to intangible assets$60$73$79
Based on current intangibles subject to amortization, estimated amortization expense for each of the succeeding years is as follows: 2016 – $46 million, 2017– $44 million, 2018 – $35 million, 2019–$33 million, 2020 – $32 million, and cumulatively thereafter – $199 million.
NOTE 10 INCOME TAXES
The components of International Paper’s earnings from continuing operations before income taxes and equity earnings by taxing jurisdiction were as follows:
In millions 2015 2014 2013
Earnings (loss)
U.S.$1,147$565$775
Non-U.S.119 307 453
Earnings (loss) from continuing operations before income taxes and equity earnings$1,266$872$1,228
The provision (benefit) for income taxes (excluding noncontrolling interests) by taxing jurisdiction was as follows:
In millions 2015 2014 2013
Current tax provision (benefit)
U.S. federal$62$175$(663)
U.S. state and local 12 9(98)
Non-U.S.111 74 95
$185$258$(666)
Deferred tax provision (benefit)
U.S. federal$321$(67)$206
U.S. state and local 30 5(18)
Non-U.S.(70)(73)(20)
$281$(135)$168
Income tax provision (benefit)$466$123$(498)
The Company’s deferred income tax provision (benefit) includes a $3 million provision, a $13 million benefit and a $7 million provision for 2015, 2014 and 2013, respectively, for the effect of changes in non-U.S. and U.S. state tax rates.
International Paper made income tax payments, net of refunds, of $149 million, $172 million and $291 million in 2015, 2014 and 2013, respectively.
A reconciliation of income tax expense using the statutory U.S. income tax rate compared with the actual income tax provision follows:
In millions 2015 2014 2013
Earnings (loss) from continuing
operations before income taxes
and equity earnings$1,266$872$1,228
Statutory U.S. income tax rate 35%35%35%
Tax expense (benefit) using statutory U.S. income tax rate 443 305 430
State and local income taxes 27 10(2)
Tax rate and permanent differences on non-U.S. earnings(44)(72)(90)
Net U.S. tax on non-U.S. dividends 12 16(15)
Tax benefit on manufacturing activities(14)(46)(27)
Non-deductible business expenses 8 7 4
Non-deductible impairments 109 35 37
Sale of non-strategic assets(61)——
Tax audits——(770)
Subsidiary liquidation—(85)—
Retirement plan dividends(5)(5)(5)
Tax basis adjustments——(33)
Tax credits(15)(34)(23)
Other, net 6(8)(4)
Income tax provision (benefit)$466$123$(498)
Effective income tax rate 37%14%(41)%
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The tax effects of significant temporary differences, representing deferred income tax assets and liabilities at December 31, 2015 and 2014, were as follows:
In millions 2015 2014
Deferred income tax assets:
Postretirement benefit accruals$172$189
Pension obligations 1,403 1,517
Alternative minimum and other tax credits 283 342
Net operating and capital loss carryforwards 732 672
Compensation reserves 265 280
Other 244 266
Gross deferred income tax assets 3,099 3,266
Less: valuation allowance(430)(415)
Net deferred income tax asset$2,669$2,851
Deferred income tax liabilities:
Intangibles$(271)$(316)
Plants, properties and equipment(2,727)(2,707)
Forestlands, related installment sales, and investment in subsidiary(2,253)(2,290)
Gross deferred income tax liabilities$(5,251)$(5,313)
Net deferred income tax liability$(2,582)$(2,462)
Deferred income tax assets and liabilities are recorded in the accompanying consolidated balance sheet under the captions Deferred income tax assets, Deferred charges and other assets, Other accrued liabilities, and Deferred income taxes. There is a decrease in deferred income tax assets principally relating to the tax impact of changes in qualified pension liabilities and the utilization of tax credits. Deferred tax liabilities decreased primarily due to a reduction in the intangibles deferred tax liability. Of the $2.3 billion forestlands, related installment sales, and investment in subsidiary deferred tax liability, $1.4 billion is attributable to an investment in subsidiary and relates to a 2006 International Paper installment sale of forestlands and $840 million is attributable to a 2007 Temple-Inland installment sale of forestlands (see Note 12). Certain tax attributes reflected on our tax returns as filed differ from those reflected in the deferred income tax accounts due to uncertain tax benefits.
The valuation allowance for deferred income tax assets as of December 31, 2015, 2014 and 2013 was $430 million, $415 million and $413 million, respectively. The net change in the total valuation allowance for the years ended December 31, 2015 and 2014 was an increase of $15 million and an increase of $2 million, respectively.
A reconciliation of the beginning and ending amount of unrecognized tax benefits for the years ended December 31, 2015, 2014 and 2013 is as follows:
In millions 2015 2014 2013
Balance at January 1$(158)$(161)$(972)
(Additions) reductions based on tax positions related to current year(6)(15)(22)
Additions for tax positions of prior years(6)(1)(29)
Reductions for tax positions of prior years 7 9 824
Settlements 2—26
Expiration of statutes of
limitations 4 2 11
Currency translation adjustment 7 8 1
Balance at December 31$(150)$(158)$(161)
Included in the balance at December 31, 2015, 2014 and 2013 are $1 million, $1 million and $1 million, respectively, for tax positions for which the ultimate benefits are highly certain, but for which there is uncertainty about the timing of such benefits. However, except for the possible effect of any penalties, any disallowance that would change the timing of these benefits would not affect the annual effective tax rate, but would accelerate the payment of cash to the taxing authority to an earlier period.
The Company accrues interest on unrecognized tax benefits as a component of interest expense. Penalties, if incurred, are recognized as a component of income tax expense. The Company had approximately $34 million and $41 million accrued for the payment of estimated interest and penalties associated with unrecognized tax benefits at December 31, 2015 and 2014, respectively.
The major jurisdictions where the Company files income tax returns are the United States, Brazil, France, Poland and Russia. Generally, tax years 2003 through 2014 remain open and subject to examination by the relevant tax authorities. The Company is typically engaged in various tax examinations at any given time, both in the United States and overseas. In 2013, the Company concluded its examination with the U.S. Internal Revenue Service for the tax years 2006 through 2009 for both International Paper Company and Temple-Inland. As a result of the completion of the examinations, the Company reduced its unrecognized tax benefits by approximately $844 million. Other pending audit settlements and the expiration of statute of limitations could further reduce the uncertain tax positions by $39 million during the next twelve months. While the Company believes that it is adequately accrued for possible audit adjustments, the final resolution of these examinations cannot be determined at this time and could result in final settlements that differ from current estimates.
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Included in the Company’s 2015, 2014 and 2013 income tax provision (benefit) are $(121) million, $(453) million and $(869) million, respectively, related to special items. The components of the net provisions related to special items were as follows:
In millions 2015 2014 2013
Special items$(84)$(372)$(95)
Tax-related adjustments:
Return to accrual 23——
Internal restructurings(62)(90)(4)
Settlement of tax audits and legislative changes—10(770)
Medicare D deferred income tax write-off———
Other tax adjustments 2(1)—
Income tax provision (benefit) related to special items$(121)$(453)$(869)
Excluding the impact of special items and nonoperating pension expense, the 2015, 2014 and 2013 income tax provisions were $687 million, $659 million and $497 million, respectively, or 33%, 31% and 26%, respectively, of pre-tax earnings before equity earnings.
The following details the scheduled expiration dates of the Company’s net operating loss and income tax credit carryforwards:
In millions 2016
Through
2025 2026
Through
2035 Indefinite Total
U.S. federal and non-U.S. NOLs$76$—$519$595
State taxing jurisdiction NOLs 147 57—204
U.S. federal, non-
U.S. and state tax credit carryforwards 144 32 241 417
U.S. federal and state capital loss carryforwards 23——23
Total$390$89$760$1,239
Deferred income taxes are not provided for temporary differences of approximately $5.7 billion, $5.2 billion and $5.1 billion as of December 31, 2015, 2014 and 2013, respectively, representing earnings of non-U.S. subsidiaries intended to be permanently reinvested. Computation of the potential deferred tax liability associated with these undistributed earnings and other basis differences is not practicable.
The American Taxpayer Relief Act of 2012 (the “Act”) was signed into law on January 2, 2013. The Act retroactively restored several expired business tax provisions, including the research and experimentation credit and the Subpart F controlled foreign corporation look-through exception. Because a change in tax law is accounted for in the period of enactment, the retroactive effect of the Act on the Company's U.S.
federal taxes for 2012 of a benefit of approximately $32 million was recognized in the first quarter of 2013.
NOTE 11 COMMITMENTS AND CONTINGENT LIABILITIES
PURCHASE COMMITMENTS AND OPERATING LEASES
Certain property, machinery and equipment are leased under cancelable and non-cancelable agreements.
Unconditional purchase obligations have been entered into in the ordinary course of business, principally for capital projects and the purchase of certain pulpwood, logs, wood chips, raw materials, energy and services, including fiber supply agreements to purchase pulpwood that were entered into concurrently with the Company’s 2006 Transformation Plan forestland sales and in conjunction with the 2008 acquisition of Weyerhaeuser Company’s Containerboard, Packaging and Recycling business.
At December 31, 2015, total future minimum commitments under existing non-cancelable operating leases and purchase obligations were as follows:
In millions 2016 2017 2018 2019 2020 Thereafter
Lease obligations$118$95$72$55$41$128
Purchase obligations (a)3,001 541 447 371 358 1,579
Total$3,119$636$519$426$399$1,707
(a)Includes $2.1 billion relating to fiber supply agreements entered into at the time of the Company’s 2006 Transformation Plan forestland sales and in conjunction with the 2008 acquisition of Weyerhaeuser Company’s Containerboard, Packaging and Recycling business.
Rent expense was $170 million, $154 million and $168 million for 2015, 2014 and 2013, respectively.
GUARANTEES
In connection with sales of businesses, property, equipment, forestlands and other assets, International Paper commonly makes representations and warranties relating to such businesses or assets, and may agree to indemnify buyers with respect to tax and environmental liabilities, breaches of representations and warranties, and other matters. Where liabilities for such matters are determined to be probable and subject to reasonable estimation, accrued liabilities are recorded at the time of sale as a cost of the transaction.
ENVIRONMENTAL PROCEEDINGS
CERCLA and State Actions
International Paper has been named as a potentially responsible party in environmental remediation actions under various federal and state laws, including the
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Comprehensive Environmental Response, Compensation and Liability Act (CERCLA). Many of these proceedings involve the cleanup of hazardous substances at large commercial landfills that received waste from many different sources. While joint and several liability is authorized under CERCLA and equivalent state laws, as a practical matter, liability for CERCLA cleanups is typically allocated among the many potential responsible parties. Remediation costs are recorded in the consolidated financial statements when they become probable and reasonably estimable. International Paper has estimated the probable liability associated with these matters to be approximately $93 million in the aggregate as of December 31, 2015.
Cass Lake: One of the matters included above arises out of a closed wood treating facility located in Cass Lake, Minnesota. During 2009, in connection with an environmental site remediation action under CERCLA, International Paper submitted to the United States Environmental Protection Agency (EPA) a remediation feasibility study. In June 2011, the EPA selected and published a proposed soil remedy at the site with an estimated cost of $46 million. The overall remediation reserve for the site is currently $47 million to address the selection of an alternative for the soil remediation component of the overall site remedy. In October 2011, the EPA released a public statement indicating that the final soil remedy decision would be delayed. In the unlikely event that the EPA changes its proposed soil remedy and approves instead a more expensive clean-up alternative, the remediation costs could be material, and significantly higher than amounts currently recorded. In October 2012, the Natural Resource Trustees for this site provided notice to International Paper and other potentially responsible parties of their intent to perform a Natural Resource Damage Assessment. It is premature to predict the outcome of the assessment or to estimate a loss or range of loss, if any, which may be incurred.
Other Remediation Costs
In addition to the above matters, other remediation costs typically associated with the cleanup of hazardous substances at the Company’s current, closed or formerly-owned facilities, and recorded as liabilities in the balance sheet, totaled approximately $46 million as of December 31, 2015. Other than as described above, completion of required remedial actions is not expected to have a material effect on our consolidated financial statements.
LEGAL PROCEEDINGS
Environmental
Kalamazoo River: The Company is a potentially responsible party (PRP) with respect to the Allied Paper, Inc./Portage Creek/Kalamazoo River Superfund Site in Michigan. The EPA asserts that the site is contaminated by polychlorinated biphenyls (PCBs) primarily as a
result of discharges from various paper mills located along the Kalamazoo River, including a paper mill formerly owned by St. Regis Paper Company (St. Regis). The Company is a successor in interest to St. Regis. Although the Company has not received any orders from the EPA, in December 2014, the EPA sent the Company a letter demanding payment of $19 million to reimburse the EPA for costs associated with a Time Critical Removal Action of PCB contaminated sediments from a portion of the site. The Company’s CERCLA liability has not been finally determined with respect to this or any other portion of the site and we have declined to reimburse the EPA at this time. As noted below, the Company is involved in allocation/apportionment litigation with regard to the site. Accordingly, it is premature to predict the outcome or estimate our maximum reasonably possible loss with respect to this site. However, we do not believe that any material loss is probable.
The Company was named as a defendant by Georgia-Pacific Consumer Products LP, Fort James Corporation and Georgia Pacific LLC in a contribution and cost recovery action for alleged pollution at the site. The suit seeks contribution under CERCLA for costs purportedly expended by plaintiffs ($79 million as of the filing of the complaint) and for future remediation costs. The suit alleges that a mill, during the time it was allegedly owned and operated by St. Regis, discharged PCB contaminated solids and paper residuals resulting from paper de-inking and recycling. NCR Corporation and Weyerhaeuser Company are also named as defendants in the suit. In mid-2011, the suit was transferred from the District Court for the Eastern District of Wisconsin to the District Court for the Western District of Michigan. The trial of the initial liability phase took place in February 2013. Weyerhaeuser conceded prior to trial that it was a liable party with respect to the site. In September 2013, an opinion and order was issued in the suit. The order concluded that the Company (as the successor to St. Regis) was not an “operator,” but was an “owner,” of the mill at issue during a portion of the relevant period and is therefore liable under CERCLA. The order also determined that NCR is a liable party as an "arranger for disposal" of PCBs in waste paper that was de-inked and recycled by mills along the Kalamazoo River. The order did not address the Company's responsibility, if any, for past or future costs, which is the subject of a separate trial, in which trial testimony was given between September and December 2015 and post-trial briefing is currently scheduled to be completed in March 2016. The Court has not yet ruled to what extent it will decide responsibility for future costs. We are unable to predict the outcome or estimate our maximum reasonably possible loss. However, we do not believe that any material loss is probable.
Harris County: International Paper and McGinnis Industrial Maintenance Corporation, a subsidiary of
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Waste Management, Inc., are PRPs at the San Jacinto River Waste Pits Superfund Site (San Jacinto River Superfund Site) in Harris County, Texas, and have been actively participating in investigation and remediation activities at this Superfund Site. In December 2011, Harris County, Texas filed a suit against the Company seeking civil penalties with regard to the alleged discharge of dioxin into the San Jacinto River from waste impoundments that are part of the San Jacinto River Superfund Site. In November 2014, International Paper secured a zero liability jury verdict. Harris County appealed the verdict in April 2015, and that appeal is pending. The Company is also defending an additional lawsuit related to the San Jacinto River Superfund Site brought by approximately 400 individuals who allege property damage and personal injury. Because this case is still in the discovery phase, it is premature to predict the outcome or to estimate a loss or range of loss, if any, which may be incurred.
Antitrust
Containerboard: In September 2010, eight containerboard producers, including International Paper and Temple-Inland, were named as defendants in a purported class action complaint that alleged a civil violation of Section 1 of the Sherman Act. The suit is captioned Kleen Products LLC v. International Paper Co. (N.D. Ill.). The complaint alleges that the defendants, beginning in February 2004 through November 2010, conspired to limit the supply and thereby increase prices of containerboard products. The class is all persons who purchased containerboard products directly from any defendant for use or delivery in the United States during the period February 2004 to November 2010. The complaint seeks to recover an unspecified amount of treble actual damages and attorneys' fees on behalf of the purported class. Four similar complaints were filed and have been consolidated in the Northern District of Illinois. In March 2015, the District Court certified a class of direct purchasers of containerboard products; in June 2015, the United States Court of Appeals for the Seventh Circuit granted the defendants' petition to appeal and the class certification issue is now pending in that court. In June 2015, International Paper and Temple-Inland were named as defendants in a lawsuit captioned Del Monte Fresh Products N.A., Inc. v. Packaging Corporation of America (S.K. Fl.), in which the Plaintiff asserts substantially similar allegations to those raised in the Kleen Products LLC action. Pursuant to a tolling agreement signed by all parties, the case was voluntarily dismissed without prejudice in November 2015. Moreover, in January 2011, International Paper was named as a defendant in a lawsuit filed in state court in Cocke County, Tennessee alleging that International Paper violated Tennessee law by conspiring to limit the supply and fix the prices of containerboard from mid-2005 to the present. Plaintiffs in the state court action seek certification of a class of
Tennessee indirect purchasers of containerboard products, damages and costs, including attorneys' fees. No class certification materials have been filed to date in the Tennessee action. The Company disputes the allegations made and is vigorously defending each action. However, because the Kleen Products LLC action is in the discovery stage and the Tennessee action is in a preliminary stage, we are unable to predict an outcome or estimate a range of reasonably possible loss.
Gypsum: Beginning in late December 2012, certain purchasers of gypsum board filed a number of purported class action complaints alleging civil violations of Section 1 of the Sherman Act against Temple-Inland and a number of other gypsum manufacturers. The complaints were similar and alleged that the gypsum manufacturers conspired or otherwise reached agreements to: (1) raise prices of gypsum board either from 2008 or 2011 through the present; (2) avoid price erosion by ceasing the practice of issuing job quotes; and (3) restrict supply through downtime and limiting order fulfillment. On April 8, 2013, the Judicial Panel on Multidistrict Litigation ordered transfer of all pending cases to the U.S. District Court for the Eastern District of Pennsylvania for coordinated and consolidated pretrial proceedings, and the direct purchaser plaintiffs and indirect purchaser plaintiffs filed their respective amended consolidated complaints in June 2013. The amended consolidated complaints alleged a conspiracy or agreement beginning on or before September 2011. The alleged classes were all persons who purchased gypsum board directly or indirectly from any defendant. The complainants seek to recover unspecified treble actual damages and attorneys' fees on behalf of the purported classes. In February 2015, we executed a definitive agreement to settle these cases for an immaterial amount, and this settlement received final court approval and was paid in the third quarter of 2015.
In March 2015, several homebuilders filed an antitrust action in the United States District Court for the Northern District of California alleging that they purchased gypsum board and making similar allegations to those contained in the above settled proceeding. That lawsuit was transferred by the Judicial Panel on Multidistrict Litigation to the Eastern District of Pennsylvania. The homebuilders filed a notice to opt out of the class settlements and recently amended their complaint to assert that the alleged conspiracy or conspiracies continued into 2015. The Company intends to dispute the allegations made and to vigorously defend that lawsuit.
In addition, in September 2013, similar purported class actions were filed in courts in Quebec, Canada and Ontario, Canada, with each suit alleging violations of the Canadian Competition Act and seeking damages and injunctive relief. In April 2015, a similar class action
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was filed in British Columbia, Canada. In May 2015, we reached an agreement in principle to settle these Canadian cases for an immaterial amount. In November 2015, a definitive settlement agreement was executed and is subject to court approval.
Tax
On October 16, 2015, the Company was notified of a $92 million tax assessment issued by the state of Sao Paulo, Brazil for tax years 2011 through 2013. The assessment pertains to invoices issued by the Company related to the sale of paper to the editorial segment, which is exempt from the payment of ICMS value-added tax. This assessment is in the preliminary stage.The Company does not believe that a material loss is probable.
General
The Company is involved in various other inquiries, administrative proceedings and litigation relating to environmental and safety matters, labor and employment, contracts, sales of property, intellectual property, personal injury and other matters, some of which allege substantial monetary damages. While any proceeding or litigation has the element of uncertainty, the Company believes that the outcome of any of these lawsuits or claims that are pending or threatened or all of them combined (other than those that cannot be assessed due to their preliminary nature) will not have a material effect on its consolidated financial statements.
NOTE 12 VARIABLE INTEREST ENTITIES
In connection with the 2006 sale of approximately 5.6 million acres of forestlands, International Paper received installment notes (the Timber Notes) totaling approximately $4.8 billion. The Timber Notes, which do not require principal payments prior to their maturity which was originally August 2016, are supported by irrevocable letters of credit obtained by the buyers of the forestlands.
During 2006, International Paper contributed the Timber Notes to newly formed special purpose entities (the Borrower Entities) in exchange for Class A and Class B interests in these entities. Subsequently, International Paper contributed its $200 million Class A interests in the Borrower Entities, along with approximately $400 million of International Paper promissory notes, to other newly formed special purpose entities (the Investor Entities, and together with the Borrower Entities, the Entities) in exchange for Class A and Class B interests in these entities, and simultaneously sold its Class A interest in the Investor Entities to a third party investor. As a result, at December 31, 2006, International Paper held Class B
interests in the Borrower Entities and Class B interests in the Investor Entities valued at approximately $5.0 billion. International Paper did not provide any financial support that was not previously contractually required for the years ended December 31, 2015, 2014, or 2013.
Following the 2006 sale of forestlands and creation of the Entities discussed above, the Timber Notes were used as collateral for borrowings from third party lenders, which effectively monetized the Timber Notes.
Also during 2006, the Entities acquired approximately $4.8 billion of International Paper debt obligations for cash, resulting in a total of approximately $5.2 billion of International Paper debt obligations held by the Entities at December 31, 2006. The various agreements entered into in connection with these transactions provided that International Paper had, and intended to effect, a legal right to offset its obligation under these debt instruments with its investments in the Entities. Accordingly, for financial reporting purposes, International Paper had offset approximately $5.2 billion of Class B interests in the Entities against $5.3 billion of International Paper debt obligations held by these Entities at December 31, 2014, and despite the offset treatment, these remained debt obligations of International Paper. Remaining borrowings of $50 million are included in Long-term debt in the accompanying consolidated balance sheet at December 31, 2014. Additional debt related to the above transaction of $107 million is included in Notes payable and current maturities of long-term debt at December 31, 2014.
The use of the Entities facilitated the monetization of the credit enhanced Timber Notes in a cost effective manner by increasing borrowing capacity and lowering the interest rate, while providing for the offset accounting treatment described above. Additionally, the monetization structure preserved the $1.4 billion tax deferral that resulted from the 2006 forestlands sales.
Based on an analysis of the Entities under ASC 810, "Consolidation," that considers the potential magnitude of the variability in the structures and which party has a controlling financial interest, International Paper determined that it was not the primary beneficiary of the Entities at December 31, 2014, and therefore, did not consolidate its investments in the Entities. The Company also determined that the source of variability in the structures is the value of the Timber Notes, the assets most significantly impacting the structures' economic performance. The credit quality of the Timber Notes is supported by irrevocable letters of credit obtained by the Timber Note issuers. International Paper analyzed which party had control over the economic performance of each Entity, and concluded International Paper did not have control over significant decisions surrounding the Timber Notes and letters of
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credit and therefore was not the primary beneficiary at December 31, 2014. The Company’s maximum exposure to loss at December 31, 2014 equaled the principal amount of the Timber Notes; however, an analysis performed by the Company concluded the likelihood of this exposure was remote.
During the third quarter of 2015, we initiated a series of actions in order to extend the 2006 monetization structure and maintain the long-term nature of the $1.4 billion deferred tax liability.First, International Paper acquired the Class A interests in the Investor Entities from a third party for $198 million in cash. As a result, International Paper became the owner of all of the Class A and Class B interests in the Entities and became the primary beneficiary of the Entities. The assets and liabilities of the Entities, primarily consisting of the Timber Notes and third party bank loans, were recorded at fair value as of the acquisition date of the Class A interests.
Subsequent to purchasing the Class A interests in the Investor Entities, International Paper restructured the Entities, which resulted in the formation of wholly-owned, bankruptcy-remote special purpose entities (the 2015 Financing Entities). As part of the restructuring, the Timber Notes held by the Borrower Entities, subject to the third party bank loans, were contributed to the 2015 Financing Entities along with approximately $150 million in International Paper debt obligations, approximately $600 million in cash and approximately $130 million in demand loans from International Paper, and certain Entities were liquidated. As a result of these transactions, International Paper began consolidating the 2015 Financing Entities during the third quarter of 2015. Also, during the third quarter of 2015, the 2015 Financing Entities used $630 million in cash to pay down a portion of the third party bank loans and refinanced approximately $4.2 billion of those loans on nonrecourse terms (the 2015 Refinance Loans).
During the fourth quarter of 2015, International Paper extended the maturity date on the Timber Notes for an additional five years. The Timber Notes are shown in Financial assets of special purpose entities on the accompanying consolidated balance sheet and mature in August 2021 unless extended for an additional five years. These notes are supported by approximately $4.8 billion of irrevocable letters of credit. In addition, the Company extinguished the 2015 Refinance Loans scheduled to mature in May 2016 and entered into new nonrecourse third party bank loans totaling approximately $4.2 billion (the Extension Loans). Provisions of loan agreements related to approximately $1.1 billion of the Extension Loans require the bank issuing letters of credit supporting the Timber Notes pledged as collateral to maintain a credit rating at or above a specified threshold. In the event the credit
rating of the letter of credit bank is downgraded below the specified threshold, the letters of credit must be replaced within 60 days with letters of credit from a qualifying financial institution. The Extension Loans are shown in Nonrecourse financial liabilities of special purpose entities on the accompanying consolidated balance sheet and mature in the fourth quarter of 2020. The extinguishment of the 2015 Refinance Loans of approximately $4.2 billion and the issuance of the Extension Loans of approximately $4.2 billion are shown as part of reductions of debt and issuances of debt, respectively, in the financing activities of the consolidated statement of cash flows.
The Extension Loans are nonrecourse to the Company, and are secured by approximately $4.8 billion of Timber Notes, the irrevocable letters of credit supporting the Timber Notes and approximately $150 million of International Paper debt obligations. The $150 million of International Paper debt obligations are eliminated in the consolidation of the 2015 Financing Entities and are not reflected in the Company’s consolidated balance sheet.
The purchase of the Class A interests and subsequent restructuring described above facilitated the refinancing and extensions of the third party bank loans on nonrecourse terms. The transactions described in these paragraphs result in continued long-term classification of the $1.4 billion deferred tax liability recognized in connection with the 2006 forestlands sale.
As of December 31, 2015, the fair value of the Timber Notes and Extension Loans is $4.68 billion and $4.28 billion, respectively. The Timber Notes and Extension Loans are classified as Level 2 within the fair value hierarchy, which is further defined in Note 14.
Activity between the Company and the 2015 Financing Entities (the Entities prior to the purchase of the Class A interest discussed above) was as follows:
In millions 2015 2014 2013
Revenue (a)$43$38$45
Expense (a)81 72 79
Cash receipts (b)21 22 33
Cash payments (c)71 73 84
(a)The net expense related to the Company’s interest in the Entities is included in the accompanying consolidated statement of operations, as International Paper has and intends to effect its legal right to offset as discussed above. After formation of the 2015 Financing Entities, the revenue and expense are included in Interest expense, net in the accompanying consolidated statement of operations.
(b)The cash receipts are equity distributions from the Entities to International Paper prior to the formation of the 2015 Financing Entities. After formation of the 2015 Financing Entities, cash receipts are interest received on the Financial assets of special purpose entities.
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(c)The cash payments are interest payments on the associated debt obligations discussed above. After formation of the 2015 Financing Entities, the payments represent interest paid on Nonrecourse financial liabilities of special purpose entities.
In connection with the acquisition of Temple-Inland in February 2012, two special purpose entities became wholly-owned subsidiaries of International Paper.
The use of the two wholly-owned special purpose entities discussed below preserved the tax deferral that resulted from the 2007 Temple-Inland timberlands sales. The Company recognized an $840 million deferred tax liability in connection with the 2007 sales, which will be settled with the maturity of the notes in 2027.
In October 2007, Temple-Inland sold 1.55 million acres of timberland for $2.38 billion. The total consideration consisted almost entirely of notes due in 2027 issued by the buyer of the timberland, which Temple-Inland contributed to two wholly-owned, bankruptcy-remote special purpose entities. The notes are shown in Financial assets of special purpose entities in the accompanying consolidated balance sheet and are supported by $2.38 billion of irrevocable letters of credit issued by three banks, which are required to maintain minimum credit ratings on their long-term debt. In the third quarter of 2012, International Paper completed its preliminary analysis of the acquisition date fair value of the notes and determined it to be $2.09 billion. As of December 31, 2015 and 2014, the fair value of the notes was $2.10 billion and $2.27 billion, respectively. These notes are classified as Level 2 within the fair value hierarchy, which is further defined in Note 14.
In December 2007, Temple-Inland's two wholly-owned special purpose entities borrowed $2.14 billion shown in Nonrecourse financial liabilities of special purpose entities. The loans are repayable in 2027 and are secured only by the $2.38 billion of notes and the irrevocable letters of credit securing the notes and are nonrecourse to us. The loan agreements provide that if a credit rating of any of the banks issuing the letters of credit is downgraded below the specified threshold, the letters of credit issued by that bank must be replaced within 30 days with letters of credit from another qualifying financial institution. In the third quarter of 2012, International Paper completed its preliminary analysis of the acquisition date fair value of the borrowings and determined it to be $2.03 billion. As of December 31, 2015 and 2014, the fair value of this debt was $1.97 billion and $2.16 billion, respectively. This debt is classified as Level 2 within the fair value hierarchy, which is further defined in Note 14.
Activity between the Company and the 2007 financing entities was as follows:
In millions 2015 2014 2013
Revenue (a)$27$26$27
Expense (b)27 25 29
Cash receipts (c)7 7 8
Cash payments (d)18 18 21
(a)The revenue is included in Interest expense, net in the accompanying consolidated statement of operations and includes approximately $19 million, $19 million and $19 million for the years ended December 31, 2015, 2014 and 2013, respectively, of accretion income for the amortization of the purchase accounting adjustment on the Financial assets of special purpose entities.
(b)The expense is included in Interest expense, net in the accompanying consolidated statement of operations and includes approximately $7 million, $7 million and $7 million for the years ended December 31, 2015, 2014 and 2013, respectively, of accretion expense for the amortization of the purchase accounting adjustment on the Nonrecourse financial liabilities of special purpose entities.
(c)The cash receipts are interest received on the Financial assets of special purpose entities.
(d)The cash payments are interest paid on Nonrecourse financial liabilities of special purpose entities.
NOTE 13 DEBT AND LINES OF CREDIT
In 2015, International Paper issued $700 million of 3.80% senior unsecured notes with a maturity date in 2026, $600 million of 5.00% senior unsecured notes with a maturity date in 2035, and $700 million of 5.15% senior unsecured notes with a maturity date in 2046. The proceeds from this borrowing were used to repay approximately $1.0 billion of notes with interest rates ranging from 4.75% to 9.38% and original maturities from 2018 to 2022, along with $211 million of cash premiums associated with the debt repayments. Additionally, the proceeds from this borrowing were used to make a $750 million voluntary cash contribution to the Company's pension plan. Pre-tax early debt retirement costs of $207 million related to the debt repayments, including the $211 million of cash premiums, are included in restructuring and other charges in the accompanying consolidated statement of operations for the twelve months ended December 31, 2015.
During the second quarter of 2014, International Paper issued $800 million of 3.65% senior unsecured notes with a maturity date in 2024 and $800 million of 4.80% senior unsecured notes with a maturity date in 2044. The proceeds from this borrowing were used to repay approximately $960 million of notes with interest rates ranging from 7.95% to 9.38% and original maturities from 2018 to 2019. Pre-tax early debt retirement costs of $262 million related to these debt repayments, including $258 million of cash premiums, are included in Restructuring and other charges in the accompanying consolidated statement of operations for the twelve months ended December 31, 2014.
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Amounts related to early debt extinguishment during the years ended December 31, 2015, 2014 and 2013 were as follows:
In millions 2015 2014 2013
Debt reductions (a)$2,151$1,625$574
Pre-tax early debt extinguishment costs (b)207 276 25
(a)Reductions related to notes with interest rates ranging from 2.00% to 9.38% with original maturities from 2014 to 2031 for the years ended December 31, 2015, 2014 and 2013. Includes the $630 million payment for a portion of the Special Purpose Entity Liability (see Note 12 Variable Interest Entities).
(b)Amounts are included in Restructuring and other charges in the accompanying consolidated statements of operations.
A summary of long-term debt follows:
In millions at December 31 2015 2014
8.7% note – due 2038$264$264
9 3/8% note – due 2019 295 420
7.95% debentures – due 2018 648 903
7.5% note – due 2021 603 979
7.3% notes – due 2039 721 721
6 7/8% notes – due 2023 – 2029 131 131
6.65% note – due 2037 4 4
6.4% to 7.75% debentures due 2025 – 2027 142 142
6 3/8% to 6 5/8% notes – due 2016 – 2018 185 358
6.0% notes – due 2041 585 585
5.25% to 5.3% notes – due 2015 – 2016 261 457
5.00% to 5.15% – due 2035 – 2046 1,280—
4.8% notes - due 2044 796 796
4.75% notes – due 2022 817 896
3.65% to 3.80% notes – due 2024 – 2026 1,490 797
Floating rate notes – due 2015 – 2025 (a)438 271
Environmental and industrial development
bonds – due 2015 – 2035 (b)594 950
Short-term notes (c)5 424
Other (d)67 275
Total (e)9,326 9,373
Less: current maturities 426 742
Long-term debt$8,900$8,631
(a)The weighted average interest rate on these notes was 2.9% in 2015 and 2.8% in 2014.
(b)The weighted average interest rate on these bonds was 5.8% in 2015 and 5.7% in 2014.
(c)The weighted average interest rate was 2.2% in 2015 and 2.6% in 2014. Includes $5 million at December 31, 2015 and $91 million at December 31, 2014 related to non-U.S. denominated borrowings with a weighted average interest rate of 2.2% in 2015 and 7.2% in 2014.
(d)Includes $8 million at December 31, 2015 and $20 million at December 31, 2014 related to the unamortized gain on interest rate swap unwinds (see Note 14 Derivatives and Hedging Instruments).
(e)The fair market value was approximately $9.9 billion at December 31, 2015 and $10.6 billion at December 31, 2014.
Total maturities of long-term debt over the next five years are 2016 – $426 million; 2017 – $43 million; 2018 – $811 million; 2019 – $427 million; and 2020 – $183 million.
At December 31, 2015, International Paper’s credit facilities (the Agreements) totaled $2.1 billion. The Agreements generally provide for interest rates at a floating rate index plus a pre-determined margin dependent upon International Paper’s credit rating. The Agreements include a $1.5 billion contractually committed bank facility that expires in August 2019 and has a facility fee of 0.15% payable annually. The liquidity facilities also include up to $600 million of uncommitted financings based on eligible receivables balances (approximately $600 million available as of December 31, 2015) under a receivables securitization program that expires in December 2016. At December 31, 2015, there were no borrowings under either the bank facility or receivables securitization program.
Maintaining an investment grade credit rating is an important element of International Paper’s financing strategy. At December 31, 2015, the Company held long-term credit ratings of BBB (stable outlook) and Baa2 (stable outlook) by S&P and Moody’s, respectively.
NOTE 14 DERIVATIVES AND HEDGING ACTIVITIES
International Paper periodically uses derivatives and other financial instruments to hedge exposures to interest rate, commodity and currency risks. International Paper does not hold or issue financial instruments for trading purposes. For hedges that meet the hedge accounting criteria, International Paper, at inception, formally designates and documents the instrument as a fair value hedge, a cash flow hedge or a net investment hedge of a specific underlying exposure.
INTEREST RATE RISK MANAGEMENT
Our policy is to manage interest cost using a mixture of fixed-rate and variable-rate debt. To manage this risk in a cost-efficient manner, we enter into interest rate swaps whereby we agree to exchange with the counterparty, at specified intervals, the difference between fixed and variable interest amounts calculated by reference to a notional amount.
Interest rate swaps that meet specific accounting criteria are accounted for as fair value or cash flow hedges. For fair value hedges, the changes in the fair value of both the hedging instruments and the underlying debt obligations are immediately recognized in interest expense. For cash flow hedges, the effective portion of the changes in the fair value of the hedging
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instrument is reported in Accumulated other comprehensive income (“AOCI”) and reclassified into interest expense over the life of the underlying debt. The ineffective portion for both cash flow and fair value hedges, which is not material for any year presented, is immediately recognized in earnings.
FOREIGN CURRENCY RISK MANAGEMENT
We manufacture and sell our products and finance operations in a number of countries throughout the world and, as a result, are exposed to movements in foreign currency exchange rates. The purpose of our foreign currency hedging program is to manage the volatility associated with the changes in exchange rates.
To manage this exchange rate risk, we have historically utilized a combination of forward contracts, options and currency swaps. Contracts that qualify are designated as cash flow hedges of certain forecasted transactions denominated in foreign currencies. The effective portion of the changes in fair value of these instruments is reported in AOCI and reclassified into earnings in the same financial statement line item and in the same period or periods during which the related hedged transactions affect earnings. The ineffective portion, which is not material for any year presented, is immediately recognized in earnings.
The change in value of certain non-qualifying instruments used to manage foreign exchange exposure of intercompany financing transactions and certain balance sheet items subject to revaluation is immediately recognized in earnings, substantially offsetting the foreign currency mark-to-market impact of the related exposure.
COMMODITY RISK MANAGEMENT
Certain raw materials used in our production processes are subject to price volatility caused by weather, supply conditions, political and economic variables and other unpredictable factors. To manage the volatility in earnings due to price fluctuations, we may utilize swap contracts or forward purchase contracts.
Derivative instruments are reported in the consolidated balance sheets at their fair values, unless the derivative instruments qualify for the normal purchase normal sale ("NPNS") exception under GAAP and such exception has been elected. If the NPNS exception is elected, the fair values of such contracts are not recognized on the balance sheet.
Contracts that qualify are designated as cash flow hedges of forecasted commodity purchases. The effective portion of the changes in fair value for these instruments is reported in AOCI and reclassified into earnings in the same financial statement line item and in the same period or periods during which the hedged transactions affect earnings. The ineffective and non-qualifying portions, which are not material for any year presented, are immediately recognized in earnings.
The change in the fair value of certain non-qualifying instruments used to reduce commodity price volatility is immediately recognized in earnings.
The notional amounts of qualifying and non-qualifying instruments used in hedging transactions were as follows:
In millions December 31, 2015 December 31, 2014
Derivatives in Cash Flow Hedging Relationships:
Foreign exchange contracts (Sell / Buy; denominated in sell notional): (a)
Brazilian real / U.S. dollar - Forward—166
British pounds / Brazilian real - Forward—5
European euro / Brazilian real - Forward—9
European euro / Polish zloty - Forward 260 280
Mexican peso / U.S. dollar - Forward 136—
U.S. dollar / Brazilian real - Forward—125
Derivatives in Fair Value Hedging Relationships:
Interest rate contracts (in USD)17 230
Derivatives Not Designated as Hedging Instruments:
Electricity contract (in Megawatt Hours)1—
Foreign exchange contracts (Sell / Buy; denominated in sell notional):
European euro / British pounds 25—
Indian rupee / U.S. dollar 49 43
Mexican peso / U.S. dollar 131 187
U.S. dollar / Brazilian real—11
Interest rate contracts (in USD)38—
(a)These contracts had maturities of three years or less as of December 31, 2015.
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The following table shows gains or losses recognized in AOCI, net of tax, related to derivative instruments:
Gain (Loss)
Recognized in AOCI on Derivatives
(Effective Portion)
In millions 2015 2014 2013
Foreign exchange contracts$(3)$10$—
Total$(3)$10$—
During the next 12 months, the amount of the December 31, 2015 AOCI balance, after tax, that is expected to be reclassified to earnings is a gain of $3 million.
The amounts of gains and losses recognized in the consolidated statement of operations on qualifying and non-qualifying financial instruments used in hedging transactions were as follows:
Gain (Loss)
Reclassified from
AOCI
into Income
(Effective Portion)Location of Gain
(Loss)
Reclassified
from AOCI
into Income
(Effective Portion)
In millions 2015 2014 2013
Derivatives in Cash Flow Hedging Relationships:
Foreign exchange contracts$(12)$4$7 Cost of products sold
Total$(12)$4$7
Gain (Loss)
Recognized
in Income Location of Gain (Loss)
in Consolidated Statement of
Operations
In millions 2015 2014 2013
Derivatives in Fair Value Hedging Relationships:
Interest rate contracts$3$1$(1)Interest expense, net
Debt(3)(1)1 Interest expense, net
Total$—$—$—
Derivatives Not Designated as Hedging Instruments:
Electricity Contracts$(7)$(2)$4 Cost of products sold
Embedded derivatives——(1)Interest expense, net
Foreign exchange contracts(4)(1)(5)Cost of products sold
Interest rate contracts 13(a)12(b)21 Interest expense, net
Total$2$9$19
(a)Excluding gain of $3 million related to debt reduction recorded to Restructuring and other charges.
(b)Excluding gain of $7 million, net related to debt issuance and debt reduction recorded to Restructuring and other charges.
The following activity is related to fully effective interest rate swaps designated as fair value hedges:
2015 2014
In millions Issued Terminated Undesignated Issued Terminated Undesignated
Second Quarter$—$175$38$—$—$—
First Quarter———55——
Total$—$175$38$55$—$—
Fair Value Measurements
International Paper’s financial assets and liabilities that are recorded at fair value consist of derivative contracts, including interest rate swaps, foreign currency forward contracts, and other financial instruments that are used to hedge exposures to interest rate, commodity and currency risks. In addition, a consolidated subsidiary of
International Paper has an embedded derivative. For these financial instruments and the embedded derivative, fair value is determined at each balance sheet date using an income approach.
The guidance for fair value measurements and disclosures sets out a fair value hierarchy that groups
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fair value measurement inputs into the following three classifications:
Level 1: Quoted market prices in active markets for identical assets or liabilities.
Level 2: Observable market-based inputs other than quoted prices included within Level 1 that are observable for the asset or liability, either directly or indirectly.
Level 3: Unobservable inputs for the asset or liability reflecting the reporting entity’s own assumptions or external inputs from inactive markets.
Transfers between levels are recognized at the end of the reporting period. All of International Paper’s derivative fair value measurements use Level 2 inputs.
Below is a description of the valuation calculation and the inputs used for each class of contract:
Interest Rate Contracts
Interest rate contracts are valued using swap curves obtained from an independent market data provider. The market value of each contract is the sum of the fair value of all future interest payments between the contract counterparties, discounted to present value. The fair value of the future interest payments is determined by comparing the contract rate to the derived forward interest rate and present valued using the appropriate derived interest rate curve.
Foreign Exchange Contracts
Foreign currency forward contracts are valued using foreign currency forward and interest rate curves obtained from an independent market data provider. The fair value of each contract is determined by comparing the contract rate to the forward rate. The fair value is present valued using the applicable interest rate from an independent market data provider.
Electricity Contract
The electricity contract is valued using the Mid-C index forward curved obtained from the Intercontinental Exchange. The market value of the contract is the sum of the fair value of all future purchase payments between the contract counterparties, discounted to present value. The fair value of the future purchase payments is determined by comparing the contract price to the forward price and present valued using International Paper's cost of capital.
Embedded Derivative
Embedded derivatives are valued using a hypothetical interest rate derivative with identical terms. The hypothetical interest rate derivative contracts are fair valued as described above under Interest Rate Contracts.
Since the volume and level of activity of the markets that each of the above contracts are traded in has been normal, the fair value calculations have not been adjusted for inactive markets or disorderly transactions.
The following table provides a summary of the impact of our derivative instruments in the consolidated balance sheet:
Fair Value Measurements
Level 2 – Significant Other Observable Inputs
Assets Liabilities
In millions December 31, 2015 December 31, 2014 December 31, 2015 December 31, 2014
Derivatives designated as hedging instruments
Foreign exchange contracts – cash flow$5(a)$16(b)$1(c)$14(c)
Total derivatives designated as hedging instruments$5$16$1$14
Derivatives not designated as hedging instruments
Electricity contract$—$—$7(d)$2(c)
Foreign exchange contracts—1(a)—2(c)
Total derivatives not designated as hedging instruments$—$1$7$4
Total derivatives$5$17$8$18
(a)Included in Other current assets in the accompanying consolidated balance sheet.
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(b)Includes $14 million recorded in Other current assets and $2 million recorded in Deferred charges and other assets in the accompanying consolidated balance sheet.
(c)Included in Other accrued liabilities in the accompanying consolidated balance sheet.
(d)Includes $4 million recorded in Other accrued liabilities and $3 million recorded in Other liabilities in the accompanying consolidated balance sheet.
The above contracts are subject to enforceable master netting arrangements that provide rights of offset with each counterparty when amounts are payable on the same date in the same currency or in the case of certain specified defaults. Management has made an accounting policy election to not offset the fair value of recognized derivative assets and derivative liabilities in the consolidated balance sheet. The amounts owed to the counterparties and owed to the Company are considered immaterial with respect to each counterparty and in the aggregate with all counterparties.
Credit-Risk-Related Contingent Features
International Paper evaluates credit risk by monitoring its exposure with each counterparty to ensure that exposure stays within acceptable policy limits. Credit risk is also mitigated by contractual provisions with the majority of our banks. Certain of the contracts include a credit support annex that requires the posting of collateral by the counterparty or International Paper based on each party’s rating and level of exposure. Based on the Company’s current credit rating, the collateral threshold is generally $15 million.
If the lower of the Company’s credit rating by Moody’s or S&P were to drop below investment grade, the Company would be required to post collateral for all of its derivatives in a net liability position, although no derivatives would terminate. The fair values of derivative instruments containing credit-risk-related contingent features in a net liability position were $1 million as of December 31, 2015 and December 31, 2014, respectively. The Company was not required to post any collateral as of December 31, 2015 or 2014.
NOTE 15 CAPITAL STOCK
The authorized capital stock at both December 31, 2015 and 2014, consisted of 990,850,000 shares of common stock, $1 par value; 400,000 shares of cumulative $4 preferred stock, without par value (stated value $100 per share); and 8,750,000 shares of serial preferred stock, $1 par value. The serial preferred stock is issuable in one or more series by the Board of Directors without further shareholder action.
The following is a rollforward of shares of common stock for the three years ended December 31, 2015, 2014 and 2013:
Common Stock
In thousands Issued Treasury
Balance at January 1, 2013 439,894 13
Issuance of stock for various plans, net 7,328(533)
Repurchase of stock—11,388
Balance at December 31, 2013 447,222 10,868
Issuance of stock for various plans, net 1,632(4,668)
Repurchase of stock—22,534
Balance at December 31, 2014 448,854 28,734
Issuance of stock for various plans, net 62(4,230)
Repurchase of stock—12,272
Balance at December 31, 2015 448,916 36,776
NOTE 16 RETIREMENT PLANS
International Paper sponsors and maintains the Retirement Plan of International Paper Company (the “Pension Plan”), a tax-qualified defined benefit pension plan that provides retirement benefits to substantially all U.S. salaried employees and hourly employees (receiving salaried benefits) hired prior to July 1, 2004, and substantially all other U.S. hourly and union employees who work at a participating business unit regardless of hire date. These employees generally are eligible to participate in the Pension Plan upon attaining 21 years of age and completing one year of eligibility service. U.S. salaried employees and hourly employees (receiving salaried benefits) hired after June 30, 2004 are not eligible to participate in the Pension Plan, but receive a company contribution to their individual savings plan accounts (see Other U.S. Plans); however, salaried employees hired by Temple Inland prior to March 1, 2007 also participate in the Pension Plan. The Pension Plan provides defined pension benefits based on years of credited service and either final average earnings (salaried employees and hourly employees receiving salaried benefits), hourly job rates or specified benefit rates (hourly and union employees).
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In connection with the Temple-Inland acquisition in February 2012, International Paper assumed administrative responsibility for the Temple-Inland Retirement Plan, a defined benefit plan which covers substantially all employees of Temple-Inland. The Temple-Inland Retirement Plan merged with the Retirement Plan of International Paper Company on December 31, 2014.
The Company also has three unfunded nonqualified defined benefit pension plans: a Pension Restoration Plan available to employees hired prior to July 1, 2004 that provides retirement benefits based on eligible compensation in excess of limits set by the Internal Revenue Service, and two supplemental retirement plans for senior managers (SERP), which is an alternative retirement plan for salaried employees who are senior vice presidents and above or who are designated by the chief executive officer as participants. These nonqualified plans are only funded to the extent of benefits paid, which totaled $62 million, $38 million and $28 million in 2015, 2014 and 2013, respectively, and which are expected to be $22 million in 2016.
The Company will freeze participation, including credited service and compensation, for salaried employees under the Pension Plan, the Pension Restoration Plan and the two SERP plans for all service on or after January 1, 2019.Credited service was previously frozen for the Temple Retirement Plans. This change will not affect benefits accrued through December 31, 2018.For service after this date, employees affected by the freeze will receive Retirement Savings Account contributions as described later in this Note 16.
Many non-U.S. employees are covered by various retirement benefit arrangements, some of which are considered to be defined benefit pension plans for accounting purposes.
OBLIGATIONS AND FUNDED STATUS
The following table shows the changes in the benefit obligation and plan assets for 2015 and 2014, and the plans’ funded status. The U.S. combined benefit obligation as of December 31, 2015 decreased by $302 million, due to an increase in the discount rate assumption used in computing the estimated benefit obligation partially offset by updated demographic assumptions. Our mortality assumption for the year ended December 31, 2014 reflects adoption of the newly issued Society of Actuaries longevity improvement sale, with Company specific adjustments. U.S. plan assets increased by $5 million, primarily reflecting a $750 million qualified pension contribution in 2015 offset by benefit payments.
2015 2014
In millions U.S.
Plans Non-
U.S.
Plans U.S.
Plans Non-
U.S.
Plans
Change in projected benefit obligation:
Benefit obligation, January 1$14,741$233$12,903$228
Service cost 161 6 145 5
Interest cost 597 10 600 13
Curtailments———(4)
Settlements(43)(12)——
Actuarial loss (gain)(254)(1)1,755 12
Divestitures——(23)—
Other———12
Plan amendments——133—
Benefits paid(764)(7)(772)(13)
Effect of foreign currency exchange rate movements—(25)—(20)
Benefit obligation, December 31$14,438$204$14,741$233
Change in plan assets:
Fair value of plan assets, January 1$10,918$180$10,706$181
Actual return on plan assets(1)4 593 13
Company contributions 813 9 391 8
Benefits paid(764)(7)(772)(13)
Settlements(43)(12)——
Other———6
Effect of foreign currency exchange rate movements—(19)—(15)
Fair value of plan assets, December 31$10,923$155$10,918$180
Funded status, December 31$(3,515)$(49)$(3,823)$(53)
Amounts recognized in the consolidated balance sheet:
Non-current asset$—$7$—$8
Current liability(22)(2)(62)(3)
Non-current liability(3,493)(54)(3,761)(58)
$(3,515)$(49)$(3,823)$(53)
Amounts recognized in accumulated other comprehensive income under ASC 715 (pre-tax):
Prior service cost$166$—$209$—
Net actuarial loss 4,899 42 4,812 40
$5,065$42$5,021$40
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The components of the $44 million and $2 million increase related to U.S. plans and non-U.S. plans, respectively, in the amounts recognized in OCI during 2015 consisted of:
In millions U.S.
Plans Non-
U.S.
Plans
Current year actuarial (gain) loss$530$5
Amortization of actuarial loss(428)(1)
Amortization of prior service cost(43)—
Settlements(15)—
Effect of foreign currency exchange rate movements—(2)
$44$2
The accumulated benefit obligation at December 31, 2015 and 2014 was $14.3 billion and $14.6 billion, respectively, for our U.S. defined benefit plans and $189 million and $208 million, respectively, at December 31, 2015 and 2014 for our non-U.S. defined benefit plans.
The following table summarizes information for pension plans with an accumulated benefit obligation in excess of plan assets at December 31, 2015 and 2014:
2015 2014
In millions U.S.
Plans Non-U.S.
Plans U.S.
Plans Non-U.S.
Plans
Projected benefit obligation$14,438$182$14,741$196
Accumulated benefit obligation 14,282 168 14,559 176
Fair value of plan assets 10,923 126 10,918 135
ASC 715, “Compensation – Retirement Benefits” provides for delayed recognition of actuarial gains and losses, including amounts arising from changes in the estimated projected plan benefit obligation due to changes in the assumed discount rate, differences between the actual and expected return on plan assets and other assumption changes. These net gains and losses are recognized prospectively over a period that approximates the average remaining service period of active employees expected to receive benefits under the plans to the extent that they are not offset by gains in subsequent years. The estimated net loss and prior service cost that will be amortized from AOCI into net periodic pension cost for the U.S. plans during the next fiscal year are expected to be $374 million and $41 million, respectively.
NET PERIODIC PENSION EXPENSE
Service cost is the actuarial present value of benefits attributed by the plans’ benefit formula to services rendered by employees during the year. Interest cost represents the increase in the projected benefit obligation, which is a discounted amount, due to the passage of time. The expected return on plan assets reflects the computed amount of current-year earnings from the investment of plan assets using an estimated long-term rate of return.
Net periodic pension expense for qualified and nonqualified U.S. and non-U.S. defined benefit plans comprised the following:
2015 2014 2013
In millions U.S.
Plans Non-
U.S.
Plans U.S.
Plans Non-
U.S.
Plans U.S.
Plans Non-
U.S.
Plans
Service cost$161$6$145$5$188$4
Interest cost 597 10 600 13 576 11
Expected return on plan assets(783)(11)(762)(14)(738)(11)
Actuarial loss / (gain)428 1 374—485 1
Amortization of prior service cost 43—30—34—
Curtailment gain———(4)——
Settlement loss 15—————
Net periodic pension expense (a)$461$6$387$—$545$5
(a) Excludes $1 million in curtailments in 2014 related to the pension freeze remeasurement that were recorded in restructuring and other charges.
The increase in 2015 pension expense reflects a decrease in the discount rate from 4.65% in 2014 to 4.10% in 2015, updated mortality assumptions, higher amortization of unrecognized actuarial losses and a settlement charge in 2015.
ASSUMPTIONS
International Paper evaluates its actuarial assumptions annually as of December 31 (the measurement date) and considers changes in these long-term factors based upon market conditions and the requirements for employers’ accounting for pensions. These assumptions are used to calculate benefit obligations as of December 31 of the current year and pension expense to be recorded in the following year (i.e., the discount rate used to determine the benefit obligation as of December 31, 2015 was also the discount rate used to determine net pension expense for the 2016 year).
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Major actuarial assumptions used in determining the benefit obligations and net periodic pension cost for our defined benefit plans are presented in the following table:
2015 2014 2013
U.S.
Plans Non-
U.S.
Plans U.S.
Plans Non-
U.S.
Plans U.S.
Plans Non-
U.S.
Plans
Actuarial assumptions used to determine benefit obligations as of December 31:
Discount rate 4.40%4.64%4.10%4.72%4.90%5.07%
Rate of compensation increase 3.75%4.12%3.75%4.03%3.75%4.13%
Actuarial assumptions used to determine net periodic pension cost for years ended December 31:
Discount rate (a)4.10%4.72%4.65%5.07%4.10%4.96%
Expected long-term rate of return on plan assets (b)7.75%6.64%7.75%7.53%8.00%7.04%
Rate of compensation increase 3.75%4.03%3.75%4.13%3.75%3.17%
(a) Represents the weighted average rate for 2014 due to the remeasurement in the first quarter of 2014.
(b)Represents the expected rate of return for International Paper's qualified pension plan for 2014 and 2013. The weighted average rate for the Temple-Inland Retirement Plan was 7.00% and 6.16% for 2014 and 2013, respectively.
The expected long-term rate of return on plan assets is based on projected rates of return for current and planned asset classes in the plan’s investment portfolio. Projected rates of return are developed through an asset/liability study in which projected returns for each of the plan’s asset classes are determined after analyzing historical experience and future expectations of returns and volatility of the various asset classes.
Based on the target asset allocation for each asset class, the overall expected rate of return for the portfolio
is developed considering the effects of active portfolio management and expenses paid from plan assets. The discount rate assumption was determined from a universe of high quality corporate bonds. A settlement portfolio is selected and matched to the present value of the plan’s projected benefit payments. To calculate pension expense for 2016, the Company will use an expected long-term rate of return on plan assets of 7.75% for the Retirement Plan of International Paper,
a discount rate of 4.40% and an assumed rate of compensation increase of 3.75%. The Company estimates that it will record net pension expense of approximately $364 million for its U.S. defined benefit plans in 2016, with the decrease from expense of $461 million in 2015 reflecting an increase in the discount rate to 4.40% in 2016 from 4.10% in 2015, updated demographic assumptions, and lower amortization of unrecognized losses.
For non-U.S. pension plans, assumptions reflect economic assumptions applicable to each country.
The following illustrates the effect on pension expense for 2016 of a 25 basis point decrease in the above assumptions:
In millions 2016
Expense/(Income):
Discount rate$36
Expected long-term rate of return on plan assets 27
Rate of compensation increase(2)
PLAN ASSETS
International Paper’s Board of Directors has appointed a Fiduciary Review Committee that is responsible for fiduciary oversight of the U.S. Pension Plan, approving investment policy and reviewing the management and control of plan assets. Pension Plan assets are invested to maximize returns within prudent levels of risk.
The Pension Plan maintains a strategic asset allocation policy that designates target allocations by asset class. Investments are diversified across classes and within each class to minimize the risk of large losses. Derivatives, including swaps, forward and futures contracts, may be used as asset class substitutes or for hedging or other risk management purposes. Periodic reviews are made of investment policy objectives and investment manager performance. For non-U.S. plans, assets consist principally of common stock and fixed income securities.
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International Paper’s U.S. pension allocations by type of fund at December 31, and target allocations were as follows:
Asset Class 2015 2014 Target
Allocations
Equity accounts 48%47%43%-54%
Fixed income accounts 33%33%25%-35%
Real estate accounts 10%10%7%-13%
Other 9%10%8%-17%
Total 100%100%
The 2014 actual allocations shown represent a weighted average of International Paper and Temple-Inland plan assets as the TIN plan was fully merged into the IP plan by 2015.
The fair values of International Paper’s pension plan assets at December 31, 2015 and 2014 by asset class are shown below. Plan assets included an immaterial amount of International Paper common stock at December 31, 2015 and 2014. Hedge funds disclosed in the following table are allocated equally between equity and fixed income accounts for target allocation purposes.
Fair Value Measurement at December 31, 2015
Asset Class Total Quoted
Prices in
Active
Markets
For
Identical
Assets
(Level 1)Significant
Observable
Inputs
(Level 2)Significant
Unobservable
Inputs
(Level 3)
In millions
Equities – domestic$2,150$1,382$768$—
Equities – international 2,563 1,818 745—
Corporate bonds 1,286—1,286—
Government securities 518—518—
Mortgage backed securities 217—217—
Other fixed income 275—265 10
Commodities 118—118—
Hedge funds 894——894
Private equity 492——492
Real estate 1,094——1,094
Risk parity funds 341—1 340
Cash and cash equivalents 975 975——
Total Investments$10,923$4,175$3,918$2,830
Fair Value Measurement at December 31, 2014
Asset Class Total Quoted
Prices
in
Active
Markets
For
Identical
Assets
(Level 1)Significant
Observable
Inputs
(Level 2)Significant
Unobservable
Inputs
(Level 3)
In millions
Equities – domestic$2,268$1,380$888$—
Equities – international 2,397 1,815 582—
Corporate bonds 1,230—1,230—
Government securities 1,282—1,282—
Mortgage backed securities 172—172—
Other fixed income 207—197 10
Commodities 170—170—
Hedge funds 867——867
Private equity 519——519
Real estate 1,101——1,101
Risk parity funds 376——376
Cash and cash equivalents 329 329——
Total Investments$10,918$3,524$4,521$2,873
Equity securities consist primarily of publicly traded U.S. companies and international companies. Publicly traded equities are valued at the closing prices reported in the active market in which the individual securities are traded.
Fixed income consists of government securities, mortgage-backed securities, corporate bonds and common collective funds. Government securities are valued by third-party pricing sources. Mortgage-backed security holdings consist primarily of agency-rated holdings. The fair value estimates for mortgage securities are calculated by third-party pricing sources chosen by the custodian’s price matrix. Corporate bonds are valued using either the yields currently available on comparable securities of issuers with similar credit ratings or using a discounted cash flows approach that utilizes observable inputs, such as current yields of similar instruments, but includes adjustments for certain risks that may not be observable, such as credit and liquidity risks. Common collective funds are valued at the net asset value per share multiplied by the number of shares held as of the measurement date.
Commodities consist of commodity-linked notes and commodity-linked derivatives. Commodities are valued at closing prices determined by calculation agents for outstanding transactions.
Hedge funds are investment structures for managing private, loosely-regulated investment pools that can pursue a diverse array of investment strategies with a wide range of different securities and derivative instruments. These investments are made through funds-of-funds (commingled, multi-manager fund structures) and through direct investments in individual hedge funds. Hedge funds are primarily valued by each
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fund’s third-party administrator based upon the valuation of the underlying securities and instruments and primarily by applying a market or income valuation methodology as appropriate depending on the specific type of security or instrument held. Funds-of-funds are valued based upon the net asset values of the underlying investments in hedge funds.
Private equity consists of interests in partnerships that invest in U.S. and non-U.S. debt and equity securities. Partnership interests are valued using the most recent general partner statement of fair value, updated for any subsequent partnership interest cash flows.
Real estate includes commercial properties, land and timberland, and generally includes, but is not limited to, retail, office, industrial, multifamily and hotel properties. Real estate fund values are primarily reported by the fund manager and are based on valuation of the
underlying investments which include inputs such as cost, discounted cash flows, independent appraisals and market based comparable data.
Risk Parity Funds are defined as engineered beta exposure to a wide range of asset classes and risk premia, including equity, interest rates, credit, and commodities. Risk parity funds seek to provide high risk-adjusted returns while providing a high level of diversification relative to a traditional equity/fixed income portfolio.These funds seek to achieve this objective with the use of modest leverage applied to lower-risk, more diverse asset classes. Investments in Risk parity funds are valued using monthly reported net asset values. Also included in these funds are related derivative instruments which are generally employed as asset class substitutes for managing asset/liability mismatches, or bona fide hedging or other appropriate risk management purposes. Derivative instruments are generally valued by the investment managers or in certain instances by third-party pricing sources.
The fair value measurements using significant unobservable inputs (Level 3) at December 31, 2015 were as follows:
Fair Value Measurements Using Significant Unobservable Inputs (Level 3)
In millions Other
fixed
income Hedge
funds Private
equity Real
estate Risk parity funds Total
Beginning balance at December 31, 2014$10$867$519$1,101$376$2,873
Actual return on plan assets:
Relating to assets still held at the reporting date—27 27 41(39)56
Relating to assets sold during the period—3(9)27(7)14
Purchases, sales and settlements—(3)(45)(75)10(113)
Transfers in and/or out of Level 3——————
Ending balance at December 31, 2015$10$894$492$1,094$340$2,830
FUNDING AND CASH FLOWS
The Company’s funding policy for the Pension Plan is to contribute amounts sufficient to meet legal funding requirements, plus any additional amounts that the Company may determine to be appropriate considering the funded status of the plans, tax deductibility, cash flow generated by the Company, and other factors. The Company continually reassesses the amount and timing of any discretionary contributions. Contributions to the qualified plan totaling $750 million, $353 million and $31 million were made by the Company in 2015, 2014 and 2013, respectively. Generally, International Paper’s non-U.S. pension plans are funded using the projected benefit as a target, except in certain countries where funding of benefit plans is not required.
At December 31, 2015, projected future pension benefit payments, excluding any termination benefits, were as follows:
In millions
2016$782
2017 792
2018 803
2019 818
2020 832
2021 – 2025 4,365
OTHER U.S. PLANS
International Paper sponsors the International Paper Company Salaried Savings Plan and the International Paper Company Hourly Savings Plan, both of which are tax-qualified defined contribution 401(k) savings plans.
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Substantially all U.S. salaried and certain hourly employees are eligible to participate and may make elective deferrals to such plans to save for retirement. International Paper makes matching contributions to participant accounts on a specified percentage of employee deferrals as determined by the provisions of each plan. For eligible employees hired after June 30, 2004, the Company makes Retirement Savings Account contributions equal to a percentage of an eligible employee’s pay.
The Company also sponsors the International Paper Company Deferred Compensation Savings Plan, which is an unfunded nonqualified defined contribution plan. This plan permits eligible employees to continue to make deferrals and receive company matching contributions when their contributions to the International Paper Salaried Savings Plan are stopped due to limitations under U.S. tax law. Participant deferrals and company matching contributions are not invested in a separate trust, but are paid directly from International Paper’s general assets at the time benefits become due and payable.
Company matching contributions to the plans totaled approximately $100 million, $112 million and $120 million for the plan years ending in 2015, 2014 and 2013, respectively.
NOTE 17 POSTRETIREMENT BENEFITS
U.S. POSTRETIREMENT BENEFITS
International Paper provides certain retiree health care and life insurance benefits covering certain U.S. salaried and hourly employees. These employees are generally eligible for benefits upon retirement and completion of a specified number of years of creditable service. Excluded from company-provided medical benefits are salaried employees whose age plus years of employment with the Company totaled less than 60 as of January 1, 2004. International Paper does not fund these benefits prior to payment and has the right to modify or terminate certain of these plans in the future.
In addition to the U.S. plan, certain Brazilian and Moroccan employees are eligible for retiree health care and life insurance benefits.
The components of postretirement benefit expense in 2015, 2014 and 2013 were as follows:
In millions 2015 2014 2013
U.S.
Plans Non-
U.S.
Plans U.S.
Plans Non-
U.S.
Plans U.S.
Plans Non-
U.S.
Plans
Service cost$1$1$1$1$2$2
Interest cost 11 5 14 6 14 5
Actuarial loss 6 1 5 1 7—
Amortization of prior service credits(10)(2)(13)(1)(24)—
Net postretirement (benefit) expense (a)$8$5$7$7$(1)$7
(a) Excludes $7 million of curtailment gains in 2013 related to the sale of Building Products that were recorded in Net (gains) losses on sales and impairments of businesses in the consolidated statement of operations.
International Paper evaluates its actuarial assumptions annually as of December 31 (the measurement date) and considers changes in these long-term factors based upon market conditions and the requirements of employers’ accounting for postretirement benefits other than pensions.
The discount rates used to determine net U.S. and non-U.S. postretirement benefit cost for the years ended December 31, 2015, 2014 and 2013 were as follows:
2015 2014 2013
U.S.
Plans Non-
U.S.
Plans U.S.
Plans Non-
U.S.
Plans U.S.
Plans Non-
U.S.
Plans
Discount rate 3.90%11.52%4.50%11.94%3.70%8.43%
The weighted average assumptions used to determine the benefit obligation at December 31, 2015 and 2014 were as follows:
2015 2014
U.S.
Plans Non-
U.S.
Plans U.S.
Plans Non-
U.S.
Plans
Discount rate 4.20%12.23%3.90%11.52%
Health care cost trend rate assumed for next year 7.00%11.41%7.00%11.38%
Rate that the cost trend rate gradually declines to 5.00%5.94%5.00%6.11%
Year that the rate reaches the rate it is assumed to remain 2022 2026 2022 2025
A 1% increase in the assumed annual health care cost trend rate would have increased the U.S. and non-U.S. accumulated postretirement benefit obligations at December 31, 2015 by approximately $11 million and $7 million, respectively. A 1% decrease in the annual trend rate would have decreased the U.S. and non-U.S. accumulated postretirement benefit obligation at December 31, 2015 by approximately $10 million and $6 million, respectively. The effect on net postretirement
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benefit cost from a 1% increase or decrease would be approximately $1 million for both U.S. and non-U.S. plans.
The plan is only funded in an amount equal to benefits paid. The following table presents the changes in benefit obligation and plan assets for 2015 and 2014:
In millions 2015 2014
U.S.
Plans Non-
U.S.
Plans U.S.
Plans Non-
U.S.
Plans
Change in projected benefit obligation:
Benefit obligation, January 1$306$59$322$72
Service cost 1 1 1 1
Interest cost 11 5 14 6
Participants’ contributions 12—15—
Actuarial (gain) loss—(1)14 19
Other———(26)
Plan amendments—1—(7)
Benefits paid(57)(1)(62)(1)
Less: Federal subsidy 2—2—
Currency Impact—(19)—(5)
Benefit obligation, December 31$275$45$306$59
Change in plan assets:
Fair value of plan assets, January 1$—$—$—$—
Company contributions 45 1 47 1
Participants’ contributions 12—15—
Benefits paid(57)(1)(62)(1)
Fair value of plan assets, December 31$—$—$—$—
Funded status, December 31$(275)$(45)$(306)$(59)
Amounts recognized in the consolidated balance sheet under ASC 715:
Current liability$(29)$(2)$(33)$(2)
Non-current liability(246)(43)(273)(57)
$(275)$(45)$(306)$(59)
Amounts recognized in accumulated other comprehensive income under ASC 715 (pre-tax):
Net actuarial loss (gain)$42$15$44$23
Prior service credit(12)(2)(22)(5)
$30$13$22$18
The non-current portion of the liability is included with the postemployment liability in the accompanying consolidated balance sheet under Postretirement and postemployment benefit obligation.
The components of the $8 million and ($5) million increase and decrease in the amounts recognized in OCI during 2015 for U.S. and non-U.S. plans, respectively, consisted of:
In millions U.S.
Plans Non-
U.S.
Plans
Current year actuarial gain$4$—
Amortization of actuarial (loss) gain(6)(1)
Current year prior service cost—1
Amortization of prior service credit 10 2
Currency impact—(7)
$8$(5)
The portion of the change in the funded status that was recognized in either net periodic benefit cost or OCI for the U.S. plans was $17 million, $33 million and $63 million in 2015, 2014 and 2013, respectively. The portion of the change in funded status for the non-U.S. plans was $0 million, $14 million, and $19 million in 2015, 2014 and 2013, respectively.
The estimated amounts of net loss and prior service credit that will be amortized from OCI into net U.S. postretirement benefit cost in 2016 are expected to be $6 million and $(4) million, respectively. The estimated amounts for non-U.S. plans in 2016 are expected to be $1 million and $(2) million, respectively.
At December 31, 2015, estimated total future postretirement benefit payments, net of participant contributions and estimated future Medicare Part D subsidy receipts, were as follows:
In millions Benefit
Payments Subsidy Receipts Benefit
Payments
U.S.
Plans U.S.
Plans Non-
U.S.
Plans
2016$31$1$2
2017 28 1 2
2018 27 1 2
2019 25 1 2
2020 24 1 3
2021 – 2025 98 6 21
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NOTE 18 INCENTIVE PLANS
International Paper currently has an Incentive Compensation Plan (ICP) which, upon the approval by the Company’s shareholders in May 2009, replaced the Company’s Long-Term Incentive Compensation Plan (LTICP). The ICP authorizes grants of restricted stock, restricted or deferred stock units, performance awards payable in cash or stock upon the attainment of specified performance goals, dividend equivalents, stock options, stock appreciation rights, other stock-based awards, and cash-based awards at the discretion of the Management Development and Compensation Committee of the Board of Directors (the Committee) that administers the ICP. Additionally, restricted stock, which may be deferred into RSU’s, may be awarded under a Restricted Stock and Deferred Compensation Plan for Non-Employee Directors.
STOCK OPTION PROGRAM
International Paper accounts for stock options in accordance with guidance under ASC 718, “Compensation – Stock Compensation.” Compensation expense is recorded over the related service period based on the grant-date fair market value. Since all outstanding options were vested as of July 14, 2005, only replacement option grants are expensed.
During each reporting period, diluted earnings per share is calculated by assuming that “in-the-money” options are exercised and the exercise proceeds are used to repurchase shares in the marketplace. When options are actually exercised, option proceeds are credited to equity and issued shares are included in the computation of earnings per common share, with no effect on reported earnings. Equity is also increased by the tax benefit that International Paper will receive in its tax return for income reported by the employees in their individual tax returns.
Under the program, upon exercise of an option, a replacement option may be granted under certain circumstances with an exercise price equal to the market price at the time of exercise and with a term extending to the expiration date of the original option.
The Company has discontinued the issuance of stock options for all eligible U.S. and non-U.S. employees. In the United States, the stock option program was replaced with a performance-based restricted share program to more closely tie long-term incentive compensation to Company performance on two key performance drivers: return on invested capital (ROIC) and total shareholder return (TSR). All outstanding options expired on March 15, 2015.
The following summarizes the status of the Stock Option Program and the changes during the three years ending December 31, 2015:
Options
(a,b)Weighted
Average
Exercise
Price Weighted
Average
Remaining
Life
(years)Aggregate
Intrinsic
Value
(thousands)
Outstanding at December 31, 2012 9,136,060$38.79 1.15$1,077
Granted 4,744 48.11
Exercised(7,317,825)38.57
Expired(70,190)37.15
Outstanding at December 31, 2013 1,752,789 39.80 0.67 16,175
Granted 3,247 49.13
Exercised(1,634,858)39.80
Expired(49,286)41.50
Outstanding at December 31, 2014 71,892 39.03 0.18 1,046
Granted——
Exercised(62,477)39.05
Expired(9,415)38.92
Outstanding at December 31, 2015—$—0.00$—
(a)The table does not include Continuity Award tandem stock options described below. No fair market value is assigned to these options under ASC 718. The tandem restricted shares accompanying these options are expensed over their vesting period.
(b)The table includes options outstanding under an acquired company plan under which options may no longer be granted.
PERFORMANCE SHARE PLAN
Under the Performance Share Plan (PSP), contingent awards of International Paper common stock are granted by the Committee. The PSP awards are earned evenly over a three-year period. PSP awards are earned based on the achievement of defined performance rankings of ROIC and TSR compared to ROIC and TSR peer groups of companies. Awards are weighted 75% for ROIC and 25% for TSR for all participants except for officers for whom the awards are weighted 50% for ROIC and 50% for TSR. The ROIC component of the PSP awards is valued at the closing stock price on the day prior to the grant date. As the ROIC component contains a performance condition, compensation expense, net of estimated forfeitures, is recorded over the requisite service period based on the most probable number of awards expected to vest. The TSR component of the PSP awards is valued using a Monte Carlo simulation as the TSR component contains a market condition. The Monte Carlo simulation estimates the fair value of the TSR component based on the expected term of the award, a risk-free rate, expected dividends, and the expected volatility for the Company and its competitors. The expected term is estimated based on the vesting period of the awards, the risk-free rate is based on the yield on U.S. Treasury securities matching the vesting period, and the volatility is based on the Company’s historical volatility over the expected term.
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PSP grants are made in performance-based restricted stock units. The 2012 PSP awards issued to certain members of senior management were accounted for as liability awards, which were remeasured at fair value at each balance sheet date. The valuation of these PSP liability awards was computed based on the same methodology as the PSP equity awards. On December 8, 2014, IP eliminated the election for executives to withhold more than the minimum tax withholding for the 2013 and 2014 grants making them equity awards.
The following table sets forth the assumptions used to determine compensation cost for the market condition component of the PSP plan:
Twelve Months Ended December 31, 2015
Expected volatility 19.01%-36.02%
Risk-free interest rate 0.21%-1.10%
The following summarizes PSP activity for the three years ending December 31, 2015:
Share/Units Weighted
Average
Grant Date
Fair Value
Outstanding at December 31, 2012 8,660,855$28.37
Granted 3,148,445 40.76
Shares issued(3,262,760)32.48
Forfeited(429,051)34.58
Outstanding at December 31, 2013 8,117,489 31.20
Granted 3,682,663 46.82
Shares issued(4,025,111)37.18
Forfeited(499,107)43.10
Outstanding at December 31, 2014 7,275,934 34.98
Granted 1,863,623 53.25
Shares issued(2,959,160)37.09
Forfeited(322,664)53.97
Outstanding at December 31, 2015 5,857,733$38.69
EXECUTIVE CONTINUITY AND RESTRICTED STOCK AWARD PROGRAMS
The Executive Continuity Award program provides for the granting of tandem awards of restricted stock and/or nonqualified stock options to key executives. Grants are restricted and awards conditioned on attainment of a specified age. The awarding of a tandem stock option results in the cancellation of the related restricted shares. The final award under this program was paid in 2013.
The service-based Restricted Stock Award program (RSA), designed for recruitment, retention and special recognition purposes, also provides for awards of restricted stock to key employees.
The following summarizes the activity of the Executive Continuity Award program and RSA program for the three years ending December 31, 2015:
Shares Weighted
Average
Grant Date
Fair Value
Outstanding at December 31, 2012 151,549$30.49
Granted 67,100 44.41
Shares issued(88,775)32.30
Forfeited(17,500)37.75
Outstanding at December 31, 2013 112,374 36.24
Granted 89,500 48.19
Shares issued(83,275)33.78
Forfeited(4,000)45.88
Outstanding at December 31, 2014 114,599 47.03
Granted 36,300 50.06
Shares issued(27,365)45.35
Forfeited(3,166)50.04
Outstanding at December 31, 2015 120,368$48.24
At December 31, 2015, 2014 and 2013 a total of 16.2 million, 16.3 million and 17.8 million shares, respectively, were available for grant under the ICP.
Stock-based compensation expense and related income tax benefits were as follows:
In millions 2015 2014 2013
Total stock-based compensation expense (included in selling and administrative expense)$114$118$137
Income tax benefits related to stock-based compensation 88 92 74
At December 31, 2015, $126 million of compensation cost, net of estimated forfeitures, related to unvested
restricted performance shares, executive continuity awards and restricted stock attributable to future performance had not yet been recognized. This amount will be recognized in expense over a weighted-average period of 1.6 years.
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NOTE 19 FINANCIAL INFORMATION BY INDUSTRY SEGMENT AND GEOGRAPHIC AREA
International Paper’s industry segments, Industrial Packaging, Printing Papers and Consumer Packaging Businesses, are consistent with the internal structure used to manage these businesses. All segments are differentiated on a common product, common customer basis consistent with the business segmentation generally used in the Forest Products industry.
For management purposes, International Paper reports the operating performance of each business based on earnings before interest and income taxes (EBIT). Intersegment sales and transfers are recorded at current market prices.
External sales by major product is determined by aggregating sales from each segment based on similar products or services. External sales are defined as those that are made to parties outside International Paper’s consolidated group, whereas sales by segment in the Net Sales table are determined using a management approach and include intersegment sales.
The Company also holds a 50% interest in Ilim that is a separate reportable industry segment. The Company recorded equity earnings (losses), net of taxes, of $131 million, $(194) million and $(46) million in 2015, 2014, and 2013, respectively, for Ilim. Equity earnings (losses) includes an after-tax foreign exchange gain (loss) of $(75) million, $(269) million and $(32) million in 2015, 2014 and 2013, respectively, primarily on the remeasurement of U.S. dollar-denominated net debt.
Summarized financial information for Ilim which is accounted for under the equity method is presented in the following table.
Balance Sheet
In millions 2015 2014
Current assets$455$458
Noncurrent assets 968 1,223
Current liabilities 665 899
Noncurrent liabilities 715 742
Noncontrolling interests 21 15
Income Statement
In millions 2015 2014 2013
Net sales$1,931$2,138$1,897
Gross profit 971 772 562
Income from continuing operations 254(387)(76)
Net income attributable to Ilim 237(360)(71)
At December 31, 2015 and 2014, the Company's investment in Ilim was $172 million and $170 million, respectively, which was $161 million and $158 million, respectively, more than the Company's proportionate share of the joint venture's underlying net assets. The differences primarily relate to purchase price fair value adjustments and currency translation adjustments. The Company is party to a joint marketing agreement with Ilim, under which the Company purchases, markets and sells paper produced by Ilim. Purchases under this agreement were $170 million, $200 million and $114 million for the years ended December 31, 2015, 2014 and 2013, respectively.
INFORMATION BY INDUSTRY SEGMENT
Net Sales
In millions 2015 2014 2013
Industrial Packaging$14,484$14,944$14,810
Printing Papers 5,031 5,720 6,205
Consumer Packaging 2,940 3,403 3,435
Corporate and Intersegment Sales(90)(450)(967)
Net Sales$22,365$23,617$23,483
Operating Profit
In millions 2015 2014 2013
Industrial Packaging$1,853$1,896$1,801
Printing Papers 533(16)271
Consumer Packaging(25)178 161
Operating Profit 2,361 2,058 2,233
Interest expense, net(555)(601)(612)
Noncontrolling interests / equity earnings adjustment (a)(8)(2)1
Corporate items, net(36)(51)(61)
Restructuring and other charges(238)(282)(10)
Net gains (losses) on sales and impairments of businesses—(38)—
Non-operating pension expense(258)(212)(323)
Earnings (Loss) From Continuing Operations Before Income Taxes and Equity Earnings$1,266$872$1,228
Restructuring and Other Charges
In millions 2015 2014 2013
Industrial Packaging$—$7$(2)
Printing Papers—554 118
Consumer Packaging 10 8 45
Corporate 242 277(5)
Restructuring and Other Charges$252$846$156
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Assets
In millions 2015 2014
Industrial Packaging$14,483$14,852
Printing Papers 4,696 5,393
Consumer Packaging 2,115 3,249
Corporate and other (b)9,293 5,190
Assets$30,587$28,684
Capital Spending
In millions 2015 2014 2013
Industrial Packaging$858$754$629
Printing Papers 361 318 294
Consumer Packaging 216 233 208
Distribution (c)——9
Subtotal 1,435 1,305 1,140
Corporate and other (b)52 61 58
Total$1,487$1,366$1,198
Depreciation, Amortization and Cost of Timber Harvested (d)
In millions 2015 2014 2013
Industrial Packaging$725$775$805
Printing Papers 307 367 446
Consumer Packaging 215 223 206
Corporate 47 41 74
Depreciation and Amortization$1,294$1,406$1,531
External Sales By Major Product
In millions 2015 2014 2013
Industrial Packaging$14,421$14,837$14,729
Printing Papers 4,919 5,360 5,443
Consumer Packaging 2,907 3,307 3,311
Other 118 113—
Net Sales$22,365$23,617$23,483
INFORMATION BY GEOGRAPHIC AREA
Net Sales (e)
In millions 2015 2014 2013
United States (f)$16,554$16,645$16,371
EMEA 2,770 3,273 3,250
Pacific Rim and Asia 1,501 1,951 2,114
Americas, other than U.S.1,540 1,748 1,748
Net Sales$22,365$23,617$23,483
Long-Lived Assets (g)
In millions 2015 2014
United States$9,683$9,476
EMEA 827 926
Pacific Rim and Asia 353 897
Americas, other than U.S.1,085 1,553
Corporate 398 383
Long-Lived Assets$12,346$13,235
(a)Operating profits for industry segments include each segment’s percentage share of the profits of subsidiaries included in that segment that are less than wholly-owned. The pre-tax noncontrolling interests and equity earnings for these subsidiaries is added here to present consolidated earnings from continuing operations before income taxes and equity earnings.
(b)Includes corporate assets and assets of businesses held for sale.
(c)The xpedx business, which historically represented the Company's Distribution reportable segment, was spun off July 1, 2014.
(d)Excludes accelerated depreciation related to the closure and/or repurposing of mills.
(e)Net sales are attributed to countries based on the location of the seller.
(f)Export sales to unaffiliated customers were $2.0 billion in 2015, $2.3 billion in 2014 and $2.4 billion in 2013.
(g)Long-Lived Assets includes Forestlands and Plants, Properties and Equipment, net.
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INTERIM FINANCIAL RESULTS (UNAUDITED)
In millions, except per share amounts and stock prices 1st
Quarter 2nd
Quarter 3rd
Quarter 4th Quarter Year
2015
Net sales$5,517$5,714$5,691$5,443$22,365
Gross margin (a)1,673 1,746 1,800 1,678 6,897
Earnings (loss) from continuing operations before income taxes and equity earnings 406 266(b)329(b)265(b)1,266(b)
Gain (loss) from discontinued operations—————
Net earnings (loss) attributable to International Paper Company 313 227(b,c)220(b,c)178(b,c)938(b,c)
Basic earnings (loss) per share attributable to International Paper Company common shareholders:
Earnings (loss) from continuing operations$0.74$0.54(b)$0.53(b)$0.43(b)$2.25(b)
Gain (loss) from discontinued operations—————
Net earnings (loss)0.74 0.54(b,c)0.53(b,c)0.43(b,c)2.25(b,c)
Diluted earnings (loss) per share attributable to International Paper Company common shareholders:
Earnings (loss) from continuing operations 0.74 0.54(b)0.53(b)0.43(b)2.23(b)
Gain (loss) from discontinued operations—————
Net earnings (loss)0.74 0.54(b,c)0.53(b,c)0.43(b,c)2.23(b,c)
Dividends per share of common stock 0.4000 0.4000 0.4000 0.4400 1.6400
Common stock prices
High$57.90$56.49$49.49$44.83$57.90
Low 51.35 47.39 37.11 36.76 36.76
2014
Net sales$5,724$5,899$6,051$5,943$23,617
Gross margin (a)1,690 1,839 1,996 1,838 7,363
Earnings (loss) from continuing operations before income taxes and equity earnings(139)(d)152(d)552(d)307(d)872(d)
Gain (loss) from discontinued operations(7)(e)(13)(e)16(e)(9)(e)(13)(e)
Net earnings (loss) attributable to International Paper Company(95)(d-f)161(d-f)355(d-f)134(d-f)555(d-f)
Basic earnings (loss) per share attributable to International Paper Company common shareholders:
Earnings (loss) from continuing operations$(0.20)(d)$0.40(d)$0.80(d)$0.34(d)$1.33(d)
Gain (loss) from discontinued operations(0.01)(e)(0.03)(e)0.04(e)(0.02)(e)(0.03)(e)
Net earnings (loss)(0.21)(d-f)0.37(d-f)0.84(d-f)0.32(d-f)1.30(d-f)
Diluted earnings (loss) per share attributable to International Paper Company common shareholders:
Earnings (loss) from continuing operations(0.20)(d)0.40(d)0.79(d)0.34(d)1.31(d)
Gain (loss) from discontinued operations(0.01)(e)(0.03)(e)0.04(e)(0.02)(e)(0.02)(e)
Net earnings (loss)(0.21)(d-f)0.37(d-f)0.83(d-f)0.32(d-f)1.29(d-f)
Dividends per share of common stock 0.3500 0.3500 0.3500 0.4000 1.4500
Common stock prices
High$49.71$50.65$51.98$55.73$55.73
Low 44.43 44.24 46.77 44.50 44.24
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Note: Since basic and diluted earnings per share are computed independently for each period and category, full year per share amounts may not equal the sum of the four quarters. In addition, the unaudited selected consolidated financial data are derived from our audited consolidated financial statements and have been revised to reflect discontinued operations.
Footnotes to Interim Financial Results
(a)Gross margin represents net sales less cost of products sold, excluding depreciation, amortization and cost of timber harvested.
(b)Includes the following pre-tax charges (gains):
2015
In millions Q1 Q2 Q3 Q4
Riegelwood mill conversion costs, net of proceeds from sale of the Carolina Coated Bristols brand$—$(14)$7$15
Timber monetization restructuring——17(1)
Early debt extinguishment costs—207——
Refund and state tax credits—(4)——
IP-Sun JV impairment——186(12)
Legal reserve adjustment———15
Impairment of Orsa goodwill and trade name intangible———137
Other items—1 1 4
Total$—$190$211$158
(c) Includes the following tax expenses (benefits):
2015
Q1 Q2 Q3 Q4
Tax expense for cash pension$—$23$—$—
Tax benefit related to IP-Sun JV——(67)—
Other items—5—2
Total$—$28$(67)$2
(d) Includes the following pre-tax charges (gains):
2014
Q1 Q2 Q3 Q4
Temple-Inland integration$12$2$1$1
Courtland mill shutdown 495 49 3 7
Early debt extinguishment costs—262 13 1
India legal contingency resolution——(20)—
Multi-employer pension plan withdrawal liability——35—
Foreign tax amnesty program——32—
Asia Industrial Packaging goodwill impairment———100
Loss on sale by investee and impairment of investment———47
Other items 4(4)13(1)
Total$511$309$77$155
(e) Includes the after-tax operating earnings of the xpedx business prior to the spin-off and the following after-tax charges (gains):
2014
Q1 Q2 Q3 Q4
xpedx spinoff$10$20$(14)$—
Building Products divestiture 2—(2)9
xpedx restructuring—(1)——
Total$12$19$(16)$9
(f) Includes the following tax expenses (benefits):
2014
Q1 Q2 Q3 Q4
State legislative tax change$10$—$—$—
Internal restructuring———(90)
Other items(1)———
Total$9$—$—$(90)
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ITEM 9. CHANGES IN AND DISAGREEMENTS WITH ACCOUNTANTS ON ACCOUNTING AND FINANCIAL DISCLOSURE
None.
ITEM 9A. CONTROLS AND PROCEDURES
EVALUATION OF DISCLOSURE CONTROLS AND PROCEDURES
We maintain disclosure controls and procedures that are designed to ensure that information required to be disclosed by us in the reports we file or submit under the Securities and Exchange Act of 1934, as amended (the “Exchange Act”), is recorded, processed, summarized and reported within the time periods specified in the SEC’s rules and forms, and that such information is accumulated and communicated to management, including our principal executive officer and principal financial officer, as appropriate, to allow timely decisions regarding required disclosure. As of December 31, 2015, an evaluation was carried out under the supervision and with the participation of the Company’s management, including our principal executive officer and principal financial officer, of the effectiveness of our disclosure controls and procedures, as defined by Rule 13a-15 under the Exchange Act. Based upon this evaluation, our principal executive officer and principal financial officer have concluded that the Company’s disclosure controls and procedures were effective as of December 31, 2015.
MANAGEMENT’S REPORT ON INTERNAL CONTROL OVER FINANCIAL REPORTING
Our management is responsible for establishing and maintaining adequate internal control over our financial reporting. Internal control over financial reporting is the process designed by, or under the supervision of, our principal executive officer and principal financial officer, and effected by our Board of Directors, management and other personnel, to provide reasonable assurance regarding the reliability of financial reporting and the preparation of financial statements for external purposes in accordance with accounting principles generally accepted in the United States (GAAP). Our internal control over financial reporting includes those policies and procedures that:
•pertain to the maintenance of records that, in reasonable detail, accurately and fairly reflect the transactions and dispositions of our assets;
•provide reasonable assurance that transactions are recorded as necessary to allow for the preparation of financial statements in accordance with GAAP, and that our receipts and expenditures are being made only in accordance with authorizations of our management and directors;
•provide reasonable assurance regarding prevention or timely detection of unauthorized acquisition, use or disposition of our assets that could have a material effect on our consolidated financial statements; and
•provide reasonable assurance as to the detection of fraud.
All internal control systems have inherent limitations, including the possibility of circumvention and overriding of controls, and therefore can provide only reasonable assurance of achieving the designed control objectives. The Company’s internal control system is supported by written policies and procedures, contains self-monitoring mechanisms, and is audited by the internal audit function. Appropriate actions are taken by management to correct deficiencies as they are identified.
As of December 31, 2015, management has assessed the effectiveness of the Company’s internal control over financial reporting. In a report included on pages 40 and 41, management concluded that the Company’s internal control over financial reporting was effective as of December 31, 2015.
In making this assessment, we used the criteria described in “Internal Control – Integrated Framework (2013)” issued by the Committee of Sponsoring Organizations of the Treadway Commission.
Our independent registered public accounting firm, Deloitte& Touche LLP, with direct access to our Board of Directors through our Audit and Finance Committee, has audited the consolidated financial statements prepared by us. Deloitte& Touche LLP has also issued an attestation report on our internal control over financial reporting. Their report on the consolidated financial statements and attestation report are included in Part II, Item 8 of this Annual Report under the heading “Financial Statements and Supplementary Data.”
MANAGEMENT’S PROCESS TO ASSESS THE EFFECTIVENESS OF INTERNAL CONTROL OVER FINANCIAL REPORTING
To comply with the requirements of Section 404 of the Sarbanes-Oxley Act of 2002, we followed a comprehensive compliance process across the enterprise to evaluate our internal control over financial reporting, engaging employees at all levels of the organization. Our internal control environment includes an enterprise-wide attitude of integrity and control consciousness that establishes a positive “tone at the top.” This is exemplified by our ethics program that includes long-standing principles and policies on ethical business conduct that require employees to maintain the highest ethical and legal standards in the conduct of our business, which have been distributed to all employees; a toll-free telephone helpline whereby any
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employee may report suspected violations of law or our policy; and an office of ethics and business practice. The internal control system further includes careful selection and training of supervisory and management personnel, appropriate delegation of authority and division of responsibility, dissemination of accounting and business policies throughout the Company, and an extensive program of internal audits with management follow-up. Our Board of Directors, assisted by the Audit and Finance Committee, monitors the integrity of our financial statements and financial reporting procedures, the performance of our internal audit function and independent auditors, and other matters set forth in its charter. The Committee, which consists of independent directors, meets regularly with representatives of management, and with the independent auditors and the Internal Auditor, with and without management representatives in attendance, to review their activities.
CHANGES IN INTERNAL CONTROL OVER FINANCIAL REPORTING
There have been no changes in our internal control over financial reporting during the quarter ended December 31, 2015, that have materially affected, or are reasonably likely to materially affect, our internal control over financial reporting.
ITEM 9B. OTHER INFORMATION
None.
PART III.
ITEM 10. DIRECTORS, EXECUTIVE OFFICERS AND CORPORATE GOVERNANCE
Information concerning our directors is hereby incorporated by reference to our definitive proxy statement that will be filed with the Securities and Exchange Commission (SEC) within 120 days of the close of our fiscal year. The Audit and Finance Committee of the Board of Directors has at least one member who is a financial expert, as that term is defined in Item 401(d)(5) of Regulation S-K. Further information concerning the composition of the Audit and Finance Committee and our audit committee financial experts is hereby incorporated by reference to our definitive proxy statement that will be filed with the SEC within 120 days of the close of our fiscal year. Information with respect to our executive officers is set forth on pages 5 and 6 in Part I of this Form 10-K under the caption, “Executive Officers of the Registrant.”
Executive officers of International Paper are elected to hold office until the next annual meeting of the Board of Directors following the annual meeting of shareholders and, until the election of successors, subject to removal by the Board.
The Company’s Code of Business Ethics (Code) is applicable to all employees of the Company, including the chief executive officer and senior financial officers, as well as the Board of Directors. We disclose any amendments to our Code and any waivers from a provision of our Code granted to our directors, chief executive officer and senior financial officers on our Internet Web site within four business days following such amendment or waiver. To date, no waivers of the Code have been granted.
We make available free of charge on our Internet Web site at www.internationalpaper.com, and in print to any shareholder who requests them, our Corporate Governance Principles, our Code of Business Ethics and the Charters of our Audit and Finance Committee, Management Development and Compensation Committee, Governance Committee and Public Policy and Environment Committee. Requests for copies may be directed to the corporate secretary at our corporate headquarters.
Information with respect to compliance with Section 16(a) of the Securities and Exchange Act and our corporate governance is hereby incorporated by reference to our definitive proxy statement that will be filed with the SEC within 120 days of the close of our fiscal year.
ITEM 11. EXECUTIVE COMPENSATION
Information with respect to the compensation of executives and directors of the Company is hereby incorporated by reference to our definitive proxy statement that will be filed with the SEC within 120 days of the close of our fiscal year.
ITEM 12. SECURITY OWNERSHIP OF CERTAIN BENEFICIAL OWNERS AND MANAGEMENT AND RELATED STOCKHOLDER MATTERS
A description of the security ownership of certain beneficial owners and management and equity compensation plan information is hereby incorporated by reference to our definitive proxy statement that will be filed with the SEC within 120 days of the close of our fiscal year.
ITEM 13. CERTAIN RELATIONSHIPS AND RELATED TRANSACTIONS, AND DIRECTOR INDEPENDENCE
A description of certain relationships and related transactions is hereby incorporated by reference to our definitive proxy statement that will be filed with the SEC within 120 days of the close of our fiscal year.
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ITEM 14. PRINCIPAL ACCOUNTANT FEES AND SERVICES
Information with respect to fees paid to, and services rendered by, our principal accountant, and our policies and procedures for pre-approving those services, is hereby incorporated by reference to our definitive proxy statement that will be filed with the SEC within 120 days of the close of our fiscal year.
PART IV.
ITEM 15. EXHIBITS AND FINANCIAL STATEMENT SCHEDULES
(1)Financial Statements – See Item 8. Financial Statements and Supplementary Data.
(2)Financial Statement Schedules – The following additional financial data should be read in conjunction with the consolidated financial statements in Item 8. Schedules not included with this additional financial data have been omitted because they are not applicable, or the required information is shown in the consolidated financial statements or the notes thereto.
Additional Financial Data
2015, 2014 and 2013
Consolidated Schedule: II-Valuation and Qualifying Accounts.90
(3.1)Restated Certificate of Incorporation of International Paper Company (incorporated by reference to Exhibit 3.1 to the Company’s Current Report on Form 8-K dated May 13, 2013).
(3.2)By-laws of International Paper Company, as amended through February 9, 2016
(incorporated by reference to Exhibit 3.1 to the Company’s Current Report on Form 8-K dated February 8, 2016).
(4.1)Indenture, dated as of April 12, 1999, between International Paper and The Bank of New York, as Trustee (incorporated by reference to Exhibit 4.1 to the Company’s Current Report on Form 8-K dated June 29, 2000).
(4.2)Supplemental Indenture (including the form of Notes), dated as of June 4, 2008, between International Paper Company and The Bank of New York, as Trustee (incorporated by reference to Exhibit 4.1 to the Company’s Current Report on Form 8-K dated June 4, 2008).
(4.3)Supplemental Indenture (including the form of Notes), dated as of May 11, 2009, between International Paper Company and The Bank of New York Mellon, as trustee (incorporated by reference to Exhibit 4.1 to the Company's Current Report on Form 8-K dated May 11, 2009).
(4.4)Supplemental Indenture (including the form of Notes), dated as of August 10, 2009, between International Paper Company and The Bank of New York Mellon, as trustee (incorporated by reference to Exhibit 4.1 to the Company's Current Report on Form 8-K dated August 10, 2009).
(4.5)Supplemental Indenture (including the form of Notes), dated as of December 7, 2009, between International Paper Company and The Bank of New York Mellon Trust Company, N.A., as trustee (incorporated by reference to Exhibit 4.1 to the Company's Current Report on Form 8-K dated December 7, 2009).
(4.6)Supplemental Indenture (including the form of Notes), dated as of November 16, 2011, between the Company and The Bank of New York Mellon Trust Company, N.A., as trustee (incorporated by reference to Exhibit 4.1 to the Company's Current Report on Form 8-K dated November 16, 2011).
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(4.7)Supplemental Indenture (including the form of Notes), dated as of June 10, 2014, between the Company and The Bank of New York Mellon Trust Company, N.A., as trustee (incorporated by reference to Exhibit 4.1 to the Company's Current Report on Form 8-K dated June 10, 2014).
(4.8)Supplemental Indenture (including the form of Notes), dated as of May 26, 2015, between the Company and The Bank of New York Mellon Trust Company, N.A., as trustee (incorporated by reference to Exhibit 4.1 to the Company's Current Report on Form 8-K dated May 26, 2015).
(4.9)In accordance with Item 601 (b)(4)(iii)(A)of Regulation S-K, certain instruments respecting long-term debt of the Company have been omitted but will be furnished to the Commission upon request.
(10.1)Amended and Restated 2009 Incentive Compensation Plan (ICP) (incorporated by reference to Exhibit 99.1 to the Company's Current Report on Form 8-K dated February 10, 2014). +
(10.2)2015 Management Incentive Plan (incorporated by reference to Exhibit 10.2 to the Company’s Annual Report on Form 10-K for the fiscal year ended December 31, 2014). +
(10.3)2016 Management Incentive Plan
(incorporated by reference to Exhibit 99.1 to the Company’s Current Report on Form 8-K dated February 8, 2016) +
(10.4)Amended and Restated 2009 Executive Management Incentive Plan, including 2015 Exhibits thereto (incorporated by reference to Exhibit 10.4 to the Company’s Annual Report on Form 10-K for the fiscal year ended December 31, 2014). +
(10.5)2016 Exhibits to the Amended and Restated 2009 Executive Management Incentive Plan. +
(10.6)Restricted Stock and Deferred Compensation Plan for Non-Employee Directors, Amended and Restated as of May 10, 2010 (incorporated by reference to Exhibit 10.1 to the Company’s Quarterly Report on Form 10-Q for the quarter ended June 30, 2010).+
(10.7)Form of Restricted Stock Award Agreement. (incorporated by reference to Exhibit 10.8 to the Company’s Annual Report on Form 10-K for the fiscal year ended December 31, 2013). +
(10.8)Form of Restricted Stock Unit Award Agreement (cash settled). (incorporated by reference to Exhibit 10.9 to the Company’s Annual Report on Form 10-K for the fiscal year ended December 31, 2013). +
(10.9)Form of Restricted Stock Unit Award Agreement (stock settled). (incorporated by reference to Exhibit 10.10 to the Company’s Annual Report on Form 10-K for the fiscal year ended December 31, 2013). +
(10.10)Form of Performance Share Plan award certificate. +
(10.11)Pension Restoration Plan for Salaried Employees (incorporated by reference to Exhibit 10.1 to the Company’s Quarterly Report on Form 10-Q for the quarter ended March 31, 2009). +
(10.12)Unfunded Supplemental Retirement Plan for Senior Managers, as amended and restated effective January 1, 2008 (incorporated by reference to Exhibit 10.21 to the Company’s Annual Report on Form 10-K for the fiscal year ended December 31, 2007). +
(10.13)Amendment No. 1 to the International Paper Company Unfunded Supplemental Retirement Plan for Senior Managers, effective October 13, 2008 (incorporated by reference to Exhibit 10.3 to the Company’s Current Report on Form 8-K dated October 17, 2008). +
(10.14)Amendment No. 2 to the International Paper Company Unfunded Supplemental Retirement Plan for Senior Managers, effective October 14, 2008 (incorporated by reference to Exhibit 10.5 to the Company’s Current Report on Form 8-K dated October 17, 2008). +
(10.15)Amendment No. 3 to the International Paper Company Unfunded Supplemental Retirement Plan for Senior Managers, effective December 8, 2008 (incorporated by reference to Exhibit 10.20 to the Company’s Annual Report on Form 10-K for the fiscal year ended December 31, 2008). +
(10.16)Amendment No. 4 to the International Paper Company Unfunded Supplemental Retirement Plan for Senior Managers, effective January 1, 2009 (incorporated by reference to Exhibit 10.1 to the Company’s Quarterly Report on Form 10-Q for the quarter ended September 30, 2009). +
(10.17)Amendment No. 5 to the International Paper Company Unfunded Supplemental Retirement Plan for Senior Managers, effective October 31, 2009 (incorporated by reference to Exhibit 10.17 to the Company’s Annual Report on Form 10-K for the fiscal year ended December 31, 2009). +
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(10.18)Amendment No. 6 to the International Paper Company Unfunded Supplemental Retirement Plan for Senior Managers, effective January 1, 2012 (incorporated by reference to Exhibit 10.21 to the Company's Annual Report on Form 10-K for the fiscal year ended December 31, 2011). +
(10.19)Form of Non-Competition Agreement, entered into by certain Company employees (including named executive officers) who have received restricted stock (incorporated by reference to Exhibit 10.22 to the Company’s Annual Report on Form 10-K for the fiscal year ended December 31, 2008). +
(10.20)Form of Non-Solicitation Agreement, entered into by certain Company employees (including named executive officers) who have received restricted stock (incorporated by reference to Exhibit 10.5 to the Company’s Quarterly Report on Form 10-Q for the quarter ended March 31, 2006). +
(10.21)Form of Change-in-Control Agreement - Tier I, for the Chief Executive Officer and all "grandfathered" senior vice presidents elected prior to 2012 (all named executive officers) - approved September 2013 (incorporated by reference to Exhibit 10.1 to the Company’s Quarterly Report on Form 10-Q for the quarter ended September 30, 2013). +
(10.22)Form of Change-in-Control Agreement - Tier II, for all future senior vice presidents and all "grandfathered" vice presidents elected prior to February 2008 - approved September 2013 (incorporated by reference to Exhibit 10.2 to the Company’s Quarterly Report on Form 10-Q for the quarter ended September 30, 2013). +
(10.23)Form of Indemnification Agreement for Directors (incorporated by reference to Exhibit 10.13 to the Company’s Annual Report on Form 10-K for the fiscal year ended December 31, 2003). +
(10.24)Board Policy on Severance Agreements with Senior Executives (incorporated by reference to Exhibit 10.1 to the Company’s Current Report on Form 8-K filed on October 18, 2005).+
(10.25)Board Policy on Change of Control Agreements (incorporated by reference to Exhibit 10.2 to the Company’s Current Report on Form 8-K filed on October 18, 2005). +
(10.26)Time Sharing Agreement, dated October 17, 2014 (and effective November 1, 2014), by and between Mark S. Sutton and International Paper Company (incorporated by reference to Exhibit 99.1 to the Company’s Current Report on Form 8-K dated October 14, 2014).+
(10.27)Five-Year Credit Agreement dated as of August 5, 2014, among International Paper Company, JPMorgan Chase Bank, N.A., individually and as administrative agent, and certain lenders (incorporated by reference to Exhibit 10.1 to the Company’s Quarterly Report on Form 10-Q for the quarter ended September 30, 2014).
(10.28)Equity Transfer Agreement dated October 7, 2015, between International Paper Investment (Shanghai) Co., Ltd. and Shandong Sun Holding Group Co., Ltd.
(11)Statement of Computation of Per Share Earnings.
(12)Computation of Ratio of Earnings to Fixed Charges and Preferred Stock Dividends.
(21)List of Subsidiaries of Registrant.
(23)Consent of Independent Registered Public Accounting Firm.
(24)Power of Attorney (contained on the signature page to the Company’s Annual Report on Form 10-K for the year ended December 31, 2015).
(31.1)Certification by Mark S. Sutton, Chairman and Chief Executive Officer, pursuant to Section 302 of the Sarbanes-Oxley Act of 2002.
(31.2)Certification by Carol L. Roberts, Chief Financial Officer, pursuant to Section 302 of the Sarbanes-Oxley Act of 2002.
(32)Certification pursuant to 18 U.S.C. Section 1350, as adopted pursuant to Section 906 of the Sarbanes-Oxley Act of 2002.
(101.INS)XBRL Instance Document
(101.SCH)XBRL Taxonomy Extension Schema
(101.CAL)XBRL Taxonomy Extension Calculation Linkbase
(101.DEF)XBRL Taxonomy Extension Definition Linkbase
(101.LAB)XBRL Taxonomy Extension Label Linkbase
(101.PRE)XBRL Extension Presentation Linkbase
Management contract or compensatory plan or arrangement.
Filed herewith
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SCHEDULE II – VALUATION AND QUALIFYING ACCOUNTS
INTERNATIONAL PAPER COMPANY AND CONSOLIDATED SUBSIDIARIES
SCHEDULE II – VALUATION AND QUALIFYING ACCOUNTS
(In millions)
For the Year Ended December 31, 2015
Balance at
Beginning
of Period Additions
Charged to
Earnings Additions
Charged to
Other
Accounts Deductions
from
Reserves Balance at
End of
Period
Description
Reserves Applied Against Specific Assets Shown on Balance Sheet:
Doubtful accounts – current$82$11$—(23)(a)$70
Restructuring reserves 16 5—(11)(b)10
For the Year Ended December 31, 2014
Balance at
Beginning
of Period Additions
Charged to
Earnings Additions
Charged to
Other
Accounts Deductions
from
Reserves Balance at
End of
Period
Description
Reserves Applied Against Specific Assets Shown on Balance Sheet:
Doubtful accounts – current$109$11$—(38)(a)$82
Restructuring reserves 51 41—(76)(b)16
For the Year Ended December 31, 2013
Balance at
Beginning
of Period Additions
Charged to
Earnings Additions
Charged to
Other
Accounts Deductions
from
Reserves Balance at
End of
Period
Description
Reserves Applied Against Specific Assets Shown on Balance Sheet:
Doubtful accounts – current$119$38$—(48)(a)$109
Restructuring reserves 17 46—(12)(b)51
(a)Includes write-offs, less recoveries, of accounts determined to be uncollectible and other adjustments.
(b)Includes payments and deductions for reversals of previously established reserves that were no longer required.
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SIGNATURES
Pursuant to the requirements of Section 13 or 15(d) of the Securities Act of 1934, the Registrant has duly caused this report to be signed on its behalf by the undersigned, thereunto duly authorized.
INTERNATIONAL PAPER COMPANY
February 25, 2016
By:/S/ S HARON R. R YAN
Sharon R. Ryan
Senior Vice President, General Counsel
and Corporate Secretary
POWER OF ATTORNEY
KNOW ALL MEN BY THESE PRESENTS, that each person whose signature appears below constitutes and appoints Sharon R. Ryan and Deon Vaughan as his or her true and lawful attorney-in-fact and agent, acting alone, with full power of substitution and resubstitution for him or her and in his or her name, place and stead, in any and all capacities, to sign any or all amendments to this annual report on Form 10-K, and to file the same, with all exhibits thereto and other documents in connection therewith, with the Securities and Exchange Commission, granting unto said attorney-in-fact full power and authority to do and perform each and every act and thing requisite or necessary to be done, hereby ratifying and confirming all that said attorney-in-fact and agent, or her substitute or substitutes, may lawfully do or cause to be done by virtue hereof.
Pursuant to the requirements of the Securities Exchange Act of 1934, this report has been signed below by the following persons on behalf of the registrant and in the capacities and on the dates indicated:
Signature Title Date
/S/M ARK S. S UTTON Chairman of the Board & Chief Executive Officer and Director February 25, 2016
Mark S. Sutton
/S/D AVID J. B RONCZEK Director February 25, 2016
David J. Bronczek
/S/W ILLIAM J. B URNS Director February 25, 2016
Willliam J. Burns
/S/A HMET C. D ORDUNCU Director February 25, 2016
Ahmet C. Dorduncu
/S/I LENE S. G ORDON Director February 25, 2016
Ilene S. Gordon
/S/J AY L. J OHNSON Director February 25, 2016
Jay L. Johnson
/S/S TACEY J. M OBLEY Director February 25, 2016
Stacey J. Mobley
/S/J OAN E. S PERO Director February 25, 2016
Joan E. Spero
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/S/J OHN L. T OWNSEND III Director February 25, 2016
John L. Townsend III
/S/W ILLIAM G. W ALTER Director February 25, 2016
William G. Walter
/S/J. S TEVEN W HISLER Director February 25, 2016
J. Steven Whisler
/S/R AY G. Y OUNG Director February 25, 2016
Ray G. Young
/S/C AROL L. R OBERTS Senior Vice President and Chief Financial Officer February 25, 2016
Carol L. Roberts
/S/T ERRI L. H ERRINGTON Vice President – Finance and Controller February 25, 2016
Terri L. Herrington
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APPENDIX I
2015 LISTING OF FACILITIES
(all facilities are owned except noted otherwise)
PRINTING PAPERS Paulinia, São Paulo, Brazil Stone Mountain, Georgia leased
Yanzhou City, China (2)Tucker, Georgia
Uncoated Papers and Pulp Veracruz, Mexico Aurora, Illinois (3 locations)
U.S.:Kenitra, Morocco Bedford Park, Illinois (2 locations) 1 leased
Selma, Alabama (Riverdale Mill)Edirne, Turkey Belleville, Illinois
Cantonment, Florida (Pensacola Mill)Corum, Turkey (1)Carroll Stream, Illinois
Ticonderoga, New York Des Plaines, Illinois
Riegelwood, North Carolina Corrugated Container Lincoln, Illinois
Eastover, South Carolina U.S.:Montgomery, Illinois
Georgetown, South Carolina Bay Minette, Alabama Northlake, Illinois
Sumter, South Carolina Decatur, Alabama Rockford, Illinois
Franklin, Virginia Dothan, Alabama leased Butler, Indiana
Huntsville, Alabama Crawfordsville, Indiana
International:Conway, Arkansas (2 locations)Fort Wayne, Indiana
Luiz Antônio, São Paulo, Brazil Fort Smith, Arkansas (2 locations)Hammond, Indiana
Mogi Guacu, São Paulo, Brazil Russellville, Arkansas (2 locations)Indianapolis, Indiana (3 locations)
Três Lagoas, Mato Grosso do Sul, Brazil Phoenix (Tolleson), Arizona Saint Anthony, Indiana
Saillat, France Yuma, Arizona Tipton, Indiana
Kadiam, India Anaheim, California Cedar Rapids, Iowa
Rajahmundry, India Buena Park, California leased Waterloo, Iowa
Kwidzyn, Poland Camarillo, California Garden City, Kansas
Svetogorsk, Russia Carson, California Bowling Green, Kentucky
Compton, California Lexington, Kentucky
INDUSTRIAL PACKAGING Elk Grove, California Louisville, Kentucky
Exeter, California Walton (Richwood), Kentucky
Containerboard Gilroy, California (2 locations)Bogalusa, Louisiana
U.S.:Los Angeles, California leased Lafayette, Louisiana
Pine Hill, Alabama Modesto, California Shreveport, Louisiana
Prattville, Alabama Ontario, California Springhill, Louisiana
Cantonment, Florida (Pensacola Mill)Salinas, California Auburn, Maine
Rome, Georgia Sanger, California Three Rivers, Michigan
Savannah, Georgia San Leandro, California leased Arden Hills, Minnesota
Cayuga, Indiana Santa Fe Springs, California (2 locations)Austin, Minnesota
Cedar Rapids, Iowa Stockton, California Fridley, Minnesota
Henderson, Kentucky Tracy, California Minneapolis, Minnesota leased
Maysville, Kentucky Golden, Colorado Shakopee, Minnesota
Bogalusa, Louisiana Wheat Ridge, Colorado White Bear Lake, Minnesota
Campti, Louisiana Putnam, Connecticut Houston, Mississippi
Mansfield, Louisiana Orlando, Florida Jackson (Richland), Mississippi
Vicksburg, Mississippi Plant City, Florida Magnolia, Mississippi leased
Valliant, Oklahoma Tampa, Florida leased Olive Branch, Mississippi
Springfield, Oregon Columbus, Georgia Fenton, Missouri
Orange, Texas Forest Park, Georgia Kansas City, Missouri
Griffin, Georgia Maryland Heights, Missouri
International:Kennesaw, Georgia leased North Kansas City, Missouri leased
Franco da Rocha, São Paulo, Brazil Lithonia, Georgia St. Joseph, Missouri
Nova Campina, São Paulo, Brazil Savannah, Georgia St. Louis, Missouri
A-1
Table of Contents
Omaha, Nebraska Amarillo, Texas Bellusco, Italy
Barrington, New Jersey Carrollton, Texas (2 locations)Catania, Italy
Bellmawr, New Jersey Edinburg, Texas Pomezia, Italy
Milltown, New Jersey El Paso, Texas San Felice, Italy
Spotswood, New Jersey Ft. Worth, Texas leased Kuala Lumpur, Malaysia
Thorofare, New Jersey Grand Prairie, Texas Juhor, Malaysia
Binghamton (Conklin), New York Hidalgo, Texas Apodaco (Monterrey), Mexico leased
Buffalo, New York McAllen, Texas Ixtaczoquitlan, Mexico
Rochester, New York San Antonio, Texas (2 locations)Juarez, Mexico leased
Scotia, New York Sealy, Texas Los Mochis, Mexico
Utica, New York Waxahachie, Texas Puebla, Mexico leased
Charlotte, North Carolina (2 locations)Lynchburg, Virginia Reynosa, Mexico
1 leased Petersburg, Virginia San Jose Iturbide, Mexico
Lumberton, North Carolina Richmond, Virginia Santa Catarina, Mexico
Manson, North Carolina Moses Lake, Washington Silao, Mexico
Newton, North Carolina Olympia, Washington Villa Nicolas Romero, Mexico
Statesville, North Carolina Yakima, Washington Zapopan, Mexico
Byesville, Ohio Fond du Lac, Wisconsin Agadir, Morocco
Delaware, Ohio Manitowoc, Wisconsin Casablanca, Morocco
Eaton, Ohio Singapore, Singapore
Kenton, Ohio International:Almeria, Spain
Madison, Ohio Manaus, Amazonas, Brazil Barcelona, Spain
Marion, Ohio Paulinia, São Paulo, Brazil Bilbao, Spain
Marysville, Ohio leased Rio Verde, Goias, Brazil Gandia, Spain
Middletown, Ohio Suzano, São Paulo, Brazil Madrid, Spain
Mt. Vernon, Ohio Las Palmas, Canary Islands Bangkok, Thailand
Newark, Ohio Tenerife, Canary Islands Adana, Turkey
Streetsboro, Ohio Rancagua, Chile Bursa, Turkey
Wooster, Ohio Baoding, China Corlu, Turkey
Oklahoma City, Oklahoma Beijing, China Corum, Turkey
Beaverton, Oregon (2 locations)Chengdu, China Gebze, Turkey
Hillsboro, Oregon Dalian, China Izmir, Turkey
Portland, Oregon Dongguan, China
Salem, Oregon leased Guangzhou, China (2 locations)
Biglerville, Pennsylvania (2 locations)Hohhot, China Recycling
Eighty-four, Pennsylvania Nanjing China U.S.:
Hazleton, Pennsylvania Shanghai, China (2 locations)Phoenix, Arizona
Kennett Square (Toughkenamon),
Pennsylvania Shenyang, China Fremont, California
Lancaster, Pennsylvania Suzhou, China Norwalk, California
Mount Carmel, Pennsylvania Tianjin, China (2 locations)West Sacramento, California
Georgetown, South Carolina Wuhan, China Denver, Colorado (1)
Laurens, South Carolina Arles, France Itasca, Illinois
Lexington, South Carolina Chalon-sur-Saone, France Des Moines, Iowa
Ashland City, Tennessee leased Creil, France Wichita, Kansas
Cleveland, Tennessee LePuy, France (Espaly Box Plant)Roseville, Minnesota
Elizabethton, Tennessee leased Mortagne, France Omaha, Nebraska
Morristown, Tennessee Guadeloupe, French West Indies Charlotte, North Carolina
Murfreesboro, Tennessee Batam, Indonesia Beaverton, Oregon
A-2
Table of Contents
Eugene, Oregon leased DISTRIBUTION
Memphis, Tennessee leased (1)
Carrollton, Texas IP Asia
Salt Lake City, Utah International:
Richmond, Virginia China (8 locations)
Kent, Washington Malaysia
Taiwan
International:Thailand
Monterrey, Mexico leased Vietnam
Xalapa, Veracruz, Mexico leased
FOREST PRODUCTS
Bags
U.S.:Forest Resources
Buena Park, California International:
Beaverton, Oregon Approximately 335,000 acres in Brazil
Grand Prairie, Texas
CONSUMER PACKAGING
Coated Paperboard
Augusta, Georgia
Riegelwood, North Carolina
Prosperity, South Carolina
Texarkana, Texas
Foodservice
U.S.:
Visalia, California
Shelbyville, Illinois
Kenton, Ohio
International:
Shanghai, China
Beijing, China
Bogota, Colombia
Cheshire, England leased
(1) Closed March 2015
(2) Closed October 2015
A-3
Table of Contents
APPENDIX II
2015 CAPACITY INFORMATION
CONTINUING OPERATIONS
(in thousands of short tons)U.S.EMEA Americas,
other
than U.S.Asia India Total
Industrial Packaging
Containerboard (a)13,131 48 360——13,539
Printing Papers
Uncoated Freesheet 1,808 1,150 1,135—258 4,351
Bristols 165————165
Uncoated Papers and Bristols 1,973 1,150 1,135—258 4,516
Dried Pulp 1,335 346 140——1,821
Newsprint—124———124
Total Printing Papers 3,308 1,620 1,275—258 6,461
Consumer Packaging
Coated Paperboard 1,568 379—1,413(b)—3,360
(a) In addition to Containerboard, this also includes saturated kraft, kraft bag, and gypsum.
(b) The Company's ownership interest in the Asian Coated Paperboard business was sold in October 2015.
Forest Resources
We own, manage or have an interest in approximately 1.4 million acres of forestlands worldwide. These forestlands and associated acres are located in the following regions:(M Acres)
Brazil 335
We have harvesting rights in:
Russia 1,047
Poland—
Total 1,382
A-4 |
189979 | https://artofproblemsolving.com/wiki/index.php/Conditional_probability?srsltid=AfmBOooApyixYJVJu4jKXiUZn49oMYL56noWDootAn6iR29aoZZPlUsu | Art of Problem Solving
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Conditional probability
Conditional probability is the probability of an event occurring, assuming that another event has already occurred. is said as the probability of event B given A
Contents
[hide]
1 Example
2 Solution
3 Formula
4 Different Problem
5 Different approaches
6 Let's make a deal
7 Review Problems
8 See also
Example
Let us say that fair sided dice are rolled and their face up values sum is . What is the probability that the face up value of the one dice is ?
Solution
Let call the first dice and the second one . There are 5 ways for and 2 of those ways (distinct) includes a 2. Therefore, our answer is .
Formula
The formula for conditional probability is where represents the conditional probability. is also said as the probability of event B occurring given event A occurs. is the probability . We can also represent as
Different Problem
A fair standard die is tossed 3 times. Given that the sum of the first two tosses equals the third, what is the probability that at least one 2 is tossed? (Source AMC)
Solution: The probability that the sum of the first two die is equal to the 3rd die is just another way to also equivalent to the probability that the sum of the numbers on the first two die is less than 7, which is . The probability that a two is rolled and that it meets the first condition is . dividing gets us .
Different approaches
In the diagram above, it represents a diagram of the probabilities of A and B occurring, where event A doesn't have to happen, so that event B happens. For example, if we have a group of cats and dogs, and we pick two animals, and the first one is a dog, what is the probability that we pick a cat? Geometrically speaking, would be the ratio of to , since is .
In this diagram the above (grey part) has to occur so that can occur. Let's take the cat and dog example, but instead we are given that the first pet chosen is a cat, and we want the probability of choosing a cat for the second one. Geometrically speaking, would be the since is .
Let's make a deal
The Monty Hall problem is related to conditional probability. Monty Hall runs the show, "Let's Make a Deal." The problem states, Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, and the host, who knows what's behind the doors, opens another door, which has a goat. He then says to you, "Do you want to pick door change your door?" Is it to your advantage to switch your choice?
The short answer is yes. The host doesn't just choose a door at random to reveal. He follows three rules, he doesn't reveal the door with a car, he doesn't reveal your door, and he always gives the option to switch. We start with a chance of winning the car, and the host knowingly reveals a goat, which he always has the option to do, so why should your chance of winning increase to . This is a bit skeptical, but we can use a sample space to see this.
Remember that the host can always reveal a goat. If we choose the car correctly on the first try, which happens at chance, swapping would lose, but if we choose the goat on the first try, which happens at chance, swapping would win.
But how can we solve this using conditional probability? Let's look at 60 different cases with equal probability, where we choose Door 1. (We label the goat and ) There are 20 cases where we choose and the host reveals the other goat. There are 10 cases where we choose the car and the host reveals . There are 10 cases where we choose the car and the host reveals . There are 20 cases where we choose and the host reveals the other goat. will be the probability that we win by switching, and will be the probability that we choose Door 1, which makes constant at . Since there are 40 cases out of 60 cases, or that we win, by choosing Door 1, by conditional probability, we have .
Review Problems
Two fair coins are simultaneously flipped. This is done repeatedly until at least one of the coins comes up heads, at which point the process stops, what is the probability that the other coin also comes up heads on the last flip? (Source HMMT)
The probability that event A occurs is , and the probability that event B occurs is . What are the minimum and maximum possible values of ? What are the minimum and maximum possible values of and ? (Source AMC)
What fraction of all permutations of the numbers such that the first term is not and the third has ? (Source AMC)
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Write sqrt(-3) in simplest radical form.
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$$\sqrt{3} i$$3i
Explanation
Rewrite $$\sqrt {-a}$$−a as $$\sqrt{a}\cdot\sqrt{-1}$$a⋅−1:$$\sqrt{3} \times \sqrt{-1}$$3×−1
Rewrite by definition $$i=\sqrt{-1}$$i=−1:$$\sqrt{3} i$$3i
Answer: $$\sqrt{3} i$$3i
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189981 | https://germanna.edu/sites/default/files/2022-03/How%20to%20Graph%20Trigonometric%20Functions.pdf | Provided by The Academic Center for Excellence 1 How to Graph Trigonometric Functions Created September 2013 How to Graph Trigonometric Functions This handout includes instructions for graphing processes of basic, amplitude shifts, horizontal shifts, and vertical shifts of trigonometric functions. You can navigate to specific sections of this handout by clicking the links below. The Unit Circle and the Values of Sine and Cosine Functions: pg. 1 Graphing Sine and Cosine Functions y = sin x y = cos x: pg. 3 Graphing the Tangent Function y = tan x: pg. 6 The Form y = A sin (Bx + C) + D: pg. 7 Amplitude Shifts of Trigonometric Functions: pg. 8 Horizontal Shifts of Trigonometric Functions: pg. 8 Period Compression or Expansion of Trigonometric Functions: pg. 9 Vertical Shifts of Trigonometric Functions: pg. 10 Strategies, Summary, and Exercises: pg. 11 The Unit Circle and the Values of Sine and Cosine Functions The unit circle is a circle with a radius that equals 1. The angle θ is formed from the φ (phi) ray extending from the origin through a point p on the unit circle and the x-axis; see diagram below. The value of sin θ equals the y-coordinate of the point p and the value of cos θ equals the x- coordinate of the point p as shown in the diagram below. Provided by The Academic Center for Excellence 2 How to Graph Trigonometric Functions Created September 2013 (0,1) p = (cos θ, sin θ) θ (-1,0) (1,0) (0,-1) This unit circle below shows the measurements of angles in radians and degrees. Beginning at 0π, follow the circle counter-clockwise. As angle θ increases to π 2 radians or 90°, the value of cosine (the x-coordinate) decreases because the point is approaching the y-axis. Meanwhile, the value of sine (the y-coordinate) increases. When one counter-clockwise revolution has been completed, the point has moved 360° or 2π. π or 90° 2 0π or 0° π or 180° 2π or 360° 3𝜋𝜋 2 or 270° Provided by The Academic Center for Excellence 3 How to Graph Trigonometric Functions Created September 2013 Graphing Sine and Cosine Functions y = sin x and y = cos x There are two ways to prepare for graphing the basic sine and cosine functions in the form y = sin x and y = cos x: evaluating the function and using the unit circle. To evaluate the basic sine function, set up a table of values using the intervals 0π, π 2 , 3π 2 , and 2π for x and calculating the corresponding y value. f(x) or y = sin x f(x) or y x 0 0π 1 π 2 0 π -1 3π 2 0 2π To use the unit circle, the x-coordinates remain the same as within the list above. To find the y- coordinate of the point to graph, first locate the point p on the unit circle that corresponds to the angle θ given by the x-coordinate. Then, use the y-coordinate of the point p as the y value of the point to graph. To draw the graph of one period of sine or y = sin x, label the x-axis with the values 0π, π 2, π, 3π 2 , and 2π. Then plot points for the value of f(x) or y from either the table or the unit circle. Provided by The Academic Center for Excellence 4 How to Graph Trigonometric Functions Created September 2013 Other points may be added for the intermediate values between those listed above to obtain a more complete graph, and a best fit line can be drawn by connecting the points. The figure below is the completed graph showing one and a half periods of the sine function. The graph of the cosine function y = cos x is drawn in a similar manner as the sine function. Using a table of values: f(x) or y = cos x f(x) or y x 1 0π 0 π 2 -1 π 0 3π 2 1 2π 1 0π -1 π 2 π 3π 2 2π 1 0π -1 π 2 π 3π 2 2π 5π 2 3π One period y = sin x Provided by The Academic Center for Excellence 5 How to Graph Trigonometric Functions Created September 2013 To use the unit circle, the x-coordinate remains the same as the list on the previous page. To find the y-coordinate of the point to graph, first locate the point p on the unit circle that corresponds to the angle θ given by the x-coordinate. Then, use the x-coordinate of the point p as the y value of the point to graph. To draw the graph of one period of cosine or y = cos x, label the x-axis with the values 0π, 𝜋𝜋 2, π, 3𝜋𝜋 2 , and 2π. Then plot points for the value of f(x) or y from either the table or the unit circle. Add other points as required for the intermediate values between those above to obtain a more complete graph, and draw a best fit line connecting the points. The graph below shows one and a half periods. 1 0π π 2 -1 π 3π 2π 2 One period 5π 3π 2 y = cos x 1 0π -1 π 2 π 3π 2 2π Provided by The Academic Center for Excellence 6 How to Graph Trigonometric Functions Created September 2013 Graphing the Tangent Function y = tan x The tangent value at angle θ is equal to the sine value divided by the cosine value ቀ 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉ቁ of the same angle θ. The value of tangent at 0π for the unit circle is 0 1, which is equivalent to 0. The value of tangent at 𝜋𝜋 2 is 1 0. This yields a divide by 0 error or undefined (try this in your calculator). Therefore, the tangent function is undefined at 𝜋𝜋 2. This is illustrated by drawing an asymptote (vertical dashed line) at 𝜋𝜋 2. See the figure below. The value of tangent at π is 0 1, which results in 0. To determine how the tangent behaves between 0π and the asymptote, find the sine and cosine values of π 4, which is half way between 0π and π 2. Looking at the handout Common Trigonometric Angle Measurements, the tangent of π 4 is √2 2 (sine) divided by √2 2 (cosine). Flipping the cosine value and multiplying gives √2 2 × 2 √2 which simplifies to 1. The value of tangent at π 4 is therefore 1. These points have been added to the graph below. 1 0π -1 π 2 π 3π 2 2π 1 0π -1 π 4 π 2 π 3π 2 2π Provided by The Academic Center for Excellence 7 How to Graph Trigonometric Functions Created September 2013 1 0 -1 π 4 π 2 3π 4 π 5π 4 3π 7π 2 4 2 Next, calculate the value of tangent for 3𝜋𝜋 4 . Consulting with Common Trigonometric Angle Measurements, the tangent of 3𝜋𝜋 4 is √2 2 (sine) divided by – √2 2 (cosine). This simplifies to a tangent value of -1. Now, draw the tangent function graph so that the line approaches the asymptote without touching or crossing it. The image on the next page shows the completed graph of one and a half periods of the tangent function. One period The period of the basic tangent function is π, and the graph will repeat from π to 2π. The Form y = A sin(Bx + C) + D The form y = A sin(Bx + C) is the general form of the sine function. From this general form of the sine function, the amplitude, horizontal, phase, and vertical shifts from the basic trigonometric forms can be determined. A : modifies the amplitude in the y direction above and below the center line B : influences the period and phase shift of the graph C : influences the phase shift of the graph D : shifts the center line of the graph on the y-axis y = tan x Provided by The Academic Center for Excellence 8 How to Graph Trigonometric Functions Created September 2013 Amplitude Shifts of Trigonometric Functions The basic graphs illustrate the trigonometric functions when the A value is 1. This A = 1 is used as an amplitude value of 1. If the value A is not 1, then the absolute value of A value is the new amplitude of the function. Any number |A| greater than 1 will vertically stretch the graph (increase the amplitude) while a number |A| smaller than 1 will compress the graph closer to the x-axis. Example: Graph y = 3 sin x. Solution: The graph of y = 3 sin x is the same as the graph of y = sin x except the minimum and maximum of the graph has been increased to -3 and 3 respectively from -1 and 1. Horizontal Shifts of Trigonometric Functions A horizontal shift is when the entire graph shifts left or right along the x-axis. This is shown symbolically as y = sin(Bx – C). Note the minus sign in the formula. To find the phase shift (or the amount the graph shifted) divide C by B ቀ C 𝐵𝐵ቁ. For instance, the phase shift of y = cos(2x – π) can be found by dividing π (C) by 2 (B), and the answer is 𝛑𝛑 𝟐𝟐. Another example is the phase shift of y = sin(-2x – π) which is –π (C) divided by -2 (B), and the result is 𝛑𝛑 𝟐𝟐. Be careful when dealing 3 2 1 Amplitude is now 3 up 0 -1 π 2 Amplitude is now 3 down π 3π 2 2π 5π 2 3π -2 -3 y = 3 sin x Provided by The Academic Center for Excellence 9 How to Graph Trigonometric Functions Created September 2013 with the signs. A positive sign takes the place of the double negative signs in the form y = sin(x + π). The C is negative because this example is also written as y = sin(x-(- π)), which produces the negative π phase shift (graphed below). It is important to remember a positive phase shift means the graph is shifted right or in the positive direction. A negative phase shift means the graph shifts to the left or in the negative direction. Period Compression or Expansion of Trigonometric Functions The value of B also influences the period, or length of one cycle, of trigonometric functions. The period of the basic sine and cosine functions is 2π while the period of the basic tangent function is π. The period equation for sine and cosine is: Period = 𝟐𝟐𝛑𝛑 |𝐁𝐁|. For tangent, the period equation is: Period = 𝛑𝛑 |𝐁𝐁|. Period compression occurs if the absolute value of B is greater than 1; this means the function oscillates more frequently. Period expansion occurs if the absolute value of B is less than 1; this means the function oscillates more slowly. The starting point of the graph is determined by the phase shift. To determine the key points for the new period, divide the period into 4 equal parts and add this part to successive x values beginning with the starting point. Phase shift = -π 1 -π -π 2 0 π 2 π 3π 2 2π -1 y = sin(x + π) Provided by The Academic Center for Excellence 10 How to Graph Trigonometric Functions Created September 2013 Example: Graph y = sin(2x – π) Solution: A = 1, B = 2, C = π, D = not written so 0 The amplitude of A is 1 The horizontal (phase) shift is 𝑪𝑪 𝑩𝑩 = 𝛑𝛑 𝟐𝟐 = 𝛑𝛑 𝟐𝟐 The period changes to 𝟐𝟐𝛑𝛑 |𝐁𝐁| = 𝟐𝟐𝛑𝛑 |𝟐𝟐| = π The starting point is π 2. To find the 5 key points, divide π by 4 to obtain the value of the 4 equal parts: π 4. Then add π 4 to the starting point and each successive value of x to find the remaining four points. The 5 key points are: π 2, 3π 4 , π, 5π 4 , 3π 2 . Once the key points are determined, evaluate the equation at these values and plot the resulting points. Vertical Shifts of Trigonometric Functions A vertical shift occurs when the entire graph shifts up or down along the y-axis. This is shown symbolically as y = sin (x) + D. This is different from horizontal shifts because there are no parentheses around the D, and D is always a constant. In these examples, the graph of y = sin (x) + 2 would shift the centerline, thus the entire graph, up 2 units, and y = tan (x) – 1 would shift the entire graph down 1 unit.
1 Phase shift = π 2 0π -1 π 2 3π 4 π 3π 2 7π 4 2π 5π 2 9π 4 3π 5π 4 y = sin(2x – π) Provided by The Academic Center for Excellence 11 How to Graph Trigonometric Functions Created September 2013 Strategies, Summary, and Exercises By using the following guidelines, it will make trigonometric functions easier to graph: 1) Recall the values of sine and cosine on the unit circle. 2) Identify the amplitude shift value A, in y = A sin x, y = A cos x, etc. 3) Identify the horizontal shift value 𝑪𝑪 𝑩𝑩, in y = sin(Bx – C), y = cos(Bx – C), etc. 4) The period compression or expansion of the graph is determined by dividing the period of the basic function by the absolute value of B. 5) Identify the vertical shift value D, in y = sin (x) + D, y = cos (x) + D, etc. 6) The phase shift (if any) is the starting point to graphing the function. Divide the period by 4 to determine the 5 points to graph on the x-axis. 7) Use the 5 points to determine corresponding y-coordinate. Exercises: Find the amplitude shift, horizontal shift, period, and vertical shift of the following and graph one period. 1) y = 1 4 sin x + 1 2) y = 2 sin x 3) y = 1 2 cos(– x + π) – 1 1 y = tan x – 1 0π -1 π 4 π 2 3π 4 π 5π 4 3π 2 2π -2 Provided by The Academic Center for Excellence 12 How to Graph Trigonometric Functions Created September 2013 Solutions to Exercises 1) y = 1 4sin x + 1 First find the values of A, B, C, and D A = 1 4, B = 1, C = not written which means 0, and D = 1 Amplitude shift is A, A = 1 4 Horizontal shift is 𝐶𝐶 𝐵𝐵 = 0 1 = 0 Period is 2𝜋𝜋 |1| = 2π Vertical shift is D = 1 Graph of one period of y = 1 4 sin x + 1 2) y = 2 sin x First find the values of A, B, C, and D A =2, B = 1, C = not written which means 0, and D = 0 Amplitude shift is A, A = 2 Horizontal shift is 𝐶𝐶 𝐵𝐵 = 𝟎𝟎 𝟏𝟏= 0 Period is 𝟐𝟐𝟐𝟐 |𝟏𝟏| = 2π Vertical shift is D = 0 1 ¼ 1 ¾ ½ ¼ 0π -¼ π 2 π 3π 2 2 Provided by The Academic Center for Excellence 13 How to Graph Trigonometric Functions Created September 2013 Graph of one period of 2sin x 3) y = 1 2cos(– x + π) – 1 First find the values of A, B, C, and D A = 1 2, B = –1, C = –π, and D = –1 Amplitude shift is A, A = 1 2 Horizontal shift is 𝐶𝐶 𝐵𝐵 = –π −𝟏𝟏= 𝜋𝜋 Period is 𝟐𝟐𝟐𝟐 |−𝟏𝟏| = 2π Vertical shift is D = –1 Graph of one period of y = 1 2cos(– x + π) – 1 2 1 0π π -1 π 2 3π 2 2π -2 -½ -1 0π π 2 π 3π 2 2π 5π 4 3π |
189982 | https://www.youtube.com/watch?v=hel6vCjmTgE | The Movement of Water and Solutes in Plants | Chapter 30 - Raven Biology of Plants
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Posted: 10 Sep 2025
Chapter 30 of Raven Biology of Plants (Eighth Edition) examines how plants transport water, inorganic nutrients, and organic assimilates through their bodies, focusing on the xylem and phloem as the vascular highways of plant physiology.
The chapter begins with transpiration, the loss of water vapor through leaves, described as an “unavoidable evil” since CO₂ uptake for photosynthesis inevitably causes water loss. Stomata regulate this exchange through turgor-driven movements of guard cells, controlled by blue-light photoreceptors, potassium and sucrose fluxes, and the radial orientation of cellulose microfibrils. Environmental factors—including temperature, humidity, and air currents—influence transpiration rates, while adaptations such as cuticles, sunken stomata, and crassulacean acid metabolism (CAM) help reduce water loss.
The cohesion-tension theory explains how water ascends tall trees: evaporation from mesophyll cells creates a negative pressure that pulls water upward through xylem conduits. Cohesion between water molecules and adhesion to cell walls sustain this tension. Cavitation and embolism (air bubbles in xylem) can block transport, but pit membranes and torus-margo structures localize damage. Experiments using pressure chambers, sap flow measurements, and dendrometer readings provide strong evidence for this mechanism. Root pressure contributes to water movement under low transpiration, leading to guttation, while hydraulic redistribution allows deep-rooted plants to move water upward or laterally to drier soils, benefiting neighboring plants and entire ecosystems.
Water and ion uptake by roots occurs mainly through root hairs, with three pathways—apoplastic, symplastic, and transcellular. Casparian strips in the endodermis enforce selective uptake, and ion absorption requires metabolic energy for both epidermal uptake and secretion into xylem vessels. Mycorrhizal fungi greatly enhance nutrient acquisition, especially phosphorus, zinc, and copper.
The chapter then turns to phloem transport (translocation), the movement of sugars and other assimilates from sources (photosynthesizing leaves, storage organs) to sinks (growing tissues, fruits, seeds). Evidence from girdling experiments, radioactive tracers, and aphid stylet studies confirms that sieve tubes transport sucrose-rich sap at rapid rates (50–100 cm per hour). The widely accepted pressure-flow hypothesis, first proposed by Ernst Münch, explains assimilate movement as osmotically driven: sugar loading lowers sieve tube water potential, drawing in xylem water and generating turgor pressure that drives flow toward sinks.
Phloem loading occurs either apoplastically (requiring active sucrose-proton symport) or symplastically. Symplastic loaders may use polymer trapping, converting sucrose into raffinose and stachyose in intermediary cells to maintain concentration gradients. Passive symplastic loading occurs in many trees, where high mesophyll sugar concentrations alone drive diffusion into phloem. Phloem unloading at sinks can be symplastic or apoplastic, depending on tissue type, with energy-dependent steps required for sugar storage in sink cells. Alongside sucrose, phloem transports amino acids, proteins, RNAs, hormones, and inorganic ions, linking the transport of water, nutrients, and information across the plant body.
The essay Green Roofs: A Cool Alternative highlights how urban green roofs use transpiration and water cycling to cool air, reduce stormwater runoff, insulate buildings, and improve urban biodiversity—an applied example of plant transport processes benefiting human environments.
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Raven Biology of Plants Chapter 30 summary, plant water and solute transport explained, transpiration stomata guard cell turgor blue light K+ sucrose, cohesion-tension theory xylem ascent of sap embolism cavitation pit membranes torus margo, root hairs water absorption Casparian strip apoplastic symplastic transcellular pathways, ion uptake active transport root pressure guttation hydraulic redistribution, mycorrhizas phosphorus zinc copper absorption symbiosis, phloem transport assimilates translocation pressure-flow hypothesis Münch, source to sink movement sucrose osmosis turgor gradient, phloem loading apoplastic sucrose-proton symport symplastic polymer trapping raffinose stachyose intermediary cells, phloem unloading apoplastic symplastic sink tissues, sieve tube elements companion cells aphid stylet experiments radioactive tracers, phloem sap composition amino acids proteins RNAs hormones inorganic ions, CAM plants nocturnal transpiration xerophytes water-saving adaptations, environmental influences transpiration temperature humidity wind
Transcript:
Welcome to the Deep Dive, your shortcut to being well-informed. We cut through the noise and get straight to the core of fascinating topics. Today, we're uh digging into something fundamental but maybe overlooked. How plants actually move stuff around inside them. Water, nutrients, you name it. We're pulling the key insights from a chapter in the classic raven biology of plants. And our goal is simple. Give you a clear grasp of these amazing systems. No textbook needed. I mean, think about it. Plants seem so still, right? But inside they're like these incredible hydraulic engineers. They're moving huge amounts of water and nutrients, sometimes over enormous distances. Imagine a giant redwood pulling water hundreds of feet up. We'll be focusing on the two main transport tissues, xyllem and phum. And their jobs are actually more connected and uh surprising than you might think. Yeah, it's fascinating how long it took us to really figure these processes out. For centuries, people saw SAP moving, but the how was a real puzzle. It wasn't really until, you know, the late 19th, early 20th century that the pieces started fitting together. How water defies gravity, how sugars get distributed. It shows how much we've learned. Absolutely. And some of those early observations are just incredible. Let's get into the water and nutrient movement part first. Picture this. Back in the 1700s, Steven Hails, an English physician, found a sunflower could lose water vapor. He called it persspiring 17 times faster than a human over 24 hours. 17 times. Yeah. And that's just one sunflower. A single large tree can lose hundreds of liters, maybe 200 to 400 in just one day. This process, losing water vapor from leaves, mostly is transpiration. So why why lose all that water? It seems wasteful, but it's actually uh fundamentally linked to photosynthesis. It's sometimes called an unavoidable evil. See, plants need big leaves for sunlight, and they need moist surfaces inside those leaves so CO2 can dissolve and get into the cells. But if you have water exposed to air that isn't saturated, you get evaporation. It's simple physics. So getting CO2 in means water inevitably gets out. Too much loss though and the plant is in trouble, right? But plants have tricks up their sleeves. Yeah. Adaptations to cut down on that loss like the waxy coating on leaves, the cuticle. That's a pretty good barrier. It is. Yeah. Very effective. But the real control points are these tiny pores called stamata. They're microscopic, maybe only 1% of the leaf surface, but they account for like over 90% of the water lost through transpiration. Exactly. It happens in two steps. Water evaporates from cell surfaces inside the leaf, turning into vapor in the air spaces there. Then that vapor diffuses out into the drier outside air through those stamatal pores. It's all driven by differences in water vapor concentration. And the way these stamata open and close is just ingenious. Each pore has two guard cells around it and it all comes down to changes in the shape of those guard cells controlled by something called tur pressure. Tur pressure like water pressure inside the cell pretty much when stamata need to open the guard cells actively pump in solutes mostly potassium ions K plus N to start and later sucrossse plays a role too. This makes the inside of the guard cell more concentrated lowering its water potential makes it thirstier essentially. Ah so water flows in. Exactly. Water rushes in by osmosis, inflating the guard cells. They swell up, increase their internal pressure, their tur. Because of how they're built, as they inflate, they bow outwards, opening the pore between them. Think of like two slightly curved balloons inflating side by side. Okay, I can picture that. So closing is just the reverse. Yep. Salutes leave, water follows, the cells lose tur, become flaccid, and the pore closes up. It's often a daily cycle. K+ in the morning, sucrossse later on. And you mentioned how they're built. The cell walls matter too, right? Oh, absolutely. Crucial. The cellulose microfibals, the tiny reinforcing rods in the cell walls are arranged radially, kind of like spokes on a wheel around the guard cell. This means when the cell takes up water, it gets longer and bends outwards, opening the pool, rather than just swelling sideways like a regular balloon might. That structure is key. Clever. And what about the environment? Does that affect the stamata? Definitely. CO2 concentration is a big one. Higher CO2 inside the leaf usually triggers closure. Plants don't need to keep the gates wide open if CO2 is plentiful. Temperature also has an effect. Generally between 10 and 25 Celsius, it's not huge, but above 30 35° C, many plants start closing their stamata. Why is that? Just the heat. It's partly because higher temperatures increase respiration, which releases CO2 inside the leaf. That buildup signals closure. It's a water-saving move in hot conditions, often leading to midday closure. And plants have internal clocks, too. Circadian rhythms. They do. Many plants will open and close their stamata on a roughly 24-hour cycle. Even if you keep them in constant light or darkness, it's built-in programming. Okay, now this is cool. See plants, cacti, pineapples. Sorry. They do things differently, right? They really do. Craillation, acid metabolism. CM. It's a fantastic adaptation for deserts. They flip the whole thing. They open their stamata at night when it's cooler and more humid. So they lose less water. They take in CO2 then and store it chemically. Yeah. As organic acids. And then during the day stamata are shut tight, but they use the CO2 they stored overnight for photosynthesis powered by the sunlight. It's a brilliant water saving strategy. Amazing. Are there other plants that do weird things at night? Well, some regular C3 plants actually don't fully close their stamata at night. This is called nocturnal transpiration. They lose some water, yes, but the thought is it might help them pull in more mineral nutrients from the soil when evaporation isn't as intense. Interesting trade-off. So, besides stamata, what else affects how fast a plant loses water? Several things. Temperature. Obviously, evaporation increases with heat, roughly doubling for every 10° C rise. Though, the evaporation itself does cool the leaf a bit like sweating kind of. Yeah, humidity is huge, too. If the air outside is already very humid, the gradient driving water vapor out is much smaller, so transpiration slows down. Explains why rainforest plants can have those huge leaves. Exactly. They can afford to maximize light capture. Whereas in dry, windy grasslands, you see narrower leaves, thicker cuticles, everything geared towards conserving water and wind. Does that matter? Oh, yes. Wind blows away the little layer of human air right next to the leaf surface called the boundary layer. This steepens the gradient between the inside of the leaf and the outside air, so transpiration usually increases. Okay, so plants are losing all this water, which brings us back to that redwood. How on earth does water get from the roots way down there all the way to the top leaves hundreds of feet up? That's the million-dollar question, isn't it? We know water moves in the xyllem, those bundles of pipelike cells. Simple experiments like putting a cut stem in colored water show the dye moving up through the xyllem vessels and trait guides. But how is it pushed from the bottom? For a long time, people thought maybe root pressure pushed it up, but that's definitely not strong enough for tall trees. The dominant theory now is the cohesion tension theory. Cohesion tension sounds sticky. It kind of is. The core idea is that water is pulled from the top, not pushed from below. It starts with transpiration evaporation from the cells and the leaves. This loss of water lowers the water potential inside those leaf cells. Makes them thirstier again. Exactly. This creates a negative pressure, a tension or a pull on the water remaining in the leaf. Now, water molecules have this amazing property called cohesion. They stick to each other really strongly like tiny magnets because of hydrogen bonds. Is that a chain? Precisely. So the tension created in the leaves pulls on this continuous chain or column of water molecules extending all the way down the xylem through the stem right down to the roots. And there's also adhesion water molecules sticking to the xyllem walls which helps counteract gravity and keep the column intact. So maybe cohesion adhesion tension theory is more accurate. Wow. So the pulling force comes from evaporation at the top. Yes. And the energy for this entire massive water transport system ultimately comes from the sun driving that evaporation. The plant just provides the plumbing. There are experiments showing water uptake closely follows transpiration rates and models using things like porous clay cups attached to mercury columns that demonstrate this physical pulling effect. That's incredible. Yeah, but it sounds risky. What if that chain of water breaks? That's the weak point. The Achilles heel. You could say air bubbles. If the tension becomes too great or if stresses like freezing occur, the water column can break. That's called cavitation. If the conduit then fills with air or water vapor, that's an embolism. And once that happens, that particular xyllem vessel or trade is basically out of commission. It can't transport water anymore. So, how do plants deal with that? It must happen all the time, especially in dry or cold conditions. They have several coping mechanisms. First, the conduits are very narrow, which helps maintain the cohesive forces. More importantly, the connections between adjacent conduits have these structures called pit membranes. They're porous, allowing water through, but the pores are tiny. The surface tension of water across these tiny pores is usually strong enough to prevent an air bubble in one embleized conduit from spreading or seeding into its neighbors. It contains the damage like fire doors. That's a great analogy. Conifers have an even fancier system with a structure called a Taurus in their pits that can actually act like a plug to seal off a damaged tracheide. So plants can tolerate some embolisms. Yes, there's redundancy. But widespread embolism caused by severe drought or freeze thaw cycles can be fatal. How do scientists even know this tension exists? You can't exactly stick a pressure gauge into a single xyllem cell. True. They use clever methods. One is the pressure chamber. sometimes called a pressure bomb. You cut off a leafy twig, put it in a sealed chamber with the cut end sticking out, and then pressurize the chamber with gas. When you cut the twig, the water column under tension snaps back into the xyllem. The pressure needed in the chamber to force the water back out to the cut surface equals the negative pressure or tension the water was under originally. Ah, okay. Indirect measurement, right? They can also measure sap flow velocity directly, maybe using heat pulses. They gently heat the sap in one spot in time, how long it takes for the warmer sap to reach a sensor further up. And these measurements confirm the flow starts in the twigs in the morning following the sun, then moves down the trunk, consistent with a pole from above. Any other evidence? Yeah, another cool one is measuring tiny changes in tree trunk diameter. Trunks actually shrink slightly during the day when transpiration and tension are highest because the tension pulls the xyllem walls inward. Then they swell slightly at night when tension relaxes. It's minuscule but measurable. Wow. So, that all points back to that solarp powered cohesion-driven pole. It really does. It's a testament to the unique physical properties of water and the elegant structures plants have evolved. But even this amazing system have limits, right? You mentioned redwoods earlier. Is there a maximum height? It seems so. Studies on the tallest trees suggest the tension in their upper xyllem is getting dangerously close to the point where widespread cavitation would occur. The sheer effort of pulling water that high also creates water stress that likely limits leaf growth and photosynthesis at the very top. The current thinking is the maximum height is probably somewhere around 122 to 130 m. Gravity and physics set a limit. Okay, so the water gets pulled up, but how does it get into the roots from the soil in the first place? Roots are the foundation literally and figuratively. They anchor the plant, of course, but they also have to absorb enormous amounts of water to replace what's lost through transpiration. And the key players there are the root hairs. Absolutely. These aren't separate roots, but tiny finger-like extensions of the roots epidermal cells. They massively increase the surface area for absorption. A single rye plant might have billions of them, creating a huge interface with the soil. So, water soaks in through these hairs. Then, where does it go? It has to cross the root to get to the central vascular cylinder where the xyllem is located. It can travel between cells through the cell walls. That's the opalast pathway. Or it can move through cells either crossing membranes from cell to cell the transcellular pathway or moving from cytoplasm to cytoplasm through little connections called plasmodmatada the simplastic pathway. So multiple routes. Yes. But there's a critical checkpoint. A layer of cells called the endodmis surrounds the vascular cylinder. And the walls of these endodermal cells have a waterproof band embedded in them, the casparian strip. Think of it like mortar between bricks. What does that do? It completely blocks the apoplus pathway, the route through the cell walls. So all water and dissolved minerals must cross a living plasma membrane of an endodermal cell to enter the vascular tissue. Ah, so the plant gets control. It forces everything through a membrane gatekeeper. Precisely. It allows the plant to selectively control what ions get into the xyllem stream. Some roots might also have a similar layer, the exodermis, further out. Okay. We established water is mostly pulled. But uh you mentioned root pressure earlier, a push from the bottom. Yes. Under certain conditions. When transpiration is very low, like at night, and soil moisture is high, roots can actively pump mineral ions into the xylem in the vascular cylinder. This makes the xylem sap more concentrated, lowers its water potential, and water flows in from the surrounding root cells by osmosis, building up a positive pressure from below, and this causes gutation. Exactly. Gutation is when you see little droplets of water on the tips or edges of leaves, usually in the morning. It looks like dew, but it's actually xyllem sap being forced out through special pores called hythodess by this positive root pressure. So, it's internal water being pushed out. Correct. But again, this pressure is relatively weak. It can't push water up a tall tree. And many plants, especially conifers, don't really do it. It's more common in smaller herbaceous plants when conditions are right. Fascinating. What about this other root thing, hydraulic redistribution? Sounds complex. It sounds complex, but the idea is simple. Water moving passively through roots from water soil areas to drier soil areas. It typically happens at night or when transpiration is low. If a plant has deep roots and moist soil and shallow roots and dry soil, water absorbed by the deep roots can move up into the shallow roots and then actually exit into the drier surface soil. That's hydraulic lift. So the plant is watering its own surface roots or even its neighbors potentially. Yes. It can also happen downward hydraulic descent or sideways. The benefits can be significant. It can keep surface roots and their associated microbes like beneficial microisal fungi hydrated during dry periods. It might share water with neighboring plants like sugar maples sharing deeper water with shallower rooted plants. And on a large scale like in the Amazon, it's thought to significantly boost water cycling and influence regional climate. Wow. Roots doing plumbing work outside the plant, too. Okay, let's switch slightly to the minerals, the inorganic nutrients. How do they get taken up? Is it just dissolved in the water? Some comes along with the water flow, but uptake is primarily an active energy requiring process. Plants have to pump ions in, often against a steep concentration gradient. You mean there are way more nutrients inside the root cells than in the soil? Often, yes. Like pea roots might have 75 times more potassium inside than in the surrounding soil solution. Rudabagger root vacules can concentrate potassium 10,000fold. That doesn't happen passively. So, the plant spends energy on this. Definitely there are two main energy dependent steps. Actively absorbing ions from the soil across the membranes of the outer root cells especially root hairs and then actively secretreting those ions from adjacent parankma cells into the xylem vessels. If you deprive roots of oxygen stopping respiration and energy production mineral uptake plummets and those fungi you mentioned my coriza they help here too. Hugely important, especially for nutrients that don't move easily in the soil, like phosphorus, zinc, and copper. The fungal threads extend far out from the root, acting like an extension cord, accessing nutrients the root couldn't reach and transporting them back to the plant. It's a critical symbiosis for most plants. Once the ions are in the xylem, do they just go straight up? Mostly, yes, carried by the transpiration stream, but there's also exchange happening between the xyllem and the phe, the other transport system. Ions can move sideways between the two streams. So nutrients can circulate around. Yes, to some extent. Some ions like potassium, phosphate, and chloride are quite mobile in the floam. This means they can be easily moved from older leaves, say to newer growing areas where they're needed more. Others like calcium, boron, and iron are relatively immobile in the phe. Once the xyllem delivers them to a leaf, they tend to stay put. This difference is important for understanding deficiency symptoms and even for things like folure fertilization spraying nutrients on leaves. Okay, that brings us perfectly to the floam. We've done water and minerals up the xyllem. What about the food, the sugars? That's the floam's job, right? Exactly. Floam transport technically called transllocation moves the sugars produced during photosynthesis. The assimilates around the plant. It follows a source to sync pattern. Source to sync. The source is any part of the plant that's producing or releasing more sugar than it needs for itself. Usually mature photosynthesizing leaves are the main sources. Storage organs releasing stored reserves like a potato tuber sprouting can also be sources. A sink is any part that needs sugars for growth or storage. Think root tips, chute tips, developing flowers, fruits, seeds or storage organs that are filling up like that potato tuber later in the season. So the direction can change. Absolutely. A leaf might be a source when it's mature, but it was a sink when it was young and growing. Roots might be sinks most of the time, but could become sources if they're exporting stored sugar. Developing fruits are very strong sinks, often drawing sugars from many different leaves. How do we know sugars move specifically in the phe tubes? Classic experiments like girdling or ringing a tree are key evidence. If you remove a ring of bark, which includes the floam, down to the wood, the violum, you see sugars accumulate above the ring causing swelling. Things below the ring are starved because you cut the downward pipeline. Precisely. We can also use radioactive tracers. If you feed a leaf radioactive carbon dioxide, 14 CO2, it gets incorporated into sugars during photosynthesis. You can then track where that radioactive sugar moves and auto radiography shows it traveling within the PHM civ tubes. Clever. And what about aphids? You mentioned them helping with xylem research. Wait, was it xyllem or phum? Phum. Aphids are fantastic natural tools for studying phe. They have these incredibly fine needle-like mouth parts called stylots that they can insert directly into a single civ tube cell. Wow. Precision insects. Totally. And the pressure inside the civ tube is actually quite high. So the floam sap is forced out through the stylot and into the aphid. Sometimes it comes out the other end as sugary honeydew. So scientists collect the aphid's output or even just the sap exuding from the stylite if the aphid is carefully detached. Analyzing this sap confirms it's mostly sucrossse often 10 25% plus amino acids hormones some minerals and the flow rates measured this way can be really fast 50 to 100 cm much faster than diffusion. Okay so how does it move that fast? not diffusion. What's the mechanism? The widely accepted explanation is the pressure flow hypothesis first proposed by Ernst Mench back in 1927. It's based entirely on osmosis creating a pressure gradient. Osmosis again seems important in plants. Fundamental. Here's how pressure flow works step by step. One, PHM loading. At the source, say a leaf cell making sugar. Sucrose is actively loaded into the phum sift tubes. This requires energy. Pumping sugar into the civ tube makes the sap inside very concentrated. Okay, high sugar concentration in the civ tube at the source. Two, water enters. This high sugar concentration lowers the water potential inside the civ tube compared to the nearby xyllem. So water moves by osmosis from the xyllem into the civ tube at the source end inflating it like the guard cells. Exactly. This influx of water builds up high trigger pressure within the civ tube at the source. Three, bulk flow. Now you have high pressure at the source end and lower pressure at the sink end where sugar is being removed. This pressure difference drives the entire solution water and dissolved sugars through the sift tubes from source to sink. It's a bulk flow like water through a pipe land at the sink. Four flow unloading at the sink. Say a growing root tip. Sugars are removed from the sift tube either passively or actively to be used or stored. This makes the sap in the sift tube less concentrated. Five. Water exits with the sugar removed. The water potential inside the civ tube at the sink becomes higher than in the surrounding tissues. Often the xyllem again. So water moves by osmosis out of the civ tube at the sink and that water can go back into the xyllem circulation as a whole cycle. It is water circulates between xylem and phum facilitating the pressure gradient that drives sugar transport. Critically the civ tubes themselves are largely passive conduits for this long-distance flow. But the loading at the source and often the unloading at the sink require active transport and metabolic energy from the plant. You mentioned loading requires energy. How does that sugar actually get from the leaf cell into the civ tube? There are a couple of main routes in many herbaceous plants. It's epiplastic loading. Sucrose moves out into the cell wall space the aoplast near the phe and then it's actively pumped into this civ tube companion cell complex. Companion cells are closely associated helper cells for civ tubes. So it takes a detour outside the cells first. Yes. And that pumping step often involves a proton pump using ATP energy creating a proton gradient and then a simporter protein that brings sucrose into the phe along with a proton flowing back down its gradient like a revolving door powered by protons. Yeah. What's the other? Simplastic loading. Here sugars move entirely through plasmmoismata. those cellto cell connections from the photosynthetic cells all the way into the civ tubes. Some plants using this route have a neat trick called polymer trapping. In specialized intermediary cells next to the civ tubes, sucrose is converted into larger sugars like rafinos or stachios. Why bigger sugars? Because these larger sugars are too big to diffuse back out through the narrow plasmo they came in through, but they can move forward into the larger civ tube connections. It effectively traps the sugar near the floam and maintains a concentration gradient driving diffusion towards the sift tube. Still requires energy but indirectly and some just diffuse passively. Yes, some trees seem to use passive some plastic loading. Sugars just move down their concentration gradient from the leaf cells into the floam where the flow itself maintains a lower concentration. No active pumping needed for loading itself. So different strategies for loading. What about unloading at the sink? Is that active too? It can be passive initially, just diffusing out down a concentration gradient, but often the subsequent steps moving the sugar into the actual storage or growth cells of the sink tissue require energy. For example, to store huge amounts of sugar in a sugarbeat root, the plant has to actively pump sucrose into the storage vacules, definitely requiring energy. In growing tissues, energy is needed just to maintain the gradient by quickly using up the arriving sugars. Okay, let's try and wrap this up. We've gone from uh water loss being unavoidable for photosynthesis inspiration. Yeah. To the absolutely mindbending cohesion tension theory pulling water up hundreds of feet powered by the sun and water's own stickiness. Then into the roots, how they absorb water and actively grab nutrients, sometimes with fungal help. The Casparian strip checkpoint, root pressure, hydraulic redistribution. Lots going on down there. And finally, the PHM's pressure flow system. Shipping sugars from source to sink using osmotic gradients. Loading, flowing, unloading. An elegant system. It really is. From tiny pores opening and closing to these massive flows throughout the whole plan. It definitely makes you look at plants, even common ones, with a bit more awe, doesn't it? Master engineers. Truly. Absolutely. And perhaps a final thought to leave you with. Consider this intricate network largely powered by simple physics evaporation, cohesion, osmosis, and solar energy. How universal might these physical principles be for transport systems and life elsewhere? If life evolved under different conditions, different atmospheres, maybe without trees like ours, would similar physical solutions emerge? It makes you ponder the fundamental constraints and opportunities that physics places on biology. That's a great question to ponder. Thank you for joining us on this deep dive into the hidden dynamic world inside plants. |
189983 | https://www.chegg.com/homework-help/questions-and-answers/lab-triangle-area-comparison-classes-use-python-idle-given-class-triangle-complete-program-q87229429 | Your solution’s ready to go!
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Question: LAB: Triangle area comparison (classes) - use Python/IDLE Given class Triangle, complete the program to read and set the base and height of triangle1 and triangle2, determine which triangle's area is larger, and output the larger triangle's info, making use of Triangle's relevant methods. You should implement the method get_area(self) to return the area
LAB: Triangle area comparison (classes) - use Python/IDLE
Given class Triangle, complete the program to read and set
the base and height of triangle1 and triangle2, determine which
triangle's area is larger, and output the larger triangle's info,
making use of Triangle's relevant methods. You should
implement the method get_area(self) to return the area of
the rectangle using formula A = bh/2
Ex: If the input is:
where 3.0 is triangle1's base, 4.0 is triangle1's height, 4.0 is
triangle2's base, and 5.0 is triangle2's height, the output is:
CURRENT CODE:
class Triangle:
def __init__(self):
self.base = 0
self.height = 0
def set_base(self, user_base):
self.base = user_base
def set_height(self, user_height):
self.height = user_height
def get_area(self):
TODO: Implement this method to calculate the area
as a = bh/2 and return the value
def print_info(self):
print('Base:
{:.2f}'.format(self.base))
print('Height:
{:.2f}'.format(self.height))
print('Area:
{:.2f}'.format(self.get_area()))
if __name__ == "__main__":
triangle1 = Triangle()
triangle2 = Triangle()
TODO: Read and set base and height for triangle1
(use set_base() and set_height())
TODO: Read and set base and height for triangle2
(use set_base() and set_height())
TODO: Determine larger triangle (use
get_area())
print('Triangle with larger area:')
TODO: Output larger triangle's info (use
print_info())
Fill in the blanks
The function plan for function get_area() is as follows:
Define a class Triangle:
Return the area of t...
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189984 | http://math2.org/math/trig/hyperbolics.htm | Hyperbolic Trigonomic Identities
Math2.org Math Tables: Hyperbolic Trigonometric Identities
(Math)
Hyperbolic Definitions
sinh(x) = ( e x - e-x )/2
csch(x) = 1/sinh(x) = 2/( e x - e-x )
cosh(x) = ( e x + e -x )/2
sech(x) = 1/cosh(x) = 2/( e x + e-x )
tanh(x) = sinh(x)/cosh(x) = ( e x - e-x )/( e x + e-x )
coth(x) = 1/tanh(x) = ( e x + e-x)/( e x - e-x )
cosh 2(x) - sinh 2(x) = 1
tanh 2(x) + sech 2(x) = 1
coth 2(x) - csch 2(x) = 1
Inverse Hyperbolic Defintions
arcsinh(z) = ln( z + (z 2 + 1) )
arccosh(z) = ln( z (z 2 - 1) )
arctanh(z) = 1/2 ln( (1+z)/(1-z) )
arccsch(z) = ln( (1+(1+z 2) )/z )
arcsech(z) = ln( (1(1-z 2) )/z )
arccoth(z) = 1/2 ln( (z+1)/(z-1) )
Relations to Trigonometric Functions
sinh(z) = -i sin(iz)
csch(z) = i csc(iz)
cosh(z) = cos(iz)
sech(z) = sec(iz)
tanh(z) = -i tan(iz)
coth(z) = i cot(iz) |
189985 | https://www.orthobullets.com/pathology/8075/glomus-tumor | Glomus Tumor - Pathology - Orthobullets
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89 DAYS BEFORE EVENT
10th Annual Baseball Sports Medicine: Game-Changing Concepts
Dec 03, 2025 — Dec 05, 2025
Christopher S. Ahmad Columbia University Medical CenterAnthony Romeo Chicago, IllinoisRegister
89 Days Left!
Description
We invite you to join us as we celebrate the 10th anniversary of the Baseball Sports Medicine: Game-Changing Concepts course.
In recognition of this milestone, we're expanding the course: Baseball Sports Medicine: Game-Changing Conceptswill take place over 3 days instead of 2, on December 3-5, 2025.
Once again, the course will be held at the heart of baseball in the US, the headquarters of MLB's Office of the Commissioner of Baseball in New York City.
Through a mix of didactic presentations, video and technique spotlights, and case-based discussions, the course’s expert faculty will help attendees navigate key issues in the management of injuries in baseball athletes at all levels of play.
Many of the faculty are physicians and trainers from MLB teams, and they will offer practical, evidence-based solutions for managing:
Shoulder injuries,including rotator cuff tears, latissimus injuries, and SLAP tears
Upper extremity nerve and vascular issues, including suprascapular nerve injury, aneurysms and thrombosis, and neurogenic TOS
Non-UCL elbow injuries, including olecranon stress fractures, valgus extension overload, and ulnar nerve issues
UCL injuries,including surgical techniques, outcomes, complications, and revision
Batting injuries,including batter’s shoulder, hand and wrist fractures, and oblique strain
Knee, hip, spine, and hamstring issues, including FAI, athletic pubalgia, and ACL injuries
They’ll also share theirrehabilitationprotocols, including soft tissue modalities, functional assessment criteria for initiation of throwing, and neurocognitive training.
Ample time has been allotted during each session to give you the opportunity to ask the faculty questions. Be sure to come to the meeting with the questions you need answered to help enhance your practice!
more
21 Faculty
Christopher Camp, MD Mayo Clinic, US Eric Wagner, MD Emory Musculoskeletal Institute, US Peter Chalmers, MD University of Utah, Salt Lake City, US Michael Freehill, MD Redwood City, CA, US Christopher Ahmad, MD The Center for Shoulder, Elbow and Sports Medicine, New York, US Justin Greisberg, MD NewYork-Presbyterian Hospital, New York, US Mark Schickendantz, MD Cleveland Clinic Sports Health Center, US Anthony Romeo, MD Chicago , US Steven Cohen, MD Philadelphia, US Timothy Griffith, MD Suwanee, US
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Updated: Jun 22 2021
Glomus Tumor
Patrick O'Donnell MD/PhD
Topic Podcast
Glomus Tumor
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summary
Glomus Tumors are rare benign tumors of the glomus body, often occurring in the subungual region.
The condition is typically seen in patients between the ages of 20 and 40 who present with a painful subungal mass with bluish discoloration.
Diagnosis is made with a biopsy showing a well-defined lesion lacking cellular atypia or mitotic activity with the presence of small round cells with dark nuclei.
Treatment is usually marginal excision.
Epidemiology
Demographics
occurs in patients 20 to 40 years of age
Anatomic ocation
75% occur in hand
50% are subungual
50% have erosions of distal phalanx (primary involvement of bone being very rare)
less common locations: palm, wrist, forearm, foot
Etiology
Forms
may involve either the soft tissue and/or bone
Frequently associated with a delay in diagnosis
Anatomy
Glomus body
the glomus body is a perivascular temperature regulating structure frequently located at the tip of a digit or beneath the nail
Presentation
Symptoms (classic triad)
paroxysmal pain
exquisite tenderness to touch
cold intolerance
Physical exam
small bluish nodule
often difficult to see, especially in the subungual location
nail ridging or discoloration is common
Love test
pressure to the area with a pinhead elicits exquisite pain
100% sensitive, 78% accurate
Hildreth test
tourniquet inflation reduces pain/tenderness and abolishes tenderness to the Love test
92% sensitive, 91% specific
Imaging
Radiographs
glomus tumors can produce a pressure erosion of the underlying bone and an associated deformity of the bone cortex
MRI
helpful to establish diagnosis
present as a low T1 signal and high T2 signal
Studies
Histology
well-defined lesion lacking cellular atypia or mitotic activity
small round cells with dark nuclei
associated small vessels in a hyaline/myxoid stroma
can show gland-like or nest structures, separated by stromal elements
Treatment
Operative
marginal excision is curative
indications
symptoms affecting quality of life
outcomes
due to the benign nature of this disease, recurrence is uncommon
several cases of malignant glomus tumors have been reported in the literature
reconstruction of nail bed contour with autologous fat graft
indications
for large defects after resection
Complications
Recurrence
20%
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189987 | http://www.hexagon.edu.vn/images/resources/upload/48f705559d5da29b51bb76fc555d8c3c/problems-from-the-book-1-pdf_1380724174.pdf | Problems from the Book Titu Andreescu Gabriel Dospinescu Problems from the Book XYZP,s, Titu Andreescu University of Texas at Dallas School of Natural Sciences and Mathematics 800 W Campbell Road Richardson, TX 75080 USA titu.andreescuOutdallas.edu Gabriel Dospinescu Ecole Normale Superieure Departement de Mathematiques 45, rue d'Ulm Paris, F-75230 France gdospi2002@yahoo.com Library of Congress Control Number: 2008924026 ISBN-13: 978-0-9799269-0-7 © 2008 XYZ Press, LLC All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (XYZ Press, LLC, 1721 Monaco Drive, Allen TX, 75002, USA) and the authors except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of tradenames, trademarks, service marks and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. 9 8 7 6 5 4 3 2 1 www.awesomemath.org Cover design by Iury Ulzutuev Math isn't the art of answering mathematical questions, it is the art of asking the right questions, the questions that give you insight, the ones that lead you in interesting directions, the ones that connect with lots of other interesting questions — the ones with beautiful answers. — G. Chaitin vii Preface What can a new book of problems in elementary mathematics possibly con-tribute to the vast existing collection of journals, articles, and books? This was our main concern when we decided to write this book. The inevitability of this question does not facilitate the answer, because after five years of writ-ing and rewriting we still had something to add. It could be a new problem, a comment we considered pertinent, or a solution that escaped our rationale until this predictive moment, when we were supposed to bring it under the scrutiny of a specialist in the field. A mere perusal of this book should be sufficient to identify its target audi-ence: students and coaches preparing for mathematical Olympiads, national or international. It takes more effort to realize that these are not the only potential beneficiaries of this work. While the book is rife with problems collected from various mathematical competitions and journals, one cannot neglect the classical results of mathematics, which naturally exceed the level of time-constrained competitions. And no, classical does not mean easy! These mathematical beauties are more than just proof that elementary mathematics can produce jewels. They serve as an invitation to mathematics beyond com-petitions, regarded by many to be the "true mathematics". In this context, the audience is more diverse than one might think. Even so, as it will be easily discovered, many of the problems in this book are very difficult. Thus, the theoretical portions are short, while the emphasis is squarely placed on the problems. Certainly, more subtle results like quadratic reciprocity and existence of primitive roots are related to the basic results in linear algebra or mathematical analysis. Whenever their proofs are par-ticularly useful, they are provided. We will assume of the reader a certain familiarity with classical theorems of elementary mathematics, which we will use freely. The selection of problems was made by weighing the need for rou- viii tine exercises that engender familiarity with the joy of the difficult problems in which we find the truly beautiful ideas. We strove to select only those problems, easy and hard, that best illustrate the ideas we wanted to exhibit. Allow us to discuss in brief the structure of the book. What will most likely surprise the reader when browsing just the table of contents is the absence of any chapters on geometry. This book was not intended to be an exhaustive treatment of elementary mathematics; if ever such a book appears, it will be a sad day for mathematics. Rather, we tried to assemble problems that enchanted us in order to give a sense of techniques and ideas that become leitmotifs not just in problem solving but in all of mathematics. Furthermore, there are excellent books on geometry, and it was not hard to realize that it would be beyond our ability to create something new to add to this area of study. Thus, we preferred to elaborate more on three important fields of elementary mathematics: algebra, number theory, and combinatorics. Even after this narrowing of focus there are many topics that are simply left out, either in consideration of the available space or else because of the fine existing literature on the subject. This is, for example, the fate of functional equations, a field which can spawn extremely difficult, intriguing problems, but one which does not have obvious recurring themes that tie everything together. Hoping that you have not abandoned the book because of these omissions, which might be considered major by many who do not keep in mind the stated objectives, we continue by elaborating on the contents of the chapters. To start out, we ordered the chapters in ascending order of difficulty of the mathematical tools used. Thus, the exposition starts out lightly with some classical substitution techniques in algebra, emphasizing a large number of examples and applications. These are followed by a topic dear to us: the Cauchy-Schwarz inequality and its variations. A sizable chapter presents ap-plications of the Lagrange interpolation formula, which is known by most only through rote, straightforward applications. The interested reader will find some genuine pearls in this chapter, which should be enough to change his or her opinion about this useful mathematical tool. Two rather difficult chapters, in which mathematical analysis mixes with algebra, are given at the end of the book. One of them is quite original, showing how simple consideration of ix integral calculus can solve very difficult inequalities. The other discusses prop-erties of equidistribution and dense numerical series. Too many books consider the Weyl equidistribution theorem to be "much too difficult" to include, and we cannot resist contradicting them by presenting an elementary proof. Fur-thermore, the reader will quickly realize that for elementary problems we have not shied away from presenting the so-called non-elementary solutions which use mathematical analysis or advanced algebra. It would be a crime to con-sider these two types of mathematics as two different entities, and it would be even worse to present laborious elementary solutions without admitting the possibility of generalization for problems that have conceptual and easy non-elementary solutions. In the end we devote a whole chapter to discussing criteria for polynomial irreducibility. We observe that some extremely efficient criteria (like those of Peron and Capelli) are virtually unknown, even though they are more efficient than the well-known Eisenstein criterion. The section dedicated to number theory is the largest. Some introductory chapters related to prime numbers of the form 4k + 3 and to the order of an element are included to provide a better understanding of fundamental results which are used later in the book. A large chapter develops a tool which is as simple as it is useful: the exponent of a prime in the factorization of an inte-ger. Some mathematical diamonds belonging to Paul Eras and others appear within. And even though quadratic reciprocity is brought up in many books, we included an entire chapter on this topic because the problems available to us were too ingenious to exclude. Next come some difficult chapters concern-ing arithmetic properties of polynomials, the geometry of numbers (in which we present some arithmetic applications of the famous Minkowski's theorem), and the properties of algebraic numbers. A special chapter studies some ap-plications of the extremely simple idea that a convergent series of integers is eventually stationary! The reader will have the chance to realize that in mathematics even simple ideas have great impact: consider, for example, the fundamental idea that in the interval (-1, 1) the only integer is 0. But how many fantastic results concerning irrational numbers follow simply from that! Another chapter dear to us concerns the sum of digits, a subject that always yields unexpected and fascinating problems, but for which we could not find a unique approach. x Finally, some words about the combinatorics section. The reader will imme-diately observe that our presentation of this topic takes an algebraic slant, which was, in fact, our intention. In this way we tried to present some unex-pected applications of complex numbers in combinatorics, and a whole chapter is dedicated to useful formal series. Another chapter shows how useful linear algebra can be when solving problems on set combinatorics. Of course, we are traditional in presenting applications of Turan's theorem and of graph theory in general, and the pigeonhole principle could not be omitted. We faced diffi-culties here, because this topic is covered extensively in other books, though rarely in a satisfying way. For this reason, we tried to present lesser-known problems, because this topic is so dear to elementary mathematics lovers. At the end, we included a chapter on special applications of polynomials in num-ber theory and combinatorics, emphasizing the Combinatorial Nullstellensatz, a recent and extremely useful theorem by Noga Alon. We end our description with some remarks on the structure of the chapters. In general, the main theoretical results are stated, and if they are sufficiently profound or obscure, a proof is given. Following the theoretical part, we present between ten and fifteen examples, most from mathematical contests or from journals such as Kvant, Komal, and American Mathematical Monthly. Others are new problems or classical results. Each chapter ends with a series of problems, the majority of which stem from the theoretical results. Finally, a change that will please some and scare others: the end-of-chapter problems do not have solutions! We had several reasons for this. The first and most practical consideration was minimizing the mass of the book. But the second and more important factor was this: we consider solving problems to necessarily include the inevitably lengthy process of trial and research to which the inclusion of solutions provides perhaps too tempting of a shortcut. Keeping this in mind, the selection of the problems was made with the goal that the diligent reader could solve about a third of them, make some progress in the second third and have at least the satisfaction of looking for a solution in the remainder. We come now to the most delicate moment, the one of saying thank you. First and foremost, we thank Marin Tetiva and Paul Stanford, whose close reading of the manuscript uncovered many errors that we would not have xi liked in this final version. We thank them for the great effort they put into reviewing the book. All of the remaining mistakes are the responsibility of the authors, who would be grateful for reports of errors so that in a future edition they will disappear. Many thanks to Radu Sorici for giving the book the look it has now and for the numerous suggestions for improvement. We thank Adrian Zahariuc for his help in writing the sections on the sums of digits and graph theory. Several solutions are either his own or the fruit of his experience. Special thanks are due to Valentin Vornicu for creating Mathlinks, which has generated many of the problems we have included. His website, mathlinks ro, hosts a treasure trove of problems, and we invite every passionate mathematician to avail themselves of this fact. We would also like to thank Ravi Boppana, Vesselin Dimitrov, and Richard Stong for the excellent problems, solutions, and comments they provided. Lastly, we have surely forgotten many others who helped throughout the writing process; our thanks and apologies go out to them. Titu Andreescu Gabriel Dospinescu titu.andreescuAutdallas.edu gdospi2002©yahoo.com Contents 1 Some Useful Substitutions 1.1 Theory and examples 1.2 Problems for training 1 3 20 2 Always Cauchy-Schwarz... 25 2.1 Theory and examples 27 2.2 Problems for training 43 3 Look at the Exponent 47 3.1 Theory and examples 49 3.2 Problems for training 67 4 Primes and Squares 73 4.1 Theory and examples 75 4.2 Problems for training 89 5 T2's Lemma 93 5.1 Theory and examples 95 5.2 Problems for training 111 xiv CONTENTS 6 Some Classical Problems in Extremal Graph Theory 115 6.1 Theory and examples 117 6.2 Problems for training 128 7 Complex Combinatorics 131 7.1 Theory and examples 133 7.2 Problems for training 148 8 Formal Series Revisited 153 8.1 Theory and examples 155 8.2 Problems for training 173 9 A Brief Introduction to Algebraic Number Theory 179 9.1 Theory and examples 181 9.2 Problems for training 200 10 Arithmetic Properties of Polynomials 205 10.1 Theory and examples 207 10.2 Problems for training 227 11 Lagrange Interpolation Formula 233 11.1 Theory and examples 235 11.2 Problems for training 259 12 Higher Algebra in Combinatorics 263 12.1 Theory and examples 265 12.2 Problems for training 282 13 Geometry and Numbers 289 13.1 Theory and examples 291 13.2 Problems for training 309 14 The Smaller, the Better 313 14.1 Theory and examples 315 14.2 Problems for training 327 CONTENTS xv 15 Density and Regular Distribution 333 15.1 Theory and examples 335 15.2 Problems for training 350 16 The Digit Sum of a Positive Integer 353 16.1 Theory and examples 355 16.2 Problems for training 369 17 At the Border of Analysis and Number Theory 375 17.1 Theory and examples 377 17.2 Problems for training 394 18 Quadratic Reciprocity 399 18.1 Theory and examples 401 18.2 Problems for training 419 19 Solving Elementary Inequalities Using Integrals 425 19.1 Theory and examples 427 19.2 Problems for training 445 20 Pigeonhole Principle Revisited 451 20.1 Theory and examples 453 20.2 Problems for training 473 21 Some Useful Irreducibility Criteria 479 21.1 Theory and examples 481 21.2 Problems for training 501 22 Cycles, Paths, and Other Ways 505 22.1 Theory and examples 507 22.2 Problems for training 519 23 Some Special Applications of Polynomials 523 23.1 Theory and examples 525 23.2 Problems for training 543 xvi CONTENTS Bibliography 547 Index 553 THEORY AND EXAMPLES 3 1.1 Theory and examples We know that in most inequalities with a constraint such as abc = 1 the substitution a = — x , b = z c = — simplifies the solution (don't kid yourself, z x y not all problems of this type become easier!). The use of substitutions is far from being specific to inequalities; there are many other similar substitutions that usually make life easier. For instance, have you ever thought of other conditions such as xyz = x + y + z+ 2; xy + yz + zx + 2xyz =1; x2 +y2 + z2 + 2xyz = 1 or x2 + y2 + z2 = xyz + 4? The purpose of this chapter is to present some of the most classical substitutions of this kind and their applications. You will be probably surprised (unless you already know it...) when finding out that the condition xyz = x + y + z + 2 together with x, y, z > 0 implies the existence of positive real numbers a, b, c such that b+c c+ a a + b x= y= z= a b Let us explain why. The condition xyz=x+y+z+ 2 can be written in the following equivalent way: 1 1 1 1+x 1+y + 1+z 1. Proving this is just a matter of simple computations. Now take 1 1 1 a = b= 1+y, c= 1+ x' 1+z Then a + b + c = 1 and x = 1— a = b+ c. Of course, in the same way a a + a + c c + a a + b we find y = c z = a+ b . The converse (that is, b b ' a ' b ' c satisfy xyz = x + y + z + 2) is much easier and is settled again by basic computations. Now, what about the second set of conditions, that is x, y, z > 0 4 1. SOME USEFUL SUBSTITUTIONS and xy+yz+zx+2xyz = 1? If you look carefully, you will see that it is closely related to the first one. Indeed, x, y, z > 0 satisfy xy + yz + zx + 2xyz = 1 if 1 1 1 1 and only if 111 —, — verify = + + + 2, so the substitution here is x y z xyz x y z a b c x = , z = b-Fc' Y c+a a+b • Now, let us take a closer look at the other substitutions mentioned at the beginning of the chapter, namely x2 + y2 + z2 + 2xyz = 1 and x2 + y2 + z2 = xyz +4. Let us begin with the following question, which can be considered an exercise, too: consider three real numbers a, b, c such that abc = 1 and let 1 1 1 x = a + y =- b + — b , z = c + — a The question is to find an algebraic relation between x, y, z, independent of a, b, c. An efficient way to answer this question (that is, without horrible computations that result from solving the quadratic equations) is to observe that xyz ( a+ 1) (b+ 1 (c+ 1 (a2 + a2 + (b2 + b2 + (c2 + c2 + 2 (x2 2) + (y2 2) + (z2 2) + 2. Thus X2 ± y2 ± Z2 - xyz = 4. THEORY AND EXAMPLES 5 Because 'a + a > 2 for all real numbers a, it is clear that not every triple (x, y, z) satisfying (1.2) is of the form (1.1). However, with the extra-assumption minflx1,1Y1,1z11 > 2 things get better and we do have the converse, that is if x, y, z are real numbers with min{ Ix', lyl, lz1} > 2 and satisfying (1.2), then there exist real numbers a, b, c with abc = 1 satisfying (1.1). Actually, it suf-fices to assume only that max(14 I I , Izi) > 2. Indeed, we may assume that I x 1 > 2. Thus there exists a nonzero real number u such that x = u + 1 . Now, let us regard (1.2) as a quadratic equation with respect to z. Because the discriminant is nonnegative, it follows that (x2 – 4) (y2 – 4) > 0. But since Ix I > 2, we find that y2 > 4 and so there exist a non-zero real number v for which y = v + 1 – v. How do we find the corresponding z? Simply by solving the second degree equation. We find two solutions: 1 U V Zi = UV + , Z2 = -+-uv V U and now we are almost done. If z = uv + 1 — V U we take (a, b, c) = C u,v v, ) uv uv u v 1 u and if z = – + –, then we take (a, b, c) = (–, v, – . U V Inspired by the previous equation, let us consider another one, X2 + y2 + Z2 + xyz = 4 where x, y, z > 0. We will prove that the set of solutions of this equation is the set of triples (2 cos A, 2 cos B, 2 cos C), where A, B, C are the angles of an acute triangle. First, let us prove that all these triples are solutions. This reduces to the identity cos2 A + cos2 B + cos2 C + 2 cos A cos B cos C = 1. 6 1. SOME USEFUL SUBSTITUTIONS This identity can be proved readily by using the sum-to-product formulas. For the converse, we see first that 0 < x, y, z < 2, hence there are num- bers A, B E (0, -2 ) such that x = 2 cos A, y = 2 cos B. Solving the equa- tion with respect to z and taking into account that z E (0, 2) we obtain z = —2 cos(A + B). Thus we can take C = 7r — A — B and we will have (x, y, z) = (2 cos A, 2 cos B, 2 cos C). Let us summarize: we have seen some nice substitutions, with even nicer proofs, but we still have not seen any applications. We will see them in a moment... and there are quite a few problems that can be solved by using these "tricks". First, an easy and classical problem, due to Nesbitt . It has so many extensions and generalizations that we must discuss it first. [Example 1. Prove that a b c 3 b+c + c+a > a+b 2 for all a, b,c> 0. Solution. With the "magical" substitution, it suffices to prove that if x, y, z > 0 satisfy xy + yz + zx + 2xyz = 1, then x+ y + z 3 Let us suppose that 5_ (x + y + z)2 3 3 , we also have 4 4 4 , a contradiction, so we are done. Let us now increase the level of difficulty and make an experiment: imagine that you did not know about these substitutions and try to solve the following problem. Then look at the solution provided and you will see that sometimes a good substitution can solve a problem almost alone. 3 this is not the case, i.e. x+ y + z < — 2. Because xy + yz + zx we must have xy + yz + zx < 1 — 4 3 and since xyz <(x+ y+z ) 3 2xyz < —. It follows that 1 = xy+yz+zx+2xyz < — 3 +1 = 1 THEORY AND EXAMPLES 7 Example 2. Let x, y, z > 0 be such that xy + yz + zx + 2xyz = 1. Prove that 1 -1+- + -1 1 > 4(x +y+z). x y z [Mircea Lascu] Solution. With our substitution the inequality becomes b+c c+a a+b a > 4 a b+c + c+a + a+b But this follows from 4a a a 4b b b 4c c c , , - . b+c - b c c+a c a a+b - a b Simple and efficient, these are the words that characterize this substitution. Here is a geometric application of the previous problem. Example 3. Prove that in any acute-angled triangle ABC the following inequality holds cos2 A cos2 B + cos2 B cos2 C + cos2 C cos2 A 1 < -4 (cos2 A + cos2 B cos2 C). [Titu Andreescu] Solution. We observe that the desired inequality is equivalent to cos A cos B cos B cos C cos A cos C cos C cos A cos B 1 ( cos A cos B cos C - 4 cos B cos C cos C cos A cos A cos B 8 1. SOME USEFUL SUBSTITUTIONS Setting x= cos B cos C y = cos A cos C cos A cos B z= cos B cos C cos A the inequality reduces to 1 1 1 4(x + y + z) < —+—+ —. x y z But this is precisely the inequality in the previous example. All that remains is to show that xy + yz + zx + 2xyz = 1. This is equivalent to cos2 A + cos2 B + cos2 C + 2 cos A cos B cos C = 1, which we have already discussed. The following problem is a nice characterization of the equation (1.2) by poly-nomials and also teaches us some things about polynomials, in two or three variables. Example 4.] Find all polynomials f (x, y, z) with real coefficients such that f(a+-1,b+— 1,c+ 1 ) =0 a c whenever abc = 1. [Gabriel Dospinescu] Solution. From the introduction, it is now clear that the polynomials divisible by x2 + y2 + z2 — xyz — 4 are solutions to the problem. But it is not obvious why any desired polynomial should be of this form. To show this, we use the classical polynomial long division. There are polynomials g(x, y, z), h(y, z), k(y, z) with real coefficients such that f (x, y, z) (x2 + y2 + z2 — xyz — 4)g (x , y, z) + xh(y, z) + k(y , z) THEORY AND EXAMPLES 9 Using the hypothesis, we deduce that 0= (a+-1)h(b+-1 c+-1) +k(b+-1 c+-1) a b' c b' c whenever abc = 1. Well, it seems that this is a dead end. Not exactly. Now 1 we take two numbers x, y such that min{ > 2 and we write x = b + b — y y2 4 y = c + 1 — c with b = x+x2- 4 , c = 2 2 Then it is easy to compute a + 1 . Itis exactly a xy + -V(x2 — 4)(y2 — 4). So, we have found that (xy + V(x2 — 4)(y2 — 4))h(x, y) + k(x, y) = 0 whenever min{ } > 2. And now? The last relation suggests that we should prove that for each y with > 2, the function x \/x2 — 4 is not () rational, that is, there are not polynomials p, q such that N/x2 4 = p x q(x) But this is easy because if such polynomials existed, than each zero of x2 — 4 should have even multiplicity, which is not the case. Consequently, for each y with > 2 we have h(x,y) = k(x,y) = 0 for all x. But this means that h(x, y) = k(x, y) = 0 for all x, y, that is our polynomial is divisible by x2 + y2 ± z2 — xyz — 4. The level of difficulty continues to increase. When we say this, we refer again to the proposed experiment. The reader who will try first to solve the prob-lems discussed without using the above substitutions will certainly understand why we consider these problems hard. Example 5. Prove that if x, y, z > 0 and xyz = x + y + z + 2, then 2(Vxy + Vyz + -‘ 51 -) x + y + z + 6. 10 1. SOME USEFUL SUBSTITUTIONS Solution. This is tricky, even with the substitution. There are two main ideas: using some identities that transform the inequality into an easier one and then using the substitution. Let us see. What does 2(./Ty - + \/yz + ./zx) suggest? Clearly, it is related to (N/Y + -N5 + 15)2 - (X ± + z). Consequently, our inequality can be written as + N/V + A/iz < N/2(x + y + z + 3). The first idea that comes to mind (that is using the Cauchy-Schwarz inequality in the form Vi + .‘ 5 + < V3(x + y + z) < V2(x + y + z + 3)) does not lead to a solution. Indeed, the last inequality is not true: setting x + y + z = we have 3s < 2(s + 3). This is because the AM-GM inequality implies that 3 xy z < — 83 , so 8 > + 2, which is equivalent to (s — 6)(8 + 3)2 > 0, implying 27 27 — s > 6. Let us see how the substitution helps. The inequality becomes lb+c a c+a lab c 2 (b+c c+a a+b + 3) a The last step is probably the most important. We have to change the expres-sion c c+a a+b sion + + + 3 a little bit. a b c We see that if we add 1 to each fraction, then a+ b+c will appear as a common factor, so in fact b+c c+a a+b , 1 1 1 a +3 =(a+b+c)G+ And now we have finally solved the problem, amusingly, by employing again the Cauchy-Schwarz inequality: \lb+c a c+a lab c (b+c+c+a+a+b)(-1 +-1). THEORY AND EXAMPLES 11 We continue with a difficult 2003 USAMO problem. There are numerous proofs for this inequality, none of them easy. The following solution is again not simple, but seems natural for someone familiar with such a substitution. Example 6.1 Prove that for any positive real numbers a, b, c the following inequality holds (2a + b + c)2 (2b + c a) 2 (2c + a + b)2 2a2 + + c)2 2b 2 + (c + a)2 2c 2 + (a + b)2 < 8 [Titu Andreescu, Zuming Feng] USAMO 2003 Solution. The desired inequality is equivalent to (2+ b + cV ( b 2+ c+ ay ( 2 + a+bV a ) c ) 2 + (b+c + 2 + (c+ a) 2 + 2 + (a +b)2 al " b 2 a I c) Taking our substitution into account, it suffices to prove that if xyz = x + y+ z + 2, then (2 + x)2 + (2 + y)2 (2 + z)2 < 8. 2 + x2 This is in fact the same as 2x + 1 2 + y2 2y + 1 2 + z2 — 2z + 1 < 5 x2 + 2 y2 + 2 + z2 + 2 — 2' Now, we transform this inequality into (x — 1)2 (y — 1)2 (z — 1)2 1 x2 + 2 y2 + 2 z2 + 2 — 2 This last form suggests using the Cauchy-Schwarz inequality to prove that (x — 1)2 (y — 1)2 (z — 1)2 > (x + y + z — 3)2 X2 ± 2 y2 +2 z2 2 x2 ± y2 ± z2 6 12 1. SOME USEFUL SUBSTITUTIONS So, we are left with proving that 2(x + y + z — 3)2 > x2 + y2 + z2 + 6. But this is not difficult. Indeed, this inequality is equivalent to 2(x + y + z — 3)2 > (x + y + z)2 — 2(xy + yz + zx) + 6. Now, from xyz > 8 (recall who x, y, z are and use the AM-GM inequality three times), we find that xy+yz +zx > 12 and x +y+z > 6 (by the same AM-GM inequality). This shows that it suffices to prove that 2(s — 3)2 > s 2 — 18 for all s > 6, which is equivalent to (s — 3) (s — 6) > 0, clearly true. And this difficult problem is solved! The following problem is also hard. Yet there is an easy solution using the substitutions described in this chapter. [Example 7.1 Prove that if x, y, z > 0 satisfy xy + yz + zx + xyz = 4 then x + y + z xy + yz + zx. India 1998 Solution. Let us write the given condition as xy ± yz zx xyz 2 2 + 2 • 2 + 2 • 2 +2 2 • 2 • 2 1. Hence there are positive real numbers a, b, c such that 2a 2b 2c b+c' Y c+a' x = z= a+b. But now the solution is almost over, since the inequality x+y+z>xy+yz+zx is equivalent to a b c 2ab 2bc 2ca b+c + c+a + a+b (c+a)(c+b) + (a+b)(a+c) + (b+a)(b+c)• THEORY AND EXAMPLES 13 After clearing denominators, the inequality becomes a(a + b) (a + c) + b(b + a)(b + c) + c(c + a)(c + b) > > 2ab(a + b) + 2bc(b + c) + 2ca(c + a). After basic computations, it reduces to a(a — b) (a — c) + b(b — a) (b — c) + c(c — a)(c — b) > 0. But this is Schur's inequality! Here is a difficult problem, in which the substitution described plays a key role, but cannot solve the problem alone. Elkample S. Prove that if x, y, z > 0 satisfy xyz = x + y + z + 2, then xyz(x — 1)(y — 1)(z — 1) < 8. [Gabriel Dospinescu] Solution. Using the substitution b + c c + a a + b x = -= z = a b the inequality becomes (a + b)(b + c)(c + a)(a + b — c)(b + c — a)(c + a — < 8a2b2c2 (1.4) for any positive real numbers a, b, c. It is readily seen that this form is stronger than Schur's inequality (a + b — c)(b + c — a)(c + a — < abc. 14 1. SOME USEFUL SUBSTITUTIONS First, we may assume that a, b, c are the sides of a triangle ABC, since other-wise the left-hand side in (1.4) is negative. This is true because no more than one of the numbers a + b — c, b + c — a, c+ a — b can be negative. Let R be the center of the circumcircle of triangle ABC. It is not difficult to deduce the following identity (a+b—c)(b+c—a)(c+a b)= (a+b+c)R2. Consequently, the desired inequality can be written as (a + b + c)R2 > (a + b)(b + c)(c + a) 8 But we know that in each triangle ABC, 9R2 > a2 + b2 + c2. Hence it suffices to prove that 8(a + b + c) (a2 + b2 + c 2) > 9(a + b)(b + c)(c + a). This inequality is implied by the following ones: 8(a + b + c)(a2 +b 2 + c 2) > 8(a + b + c)3 and 9(a+b)(b+c)(c+a)< 8(a+b+c)3. The first inequality reduces to a2b2c2 a2 ± b2 ± e2 > 1 _ (a + b + c)2 , 3 while the second is a consequence of the AM-GM inequality. By combining these two results, the desired inequality follows. Of a different kind, the following problem and the featured solution prove that sometimes an efficient substitution can help more than ten complicated ideas. THEORY AND EXAMPLES 15 Example 9. J Let a, b, c > 0. Find all triples (x, y, z) of positive real numbers such that { x+y+z=a+b+c a2x + b2y + c2z + abc = 4xyz [Titu Andreescu] IMO Shortlist 1995 Solution. We try to use the information given by the second equation. This equation can be written as a2 b2 c2 abc —+++ = 4 yz zx xy xyz and we already recognize the relation U2 ± V2 ± W2 + UVW = 4 where u = V a yz , v = V b zx , w = V c xy. According to example 3, we can find an acute-angled triangle ABC such that u = 2 cos A, v = 2 cos B, w = 2 cos C. We have made use of the second condition, so we use the first one to deduce that x + y + z = 2.Vxy cos C + 2Vyz cos A + 2.\/zx cos B. Trying to solve this as a second degree equation in VY, we find the discriminant —4(5 sin C — AFzsinB)2. Because this discriminant is nonnegative, we infer that N5 sin C = -Vi sin B and -VY = N5 cos C + Nii cos B. Combining the last two relations, we find that sin A sin B sin C 16 1. SOME USEFUL SUBSTITUTIONS Now we square these relations and we use the fact that cos A =- The conclusion is: 2. V a yz 2A/zx cos B = cosC = 2.\/xy . b+ c c+ a a + b x =- z = 2 y= 2 2 and it is immediate to see that this triple satisfies both conditions. Hence there is a unique triple that is solution to the given system. And now, we come back to an earlier problem, this time with a solution based on geometric arguments. [Example 10. Prove that if the positive real numbers x, y, z satisfy xy + yz + zx + xyz = 4, then x+y+z>xy+yz+zx. India 1998 Solution. The relation given in the hypothesis of the problem is not an im-mediate analogue of the equation (1.3) Let us write the condition xy + yz + zx + xyz = 4 in the form vxy2 z V y 2 V ZX 2 Vxy • Vyz • -fzx = 4. Now, we can use the result from example 3 and we deduce the existence of an acute-angled triangle ABC such that { yz = 2 cos A A/zx = 2 cos B .\/xy = 2 cos C. x = \/ 2 co s B cos C = \I 2 cos A cos C cos A cos B z = \/2 cos A cos B cos C THEORY AND EXAMPLES 17 We solve the system and we find the triplet 2 cos B cos C 2 cos A cos C 2 cos A cos B (x, y, z) = cos A cos B cos C Hence we need to prove that cos B cos C cos A cos C cos A cos B cos A cos B cos C > 2(cos2 A + cos2 B + cos2 C). This one is a hard inequality and it follows from a more general result. Lemma 1.1. If ABC is a triangle and x, y, z are arbitrary real numbers, then x2 + y2 + z2 > 2yz cos A + 2zx cos B + 2xy cos C. Proof. Let us consider points P, Q, R on the lines AB, BC, CA, respectively, such that AP = BQ = CR = 1 and P, Q, R do not lie on the sides of the triangle. Then we see that the inequality is equivalent to (x•AP + y BQ + z • CR)2 > 0, which is obviously true. Note that the condition x + y + z = 2Vxy cos C + 2 \/yz cos A + 2Vzx cos B is the equality case in the lemma. It offers another approach to Example 9. The lemma being proved, we just have to take in the above lemma and the problem will be solved. And finally, an apparently intricate recursive relation. 18 1. SOME USEFUL SUBSTITUTIONS Example 11. Let (an)n>o be a non-decreasing sequence of positive integers such that ao = al = 47 and 4_1 ±an 2 ±a2 nr+i _ an_i an and- 4 for n > 1. Prove that 2 + an and 2 + -V2 + an are perfect squares for all n > 0. [Titu Andreescu] Solution. Let us write an = xn + — 1 , with xn-> 1. Then the given condition xn becomes xn+i = xnxn-i (we have used here explicitly that xn > 1), which shows that (ln xn)n>0 is a Fibonacci-type sequence. Since xo = xi, we deduce that xn = xrn, where F 0 = F1 = 1, Fn+i = Fn F n_1. Now, we have to do 47 + V472 - 1 more: what is xo? And the answer xo = 2 won't suffice. Let us remark that (\i5+ from where we find that 1 r_)2 =49 v X0 1 = 7. 21 Similarly, we obtain that 1 = 3. + 'XI Solving the equation, we obtain 2 = A I ix7i = (1+ 2v5) 2) that is So = A8. And so we have found the general formula an = A8Fm + A-8F,L. And now the problem becomes easy, since (A4F, A-4F,N2 (A2Fn A-2F,)2. an + 2 = ) and 2 + -V2 + ari = THEORY AND EXAMPLES 19 1 All we are left to prove is that A2k + — A2k E N for all k E N. But this is not difficult, since A2 ± A -1 N, A4 + E N z A4 and 1 A2 k+1 (A2 + 1 ) (A2k 1 ) A2 k-1 1 A2(k+1) - A2 A2k A2(k-1) 20 1. SOME USEFUL SUBSTITUTIONS 1.2 Problems for training 1. Find all triples x, y, z of positive real numbers, solutions to the system: { x2 + y2 + Z2 = xyz + 4 xy + yz + zx = 2(x + y + z) 2. Prove that if x, y, z > 0 satisfy xy + yz + zx + 2xyz = 1, then 1 3 xyz < — 8 and xy + yz + zx > — 4. 3. Prove that for any positive real numbers a, b, c the following inequality holds b+c c+a a+b a b c 9 + + > a b c — b+c ± c+a+ a+b + 2 . J. Nesbitt 4. Let a, b,c> 0 such that a2 + b2 + c 2 + abc = 4. Prove that \/ (2 — a)(2 — b) + (2 + a)(2 + b) (2 — b)(2 — c) ± (2 + b)(2 + c) (2 — c)(2 — a) 1 • (2+c)(2+a) Cristinel Mortici, Romanian Inter-county Contest 5. Prove that if a, b,c> 0 satisfy the condition la2 + b2 + c2 41 = abc, then (a — 2)(b — 2) + (b — 2)(c — 2) + (c— 2)(a — 2) > 0. Titu Andreescu, Gazeta Matematica 6. Prove that if x, y,z > 0 and xyz = x + y + z + 2, then xy + yz + zx > 2(x + y + z) and N5 + N/Y+ V7z 5 .\/xyz. PROBLEMS FOR TRAINING 21 7. Let x, y, z > 0 such that xy + yz + zx = 2(x + y + z). Prove that xyz 4 — (3 + cos(A — B) + cos(B — C) + cos(C — A)). — Titu Andreescu 9. Prove that in every acute-angled triangle ABC, (cos A + cos B)2 + (cos B + cos C)2 + (cos C + cos A)2 < 3. 10. Find all triples (a, b, c) of positive real numbers, solutions to the system f a2 + b2 + c 2 + abc = 4 a + b + c = 3 Cristinel Mortici, Romanian Inter-county Contest 11. Find all triplets of positive integers (k,l, m) with sum 2002 and for which the system x - + — = k y x z — + — = / z y z x — + — = m x z has real solutions. Titu Andreescu, proposed for IMO 2002 22 1. SOME USEFUL SUBSTITUTIONS 12. Prove that in any triangle the following inequality holds A B C < cost 2 ( sin — 2 + sin — 2 + sin — —A + cos .13 + cos2 — C. 2 2 2 13. Find all functions f : (0, oo) —+ (0, oo) with the following properties: a) 1(x) + f (y) + f (z) + f (xyz) = f (/) f (Vyz)f ( N/zx) for all x, y, z; b) if 1 < x < y then f (x) < f (y). Hojoo Lee, IMO Shortlist 2004 14. Prove that if a, b, c > 2 satisfy the condition a2 + b2 + c2 = abc + 4, then a + b + c + ab + ac + bc > 2,\/(a + b + c + 3)(a2 ± b2 + c2 3). Marian Tetiva 15. Let x, y, z > 0 such that xy + yz + zx + xyz = 4. Prove that 1 1 1 2 3( +—+ ) > (X+2)(Y+2)(Z+2). Vi N5 "V i Gabriel Dospinescu 16. Prove that in any acute-angled triangle the following inequality holds ( cos A 1 2 C COS B 2 + (COS C) 2 cos B ) cos C ) cos A +8cosAcosBcosC> 4. Titu Andreescu, MOSP 2000 PROBLEMS FOR TRAINING 23 17. Solve in positive integers the equation (x + 2)(y + 2)(z + 2) = (x + y + z + 2)2. Titu Andreescu 18. Let n > 4 be a given positive integer. Find all pairs of positive integers (x, y) such that (x + y)2 xy = n - 4. n Titu Andreescu 19. Let the sequence (an)n>o, where ao = al = 97 and an+1 = an-Ian + ./(a,2, — 1)(an 2 1 — 1) for all n > 1. Prove that 2 + -V2 + 2ar, is a perfect square for all n > 0. Titu Andreescu 20. Prove that if a, b, c > 0 satisfy a2 + b2 + c2 + abc = 4 then 0 < ab + bc + ca - abc < 2. Titu Andreescu, USAMO 2001 21. Prove that if a, b, c > 0 and x = a + b -1 ' a y = b + -1, z = c + -1, then xy+yz+zx > 2(x+y+z). Vasile Cartoaje 22. Prove that for any a, b, c > 0, (b + c - a)2 (c + a — b)2 ± (a + b - c)2 > 3 + a2 (c a)2 b2 (a + b)2 + c 2 Japan 1997 THEORY AND EXAMPLES 27 2.1 Theory and examples In recent years the Cauchy-Schwarz inequality has become one of the most used results in contest mathematics, an indispensable tool of any serious problem solver. There are countless problems that reduce readily to this inequality and even more problems in which the Cauchy-Schwarz inequality is the key idea of the solution. In this unit we will not focus on the theoretical results, since they are too well-known. Yet, examples that show the Cauchy-Schwarz inequality at work are not as readily available. This is the reason why we will see this inequality in action in several simple examples first, gradually leading to uses of the Cauchy-Schwarz inequality in some of the most difficult problems. Let us begin with a very simple problem. Though it is a direct application of the inequality, it underlines something less emphasized: the analysis of the equality case. Example Prove that the finite sequence ao, al, , an of positive real numbers is a geometrical progression if and only if (a( 2 )±4±• • •+an_1)(4 .-Fd-F• • .-Fari 2 ) = (aoai +• • •+an—lan)2 . Solution. We see that the relation given in the problem is in fact the equality case in the Cauchy-Schwarz inequality. This is equivalent to the proportion-ality of the n-tuples (ao, al, . , an_i) and (al, a2, , an), that is ao al an—i - = - = • • • = al a2 an But this is just actually the definition of a geometrical progression. Hence the problem is solved. Note that Lagrange's identity allowed us to work with equivalences. Another easy application of the Cauchy-Schwarz inequality is the following problem. This time the inequality is hidden in a closed form, which suggests using calculus. There exists a solution that uses derivatives, but it is not as elegant as the one featured: 28 2. ALWAYS CAUCHY-SCHWARZ... Example 2.1 Let p be a polynomial with positive real coefficients. Prove that p(x2)p(y2) > p2(xy) for any positive real numbers x, y. Russian Mathematical Olympiad Solution. If we work only with the closed expression p(x2)p(y2) > the chances of seeing a way to proceed are small. So, let us write ao + aix + • • • + anxn . The desired inequality becomes (ao + aix2 + • • • + anx2n)(ao + aiy2 + • • any2n) > (ao + aixy + • • • + anxnyn)2. P2 (xY), p(x) = And now the Cauchy-Schwarz inequality comes into the picture: (ao + aixy + • • • + anxnYn)2 -= (Vao • Vao Vaix2 • Vaiy2 + • • + anxn • anyn)2 02 + any 2n). < (ao + aix2 + • • • + anx2n)(ao a + And the problem is solved. Moreover, we see that the conditions x, y > 0 are redundant, since we have of course p2(xy) < p2(1xyl). Additionally, note an interesting consequence of the problem: the function f : (0, co) (0, co), f(x) = lnp(ex) is convex, that is why we said in the introduction to this problem that it has a solution based on calculus. The idea of that solution is to prove that the second derivative of this function is nonnegative. We will not prove this here, but we note a simple consequence: the more general inequality p(x/Dp(4) . . . p(4) > Pk (X1X2 Xk), which follows from Jensen's inequality for the convex function f (x) = lnp(ex). Here is another application of the Cauchy-Schwarz inequality, though this time you might be surprised why the "trick" fails at a first approach: THEORY AND EXAMPLES 29 Prove that if x, y, z> > 0 satisfy — 1 + — 1 + — 1 = 2, then x y z Example 3. — + — 1 + — 1 < + y + z. Iran 1998 Solution. The obvious and most natural approach is to apply the Cauchy-Schwarz inequality in the form Vx-1+ — 1 + N/z — 1 < V3(x + y + z — 3) and then to try to prove the inequality \/3(x + y + z — 3) < .Vx+ y + z, which is equivalent to x + y + z < — 9. Unfortunately, this inequality is not 2 true. In fact, the reversed inequality holds, that is x+ y + z > — 9, since — 2 1 1 9 2 = x + y + z 1 > x+y+z. Thus this approach fails, so we try another, using again the Cauchy-Schwarz inequality, but this time in the form .Vx — 1 + —1+ — 1 = -va • •/x — a 1 + .NA Y b l +•VC• (a + b+c)(x— a 1 y 1 z c 11 • We would like to have the last expression equal to A/x+ y + z. This encourages us to take a = x, b = y, c = z, since in this case x— a 1 y — 1 z-1 b = 1 and a+b+c-=x+y+z. Hence this idea works and the problem is solved. We continue with a classical result, the not so well-known inequality of Aczel. We will also see during our trip through the world of the Cauchy-Schwarz inequality a nice application of Aczel's inequality. z-1 30 2. ALWAYS CAUCHY-SCHWARZ... Example 4 Let al, a2, , an, b1, b2, . , bn, be real numbers and let A, B > 0 such that A2 > a? + a3 + • • • + an 2 or B 2 > b? + + • • • + bn 2 Then (A2 ? 2 — a — a 2 _ an 2)( ? B2 b — — • • • — b2 ) < (AB — aibi — a2b2 — • • • — anbn)2. [Aczel] Solution. We observe first that we may assume that A2 > a? + a3 + • • + an 2 and B 2 > b? + + • • • + bn 2 . Otherwise the left-hand side of the desired inequality is less than or equal to 0 and the inequality becomes trivial. From our assumption and the Cauchy-Schwarz inequality, we infer that aibi + a2b2 + • • • + anbn < ,Va? + 4 + • • • + an 2 , Vbi +b2 +•••+bn < AB Hence we can rewrite the inequality in the more appropriate form aibi + a2b2 + • • • + anb, + V(A2 — a)(B2 — b) < AB , where a -= a? + a3 + • • • + an and b = b? + b2 + • • • + bn 2 Now, we can apply the Cauchy-Schwarz inequality, first in the form albs + a2b2 + • • • + anbn + \/(A2 — a)(B2 — b) < fctb + \/(A2 — a)(B2 — b) and then in the form + V(A2 a)(B2 — b) <~/(a + A2 — a) (b B2 — b) = AB. And by combining the last two inequalities the desired inequality follows. As a consequence of this inequality we discuss the following problem, in which the condition seems to be redundant. In fact, it is the key that suggests using Aczel's inequality. THEORY AND EXAMPLES 31 Example 57 Let al, a2, , an, b1, b2, , bn be real numbers such that (al + • • • + an — 1)(14 . + • • • + bn 2 — 1) > (aibi + • • • + anbn — 1)2. Prove that a? + 4 + • + an > 1 and bq + b2 + • • • + bn 2 > 1. [Titu Andreescu, Dorffi Andrica] USA TST 2004 Solution. First of all, it is not difficult to observe that an indirect approach is more efficient. Moreover, we may even assume that both numbers aT + 4 + • • + an — 1 and bi + b2 + • • • + bn 2 — 1 are negative, since they have the same sign (this follows immediately from the hypothesis of the problem). Now, we want to prove that (a? + • • • + an 2 — 1)(bT + • • • + bm 2 — 1) < (aibi + • • • + anbn — 1)2 (2.1) in order to obtain the desired contradiction. And all of a sudden we arrived at the result in the previous problem. Indeed, we have now the conditions 1 > a? + a3 + • • • + an 2 and 1 > b7 . + b2 + + bn 2 , while the conclusion is (2.1). But this is exactly Aczel's inequality, with A = 1 and B = 1.The conclusion follows. The Cauchy-Schwarz inequality is extremely well hidden in the next problem. It is also a refinement of the Cauchy-Schwarz inequality, as we can see from the solution. Example 6. For given n > k > 1 find in closed form the best constant T (n, k) such that for any real numbers xi, x2, , xn the fol-lowing inequality holds: i=1 ) 2 2 i) > T (n, k) [k i=1 k i=1 2 ___ k 32 2. ALWAYS CAUCHY-SCHWARZ... 1<i<j T(n, k) (Xi — j ) . 1<i<j<k [Gabriel Dospinescu] Solution. In this form, we cannot make any reasonable conjecture about T (n, k), so we need an efficient transformation. We observe that (Xi — x3)2 1<i<j<n is nothing else than and also (Xi — Xj)2 = 1<i 0. We also observe that in the left-hand side there are n — k variables that do not appear in the right-hand side and that the left-hand side is minimal when these variables are equal. So, let us take them all to be zero. The result is THEORY AND EXAMPLES 33 which is equivalent to Now, if kT (n, k) — n > 0, we can take a k-tuple (xi, x2, . , xk) such that E x, = 0 and Ex? 0 and we contradict the inequality (2.2). Hence we i=i must have kT (n, k) — n < 0 that is T(n, k) < — k . Now, let us proceed with the converse, that is showing that n n 2 k nE4 — Exi > _ n kE4— — k i=-1 i=i i=i Exi i=i 2 1 (2.3) for all real numbers x1, x2, .. , xn. If we manage to prove this inequality, then it will follow that T (n, k) = — k . But (2.3) is of course equivalent to — ) 2 k xi 2 i=k+1 i=1 Now, we have to apply the Cauchy-Schwarz inequality, because we need xi. i=k+1 We find that n n xi > n — k L xi 2 n i=k+1 i=k+1 34 2. ALWAYS CAUCHY-SCHWARZ... and so it suffices to prove that n — n k A2 > (A + B) — 132, (2.4) k where we have taken A = xi and B = xi. But (2.4) is straightforward, 2=k-1-1 i=1 since it is equivalent to (kA — (n — k)B)2 k(n — k)B2 > 0, which is clear. Finally, the conclusion is settled: T (n, k) = — k is the best con-stant. We continue the series of difficult inequalities with a very nice problem of Murray Klamkin. This time, one part of the problem is obvious from the Cauchy-Schwarz inequality, but the second one is not immediate. Let us see. [Example 7.] Let a, b, c be positive real numbers. Find the extreme values of the expression Va2x2 b2y2 c2z2 Vb2 x2 c2y2 a2z2 \/c2x2 a2y2 b2z2 where x, y, z are real numbers such that x2 + y2 + z2 = 1. [Murray Klamkin] Crux Mathematicorum Solution. Finding the upper bound does not seem to be too difficult, since from the Cauchy-Schwarz inequality it follows that Va2x2 b2y2 c2z2 Vb2x2 c2y2 a2z2 V c2x2 a2y2 b2z2 < THEORY AND EXAMPLES 35 < V3(a2x2 b2y2 c2z2 c2y2 a2z2 c2x2 a2y2 b2z2) = ,V3(a2 b2 c2). We have used here the hypothesis x2 + y2 + z2 1. Thus, V3(a2 b2 c2) is the upper bound and this value if attained for x = y = z = 3 But for the lower bound things are not so easy. Investigating what happens when xyz = 0, we conclude that the minimal value should be a+ b+c, attained when two variables are zero and the third one is 1 or —1. Hence, we should try to prove the inequality Va2x2 b2y2 c2z2 Vb2 x2 c2y2 a2z2 1 c2x2 a2y2 b2 z2 > a + b + c. Why not square it? After all, we observe that a2x2 b2y2 c2z2 b2x2 c2y2 a2z2 c2x2 a2y2 b2z2 a2 b2 c2 so the new inequality cannot have a very complicated form. It becomes Va2x2 b2y2 c2z2 Vb2x2 c2y2 a2z2 Vb2x2 c2y2 a2z2 x2 + a2y2 b2z2 + c2x2 + a2y2 + b2z2 • Va2x2 + b2y2 + c 2z2 > ab + be + ca which has great chances to be true. And indeed, it is true and it follows from — what else, the Cauchy-Schwarz inequality: 1/a2x2 b2y2 c2z2 Vb2x2 c2y2 + a2 z 2 Z > abx2 + bcy2 + caz2 and the other two similar inequalities. This shows that the minimal value is indeed a + b + c, attained for example when (x, y, z) = (1, 0, 0). It is now time for the champion inequalities. Do not worry if the time you spend on them is much longer than the time spent for the other examples: these problems are difficult! There are inequalities where you can immediately 36 2. ALWAYS CAUCHY-SCHWARZ... see that you should apply the Cauchy-Schwarz inequality. Yet, applying it in- correctly can be very annoying. This is the case with the following example, where there is only one possibility to solve the problem using Cauchy-Schwarz: Prove that for any real numbers a, b, c, x, y, z the following in-equality holds: ax + by + cz + /(a2 ± b2 c2 (x2 + y2 ± z2) 2 , > 3 —(a+b+c)(x+y+z). — [Vasile Cartoaje] Kvant Solution. It is quite clear that a direct application of the Cauchy-Schwarz inequality for V(a2 b2 c2 ) (x2 + y2 + z2) has no chance to work. Instead, if we develop 3(a + b + c) (x + y + z) we may group a, b, c and therefore try again the same method. Let us see: 2 — 3 (a + b + c)(x + y + z)—(ax + by + cz) 2y + 2 3 z — x + b 2x + 3 2z — y + c 2x + 3 2y — z =a and the latter can be bounded by a2 + b 2 + C2 • VE ( 2x-1-2y—z I 2. 3 ) All we have 2x-I-2y—z )2 < x2 + y2 + z2 , to do now is to prove the easy inequality E ( which 3 / is actually an equality! Example 9. Prove that for any nonnegative numbers al, a2, . • • , an 1 that E, = _ 2, the following inequality holds: i=1 such THEORY AND EXAMPLES 37 1<i<j<n aia3 < n(n — 1) (1 — aj)(1 — a3) — 2(2n — 1)2 • [Vasile Cartoaje] Solution. This is a very hard problem, in which intuition is better than tech-nique. We will concoct a solution using a combination of the Cauchy-Schwarz inequality and Jensen's inequality, but we warn the reader that such a solution cannot be invented easily. Fasten your seat belts! Let us write the inequality in the form n 2 ai n(n — 1) 1 — ai i=1 ) < (1—a2)2 (2n — 1)2 We apply now the Cauchy-Schwarz inequality to find that ai (1 — ai)2) Thus, it remains to prove the inequality aa2 ai 2 n(n — 1) (1 _ ao 2=1 2 — (1 _ ao2 + (2n — 1)2 The latter can be written of course in the following form: ai(1 — 2a,) < 2n(n — 1) z=i (1 — a02 (2n — 1)2 • This encourages us to study the function x( 2x f : [0 —1 —> R, f (x) = 1 ' 2 (1 ) 2 ) and )2 and to see if it is concave. This is not difficult, for a short computation shows that f"(x) = —6x (1 _ x)4 < 0. Hence we can apply Jensen's inequality to com-plete the solution. (iES 1<i<j<n ) 2 < (a, + • • • + aj) . \ 2 38 2. ALWAYS CAUCHY-SCHWARZ... We continue this discussion with a remarkable solution, found by Claudiu Raicu, a member of the Romanian Mathematical Olympiad Committee, to the difficult problem given in 2004 in one of the Romanian Team Selection Tests. Example 10. Let al, a2, , an be real numbers and let S be a non-empty subset of {1, 2, , n}. Prove that [Gabriel Dospinescu] Romanian TST 2004 Solution. Let us define si = al + a2 + • • • + a, for i > 1 and so = O. Now, partition S into groups of consecutive numbers. Then E , is of the form iES Sil — sil Sj2 — 8i2 + • • • + Sik — Sik , with 0 < it < i2 < • • • < ik < n, jl < j2 < • • < ik and also it < ji, , ik < jk. Now, let us observe that the left-hand side is nothing other than n + (s3 2 (Sj — Si) 2 . i=1 1<i<j<n 1<i<j<n Hence we need to show that \ 2 — Sil + 2 - + • • • + Sik Sik )2 (Si - Si) . 0<i<j<n±1 Let us take al = sil, a2 = s ji , , a2k_1 = 5 4, a2k = sjk and observe the obvious (but important) inequality E (si — Si) 2 — 0<i<j<n 1<i<j<2k THEORY AND EXAMPLES 39 And this is how we arrived at the inequality (al — a2 + a3 — • • • + a2k-1 a2k)2 < ai — a:3) . (2.5) 1<i<j<2k The latter inequality can be proved by using the Cauchy-Schwarz inequality k-times: (al — a2 + a3 — " • + a2k-1 a2k)2 < k((ai — a2)2 + (a3 — a4)2 + • • • + — a2ic)2) (al — a2 + a3 — " • + a2k-1 a2k)2 < k((ai — a4)2 + (a3 — a6)2 + • • • + (a2k-1 — a2)2) (al — a2 + a3 — • • • + a2k-1 a2k)2 < k((ai — a2k)2 + (a3 — a2)2 + • • • + (a2k-1 — a2k-2)2) and by summing up all these inequalities. In the right-hand side we obtain an even smaller quantity than (ai — a3)2, which proves that (2.5) is 1<i<j<2k correct. The solution ends here. The following is a remarkable inequality in which the Cauchy-Schwarz inequal-ity is extremely well hidden. We must confess that the following solution was found after several weeks of trial and error: Example 1171 Prove that for any positive real numbers a, b, c, x, y, z such that xy + yz zx = 3, a , , b b-PcW+z)+ c±a(x+z)+ a+b(x+y)> 3. [Titu Andreescu, Gabriel Dospinescu] •\/ 3 , -4(xy + yz + zx)+ 3 4(xy + yz + zx) < 2 (b+ a 2 c) V(x+y+z)2 3 40 2. ALWAYS CAUCHY-SCHWARZ... Solution. This is probably the best example of how finding the good homo-geneous inequality simplifies the solution. In our case, it suffices to prove the homogeneous inequality b + c(y + + c + a(x + + a + b(x + y) -V3(xy + yz + zx). And now we can assume that x + y + z = 1! Let us apply then the Cauchy-Schwarz inequality: a x + c+a y + a+b z + V3(xy+ yz + zx) < ( a b + c ) 2 b+c • V x2+ Therefore, the problem will be solved if we manage to prove that ( a )2 a which is the same as ab E (a + c)(b + c) 4 3' This reduces to (a + b + c) (ab + be + ca) > 9abc which is clearly true. Finally, two classical inequalities show the power of a clever application of the Cauchy-Schwarz inequality combined with some analytic tools: Example 12. Prove that for any real numbers al, a2, an the following inequality holds: n i=1 3 2 b + c b + [Hilbert] THEORY AND EXAMPLES 41 Solution. Here is a unusual way to apply the Cauchy-Schwarz inequality: ) 2 i=1 j=1 i ° j=1 6 .6 +3 6 Vi v Jai -rjaj 4/7 aiaj x--n 0,4 Ni5ct.
1 k = + L'a z,.9-1+ ) • i,j1 i,=1 By rearranging terms in both sums, it is enough to prove that for any positive integer m A/Fri < E (M n) \n 7r. n>1 Fortunately, this is not difficult, because the inequality 1 n+1 dx (n + m +1)Vn +1 fri (X + M)fi holds as a consequence of the monotonicity of f (x) =(x+myvi. By adding up these inequalities, we deduce that E 1 < dx n>0 (n + m + 1Wn ± 1 Jo (x + m)-Vi • With the change of variable x = mu2, a simple computation shows that the last integral is Ir and this finishes the solution. We end this chapter with a remarkable inequality due to Fritz Carlson . There are many analytic methods of proving this result, but undoubtedly the follow-ing one, due to Hardy, will make you say: always Cauchy-Schwarz! 42 2. ALWAYS CAUCHY-SCHWARZ... Example 13. For any real numbers al, a2, ..., an we have +. • •±n2a2 ) > (al +• • •±0,70 72 • (a? 2 2)( 2 4 2 ma2 +- • •+an a l + a2 n — 4 [Fritz Carlson] Solution. Choose some arbitrary positive numbers x, y and use the Cauchy-Schwarz inequality in the form (al a2 an)2 < E (x + yk2)4 • k=1 1 x yk2 • k>1 Because the function f(z) = x±l yz2 is decreasing, we have 1dz k>1 x + yk2 f x + yz2 • It is immediate to check that the last integral equals 2/ T . Therefore, if we let S = 4 + 4 + + an and T = 4+ 224 + + n2an 2 , then we have for all positive numbers x, y the inequality 7r (al + a2 + + a n)2 < 2\/xy (Sx + Ty). And now, we can make a choice for x, y, so as to minimize the last quantity. It is not difficult to see that a smart choice is x = s and y = All it remains is to insert these values in the previous inequality and to take the square of this relation. PROBLEMS FOR TRAINING 43 2.2 Problems for training 1. Let a, b, c be nonnegative real numbers. Prove that (ax2 + bx + c)(cx2 + bx + a) > (a + b + c)2 x2 for all nonnegative real numbers x. Titu Andreescu, Gazeta MatematicA 2. Let p be a polynomial with positive real coefficients. Prove that if p ( 1 1 i — x) > p(x) i s true for x = 1, then it is true for all x > 0. Titu Andreescu, Revista Matematica Timi§oara 3. Prove that for any real numbers a, b, c > 1 the following inequality holds: -Va —1+Vb—l+Vc—l< V a(bc + 1) . 4. For any positive integer n find the number of ordered n-tuples of integers (ai , a2, ... , an) such that ai. + a2 + • • • + an > n2 and aT . + d + • • • + an 2 < n 3 + 1. China 2002 5. Prove that for any positive real numbers a, b, c, 1 1 1 1 (a + b+ c+ -Nbc)2 > a+b + b+c + c+a + 2.I /abc — (a+b)(b+c)(c+a)• Titu Andreescu, MOSP 1999 44 2. ALWAYS CAUCHY-SCHWARZ... 6. Let al, a2, , an, bi, b2, , bn be real numbers such that E a jai > 0. 1<i<j aiaj bibj ij<n 1<i0j<n 1<i0j<n 1< ) Alexandru Lupas, AMM 7. Let n > 2 be an even integer. We consider all polynomials of the form xn + an_ixn-1 + • • • + aix + 1, with real coefficients and having at least one real zero. Determine the least possible value of a7+ a2 + • • • +an 2_1. Czech-Polish-Slovak Competition 2002 8. The triangle ABC satisfies the relation 2 A 2 c i) 2 (1 2 (Cot — 2 ) + (2 cot B) + (3 cot — 2 = 7r Show that ABC is similar to a triangle whose sides are integers, and find the smallest set of such integers. Titu Andreescu, USAMO 2002 9. Let x1, x2, ... , xn be positive real numbers such that 1 1 1 + + • + = 1. 1 + x1 1 + x2 1 + xn PROBLEMS FOR TRAINING 45 Prove the inequality 1 1 1 V--.+-\/+•••+-Vxn?(7/-1) xi x2 ( + A/xi, + ). Vojtech Jarnik Competition 2002 7r 10. Given real numbers xi, x2, , xi° E [0, --] 2 such that sin2 xi + sin2 x2 + • • • + sin2 x10 = 1. Prove that 3(sin xi + sin x2 + • • • + sin xio) < cos xi + cos x2 + • • + cos xio• Saint Petersburg, 2001 11. Prove that for any real numbers xi, x2, , xn the following inequality holds n n 2 xii) < 2(n2 _ 1) vn n v i=1 i=1 — 3 (z—, — xi 1 2) . i=1 j=1 IMO 2003 12. Let al, a2, ..., an be positive real numbers which add up to 1. Let ni be the number of integers k such that 21' > ak > 2'. Prove that < 4 + -Vlog2(n). 2' i>1 L. Leindler, Miklos Schweitzer Competition 46 2. ALWAYS CAUCHY-SCHWARZ... 13. Let n > 2 and xi, x2,... , xn be positive real numbers such that ( 1 1 1 (xi ± xz ± • • • ± xn ) (— + — +...+ —) = n2 + 1. Xi X2 xn Prove that (x + 4 + • • - ± X2 ) 1 1 i .4_ 1 -, ,2 L A , 2 n 2 2 , • • • , --- , . 1 -. 1 ( X X2 Xn n(n 1). Gabriel Dospinescu 14. Prove that for any positive real numbers al, az, .. • , an, x1, x2, • , xn such that i<i<j<n the following inequality holds xixi = (2) 7 al az + • • • + an (xz + xn) + • an + (xi --1-• • • +xn_i) > n. al + • • ± an-i Vasile Cartoaje, Gabriel Dospinescu THEORY AND EXAMPLES 49 3.1 Theory and examples Most of the time, proving divisibility reduces to congruences or to the famous theorems such as those of Fermat, Euler, or Wilson. But what do we do when we have to prove, for example, that lcm(a, b, c)2 I lcm(a, b) • lcm(b, c) • lcm(c, a) for any positive integers a, b, c? One thing is sure: the above methods fail. Yet, another smart idea appears: if we have to prove that alb, then it suffices to show that the exponent of any prime number in the prime factorization of a is at most the exponent of that prime in the prime factorization of b. For simplicity, let us denote by vp(a) the exponent of the prime number p in the prime factorization of a. Of course, if p does not divide a, then vp(a) = 0. Also, it is easy to prove the following properties of vp(a): • vp(a + b) > minfvp(a), vp(b)} • vp(ab) = vp(a) vp(b) for any positive integers a and b. Now, let us rephrase the above idea in terms of vp(a): alb if and only if for any prime p we have vp(a) < vp(b), and a =- b if and only if for any prime p, vp(a) = vp(b). • vp(gcd(ai,a2, • vp(lcm(ai, a2, n • vp(n!) -= [— pi , an)) = , an)) = n [pd max{vp n 3 LP _I minfvp(ai), vp(a2), , vp(an)}, (al), vp(a2), , vp(an)} n — sp(n). 1 3 — 1 Here, sp(n) is the sum of the digits of n when written in base p. Observe that the third and fourth properties are simple consequences of the definitions. Less straightforward is the fifth property; it follows from the fact that among the numbers 1, , n there are [— ni multiples of p, multiples of p2 and so on. p p2 The other equality is not difficult. Indeed, let us write n = ao±aip±• • • +akpk, where ao, , ak E — 11 and ak # O. Then H n 1:4 +... = ±a2p+ • akpk-1 + a2 -1-a3p± • • • + akpk-2 ± • ak, P i P 50 3. LOOK AT THE EXPONENT and now, using the formula P 1 + p + p = p —1 we find exactly the fifth property. [Example 1. 1 , 7 Let a and b be positive integers such that alb2 b3 a4 as b 6 , b71a8, .... Prove that a = b. Solution. We will prove that vp(a) = vp(b) for any prime p. The hypothesis 027 b3 a4, a51b67 b71a8,...
and on+3 a4n+4 for is the same as azin+1 1 b4n-F2 all positive integers n. But the relation a4n+11b4n+2 can be written as (4n + 1)vp(a) < (4n + 2)vp(b) for all n, so that vp(a) < lim 4n + 2vp(b) = vp(b). n--, c>o 4n + 1 Similarly, the condition b4n+3 a4n+4 implies vp(a) > vp(b) and so vp(a) = vp(b). The conclusion now follows. We have mentioned at the beginning of the discussion a nice and easy problem, so probably it is time to solve it, although you might have already done this. Example 2. Prove that lcm(a,b,c)211cm(a,b) • lcm(b, c) • lcm(c, a) for any positive integers a, b, c. Solution. Let p be an arbitrary prime number. We have vp(lcm(a, b, c)2) =- 2 max{x, y, z} and vp(lcm(a, b) • lcm(b, c) • lcm(c, a)) = max{x, y} + max{y, z} + max{z, x}, where x = vp(a), y = vp(b), z = vp(c). So we need to prove that max{x, y} + max{y, z} + max{z, x} > 2 max{x, y, z} THEORY AND EXAMPLES 51 for any nonnegative integers x, y, z. But this follows by symmetry: we may assume that x > y > z and the inequality reduces to 2x + y > 2x. It is time for some difficult problems. The ones we chose to present are all based on the observations from the beginning of the chapter. Example 3. Prove that there exists a constant c such that for any positive integers a, b,n that satisfy a! • b! In! we have a+b<n+cln n. [Paul Erdos] Solution. Of course, there is no reasonable estimation of this constant, so we should better see what happens if a! • b! In!. Then v2(a!) + v2(b!) < v2(n!), which can be also written as a — 32(a) + b — s2(b) < n — 32(n) < n. So we have found almost exactly what we needed: a + b < n + 32(a) + s2(b). Now, we need another observation: the sum of digits of a number A when written in binary is at most the number of digits of A in base 2, which is 1 + [log2 A] (this follows from the fact that 2k-1 < A < 2k, where k is the number of digits of A in base 2). Hence we have the estimations a + b < n + 32(a) + 82 (b) < n + 2 + log2 ab < n + 2 + 2 log2 n (since we have of course a, b < n). And now the conclusion is immediate. The following problem appeared in Kvant. It took quite a long time before an Olympian, S. Konyagin, found a simple solution. We will not present his solution here, but another one, even simpler. Example 4. Is there an infinite set of positive integers such that no matter how we choose some elements of this set, their sum is not a perfect power? Kvant 52 3. LOOK AT THE EXPONENT Solution. Let us take A = {2n 3n+1 in > 1} If we consider some differ-ent numbers from this set, their sum will be of the form 2' • 3x+1 y, where (y, 6) = 1. This is certainly not a perfect power, since otherwise the exponent should divide both x and x 1. Thus this set is actually a good choice. The following problem shows the beauty of elementary Number Theory. It combines diverse ideas and techniques, and the result we are about to present is truly beautiful. You might also want to try a combinatorial approach by counting the invertible matrices with entries in the field Z/2Z. Example 57 Prove that for any positive integer n, n! is a divisor of n-1 H (2n — 2k). k=0 Solution. Let us take a prime number p. We may assume that p < n. First, let us see what happens if p = 2. We have v2(n!) = n — s2(n) < n — 1 and also V2 ( n-1 ( 2 n — 2k)) k=0 n-1 k=0 v2(2n — 2k) > n — 1 (since 2n — 2k is even for k > 1). Now, let us assume that p > 2. From Fermat's theorem we have pl2P-1 — 1, so p1214P-1) — 1 for all k > 1. Now, n-1 H (2n — 2k) = 2 n(n2 1) fl (2k - 1) k=0 k=1 and from the above remarks we infer that n-1 2k - 1) k=1 THEORY AND EXAMPLES 53 > >2 , vp(2k(P-1) - 1) > card{ < k(p - 1) < n}. 1<k(p-1)<n Because we find that But card{k11 < k(p 1) n [p - n-1 V p H (2 n — 2k)) k=0 n - s (n) n - 1 n vp(n!) = p - 1 -< p - 1 p -1' [p _ n and since vp(n!) E Z, we must have vp(n!) < [p d . From these two inequalities, we conclude that (n-1 VP H (2n — 2k) > vp(n!) k=0 and the problem is solved. Diophantine equations can also be solved using the method described in this chapter. Here is a difficult one, given at a Russian Olympiad. Prove that the equation 11 1 1 = +—+ + lon ni! n2! nk• does not have integer solutions such that 1 < n1 < • • • < nk• Tuymaada Olympiad Solution. We have 10n((ni + 1) ... (nk - 1)nk + • • • + (nk_i + 1) • • • (nk - 1)nk + 1) = nk! 54 3. LOOK AT THE EXPONENT which shows that nk divides 10n. Let us write nk = 2' • 5v. Let S = (ni + 1) ... (nk - 1)nk + • + (nk-i + 1) ... (nk - 1)nk + 1. First of all, suppose that x, y are positive. Thus, S is relatively prime to 10. It follows that v2(nk!) = v5(nk!). This implies [— nk = [— nk 53 for all j (because nk we clearly have [—] > [1 -' ±1 ) and so nk < 3. A simple verification shows 2j 53 that there is no solution in this case. Next, suppose that y = 0. Then S is odd and thus v2(nk!) = n < v5(nk!). Again, this implies v2(nk!) = v5(nk!) and we have seen that this yields no solution. Thus x = 0. A crucial obser-vation is that if nk > nk_1 + 1, then S is odd and thus we find again that v2(nk!) = n < v5(nk!), impossible. Hence nk = nk_i +1. But then, taking into account that nk is a power of 5, we deduce that S is congruent to 2 modulo 4 and thus v2(nk!) = n + 1 < v5(nk!) + 1. It follows that [— nil < 1 + [— nil and 2 5 thus nk < 6. Because nk is a power of 5, we find that nk = 5, nk-1 < 4 and exhausting all of the possibilities shows that there are no solutions. A tricky APMO 1997 problem asked to prove that there is a number 100 < n < 1997 such that ni2n + 2. We will invite you to verify that 2 • 11 • 43 is a solution, and especially to find out how we arrived at this number. Yet... small verifications show that all such numbers are even. Proving this turns out to be a difficult problem and this was proved for the first time by Schinzel. Example 7.1 Prove that for any n > 1 we cannot have n1271-1 + 1. [Schinzel] Solution. Although very short, the proof is tricky. Suppose n is a solution. Let n = Hpi k' where p1 < 7,2 < • < Rs are prime numbers. The idea i=1 is to look at v2(pi - 1). Choose that pi which minimizes this quantity and write pi = 1 + 2rtrni, with mi odd. Then n 1 (mod 2ri) and we can write THEORY AND EXAMPLES 55 n — 1 = 2rit. We have 22rit —1 (mod pi), thus —1 = 22ritmi = 2(pi-1)t = 1 (mod pi) (the last congruence being derived from Fermat's little theorem). Thus pi = 2, which is clearly impossible. We continue with a very nice and difficult problem, in which the idea of look-ing at the exponents is really helpful. It seems to have appeared for the first time in AMM, but over the last few years, it has been proposed in various national and international contests. Example 8. Prove that for any integers al, a2, , an the number H ai i— j 1<i<j<n is an integer. [Armond E. Spencer] AMM E 2637 Solution. We consider a prime number p and prove that for each k > 1, there are more numbers divisible by pk in the sequence of differences (a,, — a3)1<z<3<n than in the sequence (i — j)i<z<i<n. Because ( vp fl (ai — aj) I = Npk H (ai — ai) I where Npk ({ 0;1 ‹.1 n}) is the number of terms in the sequence A that are multiples of x and vp H (i- j) 1<i<j<n k>1 N pk H j) ) , 1<i<j<n 1<i<j<n k>1 1<i<j<n, NPk H (i—j)) =131-1 ( 1+ [ 2nPi ]) () 1<i<j<n i=1 (3.1) 56 3. LOOK AT THE EXPONENT the problem will be solved if we prove our claim. Fix k > 1 and suppose that there are exactly b i indices j E {1, 2, ... , n} such that a3 i (mod pk), for each i E {0,1, ... ,pk — 1}. Then k - P —1 bi Npk (ai ai)) = E (2). 1<i<j<n i=0 Let us see what happens for ai = i. If i = 0, then the number of 1 < j < n for which j = 0 (mod pk) is . If i > 0 then any 1 < j < n for which j = i (mod pk) has the form rpk i for some 0 < r < [n p 7,1 . Thus we find 1 + [V] indices in this case. Hence By changing j = pk — 1 in (3.1), we infer that H Npk i<i<j<n j=0 so it suffices to prove that Pk -1 pk n±i E (;) E P k 2 i=o =o THEORY AND EXAMPLES 57 pk —1 pk_i Now, observe that we need to find the minimum of >2 , x 2i) , when E xi =_ n i=o i=o (it is clear from the definition of bi that pk —1 P k 1 - E bi= n=E [n j Pk ] i=0 from the definition of bi). For this, let us suppose that xo < x1 < x2 < • • • < xpk_i is the pk-tuple for which the minimum is reached (such a pk-tuple pk exists since the equation E xi = n has a finite number of solutions). If i=o xpk_i > xo + 1, then we consider the n-tuple (x0 + 1, xi, . • • , xpk_2, xpk_i — 1), where the sum of components is n, but for which (xo + 1) + (xi) ± (xpk_2) (xpk_i — 1) 2 2 2 2 < (X0) + (x 1) (x pk_2) (xpk_i 2 2 2 2 ). The last inequality is true, since it is equivalent to xpk_i > xo + 1. But this contradicts the minimality of (xo, xi, , x2, , xpk_i). So, xpk_i < xo + 1, and from here it follows that xi E {xo, xo + 1} for all i E {0, 1, 2, ... ,pk — 1}. Hence there is a j E {0, 1, 2, . ,pk — 1} such that xo = xi = • • • = x3 and x3+1 = x3+2 = • • • = xpk_i = xo + 1. Because the variables xr add up to n, we must have (j 1)XO (Pk j 1)(X0 + 1) = n, thus pk(xo + 1) = n + j + 1. Therefore Er-01 ( b ) > (j + 1) (2) + (pk — j — 1) (x021). Finally, observe that for all 0 < i < pk — 1 we have [n ptii xo + 1 + P —jT1 ] and this is equal to xo + 1 if i > j + 1 and to xo otherwise. P Therefore k —1 Ln+i E pk 1) (X 2 0) ± (pk 1) (xo + 1) ) 2 i=o 58 3. LOOK AT THE EXPONENT The next exercise is particularly difficult, but the ideas used in its solution are extremely useful when solving some other problems. Example 9. Let a and b be two distinct positive rational numbers such that for infinitely many integers n, an — bn is an integer. Prove that a and b are also integers. [Gabriel Dospinescu] Mathlinks Contest Solution. Let us start by writing a = -x , b = -z , where x, y, z are distinct positive integers with no common factor, and x y. We are given that znIx" - y" for all positive integers n in an infinite set M. Assume that z > 1 and take p a prime divisor of z. If p does not divide x, it follows that it cannot divide y. Now, we have two cases: i) If p = 2, then let n be such that 2'Ixn - y". Write n = 2unjn, where jn is odd. From the identity x2un - y2"in = (xin - yin)(xin + yin) ... (x2" -lin y2un-ljn) it follows that u„-i kin 2kin). v2(xn - y") = v2(x - yin) + 2 k=0 But xin-1 + xin-2y + • • • + xyin-2 +yin-1 is clearly odd (since jn, x, y are odd), hence v2 (xin - yin) = v2 (x - y). Similarly, we can prove that v2(x3- + yjn) = v2(x + y). Because x2kin + y2kin = 2 (mod 4), THEORY AND EXAMPLES 59 for k > 0, we finally deduce that 2un jn < v2 (xn — yTh ) < v2(x + y) + v2(x — + un — 1 (3.2) Consequently, (2un)nEm is bounded, a simple reason being the inequality 2un < v2(x + v2(x — y) + un — 1. Hence (un)nEm takes only a finite number of values, and from (3.2) it follows that (jn)nEm also takes a finite number of values, that is M is finite, a contradiction. ii) Suppose that p is odd and let d be the least positive integer k such that pixk — yk . Then for any n in M we have pixn — yn. Let x = tu, y = tv, where (u, v) = 1. Clearly, tuv is not a multiple of p. It follows that p (u — vd,u d n vn ) = u(n,d) v(n,d) I x(n,d) y(n , d ) and by the choice of d, we must have din. Therefore any element of M is a multiple of d. Take now n in M and write it in the form n = md, for some positive integer m. Let A = and B =- yd. Then Pm I Pn yn Am Bm and this happens for infinitely many m. Moreover, /AA — B. Let R be the infinite set of those m. We will prove now a very useful result in this type of problems: Theorem 3.1. Let p be an odd prime and let A, B be positive integers, not divisible by p and such that plA. — B. Then for all positive integers n we have vp(An — Bn) = Vp(72) vp(A — B). Proof. The proof of this theorem is natural, even though it is quite long and technical. Indeed, let us write n = pk • I with gcd(/,p) = 1. We will prove the result by induction on k. First, suppose that k = 0. Observe that vp(An — Bn) = vp(A — B) if and only if p does not divide An-1 + An-2B 60 3. LOOK AT THE EXPONENT ABn-2 Bn-1 If the latter does not hold, because A = B (mod p), we infer that pInAn-1 and this cannot hold because k = 0 and gcd(A,p) = 1. Suppose now that the result holds for k and take n = pk+1/ with gcd(/,p) = 1. Then, if m = pk/ we can apply the inductive hypothesis and write: vp(An — Bn) = Vp(AmP — BmP) = vp(Am — Bm)+ vp(Am(P-1) + Am(P-2) Bm + • • • + Am Bm(P-2) + Bm(P -1) ) = v p(A — B) + k + vp(Am(P -1) + Am(P -2) Bm + • • • + Am Bm(P-2) + Bm(P -1) ) So, we need to prove that vp(Am(P-1) + Am(P-2)Bm + • + AmBm(P-2) + Bm(P-1)) = 1. But this is not difficult. First, note that if we put Am = a, Bm = b, it is enough to prove that if vp(a) = vp(b) = vp(a — b) — 1 = 0, then vp (ap-1 ap-2 b abp-2 + bp-1) 1. Now, write b = a + pc for some integer c and observe that using the binomial formula we can write aP-1+ aP-2b + • • • + abP -2 + bP-1 = aP-1+ aP-2 (a + pc) + aP - 3 (a2 + 2apc)+ • • + a2 (aP-3 + (p — 3)aP-4pc) + a (aP-2 + (p — 2)aP-3pc) + aP-1 + (p 1)aP-2 pc = paP-1 + caP-2p2 P — 2 1 = paP-1 (mod p 2), which proves the inductive step and finishes the proof of the theorem. Let us come back to our problem. Using the theorem, we deduce that for infinitely many m we have m < vp(Am — Br') = v p(A — B) + vp(m) < vp(A — B) + Llogp mJ , THEORY AND EXAMPLES 61 which is clearly impossible. Hence plx and ply, in contradiction with the fact that x, y, z are relatively prime. This shows that z = 1 and a, b are integers. If you thought this is the last challenge on this chapter, you are wrong! The following problems can be called The Eras Corner. They were especially kept for the end of the chapter, because of their beauty and difficulty. Example 10.1 a) Prove that for any positive integer n there exist positive integers al < a2 < • • < an such that ai — alai for all i < j. b) Prove that there exists a positive constant c such that for any n and any sequence al < a2 < < an which satisfies the conditions of a), al > n'. [Paul Eras ] Miklos Schweitzer Competition Solution. If a) is not so difficult, b) needs culture and ingenuity. The proof of a) is of course by induction on n. For n = 1 it is enough to take al = 1. Suppose that al < a2 < < an is a good sequence and let us take b = aia2 • • • an. The sequence b, b + al, b + a2, b + an is also good and shows how the inductive step works. Now, let us discuss b). Take any prime number p < n and observe that if ai a 3 (mod p) then ai = a 3 = 0 (mod p). Therefore at most p — 1 among the numbers al, a2, an are not multiples of p. Consider the multiples of p among al, a2, an and divide them by p. We obtain another good sequence, and the previous argument shows that this new sequence has at most p — 1 terms not divisible by p. Repeating this argument yields yp(aia2 • • an) > (n — (p — 1)) + (n — 2(p — 1)) + + (n [p n (7) — 1)). A small computation shows that if p < then the last quantity exceeds 3p ' 2 Therefore aia2...an > fl p 3P . But it is clear that al > an — al, so p< V7 n, > a a > ia2 • • an 2 — 2 62 3. LOOK AT THE EXPONENT which shows that 1 3 -Frz P al > — 2 • e P < So, all we need now is to prove that there exists a constant c > 0 such that in pp > c • Inn. Actually, we will prove more, that p<n lnp = lnn + 0(1). P The tool will be again the factorization of n!. Indeed, this gives the identity ln(n!) = >vp(n!) • lnp. On the one hand, using Stirling's formula n! ( 71 e )n-V27rn, we deduce that ln(n!) = n(ln n — 1) + 0(ln n). On the other hand, 71 23 — 1 < Vp(77,!) < Therefore p<n Because the series 0 (n) . lnp lnp P(P1) is clearly convergent, it follows that n- E 29091) — P p<n And now, we will prove the following result also due to Erd6s: f p < 4n-1 if p<n n > 1. The proof of this theorem is magnificent. We use induction. For small values of n it is clear. Now, assume the inequality true for all values smaller than n and let us prove that fl p < 4m-1. If n is even, we have nothing to p<n H . prove, since H p = p < 4n-2 < 4n-1 p<n p<n-1 THEORY AND EXAMPLES 63 Now, assume that n = 2k + 1 and consider the binomial coefficient 2k+1 (k + 2) ... (2k + 1) k ) k! 217\ An application of the identity 22k+1 = Ei>0 (2k2 ) shows that (2k+1) < 4k Thus, using the inductive hypothesis, we find P P H p < 4k 4k 4n-1. p<n p<k+1 k+2<p<2k+1 This result shows that In fl p = 0(n), so using the previous estimations we p<rt can write >2, ln p = In n 0(1). p<n Here is a refinement and proof of the famous Bertrand's postulate, asserting that between n and 2n there is always a prime number if n > 1. Actually, the result proved in the next example shows that much more is true for suf-ficiently large n and also gives an effectively computable constant c < 10000 for the proof of Bertrand's postulate. Simple computations allow after that a complete proof of this result. However, we prefer the more quantitative result below: Example 11.1 For any e > 0 there exists an no such that for all n > no there are at least (2 6) iogn 2(n) primes between n and 2n. 3 [Paul Erd6s] Solution. A very good way of obtaining interesting bounds for the counting functions of prime numbers is to study the powers that divide the binomial coefficient (2n). Why is this number so special? First of all, because it is quite easy to evaluate it asymptotically. One can easily prove, for instance (2: using Stirling's formula that 4n There are, however, much more -/rn • 64 3. LOOK AT THE EXPONENT elementary estimations. For example, using the fact that (2 n n) is the largest binomial coefficient and that the sum of these binomial coefficients is 471, we easily infer the inequality (2nn) 2n4+1, which is more than enough for our modest purposes. Now, another important fact about this binomial coefficient is that the prime powers dividing it do not have large exponents. Indeed, VP [p ] < [loge 27d , k>1 ( [Pk n 0 n which shows that the largest power of p dividing (2 n n) does not exceed 2n. This implies that the exponent of any prime p> N/2n is at most 1. But the remarkable observation that Erdos had is that actually this special binomial coefficient is not a multiple of any prime between 3 and n, as you can im- mediately establish using the fact that vp((2nn)) > [27: _ 2 7 4]). So, k>1 P using all these observations, we infer that 2n + 1 — n 471 2n < H 2n • H p- H P. p< 2n I /n<p<V n<p<2n Using the result proved in the solution of the previous example, we deduce (2 1 -2+1, so if that 2n p < 4 Also, it is clear that 2n <472 n,<p<V p< 2n f(n) is the number of primes between 7/ and 2n, then 4n 1+ \/2n —1 < (2n) (2n)f (n). By taking logarithms, we finally deduce that f(n) > 3 — 0( \rn • In n) log2 n from which the conclusion follows immediately. 2n + 1 But the most subtle and difficult problem of this chapter (and probably of the whole book) is the following fascinating result, conjectured by Palfy and THEORY AND EXAMPLES 65 proved by Eras using Sylvester's theorem on prime divisors of consecutive numbers. The following marvelous solution by M. Szegedi was taken from the note "a (mod p) < b (mod p) for all primes p implies a = b", published in the second issue of the American Mathematical Monthly, 1987: Example 12. Let a, b be positive integers such that for all prime numbers p, a (mod p) < b (mod p). Then a = b. [Erdos, Palfy] Miklos Schweitzer Competition 1984 Solution. This solution will not be short, but it has the merit of being com-pletely elementary. It follows from a very subtle analysis of the prime powers dividing (a b) (for it is clear that by taking a prime p > a + b we obtain a < b). Hence suppose that a < b. Observe that if a < a then by letting c = b — a we have 0 < c < a and also c (mod p) = b (mod p) — a (mod p) < b (mod p) (because 0 < b (mod p) — a (mod p) < p). Therefore it is enough to prove that the case 0 < a < 2 is impossible. Let (a b) = where A = a! and B = b(b — 1) (b — a +1). Also, let A(pk) and B(pk) be the number of factors of A and B respectively, that are multiples of pk. It is clear that A(pk) = [ p 4] and B(pk) =114, L P 1 — k a . LP Then, by using the fact that 0 < [x + y] — [x] — [y] < 1 for all real numbers x, y, we infer that B(pk) — A(pk) is 0 or 1. Now, the crucial observation is that A(p) > B(p). Indeed, the first multiple of p that appears in the product a • (a — 1) • • • 2 • 1 is a — a (mod p), while the first multiple of p in b • (b — 1) (b — a + 2) • (b — a + 1) is b — b (mod p). Using this remark and the fact that the sequences 1,2, ... , a and b — a + 1, b — a + 2, ... , b have the same length, we infer that A(p) > B(p). But, as we have already seen, this implies A(p) = B(p). Therefore if p > a then surely A(p) = 0, so B(p) = 0 and so A(pk) = B(pk) = 0 for all positive integers k and all p > a. Therefore A = H pA(P)+A(P2)±... and B = H pB(P)+B(P2)±— , p<a p<a (b\ B = = HP B(p)—A(p)+B(732)—A(732)+•-• SO a A p<a 66 3. LOOK AT THE EXPONENT There is another crucial observation to be made: if m(p) is the largest k such that B(pk) is not zero, then using the fact that A(p) = B(p) we obtain B(p) — A(p) + B(p2) — A(p2) + • • • m(p) (B(p)) - A(p')), SO j=2 B(p) — A(p) + B(p2) A(p2) + • • • < m(p) — 1 (recall that we have established the inequality B(pk) — A(pk) < 1). Therefore (a b) is a divisor of 11 pm(P) —1 and so p<a (b a +1) • (b — a + 2) • • b prn(P) p<a is a divisor of FT! p . However, the last divisibility cannot hold for b > 2a. p<a Indeed, it is clear that a! < aa_ir(a) < (b — a +1) (b — a + 2) • • • b P fj pm(p) p<a p<a because after cancellations are made in (b a + 1) • (b — a + 2) • • b pm(P) p<a we obtain a — 7r(a) factors all equal to at least b — a +1 > a, a contradiction. PROBLEMS FOR TRAINING 67 3.2 Problems for training 1. Prove the identity lcm(a, b, c)2 gcd(a, b, c)2 lcm(a, b) • lcm(b, c) • lcm(c, a) gcd(a, b) • gcd(b, c) • gcd(c, a) for all positive integers a, b, c. USAMO 1972 2. Let a, b, c, d be positive integers such that ab = cd. Prove that gcd(a, c) • gcd(a, d) = a • gcd(a, b, c, d). Polish Olympiad 3. Let al, a2, , ak, b1, b2, , bk be positive integers such that gcd(ai, bi) = 1 for all i E {1, 2, , k}. Let m = lcm(bi, b2, , bk). Prove that (aim a2m a km) gcd gcd(ai , a2, • • • , ak)- b1 b2 • bk IMO 1974 Shortlist 4. Let n be a positive integer such that 2n-20051n!. Prove that n has at most 2005 non-zero digits when written in base 2. 5. Let 0 < al < • • • < an be integers. Find the greatest m for which we can find the integers 0 < b1 < < bm such that 2ak = E bk and H(2ak )! = bk!• k=i k-1 k=1 k=1 Gabriel Dospinescu 68 3. LOOK AT THE EXPONENT 6. Show that if it is a positive integer and a and b are integers, then n! divides a(a + b)(a + 2b) • • • (a + (n — 1)b)bn-1. IMO 1985 Shortlist 7. Prove that the product of the numbers between 21917 + 1 and 21991 — 1 is not a perfect square. Tournament of the Towns 1991 8. Let a, b, c be positive integers such that clay — V. Prove that cl aa cibc . I.Niven, AMM E 564 9. Prove the identity n (n+ 1) lcm ((n o), (1),..., (I n)) = lcm(1,2,...,n + 1) for any positive integer n. Peter L. Montgomery, AMM E 2686 10. Prove that the least common multiple of the numbers 1, 2, ... ,n equals 1), n , ... , n the least common multiple of the numbers (n (2) (n) if and only if n + 1 is a prime. Laurentiu Panaitopol, Romanian TST 1990 11. Find v2(A), where A is the numerator of 1 +A-kt + ' • • • + 2k1 -1 ' J.L.Selfridge, AMM E 1408 PROBLEMS FOR TRAINING 69 12. Let al, a2, ak be positive integers not exceeding n such that ai does not divide fl a3 for all i. Denote by 7r(n) is the number of primes not exceeding n. Prove that k < 7r(n). Erdos 13. Let al, , an > 0 be such that whenever k is a prime number or a power of a prime number, we have an {T al}+...± {T} <1 Prove that there is a unique i E {1,2,..., n} such that al + • • + an < 1 + [ad. Tache Alexandru 14. Let m be an integer greater than 1. Suppose that a positive integer n satisfies nlam - 1 for all integers a relatively prime to n. Prove that n < 4m(2m - 1). Find all cases of equality. Gabriel Dospinescu, Marian Andronache, Romanian TST 2004 15. Prove that the sequence (xn)n>i, xn being the exponent of 2 in the 2 22 2' decomposition of the numerator of -1 + + • + , goes to infinity as 2 n -p oo. Even more, prove that > 2n — n + 1. Adapted from a Kvant problem 16. Let x, y be relatively prime different natural numbers. Prove that for infinitely many primes p, the exponent of p in xP-1 - yP-1 is odd. AMM 70 3. LOOK AT THE EXPONENT 17. Find the exponent of 2 in the prime factorization of the number (22n1/ ( 2n 2n1) • J. Desmong, W.R.Hastings, AMM E 2640 18. Let n be an integer greater than 1 and let a, b be positive integers smaller than n. Prove that there exists a prime number p such that min(sp(a) + sp(n — a), sp(b) + sp(n — b)) _>. p— 1 ± sp(n). Gabriel Dospinescu 19. Prove that there exists an absolute constant c such that for any positive integers a, b, n for which a!b! = n! and 1 < a < b < n we have n < b + cln inn. Paul Era's, AMM 6669 20. Prove that the product of at most 25 consecutive integers is not a square. Narumi's Theorem 21. Prove that for all positive integers n different from 3 and 5, n! is divisible by the number of its positive divisors. Paul Eras, Miklos Schweitzer Competition 22. Let (an)n>i be a sequence of positive integers such that gcd(am, an) = agcd(m,n) for all positive integers m, n. Prove that there exists a unique sequence of positive integers (bn)n>1 such that an = fl bd. dln Marcel Tena PROBLEMS FOR TRAINING 71 23. Let f (n) be the maximum size of a subset A of {1,2, ..., n} which does not contain two elements i, j such that it2j. Prove that f (n) = s+ 0(ln n). Paul ErdOs, AMM E 3403 THEORY AND EXAMPLES 75 4.1 Theory and examples The study of the properties of prime numbers is very well-developed, yet many old conjectures and open questions are still waiting to be solved. In this chapter, we present properties of some classes of primes and also of some classical results related to representations as sum of two squares. At the end of the unit, we will discuss, as usual, some nonstandard and surprising problems. Because we will use some facts several times, we prefer to fix some notations before discussing the problems. So, we will consider the sets Pi and P3 of all prime numbers of the form 4k + 1 and 4k + 3, respectively. Also, Q2 will be the set of all numbers that can be written as the sum of two perfect squares. Our purpose is to present some classical results related to Pi, P3, Q2. The most spectacular property of the set Pi is the fact that any of its elements is the sum of the squares of two positive integers. This is not a trivial property and we will present a beautiful proof of it next. Example 1.1 Prove that Pi is a subset of Q2. [Fermat] Solution. We need to prove that any prime number of the form 4k + 1 is the sum of two squares. We will use a very nice result: Theorem 4.1 (Thue). If n is a positive integer and a is relatively prime to n, then there exist integers 0 < x, y < VT" such that xa ±y (mod n) for a suitable choice of the signs + or —. Proof. The proof is simple, but the theorem itself is a diamond. Indeed, let us consider all the values xa — y, with 0 < x, y < INFri J. So, we have a list of + 1)2 > n numbers and it follows that two numbers among them give the same remainder when divided by n, let them be axi — yi and axe — y2. It is not difficult to see that we may assume that x1 > x2 (we certainly cannot have xi = x2 or yi = y2). If we take x = x1 — x2 and Y = 1Y1 — Y21, all the conditions will be satisfied, so the theorem is proved. K 76 4. PRIMES AND SQUARES We will use now Wilson's theorem to find an integer n such that pin2 + 1. Indeed, let us write p = 4k +1 and observe that we can take n = (2k)!. Why? Because from Wilson's theorem we have —1 ==._ (p — 1)! .=-=- 1 • 2 • ... (p 2 1 P p 2 ) (p — 1) • • • 2 (-1)Y ((P 2 1 ) !) = ((2k)!)2 (mod p) and the claim is proved. Now, since pin2 + 1, it is clear that p and n are relatively prime. Hence we can apply in order to find positive integers 0 < x, y < \Fp (since Fp 0 Q) such that pin2x2 — y2. Because pin2 + 1, we find that /342 + y2 and because 0 < x, y < \i -p, we conclude that we have in fact p = x2 + y2. The theorem is proved. It is time now to study some properties of the set P3. Because they are easier, we will discuss them in a single example. [Example 2.1 Let p E P3 and suppose that x and y are integers such that pjx2 + y2. Show that plgcd(x, y). Consequently, any number of the form n2 + 1 has only prime factors that belong to Pi or are equal to 2. Conclude that P 1 is infinite and then that P3 is infinite. Solution. Let us focus on the first question. Suppose that plgcd(x, y) is not true. Then, it is obvious that xy is not a multiple of p. Because plx2 + y2, we can write x2 —y 2 (mod p). Combining this with the observation that gcd(x,p) = gcd(y,p) = 1 and with Fermat's little theorem, we find that 1 xp—i _ (_1)Yyp-1 = 1)2k+1 = —1 (mod p) (for p = 4k + 3), which is impossible. This settles the first question. The second one follows clearly from the first one. Now, it remains to prove the third assertion. Proving that P3 is infinite is almost identical with the proof that there are infinitely many primes. Indeed, suppose that Doi7A 2,9u 9 l• • • )1371 are all the elements of P3 greater THEORY AND EXAMPLES 77 than 3 and consider the odd number N = 4p1p2 + 3. Because N 3 (mod 4), N must have a prime factor that belongs to P3. But since pi is not a divisor of N for any i = 1, 2, ..., n, the contradiction is reached and thus P3 is infinite. In the same manner we can prove that P 1 is infinite, but this time we must use the second question. Indeed, we consider this time the number M = (q1q2. qm)2 + 1, where qi, q2, , qm, are the elements of P 1 and then simply apply the result from the second question. The conclusion is clear. It is not difficult now to characterize the elements of the set Q2. A number is a sum of two squares if and only if any of its prime factors that also belongs to P3 appears at an even exponent in the decomposition of that number. The proof is just a consequence of the first example and we will not insist on anything more. Having presented some basic results that we will further use in this unit, it is time to see some applications that these two examples have. As a simple consequence of the first example, we consider the following problem, which is certainly easy for someone who knows Fermat's theorem regarding the ele-ments of P1 and difficult enough otherwise. Example 3. Find the number of integers x E { —1997, ... , 1997} for which 19971x2 + (x + 1)2. India 1998 Solution. We know that any quadratic congruence reduces to the congruence x2 a (mod p). So, let us proceed and reduce the given congruence to this special form. This is not difficult, since x2 + (x + 1)2 .= 0 (mod 1997) is equiv-alent to 2x2 + 2x +1 a - 0 (mod 1997), which in turn becomes (2x +1)2 +1 0 (mod 1997). Because 1997 E P 1, the congruence n2 —1 (mod 1997) has at least one solution. More precisely, there are exactly two solutions that belong to {1, 2, ... ,1996}, because if no is a solution, then so is 1997 — no and it is clear that this equation has at most two noncongruent solutions mod 1997. Because gcd(2, 1997) = 1, the function x 2x + 1 is a permu-tation of Z/1997Z, and so the initial congruence has exactly two solutions 78 4. PRIMES AND SQUARES with x E {1, 2, , 1996}. In a similar way, we find that there are exactly two solutions with x E { —1997, —1996, ... , —1}. Therefore there are exactly four numbers x E { —1997, ... , 1997} such that 19971x2 + (x + 1 ) 2 . We continue with a much trickier problem, proposed by Romania for the 1996 IMO. Even though it uses only the elementary facts about P3 proved before, this problem is fairly difficult: Let No denote the set of nonnegative integers. Is there a bijec-tive function f : No —> No such that for all nonnegative integers m, n we have f (3mn + m + n) = 4f (m) f (n) + f (m) + f (n)? IMO 1996 Shortlist Solution. The first step is to notice that one can change the given relation into ( (3m + 1)(3n + 1) — 1) (4f (m) + 1)(4f (n) + 1) — 1 3 4 This has the advantage that after introducing the function g : 3 • No + 1 —> 4 • No + 1, g(n) = 4f ( n -3 1 ) + 1, it becomes g(mn) = g (m)g(n), which is much easier than the initial relation. Because one can easily reconstruct f from g by f (n) -=- g(3n+ 41) 1 , the question becomes: is there a bijective multiplicative function g between 3•No +1 and 4•No +1, that is are the monoids 3.No +1 and 4 • No + 1 isomorphic? Let us introduce the analogous sets T1, T2 of positive primes of the form 3k + 1 and 3k + 2. In the same way as we proved that P1, P3 are infinite, you can prove that T 1, T2 are infinite. Because they are clearly countable, there exists a bijection between P 1 and T 1 and a bijection between P3 and T2. This gives us a bijection '0 between P1 U 8 and T1 U T2 which maps P1 onto T1 and P3 onto T2 bijectively. Now, it is not difficult to construct an isomorphism g: define g(1) = 1 and if n > 1 is in 3 • No + 1 write n = p1p2 • • pk for some prime numbers pi E T1 U T2 , not necessarily distinct and define g(n) = b(pi) 0(p2) • • • 0 (pk). We need to verify that g is well-defined, multiplicative and bijective. First of all, note that there is an even THEORY AND EXAMPLES 79 number of elements of T2 among pl, P2, -.1 pk. Then there is an even number of .elements of P3 among 111(pi) and thus g(n) E 4 • No + 1. Thus g is well-defined. Clearly g is multiplicative (by the definition itself) and, using the properties of zb it is immediate to verify that g is also bijective. This proves the existence of a function f with the desired properties. From a previous observation, we know that the condition that a number is a sum of two squares is quite restrictive. This suggests that the set Q2 is rather sparse. This conclusion can be translated into the following nice problem. Example 5. Prove that Q2 does not have bounded gaps, that is there are ar-bitrarily long sequences of consecutive integers, none of which can be written as the sum of two perfect squares. AMM Solution. The statement of the problem suggests using the Chinese Remain-der Theorem, but here the main idea is to use the complete characteriza-tion of the set Q2 we have just discussed: Q2 = E if pin and p E P3, then vp(n) E 2Z}. We know what we have to do. We will take long sequences of consecutive integers, each of them having a prime factor that belongs to P3 and has exponent 1. More precisely, we take different elements of P3, let them be pl,P2, , pn (we can take as many as we need, since P3 is infinite) and then we look for a solution to the system of congruences x — 1 (mod p?) x p2 — 2 (mod /A) x pn — n (mod pn 2 ) The existence of such a solution follows from the Chinese Remainder Theorem. Thus, the numbers x +1,x + 2, ... , x n cannot be written as the sum of two perfect squares, since + i, but g does not divide x i. Because n is as large as we want, the conclusion follows. 80 4. PRIMES AND SQUARES The Diophantine equation x(x +1)(x + 2) • • • (x + n) = yk has been extensively studied by many mathematicians and great results have been obtained by Erdos and Selfridge. But these results are very difficult to prove and we prefer to present a related problem, with a nice flavor of elementary mathematics. Example 6.] For any p in P3, prove that no set of p — 1 consecutive positive integers can be partitioned into two subsets, each having the same product of the elements. Solution. Let us suppose that the positive integers x + 1, x + 2, ... , x + p — 1 have been partitioned into two classes X, Y, each of them having the same product of the elements. If at least one of the p — 1 numbers is a multiple of p, then there must be another one divisible by p (since in this case both products of elements from X and Y must be multiples of p), which is clearly impossible. Thus, none of these numbers is a multiple of p, which means that the set of the remainders of these numbers when divided by p is exactly 1, 2, ... ,p — 1. Also, from the hypothesis it follows that there exists a positive integer n such that (x + 1)(x + 2) • • • (x + p — 1) = n2. Hence n2 1 • 2 • • • (p — 1) —1 (mod p), the last congruence following from Wilson's theorem. But from the second example we know that the congruence n2 —1 (mod p) is impossible for p E P3 and this is the needed contradiction. The results in the second example are useful tools in solving nonstandard Dio-phantine equations. You can see this in the following two examples. Prove that the equation x4 = y2 + z2 + 4 does not have integer solutions. [Reid Barton] Rookie Contest 1999 Solution. Practically, we have to show that x4 — 4 does not belong to Q2. Hence we need to find an element of P3 that has an odd exponent in the prime THEORY AND EXAMPLES 81 factorization of x4 — 4. The first case is when x is odd. Using the factorization x4 — 4 = (x2 — 2)(x2 + 2) and the observation that x2 + 2 = 3 (mod 4), we deduce that there exists p E P3 such that vp(x2 +2) is odd. But since p cannot divide x2 — 2 (otherwise pl x2 +2 — (x2 — 2), which is not the case), we conclude that vp(x4 — 4) is odd, and so x4 — 4 does not belong to Q2. We have thus shown that in any solution of the equation, x is even, let us say x = 2k. Then, we must also have 4k4 — 1 E Q2, which is clearly impossible since 4k4 — 1 3 (mod 4) and thus 4k4 — 1 has a prime factor that belongs to P3 and has odd exponent. Moreover, it is worth noting that the equation x2 + y2 = 4k + 3 can be solved directly, by working modulo 4. The following problem is much more difficult, but the basic idea is the same. Yet the details are not so obvious and, most importantly, it is not clear how to begin. [Example 8.1 Let p E P3 and suppose that x, y, z, t are integers such that x2P y2P Z2P = t2P. Prove that at least one of the numbers x, y, z,t is a multiple of p. [Barry Powel] AMM Solution. Without loss of generality, we may assume that x, y, z, t are rela-tively prime. Next, we prove that t is odd. Supposing the contrary, we obtain X2P + y2P Z2P = 0 (mod 4). Because a2 (mod 4) E {0, 1}, the latter implies that x, y, z are even, contradicting the assumption that gcd(x, y, z, t) = 1. Hence t is odd. This implies that at least one of the numbers x, y, z is odd. Suppose that z is odd. We write the equation in the form X2P t2P - Z2P (t2 - z 2) y2P = t 2 -z 2 and look for a prime q E P3 with an odd exponent in the decomposition of a factor that appears in the right-hand side. The best candidate for this factor seems to be , t2P - Z2P t2 z2 (t2)p-1 (t2)p-2z2 (z2)p-1 82 4. PRIMES AND SQUARES which is congruent to 3 (mod 4). This follows from the hypothesis p E P3 and the fact that a2 1 (mod 4) for any odd number a. Hence there is a t2p _ z2p q E P3 such that v q ( t2 — z2 ) is odd. Because x2P + y2P E Q2, it follows that vq(x2P + y2P) is even and so vq(t2 — z2) is odd. In particular, qlt2 — z2 and, because ql(t2)p-1 + (e)p-2z2 + ... + (z2)p-1 , we deduce that qlpt2(P-1). If q p, then qlt, hence qlz and also q1x2P + y2P. Because q E P3, we infer that qlgcd(x, y, z, t) = 1, which is clearly impossible. Therefore q = p and so plx2P + y2P. Because p E P3, we find that plx and ply. The conclusion follows. The previous results are used in the solution of the following problem. Even if the problem is formulated as a functional equation, we will immediately see that it is pure number theory mixed with some simple algebraic manipulations. I Example 9.1 Find the least nonnegative integer n for which there exists a nonconstant function f : Z —> [0, oo) with the following prop-erties: a) PxY) = f(x).f(Y); b) 2 f (x2 + y2) — f (x) — f (y) E {0,1, ... , n} for all x, y E Z. For this n, find all functions with the above properties. [Gabriel Dospinescu] Crux Mathematicorum Solution. First of all, we will prove that for n = 1 there are functions which satisfy a) and b). For any p E P3 define: _{0 if plx otherwise Using the properties of P 1 and P3, you can easily verify that fp satisfies the conditions of the problem. Hence fp is a solution for all p E P3. THEORY AND EXAMPLES 83 We will prove now that if f is nonconstant and satisfies the conditions in the problem, then n > 0. Suppose not. Then 2f (x2 + y2) = f (x) + f (y) and hence 2f(x2 + 02) 2f (x)2 = f (x) + f (0).. It is clear that we have f (0)2 = f(0). Because f is nonconstant, we must have f(0) = 0. Consequently, 2f(x)2 = 1 , f(x) for every integer x. But if there exists x such that f(x) = 2 — then 2f(x2)2 f(x 2), contradiction. Thus, f(x) = 0 for any integer x and f is constant, contradiction. So, n = 1 is the least number for which there are nonconstant functions which satisfy a) and b). We will now prove that any nonconstant function f which satisfies a) and b) must be of the form fp: or the function sending all nonzero integers to 1 and 0 to 0. We have already seen that f(0) = 0. Since 1(1)2 = f(1) and f is nonconstant, we must have f (1) = 1. Also, 2f (x)2 — f (x) = 2 f (x2 + 02) — f (x) — f (0) E {0,1} for every integer x. Thus f (x) E {0,1}. Because f (-1)2 = f (1) = 1 and f(-1) E [0, oo), we must have f (-1) = 1 and f (—x) = f (-1)f (x) = f (x) for any integer x. Then, since f (xy) = f (x) f (y), it suffices to find f (p) for any prime p. We prove that there is exactly one prime p for which f (p) = 0. Because f is nonconstant and f is not the function sending all nonzero integers to 1, there is a prime number p for which f (p) = 0. Suppose there is another prime q for which f (q) = 0. Then 2 f (p2 + q2) E {0, 1}, which means f(p2 q2) = 0. Then for any integers a and b we must have: 0 _ 2f(a2 b2)f(p2 q2) 2f ((ap + bq)2 + (aq — bp)2). Observe that 0 < f (x) + f (y) < 2f (x2 + y2) for any x and y, so we must have f (ap + bq) = f (aq — bp) = 0. But p and q are relatively prime, so there are integers a and b such that aq — by = 1. Then 1 = f(1) = f (aq — bp) = 0, a contradiction. So, there is exactly one prime p for which f (p) = 0. Let us suppose that p = 2. Then f (x) = 0 for any even x and 2f (x2 + y2) = 0 for any odd numbers x and y. This implies that f (x) = f (y) = 0 for any odd numbers x and y and thus f is constant, contradiction. Therefore p E Pl U P3. Suppose p E P1. According example 1, there are positive integers a and b such that p = a2 + b2. Then we must have f (a) = f (b) = 0. But max{a, b} > 1 and 84 4. PRIMES AND SQUARES there is a prime number q such that ql max{a, b} and f (q) = 0 (otherwise, we would have f(max{a, b}) = 1). But it is clear that q < p and thus we have found two distinct primes p and q such that f (p) = f (q) = 0, which, as we have already seen, is impossible. Consequently, p E P3 and we have f (x) = 0 for any x divisible by p and f(x) = 1 for any x which is not divisible by p. Hence, f must be fp and the conclusion follows. We end this chapter with two beautiful problems concerning properties of prime numbers of the form 4k + 1 or 4k + 3. We saw that Q2 does not have bounded gaps. In fact, much more is true. We will show that Q2 has den-sity zero. Define the density of a set of positive integers Pi as the limit (if it exists) of the sequence P1 x(x) where P1 (x) is the counting function of the set P 1, that is Pi(x) = E 1. Before proving that Q2 has density zero, we want to prove a jewel of mathematics, the first step in analytic number theory: Example 1071 The sets P1 and P3 have Dirichlet density 1, that is lim s—>1 In 5-11 and similarly for P3. 1 ps 2 pEPi 1 [Dirichlet] (71) Solution. Let us consider s > 1 and L(s) = E Ans , where A(n) = 0 if n n>1 is even and A(n) = (-1) X 2 1 otherwise. It is clear that A(n) • A(m) = A(mn). Using this, it is not difficult to see that p2 s ) L(s) H (1+ A p +A p2 + _f 1—la(p)' 1 1 A(p). s p) P ps L(s) = H (1+ A p (P) + P A(':s) s 1 A(P) • p Ps 1 1 1 L(s) =1— + — + • • • > 0, 75 (4.1) THEORY AND EXAMPLES 85 Indeed, let us define P(x) = + + , A(P2) + • .•). It is a finite product where P1(x) is the set of positive integers having all prime divisors not ex- ceeding x. Thus the difference between the sum of the absolutely convergent A(n) A( ) series E — ns and P(x) is just the sum of A(n) taken over the set of all positive 8 n> 1 integers that have at least one prime divisor greater than x, thus it is certainly bounded in absolute value by E b. Because this converges to 0 for x oo, n>x it follows that P(x) converges to L(s) for x —> oo, so we have p<x of absolutely convergent series, so we can write P(x) = E A(n) ns nEP1(x) Now, observe that so we can take logarithms in both sides of (4.1) in order to obtain In L(s) = p In (1 A(P) I. Ps Finally, observe that there exists a constant w such that I —1n(1— x)—x I < Cx2 (1 for all 0 < x < . Indeed, the function —ln x)— x is continuous on [0, 2 1] , so x it is bounded. Therefore In L(s) — E A(p) P P 86 4. PRIMES AND SQUARES Now, let us prove that ln L(s) is bounded for s —> 1. Indeed, from 1 L(s) = (1 — 1 8) + — + " = 1 ( 1 79 5 — • it follows that for s > 1 we have ln L(s) E (ln 0). With exactly the same arguments (applied this time for the function 0(n) = 1 for odd n and 0(n) = 0 I,b for even n and Li(s) = 2 (n) ), we can prove that n>1 ln is bounded for s 1. However, it is clear that E _ ns =( 1_ s) • ((s), where c(s) = E s is the famous Riemann's function. Because ln L(s) is n>1 bounded, it follows from a previous inequality that > A(P) is also bounded p>2 near 1. Finally, we deduce from these observations that _ 1 = 0(1) ps ps pEP1 pEP3 and - + - = ln(1 — 2') + In ((s) + 0(1) PEPi pEP3 for s —f 1. A simple integral estimation shows that ln(1-2-8)+1n ((s) ln s l i for s —> 1, which finishes the proof of this beautiful theorem. Now, let us see why the set Q2 has zero density. The proof of this result will surely look very complicated. Actually, it is a motivation to give some other very useful results connected to this problem. First of all, let us start with nE2N+1 THEORY AND EXAMPLES 87 Theorem 4.2. Let P be a set of prime numbers. The set of positive integers n divisible by some prime p E P has density 1 if n (1 — = O. qEP Proof. The proof of this result is quite simple, even though in order to make it rigorous we need some technical details. It is clear that P is infinite, so let p1 < P2 < ••• be its elements. Let E be the set of the numbers n divisible by some prime p E P and let X be the set of positive integers n that are not divisible by any element of P. Also, let f(x,y) be the cardinal of the set of those numbers not exceeding x and which are relatively prime to fl q. Using qEP,q<y the Inclusion-Exclusion Principle and the fact that the number of multiples of pi1pi2...pis not exceeding x differs by at most 1 from we deduce 13 1. 1j 2 •• •.Ps that f (x, y) = x • H (1_1+0(2Y) 4E P,q<y (because in the sum appearing in the Inclusion-Exclusion Principle there are 2Y terms of the form + 0(1)). Now, by choosing y = ln x we deduce Ps Ps2 ' • • • 'Ps, f (x, ln x) = x • (1 - ) 0 (xln 2). qEP,q<ln x Because the counting function of X satisfies R(x) < f(x,y) for all x, y and because limx_,,„, - 1) = 0, it follows that R(x) = o(x), that is X qEP,q<ln x has zero density. It is clear then that E has density 1. 0 Now, using the previous theorem due to Dirichlet, we can easily establish that E 1 = 00. Because In (1 — ) _ 0 (y ) it easily follows that pEP3 H (1_1) = 0. By the previous theorem, it follows that the set of integers that pEP3 divisible by at least an element of P3 has density 1. Now, let P3(x) be the counting function of the set of positive integers that are not divisible by 4 or 88 4. PRIMES AND SQUARES by any element of P3. They are the only integers that are sums of two coprime squares. Also, we have proved that P3(x) = o(x). It is also clear that if Sq(x) is the counting function of the set of positive integers that are sums of two squares, then Sq(x) < E P3 UT) . Now, for N an arbitrary positive integer, observe that E P3 (;) < P3(N) 3 z <N because P3 UT) < P3(N) for these j and the sum has at most •VX nonzero terms. On the other hand, E 8 (-.-2 -) < supp3(t) x < 3x • sup P4(t) >N 3 t>N t 3>1 — t>N t f, 2 Everything should be clear now: for c > 0 choose N such that supt›N P3t(t) < 6 4B(N)2 Then for x > we have Sq(x) < cx, which means that Sq(x) = x. PROBLEMS FOR TRAINING 89 4.2 Problems for training 1. Prove that a positive integer can be written as the sum of two perfect squares if and only if it can be written as the sum of the squares of two rational numbers. Euler 2. a) Prove that for any real number x and any nonnegative integer N one can find integers p and q such that I qx — PI < N + 1 b) Suppose that a is a divisor of a number of the form n2 + 1. Prove that a E Q2. 3. Prove that the equation 3k = m2 + n2 + 1 has infinitely many solutions in positive integers. Saint-Petersburg Olympiad 4. Prove that each p E P 1 can be represented in exactly one way as the sum of the squares of two integers, up to the order of the terms and signs of the terms. 5. Find all positive integers n for which the equation n = x2 + y2, with 0 < x < y and gcd(x, y) = 1 has exactly one solution. 6. Prove that the number 4mn — m — n cannot be a perfect square if m and n are positive integers. IMO 1984 Shortlist 7. The positive integers a, b have the property that the numbers 15a + 16b and 16a — 15b are both perfect squares. What is the least possible value that can be taken by the smallest of the two squares? IMO 1996 90 4. PRIMES AND SQUARES 8. Find all pairs (m, n) of positive integers such that m2 — 13m + (n! — 1)m • Gabriel Dospinescu x2+2 is 9. Find all pairs (x, y) of positive integers such that the number x — y a divisor of 1995. Bulgaria 1995 10. Find all n-tuples (al, a2, , an) of positive integers such that (al! — 1)(a2! — 1) ... (an! — 1) — 16 is a perfect square. Gabriel Dospinescu 11. Prove that there are infinitely many pairs of consecutive numbers, no two of which have any prime factor that belongs to P3. 12. Ivan and Peter alternately write down 0 or 1 until each of them has written 2001 digits. Peter is a winner if the number, whose binary rep-resentation has been obtained, cannot be expressed as the sum of two perfect squares. Prove that Peter has a winning strategy whenever Ivan starts. Bulgaria 2001 13. Prove that the equation y2 = x5 — 4 has no integer solutions. Balkan Olympiad 1998 PROBLEMS FOR TRAINING 91 14. It is a long standing conjecture of Erdos that the equation '1 = l x - + l y + 1 has solutions in positive integers for all positive integers M. Prove that the set of those n for which this statement is true has density 1. 15. Let T the set of the positive integers n for which the equation n2 = a2+b2 has solutions in positive integers. Prove that T has density 1. Moshe Laub, AMM 6583 16. Find all positive integers n such that the number 2' — 1 has a multiple of the form m2 + 9. IMO 1999 Shortlist 17. Prove that the set of odd perfect numbers (that is for which a(n) = 2n, where a(n) is the sum of the positive divisors of n) has zero density. 18. Prove that the equation x8 + 1 = n! has only finitely many solutions in nonnegative integers. 19. Find all functions f : Z± —> Z with the properties: 1. f(a) > f(b) whenever a divides b. 2. for all positive integers a and b, f (ab) + f (a2 + b2) = f(a) + f (b) . Gabriel Dospinescu, Mathlinks Contest 20. Let Lo = 2, L1 = 1 and Ln+2 = Ln±i + Lr, be the famous Lucas's sequence. Then the only n > 1 for which L7-, is a perfect square is n = 3. Cohn's theorem THEORY AND EXAMPLES 95 5.1 Theory and examples T2 's lemma is clearly a direct application of the Cauchy-Schwarz inequality. Some will say that it is actually the Cauchy-Schwarz inequality and they are not wrong. Anyway, this particular lemma has become very popular among the American students who attended the training of the USA IMO team. This happened after a lecture delivered by the first author at the Mathematical Olympiad Summer Program (MOSP) held at Georgetown University in June, 2001. But what exactly does this lemma say? It says that for any real numbers al, a2, , an and any positive real numbers xl, xz, , xn, the inequality ,2 ,2 a2 (al + az + • • • + an)2 — + + • • • + n > X1 X2 Xn X1 ± X2 ' • • ± Xn holds. And now we see why calling it also the Cauchy-Schwarz inequality is natural, since it is practically an equivalent form of this inequality: (a? + 4 + a2 + (X1 + X2 + + Xn) X1 X2 Xn X1 • + 2 22 a2 + • • • + a, fxTi, X2 Xn But there is another nice proof of (5.1), by induction. The inductive step is reduced practically to the case n = 2, which is immediate. Indeed, it boils al down to (alx2 — a2x1)2 > 0 and the equality occurs if and only if — = az x1 x2 Applying this result twice it follows that 2 2 ai (al + a2 + a3)2 a2 a 3 2 > (al + a2)2 + > X1 X2 X3 X1 + X2 X3 xl + X2 + X3 96 5. T2'S LEMMA and we see that a simple inductive argument finishes the proof. With this brief introduction, let us discuss some problems. And there are plenty of them given in mathematical contests or proposed in mathematical magazines! First, an old problem, that became classical. We will see that with T 2's lemma it becomes straightforward and even more, we will obtain a refinement of the inequality. Prove that for any positive real numbers a, b, c a3 b3 c3 a + b + c a2 + ab + b2 + b2 + bc + c2 + c2 + ca + a2 > 3 Tournament of the Towns, 1998 Solution. We will change the left-hand side of the inequality so that we could apply T2's lemma. This is not difficult: we just have to write it in the form a4 b4 a(a2 + ab + b2) b(b2 + be + c2) c(c2 + ca + a2) It follows that the left-hand side is greater than or equal to (a2 b2 c2)2 a3 + b3 + c 3 + ab(a + b) + bc(b + c) + ca(c + a) But we can easily observe that a3 + b3 + c 3 + ab(a + b) + bc(b + c) + ca(c + a) = (a + b + c)(a2 + b2 + c2) , so we have proved an even stronger inequality, that is a3 b3 c3 a2 b2 e2 a2 + ab + b2 b 2 + bc + c2 c 2 + ca + a2 — a + b + c • The second example also became representative for a whole class of problems. There are countless examples of this type in numerous contests and mathe-matical magazines, so we find it necessary to discuss it at this point. THEORY AND EXAMPLES 97 Example 2 1 For arbitrary positive real numbers a, b, c, d prove the inequal-ity 2 b+2c a +3d c+2d+3a d+2a+3b+ a+2b+3c— > 3 [Titu Andreescu] IMO 1993 Shortlist Solution. If we write the left-hand side in the form a2 b2 c2 d2 a(b + 2c + 3d) + b(c + 2d + 3a) + c(d + 2a + 3b) d(a + 2b + 3c)' then the way to continue is clear, since from the lemma we obtain a b+2c+3d + c+2d+3a + d+2a+3b + a+2b+3c (a+b+c+d)2 4(ab+bc+cd+da+ac+bd). Hence it suffices to prove the inequality 3(a+b+c+d)2 >8(ab+bc+cd+da+ac+bd). But it is not difficult to see that (a+b+c+ a)2 a2 b2 e2 + a +2(ab+bc+cd+da+ac+bd), implies 8(ab+bc+cd+da+ac+bd)=4(a+b+c+d)2 4(a2 b2 c2 a2). Consequently, we are left with the inequality 4(a2 b2 e2 a2) > (a+b+c+d)2, which is just the Cauchy-Schwarz inequality for four variables. The problem below, given at the IMO 1995, was discussed extensively in many publications. It could be also solved by using the above lemma. 98 5. T2'S LEMMA Let a, b, c be positive real numbers such that abc = 1. Prove that 1 1 1 3 a3(b + c) + b3(c + a) + c3(a + b) > 2. Solution. We have: 1 1 1 1 1 1 _ a2 4_ b2 + e2 a3 (b + c) + b3 (c + a) + c3 (a + b) a(b + c) b(c + a) c(c + a) ( + + 1 1 1) = 2 > a b c) (ab + bc + ca)2 ab + bc + ca > 3 -- — 2(ab + be + ca) 2(ab + be + ca) 2 — 2' the last inequality following from the AM-GM inequality. The following problem is also not difficult, but it uses a nice combination of this lemma and the Power-Mean inequality. It is another example in which proving the intermediate inequality (that is, the inequality that remains to be proved after using the lemma) is not difficult. Let n > 2. Find the minimal value of the expression xi x2 + x3 + • • • + xn xl + x3 + • • • + xn x5 +...+ Xi ± X2 + • • ' Xn-1 7 where xl, x2, , x7, are positive real numbers satisfying ± 4 ± • • • + Xn 2 = 1. Turkey, 1997 THEORY AND EXAMPLES 99 Solution. Usually, in such problems the minimal value is attained when the variables are equal. So, we conjecture that the minimal value is 1 attained when x1 = x2 = • • • = xn = . Indeed, by using the lemma, it n follows that the left-hand side is greater than or equal to 2 Ex? Exi(xl +•••+xi_i+xi+,.+•••+ Xn) i=1 But it is not difficult to observe that n (n 2 Exi(x,+•••+xi_i+xi+,.+•••+ xn) = E xi) -1. So, proving that 5 X5 1 X5 2 + • + Xn X2 + X3 + • • • + Xn + X3 + • • • + Xn Xi + X2 + " • + Xn-1 1 n(n — 1) reduces to proving the inequality 2 T E Xi > i=1 2 n Xi) — 1 i=1 n(n — 1) But this is a simple consequence of the Power-Mean inequality. Indeed, we have n(n — 1) 100 5. T2'S LEMMA implying E 3 xi > ,Fri and E x2 < i=i i=1 The conclusion follows. In 1954, H.S.Shapiro asked whether the following inequality is true for any positive real numbers at, a2, , an: at a2 > a2 + a3 a3 + ++ al + a2 — 2 The question turned out to be extremely difficult. The answer is really unex-pected: the inequality holds for all odd integers smaller than or equal to 23 and all even integers smaller than or equal to 12, but fails for all the others. Let us examine the case it = 5, a problem proposed for MOSP 2001. Prove that for any positive real numbers al, a2, a3, a4, a5, al a2 a3 a4 a5 5 > a2 + a3 a3 + a4 a4 + a5 a5 + al al + a2 2 Solution. Again, we apply the lemma and we conclude that it suffices to prove the inequality (al + a2 + a3 + a4 + a5)2 2 [ai(a2 + a3) + a2(a3 + a4) + a3(a4 + a5) + a4(a5 + at) + a5(ai + a2)] Let us denote al + a2 + a3 + a4 + a5 = S. Then we observe that al (a2 + a3) + a2(a3 + a4) + a3(a4 + a5) + (a5 + al) + a5(ai + a2) (S — ) + a2(S — a2) + a3(S — a3) + azi(S — a4) + a5(S a5) 2 S2 — - — a3 a5 2 THEORY AND EXAMPLES 101 With this identity, we infer that the intermediate inequality is in fact _ 5 (s 2 _ a2 i _ a2 2 _ a3 2 _ a4 2 _ a5 2 ) , (al + a2 + a3 + a4 + a5)2 4 equivalent to 5(4 + 4 + 4 + ce, + 4) > 82, which is nothing else then the Cauchy-Schwarz inequality. Another question arises: is there a positive real number such that for any positive real numbers al, a2, ... , a, and any n > 3 the following inequality holds: al a2 a, + + • • • + > cn. a2 + a3 a3 + a4 al + a2 This time, the answer is positive, but finding the best such constant is an extremely difficult task. It was first solved by Drinfield (who, by the way, is a Fields' medalist). The answer is quite complicated and we will not discuss it here (for a detailed presentation of Drinfield's method the interested reader can consult the written examination given at ENS in 1997). The following problem, given at the Moldavian TST in 2005, shows that c = -\/ — 1 is such a constant (not optimal). The optimal constant is quite complicated, but an approximation is 0.49456682. For any al, a2, ... , an, and any n > 3 the following inequality holds: al a2 a, + + + > (N/2 — 1)n. a2 + a3 a3 + al al + a2 The proof is completely elementary, yet very difficult to find. An ingenious argument using, the arithmetic-geometric means inequality does the job: let us write the inequality in the form al + a2 + a3 a2 + a3 + a4 an + al + a2 + + • • + > .N,/ • n. a2 + a3 a3 + al al + a2 Now, using the AM-GM inequality, we see that it suffices to prove the stronger inequality: al + a2 + a3 a2 + a3 + azi an + al + a2 ? (An. a2 + a3 a3 + a4 al + a2 102 5. T2'S LEMMA Observe that , ad-H. _ r_ , ai+2)2 \ 2 (Cti ad±i + ai+2) = (ai --r 2 + 2 _> 4 (ai + a21) (a21 21 + ai+2) (the last inequality being another consequence of the AM-GM inequality). Thus, Thai + ai+1 + ai+2)2 > 11(2ai + ai+i) 11(2ai+2 + adH- d=1 d=i d=1 Now, the real trick is to rewrite the last products appropriately. Let us observe that 11(2ai+2 ad+i) = H(2ad-ki + ad), d=1. d=i SO H(2ai + ai+i) fl(2ai+2 + ai+1) = II[(2ai aid-1)(ai + 2ad-14)] d=i d=i > H(2(ai + ai+1)2) = 2n Bad + d=i d=i ) 2 The conclusion now follows. This lemma came handy even at the IMO 2005 (problem 3). In order to prove that for any positive real numbers x, y, z such that xyz > 1 the following inequality holds + y2 + z2 <3 L-s x5 + y2 + z2 a few students successfully used the above mentioned lemma. For example, a student from Ireland applied this result and called it "SQ Lemma". During the THEORY AND EXAMPLES 103 coordination, the Irish deputy leader explained what "SQ" stood for: "...escu". A typical solution using this lemma is as follows: x4 y4 z4 (x2 + y2 ± z2)2 X5 + y 2 + Z2 = + +—> 1/x y2 z 2 1 4. y2 + z2 X hence — 1 +y 2 +z 2 x2 + y2 + Z2 < \--"' X —2+ xy + yz + zx 5 + y2 + z2 — 2 + y2 + z2 + y2 + z2) < 3. x xyz(x2 It is now time for the champions. We begin with a difficult geometric inequal-ity for which we have found a direct solution using T2's lemma. Here it is. Example 6. Let ma, mb, mc, ra, rb, r, be the lengths of the medians and the radii of the circumscribed circles in a triangle ABC. Prove that the following inequality holds rarb rbr, rcra > 3. mamb MbMc McMa [Ji Chen] Crux Mathematicorum Solution. Of course, we start by translating the inequality into an algebraic one. Fortunately, this is not difficult, since using Heron's relation and the formulas K -V2b2 + 2c2 — a2 ra = s—a, ma= 2 and the like, the desired inequality takes the equivalent form (a+b+c)(b+c—a) (a+b+c)(c+a—b) \/2a2 + 2b2 — c2 • -V2a2 + 2c2 — b2 V2b 2 + 2a2 — c2 • \/2b2 + 2c2 — a2 (a+b+c)(a+b—c) + V2c2 + 2b2 — a2 • V2c2 + 2a2 — b2 > 3. 104 5. T2'S LEMMA In this form, the inequality is more than monstrous, so we try to see if a simpler form holds, by applying the AM-GM inequality to each denominator. So, let us try to prove the stronger inequality 2(a + b + c)(c + b — a) 2(a + b + c) (c + a — b) 4a2 ± b2 + c2 + 4b2 + c2 ± a2 2(a + b + c) (a + b — c) > 3. c+b—a c+a—b a+b—c > 3 4a2 + b2 + c2 + 4b2 + c 2 + a2 + 4c2 + a2 + b2 2(a + b + c) we see that by T2's lemma the left-hand side is at least (a + b ± c)2 (b+c—a)(4a2 -I- b2 -I- c2) ± (c + a — b)(4b2 ± a2 ± c 2) + (a ± b — c)(4c2 ± a2 + b2) • Basic computations show that the denominator of the last expression is equal to 4a2 (b + c) + 4b2 (c + a) + 4c2 (a + b) — 2 (a3 + b 3 + c 3) and consequently the intermediate inequality reduces to the simpler form 3(a3 + b 3 + c 3) + (a + b + c)3 > 6[a2 (b + c) + b2 (c + a) + c2(a + b)]. Again, we expand (a + b + c)3 and obtain the equivalent inequality 4(a3 + b3 + c 3) + 6abc > 3 [a2 (b + c) + b2 (c + a) + c2(a + b)] , which is not difficult at all. Indeed, it follows from the inequalities 4(a3 + b3 + c 3) ? 4[a2(b + c) + b2(c + a) + c2(a + b)] — 12abc and + 4c2 + a2 + b2 — Written in the more appropriate form a2(b + c) + b2(c + a) + c2(a + b) > 6abc. THEORY AND EXAMPLES 105 The first one is just an equivalent form of Schur's inequality, while the second follows immediately from the identity a2 (b + c) + b2 (c + a) + c2 (a + b) — 6abc = a(b — c)2 + b(c — a)2 + c(a — b)2. Finally, we have managed to prove the intermediate inequality, and hence the problem is solved. The journey continues with a very difficult problem, given at the Japanese Mathematical Olympiad in 1997, and which became infamous due to its dif-ficulty. We will present two solutions for this inequality. The first one uses a nice combination between the T2 lemma and the substitution discussed in the unit "Two useful substitutions". Example 7. Prove that for any positive real numbers a, b, c the following inequality holds (b + c — a)2 (c + a — b)2 (a + b — c)2 3 + c2 + (a + b)2> 5' a2 + (b + b2 (C a)2 Japan 1997 Solution. Of course, from the introduction to this problem, the reader has already noticed that it is useless to try a direct application of the lemma, since any such approach is doomed. But with the substitution b + c c + a a + b x= y= z a b c we have to prove that for any positive real numbers x, y, z satisfying xyz = x + y + z + 2, the inequality (x — 1)2 (y — 1)2 (z — 1)2 3 — x2 + 1 y2 + 1 z2 + 1 > — 5 holds. It is now time to use T2's lemma in the form (x — 1)2 (y — 1)2 (z — 1)2 (x y + z — 3)2 x2 + 1 y2 + 1 z2 + 1 x2 + y2 + z2 + 3' 106 5. T2'S LEMMA Hence it is enough to prove the inequality (x + y + z — 3)2 3 x2 ± y2 + z2 > 3 — 5 But this is equivalent to (x+y+z)2 — 15(x+y+z)+3(xy+yz+zx)+18 >O. This is not an easy inequality. We will use the proposed problem 6 from the chapter Two Useful Substitutions to reduce the above inequality to the form (x + y + z)2 — 9(x + y + z) + 18 > 0, which follows from the inequality x + y + z > 6. And the problem is solved. But here is another original solution. Alternative solution. Let us apply T 2's lemma in the following form: (b + c — a)2 (c + a — b)2 (a + b — c)2 a2 + (b + c)2 b 2 + (c + a)2 c 2 + (a + b)2 — ((b +c)2 — a(b c))2 ((c a)2 b(c + a))2 + ((a + b)2 ca + b)) 2 a2(b + c)2 + ± b2(c c)2 (c a)4 c2 (a + + (a ± b)4 4(a2 + b2 + c 2)2 a2 (b + c)2 + b2(c + a)2 + c 2(a + b)2 + (a + b)4 + (b + c)4 + (c + a)4 Consequently, it suffices to prove that the last quantity is greater than or equal to — 3. This can be done by expanding everything, but here is an elegant proof 5 using the observation that a2(b + c)2 + b2 (c + a)2 = [(a + b)2 + (b + c)2 + (c + a)2] (a2 + b2 + c 2) +2ab(a + b)2 + 2bc(b + c)2 + 2ca(c + a)2. Because c2 (a ± b)2 ± (a + + (c a)4 (a+b)2 + (b+c)2 +(c+a)2 < 4(a2 +b2+c2), THEORY AND EXAMPLES 107 we observe that the desired inequality reduces to 2ab(a + b)2 + 2bc(b + c)2 + 2ca(c + a)2 < 3(a2 b2 c2)2. But this inequality is not so difficult. Indeed, first we observe that 2ab(a + b)2 + 2bc(b + c)2 + 2ca(c + a)2 < 4ab(a2 + b2) + 4bc(b2 + c2) 4ca(c2 a 2). Then, we also find that 4ab(a2 b2) < a4 + b4 + 6a2b2, since (a — b)4 > 0. Hence 4ab(a2 + b2) + 4bc(b2 c2 4ca(c2 + a2) < 2(a2 b2 c2)2 +2(a2b2 + b2c2+ czaz) < _ 8 (az + b2 c2)2 3 and so the problem is solved. With minor changes, we can readily see that this solution even works without the assumption that a, b, c are positive. We end this discussion (which remains probably permanently open) with a series of more difficult problems, based on less obvious applications of T2 's lemma. Let al, a2, , an > 0 such that al + a2 + • • • + an = 1. Prove that: (aia2 + • • • + anal.) an a2 2 + a2 ciT + al) n+1• [Gabriel Dospinescu] +"'+ an an + — al ( an ) 2 al 108 5. T2'S LEMMA Solution. How can we get to aia2 + a2a3 + • • + anal? Probably from a 2 2 1 a2 + + an a2 a2a3 anal after we use the lemma. So, let us try the following estimation: al a2 an = a 2 1 a2 2 a2 a3 al aia2 a2a3 + • • + anal The new problem, proving that 1 aia2 + a2a3 + • • • + anal al a2 an n al a2 an + + + + al > n + 1 a2 + 3 + • • • ± — ,-, 2 _i_ ,-, a2 1 a2 + a9 al 2 ,., , v3 1 w3 seems even more difficult, but we will see that we have to make one more step in order to solve it. Again, we look at the right-hand side and we write al a2 an — + — + • • • + — as a2 a3 al (— al az an a2 a3 + — + • • • + — al al az ,+ an — + — • • • + — a2 a3 al After applying T2's lemma, we find that al a2 9 a22 + + a2 a5 + (al) 2 a2 2 an a2 a3 a3 al + 2 al + a2 + al al a2 a2 a3 1' al an — + -2 C +...+— az a3 al al a2 an, • 1 — a2 a3 al 2 al an, We are left with an easy problem: if t = — + • • • + — , then t nt a2 al 1 + t n + 1' or t > n. But this follows immediately from the AM-GM inequality. THEORY AND EXAMPLES 109 Example 9.] Prove that for any positive real numbers a, b, c the following inequality holds (a + b)2 (b + c)2 (c + a)2 c2 + ab a2 + be b2 + ca > 6. [Darij Grinberg, Peter Scholze] Solution. We do not hide from you that things become really complicated here. However, let us try to use T2's lemma again, but of course not in a direct form, since that one is doomed. Trying to make the numerators as strong as possible, we may first try the choice (a + b)4. And so, we know that the left hand side is at least (E (a + b)2)2 E (a + b)2(c2 + ab) • So, we should see whether the inequality (E (a + b)2) 2 > 6 E (a + b)2(c2 + ab) holds. However, this is not easy, at least not without computations. With some courage, we can develop everything and reach the equivalent inequality 2(a4 b4 + c4) c4\ + ab(a2 + b2) + bc(b2 + c 2) + ca(c2 + a2)+ 2abc(a + b + c) > 6(a2b2 + by + c2a2). Fortunately, this can be broken into pieces: because bc(b2 + c 2) > 2b2c2, it is enough to prove that a4 b4 4 C abc(a + b + c) > 2(a2b2 b2c2 c2a2). Now, if you know Heron's formula for the area of a triangle, 2(a2b2 + b2c2 + c 2a2) — (a4 + b4 + c4) 110 5. T2'S LEMMA should ring a bell! It is actually equal to (a+ b+ c) (a+ b c)(b + c— a)(c+a— b). So, we are left with the classical inequality (a + b — c)(b + c a)(c + a — b) < abc. If one of a + b — c, b + c — a, c + a — b is negative, we are done. Otherwise, observe that a= (a + b c) + (c + a — b) > 2V(a + b c)(c + a — b). Multiplying this and two similar inequalities easily yields the conclusion. Do you like inequalities that can be solved with identities? Here is one which combines T2's lemma with a very strange identity. Do not worry, things like that do not appear too often. Fortunately... 1111111111V Prove that if a, b, c, d > 0 satisfy abc+bcd+cda+dab = a+b+c+d, then \/a2 + 1 + \/b2 + 1 + Jc2 + 1 + jd2 + 1 <a+b+c+d. 2 2 V 2 V 2 — [Gabriel Dospinescu] Solution. The following solution is very difficult to find, but it is the only one that the authors have. The idea is to apply T2 's lemma to an identity which is almost impossible to find. We will prove that a2 + 1 b 2 + 1 C2 + 1 d 2 + 1 =a+b+c+d a+b b+c c+d d+ a and after that T2 's lemma will do the rest. To prove the identity, just observe that (a+b)(a+c)(a+d) = a2(a+b+c+d)+abc+bcd+cda+dab = (a2+1)(a+b+c+d). Use similar identities and add them up. PROBLEMS FOR TRAINING 111 5.2 Problems for training 1. Let a, b, c, d be positive real numbers with a+ b + c + d = 1. Prove that a2 b2 c2 d2 1 a+b + b+c + c+d + d+a > — 2 Ireland 1999 2. Let a, b, c, be positive real numbers with a2 b 2 c 2 3abc. Prove that a c 9 + > b2c2 c2a2 a2b2 — a+ b+ c India 3. Let xi, x2, , xn, y1, Y2, • • • , yn, be positive real numbers such that Xi + X2 + • • • ± Xn > X1Y1 X2Y2 ± • • • ± XnYn• Prove that Xi X2 Xn Xi ± X2 + • ' Xn <—+—+...+— . Y1 Y2 Yn Romeo Ilie, Romania 1999 4. For arbitrary positive real numbers a, b, c prove the inequality a b c ± + > 1. b + 2c c + 2a a + 2b — Czech-Slovak Competition 1999 5. Prove that for all positive real numbers a, b, c satisfying a + b + c = 1, the following inequality holds a b c 9 > 1 + bc 1 + ca 1 + ab — 10 112 5. T2'S LEMMA 6. Prove that for any positive real numbers a, b, c, d satisfying ab + bc+ cd+ da = 1 the following inequality is true a3 b3 c3 d3 1 + + + > b+c+d c+d+a d+a+b a+b+c — 3 . IMO 1990 Shortlist 7. Prove that if the positive real numbers a, b, c satisfy abc = 1, then a b+c+1 c+a+1 a+b+1— >1. Vasile Cirtoaje, Gazeta Matematica 8. Prove that for any positive real numbers a, b, c, a2 bc + + C2 + ab > a + b+c. b+c c+a a+b Cristinel Mortici, Gazeta Matematica 9. Prove that for any nonnegative real numbers xi, x2, xn, Xi X2 xn > 2. xn + X2 Xi + X3 xn-1 + Xi Tournament of the Towns 1982 10. Prove that for any positive real numbers a, b, c the following inequality holds ( a 2 b 2 2 c 3 a2 +b2 ± C2 +c) c+a,) a+b) 4 ab+bc+ca. Gabriel Dospinescu PROBLEMS FOR TRAINING 113 11. Prove that for any positive real numbers a, b, c, d, e satisfying abcde = 1, a+ abc b + bcd c + cde 1 + ab + abcd + 1 + be + bcde + 1 + cd + cdea d + dea e + eab 10 +1 + de + deab + 1 + ea + eabc — 3 • Waldemar Pompe, Crux Mathematicorum 12. Let n > 4 an integer and let al, a2, , an be positive real numbers such that aT + a2 + • • • + an 2 = 1. Prove that al a2 an e 4 a2 +1 + c 3 +1 + +a2+1 > (alfdT + a2V,2 + • • • + anV(T)2. Mircea Becheanu, Bogdan Enescu, Romanian TST 2002 13. Determine the best constant kn such that for all positive real numbers al, a2, , an satisfying aia2... an = 1 the following inequality holds aia2 a2a3 anal (a7 + a2)(4 + al) + a3)(a3 + a2) + ai)(a? + a2) < kn. Gabriel Dospinescu, Mircea Lascu 14. Prove that for any positive real numbers a, b, c, (2a + b + c)2 (2b + c+ a)2 (2c+ a + b)2 2a2 + + 2b2 (c (02 2e2 + (a + b)2 < 8. Titu Andreescu and Zuming Feng, USAMO 2003 15. Let n > 13 be a positive integer and suppose that the positive numbers al, a2, ..., an satisfy the relations al ±a2+ • • • +an = 1 and al +2a2+ • • • + nan = 2. Prove that (a2—ai)V2+ (a3— a2)0+ • • .± (an < 0. Gabriel Dospinescu THEORY AND EXAMPLES 117 6.1 Theory and examples You have already seen quite a few strategies and ideas, and you might say: "Enough with these tricks! When will we go to serious facts?" We will try to convince you that the following results are more than simple tools or tricks. They help to create a good base, which is absolutely indispensable for someone who enjoys mathematics, and moreover, they are the first steps to some really beautiful and difficult theorems or problems. And you must admit that the last problems discussed in the previous units are quite serious facts. It is worth mentioning that these strategies are not a panacea. This assertion is proved by the fact that every year problems that are based on well-known tricks prove to be very difficult in contests. We will "disappoint" you again in this unit by focusing on a very familiar theme: graphs without complete subgraphs. Why do we say familiar? Because there are hundreds of problems proposed in different mathematics competi- tions around the world and in professional journals that deal with this subject. And each such problem seems to add something. Before passing to the first problem, we will assume that the basic knowledge about graphs is known and we will denote by d(V) and C(V) the number, and the set of vertices adjacent to V, respectively. Also, we will say that a graph has a complete k—subgraph if there are k vertices any two of which are connected. For simplicity, we will say that G is k-free if it does not contain a complete k—subgraph. First we will discuss one famous classical result about k-free graphs, namely Tu- ran's theorem. Before that, though, we prove a useful lemma, also known as Zarankiewicz's lemma, which is the main step in the proof of Turan's theorem. If G is a k-free graph, then there exists a vertex having degree k — at most L k — 1 2 n l . [Zarankiewicz] Also, > (1 +1 k — 1 k — 2 ni (j — 1)n. L n c(V) i=1 118 6. SOME CLASSICAL PROBLEMS IN EXTREMAL GRAPH THEORY Solution. Suppose not and take an arbitrary vertex V1. Then > [k 2 so there exists V2 E COTO. Moreover, IC(V1) n C(V2)II= d(Vi) d(V2) — IC(Vi)U C(V2))I > 2 (1 + [k — 1n]/ —n > O. Pick a vertex V3 E C(Vi) n C(V2) A similar argument shows that ic(vi ) n c(v 2 ) n c(v 3)1 > 3 (1+ Lk 1k— n _II) 2n. — 1 2 Repeating this argument, we find V4 E C(Vi) n C(V 2) n C(V3) k-2 Vk-1 E n c(4). i=i This can be proved easily by induction. Thus k-1 n c(vi) (k — 1) (1+ [k 2n]) — (k — 2)n > 0, k — 1 and, consequently, we can choose k-1 Vk c n c(vi). z=1. But it is clear that VI, V2, , Vk form a complete k graph, which contradicts the assumption that G is k-free. THEORY AND EXAMPLES 119 We are now ready to prove Turan's theorem. The greatest number of edges of a k-free graph with n vertices is k — 2 n2 — r2 ( 2), r k — 1 2 + where r is the remainder left by n when divided to k — 1. [Turan] Solution. We will use induction on n. The first case is trivial, so let us assume the result true for all k-free graphs having n — 1 vertices. Let G be a k-free graph with n vertices. Using Zarankiewicz's lemma, we can find a vertex V such that d(V) < k I i ni Because the subgraph determined by the other n — 1 vertices is clearly k-free, using the inductive hypothesis we find that G has at most k — 2 k — 2 (n — 1)2 — r? (r1) L k — 1 + k — 1 2 2 edges, where ri = n — 1 (mod k — 1). Let n = q(k — 1) + r = qi(k — 1) + + 1. Then ri E {r — 1, r + k — 2} (this is because r — r1 1 (mod k — 1)) and it is easy to check that k — 2 L k — 1 ni + k — 2 (n — 1)2 — r? (r1\ k — 2 n2 — r2 (r) k — 1 2 ( 2 ) = k — 1 2 2 The inductive step is proved. Now, it remains to construct a k-free graph k — 2 n — 1 2 — r 2 2 with n vertices and + (r 2 k ) edges. This is not difficult. Just consider k — 1 classes of vertices, r of them having q + 1 elements and the rest q elements, where q(k — 1) + r = n and join the vertices situated in different k — 2 2 — 2 r groups. It is immediate that this graph is k-free, has + ( k — 1 n r 2 2) 120 6. SOME CLASSICAL PROBLEMS IN EXTREMAL GRAPH THEORY edges and also the minimal degree of the vertices is L k k — 1 n] . This graph is 2 called Turan's graph and is denoted by T(n, k). These two theorems generate numerous beautiful and difficult problems. For example, using these results yields a straightforward solution for the following Bulgarian problem. rExample 3.1 There are 2001 towns in a country, each of which is connected with at least 1600 towns by a direct bus line. Find the largest n for which it must be possible to find n towns, any two of which are connected by a direct bus line. Spring Mathematics Tournament 2001 Solution. Practically, the problem asks to find the greatest n such that any graph G with 2001 vertices and minimum degree at least 1600 is not n-free. But Zarankiewicz's lemma implies that if G is n-free, then at least one ver-tex has degree at most [n — 2 2001 . So, we need the greatest n for which n — 1 n — 2 2001 < 1600. It is immediate to see that n = 5. Thus for n = 5 any Ln — 1 such graph G is not n-free. It suffices to construct a graph with all degrees of the vertices at least 1600, is which is 6-free. We will take of course T(2001, 6), [ 4 whose minimal degree s — 52001 = 1600 and which is (as shown before) 6-free. Thus, the answer is n = 5. Here is a beautiful application of Turan's theorem in combinatorial geometry. Example 4.J Consider 21 points on a circle. Show that at least 100 pairs of points subtend an angle less than or equal to 120° at the center. Tournament of the Towns 1986 THEORY AND EXAMPLES 121 Solution. In such problems, it is more important to choose the right graph than to apply the theorem, because as soon as the graph is appropriately chosen, the solution is more or less straightforward. Here we will consider the graph with vertices at the given points and we will connect two points if they subtend an angle less than or equal to 120° at the center. Therefore we need to prove that this graph has at least 100 edges. It seems that this is a reversed form of Turan's theorem, which maximizes the number of edges in a k-free graph. Yet, the reversed form of the reversed form is the natural one. Applying this principle, let us look at the "reversed" graph, the complementary 1 ) one. We must show that it has at most ( 2 — 100 = 110 edges. But this is 2 immediate, since it is clear that this new graph does not have triangles and 2 21 — 4 1 so, by Turan's theorem, it has at most = 110 edges, and the problem is solved. At first glance, the following problem seems to have no connection with the previous examples, but, as we will immediately see, it is a simple consequence of Zarankiewicz's lemma. It is an adaptation of an USAMO 1978 problem. Anyway, this is trickier than the actual contest problem. Example 5. There are n delegates at a conference, each of them knowing at most k languages. Among any three delegates, at least two speak a common language. Find the least number n such that for any distribution of the languages satisfying the above properties, it is possible to find a language spoken by at least three delegates. Solution. We will prove that n = 2k+3. First, we prove that if there are 2k+3 delegates, then the conclusion of the problem holds. The condition "among any three of them there are at least two who can speak the same language" suggests taking the 3-free graph with vertices the persons and whose edges join persons that do not speak a common language. From Zarankiewicz's lemma, there exists a vertex whose degree is at most [-2] = k + 1. Thus, it is not 122 6. SOME CLASSICAL PROBLEMS IN EXTREMAL GRAPH THEORY connected with at least k + 1 other vertices. Hence there exists a person A and k + 1 persons A1, A2, ... , Ak±i that can communicate with A. Because A speaks at most k languages, there are two persons among A1, A2, .. • lAk-Ei that speak with A in the same language. But that language is spoken by at least three delegates and we are done. It remains to prove now that we can create a situation in which there are 2k + 2 delegates, but no language is spoken by more than two delegates. We use again Turan's graph, by creating two groups of k + 1 delegates. Assign to each pair of persons in the first group a common language, so that the language associated is different for any two pairs in that group. Do the same for the second group, taking care that no language associated with a pair in the second group is identical to a language associated with a pair in the first group. Persons in different groups do not communicate. Then it is clear that among three persons, two will be in the same group and therefore will have a common language. Of course, any lan-guage is spoken by at most two delegates. The following problem turned out to be an upset at one of the Romanian Team Selection Tests for 2004 IMO, being solved by only four contestants. The idea is even easier than in the previous problems, but this time we need a little observation that is not so obvious. Example 6. Let Ai, A2, ... , Aim be different subsets of the set {1, 2, ... , n}. Suppose that the union of any 50 subsets has more than — 50n 51 elements. Prove that among them there are three any two of which having common elements. [Gabriel Dospinescu] Romanian TST 2004 Solution. As the conclusion suggests, we should take a graph with vertices the subsets, connecting two subsets if they have common elements. Let us assume that this graph is 3-free. The main idea is not to use Zarankiewicz's lemma, but to find many vertices with small degrees. In fact, we will prove that there are at least 51 vertices all of them having degree at most 50. Suppose this is not the case, so there are at least 51 vertices whose degrees are greater than THEORY AND EXAMPLES 123 51. Let us pick such a vertex A. It is connected with at least 51 vertices, so it must be adjacent to a vertex B whose degree is at least 51. Because A and B are each connected with at least 51 vertices, there is a vertex adjacent to both, so we have a triangle, contradicting our assumption. Therefore, we can find ,21,51, all of them having degrees at most 50. Consequently, Ai, is disjoint from at least 50 subsets. Because the union of these fifty subsets has 50 more than — 50n elements, we infer that IA,' < n — — 51n = — n . In a similar 51 51 way, we obtain IA, 3 51 < — n for all j E {1,2, , 51} and so 50 IA,, U Ai2 U • • • U Azsol < Ail + • • • + Vii501 < which contradicts the hypothesis. We continue with an adaptation of a very nice and quite challenging problem from the American Mathematical Monthly. Example 7. Prove that the complement of any 3-free graph with n vertices and m edges has at least n(n — 1)(n — 5) 2( n2 - n 2 24 + n 4 triangles. [A.W Goodman] AMM Solution. Believe it or not, the number of triangles from the complementary graph can be expressed in terms of the degrees of the vertices of the graph only. More precisely, if G is a 3—free graph, then the number of triangles from the complementary graph is () 1 3 — 2 E d(x)(n — 1 — d(x)), xEX 124 6. SOME CLASSICAL PROBLEMS IN EXTREMAL GRAPH THEORY where X is the set of vertices of G. Indeed, consider all triples (x, y, z) of vertices of G. We will count the triples that do not form a triangle in the complementary graph G. Indeed, consider the sum E d(x)(n — 1 — d(x)). It xEX counts twice every triple (x, y, z) in which x and y are connected, while z is not adjacent to any of x and y: once for x and once for y. But it also counts twice every triple (x, y, z) in which y is connected with both x and z: once for x and once for z. Therefore , 2 — 1 E d(x)(n —1— d(x)) is exactly the number of (n 24 + 3) — E d(x)( n — 1 — d(x)) > n(n 1)(n — 5) 2 xEX (m n 2 - n) 2 4 Because 1: d(x) = 2m, after a few computations the inequality reduces to xEx E d2(x) > (6.1) xEX But this is the Cauchy-Schwarz inequality combined with ExEx d(x) = 2m. Finally, two chestnuts. The following problem is not directly related to our topic at first glance, but it gives a very beautiful proof of Turan's theorem: Example 8. Let G be a simple graph. To every vertex of G one assigns a nonnegative real number such that the sum of the numbers assigned to all vertices is 1. For any two vertices connected by an edge, compute the product of the numbers associated to these vertices. What is the maximal value of the sum of these products? xEX triples (x, y, z) that do not form a triangle in the complementary graph. (Here we have used the fact that G is 3-free.) Now, it is enough to prove that THEORY AND EXAMPLES 125 Solution. The answer is not obvious at all, so let us start by making a few remarks. If the graph is complete of order n then the problem reduces to finding the maximum of E xix j knowing that x1 + x2 + • • • + xn = 1. 1<i<j<n This is easy, since xix 1<i<j<n i(i n 2 < 2 \ L-d i=1 ) The last inequality is just the Cauchy-Schwarz inequality and we have equality when all variables are n. Unfortunately, the problem is much more difficult in other cases, but at least we have an idea of a possible answer: indeed, it is easy now to find a lower bound for the maximum: if H is the complete subgraph with maximal number of vertices k, then by assigning these vertices k, and to all other vertices 0, we find that the desired maximum is at least 1(1 — We still have to solve the difficult part: showing that the desired maximum is at most 1(1 — 1). Let us proceed by induction on the number n of vertices of G. If n = 1 everything is clear, so assume the result true for all graphs with at most n — 1 vertices and take a graph G with n vertices, numbered 1, 2, ..., n. Let A be a set of vectors with nonnegative coordinates and whose components add up to 1 and E the set of edges of G. Because the function f (xi, x2, ...,xn) = E xix, is continuous on the compact set A, it attains its (imEE maximum in a point (x1, x2, ..., xn). Denote by f (G) the maximum value of this function on A. If at least one of the xi is zero, then f (G) = f (G1) where Gi is the graph obtained by erasing vertex i and all edges that are incident to this vertex. It suffices to apply the induction hypothesis to G1 (clearly, the maximal complete subgraph of Gi has at most as many vertices as the maximal complete subgraph of G ). So, suppose that all xi are positive. We may assume that G is not complete, since this case has already been discussed. So, let us assume for example that vertices 1 and 2 are not connected. Choose any number 0 < a < xi and assign to vertices 1, 2, ..., n of G the numbers xi — a, x2 + a, x3, ..., xn. By maximality of f (G), we must have E xi < xi, i,c2 126 6. SOME CLASSICAL PROBLEMS IN EXTREMAL GRAPH THEORY where C1 is the set of vertices that are adjacent to vertex 2 and not adjacent to vertex 1 (the definition of C2 being clear). By symmetry, we deduce that we must actually have EE xi, iEci iEC2 which shows that f ..., xn) = f (0, x1 + x2, x3, , xii). Hence we can ap- ply the previous case and the problem is solved. Observe that the inequality in Turan's theorem follows by taking all xi to be The final problem is a very beautiful result on the number of complete sub-graphs of a graph: [Example 9.] What is the maximal number of complete maximal subgraphs that a graph on n vertices can have? [Leo Moser, J. W. Moon] Solution. Let us suppose that n > 5, the other cases being easy to check. Let f (n) be the desired number and G a graph for which this maximum is attained. Clearly, this graph is not complete, so there are two vertices x and y not connected by an edge. In order to simplify the solution, we need several notations. Let V(x) be the set of vertices that are adjacent to x, G(x) the subgraph obtained by erasing vertex x and G(x, y) the graph obtained by erasing all edges incident to x and replacing them with edges from x to any vertex in V(y) . Finally, let a(x) be the number of complete subgraphs with vertices in V(x), maximal with respect to G(x) and let c(x) be the number of complete maximal subgraphs of G that contain x. Now, we pass to serious things: by erasing edges incident to x, exactly c(x) — a(x) complete maximal subgraphs vanish, and by joining x with all vertices of V(y), exactly c(y) complete maximal subgraphs appear. So, if c(G) is the number of complete maximal subgraphs in the graph G, then we have the relation c(G(x, y)) =- c(G) + c(y) — c(x) + a(x). THEORY AND EXAMPLES 127 By symmetry, we can assume that c(y) > c(x). By maximality of c(G), we must have c(G (x , y)) < c(G), which is the same as c(y) = c(x) and a(x) = 0. Therefore G(x, y) also has f (n) complete maximal subgraphs. In the same way, we deduce that c(G (x , y)) = c(G(y, , x)) = c(G). Now take a vertex x and let xi, x2, xk be the vertices not adjacent to x. By performing the previous operations, we change G into Gi = G(xi, x), then into C2 = Gi(X2, X) and so on until Gk = Gk_i(Xk, X), by conserving the number f (n) of maximal complete subgraphs. Observe now that Gk has the property that x, xi, ..., xk are not joined by edges, yet V(x1) = V(x2) = • • = V(xk) = V(x). Now, we know what to do: if V(x) is void, we stop the process. Otherwise, consider a vertex of V(x) and apply the previous transformation. In the end, we obtain a complete multipartite graph G' whose vertices can be partitioned into r classes with ni, n2, vertices, two vertices being connected by an edge if and only if they do not belong to the same class. Because GI has f (n) maximal complete subgraphs, we deduce that f (n) = max max nin2.••Thr• r ni +7/2±.••+n,-=n (6.2) (6.2) can be easily computed. Indeed, let (ni, n2, ...,nr) the r-tuple for which the maximum is attained. If one of these numbers is at least equal to 4, let us say ni, we consider (2, ni — 2, n3, ...,nr) for which the product of the com-ponents is at least the desired maximum. So none of the ni exceed 3. Even more, since 2 • 2 • 2 < 3 • 3, there are at most two numbers equal to 2 among ni, n2, ..., nr. This shows that f (n) = 33 if n is a multiple of 3, f (n) = 4 n-4 if n — 1 is a multiple of 3 and f (n) = 2 3Y otherwise. 128 6. SOME CLASSICAL PROBLEMS IN EXTREMAL GRAPH THEORY 6.2 Problems for training 1. In a country there are 1998 cities. At least two out of each three cities are not directly connected. What is the greatest possible number of direct flights? Japan 1998 2 2. Let xi, x2, . , xn, be real numbers. Prove that there are at most — n 4 pairs (i, j) E {1, 2, ... , n}2 such that 1 < Ixi — x3 I < 2. MOSP n2 3. Prove that if n points lie on a unit circle, then at most — 3 segments connecting them have length greater than .\/. Poland 1997 4. Let A be a subset of the set S = {1, 2, ... , 1000000} having exactly 101 elements. Prove that there exist t1, t2, , till() E S such that the sets A3 = {x t 3 Ix E A} are pairwise disjoint. IMO 2003 5. Prove that a graph with n vertices and k edges has at least — k (4k — n2) 3n triangles. APMO 1989 6. We are given 5n points in a plane and we connect some of them so that 10n2 +1 segments are drawn. We color these segments in 2 colors. Prove that we can find a monochromatic triangle. PROBLEMS FOR TRAINING 129 7. There are 1999 people participating in an exhibition. Out of any 50 people, at least two do not know each other. Prove that we can find at least 41 people who each know at most 1958 other people. Taiwan 1999 8. There are n aborigines on an island. Any two of them are either friends or enemies. One day they receive an order saying that all citizens should make and wear a necklace with zero or more stones so that i) for any pair of friends there exists a color such that each of the two persons has a stone of that color; ii) for any pair of enemies there does not exist such a color. What is the least number of colors of stones required? Belarus 2001 9. Let G be a graph with no triangles and such that no point is adjacent to all other vertices. Also, if A and B are not joined by an edge, then there exists a vertex C such that AC and BC are edges. Prove that all vertices have the same degree. APMO 1990 10. Let G be a regular graph of degree k (every vertex is adjacent to k other vertices) with n vertices. Prove that G and its complementary graph n(n-1)(n-2) nk(n-k-1) • contain together at least triangles. 6 2 11. G is a finite graph such that it does not contain a complete subgraph with 5 vertices, and any two triangles have at least point in common. Show that there is a set of at most two points whose removal leaves no triangles. IMO 2001 Shortlist 130 6. SOME CLASSICAL PROBLEMS IN EXTREMAL GRAPH THEORY 12. Prove that for every n one can construct a graph with no triangles and whose chromatic number is at least n. 13. A graph with n2 +1 edges and 2n vertices is given. Prove that it contains two triangles sharing a common edge. Chinese TST 1987 14. What is the least number of edges in a connected n-vertex graph such that any edge belongs to a triangle? Paul Erd6s, AMM E 3255 15. A graph with n vertices and k edges has no triangles. Prove that we can choose a vertex such that the subgraph induced by the remaining vertices has at most k (1 – 4k — ri2 ) edges. USAMO 1995 16. Let n 1 (mod 3) be an integer greater than 3 and consider n2 points in the plane. Find the least number of segments connecting pairs of these points such that no matter how we choose n points there exist four among them any two of which are connected by a line segment. Emil Kolev, Bulgaria 2002 THEORY AND EXAMPLES 133 7.1 Theory and examples When reading the title, you will perhaps expect a difficult unit, reflecting the complexity of combinatorics. But, this was not our intention. We just wanted to discuss some combinatorial problems that can be solved elegantly by using complex numbers. At this moment, the reader will probably say that we are crazy, but we will support our idea and prove that complex numbers can play a significant role in solving counting problems, and also in problems related to tilings. They also have numerous applications in combinatorial number theory, so our purpose is to illustrate a little bit from each of these situations. After that, you will surely have the pleasure of solving the proposed problems using this technique. To avoid repetition, we will present in the beginning of the discussion a useful result Lemma 7.1. If p is a prime number and ao, al, ... , ar_i are rational numbers satisfying ao + aie + a26.2 + • • • + ap_iEP 1 = 0, where 27r 2 w E = COS — p + . sin — p =e P then ao = al = • • • = ap-l• Proof. We will just sketch the proof, which is not difficult. It is enough to observe that the polynomials ao+aiX+a2X2+• • • +ap_iXP-1 and 1+X +X2+ • • • + XP-1 are not relatively prime-because they share a common root-and since 1+X + X2 + • • • + XP-1 is irreducible over Q (you can find a proof in the chapter concerning the irreducibility of polynomials), 1+ X + X 2 + • • • + XP-1 must divide ao + a1X + a2X2 + • • • + ap_iXP-1, which can only happen if ao = al = • • • = ap_i. Therefore, the lemma is proved and it is time to solve some nice problems. K Note, in the following examples, m(A) will denote the sum of the elements of the set A. By convention m(0) = 0. 134 7. COMPLEX COMBINATORICS The first example is an adaptation from a problem given in the Romanian Contest "Traian Lalescu". Of course, there is a solution using recursive se-quences, but it is by far less elegant than the following one. Example lid How many n-digit numbers, all of whose digits are 1, 3, 4, 6, 7, or 9 have the digit sum a multiple of 7? Solution. Let C4 k) be the number of n-digit numbers, all of whose digits are 1, 3, 4, 6, 7, 9 and whose digit sum is congruent to k modulo 7. It is clear that 6 X--" ,(k) ,k c Exl-Fx2±—±xn k=0 ,x2 ,•• •,x nE{1,3,4,6,7 ,9} = (6 + E.3 e4 e6 + 67 + where E = cos 27 — + i sin 27r —. Observing that 1 + E 6 2 + • • • + E6 = 0 and 7 7 e9 = E2 helps us bring (E e3 E4 + Es + E7 69 )n to the simpler form (–E5)n. Let us assume, for example, that n is divisible by 7 (the other cases can be discussed similarly). Then 6 Ea(k)Ek _ (_1)n k=0 = • • • = aV. Let q and from the lemma we infer that be the common value. Then 7q = because exactly 6n numbers have n case we have aV = (-1)n + n 1, 2, 3, 4, 5, 6 (mod 7). – ( 7 , 3, 4, 6, 7, 9. In this 1)n We leave you with the other cases: a(0) ( 1)n 41) 6 a ) – (-1) 71 = 6 11 – (-1)n - this is k=0 digits, all equal to 1 Following this trick, here is a slightly more difficult problem, which appeared on the Balkan Olympiad Shortlist in 2005, and which was used for the selection of the Romanian IMO 2005 team: THEORY AND EXAMPLES 135 Example 2. Let (an)n>1 be a sequence of distinct positive integers such that an < 4.999n for all n. Prove that there are infinitely many n for which the sum of digits of an is not a multiple of 5. Does the result remain true if the condition is relaxed to an < 5n for all n? [Gabriel Dospinescu] Solution. Let s(x) be the sum of digits of x, and suppose that for all n > M we have 5Is(an). Let n be such that [1r 99 -9 1] > M + 3 and let A be the set of the first 10n nonnegative integers. The numbers ak with 1 < k < [ are 127 9 9 -9 1. in A because 1 < ak < 4.999k < 10n — 1 for these numbers k. It follows that A contains at least [1 4r 99 -9 1] M numbers with digit sum divisible by 5. Now fix a number 2 < i < n and observe that if x j is the number of elements of A with i digits and having digit sum congruent to j mod 5, then 6aid-a2+•••+a, = Xo XiE X2E2 X36 3 + X4E4 = 0<a2,...,ai <9 1<ai <9 (6 + 62 + + 69)(1 + + + 69)i-1 = 0. Using the lemma, and taking into account that x0 + x1 + • • • + x4 = 9 • 102-1, E 10; we deduce that there are at most 1 + 9 = 2 • 10n-1 — 1 elements of A i=2 with the digit sum a multiple of 5. Thus [1 ,r 99 -9 1] M < 2 • 10n-1 — 1 for all sufficiently large n, which is certainly impossible. For the second part of the problem, the answer is negative. Indeed, consider the sequence starting with 1 and containing the positive integers (in increasing order) whose digit sum is divisible by 5. Let us prove that an < 5n for all n. Indeed, this is clear for n = 1, 2, 3 because al = 1, a2 = 5, a3 = 14. The crucial observation is that clearly among any 10 consecutive positive integers, exactly two are terms of the sequence. Thus a2n < 10n and a2n_1 = a2n —5 < 5(2n-1). This proves that for an < 5n the statement is no longer true. 136 7. COMPLEX COMBINATORICS The same simple, but tricky, idea can offer probably the most beautiful solu-tion for the difficult IMO 1995 problem 6. It is worth mentioning that Nikolai Nikolov won a special prize for the following magnificent solution. Let p > 2 be a prime number and let A = {1, 2, ... , 2p}. Find the number of subsets of A each having p elements and whose sum is divisible by p. IMO 1995 Solution. Consider E = cos — 27r + i sin — 27r and let x j be the number of subsets p p X of A such that IXI = p and m(X) j (mod p). Then it is not difficult to see that p-1 E xi Em(B) i=0 1<c,<c2<•••<cp<2p Eci±c2±...-1-cp But EC1±C2+•••+CP is precisely the coefficient of XP in the poly- 1<ci<c2<•••<cp<2p nomial (X + e)(X + E2) (x- 6.213\ ) Because XP - 1 = (x -1)(x - (x - EP-1), we easily find that (X + E)(X + E2)... (X + E2P) = (XP + 1)2. Thus E xi = 2, and the lemma implies xo - 2 = xi = • • • = xp-i. p-1 Since J=0 there are (2P" subsetswith p elements, it follows that 2p xo ± xi ± • • • + xp—i = ( p ) THEORY AND EXAMPLES 137 Therefore xo 2 + P 1 (O P ) ) 2) , and we are done. The following problem deals with a little more general case, even though the restriction imposed on the cardinality is no longer maintained. Let f (n) be the number of subsets of 1, 2, 3, ..., n whose ele-ments sum to 0 (mod n). The empty set is included, having the element sum equal to zero. Prove that f(n) = -n • E co(d)2. din d odd Solution. Let g(X) =n(1 + X') = E ak Xk i=1 k>o t7r and let E e 2 . It is clear that f(n) = E din. On the other hand, the i>o last sum can easily be computed in terms of g(E3). Indeed, one can verify the identity n —• 4–s • g(E3)= Ea Now, let us compute g(E3). If d = gcd(j n) (that is, E is a primitive d-th root of unity), then xd — 1 = (x — E3)(x — E23)... (x — Ed3) and so j>0 (1+ 6-3)(1+623)— (1+6' 13 ) =2 138 7. COMPLEX COMBINATORICS if d is odd and 0 otherwise. This shows that g (E3 ) = 2i if d is odd and 0 otherwise. But there are exactly co(d) values of j for which E3 is a primitive d—th root of unity, so 1 1 n — • E g (E3) = — n • E yo (d)2 . . 3=1 din d odd With a somewhat different but closely related idea we can solve the following nice problem. Example 5.1 Let n > 1 be an integer and let al, a2, , am be positive inte- gers. Denote by f (k) the number of m-tuples (ci, c2, . , cm) such that 1 < ci < ai for all i and c1 + c2 + • • + cm k (mod n). Prove that f (0) = f (1) = • • • = f (n — 1) if and only if there exists an index i E {1, 2, ... , m} such that nlai. [Reid Barton] Rookie Contest 1999 Solution. Observe that n-1 E f(k)Ek = E scl±c2±.„±cm = E2 Ecti) k=-0 1 2 be a prime number and let m and n be multiples of p, with n odd. For any function f : {1, 2, , m} —> {1, 2, , n} satisfying E f (k) 0 (mod p), consider the product 11 f (k). k=1 k=1 Prove that the sum of these products is divisible by ( P — n ni [Gabriel Dospinescu] 27r i 27r Solution. Let E = cos — + ?, sin — and let xk be the sum of fl f (k) over all p p k=1 T17, functions f : {1,2, . , m} {1,2, . . , n} such that E f (i) k (mod p). It i=1 is clear that 13 1 Exk Ek = k-=0 ,C2 7 • • • ,Cra c]. ±c2 +•••+cai C C2 . . CrnE = (E ± 2E2 ± • • • ± nEn)m Recall the identity , 1 + 2x + 3x2 + • • • + nxn — nxn+1 — (n + 1)xn + 1 (X — 1)2 Plugging e in the previous identity, we find that nEn+2 (n i)En+1 E E 2E2 • ••• TtE n = ne (e — 1)2 E — 1 WTI (E — 1)7n 140 7. COMPLEX COMBINATORICS Consequently, p-1 E xkEk = k=0 On the other hand, it is not difficult to justify that Ep—i Ep-2±•••±E±1=0<=> — (E13-2 ± 2E13-3 + • • • ± (p — 2)E p — 1). Considering (XP-2 +2XP-3 + • • • + (p — 2)X +p — 1)m = b0 + b1X + • • • + bm(p_2).X17-(p-2), we have nm n)m (co + ciE + • • • + cp_iEp-1), — where ck = E (mod p) Setting r = ( --n)m, we deduce that p xo — rc0 + (xi — rCi)E ± • • • + (Xp-1 rCp-1)EP 1 = 0. From the lemma, it follows that x0 — rc0 = xi — rci = • • • = xp—i — rcp—i = k. Because clearly co, ,cp_i are integers, it remains to prove that rik. Because pk = xo + xi + • • + xp—i — r(co + + • • • + cp-1.) = (1 + 2 + • • • + n)m — r(bo + bi + • • • + bm(p-2)) ( n(n+l)r ) r p(p 2 — 1))771 2 it is clear that rlk. Here we have used the condition ) s in the hypothesis. The problem is solved. 1 E —1 THEORY AND EXAMPLES 141 It is now time to leave these kinds of problems and to talk a little bit about some nice applications of complex numbers in tilings. Before presenting some examples, let us make some conventions: consider a rectangular table with edges parallel to two fixed (orthogonal) lines Ox and Oy. An a x b rectangle is a figure consisting of ab unit squares, with edges parallel to Ox and Oy and such that the edge parallel to Ox has length a and the one parallel to Oy has length b. For instance, the rectangle with vertices (0,0), (2,0), (2, 1), (0,1) is a 2 x 1 rectangle, while the rectangle with vertices (0,0), (1,0), (1, 2), (0,2) is a 1 x 2 rectangle. Now, the idea is to put a complex number in each square of a table and then to reformulate the hypothesis and the conclusion of a particular tiling problem in terms of complex numbers. We will see how this technique works better by solving a few actual problems. First, some easy examples. Example 7. Consider a rectangle that can be tiled by a finite combination of 1 x m and n x 1 rectangles, where m, n are positive integers. Prove that it is possible to tile this rectangle using only 1 x in rectangles or only n x 1 rectangles. [Gabriel Carroll BMC Contest 2000 Solution. Let the dimensions of the initial rectangle be a x b, for the positive integers a and b. Now let us partition the rectangle into 1 x 1 squares and denote these squares by (1,1), (1,2),..., (1, b),... , (a, 1), (a, 2), ... , (a, b). Next, put the number Em in the square labeled (x, y), where 27r27r 27r 27r el = cos — + sin — , E2 = cos — . sin —. 71 The main observation is that the sum of the numbers in any 1 x m or n x 1 rectangle is 0. This is immediate, but the consequence of this simple observa- tion is really surprising. Indeed, it follows that the sum of the numbers in all 142 7. COMPLEX COMBINATORICS the squares is 0, and so 0 = ,x ,y '1'2 1<x<a 1<y<6 i=1 j=1 a Hence at least one of the numbers i=1 El and e2 is 0. But this means that i-1 70 or ml b. In either case, the conclusion of the problem follows. The idea in the previous problem is quite useful, helping many tiling problems become straightforward. Here is one more example: 1 r Example f.1 Can we tile a 13 x 13 table from which we remove the central --- unit square using only 1 x 4 or 4 x 1 rectangles? Baltic Contest 1998 Solution. Suppose such a tiling is possible, and label the squares of the table as in the previous problem. Next, associate to square (k, j) the number ik+23. Clearly, the sum of the numbers from each 1 x 4 or 4 x 1 rectangle is 0. Therefore the sum of all labels is equal to the number corresponding to the central unit square. Hence 221 (i 413 i2 il3)(i2 i4 i26) i2 i26 •3 = , i -1 i2 which clearly cannot hold. Thus the assumption we made is wrong, and such a tiling is not possible. The example we are going to discuss now is based on the same idea, and here complex numbers are even more involved. THEORY AND EXAMPLES 143 [-Example 9.1 On an 8 x 9 table we place 3 x 1 rectangles and "broken" 1 x 3 rectangles, obtained by removing their central unit square. The rectangles and the "broken" rectangles do not overlap and cannot be rotated. Prove that there exists a set S consisting of 18 squares of the table such that if 70 unit squares of the table are covered, then the remaining two belong to S. [Gabriel Dospinescu] Solution. Again, we label the squares of the table (1,1), (1, 2), , (8, 9) by starting from the upper left corner. In the square labeled (k, j) we will place the number i3 • Ek, where i2 = -1 and E2 + E +1 = 0. The sum of the numbers from any rectangle or "broken" rectangle is 0. The sum of all numbers is 8 \ 9 ( Ek) ( ij) = k=1. j=1 Let us suppose that (al, b1) and (a2, b2) are the only uncovered squares. Then ibi ±ibzea2 = Let z1 = and z2 = ib2-102 . We have Iz1 = 1z21 = 1 and z1 + z 2 = -1. It follows that — 1 + 1 = -1 and so 4 = 4 = 1. This Z1 Z2 in turn implies the equalities i3(b1-1) = i3(b2-1) = 1, from which we conclude that b1 b2 1 (mod 4). Therefore the relation z1 + z2 = -1 becomes Ea' Ea2 = -1, which is possible if and only if the remainders of al, a2 when divided by 3 are 1 and 2. Thus we can choose S to be the set of squares that lie at the intersection of the lines 1, 2, 4, 5, 7, 8 with the columns 1, 5, 9. From the above argument, if two squares remain uncovered, then they belong to S. The conclusion is immediate. LExample 10.] Let m and n be integers greater than 1 and let ai, a2, , a, be integers, none of which is divisible by ma-1. Prove that we can find integers e l, e2, , en, not all zero, such that led < m for all i and mnleiai + e2a2 + • • • + enan• IMO 2002 Shortlist f (x) = H E x3,4) — 11 n (n-1 1 — xma/ 1 — xai i=1 j=0 i=1 144 7. COMPLEX COMBINATORICS Solution. Look at the numbers j ezaz, where 0 5_ < m — 1 for all i. Observe that we have a collection of mn numbers (denote this collection by A). We can assume that this is a complete system of residues modulo mn (otherwise, the conclusion is immediate). Now, consider f (x) = E xa. Then aEA Now take e = . Since the mn numbers we previously considered form a complete system of residues modulo mn, we must have f (e) = 0. Therefore (the hypothesis ensures that Ea/ 1) H(1 — Erna') = 0. But this clearly con- i=i tradicts the fact that none of the numbers a1, a2, , an is a multiple of mn-1. Let p be a prime number and let fk(xl, x2, •••, xn) = ak1X1 ak2x2 + • • • + aknxn be linear forms with integer coefficients for k = 1,2, ...,pn. Suppose that for all systems of integers (xi, x2, •••7 xn), not all divisible by p, fl(Xl, X2, ..., Xn), f2(X1, X2, .••, Xa), •••, fpn (Xl, X2, •••, Xn) represent every remainder mod p exactly pn-1 times. Prove that {(aki , ak2, akn) I k = 1, 2, ..., pn} is equal to i2) •••, I21,•••, in = 0,1,...,p 1}. Miklos Schweitzer Competition Solution. Let e = e P and observe that the hypothesis implies the identities pn = 0 k=, THEORY AND EXAMPLES 145 for all xi, x„, not all multiples of p. Now fix ii, i2, in. By multiplying both sides of the equality by E—"x1—i2x2 insn we deduce that E E.(aki_ii).,+•..±(akm_in)xn = 0. k=i By making the sum of all these equalities corresponding to all (xi, x2, •••, xn) E {0, 1, ...,p-1}71 and by taking into account that for (xi, x2, ••., xn) = (0, 0, ..., 0) the left-hand side equals pn, we deduce that pn n p-1 pn = II ( >; Ex (7.1) k=1 j=1 xj=0 Because the sum in the right-hand side of (7.1) is not zero, at least one term is not zero. Observe however that every term of the sum equals 0 or pn. Therefore there exists an unique k such that ak3 = i 3 (mod p) for all j. This is just another way of saying that {(aki, • • ,akm)lk = 1, = \-1, • • • • in = 0,... — 1}. The following problem, communicated by Vesselin Dimitrov, is a very special one. It concerns a concept introduced by Erdos in a paper dating back to 1952: the covering systems of congruences. More precisely, the family of ordered pairs (ai, di), (a2, d2), • • • , (ak, dk), where 1 < d1 < d2 < • • • < dk is called a covering set of congruences if x a, (mod c12) is solvable for any integer x. Erdos immediately realized that this new concept can be a source of difficult questions, and that became source of intensive research. Ethos conjectured that there exists no covering set of congruences in which all the moduli are odd. This remains open. On the other hand, Erdos used covering sets (more precisely, the set (0, 2), (0, 3), (1, 4), (3, 8), (7, 12), (23, 24)) to prove the existence of an infinite arithmetic progression of odd positive integers, 146 7. COMPLEX COMBINATORICS none of which is of the form 2n p. Schinzel also studied these systems in his researches concerning the irreducibility of polynomials. Conjectured by Erd6s, the example that comes next was proved by Sun, and what is really strange is that the solution is absolutely elementary. We thank Vesselin Dimitrov for pointing out this jewel of number theory. [Example 12.] Let F be a family of k infinite arithmetic progressions ch diZ, where 1 < d1 < • < dk. Assume that F covers 2k consecutive integers (that is, there exists an integer x such that every number in the sequence x, x 1,...,x + 2k — 1 belongs to at least one member of the family F). Then F is a covering system of congruences. [Erdos-Sun] Solution. The magical idea is to rewrite the condition that a number belongs to a union of arithmetic progressions in a more algebraic way. For instance, the fact that x + t belongs to the union of members of F can be written in the form H ( 1 _ e 24' x t — ai 0. 1<j<k Now, all we have to do is to develop this product and observe that the same relation can be expressed in the form Ea/ • e(x+t),(3/ = 0, (7.2) /cS where S = {1, 2, , k} (including the void set, in case of which the term is 0), cer and Oi depending only on the cover itself. Indeed, this is clear, because ri[ (1 e (x+t—ai) ) — (_1)11] e-27:1•Eier d 3 • e 2i7r•E (x-1-t) 3 dj 1<j<k THEORY AND EXAMPLES 147 If we manage to prove that the same relation (7.2) holds for any integer y instead of x, we are done, since it would follow that any integer belongs to at least one member of F. If we consider zi = e01, then we know that t+x El aiz i = — 0 for all 0 < t < 2k — 1. Define ur, = E./ ceiz7 and observe that un satisfies a linearly recurrent relation of order 2k, the coefficient of un being nonzero. Indeed, consider the polynomial rt, (X — zI), which has degree 2k and nonzero free term (because all zi are nonzero), and write it in the form X2k + A2k_1X2k-1 + • • • + AiX + Ao. Then we know that 2k A 2k -1 + + • • + Ao = O. By multiplying this relation by ai • zy (we allow here negative exponents as well) and by adding up these relations, we obtain a recurrence relation un+2k + A2k_iun±2k_i + • " + Ao = 0. And now... we are done: from the hypothesis, 2k consecutive terms of this sequence vanish. Since the sequence satisfies a recurrence relation of order 2k with nonzero free term, it follows by a trivial induction that all terms are zero. This finishes the proof. 148 7. COMPLEX COMBINATORICS 7.2 Problems for training 1. Can we tile a 9 x 9 table from which we remove the central unit square using only 1 x 4 or 4 x 1 rectangles? 2. Three persons A, B, C play the following game: a subset with k elements of the set {1, 2, ... ,1986} is selected randomly, all selections having the same probability. The winner is A, B, or C, according to whether the sum of the elements of the selected subset is congruent to 0, 1, or 2 modulo 3. Find all values of k for which A, B, C have equal chances of winning. IMO 1987 Shortlist 3. We roll a regular die it, times. What is the probability that the sum of the numbers shown is a multiple of 5? IMC 1999 4. Let ak, bk, ck be integers, k = 1, 2, ..., n and let f (x) be the number of ordered triples (A, B, C) of subsets (not necessarily nonempty) of the set S = {1, 2, ... , n} whose union is S and for which E ai + E bi + 3 (mod x). iEs \A iEs \B ies\c Suppose that f (0) = f (1) = f(2). Prove that there exists i E S such that 3 I az ± bi Gabriel Dospinescu 5. How many 100-element subsets of the set {1, 2, ... , 2000} have the sum of their elements a multiple of 5? Qihong Xie PROBLEMS FOR TRAINING 149 6. There are 2000 white balls in a box. There is also an unlimited supply of white, green, and red balls, initially outside the box. At each step, we can replace two balls in the box by one or two balls as follows: two whites or two reds by a green; two greens by a white and a red; a white and a green by a red or a green; and a red by a white. a) After a finite number of steps, there are exactly three balls in the box. Prove that at least one of them is green. b) Is it possible that after a finite number of steps there is only one ball in the box? Bulgaria 2000 7. A 7 x 7 table is tiled by sixteen 1 x 3 rectangles such that only one square remains uncovered. What are the possible positions of this square? Tournament of the Towns 1984 8. Let k be an integer greater than 2. For which odd positive integers n can we tile a n x n table by 1 x k or k x 1 rectangles such that only the central unit square is uncovered? Gabriel Dospinescu 9. Let n > 2 be an integer. At each point (i, j) having integer coordinates we write the number i + j (mod n). Find all pairs (a, b) of positive integers such that any residue modulo n appears the same number of times on the sides of the rectangle with vertices (0, 0), (a, 0), (a, b) , (0, b) and also any residue modulo 71 appears the same number of times in the interior of this rectangle. Bulgaria 2001 150 7. COMPLEX COMBINATORICS 10. Let be the family of subsets of the set A = {1, 2, , 3n} having the sum of their elements a multiple of 3. For each member of compute the square of the sum of its elements. Compute in a closed form the sum of the numbers obtained in this way. Gabriel Dospinescu 11. Let p be an odd prime. Prove that the 2 pi 2 numbers +1 + 2 + • • • ± P21 represent each nonzero residue class mod p the same number of times. R. L. McFarland, AMM 6457 12. Suppose that A1, A2, ..., An are n sets of p integers where p is a prime such that the pn sums E ai with a9, E Ai are all distinct mod pn. Then the n sets, with appropriate ordering, can be described as follows: the elements of Ai, taken mod pi are the numbers ci+jpi-1 for a fixed integer ci and j = 0,1, ..., p — 1. S. W. Golomb, AMM 13. Prove that the number of subsets with n elements of the set of the first 2n positive integers whose sum is a multiple of n is (-1)n n 1)dco (n d) (2 d d) din 14. Let p be an odd prime and n > 2. For a permutation o- of the set {1, 2, ..., n} define S(u) = a(1) + 2o-(2) + • • • + no-(n). Let A3 be the set of even permutations o- such that S(a) = j (mod p) and B3 be the set of odd permutations o- for which S(a) = j (mod p). Prove that n > p if and only if A3 and B3 have the same number of elements for all j. Gabriel Dospinescu PROBLEMS FOR TRAINING 151 15. Let p > 3 be a prime number and let h be the number of sequences P-1 aj 3=o number of sequences (b1, b2, , bp_i) C {0,1,3}P-1 such that p I > jb3. Prove that h < k and that the equality holds only for p = 5. IMO 1999 Shortlist 16. Let p be an odd prime and let a, b, c, d be integers not divisible by p such that {7 } + + {7,} + = 2 for all integers r not divisible by p (here 6 is the fractional part). Prove that at least two of the numbers a + b, a + c, a + d,b + c,b + d, c + d are divisible by p. Kiran Kedlaya, USAMO 1999 a2, . , ap_i) C {0,1,2}P-1 such that p j . Also, let k be the p-1 j=0 THEORY AND EXAMPLES 155 8.1 Theory and examples We start with a riddle and a challenge: what is the connection between the following problems? 1. The set of nonnegative integers is partitioned into n > 1 infinite arithmetical sequences with common differences ri, r2, , r, and first terms al, a2, an. Then al a2 an n — 1 — ±+ = ri r2 r, 2 2. The vertices of a regular polygon are colored such that each set of vertices having the same color is the set of vertices of a regular polygon. Prove that there are two congruent polygons among them. The first problem was discussed during the preparation of the USA IMO team, but it seems to be a classical result. As for the second one, well, it is a famous problem given at a Russian Olympiad, proposed by N. Vasiliev. If you have no clue, then we will give you a small hint: the methods used to solve both problems are very similar and can be included into a larger field, that of formal series. What are those? Well, given a commutative ring A, we can define another ring, called the ring of formal series with coefficients in A and denoted A. An element of A is of the form ariXn , where n>0 an E A, and it is also called the generating function of the sequence (a,),>0. The addition and multiplication are the natural ones, defined as the similar operations with polynomials: (E anxn) (1 .2 bnxn) n>0 n>0 = E (an + br)xn n>0 and (Th anXn) b Xn) = E CnXn, >o n>0 n>0 where c„ = E apbq. Yet, for an entire function g(z) = E gnzn and a formal p+q=n n>0 series f (X) = E anXn we can define the formal series g(f (X)) = E br,Xn n>0 n>0 156 8. FORMAL SERIES REVISITED obtained "formally" from the formula g(f(X)) = j gn fn(X) by developing n>0 f n(X) and grouping terms according to successive powers of X . You can (and you should, if it is the first time you encounter this object) easily prove that all formulae of the type of •eg = of +9, sin( f +g) = sin( f) cos(g)+sin(g) cos( f) and so on are valid in the ring of formal series. Also, one can define a derivative on this ring, similarly defined as the usual derivative of polynomials, by f(X) = > nai,X71-1 and check that all the properties that the derivative has on the n>1 space of polynomials are preserved. Actually, all operations that are allowed on polynomials can be transferred formally to the ring of power series, and preserve their properties, as long as they are expressed purely in terms of the coefficients (this excludes of course speaking about zeros of a formal series). As we will see in what follows, formal series have some very nice applications in different fields: algebra, combinatorics, and number theory. But let's start working now, assuming familiarity with some basic analysis tools. We warn the reader that from time to time we will insist on some questions of convergence or continuity, but at other times we wilLwork only in this ring of formal series, therefore adopting only the operations of this ring, with no further reference to questions of convergence. LExample 1. Let al, , an be complex numbers such that all + • • • + an = 0 for all 1 < k < n. Then alrnumbers are equal to 0. Solution. The experienced reader has already noticed that this problem is an immediate consequence of Newton's relations. But what can we do if we are not familiar with these relations? Here is a nice way to solve the problem (and a way to prove Newton's relations, too). First of all, observe that the given condition implies ak l + a2 k + • • • + an = 0 for all positive integers k. Indeed, let f (X) = X n+ bm_00-1 + • • • + b1X + b0 = H(X — ao. i=1 Then az k + • • • + boak—n = 0 THEORY AND EXAMPLES 157 for all k > n 1. It suffices to add these relations and to prove the statement by strong induction. Now, let us consider the function f (z) = n 1 Developing it by using i=1 — zai . 1 = 1 x + x2 ± • •• (for lx1 < 1), we obtain that f(z) = n for all sufficiently small z (meaning for such z that satisfy I z1 maxi <i<n{ la, < 1). Assume that not all numbers are zero and take ai, , a3 (1 < s < n) to be the collection of numbers of maximal ab-solute value among the n numbers and let this maximal absolute value be r. By taking a sequence zp 1 — such that lzp r I < 1, we obtain a contradiction with the relation 1 = n (indeed, it suffices to observe that the left- 1 — z a i=i P t hand side is unbounded, while the right one is bounded). This shows that all numbers are equal to 0. We are going to discuss a nice number theory problem whose solution is prac-tically based on the same idea. This result is an important step in proving that the order of any finite subgroup of GL,(Z) divides (2n)!. Indeed, it is not difficult to prove that if G is a finite subgroup of GL„(Z) then !GI divides E Tr(g) (all you need is to note that E g is idempotent, which is an gEG gEG immediate consequence of the fact that in a finite group the translations are actually permutations; or, the trace of an idempotent matrix is just its rank, and thus an integer). Working with the tensorial product matrices A ® A where A E G and repeating the above argument yields IG11 E (Tr(g))k for gEG all k > 0. Now, all we need is to apply the result below in order to conclude that (n — Tr(gi))(n — Tr(g2)) • (n — Tr(g„)), 1 — x n 158 8. FORMAL SERIES REVISITED where Tr(gi), Tr(g2), Tr(gs) are the distinct traces that appear in the list (Tr(g))gEG,g/n. Because n — Tr(gz) are distinct integers between 1 and 2n, it follows that IGI divides (2n)!. Example 2.1 Let al , a2, , aq, xi, x2, . , xq and m be integers such that ml aixi + a24 + • • + aqxq k for all k > 0. Then ml ai ll(xi — xi). i=2 Solution. Consider this time the formal series f (z) = E ai 1 — zxi i=i By using the same formula as in the first problem, we obtain (q f (z) = E ai + aixi) z + • • • , i=i i=i which shows that all coefficients of this formal series are integers divisible by m. It follows that the formal series Eai 11(1 — xiz) i=1 jai also has all of its coefficients divisible by m. Now consider S z), the t-th funda-mental symmetric sum in xi (j i). Because all coefficients of ilj ,(1-1 x3z) are multiples of m, a simple computation shows that we have the divisi-bility relation ml xq-1 1 q-2 ai — x1 (i) + • • • + (-1 isq (01. i=1 i=1 i=1 THEORY AND EXAMPLES 159 This can also be rewritten as m ai (x7-1 - xr 2S1i) • • + (-1)q-1S(i) ) 1 • j=1 Now, the trivial identity (xi - xi)... (x1 - - xi+i).. • (xi - xn) = 0 gives us the not-so obvious relation x7-1 - x7-2 + + (-1)q-1 = 0 for i > 2. Therefore q ( X-1 — Xq-2 S i1) + • • • + (-1)q-1891 21 = (xi — x2) (x1 — xn) and we are done. In order to solve the problem announced at the very beginning of the presen-tation, we need a lemma, which is interesting itself, and which we prefer to present as a separate problem. Example 3. Suppose that the set of nonnegative integers is partitioned into a finite number of infinite arithmetical progressions with common differences ri,r2, • • , rn and first terms al, a2, • , an. Then 1 1 1 — + — + • • • + — = 1. ri r2 rn Solution. Let us observe that for any Ix1 < 1 we have the identity: Exaid-kri E x a2±kr2 . . . E x and-krn E x k. k>0 k>0 k>0 k>0 Indeed, all we did was to write the fact that each nonnegative integer is in exactly one of the arithmetical sequences. The above relation becomes: xal xa2 xan 1 + + • + (8.1) 1 _ xri _ xr2 1 — Xrn 1 X 160 8. FORMAL SERIES REVISITED -1 xa li Let us multiply (8.1) by 1 - x and use the fact that m x->i 1 - x the desired relation 1 1 1 — + — + • + — = 1. r1 r2 rn = a. We find It is now time to solve the first problem. We will just take a small, but far from obvious, step and we'll be done. The fundamental relation is again (8.1). rExample 4.! The set of nonnegative integers is partitioned into n > 1 infi- nite arithmetical progressions with common differences ri, rn and first terms al, a2, , an. Then at a2 an n - 1 + + • • • + = Ti rz rn 2 MOSP Solution. Let us write the relation (8.1) in the more appropriate form: xal xan , ± • • • + = 1 (8.2) 1 +x+•••+xri-1 1+x+•••+xrn- , Now, let us differentiate (8.2) and then make x 1 in the resulting expression. An easy computation, which is left to the reader, shows that It suffices now to use the result proved in example 3 in order to conclude that al a2 an n - 1 T1 — +r2 + + rn = 2 • THEORY AND EXAMPLES 161 Some comments about these two relations are necessary. First of all, using a beautiful and difficult result due to Erd6s, we can say that the relation 1 1 1 — — ± • • • ± — = 1 r1 r2 rn implies that max(ri,r2, ,rn) < 22n-'. Indeed, this remarkable theorem due to Eras asserts that if xi, x2, , xk are positive integers whose sum of reciprocals is less than 1, then 1 1 1 1 1 1 —+—+...+— < —+—+...+—, X1 X2 xk ul U2 Uk where ui = 2, n71+1 = U2, — Un + 1. But the reader can verify immediately by induction that 1 1 1 — — ± • • • + — = 1 Ul U2 Uk 1 u1u2 . • • uk Thus we can write 1 — < 1 rn 1 or, even better, rn < u1u2 • • • un_i = un — 1 (the last relation following again by a simple induction). Another inductive argument proves that un < 22n-1. Hence max(ri,r2, , rn) < 22" 1. Using the relation proved in example 4, we also deduce that max(ai, a2, • • • , an) < (n — 1) • 22n1-1. This shows that for fixed n not only is there a finite number of ways to parti-tion the set of positive integers into n arithmetical progressions, but we also have some explicit (even though huge) bound on the common differences and first terms. It is now time to solve the remarkable problem discussed at the beginning of this chapter. We will see that using the previous results proved here, the solution becomes natural. However, the problem is still really difficult. 162 8. FORMAL SERIES REVISITED l Example 5J The vertices of a regular polygon are colored in such a way that each set of vertices having the same color is the set of ver-tices of a regular polygon. Prove that there are two congruent polygons among them. [N. Vasiliev] Russian Olympiad Solution. Let us assume that the initial polygon (which we will call big from now on) has n edges, and that it is inscribed in the unit circle, the vertices 27, , having as coordinates the numbers 1, e e2 , en-1, where E = e (of course, we will not lose generality with all these restrictions). Let ni, n2, , nk be the number of edges of the monochromatic polygons, and assume that all 217r these numbers are distinct. Let Ei = e n3 and observe that the coordinates of n the vertices of each monochromatic polygon are zi, ziEj, , zjey 1 for some complex numbers zi on the unit circle. First, a technical result. 2i,r Lemma 8.1. For any complex number z and = e P we have the identity 1 1 1 1— z 1 — z( 1 — z(P-1 1— zP Proof. Proving this lemma is a simple task. Indeed, it suffices to observe that z, , z(P-1 are exactly the zeros of P(X) = XP — e. Or, observe that P'(X) 1 1 P(X) X — z X — z(P-1' thus by taking X = 1 we obtain the desired result. Now, the hypothesis of the problem and lemma allow us to write ni n2 nk = + + n 1 — (zzi)nl 1 — (zz2)n2 1 — (zzk)nk 1 — zn . Also, the simple observation n1 + n2 + • • • + nk = n yields the new identity nizi ft, ni , 2 2 nk nzn Zni + n..,2,4 zn2 + + nkzk Znk = (1) 1 — (ZZiri 1 — (ZZ2)n2 1 — (ZZOnk 1 — Zn . (ki,j i=1 i=1 n ) n n ak - ... + ( —1)71-1 E ar ii = n! Haz. E i<i<j<n THEORY AND EXAMPLES 163 Let us assume now that ni < min(n2, , nk) and divide (1) by znl. It follows that for any nonzero z we have n2z22 zn2-ni nkzk k znk-n1 = nzn-n1 .
(2) 1 - (zzi)ni 1 - (zz2)n2 1 - (zzk)nk 1 - zn We are done: it suffices to observe that if we make z —> 0 (by nonzero values) in (2), we obtain 41 = 0, which is clearly impossible, since = 1. The proof ends here. K The problem that we are going to discuss now has appeared in various contests in different forms. It is a very nice identity that can be proved in quite messy but elementary ways. Here is a magical proof using formal series. [Example 6.] For any complex numbers al, a2, , an the following identity holds: n n E ai a ) E 3 i=1 i=1(ji Solution. Consider the formal series f( z ) 11(eza, 1). i=1 We are going to compute it in two different ways. First of all, it is clear that z2a? f (z)= 11 (zai + 2! + , i=1 164 8. FORMAL SERIES REVISITED hence the coefficient of z" is ITai. On the other hand, we can write z=i n z ct .f (z) ez ,=, a, — E e , o, (_1)n-1 E eza, ( 1)n. i=1 z=i Indeed, you are right: everything is now clear, since the coefficient of z" in ekz kn ls !
. The conclusion follows. Here are two applications of this formula. The first one is a recent Putnam problem (2004), which asked competitors to prove that for any n there exists N and some rational numbers ci,c2, cN such that X1X2 • • ' Xn = cz(azi xi + ai2x2 + • • + a,nxn)n z=i holds identically in complex variables xi, x2, xn, and aid are equal to —1,0 or 1. It is clear that the above identity furnishes an answer N = 2' where we have even more, aid E {0,1}. Some twenty years before the Putnam Competi-tion, the following problem was proposed at the Saint Petersburg Olympiad: A calculator can perform the following: add or subtract two numbers, divide any number by any nonzero integer and raise any number to the tenth power. Prove that using this calculator one can compute the product of any ten num-bers. As you can immediately see, the solution follows by the above identity. Without using it, it is really difficult to solve this problem. Not only algebra problems can be solved in an elegant manner using formal series, but also some beautiful number theory and combinatorics problems. We shall focus a little more on each type of problem in the sequel. Example 7J Let 0 = a0 < a1 < a2 < ... be a sequence of nonnegative integers such that for all n the equation a, + 2a3 + 4ak = n has a unique solution (i, j, k). Find ai998. IMO 1998 Shortlist THEORY AND EXAMPLES 165 Solution. Here is the very nice answer: 9817030729. Let A = { ao, al, • • } and let bn, = 1 if n E A and 0 otherwise. Next, consider the formal series f(x) = E tone , the generating function of the set A (we can write it in a n>0 more intuitive way f (x) = xan ). The hypothesis imposed on the set A n>0 translates into f (x) f (x2) f (x4) = 1 1 x . Replace x by x2k . We obtain the recursive relation 1 _ , nk+1 _ ok+2 ) 1 — f(x' )f(x' )f(x' — x2k Now, observe that H f(x2k) = fl (f(x23k )f(x23k+1 )f (x 23k-1-2 )) n 1-r 1 - xvk k>0 k>0 k>0 and H f(x2k) = H (f(x23k±,)f(x23k+2)f(x2.+3)) =1 1 1 1 x23k-Fi k>1 k>0 k>0 Therefore (you have observed that rigor was not the strong point in establish-ing these relations), 23k+1 f (x) = 11 1 23k = 11(' x8k) x k>0 k>0 This shows that the set A is exactly the set of nonnegative integers that use only the digits 0 and 1 when written in base 8. A quick computation based on this observation shows that the magical term asked for by the problem is 9817030729. The following problem is an absolute classic. It has appeared under different forms in Olympiads from all over the world. We will present the latest one, given at the 2003 Putnam Competition: 166 8. FORMAL SERIES REVISITED [Example 8.1 Find all partitions with two classes A, B of the set of non-negative integers having the property that for all nonnegative integers n the equation x + y = n with x < y has as many solutions (x, y) EA x A as in B x B. Solution. Let f and g be the generating functions of A and B respectively. Then f (x) = anxn , g(x) = E bnxn n>0 n>0 where, as in the previous problem, an equals 1 if n E A and 0 otherwise. The fact that A and B form a partition of the set of nonnegative integers can be also rewritten as f (x) + g(x) = xn = n>0 Also, the hypothesis on the number of solutions of the equation x + y = n implies that /2(x) — f(x2) = g2(x) g(x2) Hence f(x2) g(x2) = f( - g (x) 1 — x which can be rewritten as f (x) — g(x) 1 x ' f(x2) _ g(x2) Now, the idea is the same as in the previous problems: replace x by x2k and iterate. After multiplication, we deduce that f (x) — g(x) = 1(1_x2k ) lim (f(x2n) — g(X2n )). n—>oo k>0 Let us assume without loss of generality that 0 E A. You can easily verify that lira f(x2n) = 1 and lim g(x2n) = 0, n—>oo n—>co 1 1 — x THEORY AND EXAMPLES 167 which follows from the observation that 1 < f (x) < 1 + i x x and 0 < g(x) < x ix if 0 < x < 1. This shows that actually f(x) — g(x) = — x2k) = E( i)s2(k)xk , k>0 k>0 where .52(X) is the sum of the digits in the binary representation of x. Taking into account the relation f (x) + g(x) = 1 1 we finally deduce that A and B are respectively the set of nonnegative integers having even (respectively odd) sum of digits when written in base 2. We will discuss a nice problem in which formal series and complex numbers appear in a quite spectacular way: Let n and k be positive integers such that n > 2k-1 and let S = {1, 2, , n}. Prove that the number of subsets A of S for which x m (mod 2k) does not depend on m E xE A {0, 1, , 2k — 1}. Solution. Let us consider the function (call it formal series, if you want): f (x) = 11(1 + If we prove that 1 + x + • • • + x2k-1 divides f (x), then we have certainly done the job. In order to prove this, it suffices to prove that any 2kth root of unity, except for 1, is a root of f . But it suffices to observe that for any / E {1, 2, ... , 2k-1 — 1} we have ( / 2/7 t . 2 cos 2k + sin 2k ) = —1 2k-1—v2(l) 168 8. FORMAL SERIES REVISITED and so f ( cos — 217 + i sin 217) = 0, 2k 2k which settles our claim. Finally, it is time for a tough problem, solved by Constantin Tanasescu. rExample 10.] Let S be the set of all words which can be formed using m > 1 given letters. For any w E S, let 1(w) be its length. Also, let W C S be a set of words. We know that any word in S can be obtained in at most one way by concatenating words from W. Prove that E / 1 171 [Adrian Zahariuc] Solution. Let A be the set of all words which can be obtained by concatenating words from W. Let f(x) = E xl(w) , g(x) = E 1(w) x . wEW wEA By the definition of A, g(x) =1+ f(x)+ P(x) + • • • = 1- f(x). Hence f (x)g(x) = g(x) — 1. (8.3) Now, A (and W) has at most ink elements of length k, thus g(x) < oo and 1 1 7 7 2 f (x) < oo for x < -r-ri -. Thus for all x E 0, the expression in (8.3) is less wEW 1 THEORY AND EXAMPLES 169 ( than g(x) and so f(x) < 1 for all x E 0, — 1 . All we need now is to make m x tend to Irt — 1 and we will obtain f — 1 < 1, which is precisely the desired inequality. Indeed, observe that f can be written as f (x) = > anxn for some n>0 nonnegative real numbers an. Fix a positive integer N. Because akxk < f (x) < 1, k=o for all 0 < x < 1 by continuity of the polynomials it follows that E< 1 mk 1, k=0 and because N is arbitrary, we have E < 1, that is f (1) < 1. k>0 There is a very short solution for the following result using group theory. However, this is not the natural approach. The following solution may seem very involved and technical, but it was written in order to convince the reader that from time to time we need to work with composition of formal series, not merely with their sum and product. Example 11. 1 Let c(a) be the number of cycles (including those of length 1) in the decomposition of a into disjoint cycles. Prove that 1 E nc(a) (m + n — I) , m! o-ESm where 8, is the set of permutations of the set {1, 2, ..., m}. [Marvin Marcus] AMM 5751 Solution. Let us start with a Lemma: Lemma 8.2. For given nonnegative integers k1, k2, , kn such that kJ_ +2k2+ • • +nkn = n, the number of permutations of {1, 2, , n} which have ki cycles of length i for all i is n! ki!k2! • • • knqki 2k2 . nkn 170 8. FORMAL SERIES REVISITED Proof. Indeed, there are n! ways to fill in the elements of all cycles, but observe that every cycle of length j can be rotated around j ways and be the same cycle (so we must divide n! by jk3) and also there are kj ! ways to permute the cycles of length j in order to obtain the same permutation. All these operations being independent, the statement of the lemma follows. K Thus the sum we need to evaluate is m! n, !
k2! km!nki+k2+•-•+k mk m ki+2k2+•••+mk,n=7Th You will probably say that this is much more difficult than the initial problem, but you are not right, because the latter sum can also be written as 1 /3! !
• • • km! ( n n ) . . . P! ki!k2! 2 ) k2 m) 1cm, p +2k2+.••+mkm=rn +k2 +. +km = - p Now, observe that the multinomial formula implies that Ici!k2! 1 3! • •km! (n i )ki (n)k2 (n)km. . _ 2 ki+2k2+---1-mkm=m -f-k2 +..•+km=p Xm is the coefficient of Xm in the formal series (nX 1 + n 2 X 2 _ L n m Therefore the sum to be evaluated is the coefficient of Xm in the formal series m! • P ! 1 1 (nX nX2 - 2 + • • • + nXm rn! e 71 1 ) . 0 +10( 2 2 ± 71XM m Finally, observe that n2 2 + nX: _ n ln(1 — X), so nX +n2 +. .+nxm + = 1 e 1 2 (1 )7/ m! THEORY AND EXAMPLES 171 But using the binomial formula for (1 — x)-n we easily find the coefficient of X' in (1 —x 1 )n rn to be (n+m-1). This finishes the solution. We should also mention the beautiful solution using group theory. Remember that when a group G is acting on a set Y (that is, we can define for all g E G and x E Y an element g•x E Y such that for all g, h, x we have g•(h•x) = (gh)•x and 1•x = x), the number of orbits for the action of G on Y, that is the number of distinct sets of the form {g E G}, is equal to IGI gEG 1 IFix(g)I, where Fix(g) is the set of x E Y such that g • x = x. This is called Burnside's lemma and it is very useful, even though its proof is really simple: all you need to do is to count in two ways the pairs (g, x) such that g-x = x. Now, consider Y the set of the first m positive integers, and G the set of permutations of its elements. G acts obviously on the set of colorings of Y with n colors CI., C2, ..., C, (that is, on the set Y of functions from Y to {1, 2, ..., n}). The number of orbits is just the number of pairwise inequivalent classes of colorings, where two colorings are equivalent if they can be obtained by a permutation of G. Clearly, there are (n+7 7-1) such classes of equivalence (because they are determined by the nonnegative integers (k1, k2, kn) which add up to m, where lc, is the number of objects colored with the color CC; there are (n+77 7-1) solutions of the equation k1 +k2+- • • kn = m in nonnegative integers). On the other hand, we can use Burnside's lemma to count these pairwise inequivalent colorings. Observe that a permutation g fixes a coloring if and only if the numbers belonging to the cycles of g have the same color. Therefore, Fix(g) is the set of colorings which are constant on each cycle of g. There are n°(g) such colorings. Thus, there are 1 E nc(g) classes of colorings, and this finishes the proof of the identity. M! gEG In order to see whether you understood this type of argument, try to show 172 8. FORMAL SERIES REVISITED n (using this technique) that n divides E Ngcd(k'n) for all integers N. (Hint: k=1 count the number of classes of colorings of the vertices of a regular n-gon, two colorings being equivalent if they are obtained by a rotation keeping the polygon invariant.) PROBLEMS FOR TRAINING 173 8.2 Problems for training 1. Let zl, z2, ... , zr, be arbitrary complex numbers. Prove that for any s > 0 there are infinitely many numbers k such that 1 ,04 + 4 + • • • + ; k J > MaX(124111Z21) • • • )1 2 '4 — E.' 2. Find the general term of the sequence (xn)n>i given by In-Fk = aiXn+k-1 + ' • ' + alcxn with respect to xi, ... , xk. Here al, ... , ak and xl, .. • , xk complex numbers. are arbitrary 3. Let al, a2, ... , an be relatively prime positive integers. Find in closed form a sequence (xk)k>1 such that if yk is the number of positive integral solutions to the equation aixi + a2x2 + • • • + anxn = k, then urn — k = 1. k—>cx:, yk 4. Prove that if we partition the set of nonnegative integers into a finite number of infinite arithmetical sequences, then there will be two of them having the same common difference. 5. Is there an infinite set of nonnegative integers such that all sufficiently large integers can be represented in the same number of ways as the sum of two elements of the set? D. Newman 6. How many polynomials P with coefficients 0, 1, 2, or 3 satisfy P(2) = n, where n is a given positive integer? Romanian TST 1994 174 8. FORMAL SERIES REVISITED 7. Prove that for each positive integer k, 1 n1n2 • • • nk(ni. + nz + • • • + nk) = where the summation is taken after all k-tuples (ni, n2, ..., nk) of positive integers with no common divisor except 1. D. J. Newman, AMM 5336 8. Let n and k be positive integers. For any sequence of nonnegative inte-gers (al, a2, ak) which adds up to n, compute the product a1a2 • • ak. Prove that the sum of all these products is n(n2 — 12)(n2 22) (nz (k 1)2) (2k — 1)! 9. In how many different ways can we parenthesize a non-associative prod-uct aia2 an? Catalan 10. Let A be a finite set of nonnegative integers and define a sequence of sets by Ao = A and for all n > 0, an integer a is in An+1 if and only if exactly one of the integers a —1 and a is in An. Prove that for infinitely many positive integers k, Ak is the union of A with the set of numbers of the form k + a with a E A. Putnam Competition 11. Let Ai = 0, = {0} and An+i = {1+ xl X E Bn-Fi (An \ B n) U (Bn \ An). Find all positive integers n such that Bn = {0}? AMM PROBLEMS FOR TRAINING 175 12. For which positive integers n can we find real numbers al, a2, . . an such that {lai — ail I 1<i < <n} = {1,2,..., (n 2)}? Chinese TST 2002 13. Let f(r,n) be the number of partitions of n of the form n = bo + • • + bs where bi > rbi±i for all 0 < i < s — 1, and let g(r,n) be the number of partitions of n where each part has the form 1 + r + • • • + ri for some nonnegative integer i. Prove that f(r,n) = g(r, n) for all r and n. D. R. Hickerson, AMM 14. Is it possible to partition the set of all 12-digit numbers into groups of four numbers such that the numbers in each group have the same digits in 11 places and four consecutive digits in the remaining place? St. Petersburg Olympiad 15. Determine whether there is a subset X of the integers with the following property: for any integer n there is exactly one solution of a + 2b = n with a, b E X. Richard Stong, USAMO 1996 16. Let F(n) be the number of functions f : {1, 2, ..., n} —> {1, 2, ..., n} with the property that if i is in the range of f, then so is j for all j < i. Prove that kn F(n) 2k+1. k>0 L. Lovasz, Miklos Schweitzer Competition 176 8. FORMAL SERIES REVISITED 17. Suppose that every integer is colored using one of 4 colors. Let m, n be distinct odd integers such that m + n 0. Prove that there exist integers a, b of the same color such that a — b equals one of the numbers m, n, m — n, m + n. IMO 1999 Shortlist 18. Find all positive integers n with the following property: for any real numbers al, a2, , an, knowing the numbers a, + a3, i < j, determines the values al, a2, , an uniquely. Eras and Selfridge 19. Suppose that ao = al = 1 and (n + 3)an±i = (2n + 3)an + 3nan_i for n > 1. Prove that all terms of this sequence are integers. Kornai 20. Let x and y be noncommutative variables. Express in terms of n the constant term of the expression (x + y + x-1 + y-1)n. M. Haiman, D. Richman, AMM 6458 21. Consider (bn)n>1 a sequence of integers such that b1 = 0 and define al = 0 and an = nbn + aibn-i + • • • + an-1b1 for all n > 2. Prove that pap for any prime number p. 22. Prove that there exists a subset S of {1, 2, ..., n} such that 0, 1, 2, ..., n —1 all have an odd number of representations as x — y with x, y E S, if and only if 2n — 1 has a multiple of the form 2 • 4k — 1. Miklos Schweitzer Competition PROBLEMS FOR TRAINING 177 23. Suppose that (a„),>1 is a linearly recursive sequence of integers (that is, there exist integers r and xi, x2, ..., xr such that an-Fr = xian+r—i + x2an±,-2 + • • • + xran for all n) such that n divides an for all positive integers n. Prove that (n) is also a linearly recursive sequence. Polya 24. A set A of positive integers has the property that for some positive integers b2 j ci, the sets bzA + cz, 1 < i < n, are disjoint subsets of A. Prove that IMO 2004 Shortlist 25. Let f (n) be the number of partitions of n into parts taken from its divisors. Prove that (1 + 0(1)) r2) 1) lnn < lnf(n) < (1 + o(1)) 7 ( 2 n) lnn, where r(n) is the number of divisors of n. D. Bowman, AMM 6640 THEORY AND EXAMPLES 181 9.1 Theory and examples We have already seen some topics where algebra, number theory and combi-natorics were mixed in order to obtain some beautiful results. We are aware that such topics are not so easy to digest by the unexperienced reader, but we also think that it is fundamental to have a unified vision of elementary mathematics. This is why we have decided to combine algebra and number theory in this chapter. Your effort and patience will be tested again. The purpose of this chapter is to survey some classical results concerning algebraic numbers and their applications, as well as some connections between number theory and linear algebra. First, we recall some basic facts about matrices, determinants, and systems of linear equations. For example, the fact that any homogeneous linear system { anxi + ai2x2 + • • • + ainxn = 0 a2ix1 + a22x2 + • • • + a2nxn = 0 an 1 X 1 + an2X2 + • " a nnX n =0 in which all . . . and a12 a22 . . . ant aln a2n ann 0 has only the trivial solution. Second, we need Vandermonde's identity 1 Xi 2 X1 xi-1 1 x2 X2 2 x7 2 1-1 = H (xi - ) . (9 . 1 ) 1 Xn X2 n Xn n-1 1<i<j<n 182 9. A BRIEF INTRODUCTION TO ALGEBRAIC NUMBER THEORY Finally, when studying the algebraic numbers, we will need two more specific results. The first one is due to Hamilton and Cayley, while the second one is known as the fundamental theorem of symmetric polynomials Theorem 9.1. For any field F and any matrix A E Mn(F), if pA is the char-acteristic polynomial of A: pA(X) = det(XIn — A), then pA(A) = On. Theorem 9.2. Let A be a ring and let f E A[Xl, X2, ..., Xn] be a symmet- ric polynomial with coefficients in A, that is for any permutation a E Sn we have f X2, Xn) = f ()Cam, x0-(2), • • •, X,-(n)). Then we can find a poly- nomial g E A[Xi, X2, ..., Xn] such that f X2, • • •, Xn) = g(X1 + X2 + • • • + Xn, X1X2 X1X3 + • • • + Xn-lXn, •••, X1X2 • • Xn). This means that any symmetric polynomial with coefficients in a ring is a polynomial (with coefficients in the same ring) in the symmetric fundamental sums: Sk(Xj., Xn) = Xil • • Xik . 1<ii<i2<...<4<n As usual, we start with some easy examples. Here is a nice (and direct) application of theorem 2: Example ld Given a polynomial with complex coefficients, can one decide if it has a double zero only by performing additions, multipli-cations, and divisions on its coefficients? Solution. Yes, one can, even though at first glance this does not seem natural. Let f (x) = ao + aix + • • + anxn. Then this polynomial has a double zero if and only if = 0, where F(xi, x2, , xn) = fl (x„, — x i)2 and xi, x2, , xn are the zeros 1<z<j<n of the polynomial. At the same time F(xi,x2, • . • , xn) THEORY AND EXAMPLES 183 is symmetric with respect to xi, x2, xn, so by theorem 2 it is a polynomial in the fundamental symmetric sums in xi, x2, xn. By Vieta's formulas, these fundamental sums are just the coefficients of f (up to a sign), so F(xi, x2, .. • , xn) is a polynomial on the coefficients of f . Consequently, we can decide whether F(x1,x2,...,xn) = 0 only by using the operations on the coefficients of the polynomial mentioned in the hypothesis. This shows that the answer to the problem is positive. You may know the following classical problem: if a, b, c E Q satisfy a + b.n = 0, then a = b = c = 0. Have you ever thought about the general case? This cannot be done with only simple tricks. We need much more. Of course, there is a direct solution using Eisenstein's criterion applied to the polynomial f (X) = Xn — 2, but here is a beautiful proof using linear algebra. This time we need to be careful and work in the most appropriate field. Example 2. Prove that if ao, al, . • • , E Q satisfy ao + al 2 + • • • + ' .I /2n-1 = 0, then ao = ai = • • • = an_i = O. Solution. If ao + al 2 + • • • + an-1 V2n-1 = 0, then kao + kal + • + kan_i 1 =0 for any real number k. Hence we may assume that ao, al, , an_i E Z. The idea is to choose n values for k to obtain a system of linear equations having nontrivial solutions. Then the determinant of the system must be zero, and this will imply ao = ai = • • • = = 0. Now, let us fill in the blanks. What 184 9. A BRIEF INTRODUCTION TO ALGEBRAIC NUMBER THEORY are good values for k? This can be seen by noticing that 1Y271-1 • 7 0 . = 2 E Z. So, the values (k1, k2, , kri) = (1, 0, , 1/2n-1) are good, and the system becomes ao + al • + • • • + an_i • 1/2n-1 =0 ao • 0 + ai • n -c/P + • • • + 2an_i = 0 ao • 1/2n-1 - + 2a1 + • • + 2an_1 • V 2n-2 = 0. Viewing (1,Nr n 2, ... , .V2-1) as a nontrivial solution to the system, we conclude that ao 2an_i 2a1 al ao 2a2 an-i an-2 ao =0. But what can we do now? Expanding the determinant leads nowhere. As we said before passing to the solution, we should always work in the most appropriate field. This time the field is Z/2Z, since in this case the determinant can be easily computed; it equals ao = 0, where x means the residue class of the integer x modulo 2. Hence ao must be even, that is ao = 2b0 and we have bo an_i al ai ao 2a2 an-i an-2 a() =0. Now, we interchange the first two lines of the determinant. Its value remains 0, but when we expand it in Z2, it yields Ze il = 0. Similarly, we find that all a, are even. Let us write a2 = 2b2. Then we also have b0 + b1 + • • • + bn-1 • n-1/2n-1 = 0 and with the same reasoning we conclude that all b, are even. But of course, we can repeat this as long as we want. By the method of infinite descent, we find that ao = al = • • • = = 0. The above solution might seem exaggeratedly difficult compared with the one using Eisenstein's criterion, but the idea was too nice not to be presented here. The following problem can become a nightmare despite its apparent simplicity. THEORY AND EXAMPLES 185 Example 3. Let A = {a3+b3+c3-3abci a, b, c E Z}. Prove that if x, y E A, then xy E A. Solution. The observation that 3 3 3 r. a + b + c — oak = a c b b a c c b a leads to a quick solution. Indeed, it suffices to note that ( a c b (x b a c y x z = c b a z y x (ax+cy+bz az+by+cx ay+bx+cz = ay+bx+cz ax+cy+bz az+by+cx az+by+cx ay+bx+cz ax+cy+bz and thus (a3 + b3 + c 3 — 3abc)(x2 + y3 + z3 — 3xyz) = A3 + B 3 + C3 — 3ABC, where A = ax + bz + cy, B = ay + bx + cz, C = az + by + cx. You see, identities are not so hard to find... We all know the famous Bezout's theorem, stating that if al, a2, , an are relatively prime, then one can find integers kl, k2, , kn such that kiai + k2a2 + • • • + knan = 1. The following problem claims more, at least for n = 3. Example 4.1 Prove that if a, b, c are relatively prime integers, then there are integers x, y, z, u, v, w such that a(yw — zv) + b(zu — xw) + c(xv — yu) = 1. 186 9. A BRIEF INTRODUCTION TO ALGEBRAIC NUMBER THEORY Solution. The given condition can be written in the form det A = 1, where a x u A= b y v . C Z W So, let us prove a much more general result. Theorem 9.3. Any vector v whose integer components are relatively prime is the first column of an integral matrix with determinant equal to I. Proof. We induct on the dimension n of the vector v. Indeed, for n = 2 it is exactly Bezout's theorem. Now, assume that it is true for vectors in zn-1 and take v = (vi, v2, , vn) such that yi are relatively prime. Con- sider V Vn-i sider the numbers — , where g is the greatest common divisor of 7 7 • ' • 1 g g v1 V2, • • • , Vn-l• They are relatively prime and the matrix a12 an-1,2 has determinant equal to 1. We can find a, such that ag f3vn = 1 and matrix has integral entries and determinant 1: verify that the following u1 a12 al,n-1 ( 1)n-1 el g Vn-1 an-1,2 vn 0 • • • a n - 1 , n - 1)n-1,en-1 9 0 1)n-ia In the chapter Look at the Exponent we have seen a rather complicated solution for the following problem. This one is much easier, but difficult to find: I THEORY AND EXAMPLES 187 [Example 5. Prove that for any integers al, a2,••• n a , the number H aj — ai 1<i<j<n is an integer. [Armond Spencer] AMM E 2637 Solution. With this introduction, the way to proceed is clear. What does the expression H (aj — ai) suggest? It is the Vandermonde's identity (9.1), 1<i<j<n associated with al, a2, , an. But we have a hurdle here. We might want to use the same formula for the expression H (j - i). This is a dead end. 1<z<j<n But it is easy to prove that H (j - i) equals (n — 1)!(n — 2)! • • • 1!. Now, i<i<j<n we can write 1 -1 — 1 1<i<j<n 1! • 2! • • • (n — 1)! 1 1 1 1 al a2 a3 an • • • • • • • • • n-1 n-1 n-1 al a2 a3 an rt As usual, the last step is the most important. The above formula can be rewritten as 1 1 1 1 al a2 a3 an 1! 1! 1! 1! n-1 n-1 n-i a1 1 a2 a3 an (n — 1)! n — 1)! (n — 1)! (n — 1)! II aj — ai . . = 1<i<j<n 3 - 188 9. A BRIEF INTRODUCTION TO ALGEBRAIC NUMBER THEORY And now we recognize the form H 1<i<j<n 1 1 1 ( i ) (a2) 1 an ( 1 ) ai — (a1 2 ) (a 2 2) (a 2 n) ( al an — 1) (nag 1) (n — 1) which can be proved easily by subtracting lines. Because each number (a, is an integer, the determinant itself is an integer and the conclusion follows. At this point, you might be disappointed because we did not keep our promise: no trace of algebraic numbers appeared until now! Yet, we considered that a small introduction featuring easy problems and applications of linear algebra in number theory was absolutely necessary. Now, we can pass to the real purpose of this chapter, a small study of algebraic numbers. But what are they? Let us start with some definitions: we say that a complex number x is algebraic if it is a zero of a polynomial with rational coefficients. The monic polynomial of least degree, with rational coefficients and having x as a zero is called the minimal polynomial of x. Its other complex zeros are called the conjugates of x. Using the division algorithm, it is not difficult to prove that any polynomial with rational coefficients which has x as zero is a multiple of the minimal polynomial of x. Also, it is clear that the minimal polynomial of an algebraic number is irreducible in Q[X]. We say that the complex number x is an algebraic integer if it is zero of a monic polynomial with integer coefficients. You can prove, using Gauss's lemma, that an algebraic number is an algebraic integer if and only if its minimal polynomial has integer coefficients. In order to avoid confusion, we will call the usual integers "rational" integers in this chapter. There are two very important results concerning algebraic integers that you should know: THEORY AND EXAMPLES 189 Theorem 9.4. The sum or product of two algebraic numbers is algebraic. The sum or product of two algebraic integers is an algebraic integer. Proof. This result is extremely important, because it shows that the algebraic integers form a ring. Denote this ring by AI. None of the known proofs is really easy. The one that we are going to present first uses the funda-mental theorem of symmetric polynomials. Consider two algebraic numbers x and y and let xi, x2, ..., xn and yi, y2, ..., yrn be the conjugates of x and y n m respectively. Next, look at the polynomial f (x) = fJ 11 (X — xi — yi). We claim that it has rational coefficients. (The fact that x y is a zero of f being obvious.) This follows from the fundamental theorem of symmetric polynomials applied twice. Let R = Z[yi, y2, ..., yrn] be the ring considered in the statement of the Theorem 9.2. Because the coefficients of f are sym-metric polynomials in xi, x2, ..., xn, it follows that every coefficient of f is of the form B( \ai, az, •••, Grn) Y1) Y2) • • • ) Ym), where ai are the symmetric sums in xi, x2, ..., xri, and B is a polynomial with rational (respectively integer, if x, y are algebraic integers) coefficients. But the coefficients of f are also symmetric in Yi, Y2) • • • Y771 I so by taking R = Z[o-i,o -2,...,a-n] in Theorem 9.2, we deduce that A is a polynomial with rational (or integer) coefficients in the symmetric sums in xi, x2, ..., xn and yi, y2, ..., yrn. Thus f has rational coefficients if x, y are algebraic and f has integer coefficients if x, y are algebraic integers. There is also a solution which uses only the most elementary linear alge-bra! Indeed, we claim that a complex number z is an algebraic integer if and only if there exists a finitely generated commutative subring of C containing z. Indeed, if z is an algebraic integer, the division algorithm immediately shows that Z[z] is a finitely generated commutative subring of C. Now, sup-pose that R is a commutative subring of C which is finitely generated and contains z. Take vi, vz, vn that generate R and observe that the num-bers zvi, zv2, zvn are in R, thus they are linear combinations with integer coefficients of vi, v2, ..., vn. Let zvi = adz)]. + az2v2 + + ainvn for some in-tegers and and let A be the matrix with entries a,9. The above system of 190 9. A BRIEF INTRODUCTION TO ALGEBRAIC NUMBER THEORY equations can be written as (z/n — A)v = o, where v is the vector whose coordinates are v1, v2, ..., vn. Because v is not zero, the last relation implies det(z/n — A) = 0 and thus z is a root of the characteristic polynomial of A, (which is unitary and has integer coefficients), because so does A. This proves the claim. Now, consider two algebraic integers x, y. By the previous characterization and the fact that clearly y is an algebraic integer over Z[x], it follows that Z[x, y] -= (Z[x])[y] = Z[x]vi + • • + Z[x]vin, and since x is an algebraic integer there exist Th., .., up such that Z[x] = Zu1 + • • • + Zup. There-fore Z[x, y] c zukvi. Because Z[x + y] and Z[xy] are subsets of 1<k<p,1</<Tri Z[x, y], by applying the characterization again it follows that x y and xy are algebraic integers. Note however (and it is very important) that the set of algebraic integers is not a field (the following theorem will make this state-ment obvious), while the set of algebraic numbers is a field: if P(x) = 0 for some non-zero polynomial with integer coefficients P, then Q (1) = 0, where Q(X) = Xdeg(P) • P W. The next result is also very important, and we will see some of its applications in the following examples. Theorem 9.5. The only rational numbers which are also algebraic integers are the rational integers. Proof. The proof of this result is much easier. Indeed, suppose that x = P is a rational number (with gcd(p, q) = 1) which is also a zero of the monic polynomial with integer coefficients f (X) = Xn + an_iXn-1 + • • • + ctiX + ao. Then pn + an_ ipn—i q aipqn— aoqn = 0. Therefore q divides pn and since gcd(q,pn) = 1, we must have q = +1, which shows that x is a rational integer. Clearly, any rational integer x is an algebraic integer. Here is a very nice and difficult problem that appeared in AMM in 1998, and which is a consequence of these results. We prefer to give two solutions, one using the previous results and another one using linear algebra. A variant of THEORY AND EXAMPLES 191 this problem was given in 2004 at a Team Selection Test in Romania, and it turned out to be a surprisingly difficult problem. Example 6.1 Consider the sequence (xn)n>0 defined by x0 = 4, xi = x2 = 0, x3 = 3 and xn+4 = xn+I. + xn. Prove that for any prime p the number xp is a multiple of p. AMM Solution 1. Naturally, we start by considering the characteristic polynomial of the recursive relation: X4 — X — 1. It is easy to see that it cannot have a double zero. Using the theory of linear recursive sequences, it follows that the general term of the sequence is of the form Arl' + Br2 + Cr3 + Dr4 for some constants A, B, C, D. Here ri are the distinct zeros of the caracteristic polynomial. Because this polynomial has no rational zero, it is natural to suppose that Arr ii + Br2 + Cr7 3 1 + Dr4 is symmetric in r1, r2, r3, r4 and thus A=B=C=D. Because xo = 4, we should take A=B=C=D= 1. Now, let us see whether we can prove that xr, = ri + r7 2 2 + r7 3 ' + r4 for all n. Using Viete's formulae, we can check that this holds for n less than 4. But +4 since rim 7,7 7 +1 + an inductive argument shows that the formula is true for any n. Hence we need to prove that p divides q+ r2 + r3 + r4 for any prime number p. This follows from the more general result (which is also a generalization of Fermat's little theorem): Theorem 9.6. Let f be a monic polynomial with integer coefficients and let r1, r2, rn be its zeros (not necessarily distinct). Then A = (ri + r2 + • • + rn)P — (ri) . + r2 + • + r1 7,) ,) is a rational integer divisible by p for any prime number p. Proof. Theorem 9.2 shows that A is a rational integer because it is a symmetric polynomial in ri, r2, rn, and thus a polynomial with integer coefficients in the coefficients of f . The difficulty is to prove that it is a multiple of p. First 192 9. A BRIEF INTRODUCTION TO ALGEBRAIC NUMBER THEORY of all, let us prove by induction that if al, a2, ..., an are algebraic integers then 1 • ((al + a2 + • • • + ari)P - + 4 + • • • + al ,,;,)) is also an algebraic integer. For n = 2, this follows from the binomial formula 1 • ((a + b)P - aP - bP) = p-1 E P • P • (P) • aP-2bi . Indeed, 1- (1?) is an integer, and we obtain a sum of products t=i of algebraic integers, which is an algebraic integer. Now, if the assertion is true for n-1, consider al, a2, ..., an algebraic integers. By the inductive hypothesis, (al + a2 + • • • + an_l)P -(aP + + • • + aP n_1) E p • AI. The case n = 2 shows that (al + a2 + • • • + an)P - (al + a2 + • • • + an_i)P - E p • AI. Therefore, (al + az + • • • + an)P - + 4 + • • • + an) E p • AI (as being the sum of the above expressions), which is exactly what we needed to finish the inductive step. Now, finishing the proof of the theorem is easy: we know that 1 • ((al + a2 + • • • + an)P - + 4 + • • • + 4)) is a rational number which is also an algebraic integer. By theorem 4, it must be a rational integer. Solution 2. Let us consider the matrix ( 0 1 A = 0 0 0 0 1 0 0 0 0 1 1 1 0 0 and let Tr(X) be the sum of the entries of the main diagonal of the matrix X. We will first prove that xn = Tr(An) (here A° = /4). This is going to be the easy part of the solution. Indeed, for n = 1, 2, 3 it is not difficult to verify it. Now, assume that the statement is true for all i = 1, 2, ... , n - 1 and prove that it is also true for n. This follows from Xn = Xn_4 Xn_3 = Tr(An-4) Tr(An-3) = Tr(An-4(A + /4)) = Tr(An). We have used here the relation A4 = A + /4, which can be easily verified by a simple computation. Hence the claim is proved. Now, let us prove an important result-that is, Tr(AP) Tr(A) (mod p) for any integral matrix and any prime p. The proof is not trivial at all. A THEORY AND EXAMPLES 193 possible advanced solution is to start by considering the matrix A obtained by reducing all entries of A modulo p, then by working in a field in which the characteristic polynomial of A has all its zeros A1, A2, ... , An. This field clearly has characteristic p (it contains Zp) and so we have (using the binomial (P formula and the fact that all coefficients ' 1 < k < p — 1 are multiples of P) n n p Tr(AP) = =- (E Az)= (TrA)P, i=1 from where the conclusion is immediate, using Fermat's little theorem. But there is a beautiful elementary solution. Let us consider two integral matrices A, B, and write (A + B)P = E A1A2...Ap. A1,. .,ApE{A,13} Observe that for any A, B we have Tr(AB) = Tr(BA), and, by induction, for any Xi, X2,... , Xn and any cyclic permutation a, Tr (X X2 . • . Xn ) = Tr (X 0-(1)X0-(2) • X (n)) • Now, note that in the sum AlA2...Ap we can form 2P — 2 A1,...,ApE{A,B} groups of p-cycles and that we have two more terms, AP and B. Thus I Tr(A1A2...Ap) _= Tr(AP) + Tr(BP) A1,...,ApE{A,B} modulo p (you have already noticed that Fermat's little theorem comes in handy once again), since the sum of Tr(A1A2 Ap) is a multiple of p in any cycle. Thus we have proved that Tr(A + B)P Tr(AP) + Tr(BP) (mod p) and by an immediate induction we also have Tr(Ai + + A k)P Tr(AD + • • + Tr(4) (mod p) . 194 9. A BRIEF INTRODUCTION TO ALGEBRAIC NUMBER THEORY Next, consider the matrices Ei3 that have 1 in the position (i, j) and 0 else-where. For these matrices we have Tr(AP) =_ Tr(A) (mod p) and by using the above result we can write (using Fermat's little theorem one more time): TrAP = i,j E Tr(AP E aiiTrEij = TrA (mod p). The result is proved, and with it the fact that xp is a multiple of p. The example we are about to discuss next generated a whole mathematical theory and even an important area of research in transcendental number the-ory. Let us start by introducing a definition: for a complex polynomial f (X) = ariXn + an_iXn-1 + • • • +aiX +ao = ari(X —xi)• (X — x2) • • • (X — xn) define the Mahler measure of f to be MU) = land • max(1,1xil) • • • max(1,1xn1)- You can immediately see that M(f g) = M(f)- M(g) for all polynomials f and g. Using complex analysis, we can prove the following identity: MU) efol if(e2s7rt)Idt The next problem shows that a monic polynomial with integer coefficients and Mahler measure 1 has all of its zeros roots of unity. That is, the only algebraic integers all of whose conjugates lie on the unit disk of the complex plane are roots of the unity. This result is the celebrated Kronecker's theorem. [Example 7.] Let f be a monic polynomial with integer coefficients such that f (0) 0 and M(f) = 1. Then for each zero z of f there exists an n such that zn = 1. [Kronecker] THEORY AND EXAMPLES 195 Solution. What you are going to read now is one of those mathematical jew-els that you do not come across every day, so enjoy the following proof. Let f (X) = (X — xi) • (X — x2) • • • (X — xn) be the factorization of f in C[X]. Consider now the polynomials fk(X) = (X — x1 1) • (X — 4) • • • (X — xn). The coefficients of these polynomials are symmetric polynomials in x1, x2, •-, xn, and since all symmetric fundamental sums of x1, x2, ..., xn, are integers, all fk have integer coefficients (we used Theorem 9.2 here). What is really awesome is that there is a uniform bound on the coefficients of fk. Indeed, because all xi have absolute values at most 1, all symmetric fundamental sums in xk i,x2 k ...,xn have absolute values at most ( ) . Therefore, all coefficients of all polynomials fk are integers between — 2 ) and ( ). This shows that there are two identical polynomials among fi,12,13, .... Let i > j be such that f, = f j. Consequently, there is a permutation a of 1, 2, ..., n such that xi = Xa(i), = X afro. An easy induction shows that xi ir = x o.
3 ,.(1) for all r > 1. Because an! (1) = 1, we deduce that xi ini-a = 1 and so x1 is a root of the unity. Clearly, we can similarly prove that x2, x3, ..., xn are roots of the unity. After this example, a natural question appears: are there algebraic integers on the unit circle that are not roots of unity? The answer is yes, as the following example shows. Actually, part a) was known much before its publication in AMM. We invite the reader to take a look at the last chapter of this book for a proof of this more general result. Burnside proved a much more general result, which is left as exercise in the Problems for Training section, as a lemma in his famous theorem stating that any group whose cardinality is of the form pa qb for some primes p, q and some positive integers a, b is solvable. [Example 8. a) If a is a root of unity whose real part is an algebraic integer, then a4 = 1. b) There are algebraic integers of absolute value 1 and which are not roots of the unity. [H. S. Shapiro] AMM 4656 2 +a- 1 Solution. The proof of a) is very ingenious. Let b = Re(a) = a be the 196 9. A BRIEF INTRODUCTION TO ALGEBRAIC NUMBER THEORY real part of a, and consider ai, a2, ak the conjugates of a. We claim that the conjugates of b are distinct numbers among Re(ai), Re(a2), Re(ak). In- deed, the polynomial (x has b as a zero and its coefficients are 2 3=1 symmetric polynomials in ai (because a3 N = 1 for a suitable N), and rational by the theorem of symmetric polynomials. Thus all conjugates of b are among the zeros of this polynomial. On the other hand, if a4 1 then a4 1 for all j and so 0 < IRe(aj)I < 1, which means that the absolute value of the product of all conjugates of b is smaller than 1. Let h be the minimal polynomial of b over Q. Because b is an algebraic integer, h has integer coefficients, thus h(0) is an integer. But 11(0)1 is also the absolute value of the product of all conjugates of b, which is smaller than 1. It follows that h(0) = 0, and because h is irreducible in Q[X], it follows that h(X) = X and so b = 0, which is impossible if a4 1. Now b) is not so difficult. We will take a a zero of a polynomial of the form (X + 1)4 — uX2 for some integer u. We need to have I a I = 1 and also Re(a) needs to be an algebraic integer. If we also manage to ensure that a4 1, then we are done by a). You can easily check that by taking u = 8 all conditions are satisfied, and so 0 — 1 + iN/20 — 2 is an algebraic integer on the unit circle which is not a root of the unity. Some more comments on the previous examples are needed. First of all, it is not difficult to deduce from this result that the only monic polynomi-als with integer coefficients whose Mahler measure is 1 are products of X and some cyclotomic polynomials. A famous conjecture of Lehmer says that there exists a constant c > 1 such that if a polynomial with integer coeffi-cients has Mahler measure greater than 1, then its Mahler measure is actually greater than c. The polynomial with least Mahler measure found up to now is X19 + X9 X 7 X 6 X 5 X 4 X 3 + X + 1, whose Mahler measure is about 1.176. For some upper bounds of the Mahler measure in terms of the coefficients of the polynomial, we refer the reader to example 16 of chapter Pigeonhole Principle Revisited. Showing that a sum of square roots of positive integers is not a rational number is not difficult as long as the number of square roots is less than 3. Otherwise, THEORY AND EXAMPLES 197 this is much more complicated. Actually, one can prove the very beautiful result that if al, ..., an are positive integers such that /57 1 + • • • + Van is a rational number, then all a, are perfect squares. The following problem claims much less, but is still not simple. We will see how easy it becomes in the framework of the above results. Example 9. ] Prove that the number V10012 + 1 + V10022 + 1 + • • • + .V20002 + 1 is irrational. Chinese TST 2005 Solution. Let us suppose that the number is rational. Because it is a sum of algebraic integers, it is also an algebraic integer. By theorem 4, it follows that V10012 + 1 + V10022 + 1 + • • • + -V20002 + 1 is a rational integer. Hence V10012 + 1 + -V10022 + 1 + • • • + .V20002 + 1 — (1001 + 1002 + • • • + 2000) is a rational integer. But this cannot hold, because V10012 + 1 + V10022 + 1 + • • • + -V20002 + 1 — (1001 + 1002 + • • • + 2000) = 1 1 1 • 1001 + -V10012 + 1 + + + 1002 + -V10022 + 1 2000 + V20002 + 1 is greater than 0 and smaller than 1. The following example is very elegant, and you can easily check that the result is sharp: Example 10. Let x, y be complex numbers such that the expression x'-Y' x-y is an integer for some 4 consecutive positive integers n. Prove that it is an integer for any positive integer n. [Clark Kimberling] AMM E 2998 198 9. A BRIEF INTRODUCTION TO ALGEBRAIC NUMBER THEORY Solution. Let an be the given expression and let S = x+ y, P = xy. Observe that ari+2 - San+i + Pan = 0 for all n. Also, it is not difficult to prove that - a2 = —pa-1 art+ 1 an- 1 . Thus if an_i, an, and-1, ar,+2 are all integers, so are P71-1 and Pn. Thus P is an algebraic integer which is also rational (because P = p5-5 15 so that P is an integer. On the other hand, it is immediate to - prove by induction that an = fn(S) for some monic polynomial fn with integer coefficients, of degree n - 1. This shows that S is a zero of the monic polyno-mial with integer coefficients fn (X) - an, so S is an algebraic integer. Because S = an±2±Pan , S is also rational. Thus S is an integer, and in this case it and-1 is obvious that all terms of the sequence are integers, by the recursive relation. Here is a beautiful and difficult problem, where properties of algebraic integers come to the spotlight. Example 11.1 Let al, a2, ak be positive real numbers such that Y \l/a + n a2 + • • • +' •= v/ik is a rational number for all 7/ > 2. Prove that ai = a2 = ••• = ak = 1. Solution. First of all, we will prove that al, a2, ak are algebraic numbers and that al • a2 • • • ak = 1. Take an integer N > k and put X1 = Nyai, x2 = NVF /2, N!/— xk = ak. Then clearly xi + 4 + • • • + x k is rational for all 1 < j < N. Using New- ton's formulae, we can easily deduce that all symmetric fundamental sums of xi, x2, xk are rational numbers. Hence xi, x2, ..., xk are algebraic numbers, and so al = 4 11, a2 = •.., ak = are algebraic numbers as well. Also, by the argument above, we know that xi • x2 • • xk = Val • a2 • • • ak is rational, and this happens for all N > k. This implies immediately that al • a2 • • • ak = 1. Now let f(x) = brXr +br-iXr-1 + • • • + b0 be a polynomial THEORY AND EXAMPLES 199 with integer coefficients which vanishes at al, a2, •••, ak. Clearly, brat, •••, brak are algebraic integers. But then br( VF LI + Va2 + • • • + c/F tk) = • ( Vbral + Vbra2 + • • • + Vbrak) is also an algebraic integer. Because it is also a rational number it follows that it is a rational integer. Consequently, (br( +iy(a,2 + • • • +" -‘,X))7i>i is a sequence of positive integers. Because it converges to kbr, it eventu-ally becomes equal to kbr (from a rank). Thus there is n such that + + • • • + VaT, = k. Since al • a2 • • • ak = 1, the AM-GM inequality implies al = a2 = • • • = ak = 1 and the problem is solved. 200 9. A BRIEF INTRODUCTION TO ALGEBRAIC NUMBER THEORY 9.2 Problems for training 1. Let F1 = 1, F2 = 1 and F ri = Fn-1+ Fn-2 for all n > 3 be the Fibonacci sequence. Prove that Fn±,Fn_l— = (-1)Th and Fm+n = FnFrn-i Fn±iFm. 2. Compute the product H (63 - Ei)2, where 0<l<3<n-1 2 Ek = cos 2k71- ± 'I sin k n n for all k E {0 , 1, , n - 1}. 3. Let al, a2, , an E R. A move is transforming the n-tuple (x1, x2, . • • , xn) into the n-tuple (X1 + X2 X2 + X3 Xn-1 Xn X n + Xi Prove that if we start with an arbitrary n-tuple (al, a2) • • • an), after finitely many moves we obtain an n-tuple (A1, A2, , An) such that max lAr - Aj I < 22005 • i<i<j<n 4. Let a, b, c be relatively prime nonzero integers. Prove that for any rela-tively prime integers u, v, w satisfying au+bv+cw = 0, there are integers m, n, p such that a = nw - pv, b = pu - mw , c = mv - nu. 2 2 2 2 1 Octavian Stgna§ila, Romanian TST 1989 PROBLEMS FOR TRAINING 201 5. Prove that for any integers al, a2, , an the following number lcm(ai, a2, • • • an) a1a2 • • an H (ai — ai) 1<i<j<n is an integer divisible by 1!2! • • • (n — 2)!. Moreover, we cannot replace 1!2! • • • (n — 2)! by any other multiple of 1!2! • • (n — 2)!. 6. Let A, B, C be lattice points such that the angles of triangle ABC are rational multiples of 7r. Prove that triangle ABC is right and isosceles. 7. Let P be a regular polygon with 2001 sides, inscribed in the unit circle. Prove that the lengths of all sides and diagonals of P are irrational. AMM 8. Find all non-isosceles triangles with at least two sides rational numbers and having all angles rational numbers, when measured in degrees. Ron Evans, AMM E 2668 9. Let p be a prime and let al, a2, , ap+i be real numbers such that no matter how we eliminate one of them, the rest of the numbers can be divided into at least two nonempty pairwise disjoint subsets each having the same arithmetic mean. Prove that al = a2 = • • • = ap+i.• Marius RAdulescu, Romanian TST 1994 10. Let a, b, c be integers. Define the sequence (xn)n>0 by xo = 4, x1 = 0, x2 = 2c, x3 = 3b and xn+3 = axn—i + bxn cxn±i. Prove that for any prime p and any positive integer m, the number xpm is divisible by p. CalM Popescu, Romanian TST 2004 202 9. A BRIEF INTRODUCTION TO ALGEBRAIC NUMBER THEORY 11. Let a, b be two positive rational numbers such that for some n > 2 the number 7 , + vi) is rational. Prove that is also rational. Marius Cavachi, Gazeta Matematic6 12. Prove that the polynomial X' — 1 is divisible by a cubic monic polyno-mial, with integer coefficients if and only Marcel Tena, Gazeta Matematicg Contest 13. Prove that each of the numbers Vn + 1 — AF7, for positive integers n can be the written in the form 2 cos ( 7r) for some integers k, m. Chinese Olympiad 14. Prove that if al, a2, ..., am are positive integers, none of which is di-visible by the square of a prime number, and all having prime divi-sors in the set S = {Pi, P2 then there exist nonzero integers ci , c2, ce, d1, d2, ..., de such that > 0, all prime divisors of ci C2 • Ce are in S and (di + d2 c2 + • " + de AX) + b2 %/a2 + " • + bn Van) is a nonzero integer. Thus, the set of square roots of the square-free pos-itive integers is linearly independent over the field of rational numbers. Kvant 15. Let a be an algebraic number, and denote by K = Q[a] the field gen-erated by a over Q. Prove that there exists a positive integer m and algebraic integers al, a2, ..., am E K such that any other algebraic inte-ger in K is a linear combination, with rational integer coefficients, of al, a2, am,. if 31n or 41n. PROBLEMS FOR TRAINING 203 16. Let at, a2, ..., an be algebraic integers linearly independent over the field of rational numbers. Prove that there exists a positive constant C and a positive integer N such that for all rational numbers qi, qz, ..., qn, not all zero, C lqiai + q2a2 + • . • + qnanl ?: (km +1(721+ ... ± ', Inv- John Mc Carthy, AMM 4798 17. Let m, 11 be relatively prime numbers and x > 1 be a real number such that xm+2 -- and xn+I F-, are integers. Prove that x +1 is also an integer. 18. Consider 5 roots of order n of unity whose sum is not 0. Prove that the absolute value of their sum is at least 5'. Gerald Myerson, AMM 19. Find all solutions in rational numbers to the equation a2b + b2c + c2d + d2a = 0. THEORY AND EXAMPLES 207 10.1 Theory and examples Another topic with old tricks... you will probably say. Yet we might spend time on a problem just because we ignore obvious clues or basic aspects of it. This is why we think that talking about these "old fashioned tricks" is not because of lack of imagination, but rather an imperious need. In this note we combine some classical arithmetic properties of polynomials. This is just an introduction to this field, but some basic things should become second nature, and among them there will be some problems we discuss further. As usual, we keep some chestnuts for the end of the chapter, hoping that the hard-core solver will appreciate these extremely difficult problems. Recall that if f E Z[X] and a, b are integers, then a —b divides f (a)—f (b). This is the essential result which we will use relentlessly. Here are two applications: Let f, g be relatively prime polynomials with integer coeffi-cients. Define the sequence an = gcd(f (n), g(n)). Prove that this sequence is periodic. AMM Solution. As we have seen in previous problems, there exist polynomials F, G with integer coefficients and a positive integer A such that f F + gG = A. Thus an is a divisor of A for all n. Actually, we will prove that A is a period for the sequence (an)n>1. Let us prove that an a n,±A. We know that f (n + A) = f (n) (mod A), and since an divides A and f (n), it will also divide f (n + A). Simi-larly, an divides g(n + A) and so an I an+A. But the same relations show that an+A divides an and so an = an+A• Example 2. Let p E Z[x] be such that deg p > 1, and let A = {p(n) In E Z}. Prove that there exists an infinite arithmetical sequence none of whose terms can be expressed in the form f(x) for some integer x. 208 10. ARITHMETIC PROPERTIES OF POLYNOMIALS Solution. We will argue by contradiction: suppose that for all d > 2 and all n at least one of the numbers f (x) with x integer gives remainder n when divided by d. This means that for all n and d, the numbers p(n),p(n+1), p (n,+ d —1) give all remainders mod d. Indeed, because n, n +1, n+d— 1 are a complete system mod d, it follows that for any x, p(x) gives the same remainder mod d as one of p(n),p(n + 1), p(n + d — 1). In particular, any residue mod d appears as a residue mod d of one of the numbers p(n), p(n +1), ..., p(n+ d —1). Because deg(p) > 1, there exists n such that d = p(n + 1) — p(n) > 2. In this case, p(n) = p(n +1) (mod d) and so the numbers p(n), p(n+ 1), p(n + d — 1) give at most d — 1 distinct remainders mod d, which is a contradiction. We continue with an important result, due to Schur, that appears in many variations in contests. Even though in the topic At the Border of Analy-sis and Number Theory we prove an even more general result based on a nice analytical argument, we prefer to present here a purely arithmetical proof. Example 3.1 Let f E Z[X] be a non-constant polynomial. Then the set of prime numbers dividing at least one nonzero number among f (1), f (2), , f (n), is infinite. [Schur] Solution. First, suppose that f (0) = 1 and consider the numbers f (n!). For sufficiently large n, they are nonzero integers. Moreover, f (n!) 1 (mod n!) and so if we pick a prime divisor of each of the numbers f (n!), the conclusion follows (since in particular any such prime divisor is greater than n). Now, if f (0) = 0, everything is clear because in this case n divides f (n) for all n. x f (0 , Suppose that f (0) 0 and consider the polynomial g(x) f (f (0) )) . Clearly g E Z[X] and g(0) = 1. Applying now the first part of the solution, the problem is solved. THEORY AND EXAMPLES 209 This result has, as we have already said, important consequences. Here is a nice application. Example 4. I Suppose that f, g E Z[X] are monic nonconstant irreducible polynomials such that for all sufficiently large n, f (n) and g(n) have the same set of prime divisors. Then f = g. Solution. Indeed, by Gauss's lemma, the two polynomials are irreducible in Q[X]. In addition, if they are not equal, then the above remark and the fact that they have the same leading coefficient implies that the two polynomials are relatively prime in Q[X]. Using Bezout's theorem we conclude that there is a nonzero integer N and P, Q E Z[X] such that fP + gQ = N. This shows, that for n large enough, all prime factors of f (n) divide N. But, of course, this contradicts Schur's result. The result of Example 2 remains true if we assume the same property is valid for infinitely many numbers n. Yet the proof uses some highly non-elementary results of Erdos. The interested reader will find rich literature on this field. A refinement of Schur's theorem is discussed in the following example. The key additional ingredient is the Chinese remainder theorem. Example 5. 1 f E Z[X] be a non constant polynomial and let n, k be L_ positive integers. Prove that there exists a positive integer a such that each of the numbers f (a), f (a + 1), . , f (a + n — 1) has at least k distinct prime divisors. Bulgarian Olympiad Solution. Let us consider an array of distinct prime numbers (pii)i<i,j<k such that f (xis) 0 (mod pig) for some positive integers xis. This is just a direct consequence of Schur's theorem. Now, using the Chinese remainder theorem, we can find a positive integer a such that a + i — 1 (mod pig) for all 210 10. ARITHMETIC PROPERTIES OF POLYNOMIALS indices i and j. Using the fundamental result mentioned in the beginning (namely that f (a) — f (b) is always divisible by a — b), it follows that each of the numbers f (a), f (a+1), . . , f (a+n —1) has at least k distinct prime divisors. We continue with two more difficult examples of problems whose solutions are based on combinations of Schur's theorem with various classical arguments. [Example 6.1 For integral m, let p(m) be the greatest prime positive divisor of m. By convention we set p(1) = p(-1) = 1 and p(0) = co. Find all polynomials f with integer coefficients such that the sequence (p(f (n2)) — 2n),,>0 is bounded above. [Titu Andreescu, Gabriel Dospinescu] USAMO 2006 Solution. When searching for the possible answer, one should start with easy examples. Here, the quadratic polynomials might give an insight. Indeed, observe that if u is an odd integer then the polynomial f(X) = 4X — u2 is a solution to the problem. This suggests that any polynomial of the form c(4X — ai)(4X — 4)...(4X — al) is a solution if c is a nonzero integer and al, a2, a/ are odd integers. Indeed, any prime divisor p of f (n2) is either a divisor of c (and thus in a finite set) or a divisor of some (2n — a3) (2n + a3). In this case p — 2n < max(ai, a2, ak) and so f is a solution of the problem. We deal now with the much more difficult part: showing the converse. Take f a polynomial that satisfies the conditions of the problem, and suppose that p(f (n2)) — 2n < 2A for some constant A. Using Schur's theorem for the poly-nomial f (X 2), we deduce the existence of a sequence of different prime num-bers p3 and nonnegative integers k3 such that p3 Define the sequence r3 = min(k3 (mod p3), 1)3 — k3 (mod p3)) and observe that p3 divides f (7-3 2) and also that 0 < r3 < P32 1. Hence 1 < p3 — 2r3 < A and so the sequence (p3 — 2r3 )3>1 must take some value al infinitely many times. Let p3 — 2r = al for j in an infinite set X. Then, if m = deg( f), we have p3 14m f ((P3 2a1)2) for all j E X and also the polynomial 4772 • f((x-2a1\2\ ) ) has integer coefficients. a2 a2 i This shows that pl divides 4m • f (-,1-) for infinitely many j. Hence -, 1 1 s a root THEORY AND EXAMPLES 211 of f . Because f (n2) does not vanish, al must be odd. This means that there exists a polynomial g with integer coefficients and a rational number r such that f (X) -= r(4X — aT)g(X). Of course, g has the same property as f, and applying the previous arguments finitely many times we deduce that f must be of the form c(4X — a?)(4X — 4)...(4X — ak) for a certain rational number c and odd integers al, a2, ak. But do not forget that all coefficients of f are integers! Therefore the denominator of c is a divisor of both 4m and 4 .4...4, thus it is 1. This shows that c is an integer and the solution finishes here. The next problem, which uses Schur's theorem, also needs a classical result, a very particular case of Hensel's lemma. Let us first state and prove this result and then concentrate on the following problem. So, let us first prove the following: Lemma 10.1 (Hensel's lemma). Let f be a polynomial with integer coeffi-cients, p a prime number and n an integer such that p divides f (n) and p does not divide f (n). Then there exists a sequence (nk)k>1 of integers such that ni = n, pk divides nk+1 — nk and pk divides f (nk). Proof. The proof is surprisingly simple. Indeed, let us suppose that we have found i and search for ni+i = ni+b.pi such that pi+1 divides f (ni+i). Because 2i > i + 1, using the binomial formula yields f (ni + b • pi) = f (ni) + bpi f i (ni) (mod pz+1). Let f (ni) = cpz for some integer c. Because 74 n (mod p), we have rni) (n) (mod p) and so rni) is invertible modulo p. Let m be the inverse of rn.,) modulo p. It is enough to choose b = —mc in order to finish the inductive step. We can now discuss a difficult problem used for the preparation of the Iranian IMO team: 212 10. ARITHMETIC PROPERTIES OF POLYNOMIALS [Example 7. Find all polynomials f with integer coefficients such that nlm whenever f (n)I f (m). [Mohsen Jamali] Iranian TST Solution. [Adrian Zahariuc] With this preparation, the solution will be short, which does not mean that the problem is easy (as we already said). First of all, observe that for a nonconstant polynomial with integer coefficients such that f (0) 0 and for any k there are infinitely many prime numbers p such that PkIi(n) for some integer n. Indeed, by working with an irreducible divisor of f , we can assume that f is irreducible. Thus f and f' are relatively prime in the ring of polynomials with rational coefficients. Bezout's theorem shows in this case that there exist integer polynomials S, Q and an integer A 0 such that Sf Qf' =- A. Therefore, if p is a sufficiently large prime such that plf(n) for some 71 (the existence of infinitely many such primes follows from Schur's theorem), p will not divide fi(n), and we can apply Hensel's lemma to finish the proof of this result. Next, observe that XI f (X). Indeed, we have f(n)I f(n f (n)) for all n, so nIn + f (n) for all it which easily implies f (0) = 0. So, let us write f (X) = Xk g(X) with g(0) 0. Assume that g is nonconstant. By the previous result, there exists a prime p such that p > 19(0)1 and pk1g(m) for some integer m. Clearly, p does not divide g(p), so by the Chinese remain-der theorem there exists an integer n such that n m (mod pk) and n p (mod g(p)). Thus pk Ig(n) and g (p)Ig (n), and thus f (p)1 f (n). This implies that pin, and this is impossible, because it would follow that plg(0). Thus g is constant and the answer is: all polynomials of the form can. Here is another application of Hensel's lemma. The example below is quite a difficult problem, especially because examples of small degree cannot be found: Example 8:1 Is there a polynomial f with integer coefficients that has no rational zeros, but has a zero modulo any positive integer? Kornai THEORY AND EXAMPLES 213 Solution. The answer is yes, but it is not obvious why such polynomials exist. A very difficult theorem of Chebotarev implies that such polynomials with small degree (smaller than 5) do not exist. It can be proved that there are such polynomials of degree 5, but the example we have chosen has degree 6: define f (X) = (X2 3)(x2 _ 13)(x2+ 39). We will prove that for any n there exists m such that n1 f (m). Observe first of all that it is enough to prove this if n is a power of a prime. Indeed, if we can find ml, m2 such that ni If (mi) and n2I f (m2) for some relatively prime integers ni, n2 then by taking m such that m = m1 (mod ni) and m = m2 (mod n2) (which is possible by the Chinese remainder theorem) we have an m such that nin2If (m). Let us now deal with powers of 2. We will prove by induction the existence of a sequence xn such that 2n1xn 2 + 39. For n = 1 we take x1 = 1, for n = 2 we take x2 = 1, as well as x3 = 1. Now assume that xn 2 + 39 = 2n • k for some integer k and n > 3. Then (2n-1x x n)2 + 39 = 2n(xxn + k) (mod 2n+1). If k is even we define Xn+1 = xn• Otherwise, we define x = 1 and so xn+1 = xn + k. In either case, 2n+11xn 2+1 + 39. Now, let us deal with powers of 3 and 13; actually, this case follows immediately from Hensel's lemma applied to the polynomials X2 - 13 and X2 +3, with n = 1 and n = 6 respectively. Finally, take p a prime number different from 2, 3, 13 and observe that the identity (-39) • (13) • (1 = 1 P P P (where M denotes the Legendre symbol) implies that one of the numbers (-39)(la) and (= 3 -) equals 1. This shows the existence of an integer m P such that for some a equal to 3, -13, 39 we have plm2 + a. It is now enough to apply Hensel's lemma for the polynomial X2 + a in order to obtain a sequence xn such that PnI4, any prime p and any positive integer n, and by the remark in the beginning of the solution this polynomial is a solution of the problem. [Example 9. Find all polynomials f with integer coefficients and the follow-ing property: for any relatively prime positive integers a, b, the sequence (f (an + b)),>0 contains an infinite number of terms, any two of which are relatively prime. + a for all n. This shows that f has a root modulo pn for [Gabriel Dospinescu] 214 10. ARITHMETIC PROPERTIES OF POLYNOMIALS Solution. Clearly, constant polynomials can be eliminated. We will prove that the only polynomials with this property are those of the form X" and —X', with n a positive integer. Because changing f with its opposite does not modify the property of the polynomial, we can assume that the leading coefficient of f is positive. Hence there exists a constant M such that f (n) > 2 for all n > M. From now on, we consider only n > M. Let us prove that we have gcd(f(n), n) 1 for any such n. Suppose that there is an n > M such that gcd(f (n), n) = 1. The sequence (f (n + k f (n)))k>0 would contain at least two relatively prime numbers. Let them be s and r. Because f (n) kf(n) = kf(n) + n — n I f(kf(n)+n)— f(n), we have f(n) f(n+kf(n)) for any positive integer k. It follows that s and r are both multiples of f (n) > 2, which is impossible. We have shown that gcd(f(n), n) 1 for any n > M. Thus for any prime p > M we have pj f (p) and so pl f (0). Because any nonzero integer has a finite number of divisors, we conclude that f (0) = 0. Hence there is a polynomial q with integer coefficients such that f (X) = X q(X). It is clear that q has positive leading coefficient and the same property as f . Repeating the above argument, we infer that if q is nonconstant, then q(0) = 0 and q(X) = X h(X). Because f is nonconstant, the above argument cannot be repeated infinitely many times, and thus one of the polynomials g and h must be constant. Consequently, there are positive integers n, k such that f (X) = kX". But since the sequence (f (2n + 3))„>0 contains at least two relatively prime integers, we must have k = 1. We obtain that f is of the form X'. Because f is a solution if and only if —f is a solution, we infer that any solution of the problem is a polynomial of the form ±Xn. Now let us prove that the polynomials of the form X', —X' are solutions. It suffices to prove it for X' and even for X; but this follows from Dirichlet's the-orem. There is another more elementary approach. Suppose that xi, x2, , xp are chosen such that the numbers ax, + b are pairwise relatively prime. We prove that we can add xp+i so that axl + b, ax2 + b, . . . , axp±i + b are pairwise relatively prime. Clearly, ax1 + b, ax2 + b, . . . , axp + b are relatively prime to a, so we can apply the Chinese remainder theorem to find an xp±i greater than xi, x2, , xp, such that xp+1 (1 — b)az -1 (mod axi + b), i E {1, 2, ... ,p}, where ai -1 is a's inverse in Zx,±1). Then gcd(axp±i + b, ax, + b) = 1 for i E {1,2, , p} and thus we can add xp-Fi. THEORY AND EXAMPLES 215 Example 10.] Find all polynomials f with integer coefficients such that f (n) Inn-1 — 1 for all sufficiently large n. [Gabriel Dospinescu] Solution. Clearly, f (X) = X — 1 is a solution, so let us consider an arbitrary solution and write it in the form f (X) = (X —1)7 g(X) with r > 0 and g E Z[X] with g(1) 0. Thus there exists M such that g(n)Inn-1 — 1 for all n > M. We will prove that g is constant. Assuming the contrary, we may assume without loss of generality that the leading coefficient of g is positive. Thus there is k > M such that g(n) > 2 and g(n)Inn-1 — 1 for all n > k. Now, since n + g(n) — nig (n + g(n)) — g(n), we deduce that g(n)1g(n + g(n)) for all n. In particular, for all n > k we have g(n)Ig(n + g(n))1(n + g(n))n±g(n)-1 — 1 and g(n)10-1 — 1. Of course, this implies that — g(n) Inn±g(n) 1 1 =(nn — )rtg(n) n g(n) — 1, that is g(n)Ing(n) — 1 for all n > k. Now, let us consider a prime number p > k and let us look at the smallest prime divisor q of g(p + 1) > 2. We clearly have qlg(p +1)1(p +1)9(7)+1) — 1 and gl(p + 1)q-1 — 1. Since gcd(g(p + 1), g — 1) = 1 (by minimality) and , gcd((p + 1)9(7)+1-) — 1, (p + 1)q-1 1) 1)gcd(g(P+1),q-1) 1 = p it follows that we actually have p = q. This shows that plg(p + 1) and thus (again using the fundamental result) plg(1). Because this occurs for any prime number p > k, we must have g(1) = 0. This contradiction shows that g is indeed constant. Let g(X) = c. Thus cl2n(2n-1) — 1 for all n > M. Given that gcd(2a — 2b — 1) = 2gcd(a,b) 1, in order to show that 1cl = 1, it suffices to exhibit k < m < n such that gcd(m(2m — 1), n(2n — 1)) = 1. This is easy to achieve. Indeed, it suffices to take a prime number m greater than M, k and to choose 216 10. ARITHMETIC PROPERTIES OF POLYNOMIALS a prime number n greater than m(2m - 1). A simple argument shows that gcd(m(277' - n(2n - 1)) = 1 and so = 1. Finally, let us prove that r < 2. Assuming the contrary, we deduce that (n - 1)31nn-1 - 1 <=> (n - 1)2 n i n-2 + nn-3 n +1 for all sufficiently large n, and since nn-2 nn-3 ± • • • +n+1= = (n - 1)[nn-3 + 2nn-4 + • • • + (n - 3)n + (n - 2) + we obtain n - 1 lnn 3 + 20-4 + • • + (n - 3)n + (n - 2) + 1 for all sufficiently large n, which is clearly impossible, since n'3 + 2n' 4 + • • • + (n - 3)n + (n - 2) + 1 1 + 2 + • • • + (n - 2) + 1 (n - 1)(n - 2) + 1 (mod n - 1). 2 Hence r < 2. The relation nn-1 — 1= (n - 1)2[nn-3 + 2e-4 + • • • + (n - 3)n + (n - 2) + 1] shows that (n - 1)2inn-1 - 1 for all n > 1 and allows us to conclude that all solutions are the polynomials +(X - 1)r, with r E {0,1, 2}. After reading the solution of the following problem, you might think that the problem is very simple. Actually, it is extremely difficult. There are many possible approaches that fail and the time spent for solving such a problem can be significant. Example 11d Let f E Z[X] be a nonconstant polynomial, and let k > 2 be an integer such that 1 0(n) E Z for all positive integers n. Then there exists a polynomial g E Z[X] such that f = g1. THEORY AND EXAMPLES 217 Solution. Let us assume the contrary, and let us factor f = pil ...ps icsgk where 1 < k2 < k, and pi are different irreducible polynomials in Q[X]. Suppose that s > 1 (which is the same as negating the conclusion). Because p1 is irreducible in Q[X], it is relatively prime with /42 p, and thus (using Bezout's theo-rem and multiplication by integers) there exist polynomials Q, R with integer coefficients and a positive integer c such that Q(x)pi(x) + R(x)pi i(x)p2(x) ps(x) =- c. Now, using the result from Example 1, we can take a prime number q > I cl and a number n such that qlpi(n) 0. We have of course qlpi(n + q) (since pi (n q) = pi (n) (mod q)). The choice q > 1cl ensures that q does not divide P2(n) • • .ps(n) and so vq( f (n)) = vq(pi(n)) + kvq(g(n)). But the hypothesis implies that k vq(f(n)), so vq(pi(n)) > 2. In a similar manner we obtain vq(pi(n + q)) > 2. Yet, using the binomial formula, Pi (n + q) = Pi (n) (n) (mod q2). Hence we must have qlpi(n), which contradicts the fact that q > I cl and Q(x)pi(x) + R(x)pi i(x)p2(x) ps(x) = c. This contradiction shows that s > 1 is false and the result follows. The next problem was given at the USA TST 2005 and uses a nice combination of arithmetic considerations and complex number computations. We take advantage of many arithmetical properties of polynomials in this problem, although the problem itself is not so difficult (if we find a good way to solve it, of course...). Example 12.1 A polynomial f E Z[X] is called special if for any positive in-teger k > 1, the sequence 1(4 f (2), f (3), . contains num-bers which are relatively prime to k. Prove that for any n > 1, at least 71% of all monic polynomials of degree n with coefficients in the set {1,2, , n!} are special. [Titu Andreescu, Gabriel Dospinescu] USA TST 2005 218 10. ARITHMETIC PROPERTIES OF POLYNOMIALS Solution. Of course, before counting such polynomials, it would be better to find an easier characterization for them. Let pi, p2, , pr be all the prime numbers not exceeding n, and consider the sets Ai = {f E MI Pilf(m), d m E N}, where M is the set of monic polynomials of degree n with coefficients in the set {1, 2, ... , n!}. We will prove that the set T of special polynomials is exactly M \ U Ai. Clearly, i=i T C M \ U Ai. The converse, however, is not that easy. Let us suppose that i<r f E Z[X] belongs to M \ U A, and let p be a prime number greater than 7,=-1 n. Because f is monic, Lagrange's theorem ensures that we can find m such that p is not a divisor of f(m). It follows that for any prime number q at least one of the numbers f (1), f (2), f (3), . . . is not a multiple of q. Let k > 1 and let qi, q2,... , q, be its prime divisors. Then we can find ul, , us such that q, does not divide f (u,). Using the Chinese remainder theorem, there is a positive integer x such that x u, (mod qi). Consequently, f(x) gui) (mod qi) and thus q, does not divide f (x), so gcd(f (x), k) = 1. The equality of the two sets is now proved. Using a raw estimation, we obtain r U Ai i=i ?_ Iml - lAil. i=1 (n!)n Let us now compute IA ; I. Actually, we will show that IN = a monic polynomial in Ai, f (X) = X' + an_iXn-1 + • • • + a1X + Let f be Then, for any m > 1, 0 f (m) + (al + ap + a2p_1 + a3p-2 • . . )ni +(a2 ap+1 a2p + ... )m2 + • • + (ap—i + a2p-2 + a3p-3+ • • • )mP 1 (mod p), THEORY AND EXAMPLES 219 where, for simplicity, we put p = A. Again, using Lagrange's theorem it follows that p i ao, p I al + ap + a2p-1 + • • • • • • 1p I ap-1 a2p-2 + • • • We are going to use this later, but a small observation is still needed. Let us count the number of s-tuples (x1, x2, , x3) E {1, 2, , n!}3 such that xl + x2 + • • • + xs u (mod p), where u is fixed. Let 27r27r E = cos — + sin — p p and observe that 0= (E E2 En! p-1 1{(xl, X2, Xs) E {1) 21 • • • nfts 1 + • • + Xs k (mod P)}1. k=0 A simple argument related to the irreducibility of the polynomial 1+ X + X2 + • + XP-1 shows that all numbers that appear in the above sum are equal, and that their sum is (n!)8, thus each number equals (n!)S p We are now ready to finish the proof. Assume that among the numbers ai, ap, a2p_i, ... there are exactly vi numbers, and so on, finally there are vp_i numbers among ap_i, a2p_2, .... Using the above observations, it follows that n! (n!r1 (n!)vP-1 - (n!)fl I Ai I = p p p PP Hence I T1 (on - (n!)n p prime PP But 1 1 1( 1 1 1 1 55 ± 7 7 + < 55 + 5 + 52 + • • • < 1000 and so the percent of special polynomials is at least 100 ( 1 - - 1 ) = 75 100 1> 71. 4 27 1000 27 10 220 10. ARITHMETIC PROPERTIES OF POLYNOMIALS Just a few more observations about this problem. The authors discovered (af-ter the problem was submitted and given in the TST) that this question was the object of Jan Turk's article The fixed divisor of a polynomial published in the fourth issue of the American Mathematical Monthly, 1986. In this arti-cle, with a completely different idea and technique, much more involved and precise estimations are obtained. For instance, the author proves that the probability for a random polynomial with integer coefficients to be special is H (1 - 1), which is approximately 0.722. This shows that even though our PP estimations were very elementary, they were not far from reality. We invite the reader to read this fascinating article. Suppose that a polynomial f with integer coefficients has no double zeros. Then for any positive integer r there exists an n such that in the prime decomposition of f (n) there are at least r distinct prime divisors, all of them with exponent 1. Iranian Olympiad Solution. Already for r = 1 the problem is in no way obvious. So let's not attack the general case directly, but rather concentrate first on the case r = 1. Suppose the contrary, that is for all n the prime divisors of f (n) have exponent at least 2. Because f has no double zero, gcd(f, = 1 in C[X] and thus also in Q[X] (because of the division algorithm and Euclid's algorithm). Using Bezout's theorem in Q[X], we can find polynomials P, Q with integer coefficients such that P (n) f (n) + Q(n) f' (n) = c for some positive integer c. Using the result in the first example, we can take q > c a prime divisor of some f (n). Our hypothesis ensures that q2 lf (n). But then, also, ql f (n + q) and so q2 jf (n + q). Using Newton's binomial formula, we deduce immediately that f (n + q) = f (n) + qf y (n) (mod q2). We finally find ql (n) and so q1c, which is impossible, since our choice was q > c. Thus the case r = 1 is proved. Let us now try to prove the property by induction and suppose it is true for r. Of course, the existence of P, Q such that P (n) f (n) + Q (n) (n) = c for some positive integer c did not depend on r, so we keep the above notations. By 13. THEORY AND EXAMPLES 221 the inductive hypothesis, there is n such that at least r prime divisors of f (n) have exponent 1. Let these prime factors beP1, P2, , pr. But it is clear that n + kp7p3 p2 , has the same property: all prime divisors pi, P2, • • • , Pr have exponent 1 in the decomposition of f (n+kpiA . . . p7.
2). Because at most a finite number among them can be zeros of f , we may assume from the beginning that n is not a zero of f. Consider now the polynomial g (X) = f (n+ (pi . 1302 X), which is obviously nonconstant. Thus using again the result in Example 1, we find a prime number q > maxficl, pi, ... ,Pr, IP(n)11 and a number u such that qlg(u). If vq(g(u)) = 1, victory is ours, since a trivial verification shows that q, pi, are different prime numbers whose exponents in f (n+ (pi p r)2u) are all 1. The difficult case is when vq(g(u)) > 2. In this case, we will consider the number N = n + u(pi pr)2 + uq(pi p r)2 Let us prove that in the decomposition of f (N), all prime numbers q, pi, , pr have exponent 1. For any pi, this is true since f (N) f (n) (mod (p1 ...p r)2). Using once again the binomial formula, we obtain f (N) = f (n + (p1 p r)2u) + uq(pi pr.)21(N) (mod q2). Now, if vq( f (n)) > 2, then since vq(f (n + (pi pr)2u)) = vq(g(u)) > 2, we have qiu(pi /302 (N). Recall that the choice was q > maxficl, pi, , pr, Ip(n)l} so necessarily qlu (if qlf (N) ql(f (N), (N))1c = q < contradic-tion). But since gig (u), we have qlg(0) = f (n). Fortunately, we ensured that n is not a zero of our polynomial and also that q > max{ ... ,pr, Ip(n)1} so the last divisibility cannot hold. This finishes the inductive step and solves the problem. Did you like ErdOs's Corner in chapter Look at the Exponent? We repeat the experience, with a series of difficult problems related to prime divisors of polynomials. When we say difficult, we say however solvable, because one should know that most of the problems concerning quantitative estimates for prime divisors of polynomials are still unsolved and will probably remain so for very long time. Let us recall a few terrible results that have been obtained so far, of course without proofs. Let P(n) be the greatest prime divisor of n. Even the fact that P(f (n)) tends to oo for any polynomial f of degree at least 222 10. ARITHMETIC PROPERTIES OF POLYNOMIALS 2 is a very difficult result (even the case deg( f) = 2 requires the Thue-Siegel theorems). An extremely difficult theorem of Erd6s shows that the largest prime divisor of f (1)f (2)... f (n) is greater than n • e(ln n) c for some absolute constant c > 0. All these results require very deep results in algebraic and ana-lytic number theory. Another is the famous open question of prime-producing polynomials: any polynomial f without a fixed divisor should produce prime numbers infinitely many times. All these questions are far beyond the known results. But, of course, we will discuss just a few results with elementary (more or less) solutions. The first problem investigates Schur's theorem for a family of polynomials. The following solution was suggested to us by Vesselin Dimitrov. The beauty of the result can be easily seen when studying the second part of the problem, where we prove by elementary means a result that usually was proved using Galois theory. Even though we haven't found the first article studying this problem, we did find one signed by T. Nagell, so we will call this Nagell's theorem. [Example 14. a) Let fi , 12, • • • , fn be nonconstant polynomials with inte-ger coefficients. Prove that there are infinitely many primes numbers p with the property that fl , f2, • • • , fn have a zero in Z/p7L (that is, there exist integers k1 , k2, • • • , k,, such that pifi(ki) for all i). b) Prove that for any nonconstant polynomial f with inte-ger coefficients and any positive integer k there are infinitely many primes of the form 1 + qk that divide at least one of the numbers f (1), f (2), f (3), .... Nagell's theorem Solution. For n = 1, a) is just Schur's theorem. Actually, the idea is to reduce the study to this special case, by proving the existence of polynomials gi, g2, , gn such that fi(gi(X)) have a common nontrivial divisor. This is not immediate, however. Let us see what we are asking for: of course, if there THEORY AND EXAMPLES 223 exists a common nontrivial divisor, it must have a complex root z, so first of all we should see whether we can find g2 with rational coefficients and some z such that fi(g,(X)) have common root z. In this case, g2(z) would be all the zeros of L, so it is more than natural to start by fixing some roots xi, x2, . , xn of f2, , f,,, respectively and trying to find some z and some gi with g2(z) = xi. And now, a very useful theorem from algebraic number theory (but whose proof is completely elementary) helps us: actually, any finite extension of the field of rational numbers is generated by one element. T hat is, if al, a2, . • • , ak are algebraic numbers (over the field of rational numbers), then there exists an algebraic number a such that Q(ai, a2, , ak) = Q(a). We will leave the proof of this theorem as a beautiful exercise for the reader (in case you do not manage to solve it alone, any introductory book to algebraic number theory gives a proof of this result). Now, xi are clearly algebraic, since they are roots of f„. Thus there exists some algebraic number z for which Q(xi, x2, . , xn) = Q(z). By multiplying z by a suitable integer, we may assume that z is actually an algebraic integer. This means that each xi can be written in the form gi(z) for some polynomial gi with rational coefficients. Of course, there exists some integer N for which hi = Ng, have integer coefficients and there exists some large d for which F., (X) = Nd fi ( 11'( N x)) also has integer coefficients. Now, all Fi are divisible by P, the minimal polynomial of z in Q[X]. Because z is an algebraic integer, P is a monic polynomial with integer coefficients, and thus primitive. From Gauss's lemma, it follows that F, are divisible by P in Z[X]. Finally, let us apply Schur's theorem to this polynomial. There are infinitely many prime p > N for which F has a zero np in Z/pZ. Fix such a prime p and note that x = np. Let f2 (X) = AsXs + As_1Xs-1 + • • • + Ao. We know that p divides AsNd—shz(x)s + As iNd—s+ihi(x)s—i + • • • + AoNd. Of course, p is relatively prime to N, so p will actually divide Ashi(x)s + Thus, if N' is the inverse of N in Z/pZ, N'hi(x) is a zero of L modulo p. Since i was arbitrary, it follows that all fi have a zero in Z/pZ for any such prime p. The conclusion follows. 224 10. ARITHMETIC PROPERTIES OF POLYNOMIALS Part b) is actually a fairly immediate consequence of a). The idea is that for n > 1, any prime divisor of cbri(a), the nth cyclotomic polynomial, is either congruent to 1 modulo p or divides n. The proof of this result is not very difficult. Indeed, consider p such a prime divisor. Then plan —1 and thus, if d is the order of a modulo p, we have din and dlp — 1. Clearly, if d = n, we are done, so assume that d < n. Then since pad — 1 - = 11 kid Ok(a), there exists a divisor k of d such that plcbk(a). However, X' — 1 is the product of all cyclo-tomic polynomials whose orders divide n, so it is a multiple of Ok(X) • cbn(X). Therefore, X' — 1 will have a as a double root in Z/p.Z. This is impossible unless pin, because in this case a would be a root of nXn-1 and thus pin (since p is not a divisor of a). This proves the claim Now, using a) for the polynomials cbk (X) and f(X), we know there are infinitely many primes p such that both these polynomials have roots in the field with p elements. But the observation made in the beginning of b) shows that only finitely many of these prime numbers are not congruent to 1 modulo k. Thus, infinitely many are of the form 1 + kq and the proof finishes here. The next example concerns the very classical problem of square free numbers among polynomial values. More generally, one defines k-free numbers as non-zero integers which are not divisible by any k-th power of a prime. One can prove (the idea is exactly the same as in the problem that we will discuss) that if f is a primitive polynomial of degree d and if f is not the d-th power of a linear polynomial, then a positive proportion of positive integers n have the property that f (n) is d-free. A more difficult result was proved by Eras: under some natural conditions imposed on f , there are infinitely many n for which f (n) is d—free. Needless to say, the proof if highly nontrivial. We will discuss a closely related problem concerning square free numbers of a special form. The next result is a lot stronger than the one proved by Laurentiu Panaitopol, stating that there are infinitely many triples of consecutive numbers, all square free. The solution is adapted from a beautiful argument due to Ravi Boppana. Before passing to this problem, let us give a definition: we say that a set A of positive integers has positive density if there exists a constant c > 0 such that for all sufficiently large x there are at least cx elements of A less than x. THEORY AND EXAMPLES 225 Example 15. Prove that the set of positive integers n such that 1 — 2n(n + 1)(n + 2)(n2 + 1) is square free has positive density. [Vesselin Dimitrov] Solution. Let us search for such numbers of the form n = 180k + 1 for some positive integer k. By this choice, ln(n + 1)(n + 2)(n2 + 1) is not divisible by 4 or 9 or 25. So we can ignore the prime factors 2, 3 and 5. Let p be a prime greater than 5. There is exactly one k (mod p2) such that n = 180k + 1 is divisible by p2, exactly one k (mod p2) such that n + 1 is divisible by p2, and exactly one k (mod p2) such that n + 2 is divisible by p2. Also, there are at most two k (mod p2) such that n2 + 1 is divisible by p2. Indeed, if p21a2 + 1 and P2 Ib2 + 1, then p21(a — b) (a + b) . Then p2 la — b or p2 la + b (otherwise, p divides a — b and a + b, thus it divides a too, which is clearly impossible). Altogether there are at most five k (mod p2) such that one of n, n + 1, n + 2, or n2 + 1 is divisible by p2. Let N be a large positive integer. By the previous observation, there are at most 5 1 19 1 2 . -11 values of k between 1 and N such that n, n+ 1, n +2, or n2 +1 is divisible by p2. If p > 180N + 1, then p is too large for n, n + 1, n + 2, or n2 + 1 to be divisible by p2. Altogether the number of k between 1 and N such that one of n, n + 1, n + 2, or n2 + 1 is not square free is at most z_, v,1 p N 80+1 5 [N] 7 • We can bound the last sum by 180N+1 E (5 + 5 p N) < 57(180N + 1) + 5N E p2 2 P=7 p>7 and since \--I 1 1 1 (1 1 1 1 1 P P 2 >3 m +1)(2m — 1) < 2 7 + 7 9 ± ) 10' m 226 10. ARITHMETIC PROPERTIES OF POLYNOMIALS we infer that the number of "bad" k is at most / 4 + o(N). We used here the classical fact that 7r(x) = o(x), where 7(x) = E 1 is the counting function of p<x the prime numbers (for a proof of this result, see the chapter At the Border between Analysis and Number Theory). Therefore, the number of 1 < k < N for which all numbers n, n+1, n+2, n2+1 (where n = 180k +1) are squarefree is at least 2 + o(N). For any such number k, In(n + 1)(n + 2)(n2 + 1) is squarefree (the only common prime divisors of two numbers among n, n+ 1, n+ 2, n2 +1 are 2, 3, 5 and we saw that the choice of n ensures that 4, 9, 25 are not divisors of n(n+1) (n+2) (n2 + 1)). Thus, the number of n < 181N such that .n(n + 1)(n + 2)(n2 + 1) is squarefree is at least / 4 + o(N), which means that the set of n for which ln(n + 1)(n + 2) (n2 + 1) is squarefree has positive density. PROBLEMS FOR TRAINING 227 10.2 Problems for training 1. Given a finite family of polynomials with integer coefficients, prove that for infinitely many integers n they assume at n only composite numbers. 2. Let f and g E Z[X] be nonzero polynomials. Consider the set Df,g = {gcd(f (n), g(n)) I n E N}. Prove that f and g are relatively prime in Q[X] if and only if D f ,g is finite. Gazeta Matematical 1985 3. Prove that there are no polynomials f E Z[X] with the property that there exists an n > 3 and integers xi, , xn such that f(xi) = i = 1, n (indices are taken mod n). 4. Let f E Z[X] be a polynomial of degree n > 2. Prove that the polynomial f (f (X)) — X has at most n integer zeros. Gh. Eckstein, Romanian TST 5. Find all integers n > 1 for which there is a polynomial f E Z[X] such that for any integer k we have f (k) congruent with either 0, or 1 modulo n and both these congruences have solutions. 6. Find all polynomials f with rational coefficients such that f (n)I2n — 1 for all positive integer n. Polish Olympiad 7. Let f be a polynomial with integer coefficients and let a0 = 0 and an = f(an_i) for all n > 1. Prove that (an)n>0 is a Mersenne sequence, that is gcd(am, an) = agcd(m,n) for all positive integers m and n. 228 10. ARITHMETIC PROPERTIES OF POLYNOMIALS 8. Let p be a prime number. Find the greatest degree of a polynomial f E Z[X] having coefficients in the set {0,1, ,p — 1}, such that its degree is at most p and if p divides f(m) — f (n) then it also divides m — n. 9. Find all integers k such that if a polynomial with integer coefficients f satisfies 0 < f (0) , f (1), f (k) < k then f (0) = f (1) = • • • = f (k). IMO 1997 Shortlist 10. Let f be a polynomial with integer coefficients. Prove the equivalence of the following two properties: i) for any integer n one has f (n) E Z; ii) There exist integers n and ao, al, a2, ..., an such that f (X) = ao + al X + X (X -1) X(X-1).••(X-n+1) a2 2 + + a n n! 11. Let n be a positive integer. What is the least degree of a monic polyno-mial f with integer coefficients such that ni f (k) for any integer k. 12. Let f be a polynomial with rational coefficients such that f (n) E Z for all n E Z. Prove that for any integers m, n the number lcm[1,2, ..., deg( f)] f (m) - f (n) is an integer. MOSP 2001 13. Let P(Xl, X2, ..., X1) be a polynomial with real coefficients. Give a nec-essary and sufficient condition for P to send Z1 in Z. Deduce that for any integers al, a2, ..., an the number n a3 — I 3 is integer. <2<3 <n m — n PROBLEMS FOR TRAINING 229 14. Let f(x) = ao + aix + • • • ± amxm, with m > 2 and am 0, be a polynomial with integer coefficients. Let n be a positive integer such that: i) a2, a3, , a, are divisible by all prime factors of n; ii) al and n are relatively prime. Prove that for each positive integer k there is a positive integer c such that f (c) is divisible by Thk . Romanian TST 2001 15. Find all quadratic polynomials f E Z[X] with the property that for any relatively prime integers m, n, the numbers f (m), f (n) are also relatively prime. St. Petersburg Olympiad 16. Let d, r be positive integers with d > 2. Prove that for any nonconstant polynomial f with real coefficients of degree smaller than r, the numbers f (0) , f (1), f (dr — 1) can be divided into d disjoint sets such that the sum of the elements of each set is the same. J. 0. Shallit, AMM E 3032 17. Let a, b, c, d, k, m be integers such that a, c > 0. Suppose that m is rel-atively prime to both c and k. Prove that there exists a polynomial f of degree d with integer coefficients such that f (n) k • can±b (mod m,) for all nonnegative integers n if and only if m is a divisor of (ca — w+d. 18. Let f E Z[X] be a nonconstant polynomial. Prove that the sequence f (3') (mod n) is not bounded. 230 10. ARITHMETIC PROPERTIES OF POLYNOMIALS 19. a) Prove that for each positive integer n there is a polynomial f E Z[X] such that all numbers f (1) < f (2) < • • • < f (n) are prime numbers. b) As above, except the numbers now need to be powers of 2 rather than primes. 20. Are there polynomials p, q, r with positive integer coefficients such that 2 p(x) + (x2 — 3x + 2)q(x) and q(x) = (20 15 + 12) rx). Vietnamese Olympiad 21. Let (an)n>i be an increasing sequence of positive integers such that for some polynomial f E Z[X] we have an < f (n) for all n. Suppose also that m — nlarn — an for all distinct positive integers m, n. Prove that there exists a polynomial g E Z[X] such that an = g(n) for all n. USAMO 1995 22. We call a sequence of positive integers (an)n>1 pairwise relatively prime if gcd(ani, an) = 1 for any different positive integers m, n. Find all integer polynomials f E Z[X] such that for any positive integer c, the sequence (f (c))7i>1 is pairwise relatively prime. Here f [n] is the composition of f with itself taken n times. Leo Mosser 23. Suppose that f E Z[X] is a nonconstant polynomial. Also, suppose that for some positive integers r, k, the following property holds: for any pos-itive integer n, at most r prime factors of f (n) have exponent at most equal to k. Does it follow that any zero of this polynomial has multi-plicity at least k + 1? PROBLEMS FOR TRAINING 231 24. Prove that for all n there exists a polynomial f with integer coefficients and degree not exceeding n such that 2n divides f (x) for all even integers x and 2' divides f (x) — 1 for all odd integers x. P. Hajnal, Komal 25. Find all polynomials f with integer coefficients and such that f (p)12P —2 for any prime number p. Gabriel Dospinescu, Peter Schoelze 26. Prove that for any c > 0 there are infinitely many n such that the largest prime divisor of n2 + 1 is greater than cn. Nagell THEORY AND EXAMPLES 235 11.1 Theory and examples Almost everyone knows the Chinese Remainder Theorem, which is a remark-able tool in number theory. But does everyone know the analogous form for polynomials? Stated like this, this question may seem impossible to answer. Then, let us make it easier and also reformulate it: is it true that given some pairwise distinct real numbers xo, xi, x2, , xn and some arbitrary real num- bers ao, al, a2, , an, we can find a polynomial f with real coefficients such that f (xi) = ai for i E {0,1, , n}? The answer turns out to be positive, and a possible solution to this question is based on Lagrange's interpolation formula. It says that an example of such polynomial is n f (x) = ai H x _ xi . xi _ xi i=0 0<j<n j 2 (In what follows along this unit, a product like the above one will be written, for simplicity, just as x—x, .) x1 —x) Indeed, it is immediate to see that f (xi) = ch for i E {0, 1, , n}. Also, the above expression shows that this polynomial has degree less than or equal to n. Is this the only polynomial with this supplementary property? Yes, and the proof is not difficult at all. Just suppose we have another polynomial g of degree less than or equal to n and such that g(xi) = ai for i E {0, 1, , n}. Then the polynomial g— f also has degree less than or equal to n and vanishes at x0, xi, , xn. Thus, it must be null, and the uniqueness is proved. What is Lagrange's interpolation theorem good for? We will see in the follow-ing problems that it helps us to immediately find the value of a polynomial in a certain point if we know the values in some given points. And the reader may have already noticed that this follows directly from the formula (1), which shows that if we know the value in 1+ deg f points, then we can find the value in any other point without solving a complicated linear system. Also, we will 236 11. LAGRANGE INTERPOLATION FORMULA see that it helps in establishing some inequalities and bounds for certain spe-cial polynomials, and will even help us in finding and proving some beautiful identities. Now, let us begin the journey through some nice examples of prob-lems where this idea can be used. As promised, we will first see how we can rapidly compute the value in a certain point for some polynomials. This was one of the favorites problems in the old Olympiads, as the following example illustrates. Let F 1 = F2 = 1, Fn+2 = Fn Fn,-ki and let f be a polynomial of degree 990 such that f(k) = Fk for k E {992, , 1982}. Show that f (1983) = F1983 — 1. [Titu Andreescu] IMO 1983 Shortlist Solution. So, we have f (k + 992) = Fk+992 for k = 0, 1, ..., 990 and we need to prove that f (992+991) = F1983 —1. This simple observation shows that we don't have to bother too much with k + 992, since we could work as well with the polynomial g (x) = f (x+992), which also has degree 990. Now, the problem becomes: if g(k) = Fk+992, for k = 0, 1, ..., 990, then g(991) = F1983 — 1. But we know how to compute g(991). Indeed, looking again at the interpolation formula (as we name equation (1)), we find that 990 990 991 .9(991) = g(k)( 991 k ) (-1)k =LJ k J Fk+992(-1)k k=0 k=0 which shows that we need to prove the identity (99l) k Fk+992 — 1 )k = F1983 — 1 . k=0 We know that 990 an — bn F Ti- THEORY AND EXAMPLES 237 +1 where a = 2 and b = try a direct approach: 1 — .\/-g Bearing this in mind, we can of course 2 • 990 09, Fk+992( — 1 )k k=0 [ 990 E (7) ak+992 i)k (99 1) bk+992( i)ki k=0 k=0 But using the binomial theorem, the above sums vanish: • (99 ▪ k 1) ak+992 ( 1)k = a992 990 E (991) k=0 ) k=0 (_a)k = a992 [(1 a)991 a991]. Since a2 = a + 1, we have a992[(1 — a)991 + a991] = a(a a2 \ ) 991 + a1983 = —a + a1983. Since in all this argument we have used only the fact that a2 = a +1 and since b also satifies this relation, we find that 990 i (9k 91) /1+992,- , , \ k = 1 (a1983 b1983 a + b) k=0 a1983 b1983 a — b = F1983 — 1. And this is how, with the help of a precious formula and with some smart computations, we could solve this problem and also find a nice property of the Fibonacci numbers. The following example is a very nice problem proposed for IMO 1997. Here, the steps following the use of Lagrange's interpolation formula are even better hidden in some congruences. It is the typical example of a good Olympiad problem: no matter how much the contestant knows in that field, one may have great difficulties in solving it. 990 p-1 P-1 E(_i)k 1)1(k) k=0 f (k) (mod p). k=0 238 11. LAGRANGE INTERPOLATION FORMULA Example 2.1 Let f be a polynomial with integer coefficients, and let p be a prime such that f(0) = 0, f(1) = 1 and f(k) is congruent to either 0 or 1 modulo p, for all positive integers k. Show that the degree of f is at least p — 1. IMO 1997 Shortlist Solution. Such a problem should be solved indirectly, arguing by contradic-tion. So, let us suppose that deg f < p — 2. Then, using the Interpolation Formula, we find that p-1 f (x) = E f (k) H x -j . k — k=0 Now, since deg f < p — 2, the coefficient of xP-1 in the right-hand side of the identity must be zero. Consequently, we have P-1 (_1)p-k-1 E k!(p — 1 — k)! f (k k=0 ) From here we have one more step. Indeed, let us write the above relation in the form p-1 E(-1)k ( P k 1 ) f(k) = 0 k=0 and let us take this equality modulo p. Since k! (1) k 1 ) = — k)(p — k + 1) . . . (p — 1) = (-1)k k! (mod p) we find that -k 1) = (-1)k (mod p) and so 0. THEORY AND EXAMPLES 239 Thus, E f (k) 0 (mod p), k=0 which is impossible, since f(k) 0,1 (mod p) for all k and not all of the numbers f (k) have the same remainder modulo p (for example, f (0) and f (1)). This contradiction shows that our assumption was wrong and the conclusion follows. It's time now to see how some formidable identities are simple consequences of the Lagrange interpolation formula, although in these problems polynomials do not appear at first sight. [Di ...ample 3. Let al, a2, , an be pairwise distinct positive integers. Prove al : that for any positive integer k the number E is i=1 1(a i — aj) an integer. United Kingdom Solution. Just by looking at the expression, we recognize the Lagrange In-terpolation formula for the polynomial f(x) = xk. But we may have some problems when the degree of this polynomial is greater than or equal to n. This can be solved by working with the remainder of f modulo g(x) = (x—ai)(x—a2) . . . (x—an). So, let us proceed by writing f (x) = g(x)h(x)+r(x), where r is a polynomial of degree at most n — 1. This time we don't have to worry since the formula works, and we obtain n r(ai) TT x ai ai — aj i=1 Now, we need three observations. The first one is r(ai) = al ic, the second one is that the polynomial r has integer coefficients, and the third one is that 240 11. LAGRANGE INTERPOLATION FORMULA E a2 is just the coefficient of xn-1 in the polynomial i=i ii(ai — ai ) jOi E r(ai) f x ai . ai — ai i=1 jai Combining these observations, we find that E is the coefficient j=1 I1(ai — a 3) 3 2 of xn-1 in r, which is an integer. Thus, not only have we solved the problem, but we have also found a rapid way to compute the sums of the form E FF ai i=i ai) The following two problems concern some combinatorial sums. If the first one is relatively easy to prove using a combinatorial argument (it is a very good exercise for the reader to find this argument), for the second problem a combinatorial approach is much more difficult. But we will see that both are immediate consequences of the interpolation formula. Let f (x) = Eakxn-k. Prove that for any non-zero real num- k--=0 ber h and any real number A we have Example 4. E( 1)' k f (A + kh) = a() • n! • hn. k=0 Solution. Since this polynomial has degree at most n, we have no problems in applying the interpolation formula -v-r x - f (x) = E f (A + kh) 11 (k A-jh — j)h k=0 jk THEORY AND EXAMPLES 241 Now, let us identify the leading coefficients in both polynomials that appear in the equality. We find that 71 kh) TT 1 f ( \ il[(k — j)h] n!hn n—k 1) f + kh), k=0 .7 1c which is exactly what we had to prove. Simple and elegant! Notice that the above problem implies the well-known combinatorial identities E( ( k) kp 0 k=0 for all p E {0, 1, 2, . . . , n — 1} and k=0 (-1 n—k (n) kn = n!. (Notice that the identity remains valid for h = 0, too!) As we promised, we will discuss a much more difficult problem. The reader might say after reading the solution: but this is quite natural! Yes, it is natural for someone who knows the Lagrange Interpolation formula very well and especially for someone who thinks that using it could lead to a solution. Unfortunately, this isn't always so easy. Example 5. Prove the identity E(_i)n—k (n) kn+1 n(n + 1)!
— 2 • k=o Solution. We take the polynomial f (x) = xn . (Why don't we take the poly- nomial f (x) = xn+1? Simply because (-1)' (n appears when writing the formula for a polynomial of degree at most n.) We write the Interpolation Formula n ao = k=0 xn x(x — 1) (x — k — 1)(x — k ± 1) • • • (x n) 1) n—k (n — k)!k! k=0 242 11. LAGRANGE INTERPOLATION FORMULA Now, we identify the coefficient of xn-1 in both terms. We find that 0 = E(-1)1" (1 km(1 + 2 + • • • + n — k). And now the problem is solved, since we found that E(_i)n-k (n) kn+1 n(n + 1) E n ( (n) kn k) 2 k=0 k=0 and we also know that E( i)n-k (n) kn = nl k=0 from the previous problem. If the Lagrange interpolation formula was good only to establish identities and to compute values of polynomials, it would not have been such a great discov-ery. Of course this is not the case—it plays a fundamental role in analysis. However, we are not going to enter this field, and we prefer to concentrate on another elementary aspect of this formula and see how it can help us establish some remarkable inequalities. And some of them will be really tough. We begin with a really difficult inequality, originally published by H.S. Shapiro in the American Mathematical Monthly, in which the interpolation formula is really well hidden. But denominators can give valuable indications from time to time. ;Example 6. Prove that for any real numbers xi, x2, . , x7, E [ —1, 1] the following inequality is true: 1 E > 2n-2. i=1 H — Xi I k=0 [H. S. Shapiro] Iranian Olympiad THEORY AND EXAMPLES 243 Solution. The presence of f xj - xi i is the only hint to this problem. But 3 even if we know it, how do we choose the polynomial? The answer is simple: we will choose it to be arbitrary, and only in the end we will decide which n-1 one is optimal. So, let us proceed by taking f(x) = )' akx an arbitrary k=0 polynomial of degree n — 1. Then we have n f(x) = f(xo _FT x-x j. k=1 xk — Xi .j1c Combining this with the triangle inequality, we get If (x)I E If (x01 11 k=1 X — X j Xk — Xi Only now comes the beautiful idea, which is in fact the main step. From the above inequality we find that I an- 71 If (X1c)1 k=1 11 ixk - xj jk H (1 _ x .j1c f(x) xn-1 and since this is true for all non-zero real numbers x, we may take the limit when x —> oo and the result is pretty nice: If (x0I L. k= 111Xk 30k This is the right moment to decide what polynomial to take. We need a poly-nomial f such that I f(x)1 < 1 for all x E [-1, 1] and such that the leading coefficient is 2n-2. This time our mathematical culture will decide. And it says that Chebyshev polynomials are the best, since they are the polynomials with the minimum deviation on [-1, 1] (the reader will wait just a few seconds and 244 11. LAGRANGE INTERPOLATION FORMULA will see a beautiful proof of this remarkable result using Lagrange's interpola-tion theorem). So, we take the polynomial defined by f (cos x) = cos(n — 1)x. It is easy to see that such a polynomial exists, has degree n — 1, and leading coefficient 2n-2, so this choice solves our problem. Note also that the inequality lan_il < If (xk )1 can be proved by k=1 ixk - X3I identifying the leading coefficients in the identity f (x) k=1 f (xk) vr x - Xj xk - xj and then using the triangle inequality. The following example is a fine combination of ideas. The problem is not simple at all, since many possible approaches fail. Yet, in the framework of the previous problems and with the experience of Lagrange's interpolation formula, it is not so hard after all. [ Example 7.1 Let f E R[X] be a polynomial of degree n with leading coef-_ ficient 1, and let xo < xi < x2 < • • • < xr, be some integers. Prove that there exists k E {0,1, . . . , n} such that If(x01 > - 2n Crux Matematicorum Solution. Naturally (but would this be naturally without having discussed so many related problems before?), we start with the identity f(xo n- x - Xj xk - xj k=0 jk THEORY AND EXAMPLES 245 Now, repeating the argument in the previous problem and using the fact that the leading coefficient is 1, we find that E n if(xol > 1. k=0 H ixk - j0k It is time to use the fact that we are dealing with integers. This will allow us to find a good lower bound for H ixk - x3 I This is easy, since 30k H I Xk — xjI = I (Xk — X0)(Xk — X1) • " (Xk Xk-1)(Xk+1 Xk) • • • (Xn Xk)I jr/k > k(k — 1)(k — 2) • • • 2 • 1 • 1 • 2 • • • (n — k) = k!(n — k)!. And yes, we are done, since using these inequalities, we deduce that 71, .f(x > 1. k!(n — k)! k=0 Now, since >2, k!(n — k)! (k n! n) 2n k=0 k=0 it follows that If (x01 for some 0 < k < n. The following example is an answer to a conjecture of F. J. Dyson (1962). The elegant proof presented here, based on an identity obtained by Lagrange's in-terpolation formula, is due to I. J. Good (1970): 246 11. LAGRANGE INTERPOLATION FORMULA [Example 8.1 Let ai, a2, ..., an be nonnegative integers and let f (a1, a2, .••, an) be the constant term of the "polynomial" n (1 _ xi 1<i,j<n 3 jA3 Prove that , (al + a2 + • • • + an)! f a2 • • • an) = a1!a2!...an! Solution. Define g(ai, a2, .., an) = a1!a2!...an! We will prove by induction on al + a2 + • • • + an that f(ai, a2, •••, an) = g(ai, a2, ..., an). If al = a2 = • • • = an = 0 the claim is obviously true. Now, observe that g(ai,a2,..., an) = g(ai — 1, a2,..., an) + • • • +g(ai,a2,...,a„ — 1) if all a, are positive and an) = if ak = 0. Therefore it would be enough to prove the same relation for f. If ak = 0 it is clear that f (ai, a2, '.. 5 an) = f (al, a2, ••-, ak+17 •••, an), so assume that all a, are positive. In order to prove that f(ai, a2, an) = f (ai — 1, a2, ..., an) + • • • + f (al , a2, ..., — 1) it is enough to prove the identity 1<i,j<n 3 i=1 i x3 X •) 3 a2 xi 11 (1 — =- (1 Xi ) H _ (al + a2 + ••• + an)! --1 THEORY AND EXAMPLES 247 which reduces of course to 1 = (1 - xj i=1 joi But this is just Lagrange's interpolation formula written for the polynomial f (X) = 1 with nodes xi, x2, xn and evaluated at x = 0. We will discuss one more problem before embarking on a more detailed study of Chebyshev polynomials and their properties. This was given in the Ro-manian Mathematical Olympiad and is a very nice application of Lagrange's interpolation formula. It is sufficient to say that it follows trivially using a little bit of integration theory and Fourier series. Prove that for any polynomial f of degree n and with leading coefficient 1 there exists a point z such that izi = 1 and If (z)I > 1. [Marius Cavachi] Romanian Olympiad Solution. Of course, the idea is always the same, but this time it is necessary to find the good points for which we should write the interpolation formula. As we did before, we will be blind at the beginning and we will try to find these points in the end. Until then, let us call them xo, xi, x2, . . , xn and write E n If(X01 > 1. This inequality was already proved in Example 6. Now, consider the polyno-mial 9(x) = fJ(x — xi). i=o k=0 11 iXk — X ji jOk 248 11. LAGRANGE INTERPOLATION FORMULA We have then Ig' (xi) I = n Now, we would like, if possible, to have I xiI = 1 and also k=0191 ( 1 X01 < 1. In n k=0 II1Xk — X numbers If (xk)1 is greater than or equal to 1 and the problem would be solved. Thus, we should find a monic polynomial g of degree n + 1 with all roots of modulus 1 and such that E 1 < 1. This is trivial: it suffices, of course, k=0 Igi(Xk) n to consider g(x) -= xn+ 1 — 1. The conclusion follows. We have an explanation to give: we said the problem follows trivially with a n little bit of integration theory tools. Indeed, if we write f (x) = akxk then k=-0 one can check with a trivial computation that 1 27r ak = 27r f (eit)e—iktdt f o and from here the conclusion follows since we will have 27r 27r f(eit)e—tradt < rteitNi )dt 5_ 27r max I f (z)1. 0 lz1=1 Of course, knowing this already in 10th grade (since the problem was given to 10th grade students) is not something common... Before passing to the next more computational problem (which does not mean less interesting, of course), let us recall some properties of the Chebyshev's polynomials of the first kind. They are defined by Tn(x) = cos(n arccos(x)), or, equivalently, T n(cos x) = cos(nx). You can easily check by induction, us-ing the obvious relation Tn+i(x) = 2xTn(x) — Tn_1(x) that this gives you a this case it would follow from > 1 that at least one of the 27r = f o THEORY AND EXAMPLES 249 polynomial of degree n, having leading coefficient 2n-1 and all of whose coef-ficients are integers. Among hundreds of interesting and useful properties of these polynomials, let us state a few, the proof of which is left as a very useful exercise for the interested reader. Theorem 11.1. The polynomials TT, have the following properties: • An explicit formula for Tn is T n (x) = (x + Jx2 — 1)n + (x — x2 — 1)n 2 • The polynomials Tn and T, commute, that is Tn(Tm(x)) = Tm(Tn(x)) for all m, n and all x. • The generating function of these polynomials is given by: z(x — z) 1 — 2zx + z2 • They form an orthogonal system on the interval [-1, 1] for the weight v(x) = Jll x2 j that is for all i j positive integers the following relation holds f 1 T i(x)T3 (x) J-1 — x The following problems will be based on a very nice identity that will allow us to prove some classical results about norms of polynomials, to find the polynomials having minimal deviation on [-1, 1], and also to establish some new inequalities. In order to do all this, we need two quite technical lemmas, which are not difficult to establish, but very useful. E Tn(x)zn = n>1 for all <1 and lx1 <1. = 0. 2n\/x2 – 1 Using this formula and de Moivre's formula we easily deduce i). (x) = ' 2 1 4 i [(x + x2 — 1)n + (x – 2 — 1)71+ X [(x + 1/42 — 1)n — (x — VX2 — 1)n]. 250 11. LAGRANGE INTERPOLATION FORMULA Lemma 11.2. If we let tk = cos — k7r , 0 < k < n, then Vx2 — f (x) = ll(x - tk) = [(x + 1/42 — 1)72 — (x — — 1)nl• k=0 2n 1 Proof. The proof is simple. Indeed, if we consider Vx2 – g(x) = 27, 1 [(x X2 — 1)n — (X — X2 — 1)n], using the binomial formula we can establish immediately that it is a polyno- mial. Moreover, from the obvious fact that lim g(x) = 1, we deduce that it x–,00 xn-F1 is actually a monic polynomial of degree n 1. The fact that g(tk) = 0 for all 0 < k < n is easily verified using de Moivre's formula. All these prove the first lemma. K A little bit more computational is the second lemma. Lemma 11.3. The following relations are true: i) FIN - ti) = 2n - if 1 < k < n – 1; _1)kn 1 j n ii) fJ(to – ti) = 2n-2' j=1 n-1 - iii) ll(t, - t j ) = ( 2n-1) 2n j=0 Proof. Simple computations, left to the reader, allow us to write: THEORY AND EXAMPLES 251 To prove ii) and iii) it suffices to compute lim f'(x) and lim f'(x), using the x->1 above formula; we let the reader to do the dirty job. K Of course, you hope that all these computations have an honorable purpose; and you're right, since these lemmas will allow us to prove some very nice results. The first one is a classical theorem of Chebyshev, about minimal de-viation of polynomials on [-1, 1]. LExample 10.1 (Chebyshev's theorem) Let f E R[X] be a monic polynomial of degree n. Then 1 max If(x)1 > xE[-1,1] 2n-1 and this bound cannot be improved. Solution. Using again the observation from Problem 7, we obtain the identity: Egto 11 4, 1 = 1. k=0 Thus, we have —1 1 < max If(tic)1E - 0<k<n (tk - t jk k=0 ) Now, it suffices to apply Lemma 2 to conclude that we actually have —1 II(tk tj) 2n-1. j$k This shows that max1] 2n 1 -1 If(x)1 > and so the result is proved. To prove — that this bound is optimal, it suffices to use the polynomial T n. Then the k=0 252 11. LAGRANGE INTERPOLATION FORMULA 1 polynomial 2n-1Tn is monic of degree n and max xE[-1,1] 1 2n-1 Tn(x) 1 (11.2) 2n-1 • There are many other proofs of this result, a lot of them are much easier, but we chose this one because it shows the power of Lagrange interpolation theory. Not to mention that the use of the two lemmas allowed us to prove that the inequality presented in Example 7 is actually the best. Some years ago, Walther Janous presented in Crux the following as an open problem. It is true that it is a very difficult one, but here is a very simple solution using the results already achieved. Example 11. I Suppose that ao, , an are real numbers such that for all x E [-1, 1] we have iao + aix + • • • + anxn1 < 1. Then for all x E [-1, 1] we also have Ian + an_ix + • • + aoxml < 2n-1. [Walther Janous] Crux Matematicorum Solution. Actually, we are going to prove a stronger result, that is: Lemma 11.4. Denote, for f E R[X], Ilf 11 = xr1 ,11 If (x)1. < 1 luln11f11 -Fr 1—tilt II— tj k=0 jk n THEORY AND EXAMPLES 253 Then for any polynomial f E R[X] of degree n the following inequality is satisfied: (x)1 iTn(x)Hif for all lx1 > 1. Proof. Using Lagrange's interpolation formula and the triangle inequality, we deduce that for all u E [-1, 1], u 0, we have: The brilliant idea is to use the Lagrange interpolation formula again, this time for the polynomial Tn. We shall then have (also for u E [-1, 1] , u 0) -Fr 1— ut i 'Ulm k=0 j 11 k til (the last identity being ensured by lemma 11.2). By combining the two results, we obtain Lill for all u E [-1, 1], u # 0 and the conclusion follows. K Coming back to the problem and considering the polynomial f (x) = kX k the hypothesis says that 11f11 < 1 and so by the lemma we have k=0 I f (x)I < ITn(x)I for all lx1 > 1. We will then have for all x E [-1,1], x 0: Ian + an-ix + • • • + aoei = xnTn (-1) 254 11. LAGRANGE INTERPOLATION FORMULA It suffices to prove that xnTii (-1) x < 2n-1, which can be also written as (1 + -V1 — x2)n + (1 — .V1 — x2)n < 2n. But this inequality is very easy to prove: just set a = -V1 — x2 E [0, 1] and observe that h(a) = (1 — a)n + (1 + a)n is a convex function on [0, 1], thus its superior bound is attained at 0 or 1 and there the inequality is trivially verified. Therefore we have Ian + an_ix + • • • + aoxn I < 2n-1 and the problem is solved. Since we are here, why not continuing with some classical, but very important results of Riesz, Bernstein and Markov? We unified these results in a single example because they have the same idea, and moreover they follow one from another. We must mention that a) is a result of M. Riesz, while b) was obtained by S. Bernstein and finally c) is a famous theorem of A. Markov, the real equivalent of an even more celebrated Bernstein's complex theorem (whose proof you will surely enjoy: it is among the training problems). Example 12.1 a) Let P be a polynomial with real coefficients of degree at most n — 1 such that V1 — x21P(x)1 < 1 for all x E [-1, 1]. Prove that IP(x)I < n for all x E [-1, 1]. b) Let n 1(x) = (ak cos(kx) + bk sin(kx)) k=0 be a trigonometric polynomial of degree n with real coeffi-cients. Suppose that If (x)I < 1 for all real numbers x. Prove that If(x)i < n for all real numbers x. c) Prove that if P has degree n and real coefficients and more-over IP(x)I < 1 for all x E [-1,1] then 1 P' (x)I — x E [-1, 1]. < n2 for all Ti (-1)z-1 X — Xi x? • P (xi) Tn(x) p (x) 1=1 THEORY AND EXAMPLES 255 Solution. a) Let us write the Lagrange interpolation formula for P with the points xi, x2, ..., xn, where x3 = cos( (232-771)7r) are the zeros of the nth Cheby-shev's polynomial Tn. We obtain the important identity Take now x E [-1, 1]. Observe that if x E [xn,xi.] = [ —xi,xi] then by the hypothesis IP(x)I < 1 < 1 < n' the last inequality being equivalent N / ix? to sin() > which is clear by a convexity argument. So, assume that x > x1, the case x < —xi being identical. In this case the triangle inequality applied to the previous identity shows that I P(x ) 1 But the last sum is exactly (x). Because Tn(cos u) = cos(nu), we have Tn ' (cos u) = nsi SI n.(nu) However, an easy induction shows that I sin(nu) I < n < R U sM ul for all u and all positive integers n. This implies that IT7c(x)i < n2 for all x E [ —1, 1]. Combining this with the inequality IP(x)I < • Tr, ' (x) we deduce that IP(x)I < n for all x > x1. This finishes the proof of the first part. b) First of all, let us see what happens when all a, are zero, that is f (x) = b1 sin x + b2 sin(2x) + • • • + bn sin(nx). Observe that in exactly the same way as you could have proved the existence of the polynomial Tn (that is, by induction), you can prove the existence of a polynomial 11„, of degree n — 1 such that Rn(cos x) = sin sin x (nx) Therefore there exists apolynomial P of degree at most n — 1, with real coefficients, and such that = P(cos x). Observe that this polynomial satisfies the conditions sin x of a), because I sin x • P(cos x)I < 1 for all real x. Therefore we can apply a) to deduce that IP(x)I < n for all x E [-1, 1], that is If(x)I < n • I sin xl for all x. Dividing by x and letting x —> 0 we deduce that f'(0)1 < n. Now, 256 11. LAGRANGE INTERPOLATION FORMULA let us come back to the general problem and fix a real number x0. Define g(x) f(x+x0)2f(x-xo) . Using standard trigonometric formulae, we deduce that g(x) is of the form ci sin x + c2 sin(2x) + • • • + cn sin(nx) for some real numbers c3. The triangle inequality also ensures that Ig(x)I < 1 for all real numbers x. Thus, by the result that we have just obtained, we must have 1.V MI < n. Because g'(0) = f(x0) and because xo was arbitrary, b) is proved. c) Let us consider this time f (x) = P(cos x). An immediate induction based on the most elementary product formulae for trigonometric functions shows that (cos x)3 is a trigonometric polynomial of degree at most n. Thus by b) we must have f' (x) < n for all x. This means that Isin x • P' (cos x)I < n for all x, which shows that the polynomial satisfies the conditions of a). Thus it has values not exceeding n on [-1,1], which means that P' does not exceed n2 on [-1,1], and this finishes the proof of this beautiful theorem. We end this topic with a very difficult problem, which refines an older one given in a Japanese mathematical Olympiad in 1994. The problem has a nice story: given initially in an old Russian Olympiad, it asked to prove that n 7a max TI ai < los - max 11 I x — ail xe[0,2] — .E[0,1] i=1 i=1 for any real numbers al, a2, , an. The Japanese problem asked only to prove the existence of a constant that could replace 108. A brute force choice of points in the Lagrange interpolation theorem gives a better bound of approx-imately 12 for this constant. Recent work by Alexandru Lupa§ reduces this bound to 1 + 2.\/g. In the following, we present the optimal bound. [Example 1371 For any real numbers al, a2, , an, the following inequality j holds: (3 + 2 4 4 n + (3 — 2n max nix ail < max H lx-ai l. xE[o,2] i=1 2 xe[om i=1 [Gabriel Dospinescu] THEORY AND EXAMPLES 257 Solution. Let us denote IlfIl[a,b] = xn el[a , , ,)f ] If (x)I for a polynomial f and let, for simplicity, (3 + 24n + (3 — 2-\)n en = 2 We thus need to prove that I f II [0,2] cnIlf II Ail where f (x) = 11(x — ai) . i=i We shall prove that this inequality is true for any polynomial f , which allows us to suppose that VII [0,i] = 1. We shall prove that for all x E [1,2] we have 1 + tk f (x)I < cn. Let us fix x E [1,2] and consider the numbers xk = 2 where tk's are as in Lemma 11.2. Using the Lagrange interpolation formula, we deduce that If (x)I E k=0 Fr X — Xj " Xk — X j jec = iok x _ xi _ xi i k=o 5- E n H 2 — E n II 3 — ti k=_O.W I c Xk — Xi I k=0 j 1 Ok ti Using Lemma 11.3, we can write n 3 — t 2n-i n-1 EH = n 11(3 ti)+ k=0 jk k=1 j = 0 j=1 258 11. LAGRANGE INTERPOLATION FORMULA Based on the expression of the derivative from the proof of Lemma 11.3, we obtain: n-1 n-1 EH(3_t3)+ — t3) + 11(3 — t3) = k=1j#k j=0 j=1 = -7 -1 2n -[(3 + 2An + (3 — 2VJ)n] 3 2n 0 [(3 2A n — (3 — 2V2)Th]. +1- All we have to do now is to compute n-1 n n-1 11(3 — t3)+ H(3 — t3) = 6 H(3 — t3). j=0 j=1 j=1 But, according to Lemma 11.2, we deduce immediately that n-1 11(3- t3) 3=1 1 2n+i [(3 ± 2An (3 _ 2.‘5 v-- )71. Putting all these observations together and making a small computation, that we leave to the reader, we easily deduce that I f(x)I < cn. This proves that 11f11 [0,2] C crillf11[0,1] and solves the problem. PROBLEMS FOR TRAINING 259 11.2 Problems for training 1. A polynomial of degree 3n takes the value 0 at 2,5,8, . . . , 3n — 1, the value 1 at 1,4,7, .. . , 3n — 2 and the value 2 at 0,3,6, .. . , 3n and it's value at 3n + 1 is 730. Find n. USAMO 1984 2. A polynomial p of degree n satisfies p(k) = 2k for all 0 < k < n. Find its value at n + 1. Murray Klamkin 3. Prove that for any real number a we have the following identity E(— 1)k ( 71) (a — = n!. k=0 Tepper's identity 4. Find E(_i)k (n ) kn+2 and E(-1)k (n k) 0+s. k=0 k=0 AMM 5. Prove that jk and evaluate n+2 Xk \ • k=0 (Xk Xj) jOk 260 H. LAGRANGE INTERPOLATION FORMULA 6. Prove the identity (n k) , E( 1)k_1 (n kr = E±. k=1 k=2 Peter Ungar, AMM E 3052 7. Let a, b, c, d E R such that lax3 + bx2 + cx + dl < 1 for all x E [-1, 1]. Prove that + Ibl + 1c1+1d1 < 7. IMO Shortlist 1996 8. Define F (a, b, c) = max 1x3 — ax2 — bx — What is the least possible xE[0,3] value of this function over R3? Chinese TST 2001 9. Let a, b, c, d E R such that lax3 + bx2 + cx + dl < 1 for all x E [-1, 1]. What is the maximal value of 1cl? For which polynomials is the maximum attained? Gabriel Dospinescu 10. Let a > 3 be a real number and p be a real polynomial of degree n. Prove that max lai — p(i)1 > 1. i=0,1,...,n+1 11. Find the maximal value of the expression a2 + b2 + c2 if lax2 + bx cl < 1 for all x E Laurentiu Panaitopol PROBLEMS FOR TRAINING 261 12. Let f E R[X] a polynomial of degree n that verifies I f(x)1 < 1 for all x E [0, 1], then < 2n+1 1. 13. Let a, b, c be real numbers and let f(x) = ax2 + bx + c such that max{lf(±1)1,1f(0)11 5_ 1. Prove that if Ix1 < 1 then 5 IA4 and < 2. Spain 1996 14. Let A = fp E R[X]l degp < 3, Ip(±1)1 5 1, Find sup max 1p"(x)I. PEA lx1<1 p (±) 2 IMC 1998 15. a) Prove that for any polynomial f having degree at most n, the following identity is satisfied: xf'(x) = f (x) + „ 2Zk „ 2_, f (xzk) (1 _ zo2 k=1 where zk are the roots of the polynomial X' + 1. b) Deduce Bernstein's inequality: II f ' II < where = If(x)1. P. J. O'Hara, AMM 262 11. LAGRANGE INTERPOLATION FORMULA 16. Let f be a complex polynomial of degree at most n, and let zo, zi, ..., zd be the zeros of the polynomial Xd+1 — 1, where d > n. Define 11f11 as the maximum of f (z)1 over all complex numbers z on the unit circle of the complex plane. a) Prove that if there exist n 1 pairwise distinct zeros xo, xi, •••,xn among zo,zi,...,zd such that f (xi)1 < a d then 11 f H < 1. b) Deduce that 11f11 110 < 4d • 11/911- Gelfand 17. Let f be a complex polynomial of degree n such that f (x)1 < 1 for all x E [-1, 1]. Prove that for all k and all real numbers x such that lx1 > 1, 1.0)(x)i < IT(k)(x)1. Prove that Chebyshev's theorem is a consequence of this result. W. W. Rogosinski THEORY AND EXAMPLES 265 12.1 Theory and examples It is probably time to see the contribution of non-elementary mathematics in combinatorics. It is quite difficult to imagine that behind a simple game such as football, for example, or behind a day-to-day situation such as handshakes, there exists such a complicated machinery. But this sometimes happens, as will be soon demonstrated. In the beginning of the discussion, the reader does not need any special knowledge, just imagination and the most basic properties of matrices, but, as soon as we advance, things may change. Anyway, the most important fact is not the knowledge, but the ideas and, as we will see, it is not always easy to discover that non-elementary fact that hides behind a completely elementary problem. Now that we have clarified what is the purpose of the unit, we can begin. The first problem we are going to discuss is not classical, but it is relatively easy and shows how a very nice application of linear-algebra can solve elemen-tary problems. Let n > 3 and let An, Bn be the sets of all even, respectively odd, permutations of the set {1, 2, ... , n}. Prove that n n E E li — 0-col = E 1 0-(i)i. crEA, i=1 crE13, i=1 [Nicolae Popescu] Gazeta Matematica Solution. Writing the difference n n E E li — awl — E E Ii — um 0-EAn i=1 aEBn i=1 as n E 6(a) awl, ,Esn i=1 266 12. HIGHER ALGEBRA IN COMBINATORICS where E(a) J +1, if a- E A, 1 if a- E Bn reminds us about the formula det A = (a-)ai,(0a20-(2) • • • anu(n) ES, We have taken here Sr, = An U By,. But we have no product in our sum! This is why we take an arbitrary positive number x and consider the matrix A = (xli-31)1<,,i<n. We have det A = CIES n (_ r(a) x I 1 (1)I x I rt-a (n)I = li-a(i)l li-a(i)1 xj=i xj=1 o-E An a- E This is how we obtain the identity 1 x x2 Xn-2 Xn-1 X 1 X xn-3 xn-2 x2 X I. Xn-4 Xn-3 • ' • Xn-1 Xn-2 x 1 x i= 1 Ii-c(i)1 x i= 1 (12.1) o- even a odd Anyway, we do not have the desired difference yet. The most natural way is to differentiate the last relation, which is nothing other than a polynomial THEORY AND EXAMPLES 267 identity, and then to take x = 1. Before doing that, let us observe that the polynomial 1 x x2 Xn-2 X71-1 X 1 X Xn-3 Xn-2 X2 X 1 Xn-4 e-3 Xn-1 Xn-2 X 1 is divisible by (x — 1)2. This can be easily seen by subtracting the first line from the second and the third one and taking from each of these lines x — 1 as a common factor. Thus, the derivative of this polynomial is a polynomial divisible by x — 1, which shows that after we differentiate (12.1) and take x = 1, the left-hand side vanishes, while the right-hand side becomes n E 0-(i)1 a(i)l. aEAn i=1 aEBn i=1 This completes the proof. Here is another nice application of this idea. You probably know how many permutations do not have a fixed point. The question that arises is how many of them are even. Using determinants provides a direct answer to the question. [Example 2. Find the number of even permutations of the set {1, 2, ... , n} that do not have fixed points. Solution. Let Cn and Dn be the sets of even and odd permutations of the set {1, 2, ... , n}, that do not have any fixed points, respectively. You may recall how to find the sum 1C711 + IDn 1: using the inclusion-exclusion principle, it is not difficult to establish that it is equal to 1 1 (-1)n) n! (1 — • • + 1! 2! n! Hence if we manage to compute the difference ICn1 — answer to the question. Write ICI = E 1— E 1 crEA, crEBn a(i)#i we will be able to 268 12. HIGHER ALGEBRA IN COMBINATORICS using An, Bn from Example 1, observe that this reduces to computing the determinant of the matrix T = (tu)i<i,3<n, where f 1, if i j 1 0, if i = j t That is, — DnI = 1 1 1 1 1 0 0 1 1 0 But computing this determinant is not difficult. Indeed, we add all columns to the first and factor n — 1, then we subtract the first column from each of the other columns. The result is 1Cml — IDn = (-1)n-1 (n — 1), and the conclusion is: 1 1 1) n-2 cnl ) - + (-1)n 1)] . = [n! (1 3! ( (n 2)! In the following problems we will focus on a very important combinatorial tool, that is the incidence matrix. Suppose we have a set X = {x1, x2, . • , and X1, X2, . , Xk a family of subsets of X. Now, define the matrix A = (aij)i<i<n, where 1<j<k f 1, if xi E X3 a" _ 0, if xi cl Xi This is the incidence matrix of the family Xi, X2, . , Xk and the set X. In many situations, computing the product AT • A helps translate the conditions and the conclusion of certain problems. From this point, we turn on this machinery, and solving the problem is on its way. Let us discuss first a classical problem. It appeared at the USAMO 1979, Tournament of the Towns 1985 and in the Bulgarian Spring Mathematical Competition 1995. This says something about the classical character and beauty of this problem. THEORY AND EXAMPLES 269 Example 3. Let A1, A2, . . . , An+i be distinct subsets of the set {1, 2, ... , n}, each having exactly three elements. Prove that there are two subsets among them that have exactly one common element. Solution. We argue by contradiction and suppose that IA, n A31 E 0,21 for all i j. Now, let T = (tip) 1<2<n, be the incidence matrix of the family 1<j<n+1 A1, A2, . . . , An+1 and compute the product / n n n z t tk,ltk,2 ,, tk,ltk,n+1 k=1 k=1 k=1 tT • T = ... ... ... n n n tk,n+ltk,1 E tk,n+ltk,2 tz,n, k=1 k=1 k=1 / But Etki=Ail = 3 and E tkitki = Ai n Aj I E {0,2}. k=1 k=1 Thus, considered in the field Z/2Z, we have C:1 tT • T = (... where X is the matrix having as elements the residues classes of the elements of the matrix T. Because det X = det X, the previous relation shows that det tT • T is odd, hence nonzero. This means that tTT is an invertible matrix of order n + 1, thus rank(tT • T) = n + 1 which contradicts the inequality rank(tT • T) < rank(T) < n. This shows that our assumption is wrong and there indeed exist indices i j such that 14, n A31 = 1. The following problem is very difficult to solve by elementary means, but the solution using Linear Algebra is straightforward. [Example 4.1 Let n be even and let A1, A2, , An be distinct subsets of the set {1, 2, , n}, each of them having an even number of elements. Prove that among these subsets there are two having an even number of common elements. o I tT •T = (. . . 1 1 1 0 1 1 270 12. HIGHER ALGEBRA IN COMBINATORICS Solution. Indeed, if T is the incidence matrix of the family A1, A2, , An, we obtain as in the previous problem the following relation n A21 IA1 n Anl tT • T = (12.2) IAnnAll 1An n A21 lAn1 Now, let us suppose that all the numbers IAi n I are odd and interpret the above relation in the field Z/2Z. We find that which means again that det tT T is odd. Indeed, if we work in Z/2Z, we obtain 1 1 1 0 = 1. The technique used is exactly the same as in the previous example. Note that this is the moment when we use the hypothesis that n is even. Now, since det tT • T = det2 T, we obtain that det T is also an odd number. Hence we should try to prove that in fact det T is an even number and the problem will be solved. Just observe that the sum of elements of the column i of T is 'Ai l, hence an even number. Thus, if we add all lines to the first, we will obtain only even numbers on the first line. Because the value of the determinant does not change under this operation, it follows that det T is an even number. But a number cannot be both even and odd, so our assumption is wrong and the problem is solved. THEORY AND EXAMPLES 271 Working in a simple field such as Z/2Z can allow us to find quite interesting solutions. For example, we will discuss the following problem, used in the preparation of the Romanian IMO team in 2004. Example 5. The squares of an n x n table are colored white and black. Suppose that there exists a nonempty set of rows A such that any column of the table has an even number of white squares that also belong to A. Prove that there exists a nonempty set of columns B such that any row of the table contains an even number of white squares that also belong to B. [Gabriel Dospinescu] Solution. This is just the combinatorial translation of the well-known fact that a matrix T is invertible in a field if and only if its transpose is also invertible in that field. But this is not so easy to see. In each white square we write the number 1 and in each black square we put a 0. We thus obtain a binary matrix T = (43)1<,,3<n• From now on, we work only in Z/2Z. Suppose that A contains the rows al, a2, , ak. It follows that tai = 0 for all j = 1, 2, ..., n. Now, let us take xi = { 1, if i E A 0, f i A i It follows that the system tilZi t21Z2 + • • • + tn1Zn =0 t12Z1 t22Z2 + • • ' tn2Zn = 0 tinzi + t2nz2 + • • • + tnnzn = 0 has the nontrivial solution (xi, x2, ... , xn). Thus, det T = 0 and consequently det tT = 0. But this means that the system { him, ± ti2Y2 + " • + tlnYn t21y1 + t22y2 + " • + t2nYn =0 =0 tnlyi tn2Y2 ' tnnYn = 0 272 12. HIGHER ALGEBRA IN COMBINATORICS also has a nontrivial solution in Z/2Z. Now, we take B = y2 01 and we clearly have B 0 and E .ix = 0, i = 1,2, ..., n. But this means that any x B line of the table contains an even number of white squares that also belong to B, and the problem is solved. The cherry on the cake is the following very difficult problem, where just knowing the trick of computing to • A does not suffice. It is true that it is one of the main steps, but there are many more things to do after we compute tA•A. And if for these first problems we have used only intuitive or well-known properties of the matrices and fields, this time we need a more sophisticated arsenal: the properties of the characteristic polynomial and the eingenvalues of a matrix. It is exactly the kind of problem that knocks you down when you feel most confident. Note that the problem can also be reformulated in a more down-to-earth way: for which m, n is there a directed graph with n vertices in which every pair of vertices is connected by exactly m paths of length 2? Example 6. Let S = {1, 2, ... , n} and let A be a family of pairs of elements in S with the following property: for any i, j E S there exist exactly m indices k E S for which (i, k),(k,j) E A. Find all possible values of m and n for which this is possible. [Gabriel Carrol] Solution. It is not difficult to see what hides behind this problem. Indeed, if we take T = (ti3)].<2,3<n, where tij 1, if (i, i) E A = (), otherwise the existence of the family A reduces to THEORY AND EXAMPLES 273 So we must find all values of m and n for which there exist a binary matrix T such that m m m T2 = m m m Let us consider m m m m, B =( m m m and find the eigenvalues of B. This is not difficult, since if x is an eingenvalue, then m—x m m m m—x m =0 m m 772 — X If we add all columns to the first and then take the common factor mn — x, we obtain the equivalent form (mn — x) 1 1 m — x =0. 1 m m — x In this last determinant, we subtract from each column the first column mul- tiplied by m and we obtain in the end the equation xn-1(mn — x) = 0, which shows that the eigenvalues of B are precisely 0, 0, ... 0, mn. But these are n-1 times exactly the squares of the eigenvalues of T. Hence T has the eigenvalues 0, 0, , 0, \/mn, because the sum of the eigenvalues is nonnegative (being n-1 equal to the sum of the elements of the matrix situated on the main diagonal). Since Tr(T) E Z, we find that mn must be a perfect square. Also, because Tr(T)< ?I, we must have m < n. 274 12. HIGHER ALGEBRA IN COMBINATORICS Now, let us prove the converse. Suppose that m < n and mn is a perfect square and write m = du2, n = dv2. Let us take the matrices I = (1 . . .1), 0 ( ). dv times dv times Now, let us define the circulant matrix / I...ro...o I U 11-U Oar u v — u —1 I . . 0 . . . 0 .I \ v-u E Mv,n({0, 1}). S= Finally, we take S S E Mn({0,1}). S It is not difficult to see that m m m A2 = m m m m m m which concludes the proof. The last idea that we present here (but certainly these are not all the methoG of higher mathematics applied to combinatorics) is the use of vector spaces. Again, we will not insist on complicated concepts from the theory of vector spaces, just the basic facts and theorems Maybe the most useful fact is that if V is a vector space of dimension n (that is, V has a basis of cardinality n), then any n + 1 or more vectors are linearly dependent. As a direct application, we will discuss the following problem, which is very difficult to solve by means A = THEORY AND EXAMPLES 275 of elementary mathematics. Try first to solve it without vectors and you will see how hard it is. The following example is classical too, but few people know the trick behind it. [Example 7.] Let n be a positive integer and let A1, , An+1 be nonempty subsets of the set {1, 2, ... , n}. Show that there exist nonempty and disjoint index sets I = {i1,i2,... , ip} and J = , jql such that U Ai2 U • • • U = Ai 2 • • U Chinese Olympiad Solution. Let us assign to each subset Ak a vector vk E Rn, where vk = (x/ 1 0 , xri) and f 0, if / E Ak Xk 1 1, if / E Ak Because dim Rn = n, the vectors we have just constructed must be linearly dependent. So, we can find ai, a2, , an+i E R, not all of them 0, such that aivi + a2v2 + • " + an±ivn±i = 0. Now take / = {i, E {1,2,... ,n+1}1ai > 0} and J = E {1,2, ..., n+1}1 aj < 0}. It is clear that I and J are nonempty and disjoint. Let us prove that U Ai = U Ai and the solution will be complete. Take h E U Ai and sup-iEI iEI pose that h U A3. Then the vectors vi with j E J have zero on their hth jEJ component, so the hth component of the vector aivi + a2v2 + • • • + ari+ivn+1 is E , > 0, which is impossible, since aivi + a2V2 + • • • + an-FiVn,±1 = 0. This xEA, iEI shows that U Ai c U A 3. The reversed inclusion can be proved in exactly iEI jEJ 276 12. HIGHER ALGEBRA IN COMBINATORICS the same way, And probably l Example 8. so we conclude that U A, = U A3' iEI 3EJ an even more difficult result: Let S be a finite subset of [0,1] containing 0 and 1 and such that every distance that occurs between pairs of elements oc-curs at least twice, except for the distance 1. Prove that S contains only rational numbers. [E.G.Strauss] Iran 1998 Solution. Let (et, e2, ..., en) be a basis of the linear space spanned by S over the field of rational numbers; this basis can be chosen such that en = 1. Now, write each element x, of S in this basis: x, = aiiei + az2e2 + • • • + aznen• We can define an order relation on the set of these vectors, by saying that x,> x3 if there exists a position inn which the two vectors differ and x, has a larger coordinate in the first position where they differ. This (lexicographic) order is total, so we can choose the maximal and minimal elements for it to be x, and x3 respectively. We know that xi — xi = xk — xi for some k,I. Thus using the maximality and minimality of x, and xi respectively, we deduce that xi = (0,0, ..., 0,1) and x3 = (0,0, ..., 0). Because any other vector xr is less than x, but greater than x3, we deduce that all vectors have the first n — 1 coordinates zero, which is equivalent to the fact that all elements of S are rational. We conclude this discussion with another problem, proposed for the TST 2004 in Romania, whose idea is also related to vector spaces. Example 9. Thirty boys and twenty girls are training for the Team Se-lection Test. They observed that any two boys have an even number of common acquaintances among the girls and exactly nine boys know an odd number of girls. Prove that there ex-ists a group of sixteen boys such that any girl attending the training is known by an even number of boys from this group. [Gabriel Dospinescu] THEORY AND EXAMPLES 277 Solution. Let us consider the matrix A = (aii) where { 1, if Bi knows Fi We have considered here that B1, B2, , B30 are the boys and F1, F 2, , F20 are the girls. Now, consider the matrix T = A • At. Observe that all the elements of the matrix T, except those from the main diagonal, are even 20 (because ti3 = E aikaik is the number of common acquaintances among the k=1 girls of the boys Bi, B3). Each element on the main diagonal of T is precisely the number of girls known by the corresponding boy. Thus, if we consider the matrix T in (Z/2Z, •), it will be diagonal, with exactly nine nonzero elements on its main diagonal. From now on, we will work only in Z/2Z. We have seen that rank(T) = 9. Using Sylvester's inequality, we have 9 = rank(T) > rank(A) + rank (At) — 20 = 2 • rank (At) — 20 hence r = rank (At) < 14. Let us consider now the linear system in (Z/2Z, +, •): The set of solutions of this system is a vector space of dimension 30 — r > 16. This is why we can choose a solution (x1, x2, , x30) of the system, hav- ing at least 16 components equal to 1. Finally, consider the set M = E {1, 2, ... , 30}1 xi = -IT We have proved that 'MI > 16 and also E a3, = 0 jEm for all i = 1, 2, 20. But observe that aji is just the number of boys Bk jEM with k E M such that Bk knows Fi. Thus, if we choose the group of those boys Bk with k E M, then each girl is known by an even number of boys from this group, and the problem is solved. aii = 0, otherwise { allxi. ± a2ix2 + " • + a30,1x3o =0 auxi + a22x2 + • • • + a30,2X30 = 0 a1,20X1 a2,20X2 ' • • + a30,20X30 = 0 278 12. HIGHER ALGEBRA IN COMBINATORICS A famous result of Sylvester (proved by Gallai and then by many other math-ematicians) states that if A is a finite set of points in the plane such that there is no line which contains A, then there exists a line passing through exactly two points of A. The following example is a refinement of this result and the proof is almost magical: Example 10:1 Prove that n distinct points, not all of them lying on a line, determine at least n distinct lines. [Paul Era's] Solution. Number the points with 1, 2, ..., n. Let X be the set of distinct lines passing through two of the n points and let Ai be the set of those lines in X that contain the point i. Then any two of the sets Ai, Ai have exactly one common element. We need to prove that their union, X has at least n elements. Suppose the contrary, namely that there are only p < n such elements of X (and let /1, , /3, be these lines). Then because any homogeneous linear system with p equations and more than p unknowns has a nontrivial solution, it follows that we can assign numbers xi, x2, ..., xn, not all 0, to the points such that the sum of the numbers on each line of X is 0. Then E xi = 0 for iEt, all j. Therefore 0 = xi)2. 3=1 TE/, However, observe that in the last sum every x2 appears at least twice (since not all the points are on the same line), yet every product 2xix3 with i j appears only once (this is where we use the fact that any two sets among Ai, A2, ..., An have exactly one common point). We therefore obtain xi + xz + • + xn 2 + (x1 + x2 + • • • + xn)2 < 0, which forces all xi to be zero, which contradicts the choice of xi, x2, • • In the framework of the previous problem, the next example should not be very difficult to solve. However, it is worth saying that this problem has THEORY AND EXAMPLES 279 no combinatorial proof until now: this is the famous Graham-Pollak theorem. The solution, due to Tverberg, is taken from the excellent work Proofs from The Book. Example 11. There exists no partition of the complete graph on n vertices with fewer than n-1 complete bipartite subgraphs (such that every edge belongs to exactly one subgraph). [R. Graham, 0. Pollak] Solution. Denote by 1, 2, ..., n the vertices of the complete graph on n ver-tices and suppose that B1, B2, ..., Bm is a partition of this graph with complete bipartite subgraphs. Every such subgraph Bk is defined by two sets of ver-tices Lk and Rk. Put a real number x, in each vertex of the complete graph Kn. The hypothesis implies that E xix, = E [(E xi) • (E x j)1 . i<i<j<n k=1 iELk iERk The idea is (like in the previous problem) that if m < n — 1 then we can choose the real numbers x1, x2, xn such that not all of them are zero, x1 + x2 + • • • + xn = 0 and E xi = 0 for all k. Indeed, this linear sys-ieLk tem has a nontrivial solution, because the number of unknowns exceeds the number of equations. Using the above identity and the fact that E xF = i=i (E x j)2 — 2 E xixi, we infer that x7 + 4 + • • • + xn 2 = 0, which contra- i=1 1<i<j<n dicts the choice of x1, x2, • • • , xn• We end this chapter with a very tricky problem, which became classical: we found traces of it and variants in AMM, Mathematics Magazine, as well as Iranian, Russian, and German Olympiads: 280 12. HIGHER ALGEBRA IN COMBINATORICS Example Let G be a simple graph, all of whose vertices are colored in white. A legal operation consists of choosing a vertex and changing the color of that vertex and of all of its neighbors (vertices connected to it) to the opposite color (black in white and white in black). Prove that one can make all vertices of the graph black in a finite number of legal operations. Solution. We will assume by convention that any vertex is connected to itself, so that the adjacency matrix A of the graph (defined by az3 = 0 if i and j are not connected and 1 otherwise) is symmetric and has only 1 on the main diagonal. The idea is to prove the existence of a set S of vertices of the graph such that any vertex of G is connected to an odd number of vertices of S. In this case, all we need is to perform legal operations on the vertices of S in order to change the color of all vertices of the graph. Now, observe that if we find a vector v = (v1, v2, ..., vn) with integer coefficients such that Av has all coordinates odd numbers we are done: it is enough to choose S the set of those i such that vi is odd. Thus, the problem reduces to proving that for any binary symmetric matrix A with diagonal (1, 1, ..., 1), there exists a vector v such that Av has all coordinates odd numbers. Translated in the field F = Z/2Z, this comes down to proving that for any symmetric matrix A e 111„(F), there exists a vector v E Fn such that Av = (1, 1, ..., 1). By a classical argument, it is enough to show that the orthogonal vector space of Im(A) is a subset of the orthogonal vector space of (1,1, ..., 1). But if x is orthogonal to Im(A), then we must have Xi E aijy, 0 i=i i=i for all yi, yn E F, which means that E azixi = 0 for all j. Thus i=i n n E x j E a zjxz = 0, THEORY AND EXAMPLES 281 which can be also written as E (aii + aii)xixi + E aii4 = 0. 1<i<j<n, i=1 The matrix is symmetric, so the first sum is 0. Also, we have 4 = xi and aii = 1, so we infer that x1 + x2 + • • + xn = 0, which means that x is orthog-onal to v. This finishes the proof. 282 12. HIGHER ALGEBRA IN COMBINATORICS 12.2 Problems for training 1. The squares of an n by n board are filled with 0 or 1 such that any two lines differ in exactly -7-1 positions. Prove that there are at most n • ones on the board. 2. Consider 2n + 1 real numbers with the property that no matter how we eliminate one of them, the rest can be divided into two groups of n numbers, the sum of the numbers in the two groups being the same. Then all the numbers must be equal. 3. A handbook classifies plants by 100 attributes (each plant either has a given attribute or does not have it). Two plants are dissimilar if they differ in at least 51 attributes. Show that the handbook cannot give 51 plants all dissimilar from each other. Tournament of the Towns 1993 4. Let A1, A2, , Am be distinct subsets of a set A with n > 2 elements. Suppose that any two of these subsets have exactly one element in com-mon. Prove that m < n. 5. The edges of a regular 24-gon are colored red and blue. A step consists of recoloring each edge which has the same color as both of its neighbors in red, and recoloring each other edge in blue. Prove that after 2n-1 steps all of the edges will be red and that need not hold after fewer steps. Iran Olympiad 1998 6. Is there in the plane a configuration of 22 circles and 22 points on their union (the union of their circumferences) such that any circle contains at least 7 points and any point belongs to at least 7 circles? Gabriel Dospinescu, Moldova TST 2004 PROBLEMS FOR TRAINING 283 7. Let p be an odd prime and let 71 > 2. For any permutation a E Sn, we consider 5(a) = > k (k). k=1 Let A3 and B3 be the set of even and odd permutations a for which S(a) j (mod p) respectively. Prove that n > p if and only if A3 and B3 have the same number of elements for all j E {0, 1, ,p —1}. Gabriel Dospinescu 8. A number of teams compete in a tournament, and each team plays against any other team exactly once. In each game, 2 points are given to the winner, 1 point for a draw, and 0 points for the loser. It is known that for any subset S of teams, one can find a team (possibly in S) whose total score in the games with teams in S is odd. Prove that n is even. D. Karpov, Russian Olympiad 1972 9. Let n > 2. Find the greatest p such that for all k E {1,2, ... ,p} we have (n E Eif(i)) = E (Eif(i)) crEA, i=-1 crE.13,, i=1 where An, Bn are the sets of all even and odd permutations of the set {1,2, . , n} respectively. Gabriel Dospinescu 10. Let r be the number of disjoint cycles in the decomposition of a per-mutation a of the set {1, 2, ..., n}. Prove that a cannot be written as the product of fewer than n — r transpositions. Determine the minimal number of transpositions that generate the symmetric group of order n. 284 12. HIGHER ALGEBRA IN COMBINATORICS 11. A simple graph has the property: given any nonempty set H of its vertices, there is a vertex x of the graph such that the number of edges connecting x with the points in H is odd. Prove that the graph has an even number of vertices. 12. In an m by n table, real numbers are written such that for any two lines and any two columns, the sum of the numbers situated in the opposite vertices of the rectangle formed by them is equal to the sum of the numbers situated in the other two opposite vertices. Some of the numbers are erased, but the remaining ones allow us to find the erased numbers using the above property. Prove that at least 71+771-1 numbers remained on the table. Russian Olympiad 1971 13. Find the least m such that it is possible to decompose the complete graph on n vertices into m complete subgraphs such that every edge belongs to exactly one such subgraph. 14. Let Al, A2, ..., An, B1, B2, ..., Bn be subsets of {1,2, , n} such that: a) for any nonempty subset T of A, there is an i E A such that Ai n is odd, and b) for any i, j E A, Ai and Bi have exactly one common element. Prove that B1 = B2 = • • • = Gabriel Dospinescu 15. Let xi, x2, , xn be real numbers and suppose that the vector space spanned by xi — x3 over the rationals has dimension m. Then the vector space spanned only by those xi — xi for which xi —x3 xk —x/ whenever (i, j) (k,l) also has dimension m. Strauss's theorem PROBLEMS FOR TRAINING 285 16. Light bulbs L1, L2, ..., Li, are controlled by switches Si, 82, S. Switch S i changes the on/off status of light Li and possibly the status of some other lights. Suppose that if Si changes the status of L3 then Sj changes the status of Li. Initially all lights are off. Is it possible to operate the switches in such a way that all the lights are on? Uri Peled, AMM 10197 17. Let s be a function defined by s(ai, a2, .••, ar) = (Iai —a21, I a2 — a3 1/ • • • ar — au I ). Prove the equivalence of the following statements: i) for all non- negative integers al, a2, ar, there exists n such that the n-th iterate of s evaluated at (al, a2, ar) is (0,0, ..., 0); ii) r is a power of 2. Ducci's problem 18. Let A be a finite set of real numbers between 0 and 1 such that for all x E A there exist a, b different from x, which belong to A or which are equal to 0 or 1, such that x = az 6. Prove that all elements of A are rational. Bay Area Competition 19. Let X be a set of n prime numbers and let m be a positive integer. Find the number of subsets A of X having the following properties: i) A has m elements, all of them being square-free. ii) the product of the elements in any subset of A is not a perfect square. iii) any prime divisor of an element of A is in A. Iran 1998 20. In a contest consisting of n problems, the jury defines the difficulty of each problem by assigning it a positive integral number of points (the same number of points may be assigned to different problems). Any 286 12. HIGHER ALGEBRA IN COMBINATORICS participant who answers the problem correctly receives that number of points for the problem; any other participant receives 0 points. After the participants submitted their answers, the jury realizes that given any ordering of the participants (where ties are not permitted), it could have defined the problems' difficulty levels to make that ordering coincide with the participants' ranking according to their total scores. Determine, in terms of n, the maximum number of participants for which such a scenario could occur. Russian Olympiad 2001 21. In a society, acquaintance is mutual and even more, any two persons have exactly one common friend. Then there is a person who knows all the others. Universal friend theorem 22. Let A1, A2, , Am be subsets of {1, 2, , n}.Then there are disjoint sets I, J such that U Ai = U Ai and n, = n iEI jEJ iEI jEJ Lindstrom's theorem 23. On an m x n sheet of paper a grid dividing the sheet into unit squares is drawn. The two sides of length n are taped together to form a cylinder. Prove that it is possible to write a real number in each square, not all zero, so that each number is the sum of the numbers in the neighboring squares, if and only if there exist integers k, I such that 7/ + 1 does not divide k and 2/7r/c7r 1 17/ cos — + cos n + 1 = 2 Ciprian Manolescu, Romanian TST 1998 PROBLEMS FOR TRAINING 287 24. Let A1, A2, . , Am and B1, B2, . , B,, be subsets of {1, 2, . , n} such that A2, fl B, is an odd number for all i and j. 'Then mp < 2n-1. Benny Sudakov 25. A figure composed of 1 by 1 squares has the property that if the squares of a fixed m by n rectangle are filled with numbers the sum of all of which is positive, the figure can be placed on the rectangle (possibly after being rotated by a multiple of 2) so that the numbers it covers also have positive sum (however, the figure may not have any of its squares outside the rectangle). Prove that a number of such figures can be placed on the rectangle such that each square is covered by the same number of figures. THEORY AND EXAMPLES 291 13.1 Theory and examples It may seem weird, but geometry is really useful in number theory, and some-times it can help in proving difficult results with some extremely simple argu-ments. In the sequel we are going to exhibit a few applications of geometry in number theory, almost all of them revolving around the celebrated Minkowski's theorem. This theorem will give a very efficient criterion for a centrally sym-metric convex body to contain a nontrivial lattice point. The existence of this point has important consequences in the theory of representation of numbers by quadratic forms, and in the approximation of real numbers by rational numbers. As usual, we will present only a mere introduction to this extremely well-developed field of mathematics. You will surely have the pleasure of con-sulting some reference books about this fascinating area of research. First of all, let us define the notion of convex body (or convex set; in what follows we will call bodies sets in lir). A subset A of Rn will be called a convex body if it is convex, that is A contains the segment {tx-F(1—t)y 10 < t < 1} once it contains two points x, y. A is called centrally symmetric if it is symmetric with respect to the origin, that is —x E A if x E A. We will take for granted that convex bodies have volumes (this is more delicate than it seems, actually). We start by proving the celebrated Minkowski's theorem. Theorem 13.1 (Minkowski). Suppose that A is a bounded centrally symmetric convex body in Rn having volume strictly greater than 2n. Then there is a lattice point in A different from the origin. Proof. The proof is surprisingly simple. Indeed, begin by making a partition of Rn into cubes of edge 2, having as centers the points that have all coordinates even integers. It is clear that any two such cubes have disjoint interiors and that they cover all space. That is why we can say that the volume of A is equal to the sum of the volumes of the intersections of A with each cube (because A is bounded, it is clear that the sum will be finite). But of course, one can bring any cube into the cube centered around the origin by using a translation by a vector all of whose coordinates are even. Since translations preserve volume, we will have now an agglomeration of bodies in the central 292 13. GEOMETRY AND NUMBERS cube (the one centered at the origin), and the sum of volumes of all these bodies is greater than 2Th. It follows that there are two bodies which intersect at a point X. Now, look at the cubes where these two bodies where taken from and look at the points in these cubes whose image under these translations is the point X. We have found two different points x, y in our convex body such that x — y E 2Zn. But since A is centrally symmetric and convex, it follows x 2 i y that s a lattice point different from the origin and belonging to A. The theorem is proved. 0 Here is a surprising result that follows directly from this theorem. rExample 1. Suppose that at each lattice point in space except for the origin one draws a ball of radius r > 0 (common for all the balls). Then any line that passes through the origin will intercept some ball. Solution. Let us suppose the contrary and consider a cylinder having as axis r — 2 that very line and base a circle of radius . We choose it sufficiently long to ensure that it has a volume greater than 8. This is clearly a bounded centrally symmetric convex body in space and using Minkowski's theorem we deduce the existence of a nontrivial lattice point in this cylinder (or on the border or the corresponding sphere). This means that the line will intercept the ball centered around this point. Actually, the theorem proved before admits a more general formulation: Theorem 13.2 (Minkowski). Let A be a convex body inRn and let v1, v2, • • • , vn be linearly independent vectors in Rn. Consider the fundamental parallelepiped n P = E 0 < xi < 1 and denote by Vol(P) its volume. If A has a vol- i=i ume greater than 2n • Vol(P), A must contain at least one point of the lattice L = Zvi + • • • + Zvn, different from the origin. THEORY AND EXAMPLES 293 Proof. With all these terms, it would seem that this is extremely difficult to prove. Actually, it follows trivially from the first theorem. Indeed, by con-sidering the linear transformation f sending vi into e, = (0, 0, , 1, 0, , 0), one can easily see that P is sent into the "normal" cube in IV (that is, the set of vectors all of whose components are between 0 and 1), and that f maps L into Z. Because the transformation is linear, it will send A into a bounded Vol(A) centrally symmetric convex body of volume > 2n Vol(P) . It suffices to apply the first theorem to this bounded centrally symmetric convex body and to look at the preimage of the lattice point (in Zn), in order to find a nontrivial point of A n L. This finishes the proof of the second theorem. K In the chapter Primes and Squares we proved that any prime number of the form 4k +1 is the sum of two squares. Let us prove it differently, this time using Minkowski's theorem. Example 2.] Any prime number of the form 4k +1 is the sum of two squares. Solution. We have already proved that for any prime number of the form 4k + 1, call it p, we can find an integer a such that p1 a2 + 1. Consider then vi =- (p, 0) and v2 = (a, 1). Clearly, they are linearly independent and moreover for any point (x, y) in the lattice L = Zv1 + Zv2 we have plx2 + y2. Indeed, there are m, n E Z such that x = mp + na, y = n and thus x2 +y 2 = n2(a2 +1) 0 (mod p). In addition, the area of the fundamen-tal parallelogram is 11 vi A v211 = p. Next, consider as convex body (when the context is clear, we will no longer add bounded centrally symmetric convex body, just convex body) the disc centered at the origin and having radius . Clearly, its area is strictly greater than four times the area of the fundamental parallelogram. Thus, there is a point (x, y) different from the origin that lies in this disc and also in the lattice L = Zv1 + Zv2. For this point we have plx2 + y2 and x2 + y2 < 2p, which shows that p= x2 + y2. Proving that some Diophantine equation has no solution is a classical prob- lem, but what can we do when we are asked to prove that some equation has 294 13. GEOMETRY AND NUMBERS solutions? Minkowski's theorem and, in general, the geometry of numbers give responses to such problems. Here is an example: [Example Consider positive integers a, b, c such that ac = b2 + b + 1. Prove that the equation ax2 — (2b + 1)xy + cy2 = 1 has integer solutions. Polish Olympiad Solution. Here is a very quick approach: consider in R2 the set of points satisfying ax2 — (2b + 1)xy + cy2 < 2. A simple computation shows that it is an elliptical disc having area — 47 > 4. An elliptical disc is obviously Nid a convex body and, even more, it certainly is symmetric about the origin. Thus by Minkowski's theorem there is a point in this region different from the origin. Since ac = b2 + b + 1, we have for all x, y not both equal to 0 the inequality ax2 — (2b + 1)xy + cy2 > 0. Thus for (x, y) E Z2 \ {(0, 0)}, we have ax2 — (2b+1)xy + cy2 = 1 and the existence of a solution of the given equation is proved. The following problem (like the one above) has a quite difficult elementary solution. The solution using geometry of numbers is more natural, but it is not at all obvious how to proceed. Yet... the experience gained by solving the previous problem should ring a bell. Example 4. Suppose that n is a positive integer for which the equation x2 + xy + y2 = n has rational solutions. Then this equation has integer solutions as well. Kemal Solution. Of course, the problem reduces to: if there are integers a, b, c such that a2 + ab + b2 = c2n, then x2 + xy + y2 = n has integer solutions. We will THEORY AND EXAMPLES 295 assume that a and b are nonzero (otherwise the conclusion follows trivially); and more, a classical argument allows us to assume that a and b are each relatively prime (which implies that a and b are each relatively prime to n, too). We try again to find a pair (x, y) E Z2 \{(0, 0)} such that x2+xy+y2 < 2n and such that n divides x2 + xy + y2. In this case we will have x2 + xy + y2 = n and the conclusion follows. First, let us look at the region defined by x2 + xy + y2 < 2n. Again, simple computations show that it is an elliptical disc of area — 47 n. Next, consider the lattice formed by the points (x, y) such that n divides ax - by. The area of the fundamental parallelogram is clearly at most n. By Minkowski's theorem, we can find (x, y) E Z2 \ {(0, O)} such that x2 + xy + y2 < 2n and n divides ax - by. We claim that this yields an integer solution to the equation. Ob-serve that ab(x2 + xy + y2) = c2xyn + (ax - by) (bx - ay) and so n also divides x2+xy+y2 (since n is relatively prime with a and b) and the conclusion follows. Before continuing with some more difficult problems, let us recall that for any symmetric real matrix A such that E aijxixj > 0 i<753 <n for all x = (xi, x2, . , xn) E Rn \ {0} the set of points satisfying E ai <1 i<i,j<n Vol(Bn) where has volume equal to \Met A Vol(Bn) = n F(1+ 2) cc where Bn is the nth dimensional Euclidean ball (and F(x) = J e-ttx-idt is Euler's gamma function). There are explicit formulae for F(1 + 3) because 71-3 296 13. GEOMETRY AND NUMBERS F(n) = (n — 1)! for positive integers n (so this takes care of the case n even) and F (1 + ) = 2 3/1-1-1 2 2 x ((n — 1)/2)! for odd n. The proof of this result is not elementary and we invite you to read more about it in any decent book of multivariate integral calculus. In particular, you should notice that these results can be applied to previous problems to facilitate the computations of different areas and volumes. With these fact in mind, let us attack some serious problems. If we talked about squares, why not present the beautiful classical proof of Lagrange's theorem on representations using four squares? 1 Example 5.1 Any positive integer is a sum of four perfect squares. [Lagrange] Solution. This is going to be much more complicated, but the idea is always the same. The main difficulty is finding the appropriate lattice and centrally symetric convex body. First of all, let us prove the result for prime numbers. Let p be an odd prime number (for the prime 2 the result is obvious) and consider the sets A = {a21 a E Z/pZ}, B = {—b2 — ll b E Z/pZ}. Since there are p ± 1 distinct squares in Z/pZ (as we have already seen in previous 2 chapters), these two sets cannot be disjoint. In particular, there are integers x and y such that 0 < x, y < p — 1 and plx2 + y2 + 1. This is the observation that will enable us to find a good lattice. Consider now the vectors vi = (P, 0, 0, 0), v2 = (0,p, 0, 0), v3 = (x, y, 1, 0), v4 = (y, —x, 0,1) and the lattice L generated by these vectors. A simple computation (using the above formulas) allows us to prove that the volume of the fundamen- tal parallelepiped is p2. Moreover, one can easily verify that for each point (t, u, v, w) E L we have plt2 + u2 + v2 + w2. Even more, we can also prove (by employing the non-elementary results stated before) that the volume of \ Fr x n! THEORY AND EXAMPLES 297 the convex body C = {(t, u, v, w) E R41 t2 + u 2 + v2 + w2 < 2p} is equal to 272p2 > 16Vol(P). Thus C n L is not empty. It suffices then to choose a point (t, u, v, w) E C n L and we will clearly have t2 + u2 + v2 + w2 = p. This finishes the proof for prime numbers. Of course, everything would be nice if the product of two sums of four squares is always a sum of four squares. Fortunately, this is the case, but the proof is not obvious at all. It follows form the miraculous identity: (a2 + b2 + C2 + d2) (X2 + y2 + z2 + t2) = (ax + by + cz + dt)2 +(ay — bx + ct — dz)2 + (az — bt + dy — cx)2 + (at +bz — cy — dx)2. This is very nice, but how could one answer the eternal question: how on earth should I think of such an identity? Well, this time there is a very nice reason: instead of thinking in eight variables, let us reason only with four. Consider the numbers z1 = a + bi, z2 = c+ di, z3 =- x + yi, z4 = z + ti and introduce the matrices Zi Z2 ( Z3 4 ) M = z N = z 2 zi 4 Z3 We have det(M) = Izi 12 + 1z212 — a2 + b2 + c 2 + d2 and similarly det(N) = X2 + y2 + z2 + t2. It is then natural to express (a2 + b2 + c 2 + d2)(x2 + y2 + z2 + t2) as det(MN). But surprise! We have MN = ( ziz3 — z2z4 ziz4+ z2z3 ) and so det(MN) is again a sum of four squares. The identity is now motivated. Let us concentrate a little bit more on approximations of real numbers. We have some beautiful results of Minkowski that deserve to be presented after this small introduction to the geometry of numbers. The following one is ex-tremely important while studying algebraic number fields. — Z1Z4 Z2Z3 Z1Z3 Z2Z4 298 13. GEOMETRY AND NUMBERS Example 6.1 Let A = (ai3) be an n x n invertible matrix with real en-tries, and let Cl, , cn be positive real numbers such that c1c2 • • • cn > Idet Al. Then there are integers xi, x2, .. • xn, not all 0, such that E ,x3 < cz for all i = 1, , n. 3=1 Minkowski's linear forms theorem Solution. We need to prove that there exists a nonzero integer vector X that also belongs to the region {Y E Rni (AY)21 < Cz, i = 1, n} (here (AY)i denotes the i-th coordinate of AY). But {Y E Rni RAY)21 < ci, i = 1, . . . , n} is exactly the image through A-1 of the parallelepiped {Z E Rn < ZZ < c2 i i = 1, . . . , n} (which has volume 2nci cn). Thus {Y E Rni 1(AY),1 < ci, i =1,...,n} is a centrally symetric convex body of volume 1 I det AI 2nc1 cn > 2n. By Minkowski's theorem, this body will contain a nonzero lattice point, which satisfies the conditions of the problem. Actually, there exists a very useful sharpening of the last result: if we suppose only that ci c2 • • • cn > Idet Al, then the integers xi, x2, , xn, not all 0, can be chosen such that E ai3 x3 < ci and E azi x j < a, for all i > 2. The proof is j=i not difficult at all, once example 6 is proved. Indeed, note that if c > 0 then by the previous result there are integers xi (c), x2(c), , xn(c), not all 0 and such that (c) < c2 for i = 2, 3, n and E ai3x3(6) < ci(l+E). Because 3-1 j=i the matrix A is invertible, there exist only finitely many (xi (c), x2(E), xn(E)) with these properties for fixed E. Indeed, the condition says that the vector Ax(E) is bounded where x(c) is the vector with components xi(c). Thus the vector x(c) is also bounded in Rn. This shows that it is possible to construct a sequence Ek that converges to 0 and such that x3 = xj(ck) does not depend on k for all j. All we need is then to make k oo in the above inequalities. THEORY AND EXAMPLES 299 Now, this theorem implies Dirichlet's approximation theorem (also discussed in the chapter Density and regular distribution: for all real numbers al, a2, ..., an and all positive integers M there exist integers ml, m2, p such that < Mn and lmi — pad < - for all i). Indeed, all we need is to apply the above result to the (n + 1) x (n + 1) matrix / 1 0 0 0 —al \ 0 1 0 0 —a2 0 0 1 0 —a3 0 0 0 1 —an \ 0 0 0 0 1 j And here is a nice consequence of the previous example. Our last example of Diophantine approximation that can be obtained using Minkowski's theorem will imply the product theorem for homogeneous linear forms: l Example 7. Let A = (aid) be an n x n invertible matrix with real entries (n > 2). Show the existence of integers xi, x2, xn, not all 0, for which E aiixi + ai2x2 + - • • + ainxrd < Vn! IdetAl. i=1 Solution. Let us start by computing the volume of the figure 0(x, n) consist-ing of all points (xi, x2, ...,xn) such that 1x11+ lx21+ • • • + Ixn1 < x. For n = 1 it is certainly 2x. Now, using Fubini's theorem we can write Vol(0(x, n)) = dxidx2...dxn = dxi...dxn—i f fixr,1<x = Vol(O(x — Ixrd,n — 1))dxn = Vol(0(1,n-1))/ (x lx”.1<x I 300 13. GEOMETRY AND NUMBERS = Vol(0(1,n - 1)) 2xn 71 2r An immediate induction shows that Vol(O(x, n)) = ( ; Now, the problem asks to prove that there exists some nonzero integer vector X such that AX lies in the figure O( /n! • I det AI, n). Stated otherwise, we need to prove that the image of this figure by the linear application determined by A-1 contains a nonzero lattice point. But the volume of this centrally symetric convex body is 2n (just replace x by /n! • I det AI). Unfortunately, we cannot directly ap-ply Minkowski's theorem, because this volume is not strictly greater than 2n. However, we can imitate the argument used after the solution of the previous exercise in order to obtain the desired result: for all E > 0 we know that the octahedron O( /n! • I det AI + E, n) contains some AX, (where X, is a nonzero integer vector). One shows that these vectors X, are uniformly bounded, then extracts a constant family of vectors and takes the limit. We leave to the reader the details. The highlight of the IMO 1997, the very beautiful problem 6 also has a mag-nificent solution using geometry of numbers. Actually, we will prove much more than the result asked in the contest, which shows that, for large values of n, one of the bounds asked by the IMO problem is very weak: I Example CI For each positive integer n, let f (n) denote the number of ways of representing it as a sum of powers of two with nonnegative integer coefficients. Representations that differ only in the ordering of their summands are considered to be the same. For instance, f (4) = 4. Prove that there are two constants a, b such that n2 2-y n2 —nlog2(n)—an < f(2n) < 22—n1og2(n)—bn for all sufficiently large n. Adapted after IMO 1997 THEORY AND EXAMPLES 301 Solution. It is clear that f (2n) is just the number of nonnegative integer solutions of the equation ao + 2a 1 + • • • + 2nan = 2n, which is the same as the number of solutions in nonnegative integers of the inequation 2ai + 4a2 + • • • + Tian < 2n. For any such solution different from (0, 0, ..., 0,2n) we have an = 0 and we will consider the hypercube H(ai, a2, •, an_i) = [al, ai + 1) x [a2, a2 + 1) x x [an_i, an_i +1). It is clear that these hypercubes are pairwise disjoint for distinct solutions (al, a2, an_i). So the number of solutions of the inequation is the total volume of these hypercubes. Now, observe that any such hypercube is included in the set of points (xi, x2, xn_i) with xi > 0 and n-1 E 22(x, - 1) < 2Th. Also, the union of these cubes covers the region consisting i=1 n-1 of those points (xi, x2, ..., xn_i) with xi > 0 and E 2'xi < 2Th. Indeed, take a i=1 point (x1, x2, ..., xn_i) in this region. Then ([x1], [x2], ..., [xn_i], 0) is a solution of the inequation and the point belongs to the corresponding hypercube. Now, let us consider more generally the region R(ai, a2, ..., an, A) defined by the inequations xi > 0 and aixi + a2x2 + • • • + anxn < A. Its volume is Vol(R(ai, .••, an, A)) = dxidx2...dxn = fx,>0,aixi+•••+anx,<A i0<.,,,<2- xi,.•.,.,-1>0,aixi±-±an-ixn-i <A —anxn dxi...dxn-i A -= I an Vol(R(ai, an_i, A - anxn))dxn = A = VOI(R(al, •••) an-1) 1)) f (A - anxn)n-ldxn = An Vol(R(ai, an-1,1))• Tian This relation easily implies by induction that Vol(R(ai, a2, •.•, an)) = ni.a1an 2 ...an • Thus, because the sum of the volumes of the hypercubes is between the vol- ume of R(2,4, ..., 2n-1, 2n) and R(2,4, ..., 271-1,2 + 22 + 2n-1 + 2n) = 0 302 13. GEOMETRY AND NUMBERS R(2, 4, ..., 2n-1, 2n+1- 2), counting the solution (0, 0, ..., 0, 2Th), we deduce that the number of solutions satisfies the inequalities 2 2 < f (2n) < 1 4_ (27+1 2)n-1 1 + (n 1)! 2 Th2Z 2 n • (n - 1)! Now, note that 2 n 2 -n 2 21+0(n)-log2((n-1)!) (n - 1)! and that 1 log2((n - 1)!)= — 1n2 ((n 1) ln(n - 1) + 0(n)) = n log2(n) + 0(n) by Stirling's formula. Similarly, (2n+1 2 ) n-1 n2 = n log2 (n) + 0(n). The existence of the two constants is now obvious. We end this chapter with some difficult problems concerning representations of solutions of some Diophantine equations. We will show, using Minkowski's theorem, that if n < 4 and A is a symmetric and positive matrix (that is, txAx > 0 for all vectors x E Rn) in SL,(Z) (the set of integer n x n matrices with determinant 1), then there exists a matrix B with integer coefficients such that A = B • Bt (a result which actually holds for n < 7, for n = 8 being false). This will have some nice applications in the study of some Diophantine equations. Let us start with some notations and easy observations. A bases of Zn will be a family B = (vi, v2, ...,vp) of vectors in Z'2, such that any vector x E Zn can be uniquely expressed as kivi k2v2 + • • • + kpvp for some integers k1, k2, kp. For instance, it is clear that the canonical bases (el, e2, en) of Rn is a basis of Zn, where ez is the vector which has 1 on position i and 0 otherwise. But there are many other bases of Zn. Actually, in the chapter A Little Introduction to Algebraic Number Theory we proved that n2 2 2 • (n - 1)! 2 THEORY AND EXAMPLES 303 any integer vector whose coordinates are relatively prime can be completed to a basis of Zn. We can easily prove that any two bases B1 = (v1, v2, •-•, vp) and B2 = (W1, W2, ..., wq) have the same number of elements. Indeed, let us write v, = aiiwi + ai2W2 + • • • + atqWqr and w, = bay' + bi2v2 + • • • + bzpvp for some integers au, bu. Then if A and B are the matrices with entries au and bu, we have AB = _r„ (just replace in v, = aiiwi + ai2w2 + • • + aiqwq each w, by bay]. + bi2v2 + • • • + b2pvp and then use the linear independence of vi). Thus p = rank(AB) < rank(A) < q and by symmetry we also have q < p, so q = p. (Now one can see that, due to the existence of the canonical basis mentioned above, any basis of Zn has n elements.) Now, for an n x n matrix A with integer coefficients we can define a bilinear form gA(x, y) = E aijxiyj = xtAy, where x = xi el + x2e2 + • • • + xnen and 1<i<j<n y = y1e1 + y2e2 + • • • + ynen. Let fA be the quadratic form associated with this bilinear form, that is fA(x) = g A(x , x). Now, take B = (v1, v2, vn) a basis in Zn and suppose that vi = viiei + v2,e2 + • + vnien. By the previous argument (showing that two basis have the same cardinality) we know that V = (vu) is invertible. For an integer vector whose coordinates are xi in the canonical basis and x", in B, we have x = Vx', and a short computation shows that g A(x , y) = xit(VtAV)V. On the other hand, a direct computation shows that g A(x , y) = xitGy' where G = (g A(u,, v3)), and this shows that G = VtAV. lExample 9.1 If A E SLn(Z), n < 4, is a symmetric positive matrix, then there is a matrix B with integer entries such that A = BtB. Solution. We will keep the notations used in the introduction to this exam-ple. Let us start with a very modest result, but one which, as we will see immediately, is the key idea for solving the problem. Lemma 13.3. There exists a vector vi E Zn such that fA(vi) = 1. Proof. The proof is a direct consequence of Minkowski's theorem. Indeed, we have seen that the volume of the ellipsoid defined by fA(x) < 2 is equal to 304 13. GEOMETRY AND NUMBERS r(i+3). For n < 4, using the fact that F(n) = (n — 1)! and F(n + =- 1-3....•(2n-1) \Fr, you can easily verify that (3)71 > (F (1 + "02. This shows 2" that the volume of the ellipsoid is greater than 2n and thus it contains a nontrivial lattice point, which we call v1. Because fA(vi) > 0 (A is invertible and positive) and fA(vi) E Z, it is clear that vi is a good choice. Now, we will extend this vector v1 to a basis B = (v1,v2,...,vn) so as to have the first line and column of G constructed: Lemma 13.4. Let v1 be a vector as found in lemma 13.1. Then there exist integer vectors v2,v3,...,vn such that B = (vi,v2,...,vn) is a basis of Zn and gA(vi,vi) = 0 for all i > 2. Proof. The proof is very beautiful. Consider H = {x E Zn IgA(vi, x) = 01. Clearly, H is a submodule of Zn, thus it is of the form 7Gv2 + • • • + Zvr for some linearly independent integer vectors v2, v3, ..., vr. We claim that B = (v1, v2, ..., vr) is a basis of Zn. Indeed, take x E Zn. We need to study the equation x = kivi + v , where v E H. All we need is gA(vi, x — kiv 1) = 0, which is the same as ki fA(vi) = gA(vl , x), thus k1 = gA(vl, x). Thus k1 exists and is uniquely determined. This means (because v2, ..., yr are linearly independent) that there exist unique integers k1, k2 , k r such that x = k1v1 k2v2 + • • • + krvr. Thus B is a basis of Zn, and consequently we also have r = n. This finishes the proof of lemma 2. 1=1 Now, we can proceed to an inductive proof. We will prove that the assertion holds for n > 1 by induction. Of course, the case n = 1 is trivial, so assume that the result holds for n — 1. Using lemma 1 and lemma 2, we know that for 1 0 some matrix S with integer coefficients we have A = St • 0 A' S, where clearly A' is a symmetric positive matrix in SLn_1(Z). Applying the inductive hypothesis, we can write A' = BitB' for some matrix B' with integer entries. THEORY AND EXAMPLES 305 ( 1 0 ' ) Therefore A =134B where B = B • S. 0 Let us now discuss two beautiful applications of this result. The first is quite classical; it was among the results obtained by Fermat. However, it appeared as an old proposal for the IMO, as well as in the Iranian Olympiad in 2001. Example 10. Let x, y, z be positive integers such that xy = z2 + 1. Prove that there exist integers a, b, c, d such that x = a2 + b2, y = c2 + d2 and z = ac + bd. z Solution. Let us consider the matrix A = x y . Then A E SL2(Z) z because xy = z2 + 1. Also, tr(A) = x + y > 2, thus A has positive eigenvalues, so A is symmetric and positive. (This could have been established directly, too, by showing that 2 (XU zv)2 + v2 XU2 2zuv + yv = > 0 x for all u, v not both equal to 0.) By the previous result, A can be written as B • Bt for some matrix B = ( a d b . By identifying entries in the equality A = B • Bt, we deduce the desired representation. Note that the last example implies a famous theorem of Fermat: each prime number of the form 4k + 1 is a sum of two squares. Indeed, we saw that for such a prime p there is always an n such that pin2 + 1. However, the last theorem shows that any divisor of a number of the form n2 + 1 is a sum of two squares. Next, let us see a very difficult Diophantine equation, whose solution follows in a few lines from the important result proved above. TExample 11. Find all integers a, b, c, x, y, z such that axe + by2 + cz2 = abc + 2xyz — 1, ab + be + ca > x2 + y2 + z2, and a, b, c > 0. [Gabriel Dospinescu] Mathematical Reflections 306 13. GEOMETRY AND NUMBERS a z y Solution. Let us consider the matrix M = z ( b x Clearly, M is y x c symmetric. The condition ax2 by2 cz2 = abc + 2xyz — 1 implies that M E SL3(Z). Now, let us prove that A is positive. Because M is symmetric and invertible, it is enough to prove that its eigenvalues are positive. Let these eigenvalues be u, v, w. Then we know that u, v, w are real numbers (because M is symmetric), that uvw = 1 and u+v+w = tr(M) = a+b+c > 0. On the other hand, it is not difficult to see that uv + vw + wu = ab — z2 + bc — x2 + ac — y2 > 0, the sum of the principal second-order minors. Thus u, v, w are zeros of a polynomial of the form X3 — UX2 + VX — 1 for some nonnegative U, V. Clearly, such a polynomial can have only nonnegative zeros, thus u, v, w > 0. Because det(M) = 1, it follows that M satisfies all conditions of the previous theorem, so M is of the form tNN for some integer matrix N. If we write al a2 a3 N = b1 b2 b3 , we deduce that a = 11A112 b = 11B112, c = 11C112 C1 C2 C3 z = (A, B), y = (A, C) and finally x = (B, C) for some integer vectors A, B ,C (here II 11 and (•) are the Euclidean norm and inner product respectively) that form a basis of Z3 (they are the rows of the matrix N). All these triples found are actually solutions. Indeed, if A, B ,C form a basis in Z3, then the matrix N whose rows are A, B ,C is in GL3(Z), that is its determinant is —1 or 1, so det(tNN) = (det(N))2 = 1. Thus det(A) = 1 and ax2 + by2 + cz2 = abc + 2xyz — 1. Also, X2 + y2 + Z2 < ab bc ca is a consequence of the Cauchy-Schwarz inequality, because x2 = (13 7 1113112 dr112. THEORY AND EXAMPLES 307 Thus these are the solutions of the Diophantine equation. And last but not least, let us prove a beautiful theorem which, although not related to Minkowski's theorems, has strong connections with the geometry of numbers. You will notice, if you know the three-squares theorem, that the problem is trivial. But if not, what would you do? Without such an advanced result, the problem is not easy at all, but as we will see, a good geometric argument is the key of a very elementary solution: [Example 12.] Prove that any integer which can be written as the sum of the squares of three rational numbers can also be written as the sum of the squares of three integers. [Davenport-Cassels] Solution. Let us suppose by contradiction that the property does not hold. We will use a geometric argument combined with the extremal principle. Let S be the sphere of radius Nrn in R3 and suppose that a E S has all coordinates rational numbers. There exists an integer vector v E Z3 and an integer d > 1 such that a = d. Choose the pair (a, v) for which d is minimal. We claim that there exists a vector b E Z3 such that Ha — b11 < 1, where 11x11 is the Euclidean norm of the vector x. Indeed, it is enough to write a = (x, y, z) and to consider b = (X, Y, Z), where integers X, Y, Z are such that 1 1X — x1 < 2 Yi 1 — 2 , 1Z — ,z1 < 2 — 1. Now, since a is assumed to have at least one non-integer coordinate, a b. Consider the line ab. It will cut the sphere S in a and another point c. Let us determine precisely this point. Writing c = b + A • (a — b) and imposing the condition 11c11 2 = n yields a quadratic equation in A, with an obvious solution A = 1. Using Viete's formula for this equation, we deduce that another solution 11b11 Hab11 2—n 2 is A = On the other hand, the identity Ha b112 = n +1142 (b,v) 308 13. GEOMETRY AND NUMBERS and the fact that 0 < I la — bl 1 < 1 show that I la — bl 12 = 4 for a cer- tain positive integer A smaller than d. Therefore, A = 1(1Ib112 — n) and c = b + Ilbl A l 2' (v db) = "t i' , A for an integer vector w. This shows that the pair (c, w) contradicts the minimality of (a, v) and proves the result. PROBLEMS FOR TRAINING 309 13.2 Problems for training 1. Consider a lamp (a point) in space. Prove that no matter how we place a finite number of closed spheres of equal radius, the light of this lamp will be able to go to infinity (that is, there exists a direction in which the light will not hit any of these spheres). The spheres must not touch. Iran 2003 2. Suppose that a and b are rational numbers such that the equation axe + bye = 1 has at least one rational solution. Then it has infinitely many rational solutions. Kurschak Competition 3. Is there a sphere in R3 which has exactly one point with all coordinates rational numbers? Tournament of the Towns 4. In the plane consider a polygon of area greater than n. Prove that it contains n + 1 points Aa(xi, yi) such that xi — x3, yi — y3 E Z for all 1 < j < n + 1. Chinese TST 1988 5. Let S = {pi ., P2, pn} be a set of prime numbers and let f (S, x) be the number of integers not exceeding x, all of whose prime divisors are in S. Prove that f (S, x) (ln n! • lnpi • In p2 • • • lnpn Deduce that there are infinitely many primes. Michael Rubinstein 310 13. GEOMETRY AND NUMBERS 6. Suppose that a, b, c are positive integers such that ac = b2 + 1. Prove that the equation axe + 2bxy + cy2 = 1 is solvable in integers. 7. Let A = (au) where au (1 < i, j < n) are rational numbers such that aiixixi > 0 for all x = (xi, . , xn) E Rn \ {0}. Prove that there l<z,3<n are integers. xi, , xr, (not all zero) such that au xixi < n ' \Ydet A. 1<i,j<n Minkowski 8. Prove that if A = (au) is an n x n invertible matrix with real entries then there exist integers xl, x2, ..., xn, not all zero, such that < — n! • IdetAl. nn Product Theorem for Homogeneous Linear Forms 9. Consider a disc of radius R. At each lattice point of this disc, except for the origin, one plants a circular tree of radius r. Suppose that r is optimal with respect to the following property: if you look from the origin, you can see at least one point situated at the exterior of the disc. Prove that 1 1 < r <—. \/R2 + 1 R George Polya, AMM 10. Let f : Rn R be an even function such that f (x) > 0 for all x E Rn, different from 0, f (ax) = a f (x) for all a > 0 and all x E Rn, and f (x + y) < f (x) + f(y) for all x, y. Prove that there exists an open, bounded, convex set B = {x E Rnif(x) < 1} such that f (x) is the gauge of B, that is f (x) = inf{A > 0j E B}. n aijxj =1 PROBLEMS FOR TRAINING 311 11. Let B be a convex, open and bounded set in Rn, and let f be its gauge (defined in the previous problem). Prove that minxEzn,x00 f (x) < 2 where V (B) is the volume of B. ii/v(s)' Minkowski 12. Let a, b, c, d be positive integers such that there are 2004 pairs (x, y) with x, y E [0,1] for which ax + by, cx + dy E Z. If gcd(a, c) = 6, find gcd(b, d). Nikolai Nikolov, Bulgarian Olympiad 2005 13. Prove that there is no position in which an n by n square can cover more than (n 1)2 integral lattice points. D.J.Newman, AMM E 1954 14. Let n > 5 and let al, , an, b1, , bri be integers such that that all pairs (ai, bi) are different and laibi+i — ai±ibil = 1, 1 < i < n (here (an+i, bn+1) = (al, b1)). Prove that we can find 1 < i — j < n — 1 such that laibi — = 1. Korean TST 15. Let us denote by A(C, r) the set of points w on the unit sphere in Rn with the property that lw kl > for any nonzero vector k E Z — PP' (here w • k is the usual dot product and I kII is the Euclidean norm of the vector k E Zn). Prove that if r > n — 1 there exists C > 0 such that A(C, r) is nonempty, but if r < n — 1 there is no such C. Mathlinks Contest (after an ENS entrance exam problem) 312 13. GEOMETRY AND NUMBERS 16. Prove that for a positive integer n the following assertions are equivalent: a) n is the sum of three squares of integers; b) the set of points with all coordinates rational on the sphere centered at the origin and having radius \Ft is dense in this sphere. 17. Suppose that xi, x2, , xn are algebraic integers such that for each 1 < i < n there is at least one conjugate of x, which is not among x1, x2, • • • , xn. Prove that the set of n-tuples (f (xi), f (x2), • • • , f (xn)) with f E Z[X] is dense in RTh . 18. Let f (X) = (X — xi)(X — x2) • • • (X — xn) be an irreducible polynomial over the field of rational numbers, with integer coefficients and real zeros. Prove that nn H ixi - xi l 1<i<j<ri 71. Siegel THEORY AND EXAMPLES 315 14.1 Theory and examples Quite often, a collection of simple ideas can make very difficult problems look easy. We have seen or will see a few such examples in our journey through the world of numbers: congruences that readily solve Diophantine equations, properties of the primes of the form 4k + 3, or even facts about complex numbers, analysis or higher algebra, cleverly applied. In this unit, we will discuss a fundamental concept in number theory, the order of an element. It may seem contradictory for us to talk about simple ideas and then say "a fundamental concept". Well, what we are going to talk about is the bridge between simplicity and complexity. The reason for which we say it is a simple idea can be guessed easily from the definition: given an integer n > 1 and an integer a such that gcd(a, n) = 1, the least positive integer d for which nlad — 1 is called the order of a modulo n. The definition is correct, since from Euler's theorem we have n1 aP(n) — 1, so such numbers d exist. The complexity of this concept will be illustrated in the examples that follow. We will denote by on(a) the order of a modulo n. A simple property of an(a) has important consequences: if k is a positive integer such that nlak — 1 and d = on(a), then dlk. Indeed, because nlak — 1 and nlad — 1, it follows that niagcd(k,d) 1. But from the definition of d we have d < gcd(k, d) , which cannot hold unless dlk. Nice and easy. But could such a simple idea be of any use? The answer is yes, and the solutions of the problems to come will vouch for it. But before that we note a first application of this simple observation: on(a)fr,o(n). This is a consequence of the above property and of Euler's theorem. Now an old and nice problem, which may seem really trivial after this intro-duction. It appeared in Saint Petersburg Mathematical Olympiad and also in Gazeta Matematica. [Example 1.1 Prove that nic,o(an — 1) for all positive integers a and n. Solution. What is oan_i (a)? It may seem a silly question, since of course oan_1(a) = n. (because if ak — 1 is a multiple of an — 1, then ak — 1 > an —1 316 14. THE SMALLER, THE BETTER and so k > n-we assumed a > 1, otherwise the conclusion is clear; thus the order is at least n and on the other hand it obviously divides n) Using the observation in the introduction, we obtain exactly nico(an — 1). Here is another beautiful application of the order of an element. It is the first case of Dirichlet's theorem that we intend to discuss, a classical property. Example 2.] Prove that any prime factor of the n-th Fermat number 22n +1 is congruent to 1 modulo 2n+1. Then show that there are infinitely many prime numbers of the form 2nk + 1 for any fixed n. Solution. Let us consider a prime p such that pl22n + 1. Then p divides (22n + 1)(22n — 1) = 22n+1 — 1 and consequently op(2)12n+1. This ensures the existence of a positive integer k < n + 1 such that op(2) = 2k. We will prove that in fact k = n + 1. Indeed, if this is not the case, then op(2)12n, and so PI2°P(2) — 1122n — 1. But this is impossible, since OP + 1 and p is odd. Hence we found that op(2) = 2n+1 and we need to prove that op(2)1p — 1 to finish the first part of the question. But this follows from the introduction of this chap-ter. The second part is a direct consequence of the first. Indeed, it is enough to prove that there exists an infinite set of pairwise relatively prime Fermat's numbers (22nk 11 ink>n- Then we could take a prime factor of each such number and apply the first part to obtain that each such prime is of the form 2nk + 1. But not only is it easy to find such a sequence of pairwise relatively prime numbers, but in fact, any two different Fermat numbers are relatively prime. Indeed, suppose that digal(22n + 1,22n+k + 1). Then d122n+1 — 1 and so c/122n+k — 1. Combining this with d122n+k + 1, we obtain a contradiction. Hence both parts of the problem are solved. We continue with another special case of the well-known and difficult theo-rem of Dirichlet on arithmetical sequences. Though classical, the following problem is not straightforward, and this probably explains its presence on a Korean TST in 2003. THEORY AND EXAMPLES 317 Example 3. For a prime p, let fp(x) = xP-1+ XP-2 + • • • + x + 1. a) If plm, prove that any prime factor of fp(m) is relatively prime to m(m — 1). b) Prove that there are infinitely many positive integers n such that pn +1 is prime. Solution. a) Take a prime divisor q of fp(m). Because q11 + m + • • + mP-1, it is clear that gcd(q,m) = 1. Moreover, if gcd(q, m — 1) 1, then qlm — 1 and because ql1 + m + • • • + mP-1, it follows that qlp. But plm and we find that qlm, which is clearly impossible. More difficult is b). We are tempted to use a) and explore the properties of fp(m), just like in the previous problem. So, let us take a prime qlfp(m) for a certain positive integer m that is divisible by p. Then we have qimP — 1. But this implies that oq(m)lp and consequently oq(m) E {1,74. If oq(m) = p, then q 1 (mod p). Otherwise, qlm — 1, and because qlf p(m), we deduce that qlp. Hence q = p. But, while solving a), we have seen that this is not possible, so the only choice is plq — 1. Now, we need to find a sequence (mk)k>1 of multiples of p such that f p(mk) are pairwise relatively prime. This is not as easy as in the first example. Anyway, just by trial and error, it is not too difficult to find such a sequence. There are many other approaches, but we like the following one: take ml = p and mk = Pfp(Tri1)ip(n12) • • • ip(rnk—i)• Let us prove that fp(mk) is relatively prime to fp(mi), f p (7112) • • • fp(mk-i)• But this is easy, since fp(m1)fp(m2) • • • fp(mk—t)i.ip(mk) — f p(0) = fp(mk)— 1. Let us use this special case of Dirichlet's theorem to prove the following non-trivial result: Example 4. Let k > 2 be an integer. Prove that there are infinitely many composite numbers n with the property that nian—k — 1 for all integers a relatively prime to n. [A.Makowski] Solution. Let us choose these numbers of the form n = kp for some suitable prime number p. We need plan—k -1, so it is enough to have k, which 318 14. THE SMALLER, THE BETTER is clearly true. Next, we need klan-k — 1, which (by Euler's theorem) is true if n — k is divisible by co (k). So, it would be enough to have co(k)lp — 1 and to be sure that p > k so that gcd(p, k) = 1. But from the previous problem there are infinitely many prime numbers p 1 (mod co (k)) and those numbers greater than k furnish infinitely many good numbers n. The following problem has become a classic, and variants of it appeared in mathematics competitions. It seems to be a favorite Olympiad problem, since it uses only elementary facts and the method is nothing less than beautiful. [Example 5.1 Find the least n such that 22005117n — 1. Solution. The problem actually asks for 022005(17). We know that , 022005 (17)1(p (220135) = 22004 so 022005(17) = 2k, for some k E {1, 2, ... , 2004}. The order of an element has done its job. Now, it is time to work with exponents. We have 220051172k — 1. Using the factorization 172k — 1 = (17 — 1)(17 + 1)(172 ± (1721,1 ± 1) we proceed by finding the exponent of 2 in each factor of this product. But this is not difficult, because for all i > 0 the number 1721 + 1 is a multiple of 2, but not a multiple of 4. Hence v2(172k — 1) = 4 + k and the order is found by solving the equation k + 4 = 2005. Thus 022005 (17) = 22001. Another simple, but not straightforward, application of the order of an element is the following divisibility problem. Here, we also need some properties of the prime numbers. Find all prime numbers p and q such that p2 + 112003g +1 and q2 + 112003P + 1. [Gabriel Dospinescu] THEORY AND EXAMPLES 319 Solution. Without loss of generality, we may assume that p < q. We discuss first the trivial case p = 2. In this case, 5120034 + 1 and it is easy to deduce that q is even, hence q = 2, which is a solution to the problem. Now, suppose that p > 2 and let r be a prime factor of p2 + 1. Because r1200324 — 1, it follows that or(2003)I2q. Suppose that gcd(q, or(2003)) = 1. Then or(2003)I2 and rI20032 — 1 = 23 . 3 . 7 11 . 13. 167. It seems that this is a dead end, since there are too many possible values for r. Another simple observation narrows the number of possible cases: because 7-1/92 +1, r must be of the form 4k +1 or equal to 2, and now we do not have many possibilities: r E {2,13}. The case r = 13 is also impossible, because 2003q + 1 2 (mod 13) and rI20034 + 1. So, we have found that for any prime factor r of p2 + 1, we have either r = 2 or qlor(2003), which in turn implies qlr — 1. Because p2 + 1 is even but not divisible by 4, and because any odd prime factor of it is congruent to 1 modulo q, we must have p2 + 1 —= 2 (mod q). This implies that ql(p — 1)(p + 1). Com-bining this with the assumption that p < q yields 4+1 and in fact q = p +1. It follows that p = 2, contradicting the assumption p > 2. Therefore the only solution is p = q = 2 . A bit more difficult is the following 2003 USA TST problem. Example 7. Find all ordered triples of primes (p, q, r) such that plqr + 1, qlrP + 1,rIpq + 1. [Reid Barton] USA TST 2003 Solution. It is quite clear that p, q, r are distinct. Indeed, if for example p = q, then the relation plqr + 1 is impossible. We will prove that we cannot have p, q, r > 2. Suppose this is the case. The first condition p qr + 1 implies p I qtr 1 and so op(q) 12r. If op(q) is odd, it follows that kr — 1, which combined with plqr + 1 yields p = 2, which is impossible. Thus, op(q) is either 2 or 2r. Could we have op(q) = 2r? No, since this would imply that 2rIp — 1 and so 0 pq + 1 (mod r) 2 (mod r), that is r = 2, false. Therefore, the only possibility is op(q) = 2 and so plq2 — 1. We cannot have plq — 1, because 320 14. THE SMALLER, THE BETTER plqr +1 and p 2. Thus, plq + 1 and in fact 2p1q + 1. In the same way, we find that 2q1r + 1 and 2r1p + 1. This is clearly impossible, just by looking at the greatest among p, q, r. So, our assumption is wrong, and one of the three primes must equal 2. Suppose without loss of generality that p = 2. Then q is odd, q1r2 + 1 and r12q + 1. Similarly, or(2)12q. If qlor(2), then qtr — 1 and so q1r2 + 1 — (r2 — 1) = 2, which contradicts the already established result that q is odd. Thus, or(2)12 and r13. As a matter of fact, this implies that r = 3 and q = 5, yielding the triple (2, 5, 3). It is immediate to verify that this triple satisfies all conditions of the problem. Moreover, all solutions are given by cyclic permutations of this triple. Can you find the least prime factor of the number 225 + 1? Yes, with a large amount of work, you will probably find it. But what about the number 12215 + 1? It has more than 30000 digits, so you will probably be bored before finding its least prime factor. But here is a beautiful and short solution, which does not need a single division. Example 8. Find the least prime factor of the number 12215 + 1. Solution. Let p be this prime number. Because p divides (12215 + 1) • (12215 — 1) = 12216 — 1, we find that op(12)1216. Exactly as in the solu- tion of the first example, we find that op(12) = 216 and so 2161p —1. Therefore p > 1 + 216. But it is well-known that 216 + 1 is a prime (and if you do not believe it, you can check it!). So, we might try to see if this number divides 12215 + 1. Let q = 216 + 1. Then 12215 + 1 = 2q-1 • 3Y + 1 = 3Y + 1 ( (mod q). It remains to see whether 3 = —1. But this is done in the chap-q ter Quadratic reciprocity and the answer is positive, so indeed 3 g, 2 +1 = 0 (mod q) and 216 + 1 is the least prime factor of the number 12215 + 1. OK, you must be already tired of this old fashioned idea that any prime factor of 22' + 1 is congruent to 1 modulo 271+1. Yet, you might find the energy to devote attention to the following interesting problems. THEORY AND EXAMPLES 321 Example 9.1 Prove that for any n > 2 the greatest prime factor of 22n + 1 is greater than or equal to n 2n+2 + 1. Chinese TST 2005 Solution. You will not imagine how simple this problem really is. If the start ± is right... Indeed, let us write 22n 1 = pk ii p22 where p1 < • • • < pr are prime numbers. We know that we can find odd positive integers qi such that pi = 1+2 11+1qi. Now, reduce the relation 22n +1 =pi 1p22 modulo 22n+2. It follows that 1 1 + 2n+1 E kiqi (mod 22Th+2) and so E kiqi > 2n+1. But i=i i=1 then gr E ki > 2n+1. Now everything becomes clear, since i=i + 1 > (1 + 2n+1)ki±k2+•••+k, > 2(n-1-1)(ki+k2+•••4-kr) 2n and so k1 + k2 + • • • + kr< n +1. Then (k. > 2(n + 1) and we are done. Example 10.1 It is not known whether there are infinitely many primes of the form 22n + 1. Yet, prove that the sum of the reciprocals of the proper divisors of 22n + 1, converges to 0. [Paul Erdos] AMM 4590 Solution. Note that the sum of the reciprocals of all the divisors of n is a(nn), where o-(n) is the sum of all the divisors of n. It suffices to prove (22n+i) that 0-2 converges to 1. Let pit • • • pr ic' be the prime factorization of 2' 22n + 1 and observe that 1 < a(22n+1) < 1 i Because 22n +1 ' < (1—M' • 11 (1—k . ) i=1 P% 22n + 1 > 2n(k1±•±k •) > 2", r = 0 (2 i) and so (1_1-m, converges to 1 ( + for n —> oo. From the above inequality, 0 222 converges to 1 and the con- 2' +1 1) con- clusion follows. 322 14. THE SMALLER, THE BETTER We have seen that the order of a modulo n is a divisor of co(n). Therefore a natural question appears: given a positive integer n, can we always find an integer a whose order modulo n is exactly ( 19(n)? We call such a number a a primitive root modulo n. The answer to this question turns out to be negative, but in some cases primitive roots exist. We will prove here that primitive roots mod pn exist whenever p > 2 is a prime number and n is a positive integer. The proof is quite long and complicated, but breaking it into smaller pieces will make it easier to understand. So, let us start with a lemma due to Gauss: Lemma 14.1. For each integer n > 1, E yo(d) = n. din Proof. One of the (many) proofs goes like this: imagine that you are trying to reduce the fractions -n 1 , , 1 77, in lowest terms. The denominator of any new fraction will be a divisor of n and it is clear that for any divisor d of n we obtain co (d) fractions with denominator d. By counting in two different ways the total number of fractions obtained, we can conclude. K Take now p > 2 a prime number and observe that any element of Z/pZ has an order which divides p — 1. Consider d a divisor of p — 1 and define f (d) to be the number of elements in Z/pZ that have order d. Suppose that x is an element of order d. Then 1, x, ... Xd-1 are distinct solutions of the equation ud = 1, an equation which has at most d solutions in the field Z/pZ. Therefore 1, x, ..., Xd-1 are all solutions of this equation and any element of order d is among these elements. Clearly, x2 has order d if and only if gcd(i, d) = 1. Thus at most (p(d) elements have order d, which means that f(d) < (p(d) for all d. But since any nonzero elements has an order which divides p — 1, we deduce that f (d) = p — 1 = > yo(d) dip-1 dip-1 (we used in the last equality the lemma above). This identity combined with the previous inequality shows that f (d) = co(d) for all dlp — 1. We have thus proved the following: THEORY AND EXAMPLES 323 Theorem 14.2. For any divisor d of p -1 there are exactly c,o(d) elements of order d in Z/pZ. The above theorem implies the existence of primitive roots modulo any prime p (the case p = 2 being obvious). If g is a primitive root mod p, then the p elements 0, 1, g, g2, gP-2 are distinct and so they represent a permutation of Z/pZ. Let us fix now a prime number p > 2 and a positive integer k and show the existence of a primitive root mod pk. First of all, let us observe that for any j > 2 and any integer x we have (1 + xp)P3 2 -= 1 + xpi-1 (mod pj). Establishing this property is immediate by induction on j and the binomial formula. With this preparatory result, we will prove now the following: Theorem 14.3. If p is an odd prime, then for any positive integer k there exists a primitive root mod pk . Proof. Indeed, take g a primitive root mod p. Clearly, g +p is also a primitive root mod p. Using again the binomial formula, it is easy to prove that one of the two elements g and g + p is not a root of XP-1 - 1 mod p2. This shows that there exists y a primitive root mod p for which yP-1 # 1 (mod p 2). Let yP-1 = 1 + xp. Then by using the previous observation we can write yp ( p_i) (1 xp)pk-2 1 + Xpk-1 (mod pk ) and so pk does not divide pk2(p -1) 1. Thus the order of y mod pk is a multiple of p - 1 (because y is a primitive root mod p) which divides pk-1(p - 1) but does not divide pk-2(p 1). So, y is a primitive root mod pk. In order to finish this (long) theoretical part, let us present a very efficient criterion for primitive roots modulo pk: Theorem 14.4. Each primitive root mod p and p2 is a primitive root modulo any power of p. Proof. Let us prove first that if g is a primitive root mod p and p2 then it is also a primitive root mod p3. Let k be the order of g mod p3. Then k is a divisor of p2(p-1). Because p2 divides gk -1, k must be a multiple of p(p-1). It remains 324 14. THE SMALLER, THE BETTER to prove that k is not p(p — 1). Supposing the contrary, let gp-i = 1 +rp, then we know that p3 (1 + rp)P 1. Using again the binomial formula, we deduce that p divides r and so p2 divides gp-1 - 1, which contradicts the fact that g is a primitive root mod p2. Now, we use induction. Suppose that n > 4 and that g is a primitive root mod pn-1. Let k be the order of g mod IP. Because pn-1 divides gk —1, k must be a multiple of pn-2(p — 1). Also, k is a divisor of pri-1(p —1) = cp(pn). So, all we have to do is to prove that k is not pn-2 (p — 1). Otherwise, by Euler's theorem we can write gP'-3(P-1) = 1 + rpn-2 and from the binomial formula it follows that r is a multiple of p and so pn-1 divides g/P-3(P-1) —1, contradicting the fact that g has order pn-2 (ft 1) modulo pn-1. The theorem is thus proved. 1=1 It is important to note that the previous results allow us to find all positive integers that have primitive roots. First off all, observe that such a number n cannot be written in the form n = nln2 with gcd(ni, n2) = 1 and ni, n2 > 2. ( ) Indeed, if gcd(g, n) = 1 then g yo 2 n) (gca(ni)) 2= 1 (mod ni) and similarly g 2 1 (mod n2). Thus n divides g ( 2 n) — 1 and g cannot have order co(n). Also the fact that g2k-2 1 (mod 2k) for any odd integer g and any k > 3 (whose the proof is immediate by induction) shows that there are no primitive roots mod 2k, for k > 3. This shows that the only candidates are 2,4, pk and 2pk for an odd prime number p. And these numbers have primitive roots. For 2 and 4 it is obvious, while for powers of odd primes it has been proved above. For 2pk observe that w(2pk) = co(pk), so the odd number among g, g pk (where g is a primitive root mod pk) is a primitive root mod 2pk. Now, let us solve some problems. However, make sure you correctly remember Fermat's little theorem before attempting to solve the following problem. Example 11.1 Find all positive integers n such that nlan+1 — a for all a E Z. Solution. Consider such an integer n > 1 and observe first that is must be squarefree. Indeed, if p is a prime divisor of n, just choose a = p. Next, write n = p1p2...pk for some pairwise distinct prime numbers p1, p2, ...,pk. Fix some THEORY AND EXAMPLES 325 1 < i < k and choose a a primitive root modulo pi. Then clearly the condition nlart+1 — a implies that n is a multiple of pi — 1. Now, it is very easy to determine all such numbers n. Assume that /31 < p2 < • • • < pk and observe that /31 = 2 (because /3 1 — 1 divides n), then /32 — 112 (the same argument), thus p2 = 3. Continue in this manner to obtain p3 = 7,p4 = 43. And things change after this, because we would find that p5 —1 divides 1806 and it is easy to see that this is not possible, because the only divisors d of 1806 such that d +1 is a prime are 1, 2, 6, 42, which is not a prime number. Therefore k < 4 and such numbers are 1, 2, 6, 42,1806. A very beautiful and difficult problem comes now. We will see that using the previous results on primitive roots we can obtain a quick and elegant solution. Example 12. Find all positive integers n such that n2 2n + 1. [Laurentiu Panaitopol] IMO 1990 Solution. It is clear that any solution must be odd and that 1 and 3 are solutions, so assume that n > 5. Because 2 is a primitive root mod S and mod 9 (as you can immediately check), it follows from the above results that 2 is a primitive root mod 3k for all k. In particular, if 3k inn z + 1 then 3k122n — 1 and because the order of 2 mod 3k is 2 3k-1, we deduce that 3k-11n. This shows that v3(2n + 1) < v3(n) + 1 for all n. In particular, for any solution n of the problem we have 2v3(n) = v3(n2) < v3(2n + 1) < 1 + v3(n), so v3(n) < 1. Let us prove that we actually have v3(n) = 1, if n > 1. Let p be the smallest prime divisor of 71. Then p122n — 1, so o2(2)12n and op(2)1p — 1. By the definition of p we have gcd(2n,p — 1) = 2, so p13 and thus p = 3 and 31n. This shows that we can write n = 3a where gcd(3, a) = 1. Now, we would like to prove that a = 1 (therefore, the only solution of the problem which is greater than 1 is n = 3). Assuming the contrary, let q be its smallest prime divisor. Then q12n + 1 and q126a — 1. As above, we deduce that oq(2) is a divisor of 6a and q-1, and because gcd(a, q-1) = 1, it follows that oq(2)16 and so q163. Because gcd(a, 3) = 1, the only possibility is q = 7. But then 712a +1 = 8a + 1, which is clearly impossible. This shows that a = 1 and n = 3, a contradiction with n > 5. Hence 1 and 3 are the only solutions of the problem. 326 14. THE SMALLER, THE BETTER Finally, a chestnut from the celebrated contest Miklos Schweitzer, which uses the previous theoretical results as well as a large dose of creativity. [Example 13.1 Let p = 3 (mod 4) be a prime number. Prove that p+, H (x2 + y2) =- (-1)[ s J 1<x<y< (mod p). P.Suranyil Miklos Schweitzer Competition Solution. Let p = 4k + 3 and take g to be a primitive root modulo p and x = g2. Then the squares of the residues mod p are exactly 1, x, x2, x2k, so the product we need to evaluate is H (xi + x3) (mod p). Therefore, if o<s<a<zk P is the desired product, we have P • IT (xi — ) -=- H (x2i — x2i) (mod p). 0<i<j<2k 0<i<j<2k Observe that each of the two products is actually a Vandermonde determinant and because x2k+1 = 1, the generators of the second determinant are exactly x2k x, x 3 , x2k-1 1, x2, ..., . Hence the second determinant is obtained from the first one by k — 1) + • • • + 2 + 1 transpositions of the lines and so k(k+1) k(k+1) [p+1] P (-1) 2 (mod p). An easy case examination shows that 2 8 is even and the conclusion follows. PROBLEMS FOR TRAINING 327 14.2 Problems for training 1. Let a, n > 2 be integers such that nlan-1 — 1 and n does not divide any of the numbers ax — 1, where x < n — 1 and x — 1. Prove that n is a prime number. 2. Let p be a nonzero polynomial with integral coefficients. Prove that there are at most finitely many numbers n for which p(n) and 22n + 1 are not relatively prime. P + 1 i 2 3. Let p > 3 be a prime. Prove that any positive divisor of is of the 3 form 2kp + 1. Fermat 4. Find all positive integers m, n for which nim2•3n m 3n + 1. Bulgaria 1997 5. Find the least multiple of 19 all of whose digits are 1. Gazeta Matematica 6. Let q be a prime such that q2 divides at least one Mersenne number 2P — 1 with p a prime number. Prove that q > 3 • 109. You may take it for granted that the only primes q such that q2124-1 — 1 and which are smaller than 3 • 109 are 1093 and 3511. 7. Prove that there exists a function f with integer values such that 2n 119f (n) — 97 for any positive integer n. Vietnamese TST 1997 328 14. THE SMALLER, THE BETTER 8. Let p be a prime and let q > 5 be a prime factor of 2P + 3P. Prove that q > p. Laurentiu Panaitopol, Romanian TST 9. Prove that 3 is a primitive root mod p for any prime p of the form 2' + 1. 10. Let p be a prime number and let d be a positive divisor of p — 1. Prove that there is a positive integer n such that op(n) = d. 11. Prove that for any prime number p > 3 we have (2 ;) = 2 (mod p3). 12. Let m > 1 be an odd number. Find the least n such that 219891m" — 1. IMO 1989 Shortlist 13. Let m, n be two positive integers. Prove that the remainders of the numbers ln, 2n, ..., mn modulo m are pairwise distinct if and only if m is square-free and n is relatively prime to co(m). 14. Find all positive integers n with the property that for any positive inte-gers a, b such that nla2b + 1 we also have nla2 + b. Bulgaria 15. Let a be an integer greater than 1. Prove that the function p — 1 f : {2,3,5, 7, 11,... } —>N, f(p) = op (a) is unbounded. Jon Froemke, Jerrold W Grossman, AMM E 3216 PROBLEMS FOR TRAINING 329 16. Let f (n) be the greatest common divisor of the numbers 2n — 2, 3n — 3, 4n — 4, .... Determine f (n) and prove that f (2n) = 2. AMM 17. Let f be a polynomial with integer coefficients such that for some prime number p we have f (i) = 0 (mod p) or f (i) = 1 (mod p) for any integer i. If f (0) = 0 and f (1) = 1, prove that deg( f) > p — 1. IMO 1997 Shortlist 18. Let f be a polynomial with integer coefficients, having degree p — 1, where p > 2 is a prime number. Suppose that for all integers a, b, if a — b is not a multiple of p, then f (a) — f (b) has the same property. Prove that p divides the leading coefficient of f . 19. A Carmichael number n satisfies TO' — a for all integers a. Find all Carmichael numbers of the form 3pq with p, q prime numbers. Romanian TST 1996 20. Using the existence of Carmichael composite numbers, prove that there are infinitely many pseudo-primes, that is composite numbers n such that ni2n — 2. 21. Find all prime numbers p, q such that pql2P + 2q. 22. Find the sum of the m-th powers of the primitive roots mod p for a given prime p and a positive integer m. 330 14. THE SMALLER, THE BETTER 23. Let N > 2 be an integer and suppose that N —1 = RF, where F = Ca'? (q, being distinct primes). Suppose that (R,F) = 1 and R < F. If there exists a positive integer a such that aN-1 1 (mod N) N-1 and a 4i — 1 is relatively prime to N for all i, then N is prime. Proth, Pocklington, Lehmer Test 24. Let p be a prime number and m, n be integers greater than 1 such that inp(n-1) 1. Prove that gcd(mn-1 1, n) > 1. MOSP 2001 25. Let n be a positive integer, and let An be the the set of all a such that ni(an H- 1), 1 < a < n and a E Z. a) Find all n such that An 0. b) Find all n such that An1 is even and non-zero. c) Is there n such that lAnI = 130? Italian TST 2006 26. Let n be an odd integer and let C(n) be the number of cycles of the permutation f of {0,1, ..., n —1} sending i to 2i (mod n) for all i. Prove that C(3(2n — 1)) = C(5(2n — 1)) for all odd positive integers rt. James Propp, Mathematics Magazine 27. Let A be a finite set of prime numbers and let a be an integer greater than 1. Prove that there are only finitely many positive integers n such that all prime factors of an — 1 are in A. Iran Olympiad PROBLEMS FOR TRAINING 331 28. Prove that for any prime p there is a prime q that does not divide any of the numbers nP — p, with n > 1. IMO 2003 29. Let a be an integer greater than 1. Prove that for infinitely many n the greatest prime factor of an — 1 is greater than n loge n. Gabriel Dospinescu 30. Let p be an odd prime. Prove the existence of a positive integer k < p-1 which is a primitive root mod p and which is also relatively prime to p-1. Richard Stanley, AMM E 2488 31. Let E > 0. Prove the existence of a constant c such that for all odd primes p there exists a primitive root mod p smaller than cpI±E Vinogradov THEORY AND EXAMPLES 335 15.1 Theory and examples Recall that the sequence ({nal),>1 is dense in [0,1] if a is an irrational number, a classical theorem of Kronecker. Various applications of this nice result have appeared in different contests and will probably make the object of many more Olympiad problems. Yet, there are some examples in which this result is inef-ficient. A simple one is as follows: using Kronecker's theorem, one can prove that for any positive integer a that is not a power of 10 there exists a positive integer n such that an begins with 2008. The natural question – what fraction of numbers between 1 and n have this property (speaking here about large values of 7/) – is much more difficult, and to answer it we need some stronger tools. This is the reason we now discuss some classical approximation theo-rems, particularly the very effective Weil criterion and its consequences. The proofs of these results are nontrivial and require some heavy duty analysis. Yet, the consequences that will be discussed here are almost elementary. Of course, one cannot start a topic about approximation theorems without talk-ing first about Kronecker's theorem. We skip the proof, not only because it is very well-known, but because we will prove a much stronger result about the sequence ({nal),>1. Instead, we will discuss two beautiful problems, corollar-ies of this theorem. [Example 1.1 Prove that the sequence ( [nV2003] )72>1 contains arbitrarily long geometric progressions with arbitrarily large ratio. [Radu Gologan] Romanian TST 2003 Solution. Let p be any positive integer. We will prove that there are arbitrar-ily long geometric sequences with ratio p. Given n > 3, we will find a positive integer m such that [pkm,/2003] = pk [m\/2003] for all 1 < k < n. If the ex-istence of such a number is proved, then the conclusion is immediate. Observe that [pk m V20031 = pk [mV2003] is equivalent to ]pk { mV2003}] = 0, or to {m-V2003} < — 1 . The existence of a positive integer m with the last property Pn 336 15. DENSITY AND REGULAR DISTRIBUTION is ensured by Kronecker's theorem. Here is a problem that is apparently very difficult, but which is again a simple consequence of Kronecker's theorem. Example 2. Consider a positive integer k and a real number a such that log a is irrational. For each n > 1 let xn be the number formed by the first k digits of [an]. Prove that the sequence (xn)n>i is not eventually periodical. [Gabriel Dospinescu] Mathlinks Contest and, since p = 1 + [log m J , the claim is proved. Now consider the claim false: thus there is some T for which xn+T = xn for any large enough n. Another observation is the following: there is a positive integer r such that xr3-, > 10k-1. Indeed, assuming the contrary, we find that for all r > 0 we have xri , = 10k-1. Using the first observation, it follows that k -1+ {log [arT j} < log(1 + 10k-1) for all r. Thus log (1 + 101-1 > log [(en - [log Lain > log(arT - 1) - [log arTi arT = {rT log a} - log arT i• It suffices now to consider a sequence of positive integers (rn) such that 1- -1 < {rnT log a} (the existence is a direct consequence of Kronecker's theorem) and we deduce that 1 -i ) n 1 ar T log 1 + 10k + + log 1 > 1 for all n. aT Solution. First of all, the number formed with the first k digits of a number m is [10k-1±{l'gin}i The proof of this claim is not difficult. Indeed, let us write m = ala2 . ap, with p > k. Then m = al ak • 10" ± ak±i ap, hence al ak •10P k < Tit < (al ... ak +1) • 10P-k. It follows that al ak = [ m 10P-k THEORY AND EXAMPLES 337 The last inequality is clearly impossible. Finally, assume the existence of such an r. It follows that for n > r we have XnT = XrT thus {log LanTi > log (1 + 101, 1_1) This shows that log (1 + 101-1 ) 5_ log [an"' ] — [log LanTi < nT log a — [log anTi = {nT log a} for all n > r. In the last inequality, we used the fact that [log [x] j = [log x j, which is not difficult to establish: indeed, if [log x ] = k, then 10k < x < 10k+1, and thus 10k < [xi < 10k+1, which means that [log [x]] =- k. Finally, note that the relation log (1 + 10k_1)< {nT log a} contradicts Kronecker's theorem. This finishes the proof. We continue with two subtle results, based on Kronecker's lemma. Example 3.] For a pair (a, b) of real numbers let F(a, b) denote the sequence of general term cn = [an + b]. Find all pairs (a, b) such that F (x , y) = F(a, b) implies (x, y) = (a, b). [Roy Streit] AMM E 2726 Solution. Let us see what happens when F (x , y) = F(a, b). We must have [an + bJ = [xn y_1 for all positive integers n. Dividing this equality by n and taking the limit, we infer that a = x. Now, if a is rational, the sequence of fractional parts of an + b takes only a finite number of values, so if r is chosen sufficiently small (but positive) we will have F(a, b r) = F(a, b), so no pair (a, b) can be a solution of the problem. On the other hand, we claim that any irrational number a is a solution for any real number b. Indeed, take xi < x2 and a positive integer n such that na + xi < m < na + 338 15. DENSITY AND REGULAR DISTRIBUTION for a certain integer rn. The existence of such an n follows immediately from Kronecker's theorem. But the last inequality shows that F(a, xi) F(a, x2) and so a is a solution. Therefore the answer is: all pairs (a, b) with a irrational. Finally, an equivalent condition for the irrationality of a real number: Example 4.1 Let r be a real number in (0,1) and let S(r) be the set of positive integers n for which the interval (nr, nr + r) contains exactly one integer. Prove that r is irrational if and only if for all integers M there exists a complete system of residues modulo M, contained in S(r). [Klark Kimberling] Solution. One part of the solution is very easy: if r is rational, let M be its denominator. Then clearly if 71 is a multiple of M there is no integer k in the desired interval. Now, suppose that r is irrational and take integers m, M such that 0 < m < M. By Kronecker's theorem, the integer multiples of : 7 : form a dense set modulo M. So, there exists an integer k such that the image of Tis in (m, m+1), that is for a certain integer s we have sM+m < T < sM+m+1. It is then clear that if we take n = sM + m we have 71 m (mod M) and nr < k < nr + r. This finishes the solution. Before getting into the quantitative results stated at the beginning of this chapter, we must talk about a surprising result, which turns out to be very useful when dealing with real numbers and their properties. Sometimes, it will help us reduce a complicated problem concerning real numbers to integers, as we will see in one of the examples. But first, let us state and prove this result. Example el Let xi, x2, . , xk be real numbers and let E > 0. There exists a positive integer n and integers pi, p2, , pk such that Inxi — pi < E for all i. [Dirichlet] THEORY AND EXAMPLES 339 Solution. We need to prove that if we have a finite set of real numbers, we can multiply all its elements by a suitable integer such that the elements of the new set are as close to integers as we want. Let us choose an integer N > — 1 and partition the interval [0, 1) into N inter-vals, [0, 1) = U Js, J, = N N ) s=1 Now, choose n = Nk +1 and assign to each q in the set {1, 2, ... , n} a sequence of k positive integers al, a2, , ak, where a2 = s if and only if fqxil E Js. We obtain at most Nk sequences corresponding to these numbers, and so by the pigeonhole principle we can find 1 < u < v < n such that the same sequence is assigned to u and v. This means that for all 1 < i < k we have Taxi} — {V Xi}l < 1 < E (15.1) It suffices to pick n = v — u, pi = [vxii — [uxi j. And here is how we can use this result in problems where it is more com-fortable to work with integers. But don't kid yourself, there are not many such problems. The one we are going to discuss next has meandered between world's Olympiads: proposed at the 1949 Moscow Olympiad, it appeared next at the W.L. Putnam Competition in 1973 and later on in an IMO Shortlist, proposed by Mongolia. Example 6. Let xi, x2, , x2n+1 be real numbers with the property: for any 1 < i < 2n + 1 one can make two groups of 71 numbers by using all the x3, j i, such that the sum of the numbers in each group is the same. Prove that all the numbers must be equal. 340 15. DENSITY AND REGULAR DISTRIBUTION Solution. For integers the solution is well-known and not difficult: it suffices to note that in this case all numbers xi have the same parity, and the use of infinite descent solves the problem (either they are all even and in this case we divide each number by 2 and obtain a new set with smaller sum of magnitudes and the same properties; otherwise, we subtract 1 from each number and then divide by 2). Now, assume that they are real numbers, which is definitely a more subtle case. First of all, if they are all rational, it suffices to multiply by their common denominator and apply the first case. Suppose at least one of the numbers is irrational. Consider E > 0, a positive integer m, and some integers p1,p2,...,p2,41 such that Imxi – pil < E for all i. We claim that if E > 0 is small enough, then Pi, P2, • • • P2n+1 have the same property as xl, X2, . X2n+1. Indeed, take some i and write the given condition as E aijmxi = 0 or — pi) = —EaiiPi i0i i0i (where au E {-1, 1}). Then a z3 – Pi) < 2ne. 1 Thus if we choose E < 2rn , then aijpj = 0 and so P1, P2, • • ,P2n-F1 have the same property. Because they are all integers, pi, p2, ...,p2n+i must be all equal (again, because of the first case). Hence we have proved that for any 1 N > 2m there are integers nN,pN such that InNxi – PNI < N Because at least one of the numbers x1, x2, • • • , X2n+1 is irrational, it is not difficult to prove that the sequence (nN)N>2m is unbounded. But 2 — InNImaxIxi xi', hence maxi j I ixi xi' = 0 and the problem is solved. If you thought the last problem was too classical, here is another one, a little bit less known, but with the same flavor: THEORY AND EXAMPLES 341 r Example 7.1 Let ai, a2, •••, a2007 be real numbers with the following prop-erty: no matter how we choose 13 numbers among them, there exist 8 numbers among the 2007 which have the same arith-metic mean as the 13 chosen ones. Prove that they are all equal. Solution. Note (again) that the problem is quite easy for integers. Indeed, the assumption implies that the sum of any 13 numbers is a multiple of 13. Let ai , ai be among the 2007 numbers and let x1, x2, ..., x12 be some ak with k i and k j. Then ai 4- xi + x2 + • • • + x12 and ai + xi + x2 + • • • + x12 are multiples of 13, so ai ai (mod 13). Thus all numbers give the same remainder r modulo 13. It suffices to subtract r from all ai , to divide by 13 in order to obtain a new collection of 2007 integers, with smaller absolute values and still satisfying the property given in the statement of the problem. Repeating this procedure, we will finally obtain a collection of zeros, which means that the initial numbers were all equal. Now, let us pass to the case when all numbers are known only to be real. The idea is the same as in the previous example: we will approximate, using Dirichlet's theorem, all numbers by rational numbers with a common denom-inator. Explicitly, take some c > 0 and n and pi some integers (with n > 0) such that Inai < c for all i. Take some indices i1, i2, •••, ii3. We know that for some indices j1, j2, j8 we have nail + nai, + • • • + nai„ nail + na32 + • • naj, 13 8 • If xi = nai — pi, it follows that + Piz ' • • + P113 Pii + Pj2 + • • - pis 13 8 because I xi < E. Now, observe that if P11 + pi2 + • • • + Pio 1 3 1+ Pj2 + • + Pia 13 8 is nonzero, it is at least equal to 813. Thus, if we take c < 161 43, we know that the corresponding pi have the same property as ai. By the first case, we must < 2c, 342 15. DENSITY AND REGULAR DISTRIBUTION therefore have p1 = P2 = • • = P2007. Thus 2e > — a31 for all i, j and all E < 161 13 Clearly, this implies al = a2 = • • • = a2007 and finishes the proof. . • Now, let us turn to more quantitative results about the set of fractional parts of natural multiples of different real numbers. The following criterion, due to Weyl, deserves to be discussed because of its beauty and apparent simplicity. Theorem 15.1 (Weyl's theorem). Let (an)n>1 be a sequence of real numbers from the interval /0,1]. Then the following statements are equivalent: a) For any real numbers 0 < a < b < 1, 1 < n, E [a, 1)] } lim b a; n—>oo b) For any continuous function f : [0, 1] IR, n 1 n—>co nE Pak) = f f (x)dx; k=1 c) For any positive integer r > 1, n lim -I- diirrak = O. n—>oo n k=1 In this case we will say that the sequence is equidistributed. Proof. We will present just a sketch of the solution, but containing all the necessary ingredients. First, we observe that a) says precisely that b) is true for the characteristic function of any subinterval of [0,1]. By linearity, this remains true for any piecewise constant function. Now, there is a well-known and easy to verify property of continuous functions: they can be uniformly approximated with piecewise constant functions. That is, given E > 0, we can find a piecewise constant function g such that Ig(x) — f (x)1 < e for all x E [0, 1]. But then if we write f (ak) — f f (x)dx 1 < — n (ak) — g(ak)i+ f (x) — g(x)1dx k=1 THEORY AND EXAMPLES 343 1 1 -n Eg(ak)— f g(x)dx 0 k=1 and apply the result in b) for the function g, we easily deduce that b) is true for any continuous function. The fact that b) implies c) is immediate. More subtle is that b) implies a). Let us consider the subinterval I = [a, b] with 0 < a < b < 1. Next, consider two sequences of continuous functions fk,gk such that fk is zero on [0, a], [b,1] and 1 on [ — k 1 a + ' b — — k (being affine 1 otherwise), while gk has "the same" properties but is greater than or equal to A/ (the characteristic function of I = [a, b]). Therefore Ifil 1 n ai E [a, b]}1 E fk( ) < _ 9k (a3). 3=1 But from the hypothesis, fk(a3) —> f fk(x)dx = b — a — o and 1 n -1 E gk(a3) —> f gk(x)dx = b — a+ — 1. 0 3= Now, let us take E > 0 and k sufficiently large. The above inequalities show that actually for all sufficiently large positive integers n IN 1 < i < rt, az E ia, b+ a < 2E and the conclusion follows. You have already seen how to adapt this proof for the case a = 0 or b = 1. Finally, let us prove that c) implies b). Of course, a linearity argument allows us to assume that b) is true for any trigonometric polynomial. Because any continuous function f : [0,1] R satisfying f(0) = f(1) can be uniformly approximated by trigonometric polynomials (this is a really nontrivial result due to Weierstrass), we deduce that b) is true for 1 1 k 344 15. DENSITY AND REGULAR DISTRIBUTION continuous functions f for which f (0) = f (1). Now, given a continuous f : [0, 1] —> R , it is immediate that for any E > 0 we can find two continuous functions g, h, both having equal values at 0 and 1 and such that (x) — g(x)I < h(x) and f h(x)dx E. 0 Using the same arguments as those used to prove that b) implies a), one can easily see that b) is true for any continuous function. 111 Before presenting the next problem, we need another definition: we say that the sequence (an)n>i is uniformly distributed mod 1 if the sequence of frac-tional parts of an is equidistributed. We invite the reader to find an elementary proof for the following problem in order to appreciate the power of Weyl's cri-terion. So, here is the classical example. Let a be an irrational number. Then the sequence (na)n>i is uniformly distributed mod 1. Solution. Well, after so much work, you deserve a reward: this is a simple consequence of Weyl's criterion. Indeed, it suffices to prove that c) is true, which reduces to proving that for all integers p > 1. But this is just a geometric series!!! A one-line compu-tation shows that (15.1) is satisfied and thus we obtain the desired result. It is probably time to solve the problem mentioned at the very beginning of this note: how to compute the density of those numbers n for which 2' begins with (for example) 2006. Well, again a reward: this is going to be equally easy (of course, you need some rest before looking at some deeper results). THEORY AND EXAMPLES 345 Example 9. What is the density of the set of positive integers n for which 2n begins with 2006? Solution. 2n begins with 2006 if and only if there is a p > 1 and some digits al, a2, , ap E {0, 1, , 9} such that 2' = 2006a1a2... ap, which is clearly equivalent to the existence of p > 1 such that 2007 • 10P > > 2006 • 10P. This can be rewritten as log 2007 + p > n log 2 > log 2006 + p, 2007 2006 implying [n log 2] = p + 3. Hence log 1000 > In log 21 > log 1000 and the 1 density of our set is the density of the set t t positive integers n satisfying log 2007 > {n log 2} > log 2006. 1000 1000 2007 From Example 8, the last set has density log 2006 and this is the answer to our problem. We saw a beautiful proof of the fact that if a is irrational, then (na)n>i is uniformly distributed mod 1. Actually, much more is true, but this is also much more difficult to prove. The following two examples are important theorems. The first is due to Van der Corput and shows how a brilliant combination of algebraic manipulations and Weyl's criterion can yield difficult and important results. [-Example 10. Let (xn) be a sequence of real numbers such that the se-quences (xn+p — xn),>i are equidistributed for all p > 1. Then (xn) is also equidistributed. [Van der Corput] 346 15. DENSITY AND REGULAR DISTRIBUTION Solution. This is not an Olympiad problem!!! But mathematics is not just about Olympiads and from time to time (in fact, from a certain time on) one should try to discover what is behind such great results. This is the reason we present a proof of this theorem based on a technical lemma of Van der Corput, which turned out to be fundamental in studying exponential sums. Lemma 15.2 (Van der Corput). For any complex numbers zi, z2, , zn and any h E {1, 2, ... , n}, the following inequality is true (with the convention that zi = 0 for any integer i not in {1,2, ...,n}): 2 h-1 (n-r < (n h — 1) [2E(h — r)Re E ZiZi+r) r=1 i=1 h2 h n i=1 Proof. The simple observation that n+h-1 h-1 h E zi i=i i=1 j=0 allows us to write (via Cauchy Schwarz's inequality): 2 2 n+h-1 < (n h — 1) E i=i h2 E zi i=i 2 And next? Well, we expand h-1 E zi_, i=o and see that it is nothing other than h-1 2 E (h — r)Re ZiZi+r h r=1 (n-r i=1 i=1 zi12 . We will now prove Van der Corput's theorem, by using this lemma and Weyl's criterion. (n—i j=1 j z+3 THEORY AND EXAMPLES 347 Of course, the idea is to show that Ern — E e2i7rpxk = 0 n—>oo n k=1 for all p > 1. Fix such a p and take for the moment a positive real number h and E E (0,1) (h may depend on E). Setting z3 = e 2i"x3, we have 2i7rp(xi —xi+j) < En. 2 1 n + h — 1 [ h-1 n2 h2 hn + 2 (h — i)Re i=1 -E zj j=1 Now, observe that Re (n—i zj • Zi+j = Re (n—i e2i7rp(xi —xj+i) j=1 j=1 e2i7rp(xi —xi+j) Using Weyl's criterion for the sequences (xn+, — xn)n>i for i = 1,2, ... , h — 1, we deduce that for all sufficiently large n we have n—i i= Therefore 2 1 n + h — 1 [ hn + 2En n2 h2 h-1 i=1 n + h — 1 (1 + E) < 2(1 + E) < < 6'2 nh h 2(1 + 6) for n large enough. Now, by choosing h > we deduce that for all E2 , sufficiently large n we have < E. Hence Weyl's criterion is satisfied and thus (xn)n>1 is equidistributed. K 348 15. DENSITY AND REGULAR DISTRIBUTION This was surely the most difficult result of this chapter, but why not take one more step once we are already here? Let us prove the following weaker (but as the reader will probably agree, absolutely nontrivial) version of a famous theorem of Weyl. It is related to the equidistribution of the sequence (f (n)),,>1 where f is a real polynomial having at least one irrational coefficient other than the constant term. We will not prove this here, but focus on the following result. If f is a polynomial with real coefficients and irrational lead- ing coefficient, then the sequence (f (n)),,>1 is equidistributed. [Weyl] Solution. You have probably noticed that this is an immediate consequence of Van der Corput's theorem (but just imagine the amount of work done to arrive at this conclusion!!!): the proof by induction is immediate. Indeed, if f has degree 1, then the conclusion is clear (see example 5). Now, if the result holds for polynomials of degree at most k, it suffices (by Van der Corput's theorem) to prove that for all positive integers p, the sequence ( f (n + p) — f (n)),>1 is equidistributed. But this is exactly the induction hypothesis applied to the polynomial f (X + p) — f (X) (whose leading coefficient is clearly irrational). The proof by induction finishes here. The solution of the following problem, which is a consequence of Weyl's crite-rion, is due to Marian Tetiva: Example 12 -1 .1 If a is an irrational number and P is a nonconstant polyno-mial with integer coefficients, then there are infinitely many pairs (m, n) of integers such that P(m) = Lnai [H. A. ShahAli] Solution. Of course, we can assume that a > 0. If a < 1, no heavy ma- chinery is required: all we need is to note that for all integers in the interval THEORY AND EXAMPLES 349 ( P(m) P(m)+1) has length greater than 1, thus it contains an integer nm. a a It is clear that Lnm • a] = P(m), so we have infinitely many solutions (at least one for each m). The difficult part is when a > 1. Let us consider = as 1 . By a well-known result of Beaty, the sets A = { Lnai n > 1} and B = {Ln0_1 n > 1} give a partition of the set of positive integers. A second of observation shows that it is enough to prove the statement for polynomials P whose leading coefficient is positive. Thus starting from a certain point mo, P(m) is a positive integer, thus belonging to A or to B. Suppose that the equation P(m) = Lna I has finitely many solutions, that is for all sufficiently large m, P(m) E B. Hence for some N we have the existence of a sequence of positive integers (nm)m>.N such that P(m) = Ln,13_1. This clearly implies [PQ )] = nm 0 1, that is the fractional part of P(m) is in (1 — 1' 1) for all 0 sufficiently large m. Or, hP clearly satisfies the conditions of Weyl's criterion, so the sequence of fractional parts of P( i3 m) is dense in [0,1], which is impos-sible, because all but finitely many terms are in (1 — 16, 1). This finishes the proof of the case a > 1 and ends the solution. 350 15. DENSITY AND REGULAR DISTRIBUTION 15.2 Problems for training ( 1. Evaluate sup min I p — q01 . n>1 MEN p+q=n Putnam Competition 2. Find all integers a with the property that for infinitely many positive integers n, 2n2 [ [n • = [7/A/1 + a. Radu Gologan 3. Prove that by using different terms of the sequence [n2 x/2006 J one can construct geometric sequences of any length. 4. Let x be an irrational number and let f (t) = mina {1—t}). Prove that given any e > 0 one can find a positive integer n such that f (n2 x) < E. Iran 2004 5. Suppose that A = {ni, n2, ... } is a set of positive integers such that the sequence (cos nk)k>i is convergent. Prove that A has zero density. Marian Tetiva 6. Prove that for every k one can find distinct positive integers ni, n2, . • • , nk such that [nif21 , Ln2f21 , • • . , Lnk4 and Lni01 , [7/20] , • • • , [nk are both geometrical sequences. After a Romanian TST problem PROBLEMS FOR TRAINING 351 7. Does the sequence sin(n2) + sin(n3) converge? 8. A flea moves in the positive direction of an axis, starting from the origin. It can only jump over distances equal to and V2005. Prove that there exists an no such that the flea will be able to arrive in any interval [n, n + 1] for each n > no. Romanian Contest, 2005 9. Let z1, z2, , zn be arbitrary complex numbers. Prove that for any E > 0 there are infinitely many positive integers n such that + V Izi ± • • + 4,1 > rflaXILZ1111Z21) • • I Znif 10. Prove that the sequence consisting of the first digit of 2Th + 3' is not periodical. Tuymaada Olympiad 11. Suppose that f is a real, continuous, and periodical function such that ( E if ( kk)I) k=1 n>1 positive integers k. Give a necessary and sufficient condition ensuring n I( the existence of a constant c> 0 such that E fk) > clnn for all n. k=1 Gabriel Dospinescu 12. Let f be a polynomial with integral coefficients and let a be an ir-rational number. Can all numbers f(k), k = 1,2, ... be in the set A = { Lnaj I n > 11? Is it true that any set of positive integers with positive density contains an infinite arithmetical sequence? the sequence is bounded. Prove that f(k) = 0 for all 352 15. DENSITY AND REGULAR DISTRIBUTION 13. Let a, b be positive real numbers such that {na} + {rib} < 1 for all n. Prove that at least one of them is an integer. 14. Let a, b, c be positive real numbers. Prove that the sets A = {[nai I n > 1}, B -= {[nb_I I n > 11, C = {[nc] I n > 1} cannot form a partition of the set of positive integers. Putnam Competition 15. Let x > 1 be a real number and an = Lxn]. Can the number S = 0.ala2a3... be rational? The expansion is formed by writing down the decimal digits of ai, a2, ... in turn. Mo Song-Qing, AMM 6540 16. Let xi, x2, ... be a sequence of numbers in [0,1) such that at least one of its sequential limit points is irrational. For 0 < a < b < 1, let Nn(a, b) be the number of n-tuples (al, a2, ..., an) E {±1}Th such that a,b) ( aixi + a2x2 + • • + anxn E [a,b). Prove that Nn 2 converges to b — a. Andrew Odlyzko, AMM 6542 17. Let a be a nonzero rational number and b an irrational number. Prove that the sequence nb [na] is uniformly distributed mod 1. L.Kuipers THEORY AND EXAMPLES 355 16.1 Theory and examples Problems about the sum of the digits of a positive integer often occur in mathematical contests because of their difficulty and lack of standard ways to tackle them. This is why a synthesis of the most frequent techniques used in such problems is useful. We have selected several representative problems to illustrate how the main results and techniques work and why they are so important. We will only work in base 10 and will denote the decimal sum of the digits of the positive integer x by s(x). The following "formula" can be easily checked: s(n) = n — 10k J (16.1) k>1 From (16.1) we can deduce immediately some well-known results about s(n), such as s(n) --== n (mod 9) and s(m + n) < s(m) + s(n). Unfortunately, (16.1) is a clumsy formula, which can hardly be used in applications. On the other hand, there are several more or less known results about the sum of the digits, results which may offer simple ways to attack harder problems. The easiest of these techniques is probably just the careful analysis of the structure of the numbers and their digits. This can work surprisingly well, as we will see in the following examples. [Example 1..] Prove that among any 79 consecutive numbers, we can choose at least one whose sum of digits is a multiple of 13. Baltic Contest 1997 Solution. [Adrian Zahariuc] Note that among the first 40 numbers, there are exactly four multiples of 10. Also, it is clear that the next to last digit of one of them is at least 6. Let x be this number. Clearly, x, x +1,..., x + 39 are 356 16. THE DIGIT SUM OF A POSITIVE INTEGER among our numbers, so s(x), s(x) + 1,...,s(x) + 12 occur as sums of digits in some of our numbers. One of these sums is a multiple of 13 and we are done. We will continue with two harder problems, which still do not require any special result or technique. I Example 2. Find the greatest N for which there are N consecutive posi-tive integers such that the sum of digits of the k-th number is divisible by k, for k = 1, 2, ..., N. Tournament of Towns 2000 Solution. [Adrian Zahariuc] The answer, which is not trivial at all, is 21. The main idea is that among s(n + 2), s(n + 12) and s(n + 22) there are two consecutive numbers, which is impossible since all of them should be even. Indeed, we carry over at a + 10 only when the next to last digit of a is 9, but this situation can occur at most once in our case. So, for N > 21, we have no solution. For N = 21, we can choose N +1, N + N + 21, where N = 291 10111 — 12. For i = 1 we have nothing to prove. For 2 < i < 11, s(N + i) = 2 + 9 + 0 + 9(11! — 1) + i — 2 = i + 9 11! while for 12 < i < 21, s(N + i) = 2 + 9 + 1 + (i — 12) = i, so our numbers have the desired property. !Example 3.] How many positive integers 7/ < 102005 can be written as the sum of two positive integers with the same sum of digits? [Adrian Zahariuc] Solution. Answer: 102005 — 9023. At first glance, it is seemingly impossible to find the exact number of positive integers with this property. In fact, the following is true: a positive integer cannot be written as the sum of two numbers with the same sum of digits if and only if all of its digits except for the first are 9 and the sum of its digits is odd. THEORY AND EXAMPLES 357 Let n be such a number. Suppose there are positive integers a and b such that n = a + b and s(a) = s(b). The main fact is that when we add a + b = n, there are no carry overs. This is clear enough. It follows that s(n) = s(a) + s(b) = 2s(a), which is impossible since s(n) is odd. Now we will prove that any number n which is not one of the numbers above, can be written as the sum of two positive integers with the same sum of digits. We will start with the following: Lemma 16.1. There is a < n such that s(a) s(n — a) (mod 2). Proof. If s(n) is even, take a = 0. If s(n) is odd, then n must have a digit which is not the first one and is not equal to 9, otherwise it would have one of the forbidden forms mentioned in the beginning of the solution. Let c be this digit and let p be its position (from right to left). Choose a = 10P-1(c + 1). In the addition a + (n — a) = n there is exactly one carry over, so s(a) + s(n — a) = 9 + s(n) 0 mod 2 s(a) s(n — a) mod 2 which proves our claim. K Back to the original problem. All we have to do now is take one-by-one a "unit" from a number and give it to the other until the two numbers have the same sum of digits. This will happen because they have the same parity. So, let us do this rigorously. Set n — a = bib2 • • • bk• a = aia2 • lc, Let I be the set of those 1 < i < k for which a, + bi is odd. The lemma shows that the number of elements of I is even, so it can be divided into two sets with the same number of elements, say I1 and / 2. For i = 1, 2, ..., k define A, = L' -0 if i ¢ /, '4+2 14+1 if i E /1 or as+2 bi-1 if i E /2 and B, = a, + bi - Ai. It is clear that the numbers A = A1A2•••Ak• B = B1B2.••Bk 358 16. THE DIGIT SUM OF A POSITIVE INTEGER have the properties s(A) = s(B) and A + B = n. The proof is complete. We have previously seen that s(n) n (mod 9). This is probably the most well-known property of the function s and it has a series of remarkable ap-plications. Sometimes it is combined with simple inequalities such as s(n) < 9([log n] + 1). Some immediate applications are the following: Example 4. Find all n for which one can find a and b such that s(a) = s(b) = s(a + b) = n. [Vasile Zidaru, Mircea Lascu] Solution. We have a b=a+b - n (mod 9), so 9 divides n. If n = 9k, we can take a = b = 10k —1 and we are done, since s(10k —1) = s(2.10k —2) = 9k. [Example 5J Find all the possible values of the sum of the digits of a perfect square. Iberoamerican Olympiad 1995 Solution. What does the sum of the digits have to do with perfect squares? Apparently, nothing, but perfect squares do have something to do with re-mainders mod 9. In fact, it is easy to prove that the only possible values of a perfect square mod 9 are 0, 1, 4 and 7. So, we deduce that the sum of the digits of a perfect square must be congruent to 0, 1, 4, or 7 mod 9. To prove that all such numbers work, we will use a small and very common (but worth remembering!) trick: use numbers that consist almost only of 9-s. We have the following identities: 99.99 2 = 99...99 8 00...00 1 s(99...99 2) = 9n n n-1 n-1 THEORY AND EXAMPLES 359 9991 2 -=- 99...99 82 00...00 81 = s(99..9912) = 9n +1 n-1 n-2 n-2 n-1 99...99 22 = 99...99 84 00...00 64 s (99..99 22) = 9n + 4 n-1 n-2 n-2 n-1 99942 = 99...99 88 00...00 36 8(99..99 42) = 9n + 7 n-1 n-2 n-2 n-1 and since s(0) = 0, s(1) = 1, s(4) = 4 and s(16) = 7 the proof is complete. Example 6 ± .] Compute s (s(s (44444444))). IMO 1975 Solution. Using the inequality s(n) < 9([log n] + 1) several times we have 8(44444444) < 9( [log 44444444] + 1) < 9 • 20000 = 180000; s(s(44444444)) < 9(Llogs(44444444)] + 1) < 9(log 180000 + 1) < 63, so s(s(s(44444444))) < 14 (indeed, among the numbers from 1 to 63, the max-imum value of the sum of digits is 14). On the other hand, s(s(s(n))) s(s(n)) s(n) n (mod 9) and since 44444444 74444 = 7 734481 7(mod9), the only possible answer is 7. Finally, we present two beautiful problems which appeared in the Russian Olympiad and, later, in Kvant. lExample 7. Prove that for any N there is an n > N such that 8(3n) > s(3n+1). 360 16. THE DIGIT SUM OF A POSITIVE INTEGER Solution. Suppose by way of contradiction that there is an N > 2 such that 3(3'41) - s(3n) > 0 for all n > N. But, for n > 2, s(3n+1) s(3n) 0 (mod 9), so S(3n+1) — On) > 9 for all n > N. It follows that E (s (3 k+1) — s (3k)) 9(n - N) s(3n+1) > 9(n - N) k=N+1 for all n > N +1. But 8(371+1) < 9( [log 3n+1] + 1), so 9n - 9N < 9 + 9(n + 1) log 3, for all n > N + 1. This is obviously a contradiction. [Example 8.1 Find all positive integers k for which there exists a positive constant ck such that s(( kN N )) > ck for all positive integers N. For any such k, find the best ck. [I. N. Bernstein] Solution. It is not difficult to observe that any k of the form 2' • 5q is a solution of the problem. Indeed, in that case we have (by using the properties presented in the beginning of the chapter): s(N) = s(109-PqN) < s(2q • 5r)s(kN) = — 1 s(kN) where clearly ck = ,(29.5,) is the best constant (we have equality for N = 2q • 5'). Now, assume that k = 2' • 5q • Q with Q > 1 relatively prime to 10. Let m (p(Q) and write 10m - 1 = QR for some integer R. If Rn = R(1 + 10m + • • • + 10m(n-1)) then 10' -1 = QR, and so s(Q(Rn +1)) = s(10"+Q -1) = s(Q) and s(Rn + 1) > (n - 1)s(R) (note than the condition Q > 1, which is the same as R < 10m -1, is essential for this last inequality, because it guarantees that R +1 has at most m digits and thus when adding R +1 and 10m • R, we obtain the digits of R followed by the digits of R + 1; if we proceed in the same manner for each addition, we see that Rn + 1 has among its digits at THEORY AND EXAMPLES 361 least n-1 copies of the sequence of digits of R). By taking n sufficiently large, we conclude that for any € > 0 there exists N = 1?„,, +1 such that s(kN) < s(2r • 59)s(Q) < 6 s(N) (n — 1)s(R) This shows that the numbers found in the first part of the solution are the only solutions of the problem. If so far we have studied some remarkable properties of the function s, which were quite well-known, it is time to present some problems and results which are less familiar, but interesting and hard. The first result is the following: Lemma 16.2. If 1 < x < 10n, then s(x(lOn — 1)) = 9n. Proof The idea is very simple: all we have to do is write x = aia2...a with a3 0 (we can ignore the trailing O's of x) and note that x(lOn — 1) = aia2...a3_1(a3 — 1)99..99(9 — — a3_1)(10 — a3), n-3 which obviously has the sum of digits equal to 9n. K The previous result is by no means hard, but we will see that it can be the key in many situations. A first application is: Example 9.1 Evaluate s(9 • 99 • 9999 • ... • 99...99). 2" USAMO 1992 362 16. THE DIGIT SUM OF A POSITIVE INTEGER Solution. The problem is trivial if we know the previous result. We have . N = 9 • 99 • 9999 • ... • 99..99 < 101+2 +. .+2n-1 < 102" - 1 2"-1 so s(99...99 N) = 9 • 2n. 2' However, there are very hard applications of this apparently unimportant re- sult, such as the following problem. Example 10. Prove that for each n there is a positive integer with n nonzero digits, that is divisible by the sum of its digits. IMO 1998 Shortlist Solution. Just to assure our readers that this problem did not appear on the IMO Shortlist out of nowhere, such numbers are called Niven numbers and they are an important research source in number theory. Now, let us solve it. We will see that constructing such a number is difficult. First, we will dispose of the case n = 3k, when we can take the number 11.11 (it can 3k be easily proved by induction that 3k+21103k — 1). From the idea that we should search numbers with many equal digits and the last result, we decide that the required number p should be of the form aa...aa b • (lot — 1) , with a..aa b < 10t — 1. This number has s t 1 digits and its sum of digits is 9t. Therefore, we require s t = n — 1 and 9tlaa...aa b • (10t — 1). We now use the fact that if t is a power of 3, then 9t110t — 1. So, let us take t = 3k where k is chosen such that 3k < n < 3 k+1. If we also take into account the condition aa...aa b < 10 — 1 it is natural to pick p = 11.11 2(103k — 1) when n-3" —1 n < 2 • 3k and p = 22...22(102'3k — 1) otherwise. 2.3" THEORY AND EXAMPLES 363 We continue our investigations of finding suitable techniques for problems in-volving sum of digits with a very beautiful result, which has several interesting and difficult consequences. Lemma 16.3. Any multiple of 99...99 has sum of its digits at least 9k. k Proof. We will use the extremal principle. Suppose by way of contradiction that the statement is false, and take M to be the smallest multiple of a such that s(M) < 9k, where a = 99...99. Clearly, M > 10k, hence M = apap_i...ao, with p > k and ap 0. Take N = M —10P—ka, which is a multiple less than M of a. We will prove that s(N) < 9k. Observe that N = M —10P +10" = (ap-1)•10P+ap_110P-1+• • •+(ap_k+1)10" +- • •+ao, so that we can write s(N) < ap — 1 + ap_i + • • • + (ap_k + 1) + • • • + ao -=- s(M) < 9k. In this way, we contradict the minimality of M and the proof is completed. We will show three applications of this fact, which might seem simple, but seemingly unsolvable without it. But before that, let us insist a little bit on a very similar (yet more difficult) problem proposed by Radu Todor for the 1993 IMO: if b > 1 and a is a multiple of bn —1, then a has at least n nonzero digits when expressed in base b. The solution uses the same idea, but the details are not obvious, so we will present a full solution. Arguing by contradiction, assume that there exists A, a multiple of bn —1 with less than n nonzero digits in base b, and among all these numbers consider that number A with minimal number of nonzero digits in base b and with minimal sum of digits in base b. Suppose that a has exactly s nonzero digits (everything is in base b) and let A = al bnl + a2bn2 + • • • + as bns with ni > n2 > • > ns. We claim that s = n. First of all, we will prove that any two numbers among nl, n2, ..., ris are not 364 16. THE DIGIT SUM OF A POSITIVE INTEGER congruent mod n. It will follow that s < n. Indeed, if ni n3 (mod n) let 0 < r < n — 1 be the common value of 74 and ni modulo n. The number B = A — aibn — ajbn3 + + a3)bnni-Er is clearly a multiple of bn — 1. If ai + < b then B has s — 1 nonzero digits, which contradicts the minimality of s. So b < ai < 2b. If q =- a, + — b, then B bnn1-Fr-F1 qbnni -Fr + aibnl + • • • + +a+1 1 + • • + + a3±1bn3+1 + • • • + asbns. Therefore the sum of digits of B in base b is al + a2 + • • • +as + 1 + q— (a, +a3) < al + a2 + • • • + a,. This contradiction shows that ni, n2, •••, ns give distinct remainders ri, r2, when divided by n. Finally, suppose that s < 7/ and consider the number C = aibr1 + • • • +60". Clearly, C is a multiple of bn —1. But C < bn — 1! This shows that s = n and finishes the solution. Example 1 • Prove that for every k, we have s(n!) lim (ln(ln(n)))k = co. Solution. Due to the simple fact that 101-1°gn-I — 1 < n 1011'gni — lin!, we have s(n!) > [log n J , from which our conclusion follows. r Example 12.i Let S be the set of positive integers whose decimal represen-tation contains at most 1988 ones and the rest zeros. Prove that there is a positive integer which does not divide any element of S. Tournament of Towns 1988 Solution. Again, the solution follows directly from our result. We can choose the number 101989 — 1, whose multiples have sum of digits greater than 1988. THEORY AND EXAMPLES 365 Example 13. Prove that for each k > 0, there is an infinite arithmetical progression with a common difference relatively prime to 10, such that all its terms have the sum of digits greater than k. IMO 1999 Shortlist Solution. Let us remind you that this is the last problem from IMO 1999 Shortlist, so it is one of the hardest. The official solution seems to confirm this. But, due to the above lemma we can chose the sequence an = n(10' —1), where m > k and we are done. Now, as a final proof of the utility of these two results, we will present a hard problem from the USAMO. 1 Example 14.1 Let n be a fixed positive integer. Denote by f(n) the smallest k for which one can find a set X of n positive integers with the property s k xEY for all nonempty subsets Y of X. Prove that Ci log n < f(n) < C2 log n for some constants C1 and C2. [Titu Andreescu, Gabriel Dospinescu] USAMO 2005 Solution. We will prove that + [log(n + 1)] < f(n) < 9 log [n(n 2 1) + 1 i , which is enough to establish our claim. Let 1 be the smallest integer such that n 101 — 1 > (n + 1) 2 366 16. THE DIGIT SUM OF A POSITIVE INTEGER Consider the set X = {j(101 — 1) :1 < j < n}. By the previous inequality and our first lemma, it follows that s E x 9/ xEY for all nonempty subsets Y of X, so f (n) < 9/, and the upper bound is proved. Now, let m be the greatest integer such that n > 10' — 1. We will use the following well-known lemma: Lemma 16.4. Any set M = fai,a2,...,aml has a nonempty subset whose element sum is divisible by m. Proof. Consider the sums al, al +az, • • • , al + a2 + • • • ± an,. If one of then is a multiple of m, them we are done. Otherwise, there are two of them congruent mod m, say the i-th and the j-th. Then, + a2+2 + • • • + a3 and we are done. K From the lemma, it follows that any n-element set X has a subset Y whose element sum is divisible by 10' — 1. By our second lemma, it follows that s ( Ex > m f (n) > m, xEY and the proof is complete. The last solved problem is one we consider to be very hard, and which uses different techniques than the ones we have mentioned so far. Example 15. Let a and b be positive integers such that s(an) = s(bn) for all n. Prove that log bis an integer. [Adrian Zahariuc, Gabriel Dospinescu] THEORY AND EXAMPLES 367 Solution. We start with an observation. If gcd(max{a, b}, 10) = 1, then the problem becomes trivial. Indeed, suppose that a = max{a, b}. Then, by Euler's theorem, al 109'(a) - 1, so there is an n such that an = 10P(a) - 1, and since numbers consisting only of 9-s have a digit sum greater than all previous numbers, it follows that an = bn, so a = b. Let us now solve the harder problem. For any k > 1 there is an nk such that 10k < ank < 10k + a - 1. It follows that s(ank) is bounded, so s(bnk) is bounded. On the other hand, 10- b -a < bnk <10- -a + b, so, for sufficiently large k, the first p nonzero digits of a are exactly the same as the first p digits of bnk. This means that the sum of the first p digits of -61 is bounded, which could only happen when this fraction has finitely many a decimals. Analogously, we can prove the same result about 6. Let a = 2x5Ym and b = 2z56mi, where gcd(m, 10) = gcd(rni, 10) = 1. It follows that mlm' and m'im, so m = m'. Now, we can write the hypothesis as s(2z5umn2c-x5c-y) = s(2x5Ymn2c_x5c_) = s(mn) for all c > max{x, y}. Now, if p max {z +c-x,u+c- y} - min {z + c - x , u + c - y} , we find that there is a k E {2, 5} such that s(mn) = s(mkP n) for all positive integer n. It follows that s(mn) = s(kPmn) = s(k2Pmn) = s(k3Pmn) = • • • Let t = aP, so log t E R - Q unless p = 0. Now, we will use the following: Lemma 16.5. If log t E R - Q, then for any sequence of digits, there is a positive integer n such that trim starts with the selected sequence of digits. Proof. If we prove that {{log E Z+} is dense in (0,1), then we are done. But log trim = n log t + m and by Kronecker's theorem {{n log t}In E Z+} is dense in (0,1), so the proof of the lemma is complete. K The lemma implies the very important result that s(tnm) is unbounded for p 0, which is a contradiction. Hence p = 0 and z+c-x=u+c- y, so 368 16. THE DIGIT SUM OF A POSITIVE INTEGER a = 10x—zb. The main proof is complete. This problem can be nicely extended to any base. The proof of the general case is quite similar, although there are some very important differences. The aforementioned methods are just a starting point in solving such problems since the spectrum of problems involving the sum of the digits is very large. The techniques are even more useful when they are applied creatively. PROBLEMS FOR TRAINING 369 16.2 Problems for training 1. We start with a perfect number (which is equal to the sum of its divisors, except itself) different from 6 and calculate its sum of digits. Then, we calculate the sum of digits of the new number and so on. Prove that we will eventually get 1. 2. Prove that for any positive integer n there are infinitely many numbers m not containing any zero, such that s(n) = s(mn). Russian Olympiad 1970 3. Prove that among any 39 consecutive positive integers there is one whose digit sum is divisible by 11. Russian Olympiad 1961 4. Prove that E s(n) = ln 10 n(n + 1) 9 . >1 0. Shallit, AMM 5. Are there positive integers n such that s(n) = 1000 and s(n2) = 1000000? Russian Olympiad 1985 6. Prove that there are infinitely many positive integers n such that s(n) + s(n2) = s(n3). Gabriel Dospinescu 370 16. THE DIGIT SUM OF A POSITIVE INTEGER 7. If s(n) = 100 and s(44n) = 800, find s(3n). Russia 1999 8. Let a and b be positive integers. Prove that the sequence saan + b]) contains a constant subsequence. Laurentiu Panaitopol, Romanian TST 2002 9. Find explicitly a Niven number with 100 digits. St. Petersburg 1990 10. Are there arbitrarily long arithmetic sequences whose terms have the same digit sum? What about infinite arithmetic sequences? 11. Let a be a positive integer such that s(an + n) = 1 + s(n) for any sufficiently large n. Prove that a is a power of 10. Gabriel Dospinescu 12. Are there 19 positive integers with the same digit sum, which add up to 1999? Rusia 1999 13. Call a positive integer m special if it can be written in the form n + s(n) for a certain positive integer n. Prove that there are infinitely many positive integers that are not special, but among any two consecutive numbers, at least one is special. Kvant PROBLEMS FOR TRAINING 371 14. Find all x such that s(x) = s(2x) = s(3x) = • • • = s(x2). Kurschak Competition 1989 15. Let a, b, c, d be prime numbers such that 2 < a < c and a b. Suppose that for sufficiently large n, the numbers an+ b and cn+d have the same digit sum in any base between 2 and a — 1. Prove that a = c and b = d. Gabriel Dospinescu 16. Let (an)n>i be a sequence such that s(an) > n. Prove that for any n the following inequality holds 1 1 1 — + — + • • • + — < 3.2. al a2 an Can we replace 3.2 by 3? Laurentiu Panaitopol 17. Prove that one can find ni < n2 < • • • < n50 such that nl + s(ni) = n2 + s(n2) = • • • = n5o + s(n5o)• Poland 1999 18. Let S be a set of positive integers such that for any a E — Q, there is a positive integer n such that Lan] E S. Prove that S contains numbers with arbitrarily large digit sum. Gabriel Dospinescu 19. Find the smallest positive integer which can be expressed at the same time as the sum of 2002 numbers with the same digit sum and as the sum of 2003 numbers with the same digit sum. 372 16. THE DIGIT SUM OF A POSITIVE INTEGER 20. Are there polynomials p E Z[X] such that lim s(p(n)) = 00? n—>oo 21. Prove that there exists a constant c > 0 such that for all n we have s(r) > clnn. 22. Prove that there are arbitrarily long sequences of consecutive numbers which do not contain any Niven numbers. Mathlinks Contest 23. Define f (n) = n + s(n). A number m is called special if there is a k such that f (k) = m. Prove that there are infinitely many special numbers of the form 10n + b if and only if b — 1 is special. Christopher D. Long 24. Let k be a positive integer. Prove that there is a positive integer m such that the equation n + s(n) = m has exactly k solutions. Mihai Manea, Romanian TST 2003 25. Let x7, be a strictly increasing sequence of positive integers such that v2(xn) — v5(x,i) has the limit oo or —oo. Prove that s(x„,) tends to oo. Bruno Langlois 26. Is there an increasing arithmetic sequence with 10000 terms such that the digit sum of its terms forms again an increasing arithmetic sequence? Tournament of the Towns PROBLEMS FOR TRAINING 373 27. Prove that the sum of digits of 9n is at least 18 for n > 1. AMM 28. Prove that there exists a constant C such that for all N, the number of Niven numbers smaller than N is at most C (ln x)2/3 29. Is there an infinite arithmetic progression containing no Niven numbers? Gabriel Dospinescu THEORY AND EXAMPLES 377 17.1 Theory and examples "Olympiad problems can be solved without using concepts from analysis (or linear algebra)" is a sentence often heard when talking about problems given at various mathematics competitions. This is true, but the essence of some of these problems lies in analysis, and this is the reason that such problems are always the highlight of a contest. Their elementary solutions are very tricky and sometimes extremely difficult to design, while when using analysis they can fall apart rather quickly. Well, of course, "quickly" only if you see the right sequence (or function) that hides behind each such problem. Practically, in this chapter our aim is to exhibit convergent integer sequences. Clearly, these sequences must eventually become constant, and from here the problem becomes much easier. The difficulty lies in finding those sequences. Some-times this is not so challenging, but most of the time it turns out to be a very difficult task. We develop skills in "hunting" for these sequences first by solving some easier questions, and after that we tackle the real problem. As usual, we begin with a classical and beautiful problem, which has many applications and extensions. Example 1 d Let f, g E Z[X] be two nonconstant polynomials such that f(n)Ig(n) for infinitely many n. Prove that f divides g in ((2[X]. Solution. Indeed, we need to look at the remainder of g when divided by f in Q[Xl. Let us write g = f • q r, were q, r are polynomials in Q[X] with deg r < deg f. Now, multiplying by the common denominator of all coefficients of the polynomials q and r, the hypothesis becomes: there exist two infinite integer sequences (an)n>i, (bn)n>i and a positive integer N such that bn = N f r(an) (we could have some problems with the zeros of f, but they (an) are only finitely many, so for 71 large enough, an is not a zero of f). Because r(an) deg r < deg f, it follows that f (an) —> 0, thus (bn)n>1 is a sequence of integers 378 17. AT THE BORDER OF ANALYSIS AND NUMBER THEORY that converges to 0. This implies that this sequence will eventually become the zero sequence. Well, this is the same as r(an) = 0 from a certain point no, which is practically the same as r = 0 (do not forget that any nonzero polynomial has only finitely many zeros). The problem is solved. The next problem is a special case of a much more general and classical result: if f is a polynomial with integer coefficients, k is an integer greater than 1, and .0 (n) E Q for all n, then there exists a polynomial g E Q[X] such that f (x) = gk (x). We will not discuss here this general result (the reader will find a proof in the chapter Arithmetic Properties of Polynomials). Let a, b, c be integers with a 0 such that an2 + bn + c is a perfect square for any positive integer n. Prove that there exist integers x and y such that a = x2, b = 2xy, c = y2. Solution. Let us begin by writing an2 bn + c = xri for a certain sequence (xn)n>1 of nonnegative integers. We would expect that xn — nVa, converges. And yes, it does, but it is not a sequence of integers, so its convergence is more or less useless. In fact, we need another sequence. The easiest way is to work with (x,,,+i — Xn)n>1) since this sequence certainly converges to -Va (you have already noticed why it was not useless to find that xn — nfci, is convergent; we used this to establish the convergence of (xn+i — xn)n>1). This time, the sequence consists of integers, so it is eventually constant. Hence we can find a positive integer M such that xn±i = x,n + .va for all n > M. Thus a must be a perfect square, that is a = x2 for some integer x. A simple induction shows that xn = xm + (n — M)x for n > M and so (xM — Mx + nx)2 = x2n2 + bn + c for all n > M. Identifying the coefficients finishes the solution, since we can take y = xm — Mx. Even this very particular case is interesting. Indeed, here is a very nice appli-cation of the previous problem: THEORY AND EXAMPLES 379 Example 3. Prove that there cannot exist three polynomials P, Q, R with integer coefficients, of degree 2, and such that for all integers x, y there exists an integer z such that P(x) + Q(y) = R(z). Tuymaada Olympiad Solution. Using the above result, the problem becomes straightforward. In-deed, suppose that P(X) = aX2 + bX + c,Q(X) = dX2 + eX + f and R(X) = mX2 + nX + p are such polynomials. Fix two integers x, y. Then the equation mz2 + nz + p — P(x) — Q(y) = 0 has an integer solution, so the discriminant is a perfect square. It means that m(4P(x)+4Q(y)-4p)+n2 is a perfect square and this for all integers x, y. Now, for a fixed y, the polynomial of second degree 4mP(X) + m(4Q(y) — 4p) + n2 transforms all integers into perfect squares. By the previous problem, it is the square of a polynomial of first degree. In particular, its discriminant is zero. Because y is arbitrary, it follows that Q is constant, which is not possible because deg(Q) = 2. Another easy example is the following problem, in which finding the right convergent sequence of integers in not difficult at all. But, attention must be paid to details! LExample 4.1 Let al, a2, , ak be positive real numbers such that at least one of them is not an integer. Prove that there exit infinitely many positive integers n such that n and [aim] + La2n + • • • + l_akni are relatively prime. [Gabriel Dospinescu] Solution. The solution to such a problem needs to be indirect. So, let us as-sume that there exists a number M such that n and [am n] + La2n] +• • • + Lakni are not relatively prime for all n > M. Now, what are the most efficient num-bers n to be used? They are the prime numbers, since if n is prime and it is not relatively prime with Lain.] + [a2n] + • • • + [akn] , then it must divide 380 17. AT THE BORDER OF ANALYSIS AND NUMBER THEORY Lain] + La2n] + • • • + [akni. This suggests considering the sequence of prime numbers (pn)n>1. Since this sequence is infinite, there is N such that pn > M for all n > N. According to our assumption, this implies that for all n > N there exist a positive integer xn such that Laipni [a2pn j +• •+ [akpn j = xnPn• And now, you have already guessed what is the convergent sequence! Yes, it [aoni + [a2Prii + • • • + [akPni is (xn)n>N. This is clear, since converges to pn al + a2 + • • • + ak. Thus we can find P such that xn = al + a2 + • • • + ak for all n > P. But this is the same as { alpn} {a2pn} + • • • + { akpn} = 0. This says that aipn are integers for all i = 1, 2, ..., k and n > P and so ai are integers for all i, contradicting the hypothesis. Step by step, we start to build some experience in "guessing" the sequences. It is then time to solve some more difficult problems. The next one may seem obvious after reading its solution. In fact, it is just that type of problem whose solution is very short, but difficult to find. Let a and b be integers such that a • 2n + b is a perfect square for all positive integers n. Prove that a = 0. Polish TST Solution. Suppose that a 0. Then a > 0, otherwise for large values of n the number a • 2n + b is negative. From the hypothesis, there exists a sequence of positive integers (xn)n>1 such that xri = \fa • 2n + b for all n. A direct computation shows that lim (2xn — xn+2) = 0. This implies the existence of n—>co a positive integer N such that 2xn = xn+2 for all n > P. But 2xn = xn+2 is equivalent to b = 0. Then a and 2a are both perfect squares, which is impos-sible for a 0. This shows that our assumption is wrong, and so a = 0. Schur proved that if f is a non constant polynomial with integer coefficients, then the set of primes dividing at least one of the numbers f (1), f (2), . . . is infinite. The following problem is an extension of this result. THEORY AND EXAMPLES 381 Example 6. Suppose that f is a polynomial with integer coefficients and that (an) is a strictly increasing sequence of positive integers such that an < f (n) for all n. Then the set of prime numbers dividing at least one term of the sequence (an) is infinite. Solution. The idea is very nice: for any finite set of prime numbers pl, and any k > 0, we have ,kai 1 k(aN < 00. ' • • ,, N Indeed, it suffices to observe that we actually have 1 Pi nkai PN k ki II cti,a25•••,ceN>0 2-1 • • • N j=1 i>0 Pi j=1 3 1 On the other hand, by taking k = we have 2 deg( f) 1 =oo (.f (n))k n>1 Thus, if the conclusion of the problem is not true, we can find pl, such that any term of the sequence is of the form pia' ...pk ir and thus kai „ka N < DD. n>1 n oa2,•••,aN>0 P1 ...PN On the other hand, 1 • >1 n ak f (n))k = C°' a contradiction. The same idea is used in the following problem. 1 ak 1 382 17. AT THE BORDER OF ANALYSIS AND NUMBER THEORY Example 7. Let a and b be integers greater than 1. Prove that there is a multiple of a which contains all digits 0, 1, , b — 1 when written in base b. Adapted after a Putnam Competition problem Solution. Let us suppose the contrary. Then any multiple of a misses at least one digit when written in base b. Since the sum of inverses of all multiples of 1 1 a diverges (because 1 + — 2 + — 3 + • • • = co), it suffices to show that the sum of inverses of all positive integers missing at least one digit in base b is conver-gent, and we will reach a contradiction. But of course, it suffices to prove it for a fixed (but arbitrary) digit j. For any n > 1, there are at most (b — 1)n numbers which have n digits in base b, all different from j. Thus, since each one of them is at least equal to bn-1, the sum of inverses of numbers that miss n ( b ' the digit j when written in base b is at most equal to Eb b 1 which converges. The conclusion follows. The following example generalizes an old Kvant problem. Example 8.1 Find all polynomials f with real coefficients such that if n is a positive integer which is written in base 10 only with ones, then f(n) has the same property. [Titu Andreescu, Gabriel Dospinescu] Putnam 2007 Solution. Let f be such a polynomial and observe that from the hypothesis it follows that there exists a sequence (an,),,>1 of positive integers such that f 1091) = ur9n-1 . But this sequence (an)n>1 cannot be really arbitrary: actually we can find precious information from an asymptotic study. Indeed, suppose that deg(f) = d > 1. Then there exists a nonzero number A such that f(x) ti Aid for large values of x. Therefore f (10n9— 6,4 1) A lOnd. Thus • 10an Th . lO nd. This shows that the sequence (an — nd)n>i converges to a limit 1 such that A = 9d-1 10 /. Because this sequence consists of integers, it n>1 THEORY AND EXAMPLES 383 becomes eventually equal to the constant sequence 1. Thus from a certain point f ( i.on 9-1) = 1.13 9 + —1 lcr 9— we have If xn = 1 we deduce that the equation f (x) _ (9x+1)9 d.101 -1 (9x+i)d-io'-1 has infinitely many solutions, so f (X) = 9 Thus this is the general term of these polynomials (not including here obvious constant solutions), being clear that all such polynomials satisfy the condi-tions of the problem. We return to classical mathematics and discuss a beautiful problem that ap-peared in the Tournament of the Towns in 1982, in a Russian Team Selection Test in 1997, and also in the Bulgarian Olympiad in 2003. Its beauty explains why the problem was so popular among the exam writers. Example 9. Let f be a monic polynomial with integer coefficients such that for any positive integer n the equation f(x) = 2n has at least one positive integer solution. Prove that deg(f) = 1. Solution. The problem states that there exists a sequence of positive integers (xn)n>1 such that f (xn) = 2n. Let us suppose that deg(f) = k > 1. Then, for large values of x, f (x) behaves like Xk . So, trying to find the right convergent sequence, we could try first to "think big": we have xn = r, that is for large n, xn behaves like 2T.. Then, a good possible convergent sequence could be xn+k - 2xn. Now, the hard part: proving that this sequence is indeed convergent. First, we will show that Xn-Fk converges to 2. This is easy, since xn the relation f (xn+k) = 2k f (xn) implies f (Xn+k) (Xn+k) k = 2k f(xn) Xn-Fk xn J xn and since lim 1(k ) = 1 and lim xn = co, we find that indeed lim x "+'` = 2. x--•00 x n—>oo We see that this will help us a lot. Indeed, write f (x) = Xk aixi • 384 17. AT THE BORDER OF ANALYSIS AND NUMBER THEORY Then f (xn+k) = 2k f (xn,) can be also written as k-1 a,(2kxi n — Xi n+k) Xn+k — 2xn = k—i (2X )iXri k— ±i k-1 i=0 • Xn±k But from the fact that urn = 2, it follows that the right-hand side of n—>oo xn the above relation is also convergent. Hence (xn+k — 2xn)n>i converges and so there exist M, N such that for all n > M we have Xn-Fk = 2xn+N . But now the solution is almost over, since the last result combined with f (xn+k) = 2k f (xn) yields f(2xn + N) = 2k f (x, i) for n > M, that is f (2x + N) = 2k f (x). So, an arithmetical property of the polynomial turned into an algebraic one by using analysis. This algebraic property helps us to finish the solution. Indeed, we see that if z is a complex zero of f, then 2z + N, 4z + 3N, 8z + 7N,... are all zeros of f . Since f is nonzero, this sequence must be finite and this can happen only for z = —N. Because —N is the only zero of f, we deduce that f(x) = (x + N)k . But since the equation f(x) = 22k+1 has positive integer roots, we find that 2-k - E Z, which implies k = 1, a contradiction. Thus, our assumption was wrong and deg( f) = 1. The following problem generalizes the problem above. Example 16.1 Find all complex polynomials f with the following property: there exists an integer a greater than 1 such that for all suffi-ciently large positive integer n, the equation f(x) = an2 has at least one solution in the set of positive integers. [Gabriel Dospinescu] Mathlinks Contest Solution. From the beginning we exclude the constant polynomials, so let f be a solution of degree d > 1. Let (x,),,>no be a sequence of positive integers such that f(xn) = an2 for some integer a greater than 1. Now, observe that THEORY AND EXAMPLES 385 we can choose A such that the polynomial g(X) = f (x + A) has no term of degree d -1. Define yn = xn - A and observe that g(yn) = an2. Now, what really interests us is the asymptotic behavior of the sequence yn. This boils down to finding the behavior of the solution of the equation g(y) = z when z is very large. In order to do this, put g(y) = Byd + Cy' + • • • with B > 0 (the fact that B > 0 is obvious because g(x) remains positive for arbitrarily large values of x). Now, suppose that C 0. The choice of A ensures that e < d - 2. Therefore, if we define z = Ud and Byd = vd, E = and finally m = d - e, then we have ud = '0(1+ EV-771 o(v')). Thus u = v(1 + Ey' + o(v-m))/ = v (1 — Ev-m + o(v-m)) = v + —Evl-'m. + o(vi-m). This shows that u v, and combining this observation with the previous result gives v = u - -Ed til-m 0(Iti—m). Coming back to our notations, we infer that My = z j a - + o(z- fi) where p = m -1. Finally, this can be written in the form y = Fz d + Gz-a + o(z') (the definitions of F, G and a are obvious from the last formula). Coming back to the relation g(yn) = an2 n2 we deduce that yn = Fad + Ga-an2 + o(a-" 2 ). Therefore Yd+n 2n-Fd 2n+d—an2 \ ) = Fa d n2 a + o/a . +d 2n yn This shows that if we define zn = Yn+d - a then zn = o(1). On the other hand, by definition of yn we obtain that an+1 = zn + A(1 - a2n+2+d) is an integer. Therefore, the relation Zn+1 a2 zn = an±i — a2 a n A(a 2 - 1) and the fact that zn = o(1) shows that an±i - a2an is eventually constant, equal to A(1 - a2). Thus for sufficiently large n we have zn±i = a2zn, so we have proved the existence of a constant K such that zn = Ka2n for sufficiently large n. Because a > 1 and zn = o(1), it follows that K = 0 and thus zn = 0 for sufficiently large n. But the assumption C # 0 implies that G 0 and 386 17. AT THE BORDER OF ANALYSIS AND NUMBER THEORY 2 , by one of the previous relations we also have zn Ga2n+d—anwhich is not true if zn = 0 from a certain point on. This contradiction shows that f (X) = B (X — A)d for some rational numbers A, B (because f takes rational values for infinitely many rational values of the variable, it is equal to its Lagrange interpolation polynomial, thus it has integer coefficients). Let B = 9 and A = s. Then p(sxn — r)d= qsdan2. By taking n a multiple of d greater than no we obtain the existence of integers pi, qi such that p= pl, q = 4. Thus n2 Pi (sxn — r) = , which shows that ant is a d-th power for all sufficiently large n. This implies the existence of an integer b such that a = bd. Now, by taking pi, qi, s > 0 (we can do that, without loss of generality), we deduce that for some ni (which we will identify with no from now on, by eventually r 2 in enlarging no) we have sxn = qiSb Let a = gcd(s, pi) and write s = 2 au, pi = av with gcd(u, v) = 1. Then auxn = r qi u vr thus v 10'2 and so 2 for all n > no, auxn = r+ qibn0 ubn2 no . By taking n = no we deduce that ulr. Because uls, it follows that u = 1 and so sxn -= r+ql, . Note that gcd(v, qi) = 2 1 because v In., so vlbno. Let bno 2 = my. Thus sxn = r + mqibn2 —no By taking 2 again n = no, we obtain that mqi —r (mod s), so r(1 — bn2 —no) 0 (mod s) and so bn2—n° 1 (mod s) for all n > no. Applying this relation to n + 1 and making the division in the group of invertible residues mod s, we infer that b2n+1 = 1 (mod s) for all sufficiently large n. Repeating this procedure, we deduce that b2 —= 1 (mod s) and so b 1 (mod s). This implies my = bng 1 (mod s) and since r —mqi (mod s) and gcd(s, v) = 1, we finally obtain the necessary condition ry (mod s). Now, let us show that the conditions gcd(pi, qi) = gcd(r, s) = 1 and pi = sv, gcd(s, v) = 1, ry (mod s) are sufficient for the polynomial f (X) = ( 1 .(X — I s )) to be a solution of the problem. Indeed, using the Chinese Remainder Theorem, we can choose b such that b 0 (mod v) and b 1 (mod s). Thus vIry + q1bn2 and also slry + qibn . Because gcd(s, v) = 1 it follows that there exists a sequence xn of positive integers such that ry + q1bn2 = syxn. Thus f (xn) = bdn2 and the problem is finally solved. The idea behind the following problem is so beautiful that any reader who THEORY AND EXAMPLES 387 attempts to solve it will feel generously rewarded by discovering this mathe-matical gem either by herself or himself, or in the solution provided. [Example 11.] Let 7r(n) be the number of prime numbers not exceeding n. Prove that there exist infinitely many n such that 7r(n)In. [S. Golomb] AMM Solution. Let us prove the following result, which is the key to the problem. Lemma 17.1 that lim — an n—>oo n particular, an . For any increasing sequence of positive integers (ari)n>i such = 0, the sequence n contains all positive integers. In an n>1 divides n for infinitely many n. Proof. Even if it seems unbelievable, this is true. Moreover, the proof is ex-tremely short. Let m be a positive integer. Consider the set A=(n>l amn ll mn 1 — m} • m a n This set contains 1 and it is bounded, since lim = 0. Thus it has n—>00 mn amk a maximal element k. If = — 1 — n , then m is in the sequence mk an) n>1 Otherwise, we have am(k+i) > amk > k +1, which shows that k +1 is also in the set, in contradiction with the maximality of k. The lemma is proved. 0 Thus, all we need to show now is that lim 71-(n)= 0. Fortunately, this is well n—>co n known and not difficult to prove. There are easier proofs than the following one, but we prefer to deduce it from a famous and beautiful result of Eras: p < 4n-i. This was proved in chapter Look at the Exponent, but really p<72 we expect you to know how to prove it (it is one of those marvelous proofs 388 17. AT THE BORDER OF ANALYSIS AND NUMBER THEORY that cannot be forgotten). Now, the fact that lim 7r(n)= 0 follows easily. n—>co n Indeed, fix k > 1. We have for all large n the inequality (n — 1) log 4 > j log p > (7r(n) — 7r(k)) log k, k<p<n which shows that (n — 1) log 4 7r(n) < 7r(k) + . log k 7r(n) This proves that lim = 0. The problem is finally solved. n—>oc n A somewhat tricky, but less technical problem follows now. A special case of it was proposed by the USA for IMO 1990: Example 12.1 Let f be a polynomial with rational coefficients, of degree at least 2, and let (an)n>i be a sequence of rational numbers such that f (an+1) = an for all n. Prove that this sequence is periodic. [Bjorn Poonen] AMM 10369 Solution. First of all, it is clear that the sequence is bounded. Indeed, because deg( f) > 2 there exists M such that If (x)1 > lxi if Ix1 > M. By taking M suffi-ciently large one can also assume that M > lai 1. Then an immediate induction shows that lanl < M for all n. We will now prove that for some positive integer N we have Nan e Z for all n. Indeed, let al = 9 for some integers p, q and let k be a positive integer such that k f = f sXs + • • • + fiX + fo E Z[X]. Define N = qh. Then Nal = pfs E Z, and clearly if Nan E Z then Nan+i is a ratio-nal zero of the monic polynomial with integer coefficients kN3 (f (N x ) — an), f, so it is an integer. This shows that (Nan)n>i is a bounded sequence of in- tegers, therefore it takes only a finite number of values. Suppose that the sequence (an)n>i takes at most m different values. Consider (m + 1)-tuples THEORY AND EXAMPLES 389 (64, ai+m) for positive integers i. There are at most mni+1 such (m+1)- tuples that can be formed and in each such (m + 1)-tuple there exists a value taken at least twice. Therefore there exists a pattern that is repeated infinitely many times, which means that there exists k such that for all positive integers n there exists j > n for which a3 = a 3+k. But applying fr to this last relation and taking into account that fr(an+r) = an shows that an = a n±k for all n. That is, the sequence is periodical. A fine concoction of number theory and analysis is used in the solution of the next (very) difficult problem. We will see one of the thousands of unexpected applications of Pell's equation: [Example 13.] Find all polynomials p and q with integer coefficients such that p(X)2 = (X2 + 6X + 10)q(X)2 — 1. Vietnamese TST 2002 Solution. One easy step is to notice that X2 + 6X + 10 = (X + 3)2 + 1, so by taking f (X) = p(X — 3) and g(X) = q(X — 3) the problem "reduces" to solving the equation (X2 + 1)f (X)2 = g(X)2 + 1 in polynomials with integer coefficients. Of course, we may assume that the leading coefficients of f and g are positive and also that both polynomials are nonconstant. Therefore there exists an M such that f (n) > 2, g (n) > 2 for all n > M. As it is well known, the solutions in positive integers to the Pell equation x2 + 1 = 2y2 are (xn, yn) where (1 .0\ 2n-1 )(1 — 0)2n-1 (1 + 42n-1 )2n-1 2 xn = 2 Yn — • Observe that g2(xn) + 1 = 2(ynf(xn))2. There exist two sequences of positive integers (an)n>m and (bn)n>m such that g(xn) = xar, and yn f(xn) = ybn. Let k = deg(g) and m = deg(f). Because the sequence xn k 2g(xn) xn k (1 + 0)2n-1 g(X) = 2 (x + Vx2 + 1)k + (X VX2 + i)k 390 17. AT THE BORDER OF ANALYSIS AND NUMBER THEORY 2x., 271-1) clearly converges to a nonzero limit, so does the sequence (1± ) k( and therefore the sequence (1 + 0)2an-1—k(2n-1) converges to a nonzero limit. This sequence having integer terms, it becomes constant from a certain point. Hence there exists no > M and an integer u such that 2an - 1- k(2n - 1) = u x ) xk(1±.Au±( 1 )k(i ' for all n > no. Thus g - = 2 holds for all x of the form (1+ 0)2n-1. Because this equality between two rational functions holds for infinitely many values of the argument, it follows that it is actually true for all x. By looking at the leading coefficient in both sides of the equality (after multiplication by X k) we deduce that (1+ -\/)"-' is rational, which cannot hold unless u = 0. Thus The expression in the right-hand side of the last equality is a polynomial with integer coefficients only for odd values of k. This also gives the expression of f: 2 N/X2 + 1 The solutions of the original problem are easily deduced from f and g by a translation. The previous example deserves a little digression. Actually, one can find all polynomials with real coefficients that satisfy (X2 + 1)f(X)2 = g(X)2 + 1. Indeed, it is clear that f and g are relatively prime. By differentiation, the last relation can be written as (X2 + 1)f(X)f (X) + X f 2(X) = g(X)g'(X). Thus f divides gg', and by Gauss's lemma we deduce that fig'. The relation (X2 + 1)f2(X) = g(X)2 + 1 also shows that deg(f) = deg(g') and so there exists a constant k such that f (X) = kg' (X). Therefore k2(X2 + 1)g'(X)2 = g(X)2 + 1. By identifying the leading coefficient of g in the two sides, we im-mediately find that k2 = n 2. This shows that ig:g (()()2 = i±x2 . By changing g and -g we may assume that g'(X) > 0 for sufficiently large x and thus for such values of the variable we have ,gi(x) - . This shows that the V/9(x)2+1 x2+1 f (X) = (X + -VX2 +1)k + (-X + VX2 +1)k THEORY AND EXAMPLES 391 function In 9(x)+V9(x)2+1 is constant in a neighborhood of infinity. This (x+,/x2+1),, allows us to find g in such a neighborhood and thus to find g on the whole real line. It is time now for the last problem, which is, as usual, very hard. We do not exaggerate when we say that the following problem is exceptionally difficult. Example 14. Let a and b be integers greater than 1 such that an – 1 bn – 1 for every positive integer n. Prove that b is a natural power of a. [Marius Cavachi] AMM Solution. This time we will be able to find the right convergent sequence only after examining a few recursive sequences. Let us see. So, initially we are ( given that there exists a sequence of positive integers (xn 1) )n>i such that xn (1) =- n bn — 1 . Then, 41) — b for large values of n. So, we could expect that an – 1 a ( the sequence (xn 2) )n>l, 4 2) = bx$, 1) – axS,111, to be convergent. Unfortunately, bn+1 (a – 1) – 0/11+1 (b – 1) a – b (an – 1)(0+1 - 1) which is not necessarily convergent. But... if we look again at this sequence, we see that for large values of n it grows like ( — b , so much slower. And a2 this is the good idea: repeat this procedure until the final sequence behaves like a k-I- b 1 , where k is chosen such that ak < b < ak+1. Thus the final ( sequence will converge to 0. Again, the hard part has just begun, since we have to prove that if we define xS., i ,+1) = bx, ) – aix() +1 then lim x ik+1) = 0. n—>oo This is not easy at all. The idea is to compute xn (3) and after that to prove 392 17. AT THE BORDER OF ANALYSIS AND NUMBER THEORY (i the following statement: for any i > 1 the sequence (x ) )rt>1 has the form cibn + • • • + clan + co (an+i-1 — 1)(an+i-2 — 1) ... (an — 1) for some constants co, ci, , ci. Proving this is not so hard, the hard part was to think of it. How can we prove the statement other than by induc-tion? And induction turns out to be quite easy. Supposing that the state-ment is true for i, then the corresponding statement for i 1 follows from 4+1) -= bxW — aix(i) ±1 directly (note that in order to compute the difference, we just have to multiply the numerator cibn ci_ a (i-1)n c an + co by b and an+i — 1. Then, we proceed in the same way with the second fraction and the term bn±lan+i will vanish). So, we have found a formula which shows that as soon as ai > b we have lim x, i) = 0. So, lim xi lc+1) = 0. Another n—>oo n—>oo step of the solution is to take the minimal index j such that lim 4) = 0. n—>oo Clearly, j > 1 and the recursive relation x i ,+1) = bx,•: ;) — aix() +1 shows that 4j) E Z for all n and i. Thus, there exists an M such that whenever n > M we have xn (i) = 0. This is the same as bxSij 1) = aixn (3 ± . 1 1) for all n > M, which n—M implies x2-11 = ) Xm Ci ') for all n > M. Let us suppose that b is not ( ai a multiple of a. Because ( ) n—M Xm (i-1) E Z for all n > M, we must have a3 u-1) xm (i-1) = 0 and so xn = 0 for n > M, which means lim x j-1) = 0. But n—>oo this contradicts the minimality of j. Thus we must have alb. Let us write b = ca. Then, the relation an —11bn — 1 implies an — lIcn — 1. And now we are finally done. Why? We have just seen that an — 11cn — 1 for all n > 1. But our previous argument applied to c instead of b shows that alc. Thus, c = ad and we deduce again that ald. Since this process cannot be infinite, b must be a power of a. It is worth saying that an even stronger result holds: it is enough to suppose that an — 1 Ibn 1 for infinitely many n. But this is a much more difficult problem and it follows from a 2003 result of Bugeaud, Corvaja and Zannier: THEORY AND EXAMPLES 393 If integers a, b > 1 are multiplicatively independent in Q (that is logo b cl Q or an bm for n, m 0), then for any 6 > 0 there exists no = no(a, b, E) such that gcd(an — 1, bn — 1) < 2" for all n > no. Unfortunately, the proof is too advanced to be presented here. 394 17. AT THE BORDER OF ANALYSIS AND NUMBER THEORY 17.2 Problems for training 1. Let (an)n>1 be an increasing sequence of positive integers such that an lai + a2 + • • • + an_i for all n > 2002. Prove that there exists no such that an = al + a2 + • • • + an_i for all n > no. Tournament of the Towns 2002 2. Let f E Z[X] be a polynomial of degree k such that /f (n) E Z for all n. Prove that there exist integers a and b such that f (x) = (ax + b)k . 3. Find all arithmetical sequences (an)n>i of positive integers (an)n>1 such that al + a2 • + an is a perfect square for all n > 1. Laurentiu Panaitopol, Romanian Olympiad 1991 4. Prove that any infinite arithmetical sequence contains infinitely many terms that are not perfect powers. 5. Let a, b, c > 1 be positive integers such that for any positive integer n there exists a positive integer k such that ak b k = 2cn. Prove that a -=- b. 6. Let p be a polynomial with integer coefficients such that there exists a sequence of pairwise distinct positive integers (an)n>i such that p(ai) = 0, p(a2) = al, p(a3) = a2, .... Find the degree of this polynomial. Tournament of the Towns 2003 7. Find all pairs (a, b) of positive integers such that an + b is triangular if and only if n is triangular. After a Putnam Competition problem PROBLEMS FOR TRAINING 395 8. Let a and b be positive integers such that for any 71, the decimal repre-sentation of a + bn contains a sequence of consecutive digits which form the decimal representation of n (for example, if a = 600, b = 35, n = 16 we have 600 + 16 • 35 = 1160). Prove that b is a power of 10. Tournament of the Towns 2002 9. Let a and b be integers greater than 1. Prove that for any given k > 0 there are infinitely many numbers n such that co(an + b) < kn, where ( i9 is the Euler totient function. Gabriel Dospinescu 10. Let A, B be two finite sets of positive real numbers such that { EXn171EN}C{EXTh 171ENT}. xEA xEB Prove that there exists a k ER such that A= {xi' 1 x E B}. Gabriel Dospinescu 11. Suppose that a is a positive real number such that all numbers 1', 2a, 3a, . . . are integers. Prove that a is also integer. Putnam Competition 12. Find all a, b, c such that a • 4n + b • 6n + c • 9n is a perfect square for all sufficiently large n. Vesselin Dimitrov 396 17. AT THE BORDER OF ANALYSIS AND NUMBER THEORY 13. Let f and g be two real polynomials of degree 2 such that for any real number x, if f (x) is integer, then so is g(x). Prove that there are integers m, n such that g(x) = m f (x) + n for all x. Bulgarian Olympiad 14. Let b be an integer greater than 4 and define the number in base b. Prove that xn and only if b = 10. xn -,--- 11 22 . .. 2 5 ....„—.,....,--, n-1 n is a perfect square for all sufficiently large n if Laureniiu Panaitopol, IMO 2004 Shortlist 15. Let A be a set of positive integers containing at least one number among any 2006 consecutive positive integers, and let f be a nonconstant poly-nomial with integer coefficients. Prove that for sufficiently large n there are at least On In n different primes dividing the number ri f (k). 1<k1 of positive integers satis-fying an < 100n for all rt, one can find infinitely many terms containing at least 1986 consecutive l's. Kvant 17. Find all triplets (a, b, c) of integers such that a • 2n + b is a divisor of cn + 1 for any positive integer n. Gabriel Dospinescu, Mathematical Reflections PROBLEMS FOR TRAINING 397 18. Let f be a complex polynomial such that for all positive integers n, the equation f (x) = n has at least one rational solution. Prove that f has degree at most 1. Mathlinks Contest 19. Let f be a polynomial with rational coefficients such that f(2n) is a perfect square for all positive integers n. Prove that there exists a poly-nomial g with rational coefficients such that f = g2 . Gabriel Dospinescu 20. Suppose that b1, b2, , bm, are rational numbers and bo, b_1, b_2, ... are real numbers such that the series binzm + • • • + biz + b0 + _z1 622 converges outside some circle and takes integral values for infinitely many integers z. Prove that b0 is rational and bi = 0 for all i < 0. Skolem 21. a) Let b1, b2, bm, and b0, b_1, b 2, ... be real numbers such that the ± • • • ± biz ± bo _ bv bz -22 is series f (z) = brnen + not everywhere divergent and represents integers for all sufficiently large integers z. Prove that f (z) is a polynomial. b) Deduce that a polynomial f with the properties that f(Z) C Z and f (n) is a k-th power of an integer for all sufficiently large integers n is the k-th power of a polynomial with rational coefficients. 22. a) Find all increasing functions defined on the set of positive integers, with real values and such that f (ab) = f (a) f (b) for all a and b. b) The same questions if we assume only that f (ab) = f (a) f (b) for all relatively prime positive integers a and b. Paul ErdOs 398 17. AT THE BORDER OF ANALYSIS AND NUMBER THEORY 23. Let f, g be two polynomials with real coefficients such that f (Q) = g (Q) Prove that there exist rational numbers a, b such that f (X) = g (aX b). Miklos Schweitzer Competition 24. Let al, a2, an and b1, b2, bm be positive integers such that any inte- ger x satisfies at least one congruence x az (mod bi) for some i. Prove that there exists a nonempty subset I of {1, 2, ..., n} such that EzE/ b is an integer. M. Zhang 25. Suppose that f, g are two nonconstant rational functions such that if f (zo) is integer for some complex number zo then so is g(zo). Show that there exists a polynomial with rational coefficients P such that g(z) = P(f (z)). (This is for the die-hards!) Gary Gundersen, Steve Osborn, AMM 6410 THEORY AND EXAMPLES 401 18.1 Theory and examples For a prime p, define the function (— a P ) : Z —> {-1,1} by (— a P) = 1 if the ( equation x2 = a has at least one solution in Z/pZ and a) = —1 otherwise. P In the first case, we say that a is a quadratic residue modulo p; otherwise we say that it is a quadratic non-residue modulo p. This function is called Legen-dre's symbol and plays a fundamental role in number theory. We will unfold some easy properties of Legendre's symbol first, in order to prove a highly non-trivial result, Gauss's famous quadratic reciprocity law. First, let us present a ( useful theoretical (but not very practical) way of computing a ) due to Euler. P Theorem 18.1. The following identity is true provided p a: In particular, we have (a —) a 2 = (-1)2V. (mod p). Proof. We will prove this result and many other simple facts concerning quadratic residues in what follows. First, let us assume that (— a = 1, and let x be a P solution to the equation x2 = a in Z/pZ. Using Fermat's little theorem, we ( find that a Y = xP-1 = 1 (mod p). Thus the equality — a = a Y (mod p) P holds for all quadratic residues a modulo p. In addition, for any quadratic residue we have a Y = 1 (mod p). Now, we will prove that there are exactly p — 1 quadratic residues in Z/pZ \ {0}. This will enable us to conclude that 2 quadratic residues are precisely the zeros of the polynomial X p-1 2 -1 and also that non quadratic residues are exactly the zeros of the polynomial X 2 + 1 (from Fermat's little theorem). Note that Fermat's little theorem implies that 402 18. QUADRATIC RECIPROCITY the polynomial XP-1 - 1 = (X p 2 1 - 1)(X y 2 1 + 1) has exactly p - 1 zeros in the field Z/pZ. But in a field, the number of different zeros of a polynomial cannot exceed its degree. Thus each of the polynomials X p 2 -1 and XY +1 has at most p 2 1 zeros in Z/pZ. These two observations show that in fact - each of these polynomials has exactly p 2 1 zeros in Z/pZ. Let us observe next that there are at least p 2 1 quadratic residues modulo p. Indeed, all numbers i2 (mod p) with 1 < i < P 2 1 are quadratic residues and they are - all different (modulo p). This shows that there are exactly p 2 1 quadratic residues in Z/pZ \ {0} and also proves Euler's criterion. K Euler's criterion is a very useful result. Indeed, it allows a very quick proof of the fact that (-a) : Z {-1, 1} is a group morphism. Indeed, (ab — p ) (ab) 2 = a 2 = (-) (-) P P (mod p). The relation (— ab = a b ) P P shows that while studying Legendre's sym- bol, it suffices to focus on the prime numbers only. Also, the same Euler's criterion implies that (-a) = (-b) whenever a b (mod p). It is now time to discuss Gauss's celebrated quadratic reciprocity law. First of all, we will prove a lemma (also due to Gauss). Lemma 18.2. Let p be an odd prime and let a E Z such that gcd(a, p) = 1. De-fine the least residue of a (mod n) as the integer a' such that a a' (mod n) and -3 < a' < Let a j be the least residue of aj (mod p) and I be the number of integers 1 < j < P21 for which aj < 0. Then (7 )) = (-1)1. THEORY AND EXAMPLES 403 Proof The proof is not difficult at all. Observe that the numbers la3 I for 1 < j < r = P21 are a permutation of the numbers 1,2, ..., r. Indeed, we have 1 < la31 < r and la31 laid (otherwise, we have either pla(j + k) or pla(j — k) which is impossible because gcd(a, p) = 1 and 0 < j + k < p). Therefore a1a2 • • • ar = (-1)1lailla21• • • larl = (-1)1r!. By the definition of a3 we also have a1a2 • • ar arr! (mod p) and so ar (-1)/ (mod p). Using Euler's criterion, we deduce that ((,) = (-1)/. K Using Gauss's lemma, the reader will enjoy the proof of the following classical results. Theorem 18.3. The identity (-2) = (-1) P28 8 holds for any odd prime p. p Proof. Let us take a = 2 in Gauss's lemma and observe that 1 = P21 . Indeed, we have a3 = 2j if 1 < j < Hi and a3 = 2j — p if H i] < j < P21. Now, the conclusion follows, because 1 = P21 [land P 2-8 1 have the same 4 parity, as you can easily check. El But perhaps the most striking consequence of Gauss's lemma is the famous: Theorem 18.4 (Quadratic reciprocity law). For all distinct odd primes p and q, the following identity holds: (1 1 ( P g) = ( — 1)Y Proof The proof is a little bit more involved than that of the previous result. Consider R the rectangle defined by 0 < x < 2 and 0 < y < 2 , and let = (-1)/ and (1) = (-1)m, where 1, m are defined as in Gauss's lemma. Observe that 1 is the number of lattice points (x, y) such that 0 < x < z and < px — qy < 0. These inequalities force y < P+ 21 and because y is an 404 18. QUADRATIC RECIPROCITY integer, it follows that y < 2. Therefore 1 is the number of lattice points in R that satisfy —1 < px — qy < 0 and similarly m is the number of lattice points in R that satisfy —5 < qy — px < 0. Using Gauss's lemma, it is enough to prove that (P-14 (q-1) (1 + m) is even. Because (73 1) 1q 1) is the number of lattice points in R, (P-1) 4 (q-1) (1 + m) is the number of lattice points in R that satisfy px — qy < —1 or qy — px < — 22 These points determine two regions in R, which are clearly disjoint. Moreover, they have the same number of lattice points because x =- 21 x', y = P+ 21 y' gives a one-to-one correspondence between the lattice points in the two regions. This shows that (P-1) q-1) (1 + m) is even and finishes the proof of this celebrated theorem. Using this powerful arsenal, we are now able to solve some interesting prob-lems. Most of them are merely direct applications of the above results, but we think that they are still worthy, not necessarily because they appeared in various contests. Example 1. Prove that the number 2n. + 1 does not have prime divisors of the form 8k — 1. Vietnamese TST 2004 Solution. For the sake of contradiction, assume that p is a prime of the form 8k — 1 that divides 2' + 1. Of course, if n is even, the contradiction is imme-diate, since in this case we have —1 (23)2 (mod p) and so —1 = (-1)Y 1. Now, assume that n is odd. Then —2 (2n ± 1 )2 (mod p) and ( 2) = 1. This can be also written in the form — 1 2 1, or P P P ”2 i (-1)2=4- = 1. But if p is of the form 8k — 1 the latter cannot hold and this is the contradiction that solves the problem. THEORY AND EXAMPLES 405 Using the same idea and a bit more work, we obtain the following result. Example 2.] Prove that for any positive integer n, the number 23n + 1 has at least n prime divisors of the form 8k + 3. [Gabriel Dospinescu] Solution. Using the result of the previous problem, we deduce that 2n + 1 does not have prime divisors of the form 8k + 7. We will prove that if n is odd, then it has no prime divisors of the form 8k + 5 either. Indeed, let p be a prime divisor of 2n +1. Then 2" 1 —1 (mod p) and so —2 (2 n+1 ) 2 (mod p) . Using the same argument as the one in the previous problem, we deduce that p 8 2 — 1 2 p — 1 is even, which cannot happen if p is of the form 8k + 5. Now, let us solve the proposed problem. We assume n > 2 (otherwise the verification is trivial). The essential observation is the identity 23n + 1 = (2 + 1)(22 — 2 + 1)(22.3 — 23 + 1) (22.3n — 23n 1 + 1) Now, we prove that for all 1 < i < j < n-1, gcd(22•3' —23' +1, 22•33 —233 +1) = 3. Indeed, assume that p is a prime number dividing gcd(22•3' — 23' + 1, 22.33 — 23' + 1) We then have p1232+' + 1. Thus, 2 (23'41)33-i-1 = (-1)33-'-1 -= —1 (mod p), implying 0 = 22'33 — 233 + 1 1 — (-1) + 1 = 3 (mod p) This cannot happen unless p = 3. But since v3(gcd(223" — 23' + 1, 22.33 — 233 + 1)) = 1, as you can immediately check, it follows that gcd(22.32 — 23' + 1, 22.33 — 233 + 1) = 3 406 18. QUADRATIC RECIPROCITY and the claim is proved. It remains to show that each of the numbers 22'3' — 23' + 1, with 1 < i < n — 1 has at least a prime divisor of the form 8k + 3, different from 3. From the previous remarks, it will follow that 23" + 1 has at least n — 1 distinct prime divisors of the form 8k + 3, and since it is also divisible by 3, the solution will be complete. Fix i E {1, 2, ... , n — 1} and observe that any prime factor of 22'3' — 23' +1 is also a prime factor of 23' + 1. Thus, from the first remark, this factor must be of one of the forms 8k + 1 or 8k + 3. Because v3(22°3' — 23' + 1) = 1, all prime divisors of 22•3' — 23' + 1 except for 3 are of the form 8k + 1, so 22'3' — 23' + 1 -=- 8 (mod 8), which is clearly impossible. Thus at least a prime divisor of 22.3" — 23' + 1 is different from 3 and is of the form 8k+3. The claim is proved and the conclusion follows. We have seen a beautiful proof of the following result in the chapter Geom-etry and Numbers. But there is another way to solve it, probably more natural and which turns out to be very useful in some other problems, too: Example 3. Let n be a positive integer such that the equation X2-Exy-Ey2 = n has a solution in rational numbers. Prove that this equation also has a solution in integers. Komal Solution. This looks quite familiar, especially after the discussion in chapter Primes and Squares. Indeed, let us start with a natural question: which primes can we expressed in the form x2 + xy + y2 for some integers x, y? Suppose p is such a prime number. Then 4p = (2x + y)2 + 3y2. This shows that (2x + y)2 = —3y2 (mod p). Now, if p # 3 then y 0 (mod p) because otherwise x 0 (mod p) and so p21p, clearly false. The last relation implies therefore that () = 1. Using the quadratic reciprocity law, we easily infer that this is equivalent to (5) = 1 and this happens precisely when p 1 (mod 3). Therefore the primes that can be expressed as x2 + xy + y2 are 3 and p 1 (mod 3). We are not done yet, because we need to prove that all such primes can be written like that. For 3, there is no problem, but this is not THEORY AND EXAMPLES 407 the case with arbitrary p 1 (mod 3). Take such a prime number p. From the above arguments we know that (=1 1, = 1, which means that there exists a such that a2 = —3 (mod p). Now, recall Thue's lemma proved in chapter Primes and Squares: there exist integers 0 < x, y < 05 not both zero such that a2x2 y 2 (mod p). Therefore pl3x2 + y2. Because 0 < 3x2 + y2 < 4p, we deduce that 3x2 + y2 is one of the numbers p, 2p, 3p. If it is p, then we obtain p = (y — x)2 + (y — x) • 2x + (2 • x)2. If it is 3p then y must be a multiple of 3, say y = 3z and then p = x2 + 3z2, thus we get the previous case. Finally, suppose that 2p = x2 + 3y2. Then clearly x, y have the same parity. But then x2 + 3y2 is a multiple of 4, contradiction, because 2p is not divisible by 4. Thus this case is excluded and the proof of the first part is finished. Now, we can attack the problem. Suppose that the equation x2 + xy + y2 = n has rational solutions, that is the equation a2 + ab + b2 = c2n has integer solutions with gcd(a, b, c) = 1. Take p a prime divisor of n and assume that vp(n) is odd. We claim that p 3 or p 1 (mod 3). If not then pl a and plb by the previous arguments, thus we can simplify by p2 both members of the equation. Repeating this operation, we deduce in the end that plc, which contradicts the fact that gcd(a, b, c) = 1. Thus all prime divisors of the form 3k + 2 of n appear with even exponent. As we have already seen, all prime divisors of n not of the form 3k + 2 are of the form u2 + uv + v2. Thus, all we need to prove now is that the product of two numbers of the form u2 + uv + v2 is of the same form. But this is not difficult, because if f = cm- then (u2 + uv + v2)(w2 + wt + t2) = (u — ev)(u — € 2 v) (w — et) (w — E2t) that is (A — €B)(A — €2B) for A = uw — vt,B = ut + vw + vt and we are done. If you did find the above solution cumbersome, you are right! At first glance, the following problem seems trivial. It is actually very tricky, because brute force takes us nowhere. Yet, in the framework of the above results, this should not be so difficult. Example 4d Find a number n between 100 and 1997 such that On + 2. APMO 1997 408 18. QUADRATIC RECIPROCITY Solution. We will fail if we try to search for odd numbers (actually, this result was proved in the topic Look at the Exponent! and is due to Schinzel). So let us search for even numbers. The first attempt is to chose n = 2p, for some prime p. Unfortunately, this choice is ruled out by Fermat's little theorem. So let us set n = 2pq, for some different primes p and q. We need pql22Pq-1 + 1 and so we must have (--2) = = 1. Also, using Fermat's little theo-rem, p122q-1 + 1 and ql22P-1 + 1. A simple case analysis shows that q = 3, 5, 7 are not good choices, so let us try q = 11. We find p = 43 and so it suffices to show that pq122N-1 + 1 for q = 11 and p = 43. This is not very hard: we have p122q-1 + 1, implying p12P(2q-1) + 1 = 22pq-p 1. Then p122P" + 2P-1 and using Fermat's theorem (pl2P-1 - 1) we get p122P" + 1 and an analogous reasoning shows that q122P" + 1, finishing the proof. Are we wrong to present the following example? It apparently has no connec-tion with quadratic reciprocity, but let us take a closer look. 7 5ca.m ---57 Let f, g : N N be functions with the properties: i) g is surjective; ii) 2f(n)2 = n2 + g \ 2 m for all positive integers n; iii) f (n) - n < 2004 \Fri for all n. Prove that f has infinitely many fixed points. [Gabriel Dospinescu] Moldova TST 2005 Solution. Let pn be the sequence of prime numbers of the form 8k + 3 (the fact that there are infinitely many such numbers is a trivial consequence of Dirichlet's theorem, but we invite the reader to find an elementary proof). It is clear that for all n we have C2 / ) = Pn Using the condition i) we can find xn, such that g(xn) = pn, for all n. It follows that 2f (xn)2 = xn 2 +p n 2 , which yields 2 f (xn)2 x n 2 (mod pa). Because exist sequences of positive integers an, bn such that xn = a npn and f (xn) = bnpn for all n. Clearly, ii) implies the relation 2b?, = an + 1. Finally, using the property If (n) – n1 < 2004Iii we have ( 2 pn ) Pni f (xn). Thus there = –1, the last congruence shows that pn Ix, and f (xn) xn That is Va2 + 1 lim n n–>oo an bn — – 1 an = 2004 .Vxn THEORY AND EXAMPLES 409 The last relation implies lim an =- 1. Therefore, starting from a certain point, n—too we have an = 1 = bn, that is f (Pn) = Pn and the conclusion follows. We continue with a difficult classical result that often proves very useful. It characterizes the numbers that are quadratic residues modulo all sufficiently large prime numbers. Of course, perfect squares are such numbers, but how to prove that they are the only ones? Actually, this result has been extensively generalized, but all proofs are based on class field theory, a difficult series of theorems in algebraic number theory, that are far beyond the scope of this elementary book. ( Suppose that a is a non-square positive integer. Then – a = P –1 for infinitely many prime numbers p. Solution. One may assume that a is square-free. Let us write a = 2'qi q2 • • • qn, where q, are different odd primes and e E {0,1}. Let us assume first that n > 1 (that is a 2) and consider some odd distinct primes ri, r2, ... , rk, each of them different from qi, q2, . . . , qn. We will show that there is a prime p, differ-ent from r1, r2, ... , rk, such that – a (= –1. Let s be a quadratic non-residue P modulo qn. 410 18. QUADRATIC RECIPROCITY Using the Chinese remainder theorem, we can find a positive integer b such that b-1 (mod ri), 1 < i < k b = 1 (mod 8), b-1 (mod qi), 1 < i < n b s (mod qn). Now, write b = /31 • p2 • • • pm, with pi odd primes, not necessarily distinct. Using the quadratic reciprocity law, it follows that rn 2 ( P 2 Pi -1 62-1 1(-1) 8 = ( -1) 8 = 1 i ) = i=i and nz m (,) 11( 3.)v 5 y . ( pi) ( (b) (b qi) j=1 for all i e {1, 2, ... ,n}. Hence e n m (pi) = n (19 )1 rin (g=i 19 i=1 j=1 = Fr n z qi ) qn ( ) qn ) '- 1 (We used the following observations in the above equalities: for any odd num-bers b1, , bm, if b = bib2 • • bm then the numbers bi -1 b2 — 1 8 8 E b i ;1 b — 1 2 i=1 i=1 and THEORY AND EXAMPLES 411 are even. We leave to the reader this easy exercise, which can be handled by induction for instance.) ( Thus, there exists i E {1, 2, ... , m} such that — a = —1. Because b -,- 1 pi (mod rz), 1 < i < k, we also have pi E {1, 2, ... } \ {ri, r2, ... , rk} and the claim is proved. The only case left is a = 2. But this is very simple, since it suffices to use p 8 2 — 1 Dirichlet's theorem to find infinitely many primes p such that is odd. As in other units, we will now focus on some special cases. This time it is a problem almost trivial with the above framework but seemingly impossible to solve otherwise (we say this because there is a beautiful, but very difficult, solution using analytical tools, which we will not present here). Example 7. Suppose that al, a2, . . a2004 are nonnegative integers such that ay + a2+ • • • + a2004 is a perfect square for all positive in-tegers n. What is the least number of such integers that must equal 0? [Gabriel Dospinescu] Mathlinks Contest Solution. Suppose that al, a2, ... , ak are positive integers such that ay. +a2+ • • • +ar k ' is a perfect square for all n. We will show that k is a perfect square. In ( order to prove this, we will use the above result and show that — k = 1 for all P sufficiently large primes p. This is not a difficult task. Indeed, consider a prime p, greater than any prime divisor of aia2 ... ak. Using Fermat's little theorem, p-1 p-1 p-1 ___ 7 p-1 p-1 p-1 • al + a 2 + • • • + ak = ic (mod p), and since al + a 2 + • • • + ak is a ( perfect square, it follows that k ) = 1. Thus k is a perfect square. And now P the problem becomes trivial, since we must find the greatest perfect square less than 2004. A quick computation shows that this is 442 = 1936 and so the desired minimal number is 68. 412 18. QUADRATIC RECIPROCITY Here is another nice application of this idea. It is adapted after a problem given at the Saint Petersburg Olympiad. Actually, much more is true: Con-sider f a monic polynomial with integer coefficients, irreducible over Q and having degree greater than 1. Then there are infinitely many prime numbers p such that f has no root modulo p. For a proof of this result using (the difficult) Chebotarev's theorem and an elementary theorem of Jordan, as well as for many other aspects of this problem, the reader can consult Serre's beau-tiful paper On a theorem of Jordan, Bull.A.M.S 40 (2003). Suppose that f E Z[X] is a second degree polynomial such that for any prime p there is at least one integer n for which pl f (n). Prove that f has rational zeros. Solution. Let f (x) = ax2 + bx + c be this polynomial. It suffices to prove that b2 — 4ac is a perfect square. This boils down to proving that it is a quadratic residue modulo any sufficiently large prime. Pick a prime number p and an integer n such that pl f (n). Then b2 — 4ac (2an b)2 (mod p) and so (b2 — 4ac) = 1. This shows that our claim is true and finishes the solution. Some of the properties of Legendre's symbol can also be found in the following problem. Example 9.] Let p be an odd prime and let f (x) = THEORY AND EXAMPLES 413 a) Prove that f is divisible by X — 1 but not by (X — 1)2 if and only if p 3 (mod 4); b) Prove that if p 5 (mod 8) then f is divisible by (X — 1)2 and not by (X — 1)3. [CalM Popescu] Romanian TST 2004 Solution. The first question is not difficult at all. Observe that p-1 f (1) = ( P = 0 — by the simple fact that there are exactly P 2 1 quadratic residues modulo p — and P 2 1 quadratic non-residues in {1, 2, p — 1}. Also, f(i) = (i _ 1) () = Ei p-1 . p-1 i=1 i=1 because f (1) = 0. The same idea of summing up in reversed order allows us to write: p-1 Ei z i) (P i) P i=i i=i P-1 p-1 — ( — (-1) 2 p=1. f'(1) i=i (we used again the fact that f (1) = 0). Hence for p 1 (mod 4) we must also have 1(1) = 0. In this case f is divisible by (X — 1)2. On the other hand, if p 3 (mod 4), then p-1 p-1 . f'(1) = i P p-1 i=1 P(P — 1) 1 (mod 2) 2 p-1 p-1 (2i — 1)2 (2i — 1) t (2i — 1) i=1 i=1 (mod 8). 414 18. QUADRATIC RECIPROCITY and so f is divisible by X — 1 but not by (X — 1)2. The second question is much more technical, even though it uses the same main idea. Observe that p-1 p-1 p-1 f"(1) = E(i2 — 3i + 2) (—) — E i2 ( i) — — P i=i i=i (P z (once again we used the fact that f (1) = 0). Observe that the condition p 5 (mod 8) implies, by a), that f is divisible by (X — 1)2, so actually p-1 f"(1) = i2 (:) Let us break this sum into two pieces and treat each of them independently. We have p-i p-1 2 2 2 02 = 4 C) Ei ( E(2 i=1 i=1 Note that P-1 P-1 2 2 2 „2 E i2 (p i ) >2 , i2 =Ei F 1 (mod 2), i=i i=i i=i 8 SO p-1 E(202 ( 2i) +4 (mod 8) i=1 (actually, using the fact that (-2) = (-1), we obtain that its value is —4). On the other hand, THEORY AND EXAMPLES 415 If we prove that the last quantity is a multiple of 8, then the problem will be solved. But note that f (1) = 0 implies P-1 p-1 o = E (2i) +._, (2i 1) i=i P P i=i ) Also, (ft (p — 1 P-1 P-23 2 2 r: — 1 + E i=i i=i (13) — 1 ± E i=i( p-3 p-1 + (2i + 1 2 (2i — 1) P P i=i J i=i Therefore (2i — 1 ) = 0 and the problem is finally solved. P There are more than 100 different proofs of the quadratic reciprocity law, each of them having a truly beautiful underlying idea. We decided not to present the proof using Gauss sums, which is probably the shortest one, as it needs some preparations concerning finite fields and their extensions. Instead, we present the following proof, which greatly simplifies the approach of V.A. Lebesgue. Example 10.1 Let p and q be distinct odd primes. Prove that the equation x2 1 x2 2 ± x3 2 x4 2 ± xp 2 = has qP-1 + q p1 2 solutions in (Z/qZ)P. Deduce a new proof of the quadratic reciprocity law. [Wouter Castryck] 416 18. QUADRATIC RECIPROCITY Solution. For an odd number n let us define Nn, the number of solutions of the equation xi — 4 + x3 — xi + + xn 2 = 1 in (Z/qZ)n. By replacing xi with xi + x2 we obtain an equation with the same number of solutions: Xj . — • • • + Xn 2 — 1 = —2X1X2. There exist two kinds of solutions of this last equation: those in which xi 0 and those in which x1 = 0. The first case is very easy, because for any choice of x1 0 and any choice of x3, ..., xn there is precisely one x2 such that (xi, x2, xn) is a solution. Thus the first case gives qn 2 (q — 1) solutions of the equation. The second case is even easier, because the equation reduces to the corresponding one for n — 2, so this second case gives qNn_2 new solutions (the factor q comes from the fact that any solution of 2 X3 — X4 + • • • Xn 2 -= gives q solutions of + — • • • + Xn 2 — 1 = —2X1X2) x2 being arbitrary). Therefore Nn = qn-2(q — 1) + On-2 and an immediate induction shows that .Nn = qn-1 + gn 2 1 . The first part of the problem is now clear. It is pretty clear that Np can be written as Np= (1+ (61-)) • (1+ 2 )) • • • (1 + ai±a2-1-•••-1-ap=1 because the equation x2 = a has 1 + (P) solutions in Z/pZ by definition of Legendre's symbol. On the other hand, imagine that we develop each product in the previous sum and collect terms. There will be a contribution of qP-1 THEORY AND EXAMPLES 417 coming from 1 (because there are qP-1 solutions of the equation al + a2 + • • • + ap = 1) and another contribution coming from the last product, namely ( (-1)Y E (alai • • ap). q ai+.••+ap=1 All other contributions are zero because EX (P) = 0. Thus N = qP- + ai+.••-fap=1 Now, those p-tuples (al, az, ap) with al + a2 + • • • + ap = 1 and not all ai equal to p-1 can be collected in groups of size p and so, modulo p, the last quantity equals 1 + (( 1) q (Pq P ), which reduces to 1 + (-1)x21.921 • (2) (everything is taken mod p). On the other hand, the explicit value of NP ob- tained in the first part shows that NP is congruent to 1+ (p 1) modulo p. Thus the two quantities must be equal modulo p, and since their values are -1 or 1 they are actually equal, which implies the quadratic reciprocity law. Finally, a difficult problem. Find all positive integers n such that 2' - 1I3" - 1. [J. L. Selfridge] AMM Solution. We will prove that n = 1 is the only solution to the problem. Sup-pose that n > 1 is a solution. Then 2n — 1 cannot be a multiple of 3, hence n is odd. Therefore, 2' = 8 (mod 12). Because any odd prime different from 3 is of one of the forms 12k ± 1 or 12k ± 5 and since 2n — 1 = 7 (mod 12), it follows that 2n — 1 has at least a prime divisor of the form 12k ± 5, call it ( p. We must have 3 = 1 (since 3n 1 (mod p) and n is odd) and using P the quadratic reciprocity law, we finally obtain (3) = (-1) 2 . On the other ((-1)Y) E 418 18. QUADRATIC RECIPROCITY hand, (12) = (— ±2) = —(±1). Consequently, —(±1) = (-1)Y = +1, which 3 3 is the desired contradiction. Therefore the only solution is n = 1. PROBLEMS FOR TRAINING 419 18.2 Problems for training 1. Let p = 2 (mod 3) be a prime number Prove that the equation xl i) . + 4 + • • • + + 1 = (xi + x2 + • • xn)2 has no integer solutions. Laurentiu Panaitopol, Gazeta Matematica 2. Let xi = 7 and xn±i = 2xn 2 — 1, for n > 1. Prove that 2003 does not divide any term of the sequence. Valentin Vornicu, Mathlinks Contest 3. Prove that for any odd prime p, the least positive quadratic non-residue modulo p is smaller than 1 + 4. Prove that the number 3Th + 2 does not have prime divisors of the form 24k + 13. Laurentiu Panaitopol, Gazeta Matematica 5. Let k = 22n + 1 for some positive integer n. Prove that k is a prime if and only if k is a factor of 3—T- 1 + 1. Taiwanese Olympiad 1997 6. What is the number of solutions to the equation a2 + b2 = 1 in Z/pZ x Z/pZ? What about the equation a2 — b2 = 1? 7. Find all prime numbers q such that 19931(q — 1)4 + 1. Serban Nacu, Gazeta Matematica 420 18. QUADRATIC RECIPROCITY 8. Let p —1 (mod 8) be a prime number and let m, n be positive integers such that .VP > 7 7 1. Prove that Vfo > + Radu Gologan 9. Let a and b be integers relatively prime to an odd prime p. Prove that p-1 . + bi) — (a z=1 (az2 10. Let p be a prime of the form 8k + 7. Evaluate the following sum 2 1 L p 2] . k=1 Calin Popescu, AMM 11. Let A be the set of prime numbers dividing at least one of the numbers 2n2+1 — 3n. Prove that both A and N\ A are infinite. Gabriel Dospinescu 12. Let p be a prime number. Prove that the following statements are equiv-alent: i) there is a positive integer n such that pin2 — n + 3; ii) there is a positive integer m such that plm2 — in + 25. Polish Olympiad 13. Let p be a prime of the form 4k + 1. Evaluate 2k p2] 2 [k ,:j —1 L PROBLEMS FOR TRAINING 421 14. Suppose that p is an odd prime and that A and B are two different non-empty subsets of {1, 2, ... ,p — 1} for which i) A U B = {1,2,...,p — 1}; ii) If a, b are both in A or both in B, then ab (mod p) E A; iii) If a E A, b E B, then ab E B. Find all such subsets A and B. Indian Olympiad 15. Let m, n be integers greater than 1 with n odd. Suppose that n is a quadratic residue mod p for any sufficiently large prime number p —= —1 (mod 2'). Prove that n is a perfect square. Ron Evans, AMM E 2627 16. Let a, b, c be positive integers such that b2 — 4ac is not a perfect square. Prove that for any n > 1 there are n consecutive positive integers, none of which can be written in the form (ax2 + bxy + cy2)z for some integers x, y, z with z > 0. Gabriel Dospinescu 17. Prove that if n is a positive integer such that the equation x3-3xy2d-y3 = n has an integer solution (x, y), then it has at least three such solutions. IMO 1982 18. Suppose that for a certain prime p a polynomial with integral coefficients f (x) = ax2 + bx + c takes values at 2p —1 consecutive integers which are all perfect squares. Prove that plb2 — 4ac. IMO Shortlist 422 18. QUADRATIC RECIPROCITY 19. Suppose that 0(57" — 1) = 5' — 1 for a pair (m, n) of positive integers. Here 0 is Euler's totient function. Prove that gcd(m, n) > 1. Taiwanese TST 20. Let a and b be positive integers such that a > 1 and a b (mod 2). Prove that 2a — 1 is not a divisor of 3b — 1. J.L.Selfridge, AMM E 3012 +3)n+1 i 21. Let m, n be positive integers such that A = (m 3m is an integer. Prove that A is odd. Bulgaria 1998 22. Prove that the numbers 3n + 1 have no divisor of the form 12k + 11. Fermat 23. Let p —1 (mod 8) be a prime number Prove that there exists an integer x such that x2; 2 is the square of an integer. 24. Let p = 4k +3 be a prime number. Find the number of different residues mod p of (x2 + y2)2 where gcd(x,p) = gcd(y,p) = 1. Bulgarian TST 2007 25. Let p be a prime of the form 4k + 1 such that p2I2P — 2. Prove that the greatest prime divisor q of 2P — 1 satisfies the inequality 2q > (6p)P. Gabriel Dospinescu PROBLEMS FOR TRAINING 423 26. Find all positive integers a, b, c, d such that a + b d2 = 4abc. Vietnamese TST 27. Let p be a prime number of the form 4k + 1. Prove that p-1 4 2 1 E [Vi p] = P 12 THEORY AND EXAMPLES 427 19.1 Theory and examples Why are integrals pertinent for solving inequalities? When we say integral, we say in fact area which is a measurable concept, a comparable one. That is why there are plenty of inequalities which can be solved with integrals, some of them with a completely elementary statement. They seem elementary, but sometimes finding elementary solutions for them is a real challenge. Instead, there are beautiful and short solutions using integrals. The hard part is to find the integral that hides behind the elementary form of the inequality (and to be honest, the idea of using integrals to solve elementary inequalities is practically nonexistent in Olympiad books). Recall some basic properties. • For all integrable functions f, g : [a, b] R and all real numbers a,13, fa b(af(x)+/3g(x))dx = a f f (x)dx + /3 f b g(x) (linearity of integrals). a a • For all integrable functions f, g : [a, b] --+ R such that f < g we have b f (x)dx < f g(x)dx (monotonicity for integrals). a • For all integrable function f : [a, b] R we have b f 2 (X)dX > O. Also, the well-known elementary inequalities of Cauchy-Schwarz, Chebyshev, Minkowski, Holder, Jensen, and Young have corresponding integral inequal-ities, which are derived immediately from the algebraic inequalities (indeed, one just has to apply the corresponding inequalities for the numbers f (a + — k (b — a)) , g (a + Ti k (b — a)) , . . . where k E {1,2,...,n} and to use the fact that b fa f a f (x)dx = liM a+ Ti (b — a)) I. 428 19. SOLVING ELEMENTARY INEQUALITIES USING INTEGRALS It seems at first glance that this is not a very intricate and difficult theory. Totally false! We will see how powerful this theory of integration is, and especially how hard it is to look beneath the elementary surface of a problem. To convince yourself of the strength of the integral, take a look at the following beautiful proof of the AM-GM inequality using integrals. This proof was found by H. Alzer and published in the American Mathematical Monthly. Example 1. Prove that for any al, a2, , an > 0 we have the inequality al + az ' • • + an > .Va1a2 • • • an. n Solution. Let us suppose that al < a2 < • • • < an and let + a2 + • • • + an A = , G= • an. Of course, we can find an index k E {1, 2, ... , n — 1} such that ak < G < ak-o.• Then it is immediate to see that A G 1 1 1 n raj 1 1 — — 1 — f _ dt + — n j Gj t z-1 ai i=k+1 ldt and the last quantity is clearly nonnegative, since each integral is nonnegative. Truly wonderful, is not it? This is also confirmed by the following problem, an absolute classic whose solution by induction can be a real nightmare. r Example 2 : 1 Let al, at, , an be real numbers. Prove that ctiai > O. = 1 i+ 3 = 1 Polish Mathematical Olympiad THEORY AND EXAMPLES 429 Solution. Now we will see how easy this problem is if we manage to handle integrals. Note that = f azajti+j-1 dt. + o We have translated the inequality into the language of integrals. The inequal-ity ajai > 0 is equivalent to E fl a iajti+j —ldt > 0, i,j=i ° or, using the linearity of the integrals, to n E ajaiti+j-1 dt > 0. Jo i,j=i This suggests finding an integrable function f such that n f 2 (t) -a 3 -ti+j-1dt. i,j=1 This is not difficult, because the formula 2 n aixi) = E .a-x.x- z 3 3 a i=1 i,j=1 solves the task. We just have to take f (x) a2 .a and we are done. 430 19. SOLVING ELEMENTARY INEQUALITIES USING INTEGRALS We continue the series of direct applications of classical integral inequalities with a problem which may also present serious difficulties if not attacked appropriately. rExample 3. Let t > 0 and define the sequence (x,i)n>i by 1+t +•••+ tn xn = - n + 1 Prove that X1 < -Vx2 < a x3 < 1 1 < • • • [Walther Janous] Crux Mathematicorum Solution. It is clear that we have t — 1 t 1 ,A f undo = xn = 1 1 — t jt undu Using the first of these forms for t> 1 and the second for t < 1 the inequality to be proved (clear for t = 1) reduces to the more general inequality k 1 b f k (X)dX < k+1 b — a a 1 fb fk±i(x)dx b — a a for all k > 1 and any nonnegative integrable function f : [a, b] -4 R. And yes, this is a consequence of the Power Mean Inequality for integral functions. The following problem has a long and quite complicated proof by induction. Yet using integrals it becomes trivial. lExample 4 2 1 Prove that for any positive real numbers x, y and any positive integers m, n (n — 1)(m — 1)(xm±n ym+n) (M, n — 1)(xmyn + xThym) > mn(xm+n- ly ym+n—lx). THEORY AND EXAMPLES 431 Solution. We transform the inequality as follows: +n-1 m+n- ym)(xn yn) <=> mn(x — y)(xrn — Y 1) > (m + n — 1)(xm — xm+n-1 Ym+n-1 xm ym xn yn (m + n — 1)(x — y) > m(x — y) n(x — y) (we have assumed that x > y). The last relations can immediately be trans-lated with integrals in the form x (x — y) f t m+n-2 dt ? f _ tm-1 dt f to-1 dt. And this follows from the integral form of Chebyshev inequality. A nice blending of the arithmetic and geometric inequalities as well as integral calculus allows us to give a beautiful short proof of the following inequality. Example 5.1 Let xi, x2, .. , xk be positive real numbers with xix2 • • • xn < 1 and m, n positive real numbers such that n < km. Prove that m(x7 + x2 + • • • + — k) > n(xl 714 1 . . . xi kn — 1). IMO Shortlist 1985 Solution. Applying the AM-GM inequality, we find that m(x7 + + Xr k i — k) > m(k V (xlx2 xk)n — k). P = ,c/xix2 < 1. mkPn — mk > nPrnk — n, which is the same as Let We have to prove that Pn -1 Pmk -1 n — mk 432 19. SOLVING ELEMENTARY INEQUALITIES USING INTEGRALS This follows immediately from the fact that Ps — 1 p xln P xtdt is decreasing as a function of positive x (for P < 1). We have seen a rapid but difficult proof for the following problem, using the Cauchy-Schwarz inequality. Well, the problem originated by playing around with integral inequalities, and the following solution will show how one can create difficult problems starting from trivial ones. Prove that for any positive real numbers a, b, c such that a + b + c = 1 we have (ab + bc + ca) (1)2 + b + c 2 + c + a 2 + a) > 4 [Gabriel Dospinescu] Solution. As in the previous problem, the most important aspect is to trans-late the expressionb2 + b + c2 + c + a 2 + a in the integral language. Fortu-nately, this isn't difficult, since it is just a Jo (x + b)2 (x + c)2 (x + a)2) Now, using the Cauchy-Schwarz inequality, we infer that (do not forget about a + b + c = 1): a b c ab c (x+b)2+ (x+c)2+ (x+a)2—x+b + + x+c x+a Using the same inequality again, we compare a + b c + with x+b x+c a+x 1 x + ab + be + ca. Consequently, a b c 1 + + > (x + b)2 (x + c)2 (x + a)2 — (x + ab + be + ca)2 ' dx. ) 2 THEORY AND EXAMPLES 433 and we can integrate this to find that a b c 1 b2 + b + c2 + c a2 + a (ab bc ca)(ab bc ca + 1) • Now, all we have to do is to notice that ab be + ca + 1 < 4 — — 3 It seems a difficult challenge to find and prove a generalization of this inequality to n variables. There is an important similarity between the following problem and example 2, yet here it is much more difficult to see the relation with integral calculus. Example 771 Let n > 2 and let S be the set of all sequences (al, a2, . , an) C [0, 00) which satisfy n 1 — aia >0. • i + j — n n ,± i a E Find the maximum value of the expression E + over i=1 j=1 all sequences from S. [Gabriel Dospinescu] Solution. Consider the function f : R R, f (x) = al + azx + • • • + a x Let us observe that n n E, E ai j i=1 j=1 i i=1 n a • 1 ) f (x)dx + ° 434 19. SOLVING ELEMENTARY INEQUALITIES USING INTEGRALS 1 = I f (x) E aixz-1) dx = f x f 2 (x)dx. 1 So, if we denote M = E i+j , we infer (using the hypothesis) that 1<i,j<n fl x f 2(x)dx < M. On the other hand, we have the identity n n x--. x--I ai + a3 9 al + ..4- j ± an al + + an i-1- j ' 2 n + 1 1 n+1 1 2n i=1 j=1 =2 I (x + x2 + • • • + xn) f (x)dx. Now, the problem becomes easy, since we must find the maximal value of 1 2 J (x + x2 + • • • + xn)f(x)dx 0 where f 1 o x f 2(x)dx < M. The Cauchy-Schwarz inequality for integrals is the way to go: fo1 ( + X2 ± • • • ± xn)f(x)dx = ( 1 2 fVXf 2(X) VX(1 ± X ± • • • ± Xn-1)2dX) = I xf 2(x)dx f (1 + x + • • • +xn-1)2dx < M2. THEORY AND EXAMPLES 435 n n ai + a This shows that i± < 2M and now the conclusion easily follows: i=1 j=1 the maximal value is 2 E 1 2+j, attained for ai = a2 = • • • = an = 1. 1 x2y +y2 x+y2 z+z2y + z2x + x2 z where x = ta " — y = tb-A , z = tc-1 and integrate the inequality as t ranges between 0 and 1. And surprise... since what we get is exactly the desired inequality. In the same category, here is another application of this idea. Prove that for any positive real numbers a, b, c the following inequality holds: 1 1 1 ( 1 1 1 ———+2 3a + 3b + 3c 2a + b + 2b+c 2c+a 436 19. SOLVING ELEMENTARY INEQUALITIES USING INTEGRALS 1 1 1 > 3 (a+2b + b+2c + c+2a) . [Gabriel Dospinescu] Solution. If the previous problem could be solved using bunching (or not? anyway, we haven't tried), this one is surely impossible to solve in this manner. With the experience from the previous problem, we see that the problem asks us in fact to prove that x3 + y3 + z3 2(x2y y2z z2x) > 3(xy2 yz2 zx2) for any positive real numbers x, y, z. Let us assume that x = min(x, y, z) and write y = x m, z = x n for some nonnegative real numbers m, n. Simple computations show that the inequality is equivalent to 2x(m2 — mn + n2) + (n — m)3 + m3 > (n — m)m2. Therefore, it suffices to prove that (n — m)3 + M3 > (n — m)m2, n — m ), At the start of this topic we said that there is a deep relation between integrals and areas, but in the sequel we seemed to neglect the last concept. We ask the reader to accept our apologies and bring to their attention two mathematical gems, in which they will surely have the occasion to play around with areas. If only this was easy to see... In fact, these problems are discrete forms of Young and Steffensen inequalities for integrals. [Example 10. Let al > a2 > • • • > an±i = 0 and let b1, b2, . , bn E [0,1]. Prove that if k =[ Eb2+1, i=i which is the same as t3 +1 > t for all t> —1 (via the substitution t = m which is immediate. THEORY AND EXAMPLES 437 then aibi < E a i=1 i=1 St. Petersburg Olympiad, 1996 Solution. The very experienced reader will have already seen a resemblance to Steffensen's inequality: for any continuous functions f , g : [a, b] R such that f is decreasing and 0 < g < 1 we have fa a+k f (x)dx > f b f(x)g(x)dx, where k = g(x)dx. a So, probably an argument using areas (this is how we avoid integrals and argue with their discrete forms, areas!!!) could lead to a neat solution. Let us con- sider a coordinate system XOY and let us draw the rectangles R1, R2, . • • , Rn such that the vertices of Ri are the points (i — 1,0), (i, 0), (i — 1, (i, ai) (we need n rectangles of heights al, a2, , an, and horizontal sides 1, so as to view ai as a sum of areas) and the rectangles Si, S2, . , Sn, where the ver- i=1 i—i tices of Si are the points ( j=i b3, 0) , bi, ai) , (E bi , ai) j=1 (where = 0). We have made this choice because we need two sets j=1 of pairwise disjoint rectangles with the same heights and areas al, a2, • , an and albs, a2b2, ••• anbn respectively, so that we can compare the areas of the unions of the rectangles in the two sets. Thus, we have to show that the set of rectangles Si, 82, Sn can be covered by the rectangles R1, R2, • • • Rk+1. This is quite obvious, by drawing a picture, but let us make it rigorous. Since 438 19. SOLVING ELEMENTARY INEQUALITIES USING INTEGRALS the width of the union of 81, , Sn is E b, < k + 1 (and the width of 3=1 R1, R2, , Rk+i is k+1), it is enough to prove this for any horizontal line. But if we consider a horizontal line y = p and an index r such that ar > p > ar+1, then the corresponding width for the set R1, R2, • • , Rk+1 is p, which is at least b1 + b2 + • • • + bp, the width for Si, 82, , Sn. And the problem is solved. And now the second problem, given this time in a Balkan Mathematical Olympiad. Example 11.1 Let (xn)n>0 be an increasing sequence of nonnegative integers such that for all k E N the number of indices i E N for which xi < k is yk < Do. Prove that for all m, n E N, n j=0 y3 > (m + 1)(n + 1). Balkan Mathematical Olympiad 1999 Solution. Again, the experienced reader will see immediately a similarity with Young's inequality: for any strictly increasing one-to-one map f : [0, A] [0, B] and any a E (0, A), b E (0, B) we have the inequality 1. a f (x)dx + f f -1(x)dx ab. Indeed, it suffices to take the given sequence (xn)n>0 as the one-to-one in- creasing function in Young's inequality and the sequence (Yn)Th>0 as the in- m m verse of f . Just view x, and y3 as the corresponding integrals, and z=0 3=o the similarity will be obvious. Thus, probably a geometrical solution is hiding behind some rectangles again. Indeed, consider the vertical rectangles with width 1 and heights xo, xi, , xin and the rectangles with width 1 and heights yo, Yi, • • • , yn. Then in a similar way one can prove that the set of these rect-angles covers the rectangle of sides m + 1 and n + 1. Thus the sum of their areas is at least the area of this rectangle. THEORY AND EXAMPLES 439 It will be difficult to solve the following problems using integrals, since the idea is very well hidden. Yet there is such a solution, and it is more than beautiful. Prove that for any al, a2, . , an and b1, b2, . , bn > 0 the following inequality holds Poland, 1999 Solution. Let us define the functions fi, g, : [0, oo) R, t E , , fi(x) = r [0 ad 0, t > ai Also, let us define and gi(x) = 1, x E [0, bib 0, x > bi. n f (x) = fi(x), g(x) = i=1 i=1 co Now, let us compute I f (x)g(x)dx. We see that fp ' f (x)g(x)dx = i<,j<n E i fi(x)gi (x)) dx 0 — E foo fi(x)gi(x)dx = 1<i,j<n ° A similar computation shows that <i,j <n min(ai, bi). 00 f 2 (x)dx = E min(ai, <i,j 2 foc- f (x)g(x)dx, we find that i min(ai, ai) + > min(bi, bi) > 2 min(ai, kJ). 1<i,j<n 1<i,j<n 1<i,j<n Now, remember that 2 min(x, y) = x + y — Ix — yl and the last inequality becomes lai ail + E Ibi — bjI < 2 E iai — bjl 1<i,j<n 1<i,j<n 1<i,j<n and since 1<i,j<n the problem is solved. ai —a3I=2 — 1<i<j 0 and let x1, x2, , xn be real numbers such that E aixi = 0. i=i a) Prove that the inequality xixilai — aj < 0 holds; 1<i<j R, Now, let us compute f 2 (x)dx = x x Lc() A[0,a21(x)A[0,a,j(x)dx 1<i,j2 , xixj min(ai, aj)• 1<i,j 0. 1<i,j<n min(ai aj) = + aj — ai — aj I , 2 and 1<z,j<n we conclude that xixj(ai + aj) -= n n ai xi = 0, xixi I ai — ai I <0. 1<i 0 and let Ai = E {1, 2, , a3 = Then A1, A2, , Ak is a partition of the set {1, 2, , n} and we also have ( E x3) A[0,bi] = 0 z=1 JEAi almost anywhere, from which we easily conclude that E x3 = 0 for alli E {1,2,...,k}. zeit, The conclusion follows. Because we have proved the nice inequality xixi min(ai, a3) > 0 1<i,j 0 let us take a further step and give the magnificent proof found by Ravi Boppana for one of the most difficult inequal-ities ever given in a contest. The solution is based on the above result. Example 14. Prove the following inequality k min(ajai, bibi) < i b, and a3 > b3, which leaves us with only two cases. The first one is when at least one of the two inequalities a, > bi and a3 > b3 becomes an equality. This case is trivial, so let us assume the contrary. Then xix j min(ri, ri) = bibi min ( — bi bi 1 a 1 = bibi min c -12 - a = 7 1 ' b3 ai = min(aibi, aibi) — bibj = min(aib3, bi) — min(aia , bibi). Now, we can write min(aib3, 1<i,j<n the last inequality being the main ingredient of the preceding problem. Finally, a problem, which is a consequence of this last hard inequality. Con- sider this a hint and try to solve it, since otherwise the problem is really hard. Example 15. I Let xi, x2, , xn be some positive real numbers such that — XiXil = 1<i,j<n Prove that x = n. i=1 Ixi — xil. <i,j 0, i<i,j E min(1, xixi). 1<i,j<n 1<i,j<n u+ v — lu—v1 Now, use the formula min(u,v) = 2 and rewrite the above in- equality in the form 2 2n E xi — E xi - n2 + xi — E - xixii. i=1 1<i,j<n i=1 1<i,j<n Taking into account that E - xixi i = 1<i,j<n we obtain 1<i,j n2 + n i=1 i=1 which can be rewritten as (En X — n)2 < 0 Therefore Ei n =i Xi = n. ) 2 PROBLEMS FOR TRAINING 445 19.2 Problems for training 1. Let a, b, n be positive integers with a < b. Prove that In +• • + — (bn, +1 1 1 1 < In — a. an + 1) an + 1 an +2 2 bn 2. Prove that for any a > 0 and any positive integer n the inequality (n + 1)a±1 — 1 la ± 2a ± na < holds. Also, for a E (-1, 0) we have the reversed inequality. 3. Prove that for any real number x 2k-1 n E x2k (n + 1) x . k=0 k=1 Harris Kwong, College Math. Journal 4. For any positive real number x and all positive integers n we have: (271) C o n) (2 1 n) (2 2 n) 2n,) x I + . x +1 x + 2 x + 2n> 0 Kornai 5. Let n E N, xo = 0, xi > 0(i = 1,2, ..., n), xi = 1. Prove that 1 n E + xo + x1 + • • • xi + + • • • ± xn i=, a + 1 < China 1996 446 19. SOLVING ELEMENTARY INEQUALITIES USING INTEGRALS 6. Let f be a continuous and monotonically increasing function f : [0,1] R such that f (0) = 0 and f (1) = 1. Prove that 9 10 f ()± -, f 1() < 99 — 10. k=1 k=1 St. Petersburg 1991 7. Prove that the function f : [0,1) —+ R defined by f(x)=log2(1—x)+x+x2 -Ex4 +x8 +"• is bounded. 8. Let 0 = x1 < < X2n+1 = 1 be some real numbers. Prove that if xj+1 — xi < h for all 1 < i < 2n then 1 — h 2n, 1 2 + h 2 < Ex2i(x2i+i x2i_i) < • i=1 Turkish TST 1996 9. Prove that for any real numbers al, a2, .. • , an n 2 i+ j_iaja3 > ij i) • i,3=1 i=1 10. Let k E N, al, a2, , E R with ar,H_i = al. Prove that k-1 k- j-1 n k a a — k-2 (E al) 11+ a b c ab+bc+ca Marius and Sorin Radulescu 12. Prove that we can find a constant c such that for any x > 1 and any positive integer n we have n kx 1 (k2 + x)2 2 k=1 IMC 1996 13. Prove that for all al, a2, .. • , an, b1, b2, • • • , bn > 0 the inequality holds ( E min(ai, ai)) ( E min(bi, bi)) > ( E min(ai, bi)) . 1< i,j<n 1<i,j<n 1<i,j<n Don Zagier 14. Consider vectors al, a2, an and b1, b2, bm in line through the origin, let the projection of the A1, A2, ..., An and B1, B2, ..., Bm. Suppose that the plane, and for every vectors onto the line be we always have IAII + IA21+ • • • + lAn1 + IB21+ • • • + Prove that fall + la21+ •-• + 1b21+ • + Here, Ivi is the length of the vector v. < x 448 19. SOLVING ELEMENTARY INEQUALITIES USING INTEGRALS 15. Prove that for any x1 > x2 > • • • > xn > 0 we have 22 Xi+1 ± • • • + Xn < 7r E xi. i=i Adapted after an IMC 2000 problem 16. Let co be Euler's totient function, with cp(1) = 1. Prove that 1 (p(k) k 1 — — < E ln 2n 2k2 <1. k=1 Gabriel Dospinescu 17. Let ri,r2, , rn be some positive numbers which add up to 1 and x1, x2, , xn some positive real numbers. Also let A = E rixi and G = H xi i=i a) Let us denote tdt I (x, a) = f occ ( 1 t)( x at) 2 Prove that ri(xi — A)2/ (xi, A), i=1 and hence deduce the arithmetic-geometric inequality. b) Suppose that xi < 2 and define A', G' to be the corresponding means for 1 — xi. Prove that 4 > §. n In Oral Examination ENS PROBLEMS FOR TRAINING 449 18. Prove that for any positive real numbers xi, x2, ,x,, such that n 1 E1 + x, 2 we have the inequality 1 x, + x3 2 Gabriel Dospinescu 19. Let xi, x2, xn and yi , y2, ..., yr, be positive real numbers such that for all positive t there are at most c pairs (i, j) satisfying x, + y3 > t. Prove that (xi + X2 + • • • + xn) (Y1 + Y2 + • • • + yn) < max (x, + Yi)• 1 1 1 n i_i E a, S S + a, — ai i.i Gabriel Dospinescu 21. Let m, n be positive integers and let x j,3 E [0, 1] for i = 1,2, ...., m and j = 1,2,....,n. Prove that 450 19. SOLVING ELEMENTARY INEQUALITIES USING INTEGRALS 22. Find the best constant k such that for any n and any nonnegative real numbers xi, ...xn we have (x1 + 2x2 + • • • + nxn)(x? + • • • + xn 2 ) > k(xi + • • +xn)3. 23. Prove that for any a1, a2, , an > 0 we have the following inequality a•a • 3 < 1 (xi +x2+• • •+xn)4. Carleson's inequality 25. Prove that for any real numbers al, a2, ..., an we have: ai • ai > 0. 1 — — THEORY AND EXAMPLES 453 20.1 Theory and examples It is very difficult to imagine a completely trivial mathematical statement which has absolutely nontrivial applications. And if there is such a candidate, then surely the pigeonhole principle will be the winner: what could be easier than the observation that if we put more than n objects in n boxes, there will be a box containing at least two objects? Yet, this observation, combined with some trivial variations, turn out to be a completely revolutionary idea in math-ematics. Quantitative results such as Siegel's lemma, or the fact that the class group of a number field is finite, are fundamental results in number theory, and are all consequences of this principle. There is also an enormous quantity of difficult Ramsey-type (and other) results in combinatorics, all based on this little observation. The purpose of this chapter is to present some of these applications of the pigeonhole principle, most of them elementary. Let us begin with some combinatorial statements in which the use of the pi-geonhole principle is more or less clear. But the reader must pay attention, because what is easy to state is not necessarily easy to write! This is why even the easiest problems of this chapter will have some subtle parts, and the reader should not expect straightforward applications of the pigeonhole principle. Example 1. Let A1, A2, ..., A50 subsets of a finite set A such that any subset has more than half of the number of elements of A. Prove that there exists a subset of A with at most 5 elements that has nonempty intersection with each of the 50 subsets. Great Britain 1976 Solution. Let A = {al, a2, an} and define f (i) to be the number the subsets among A1, A2, ..., A50 that contain a2. Then clearly f(1) + f(2) + • • • + f(n) = 1A1 + 1A21 + • • + IA501 > 25n. Thus there exists an i such that f (i) > 26, which implies the existence of an ax in at least 26 subsets, let them be A25, A26, ..., A50. Working with the remain- ing 24 subsets only and using the same argument we deduce the existence of 454 20. PIGEONHOLE PRINCIPLE REVISITED an element ay which belongs to at least 13 subsets among A1, A2, ..., A24, let them be Al2, A13, ..., A24. Similarly, there exists a, which belongs to at least 6 subsets among A1, ..., A11, let them be A6, A7, ..., A11 and if we continue this process we define similarly at, and ay. It is clear that the set of ax, ay, az, a„, a, satisfies all conditions of the problem. The strange statement of the following problem should not mislead the reader: after all, we have said that all problems of this chapter are based on the pigeonhole principle, but we haven't said where this idea hides. After reading the solution, the reader will surely say: but it was obvious! Yes, it is obvious, but only if we proceed correctly... Example 2. Let A = {1, ..., 100} and let A1, A2, ..., A, be subsets of A, each with 4 elements, any two of them having at most 2 elements in common. Prove that if 171 > 40425 then there exist 49 subsets among the chosen ones such that their union is A, but the union of any 48 subsets (among the 49) is not A. [Gabriel Dospinescu] Solution. Let us consider the collection of all two-element subsets of each A1, A2, ..., A,. We obtain a collection of 6m two-element subsets of A. But the number of distinct subsets of cardinal 2 in A is 4950. Thus, by the pigeon-hole principle, there exist distinct elements x, y E A which belong to at least 49 subsets. Let these subsets be A1, A2, ..., A49. Then the conditions of the problem imply that the union of these subsets has 2 + 49 x 2 = 100 elements, so the union is A. However, the union of any 48 subsets among these 49 has at most 2 + 2 x 48 = 98 elements, so it is different of A. The following example is, in a certain sense, typical for problems involving estimations of trigonometric sums. Its presence as the last problem in an in-ternational contest for undergraduate students shows that it is more difficult than it looks, even though the solution is again a pure application of the pi-geonhole principle. THEORY AND EXAMPLES 455 rExample 3.; Let A be a subset of Zn with at most 11 7 7 1 elements. Prove that there exists a nonzero integer r such that > e z n ST> . sEA IMC 1999 (e a;•— ,7 al e akt Solution. Let A = {al, a2, ak} and define g(t) -= ) for 0 < t < n — 1. If we divide the unit circle into 6 equal arcs then these k-tuples are divided into 6k classes. Because n > 6k, there are two k-tuples in the same class, that is there exist t1 < t2 such that g(ti) and g(t2) are in the same class. Observe now that if we consider r = t2 — ti then Re (e cos (27ra, t2—ti) ) > cos (L). Therefore If(r)1 > Re(f(r)) > IAI 3 — 2 and the problem is solved. Sometimes, even the completely obvious observation that an infinite sequence taking only a finite number of values must have (at least) two equal terms (ac-tually, an infinite constant subsequence) can be really useful. This is shown by the following extension of a difficult problem given in a Romanian TST in 1996: Example 4.1 Let xi, x2, ..., xk be real numbers such that A = {cos(n7rxi) + cos(n7rx2) + • • • + cos(n7rxk)In E N} is finite. Prove that x, are all rational numbers. [Vasile Pop] Solution. The beautiful idea is that if the sequence an = cos(n7rxi)+cos(n7rx2)+ • • • + cos(n7rxk) takes a finite number of distinct values, then so does the se-quence in Rk defined by un = (an, a2n, akn). Thus there exist m < n such that an = am, a2n = a2m, akn = akm. Let us analyze these relations more closely. We know that cos(nx) is a polynomial of degree n with integer coef-ficients in cos(x). If A, = cos(n7rx,) and B2 = cos(m7rx,) then the previous relations combined with this observation, show that Al + A Z + • • • + 24.3 k = Bi + B2 + • • • + Bz for all j = 1,2, . . . , k. Using Newton's formula, we deduce 456 20. PIGEONHOLE PRINCIPLE REVISITED that the polynomials having zeros A1, A2, ..., Ak and B1, B2, ..., Bk are equal. Thus there exists a permutation a of 1,2, ..., n such that Ai, = Bo.(i). Thus cos(n7rxi) = cos(m7rx,(,)), which means that nxi —mx,(,) is a rational number for all i. This easily implies that all xi, are rational numbers. The same idea can be used with success when dealing with remainders of re-cursive sequences modulo certain positive integers. This kind of problem has become quite classical, being present in lots of mathematical competitions . Consider the sequence (an)n>1 defined by al = a2 = a3 = 1 and an+3 = an+lan,+2 + an. Prove that any positive integer has a multiple which is a term of this sequence. [Titu Andreescu, Dorel Mihet] Revista Matematica Timi§oara Solution. Consider a positive integer N and let the first term of the extended sequence to be a0 = 0. Now, look at the sequence of triples (an, an+1, an+2) re-duced mod N. This sequence takes at most N3 distinct values because there are N possible remainders mod N. Thus we can find two positive integers i < j such that a, = a3 (mod N), ai+1 a 3+1 (mod N) and (4+2 = a3+2 (mod N). Using the recursive relation, we deduce that the sequence becomes periodic mod N with period j — i. Indeed, it follows immediately from the recursive relation that ak ak.+3_, (mod N) for all k > i, and using the fact that an = an,±3 — an±lan+2 we can proceed backwards with an inductive argu-ment to prove that ak ak±i_, (mod N) for all k < i. In particular, it follows that ai_i is a multiple of N, so N divides at least one term of the sequence. A classical application of the pigeonhole principle is to prove that for any col-oring of the lattice points in a plane with a finite number of colors, there are rectangles having all vertices of the same color. We advise the reader who does not know this problem to solve it first and then to proceed to the following similar problem. THEORY AND EXAMPLES 457 I Example 6. i Let m, n be positive integers and let A be a set of lattice points in the plane such that any open disc of radius m contains at least one point of A. Prove that no matter how we color the points in A with n colors there exist four points of the same color in A which are vertices of a rectangle. Romanian TST 1996 Solution. Consider first a huge square of side-length a (to be determined later) and having sides parallel to the coordinate axes. Divide it into smaller squares of side-length 2m and inscribe a circle in each such smaller square. We find at least [4a;2 ] circles of radius m inside this huge square, and thus at least as many points of A. But these points lie on a — 1 vertical lines. By the pigeonhole principle, there exists a vertical line containing at least n +1 points of A if a is suitably chosen (for instance, any multiple of 4nm2). Again by the pigeonhole principle, two of these points have the same color. This shows that in any such huge square there exists a vertical line and two points on it that have the same color. Because there are finitely many positions of these pairs of points on a segment of finite length and because we can put infinitely many huge squares consecutively on the Ox axis, there will be two squares in which the points of the same color and on the same vertical line have identical posi-tions and same color. These points will determine a monochromatic rectangle. It is time to consider some more involved problems in which the use of the pigeonhole principle is far from obvious. Several articles in the American Mathematical Monthly were dedicated to the following problem, which shows that it is not surprising that only a few students solved it when it was pro-posed for the Putnam Competition (in a weaker form than the example below): Example 7. Let Sa be the set of numbers of the form Lna I for some positive integer n. Prove that if a, b, c are positive real numbers, then the three sets Sa, Sb, Sa cannot be pairwise disjoint. 458 20. PIGEONHOLE PRINCIPLE REVISITED Solution. Let us pick an integer N and consider the triples ({ " c 1: ,} , , { }) for i = 0,1, ..., N3. These points lie in the unit cube [0,1]3 so by the pigeon- hole principle there are two points lying in a cube of side-length N , that is there exist i > j such that for some integers m, n, p we have — 'a < a N' b —n < N. This can be written as Ii — j — mal < 1 1, I i — j — nbl < 1 and Ii — j — pcI < 1. Therefore all numbers [ma], [nb_1, are equal to i — j or i — j — 1, which shows that some two of them are equal, and so two of the sets Sa, Sb, Sc intersect. We continue with a very beautiful problem from an Iranian Olympiad, where there are some traps in applying the pigeonhole principle. Example 8. Let m be a positive integer and n = 2m+1. Consider fi, f2, f n [0, 1] [0, 1] to be increasing functions such that f,(0) = 0 and fi(x) — fi(y)I < — yl for all 1 i < n and all x, y E [0,1]. Prove that there exist 1 < i < j < n such that Ifi(x)— f3 (x) < m+ 1 1 for all x E [0, IT Iran 2001 Solution. This time, everything is clear: the solution of this problem should use the pigeonhole principle. But how? Looking at the graph of such functions, we observe that the points of a regular subdivision of [0,1] play a special role in their behavior. Therefore, let us concentrate more on these points, so let us associate to each function fi an (m + 1)-tuple (al a2 (i), ..., arn+1 (i)), where ai(i) is the smallest integer k such that fi(± m e [7:+1, t'1]. In this way, we can control the behavior of the function f, very well at all points of the regular distribution (0, 77, 1 +1, m+1, ..., 1). Because A is increasing, it is clear that ai±i(i) > ai(i). Also, the inequality A (ni + + 1 1) A (741) 5 ni1 +1 assures us that a3+1(i) < a3 (i) + 1. Furthermore, note that 1 1 0<f,(m+1)=f,(m+1)f 1 ,(0)<m + 1, THEORY AND EXAMPLES 459 so al (i) = 0 for all i. Therefore there are at most 2' such sequences that can be associated with 11, 12, ..., fn. By the pigeonhole principle, two functions fi, fi with i < j must be associated with the same (m + 1)-tuple. This shows that we can find some integers tic', b n.,±1 and two indices i < j such that fi(mk +1) and fi( mk +i) are both in [ni b1 4_1, M n for all k = 0, 1, m + 1. Now we are almost done, because we have found our candidates i, j. What remains to be verified is straightforward. Indeed, consider x E [0, 1] and k an [mk ±i 7 : ‘,+ + 1 1.1. integer to be such that x E j We know that 0 < bk+1 — bk < 1 by the previous observations. Now, we have two cases. First let bk+1 = bk, so (k +1) bk + 1 1 k 1 fz(x) m+1 - m+i - m+1+ firri+1) fi(x)± m+1' and by a similar argument we also obtain f3(x) < fe(x) + n.,± 1. So, assume that bk+1 = bk + 1. Then fi(x) < fi ( m k + 1) + x m + 1 — < x + m + 1 ' k bk — k +1 f3(x) > f ( k +1) 3 771+1 +X r11+1 m +1' k +1 > x + bk — k from where fi(x) — fi(x) < m ± 0.. Analogously we obtain fi(x) — fi(x) < m1 +1, which shows that in both cases I fi(x) — f3(x)I < m1 +1. In the same category of difficult (or very difficult) problems can be included the next example, too. Here it is absolutely not obvious how to use the pigeon-hole principle. The solution presented here was given by Gheorghe Eckstein: Example 9.1 49 students take a test consisting of 3 problems, marked from 0 to 7. Show that there are two students A and B such that A scores at least as many as B for each problem. while IMO 1988 Shortlist 460 20. PIGEONHOLE PRINCIPLE REVISITED Solution. Let us consider the set of triples (a, b, c) where each component can be 0,1, ...7. We define an order on these triples by saying that (a, b, c) is greater than or equal to (x, y, z) if a > x, b > y, and c > z. A similar order is defined for pairs (a, b). We need to prove that among any 49 triples there are two that are comparable. Supposing the contrary, it is clear that such a set A of triples cannot contain two triples with the same first two coordinates. Now, consider the following chains: (1) (0,0) < (0,1) < (0,2) < (0, 3) < (0,4) < (0,5) < (0,6) < (0, 7) < (1, 7) < (2, 7) < (3, 7) < (4, < (5, 7) < (6, 7) < (7, 7) (2) (1, 0) < (1,1) < (1, 2) < (1, 3) < (1, 4) < (1, 5) < (1, 6) < (2, 6) < (3, 6) < (4, 6) < (5,6) < (6, 6) < (7, 6) (3) (2, 0) < (2, 1) < (2, 2) < (2, 3) < (2, 4) < (2, 5) < (3, 5) < (4, 5) < (5, 5) < (6,5) < (7,5) (4) (3,0) < (3,1) < (3,2) < (3,3) < (3,4) < (4,4) < (5,4) < (6,4) < (7,4). Note that no such chain can contain more than 8 pairs of the first two co-ordinates of some triples in A (otherwise there are two with the same last coordinate among them and so they are comparable). On the other hand, there are 48 pairs (a, b) with 0 < a, b < 7 covered by these four chains. There-fore there are 64 - 48 = 16 remaining pairs of two elements which are not covered by the chains. Each such pair corresponds to at most one element of A. Therefore A has at most 4 x 8 16 = 48 elements, a contradiction. Note that the above construction shows that the property fails with only 48 students. A highly nontrivial example of how the pigeonhole principle can be used in combinatorial problems is the following example. The solution was given by Andrei Jorza. Example 10.1 The 2' rows of a 2" x n table are filled with all the different n-tuples of 1 and -1. After that, some numbers are replaced THEORY AND EXAMPLES 461 by zeros. Prove that there exists a nonempty set of rows such that their sum is the zero vector. Tournament of the Towns 1996 Solution. Take any numbering L1, L2, ..., L2n of the rows before the replace-ment of some numbers by 0, in such a way that L1 is the vector with all coordinates equal to 1 and Len is the vector (-1, —1, ..., —1). Define f(L) to be the new line, obtained by (possibly) replacing some numbers by 0. For any row L that now contains some zeros, let g(L) be the corresponding row in the initial table, obtained by the following rule: any 1 in L becomes the value —1 in g(L) and any 0 or —1 in L becomes a 1 in g(L). Now, define the following sequence: x0 = (0,0, ..., 0), x1 = f (Li) and xr±i = xr f (g(xr))• We claim that all terms of this sequence have all coordinates equal to 0 or 1. This is clear if n = 1. Assuming that it holds for xi., observe that the only places in which the value —1 can appear in f(g(x,)) are those on which xr has a 1, thus all coordinates of xr±i are nonnegative. Also, the places on which a 1 appears on f(g(xr)) must be among the places on which xr had a 0. This proves that xr±i also has all coordinates equal to 0 or 1. Now, it follows from the pigeonhole principle that for some i > j we have xi = xj, which can be also written as f (g(x3)) + f (g(x3+1)) + • • • + f (g(x,-1)) = 0 and this means precisely that a sum of rows in the new table is zero. There is no trace of the pigeonhole principle in the following problem. At least at first glance. However, a very clever argument based on the pigeonhole principle allows an elegant proof: Example 11: Let (an)n,>1 be an increasing sequence of positive integers such that ari±i — an < 2001 for all n. Prove that there are infinitely many pairs (i, j) with i < j such that az a 3. Solution. Let us construct an infinite matrix A with 2001 columns in the following way: the first line consists of the numbers al +1, al +2, ..., al +2001. 462 20. PIGEONHOLE PRINCIPLE REVISITED Now, suppose we constructed the first k lines and the kth line is x1 + 1, xi + 2, ..., x1 + 2001. Define the (k + 1)st line to be N + xi +1, N + xi + 2, ..., N + xi + 2001 where N = + 1)(xi + 2) • • • (xi + 2001). The way in which this matrix is constructed ensures (an inductive argument doing the job) that for any two elements situated on the same column, one divides the other. Now, pick any 2002 consecutive lines. On each line there is at least one term of the sequence, because (an)n>1 is increasing and an±i — an < 2001. Thus there are at least 2002 terms of the sequence on the matrix formed by the selected lines. By the pigeonhole principle, there exist two terms of the sequence on some of the 2001 columns. Those terms will form a good pair. Thus for each choice of 2002 consecutive lines we find a good pair. Because the numbers on each column are increasing, it is enough to apply this procedure to the first 2002 lines, then to the next 2002 lines and so on. This will produce infinitely many good pairs. The following example was taken from an article called "24 Times the Pigeon-hole Principle". We must confess we did not count exactly how many times this phrase appears in the following solution, but we do warn the reader that this will normally take a considerable amount of time. rExample 12.1 Let n > 10. Prove that for any coloring with red and blue of the edges of the complete graph with n vertices there exist two vertex-disjoint triangles having all six edges colored with the same color. [Ioan Tomescu] Solution. Have courage, this is going to be long! First, we will establish a very useful result, that will be repeatedly used in the solution: Lemma 20.1. Every coloring with two colors of the complete graph with six vertices induces a monochromatic triangle. The only coloring with two colors of the complete graph with five vertices that does not induce monochromatic triangles has the form: there exists a pentagon with edges red and diagonals blue. THEORY AND EXAMPLES 463 Proof. Consider first the case of a complete graph with five vertices. It is clear that with every vertex there are at least two incident edges having the same color. If for a vertex at least three of them have the same color, it can be easily argued that a monochromatic triangle appears. So, suppose that every vertex is incident with two red and two blue edges. Let x be an arbitrary vertex and suppose that xy and xz are red. Then yz is blue. Now, let t be a vertex distinct from x, y, z and suppose that the edge connecting y and t is red and the edge connecting x and t is blue. Let w be the fifth vertex of the graph. Then the edges wz and wt are red, while wx and wy are blue. Similarly, zt is blue and so we can consider the pentagon xytwz which has red edges and blue diagonals. The case of the complete graph with six vertices is much easier: pick a vertex x. There exist three edges having the same color (say red) leaving from x (again the pigeonhole principle). Let y, z, t be their extremities. If yzt is blue, we are done. Otherwise, assume that yz is red. Then xyz is a monochromatic triangle. The lemma is proved. El Now, choose six vertices of the graph. They clearly induce a complete subgraph with six vertices. By the lemma, there exists a monochromatic triangle xyz. If we consider six of the remaining seven vertices, we find another monochro-matic triangle uvw, whose set of vertices is disjoint from the set of vertices of xyz. If the two triangles have the same color, we are done. Otherwise, suppose that xyz is red and uvw is blue. Because there are nine edges between the two triangles, by the pigeonhole principle at least five edges have the same color, say blue. By the same principle, there exists a vertex of xyz, call it x, which is incident with at least two blue edges having the other extremity in triangle uvw. Suppose without loss of generality that these vertices are u, v. Thus two triangles xyz and xuv appear with x as a common vertex, the edges of xyz being red and the edges of xuv blue. Look at the remaining five ver-tices, which form a complete graph with five vertices. If this graph contains a monochromatic triangle, we are done. Otherwise, by the lemma the remain-ing five vertices form a pentagon abcde with red sides and blue diagonals. By the pigeonhole principle, there exist three edges among those connecting x to 464 20. PIGEONHOLE PRINCIPLE REVISITED vertices of abcde that have the same color. Now we have two cases. In the first case, vertices y and z are joined by at least three edges having the same color with vertices of abcde. If, for instance, the color corresponding to y is blue, then we can consider two blue edges joining y with abcde. Then no blue triangle with a vertex in y appears if and only if the two blue edges join y with two consecutive vertices of the pentagon, for example with a and b. But there is still a third blue edge joining y with one of c, d, e, and this shows the existence of a blue triangle with vertices y and the two extremities of a diagonal of the pentagon. So, two blue triangles with disjoint sets of vertices appear. Let us now consider the case when y and z are each joined by at least three red edges with the vertices of the pentagon. So, there is a red triangle with vertices x and two neighboring vertices of the pentagon, say a and b. Consider now y, z, c, d, e. If the induced complete graph with five vertices contains a monochromatic triangle, we are done, because we still have the red triangle xab and the blue triangle xuv. Otherwise, again using the lemma, yz, cd and de are red, so either ze, yc are red or zc, ye are red. In both cases all other edges of the complete graph induced are blue. Let us consider just the first subcase (ze, yc red), the second one being similarly treated. Then y is joined by at least three red edges with vertices of abcde, and since yd and ye are diagonals in ycdez (thus they are blue), it follows that ya and yb are red. Similarly we find that za, zb are red and so we have two good triangles zae and xyb. Finally, let us consider the second case. Actually, all we have to do is to argue as in the first case, by considering vertices u, v joined each by at least three edges of the same color with vertices of abcde. So we are done. The following problems are more computational, but contain much more math-ematics than the previous examples. The first one is a famous example due to Behrend, concerning subsets with large cardinality containing no three el-ements in arithmetic progression. This is related to an even more famous (but notoriously difficult) theorem of Roth: the maximum cardinality of a subset of {1, 2, ..., n} having no three elements in arithmetic progression is at most C ln(l n ) for an absolute constant C. This was refined by Bourgain to n n THEORY AND EXAMPLES 465 (l) V lnn Cn The proofs of these results are very deep, but finding a lower ln rt bound for the maximum cardinality of such a set is is not so difficult, if you use the pigeonhole principle. Only easy when compared to the proofs of the mentioned theorems, of course... [Example fid There exists an absolute constant c > 0 such that for all suffi-ciently large integers N there exists a subset A of {1, 2, ..., N} with at least Ne —cin N elements and such that no three ele-ments of A form an arithmetic progression. Behrend's theorem Solution. The beautiful idea that provides an elegant proof of this result is the observation that a line cuts a sphere of R in at most two points. For n = 3, this is immediate geometrically, and for larger n this follows from the Cauchy-Schwarz inequality: if 11x11 = 11Y11 = ax + (1— a)YI I = r for some E (0,1), it easily follows by squaring the last relation that (x, y) = 11111 • 11Y1I, where () is the natural inner product and II•11 the Euclidean norm. By the Cauchy-Schwarz inequality, the last relation implies that x, y are colinear and from here the conclusion easily follows. Now, define F(n, M, r) to be the set of vectors x all of whose coordinates xi, x2, xn are in {1, 2, ..., M} and such that xi + 4 + • xn 2 = r2. Fix n, M and observe that as r2 varies from n to nM2, the sets F(n, M, r) cover the set of vectors with all coordinates in {1, 2, ..., M}. Using the pigeonhole principle it follows that there exists some r n such that ,\Ft, < r < M/ for which F(n, M, r) has at least n(m m , n 1) > m 2 elements. Let us now define the function f from F(n, M, r) to {1, 2, ..., N} by f (xi, x2, ..., xn) = E (2M)i—lxi. We claim that if f (x), f (y), f (z) form an i=1 arithmetic progression, then x = y = z. Indeed, it follows that (xi + yi — 2z0(2M)z-1 = 0. i=1 Put cti = xi + yi — 2zi. Then jail < 2M —1 and the last relation easily implies n-1 that ai = 0 for all i (indeed, I E ai(2M)z-11 < (2M)n-1, so an = 0; now, i=1 466 20. PIGEONHOLE PRINCIPLE REVISITED use an inductive argument to finish the proof). Therefore x + y = 2z and because x, y, z lie on a sphere, the observation made in the beginning of the solution shows that x = y = z. Also, f is injective: if f (x) = f (y), then f(x) + f(y) = 2f (y) and from the above argument, x + y = 2y, thus x = y. Finally, Ii(xi, x2, xn)I < M (2M)n 1 < (2M)Th. 2M — 1 — Therefore, if we consider M the largest integer such that (2M)n < N, then f (F(n, M, r)) is a subset of {1, 2, ..., N} which has no arithmetic progressions of length three. Now we need to choose some n as to obtain an optimal cardi-nality for f (F(n, M, r)). But this cardinality is the same as that of F(n, M, r) n 2 (b n ecause f is injective), which is at least ivi by the choice of r. But Nn2 Mn-2 4n-2n • So choose n the integer part of On N to see that f (F(n, M, r)) has at least Ne—cln N elements and has no three elements in arithmetic progression. We now pass to another revolutionary result, the famous Siegel's lemma. The applications of this theorem are so numerous and important that they would fill a book by themselves. We leave the interested reader to search in the huge literature of transcendental number theory for variations of the following result and for applications, among them the difficult Thue-Siegel-Roth theorem (do not kid yourselves, these require much more than Siegel's lemma alone!). Let 1 < m < n be integers and let A = (aii)i<2<n,1<j<rn be a matrix with integer entries. Suppose that for all 1 < j < m, the number Ai = E laiii is nonzero. Prove that i=1 there exist integers xi, x2, ..., xn, not all zero, such that Ixil < n-'-VA1A2 ...Am for all 1 < i < n and E aiixi = 0 for all 1 < j < m. Siegel's lemma THEORY AND EXAMPLES 467 Solution. The idea is the following: for a nonnegative real number M, we will prove that the quantity E aijx, cannot take too many distinct values when (xi, x2, ..., xTh) goes through the set of vectors with integer coordinates, all of them between 0 and M. It will follow that the image of the function ( n n f (xi, x2, ..., xn) = E anxi, ... aimxi i=i i=i is not too big and we will be able to use the pigeonhole principle as long as ([M] + 1)n is greater than the image of f. Consider integers ai, a2, ..., an and suppose that al, a2, % are nonnegative and ap+l, a p+2, ..., an are negative. Then it is clear that for any integers x, such that 0 < x, < M we have [M j (ap±i + • • • + an) < aixi + a2x2 + • • • + anxn < (al +•• • + ap) [M j Thus there are at most 1+ ( I ai I + a2 + • • • + Ian I) values taken by al xi + a2x2 + • • • + anxn, which means that the image of f has at most (1 + I_MjAi)(1+ [M] A2) • • • (1 + [] Am) < A1A2 • • • Am(1 + I_Mpm elements. Because there are (1 + [M]Dn vectors in Zn all of whose coor-dinates are between 0 and M, it follows that f is not injective if we take M = n-T/ AiA2 • • • Am. Thus there exist two distinct vectors x, y such that f(x) = f (y). It is clear that the vector v = x — y satisfies all the desired conditions. And here is a surprising, yet very challenging, application of Siegel's lemma, inspired by a USAMO problem: [Example 15.] Let C > 0 and A < e 11 be two real numbers and let f : {1, 2, ...} —> {1, 2, ...} be a function satisfying f (n) < C An for all n. Suppose that f (n + p — 1) — f (n) is a multiple of 468 20. PIGEONHOLE PRINCIPLE REVISITED p for any prime number p and any n. Prove that there are integers ri, r2, T., not all zero such that for all n we have rif (n) + r2f (n + 1) + • + rsf (n + s — 1) =0. [Gabriel Dospinescu, Vesselin Dimitrov] Solution. Let us consider positive integers m, n such that in = [3-] (this is usually the best choice in Siegel's lemma) and let us define (43 = f (i + j) for 1 < i < n and 1 < j < m. We claim that we can choose some m such that if x1, x2, ..., xn is a solution of the system given by Siegel's lemma, then xi f (j + 1) + x2 f + 2) + • • • + xn f (j + n) = 0 holds for all positive integers j. For this, we will need some preparation, which will be done in the next paragraph. Take x to be any solution given by Siegel's lemma, and observe that the desired relation holds for j < m. Assume that it fails for some k > m and let k + 1 be the smallest index for which it fails (thus it holds for all j < k and k > m). Consider p any prime smaller than k + 2. Then 1 < k + 2 — p < k and so xi f (1 + k + 2 — p) + + xn f (rt + k + 2 — p) = 0. But this last sum is congruent (mod p) to A = xif(1+ (k + 1)) + • • + xn f (n + (k + 1)) (20.1) which is nonzero by the choice of k. This shows that the last quantity A is actually a multiple of the product of all primes up to k + 1. The desired contradiction will follow from the fact that Siegel's lemma and the hypothesis on f ensure that A is small enough and thus cannot be divisible by the product of all p with p < k + 1. Let us estimate first the growth of x3. Using the notations of Siegel's lemma, we have A < C(A3+1 + • • • + Al+n) < C1An+3, THEORY AND EXAMPLES 469 where C1 > 1 depends only on A, C. Thus 1 mn m(rn-I-1) IX3 1 < (Ai • ..AIn )n—m < C 1 11' . A n—rn 2(n—m) for some C2 > 0 depending only on A, C. Therefore < C2A5n14, Ixif(1 + (k +1)) + • • • + xn f (n + (k +1))1< (max(Ixj) • C(Ak+l+l + • • • + An+k+1) < c3A9n/ 4+k where C3 is again a constant depending only on A, C. Now, we can prove the claim and thus end the solution. Suppose that the statement does not hold, so for infinitely many k (remember that for each m the corresponding k was at least m) we will have Hp < C3A9701±k. p m > n/2 — 1, we have A9n/4+kc3 < A11k/2c4. Thus for infinitely many k one must have llk ln A + ln C4 > E In p 2 p e , a contradiction with the choice of A. We end this chapter with a very challenging problem concerning the growth of coefficients of divisors of a polynomial whose coefficients are 0, 1 or —1. This type of problems, concerning the multiplicity of roots of polynomials with co-efficients —1,0, 1 has been subject to extensive research, but seems to be a quite difficult problem. One estimation in the following problem can easily be obtained using the pigeonhole principle; the other requires a beautiful theorem of Landau. 470 20. PIGEONHOLE PRINCIPLE REVISITED [Example 10.1 For n > 2, let An be the set of polynomial divisors of all polynomials of degree n with coefficients in { —1, 0,1}. Let C(n) be the largest coefficient of a polynomial with integer coefficients that belongs to An. Prove that for any E > 0 there exists a k such that for all n > k, 2 . 71 < C(n) < 2n. Solution. Let us start with the left hand side inequality: C(n) > 2j e. For a polynomial f with coefficients 0 or 1 and degree at most n define the function cb(f) = (f (1), f (1), ..., fN-1(1)). Taking into account that all coefficients of f are 0 or 1, we can immediately deduce that f(3)(1) < (1 + n)3+1 for all j, thus the image of f has at most (1 + n)1+2+ •+N < (1 + n)N2 elements. On the other hand, f is defined on a set of 2n+1 elements. So, if 2n+1 > (i+n)N2 then by the pigeonhole principle two polynomials f,g will have the same image and thus their difference will have all coefficients —1, 0 or 1 and degree at most n. Also, f — g will be divisible by (X — 1)N. Thus C(n) > ( N N), because the \ . largest coefficient of (X — 1)N is 2 N N) Because ( N N) is the largest binomial 4 coefficient among (N), we have (2N\ 1 -- , 2N-F1 > 2N for N > No. By taking N = log( n n+1) [ 2 , we have (1 + n)N2 < 2n+1, thus C(n) > 2N and it is easy to see that N > rl,'-' for n large enough. The other part, C(n) < 2', is much more subtle. For a polynomial f(X) = an,Xn + • • • + a1X + ao with real coefficients (everything that follows applies verbatim for complex coefficients), define its Mahler measure by n M(f) = lamp max(1, lxii) (20.2) THEORY AND EXAMPLES 471 where xi are the roots of f . The following inequality is due to Landau: Lemma 20.2. M(f) < Va8 + a? + • • • + Proof. There are many proofs of this lemma, but we particularly like the following one, which we haven't encountered in the literature. Consider N > n and let zi, z2, zN be the N-th roots of unity. A simple computation, based on the fact that E =1 3 Zk . = N if Njk and 0 otherwise, shows that 3 E If(zi)12 = N n j=1 i=0 ) i.Z.i = E aijav • zr = N • Ea?. u,v=0 j=1 i=0 Now, applying the AM-GM inequality, we obtain that E f (zi) f (z2) f (zN)12. i=0 On the other hand, the identity (X — zi)(X — z2) • • • (X — zN) = X N — 1 and the fact that f (X) = an(X — xl)(X — x2) • • (X — xn) imply the identity 1 f (zi )f (z2) • • • f(zN) 1 = 1 an IN 11- — 4111 —x2 I... 11 — 1, which, combined to the previous inequality, implies 472 20. PIGEONHOLE PRINCIPLE REVISITED Now, it is pretty clear that limN,c.9 V11 z N I = max(1, I,z1) whenever 1. Thus the inequality is proved whenever all roots of f lie outside the unit circle. What happens in the opposite case? It really does not matter! Actu-ally, Viete's formulae show that the inequality M(f) < .\/4 + ai + .• • + an 2 reduces to an inequality involving only absolute values of polynomials in xi. If this inequality holds whenever the variables x, are not on the unit circle, it also holds in the other cases, by continuity. Therefore the lemma is proved. K The previous lemma shows that polynomials with all coefficients of absolute value at most 1 have Mahler measure at most In + 1. Take now any divisor f of a polynomial g with all coefficients —1, 0,1 and write g = hf. Suppose that f has integer coefficients. It is easy to see that M(g) = M(h)M(f) > M(f). Thus M(f) < Vn + 1. Now, observe that by Viete's formula, the triangular inequality and the obvious fact that < M(f) for all distinct i1 , ..., is and all s, we have that any coefficient of f is bounded in absolute value by the fact that ([2J/ M(f)< n+ 1 • ( 1_2J ) < 2n for sufficiently large n. Thus the conclusion follows. PROBLEMS FOR TRAINING 473 20.2 Problems for training 1. On a piece of paper 4n unit squares are marked, their edges being par-allel to the edges of the paper. Prove that there exist n pairwise disjoint squares. 2. Let k be an integer greater than 1. Prove that there exists a prime number p and a stricly increasing sequence of positive integers al, az, ••• such that all numbers p + kai, p + ka2, ... are primes. 3. Let 0 < al < az < < awl < 5050 be integers. Prove that we can choose four distinct integers ai , a3, ak, al such that 5050 divides a2 + a3 — ak — ai. Poland 1999 4. Prove that for infinitely many positive integers A the equation Lx/j + Ly.\5] = A has at least 1980 solutions in positive integers. Russia 1980 5. A positive integer is written in each square of an n2 x n 2 chess board. The difference between the numbers in any two squares sharing an edge is at most n. Prove that at least 1 + [3 j of the squares contain the same number. Hungary 1999 6. Prove that any integer k greater than 1 has a multiple less than k4 which has at most four distinct digits. IMO 1987 Shortlist 474 20. PIGEONHOLE PRINCIPLE REVISITED 7. Are there 10000 numbers with ten digits, all multiples of 7, and with the property that any one of them can be obtained from the first number by a suitable permutation of its digits? Czech-Slovak Match 1995 8. Find the greatest positive integer n for which there exist nonnegative integers xi, x2, ..., xn, not all of them equal to 0, and such that n3 does not divide any of the numbers aixi±a2x2+• • •+anxr, with al, a2, •••, an = ±1. Romanian TST 9. The complete graph with 12 vertices has its edges painted in 12 colors. Is it possible that for any three colors there exist three vertices which are joined with each other by segments having these three colors? Russia 1995 10. Prove that among any 2m + 1 integers whose absolute values do not exceed 2m — 1 one can always choose three that add up to 0. 11. Prove that any sequence of mn + 1 real numbers contains an increasing subsequence with m + 1 terms or a decreasing subsequence with n + 1 terms. Erd6s-Szekeres's theorem 12. Let A be the set of the first 2m n positive integers and let S be a subset of A with (2in — 1)n + 1 elements. Prove that there exist a0, al, ..., distinct elements of S such that a() I al ••• am. Romanian TST 2006 PROBLEMS FOR TRAINING 475 13. Can you put 18 rectangles of size 1 x 2 on a 6 x 6 board such that there exists no straight line connecting two opposite sides of the table which goes along sides of the rectangles? N.B.Vasiliev, Kvant 14. Consider a set of 2002 positive integers not exceeding 10100. Prove that this set has two nonempty disjoint subsets with the same size, the same sum of elements, and the same sum of squares of the elements. Poland 2001 15. Consider an 11 x 11 chess board whose unit squares are colored using three colors. Prove that there exists an m x n rectangle with 2 < m, n < 11 whose vertices are in squares having the same color. loan Tomescu, Romanian TST 1988 16. 50 students compete in a contest where every participant has the same 8 problems to solve. At the end, 171 correct solutions were received. Prove that at least 3 problems were solved by at least 3 students. Valentin Vornicu, Radu Gologan, Mathlinks Contest 17. Prove that given any n2 integers, we can always put them in an n x n matrix whose determinant is divisible by n[n J. Titu Andreescu, Revista Matematica Timisoara 18. Let A be the set of the first 40 positive integers. Find the least n for which one can partition A into n subsets such that a b + c whenever a, b, c (not necessarily distinct) are in the same subset. Belarus 2000 476 20. PIGEONHOLE PRINCIPLE REVISITED 19. Consider a 100 x 1997 board whose unit squares are filled with 0 or 1 in such a way that every column contains at least 75 ones. Prove that we can erase 95 rows such that there exists at most one column consisting of zeros in the remaining table. Bulgaria 1997 20. Prove that no matter how we choose more than 2' 1+1 points in Rn, all of whose coordinates are ±1, there exists an equilateral triangle with vertices in three of these points. Putnam 2000 21. Let a be a real number with 0 < a < a and let (an)n>1 be an increasing sequence of positive integers such that for all sufficienly large n there are at least n • a terms of the sequence smaller than n. Prove that for all k > a there are infinitely many terms of the sequence that can be written as the sum of at most k other terms of the sequence. Paul Eras, AMM 22. Prove that for all N there exists a k such that more than N prime numbers can be written in the form T2+k for some integer T. Generalize it to any polynomial f (T). Sierpinski 23. Let f(n) be the largest prime divisor of n, and consider (an)n>1 a strictly increasing sequence of positive integers. Prove that the set containing f(ai a 3) for all i # j is unbounded. 24. Let P0, P1, ..., Pri,1 be some points on the unit circle. Also let A1A2...An be a regular polygon inscribed on this circle. Fix an integer k, with 1 < k < Z. Prove that one can find i, j such that Azi1 3 > A1Ak > PiP3• PROBLEMS FOR TRAINING 477 25. Let k be an integer, and let al, a2, ari be integers which give at least k + 1 distinct remainders when divided by n + k. Prove that some of these n numbers add up to n + k. Kornai 26. Let S be the set of the first 280 positive integers. Find the least n such that any subset with n elements of S contains 5 numbers that are pairwise relatively prime. IMO 1991 27. For a pair a, b of integers with 0 < a < b < 1000, the subset S of {1, 2, ..., 2003} is called a skipping set for (a, b) if for every pair of ele-ments (81, s2) E 82, — s21 is different from a and b. Let f (a, b) be the maximum size of a skipping set for (a, b). Determine the maximum and minimum values of f . Zuming Feng, USA TST 2003 28. Let n > 3 and let X be a subset (with 3n2 elements) of the set of the first n3 positive integers. Prove that there exist nine distinct elements of X, al, a2, a9 and nonzero integers x, y, z such that al x + a2y + a3z = 0, a4x + + a6z = 0 and a7x a8y + a9z = 0. Marius Cavachi, Romanian TST 1996 THEORY AND EXAMPLES 481 21.1 Theory and examples It is notoriously difficult to decide whether a given polynomial is irreducible over a certain field. There exist a variety of criteria that allow us to prove that a certain polynomial is irreducible, but unfortunately they are very lim-ited, and their hypotheses are usually not satisfied. Furthermore, there are not many elementary techniques: a few classical irreducibility criteria and the study of roots of polynomials are practically the only ideas that we will dis-cuss in this chapter. But, as you can easily see, even those are not trivial, and some of the problems can be extremely difficult, even though they have elementary solutions. We will discuss a very useful irreducibility criterion, Capelli's theorem, which is really not as well known as it should be, and we will see some striking consequences of this result. Also, we will insist on the method of studying the roots of polynomials, because it gives elegant solu-tions for problems of this type: Perron's criterion and Rouche's theorem are discussed, as well as some applications. Finally, we will see that working with reductions of polynomials modulo primes can often give precious information about their irreducibility properties. In this chapter, we will assume that the reader is familiar with notions of algebraic number theory, but those will not exceed the results discussed in the chapter A Brief Introduction to Alge-braic Number Theory. We will begin the discussion with the most elementary method, which is the study of roots of polynomials. Let us observe from the beginning two quite useful results: if a monic polynomial with integer coefficients f has a nonzero free term (constant term) and exactly one root of absolute value greater than 1, then f is irreducible in Q[X]. Indeed, if f = gh for some nonconstant polynomials g, h with integer coefficients, we may assume that g has all roots of absolute value smaller than 1. Then Ig(0)1 < 1, because it is just the product of the absolute values of the roots of g. Because 19(0)1 is an integer, it follows that g(0) = 0 and thus f(0) = 0, contradiction. The second result is very similar: if f is monic and all roots of f are outside the closed unit disc and If (0)1 is a prime number, then f is irreducible in Q[X]. Indeed, with the same notations, we may assume that Ig(0)1 = 1. Because 482 21. SOME USEFUL IRREDUCIBILITY CRITERIA Ig(0)1 is the product of the absolute values of the roots of g, it follows that at least one root of g is within the unit disc. But then f has at least one root in the closed unit disc, which is a contradiction. Here are some examples, the first two extremely simple, but useful, and the others more and more difficult. 1 1 Example 171 Let f (X) = ao +aiXd- • • • +a,,,,X71 be a polynomial with integer coefficients such that a() is prime and lao 1 > 'all + 1a21 + • • • + lard- Prove that f is irreducible in Z[X]. Solution. By previous arguments, it is enough to prove that all zeros of f are outside the closed unit disk of the complex plane. But this is not difficult, because if z is a zero of f and if 1z1 < 1 then laol = a2z2 ± • • • ± anzn1 < la21 + • • • + lank which contradicts the hypothesis of the problem. The previous example may look a bit artificial, but it is quite powerful for the-oretic purposes. For example, it immediately implies the Goldbach theorem for polynomials with integer coefficients: any such polynomial can be written as the sum of two irreducible polynomials. Actually, it proves much more: for any polynomial f with integer coefficients there are infinitely many positive integers a such that f + a is irreducible in Z[X]. We have already discussed algebraic numbers and some of their properties. We will see that they play a fundamental role in proving the irreducibility of a polynomial. However, we will work with an extension of the notion of alge-braic number: for any field K C C, we say that the number z E C is algebraic over K if there exists a polynomial f E K[X] such that f(z) = 0. Exactly the same arguments as those presented for algebraic numbers over Q allow us to deduce the same properties of the minimal polynomial of an algebraic number over K. Also, a is an algebraic number, the set K[a] of numbers of the form g(a) with g E K[X] is a field included in C. The following fundamental result is frequently used. THEORY AND EXAMPLES 483 L [ Exam I 2. Example Let K be a subfield of C, p a prime number, and a E K. The polynomial XP — a is reducible in K [X] if and only if there exists b E K such that a = b P . Solution. One implication being obvious, let us concentrate on the more diffi-cult part. Suppose that XP — a is reducible in K[X], and consider a such that ceP = a. Let f be the minimal polynomial of a over K and let m = deg( f). Clearly m < p. Let f (X) = (X — ai)(X — a2)...(X — am) and introduce the numbers r1 = a, ri = for i > 2. Because f divides XP — a, we have rP = 1. Hence (-1)mf (0) = cam for some c, a root of unity of order p. Since m < p, there exist integers u, v such that urn + vp = 1. It follows that (-1)um fu (0) = cuai—vp Combining this observation with the fact that aP = a, we deduce that cu a = (-1)mu f (0)u av = b E K, thus a = aP = b P . This finishes the proof of the hard part of the problem. We continue with a very beautiful result, the celebrated Cohn's theorem. It shows how to produce lots of irreducible polynomials: just pick prime num-bers, write them in any base you want and make a polynomial with the digits in that base! Example 3.] Let b > 2 and let p be a prime number Write p = ao + alb + • • • + aribn with 0 < ai < b — 1. Then the polynomial f (X) = anXn + an_1Xn-1 + • + aiX + ao is irreducible in Q [X] . [Cohn's theorem] Solution. It is clear that gcd(ao, al, ..., am) = 1, so by Gauss's lemma it is enough to prove that f is irreducible over Z. First, we will discuss the case b > 3, the case b = 2 being, as we will see, much more difficult. Suppose that f (X) = g(X)h(X) is a nontrivial factorization of f . Because p is a prime, one of the numbers g(b) and h(b) is equal to 1 or —1. Let this number be g(b) and let x1, x2 i ..., x,, be the zeros of f . There exists a subset A of the set 484 21. SOME USEFUL IRREDUCIBILITY CRITERIA {1, 2, ..., n} such that g(X) = a fl (X — xi). We now prove a helpful result. TEA Lemma 21.1. Each complex zero of f has either nonpositive real part or an /4b-3 absolute value smaller than 1+I 2 Proof. The proof is rather tricky, but not complicated. It is enough to observe that if lzl > 1 and Re(z) > 0 then Re (1) > 0 and so by the triangle inequality an + (b — 1) + • • • + T-t„.„) z > Re (an + b —1 > 1z12 — Izi — (b — 1) . z ) 1z12 Therefore if f (z) = 0 and Re(z) > 0 then either lzl < 1 or 1z1 < 1+ 246-3 and this establishes the lemma. K f (z) zn It remains now to cleverly apply this result. We claim that for any zero x, of f we have lb — x, I > 1. Indeed, if Re(x,) < 0, everything is clear. Otherwise, lb — xil > b — xi > b 1+v 6-3 > 1, as you can easily verify if b > 3. Now, everything is clear, because this result implies that Ig(b)1> 1, a contradiction. Now let us deal with the very difficult case b = 2. We will present a very beautiful solution communicated by Alin Bostan. The idea is to prove that 2 — x, 1 > I 1 — xi I for any zero xi of f. Keeping the previous notations, we will deduce that 1 = 1g(2)1 > 1g(1)1 and so g(1) = 0. This implies f (1) = 0, which is clearly impossible. Now take x to be a zero of f and observe that if 12 — xl < 11 — xl then Re(x) > (3) , and so if y = we have ly1 < 1 and y satisfies a relation of the form yn 2 ( 1 2 _ 1) yn_i , 2f 2 1 — y +1= O. Multiplying by yn+1 and adding the two relations, we find another relation of the same type (but with n increased) and by repeating this argument we THEORY AND EXAMPLES 485 deduce that there are infinitely many N for which y satisfies the relation (1 1) , 1 1 y N + -2 ± -2 + • • • + (-2 ± -2) +1 =0. This can be also written as + 2 1 1 (y + y2 + 2 yN) (±y ± y2 ± yN) The triangle inequality implies 2 y y N + 1 2(1 - y) and this for infinitely many N. Taking into account the fact that ly1 < 1, we deduce from the above inequality that ly1 - - I y I Finally, the last inequality implies 2x - 1 1 <2 lx 1 - 1 - x - 1 and thus 12x - 11 < 12x - 21, which is impossible for Re(x) > 3. This finishes the proof of the claim and also the solution of this difficult problem. We end this part of the chapter with a very beautiful criterion due to Perron and with a difficult theorem of Selmer. Perron's criterion is quite similar to the first example, but much more difficult to prove: it states that if a coefficient is "too large", then the polynomial is irreducible. Here is the precise statement: [hxample 4. Let .f (X) = X n + an_iXn-1 + • • • + aiX + ao be a polynomial with integer coefficients. If lan_11 > 1+1ao1+1a11+ • • • + lan-21 and a() 0 then f is irreducible in Q[X]. _< I j ( V -Y N y 1+ ) 1 2 - y 1 - y [Perron] 486 21. SOME USEFUL IRREDUCIBILITY CRITERIA Solution. We will prove that f has exactly one zero outside the closed unit disk of the complex plane. This will show that f is irreducible in Z [X], and by Gauss's lemma it will also be irreducible in Q[X]. It is quite clear that no zero of f is on the unit circle, because if z is such a zero, then Ian-ll = lan_izn-11 = Izn +an-2Zn-2 • ± alZ +aol < 1+ laol+- • + lan-21 1, a contradiction. On the other hand, I f (0)1 > 1, so by Viete's formula at least one zero of f lies outside the unit disk. Call this zero xi and let x2, ..., xr, be the other zeros of f . Let (X) g(X) = xn-1 bn_2xn-2 • • + biX + bo = _A — xi By identifying coefficients in the formula f (X) = (X — xi)g(X), we deduce that an-1 = bn-2 — xi, an-2 = bn-3 bn-2X1, al = bo — bixi, ao = —boxi. Therefore the hypothesis > 1+1aol + Ia1 + • • • + lan-21 can be rewritten as 1bn_2 — x11 > 1 + 1bn-3 — bn-2xil + • • • + lboxil• Taking into account that Ibn-21 + 411 > lbn-2 — xi I and lbn-3 bn-24 — 1bn-31, • • • , Ibo — bixil we deduce that — > + + • • • + Ibn-21) and since 411 > 1, it follows that jbol + l b1 I + • • • Ibn_21 < 1. Using an argument based on the triangle inequality, similar to the one in the first example, we immediately infer that g has only zeros inside the unit disk, which shows that f has exactly one zero outside the unit disk. This finishes the proof of this criterion. The above elegant solution, due to Laurentiu Panaitopol, shows that deep theorems can be avoided even when this seems impossible. The classical proof of this criterion uses Rouche's theorem. Because this is also a very powerful tool, we prefer to prove it in a very particular, but very common, case for polynomials and circles. THEORY AND EXAMPLES 487 Theorem 21.2 (Rouche's theorem). Let P, Q be two polynomials with complex coefficients and let R be a positive real number. If P, Q satisfy the inequality 1P(z) Q(z)i < 1Q(z)i for all z on the circle of radius R, centered at the origin, then the two polynomials have the same number of zeros inside the circle, multiplicities being counted. Proof. The proof of this theorem is not elementary, but with a little bit of integral calculus it can be proved in a very elegant way. Let L be the set of all curves -y : [0, 27r] —> C which are differentiable, with continuous derivative, such that -y(0) = -y(27r) and 7 does not vanish. The index of 7 E L is defined as 1 j(2' -)/(t) -1(7) 2i7r 0 -y(t) dt (21.1) t / We claim that I(y) is an integer. Indeed, consider K(t) = ef -Y ° -Y((x x)) dx and note that K is differentiable and that K'(t) = K(t) 7'(tt) • This shows that 7() K (t s a constant function. Therefore, because -y(0) = -y(27r), we must have 7(t) ) i K(0) = K(27r), which says exactly that I(y) is an integer. The following result is essential in the proof: Lemma 21.3. The index of a curve -y E L contained in a disc that does not contain the origin is 0. Proof. Let B(x, r) be the open disc of center x and radius r > 0 and suppose that -y is contained in B(w, s), a disc that does not contain the origin (thus s < ICJI) that is l-y(t) — w < s for all t. The idea is to make a continuous deformation of 7, keeping the index unchanged, and such that at a certain moment the index of the new curve can be trivially computed. In order to do this, take u E [0, 1] and consider the application fu(t) = u-y(t) + (1 — u)w, 488 21. SOME USEFUL IRREDUCIBILITY CRITERIA defined on [0, 276. The triangle inequality shows that fu E L and also that this curve is contained in B(w, s). On the other hand, we claim that the mapping cb(u) = I(fu) is continuous. Because it takes only integer values (by the previous remark), it will be constant. Therefore, I(y) = I(f1) = .1"(f0) = 0. So, let us prove that I(fu) is continuous with respect to u. Indeed, note that .g)(t) fu(t) fv(t) w • (u — v) • -/(t) (u-y(t) + (1 — u)w)(v-y(t) (1 — v)w) < PI vI1-1(t)1 — s)2 because lu-y(t)+ (1— u)wl 172117(0-w' > IwI - S. This inequality shows by integration that I(fu) satisfies IWI f27,- li(fu)-i(fv)i c 27,(1w1- 171(t)Idt • lu — v1, which proves that /(fu) is continuous, and finishes the proof of the lemma. This lemma implies that two curves in L sufficiently close have equal index. Indeed, suppose that yl and 72 are in L and satisfy 1-yi(t) — -y2(t)1 < 1'Y2(t) for all t. Then the curve -y(t) = -r]-(t) (t) 72 satisfies 1-y(t) — l < 1 for all t. Because 17(0 — 1 is also continuous on the compact interval [0, 27r], it follows that its maximum is smaller than 1, that is, there exists a disc that does not contain the origin and which contains -y. By the lemma, -y has index 0. But a quick computation shows that I(-y) = — I (-y2). Thus yi and y2 have the same index. Finally, let us prove this particular case of Rouche's theorem. Consider the curves y1(t) = P(Reit) and -y2(t) = Q(Reit). Observe that the inequality IP(z) Q(z)I < IQ(z)I implies that yi does not vanish on [0, 27]. Thus -yi, -y2 are in L and also j-yi(t) 72(01 < 1 1-Y2(01. Thus the two curves have the same index. But for a polynomial P one can easily compute the index of the associated curve! Indeed, suppose that P(z) = a(z — zi)(z — z2) • • • (z — zn), THEORY AND EXAMPLES 489 where zi are not necessarily distinct. Then it is well known that This shows that if -y(t) = P(Reit), then R n f zir eitdt I (Y) = 27 Jo Reit — zi . j=1 Now, we have seen that Izj j R. Suppose that I zj I < R. Then /27 eitdt 1dt 27r Jo Reit — zi R Jo 1— z ie-it Indeed, zm -imt = 1 + j jam — FZ Trt>1 and the mean value of e-zmt over [0, 27r] is 0 for all m > 1. It is enough to change the order of integral and summation (which is legal, because of the uniform convergence with respect to t) in order to see that 1 f27 dt Rio 1—a = 27r • Now, in exactly the same way, you can prove that /27 eitdt Jo Reit — zi =0 if 1z31 > R. Thus I (-y) is exactly the number of zeros of P inside the circle of radius R centered at the origin. This finishes the proof of Rouche's theorem. 1 490 21. SOME USEFUL IRREDUCIBILITY CRITERIA Observe that Perron's criterion instantly solves the following old IMO prob-lem: the polynomial Xn 5x-n-1 + 3 is irreducible in Q[X], just because "five is greater than four!" Here are two nicer examples, where this criterion turns out to be extremely efficient. The solution to the first problem is due to Mikhail Leipnitski. Example 5.1 Let f1 i f2, fn be polynomials with integer coefficients. Prove that there exists a reducible polynomial g E Z[X] such that all polynomials fi + g, f2 + fn + g are irreducible in Q[X]. Iranian Olympiad Solution. Using Perron's criterion, it is clear that if M is sufficiently large and m is greater than max(deg(f1), deg(f2), deg(fn)), the polynomials Xm+1 + MXm + fi(X) are all irreducible in Q[X]. Therefore we can choose g(X) = Xm+1 MXm. rExample 6.1 Let (fn)n>0 be the Fibonacci sequence, defined by fo = 0, fi = 1 and f Jn+1 = fn + fn-1- Prove that for any n > 2 the polyno-mial Xn + fnin+1Xn-1 + • • • + f2f3X + fi f2 is irreducible in Q[X]. [Valentin Vornicu] Mathlinks Contest Solution. By Perron's criterion, it suffices to verify the inequality fn-Fi fn > + + f2f1 + 1 for all n > 3. For n = 3 it is obvious. Supposing the inequality true for n, we have fn+ifn + fnfn-1 + • • • + f2h + 1 < fn+lfn + fn+ifn < fn+2fn+1, THEORY AND EXAMPLES 491 because this is equivalent to 2.fn < fro-2 = fn-Fi + fn and this one is obvious. The inductive step is proved and so is the proof for n > 3. Finally, a very difficult example of an irreducibility problem that can be solved by studying the roots of polynomials. It generalizes a classical result stating that XP — X —1 is irreducible over the field of rational numbers if p is a prime number. Example 7.] Prove that X' — X —1 is irreducible in Q[X] for all n. Selmer's theorem Solution. Let us consider a factorization Xn — X —1 -= f(X)g(X) for some integer nonconstant polynomials f, g. It is not difficult to check that X' — X — 1 has distinct complex roots. Thus f will have some roots z1i z2, •••, •zs of X' — X — 1, which are pairwise distinct. The essential observation is the following estimation: Lemma 21.4. For each root z of X' — X —1 one has 2Re — 1 —) > i 1. z z 1 i 2 Proof. By writing z = reit, the inequality comes down to (1+2r cos t)(r2 —1) > 0. However, r2" = lz12n = 1Z + 112 = 1 + 2r cost + r2, so what we need is (7,2n — r2)(r2 1) > 0, which is clear. I=1 Using the lemma, it follows that 2Re (zi — 1 1 1 —) + 2Re (z2 — —) + • • • + 2Re (z, — — z, > zi Z2 1 1 1 > + + + s > 0, I z112 1z212 1z81 2 492 21. SOME USEFUL IRREDUCIBILITY CRITERIA by the AM-GM inequality, because the product of is just If (0)1 = 1. Thus Re (zi — 1 ) + Re (z2 — 1 ) + • • • + Re (z, — — z, >0. Z1 Z2 On the other hand, because f is monic and has integer coefficients, 1 1 Re (zi — —) + Re (z2 — —) + • • • + Re (z, — — 1 Zi Z2 Zs is an integer, so it is actually at least 1. Working similarly with g, we de- duce that Re (zi — 1 + z2 — -3 — + • • • + zn — k) > 2, where zi, z2, ..., zn are Z1 Z2 Zn all roots of Xn — X —1. However, this is impossible, because by Viete's formula i z1 — + z2 — 1 + • • • + zn — Zn — = 1. This shows that any such factorization Zi Z2 is impossible, and so Xn — X — 1 is irreducible in Z[X]. All we need now is to apply Gauss's lemma to obtain a complete proof. We pass now to a proof of the celebrated Capelli's theorem. As we will immedi-ately see, this is a very powerful criterion for the irreducibility of compositions of polynomials, even though the proof is really easy. However, this does not seem to be well known, especially in the world of mathematical competitions. We thank Marian Andronache for showing us this striking result and some of its consequences. Example 8. Let K be a subfield of C and f, g E K[X]. Let a be a complex root of f and assume that f is irreducible in K [X] and g(X) — a is irreducible in K[a][X]. Then f (g (X)) is irreducible in K[X]. Capelli's theorem Solution. Define h(X) = g (X) — a and consider a zero of the polynomial h. Because f (g (13)) = f (a) = 0, /3 is algebraic over K. Let deg( f) = n, deg(h) = m and let s be the minimal polynomial of /3 over K. If we manage to prove that deg(s) = mn, then we are done, since s is irreducible over K and s divides THEORY AND EXAMPLES 493 f (g(X)), which has degree mn. So, let us suppose the contrary. By using a re-peated division algorithm, we can write s = rn_ign-i+rn,2gn-2+. • •+rig+ro, where deg(ri) < m. Hence rn_1(0)an-i. + • • • + ri (0)a + ro(0) = 0. By group-ing terms according to increasing powers of /3, we deduce from the last relation an equation kin_1(a)0" 1-1 + • • • + ki (a)13 + ko(a) = 0. Here the polynomials Is have coefficients in K and have degree at most n - 1. Because h is irreducible in K (a)[X], the minimal polynomial of 0 over K(a) is h and thus it has degree m. Therefore the last relation implies km-i. (a) = • • • = ki (a) = ko(a) = 0. Now, because f is irreducible in K[X], the minimal polynomial of a has degree n, and since deg(ki) < n, we must have km--i = • • • = ki = ko = 0. This shows that rn_i = • • • = ri = r o = 0 and thus s = 0, which is clearly a contradiction. This shows that s has degree mn, and thus it is equal (up to a multiplicative constant) to f (g(X)) and this polynomial is irreducible. This previous proof could have been written in a much shorter and con-ceptual form, using some basic facts of extensions of fields. Namely, let 0 be a zero of g - a. Then [K (a, (5) : K(a)] = deg(g) because g - is irreducible, and thus the minimal polynomial of 0 over K(a). On the other hand, f being irreducible over K, it is the minimal polynomial of a over K. Thus [K(a) : K] = deg(f). Thus, by multiplicativity of de-grees in extensions, [K(a, 0) : K] = deg(f) • deg(g). On the other hand, a = g(0), thus K(a, /3) = K(3), so the degree of /3 over K is at least deg(f) • deg(g) = deg(f(g(X)). Because f (g(X)) has j3 as zero, it follows that it is the minimal polynomial of 0 over K and so it is irreducible over K. Using the previous result, we obtain a generalization (and a more general statement) of two difficult problems given in recent Romanian TST's: Example 9.1 Let f be a monic polynomial with integer coefficients and let p be a prime number. If f is irreducible in Z[X] and .V(_i)deg(f)f (0) is irrational, then f(XP) is also irreducible in Z,[X]. Solution. Consider a a complex zero of f and let n = deg( f) and g(X) = XP and h = g - a. Using previous results, it suffices to prove that h is irreducible 494 21. SOME USEFUL IRREDUCIBILITY CRITERIA in Q[a] [X]. Because Q[a] is a subfield of C, it suffices to prove that a is not the p-th power of an element of Q[a]. Suppose there is u E Q[X] of degree at most n — 1 such that a = uP (a). Let al, a2, ..., an be the zeros of f . Because f is irreducible and a is one of its zeros, f is the minimal polynomial of a, so f must divide uP(X) — X. Therefore al • a2 • • • an = (u(ai)•u(a2) • • • u(an))P. Finally, using the fundamental theorem of symmetric polynomials, u(a1)-u(a2) • • • u(an) is rational. But al • a2 • • an = (-1)nf (0), implies ,V(-1)"f (0) E Q, a contradiction. A direct application of Capelli's theorem solves the following problem, which is not as easy otherwise: Example 10. I Prove that for each positive integer n the polynomial f (X) = (X2 + 12)(X2 + 22) (X2 + n2) + 1 is irreducible in Z[X]. Japan 1999 Solution. Consider the polynomial g(X) = (X + 12)(X + 22)...(X + n2) + 1. Let us prove first that this polynomial is irreducible in Z[X]. Suppose that g(X) =- F(X)G(X) with F, G E Z[X] nonconstant. Then F(—i2)G(—i2) = 1 for any 1 < i < n. Therefore F(—i2) and G(—i2) are equal to 1 or —1 and since their product is 1, we must have F(—i2) = G(—i2) for all 1 < i < n. This means that F — G is divisible by (X +12)(X + 22)...(X + n2) and because it has degree at most n — 1, it must be the zero polynomial. Therefore g = F2 and so (n!)2 + 1 = g(0) must be a perfect square. This is clearly impossible, so g is irreducible. All we have to do now is to apply the result in example 4. Sophie Germain's identity m4 + 4n4 = (m 2 — 2mn + 2n2)(m2 + 2mn 2n2) shows that the polynomial X4 + 4a4 is reducible in Z[X] for all integers a. However, finding an irreducibility criterion for polynomials of the form X' + a is not an easy task. The following result, even though very particular, shows that this problem is not an easy one. Actually, there exists a general criterion, also known as Capelli's criterion: for rational a and m > 2, the polynomial X' — a is irreducible in Q[X] if and only if va, is irrational for any prime p dividing m and also, if 41m, a is not of the form —4b4 with b rational. THEORY AND EXAMPLES 495 Example 11. L Let n > 2 be an integer and let K be a subfield of C. If the polynomial f (X) = — a E K[X] is reducible in K[X], then either there exists b E K such that a = b2 or there exists c E K such that a = —4c4. Solution. Suppose the contrary, that X2 — a is irreducible in K[X]. Let a be a zero of this polynomial. First, we will prove that X4 — a is irreducible in K[X]. Using the result in example 8, it is enough to prove that X2 — a is irreducible in K[a][X]. If this is not true, then there are u, v E K such that a = (u + av)2, which can be also written as v2a2 + (2uv — 1)a + u2 = 0. Because a2 E K and a is not in K, it follows that 2uv = 1 and u2 + av2 = 0. Thus a = —4u4 and we can take c = u, a contradiction. Therefore X2 — a is irreducible in K[a][X] and X4 — a is irreducible in K[X]. Now, we will prove by induction on n the following assertion: for any subfield K of C and any a E K not of the form b2 or —4c4 with b, c E K, the polynomial X2n — a is irreducible in K[X]. Assume it is true for n — 1 and take a a zero of X2 — a. Let Kt be the set of xt when x E K. Then with the same argument as above one can prove that a does not belong to —K2[a] (thus it is not in —4K4[a]) and it does not belong to K2[a]. Therefore X2Th — a is irreducible over K[a]. In the same way we prove that X2"-1 + a is irreducible over K (a). Now, observe that X2n — a = (X n.-1 a)(x2".-1 a) , so it has at most two irreducible factors over K. If it is not irreducible over K, then one of its ir-reducible factors over K will be X2n + a or X2n-1 — a, thus one of these polynomials would have coefficients in K. This would imply that a E K, which means that a is a square in K. This is a contradiction which finishes the proof. The following example is a notoriously difficult problem given a few years ago in a Romanian Team Selection Test. rExample 12 — .] Prove that the polynomial (X2 + X)2n + 1 is irreducible in Q[X] for all integers n > 0. [Marius Cavachi] Romanian TST 1997 496 21. SOME USEFUL IRREDUCIBILITY CRITERIA Solution. Using Capelli's theorem, it is enough to prove that if a is a root of f (X) = X2n + 1 (which is clearly irreducible in Q[X] by Eisenstein's theorem applied to f (X + 1)), then X2 + X — a is irreducible in Q[a] [X] (this is also immediate from the previous problem). But this is not difficult, because a polynomial of degree 2 (or 3) is reducible over a field if and only if it has roots in that field. Here, it is enough to prove that we cannot find a polynomial g E Q[X] such that g(a)2 + g(a) = a. Suppose by contradiction that g is such a polynomial. Then, if al, a2, a2Th are the roots of f it follows from the irreducibility of f that (g (a,) + = a, + 4 for all i. By multiplying these relations, we deduce that f (— 2 1) is the square of a rational number (the argument is always the same, based on the theorem of symmetric polynomi-als). But this means that 42"+1 is a perfect square, which is clearly impossible. A very efficient method for proving that a certain polynomial is irreducible is working modulo p for suitable prime numbers p. There are several criteria involving this idea, and Eisenstein's criterion is probably the easiest to state and verify. It asserts that if f (X) = a„Xn + an_1Xn-1 + • • aiX + a0 is a polynomial with integer coefficients for which there exists a prime p such that p divides all coefficients except an and p2 does not divide a0 then f is irreducible in Q[X]. The proof is not complicated. Observe first of all that by dividing f by the greatest common divisor of its coefficients, the resulting polynomial is primitive and has the same property. Therefore we may assume that f is primitive and so it is enough to prove the irreducibility in Z[X]. Suppose that f = gh for some nonconstant integer polynomials g, h and look at this equality in the field Z/pZ. Let f be the polynomial f reduced modulo p. We have gh = anXn (by convention, an will also denote an (mod p)). This implies that g (X) = bXr and h (X) = cX' for some 0 < r < n, with be = an. Suppose first that r = 0. Then h(X) = cXn pu(X) for a certain polynomial with integer coefficients u. Because p does not divide an, it does not divide c and so deg(h) > n, contradiction. This shows that r > 0 and similarly r < n. Thus there exist polynomials u, v with integer coefficients such that g(X) = bXr + pu(X) and h(X) = cX' +pv(X). This shows that a0 = f (0) = p2u(0)v(0) is a multiple of p2, contradiction. Before passing to the next example, note two important consequences of Eisen- THEORY AND EXAMPLES 497 stein's criterion. First, if p is a prime number, then f (X) = 1 + X + X 2 + • • • + XP-1 is irreducible in Q[X]. This follows from Gauss's lemma and the observation that f (X + 1) = +xy - 1) satisfies the conditions of Eisen-stein's criterion. Second, for all n there is a polynomial of degree n which is irreducible in Q[X]. Indeed, for Xri — 2, Eisenstein's criterion can be applied with p = 2 and the result follows from Gauss's lemma. The following example is more general than Eisenstein's criterion. And older! lExample 13 Let k = f n pg with n > 1, p a prime, and f and g polyno-mials with integer coefficients such that deg( fn) > deg(g), k is primitive, and there exists a prime p such that f is irre-ducible in Z/pZ[X] and f does not divide g. Then k is irreducible in Q[X]. [Schonemann's criterion] Solution. Suppose that k = k1k2 is a nontrivial factorization in polynomials with integer coefficients. By passing to Z/pZ[X] we deduce that kiq = (f )11. From the hypothesis and this equality, it follows that there exist nonnegative integers u, v with u + v = n and polynomials with integer coefficients gi, g2 such that k1 = fu + p91 and k2 = ft' + pg2, with deg(gi) < u deg(f) and deg(g2) < v deg(f). From here we infer that g = fug2+ fvgi+pg192. Because k1 is not identical 1, we have u > 0 and v > 0. Let us assume, without loss of generality, that u < v. From the previous relation there exists a polynomial h with integer coefficients such that g = fuh+pg192. It is enough to pass again in Z/pZ[X] this last relation to deduce that f divides g, which contradicts the hypothesis. Therefore F is irreducible. Here is an application of the above criterion, hardly approachable otherwise: rExample 14. Let p be a prime of the form 4k+3 and let a, b be integers such that min(vp(a), vp(b — 1)) = 1. Prove that the polynomial X2P + aX + b is irreducible in Z[X]. [Laurentiu Panaitopol, Doru $teanescu] 498 21. SOME USEFUL IRREDUCIBILITY CRITERIA Solution. Indeed, the fact that p = 3 (mod 4) ensures that X2 + 1 is irre-ducible in Z/pZ[X] (indeed, being of degree 2, it is enough to prove that it has no roots in Z/pZ, which was proved for instance in the chapter Primes and Squares). Let us try to write X2P + aX + b as (X2 + 1)P + pg(X), just as in the previous example. It is enough to take g(X) a x + b-1 + 1 . [(P P x 2(p-1) + x 2(p-2) + ... + P x-2 . P p p 1 2 P 1 Now it is immediate that all conditions of Schonemann's criterion are satisfied, so the problem is solved. Now let us see a beautiful proof of the irreducibility of the cyclotomic poly-nomials. This is not an easy problem, as the reader can immediately observe. But for the reader who is not so familiar with these polynomials, let us make a (very small) introduction. Let n be a positive integer. If n = 1 we define 01(X) = X - 1 and if n > 1 we put 2ikw (X — e n (21.3) gcd(k,n)=1,1<k co(d) = n (proved in the chapter The Smaller, the Better). din Now, let us prove the following important result: THEORY AND EXAMPLES 499 [Example 15.1 The polynomial 07., is irreducible in Q[X] for each positive integers n. Solution. Let a be a primitive root of unity of order n and let p be a prime number relatively prime to n. Let f and g be the minimal polynomials of a and aP over the field of rational numbers. Because a is an algebraic integer, f, g have integer coefficients. Also, because g(aP) = 0, it follows that f divides g(XP). The idea is that in Z/pZ we have g(XP) = g(X)P and so if f and g are the polynomials f, g reduced modulo p, then f divides (g)P in Z/pZ[X]. Thus, if r is a root of f in some algebraic closure of Z/pZ, then g(r) = 0. Now, suppose that f g. Both f and g divide On in Z[X], because aP is also a primitive root of unity. Because f g are irreducible, they are relatively prime and fg divides On in Z[X], thus fg divides Xn — 1 (seen as a poly-nomial in Z/pZ[X]). But this is impossible, because it would follow that r is a root of multiplicity at least 2 of the polynomial Xn — 1 modulo p, that is we also have nrn-1 = 0 in that algebraic extension. Because n and p are relatively prime, this implies that r = 0, which is impossible, because rTh = 1. The above contradiction shows that f = g, that is a and aP have the same minimal polynomial for all prime numbers p relatively prime to n. This im-mediately implies that a and Cek have the same minimal polynomial for all k relatively prime to 71. Thus, the minimal polynomial of a must have as roots all primitive roots of unity of order n and thus degree at least (,o(n), which means that it is On, that is On is irreducible. There exists another beautiful proof of this result, but which uses the difficult (and non elementary) Dirichlet's theorem on primes in arithmetic progressions. Let w be a primitive root of unity of order n and let s = cp(n) = deg(0,). Also, let f be an irreducible factor of On with integer coefficients, which has w as a zero. Then the zeros of f (which, as we have seen in chapter A Brief Introduction to Algebraic Number Theory, are called the conjugates of w) are of the form wt. Also, if On is not irreducible then the number of zeros of f is smaller than s. Now, take p to be a prime number. Because f is monic and has all zeros of absolute value 1, it follows that < 2s. But 500 21. SOME USEFUL IRREDUCIBILITY CRITERIA because f(w) = 0, it follows that f is is an algebraic integer (this result is not obvious, but it has been discussed in the same chapter). Its conjugates are also algebraic integers of the form f(wtP). Thus if we choose p > 2S, then all conjugates of the algebraic integer f(wP) are inside the unit disc of the complex plane, thus f(wP) = 0 (indeed, if x = f(u)P) and g is the minimal polynomial of x, then by Gauss's lemma g has integer coefficients, and thus the product of the absolute values of all conjugates of x is just lg(0)1; if all conjugates are inside the unit disc, then g(0) = 0 and because g is irreducible, g(X) = X, thus x = 0). Therefore, for any prime number p > 2', wP is a zero of f. All we need to observe now is that Dirichlet's theorem assures us of the existence of infinitely many primes p r (mod n) for any r such that gcd(r, n) = 1. Therefore all of with gcd(r, n) = 1 are zeros of f, which shows that deg(f) > deg(On) and proves the irreducibility of On. PROBLEMS FOR TRAINING 501 21.2 Problems for training 1. Let n be an integer greater than 2. Prove that the polynomial f(X) = X(X — (n! +1))(X — 2(n! + 1)) . - - (X — (n —1)(n! +1))+n! is irreducible in Z[X], but f(x) is composite for all integers x. 2. Let p be an odd prime and k > 1. Prove that for any partition of the set of positive integers into k classes there is a class and infinitely many polynomials of degree p — 1 with all coefficients in that class and which are irreducible in Z[X]. Marian Andronache, Ion Savu, Unesco Contest 1995 3. Find the number of irreducible polynomials of the form XP + pXc + pX1 +1, where p > 5 is a fixed prime number and k, 1 are subject to the conditions 1 < 1 < k < p —1. Valentin Vornicu, Romanian TST 2006 4. Find all integers k such that X"±1 + kXTh — 870X2 + 1945X + 1995 is reducible in Z[X] for infinitely many M. Vietnamese TST 1995 5. Let p and q be distinct prime numbers and n > 3. Find all integers a for which X' + aXn-1 + pq is reducible in Z[X]. Chinese TST 1994 6. Let n and r be positive integers. Prove the existence of a polynomial f with integer coefficients and degree n such that for any polynomial g with integer coefficients and degree at most n, if the coefficients of f — g have absolute values at most r, then g is irreducible in Q[X]. Miklos Schweitzer Competition 502 21. SOME USEFUL IRREDUCIBILITY CRITERIA 7. Prove that for any positive integer n, the polynomial (x-2 ± 2)n + 5(X2n-1. 10Xn ± 5) is irreducible in Z[X]. Laurentiu Panaitopol, Doru Stefanescu 8. Let p be a prime of the form 4k +3 and let n be a positive integer. Prove that (X2 + 1)n + p is irreducible in Z[X]. N. Popescu, Gazeta Matematicg 9. Find all positive integers n such that Xn + 64 is reducible in Q[X]. Bulgarian Olympiad 10. Let f (X) = amXm + am_iXm-1 + • • • + aiX + a0 be a polynomial of degree m in Z[X] and define H = max IN. If f (n) is prime for o H + 2 then f is irreducible in Z[X]. AMM 11. Let f be a monic polynomial of fourth degree which has exactly one real zero. Prove that f is reducible in Q[X]. MOSP 2000 12. Let a and n be integers and p be a prime such that p > 'al + 1. Prove that Xn + aX + p is irreducible in Z[X]. Laurentiu Panaitopol, Romanian TST 1999 PROBLEMS FOR TRAINING 503 13. Let p > 3 be a prime number and m, n be positive integers. Prove that X' + X' + p is irreducible in Z[X]. Laurentiu Panaitopol 14. Let p be a prime number and let k be an integer not divisible by p. Prove that XP — X + k is irreducible in Z[X]. 15. Let A be the ring of Gaussian integers Z[i] and let zi , z2, E A be such that — ziI > 2 for all i > 1. Prove that the polynomial 1 + (X — zi)(X — z2) • • • (X — zn) is irreducible in A[X]. Oral Examination ENS 16. Let f e Z[X] be a monic polynomial irreducible in Z[X], and suppose that there exists a positive integer m such that f(X") is reducible in Z[X]. Show that for any prime p dividing f (0) we have vp( f (0)) > 2. 17. Let f be a monic polynomial with integer coefficients having distinct integer roots. Prove that f2 + 1 and f4 + 1 are irreducible in Q[X]. 18. Let p, q be odd prime numbers such that p 1 (mod 8) and (7 ) = 1. Prove that the polynomial (X2 — p + q)2 — 4qX2 is irreducible in Z[X] but that it is reducible mod m for all integers m. David Hilbert 19. Prove that for all positive integers d there is a monic polynomial f of degree d such that Xn + f (X) is irreducible in Z[X] for all n. 20. Let d > 1 be an integer and let f (n) be the probability that a polyno-mial of degree n with all coefficients bounded by n in absolute value is reducible in Z[X]. Prove that f (n) = 0(in2n). 504 21. SOME USEFUL IRREDUCIBILITY CRITERIA 21. Let f be a primitive polynomial with integer coefficients of degree n for which there exist distinct integers xi, x2, ..., xr, such that 171+1 I! 0 < If(x2)1 < L n 2+1 7 1 . Prove that f is irreducible in Z[X]. Polya-Szego 22. Factor the polynomial X2005 — 2005X + 2004 over Z[X]. Valentin Vornicu, Mathlinks Contest 23. Is there a polynomial f with rational coefficients such that f(1) —1 and Xnf(X) +1 is reducible for all n > 1? Schinzel 24. Let f be an irreducible polynomial in Q[X] of degree p, where p > 2 is prime. Let xi, x2, ..., xp be the zeros of f. Prove that for any nonconstant polynomial g with rational coefficients, of degree smaller than p, the numbers g(xi),g(x2),...,g(xp) are pairwise distinct. Toma Albu, Romanian TST 1983 25. Let a be a nonzero integer. Prove that the polynomial + aXn-1 + + aX 2 + aX 1 is irreducible in Z[X]. Marian Andronache, Ion Savu, Romanian Olympiad 1990 26. Let p1, p2, ...,pn be distinct prime numbers. Prove that the polynomial f (x) = (x + eiN/Fi + e2VP2 + • • • + enN/F 971) ei,e2,•••,en=±1 is irreducible in Z[X]. THEORY AND EXAMPLES 507 22.1 Theory and examples After a very elementary chapter about extremal properties of graphs, it is time to see how the study of their cycles can give valuable information in combinatorial problems. We will assume in this chapter some familiarity with basic concepts of graph theory that can be found in practically any book of combinatorics. We prefer to do so, because recalling all definitions would re-quire a large digression and would largely diminish the quantity of examples presented. And since the topic is very subtle and the problems are in general difficult, we think it is better to present many examples. We would like to thank Adrian Zahariuc for the large quantity of interesting results and solu-tions that he communicated to us. We start with a simple, but important result. It was extended by Eras in a much more difficult to prove statement: if the number of edges of a graph on n vertices is at least (n-21)k then there exists a cycle of length at least k + 1 (if k > 1). Let us remain modest and prove the following much easier result : Example 1.1 In a graph G with n vertices, every vertex has degree at least k. Prove that G has a cycle of length at least k + 1. Solution. The shortest solution uses the extremal principle. Indeed, consider the longest chain xo, xi, ..., xi. in G and observe that this maximality property ensures that all vertices adjacent to x0 are in this longest chain. Or, the degree of x0 being at least k, we deduce that there exists a vertex xi adjacent to x0 such that k < i < r. Therefore xo, x1, ..., xi, x0 is a cycle of length at least k+1. Any graph with n vertices and at least n edges must have a cycle. The following problem is an easy application of this fact: Example 2.1 Suppose 2n points of an n x n grid are marked. Prove that there exists a k > 1 and 2k distinct marked points al, a2, •••, a2k such that for all i, a2i_1 and a2, are in the same row, while a2z and a2,4_1 are in the same column. IMC 1999 508 22. CYCLES, PATHS, AND OTHER WAYS Solution. Here it is not difficult to discover the graph to work on. It is enough to look at the n lines and n columns as the two classes of a bipartite graph. We connect two vertices if the intersection of the corresponding row and column is marked. Clearly, this graph has 2n vertices and 2n edges, so there must exist a cycle. But the existence of a cycle is equivalent (by the definition of the graph) to the conclusion of the problem. The following example is an extremal problem in graph theory, of the same kind as Turan's theorem. This type of problem can go from easy or even trivial to extremely complex and complicated results. Of course, we will discuss just the first type of problem. [Example 3.1 Prove that every graph on n > 4 vertices and m > n±n V 4 4n-3 edges has at least one 4-cycle. Solution. Let us count, in two different ways, the number of triples (c, a, b) where a, b, c are vertices such that c is connected to both a and b. For a fixed vertex c, there are d(c)2 — d(c) possibilities for the pair (a, b), where d(c) denotes the valence of c. It follows that there are at least E(d(c)2 — d(c)) triples. By the Cauchy-Schwarz inequality, if m represents the number of edges of the graph, then 2 E d(e)2 _ d(c) > 4m 2m (22.1) C Now, if there are no 4-cycles, then for fixed a and b there is at most one vertex c that appears in a triple (a, b, c). Hence we obtain at most n(n — 1) triples. It follows that 47712 2m < n 2 n, which implies that m < n±n 44n-3 a contradiction. THEORY AND EXAMPLES 509 Recall that a graph in which every vertex has degree 2 is a disjoint union of cycles. It turns out that this very innocent observation is more than helpful in some quite challenging problems. Here are some examples, taken from different contests: Example 4. A company wants to build a 2001 x 2001 building with doors connecting pairs of adjacent rooms (which are 1 x 1 squares, two rooms being adjacent if they have a common edge). Is it possible for every room to have exactly 2 doors? [Gabriel Carol]] Solution. Let us analyze the situation in terms of graphs: suppose such a situation is possible, and consider the graph G with vertices representing the rooms and connecting two rooms if there exists a door between them. Then the hypothesis says that the degree of any vertex is 2. Thus G is a union of disjoint cycles C1, C2, ..., Cp. However, observe that any cycle has even length, because the number of vertical steps is the same in both directions and the same holds for horizontal steps. Therefore the number of vertices of G, which is the sum of lengths of these cycles, is an even number, a contradiction. Reading the solution to the following problem, one might say that it is ex-tremely easy: there is no tricky idea behind it. But there there are many possible approaches that can fail, and this probably explains its presence on the list of problems proposed for the IMO 1990. Example 5.1 Let E be a set of 2n —1 points on a circle, with n > 2. Suppose that precisely k points of E are colored black. We say that this coloring is admissible if there is at least one pair of black points such that the interior of one of the arcs they determine contains exactly n points of E. What is the smallest k such that any coloring of k points of E is admissible? IMO 1990 510 22. CYCLES, PATHS, AND OTHER WAYS Solution. Consider G the graph having vertices the black points of E and join two points x, y by an edge if there are n points of E on one of the two open arcs determined by x and y. Thus the problem becomes: what is the least k such that among any k vertices of this graph at least two are adjacent? The problem becomes much easier with this statement, because of the fact that the degree of any vertex in G is clearly 2, thus G is a union of disjoint cycles. It is clear that for a single cycle of length r, the least value of k is 1 + L 2 J . Now, observe that if 2n - 1 is not a multiple of 3 then G is ac-tually a cycle (because (gcd(n + 1, 2n - 1) = 1), while in the other case G is the union of three disjoint cycles of length 2n1 . Therefore the least k is n= L 12n 2 1] +1 if 2n -1 is not a multiple of 3 and n 1 = 3 [2n 6 -1] +1 otherwise. Finally, a more involved example using the same idea, but with some compli-cation which are far from obvious. L Example 6 Consider in the plane the rectangle with vertices (0, 0), (m, 0) (0, n), (m, n), where m and n are odd positive integers. Par-tition it rectangle into triangles satisfying the following condi-tions: 1) Each triangle has at least one side (called the good side; the sides that are not good will be called bad) on a line x = j or y = k for some nonnegative integers j, k, such that the height corresponding to that side has length 1; 2) Each bad side is common for two triangles of the partition. Prove that there are at least two triangles having two good sides each. IMO 1990 Shortlist Solution. Let us define a graph G having as vertices the midpoints of the bad sides and as edges the segments connecting the midpoints of two bad sides in a triangle of the partition. Thus, any edge is parallel to one of the sides of the rectangle, being at distance k from the sides of the rectangle, for a suitable integer k. Also, it is clear that any vertex has degree at most 2, so we have three cases. The easiest is when there exists an isolated vertex. Then the THEORY AND EXAMPLES 511 triangles that have the side containing that vertex as common side have two good sides. Another easy case is when there exists a vertex x having degree 1. Then x is the end of a polygonal line formed by edges of the graph, and having the other end a point y, which is the midpoint of a side in a triangle having two good sides. The conclusion follows in this case, too. Thus, it remains to cover the "difficult" case when all vertices have degree 2. Actually, we will show that this case is impossible. Observe that until now we haven't used the hypothesis that m, n are odd. This suggests looking at the cycles of G. Indeed, we know that G is a union of disjoint cycles. If we manage to prove that the number of squares traversed by any cycle is even, it would follow that the table has an even number of unit squares, which is impossible, because mn is odd. Divide first the rectangle by its lattice points into mn unit squares. So, fix a cycle and observe that from the hypothesis it follows that the center of any square is contained in only one cycle. Now, by alternatively coloring the cells of the rectangle with white and black, we obtain a chessboard in which every cycle passes alternatively on white and black squares, so it passes through an even number of squares. This proves the claim and shows that G cannot have all vertices of degree 2. The next problem is already unobvious, and the solution is not immediate, because it requires two arguments which are completely different: a construc-tion and a proof of optimality. Starting with some special cases is often the best way to proceed, and this is indeed the key here. Example 7. i Let n be a positive integer. Suppose that n airline companies offer trips to citizens of N cities such that for any two cities there exists a direct flight in both directions. Find the least N such that we can always find a company which can offer a trip in a cycle with an odd number of landing points. Adapted after IMO 1983 Shortlist Solution. By starting with small values of n, we can guess the answer: N = 2' +1. But it is not obvious how to prove both that for 2' the assertion in the 512 22. CYCLES, PATHS, AND OTHER WAYS problem is not always true and the fact that for 272 + 1 cities the conclusion always holds. Let us start with the first claim: the result is not always true if we allow only 2n cities. Indeed, let the cities be Co, C1, C2n-i. Write every number smaller than 2n in base 2 with n digits (we allow zeros in the first positions), and let us join two cities Ci and C3 by a flight offered by an airline company Al if the first digit of i and j is different, by a flight offered by A2 if the first digits are identical, but the second digit differs in the two numbers and so on. Because the i-th digit is alternating in the vertices of a cycle for company Ai, it follows that all cycles realized by 24.7, are even. Therefore N > 2n + 1. Now, we prove by induction that the assertion holds for N > 2n + 1. For n = 1 everything is clear, so assume the result for n — 1. Suppose that all cycles in the graph of flights offered by company An are even (otherwise we have found our odd cycle). Therefore the graph of flights offered by An is bipartite, that is there exists a partition B1, B2, ..., Bni, D1, D2, ..., Dp of the cities such that any flight offered by An connects one of the cities Bi with one of the cities Dk. Because m+p = 2n + 1, we may assume that m > 2n-1+1. But then the cities B1, B2, ..., Bm are connected only by flights offered by A1, A2, ..., An_1, so by the induction hypothesis one of these companies can offer an odd cycle. This finishes the induction step and shows that N = 2n + 1 is the desired number. Here comes a very challenging problem with a very beautiful idea: Example 80 On an infinite checkerboard are placed 111 non-overlapping corners, L-shaped figures made of 3 unit squares. Suppose that for any corner, the 2 x 2 square containing it is entirely covered by the corners. Prove that one can remove each num-ber between 1 and 110 of the corners so that the property will be preserved. St. Petersburg 2000 Solution. We will argue by contradiction. Assuming that by removing any 109 corners the property is no longer preserved, it would follow that no 2 x 3 rectangle is covered by 2 corners. Now, define the following directed graph with vertices on the corners: for a fixed corner C, draw an edge from it to the THEORY AND EXAMPLES 513 corner that helps covering the 2 x 2 square containing C. It is clear that if in a certain corner there is no entering edge, we may safely remove that corner, contradiction. Therefore, in every corner there exists an entering edge and so the graph constructed has the property that every edge belongs to some cycle. We will prove that the graph cannot be a cycle of 111 vertices. Define the "center" of a corner as the center of the 2 x 2 square containing it. The first observation, that no two corners can cover a 2 x 3 rectangle, shows that in a cycle the x coordinate of the centers of the vertices are alternatively even and odd. Thus the cycle must have an even length, which shows that the graph itself cannot be a cycle. Therefore, it has at least two cycles. But then we may safely remove all the corners except those in a cycle of smallest length and the property will be preserved, thus again a contradiction. The following result is particularly nice: There are n competitors in a table-tennis contest. Any 2 of them play exactly once against each other and no draws are possible. We know that no matter how we divide them into 2 groups A and B, there is some player from A who defeated some player from B. Prove that at the end of the competition, we can sit all the players at a round table such that everyone defeated his or her right neighbor. Solution. Clearly, the problem refers to a tournament graph, that is, a di-rected graph in which any two vertices are connected in exactly one direction. We have to prove that this graph contains a Hamiltonian circuit. Take the longest elementary cycle, v1, v2, ..., vrn with pairwise distinct vertices, and take some other vertex v. Unless all edges come either out of v or into v, there is some i such that viv and vvi±i are edges. Then, vi, v2, ..., vZ v, vi+i, vim, is a longer elementary cycle, contradiction. Therefore, there are only two kinds of vertices v E V — {vi}: (type A) those for which all vv, are edges; and (type B) those for which all v.,v are edges. If there is some edge ba with a of type A and b of type B, then we can construct once again a longer circuit: b, a, v1, ..., vim,. Therefore, for any a E A and b E B, ab is an edge. Consider the partition V = B U (A U {v2}). Due to the hypothesis, since all edges between the two classes point towards B, we must have B = 0. But, once again, V = A U { 514 22. CYCLES, PATHS, AND OTHER WAYS is a forbidden partition, so A = 0. Therefore, the circuit is Hamiltonian. Before discussing the next problem, we need to present a very useful result, which is particularly easy to prove, but has interesting applications. This is why it will be discussed as a separate problem and not as a lemma: [Example 10.1 Prove that a graph is bipartite if and only if all of its cycles have even length. Solution. One part of the result is immediate: if the graph is bipartite then obviously it cannot have odd cycles, because there is no internal edge in one of the two classes of the partition. The converse is a little bit trickier. Suppose that a graph G has no odd cycles and start your "journey" with an arbitrary vertex v and color this vertex white. Continue your trip through the vertices of the graph, by coloring all neighbors of the initial vertex in black. Continue in this manner, by considering this time every neighbor of v as an initial point of a new trip and color new vertices by the described rule, avoiding vertices that are already assigned a color. We must prove that you can do your trip with no problem. But the only problem that may occur is to have two paths to a certain vertex (called a problem vertex), each leading to a different color. But this is impossible, since all cycles are even. Indeed, any two paths from v to this problem vertex must have the same parity. Therefore we have a valid coloring of the vertices of the graph, and by construction this proves that G is bipartite. And here is an application: Example 11. A group consists of n tourists. Among any 3 of them there are 2 who are not familiar. For every partition of the tourists in 2 buses, we can always find 2 tourists that are in the same bus who are familiar with each other. Prove that there is a tourist who is familiar with at most ki tourists. Bulgaria 2004 for some k. The solution ends here. THEORY AND EXAMPLES 515 Solution. Construct a graph G on n vertices corresponding to the n tourists. We construct the edge ab if and only if the tourists a and b are familiar with each other. By the hypothesis, G is not bipartite, so it must have an odd cycle. Let al, a2, a/ be the smallest odd cycle. Since 1 is odd and 1 > 3, we must have 1> 5. It is clear that there are no other edges among the a2 except aini+1. If some vertex v is connected to a, and a3, it is easy to show that the "distance" between i and j is 2, that is equals 2 or 1— 2, since otherwise we would have a smaller odd cycle. Therefore, every vertex which does not belong to the cycle is adjacent to at most 2 ai's. Even more, every vertex of the cycle is connected to exactly 2 ai's. Therefore, if c(v) is the number of edges between v and the vertices of the cycle, c(v) < 2, so At first glance, the following has nothing to do with graphs and cycles. Well, it does! Here is a beautiful solution by Adrian Zahariuc: r xample 12.1 In each square of a chessboard is written a positive real num-ber such that the sum of the numbers in each row is exactly 1. It is known that for any 8 squares, no two in the same row or column, the product of the numbers written in these squares does not exceed the product of the numbers on the main diagonal. Prove that the sum of the numbers on the main diagonal is at least 1. St. Petersburg 2000 516 22. CYCLES, PATHS, AND OTHER WAYS Solution. First, let us label the rows and the columns 1,2, ..., 8, consecutively, in increasing order. Suppose by way of contradiction that the sum of the num-bers on the main diagonal is less than 1. Then on row k there is some cell (k, j) such that the number written in it is greater than the number written in cell (j, j), that is, the one in the same column, on the main diagonal. Color (k, j) red and draw an arrow from row k to row j. Some of these arrows must form a loop. From each row belonging to the loop we choose the red cell, and from all other rows we choose the cell on the main diagonal. All these 8 cells lie in different rows and different columns and their product exceeds the product of the numbers on the main diagonal, a contradiction. Therefore our assumption is false, and the sum of the numbers on the main diagonal is at least 1. And for the die-hards, here are two very difficult problems communicated to us by Adrian Zahariuc: Example 13. There are two airline companies in Wonderland. Any pair of cities is connected by a one-way flight offered by one of the companies. Prove that there is a city in Wonderland from which any other city can be reached via airplane without changing the company. Iranian TST 2006 Solution. We would rather reformulate the problem in terms of graph theory: given a bichromatic (say, red and blue) tournament G(V, E) (i.e. a directed graph in which there is precisely one edge between any pair of vertices). We have to prove that there is a vertex v such that, for any other vertex u, there is a monochromatic directed path from v to u. Such a point will be called "strong". Let V I = n. We will prove the claim by induction on n. The base case is trivial. Suppose it is true for n — 1; we will prove it for n. Now suppose by way of contradiction that the claim fails for some G. By the inductive hypothesis we know that for each v E V there is some s(v) E V-{v} which is a strong point in G — {v}. Clearly, s(v) s(v") for all v v', since THEORY AND EXAMPLES 517 otherwise s(v) would be strong in G. Let f = s-1, i.e. s(f(v)) = v for all v. It is clear that from v we can reach all points through a monochromatic path except f(v). For each v, draw an arrow from v to f(v). These arrows must form a loop. If this loop does not contain all n vertices of the graph, by the inductive hypothesis we must have a strong vertex in this graph, which contradicts the fact that we can't reach f(v) from v. Hence, this loop is a Hamiltonian circuit v1, v2, ..., vn. Let vn+1 =- vi. From vi, we can reach all vertices except vi+1 because vi+i = f(vi). We can't reach v from u through paths of both colors since, in that case, from u we could reach all the points we could reach from v, including f (u), which is false. For v f (u), let c(uv) be the color of all paths from u to v. It is clear that c(uv) c(v f (u)). We have c(uv) c(v f (u)) c(f (u) f (v)), so c(uv) = c(f (u) f (v)) for u # v f (u). In other words, c(vkvk+m) = c(vio_ivk±„,+1)• From here, it is easy to fill in the details. Basically, we just have to take m > 1 coprime with n to get that we can travel between any two points through paths of color c(vovm) and we are done. Example 14. Does there exist a 3-regular graph (that is, every vertex has degree 3) such that any cycle has length at least 30? St. Petersburg 2000 Solution. Even though the construction will not be easy, the answer is: yes, there does. We construct a 3-regular graph G, by induction on n such that any cycle has length at least n. Take G3 = K4, the complete graph on 4 vertices. Now, suppose we have constructed Gn(V, E) and label its edges 1, 2, ..., m. Take an integer N > n2m and let V' = V x ZN. If the edge numbered k in G, is ab, we draw an edge in an±i(Vi, .E') between (a, x) and (b, x + 2k) for all x E ZN. It is clear that Gri+1 is 3-regular. We show that Gn+i has the desired property, i.e. it contains no cycle of length less than n + 1. Suppose (a1, x1), ..., (at, xt) is a cycle with t < n. Clearly, al, a2, •.., at is a cycle of Gn. 518 22. CYCLES, PATHS, AND OTHER WAYS Therefore t = n, and all ch are distinct. We have n 0 = (xi — x2) + (x2 — x3) + • • + (xn — ±2ki (mod N). (22.3) =1 This sum is nonzero since all k3 are distinct, and also it is at most n2m < N in absolute value, a contradiction. Therefore this graph has all the desired properties, and the inductive construction is complete. PROBLEMS FOR TRAINING 519 22.2 Problems for training 1. Prove that any graph on n > 3 vertices having at least 2 + ( n21) edges has a Hamiltonian cycle. Does the property remain true if 2 + (n21) is replaced by a smaller number? 2. In a group of 12 people, among any 9 persons one can find five, any two of whom know each other. Show that there are 6 people in this group, any two of whom know each other. Russia 1999 3. In a connected simple graph any vertex has degree at least 3. Prove that the graph has a cycle such that it remains connected after the edges of this cycle are deleted. Kornai 4. For a given n > 2 find the least k with the following property: any set of k cells of an n x n table contains a nonempty subset A such that in every row and every column of the table there is an even number of cells belonging to A. Poland 2000 5. In a society of at least 7 people each member communicates with three other members of the society. Prove that we can divide this society in two nonempty groups such that each member communicates with at least 2 members of their own group. Czech-Slovak Match 1997 520 22. CYCLES, PATHS, AND OTHER WAYS 6. Let n be a positive integer. Can we always assign to each vertex of a 2n-gon one of the letters a and b such that the sequences of letters obtained by starting at a vertex and reading counterclockwise are all distinct? Japan 1997 7. On an n x rt table real numbers are put in the unit squares such that no two rows are identically filled. Prove that one can remove a column of the table such that the new table has no two rows identically filled. 8. Let G be a simple graph with 2n + 1 vertices and at least 3n + 1 edges. Prove that there exists a cycle having an even number of edges. Prove that this is not always true if the graph has only 3n edges. Miklos Schweitzer Competition 9. There are 25 towns in a country. Find the smallest k for which one can set up bidirectional flight routes connecting these towns so that the following conditions are satisfied: (i) from each town there are exactly k direct routes to k other towns; (ii) if two towns are not connected by a direct route, there is a town which has direct routes to these two towns. Vietnamese TST 1997 10. Let G be a tournoment (directed graph such that between any two ver-tices there is exactly one directed edge) such that its edges are colored either red or blue. Prove that there exists a vertex of G, say v, with the property that for every other vertex u there is a monochromatic directed path from v to u. Iranian TST 2006 PROBLEMS FOR TRAINING 521 11. Some pairs of towns are connected by road. At least 3 roads leave each town. Show that there is a cycle containing a number of towns which is not a multiple of 3. Russia 12. Prove that the m [ aximal 3( number of edges in a graph of order n without n 1) an even cycle is 2 1' 13. On the edges of a convex polyhedra we draw arrows such that from each vertex at least one arrow is pointing in and at least one is pointing out. Prove that there exists a face of the polyhedra such that the arrows on its edges form a circuit. Dan Schwartz, Romanian TST 2005 14. A connected graph has 1998 points and each point has degree 3. If 200 points, no two of them joined by an edge, are deleted, show that the result is still a connected graph. Russia 1998 THEORY AND EXAMPLES 525 23.1 Theory and examples Undoubtedly, polynomials represent a powerful tool in practically any area of mathematics, simply because they manage to create a subtle link between analysis and algebra: on the one hand, considering them as formal series comes handy in arithmetic and combinatorics; on the other hand their analytic prop-erties (location of zeros, complex-analytic properties, etc) are particularly in-teresting for effective estimations. The purpose of this chapter is to present some striking applications of these ideas in number theory and combinatorics. We will merely scratch the surface, but we are convinced that even this small amount will show the reader what profound mathematical objects polynomi-als are. A particularly important result to be discussed is the revolutionary "Combinatorial Nullstellensatz" of Noga Alon, which shows perfectly well the power of algebraic methods in combinatorics. We begin, as usual, with a very easy problem. However, it is not entirely trivial because there are many approaches that can fail. A purely algebraic solution is both easy and insightful. Is there a set of points in space which cuts any plane in a finite, nonzero number of points? IMO 1987 Shortlist Solution. The idea is very simple: by taking such a set A to be the set of points of the form (f (t), g(t), h(t)), we need to find functions f, g, h such that for any a, b, c not all zero and any d, the equation a f (t) + bg(t) + ch(t) + d = 0 has a finite nonzero number of solutions. This suggests taking polynomials f, g, h. One of the many choices is f (t) = t5, g(t) = t3 and h(t) = t. Indeed, the equation at5 bt 3 ct d = 0 clearly has a finite number of solutions and has at least one, since any polynomial of odd degree has at least one real root. This shows that such a set exists. 526 23. SOME SPECIAL APPLICATIONS OF POLYNOMIALS You may very well know the classical problem stating that if 2' + 1 is a prime number, then n is a power of 2 (the reader who does not know it is urged to give it some thought before passing to the next problem!). The following example is an adaptation of this classical result, but it is not as immediate as the cited problem. Prove that if 4" 1 — 2rn + 1 is a prime number, then all prime divisors of m are smaller than 5. [S. Golomb] AMM Solution. Suppose that p is a prime divisor of m, with p > 3. Write m = np. Then en - 2m + 1 = P(-2"), where P(X) = X2P + XP + 1. We claim that P is a multiple of X2 + X +1. Indeed, X2 + X +1 has distinct complex roots and any of its roots is clearly a root of P. Therefore X2 + X +1 divides P in C[X], thus in Q[X] too. Because X2 + X +1 is monic, Gauss's lemma implies that P is divisible by X2 + X +1 in Z[X]. Therefore, P(-211) is a multiple of 4' — 2n + 1 > 1, so 4m — 2' + 1 is not a prime number. We continue with a fairly tricky problem, whose beautiful solution was com-municated by Gheorghe Eckstein. This will be a preparation for the next challenging problem. Prove that the number obtained by multiplying all 2100 num- bers of the form +1 ± '/100 is the square of an integer. Tournament of the Towns Solution. The crucial observation is that if P E Z[X] is an even polynomial, then for every positive integer k, the polynomial P(X — fi-c)P(X + /) is also an even polynomial with integer coefficients. Now, consider the polynomials -P1 (X) = X, Pk (X) = Pk-1(X - 1/k)Pk_i (X + VTC) THEORY AND EXAMPLES 527 for k > 2. By the first observation, Pioo is an even polynomial with integer coefficients. But it is clear that the desired product is just P100(1)P1oo(-1), so it is a perfect square. This finishes the solution. As we said, the next problem is very challenging. The solution presented here is due to Pierre Bornsztein, and can be adapted to prove much more: the square roots of the squarefree positive integers are linearly independent over the set of rational numbers. There are also elementary proofs of this deep result, but the following argument is simply stunning. Interested readers will find in the exercise section a much more general (and difficult) statement that can be proved using polynomial techniques, and which we strongly recommend. Example 4. Let al, a2, an be positive rational numbers such that Val + .\/t2 + + Van is a rational number. Prove that the ai are all rational numbers. Solution. If all xi = c1 ,,„ then x2 are rational numbers and the sum S of the xi's is also rational. Let us assume that x1 is not rational and consider the polynomial P(X) = H (X - + u2x2 + • • • + unxn) (23.1) Clearly, when we expand this polynomial x2, x3, ..., xn appear with even ex-ponents because the polynomial is invariant under the substitutions x2 —> —x2, ..., xn --> —xn. After expansion, the polynomial can be written as P(X) = for some polynomials with rational coefficients N and D. Because P vanishes at S, we deduce that xiD(S,xT,...,xn 2) = N(S,x7,...,xn 2), and the assump- tion that x1 is irrational implies that D(S, xi, ...,x2 n) = N(S,xT, ..., x7, 2) = 0. 528 23. SOME SPECIAL APPLICATIONS OF POLYNOMIALS But then we also have P(S, -xi, x2, ...,xn) = 0, which is impossible, since P(S, -xi, x2, ..., xii) is a product of positive numbers. This contradiction shows that xi is rational and, by induction, all xi are rational. The following problem became a classical application of polynomial techniques. It was also used in a Balkan Mathematical Olympiad and more recently in a Chinese TST. The following solution is probably a reason for its popularity. Example 5. A positive integer p is prime if an only if each equiangular polygon with p vertices and rational side-lengths is regular. Solution. We will first prove that if n is a positive integer, E = etr7, and al, a2, ..., an are positive real numbers, then there exists an n-gon with equal angles and side-lengths al, a2, ..., an if and only if al + a2E + • • • + anEn-1 = 0. This is not difficult: it is enough to consider the edges of the polygon as ori-ented vectors in clockwise direction. Clearly, their sum is 0. However, one can translate these vectors so that all of them have origin at 0, the origin of the plane. By choosing the positive semiaxis al, the complex numbers correspond-ing to the extremities of the vectors are al, a2E, anin-1, from where we find al + a2E + • • • + anEn-1 = 0. The converse is easy, because the construction follows from the previous argument. Now assume that p is a prime number, and consider a polygon with side-lengths al, a2, %, all rational numbers, and whose angles are equal. It follows that al + a2E + • • • + apEP-1 = 0 and the irreducibility of the polynomial 1 + X + • • + XP-1 over the field of rational numbers shows that al = a2 = • • • = ap, so the polygon is regular (the argument is identical to the proof of the first lemma in chapter Complex Combinatorics). For the converse, let us assume that p is not a prime and prove that there exists a non-regular poly-gon with rational side-lengths and equal angles. Let us write p = mn for some 2z, m, n > 1. Then E = e P satisfies the equation 1 and also the equation 1+ c+ +€73-1 = 0. By adding these two equations, we obtain a relation of the form al + a2€ + • • • + ap€P-1 = 0, where all a, are equal to 1 or 2 and not all of them equal. The observation in the beginning of the ± ± e(m-1)n = 0 THEORY AND EXAMPLES 529 solution shows that there exists a polygon with equal angles and side-lengths al, a2, ap. Clearly, this polygon is not regular. We continue with two difficult problems. The first one is classical, but very difficult. It belongs to a large class of additive problems in number theory and it is quite remarkable that it has a purely algebraic solution. A similar state-ment is the famous four-squares theorem, stating that any positive integer is the sum of four squares of integers, or the notoriously difficult Waring prob-lem, stating that for any k there is m such that any sufficiently large integer is a sum of at most m powers of exponent k. We leave it to the interested reader to deduce from the four squares theorem that any positive integer is the sum of 53 fourth powers! Example 6. Prove that any rational number can be written as the sum of the cubes of three rational numbers. Solution. If someone really wants to be cruel, they will just write the following identity: X3 - 3 6 3 ( X3 + 3 5X -1-- 36 3 9X2 + 35x 3 = x. ( 9X2 + 81x + 36 4 9x2 + 81x + 36 + 9x2 + 81x + 36 Well, how on earth can we come with such a thing? A natural idea would be to look for a representation of x as a sum of cubes of three rational functions. So let us try to find first two polynomials f,g such that f3 + g3 has a cubic factor. On the other hand, the factorization f3 g 3 (f g)(f zg)(f + z2g), 2%, where z = e 3 suggests a smart choice: f +zg = (X — z)3 and f +z2g = (X — z2)3. A small computation shows that f = X 3 — 3X — 1 and g = —3X2 — 3X. This already gives us the identity (x3 — 3x — 1)3 + (-3x2 — 3x)3 = (x2 + x + 1)3((x — 1)3 — 9x), 530 23. SOME SPECIAL APPLICATIONS OF POLYNOMIALS which easily implies the relation presented in the very beginning of the solu-tion, after changing 9x to x. The next problem has a particular flavor, because of the nice idea really well-hidden and of the technical difficulties that appear at all steps of the solution. Definitely not a friendly problem in a mathematical competition, but excellent spiritual food! On an m x n sheet of paper a grid dividing the sheet into unit squares is drawn. Then, the two sides of length n are taped together to form a cylinder. Prove that one can write a real number in each square, not all numbers being zero, such that each number is the sum of the numbers in the neighboring squares, if and only if there are integers k, 1 such that n + 1 does not divide k and cos (2 m 17r \ + cos ( k7T = n + 1 2 [Ciprian Manolescu] Romanian TST 1998 Solution. Number the rings 1,2, ..., n going downwards and the columns 1,2, ..., m, anticlockwise. The idea is to associate to each ring a polynomial Pi (X)= ail + ai2X + • • • + aimXm-1 and to study how the condition imposed on the numbers translates in terms of these polynomials. This is not difficult, because such numbers exist if and only if P a(X) = Pi-1(X) + + (Xm-1 + X)P i(X) (mod Xm - 1), where Po = Pn+1 = 0. This can be also written as P i±i (X) 1= - (1 - X - Xm-1)P i(X) - Pi-1(X) (mod Xm - 1) and so Pi (X)= Qi(X)P1(X), where Qi is the sequence defined by Qo = 0, Qi = 1 and Qi+i (X) = (1 - X - Xm-1)Q ,(X) - Q (X) (mod Xm - 1). THEORY AND EXAMPLES 531 The condition for all numbers to be zero becomes Pi 0. So, the condition of the problem is satisfied if and only if we can find a nonzero polynomial Pi (of course, (mod X' — 1)) such that PiQ,,,±1 = 0 (mod X' — 1), which means that Q„,+..1 and X' — 1 are not relatively prime. This is also equivalent to the existence of a number z such that zm = 1 and Qm±i(z) = 0. If xk = Qk(z), the identity satisfied by Qi becomes x0 = 0, x1 = 1 and 1) Now, if a = 1 — z — z-1, the relation becomes xi±i — ax, x j_ 1 = 0. Let ri,r2 be the roots of the equation t2 — at + 1 = 0. Then ri, r2 are nonzero, rn+1_,,,n+1 so if xr,,±1 = 0, then we surely have r1 r2 and also x72+1 = ri -r2 2 . Thus the condition on m,n is to have rrl xn+1 — 7.2 7/±1, that is there exists x such that equivalent to the existence of a nontrivial root of order n + 1 of 1, say x, such that a2x = (1 + x)2, that is 2 + 2Re(x) = (1 — 2Re(z))2. Of course, this is equivalent to the condition of the problem. This finishes the solution. Let us now turn to some combinatorial problems. We begin with a very beau-tiful result. Do not underestimate it because of its short proof — it is far from being trivial. Actually, this old conjecture of Artin plays a very important role in additive number theory and has given birth to some important theorems of Ax and Katz, which are unfortunately well beyond the scope of this book. Example 8. Let f1, f2, •••, fk be polynomials in Z/pZ[Xi, X2, X,-,] such that E deg(L) < n. Then the cardinality of the set of vec-i.---1 tors (xi, x2, ..., xn) E (Z/pZ)n such that f,(x) = 0 for all i = 1, 2, ..., k is a multiple of p. Chevalley-Warning theorem 1, x # 1 and also r2 = xr1. Using Viete's formula, this becomes Solution. The idea is that the cardinality of the set of common zeros of L 532 23. SOME SPECIAL APPLICATIONS OF POLYNOMIALS can be expressed more conveniently as (1 — fi(x)P-1)(1 — f2(x)P-1)- • • (1 — fk(x)P-1), x=(xi,•••,xn)E(Z/pZ)' where we understand by fi(x) the element fi(xi, x2, ..., xn). Indeed, this fol-lows easily from Fermat's little theorem, because the polynomial P (X) = (1 — f 1(X)P-1)(1 — f2(X)P-1) • • • (1 — fk(X)P-1) (here X = X2, ..., Xri)) has the property that P(xi, xz, xn) = 0 if and only if at least one of fi(xi, x2, ..., xn) is nonzero and 1 otherwise. Now, let us prove that E P(x) = 0. In order to do this, it is enough xe (z/pz)n to prove it for any monomial of P, of the form Xi ' x2 a2 XnnObserve that in any such monomial we have al + a2 + • • • + an < Th(P — 1), because of k the condition E deg(fi) < n. This means that there exists an i such that i=1 ai < p — 1. Observe that iLl x? xa rin H xE(Z/pZ)' 3=1 x 3 EZ/pZ and because cb,,, < p — 1, by a result proved in the chapter The Smaller, the Better, > xi ai = 0, which shows that > P(x) = 0 in Z/pZ. This x,,EZ/pZ xE(Z/pZ)n finishes the proof, because it follows that the cardinality of the set is a multiple of p. Finally, observe that if we assume that MO) = 0 for all i, it follows that fi have at least one nonzero common root in the field with p elements, which is anything but trivial! We continue with an apparently immediate application of Chevalley-Warning theorem: the famous Erdos-Ginzburg-Ziv theorem. There are many other approaches to this beautiful result, but the way in which it follows from Chevalley-Warning's theorem had to be presented. THEORY AND EXAMPLES 533 [ Example 9. Prove that from any 2n - 1 integers one can choose n whose sum is divisible by n. Erd6s-Ginzburg-Ziv theorem Solution. Let us suppose first that n = p is a prime number. As we will see, this is actually the hard part of the theorem. Consider the polynomials over Z/pZ: f1 (X1, X2, X2p-1) = Xr1 X2-1 x-2p 1 1 , f2 (X1) X2, • • • X2p-1) = aiXT1 a24 -1 + • • • + a2p-1X2p 11 where al, a2, a2p- I are the 2p - 1 numbers. Clearly, the conditions of Chevalley-Warning's theorem are satisfied and so the system fi (X) -= f2 (X) = 0 has a nontrivial solution (x,),-1,...,2p-1. Let I be the set of those 1 < i < 2p-1 such that xi 0. Then from Fermat's little theorem fi(xi, x2, •••,x2p- i) III (mod p) and f2(xl, x2, ..., x2p_1) = Eici a, (mod p) and so p divides 1/1 and > ai. Because I has at least 1 and at most 2p - 1 elements, it follows that it has exactly p elements, and the theorem is proved in this case. In order to finish the proof of the theorem, it is enough to prove that if it holds for a and b integers greater than 1, it also holds for ab. So, take 2ab - 1 integers look at the first 2a - 1 among them. There are some a whose sum is a multiple of a. Put them in a box labelled 1 and look at the remaining numbers. You have at least 2a(b - 1) -1 > 2a - 1, so you can find some other a numbers whose sum is a multiple of a. Put them in a box labelled 2. At each stage, as long as you still have at least 2a - 1 numbers which are not yet in a box, you can create another box with a numbers, the sum of which is a multiple of a. So, you can create at least 2b - 1 such boxes. Now, apply the induction hypothesis for the sums of the numbers in the first 2b - 1 boxes di-vided by a and you will obtain a collection of ab numbers the sum of which is a multiple of ab. This shows that the theorem holds for ab and finishes the proof. The next example presents a truly amazing theorem, appeared in the revo- lutionary article "Combinatorial Nullstellensatz" by Noga Alon and which is 534 23. SOME SPECIAL APPLICATIONS OF POLYNOMIALS now a must in algebraic combinatorics. The reader with background in com-mutative algebra will immediately understand the title of the article: yes, it is related to the even more famous Nullstellensatz of Hilbert, one of the basic results of algebraic geometry and probably one of the most important theo-rems in mathematics. What does the latter say? Well, the strong form says that if fl , f2, fk are polynomials with complex coefficients in n variables and if f is another such polynomial which vanishes at all common zeros of the polynomials fi, f2, fk, then some power of f can be written in the form + f292 + • + fk9k for some polynomials gi, g2, ...,gk. Note for instance that if fi, f2, fk have no common zeros then there will be gi,g2, gk such that fi9i + f292 + • • • + fk9k = 1, a fact far from being obvious! Actually, the proof of Hilbert's Nullstellensatz is difficult and really needs a fair amount of commutative algebra, so we will not present it here. The reader can find a proof in practically any book of algebraic geometry. Note however that there are substantial differences between this statement and the "Combinato-rial Nullstellensatz", and they probably explain why the latter is so well-suited for combinatorial applications. Example 10. Let F be a field, f E F[X1, X2, •-, Xri] a polynomial, and let Si, S2, ..., Sn be nonempty subsets of F. a) If f (si, 82, •.., sn) = 0 for all (51,82, sn) E Si X S2 X ... X Sn, then f lies in the ideal generated by the polyno-mials gi(Xi) (Xi — s). Moreover, the polynomials hi, h2, hn satisfying f = 91h1 + g2h2 + • • • + gnhn can be chosen such that deg(hi) < deg(f) — deg(g,) for all i. Finally, if 91, 92,..., 9n E R[Xi, X2, ..., Xn] for some subring R of F, one can choose hi with coefficients in R. b) If deg(f) = t1 + t2 + • • • + tri, where t, are nonnega-tive integers such that t, < Si and if the coefficient of Xil X 2 • X m tn is not zero, then there exist s, E Si such that f (si, 82, ..., Sn) 0. [Noga Alon] Combinatorial Nullstellensatz THEORY AND EXAMPLES 535 Solution. a) The idea is that any element si of Si satisfies an algebraic equa-tion of degree S so any power of si is a linear combination of 1 s s1siI-1 with coefficients independent of the choice of si E Si. Indeed, if 9i3Xi j=0 Isd-1 then = E giisi. This allows us to "reduce" the polynomial f by replac- 3=0 ing every V ic with a linear combination of 1, Xi, ..., Xis 1. This corresponds to subtractions from f of polynomials of the form gihi, with deg(hi) < deg(f)- n deg(gz). So we see that by subtracting a linear combination E gihi from f we i=i obtain a polynomial fi whose degree in Xi is at most Si 1l - 1 and such that 0 = s2,..., sn) = s2,..., sm) for all .5, E Si. But this immediately implies fi = 0. Indeed, fi can be written as F0 + F 1X1 + ". + for some polynomials Fi E X 7 , such that Fi has degree in Xj at most - 1. Now, for all s2 E S2, ..., Sn E S n, the polynomial Fo(s2,-,sn) 1(82, sn)X l is11-1 has at least 1.511 zeros in the field F, so it is identically zero, that is = • " = Fis,1-1(s2,•••, sn) = 0 for all (s2, sn) E S2 • • • Sn. An inductive reasoning shows that F0 = • • • = = 0 and so fi = 0. This finishes the proof of a). b) This is a di-rect consequence of a). Suppose by contradiction that f vanishes on Si x 82 x • • • x 8n. By taking subsets of S i with ti + 1 elements, we can assume that Si 1l = ti + 1. Let hi and g i be defined as in a). It follows that the coefficient of X11X22 • • • Xn tn in g1 hi + g2h2 + • • • + gnitn is not zero. Because deg(hi) < deg(f) - deg(gi), the coefficient of X11X22 • X n tn in gihi is zero: any monomial appearing in this polynomial and having degree deg(f) is a multiple of .X- z+1, contradiction. 536 23. SOME SPECIAL APPLICATIONS OF POLYNOMIALS Let us see now some applications of this result. First, some direct consequences which already show the power of the method: try to find other solutions to these problems and you will see that they are far from being trivial. This is probably also a reason for selecting the next problem as problem 6 at the International Mathematical Olympiad in 2007. Example 117.1 Let n be a positive integer and consider the set S {(x,y,z)lx,y,z E {0, 1, ...,n},x + y + z > 0} as a set of points in space. Find the minimum number of planes, the union of which contains S but does not contain (0, 0, 0). IMO 2007 Solution. Let aix+biy+ciz = d, be the equations of these planes and consider the polynomial f (X , Y, Z) =- H (aiX biY ciZ — di) — m • H (X - i)(Y — i)(Z — i), i=i where m is chosen such that f (0, 0, 0) = 0. If k < 3n, then clearly the co-efficient of XmYriZn in f is nonzero. Thus, by combinatorial Nullstellensatz there are integers x, y, z E {0, 1, ..., n} such that f (x, y, z) 0. If at least one of x, y, z is nonzero, then clearly both terms defining f are zero, a contradic-tion. Thus (x, y, z) = (0, 0, 0), which contradicts the fact that f (0, 0, 0) = 0. Therefore k > 3n and since for k = 3n an example is immediate, we deduce that this is the answer to the problem. And now a very similar statement: Example 12. Let p be a prime and let S1, S2, . Sk be sets of non-negative integers, each containing 0 and having pairwise distinct ele- ments modulo p. Suppose that Ezasil — 1) > p. Prove THEORY AND EXAMPLES 537 that for any elements al, ... , ak E Z/pZ, the equation xiai + x2a2+• • •+xkak = 0 has a solution (x1, ... , xk) E S1 X • • • X Sk other than the trivial one (0, ... , 0). Troi-Zannier's theorem Solution. [Peter Scholze] Consider the polynomial P(Xi, ..., Xk) = (aiXi + a2X2 + • • • + akX0P-1 — 1 +C fJ (yi +81) fJ (X 2 +8 2)... H (Xk +.9k) oosi Esi 00.92E52 ooskEsk where C is chosen such that P(0, ..., 0) = 0. Because of the third condition, the coefficient of xr11-1...xlso—i is s nonzero. Therefore there are ti E Si , ..., tk E Sk with P(ti, ..., tk) # 0. Since P(0, ..., 0) = 0, it is clearly not the zero solution. Thus, c H (ti+81) IT (t2 - 82 ) • • • H ( tk — 8k) 00,91ES1 0As2ES2 008k ESk must be zero, which implies that (aiti + • • • + aktk)P-1 1. It remains only to note that Fermat's little theorem gives aiti + • • • + aktk = 0. The category of deep results with short proofs is going to be represented once again, this time with a really important result of additive combinatorics, one of those mathematical fields which exploded in the twentieth century. Of course, there are many other proofs of this result, all of them very ingenious. The result itself is important: as an exercise (solved by Cauchy about two hundred years ago...), try to prove this using this Lagrange's famous theorem stating that any positive integer can be written as a sum of four squares of integers. There are very elementary arguments, as we will see, but the combinatorial Nullstellensatz also implies this result and actually much more. 538 23. SOME SPECIAL APPLICATIONS OF POLYNOMIALS [Example 13. For any subsets A, B of Z/pZ the following inequality holds IA + BI > min(p, + — 1). Cauchy-Davenport theorem Solution. There is one very simple case: lAl + > p + 1. In this case, A + B = Z/pZ, since for any x E Z/pZ, the function f (a) = x — a defined on A cannot take all its values outside B, because it is injective. The difficult case is when +1/31 < p. Let us suppose that IA+Bl < IA1+1.131— 2 and let us choose a subset C of Z/pZ containing A+ B and having (Al +1.BI — 2 ele-ments. The polynomial f X2) = 11 (X1 + X2 — c) E Z/pZ[X] has degree cEC 1.131— 2 and vanishes on A x B. In order to obtain a contradiction, it is I thus sufficient to prove that XiA —1 X2 Ii appears with a nonzero exponent in f . However, it is clear that this exponent equals (lAr Ari l-2) (mod p), which is not zero, because +1./31 — 2 < p — 2. Using the previous theorem, we obtain the desired contradiction. Before passing to the next example, let us present a truly magnificient (for its simplicity!) proof of the previous result, which is probably more natural when seeing the statement for the first time, but which is by no means as obvious as it looks! We shall prove the result by induction on Al, the case when IA = 1 being obvious. Clearly, we may assume that > 1 and also that IBI < p. Now, A having more than one element, by shifting it we may assume that it contains 0 and some x 0. Now, B is nonempty and B Z/pZ, so there must be an integer n such that nx E B but (n + 1)x does not belong to B. By shifting B this time we may suppose that 0 E B, but x is not in B. Thus, A n B is a proper nonempty subset of A and we may use the induction hypothesis for it and A U B. Because A + B contains (A n B) + (A U B) and 'An/31 +1AuB1 =1A1+1B1, the conclusion follows. Even though this proof is very beautiful and short, it should be noted that Alon's technique is much more powerful. Indeed, Alon THEORY AND EXAMPLES 539 shows in his seminal paper that his theorem implies a famous conjecture of Eras-Heilbronn, with a very similar statement, but with no elementary proof (exercise for the reader: check that the above elementary solution does not work for the following result): for any nonempty subset A of Z/pZ one has 1{a + b E A, a min(p, 21AI — 3). The next problem uses the proof given by Noga Alon for a special case of a difficult conjecture of Snevily. Again, the combinatorial nillstellensatz is well-suited, but this time it is not so clear that its hypotheses are satisfied. Actually, the most difficult part in using this powerful tool is finding the good polynomial, but there are situations when it is even more difficult to check the hypothesis, because the polynomial can have a quite complicated expression. [Example 14.1 Let p be a prime number, and let al, a2, ak E Z/pZ, not necessarily distinct. Prove that for any distinct elements b1, b2, bk of Z/pZ there exists a permutation a such that the elements al + bum, a2 + b0(2), ak + b,(k) are pairwise distinct. Alon's theorem Solution. Let B = b2, ..., bk} and suppose the contrary, that is for all choices of distinct elements xl, x2, ..., xk at least two of the elements x1 + al, ..., xk + ak are identical. That is, if x1, x2, ..., xk are distinct elements of B, we have n (x, + ai — xj — ai) = 0 in Z/pZ. We can relax the restriction 1<z<j<k of x1, x2, ..., xk being pairwise distinct by considering the polynomial f (Xi, X2, ..., Xk) = H ( X, - ) ( X , + ai — Xi — ai ). 1<i<j<k The previous remark shows that f vanishes on Bk. Clearly, f can be written as H _ xj)2±xx1,x2,...,xo 1<i<j<k 540 23. SOME SPECIAL APPLICATIONS OF POLYNOMIALS for some polynomial g of smaller degree. Also, deg(f) = k(k — 1). We will try to find t1, t2, tk such that t1+t2+• • •+tk = k(k-1), ti < k and the coefficient of XI' X 2 • • • Xk tk is nonzero in f . The first two conditions impose ti = t2 = • • • = tk = k —1, so the question is whether the coefficient of (Xi X2 • • • Xk)k-1 in f is zero. Of course, if we manage to prove that (X1X2 • • • Xk)k-1 appears with a nonzero coefficient in n - Xj)2 we can apply Alon's theorem i<i<3<k and obtain the desired contradiction. However, using the result in example 8 of the Lagrange Interpolation Formula chapter, we deduce that the free term of 11 ( 1 - ) is Id. But note that i<20j<k n xj)2 IT (1 X xi) = (-1) k(k-1) 15i<jk 1.<i0j<k 3 (X1X2 • • . )(k)k-1 (23.2) so the coefficient of (Xi X2 • • • Xk_i) is nonzero in Z/pZ because of the as-sumption k < p. This finishes the proof of the result. We have already seen examples of combinatorial problems for which it is almost impossible to find combinatorial solutions. We continue with an example, which is a quite deep result of Alon, Friedland and Katai. Needless to say, the solution using combinatorial Nullstellensatz is practically straightforward. There are, however, limits of the method, for instance one does not know if in the next result one can replace p prime by any positive integer. [Example 15.1 Let G be a graph with no loops (yet, multiple edges are al-lowed) and let p be a prime number. Assume that all vertices have degree at most 2p — 1 and the average degree of the graph is greater than 2p — 2. Prove that G has a p-regular subgraph (a subgraph in which every vertex has degree p). Solution. Let us consider the incidence matrix where v denotes a vertex and e an edge and av e = 1 if v E e and 0 otherwise. Let xe be a variable THEORY AND EXAMPLES 541 associated with each edge e and consider the polynomial The hypothesis implies that f has degree 1E1 and because the coefficient of fl x e is nonzero, Alon's Nullstellensatz implies the existence of values xe E {0,1} such that the evaluation of f at these xe is not zero. Now, clearly this is not the zero vector, thus using Fermat's little theorem we deduce that all EeEE a v,exe are 0 in Z/pZ, that is if we look at the subgraph of those edges e such that xe = 1, all vertices have degrees multiples of p, smaller than 2p. Thus this subgraph is p-regular. The reader is urged to take a look at the problems for training for many other applications of combinatorial Nullstellensatz, a theme that will surely become recurrent in algebraic combinatorics and additive number theory. We will now present a quite subtle result, based on algebraic properties of polynomials. We have already encountered this type of argument in a previous chapter, but the result and the method are too important not be presented. Example 16. i Let F be a family of subsets of a set X with n elements. Suppose that there exists a set L with s elements such that An BC E L for all distinct members A, B E F. Prove that F has at most () n (rz (7?, — 1) + • • • + elements. Frankl-Wilson theorem 542 23. SOME SPECIAL APPLICATIONS OF POLYNOMIALS Solution. Let L = {l1,12, ...,1,} and assume without loss of generality that X = {1, 2, ..., n}. Finally, call A1, A2, ..., A, the elements of F, such that 11111 < 11121 < ••. < lAml. We will associate with each set Az its characteristic vector vi = (vi 3)1<3<ri defined by: v23 = 1 if j E Ai and 0 otherwise. Observe that if (x, y) = xiyi +x2y2+• • • +xnyn is the standard euclidean inner product, then IA, n A3 = (v i, v3). Now, let us define the polynomials fi(x) = fl ((x, vi) - lk) k,lk <1Ai for i = 1, 2, ..., m. The main idea is to consider the restrictions of these polynomials to the vertices of the unit cube, that is the set Y = {0,1}n. Because x72 , = x, if x, E {0,1}, it is clear that these restrictions can be writ-ten in the form gi(xl , x„), where g, are polynomials of degree at most s and have degree at most 1 in each variable. What is remarkable is that these functions f, : Y —> R are linearly independent. This is not difficult: if Ai 8 (x) + A2 f2 (X) + • • • + fm(x) = 0 for x E Y, then by taking x = v3 for all j and using the fact that fi(v3) = 0 if j < i and fi(vi) 0 (which is obvious), we immediately deduce by induction that all A, are 0. The result follows from the fact that the vector space generated by these functions has dimension m and is a subspace of the vector space of polynomials of maximum degree at most s and partial degrees at most 1, which has dimension (n) n (n) (rt ) — 1) + • • • + ) V)). PROBLEMS FOR TRAINING 543 23.2 Problems for training 1. Let n, m be positive integers with n < m — 1 and let al, a2, a, be nonzero integers such that for all 0 < k < m we have al a2 2k am M k = 0. Prove that there are at least n 1 pairs of consecutive terms having opposite signs in the sequence al, a2, a,. Russia 1996 2. Let al, a2, aim and b1, b2, b100 be 200 distinct real numbers. Con- sider an n x n table and put the number a, bj in the (i, j) position. Suppose that the product of the entries in each column is 1. Prove that the product of the entries in each row is —1. Russian Olympiad 3. The finite sequence {ak}i n. Noga Alon 9. Let Si 82, , Sn be subsets of Z/pZ and let S = Sl x S2 x • • • x Sn. Consider polynomials fi, f2,..., fk in n variables over Z/pZ such that n (p —1) • deg(fi) i=i (lsil - 1). i=1 Prove that if the system fi(x) = f2(x) = • • • = fk(x) = 0 has a solution a E S, then it has another solution b E S. PROBLEMS FOR TRAINING 545 10. Let A be a subset of Z/pZ, where p is a prime number. Prove that among the elements a + b where abiEJ1 there are at least min(p, 21.211 — 3) distinct elements. Erdos-Heilbronn conjecture 11. Let A1, A2, ..., An be subsets of Z/pZ such that Ei<n (1 + < p. Also, let A = {ai, a2, ..., an} and B 1, B2, be of Z/pZ such that 1Bi l > (n — 1)(1 + Ad). Prove that we can select bi E Bi such that ai + + bj for i # j and (aj + bj) — (ai+ bi) does not belong to A. 12. Let F be a family of subsets of {1, 2, n} such that IAA = k whenever A E F and IA n B1 E L for all distinct members A, B E F, L being a set with s elements. Prove that F has at most (n s) elements. Frankl-Wilson 13. Let A1, A2, ..., A, and B1, B2, ..., 139„ be subsets of {1, 2, ..., n} such that there exists a set L with k elements for which n E L if i < j and nBi j does not belong to L for all i. Prove that m < (7) ± (n 2) + • • H- ( n k). 14. Let p, q be prime numbers and r a positive integer such that qlp —1, q does not divide r and p > rq-1. Let al, a2, ar. be integers such that - 1 p - 1 27 1 al q + a2 q + • + a,. q is a multiple of p. Prove that at least one of the ai's is a multiple of p. AMM 15. Prove that there exists a positive integer n such that any prime divisor of r — 1 is smaller than 21993 - 1. Komal 546 23. SOME SPECIAL APPLICATIONS OF POLYNOMIALS 16. 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Research Counsil Israel, 1961, 41-43. Fomin A.A., Kuznetsova G.M., International Mathematical Olympiads, Drofa, Moskva, 1998. Forman R., Sequences with many Primes, Amer. Math. Monthly, 99 (1992), 548-557. Freiling C., Rinne D., Tiling a square with similar rectangles, Mathe-matical Research Letters 1, 547-558, 1994. Gelca R., Andreescu T., Putnam and Beyond, Springer, 2007 Gerst I., Brillhart J., On the Prime Divisors of Polynomials, Amer. Math. Monthly, 78 (1971), 250-266. Godsil C., Tools from linear algebra, Handbook of Combinatorics, edited by R. Graham, M. Grotschel and L. Lovasz, Elsevier and M.I.T Press, 1995, 1705-1748. Graham R.L., Knuth D.E., Patashnik 0., Concrete Mathematics, 2nd edition, Addison-Wesley, 1989. Greitzer S.L., International Mathematical Olympiads 1959-1977, M.A.A., Washington, D.C., 1978. Guy, R.K., Unsolved Problems in Number Theory, Springer, 3rd edition, 2004. BIBLIOGRAPHY 551 O'Hara P.J., Another proof of Bernstein's theorem, Amer. Math. Monthly 80, 1973, 673-674. Hardy G.H, Wright E.M., An introduction to the Theory of Numbers, Oxford 1979. Hlawka E., Schoibengeier J., Taschner R., Geometric and Analytic Num-ber Theory, Springer-Verlag, 1991. Huneke C., The friendship theorem, Amer. Math. Monthly, No. 2, 2002, 192-194. Oleszkiewicz K., An elementary proof of Hilbert's inequality, Amer. Math. Monthly, 100, 1993, 276-280. Mignotte M., An Inequality about Factors of Polynomials, Mathematics of Computation, 128 (1974), 1153-1157. Mitrinovic D.S, Vasic P.M, Analytic inequalities, Springer-Verlag, 1970. Murty M.R., Prime Numbers and Irreducible Polynomials, Amer. Math. Monthly, No. 5, 2002, 452-458. Nathanson M.B., Additive Number Theory, Springer 1996. Polya G., Szego G., Problems and theorems in analysis, Springer-Verlag, 1976. Prasolov V.V, Polynomials, Algorithms and Computation in Mathemat-ics, Volume 11 Springer-Verlag, 2003. Rogosinski W.W., Some elementary inequalities for polynomials, Math. Gaz, 39, No. 327, 1955, 7-12. Roitman M., On Zgismondy Primes, Proceedings of the Amer. Math. Soc, 125 (1997), No. 7, 1913-1919. Savchev S., Andreescu T., Mathematical Miniatures, New Mathematical Library, MAA 2002. 552 BIBLIOGRAPHY Seres I., Irreducibility of Polynomials, Journal of Algebra 2, 283-286, 1963. Serre J.-P., A Course in Arithmetic, Springer-Verlag 1973. Siegel, C.L., Lectures on the Geometry of Numbers, Springer-Verlag, 1989 (notes by B.Friedman rewritten by K.Chandrasekharan with assis-tance of R.Suter). Sierpinski W., Elementary Theory of Numbers, Polski Academic Nauk, Warsaw, 1964. Sierpinski W., 250 Problems in Elementary Number Theory, American Elsevier Publishing Company, Inc., New York, Warsaw, 1970. Stanley R.P., Enumerative Combinatorics, Cambridge University Press, 2nd edition, 2000. Steele Michael J., The Cauchy-Schwarz Master Class, Cambridge Uni-versity Press, 2004. Sun Z.W., Covering the Integers by Arithmetic Sequences, Trans. Amer. Math. Soc, Vol 348, Nr 11, 1996, 4279-4320. Tomescu I., Melter R.A., Problems in Combinatorics and Graph Theory, John Wiley Sons, 1985. Tomescu I. et al., Balkan Mathematical Olympiads 1984-1994, Gil, Za-lau, 1996. Turk J., The Fixed Divisor of a Polynomial, Amer. Math. Monthly, 93 (1986), 282-286. Yaglom A.M., Yaglom I.M., Challenging Mathematical Problems with Elementary Solutions, Dover Publications, 1987. Zannier U., A note on securrent mod p sequences, Acta Arithmetica XLI, 1982. Index Aczel's Inequality, 29 Algebraic Integer, 188 Algebraic Numbers, 188 Alon's Combinatorial Nullstellensatz, 533 Alon's Permanent Lemma, 544 Alon-Friedland-Katai Theorem, 539 Behrend's Lemma, 464 Bernstein's Theorem, 254 Bertrand's Postulate, 63 Burnside's Lemma, 171 Capelli's Theorem, 492 Carlson's Inequality, 41 Cauchy-Davenport Theorem, 537 Chebyshev's Polynomials, 243 Chebyshev's Theorem, 251 Chevalley-Warning Theorem, 531 Cohn's Irreducibility Criterion, 483 Cyclotomic Polynomials, 498 Davenport-Cassels Lemma, 307 Erdos, Palfy, 65 Erdos, Sun, 146 Erdos-Ginzburg-Ziv Theorem, 532 Erd6s-Heilbronn Conjecture, 539 Euler's Criterion, 401 Formal Series Ring, 157 Frankl-Wilson Theorem, 541 Fundamental Theorem of Symmetric Polynomials, 182 Group Action, 171 Hamilton-Cayley's Theorem, 182 Hensel's Lemma, 211 Hilbert's Inequality, 40 Hilbert's Nullstellensatz, 534 Incidence Matrix, 268 Index of a Curve, 481 553 554 INDEX Kronecker's Theorem, 194 Lagrange Interpolation Formula, 235 Lagrange's Four Squares Theorem, 296 Landau's Inequality, 469 Legendre's Symbol, 401 Mahler's Measure, 194 Markov's Theorem, 254 Minimal Polynomial, 188 Minkowski's Convex Body Theorem, 291 Minkowski's Linear Forms Theorem, 298 Nagell's Theorem, 222 Nesbitt, 6 Niven Numbers, 362 Order of a mod n, 315 P-adic Valuation, 49 Pell Equation, 389 Perron's Criterion, 485 Primitive Element Theorem, 222 Primitive Root mod n, 322 Quadratic Reciprocity Law, 403 Quadratic Residue, 401 Riesz's Theorem, 254 Rouche's Theorem, 481 SchOnemann's Criterion, 498 Schur's Theorem, 208 Selmer's Theorem, 485 Shapiro's Inequality, 100 Siegel's Lemma, 466 Sophie Germain's Identity, 494 Stirling's Formula, 62 Thue's Lemma, 75 Thue's Theorem, 76 Troi-Zannier Theorem, 537 Turan's Theorem, 119 Van der Corput's Lemma, 345 Vandermonde's Identity, 187 Waring's Problem, 529 Weyl's Theorem, 342 Young's Inequality, 427 Zarankiewicz's Lemma, 117 |
189988 | https://sites.math.duke.edu/education/ccp/materials/linalg/curvefit/curvfit3.html | Curve Fitting, Part 3
Curve Fitting
Part 3: Linearization
It is common practice to try to fit non-linear models to data by first applying some transformation to the model that "linearizes" it. For example, suppose we want to fit the non-linear exponential model
y = a e bt
to the U.S. population data from Part 1. The standard trick is to linearize the model by taking logs:
ln(y) = ln(a) + b t.
Now we have a model in which the parameters A = ln(a) and b appear linearly. We can fit a least squares line to the data
(T 1, ln(Y 1) ) ), (T 2, ln(Y 2) ), ... , (T 10, ln(Y 10) ).
That is, we fit a least squares line to the semi-log plot of ln(y) versus t shown below.
For the U.S. population data, the vectors
y = ( ln(Y 1), ln(Y 2), ..., ln(Y 10) )T,
t = ( T 1, T 2, ..., T 10 )T ,
and 1 = (1, 1, ..., 1)T
are defined in your helper application worksheet. Solve the normal equations for the parameters A = ln(a) and b of best fit and then find the best fitting exponential model y = a e bt.
Plot the exponential function that you just found together with a scatter plot of the original U.S. Population data (not the logarithms). How good is the fit?
Compute the residuals y - p and the sum of squares S of the residuals. Make a residual plot. What do you learn from the plot about the goodness of fit?
Compare the fit of the exponential model to the quadratic model of Part 1. You should have already computed S for the the quadratic model. You should also make a residual plot for the quadratic model for comparison purposes.
Small Data Setty 1.0 2.5 2.0 8.0 3.0 19.9 4.0 50.0
Above is a small data set. Let's try to fit this data with a power function of the form
y = a t b
by first linearizing the model by taking logs:
ln(y) = ln(a) + b ln(t).
For the small data set above, the vectors
y = ( ln(Y 1), ln(Y 2), ln(Y 3), ln(Y 4) )T,
t = ( ln(T 1), ln(T 2), ln(T 3), ln( T 4) )T ,
and
1 = (1, 1, 1, 1)T
are defined in your helper application worksheet. Fit a line to the log-log (ln(T i),ln(Y i)) data and use the results to find the model parameters for the corresponding power function y = a t b
Plot the power function that you just found together with a scatter plot of the original small data set (not the log-log plot). How good is the fit?
Compute the residuals y - p and the sum of squares S of the residuals. Make a residual plot. What do you learn from the plot about the goodness of fit?
modules at math.duke.edu |
189989 | https://nutriverse.io/nutrisurvey/muac.html | A Guide to Implementing Nutrition and Food Security Surveys
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Child Anthropometry
1 Childhood undernutrition and anthropometry
2 Measuring weight
3 Measuring height
4 Measuring mid-upper arm circumference (MUAC)
5 Checking for oedema
6 Anthropometric measurement standardisation test
Dietary Intake
7 Measuring dietary diversity
8 Measuring food consumption
Food security
9 Household Food Insecurity Access Scale
References
Table of contents
4.1 Equipment
4.2 Personnel
4.3 Steps in measuring MUAC
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Child Anthropometry
4 Measuring mid-upper arm circumference (MUAC)
4 Measuring mid-upper arm circumference (MUAC)
4.1 Equipment
MUAC is a quick and simple way to determine whether or not a child is malnourished using a simple coloured plastic strip. There are different types of MUAC tape available. All are graduated in millimetres and some are colour coded (red, yellow and green) to indicate the nutritional status of a child or adult. The colour codes and gradations vary depending on the tape type.
The most appropriate MUAC tape to use would be the tapes that use the latest WHO Growth Standards cut-offs for acute malnutrition. These are the tapes that have three colours (red, yellow, green) with colour cut-offs at 115 mm and 125 mm. The MUAC tapes should also be precise up to 1 mm. The material for the MUAC tape needs to be flexible but non-stretchable. An example of this kind of MUAC tape is shown in Figure 4.1.
4.2 Personnel
Only a single measurer is required to measure the MUAC of a child. If with an assistant is available, he/she records the MUAC measurement.
4.3 Steps in measuring MUAC
When measuring MUAC, ensure work at eye level. Sit down when possible. Very young children can be held by their mother during this procedure. Ask the mother to remove clothing that may cover the child’s left arm.
Measurer calculates the midpoint of the child’s left upper arm by first locating the tip of the child’s shoulder (arrows 1 and 2) with finger tips. Bend the child’s elbow to make a right angle (arrow 3). Place the tape at zero, which is indicated by two arrows, on the tip of the shoulder (arrow 4) and pull the tape straight down past the tip of the elbow (arrow 5). Read the number at the tip of the elbow to the nearest centimetre. Divide this number by two to estimate the midpoint. As an alternative, bend the tape up to the middle length to estimate the midpoint. A piece of string can also be used for this purpose. Either you or an assistant can mark the midpoint with a pen on the arm (arrow 6).
Measurer straightens the child’s arm and wraps the tape around the arm at midpoint. Make sure the numbers are right side up. Make sure the tape is flat around the skin (arrow 7).
Measurer and assistant inspects the tension of the tape on the child’s arm. Make sure the tape has the proper tension (arrow 7) and is not too tight or too loose (arrows 8-9). Repeat any steps as necessary.
Assistant is on ready with the paper questionnaire or the mobile device.
When the tape is in the correct position on the arm with the correct tension, measurer reads and calls out the measurement to the nearest 0.1 cm. (arrow 10).
Assistant immediately records the measurement on the questionnaire or the mobile device and shows it to the measurer.
While the assistant records the measurement, measurer loosens the tape on the child’s arm.
Measurer checks the recorded measurement on the questionnaire or mobile device for accuracy and legibility then instructs the assistant to erase and correct any errors.
Measurer removes the tape from the child’s arm.
Common mistakes when measuring MUAC
measuring on the right arm;
estimating (rather than measuring) the mid-point of the upper arm;
bending the MUAC tape when measuring the midpoint;
not measuring the midpoint from the tip of the shoulder to the elbow bend;
pulling the MUAC tape too tight;
not pulling the MUAC tape tight enough (too slack); and,
not reading the tape accurately (to nearest millimetre).
3 Measuring height
5 Checking for oedema |
189990 | https://dspace.mit.edu/bitstream/handle/1721.1/153490/18-05-spring-2014/contents/readings/MIT18_05S14_Reading3.pdf | Conditional Probability, Independence and Bayes’ Theorem Class 3, 18.05 Jeremy Orloff and Jonathan Bloom 1 Learning Goals 1. Know the definitions of conditional probability and independence of events. 2. Be able to compute conditional probability directly from the definition. 3. Be able to use the multiplication rule to compute the total probability of an event. 4. Be able to check if two events are independent. 5. Be able to use Bayes’ formula to ‘invert’ conditional probabilities. 6. Be able to organize the computation of conditional probabilities using trees and tables. 7. Understand the base rate fallacy thoroughly. 2 Conditional Probability Conditional probability answers the question ‘how does the probability of an event change if we have extra information’. We’ll illustrate with an example. Example 1. Toss a fair coin 3 times. (a) What is the probability of 3 heads? answer: Sample space Ω = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }. All outcomes are equally likely, so P (3 heads) = 1/8. (b) Suppose we are told that the first toss was heads. Given this information how should we compute the probability of 3 heads? answer: We have a new (reduced) sample space: Ω' = {HHH, HHT, HTH, HTT }. All outcomes are equally likely, so P (3 heads given that the first toss is heads) = 1/4. This is called conditional probability, since it takes into account additional conditions. To develop the notation, we rephrase (b) in terms of events. Rephrased (b) Let A be the event ‘all three tosses are heads’ = {HHH}. Let B be the event ‘the first toss is heads’ = {HHH, HHT, HTH, HTT }. The conditional probability of A knowing that B occurred is written P (A|B) This is read as ‘the conditional probability of A given B’ or ‘the probability of A conditioned on B’ 1 2 3 18.05 class 3, Conditional Probability, Independence and Bayes’ Theorem, Spring 2014 or simply ‘the probability of A given B’. We can visualize conditional probability as follows. Think of P (A) as the proportion of the area of the whole sample space taken up by A. For P (A|B) we restrict our attention to B. That is, P (A|B) is the proportion of area of B taken up by A, i.e. P (A ∩ B)/P (B). B A A ∩B Conditional probability: Abstract visualization and coin example Note, A ⊂ B in the right-hand figure, so there are only two colors shown. The formal definition of conditional probability catches the gist of the above example and visualization. Formal definition of conditional probability Let A and B be events. We define the conditional probability of A given B as P (A ∩ B) P (A|B) = , provided P (B) (1) = 0. P (B) Let’s redo the coin tossing example using the definition in Equation (1). Recall A = ‘3 heads’ and B = ‘first toss is heads’. We have P (A) = 1/8 and P (B) = 1/2. Since A ∩ B = A, we 1/8 also have P (A ∩ B) = 1/8. Now according to (1), P (A|B) = = 1/4, which agrees with 1/2 our answer in Example 1b. Multiplication Rule The following formula is called the multiplication rule. P (A ∩ B) = P (A|B) · P (B). (2) This is simply a rewriting of the definition in Equation (1) of conditional probability. We will see that our use of the multiplication rule is very similar to our use of the rule of product in counting. In fact, the multiplication rule is just a souped up version of the rule of product. We start with a simple example where we can check all the probabilities directly by counting. Example 2. Draw two cards from a deck. Define the events: S1 = ‘first card is a spade’ ˙ and S2 = ‘second card is a spade’. What is the P (S2|S1)? answer: We can do this directly by counting: if the first card is a spade then of the 51 cards remaining, 12 are spades. P (S2|S1) = 12/51. 3 4 18.05 class 3, Conditional Probability, Independence and Bayes’ Theorem, Spring 2014 Now, let’s recompute this using formula (1). We have to compute P (S1), P (S2) and P (S1 ∩ S2): We know that P (S1) = 1/4 because there are 52 equally likely ways to draw the first card and 13 of them are spades. The same logic says that there are 52 equally likely ways the second card can be drawn, so P (S2) = 1/4. Aside: The probability P (S2) = 1/4 may seem surprising since the value of first card certainly affects the probabilities for the second card. However, if we look at all possible two card sequences we will see that every card in the deck has equal probability of being the second card. Since 13 of the 52 cards are spades we get P (S2) = 13/52 = 1/4. Another way to say this is: if we are not given value of the first card then we have to consider all possibilities for the second card. Continuing, we see that 13 · 12 P (S1 ∩ S2) = = 3/51. 52 · 51 This was found by counting the number of ways to draw a spade followed by a second spade and dividing by the number of ways to draw any card followed by any other card). Now, using (1) we get P (S2 ∩ S1) 3/51 P (S2|S1) = = = 12/51. P (S1) 1/4 Finally, we verify the multiplication rule by computing both sides of (2). 13 · 12 3 12 1 3 P (S1 ∩ S2) = = and P (S2|S1) · P (S1) = · = . QED 52 · 51 51 51 4 51 Think: For S1 and S2 in the previous example, what is P (S2|S1 c)? Law of Total Probability The law of total probability will allow us to use the multiplication rule to find probabilities in more interesting examples. It involves a lot of notation, but the idea is fairly simple. We state the law when the sample space is divided into 3 pieces. It is a simple matter to extend the rule when there are more than 3 pieces. Law of Total Probability Suppose the sample space Ω is divided into 3 disjoint events B1, B2, B3 (see the figure below). Then for any event A: P (A) = P (A ∩ B1) + P (A ∩ B2) + P (A ∩ B3) P (A) = P (A|B1) P (B1) + P (A|B2) P (B2) + P (A|B3) P (B3) (3) The top equation says ‘if A is divided into 3 pieces then P (A) is the sum of the probabilities of the pieces’. The bottom equation (3) is called the law of total probability. It is just a rewriting of the top equation using the multiplication rule. 4 18.05 class 3, Conditional Probability, Independence and Bayes’ Theorem, Spring 2014 Ω B3 B2 B1 A ∩B1 A ∩B2 A ∩B3 The sample space Ω and the event A are each divided into 3 disjoint pieces. The law holds if we divide Ω into any number of events, so long as they are disjoint and cover all of Ω. Such a division is often called a partition of Ω. Our first example will be one where we already know the answer and can verify the law. Example 3. An urn contains 5 red balls and 2 green balls. Two balls are drawn one after the other. What is the probability that the second ball is red? answer: The sample space is Ω = {rr, rg, gr, gg}. Let R1 be the event ‘the first ball is red’, G1 = ‘first ball is green’, R2 = ‘second ball is red’, G2 = ‘second ball is green’. We are asked to find P (R2). The fast way to compute this is just like P (S2) in the card example above. Every ball is equally likely to be the second ball. Since 5 out of 7 balls are red, P (R2) = 5/7. Let’s compute this same value using the law of total probability (3). First, we’ll find the conditional probabilities. This is a simple counting exercise. P (R2|R1) = 4/6, P (R2|G1) = 5/6. Since R1 and G1 partition Ω the law of total probability says P (R2) = P (R2|R1)P (R1) + P (R2|G1)P (G1) (4) 4 5 5 2 = · + · 6 7 6 7 30 5 = = . 42 7 Probability urns The example above used probability urns. Their use goes back to the beginning of the subject and we would be remiss not to introduce them. This toy model is very useful. We quote from Wikipedia: In probability and statistics, an urn problem is an idealized mental exercise in which some objects of real interest (such as atoms, people, cars, etc.) are represented as colored balls in an urn or other container. One pretends to draw (remove) one or more balls from the urn; the goal is to determine the probability of drawing one color or another, or some other properties. A key parameter is whether each ball is returned to the urn after each draw. 5 5 18.05 class 3, Conditional Probability, Independence and Bayes’ Theorem, Spring 2014 It doesn’t take much to make an example where (3) is really the best way to compute the probability. Here is a game with slightly more complicated rules. Example 4. An urn contains 5 red balls and 2 green balls. A ball is drawn. If it’s green a red ball is added to the urn and if it’s red a green ball is added to the urn. (The original ball is not returned to the urn.) Then a second ball is drawn. What is the probability the second ball is red? answer: The law of total probability says that P (R2) can be computed using the expression in Equation (4). Only the values for the probabilities will change. We have P (R2|R1) = 4/7, P (R2|G1) = 6/7. Therefore, 4 5 6 2 32 P (R2) = P (R2|R1)P (R1) + P (R2|G1)P (G1) = · + · = . 7 7 7 7 49 Using Trees to Organize the Computation Trees are a great way to organize computations with conditional probability and the law of total probability. The figures and examples will make clear what we mean by a tree. As with the rule of product, the key is to organize the underlying process into a sequence of actions. We start by redoing Example 4. The sequence of actions are: first draw ball 1 (and add the appropriate ball to the urn) and then draw ball 2. G1 R1 R2 G2 R2 G2 5/7 2/7 4/7 3/7 6/7 1/7 You interpret this tree as follows. Each dot is called a node. The tree is organized by levels. The top node (root node) is at level 0. The next layer down is level 1 and so on. Each level shows the outcomes at one stage of the game. Level 1 shows the possible outcomes of the first draw. Level 2 shows the possible outcomes of the second draw starting from each node in level 1. Probabilities are written along the branches. The probability of R1 (red on the first draw) is 5/7. It is written along the branch from the root node to the one labeled R1. At the next level we put in conditional probabilities. The probability along the branch from R1 to R2 is P (R2|R1) = 4/7. It represents the probability of going to node R2 given that you are already at R1. The muliplication rule says that the probability of getting to any node is just the product of the probabilities along the path to get there. For example, the node labeled R2 at the far left really represents the event R1 ∩ R2 because it comes from the R1 node. The multiplication rule now says 5 4 P (R1 ∩ R2) = P (R1) · P (R2|R1) = · , 7 7 6 6 18.05 class 3, Conditional Probability, Independence and Bayes’ Theorem, Spring 2014 which is exactly multiplying along the path to the node. The law of total probability is just the statement that P (R2) is the sum of the probabilities of all paths leading to R2 (the two circled nodes in the figure). In this case, 5 4 2 6 32 P (R2) = · + · = , 7 7 7 7 49 exactly as in the previous example. 5.1 Shorthand vs. precise trees The tree given above involves some shorthand. For example, the node marked R2 at the far left really represents the event R1 ∩ R2, since it ends the path from the root through R1 to R2. Here is the same tree with everything labeled precisely. As you can see this tree is more cumbersome to make and use. We usually use the shorthand version of trees. You should make sure you know how to interpret them precisely. G1 R1 R1 ∩R2 R1 ∩G2 G1 ∩R2 G1 ∩G2 P(R1) = 5/7 P(G1) = 2/7 P(R2|R1) = 4/7 P(G2|R1) = 3/7 P(R2|G1) = 6/7 P(G2|G1) = 1/7 Independence Two events are independent if knowledge that one occurred does not change the probability that the other occurred. Informally, events are independent if they do not influence one another. Example 5. Toss a coin twice. We expect the outcomes of the two tosses to be independent of one another. In real experiments this always has to be checked. If my coin lands in honey and I don’t bother to clean it, then the second toss might be affected by the outcome of the first toss. More seriously, the independence of experiments can by undermined by the failure to clean or recalibrate equipment between experiments or to isolate supposedly independent observers from each other or a common influence. We’ve all experienced hearing the same ‘fact’ from different people. Hearing it from different sources tends to lend it credence until we learn that they all heard it from a common source. That is, our sources were not independent. Translating the verbal description of independence into symbols gives A is independent of B if P (A|B) = P (A). (5) That is, knowing that B occurred does not change the probability that A occurred. In terms of events as subsets, knowing that the realized outcome is in B does not change the probability that it is in A. 7 18.05 class 3, Conditional Probability, Independence and Bayes’ Theorem, Spring 2014 If A and B are independent in the above sense, then the multiplication rule gives P (A ∩ B) = P (A|B) · P (B) = P (A) · P (B). This justifies the following technical definition of independence. Formal definition of independence: Two events A and B are independent if P (A ∩ B) = P (A) · P (B) (6) This is a nice symmetric definition which makes clear that A is independent of B if and only if B is independent of A. Unlike the equation with conditional probabilities, this definition makes sense even when P (B) = 0. In terms of conditional probabilities, we have: 1. If P (B) 0 then A and B are independent if and only if P (A|B) = P (A). 2. If P (A) 0 then A and B are independent if and only if P (B|A) = P (B). Independent events commonly arise as different trials in an experiment, as in the following example. Example 6. Toss a fair coin twice. Let H1 = ‘heads on first toss’ and let H2 = ‘heads on second toss’. Are H1 and H2 independent? answer: Since H1 ∩ H2 is the event ‘both tosses are heads’ we have P (H1 ∩ H2) = 1/4 = P (H1)P (H2). Therefore the events are independent. We can ask about the independence of any two events, as in the following two examples. Example 7. Toss a fair coin 3 times. Let H1 = ‘heads on first toss’ and A = ‘two heads total’. Are H1 and A independent? answer: We know that P (A) = 3/8. Since this is not 0 we can check if the formula in Equation 5 holds. Now, H1 = {HHH, HHT, HTH, HTT} contains exactly two outcomes (HHT, HTH) from A, so we have P (A|H1) = 2/4. Since P (A|H1) P (A) these events are not independent. Example 8. Draw one card from a standard deck of playing cards. Let’s examine the independence of 3 events ‘the card is an ace’, ‘the card is a heart’ and ‘the card is red’. Define the events as A = ‘ace’, H = ‘hearts’, R = ‘red’. (a) We know that P (A) = 4/52 = 1/13, P (A|H) = 1/13. Since P (A) = P (A|H) we have that A is independent of H. (b) P (A|R) = 2/26 = 1/13. So A is independent of R. That is, whether the card is an ace is independent of whether it’s red. (c) Finally, what about H and R? Since P (H) = 1/4 and P (H|R) = 1/2, H and R are not independent. We could also see this the other way around: P (R) = 1/2 and P (R|H) = 1, so H and R are not independent. 6.1 Paradoxes of Independence An event A with probability 0 is independent of itself, since in this case both sides of equation (6) are 0. This appears paradoxical because knowledge that A occurred certainly ̸= ̸= ̸= 8 18.05 class 3, Conditional Probability, Independence and Bayes’ Theorem, Spring 2014 gives information about whether A occurred. We resolve the paradox by noting that since P (A) = 0 the statement ‘A occurred’ is vacuous. Think: For what other value(s) of P (A) is A independent of itself? 7 Bayes’ Theorem Bayes’ theorem is a pillar of both probability and statistics and it is central to the rest of this course. For two events A and B Bayes’ theorem (also called Bayes’ rule and Bayes’ formula) says P (A|B) · P (B) P (B|A) = . (7) P (A) Comments: 1. Bayes’ rule tells us how to ‘invert’ conditional probabilities, i.e. to find P (B|A) from P (A|B). 2. In practice, P (A) is often computed using the law of total probability. Proof of Bayes’ rule The key point is that A ∩ B is symmetric in A and B. So the multiplication rule says P (B|A) · P (A) = P (A ∩ B) = P (A|B) · P (B). Now divide through by P (A) to get Bayes’ rule. A common mistake is to confuse P (A|B) and P (B|A). They can be very different. This is illustrated in the next example. Example 9. Toss a coin 5 times. Let H1 = ‘first toss is heads’ and let HA = ‘all 5 tosses are heads’. Then P (H1|HA) = 1 but P (HA|H1) = 1/16. For practice, let’s use Bayes’ theorem to compute P (H1|HA) using P (HA|A1).The terms are P (HA|H1) = 1/16, P (H1) = 1/2, P (HA) = 1/32. So, P (HA|H1)P (H1) (1/16) · (1/2) P (H1|HA) = = = 1, P (HA) 1/32 which agrees with our previous calculation. 7.1 The Base Rate Fallacy The base rate fallacy is one of many examples showing that it’s easy to confuse the meaning of P (B|A) and P (A|B) when a situation is described in words. This is one of the key examples from probability and it will inform much of our practice and interpretation of statistics. You should strive to understand it thoroughly. Example 10. The Base Rate Fallacy Consider a routine screening test for a disease. Suppose the frequency of the disease in the population (base rate) is 0.5%. The test is highly accurate with a 5% false positive rate and a 10% false negative rate. You take the test and it comes back positive. What is the probability that you have the disease? 9 18.05 class 3, Conditional Probability, Independence and Bayes’ Theorem, Spring 2014 answer: We will do the computation three times: using trees, tables and symbols. We’ll use the following notation for the relevant events: D+ = ‘you have the disease’ D− = ‘you do not have the disease T + = ‘you tested positive’ T − = ‘you tested negative’. We are given P (D+) = .005 and therefore P (D−) = .995. The false positive and false negative rates are (by definition) conditional probabilities. P (false positive) = P (T + |D−) = .05 and P (false negative) = P (T −|D+) = .1. The complementary probabilities are known as the true negative and true positive rates: P (T −|D−) = 1 − P (T + |D−) = .95 and P (T + |D+) = 1 − P (T −|D+) = .9. Trees: All of these probabilities can be displayed quite nicely in a tree. D+ D− T+ T− T+ T− .005 .995 .9 .1 .05 .95 The question asks for the probability that you have the disease given that you tested positive, i.e. what is the value of P (D+|T +). We aren’t given this value, but we do know P (T +|D+), so we can use Bayes’ theorem. P (T + |D+) · P (D+) P (D + |T +) = . P (T +) The two probabilities in the numerator are given. We compute the denominator P (T +) using the law of total probability. Using the tree we just have to sum the probabilities for each of the nodes marked T + P (T +) = .995 × .05 + .005 × .9 = .05425 Thus, .9 × .005 P (D + |T +) = = 0.082949 ≈ 8.3%. .05425 Remarks: This is called the base rate fallacy because the base rate of the disease in the population is so low that the vast majority of the people taking the test are healthy, and even with an accurate test most of the positives will be healthy people. Ask your doctor for his/her guess at the odds. To summarize the base rate fallacy with specific numbers 95% of all tests are accurate does not imply 95% of positive tests are accurate We will refer back to this example frequently. It and similar examples are at the heart of many statistical misunderstandings. 18.05 class 3, Conditional Probability, Independence and Bayes’ Theorem, Spring 2014 10 Other ways to work Example 10 Tables: Another trick that is useful for computing probabilities is to make a table. Let’s redo the previous example using a table built with 10000 total people divided according to the probabilites in this example. We construct the table as follows. Pick a number, say 10000 people, and place it as the grand total in the lower right. Using P (D+) = .005 we compute that 50 out of the 10000 people are sick (D+). Likewise 9950 people are healthy (D−). At this point the table looks like: D+ D− total T + T − total 50 9950 10000 Using P (T + |D+) = .9 we can compute that the number of sick people who tested positive as 90% of 50 or 45. The other entries are similar. At this point the table looks like the table below on the left. Finally we sum the T + and T − rows to get the completed table on the right. D+ D− total T + 45 498 T − 5 9452 total 50 9950 10000 D+ D− total T + 45 498 543 T − 5 9452 9457 total 50 9950 10000 Using the complete table we can compute |D + ∩ T + | 45 P (D + |T +) = = = 8.3%. |T + | 543 Symbols: For completeness, we show how the solution looks when written out directly in symbols. P (T + |D+) · P (D+) P (D + |T +) = P (T +) P (T + |D+) · P (D+) = P (T + |D+) · P (D+) + P (T + |D−) · P (D−) .9 × .005 = .9 × .005 + .05 × .995 = 8.3% Visualization: The figure below illustrates the base rate fallacy. The large blue area represents all the healthy people. The much smaller red area represents the sick people. The shaded rectangle represents the the people who test positive. The shaded area covers most of the red area and only a small part of the blue area. Even so, the most of the shaded area is over the blue. That is, most of the positive tests are of healthy people. 18.05 class 3, Conditional Probability, Independence and Bayes’ Theorem, Spring 2014 11 D− D+ 7.2 Bayes’ rule in 18.05 As we said at the start of this section, Bayes’ rule is a pillar of probability and statistics. We have seen that Bayes’ rule allows us to ‘invert’ conditional probabilities. When we learn statistics we will see that the art of statistical inference involves deciding how to proceed when one (or more) of the terms on the right side of Bayes’ rule is unknown. MIT OpenCourseWare 18.05 Introduction to Probability and Statistics Spring 2014 For information about citing these materials or our Terms of Use, visit: |
189991 | https://ocw.mit.edu/courses/res-6-012-introduction-to-probability-spring-2018/d9b5409a84567a002ebecf3b188dd2c7_MITRES_6_012S18_L04.pdf | ECTURE 4~ Counting o·screte uniform law -Assume n consists of n equally likely elemen · ts -Assume consists of k elements k Ten: prob= n n 1 • Applicat·ons permutat·ons number of subsets combina ·ans binom·a probabi it·es partit·ons 1 Basic counting principle 4 shirts 3 ties 2 jackets Number of possible attires? • r stages • ni choices at stage i Number of choices is: 2 Basic counting p.
rinciple examples • Number of license plates with 2 letters followed by 3 digits : • ... if repetition is prohibited: • Permutations: Number of ways of ordering n elements: • Number of subsets of {1, ... , n}: 3 Example • Find the probability that: six rolls of a (six-sided) die all give different numbers. (Assume all outcomes equally likely.) 4 ----- n!
Combinations • Definition. (~). be of k-e e e t u · e s k!(n-k)' o give n-el e t e • wo ways of constructing an ordered se uence of k distinct ·terns. Choose the k items one at a time Choose k items, then order them 5 n) n!
(k = k!(n - k)! (~) = 6 -> Binomia probabi -t-es ( ) -p • P(HTTHHH) • P(pa icular sequence) = • P(particular k-head sequence) P(k eads) 7 A coin tossing problem Assumptions: • Given that there were 3 heads in 10 tosses ' • independence what is the probability that the first two tosses were heads? • P(H) = p event A : the first 2 tosses were heads P(k heads) = (:)pk(l - p)n-k -event B : 3 out of 10 tosses were heads • First solution: P(A I B) = P(A n B) = P(B) 8 A coin tossing problem Assumptions: • Given that there were 3 heads in 10 tosses, • independence what is the probability that the first two tosses were heads? • P(H) = p event A : the first 2 tosses were heads P(k heads) = (:)pk(l - p)n-k -event B : 3 out of 10 tosses were heads • Second solution: Conditional probability law (on B) is uniform 9 Partitio s • n > 1 distinct ·terns, r ~ 1 persons give ni items to person i here n 1, ... , nr are given no negative integers -w·th n1 + · · · + nr = n • Ordering n items: -Deal ni to each person i, and then order n er of art ·o (mult nomia coefficie t) n 10 Example. 52-card deck, dealt (fa·rly) to fo r playe s. ~nd P(each payer gets an ace) • Outcomes are: -number of outcomes. • Co structing an outcome with one ace for each person: distr·bute the aces distr~bute the remaining 48 ca ds 48!
4•3 ,•2· ---- 12! 12! 2! 12! • Answer: 52. 13! 13! 13! 3! 11 Example. 52-card deck, dealt (fa·r y) to fo r playe s. A smart so ution ~nd P(each payer gets an ace) Stack the deck, aces on top Deal, one at a time, to available .. sots" A• A • A • A • • • 111II11111111 11 11 I 11 11 I I I 11111 11111 I I 11 11 I 11 11 I I 1111 I I 11 12 MIT OpenCourseWare Resource: Introduction to Probability John Tsitsiklis and Patrick Jaillet The following may not correspond to a particular course on MIT OpenCourseWare, but has been provided by the author as an individual learning resource. For information about citing these materials or our Terms of Use, visit: 13 |
189992 | https://www.reddit.com/r/skateboarding/comments/13rs7c4/best_ratio_of_height_length_of_the_flat_and/ | Best ratio of height, length of the flat and radius of the transition for a mini ramp : r/skateboarding
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Best ratio of height, length of the flat and radius of the transition for a mini ramp
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I want to build a miniramp. I have little space in a small barn. I want to use the space as perfectly as possible. Is there a rule of thumb how the ratio of height, length of the flat and radius of the transition should be? The total length is 7 meters. I or we are all no pro-skaters ;)
Would be grateful for any experience and tip
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189993 | https://www.vacuum-guide.com/vacuum_components/vacuum_viewport/usa_vacuum_viewport.htm | USA vacuum viewports
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189994 | https://math.stackexchange.com/questions/424139/related-rates-with-a-cone | calculus - Related rates with a cone - Mathematics Stack Exchange
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Related rates with a cone
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I am trying to figure out the rate the water level increases in a conical tank that is 3 m height, 2 m radius at top and water flows in at 2 m 3/minute 2 m 3/minute
I know that
(1/3)π r 2 h=V(1/3)π r 2 h=V
4 π=V 4 π=V
or at 2 seconds the volume of the water is 8.37758 8.37758
so now I have
(1/3)π r 2 h∗d h/d t=V∗d V/d t(1/3)π r 2 h∗d h/d t=V∗d V/d t
which gives an incorrect answer, I am not sure if I use t = 2 volume and height or what.
calculus
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edited Jun 19, 2013 at 14:36
user138246
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asked Jun 19, 2013 at 2:47
user138246 user138246
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So do you have an initial condition that V=4 π V=4 π when t=2 t=2?DanZimm –DanZimm 2013-06-19 02:52:25 +00:00 Commented Jun 19, 2013 at 2:52
1 Parentheses, please. Some would read the first equation with everything in the denominator. Either (1/3)π r 2 h(1/3)π r 2 h or π r 2 h/3 π r 2 h/3 or π r 2 h 3 π r 2 h 3.Ross Millikan –Ross Millikan 2013-06-19 03:04:12 +00:00 Commented Jun 19, 2013 at 3:04
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By similar triangles, observe that:
h 3=r 2⟺r=2 h 3 h 3=r 2⟺r=2 h 3
Hence, substituting into the formula for the volume of a cone will help us to avoid product rule:
V=1 3 π(2 h 3)2 h=4 π 27 h 3 V=1 3 π(2 h 3)2 h=4 π 27 h 3
Differentiating each side with respect to t t yields:
d V d t=4 π 27(3 h 2)d h d t=4 π 9 h 2 d h d t⟺d h d t=9 4 π h 2 d V d t d V d t=4 π 27(3 h 2)d h d t=4 π 9 h 2 d h d t⟺d h d t=9 4 π h 2 d V d t
Since we know that d V/d t=2 d V/d t=2, we simply need to plug in the height at the specific instant in time that the question is asking for into the final equation, and we are done. If the radius is given instead, simply multiply it by 3/2 3/2 to convert to a height.
EDIT: Alright, here's my MS Paint skills. Hopefully this diagram explains where the similar triangles came from:
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edited Jun 20, 2013 at 21:24
answered Jun 19, 2013 at 4:02
AdrianoAdriano
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I don't know what similar triangles means or where those numbers come from or why that would be useful.user138246 –user138246 2013-06-19 14:27:39 +00:00 Commented Jun 19, 2013 at 14:27
2 @Jordan: Similar triangles are ones that have the same angles. The sides are in proportion, but the overall size can be different. If you slice the water with a vertical plane through the centerline, you will get a triangle. As the cone fills, that triangle gets larger, but all the versions have the same angles and are similar. That is where he gets the first line.Ross Millikan –Ross Millikan 2013-06-19 15:21:48 +00:00 Commented Jun 19, 2013 at 15:21
I still don't follow where h/3 = r/2 came from, what do those numbers represent?user138246 –user138246 2013-06-20 20:56:12 +00:00 Commented Jun 20, 2013 at 20:56
@Jordan I've edited in a diagram. Note that h h and r r are the height and radius of the water, while 3 3 and 2 2 are the height and radius of the tank.Adriano –Adriano 2013-06-20 21:26:35 +00:00 Commented Jun 20, 2013 at 21:26
Thank you I understand that, it is really simple once I realized that the top line was basically saying 1 = 1.user138246 –user138246 2013-06-20 21:52:18 +00:00 Commented Jun 20, 2013 at 21:52
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The rate of increase depends on the depth. As you say, V=1 3 π r 2 h V=1 3 π r 2 h. The calculation of the tank volume is not useful. Then d V d t=2=1 3 π(2 r h d r d t+r 2 d h d t)d V d t=2=1 3 π(2 r h d r d t+r 2 d h d t) using the product rule. Now the dimensions of the cone give you d r d h d r d h. Can you use that to combine the two terms in the last parentheses?
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answered Jun 19, 2013 at 3:09
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No, why wouldn't the area be a useful calculation? When I get the derivative am I not left with d v d h∗v=d h d t∗h∗c d v d h∗v=d h d t∗h∗c?user138246 –user138246 2013-06-19 14:29:04 +00:00 Commented Jun 19, 2013 at 14:29
1 @Jordan: It is not useful because the tank need not be full. What you want is the slope of the side.Ross Millikan –Ross Millikan 2013-06-19 15:22:36 +00:00 Commented Jun 19, 2013 at 15:22
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Cyclic sum of adjacent positive integers are power of 2
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Problem 12272 (a) of the American Mathematical Monthly asks if under the condition that the cyclic sum of adjacent positive integers are power of 2 2, those integers could be distinct.
Proposed by H. A. ShahAli, Tehran, Iran, and Stan Wagon, Macalester College, St. Paul, MN. For which integers n n with n≥3 n≥3 do there exist distinct positive integers a 1,…,a n a 1,…,a n such that a i+a i+1 a i+a i+1 is a power of 2 2 for all i∈{1,…,n}i∈{1,…,n}? (Here subscripts are taken modulo n n, so that a n+1=a 1 a n+1=a 1.)
The answer is negative for every n≥3 n≥3. An easy contradiction can be deduced by noting that the condition yields a 1<a 2<…<a n<a 1 a 1<a 2<…<a n<a 1 or vice versa, since once a "peak" a ia i+2 a ia i+2 occurs, either a i a i or a i+2 a i+2 must be negative, WLOG a 1<a 2 a 1<a 2.
Editors comment that Yuri Ionin strengthened the conclusion, using induction to prove that positive integers a 1,…,a n a 1,…,a n chosen so that cyclically a i+a i+1 a i+a i+1 is always a power of 2 2 has at most ⌈n+1 2⌉⌈n+1 2⌉ distinct elements and that this bound is sharp.
I attempt to find a proof of this comment. A valid construction is let a 1=2 a 1=2, a i+a i+1=2 i+2 a i+a i+1=2 i+2 for i=1,2,…,⌈n+1 2⌉i=1,2,…,⌈n+1 2⌉, and a⌈n+1 2⌉+i=a⌈n+1 2⌉−i+1 a⌈n+1 2⌉+i=a⌈n+1 2⌉−i+1 for left elements. And we can apply the proof of the original problem to prove the base case. However I got stuck in the induction step. We cannot merely consider a 1,…,a n−1+a n a 1,…,a n−1+a n and I don't know how to use the induction hypothesis.
My question is: how to prove the generalization by induction or other techniques?
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We do induction on n n. Assume n≥3 n≥3, let a j=max i a i a j=max i a i. If a j=a j+1 a j=a j+1 or a i=a j−1 a i=a j−1, deleting a j a j from the sequence, then by induction hypothesis we're done. Now we assume both a j>a j−1 a j>a j−1 and a j>a j+1 a j>a j+1.
Note that for any a a, there is at most one 0<b<a 0<b<a such that a+b a+b is a power of 2 2:
a<a+b<a+a=2 a a<a+b<a+a=2 a
And there is at most one number which is a power of 2 2 in the interval (a,2 a](a,2 a].
BTW, I couldn't understand the proof for the original version you mentioned without the above key observation. This is also very intuitive to understand (especially for someone with some programming experiences) based on binary arithmetic: Given the binary expression of a a, there is exactly one way to have b<a b<a cancel all the digits of a a by addition.
Therefore since a j−1+a j a j−1+a j and a j+1+a j a j+1+a j are both powers of 2 2, we must have a j−1=a j+1 a j−1=a j+1. Now by deleting a j a j and a j−1 a j−1 from the sequence, we can apply induction hypothesis to get
|{a 1,⋯,a n}|≤1+⌈(n−2)+1 2⌉≤⌈n+1 2⌉|{a 1,⋯,a n}|≤1+⌈(n−2)+1 2⌉≤⌈n+1 2⌉
Now we show the bound is sharp. When n n is odd, we can inductively build:
⎧⎩⎨⎪⎪a 1=1 a i=2 i−a i−1 a i=a n−i+1 for 2≤i≤n+1 2 for n+1 2+1≤i≤n{a 1=1 a i=2 i−a i−1 for 2≤i≤n+1 2 a i=a n−i+1 for n+1 2+1≤i≤n
In simple words, the sequence is strictly increasing for the first half until the middle point, then starts decreasing in the second half back to a 1=1 a 1=1.
Similarly When n n is even, we may define:
⎧⎩⎨⎪⎪a 1=1 a i=2 i−a i−1 a i=a n−i+2 for 2≤i≤n 2+1 for n 2+2≤i≤n{a 1=1 a i=2 i−a i−1 for 2≤i≤n 2+1 a i=a n−i+2 for n 2+2≤i≤n
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edited May 11 at 9:44
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A formula for the foot of perpendicular
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I found the following formula using derivatives but I guess there should be a geometric verification of that.
If the parametric form of a line in R 2 R 2 or R 3 R 3 is
X=P+t V⃗X=P+t V→
then the parameter value for the foot of perpendicular from the point A A to the line is given by
t=P A→.V⃗V⃗.V⃗t=P A→.V→V→.V→
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Can one prove it without using derivative?
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edited Jan 28, 2020 at 20:04
Mohammad Riazi-KermaniMohammad Riazi-Kermani
asked Jan 26, 2020 at 22:50
Mohammad Riazi-KermaniMohammad Riazi-Kermani
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2 That is the scalar (perpendicular) projection of the vector P A→P A→ on the line. That's just vector geometry.DonAntonio –DonAntonio 2020-01-26 23:08:08 +00:00 Commented Jan 26, 2020 at 23:08
Good comment,thanks.Mohammad Riazi-Kermani –Mohammad Riazi-Kermani 2020-01-26 23:30:55 +00:00 Commented Jan 26, 2020 at 23:30
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If you draw the triangle, you need it to be pythagorean for the angle between A−(P+t V)A−(P+t V) and the line to be square. So you need
∥A−(P+t V)∥2+∥t V∥2=∥A P∥2.‖A−(P+t V)‖2+‖t V‖2=‖A P‖2.
That is, as A−(P+t V)=(A−P)+t V=A P+t V A−(P+t V)=(A−P)+t V=A P+t V,
∥A P∥2+t 2∥V∥2−2 A P⋅(t V)=∥A P∥2.‖A P‖2+t 2‖V‖2−2 A P⋅(t V)=‖A P‖2.
Solving for t t (the solution t=0 t=0 doesn't count since it gives a degenerate triangle),
t=A P⋅V V⋅V.t=A P⋅V V⋅V.
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answered Jan 26, 2020 at 23:58
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189997 | https://math.mit.edu/~goemans/18310S15/lpnotes310.pdf | 18.310A lecture notes March 17, 2015 Linear programming Lecturer: Michel Goemans 1 Basics Linear Programming deals with the problem of optimizing a linear objective function subject to linear equality and inequality constraints on the decision variables. Linear programming has many practical applications (in transportation, production planning, ...). It is also the building block for combinatorial optimization. One aspect of linear programming which is often forgotten is the fact that it is also a useful proof technique. In this first chapter, we describe some linear programming formulations for some classical problems. We also show that linear programs can be expressed in a variety of equivalent ways.
1.1 Formulations 1.1.1 The Diet Problem In the diet model, a list of available foods is given together with the nutrient content and the cost per unit weight of each food. A certain amount of each nutrient is required per day. For example, here is the data corresponding to a civilization with just two types of grains (G1 and G2) and three types of nutrients (starch, proteins, vitamins): Starch Proteins Vitamins Cost ($/kg) G1 5 4 2 0.6 G2 7 2 1 0.35 Nutrient content and cost per kg of food.
The requirement per day of starch, proteins and vitamins is 8, 15 and 3 respectively. The problem is to find how much of each food to consume per day so as to get the required amount per day of each nutrient at minimal cost.
When trying to formulate a problem as a linear program, the first step is to decide which decision variables to use. These variables represent the unknowns in the problem. In the diet problem, a very natural choice of decision variables is: • x1: number of units of grain G1 to be consumed per day, • x2: number of units of grain G2 to be consumed per day.
The next step is to write down the objective function. The objective function is the function to be minimized or maximized. In this case, the objective is to minimize the total cost per day which is given by z = 0.6x1 + 0.35x2 (the value of the objective function is often denoted by z).
Finally, we need to describe the different constraints that need to be satisfied by x1 and x2.
First of all, x1 and x2 must certainly satisfy x1 ≥0 and x2 ≥0. Only nonnegative amounts of LP-1 food can be eaten! These constraints are referred to as nonnegativity constraints. Nonnegativity constraints appear in most linear programs. Moreover, not all possible values for x1 and x2 give rise to a diet with the required amounts of nutrients per day. The amount of starch in x1 units of G1 and x2 units of G2 is 5x1 + 7x2 and this amount must be at least 8, the daily requirement of starch. Therefore, x1 and x2 must satisfy 5x1 +7x2 ≥8. Similarly, the requirements on the amount of proteins and vitamins imply the constraints 4x1 + 2x2 ≥15 and 2x1 + x2 ≥3.
This diet problem can therefore be formulated by the following linear program: Minimize z = 0.6x1 + 0.35x2 subject to: 5x1 + 7x2 ≥8 4x1 + 2x2 ≥15 2x1 + x2 ≥3 x1 ≥0, x2 ≥0.
Some more terminology. A solution x = (x1, x2) is said to be feasible with respect to the above linear program if it satisfies all the above constraints. The set of feasible solutions is called the feasible space or feasible region. A feasible solution is optimal if its objective function value is equal to the smallest value z can take over the feasible region.
1.1.2 The Transportation Problem Suppose a company manufacturing widgets has two factories located at cities F1 and F2 and three retail centers located at C1, C2 and C3. The monthly demand at the retail centers are (in thousands of widgets) 8, 5 and 2 respectively while the monthly supply at the factories are 6 and 9 respectively.
Notice that the total supply equals the total demand. We are also given the cost of transportation of 1 widget between any factory and any retail center.
C1 C2 C3 F1 5 5 3 F2 6 4 1 Cost of transportation (in 0.01$/widget).
In the transportation problem, the goal is to determine the quantity to be transported from each factory to each retail center so as to meet the demand at minimum total shipping cost.
In order to formulate this problem as a linear program, we first choose the decision variables.
Let xij (i = 1, 2 and j = 1, 2, 3) be the number of widgets (in thousands) transported from factory Fi to city Cj. Given these xij’s, we can express the total shipping cost, i.e. the objective function to be minimized, by 5x11 + 5x12 + 3x13 + 6x21 + 4x22 + x23.
We now need to write down the constraints. First, we have the nonnegativity constraints saying that xij ≥0 for i = 1, 2 and j = 1, 2, 3. Moreover, we have that the demand at each retail center must be met. This gives rise to the following constraints: x11 + x21 = 8, LP-2 x12 + x22 = 5, x13 + x23 = 2.
Finally, each factory cannot ship more than its supply, resulting in the following constraints: x11 + x12 + x13 ≤6, x21 + x22 + x23 ≤9.
These inequalities can be replaced by equalities since the total supply is equal to the total demand.
A linear programming formulation of this transportation problem is therefore given by: Minimize 5x11 + 5x12 + 3x13 + 6x21 + 4x22 + x23 subject to: x11 + x21 = 8 x12 + x22 = 5 x13 + x23 = 2 x11 + x12 + x13 = 6 x21 + x22 + x23 = 9 x11 ≥0, x21 ≥0, x31 ≥0, x12 ≥0, x22 ≥0, x32 ≥0.
Among these 5 equality constraints, one is redundant, i.e. it is implied by the other constraints or, equivalently, it can be removed without modifying the feasible space. For example, by adding the first 3 equalities and substracting the fourth equality we obtain the last equality. Similarly, by adding the last 2 equalities and substracting the first two equalities we obtain the third one.
1.2 Representations of Linear Programs A linear program can take many different forms. First, we have a minimization or a maximization problem depending on whether the objective function is to be minimized or maximized.
The constraints can either be inequalities (≤or ≥) or equalities. Some variables might be unrestricted in sign (i.e. they can take positive or negative values; this is denoted by ≷0) while others might be restricted to be nonnegative. A general linear program in the decision variables x1, . . . , xn is therefore of the following form: Maximize or Minimize z = c0 + c1x1 + . . . + cnxn subject to: ai1x1 + ai2x2 + . . . + ainxn ≤ ≥ = bi i = 1, . . . , m xj ≥0 ≷0 j = 1, . . . , n.
The problem data in this linear program consists of cj (j = 0, . . . , n), bi (i = 1, . . . , m) and aij (i = 1, . . . , m, j = 1, . . . , n). cj is referred to as the objective function coefficient of xj or, more LP-3 simply, the cost coefficient of xj. bi is known as the right-hand-side (RHS) of equation i. Notice that the constant term c0 can be omitted without affecting the set of optimal solutions.
A linear program is said to be in standard form if • it is a maximization program, • there are only equalities (no inequalities) and • all variables are restricted to be nonnegative.
In matrix form, a linear program in standard form can be written as: Max z = cT x subject to: Ax = b x ≥0.
where c = c1 .
.
.
cn , b = b1 .
.
.
bm , x = x1 .
.
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xn are column vectors, cT denote the transpose of the vector c, and A = [aij] is the m × n matrix whose i, j−element is aij.
Any linear program can in fact be transformed into an equivalent linear program in standard form. Indeed, • If the objective function is to minimize z = c1x1 + . . . + cnxn then we can simply maximize z′ = −z = −c1x1 −. . . −cnxn.
• If we have an inequality constraint ai1x1 + . . . + ainxn ≤bi then we can transform it into an equality constraint by adding a slack variable, say s, restricted to be nonnegative: ai1x1 + . . . + ainxn + s = bi and s ≥0.
• Similarly, if we have an inequality constraint ai1x1 +. . .+ainxn ≥bi then we can transform it into an equality constraint by adding a surplus variable, say s, restricted to be nonnegative: ai1x1 + . . . + ainxn −s = bi and s ≥0.
• If xj is unrestricted in sign then we can introduce two new decision variables x+ j and x− j restricted to be nonnegative and replace every occurrence of xj by x+ j −x− j .
For example, the linear program Minimize z = 2x1 −x2 subject to: x1 + x2 ≥2 3x1 + 2x2 ≤4 x1 + 2x2 = 3 x1 ≷0, x2 ≥0.
LP-4 is equivalent to the linear program Maximize z′ = −2x+ 1 + 2x− 1 + x2 subject to: x+ 1 −x− 1 + x2 −x3 = 2 3x+ 1 −3x− 1 + 2x2 + x4 = 4 x+ 1 −x− 1 + 2x2 = 3 x+ 1 ≥0, x− 1 ≥0, x2 ≥0, x3 ≥0, x4 ≥0.
with decision variables x+ 1 , x− 1 , x2, x3, x4. Notice that we have introduced different slack or surplus variables into different constraints.
In some cases, another form of linear program is used. A linear program is in canonical form if it is of the form: Max z = cT x subject to: Ax ≤b x ≥0.
A linear program in canonical form can be replaced by a linear program in standard form by just replacing Ax ≤b by Ax + Is = b, s ≥0 where s is a vector of slack variables and I is the m × m identity matrix. Similarly, a linear program in standard form can be replaced by a linear program in canonical form by replacing Ax = b by A′x ≤b′ where A′ = A −A and b′ = b −b .
2 The Simplex Method In 1947, George B. Dantzig developed a technique to solve linear programs — this technique is referred to as the simplex method.
2.1 Brief Review of Some Linear Algebra Two systems of equations Ax = b and ¯ Ax = ¯ b are said to be equivalent if {x : Ax = b} = {x : ¯ Ax = ¯ b}. Let Ei denote equation i of the system Ax = b, i.e. ai1x1 + . . . + ainxn = bi. Given a system Ax = b, an elementary row operation consists in replacing Ei either by αEi where α is a nonzero scalar or by Ei + βEk for some k ̸= i. Clearly, if ¯ Ax = ¯ b is obtained from Ax = b by an elementary row operation then the two systems are equivalent. (Exercise: prove this.) Notice also that an elementary row operation is reversible.
Let ars be a nonzero element of A. A pivot on ars consists of performing the following sequence of elementary row operations: • replacing Er by ¯ Er = 1 ars Er, • for i = 1, . . . , m, i ̸= r, replacing Ei by ¯ Ei = Ei −ais ¯ Er = Ei −ais ars Er.
LP-5 After pivoting on ars, all coefficients in column s are equal to 0 except the one in row r which is now equal to 1. Since a pivot consists of elementary row operations, the resulting system ¯ Ax = ¯ b is equivalent to the original system.
Elementary row operations and pivots can also be defined in terms of matrices. Let P be an m × m invertible (i.e. P −1 exists1) matrix. Then {x : Ax = b} = {x : PAx = Pb}. The two types of elementary row operations correspond to the matrices (the coefficients not represented are equal to 0): P = 1 ...
1 α 1 ...
1 ←i and P = 1 ...
1 β ...
1 ...
1 ←i ←k .
Pivoting on ars corresponds to premultiplying Ax = b by P = 1 −a1s/ars ...
1 −ar−1,s/ars 1/ars −ar+1,s/ars 1 ...
−ams/ars 1 ←r.
2.2 The Simplex Method on an Example For simplicity, we shall assume that we have a linear program of (what seems to be) a rather special form (we shall see later on how to obtain such a form): • the linear program is in standard form, • b ≥0, • there exists a collection B of m variables called a basis such that – the submatrix AB of A consisting of the columns of A corresponding to the variables in B is the m × m identity matrix and – the cost coefficients corresponding to the variables in B are all equal to 0.
For example, the following linear program has this required form: 1This is equivalent to saying that det P ̸= 0 or also that the system Px = 0 has x = 0 as unique solution LP-6 Max z = 10 + 20 x1 + 16 x2 + 12 x3 subject to x1 + x4 = 4 2 x1 + x2 + x3 +x5 = 10 2 x1 + 2x2 + x3 + x6 = 16 x1, x2, x3, x4, x5, x6 ≥0.
In this example, B = {x4, x5, x6}. The variables in B are called basic variables while the other variables are called nonbasic. The set of nonbasic variables is denoted by N. In the example, N = {x1, x2, x3}.
The advantage of having AB = I is that we can quickly infer the values of the basic variables given the values of the nonbasic variables. For example, if we let x1 = 1, x2 = 2, x3 = 3, we obtain x4 = 4 −x1 = 3, x5 = 10 −2x1 −x2 −x3 = 3, x6 = 16 −2x1 −2x2 −x3 = 7.
Also, we don’t need to know the values of the basic variables to evaluate the cost of the solution.
In this case, we have z = 10 + 20x1 + 16x2 + 12x3 = 98. Notice that there is no guarantee that the so-constructed solution be feasible. For example, if we set x1 = 5, x2 = 2, x3 = 1, we have that x4 = 4 −x1 = −1 does not satisfy the nonnegativity constraint x4 ≥0.
There is an assignment of values to the nonbasic variables that needs special consideration. By just letting all nonbasic variables to be equal to 0, we see that the values of the basic variables are just given by the right-hand-sides of the constraints and the cost of the resulting solution is just the constant term in the objective function. In our example, letting x1 = x2 = x3 = 0, we obtain x4 = 4, x5 = 10, x6 = 16 and z = 10. Such a solution is called a basic feasible solution or bfs. The feasibility of this solution comes from the fact that b ≥0. Later, we shall see that, when solving a linear program, we can restrict our attention to basic feasible solutions. The simplex method is an iterative method that generates a sequence of basic feasible solutions (corresponding to different bases) and eventually stops when it has found an optimal basic feasible solution.
Instead of always writing explicitely these linear programs, we adopt what is known as the tableau format.
First, in order to have the objective function play a similar role as the other constraints, we consider z to be a variable and the objective function as a constraint. Putting all variables on the same side of the equality sign, we obtain: −z + 20x1 + 16x2 + 12x3 = −10.
We also get rid of the variable names in the constraints to obtain the tableau format: −z x1 x2 x3 x4 x5 x6 1 20 16 12 -10 1 0 0 1 4 2 1 1 1 10 2 2 1 1 16 Our bfs is currently x1 = 0, x2 = 0, x3 = 0, x4 = 4, x5 = 10, x6 = 16 and z = 10. Since the cost coefficient c1 of x1 is positive (namely, it is equal to 20), we notice that we can increase z by increasing x1 and keeping x2 and x3 at the value 0. But in order to maintain feasibility, we must LP-7 have that x4 = 4−x1 ≥0, x5 = 10−2x1 ≥0, x6 = 16−2x1 ≥0. This implies that x1 ≤4. Letting x1 = 4, x2 = 0, x3 = 0, we obtain x4 = 0, x5 = 2, x6 = 8 and z = 90. This solution is also a bfs and corresponds to the basis B = {x1, x5, x6}. We say that x1 has entered the basis and, as a result, x4 has left the basis. We would like to emphasize that there is a unique basic solution associated with any basis. This (not necessarily feasible) solution is obtained by setting the nonbasic variables to zero and deducing the values of the basic variables from the m constraints.
Now we would like that our tableau reflects this change by showing the dependence of the new basic variables as a function of the nonbasic variables. This can be accomplished by pivoting on the element a11. Why a11? Well, we need to pivot on an element of column 1 because x1 is entering the basis. Moreover, the choice of the row to pivot on is dictated by the variable which leaves the basis. In this case, x4 is leaving the basis and the only 1 in column 4 is in row 1. After pivoting on a11, we obtain the following tableau: −z x1 x2 x3 x4 x5 x6 1 16 12 -20 -90 1 0 0 1 4 1 1 -2 1 2 2 1 -2 1 8 Notice that while pivoting we also modified the objective function row as if it was just like another constraint. We have now a linear program which is equivalent to the original one from which we can easily extract a (basic) feasible solution of value 90. Still z can be improved by increasing xs for s = 2 or 3 since these variables have a positive cost coefficient2 ¯ cs. Let us choose the one with the greatest ¯ cs; in our case x2 will enter the basis. The maximum value that x2 can take while x3 and x4 remain at the value 0 is dictated by the constraints x1 = 4 ≥0, x5 = 2−x2 ≥0 and x6 = 8 −2x2 ≥0. The tightest of these inequalities being x5 = 2 −x2 ≥0, we have that x5 will leave the basis. Therefore, pivoting on ¯ a22, we obtain the tableau: −z x1 x2 x3 x4 x5 x6 1 -4 12 -16 -122 1 0 1 0 4 1 1 -2 1 2 -1 2 -2 1 4 The current basis is B = {x1, x2, x6} and its value is 122.
Since 12 > 0, we can improve the current basic feasible solution by having x4 enter the basis. Instead of writing explicitely the constraints on x4 to compute the level at which x4 can enter the basis, we perform the min ratio test. If xs is the variable that is entering the basis, we compute min i:¯ ais>0{¯ bi/¯ ais}.
The argument of the minimum gives the variable that is exiting the basis. In our example, we obtain 2 = min{4/1, 4/2} and therefore variable x6 which is the basic variable corresponding to row 3 leaves the basis. Moreover, in order to get the updated tableau, we need to pivot on ¯ a34.
Doing so, we obtain: 2By simplicity, we always denote the data corresponding to the current tableau by ¯ c, ¯ A, and ¯ b.
LP-8 −z x1 x2 x3 x4 x5 x6 1 2 -4 -6 -146 1 1/2 1 -1/2 2 1 0 -1 1 6 -1/2 1 -1 1/2 2 Our current basic feasible solution is x1 = 2, x2 = 6, x3 = 0, x4 = 2, x5 = 0, x6 = 0 with value z = 146. By the way, why is this solution feasible? In other words, how do we know that the right-hand-sides (RHS) of the constraints are guaranteed to be nonnegative? Well, this follows from the min ratio test and the pivot operation. Indeed, when pivoting on ¯ ars, we know that • ¯ ars > 0, • ¯ br ¯ ars ≤ ¯ bi ¯ ais if ¯ ais > 0.
After pivoting the new RHS satisfy • ¯ br = ¯ br ¯ ars ≥0, • ¯ bi = ¯ bi −¯ ais ¯ ars ≥¯ bi ≥0 if ¯ ais ≤0 and • ¯ bi = ¯ bi −¯ ais ¯ ars = ¯ ais ¯ bi ¯ ais − ¯ br ¯ ars ≥0 if ¯ ais > 0.
We can also justify why the solution keeps improving. Indeed, when we pivot on ¯ ars > 0, the constant term ¯ c0 in the objective function becomes ¯ c0 + ¯ br ∗¯ cs/¯ ars. If ¯ br > 0, we have a strict improvement in the objective function value since by our choice of entering variable ¯ cs > 0. We shall deal with the case ¯ br = 0 later on.
The bfs corresponding to B = {1, 2, 4} is not optimal since there is still a positive cost coefficient.
We see that x3 can enter the basis and, since there is just one positive element in row 3, we have that x1 leaves the basis. We thus pivot on ¯ a13 and obtain: −z x1 x2 x3 x4 x5 x6 1 -4 -8 -4 -154 2 1 2 -1 4 0 1 -1 1 6 1 1 0 0 4 The current basis is {x3, x2, x4} and the associated bfs is x1 = 0, x2 = 6, x3 = 4, x4 = 4, x5 = 0, x6 = 0 with value z = 154. This bfs is optimal since the objective function reads z = 154 −4x1 − 8x5 −4x6 and therefore cannot be more than 154 due to the nonnegativity constraints.
Through a sequence of pivots, the simplex method thus goes from one linear program to another equivalent linear program which is trivial to solve. Remember the crucial observation that a pivot operation does not alter the feasible region.
In the above example, we have not encountered several situations that may typically occur.
First, in the min ratio test, several terms might produce the minimum.
In that case, we can arbitrarily select one of them. For example, suppose the current tableau is: LP-9 −z x1 x2 x3 x4 x5 x6 1 16 12 -20 -90 1 0 0 1 4 1 1 -2 1 2 2 1 -2 1 4 and that x2 is entering the basis. The min ratio test gives 2 = min{2/1, 4/2} and, thus, either x5 or x6 can leave the basis. If we decide to have x5 leave the basis, we pivot on ¯ a22; otherwise, we pivot on ¯ a32. Notice that, in any case, the pivot operation creates a zero coefficient among the RHS. For example, pivoting on ¯ a22, we obtain: −z x1 x2 x3 x4 x5 x6 1 -4 12 -16 -122 1 0 1 0 4 1 1 -2 1 2 -1 2 -2 1 0 A bfs with ¯ bi = 0 for some i is called degenerate. A linear program is nondegenerate if no bfs is degenerate. Pivoting now on ¯ a34 we obtain: −z x1 x2 x3 x4 x5 x6 1 2 -4 -6 -122 1 1/2 1 -1/2 4 1 0 -1 1 2 -1/2 1 -1 1/2 0 This pivot is degenerate. A pivot on ¯ ars is called degenerate if ¯ br = 0. Notice that a degenerate pivot alters neither the ¯ bi’s nor ¯ c0. In the example, the bfs is (4, 2, 0, 0, 0, 0) in both tableaus. We thus observe that several bases can correspond to the same basic feasible solution.
Another situation that may occur is when xs is entering the basis, but ¯ ais ≤0 for i = 1, . . . , m.
In this case, there is no term in the min ratio test. This means that, while keeping the other nonbasic variables at their zero level, xs can take an arbitrarily large value without violating feasibility. Since ¯ cs > 0, this implies that z can be made arbitrarily large. In this case, the linear program is said to be unbounded or unbounded from above if we want to emphasize the fact that we are dealing with a maximization problem. For example, consider the following tableau: −z x1 x2 x3 x4 x5 x6 1 16 12 20 -90 1 0 0 -1 4 1 1 0 1 2 2 1 -2 1 8 If x4 enters the basis, we have that x1 = 4 + x4, x5 = 2 and x6 = 8 + 2x4 and, as a result, for any nonnegative value of x4, the solution (4 + x4, 0, 0, x4, 2, 8 + 2x4) is feasible and its objective function value is 90 + 20x4. There is thus no finite optimum.
LP-10 2.3 Detailed Description of Phase II In this section, we summarize the different steps of the simplex method we have described in the previous section. In fact, what we have described so far constitutes Phase II of the simplex method.
Phase I deals with the problem of putting the linear program in the required form. This will be described in a later section.
Phase II of the simplex method 1. Suppose the initial or current tableau is −z x1 . . .
xs . . .
xn 1 ¯ c1 . . .
¯ cs . . .
¯ cn −¯ c0 ¯ a11 . . .
¯ a1s . . .
¯ a1n ¯ b1 ≥0 .
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¯ ar1 . . .
¯ ars . . .
¯ arn ¯ br ≥0 .
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¯ am1 . . .
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¯ amn ¯ bm ≥0 and the variables can be partitioned into B = {xj1, . . . , xjm} and N with • ¯ cji = 0 for i = 1, . . . , m and • ¯ akji = 0 k ̸= i 1 k = i.
The current basic feasible solution is given by xji = ¯ bi for i = 1, . . . , m and xj = 0 otherwise. The objective function value of this solution is ¯ c0.
2. If ¯ cj ≤0 for all j = 1, . . . , n then the current basic feasible solution is optimal.
STOP.
3. Find a column s for which ¯ cs > 0. xs is the variable entering the basis.
4. Check for unboundedness. If ¯ ais ≤0 for i = 1, . . . , m then the linear program is unbounded. STOP.
5. Min ratio test. Find row r such that ¯ br ¯ ars = min i:¯ ais>0 ¯ bi ¯ ais .
6. Pivot on ¯ ars. I.e. replace the current tableau by: LP-11 −z . . .
xs . . .
xj . . .
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0 . . .
¯ cj −¯ arj¯ cs ¯ ars . . .
−¯ c0 − ¯ br¯ cs ¯ ars .
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row r . . .
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¯ arj ¯ ars . . .
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row i . . .
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¯ aij −¯ arj¯ ais ¯ ars . . .
¯ bi − ¯ br¯ ais ¯ ars .
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Replace xjr by xs in B.
7. Go to step 2.
2.4 Convergence of the Simplex Method As we have seen, the simplex method is an iterative method that generates a sequence of basic feasible solutions. But, do we have any guarantee that this process eventually terminates? The answer is yes if the linear program is nondegenerate.
Theorem 2.1. The simplex method solves a nondegenerate linear program in finitely many itera-tions.
Proof. For nondegenerate linear programs, we have a strict improvement (namely of value ¯ br¯ cs ¯ ars > 0) in the objective function value at each iteration. This means that, in the sequence of bfs produced by the simplex method, each bfs can appear at most once. Therefore, for nondegenerate linear programs, the number of iterations is certainly upper bounded by the number of bfs. This latter number is finite (for example, it is upper bounded by n m ) since any bfs corresponds to m variables being basic3.
However, when the linear program is degenerate, we might have degenerate pivots which give no strict improvement in the objective function. As a result, a subsequence of bases might repeat implying the nontermination of the method. This phenomenon is called cycling.
2.4.1 An Example of Cycling The following is an example that will cycle if unfortunate choices of entering and leaving variables are made (the pivot element is within a box).
3Not all choices of basic variables give rise to feasible solutions.
LP-12 −z x1 x2 x3 x4 x5 x6 1 4 1.92 -16 -0.96 0 -12.5 -2 12.5 1 1 0 1 0.24 -2 -0.24 1 0 −z x1 x2 x3 x4 x5 x6 1 0.96 -8 0 -4 0 1 -12.5 -2 1 12.5 0 1 0.24 -2 -0.24 1 0 −z x1 x2 x3 x4 x5 x6 1 4 1.92 -0.96 -16 0 1 -12.5 -2 1 12.5 0 1 1 0.24 -0.24 -2 0 −z x1 x2 x3 x4 x5 x6 1 -4 0.96 0 -8 0 12.5 1 1 -2 -12.5 0 1 1 0.24 -0.24 -2 0 −z x1 x2 x3 x4 x5 x6 1 -16 -0.96 1.92 4 0 12.5 1 1 -2 -12.5 0 -2 -0.24 1 0.24 1 0 −z x1 x2 x3 x4 x5 x6 1 -8 0 -4 0.96 0 -12.5 -2 12.5 1 1 0 -2 -0.24 1 0.24 1 0 −z x1 x2 x3 x4 x5 x6 1 4 1.92 -16 -0.96 0 -12.5 -2 12.5 1 1 0 1 0.24 -2 -0.24 1 0 2.4.2 Bland’s Anticycling Rule The simplex method, as described in the previous section, is ambiguous. First, if we have several variables with a positive ¯ cs (cfr. Step 3) we have not specified which will enter the basis. Moreover, there might be several variables attaining the minimum in the minimum ratio test (Step 5). If so, we need to specify which of these variables will leave the basis. A pivoting rule consists of an entering variable rule and a leaving variable rule that unambiguously decide what will be the entering and leaving variables.
The most classical entering variable rule is: LP-13 Largest coefficient entering variable rule: Select the variable xs with the largest ¯ cs > 0. In case of ties, select the one with the smallest subscript s.
The corresponding leaving variable rule is: Largest coefficient leaving variable rule: Among all rows attaining the minimum in the min-imum ratio test, select the one with the largest pivot ¯ ars. In case of ties, select the one with the smallest subscript r.
The example of subsection 2.4.1 shows that the use of the largest coefficient entering and leaving variable rules does not prevent cycling. There are two rules that avoid cycling: the lexicographic rule and Bland’s rule (after R. Bland who discovered it in 1976). We’ll just describe the latter one, which is conceptually the simplest.
Bland’s anticycling pivoting rule: Among all variables xs with positive ¯ cs, select the one with the smallest subscript s. Among the eligible (according to the minimum ratio test) leaving variables xl, select the one with the smallest subscript l.
Theorem 2.2. The simplex method with Bland’s anticycling pivoting rule terminates after a finite number of iterations.
Proof. The proof is by contradiction. If the method does not stop after a finite number of iterations then there is a cycle of tableaus that repeats. If we delete from the tableau that initiates this cycle the rows and columns not containing pivots during the cycle, the resulting tableau has a cycle with the same pivots. For this tableau, all right-hand-sides are zero throughout the cycle since all pivots are degenerate.
Let t be the largest subscript of the variables remaining. Consider the tableau T1 in the cycle with xt leaving. Let B = {xj1, . . . , xjm} be the corresponding basis (say jr = t), xs be the associated entering variable and, a1 ij and c1 j the constraint and cost coefficients. On the other hand, consider the tableau T2 with xt entering and denotes by a2 ij and c2 j the corresponding constraint and cost coefficients.
Let x be the (infeasible) solution obtained by letting the nonbasic variables in T1 be zero except for xs = −1. Since all RHS are zero, we deduce that xji = ais for i = 1, . . . , m. Since T2 is obtained from T1 by elementary row operations, x must have the same objective function value in T1 and T2. This means that c1 0 −c1 s = c2 0 −c2 s + m X i=1 a1 isc2 ji.
Since we have no improvement in objective function in the cycle, we have c1 0 = c2 0. Moreover, c1 s > 0 and, by Bland’s rule, c2 s ≤0 since otherwise xt would not be the entering variable in T2. Hence, m X i=1 a1 isc2 ji < 0 implying that there exists k with a1 ksc2 jk < 0. Notice that k ̸= r, i.e. jk < t, since the pivot element in T1, a1 rs, must be positive and c2 t > 0.
However, in T2, all cost coefficients c2 j except c2 t are nonnegative; otherwise xj would have been selected as entering variable. Thus c2 jk < 0 and a1 ks > 0.
This is a contradiction because Bland’s rule should have selected xjk rather than xt in T1 as leaving variable.
LP-14 2.5 Phase I of the Simplex Method In this section, we show how to transform a linear program into the form presented in Section 2.2. For that purpose, we show how to find a basis of the linear program which leads to a basic feasible solution. Sometimes, of course, we may inherit a bfs as part of the problem formulation.
For example, we might have constraints of the form Ax ≤b with b ≥0 in which case the slack variables constitute a bfs. Otherwise, we use the two-phase simplex method to be described in this section.
Consider a linear program in standard form with b ≥0 (this latter restriction is without loss of generality since we may multiply some constraints by -1). In phase I, instead of solving Max z = c0 + cT x subject to: (P) Ax = b x ≥0 we add some artificial variables {xa i : i = 1, . . . , m} and consider the linear program: Min w = m X i=1 xa i subject to: Ax + Ixa = b x ≥0, xa ≥0.
This program is not in the form required by the simplex method but can easily be transformed to it. Changing the min w by max w′ = −w and expressing the objective function in terms of the initial variables, we obtain: Max w′ = −eT b + (eT A)x subject to: (Q) Ax + Ixa = b x ≥0, xa ≥0 where e is a vector of 1’s. We have artificially created a bfs, namely x = 0 and xa = b. We now use the simplex method as described in the previous section. There are three possible outcomes.
1. w′ is reduced to zero and no artificial variables remain in the basis, i.e. we are left with a basis consisting only of original variables. In this case, we simply delete the columns corresponding to the artificial variables, replace the objective function by the objective function of (P) after having expressed it in terms of the nonbasic variables and use Phase II of the simplex method as described in Section 2.3.
2. w′ < 0 at optimality. This means that the original LP (P) is infeasible. Indeed, if x is feasible in (P) then (x, xa = 0) is feasible in (Q) with value w′ = 0.
LP-15 3. w′ is reduced to zero but some artificial variables remain in the basis. These artificial variables must be at zero level since, for this solution, −w′ = Pm i=1 xa i = 0. Suppose that the ith variable ofthe basis is artificial. We may pivot on any nonzero (not necessarily positive) element ¯ aij of row i corresponding to a non-artificial variable xj. Since ¯ bi = 0, no change in the solution or in w′ will result. We say that we are driving the artificial variables out of the basis. By repeating this for all artificial variables in the basis, we obtain a basis consisting only of original variables. We have thus reduced this case to case 1.
There is still one detail that needs consideration. We might be unsuccessful in driving one artificial variable out the basis if ¯ aij = 0 for j = 1, . . . , n. However, this means that we have arrived at a zero row in the original matrix by performing elementary row operations, implying that the constraint is redundant. We can delete this constraint and continue in phase II with a basis of lower dimension.
Example Consider the following example already expressed in tableau form.
−z x1 x2 x3 x4 1 20 16 12 5 0 1 0 1 2 4 0 1 2 3 2 0 1 0 2 2 We observe that we don’t need to add three artificial variables since we can use x1 as first basic variable. In phase I, we solve the linear program: w x1 x2 x3 x4 xa 1 xa 2 1 2 2 5 4 1 0 1 2 4 1 2 3 1 2 1 0 2 1 2 The objective function is to minimize xa 1 +xa 2 and, as a result, the objective function coefficients of the nonbasic variables as well as −¯ c0 are obtained by taking the negative of the sum of all rows corresponding to artificial variables. Pivoting on ¯ a22, we obtain: w x1 x2 x3 x4 xa 1 xa 2 1 -2 -1 -2 0 1 1 2 0 4 1 2 3 1 2 -2 -1 -1 1 0 This tableau is optimal and, since w = 0, the original linear program is feasible. To obtain a bfs, we need to drive xa 1 out of the basis. This can be done by pivoting on say ¯ a34. Doing so, we get: LP-16 w x1 x2 x3 x4 xa 1 xa 2 1 0 -1 -1 0 1 -3 -2 2 4 1 -4 -2 3 2 2 1 1 -1 0 Expressing z as a function of {x1, x2, x4}, we have transformed our original LP into: −z x1 x2 x3 x4 1 126 -112 1 -3 4 1 -4 2 2 1 0 This can be solved by phase II of the simplex method.
3 Linear Programming in Matrix Form In this chapter, we show that the entries of the current tableau are uniquely determined by the collection of decision variables that form the basis and we give matrix expressions for these entries.
Consider a feasible linear program in standard form: Max z = cT x subject to: Ax = b x ≥0, where A has full row rank. Consider now any intermediate tableau of phase II of the simplex method and let B denote the corresponding collection of basic variables. If D (resp. d) is an m × n matrix (resp. an n-vector), let DB (resp. dB) denote the restriction of D (resp. d) to the columns (resp. rows) corresponding to B. We define analogously DN and dN for the collection N of nonbasic variables. For example, Ax = b can be rewritten as ABxB + ANxN = b. After possible regrouping of the basic variables, the current tableau looks as follows: xB xN −z 0 ¯ cT N −¯ c0 ¯ AB = I ¯ AN ¯ b.
Since the current tableau has been obtained from the original tableau by a sequence of elemen-tary row operations, we conclude that there exists an invertible matrix P (see Section 2.1) such that: PAB = ¯ AB = I PAN = ¯ AN and Pb = ¯ b.
LP-17 This implies that P = A−1 B and therefore: ¯ AN = A−1 B AN and ¯ b = A−1 B b.
Moreover, since the objective functions of the original and current tableaus are equivalent (i.e.
cT BxB + cT NxN = ¯ c0 + ¯ cT BxB + ¯ cT NxN = ¯ c0 + ¯ cT NxN) and xB = ¯ b −¯ ANxN, we derive that: ¯ cT N = cT N −cT B ¯ AN = cT N −cT BA−1 B AN and ¯ c0 = cT B¯ b = cT BA−1 B b.
This can also be written as: ¯ cT = cT −cT BA−1 B A.
As we’ll see in the next chapter, it is convenient to define an m-vector y by yT = cT BA−1 B . In summary, the current tableau can be expressed in terms of the original data as: xB xN −z 0 cT N −yT AN −yT b I A−1 B AN A−1 B b.
The simplex method could be described using this matrix form. For example, this optimality criterion becomes cT N −yT AN ≤0 or, equivalently, cT −yT A ≤0, i.e. AT y ≥c where yT = cT BA−1 B .
4 Duality Duality is the most important and useful structural property of linear programs.
We start by illustrating the notion on an example.
Consider the linear program: Max z = 5x1 + 4x2 subject to: x1 ≤4 (1) x1 + 2x2 ≤10 (2) 3x1 + 2x2 ≤16 (3) x1, x2 ≥0.
We shall refer to this linear program as the primal. By exhibiting any feasible esolution, say x1 = 4 and x2 = 2, one derives a lower bound (since we are maximizing) on the optimum value z∗of the linear program; in this case, we have z∗≥28. How could we derive upper bounds on z∗? Multiplying inequality (3) by 2, we derive that 6x1 + 4x2 ≤32 for any feasible (x1, x2). Since x1 ≥0, this in turn implies that z = 5x1 + 4x2 ≤6x1 + 4x2 ≤32 for any feasible solution and, thus, z∗≤32.
One can even combine several inequalities to get upper bounds. Adding up all three inequalities, we get 5x1 + 4x2 ≤30, implying that z∗≤30. In general, one would multiply inequality (1) LP-18 by some nonnegative scalar y1, inequality (2) by some nonnegative y2 and inequality (3) by some nonnegative y3, and add them together, deriving that (y1 + y2 + 3y3)x1 + (2y2 + 2y3)x2 ≤4y1 + 10y2 + 16y3.
To derive an upper bound on z∗, one would then impose that the coefficients of the xi’s in this implied inequality dominate the corresponding cost coefficients: y1+y2+3y3 ≥5 and 2y2+2y3 ≥4.
To derive the best upper bound (i.e. smallest) this way, one is thus led to solve the following so-called dual linear program: Min w = 4y1 + 10y2 + 16y3 subject to: y1 + y2 + 3y3 ≥5 2y2 + 2y3 ≥4 y1 ≥0, y2 ≥0, y3 ≥0.
Observe how the dual linear program is constructed from the primal: one is a maximization problem, the other a minimization; the cost coefficients of one are the RHS of the other and vice versa; the constraint matrix is just transposed (see below for more precise and formal rules). The optimum solution to this linear program is y1 = 0, y2 = 0.5 and y3 = 1.5, giving an upper bound of 29 on z∗. What we shall show in this chapter is that this upper bound is in fact equal to the optimum value of the primal. Here, x1 = 3 and x2 = 3.5 is a feasible solution to the primal of value 29 as well. Because of our upper bound of 29, this solution must be optimal, and thus duality is a way to prove optimality.
4.1 Duality for Linear Programs in canonical form Given a linear program (P) in canonical form Max z = cT x subject to: (P) Ax ≤b x ≥0 we define its dual linear program (D) as Min w = bT y subject to: (D) AT y ≥c y ≥0.
(P) is called the primal linear program. Notice there is a dual variable associated with each primal constraint, and a dual constraint associated with each primal variable. In fact, the primal and dual are indistinguishable in the following sense: Proposition 4.1. The dual of the dual is the primal.
LP-19 Proof. To construct the dual of the dual, we first need to put (D) in canonical form: Max w′ = −w = −bT y subject to: (D′) −AT y ≤−c y ≥0.
Therefore the dual (DD′) of D is: Min z′ = −cT x subject to: (DD′) −Ax ≥−b x ≥0.
Transforming this linear program into canonical form, we obtain (P).
Theorem 4.2 (Weak Duality). If x is feasible in (P) with value z and y is feasible in (D) with value w then z ≤w.
Proof.
z = cT x x≥0 ≤(AT y)T x = yT Ax y≥0 ≤yT b = bT y = w.
Any dual feasible solution (i.e. feasible in (D)) gives an upper bound on the optimal value z∗of the primal (P) and vice versa (i.e. any primal feasible solution gives a lower bound on the optimal value w∗of the dual (D)). In order to take care of infeasible linear programs, we adopt the convention that the maximum value of any function over an empty set is defined to be −∞while the minimum value of any function over an empty set is +∞. Therefore, we have the following corollary: Corollary 4.3 (Weak Duality). z∗≤w∗.
What is more surprising is the fact that this inequality is in most cases an equality.
Theorem 4.4 (Strong Duality). If z∗is finite then so is w∗and z∗= w∗.
Proof. The proof uses the simplex method. In order to solve (P) with the simplex method, we reformulate it in standard form: Max z = cT x subject to: (P) Ax + Is = b x ≥0, s ≥0.
Let ˜ A = (A I), ˜ x = x s and ˜ c = c 0 . Let B be the optimal basis obtained by the simplex method. The optimality conditions imply that ˜ AT y ≥˜ c LP-20 where yT = (˜ cB)T ˜ A−1 B .
Replacing ˜ A by (A I) and ˜ c by c 0 , we obtain: AT y ≥c and y ≥0.
This implies that y is a dual feasible solution. Moreover, the value of y is precisely w = yT b = (˜ cB)T ˜ A−1 B b = (˜ cB)T ˜ xB = z∗. Therefore, by weak duality, we have z∗= w∗.
Since the dual of the dual is the primal, we have that if either the primal or the dual is feasible and bounded then so are both of them and their values are equal. From weak duality, we know that if (P) is unbounded (i.e. z∗= +∞) then (D) is infeasible (w∗= +∞). Similarly, if (D) is unbounded (i.e. w∗= −∞) then (P) is infeasible (z∗= −∞). However, the converse to these statements are not true: There exist dual pairs of linear programs for which both the primal and the dual are infeasible. Here is a summary of the possible alternatives: Primal Dual z∗finite unbounded (z∗= ∞) infeasible (z∗= −∞) w∗finite z∗= w∗ impossible impossible unbounded (w∗= −∞) impossible impossible possible infeasible (w∗= +∞) impossible possible possible 4.2 The dual of a linear program in general form In order to find the dual of any linear program (P), we can first transform it into a linear program in canonical form (see Section 1.2), then write its dual and possibly simplify it by transforming it into some equivalent form.
For example, considering the linear program Max z = cT x subject to: X j aijxj ≤bi i ∈I1 (P) X j aijxj ≥bi i ∈I2 X j aijxj = bi i ∈I3 xj ≥0 j = 1, . . . , n, LP-21 we can first transform it into Max z = cT x subject to: X j aijxj ≤bi i ∈I1 (P ′) − X j aijxj ≤−bi i ∈I2 X j aijxj ≤bi i ∈I3 − X j aijxj ≤−bi i ∈I3 xj ≥0 j = 1, . . . , n.
Assigning the vectors y1, y2, y3 and y4 of dual variables to the first, second, third and fourth set of constraints respectively, we obtain the dual: Min w = X i∈I1 biy1 i − X i∈I2 biy2 i + X i∈I3 biy3 i − X i∈I3 biy4 i subject to: (D′) X i∈I1 aijy1 i − X i∈I2 aijy2 i + X i∈I3 aijy3 i − X i∈I3 aijy4 i ≥cj j = 1, . . . , n y1, y2, y3, y4 ≥0.
This dual can be written in a simplified form by letting yi = y1 i i ∈I1 yi = −y2 i i ∈I2 yi = y3 i −y4 i i ∈I3.
In terms of yi, we obtain (verify it!) the following equivalent dual linear program Min w = X i∈I biyi subject to: (D) X i∈I aijyi ≥cj j = 1, . . . , n yi ≥0 i ∈I1 yi ≤0 i ∈I2 yi ≷0 i ∈I3, where I = I1 ∪I2 ∪I3.
We could have avoided all these steps by just noticing that, if the primal program is a max-imization program, then inequalities with a ≤sign in the primal correspond to nonnegative dual LP-22 variables, inequalities with a ≥sign correspond to nonpositive dual variables, and equalities corre-spond to unrestricted in sign dual variables.
By performing similar transformations for the restrictions on the primal variables, we obtain the following set of rules for constructing the dual linear program of any linear program: Primal ← → Dual Max ← → Min P j aijxj ≤bi ← → yi ≥0 P j aijxj ≥bi ← → yi ≤0 P j aijxj = bi ← → yi ≷0 xj ≥0 ← → P i aijyi ≥cj xj ≤0 ← → P i aijyi ≤cj xj ≷0 ← → P i aijyi = cj.
If the primal linear program is in fact a minimization program then we simply use the above rules from right to left. This follows from the fact that the dual of the dual is the primal.
4.3 Complementary slackness Consider a pair of dual linear programs Max z = cT x subject to: (P) Ax ≤b x ≥0 and Min w = bT y subject to: (D) AT y ≥c y ≥0.
Strong duality allows to give a simple test for optimality.
Theorem 4.5 (Complementary Slackness). If x is feasible in (P) and y is feasible in (D) then x is optimal in (P) and y is optimal in (D) iffyT (b −Ax) = 0 and xT (AT y −c).
The latter statement can also be written as either yi = 0 or (Ax)i = bi (or both) and either xj = 0 or (AT y)j = cj (or both).
Proof. By strong duality we know that x is optimal in (P) and y is optimal in (D) iffcT x = bT y.
Moreover, (cfr. Theorem 4.2) we always have that: cT x ≤yT Ax ≤yT b = bT y.
Therefore, cT x = bT y is equivalent to cT x = yT Ax and yT Ax = yT b. Rearranging these expressions, we obtain xT (AT y −c) = 0 and yT (b −Ax) = 0.
LP-23 Corollary 4.6. Let x be feasible in (P). Then x is optimal iffthere exists y such that AT y ≥ = cj if xj = 0 xj > 0 yi ≥ = 0 if (Ax)i = bi (Ax)i < bi.
As a result, the optimality of a given primal feasible solution can be tested by checking the feasibility of a system of linear inequalities and equalities.
As should be by now familiar, we can write similar conditions for linear programs in other forms.
For example, Theorem 4.7. Let x be feasible in Max z = cT x subject to: (P) Ax = b x ≥0 and y feasible in Min w = bT y subject to: (D) AT y ≥c.
Then x is optimal in (P) and y is optimal in (D) iffxT (AT y −c) = 0.
4.4 The separating hyperplane theorem In this section, we use duality to obtain a necessary and sufficient condition for feasibility of a system of linear inequalities and equalities.
Theorem 4.8 (The Separating Hyperplane Theorem). Ax = b, x ≥0 has no solution iff∃y ∈Rm : AT y ≥0 and bT y < 0.
The geometric interpretation behind the separating hyperplane theorem is as follows: Let a1, . . . , an ∈Rm be the columns of A. Then b does not belong to the cone K = {Pn i=1 aixi : xi ≥0 for i = 1, . . . , n} generated by the ai’s iffthere exists an hyperplane {x : xT y = 0} (defined by its normal y) such that K is entirely on one side of the hyperplane (i.e. aT i y ≥0 for i = 1, . . . , n) while b is on the other side (bT y < 0).
Proof. Consider the pair of dual linear programs Max z = 0T x subject to: (P) Ax = b x ≥0 LP-24 and Min w = bT y subject to: (D) AT y ≥0.
Notice that (D) is certainly feasible since y = 0 is a feasible solution. As a result, duality implies that (P) is infeasible iff(D) is unbounded. However, since λy is dual feasible for any λ ≥0 and any dual feasible solution y, the unboundedness of (D) is equivalent to the existence of y such that AT y ≥0, y ≥0 and bT y < 0.
Other forms of the separating hyperplane theorem include: Theorem 4.9. Ax ≤b has no solution iff∃y ≥0 : AT y = 0 and bT y < 0.
5 Zero-Sum Matrix Games In a matrix game, there are two players, say player I and player II. Player I has m different pure strategies to choose from while player II has n different pure strategies. If player I selects strategy i and player II selects strategy j then this results in player I gaining aij units and player II losing aij units. So, if aij is positive, player II pays aij units to player I while if aij is negative then player I pays −aij units to player II. Since the amounts gained by one player equal the amounts paid by the other, this game is called a zero-sum game. The matrix A = [aij] is known to both players and is called the payoffmatrix. In a sequence of games, player I (resp. player II) may decide to randomize his choice of pure strategies by selecting strategy i (resp. j) with some probability yi (resp. xj). The vector y (resp. x) satisfies m X i=1 yi = 1 (resp.
n X j=1 xj = 1), yi ≥0 (resp. xj ≥0) and defines a mixed strategy.
If player I adopts the mixed strategy y then his expected gain gj if player II selects strategy j is given by: gj = X i aijyi = (yT A)j = yT Aej.
By using y, player I assures himself a guaranteed gain of g = min j gj = min j (yT A)j.
Similarly, if player II adopts the mixed strategy x then his expected loss li if player I selects strategy i is given by: li = X j aijxj = (Ax)i = eT i Ax LP-25 and his guaranteed loss4 is l = max i li = max i (Ax)i.
If player I uses the mixed strategy y and player II uses the mixed strategy x then the expected gain of player I is h = P i,j yiaijxj = yT Ax.
Theorem 5.1. If y and x are mixed strategies respectively for players I and II then g ≤l.
Proof. We have that h = yT Ax = X i yi(Ax)i ≤l X i yi = l and h = yT Ax = X j (yT A)jxj ≥g X j xj = g proving the result.
Player I will try to select y so as to maximize his guaranteed gain g while player II will select x so as to minimize l. From the above result, we know that the optimal guaranteed gain g∗of player I is at most the optimal guaranteed loss l∗of player II.
The main result in zero-sum matrix games is the following result obtained by Von Neumann and called the minimax theorem.
Theorem 5.2 (The Minimax Theorem). There exist mixed strategies x∗and y∗such that g∗= l∗.
Proof. In order to prove this result, we formulate the objectives of both players as linear programs.
Player II’s objective is to minimize l. This can be expressed by: Min l subject to: (P) Ax ≤le eT x = 1 x ≥0, l ≷0 where e is a vector of all 1’s. Indeed, for any optimal solution x∗, l∗to (P), we know that l∗= maxi(Ax∗)i since otherwise l∗could be decreased without violating feasibility.
Similarly, player I’s objective can be expressed by: Max g subject to: (D) AT y ≥ge eT y = 1 y ≥0, g ≷0 Again, any optimal solution to the above program will satisfy g∗= minj(AT y∗)j.
The result follows by noticing that (P) and (D) constitute a pair of dual linear programs (verify it!) and, therefore, by strong duality we know that g∗= l∗.
4Here guaranteed means that he’ll loose at most l.
LP-26 The above Theorem can be rewritten as follows (This explains why it is called the minimax theorem): Corollary 5.3.
max eT y=1,y≥0 min eT x=1,x≥0 yT Ax = min eT x=1,x≥0 max eT y=1,y≥0 yT Ax.
Indeed min eT x=1,x≥0 yT Ax = min j (yT A)j = g and max eT y=1,y≥0 yT Ax = max i (Ax)i = l.
Example Consider the game with payoffmatrix A = 1 −3 −2 4 .
Solving the linear program (P), we obtain the following optimal mixed strategies for both players (do it by yourself!): x∗= 7/10 3/10 and y∗= 6/10 4/10 , for which g∗= l∗= −2/10.
A matrix game is said to be symmetric if A = −AT . Any symmetric game is fair, i.e. g∗= l∗= 0.
6 Exercises Problem 1-1. A company has to decide its production levels for the 4 coming months.
The demand for those months are 900, 1100, 1700 and 1300 units respectively. The maximum pro-duction per month is 1200 units. Material produced one month can be delivered either that same month or stored in inventory and delivered at some other month. It costs the company $3 to carry one unit in inventory from one month to the next. Through additional man-hours, up to 400 additional units can be produced per month but, in this case, the company incurs a cost of $7/unit. Formulate as a linear program the problem of determining the production levels so as to minimize the total costs.
Problem 1-2. A contractor is working on a project, work on which is expected to last for a period of T weeks. It is estimated that during the jth week, the contractor will need uj man-hours of labor, j = 1 to T, for this project. The contractor can fulfill these requirements either by hiring laborers over the entire T week horizon (called steady labor) or by hiring laborers on a weekly basis each week (called casual labor) or by employing a combination of both. One manhour of steady labor costs c1 dollars; the cost is the same each week. However, the cost of casual labor may vary from week to week, and it is expected to be c2j dollars/man-hour, during week j, j = 1, . . . , T. Formulate the problem of fulfilling his labor requirements at minimum cost as a linear program.
LP-27 Problem 1-3. Transform the following linear program into an equivalent linear program in stan-dard form (Max{cT x : Ax = b, x ≥0}): Min x1 −x2 subject to: 2x1 + x2 ≥3 3x1 −x2 ≤7 x1 ≥0, x2 ≷0.
Problem 1-4. Consider the following optimization problem: Min X i ci|xi −di| subject to: Ax = b x ≥0, where A, b, c and d are given. Assume that ci ≥0 for all i. As such this is not a linear program since the objective function involves absolute values. Show how this problem can be formulated equivalently as a linear program. Explain why the linear program is equivalent to the original optimization problem. Would the transformation work if we were maximizing?
Problem 1-5. Given a set (or arrangement) of n lines (see Figure 1) in the plane (described as aix + biy = ci for i = 1, . . . , n), show how the problem of finding a point x in the plane which minimizes the sum of the distances between x and each line can be formulated as a linear program.
Hint: use Problem 1-4.
Figure 1: An arrangement of lines.
Problem 1-6. Given two linear functions over x, say cT x and dT x, show how to formulate the problem of minimizing max(cT x, dT x) over Ax = b, x ≥0 as a linear program. Would the trans-formation work if you were to maximize max(cT x, dT x)? How about minimizing the maximum of several linear functions?
LP-28 Problem 1-7. A function f : R →R is said to be convex if f(αx+(1−α)y) ≤αf(x)+(1−α)f(y) for all x, y ∈R and all 0 ≤α ≤1. It is piecewise linear if R can be partitioned into intervals over which the function is linear. See Figure 2 for an example. Show how to formulate the problem of minimizing P i fi(xi) where the fi’s are piecewise linear as a linear program.
Figure 2: A convex piecewise linear function.
Problem 1-8. What is the optimum solution of the following linear program: Min 5x1 + 7x2 + 9x3 + 11x4 + 13x5 subject to: 15x1 + 14x2 + 45x3 + 44x4 + 13x5 = 1994 xi ≥0 i = 1, . . . , 8.
Problem 2-1. Solve by the simplex method: Max z = 10 + 2x2 + 3x5 subject to: x1 −x2 + x5 = 4 3x2 + x3 −x5 = 12 x2 + x4 + 2x5 = 14 2x2 + x5 + x6 = 13 x1 ≥0, x2 ≥0, x3 ≥0, x4 ≥0, x5 ≥0, x6 ≥0.
Show all intermediate tableaux.
Problem 2-2. Solve by the simplex method using only one pivot: Max z = x1 + 4x2 + 5x3 subject to: x1 + 2x2 + 3x3 ≤2 3x1 + x2 + 2x3 ≤2 2x1 + 3x2 + x3 ≤4 x1 ≥0, x2 ≥0, x3 ≥0.
LP-29 Problem 2-3. Solve by the two-phase simplex method: Max z = 3x1 + x2 subject to: x1 −x2 ≤−1 −x1 −x2 ≤−3 2x1 + x2 ≤4 x1 ≥0, x2 ≥0.
Problem 2-4. Solve by the simplex method: Max z = x11 + 2x12 + 3x21 + 4x22 + 5x31 + 7x32 subject to: x11 + x12 ≤1 x21 + x22 ≤1 x31 + x32 ≤1 x11 + x21 + x31 ≤1 x12 + x22 + x32 ≤1 xij ≥0 i ∈{1, 2, 3}, j ∈{1, 2}.
Were you expecting the optimum solution to have all components either 0 or 1?
Problem 2-5. Find a feasible solution to the following system: x1 + x2 + x3 + x4 + x5 = 2 −x1 + 2x2 + x3 −3x4 + x5 = 1 x1 −3x2 −2x3 + 2x4 −2x5 = −4 x1, x2, x3, x4, x5 ≥ 0 Problem 2-6. Use the simplex method to show that the following constraints imply x1 +2x2 ≤8: 4x1 + x2 ≤ 4 2x1 −3x2 ≤ 6 x1, x2 ≥ 0 Problem 2-7. How are the various rules of the simplex method affected when solving a minimiza-tion problem instead of a maximization problem as described in these notes?
Problem 4-1. Write the dual to: Min z = 8x1 + 2x2 + 4x3 −4x4 subject to: x1 + x2 + x3 + x4 = 10 x1 −x2 + 3x4 ≥7 −2x1 + 3x2 + 4x3 ≥13 x1 ≥0, x2 ≥0, x3 ≷0, x4 ≷0 LP-30 Problem 4-2. Is x1 = 4, x2 = 5 and x3 = 6 an optimal solution to: Min z = 14x1 + 10x2 + cx3 subject to: x1 + x2 + x3 ≥10 x1 −x2 + +x3 ≥4 3x1 + 2x2 + x3 ≥28 −x1 −x2 + 4x3 ≥15 2x1 + x2 ≥10 x1 ≥0, x2 ≥0, x3 ≥0 1. if c=5?
2. if c=8?
Justify.
Problem 4-3. Consider the linear program Max z = 4x1 + 5x2 + 2x3 subject to: 2x1 −x2 + 2x3 ≤9 (P) 3x1 + 5x2 + 4x3 ≤8 x1 + x2 + 2x3 ≤2 x1 ≥0, x2 ≥0, x3 ≥0.
1. Find an optimal solution to (P) using the simplex method.
2. Write the dual linear program. From 1, infer an optimal dual solution. Check your answer using complementary slackness.
Problem 4-4. Prove or give a counterexample to the following statement: If the optimum solution to the primal is unique, then the optimum solution to the dual is nondegenerate.
Problem 4-5. Construct a pair of dual linear programs such that both the primal and the dual are infeasible.
Problem 4-6. Consider the one constraint LP: Max z = n X j=1 cjxj subject to: n X j=1 ajxj = b xj ≥0 for all j, where b > 0.
LP-31 1. Write its dual.
2. Develop a simple test for checking the feasibility of this problem.
3. Develop a simple test for checking unboundedness.
4. Develop a simple method for obtaining a primal optimum solution and a dual optimum solution directly.
5. In terms of the optimum dual solution, how much does the optimum value of the primal (or the dual) change when b is replaced by b + ϵ?
Problem 4-7. Suppose that you are given a “black box” procedure that, when given a system of linear inequalities, either produces a feasible solution or declares that there is no feasible solution.
Show how a single call to this black box can be used to obtain an optimal solution to the linear program Min cT x subject to: Ax = b x ≥0.
Hint: Also obtain an optimal solution to the dual linear program.
Problem 4-8. Consider the linear program Max z = cT x subject to: Ax = b x ≥0, where A is m × n. Assume that this linear program is unbounded. Prove that, if we replace b by b′ for any vector b′, the resulting linear program is either infeasible or unbounded.
Problem 4-9. Prove Theorem 4.9.
Problem 4-10. (Difficult) Prove that exactly one of the following holds: 1. There exists x ≥0 : A1x < b1 and A2x ≤b2 2. There exists (y1, y2) ≥0 : AT 1 y1 + AT 2 y2 ≥0 and, either bT 1 y1 + bT 2 y2 < 0 or (bT 1 y1 + bT 2 y2 = 0 and y1 ̸= 0).
Hint: give a system of linear inequalities (≤) which has a solution iffthe system A1x < b1, A2x ≤b2 and x ≥0 has a solution.
Problem 4-11. Given a pair of feasible dual linear programs min{cT x : Ax ≥b, x ≥0} and max{bT y : AT y ≤c, y ≥0}, prove that there exists an optimal solution x to the primal and an optimal solution y to the dual such that xj > 0 whenever (AT y)j = cj and yi > 0 when-ever (Ax)i = bi. (This is sometimes referred to as strong complementary slackness or Tucker’s complementary slackness.) Hint: use Problem 4.10.
LP-32 Problem 5-1. Consider the matrix game based on the following payoffmatrix: A = 0 −2 1 2 0 3 −1 −3 0 .
Notice that A is antisymmetric, i.e. A = −AT .
1. Write the linear programs associated with both players. Show that these linear programs are equivalent in the sense that if (x, l) is feasible for player II’s linear program then (y, g) = (x, −l) is feasible for player I’s linear program and vice versa. Prove that g∗= l∗= 0.
2. Using part 1 and using complementary slackness, find the optimal strategies for both players.
LP-33 |
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college algebra unit 8 study guides
periodic functions
8.1
Graphs of the Sine and Cosine Functions
8.2
Graphs of the Other Trigonometric Functions
8.3
Inverse Trigonometric Functions
unit 8 review
Periodic functions are mathematical marvels that repeat their values at regular intervals. They're the heartbeat of many natural phenomena, from the ebb and flow of tides to the oscillations of sound waves. Trigonometric functions like sine and cosine are prime examples.
These functions can be transformed by changing their amplitude, period, phase shift, and vertical shift. Understanding these transformations is key to modeling real-world scenarios, from seasonal temperature changes to alternating current in electrical systems. Mastering periodic functions opens doors to advanced topics in mathematics and physics.
Key Concepts
Periodic functions repeat their values at regular intervals called periods
Trigonometric functions (sine, cosine, tangent) are common examples of periodic functions
Amplitude measures the height of a periodic function's peaks and troughs from its midline
Period determines the length of one complete cycle of a periodic function
Phase shift moves a periodic function horizontally to the left or right
Vertical shift moves a periodic function up or down
Frequency is the reciprocal of the period and measures the number of cycles per unit of time
Midline is the horizontal line that runs through the center of a periodic function's graph
Trigonometric Functions
Sine function $(\sin x)$ oscillates between -1 and 1 with a period of $2\pi$
Starts at the origin, reaches a maximum at $\frac{\pi}{2}$, and a minimum at $\frac{3\pi}{2}$
Cosine function $(\cos x)$ also oscillates between -1 and 1 with a period of $2\pi$
Starts at 1, reaches a minimum at $\pi$, and a maximum at $2\pi$
Tangent function $(\tan x)$ has a period of $\pi$ and undefined values at odd multiples of $\frac{\pi}{2}$
Oscillates between positive and negative infinity
Cosecant $(\csc x)$, secant $(\sec x)$, and cotangent $(\cot x)$ are reciprocals of sine, cosine, and tangent, respectively
Trigonometric functions can be transformed using amplitude, period, phase shift, and vertical shift
Graphing Periodic Functions
Start by graphing the basic function without any transformations
Apply amplitude changes by multiplying the function by a constant $|a|$
If $|a| > 1$, the graph is stretched vertically; if $0 < |a| < 1$, the graph is compressed vertically
Modify the period by dividing the input variable by a constant $|b|$
The new period is $\frac{2\pi}{|b|}$; if $|b| > 1$, the period decreases, and if $0 < |b| < 1$, the period increases
Shift the graph horizontally by adding or subtracting a constant $c$ inside the function
Positive $c$ shifts the graph to the left, while negative $c$ shifts it to the right
Shift the graph vertically by adding or subtracting a constant $d$ outside the function
Positive $d$ shifts the graph up, while negative $d$ shifts it down
Properties and Characteristics
Periodic functions have a constant period, amplitude, and midline
Even functions are symmetric about the y-axis $(f(-x) = f(x))$, while odd functions are symmetric about the origin $(f(-x) = -f(x))$
Trigonometric functions have specific even/odd properties (sine is odd, cosine is even, tangent is odd)
Periodic functions can be continuous or discontinuous
Continuous functions have no breaks or gaps in their graphs
Discontinuous functions have breaks, gaps, or undefined points (tangent, secant, cosecant)
Periodic functions can be bounded or unbounded
Bounded functions have a maximum and minimum value (sine, cosine)
Unbounded functions have no maximum or minimum value (tangent, secant, cosecant)
Real-World Applications
Modeling seasonal changes in temperature, daylight hours, or animal populations
Describing the motion of pendulums, springs, or waves (sound, light, water)
Analyzing alternating current (AC) in electrical systems
Studying the behavior of tides, planetary orbits, or celestial bodies
Representing periodic trends in economics, such as business cycles or stock market fluctuations
Modeling the vibrations of strings in musical instruments
Describing the motion of pistons in engines or the rotation of gears in machinery
Problem-Solving Strategies
Identify the type of periodic function (sine, cosine, tangent, or their reciprocals)
Determine the basic function's period, amplitude, and midline
Identify any transformations applied to the basic function
Changes in amplitude, period, phase shift, or vertical shift
Write the equation of the transformed function using the general form:
$a \cdot f(b(x - c)) + d$, where $f$ is the basic function
Graph the transformed function by applying the transformations to the basic function's graph
Analyze the graph to find key points, such as maxima, minima, or zeros
Use the graph or equation to answer questions about the function's properties or behavior
Common Mistakes and Misconceptions
Confusing the effects of changes in amplitude, period, phase shift, and vertical shift
Remember: amplitude affects height, period affects width, phase shift moves horizontally, vertical shift moves vertically
Incorrectly applying transformations to the basic function
Changes inside the function (period and phase shift) are applied before changes outside (amplitude and vertical shift)
Misinterpreting the period as the frequency or vice versa
Period is the length of one complete cycle, while frequency is the number of cycles per unit of time
Forgetting to consider the domain restrictions of certain functions (tangent, secant, cosecant)
Mistaking even functions for odd functions or vice versa
Even functions are symmetric about the y-axis, while odd functions are symmetric about the origin
Incorrectly identifying the midline of a transformed function
The midline is affected by the vertical shift, not the amplitude
Advanced Topics
Combining periodic functions through addition, subtraction, multiplication, or composition
The resulting function may have a different period, amplitude, or shape than the original functions
Analyzing the Fourier series representation of periodic functions
Any periodic function can be expressed as an infinite sum of sine and cosine functions
Studying the relationship between trigonometric functions and the unit circle
The sine and cosine of an angle can be represented as the y and x coordinates of a point on the unit circle
Exploring the connections between periodic functions and complex numbers (Euler's formula)
Euler's formula relates exponential functions with complex arguments to trigonometric functions
Investigating the use of periodic functions in differential equations and harmonic motion
Applying periodic functions to signal processing, Fourier transforms, and frequency analysis
Examining the role of periodic functions in physics, such as wave mechanics and quantum mechanics |
189999 | https://jcp.org/aboutJava/communityprocess/jsr/tiger/enhanced-for.html | Press Room | Get Java Here |
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| JSRs by Platform JSRs by Technology JSRs by Stage JSRs by Committee List of All JSRs Register for Site Use of JCP site is subject to the JCP Terms of Use and the Oracle Privacy Policy - About JCP Program Overview JCP Process Document Calendar JCP Members Executive Committee Elections - Get Involved Overview Becoming a JCP Member Java in Education - Community Resources Overview Tutorials JCP Podcasts Spec Lead Guide TCK Tool & Info License Reference - Community News News Programs & Awards Press Releases Success Stories - FAQ - Contact Us | An enhanced for loop for the Java™ Programming Language An enhanced for loop for the Java™ Programming LanguageI. IntroductionThe current for loop construct is powerful, but somewhat ill-suited to iteration over collections. The standard idiom to iterate over a collection is somewhat verbose: for (Iterator i = c.iterator(); i.hasNext(); ) { // No ForUpdate String s = (String) i.next(); ... } The situation is made little better with the addition of generics: for (Iterator i = c.iterator(); i.hasNext(); ) { String s = i.next(); ... } We propose a second form of the for loop specifically designed for iteration over collections and arrays. Traditionally this is done using a foreach keyword, but it seems unnecessary and counterproductive to add a keyword at this late date. Under the the syntax of this proposal, the above code could be replaced by this: for (String s : c) { ... } Similarly, the following code could be used to calculate the sum of an int array. int sum = 0; for (int e : a) // e is short for element; i would be confusing sum += e; This proposal affects only the compiler; it demands no support from the VM. It does not require a new keyword and is fully compatible with all preexisting programs. The construct interacts harmoniously with existing elements of the language. For example, abrupt completion of an enhanced for statement is handled exactly like abrupt completion of any other for statement.II. SyntaxThe following description is not up to JLS standards, but should be good enough for present purposes. The new form of the for statement is governed by the following production: EnhancedForStatement: for ( Type Identifier : Expression ) Statement Expression must be an instance of a new interface called java.lang.Iterable, or an array. The interface java.util.Collection (and perhaps a few other interfaces or classes) will be retrofitted to extend (or implement) java.lang.Iterable: package java.lang; public class Iterable { / Returns an iterator over the elements in this collection. There are no guarantees concerning the order in which the elements are returned (unless this collection is an instance of some class that provides a guarantee). @return an Iterator over the elements in this collection. / SimpleIterator iterator(); } The interface java.util.Iterator will be retrofitted to implement a new interface java.lang.ReadOnlyIterator: package java.lang; public interface ReadOnlyIterator { / Returns true if the iteration has more elements. (In other words, returns true if next would return an element rather than throwing an exception.) @return true if the iterator has more elements. / boolean hasNext(); / Returns the next element in the iteration. @return the next element in the iteration. @exception NoSuchElementException iteration has no more elements. / Object next(); } These new interfaces serve to prevent the dependency of the language on java.util that would otherwise result. Note that that they rely on covariant return types, which will be introduced as a part of support for generics (JSR-14).II. Semantics(Again, not up to JLS standards.) If Expression is an instance of java.util.Collection, the enhanced for statement is shorthand for this: for ( Iterator<T> $i = Expression.iterator(); $i.hasNext(); ) { Type Identifier = $i.next(); Statement } The identifier $i is used to indicate some synthetic variable name chosen by the compiler so as not to conflict with any other natural or synthetic variable name. The type T differs depending on whether Expression has a raw type or a parameterized type. If Expression has a raw type, T is Type; if Expression has a parameterized type, T is the the type parameter of Expression. Note that the cast may generate a warning or error: if Expression has a raw type, then an unchecked assignment warning will be generated, correctly implying that the implicit cast may fail. If Expression has a parameterized type, the statement will generate a compilation error if the type parameter cannot be converted to Type by assignment conversion (JLS 5.2). If Expression is an array, the enhanced for statement is shorthand for this: { Type[] $a = Expression; L1: L2: ... Lm: for (int $i = 0; $i < $a.length; $i++) { Type Identifier = $a[ $i ] ; Statement } } The identifier $a is used to indicate some synthetic variable name chosen by the compiler so as not to conflict with any other natural or synthetic variable name. The possibly empty sequence of labels preceding the enhanced for loop is represented by L1: L2: ... Lm. This sequence is moved beyond the assignment to$a so that continue and break statements with labels will work properly. The array form of the enhanced for statement will generate a compilation error if the array element type cannot be converted to Type by assignment conversion (JLS 5.2). 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