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190300 | https://www.mathepower.com/en/turningpoint.php | Akzeptieren & schließen
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Calculate turning points
What is a turning point?
A turning point is a point where the graph of a function has the locally highest value (called a maximum turning point) or the locally lowest value (called a minimum turning point). A function does not have to have their highest and lowest values in turning points, though.
This graph e.g. has a maximum turning point at (0|-3) while the function has higher values e.g. in (2|5). This implies that a maximum turning point is not the highest value of the function, but just locally the highest, i.e. there is no higher value at least in a small area around that point.
How to find turning points?
The basic idea is that tangents in a turning point have slope .
So the basic idea of finding turning points is:
Find a way to calculate slopes of tangents (possible by differentiation).
Find when the tangent slope is . There could be a turning point (but there is not necessarily one!)
This means: To find turning points, look for roots of the derivation.
Does slope always imply we have a turning point?
No. If the slope is , we max have a maximum turning point (shown above)
or a mininum turning point
or the slope just becomes for a moment though you have no turning point. Such a point is called saddle point.
Does the slope always have to be in turning points?
Yes. This is right. But not the reversion, as seen above. So we say: Having slope is necessary but not sufficient for having a turning point.
Let's assume we have slope . How can I find out if I have a maximum / minimum turning point or a saddle point?
By using the change of signs criterion.
I have to find turning points as homework and don't know how. What can I do?
Just enter your function above. Mathepower does the calculation, with explanation and step-by-step.
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190301 | https://www.cs.utexas.edu/~mitra/csFall2017/itd105/lectures/BooleanIdentities.pdf | UIL Official List of Boolean Algebra Identities (Laws) 1 Indempotent Law for OR 2 Indempotent Law for AND 3 Commutative Law for OR 4 Commutative Law for AND 5 Associative Law for OR 6 Associative Law for AND 7 Distributive Law for AND over OR 8 Distributive Law for OR over AND 9 Law of Union 10 Law of Intersection 11 Law of Absorption 12 Law of Absorption 13 Identity Law for AND 14 Identity Law for OR 15 Double Negative Law 16 Law of Complement for OR 17 Law of Complement for AND 18 DeMorgan's Law 19 DeMorgan's Law 20 Exclusive OR (XOR) 21 Exclusive NOR (XNOR) 22 Law of the "disappearing opposite" 23 Reverse of Law #8 24 FOIL (First,Outer,Inner,Last) Distribution Note: AND will always be expressed explicitly with the operator A A A A B B A C B A C B A ) ( ) ( 1 1 A 0 0 A A A 1 A A 0 A A 1 A A 0 A A D B C B D A C A D C B A ) ( ) ( C B A C A B A ) ( ) ( B A B A A B A B A B A B A B A B A B A B A B A B A B A B A A A A A B B A C B A C B A ) ( ) ( C A B A C B A ) ( ) ( ) ( C A B A C B A A B A A ) ( A B A A st) Distribution |
190302 | https://arxiv.org/pdf/1805.02111 | Ellipse, Hyperbola and Their Conjunction
Arkadiusz Kobiera Warsaw University of Technology
Abstract
This article presents a simple analysis of cones which are used to generate a given conic curve by section by a plane. It was found that if the given curve is an ellipse, then the locus of vertices of the cones is a hyperbola. The hyperbola has foci which coincidence with the ellipse vertices. Similarly, if the given curve is the hyperbola, the locus of vertex of the cones is the ellipse. In the second case, the foci of the ellipse are located in the hyperbola’s vertices. These two relationships create a kind of conjunction between the ellipse and the hyperbola which originate from the cones used for generation of these curves. The presented conjunction of the ellipse and hyperbola is a perfect example of mathematical beauty which may be shown by the use of very simple geometry. As in the past the conic curves appear to be very interesting and fruitful mathematical beings.
1
arXiv:1805.02111v2 [math.HO] 26 Jan 2019
Introduction
The conical curves are mathematical entities which have been known for thousands years since the first Menaechmus’ research around 250 B.C. . Anybody who has attempted undergraduate course of geometry knows that ellipse, hyperbola and parabola are obtained by section of a cone by a plane. Every book dealing with the this subject has a sketch where the cone is sec-tioned by planes at various angles, which produces different kinds of conics. Usually authors start with the cone to produce the conic curve by section. Then, they use it to prove some facts about the conics. Many books focus on the curves themselves and their features. Even books which describe the conics theory in a quite comprehensive way [2, 4, 1] abandon the cone after the first couple paragraphs or go to quite complex analysis of quadratics. We may find hundreds of theorems about the curves but the relation between the cone and the conics is left to the exercise section at best or authors quickly go to more complex systems of conics in three-dimensional space . Probably the cone seems to be too simple to spent time on this topic, how-ever we will show that the cone (strictly speaking family of cones) may have interesting properties as well. Apart of pure geometry, celestial mechanics is the second field where conics are important – the orbits are conic curves. Un-fortunately, the books about celestial mechanics say only a few words about the the cone if any at al. [3, 5]. In this short paper we would like to focus on the cone and its relation to conic curves which is surprisingly omitted in books, but interesting. 2Ellipse and the Cones
Let us consider following problem: Given is an ellipse E defined by two focus points F1 and F2 and vertex A. This ellipse is created by section of the cone
S by plane ρ. It is shown in Figure 1.
Figure 1: Ellipse and the cone. Our task is to find the vertex E of the cone S. Apart from the foci, the ellipse has also two characteristics points: the vertices A and B. The distances from the vertices to one of the focus e.g. F1 will be noted as
ra = |F1A|, and rb = |F1B|. The semi-axes of this ellipse are a = |AJ | and
b = |H1J| where J is the center of the ellipse. The distance between foci is 3c = |F1F2|. The radii ra and rb define the eccentricity:
e = ra − rb
ra + rb
= ca. (1) Obviously, we may use any set of these parameters to define the ellipse E,however we will prefer the radii and focus F1.The first question is about the cone: ”Is the cone S unique?” The answer is in the following lemma:
Lemma 1 If the ellipse E lies on plane ρ and it is defined by two vertices A, B, and focus F1 (or foci F1, F2 and vertex A) then it may be generated by infinite number of cones S sectioned by the plane ρ.
Proof :The proof will be explained in a rather quite informal manner. To solve this exercise let’s reduce the three-dimensional problem to a two-dimensional problem by considering plane τ which is defined by cone’s axis and foci (or vertices) of the ellipse. It is shown in Figure 2. We put line a on plane τ .The line coincidences with the ellipse vertices A and B and the foci F1, F2
as well. The line a is also an intersection of planes τ and ρ. Note that the focus points (e.g. F1) are points of tangency of a sphere of center O with the plane ρ. This sphere is called Dandelin’s sphere and it is simultaneously tangent to the cone. The tangency points of the sphere and the cone create a circle which defines plane ω . The intersection of planes ω and ρ is line
f. We create also an additional line b on plane ω which is perpendicular to
f and goes through the axis of the cone. The intersection of the Dandelin’s sphere by the plane τ is a circle with center O. The circle is tangent to lines 4t1 and t2 which are two elements of the cone. These lines are obtained by cutting the cone by plane τ . They meet line a at points A and B.
Figure 2: Section of cone S by plane τ .The problem was reduced to a problem of finding the point E which is vertex of triangle ABE that inscribes circle O. Since the ellipse E is given, the three points A, B and F1 are fixed. Point E is a point of intersection of lines t1 and t2. These lines are defined by points A and B and the circle O
which is tangent to the lines. If the radius r is smaller than a certain limit
rmax then the two lines meet at point E (this fact seems to be quite obvious so we skip this part of the proof). The maximum radius rmax is determined by the case when the lines t1 and t2 are parallel. In such case the lines t1
and t2 become element lines of a cylinder as it is a limiting case of the cone 5when the point E goes to infinity. In this case the ellipse E is obtained as a section of the cylinder. One may show that the limiting radius is equal to the minor semi-axes of the ellipse
rmax = a√1 − e2. (2) If the radius r can be of any length between 0 and rmax then the location of point E is not unique and its position depends on radius r. Hence, one can construct an infinite number of cones which may be used to generate the ellipse E.
If the cone S is not unique, the next question is: ”What is the locus of the cone vertices E?” First, we calculate the distance from the cone vertex
E to the ellipse vertex B
|EB | = |BD | + |ED | = rb + |ED | (3) The second equality results from the fact that BD and BF 1 are tangent to circle O and they have common endpoint B. Obviously, the angles ∠F1BO
and ∠DBO are equal and right triangles OF 1B and ODB are congruent. Then segments F1B and BD are of the same length rb. One can write similar equations for segment EA
|EA | = |HA | + |EH | = ra + |ED |. (4) Here we use the fact that triangles HOE and DOE are congruent and tri-6angles HOA and F1OA are congruent as well. Comparison of the above equation leads to following proposition:
Proposition 1 If the ellipse E defined by two vertices A, B and focus F1 (or foci F1, F 2 and vertex A) is generated by section of the cones S by the plane
ρ then the locus of vertices E of the all possible cones S is a hyperbola H.The foci of the hyperbola H are points A and B, vertices are points F1 and
F2 (ellipse foci).
Prof :Let us calculate the difference of length of two segments EB and EA
|EB | − | EA | = rb + |ED | − (ra + |ED |) = rb − ra = const, (5)
|EB | − | EA | = |F1F2|. (6) This difference is a constant number because ra and rb are constant as ellipse parameters, also the distance |F1F2| is obviously constant. This directly agrees with definition of a hyperbola which foci are located in points A and
B (see Figure 3). It is also clear that vertices of this hyperbola H are points
F1 and F2.
Indeed, the hyperbola H contains all the possible locations of vertices E.The left branch contains vertices where the Dandelin’s sphere is tangent to focus F1. The upper part of the branch represents the case when the sphere is above the plane ρ, lower part is for opposite position of the sphere. The 7Figure 3: Hyperbola H.right branch is for the case where the sphere is tangent at point F2. If the radius r of the sphere vanishes to 0, the point E goes toward foci F1 or F2.If the sphere’s radius r goes to the maximum value rmax the point E goes to infinity on the hyperbola’s branches. Asymptotic lines s1, s2 are the axes of cylinders which are limiting cases of the cones with vertex in infinity.
Hyperbola and the Cones
Now we can ask reversed question: What is the locus of vertices G of cones
Z which generate the given hyperbole H. One can consider the hyperbola
H which was found in the previous part. This will not reduce generality of 8our reasoning. We will keep same plane τ where four points are defined A, B, F1 and F2. They also define the hyperbola H on the plane τ . Figure 4 shows the situation where the hyperbola is created by sectioning the cone Z
by plane τ . We state the following lemma by analogy to the case of ellipse:
Figure 4: Hyperbola H created by section of the cone Z.
Lemma 2 If the hyperbola H lies on plane τ and it is defined by two foci A, B, and vertex F1 (or two vertices F1, F 2 and focus A) then it may be generated by infinite number of cones Z sectioned by plane τ .
Proof:
The proof is analogous to proof of Lemma 1. First, we reduce the problem to planimetry by considering the plane ρ (see Figure 5). 9Figure 5: Section of the cone Z by plane ρ.By lemma’s assumption we have points A, B and F1 given, then also the point F2 is established because it is the vertex of the hyperbola. The vertex
G of the cone is defined by section of two lines t3 and t4 which are elements of cone Z. The lines lie on plane ρ and go through points F1 and F2 and are tangent to two circles O3 and O4 respectively. The circles are sections of Dandelin’s spheres (Figure 4) which are tangent to the plane τ . They are also tangent to line a at points A and B which are foci of the hyperbola H.Let r1 be the radius of the circle O3. Then the point G is not unique and its position depends on the radius r1. The radius r1 may vary from zero to infinity: 0 < r 1 < ∞. Hence, there exists infinite number of cones Z which generate the hyperbola if they are sectioned by the plane τ .10
The next step is finding the locus of vertices G. By analogy to the Propo-sition 1 we write following proposition:
Proposition 2 If the hyperbola H defined by two vertices F1, F 2 and focus A (or foci A, B and vertex F1) is generated by section of the cones Z by plane
τ then the locus of vertices G of the cones Z is an ellipse E. The foci of the ellipse E are points F1 and F2, vertices are points A and B (ellipse foci).
Proof:
We will look for the relationship between the distances from the vertex
G to the points F1 and F2. First, we will consider the right triangle O3P O 4
(Figure 5). Point P is the normal projection of point O4 onto segment AO 3.By using Pythagoras theorem we have
|O3O4|2 = |AB |2 + ( r1 − r2)2. (7) where r2 is radius of the circle O3. The triangles GKO 3 and GM O 4 are also right triangles because the points K and N are points of tangency of the lines t3 and t4 to the circles O3 and O4. Hence, one can write
|O3G|2 = |GK |2 + r21 , (8)
|O4G|2 = |GM |2 + r22 . (9) Recalling that
|O3O4| = |O3G| + |O4G| (10) 11 and substituting equations (8), (9) and (10) to equation (7) the following equation is obtained (|KG | + |M G |)2 − 2|KG || M G | + 2 |O3G|| O4G| = |AB |2 − 2r1r2 == |AB |2 − 2|O4M || O3K| (11) When both sides of equation (11) are divided by |O4M || O3K| we get (|GK | + |GM |)2
2r1r2
−
( |KG ||O3K||GM ||O4M | − |O3G||O3K||O4G||O4M |
)
= |AB |2
2r1r2
− 1. (12) The triangles O3KG and O4M G are similar because they are right trian-gles and and angles ∠KGO 3 and ∠M GO 4 are equal. The second statement is true because the triangles O4M G and O4N G are congruent and angles
∠KGO 3 and ∠N GO 4 are congurent as well (points G, O3 and O4 lie on the axis of the cone, hence the segments O3G and O4G are co-linear). Let the measure of angles ∠KO 3G and ∠M O 4G be χ. Simple trigonometrical relations based on Figure 5 yield:
|KG ||O3K| = tan χ = |GM ||O4M |, (13)
|O3G||O3K| = 1cos χ = |O4G||O4M |. (14) The second term of left hand side of equation (12) can be simplified by use of the two relationship stated above
( |KG ||O3K||GM ||O4M | − |O3G||O3K||O4G||O4M |
)
= (tan χ)2 − 1(cos χ)2 = −1. (15) 12 Finally, we get the simple equation: (|GK | + |GM |)2 = |AB |2 − 4r1r2. (16) The next step is finding the product r1r2. Let’s note that the angle ∠AF 1O3
is equal to ∠AF 1O3 and it is π/ 2 − ψ. We may say the same about ∠KF 1O3.This fact leads to conclusion that the angle ∠N F 1B is equal to
∠N F 1B = π − 2∠AF 1O3 = π − 2( π/ 2 − ψ) = 2 ψ. (17) Obviously, the line F1O4 is the bisector of this angle. Hence, the angle
∠O4F1B is equal to ψ. Triangles F1AO 3 and F1BO 4 are similar and we may write the following proportion:
|O3A||F1A| = |F1B||O4B|. (18) Length of segment O3A is r1, lenght of segment O4B is r2. We may rewrite the above equation as
r1
ra
= rb
r2
. (19) Hence, r1r2 = rarb.We successfully arrived to the conclusion that the sum of length of seg-ments GK and GM is constant (|GK | + |GM |)2 = |AB |2 − 4rarb = const. (20) 13 |AB | is equal to ra + rb then (|GK | + |GM |)2 = ( ra + rb)2 − 4rarb = ( rb − ra)2. (21) The fact that ra = |AF 1| = |F1K| = |F2M | = |BF 2| and equation (21) allows us to calculate the sum of the distances between vertex G and the foci
F1 and F2
|GF 1|+|GF 2| = |GK |+ra +|GM |+ra = ( |GK |+|GM |)+2 ra = rb +ra. (22) Hence, the sum of distances of the vertex G from foci F1 and F2 is constant (|AB | = ra + rb)
|GF 1| + |GF 2| = |AB |. (23) This equation is the simplest form of definition of the ellipse and we proved the proposition.
Conclusions
We have shown the existence of a very interesting relationship between ellipse and hyperbola by use of very simple geometry. It was shown that ellipse and hyperbola are conjugate. This conjunction is created by locus of vertices of cones which generate the two conics. Although it seems to be a very basic property of the conics, surprisingly it is not mentioned even in some books devoted to conics geometry only. On the other hand it is wonderful that 14 such simple mathematics may lead to such interesting results and express the beauty of geometry that is imperfectly shown in Figure 6.
Figure 6: Conjugate ellipse E and hyperbola H as curves generated by section of cones whose vertices are located on these curves.
References
Arseny V. Akopyan. Geometry of conics . Mathematical World vol. 26. American Mathematical Society, Providence, 2007. Julian L. Coolidge. A History of The Conic Sections and Quadric Sur-faces . Dover Publications, New York, 1968. 15 Gerhard Beutler. Methods of celestial mechanics. Vol. I, Physical, Mathe-matical, and Numerical Principles . Astronomy and Astrophysics Library. Springer, Berlin ; Heidelberg, 2005. Edward Otto. Krzywe sto˙ zkowe: zajecia fakultatywne w grupie matem-atyczno - fizycznej . Pa´ nstwowe Wydawnictwa Szkolne i Pedagogiczne, 1976. Stefan Wierzbi´ nski. Mechanika nieba . PWN, 1973. 16 |
190303 | https://www.colombohurdlaw.com/statehood-a-concept-of-international-law/ | Skip to content
Statehood (A Concept of International Law)
Sovereignty is a concept of international law in which a sovereign state exercises independent control over a particular geographic area. The modern concept of ‘statehood’ has been memorialized in the ‘Montevideo Convention on the Rights and Duties of States,’ of 1933. The Montevideo Convention defines a state as (1) having a permanent population, (2) a defined geographical territory, (3) a government, and (4) the power to enter into relations with other sovereign states.
Additionally, state recognition is an essential and highly political factor in the creation of a nation. Recognition can occur even when all four above criteria have not been met or have been met imperfectly.
See Also: Self-Determination (A Concept of International Law)
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190304 | https://math.stackexchange.com/questions/4977674/size-of-maximum-matching-in-bipartite-graph-with-degree-restrictions | Size of maximum matching in bipartite graph with degree restrictions - Mathematics Stack Exchange
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Size of maximum matching in bipartite graph with degree restrictions
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I am having trouble with the following problem:
Let G=(A,B,E)G=(A,B,E) be a bipartite graph on 22 22 vertices, of which 20 20 vertices have degree five and 2 2 vertices have degree three. Find ν(G)ν(G), where ν(G)ν(G) denotes the size of a maximum matching of G G.
By Kőnig's Theorem we have that ν(G)ν(G) equals the size of a minimum vertex cover of G G, denoted by τ(G)τ(G). I resorted to finding τ(G)τ(G). By the Handshaking Lemma we know that G G has 53 53 edges, and by the inequality τ(G)≥|E|Δ(G)τ(G)≥|E|Δ(G) (where Δ(G)Δ(G) is the maximum degree of G G) we have that τ(G)≥11 τ(G)≥11. I've tried to infer more about τ(G)τ(G), such as something involving this identity due to Gallai: τ(G)+α(G)=|V|τ(G)+α(G)=|V| (where α(G)α(G) is the size of a maximum independent set).
But I'm not sure whether this is the right approach, perhaps I'm overlooking something. Any help is appreciated.
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asked Sep 28, 2024 at 19:00
JLGLJLGL
1,055 4 4 silver badges 11 11 bronze badges
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Aren't you done already? If τ(G)≥11 τ(G)≥11, you have already covered all vertices of G G in your matching. So τ(G)τ(G) is exactly 11 11.koifish –koifish 2024-09-29 04:43:48 +00:00 Commented Sep 29, 2024 at 4:43
This is true only if |A|=|B|=11|A|=|B|=11 right? If so, I think I could infer that α(G)≤11 α(G)≤11 from the identity in my post. But since G G is bipartite α(G)=11 α(G)=11 and the rest follows.JLGL –JLGL 2024-09-29 07:27:46 +00:00 Commented Sep 29, 2024 at 7:27
@JLGL For this setup, we can indeed show that |A|=|B|=11|A|=|B|=11, and further more than A,B A,B each have exactly 1 vertex with degree 3. Hint: 5×10<5×10+3×2 5×10<5×10+3×2.Calvin Lin –Calvin Lin 2024-10-02 07:24:16 +00:00 Commented Oct 2, 2024 at 7:24
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190305 | https://dictionary.cambridge.org/us/dictionary/learner-english/liaison | Definition of liaison – Learner’s Dictionary
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liaison noun (COMMUNICATION)
liaison noun (PERSON)
liaison noun (RELATIONSHIP)
(Definition of liaison from the Cambridge Learner's Dictionary © Cambridge University Press)
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Victoria sponge
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a soft cake made with eggs, sugar, flour, and a type of fat such as butter. It is made in two layers with jam or cream, or both, between them
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190306 | https://www.scribd.com/document/404040439/Geometry-Chapter-7-Sections-7-1-and-7-2 | Geometry Chapter 7 - Sections 7.1 and 7.2 | PDF | Triangle | Area
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Geometry Chapter 7 - Sections 7.1 and 7.2
The document contains geometry problems involving the Pythagorean theorem, classifying triangles, finding lengths and areas of triangles. 2. Students are asked to determine if sets of nu…
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Geometry Worksheet 7.1 & 7.2
Write the
Pythagorean Theorem Extension of the Pythagorean Theorem
If the triangle is acute th en c
2
____ a
2
b
2
If the triangle is obtuse the n c
2
_____ a
2
b
2
Solve for x.
(Simplified radical answers)
Are the triangles below right triangles? If not, what kind o f triangles are they? 8.
9.
10.
Determine whether the numbers can represent the side lengths of a triangle. If they can, classify the triangle as
acute, right,
or
obtuse
. Show work! 11.
26, 35, 62
12.
29, 18, 14
13.
30, 78, 72
14.
17, 19, 22
Decide if each set of numbers represents a
Pythagorean triple
. (Yes or no) 15.
5, 12, 13
16.
10, 17, 24
17.
7, 14,
5 7
Find the area of each figure.
(answers rounded to nearest hundredth)
20
12 16 x x 7 5 16 x 17 25 x
8 ft
10 in 7 in 9 15 12 15 36
4 95
7 8
113
28 cm 12 cm
Name ____
adGet Scribd without ads.
Find the area AND perimeter of each figure.
(answers rounded to nearest hundredth)
Area = _ Perimeter = __ _ Area = __ Perimeter = __ __
Solve each app lication problem. Draw a sketch and show work!
(answers rounded to nearest hundredth)
23.
Find the length of the diagonal of a squa re whi ch ha s a peri meter of 12 feet.
One leg of a right triangle is twic e as long as the other leg. The area of the triangle is 49 square feet. What is the length of the hy potenuse?
25.
A shipping doc k has a ramp that is used to h elp load and u nload cargo from truc ks. The ramp is 125 inches long a nd has a base that is 120 in ches long. What is the heigh t of the ramp?
As part of you r exerci se routin e you jog around a park wh ich is shape d like a r ight isosce les triangle. One side of the park, the hypotenuse of the triangle, is 850 feet long. 26
. What are the len gths of the other tw o sides that cre ate the park?
27
. The recreation boar d would like t o place a fen ce around the park. The cost of the fencing is $3.00 per foot. How much money will it cost to fence in the park?
28.
The recreation boar d also wants to plan t new grass ove r the entire park. How many s quare feet will need to be covered with grass seed?
29.
If the cost of grass seed is $5 per square foot, how mu ch will it cost to replant grass in the entire park?
20 in 11 in 14 in
20 m
10 m 16 m
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190307 | https://www.math.uni.edu/~campbell/mdm/permcomb.html | Permutations and combinations
Factorials
Permutations
Combinations
Generalized combinations
Factorials
It is convenient to define n factorial, denoted as n! as the product of the first n positive integers. For example, 4! = 4 × 3 × 2 × 1 = 24. It is manifest that 1! = 1, and we define 0! = 1.
Permutations
A permutation is an ordering or arrangement of objects. P(n,r) denotes the number of distict arrangements of r objects from n objects. For example P(5,2) = 20 because there are 20 ordered pairs from the letters abcde, viz. ab, ac, ad, ae, ba, bc, bd, be, ca, cb, cd, ce, da, db, dc, de, ea, eb, ec, ed. This number can be obtained as 5 × 4 = 20, because there are five choices for the first letter, and after that is removed, four choices remain for the second letter, You are chooosing the first letter and the second letter, hence this is an example of the multiplication (and) rule.
Note that the number of arrangements of n distinct objects is P(n,n) = n!.
The number of ways you can choose a president, vice-president, and secretary from a class of seven students is P(7,3) = 7 × 6 × 5 = 210. This can also be written a 7!/(7-3)!. In general P(n,r) = n!/(n-r)!; This can this can be interpreted as arranging all n objects, and then removing the order of the (n-r) objects which are not chosen by dividing by the number of ways to arrange them.
Exercise: If you have 7 distinct objects, how many permutations are there of all 7?, 6 of them?, five of them? four of them? three of them? two of them? one of them? zero of them? What does permuting one object or zero objects mean?
Combinations
C(n,r) (which is read as n choose r) is the number of different unordered samples of size r which can be chosen from n distinct objects. For example, C(5,2) = 10 because ab, ac, ad, ae, bc, bd, be, cd, ce, de are the only pairs of letters which can be chosen from abcde (P(5,2) = 20 because each of these pairs can be ordered two ways as listed above). C(5,2) = 5!/((5-2)!×2!) which can be interpreted as arranging all 5 objects, and then removing order among the 3 which are not chosen and also among the 2 which are chosen. In general C(n,r) is equal to n!/((n-r)!×r!)
The classical notation for C(n,r) has n above r within parenthesis, but it is hard to do that in ascii, and it is not the notation of the text.
Generalized combinations
Often one is interested in distinguishing some, but not all, of the individuals which are chosen. For example, consider the problem of selecting a president, vice-president, and a committee of three from seven individuals. This can be done by first selecting hte president and vice-president, and then choosing the committee ot three. This line of reasoning produces the algebra: P(7,2) × C(5,3) = (7!/5!) × (5!/(2! × 3!)) = 420. This can also be written as 7!/(1!1!3!2!) = 420 which can be interpreted as arranging all 7 individuals, then removing order among the presidents (there is only 1, 1! = 1), vice-presidents (there is only one), winners (there are three), and losers (there are 2). THis reasoning can also be extended to the question of how many arrangements are there of the letters in banana. There are 6!/(1!3!21) = 60; this can be interpreted as arranging all 6 letters, then removing the order among the b, the a's, and the n's.
Competencies:How many ways can a president,vice-president, and secretary be chosen from a class of 8? How many ways can a committee of three be chosen from a class of 8? How many ways can a president, vice-president, secrtary, and a committee of three be chosen from a class of 8?
How many arrangements are there of the letters in Mississippi?
Reflection:
Challenge:How many ways can you form 4 pairs of roommates from eight men?
May 2002
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190308 | https://hal.science/hal-04822171v1/file/2307.10347v1.pdf | HAL Id: hal-04822171 Submitted on 6 Dec 2024 HAL is a multi-disciplinary open access archive for the deposit and dissemination of sci-entific research documents, whether they are pub-lished or not.
The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.
L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.
Distributed under a Creative Commons Attribution 4.0 International License On affine spaces of alternating matrices with constant rank Clément de Seguins Pazzis To cite this version: Clément de Seguins Pazzis. On affine spaces of alternating matrices with constant rank. Linear and Multilinear Algebra, 2024, 10.1080/03081087.2024.2356827. hal-04822171 Providedthatn≥r+3and|F|≥minr−1,+2,wealsodeterminethe On affine spaces of alternating matrices with constant rank Cl´ ement de Seguins Pazzis∗† July 21, 2023 Abstract Let F be a field, and n ≥ r > 0 be integers, with r even. Denote by An(F) the space of all n-by-n alternating matrices with entries in F. We consider the problem of determining the greatest possible dimension for an affine subspace of An(F) in which every matrix has rank equal to r (or rank at least r). Recently Rubei has solved this problem over the field of real numbers. We extend her result to all fields with large enough cardinality. r 2 affine subspaces of rank r matrices in An(F) that have the greatest possible dimension, and we point to difficulties for the corresponding problem in the case n ≤ r + 2. AMS MSC: 15A30; 15A03 Keywords: affine space, rank, dimension, alternating forms, skew-symmetric matrices, trivial spectrum spaces 1 Introduction Let F be a field (possibly of characteristic 2). Let V be a vector space over F with finite dimension n, and r be an even integer in [0, n ]. A bilinear form b on V is called alternating whenever b(x, x) = 0 for all x ∈ V (if the characteristic of F is not 2, this means that b is skew-symmetric, otherwise these notions are ∗Universit´e de Versailles Saint-Quentin-en-Yvelines, Laboratoire de Math´ematiques de Ver- sailles, 45 avenue des Etats-Unis, 78035 Versailles cedex, France †e-mail address: dsp.prof@gmail.com 1 r r r distinct), and it is called symplectic when it is alternating and non-degenerate. We denote by A2(V ) the vector space of all alternating bilinear forms on V , and consider the following three problems: (1) What is the greatest possible dimension d=(V ) for an affine subspace of A2(V ) in which every form has rank r? (2) What is the greatest possible dimension d≥(V ) for an affine subspace of A2(V ) in which every form has rank at least r? (3) What is the greatest possible dimension d≤(V ) for an affine subspace of A2(V ) in which every form has rank at most r? In each case, we might also inquire about the structure of the spaces that attain the greatest possible dimension, but this is very difficult in general. Problem (3) has been solved in over all fields, including an explicit description of the spaces that attain the greatest possible dimension. In the recent , Elena Rubei has solved problem (1) for arbitrary n and r but only over the field of real numbers and by using specific properties of this field. Naturally, the above problems can also be stated as problems on subspaces of alternating matrices, and it convenient to display the examples in matrix fashion. Thus, we will denote by An(F) the space of all n-by-n alternating matrices (i.e. the square matrices A ∈ Mn(F) such that XT AX = 0 for all X ∈ Fn). Following a remark of Roy Meshulam for the corresponding problems in spaces of linear operators from one vector space to another, we will see shortly that problems (1) and (2) are intimately connected with so-called trivial spec- trum spaces of endomorphisms. An endomorphism u of V has trivial spectrum if SpF(u) ⊆ {0}, i.e. it has no non-zero eigenvalue in the field F (but is allowed to have nonzero eigenvalues in algebraic extensions of F). In particular, nilpotent endomorphisms have trivial spectrum, but the converse is not true in general. Followingly, a linear subspace of End(V ) is said to have trivial spectrum when all its elements have trivial spectrum. We refer to [5, 7, 9, 10] for past work on such spaces. We mention in particular the following important result, which generalizes a famous result of Gerstenhaber on spaces of nilpotent matrices : Theorem 1 (See [5, 7]). The greatest possible dimension for a trivial spectrum linear subspace of End(V ) is dim V · 2 In , the spaces that attain the greatest possible dimension, which we call the optimal trivial spectrum subspaces, were related to the classification of 2 0 AT (potentially non-symmetric) non-isotropic bilinear forms over F (provided that |F| > 2). Now, say that we have an affine subspace B of A2(V ) in which every element has rank n, i.e. is symplectic. Take an arbitrary s0 ∈ B. Assign to every bilinear form s on V the unique endomorphism u ∈ End(V ) such that s(x, y) = s0(x, u(y)) for all (x, y) ∈ V 2. This way, we create an isomorphism Φ from A2(V ) to the space As0 of all s0-alternating endomorphisms of V (an endomorphism u is s0-alternating whenever s0(x, u(x)) = 0 for all x ∈ V ). Now, let u ∈ As0 . Then u − id is non singular if and only if s0 − Φ−1(u) is symplectic. By a simple homogeneity argument, this yields that u has trivial spectrum if and only each form in the affine subspace s0 + FΦ−1 − → (u) is symplectic. Hence, denoting by B − → the translation vector space of B, we gather that Φ( B ) has trivial spectrum. And conversely, if we have a symplectic form s0 on V together with a linear subspace L of As0 with trivial spectrum, then s0 + Φ−1(L) is an affine subspace of symplectic forms on V , with the same dimension as L. Consequently, solving the case n = r in problems (1) and (2) (they are equivalent in that case) amounts to determining the greatest possible dimension for a trivial spectrum linear subspace of As when s is an arbitrary symplectic form on an n-dimensional vector space (with n even). If F is algebraically closed, this can be obtained as a consequence of corresponding results on nilpotent linear subspaces of As (see for fields with characteristic other than 2, and for fields with characteristic 2). In this note, our first major result is a generalization to all fields of large enough cardinality: Theorem 2. Let V be an F-vector space of even dimension 2n, and s be a symplectic form on V . Assume that |F| ≥ 2n − 1. Let S be a trivial spectrum linear subspace of As. Then dim S ≤ n(n − 1). The optimality of this result (apart from the restriction on the cardinality of F) is illustrated in the following example. Let W be an optimal trivial spec- trum subspace of Mn(F) (e.g. the space NTn(F) of all strictly upper-triangular matrices). Then one sees that the set of all matrices of the form A B , with A ∈ W and B ∈ An(F), represents, in the standard basis of F2n, a space of s-alternating endomorphisms for the symplectic form s whose Gram matrix in that basis equals the standard 3 2 2 −BT s s,n−r thatM˜ hasconstantrank2s. ˜ −CT 0 r s(s+1) ifn=r+1. symplectic matrix 0 In . −In 0 Obviously, this is a trivial spectrum space with dimension 2 n = n(n − 1). 2 As an immediate corollary of Theorem 2 and of this example, we obtain: Theorem 3. Let V be a vector space of even dimension 2n, and s be a symplectic form on V . Assume that |F| ≥ 2n − 1. Then the greatest possible dimension for an affine subspace of A2(V ) consisting of symplectic forms is n(n − 1). Theorem 4. Let V be a vector space of dimension n, and r = 2s be an even integer in [0, n ]. Assume that |F| ≥ n − 1 if n is even, and |F| ≥ n − 2 if n is odd. Then d≥(V ) = s(s − 1) + r(n − r) + (n − r)(n − r − 1) = dim A2(V ) − s2. r 2 Theorem 5. Let V be a vector space of dimension n, and r = 2s be an even integer in [0, n ]. Assume that |F| ≥ max r − 1, 2 + r . Then d=(V ) = ( s (n − s − 1) if n / = r + 1 Let us immediately show that the stated dimensions can be attained in those theorems. We start with the second one. Letting M be an affine subspace of non-singular matrices of Ms(F) with dimension s(s−1) (for example, Is+NTs(F)), we take (n) A B C M ˜alt := | A ∈ A (F), B ∈ M, C ∈ M (F) ⊆ An(F). −CT (n) It is easily seen that dim Malt = s(s − 1) + s(n − 2s) = s(n − s − 1). And because of the assumption that all the matrices in M are non-singular it is also clear (n) alt Next, if n = r+1 we can take an affine subspace H ⊆ An−1(F) with dimension s(s − 1) and whose elements are all non-singular (e.g. given by the previous example), and then take the space H+ := H C | H ∈ H, C ∈ Fn−1 ⊆ An(F). 4 2 r evenintegerwithn>r+2.Assumethat||≥maxr−1,2+.LetS −CT D r s(s+1) ifn=r+1. Clearly all the matrices in H+ have rank at least n − 1, and since they are alternating their rank is even, and hence at most n − 1. And finally dim H+ = dim H + (n − 1) = s(s + 1). Finally, let us start from an affine subspace H of Ar(F) in which every element is non-singular, and with dim H = s(s − 1), and consider the space H(n) := H C | H ∈ H, C ∈ Mr,n−r(F), D ∈ An−r(F) ⊆ A (F). Again, it is clear that all the matrices in H(n) have rank at least r, and the dimension of the space is dim H(n) = dim A (F) − codimAr (F) H = n − s2. Hence, it only remains to prove the inequalities d≥(V ) ≤ dim A2(V ) − s2 and d=(V ) ≤ ( s (n − s − 1) if n / = r + 1 The proof of the first one will be deduced from Theorem 3 thanks to Meshulam’s method from (Section 4). The case n = r in the second one is given by Theorem 3. For the other cases, we will use the same strategy as in Rubei’s article to deduce the inequality from Theorem 3. As an offspring of our method, in Section 5 we will obtain the following partial result on the affine spaces that attain the greatest possible dimension in problem (1): Theorem 6. Let V be a vector space of dimension n, and r = 2s > 0 be an F r 2 be an affine subspace of A2(V ) in which every element has rank r, and with dim S = s (n − s − 1). Then there exists a basis of V and an affine subspace M ⊆ GLs(F) with s(s−1) (n) dimension 2 such that S is represented in the said basis by M ˜ a l t . Moreover, the equivalence class1 of M is uniquely determined by S. 1Two subsets X and Y of Mn,p (F) are called equivalent when there exist invertible matrices P ∈ GLn(F) and Q ∈ GLp(F) such that Y = P X Q, meaning that X and Y represent the same set of linear mappings in a different choice of bases. 5 n n 2 ˜ ˜ 0 0 B D 0 0 B D As stated earlier, the classification, up to equivalence, of the affine subspaces of Ms(F) included with GLs(F) and with dimension s(s−1) is well understood when |F| > 2 (it is connected to the one of non-isotropic quadratic forms over F). Conversely, it is easily checked that if M and M′ are equivalent affine sub- (n) ′(n) spaces of Ms(F) then Malt and M alt are congruent affine subspaces of An(F). If n = r + 2, there are examples that do not fit the result of Theorem 6: for instance, one can take an affine subspace H of dimension s(s + 1) of Ar+1(F) in which every matrix has rank r, and consider the affine space of all matrices of the form H 0×1 with H ∈ H. 1×(r+1) 0 An inspection of the proof of Theorem 6 makes us worry that the special cases n ∈ {r + 1, r + 2} are far more difficult than the case n > r + 2, and we prefer to abstain from going any further. 2 Technical lemmas Our proof techniques essentially rely on basic block-matrix results from the theory of vector spaces of bounded rank matrices. Chiefly, we will use the following result, which we call the Flanders-Atkinson lemma, and several of its corollaries. We refer to , and Section 2 of for various proofs and versions of it. Lemma 7 (Flanders-Atkinson lemma). Let n, p, r be integers with 0 < r ≤ min(n, p). Assume that |F| > r. Let Jr := Ir 0 and M = A C belong to Mn,p(F), with A ∈ Mr(F) and so on. If rk(sJr + tM ) ≤ r for all (s, t) ∈ F2, then D = 0 and BAkC = 0 for every integer k ≥ 0. Corollary 8. Let n, p, r be integers with 0 < r ≤ min(n, p). Assume that |F| > r + 1. Let Jr := Ir 0 and M = A C belong to Mn,p(F), with A ∈ Mr(F) and so on. If rk(Jr + tM ) ≤ r for all t ∈ F, then D = 0 and ∀k ≥ 0, BAkC = 0. Proof of Corollary 8. One simply uses the assumption |F| > r + 1 to gather that all the (r + 1) × (r + 1) minors of xJr + yM vanish for all (x, y) ∈ F2 (take such 6 2 2 0 0 −BT D a minor as a function of (x, y), and note that it is a homogeneous polynomial of degree r + 1 that vanishes at more than r + 1 points of the projective line of F2). Then one applies the Flanders-Atkinson lemma. Corollary 9. Let n, r be integers with 0 < r ≤ n and r even. Assume that |F| > r + 1. Let JK := K 0 and M = A B belong to An(F), with A and K in Ar(F), and so on. Assume that K is invertible. If JK + tM has rank at most r for all t ∈ F, then D = 0 and BT K−1(AK−1)kB = 0 for every integer k ≥ 0. Proof. There is a subtlety here as we have assumed that |F| > r + 1 instead of |F| > r + 1. Of course, if the latter holds then it suffices to apply Corollary 8 after right-multiplying with K−1 ⊕ In−r. Now, assume that F is finite. We can choose a field extension L of F such that | L - | > r + 1. Then we claim that JK + tM has rank at most r for all t ∈ L. The key to obtain this is to use the Pfaffian (denoted by Pf) instead of the determinant! First of all, it is critical to note that if an alternating matrix N has rank at least 2s for some integer s ≥ 1, then one of its principal 2s × 2s submatrices is invertible. This can be proved as follows: first of all, we can write r′ = rk N and pick a direct factor of the radical of N that is spanned by vectors of the standard basis. The corresponding r′ × r′ submatrix of N is then invertible. And then one proceeds by downward induction by using the development of the Pfaffian along the last row/column. Now, take an arbitrary subset I of [1, n ] with cardinality r + 2, and for N ∈ An(F) denote by NI,I the corresponding principal submatrix. The mapping t ∈ L - '→ Pf((JK + tM )I,I ) is a polynomial function of degree at most r+2 , and 2 it vanishes everywhere on F. Since |F| > r 2 + 1, it also vanishes everywhere on L - . Varying I shows, thanks to the previous remark, that JK + tM has rank at most r for all t ∈ L - . Then, applying Corollary 9 in L - yields the claimed result. As a consequence of the conclusion “D = 0” in the Flanders-Atkinson lemma, we also have the following result in terms of subspaces of linear mappings: Corollary 10. Let U and V be finite-dimensional vector spaces, and S be a linear subspace of Hom(U, V ). In S, take an element u0 of maximal rank r, and assume that |F| > r. Then every element of S maps Ker u0 into Im u0. Finally, we recall a consequence of the main result of , to be used in Section 5: 7 2 2 ˜ } ^ ^ ^ Theorem 11. Let p and r be positive integers with 1 ≤ r ≤ p. Assume that |F| > 2. Let T be an affine subspace of Mr,p(F) in which every matrix has rank r, and assume that codimM r,p (F) T ≤ r(r+1) · Then, there exists an affine subspace M of Mr(F) in which every matrix is invertible, with dim M = r(r−1) , and such that T is equivalent to the space M(p) := B C | (B, C) ∈ M × Mr,p−r(F) . Moreover, the equivalence class of M is uniquely determined by T . Proof. When r ≥ 2, Theorem 11 is the special case n = r in theorem 3 of . We wish to note that the result remains true in the special case r = 1. In that case, it is essentially an obvious result on the (affine) hyperplanes of M1,n(F) that do not contain the zero vector: simply, take such a hyperplane T , choose x1 ∈ T and then extend this vector to a basis of M1,n(F) by using a basis of the translation vector space of T . This shows that T is equivalent to the space of all row vectors with first entry equal to 1, and hence the conclusion is satisfied for M := {1} (note that the uniqueness statement is obvious in that case). A close inspection of the proof of theorem 3 of also reveals that if n = r then the case r = 1 need not be discarded. The proof of the main theorem of is difficult: it relies upon the classifica- tion of the optimal trivial spectrum subspaces of Mr(F), as given in . 3 Trivial spectrum linear subspaces of alternating en- domorphisms Here we prove Theorem 2 by induction on n. The case n = 0 is trivial, and now we assume that n ≥ 1. The idea is to use the operator-vector duality, in a way that is reminiscient to the basic idea of . For x ∈ V , we consider the linear operator x : u ∈ S '→ u(x) ∈ V. This yields a linear subspace S = {x | x ∈ V } ⊆ Hom(S, V ). Now, let x ∈ V z{0}. Since S has trivial spectrum, we have Fx ∩ Sx = {0}. But we also have Sx ⊆ {x}⊥s because S consists of s-alternating operators. And 8 ^ ^ ^ ^ ^ ^ finally x ∈ {x}⊥s since s is alternating. Therefore Fx ⊕ Sx ⊆ {x}⊥s . It follows in particular that dim(Sx) ≤ 2n − 2, that is rk x ^ ≤ 2n − 2. ^ Now, we take x ∈ V z {0} such that x has the greatest possible rank in S, denoted by r′. In particular r′ ≤ 2n − 2 and Corollary 10 yields that S maps (this is where the assumption |F| ≥ 2n − 1 comes every vector of Ker x into Im x into play). So, set S′ := Ker x = {u ∈ S : u(x) = 0}. Now, even though we might have r′ < 2n − 2, we can always embed Im x = Sx into a linear hyperplane W of {x}⊥s such that Fx ⊕ W = {x}⊥s . In particular W is s-regular. The previous remark yields that Im u ⊆ W for every u ∈ S′. Because each u ∈ S′ is s-alternating and hence s-selfadjoint, we deduce that the elements of S′ vanish everywhere on W ⊥s . And finally V = W ⊕ W ⊥s because W is s-regular. Hence, by restricting to W we obtain a linear injection from S′ to a linear subspace of AsW , where sW stands for the symplectic form induced by s on W 2. The range of the said injection obviously has trivial spectrum. Hence by induction (because |F| ≥ 2n − 1 ≥ 2(n − 1) − 1) we find dim S′ ≤ (n − 1)(n − 2). By the rank theorem, we conclude that dim S = dim S′ + dim Sx ≤ (n − 1)(n − 2) + (2n − 2) = (n − 1) n. This completes the proof of Theorem 2. 4 Affine subspaces of alternating forms with bounded rank Now, we prove Theorems 4 and 5. To start with, we let S be an affine subspace of A2(V ) in which every element has rank at least r, and we assume that |F| ≥ n−1 if n is even, and |F| ≥ n − 2 if n is odd. The case n = r has already been dealt with in Theorem 2, so we assume n > r. By downward induction on r, we can assume that S actually contains an element s0 of rank r (indeed, the dimension stated in the first part of Theorem 9 −B(b)T D(b) 2 2 5 is a non-increasing functions of r, as seen by its second expression, and the cardinality assumption on |F| garantees that |F| ≥ r0 − 1 for the least possible rank r0 of the elements of S). Let us then take an arbitrary basis of V in which the last n − r vectors span the radical of s0, and let us represent the elements of S in that basis: for each b ∈ A2(V ) we have a corresponding alternating matrix M (b) = A(b) B(b) where A(b) ∈ Ar(F), B(b) ∈ Mr,n−r(F), D(b) ∈ An−r(F). Note that B(s0) = 0 and D(s0) = 0. Set K := A(s0) ∈ GLr(F) ∩ Ar(F) and consider the affine subspace T := {b ∈ S : B(b) = 0 and D(b) = 0} (which contains s0). Every element of T has rank at least r, so A(T ) is an affine subspace of matrices of Ar(F) with constant rank r. By Theorem 3, we have dim T = dim A(T ) ≤ s(s − 1), and we conclude by the rank theorem for affine mappings that dim S ≤ dim T + r(n − r) + dim An−r(F) ≤ dim An(F) − s2. Thus Theorem 4 is now proved. In the remainder, we turn to the proof of Theorem 5: we assume that every element of S has rank r, and we modify the cardinality assumption on F: here we assume that |F| ≥ max r − 1, 2 + r . Then the argument is slightly different but we start again from the previous block form. Now, we introduce the translation vector space S of S and we apply Corollary 9, which uses the assumption that |F| > r + 1 and that every element of S has rank at most r. This yields ∀b ∈ S, D(b) = 0 and B(b)T K−1B(b) = 0. (1) From the first identity, we get dim S = dim T + dim B(S). 10 2 2 2 If n = r + 1 this is clearly enough to conclude. Now, assume that n ≥ r + 2. Then we shall prove that dim B(S) ≤ (n − r)s, which will be enough to conclude. To obtain this, we interpret the second identity in (1) as meaning that the range of K−1B(b) is totally K-singular for all b ∈ S. And the expected result will come from the following lemma: Lemma 12. Let U and U ′ be finite-dimensional vector spaces, with dim U > 1, and b be a symplectic form on U ′. Let V ⊆ HomF(U, U ′) be a linear subspace in which every element has its range totally b-singular. Then (dim U )(dim U ′) dim V ≤ · 2 Moreover, if dim U > 2 and dim V = (dim U)(dim U′) then V = HomF (U, L) for some Lagrangian2 L of (U ′, b). Proof. Set p := dim U and q := dim U ′ for convenience, and r := q · If dim Vx ≤ r for all x ∈ Fpz{0}, then we directly have dim V ≤ pr by taking a basis of U . Now, assume that dim Vx0 > r for some x0 ∈ U z {0}. Consider then the subspace V′ := {u ∈ V : u(x0) = 0}. Let x ∈ U . Let u′ ∈ V′ and u ∈ V. Then (u+u′)(x) is b-orthogonal to (u+u′)(x0) = u(x0), and u(x) is b-orthogonal to u(x0). Hence u′(x) is b-orthogonal to u(x0). Thus V′x ⊆ (Vx0)⊥b . By taking a basis of a complementary subspace of Fx0 in U , we deduce that dim V′ ≤ (p − 1) dim(Vx0)⊥b . Hence by the rank theorem dim V = dim(Vx0) +dim V′ ≤ dim U ′ +(p − 2) dim(Vx0)⊥b ≤ 2r +(p − 2) r = pr. Note that the last inequality is sharp if p − 2 > 0. Now, assume that p > 2 and dim V = pr. Hence by the last remark we must have dim Vx ≤ r for all x ∈ U z {0}. Take a basis (x1, . . . , xp) of U . Since dim V = pr, the linear mapping ψ : v ∈ V '→ (v(x1), . . . , v(xp)) ∈ Vx1 × · · · × Vxp must be surjective and all the Vxi’s must have dimension r. By the surjectivity of ψ, we deduce that Vx1, . . . , Vxp are pairwise b-orthogonal. But since they all have dimension r we have Vxi = (Vxj)⊥b for all distinct i, j in [1, p ], and since p ≥ 3 we obtain that all the Vxi’s are equal. Their common value is then a Lagrangian L of (U ′, b). Hence V ⊆ HomF(U, L) and we conclude by V = HomF(U, L) since the dimensions are equal. 2For a symplectic form b on a vector space U ′, a Lagrangian is a totally b-singular subspace of U ′ with dimension dim U′ · 11 i Σ i This complete the proof of Theorem 5. 5 Affine subspaces of alternating forms with constant rank and large dimension Here, we prove Theorem 6. We come back to the situation of the previous section. Now, we assume that n > r + 2 and that S has the critical dimension s(n − s − 1), with all the forms in S of rank r. With the previous proof, we gather in particular that dim B(S) = s(n − r) and dim T = s(s − 1), and we can use the last statement of Lemma 12 to obtain a Lagrangian L of Fr for the symplectic form (X, Y ) '→ XT KY , such that K−1B(S) is the set of all matrices of Mr,n−r(F) with range included in L. Step 1: Proving that L is totally A(s)-singular for all s ∈ S Let us take an arbitrary linear section θ : B(S) → S of the projection of S onto B(S). By Corollary 9, we find that ∀N ∈ B(S), (K−1N )T A(θ(N )) K−1N = 0. (2) For i ∈ [1, n − r ], put fi : Xi ∈ L '→ A θ 0 · · · 0 KXi 0 · · · 0 where the KXi vector appears on the i-th column. Let k ∈ [1, n − r ]. Choose distinct elements i, j in [1, n − r ] z {k} (this is possible because n > r + 2). Applying (2) at the (i, j)-spot yields ∀(X1, . . . , Xn−r) ∈ Ln−r, XT n−r fℓ(Xℓ) Xj = 0. ℓ=1 Replacing Xk with 0 and subtracting the two identities thereby obtained, we get ∀(Xi, Xj, Xk) ∈ L3, XT fk(Xk)Xj = 0. Fixing Xk and varying Xi and Xj, we obtain that L is totally fk(Xk)-singular. Varying k and Xk then yields that L is totally A(θ(N ))-singular for all N ∈ B(S). As this yields for every choice of the section θ, we conclude that L is totally A(b)-singular for all b ∈ S. Because it is also totally K-singular, we conclude that L is totally A(b)-singular for all b ∈ S. 12 −R(b)T 2 2 (n) (n) ˜ (n) ˜ suchthatQR(S)Q′=M˜(n−s).ThenP:=Q⊕(Q′)TbelongstoGLn(F)and Step 2: Preparing the reduced form Now we refine the choice of the starting basis so that L = {0} × Fs and K = 0 Is . In that case every matrix of B(S) has its rows zero starting from the −Is 0 (s + 1)-th. Note that B(S) = B(S) because B(s0) = 0. And finally the first step shows that every matrix A(b) with b ∈ S has its lower-right s × s block equal to zero. Therefore, for every b of S we now have M (b) = U (b) R(b) for some U (b) ∈ As(F) and some R(b) ∈ Ms,n−s(F). Next, note that R(S) is an affine subspace of Ms,n−s(F) and dim R(S) ≥ dim S − s(s−1) = dim Ms,n−s(F) − s(s+1). Moreover, for all b ∈ S, we see that rk M (b) ≤ 2 2 s+rk R(b) (erase the first s columns) and hence rk R(b) ≥ s. Hence every matrix in R(S) has rank s. Step 3: Concluding the reduction Noting that r + 2 > 2, we see that Theorem 11 applies to R(S). This yields Q ∈ GLs(F) and Q′ ∈ GLn−s(F) (actually, one could take Q = Is) and an affine subspace M of nonsingular matrices of Ms(F), with dimension s(s−1), PM (S)PT that ⊆ M ˜ a l t . As both spaces have dimension s(n − s − 1), we conclude PM (S)PT = M ˜ . alt This completes the proof of the first statement in Theorem 6. Step 4: Uniqueness It remains to prove that the equivalence class of M is uniquely determined by S. Towards this end, we choose a basis (e1, . . . , en) of V in which S is represented by M(n). Note that L := span(es+1, . . . , en) is totally singular for all the forms in S. The key is to prove that it is the sole such space with dimension n − s. So, we take an arbitrary subspace L′ ⊆ V with dimension n − s and which is totally singular for all the forms in S. Then clearly L′ is also totally singular for all the forms in the translation vector space S of S. First of all, we prove that dim(L ∩ L′) ≥ n − s − 1. To see this, note from the definition of Malt that S contains every alternating form on V whose radical includes L. If some pair (x, y) ∈ (L′)2 is linearly independent modulo 13 ˜ 1 s 3 n−s (n) ′(n) in somebasis(e′,...,e′).ThenL′=span(e′ ,...,e′)istotallysingularfor T (n−s) ′(n−s) 1 n s+1 n L, then among such forms we can choose one b such that b(x, y) / = 0, con- tradicting a previous statement (indeed, extend (x, y, es+1, . . . , en) into a basis (x, y, es+1, . . . , en, f1, . . . , fs−2) of V , take the dual basis (ϕ1, . . . , ϕn) and con- sider the alternating form b : (z, z′) '→ ϕ1(z) ϕ2(z′) − ϕ1(z′) ϕ2(z)). Hence the projection of L′ onto V/L has dimension at most 1, and we conclude that dim(L ∩ L′) ≥ n − s − 1. Now, assume that dim(L ∩ L′) = n − s − 1, so that dim(L + L′) = n − s + 1. Note that, for all x ∈ L ∩ L′ and all b ∈ S, the linear form b(−, x) vanishes everywhere on L + L′, whence dim{b(−, x) | b ∈ S} ≤ codimV (L + L′) = s − 1. Observing the last n − r columns in the matrices of M(n), this forces (L ∩ L′) ∩ span(er+1, . . . , en) = {0}, and since we are dealing with subspaces of L we derive that dim(L ∩ L′) ≤ (n − s) − (n − r) = s. But this would yield n ≤ 2s + 1, in contradiction with the assumption n > r + 2. This shows that L′ = L. Now we can complete the proof. Let M′ be an affine subspace of nonsingular s(s−1) ′(n) matrices of Ms(F), with dimension 2 , and assume that M ˜alt represents S all the elements of S, and hence it equals span(es+1, . . . , en) by the first part of this step. Therefore the matrix P of coordinates of (e′ , . . . , e′ ) in (e1, . . . , en) 1 n reads P = P1 with P ∈ GL (F) and P ∈ GL (F). ?×s P3 Moreover PT M ˜a l t P = M ˜alt . Extracting the upper-right blocks, this yields P1 M ˜ P3 = M ˜ . By the uniqueness statement in Theorem 11 we con- 6 Open questions and comments In this final section, we wish to make some comments on the previous results and their limitations. We have already pointed to the fact that Theorem 6 has no immediate adaptation to the case n = r + 2, and the assumption n > r + 2 14 cludethatMisequivalenttoM′.ThiscompletestheproofofTheorem6. s×s AT −(Is+A)T B 2 −(Is+A)T B was extensively used in our proof. The case n = r + 1 seems to be even more difficult. What is remarkable in our proof is that, although T was shown early on to be an affine subspace of invertible elements of Ar(F) with the greatest possible dimension (that is, s(s − 1)), we did not require any classification of such spaces, i.e. of the solution to the case n = r. We suspect that such a solution is unavoidable for the case n = r + 1, and for n = r + 2 we also have the feeling that it might also be. At this point of course we have not formulated any conjecture on the form of the optimal spaces for n = r, i.e. of the affine subspaces of Ar(F) with dimension s(s−1) in which all the elements are invertible. Actually, those spaces are known if F is algebraically closed with characteristic other than 2. In that case indeed, trivial spectrum subspaces coincide with nilpotent subspaces, and hence one of the main results from (theorem 1.9 there) yields that, for a symplectic form s0 on a 2s-dimensional vector space, and for every trivial spectrum subspace S of As0 , there exists an s0-symplectic basis (e1, . . . , es, f1, . . . , fs) of V in which S is represented by the space of all matrices of the form A B with A ∈ NTs(F) and B ∈ As(F). Hence, it follows that, up to congruence, the sole affine subspace of Ar(F) that has dimension s(s − 1) and constant rank r is s×s Is + A | A ∈ NTs(F) and B ∈ As(F) . Now, an apparently reasonable conjecture would generalize the above to trivial spectrum subspaces, as follows: Instead of NTs(F), we take an arbitrary trivial spectrum (linear) subspace V of Ms(F) with dimension s(s−1). Then the corresponding affine subspace of Ar(F) with constant rank r and dimension s(s − 1) is s×s Is + A | A ∈ V and B ∈ As(F) . The conjecture would state that every affine subspace of Ar(F) with constant rank r and dimension s(s−1) is congruent to a space of this form. Yet, in working on the present article, we discovered that this conjecture is wrong, which is seen by observing the critical case where r = 4. Indeed, if this conjecture held true, 15 then every 2-dimensional affine subspace of non-singular matrices of A4(F) would have a rank 2 matrix in its translation vector space (as seen from the lower-right B cell). Yet, this is false, as we shall now see. Indeed, in A4(F) the invertibility of matrices is controlled by the Pfaffian, which is a hyperbolic quadratic form of rank 6. And to construct a counter- example is then easy, as it suffices to construct a plane P of A4(F) that does not go through zero and that can be embedded in a 3-dimensional linear subspace of A4(F) in which all the non-zero elements are invertible. This is easy if F has its u-invariant greater than 2 (otherwise it is not possible!). We will also assume that χ(F) / = 2 for convenience, but fields with characteristic 2 could also be encompassed. So, assume that there is a nonisotropic quadratic form q over F with rank 3. Then q⊥(−q) is a hyperbolic form of rank 6, and hence it is equivalent to the 4-by-4 Pfaffian. This yields that the latter has a 3-dimensional linear subspace W on which the restriction of the Pfaffian is nonisotropic, to the effect that every nonzero element of W has rank 4. To conclude, we simply pick a 2-dimensional affine subspace P of W that does not contain 0. Then of course P consists only of rank 4 matrices, and its translation vector space is also included in W and hence contains no rank 2 matrix. Hence P is an exception to the above conjecture. In practice the above abstract construction can be used to give explicit coun- terexamples. For example, for F = R we take q = ⟨1, 1, 1⟩, which prompts us to consider the space W of all real skewsymmetric matrices of the form 0 x y z −x 0 z −y A(x, y, z) = −y −z 0 x −z y −x 0 with (x, y, z) ∈ R3 (note that Pf(A(x, y, z)) = x2 + y2 + z2). And a counterex- ample is obtained by considering the plane P = {A(x, y, 1) | (x, y) ∈ R2}. This example makes us worry that obtaining the equivalent of Theorem 6 for n = r should be very difficult. Finally, noting that the case n = r +1 in Theorem 6 is close to the equivalent of deriving Theorem 11 from the main result of , it can hardly be hoped that anything short of a complete solution to the case n = r is required for the case n = r + 1, and that, even so, deriving the case n = r + 1 from the case n = r should be very difficult. Finally, even if all those problems were solved, there will still remain the issue of the cardinality assumptions in all the results we have proved so far. The 16 techniques we used made these assumptions unavoidable. Yet, for the equivalent problems on rectangular matrices there have been results that hold for all fields [8, 9] (with the exception of the field with two elements). Achieving such a goal in the present context would require a complete revolution in the methods, and so far we have failed to come up with any valid idea on how to tackle small finite fields. References M.D. Atkinson, Primitive spaces of matrices of bounded rank II. J. Austral. Math. Soc. (Ser. A) 34 (1983) 306–315. P. Fillmore, C. Laurie, H. Radjavi, On matrix spaces with zero determinant. Linear Multilinear Algebra 18 (1985) 255–266. M. Gerstenhaber, On nilalgebras and linear varieties of nilpotent matrices (I). Amer. J. Math. 80 (1958) 614–622. R. Meshulam, On two extremal matrix problems. Linear Algebra Appl. 114- 115 (1989) 261–271. R. Quinlan, Spaces of matrices without non-zero eigenvalues in their field of definition, and a question of Szechtman. Linear Algebra Appl. 434 (2011) 1580–1587. E. Rubei, Affine subspaces of skewsymmetric matrices with constant rank. Linear Multilinear Algebra (2023) in press C. de Seguins Pazzis, On the matrices of given rank in a large subspace. Linear Algebra Appl. 435-1 (2011) 147–151. C. de Seguins Pazzis, Large affine spaces of matrices with rank bounded below. Linear Algebra Appl. 437-2 (2012) 499–518. C. de Seguins Pazzis, Large affine spaces of non-singular matrices. Trans. Amer. Math. Soc. 365 (2013) 2569–2596. C. de Seguins Pazzis, From primitive spaces of bounded rank matrices to a generalized Gerstenhaber theorem. Quart. J. Math. 65-2 (2014) 319–325. 17 C. de Seguins Pazzis, Local linear dependence seen through duality I. J. Pure Appl. Algebra 219 (2015) 2144–2188. C. de Seguins Pazzis, Affine spaces of symmetric or alternating matrices with bounded rank. Linear Algebra Appl. 504 (2016) 503–558. C. de Seguins Pazzis, The structured Gerstenhaber problem I. Linear Alge- bra Appl. 567 (2019) 263–298. C. de Seguins Pazzis, The structured Gerstenhaber problem III. Linear Algebra Appl. 601 (2020) 134–169. 18 |
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Published by alxhndz, 2016-08-30 13:07:40
DISEÑO DE ELEM MAQUINAS I
Pages:
1 - 50
51 - 100
101 - 150
151 - 158
CAPITULO II: ESFUERZOS SIMPLES EM ELEMENTOS SENCILLOS DE MÁQUINAS
EjemploN°2.14 : Analizar el eje que se muestra; si resiste o no a los esfuerzos a que
estará sometido.
Solución:
ܴଵ ܴଶ ൌ 2500 ݇݃
36ܴଵ ൌ 500ݔ30 1000ݔ21 1000ݔ12 ൌ 48000 ݇݃. ܿ݉
ܴଵ ൌ 1333.33 ݇݃ ܴଶ ൌ 1166.67݇݃
ܯଵ ൌ 1333.33ݔ6 ൌ 8000 ݇݃. ܿ݉
ܯଶ ൌ 1333.33ݔ15 െ 500ݔ9 ൌ 15500 ݇݃. ܿ݉
ܯଷ ൌ 14000 ݇݃. ܿ݉
Acero AISI 1060 ܵ௬ ൌ 39 ݇݃⁄ܿ݉ଶ ܵ௨ ൌ 80 ݇݃⁄ܿ݉ଶ
ܵ௫ ൌ ܿܯ ൌ 8000ݔ1.3 ൌ 10400 ൌ 4636.28 ݇݃⁄ܿ݉ଶ
ܫ ߨ2.6ସ 2.243
64
ܨ ܿܯ15500ݔ1.6 800 24800 800
ܵ௫ ൌ ܫെ ܣൌ ߨ3.2ସ െ ߨ3.2ଶ ൌ 5.147 െ 8.042 ൌ 4818.17 െ 99.47
64 4
ൌ 4718.70 ݇݃⁄ܿ݉ଶ
ܵ௫ ൌ ܿܯ ൌ 14000ݔ1.6 ൌ 4000 ൌ 388.56 ݇݃⁄ܿ݉ଶ
ܫ ߨ3.2ସ 10.294
ܶݎ 32 4000
ܬ 2500ݔ1.6 10.294
߬௫௬ ൌ ൌ ߨ3.2ସ ൌ ൌ 388.56 ݇݃⁄ܿ݉ଶ
32
ܵሺ୫ୟ୶ሻ ൌ 4718.70 0 ඨ൬47128.70൰ଶ 388.56ଶ ൌ 2359.35 2391.13
2
ൌ 4750.48 ݇݃⁄ܿ݉ଶ
ܵሺሻ ൌ 2359.35 െ 2391.13 ൌ െ31.78 ݇݃⁄ܿ݉ଶ
߬௫ ൌ 2391.13 ݇݃⁄ܿ݉ଶ
Hallar la flecha máxima del problema anterior a través del método gráfico.
ܯ ൌ 8000 ൌ 1.698ݔ10ିଷ݇݃. ܿ݉ିଵ
ܧ. ܫ 2.1ݔ10ݔ2.243
15500 ൌ 1.434ݔ10ିଷ
2.1ݔ10ݔ5.147
1400 ൌ 1.295ݔ10ିଷ
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO II: ESFUERZOS SIMPLES EM ELEMENTOS SENCILLOS DE MÁQUINAS
ݍൌ 4ݔ10ିଷ
3.396 1.698x10-3
1x10-3
Practiquen………
2.8.-VIGAS HIPERESTATICAS
Con cierta frecuencia, se encuentran en el proyecto de máquinas, problemas en os que
no hay suficiente información para determinar todas las reacciones desconocidas en
una viga, a partir, únicamente, de consideraciones estáticas. Esto sucede cuando el
número de incógnitas es superior al de ecuaciones de equilibrio. En el caso de vigas
hiperestáticas, no puede determinarse el momento máximo de las condiciones de
equilibrio estático, de modo que es necesario encontrar primero la deformación para
que pueda determinarse el momento.
EjemploN°2.15 : Tenemos una viga uniformemente cargada; la viga esta empotrada en
un extremo y soportada en el otro por la reacción R1. El extremo empotrado tiene las
reacciones R2 y M2. Para que el sistema este en equilibrio, debe ser igual a cero la suma
de las fuerzas verticales y la de los momentos respecto a cualquier eje. Así
obtendremos dos ecuaciones, pero como las incógnitas son tres, no son suficientes
estas condiciones.
Fig.2.11
Observado la curva elástica, emplearemos las condiciones de que la flecha es cero en
los puntos A y B y que la pendiente de la curva es cero en el punto B. Escribiremos
primero la ecuación para el valor del momento en función de una distancia cualquiera x,
medida desde el apoyo de la izquierda, y lo sustituiremos en la ecuación de momentos:
d2y = M EI d2y = MX = (R1 ⋅ x) − w ⋅ x2 …………(1)
dx2 EI dx2 2
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO II: ESFUERZOS SIMPLES EM ELEMENTOS SENCILLOS DE MÁQUINAS
Integrando la ecuación de momentos obtendremos la pendiente:
E⋅I dy = R1 ⋅ x2 − w⋅ x3 + C1 ………...……….……..…..(2)
dx 2 6
Ya que debe ser cero la pendiente en el punto B, tenemos la condición de que cuando
x=1, dx/dy=0. Cuando se sustituye esta condición en la ecuación anterior obtendremos:
C1 = w⋅ λ3 − R1 ⋅ λ2
62
Sustituyendo este valor de C1 en la ecuación (2) e integrando, tendremos:
EI ⋅ y = R1 ⋅ x 3 − w ⋅x4 + w⋅l3 ⋅x − R1 ⋅l2 ⋅ x + C2 ....(3)
6 24 6 2
La flecha debe ser cero en el punto A, de forma que y=0, cuando x=0. Sustituyendo esta
condición en la ecuación (3), ésta nos dará C2=0. La ecuación (3) se convierte entonces
en:
EI ⋅ y = R1 ⋅ x3 − w ⋅x4 + w ⋅ l3 ⋅x − R1 ⋅l2 ⋅ x …..…..(4)
6 24 6 2
La condición restante es que la deformación sea cero en el punto B, o sea, y=0 para x=l.
Haciendo esta sustitución en la ecuación (4) obtendremos:
R1 ⋅ l 3 − w ⋅l 4 + w ⋅ l 4 − R1 ⋅ l 3 = 0
6 24 6 2
R1 = 3 ⋅ w ⋅ l .......................................…..…..(5)
8
Habiendo obtenido ya una reacción, las otras dos pueden obtenerse de las condiciones
de equilibrio. De la suma de fuerzas de dirección vertical encontraremos:
R1 = 5 ⋅ w ⋅ l
8
El momento flector en el extremo fijo es:
M2 = 3wl 2 − wl 2 = wl 2
8 2 8
Ya disponemos de suficiente información para el cálculo del momento máximo que, a
partir de este punto, puede determinarse de la forma ordinaria. En la figura 2.11 se
indican los diagramas de esfuerzos cortantes y de momentos flectores. La sustitución de
R1 obtenido de la ecuación (5) en (4) nos dará la deformación elástica:
( )y = w 3lx3 − 2x4 − l 3 x
48EI
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO II: ESFUERZOS SIMPLES EM ELEMENTOS SENCILLOS DE MÁQUINAS
La flecha máxima se encuentra en el punto de pendiente cero. Para hallar la situación
de este punto, la ecuación (2) debe igualarse a cero. Por sustitución del valor de C1 y R1
tendremos:
3wlx 2 − wx 3 − wl 3 = 0
16 6 48
x = 0,421λ
RESISTENCIA DE MATERIALES
Tracción, compresión y corte (o cizalladura)
σ=F τ=F
A A
σ = Tensión de tracción o compresión, kg/cm2
τ = Tensión de corte kg/cm2
F = Carga kg
A = Área de la sección recta cm2
2.9.-TENSIONES COMBINADAS
Siempre o casi siempre en un elemento de una máquina sobre él, actúan varias
cargas de diferente clase o también debido a la geometría complicada de la
pieza una carga exterior no dé por resultado una tensión sencilla.
Es necesario investigar (averiguar) las condiciones de las tensiones para las que
el material resulte más débil, por ejemplo: el fierro fundido es menos resistente a
la tracción que a la compresión.
Fig. 2.12
En (a) se observa un elemento que ha sido separado de un cuerpo, bajo un
estado de tensiones. σ x, σ y
Tensiones de tracción τ xy
Tensiones cortantes
El elemento puede cortarse por cualquier plano mn cuya normal forme un ángulo
Ø con el eje x.
En (b) se reemplaza tensión normal σ actuando perpendicular al plano mn y
una tensión de corte τ .
Por resistencia de materiales se deduce:
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO II: ESFUERZOS SIMPLES EM ELEMENTOS SENCILLOS DE MÁQUINAS
σ = σx +σ y +σx −σ y cos2φ −τ xy − sen2φ (Ec. 2.27)
2 2
τ = σx −σ y sen2φ +τ xy cos2φ (Ec. 2.28)
2
Cuando se varía 2ø entre 0 y 360° se encuentran dos valores en que la tensión
será un máximo o un mínimo.
tg2φ = − 2τ xy (Ec. 2.29)
σx −σy
Hay otros dos valores de 2Ø en los que la tensión de corte τ es máxima.
tg2φ = − σ x −σ y (Ec. 2.30)
2τ xy
Las dos tensiones σ 1 y σ 2 se denominan tensiones principales en planos
principales
σ1,σ 2 = σx +σ y ± σ x −σ y 2 +τ 2 (Ec. 2.31)
2 2 xy
Tensión máxima de corte τ máx = ± σ x −σ y 2 +τ 2 (Ec. 2.32)
2 xy
Estas expresiones se pueden expresar gráficamente mediante el Círculo de
Mohr.
Fig. 2.13
2.9.1 TENSIONES DE TORSIÓN
Cuando una barra cilíndrica está sometida a la acción de un par T.
Las tensiones cortantes varían linealmente desde cero en el centro hasta un
máximo en la periferia.
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO II: ESFUERZOS SIMPLES EM ELEMENTOS SENCILLOS DE MÁQUINAS
La tensión en la superficie es: τ = Tr
J
τ = Tensión de corte, kg/cm2
T = Par, kg-cm
r = Radio de la barra, cm
J = Momento polar de inercia, cm4
Para una barra maciza Para una barra hueca
J = π .d 4
( )J = π de 4 − di 4
32 32
T = 71 ,700 CV
n
CV = 2πnT = F.V CV = potencia en caballos de vapor
450,000 75 T = par kg-cm
N = velocidad del eje rpm
T = 63,000HP lb − pu lg
n
2.9.2 TENSIONES DE FLEXIÓN
Fig. 2.14
La viga de la figura puede representar un eje en rotación con cojinetes en R1 y
R2 y estar sometida a las cargas F1, F2 y F3 que pueden ser causadas por algún
engranaje, polea o un elemento similar.
V = Esfuerzo cortante V = dm
M = Momento flector dx
Ec. 2.33
Cuando la carga esta uniformemente repartida es útil la siguiente relación:
dv = d 2M = −w Carga repartida
dx dx2
2.9.3 TENSIONES NORMALES
Fig. 2.15
σ = ± Mc = ± M M = Momento flector de la sección en estudio kg-cm
I = Momento de inercia cm4
I I / c I/c = Momento resistente ó modulo de la sección
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO II: ESFUERZOS SIMPLES EM ELEMENTOS SENCILLOS DE MÁQUINAS
Para una sección circular el momento resistente es πd 2 (Ec. 2.34)
32
2.9.4 TENSIONES DE CORTE
Cuando varía el momento flector a lo largo de la viga, se originan en ella unas
tensiones de corte cuyo valor es función de la ley de variación de momento.
Esta tensión viene dada por: τ = V ∫C ydA el máximo corte se alcanza
Ib YO
cuando yo = 0
Para una viga de sección rectangular
τ máx = 3V
2A
Para una viga sección circular el valor máx.
aprox. τ máx = 4V
3A
Para una sección circular hueca
Fig. 2.16 τ máx= 2V
A
2.9.5 SUPERPOSICIÓN
Cuando sobre un cuerpo actúan fuerzas que producen dos clases de tensiones,
en la misma dirección, es posible calcular independientemente las tensiones y
sumarlas después, teniendo en cuenta sus signos respectivos.
Puede utilizarse el método siempre que las cargas sean proporcionales a las
tensiones que ellas originan.
2.9.6 DEFORMACIÓN UNITARIA
Se llama deformación unitaria o simplemente deformación al alargamiento por
una unidad de longitud de la barra.
ε =δλ Ɛ = Elaboración unitaria
δ = Alargamiento total (cm)
ℓ = Longitud de la barra (cm)
Deformación de corte o cizallamiento
γ = rθ
λ
θ = Desplazamiento angular de dos secciones rectas adyacentes de una
barra circular uniforme sometida a torsión.
λ = Distancia entre las 2 secciones cm.
r = Radio de la barra cm.
γ = Deformación cortante o de cizallamiento.
∗ La elasticidad es la propiedad a algunos materiales que permiten recuperar
su forma y dimensiones originales cuando desaparece la carga.
La ley de Hooke establece que, dentro de ciertos límites, la tensión en un
material es proporcional a la deformación que origina.
∗ La condición de que la tensión sea proporcional a la deformación puede
escribirse.
σ = E.ε E y G = Constantes de proporcionalidad
E = Módulo de elasticidad kg/cm2
τ = Gγ G = Módulo elástico de cizallamiento o módulo de rigidez
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO II: ESFUERZOS SIMPLES EM ELEMENTOS SENCILLOS DE MÁQUINAS
δ = Fλ τ = Tr γ = rθ θ = Tλ
AE J λ GJ
∗ Experimentalmente, Poisson demostró que cuando un material se somete a
un esfuerzo de tracción, no solo existe una deformación axial, sino también
una deformación lateral. Estas deformaciones son proporcionales entre sí,
dentro del dominio de la ley de Hooke.
µ = deformación lateral
deformación axial
µ = Coeficiente de Poisson
Las 3 contantes elásticas se relacionan E = 2G(1+ µ ) entre sí.
2.9.7 ANALISIS DE DEFORMACIONES
Anteriormente se ha analizado las tensiones ya que los elementos de las
máquinas deben dimensionarse de forma que las tensiones nunca excedan a la
resistencia del material. Pero las piezas deben proyectarse para que sean lo
bastante rígidas como para que no aparezcan excesivas deformaciones cuando
empiecen a funcionar.
2.9.8 DEFORMACIÓN DE VIGAS 1 = M = d2y (Ec. 2.35)
Por resistencia de materiales ρ EI dx 2
ρ = Radio de curvatura de una viga deformada por un momento M.
Fig. 2.17
y = deformación o flecha d3y = V
dx3 EI
dy = θ Pendiente o inclinación Esfuerzo Cortante
dx
d2y = M Momento d2y = W Carga
dx2 EI dx4 EI
2.9.9 ENERGÍA DE DEFORMACIÓN EN LA TRACCIÓN Y COMPRESIÓN
Un elemento en movimiento tiene una energía cinética, si existe un cambio en el
movimiento del cuerpo equivale un cambio en el contenido de su energía
cinética.
Considerando que no existe la rigidez absoluta, las cargas dinámicas
(producidas por engranajes, levas, volantes, etc. se transfieren a la estructura
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO II: ESFUERZOS SIMPLES EM ELEMENTOS SENCILLOS DE MÁQUINAS
como reacciones en los cojinetes, muelles y otros puntos de conexión)
representan una absorción de energía.
De acuerdo con el principio de conservación de la energía, el trabajo externo
realizado sobre un cuerpo o estructura se almacenará en su interior como
energía de deformación.
De la figura: Puesto que la barra es deformada por F, el trabajo realizado se
transforma en energía potencial de deformación.
El trabajo realizado es igual a la energía potencial de deformación almacenada
en la barra, igual al área del triángulo OAB.
Puesto que δ = Fλ y σ = F
AE A
La energía de deformación es
∗ Por tanto la capacidad de absorción de energía depende del volumen del
material ( λA) y del módulo de elasticidad.
∗ Para que un elemento pueda absorber mayor energía tiene que ser de
longitud grande y de un módulo de elasticidad bajo.
∗ La mayor cantidad de absorción de energía ocurrirá en las zonas donde σ
sea elevado.
σ elevado – zonas de concentración de tensiones.
∗ Por esta razón, se proyectan piezas que tengan una distribución uniforme de
tensiones en toda su longitud, con objeto de absorber la máxima cantidad de
energía.
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO II: ESFUERZOS SIMPLES EM ELEMENTOS SENCILLOS DE MÁQUINAS
EjemploN°2.16 :
La figura muestra 2 diseños de pernos: Los pernos permiten una tensión de
hasta 2800 kg/cm2.
Calcular la energía que puede absorber cada perno con seguridad, despreciando
el efecto de las roscas.
A1 = πD2 =π 2.162 = 3.67cm2 A2 = π x2.54 2 = 5.08cm 2
4 4 4
La tensión máxima permisible σ = 2800kg / cm2 aplicada en A1
A1σ 1 = σ 2 A2 σ2 = σ1x A1 = 2800x 3.67 = 2,030kg / cm2
A 5.08
2
La energía que puede absorber el perno (a):
Ua = σ 2 A1λ1 + σ 2 A2 λ2 = 2,8002 x3.67x5 + 2,0302 x5.08x35 = 209kg − cm
1 2 2x2.1x106 2x2 ⋅1x106
2E 2E
En el perno (b):
Ub = σ 2 A1λ1 = 28002 x3.67x40 274kg − cm
1 2 2.1x104
2E
Comparando Ua y Ub se puede observar que la reducción del área de la espiga
permite un incremento de la carga F sin incrementar σ .
∗ Para el caso de cizallamiento o cortadura la energía está dada:
U = τ 2Aλ Para un elemento redondo
4G
2.10.-TEORÍA DE LA TENSIÓN NORMAL MÁXIMA
∗ Solo tiene importancia con objeto de hacer comparaciones ya que sus
resultados pueden ser faltos de seguridad.
∗ Esta teoría establece que el fallo se verifica cuando la tensión principal mayor
es igual al límite de fluencia o al de rotura del material.
σ 1 = S y ó σ1 = Su (según convenga aplicar)
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO II: ESFUERZOS SIMPLES EM ELEMENTOS SENCILLOS DE MÁQUINAS
Gráfico: Diagrama de Mohr para
ensayo de tracción simple.
Fig. 2.21 Diagrama de Torsión Pura.
τ = σ 1 y por consiguiente el fallo debe
suceder cuando la tensión de corte
llegue a ser igual a la resistencia a la
tracción o a la compresión.
El diagrama muestra esta teoría:
Se ha supuesto que el límite de
fluencia es igual a tracción que a
compresión.
La teoría determina que el fallo se
verificará para cualquier punto cuyas
coordenadas σ 1 y σ 2 caigan sobre
o fuera del diagrama.
Según esta teoría:
∗ Los puntos que están en el interior de la figura y en el primero y tercer
cuadrante están en la zona de seguridad, mientras que los puntos del
segundo y cuarto cuadrante pueden estar en la zona de falta de seguridad.
2.10.1 TEORÍA DE LA TENSIÓN DE CORTE MÁXIMA
∗ Esta teoría es fácil de utilizar.
∗ Siempre está en la zona de seguridad.
∗ Esta teoría establece que la fluencia empieza cuando la tensión de corte
máxima iguala a la tensión de corte correspondiente al límite de fluencia en el
ensayo de tracción simple.
La fluencia empezará τ máx = Sy
2
∗ Para un estado triaxial de tensiones las tensiones de corte máximas son:
τ = σ1 −σ2 τ = σ2 −σ3 ó τ = σ1 −σ3
2 2 2
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Fig. 2.22 Por tanto la fluencia empieza cuando
la mayor de estas tres tensiones de
corte llegue a ser igual a la mitad del
límite de fluencia a la tracción.
Esta teoría el límite de fluencia por
cizalladura de un material es a lo
mucho la mitad del límite de fluencia
a la tracción.
2.10.2 TEORÍA DE VON MISES-HENCKY (O DE LA ENERGÍA DE
DISTORSIÓN O DE LA ENERGÍA DE CIZALLADURA)
∗ Es un poco más difícil que la teoría anterior.
∗ Es la más adecuada para los materiales dúctiles.
∗ Surgió como consecuencia de que la fluencia no es en absoluto un simple
fenómeno de tracción o compresión, sino que más bien de alguna manera se
relacionaba con la deformación angular del elemento.
σ = σ x2 −σ xσ y +σ y2 + 3σ xy2
Sy2 = σ12 −σ1σ 2 + σ 2
2
σ 1 , σ 2 Tensiones principales
Para torsión pura
σ 2 = −σ 1 , τ = σ 1 Ssy = 0.577Sy
Fig. 2.23
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EjemploN°2.17 :
Las tensiones en un punto de un cuerpo son σ x = 910kg / cm 2 , σ y = 210kg / cm 2
y τ xy = 840kg / cm2 donde el material tiene un S y = 2,800kg / cm2 hallar:
a) Coeficiente de seguridad por la teoría de corte máximo.
b) Coeficiente de seguridad por la teoría de distorsión.
Solución:
a) Ssy = Sy = 2800 = 1400kg / cm2 b) − S = 14702 −1470x(−350) + 3502
2 2
S y = 1672.69kg / cm2
Ssy = 1400
Cs = τ máx 910 = 1.54kg / cm2 Cs = 2800 = 1.6739
1672.69
54%
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO II: ESFUERZOS SIMPLES EM ELEMENTOS SENCILLOS DE MÁQUINAS
EjemploN°2.18 :
Un eje de 5 cm de diámetro está cargado estáticamente por torsión pura con una
torsión cortante de 700 kg/cm2. Encontrar el coeficiente de seguridad si el
material es acero laminado en caliente 4140. Emplear la teoría de Mises –
Hencky.
σ máx = σ1 = −σ 2 = 700kg / cm2
S = 7002 −700(−700) +(700)2
A = 1212 kg / cm 2
Límite de fluencia S y = 4400 CS = 4400 = 3.63
1212
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO III: DISEÑO DE LEMENTOS POR CONCENTRACIÓN DE ESFUERZOS, CARGAS CÍCLICAS Y FATIGA
3
DISEÑO DE
ELEMENTOS POR
CONCENTRACIÓN
DE ESFUERZOS,
CARGAS CÍCLICAS Y FATIGA
“Hay dos maneras de difundir la luz...
ser la lámpara que la emite, o el espejo que la refleja.”
Filosofía china
3.0.- INTRODUCCIÓN
El concepto de concentración de esfuerzos, se refiere al estado macroscópico de
esfuerzos, y tiene un significado único para problemas en el plano que involucran la
definición de esfuerzo promedio. Entonces si se barrena un agujero en una placa
sometida a tensión, el esfuerzo presente en el elemento es constante siempre y cuando
se mida a una distancia apreciable del agujero, pero el esfuerzo tangencial en el borde
del agujero se vería incrementando considerablemente.
En ingeniería y, en especial, en ciencia de los materiales, la fatiga de
materiales se refiere a un fenómeno por el cual la rotura de los materiales bajo
cargas dinámicas cíclicas se produce más fácilmente que con cargas estáticas.
Aunque es un fenómeno que, sin definición formal, era reconocido desde la
antigüedad, este comportamiento no fue de interés real hasta la Revolución
Industrial, cuando, a mediados del siglo XIX comenzaron a producir las fuerzas
necesarias para provocar la rotura con cargas dinámicas son muy inferiores a las
necesarias en el caso estático; y a desarrollar métodos de cálculo para el diseño
de piezas confiables. Este no es el caso de materiales de aparición reciente,
para los que es necesaria la fabricación y el ensayo de prototipos
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO III: DISEÑO DE LEMENTOS POR CONCENTRACIÓN DE ESFUERZOS, CARGAS CÍCLICAS Y FATIGA
3.1.-CONCENTRACIÓN DE TENSIONES
Es difícil proyectar una máquina sin permitir algunos cambios en las secciones
rectas de los elementos. (Los ejes giratorios tienen reborde para que el cojinete
asiente adecuadamente y admita carga axial, así mismo llevan chaveteros. Un
perno tiene cambio de sección en la cabeza y en la rosca).
Cuando hay variaciones en las secciones de un elemento existen zonas de
concentración que se denominan acumuladores de tensión.
Para proyectar debemos tener en cuenta un “coeficiente de concentración de
tensiones”
K t= Tensión Valor de la tensión máxima real en la discontinuidad transversal
nominal (dada por las ecuaciones elementales de tensión para la sección
mínima)
Kt = σmáx (Ec. 3.1)
σo
Para tensión: Kt = σ máx (a) Para el corte: Kts = τ máx (b)
σo τo
Los valores de Kt y Kts dependen de la geometría de la pieza.
∗ Uso de tablas:
Fig. 3.0
Si: d = 3, D = 5, r = 0.6
r = 0.6 = 0.2 D = 5 = 1.66 σ máx = Kt P
d 3 d 3 A
Tablas Kt = 1.75
σ máx = 1.75σ o
σ máx = 1.75 P
A
Kt = Factor teórico de concentración de esfuerzos
Se determina experimentalmente con procedimiento foto elásticos.
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO III: DISEÑO DE LEMENTOS POR CONCENTRACIÓN DE ESFUERZOS, CARGAS CÍCLICAS Y FATIGA
.
EjemploN°3.1 : Barra sometida a tensión con discontinuidades en su sección
transversal.
∗ Utilizar los diagramas de R.E. Peterson, “Design Factors for Stress Concentration”.
Machine Design. 1953.
Datos: a = 76 mm r = 3.175
b = 100 mm P = 3600 kg
t = 13
¿Cuál es el esfuerzo máximo en el filete?
r = 3.175 = 0.042 D = 100 = 1.32
d 76 d 76
Por tablas kt=2.60
Esfuerzo Nominal: σ = P = 3600 kg = 364.37kg / cm 2
A 7.6x1.3cm 2
Esfuerzo real máximo: σ máx = kt σ = 2.60x364.37 = 947kg / cm2
Si Ø=13 calcular el esfuerzo en A-A
σ nom = 3600 = 318.30kg / cm2
(10.0 − 1.3)x1.3
Para ingresar a la tabla debemos calcular a = 13 = 0.13 → kt = 2.65
w 100
σ máx= 2.65x318.30kg / cm2 = 843.50kg / cm2
EjemploN°3.2 :
¿Qué carga constante P puede colocarse en la barra dibujada, sin exceder la
resistencia de cedencia del material en la muesca? El material es SAE 1050 laminado
en caliente. Sy=49,500 psi (por tablas en Apéndice).
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO III: DISEÑO DE LEMENTOS POR CONCENTRACIÓN DE ESFUERZOS, CARGAS CÍCLICAS Y FATIGA
Calculamos:
r = 118 = 0.0625 kt = 2.45
d2
D = 2.5 = 1.25
d2
σ máx= S y = 49,500 lb = Ktσ σnom = 49,500 = 20,504 lb
pu lg2 2.45 pu lg2
σnom = MC = 2P.1" = 3P
I
1"(2.5 − 0.5)3
12
3P = 20,204
P = 6,734.67 lb / pu lg2
3.2.-FATIGA
∗ Es la reducción de la resistencia de un material debido a que sobre él actúan
cargas fluctuantes (o cíclicas).
∗ Los elementos pueden fallar por acción de tensiones alternativas, aún sin
llegar a valores críticos para esfuerzos estáticos, incluso a muy inferiores al
límite de fluencia.
∗ La falla por fatiga empieza por una pequeña grieta, que se desarrolla por un
cambio de sección, un chavetero, un orificio, en las marcas de fábrica e
incluso irregularidades originadas por la mecanización.
∗ La grieta va aumentando progresivamente hasta que llega un momento en
que el área o sección neta de trabajo es tan pequeña que la pieza se rompe
repentinamente.
Su = limite de rotura. S ′n < Sy < Su
Sy = limite de fluencia.
S ′n = limite de fatiga.
Para el acero:
S ′n = 0.5Su (Ec. 3.2) sí Su < 14,000 kg / cm2
Pero si: S′n = 7000kg / cm2 entonces Su ≥ 14,000 kg / cm2
Para el Hierro fundido y bronce:
S′n = 0.4Su (Ec. 3.3)
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Para casi todos los aceros se puede determinar gráficamente el Diagrama de
Fatiga o Diagrama S-N, normalmente para estos materiales ocurre que para
cada:
103 ciclos → S = 0.9Su (a)
106 ciclos → S ′n = 0.5Su (b)
NO SE ROMPEN VIDA INFINITA
(NÚMERO DE CICLOS)
Fig.3.1 Diagrama S-N
Supóngase que estamos probando una viga giratoria que fue cargada de tal manera
que el esfuerzo está muy próximo al punto de cedencia (Su) del material, es decir, que
con un esfuerzo de esta magnitud bastarían relativamente pocos ciclos para causar la
falla. Pero si otra probeta se probara con un esfuerzo más bajo, veríamos que el número
de ciclos necesario para romper la probeta aumentaría considerablemente.
EjemploN°3.3 : Representar el diagrama S-N de un acero AISI C 1035, laminado en
caliente del cual se ha hecho una probeta tipo viga giratoria, y encontrar la resistencia a
la fatiga correspondiente a una vida de 82,000 ciclos.
Solución:
Según la tabla de propiedades el acero AISI C1035, tiene:
Sy=3,800 kg/cm2 y Su=6000 kg/cm2
Por tanto:
S′n = 0.5× Sut = 0.5× 6000 = 3000kg / cm2
S = 0.9 × Sut = 0.9x6000 = 5400kg / cm2
La resistencia a la fatiga para 82000c se obtiene a través de la siguiente
relación:
S − S ′n = S − Sx
log106 − log103
log(82000)− log103
5400 − 3000 = 5400 − Sx
6 − 3 4.914 − 3
2400 = 5400 − Sx
3 1.914
1.914(2400) = 5400 − Sx
3
Sx = 5400 −1531.05 = 3868.95kg / cm2
x = log(3868.95) = 3,5876
EjemploN°3.4 : Un acero AISI 1045 tiene una resistencia a la tensión de 95 Kpsi y una
resistencia de fluencia 74 Kpsi.
a) Determinar el límite de fatiga de la viga giratoria.
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO III: DISEÑO DE LEMENTOS POR CONCENTRACIÓN DE ESFUERZOS, CARGAS CÍCLICAS Y FATIGA
b) La resistencia a la fatiga correspondiente a 104 ciclos de duración.
c) Estimar la duración correspondiente a un esfuerzo completamente invertido de
55 kpsi.
SOLUCIÓN:
a) Para aceros
S′n = 0.5× Sut = 0.5× 95 = 47.5kpsi <
S = 0.9× Sut = 0.9x95 = 85.5kpsi
b) Para 104 ciclos la resistencia a la fatiga es:
S − S ′n = S − Sx
log106 − log103 log104 − log103
85.5 − 47.5 = 85.5 − Sx
6− 3 4−3
Sx = 72.833kpsi
c) Para un esfuerzo de 55 Kpsi
S − S ′n = S − Sx
log106 − log103 log N − log103
85.5 − 47.5 = 85.5 − 55
6 − 3 log N − 3
12.6667(log N − 3) = 30.5
log N = 2.40789 + 3 = 5.40789
N = 105.40789 = 255796.582ciclos
3.2.1 COEFICIENTES MODIFICATIVOS DEL LÍMITE DE FATIGA
El límite de fatiga de una pieza puede ser muy diferente al encontrado por el
ensayo de R.R.Moore. Debido a que la pieza no tenga la superficie pulida, que
tenga puntos de concentración de tensiones, o que opera a alta temperatura.
Por esto se ha sugerido emplear “coeficientes modificativos”, todos los
coeficientes modificativos son menores que 1, donde mi nuevo Sn estará
limitado por:
Sn=ka.kb.kc.kd.ke.kg.S`n (Ec. 3.4)
En donde:
Sn = Límite de fatiga conseguido ( Kg/cm2).
S´n = Límite de fatiga de la probeta.
ka = coeficiente modificativo de superficie.
kb = coeficiente modificativo de tamaño.
kc = coeficiente modificativo de confianza.
kd = coeficiente modificativo de temperatura.
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ke = coeficiente modificativo por concentración de tensiones (no es el mismo que
el coeficiente kt de concentración de tensiones).
Kf = coeficiente modificativo por efectos diversos.
a) ACABADO SUPERFICIAL (Ka): Tiene un efecto muy significativo sobre el
límite de fatiga.
Tabla 3.1: Factor de Corrección Ka de acabado superficial
b) EFECTOS DE TAMAÑO(Kb): El ensayo de la viga rotativa proporciona el
límite de fatiga para una probeta de 0.3” Ø
Para probetas de mayor tamaño se ha encontrado que el límite de fatiga es de
un 10 a 15% menor.
∗ Por tanto para flexión y torsión el coeficiente de tamaño es de kb=0.85.
∗ Para cargar axiales kb=1.
c) COEFICIENTE DE CONFIANZA O SEGURIDAD FUNCIONAL(Kc):
Stilen, Cummings y Schulte; establecieron que la distribución de la relación de
las resistencias a la fatiga era normal para un número fijo de ciclos.
Factor de confianza:
Kc = 1 - 0.08 D (Ec. 3.5)
Relación de supervivencia por 100 Factor de multiplicación de la desviación
D
DISEÑO DE ELEMENTOS DE MÁQUINAS I
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90 1.3
95 1.6
99 2.3
99.9 3.1
99.99 3.7
d) EFECTOS DE TEMPERATURA(Kd) :
Piezas que trabajan a temperaturas elevadas pueden fallar por “Creep” o
fluencia o por fatiga o por una combinación de ambas (o debido a una corrosión)
conocido como termofluencia.
kd = 620 (la T ° en grados Farentheit) para T > 160°F
460 + T (Ec. 3.6)
para T ≤ 160° → kd = 1
e) SENSIBILIDAD A LA ENTALLA(Ke)
Un fallo por fatiga casi siempre se origina en una discontinuidad, la grieta
empieza en una entalla, un resalte o en el borde de un orificio puede también
iniciarse en una huella de herramienta o una raya.
Hay materiales que son mucho más sensibles a la entalla que otros.
f) COEFICIENTE DE CONCENTRACIÓN DE TENSIONES EN LA
FATIGA(Kf)
Kf = límite de fluencia de probetas exentas de entalla (Ec. 3.7)
límite de fluencia de probetas entalladas
El coeficiente modificativo de concentración de tensiones Ke está relacionado
con Kf
ke = 1 (Ec. 3.8)
kf (Ec. 3.9)
La sensibilidad a la entalla q:
q = k f −1
kt − 1
Primero se halla Kt a partir de la geometría de la pieza, y luego el q:
k f = 1+ q(kt −1) (Ec. 3.10)
g) EFECTOS VARIOS(Kg)
1) Tensiones residuales: Por tratamientos térmicos o trabaja en frío. si la
tensión residual superficial es de compresión el límite de fatiga mejora (por
ejemplo endurecimiento superficial mediante perdigones, el martillado y el
laminado en frío).
2) Características direccionales del material: Las piezas laminadas, forjadas o
estiradas presentan un 10 a 20 % de reducción del límite de fatiga en
dirección transversal (que a lo largo de la dirección longitudinal).
3) Defectos internos: Inclusiones de escoria u óxidos, partículas extrañas.
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO III: DISEÑO DE LEMENTOS POR CONCENTRACIÓN DE ESFUERZOS, CARGAS CÍCLICAS Y FATIGA
4) Cementado: Pueden fallar en la capa exterior o en el núcleo.
5) Corrosión: Se debe al picado que produce la corrosión y el someter a la
pieza a tensiones aumenta la corrosión.
6) Metalizado: Como el cromado, niquelado y cadmiado reducen el límite de
fatiga hasta en un 35%.
EjemploN°3.5 :
El eje que se muestra a continuación tiene movimiento rotacional y está apoyado en
cojinetes de bolas en A y D, los radios de empalme tienen 3 mm de radio y el acero es
un AISI 1050 estirado en frío y tiene un acabado a máquina.
Se desea evaluar la duración de este elemento.
Características del material Su = 690 Mpa
Sy = 580 Mpa
Solución:
RA+RD = 6800 N
RA . 550 – 68000 x 225 = 0
RA = 2781.82 N
RD = 4018.18 N
Resistencia a la fatiga: S’n = 0.5 x 690 = 345 MPa
Por tabla (3.1) el coeficiente de superficie para un Su
= 690 MPa, es Ka=0.75
Coeficiente de tamaño Kb = 0.85 (para flexión y
torsión)
Factor de concentración de esfuerzo
D = 38 = 1.19 Kt =1.6
d 32
r = 3 = 0.094
d 32
Sensibilidad a la entalla q = 0.84
q = Kf −1
Kt −1
Kf = q(Kt −1)+1 = 0.84(1.6 −1) +1 = 1.504
Ke = 1 = 1 = 0.665
Kf 1.054
Por tanto el esfuerzo (de cálculo) para fatiga es:
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO III: DISEÑO DE LEMENTOS POR CONCENTRACIÓN DE ESFUERZOS, CARGAS CÍCLICAS Y FATIGA
Sn = 0.75x0.85x0.665x645=146.24MPa
El máximo esfuerzo está en el punto B (que es el que tiene mayor momento):
MB = 695.46 N.m.
El módulo de la sección: Z = I = π (3.2)4 = 3.217cm3
c
64 3.2
2
El esfuerzo en la sección B:
σ = 695,46N − m = N
3.217cm3 21618.278 cm2
σ = 216.2 MPa
Este esfuerzo es mayor que el límite de
fatiga σ = 216 > 146.24 = Sn por lo
tanto el elemento tiene una vida finita.
621− 146 = 621 − 216
log106 − log103 log N − log103
log N = 5.55789
N = 105.55789 = 361322.27ciclos
EjemploN°3.6 :
Imaginemos que la barra soporta tanto por la parte superior como por la parte inferior de
tal manera que la carga P pueda ser
invertida por completo. Encuéntrese el
valor numérico de la carga
completamente invertida que someterá
la barra en la muesca hasta el límite de
duración.
Material SAE 1050: Sy = 49,500 psi
Su = 90,000 psi
Solución: kt = 2.45 ; q = 0.9
Según gráfica:
k f = 1 + q(kt −1)
k f = 1 + 0.9(2.45 −1) = 2.305
S = MC = 2Px1" 3P
I 1(2.5 − 0.5)3
12
k f .σ = 2.305x3P = 6.92P
6.92P = 45,000
P = 6508lb
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO III: DISEÑO DE LEMENTOS POR CONCENTRACIÓN DE ESFUERZOS, CARGAS CÍCLICAS Y FATIGA
3.3.-TENSIONES FLUCTUANTES σmin = Tensión mínima
σmáx = Tensión máxima
(a)Esfuerzo cíclico típico o σ a = Amplitud de la tensión
Tensión Fluctuante
o Esfuerso Variable
σm = Tensión mediao promedio
σr = Recorrido de la tensión
o rango de esfuerzo
σ s = Tensiónestática(la dela
estructura)
El número de repeticiones
para originar la falla,
(b)Esfuerzo o Tensión completamente depende del rango de
invertido
esfuerzo (σ r )y que el
rango necesario de esfuerzo
para originar falla a un
número de repeticiones
dado, decrece a medida que
el esfuerzo promedio (σm )
(c) Fluctuación senoidal de aumenta.
esfuerzo o Tensión repetida
La tensión estática (σ s ) es
Fig. 3.1: diagrama de esfuerzo cíclico debida a una carga fija
previa a la pieza y
normalmente es independientemente de la parte variable de la carga.
σr = σ máx −σ mín Ec. 3.11 σm = σ máx + σ min Ec. 3.12 σa = σ máx − σ min
2 2
Ec 3.13
3.4.-RESISTENCIA A LA FATIGA BAJO TENSIONES FLUCTUANTES
Hasta aquí solo hemos estudiado la manera de encontrar la magnitud de un
esfuerzo completamente invertido que un material puede aguantar de manera
indefinida. Esto lo representamos por la onda senoidal que se muestra en la Fig.
3.1b en la que el esfuerzo promedio σm =0. La mayor parte del tiempo, una
situación de esfuerzo se asemeja a la que se describe en la Fig.3.1c en la que
σm ≠0. Se emplean generalmente dos métodos:
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∗ Diagrama modificado de Goodman: Donde muestra todos los componentes.
∗ Cuando la tensión media es de compresión, el fallo se define por dos líneas
paralelas gruesas
∗ Cuando la tensión media es de tracción al fallo se define por la tensión
máxima o por el límite de fluencia.
Esfuerzo que fluctúa para materiales dúctiles analizados por Gerber, Goodman y
Soderberg:
Si trazamos la componente del esfuerzo variable (σ a ) en el eje de las “y” y el
esfuerzo promedio (σm ) en el eje de las “x”. Obtenemos las siguientes
relaciones:
DISEÑO DE ELEMENTOS DE MÁQUINAS I
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Parabola de Gerber : σa = S n − σm 2 Ec. 3.14
1 Su Ec. 3.15
Ec. 3.16
Linea de Goodman : σa = Sn 1 − σm
Su
Linea segura de Goodman : σa = S n 1 − σm
F.S Su
Linea de Soderberg : σa = Sn 1 − σm
Sy
Linea segura de Soderberg : σa = Sn 1 − σm
F.S Sy
A flexión completamente invertida el esfuerzo promedio es cero y el esfuerzo
variable σ a =Sn, lo que concuerda con la representación en la Fig. 3.1b y por
otro lado, el esfuerzo promedio es la resistencia a la tensión, σ a =0 y tenemos la
condición de un carga aplicada sólo una vez para originar la falla.
De las relaciones que obtuvimos podemos observar en la Fig.3.4 que la relación
de Soderberg es más segura que la de Goodman y esta a su vez es más segura
que la de Gerberg. En un sentido más conservador y para estar seguros de la
certeza de los valores, en la línea de Goodman y de Soderber, tanto al Su como
al Sy pueden dividirse por un factor arbitrario de seguridad (F.S), Su y Sy
FS FS
respectivamente con lo cual nos dará una relación más segura de la línea de
Goodman y de Soderberg, obviamente el σ a calculado es menor que la
relación sin el factor de seguridad.
EjemploN°3.7 : Una parte de una máquina tiene un esfuerzo debido a flexión que fluctúa
entre un esfuerzo de tensión de 40,000 lb/pulg2 y un esfuerzo de compresión de 20,000
lb/pulg2 ¿Cuál será la resistencia a la tensión mínima del acero que podría soportar
estas fluctuaciones indefinidamente?
Aplique GERBER y GOODMAN
σ r = σ máx − σ min = 40,000 − (− 20,000) = 60,000 lb / pu lg2
σa = σ máx − σ min = 60,000 = 30,000 lb / pu lg2
2 2
σm = σ máx + σ min = 20,000 = 10,000 lbpu lg2
2 2
S’n=0.5 Su (para los aceros)
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO III: DISEÑO DE LEMENTOS POR CONCENTRACIÓN DE ESFUERZOS, CARGAS CÍCLICAS Y FATIGA
Según Gerber Según Goodman
σ m 2 σa = S 'n1 − σm
S'n1 − S u Su
σa =
1 − σm
30,000 = 0.5Su − σ 2 σa = Su Su
1 m 2
Su2 Su σm
σa 2 2
σ 2 = −
m
60,000 = Su −
Su Su = 2σ a + σ m
S u 2 −60,000Su − (10,000)2 Su = 2(30,000) + 10,000
Su = 60,000 ± 36x108 + 4x108 Su = 70,000lb / pu lg2
2
Su = 60,000 ± 63,246 = 61,623lb / pu lg2
2
Aplicando Goodman con factor de seguridad 2:
σ a = S ' n 1 − σm
F.S Su
σa = Su 1 − σm
2 2 Su
σa = Su − σm
4 2
Su = 4σ a + 2σ m = 4(30,000) + 2(10,000) = 140,000lb / pu lg2
Una forma más practica es : Su = F.S(Su) = 2(70,000) = 140,000lb / pu lg2
EjemploN°3.8 : El mismo problema anterior, con un factor de seguridad 2 aplicando
Soderberg. (Importante: Se sabe que Sy varía entre 0.55% - 0.95% de Su.)
σa = 30,000 lb σm = 10,000 lb σa = Sn 1 − σm
pu lg2 pu lg2 F.S Sy
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO III: DISEÑO DE LEMENTOS POR CONCENTRACIÓN DE ESFUERZOS, CARGAS CÍCLICAS Y FATIGA
Para S y = 0.55Su Para S y = 0.95Su
0.5Su 1 − 10,000 = 30,000 0.5Su 1 − 10,000 = 30,000
2 0.55Su 2 0.95Su
Su − 18,182 − Su = 60,000 Su − 10,526 = 60,000
2 Su 2
Su = 156,364 lb Su = 141,052 lb
pu lg2 pu lg2
EjemploN°3.9 : Una flecha de diámetro de 2” hecha de acero al carbono endurecida
hasta 200 Brinell se sujeta a una torsión que fluctúa entre 24,000 lb–pulg y -6,000 lb-
pulg ¿Cuál es el factor de seguridad por el método de Soderberg?
τ máx = T .r = 24,000x1 = 15,279 lb
J π 24 pu lg2
32
τ min = − 6,000x1 = −3820 lb
π 24 pu lg2
32
τa = τ máx − τ min = 15,279 + 3820 = 9549.5 lb
2 2 pu lg2
τm = τ máx + τ min = 15,279 − 3820 = 5729.5 lb
2 2 pu lg2
Importante: El límite de fatiga a torsión cíclica es aproximadamente la mitad del límite
de duración a flexión.
• Para aceros:
Sns = 0.5Sn = 0.25Su
S ys = 0.5S y
• Para metales y aleaciones no ferrosas:
Sns = 0.2Su
• Para fierro fundido:
Sns = 0.8Su
• Por tablas: para acero al carbono de 200 Brinel:
Su = 100,000 psi
Sy = 55,000 psi
τa = Sns 1 − τm
F.S S ys
9549.5 = 0.25Su 1 − 5729.5
FS 0.5S y
9549.5 = 25,000 1 − 5729.5
FS 27,500
1 = 9,549.5 + 5,729.5 = 0.382+ 0.208 = 0.59
FS 25,000 27,500
.FS =1.69
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO III: DISEÑO DE LEMENTOS POR CONCENTRACIÓN DE ESFUERZOS, CARGAS CÍCLICAS Y FATIGA
EjemploN°3.10 : Una barra redonda de acero AISI C1018, estirada en frío, se proyecta
para resistir una carga previa de tracción de 3600 kg y una carga fluctuante de tracción
que varía de 0 a7200 kg. Debido al proyecto de sus extremos la barra tienen un
coeficiente geométrico de concentración de tensiones de 2.10, que corresponde a un
acuerdo de 3,20mm ¿Cuál será el radio de la barra, si el margen de seguridad nunca
deberá ser menor de 100 por 100? La barra ha de proyectarse para una vida infinita.
SOLUCIÓN:
Según tablas el material tienen las siguientes propiedades mecánicas.
Sy = 4900 kg/cm2 Su = 5750 kg/cm2
Así pues el límite de fatiga será: S’n = 0.5xSu=2875 kg/cm2
El valor del coeficiente de superficie ka = 0.76
El valor del coeficiente de tamaño para cargas axiales kb = 1
El valor de la sensibilidad a la entalla q = 0.8
El valor de kf = 1+q(kt-1) = 1+0.8 (2.10 – 1) = 1.88
Dando como resultado el valor por concentración de tensiones
ke = 1 = 1 = 0.531
kf 1.88
El valor del límite de fatiga será Sn = 1.76x1x0.531x2875=1,165 kg/cm2
Determinamos las tensiones:
• La tensión estática es:
σs = Fs = 3600 = 4580 kg / cm2
A πd 2 d2
4
• Recorrido de las tensiones:
σr = Fr = 7200 = 9160 kg / cm2
A πd 2 d2
4
• Por tanto la amplitud:
σa = σr = 4580 kg / cm 2
2 d2
• En este caso la tensión
media será:
σm =σs +σa = 9160
d2 kg / cm2
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO III: DISEÑO DE LEMENTOS POR CONCENTRACIÓN DE ESFUERZOS, CARGAS CÍCLICAS Y FATIGA
σ a = 0.50
σm
σa = S n − σ m
1 S u
0.50σ m σm
= 11651 − 5750
0.50σ m = 1165 − 0.2026σ m
σ m = 1658kg / cm2
σ a = 829kg / cm2
Como nos dice que debe tener una seguridad no menor de 100%:
σa = 829 414.50kg / cm2
2
σa = 4580 = 414.50
d2
d = 3.32cm
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO IV: VIGAS CURVAS EN FLEXIÓN
4
VIGAS CURVAS
EN FLEXIÓN
“La sabiduría es un adorno en la
prosperidad y un refugio en la
adversidad.”
Aristóteles
4.0.- INTRODUCCIÓN
Entendemos por vigas, en general a aquellos elementos en los cuales una de
sus dimensiones es mucho mayor que las otras dos que lo componen.
La viga curva en flexión constituye un importante elemento estructural de
ingeniería, debido a su utilización en una amplia variedad de aplicaciones; así
por ejemplo estructuras como hélices de helicópteros, ventiladores, turbinas y
sub-sistemas de estructuras más complejas pueden ser modelados como vigas
curvas De igual manera dichas vigas son usadas de forma corriente en la
construcción de puentes. Los ejemplos anteriores permiten afirmar que el
estudio de la respuesta dinámica de este componente estructural bajo diversas
condiciones, ayudaría a entender el comportamiento de ciertas estructuras
reales de mayor complejidad sometidas a condiciones similares.
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO IV: VIGAS CURVAS EN FLEXIÓN
4.1.- ESFUERZOS EN VIGAS CURVAS EN FLEXIÓN
Para determinar la distribución del esfuerzo en un elemento curvo en flexión se
que:
La sección transversal tiene un eje de simetría en un plano a lo largo de la
longitud de la viga.
Las secciones transversales planas permanecen planas después de la flexión.
El módulo de elasticidad es igual en tracción que en compresión.
El eje neutro y el eje centroidal de una viga curva, no coinciden y el esfuerzo no
varía en forma lineal como en una viga recta.
Fig.4.1 Variación lineal de los esfuerzos en una
viga recta y su distribución hiperbólica en una
viga curva
ro = Radio de la fibra externa.
ri = Radio de la fibra interna.
rn = Radio del eje neutro.
rc = Radio del eje centroidal.
h = Altura de la sección.
co = Distancia del eje neutro a la fibra externa.
ci = Distancia del eje neutro a la fibra interna.
e = Distancia del eje neutro al eje centroidal.
M = Momento flexionante, un M positivo disminuye la curvatura.
El radio del eje neutro viene dado por: rn = A (Ec 4.1)
dA
Donde: A = Área de la sección transversal ∫
El esfuerzo se determina por: r
σ = My y)
Ae(rn − (Ec 4.2)
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO IV: VIGAS CURVAS EN FLEXIÓN
La distribución del esfuerzo es hiperbólica y los esfuerzos críticos ocurren en las
superficies interna y externa donde: y = ci y y= -co respectivamente, el momento
es positivo conforme está representado en la figura.
σi = Mci (Ec 4.3) σo = Mco
Aeri Aero
(Ec4.4)
σ i : Esfuerzo de flexión en la fibra interna.
σo : Esfuerzo de flexión en la fibra interna.
A este esfuerzo se debe añadir el esfuerzo de tracción.
EjemploN°4.1 : Grafique la distribución de los esfuerzos que actúan en toda la
sección A-A del gancho de grúa de la fig. La sección transversal es rectangular
con b=0.75” y h=4” la carga a levantar es de 5000 lb.
Solución:
Área = A = bh = 0.75 x 4 = 3”
pulg2
dA = b.dr
Se sabe que:
rn = ∫ A = bh
dA ∫ ro b.dr
r ri r
rn = h
ln ro
ri
Reemplazando valores:
rn = 4 =4 = 3.641pu lg
ln 6 1.099
2
Por tanto la excentricidad:
e = rc − rn = 4 − 3.641 = 0.359p lg
El momento M (positivo)
M = F.rc = 5000(4) = 20,000lb − p lg
El esfuerzo será:
σ = F + My y)
A
Ae(rn. −
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO IV: VIGAS CURVAS EN FLEXIÓN
σ = 5000 + 20000(3.641 − r)
3 3x(0.359)r
Sustituyendo los valores de r de 2 a 6 se puede elaborar la siguiente tabla:
Tabla 4.1.- Distribución del esfuerzo para 2 < r >6
4.2.- EJES
Son elementos que sirven para transmitir potencia y en general se llaman
árboles a los ejes sin carga torsional, la mayoría de los ejes están sometidos
durante su trabajo a cargas combinadas de torsión, flexibilidad y cargas axiales.
Los elementos de transmisión: poleas, engranajes, volantes, etc., deben en lo
posible estar localizados cerca a los apoyos.
4.3.- CÁLCULO DE EJES
El diseño de ejes consiste básicamente en la determinación del diámetro
adecuado del eje para asegurar la rigidez y resistencia satisfactoria cuando el
eje transmite potencia en diferentes condiciones de carga y operación.
Los ejes normalmente tienen sección transversal circular: macizos – huecos
Para el diseño de ejes, cuando están hechos de aceros dúctiles, se analizan por
la teoría del esfuerzo cortante máximo.
Los materiales frágiles deben diseñarse por la teoría del esfuerzo normal
máximo.
El código ASME define una tensión de corte de proyectos o permisible que es la
más pequeña de los valores siguientes:
τ d = 3.30Syt (Ec4.5) Ó τ d = 0.18Sut (Ec 4.6)
Si hay concentración de tensiones debido a un acuerdo o un chavetero, la norma
dice que hay que disminuir en un 25% la tensión de corte permisible.
La tensión de corte en un eje sometido a flexión y torsión viene dado por:
σx 2 2
2
τ máx = + τ xy (Ec 4.7)
EL ESFUERZO DE TORSIÓN:
τ xy = Tr = 16T Para ejes macizos (Ec 4.8)
J π d3
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO IV: VIGAS CURVAS EN FLEXIÓN
τ xy = 16Tde Para ejes huecos (Ec 4.9)
π (de4 − di 4 )
EL ESFUERZO DE FLEXIÓN:
σx = Mr = 32M Para ejes macizos (Ec 4.10)
I π d3 Para ejes huecos (Ec 4.11)
σx = 32Mde
π (de4 − di4 )
ESFUERZOS AXIALES (COMPRESIÓN – TRACCIÓN):
σe = 4F / πd 2 Para ejes macizos (Ec 4.12)
σe = 4F / π (de2 − di2 ) Para ejes huecos (Ec 4.13)
El código ASME da una ecuación para el cálculo de un eje hueco que combina
torsión, flexión y carga axial, aplicando la ecuación del esfuerzo cortante máximo
modificada mediante la introducción de factores de choque, fatiga y columna.
16 αFdi(1 + K 2 ) 2
p(1 − C 8
de3 = K4) f M + + (CtT )2
πσ (Ec 4.14)
Para un eje macizo con carga axial pequeña o nula.
( ) ( )d 3 = 16
πσ p 22 (Ec 4.15)
C f M + CtT
Donde: Esfuerzo cortante de torsión, psi. de = Diámetro exterior,
τ xy =
pulg. Momento flector, lb-pulg. di = Diámetro interior,
M=
pulg. Momento torsor, lb-pulg. F = Carga axial, lb.
T= di/de
K=
Tensión de corte máxima, psi.
τ máx =
σx = tensión de flexión
Cf = Factor de choque y fatiga, aplicado al momento flector.
Ct = Factor de choque y fatiga, aplicado al momento de torsión.
σf = Esfuerzo de flexión, psi.
σe = Esfuerzo axial (Tensión – Compresión), psi.
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO IV: VIGAS CURVAS EN FLEXIÓN
Tabla 4.2.- Valores de Cm y Ct Cm Ct
Para ejes estacionarios: 1.0 1.0
Carga aplicada gradualmente 1.5 a 2.0 1.5 a 2.00
Carga aplicada repentinamente
Eje en rotación: 1.5 1.0
Carga aplicada gradual o corriente 1.5 a 2.0 1.0 a 1.5
Carga repentina (choques ligeros) 2.0 a 3.0 1.5 a 3.0
Carga repentina (choques fuertes)
El código ASME indica que para ejes con especificaciones técnicas definidas el
esfuerzo permisible σ p es el 30% del límite elástico, sin sobrepasar el 18% del
esfuerzo último en tracción, para ejes sin chaveteros.
Estos valores deben reducirse en 25% si existiesen chaveteros en los ejes.
α = Factor de columna, para cargas a tracción vale igual a la unidad para
compresión, se aplica:
α = 1− 1 / k) para L/K < 115 (Ec 4.16)
0.0044(L
α = Sy L 2 para L/K > 115 (Ec 4.17)
π 2nE k
n = 1 para extremos articulados
n = 2.25 para extremos fijos
n = 1.6 para extremos restringidos parcialmente, como el caso de los cojinetes
k = Radio de giro I , pulg.
A
I = Momento de inercia, pulg4
A = Área de la sección transversal, pulg2
Sy = Esfuerzo a la fluencia, psi.
4.4.- CÁLCULO DE EJES POR RIGIDEZ
El valor permisible de giro varía desde 0.026° por centím etro para máquinas de
precisión hasta 0.33°por centímetro para ejes de tra nsmisión.
θ = TL = TL 4 = 10.19TL Para eje macizo (Ec 4.18)
GJ π ⋅d Gd 4
G
32
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO IV: VIGAS CURVAS EN FLEXIÓN
10.19TL
G de4 − di 2
( )θ = Para eje hueco (Ec. 4.19)
DISEÑO DE EJE POR RIGIDEZ LATERAL: (Ec 4.20)
2 Resolución gráfica (Ec 4.21)
dy M (Ec. 4.22)
(Ec. 4.23)
2= (Ec. 4.24)
d x EI
MOMENTO TORSOR:
T = 63,000xhp (lb − pu lg)
n(r. p.m.)
T = 71,620xCV (kg − cm)
n(r. p.m.)
Ft = 33,000HP (lb)
Vm
Ft = 4500CV (kg )
Vm
Vm = pies / min Vm =m/min
Ft : Fuerza tangencial en el radio primitivo, lb.
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO V: TORNILLOS, SUJETADORES Y UNIONES
5
TORNILLOS,
SUJETADORES
Y UNIONES
“Aprender sin pensar es inútil, pensar
sin aprender es peligroso".
(Confucio)
5.0.- INTRODUCCIÓN
Los tornillos son elementos que tienen filetes enrollados en forma de hélice
sobre una superficie cilíndrica y son unos de los elementos más utilizados en las
máquinas. Podemos clasificar los tornillos, de acuerdo con la función que
cumplen, en tornillos de unión y tornillos de potencia. Los tornillos de unión son
los que sirven para unir o asegurar dos o más partes estructurales o de
maquinaria, como es el caso de los tornillos, pernos, espárragos y tornillos
prisioneros o de fijación. Los tornillos de potencia son aquellos destinados a la
transmisión de potencia y movimiento; generalmente convierten un movimiento
de giro en un movimiento de traslación.
Los tornillos se usan en estructuras, máquinas herramientas, vehículos, prensas
y elementos de elevación, entre otros. En muchos casos, los tornillos están
sometidos a cargas variables combinadas, por lo que debe aplicarse una teoría
de falla por fatiga. Un tornillo puede fallar en el núcleo o en los filetes; se debe
tener en cuenta el diámetro del tornillo, así como el número de filetes en
contacto con la tuerca.
Los sujetadores son distintos artículos de ensamblaje que se emplean para unir
diversos componentes de una pieza. Un sujetador puede ser un perno y una
tuerca, un tornillo, un clavo e incluso una grapa. Sin embargo, la mayoría de los
sujetadores utilizados en la industria son sujetadores roscados. Estos
dispositivos por lo general permiten el ensamblar y desensamblar componentes.
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO V: TORNILLOS, SUJETADORES Y UNIONES
5.1.- SUJETADORES ROSCADOS Y TORNILLO DE POTENCIA:
Los métodos clásicos de sujeción o de unión de piezas incluyen el empleo de
elementos como pernos, tuercas, tornillos de cabeza, tornillos prisioneros,
remaches, retenes de resorte, sistema de bloqueo y chavetas. Las piezas
pueden unirse de forma permanente mediante soldadura.
5.2.- TORNILLO DE POTENCIA:
Un tornillo de potencia se usa para cambiar el movimiento angular en
movimiento lineal y también para transmitir
esfuerzos.
La base de este triángulo tiene una longitud
igual a πdm.
Ángulo de hélice.
Paso o avance del tornillo.
Fuerza de rozamiento.
Diámetro medio.
Fuerza normal
Fuerza que represente la suma de todas las fuerzas unitarias axiales que
actúan sobre el área normal de la rosca.
Fuerza necesaria con el objeto de vencer la fuerza de rozamiento y hacer
ascender la carga por el plano inclinado.
Tornillo de Potencia (rosca cuadrada)
Ecuación de Equilibrio:
P-NSenα-µNCosα=0
F+µNSenα-NCosα=0
(Ec. 5.1)
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO V: TORNILLOS, SUJETADORES Y UNIONES
Dividiendo numerador y denominador
por Cosα:
ܲ ൌ ிሺ௧∝ାఓሻ ൌ ிቀഏାఓቁ ݁ݑݍݎ ݃ݐ ∝ൌ
ଵିఓ௧∝ ଵିഏഋ గௗ
(Ec. 5.2)
ܲ ൌ ܨሺ ∝ ݃ݐߤሻ ൌ ܨ ቀߨ݀ ݉ ߤቁ
1 െ ߤ∝ ݃ݐ ∝ ݃ݐ ݁ݑݍݎൌ ߨ݀݉
1 െ ߤ
ߨ݀݉
El par necesario será:
P(dm/2) (Ec. 5.3)
Entonces:
ܶ ൌ ிௗ ቀାగఓௗቁ (Ec 5.4)
ଶ గௗିఓ
Estas ecuaciones son para roscas cuadradas, (las cargas normales son
paralelas al eje del tornillo). En roscas ACME, la carga normal está inclinada
respecto al eje en una cantidad θn (igual a la mitad del ángulo de la rosca)
Su efecto es incrementar la fuerza de rozamiento.
Por tanto la ecuación del par deben dividirse por Cos θn los términos en que
interviene el rozamiento.
ܶ ൌ ிௗ ቀାగఓௗௌఏቁ (Ec. 5.5)
ଶ గௗିఓௌఏ
5.3.- EFICIENCIA O RENDIMIENTO DE UN TORNILLO
Si µ = 0 ܶ ൌ ி
ଶగ
El rendimiento es por tanto:
(Ec. 5.6)
Fig. 5.3
Cuando se carga axialmente el tornillo debe emplearse un cojinete axial o un
collar entre el elemento giratorio y estacionario para transmitir la carga axial.
El par necesario para vencer la fuerza de rozamiento en el collar será:
TC = F.µC .dc (Ec.5.7)
2
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO V: TORNILLOS, SUJETADORES Y UNIONES
5.4.-CÁLCULO DE TORNILLOS DE POTENCIA
Fig. 5.4
Momento de giro:
ܶ ൌ ܹ ቈr ቆ ቇ௧ఈାೞഇ ݂ r (Ec. 5.8)
ଵିೞഇഀ
Momento aplicado para girar el tornillo
Carga paralela al eje del tornillo
Radio medio de la rosca
Radio efectivo de la superficie de rozamiento contra la cual se apoya la
carga, llamado radio del collar
Coeficiente de rozamiento entre las roscas del tornillo y la tuerca
Coeficiente de rozamiento en el collar
Ángulo de la hélice de la rosca en el radio medio
Ángulo entre la tangente al perfil del diente y una línea radial, medido en
un plano normal a la hélice de la rosca en el radio medio.
El momento requerido para avanzar el tornillo (o la tuerca) en el sentido de la
carga:
ܶ ൌ ܹ ቈݎ ቆ ቇି௧ఈାೞഇ ݂ ݎ (Ec. 5.9)
ଵାೞഇഀ
Este valor puede ser positivo o negativo. Si es positivo, debe efectuarse trabajo
para avanzar el tornillo. Si es negativo, la carga axial aisladamente producirá
rotación.
5.5.-EFICIENCIA DE UN MECANISMO DE TORNILLO
݂ܽ݅ܿ݊݁݅ܿ݅ܧൌ ଵௐሺ௩ሻ % ൌ ଵ௧∝ % (Ec. 5.10)
ଶగ் ൭ೌభష∝శೌೞഇೞ∝ഇ൱ାೝೝ
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO V: TORNILLOS, SUJETADORES Y UNIONES
5.6.-ESFUERZOS EN LA ROSCA
Esfuerzo cortante transversal
ܵܿ ൌ ி ൌ ௐ (Ec. 5.12)
ଶగ
5.7.-ESFUERZOS EN EL NÚCLEO
Esfuerzo cortante:
߬ ൌ ் ൌ ் ௫మ ൌ ଵ் (Ec. 5.13)
గయమర గௗయ
di = Diámetro raíz
Esfuerzo axial:
ߪ ൌ ி ൌ ி ൌ ସி (Ec. 5.14)
గௗమ
ഏమ
ర
Cuando el movimiento de rotación ha de transformarse en lineal con un gran
rendimiento, se recomienda el tornillo con tuerca de bolas recirculantes. Para
ángulos de hélice mayores a 2°el rendimiento es del 90% ( el de roscas ACME
es del 25%).
Los tornillos deben tratarse térmicamente hasta una dureza de 58 RC mínimo.
EjemploN°5.1 : El tornillo mostrado se opera por medio de un momento aplicado
al extremo inferior, la tuerca está cargada y su movimiento está restringido
median guías. Suponer que el rozamiento en el cojinete de bolas es
despreciable. El tornillo tiene un diámetro exterior de 2” y una rosca triple
ACME, de 3 filetes por pulgada. El coeficiente de rozamiento de la rosca es de
0.15. Determinar la carga que puede levantarse con un momento T de 400 lb-
pulg (sel F= 1290lb)
Solución:
ܶ ൌ ܹ ݎ െߙ݃ݐ ݂݊ߠݏ݂ߙ݃ܥݐቍ ݂ ݎ
ቌ ݊ߠݏܥ
1
Donde:
Profundidad de la rosca = 0.18”
ݎ ൌ 1 െ 0.18 ൌ 0.91"
ܽ ݁ܿ݊ܽݒ1
∝ ݃ݐൌ 2ߨݎ ൌ 2ߨ0.91 ൌ 0.175 ∝ൌ 9.92°
ߠ ൌ 14.5° ݁݉ܿܽ ܽܿݏݎ ܽݎܽ
݊ߠ݃ݐൌ ∝ ݏܥߠ݃ݐ
݊ߠ݃ݐൌ ݃ݐ14.5°ݏܥ9.92 ൌ 0.255
ߠ݊ ൌ 14.2°
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO V: TORNILLOS, SUJETADORES Y UNIONES
Determinar la presión media de contacto entre las superficies del tornillo y la
tuerca
ܨ1290
ܲ ൌ 2ߨ݄ݎ݄ ൌ 2ߨሺ6ሻ0.91ݔ0.18 ൌ 210 ݅ݏ
݈ ܽݎ݁ݎݎܽܿ ݀ݑݐ݅݃݊2
݊ ൌ ݏܽൌ 1 ൌ 6 ݏܽݐ݈݁ݑݒ
3
Se observa que la diferencia entre ߠ݊ ൌ 14.2°y ߠ ൌ 14.5° ݁݉ܿܽ ܽܿݏݎ ܽݎܽes tan
pequeña que se hubiera podido utilizar ߠ ൌ 14.5° ݁݉ܿܽ ܽܿݏݎ ܽݎܽ. Entonces:
W=1290lb
5.8.-PRETENSADO DE LOS PERNOS
Cuando se desea una conexión que pueda Fig.5.6: conexión
desmontarse y que sea lo bastante sólida como con pernos
para resistir cargas exteriores de tracción, de
cizallamiento o de una combinación de ambas,
resulta que las uniones con simples pernos, son
una buena solución. En la figura 5.6, en la que el
perno se ha estirado o tensado para producir una
carga previa inicial de tracción Fi, después de lo
cual se aplican las cargas exteriores de tracción
Fi y de cizallamiento Fs.
Para determinar la parte de la carga externa que
corresponde soportar a las piezas conectadas y la
parte que corresponde soportar al perno, es
necesario definir la expresión constante de
rigidez.
Empleando la ecuación de la deformación debida
a las cargas de tracción o compresión δ=F. ℓ
/A.E, y ordenando obtendremos:
݇ ൌ ி ൌ ா … (a)
ఋ
En donde k es la constante de rigidez en kg/cm. Con objeto de hacer la
siguiente discusión tan clara como sea posible, definiremos ahora las siguientes
magnitudes de fuerzas:
Carga de tracción externa total sobre el conjunto empernado.
Carga previa inicial sobre el perno debía solo a su tensado y que existe
antes de que se aplique Ft.
Parte de Ft correspondiente al perno.
Parte de Ft correspondiente a los elementos.
Cuando se aplica la carga externa Ft al conjunto pretensado, hay un cambio en
la deformación del perno y de los elementos conectados. Puesto que el perno
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO V: TORNILLOS, SUJETADORES Y UNIONES
está inicialmente a tracción, debe experimentar un aumento en su deformación,
que vale ∆δm = Fb/kb. El subíndice b se refiere al perno y Fb es, por tanto, la
parte de la carga externa que corresponde soportar al perno. Los elementos
conectados experimentarán una disminución en su deformación, de valor ∆δm =
Fm/km. El subíndice m se refiere a los elementos o piezas que se conectan
juntos. En la hipótesis de que los elementos no se hayan separado, el aumento
en la deformación del perno deberá igualar a la disminución en la deformación
de los elementos y, por consiguiente:
ி್ ൌ ி ሺܽሻ (Ec. 5.15)
್
Puesto que Ft = Fb + Fm, tendremos
ܨ ൌ ್ி ሺܾሻ (Ec. 5.16)
್ା
Por tanto, el esfuerzo resultante sobre el perno
ܨ ൌ ܨ ܨ ൌ ್ி ܨ ሺ7 െ 10…ሻ . (b)
್ା
Del mismo modo, la compresión resultante de los elementos conectados resulta
ser
ܨ ൌ ி െ ܨ ሺ7 െ 11ሻ
್ା …. (c)
Las ecuaciones (b) y (c) son válidas en tanto que se mantenga algo de la
compresión inicial en los elementos. Si la fuerza exterior es lo bastante grande
como para eliminar completamente esta compresión, los elementos se
separarán y la carga entera deberá ser soportada por el perno.
EjemploN°5.2 : En la figura 5.6 sea km = 4kb la rigidez de los elementos respecto a
la del perno. Si la carga previa inicial en el perno es Fi = 1,000 kg y la exterior de
tracción es Ft = 1,200 kg calcular la tracción resultante en el perno y la
compresión de los elementos.
Solución:
La tracción resultante en el perno se encuentra por medio de la ecuación:
ܨ ൌ ݇ ܨ௧ ܨ ൌ ݇ሺ1.200ሻ 1.000 ൌ 1.240 ݇݃ ܽ ݅ܿܿܽݎݐó݊
݇ ݇ ݇ 4݇
La compresión en los elementos se calcula por la ecuación:
ܨ ൌ ݇ܨ௧ െ ܨ ൌ 4݇ሺ1.200ሻ െ 1.000 ൌ െ40 ݇݃ ݀݁ ܿ݅ݏ݁ݎ݉ó݊
݇ ݇ ݇ 4݇
Esto indica que la proporción de la carga que le corresponde soportar al perno
es pequeña y que depende de la rigidez relativa de los dos materiales. Puesto
que los elementos están todavía comprimidos, no hay separación de las piezas,
aunque la carga externa, en este ejemplo, sea mayor que la pretensión del
perno.
La importancia del pretensado de los pernos no puede sobreestimarse. Tiene
los dos efectos deseables siguientes:
Mejora la resistencia a la fatiga. Cuando un conjunto empernado con
pretensión se somete a la acción de cargas de fatiga, solo se aplica al
perno una pequeña proporción del cambio total en la tensión. Por tanto,
el efecto es el de mejorar la resistencia a la fatiga del perno. Debe
señalarse que esta resistencia se debe únicamente a la pretensión y no
incluye los efectos de la concentración de tensiones o de otras
irregularidades superficiales que puedan originar el fallo.
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO V: TORNILLOS, SUJETADORES Y UNIONES
Mejora en el efecto de apriete. Se ha demostrado que una tuerca se
afloja por causa de la variación de tensiones dentro de la sujeción. El
pretensado reduce la magnitud del cambio de tensiones y mejora, por
tanto, considerablemente el efecto de apriete.
Con el objeto de obtener los beneficiosos efectos del pretensado, los elementos
sujetos deben ser rígidos y el perno debe tener una elevada carga previa. Esta
condición se obtiene a menudo cuando las piezas entran en contacto metal
contra metal, esto es sin juntas, y se empernan después. En este caso, la
rigidez de los elementos es a menudo mucho mayor que la del perno y la
proporción de la carga externa que corresponde soportar al perno puede
despreciarse.
Cuando se emplea una junta, los efectos beneficiosos del pretensado pueden
preservarse parcialmente, empleando una junta rígida. Una junta blanda o el
empleo de materiales blandos, como el aluminio o magnesio, destruirán
completamente este efecto y harán que el perno soporte prácticamente la carga
entera. Cuando se emplea una junta, puede despreciarse frecuentemente la
constante de rigidez de los elementos (ya que su rigidez es mucho mayor) y
calcularse la constante para la junta sola. Se utiliza la ecuación (a), haciendo a ℓ
igual al grosor de la junta.
5.9.-PAR DE APRIETE DEL PERNO
El pretensado de un perno es la fuerza con la que éste mantiene juntos a los
elementos, si es necesario apretar el perno exactamente hasta una pretensión
determinada, el mejor modo de hacerlo es calcular la deformación del perno
empleando la fórmula δ = Fil/AE.
ܶ ൌ ிௗ ቀగାௗగఓିௗఓ ௦௦∝∝ቁ ிఓௗ (Ec. 5.17)
ଶ ଶ
La cara que mira a la arandela de una tuerca hexagonal es 1 ½ veces el
diámetro nominal del perno. Por tanto, el diámetro medio del collar es de dc =
1,25d. La ecuación puede ahora reagruparse dando:
ܶ ൌ ቂቀௗଶௗ ቁ ቀଵ௧ିఓట௧ାఓట௦௦ఈఈቁ 0,625ߤቃ ܨ݀ (Ec. 5.18)
Definamos ahora un coeficiente de par K como
ܭൌ ቀௗଶௗ ቁ ቀଵ௧ିఓట௧ାఓట௦௦ఈఈቁ 0,625ߤ (Ec. 5.19)
Y por tanto, (Ec. 5.20)
ܶ ൌ ܨܭ݀
Para pernos sin lubricar de tipo medio, k vale alrededor de 0,20.
Los coeficientes de rozamiento de la rosca y del collar para pernos varían entre
0,10 y 0,20, dependiendo del acabado de la rosca, de su exactitud y del grado
de lubricación. Pernos y tuercas de tipo medio pueden emplearse un valor de
0,15 para µ y µc.
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO V: TORNILLOS, SUJETADORES Y UNIONES
Tabla 5-1: Coeficientes de Par
Tamaño Perno K Tamaño Perno K
¼-20 NC 0,210 9/16-12NC 0,198
¼-28 NF 0,205 9/16-18NF 0,193
5/16 -18 NC 0,210 5/8-11NC 0,199
5/16-24 NF 0,205 5/8-18NF 0,193
3/8-16 NC 0,204 ¾-10NC 0,194
3/8-24 NF 0,198 ¾-16NF 0,189
7/16-14 NC 0,205 7/8-9NC 0,194
7/16 -20 NF 0,200 7/8-14NF 0,189
½ -13 NC 0,201 1-8NC 0,193
½ -20 NF 0,195 1-12NF 0,188
El par de apriete calculado o correcto debe ser alrededor del 75 por 100 del par
medio que origina la rotura.
5.10.-RESISTENCIA DEL PERNO
Ya se ha señalado la importancia del pretensado y se ha encontrado un método
de calcular el par necesario para producir una fuerza dada de sujeción. Es,
pues, apropiado que investiguemos ahora la resistencia de los pernos y que
averigüemos qué pretensión puede resistir con éxito un perno de cierto tamaño y
material.
Tabla 5-2: Especificaciones SAE para pernos, tornillería y ésparragos.
Resistencia Carga de Dureza Diámetro
Grado a la tracción Prueba Brinell plg
SAE Material
Bhn
kg/cm2 kg/cm2
0 --- --- --- --- Bajo en Carbono sin
requisitos
1 3,86 --- 207 máx. --- Acero ordinario
2 4,85 3,86 241 máx. Hasta ½
4,5 3,65 241 máx. ½ - ¾ Bajo en carbono
3,86 --- 207 máx. ¾ - 1 ½
3 7,73 5,98 207-269 Hasta ½ Medio en carbono, trabajado
7,03 5,625 207-269 Aprox. 5/8 en frío
5 8,43 5,975 241-302 Hasta ¾ Medio en carbono, templado
y revenido
8,08 5,48 235-302 ¾-1
7,38 5,2 233-285 1 – 1 ½
6 9,84 7,73 285-331 Hasta 5/8 Medio en carbono, templado
9,35 7,38 269-331 --- y revenido
7 9,14 7,38 269-321 Hasta 1 ½ Aleado, medio en carbono,
templado y revenido
8 10,55 8,44 302-352 Hasta 1 ½ Aleado, medio en carbono,
templado y revenido
La “Society of Automotive Engineers” (SAE) ha publicado durante muchos años
especificaciones de materiales para muchos productos roscados. El proyectista,
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO V: TORNILLOS, SUJETADORES Y UNIONES
naturalmente, es libre de especificar un material escogido por él para los pernos
o especificar un perno hecho según las normas SAE.
Las especificaciones SAE comprenden todos los sujetadores roscados
exteriormente e incluyen ocho grados de aceros.
La carga de prueba de un perno es la carga máxima a tracción que un perno
puede soportar sin deformación permanente.
El área para la tensión de tracción de un elemento roscado es el área de un
círculo cuyo diámetro es la media de los diámetros del núcleo y primitivo.
En uniones metal contra metal ordinarias, la rigidez km de los elementos es tan
grande, comparada con la rigidez de los pernos kb, que, para todas las
aplicaciones, el perno resulta cargado estáticamente, aunque la carga exterior
de tracción en la conexión pueda ser del tipo de fatiga. Para estas condiciones,
la pretensión mínima en el perno debe ser el 90 por 100 de la carga de prueba.
La tensión de torsión en un perno desaparece después de su apriete.
El par aplicado a la tuerca alrededor del 50 por 100 del mismo se emplea para
vencer el rozamiento entre la cara de contacto de la tuerca y el elemento del 40
por 100 del restante se emplea para vencer el rozamiento de la rosca y el resto
produce la tracción en el perno.
5.11.-UNIONES A TRACCIÓN CON PERNOS Y JUNTAS
Frecuentemente se pueden emplear cierres herméticos en las uniones,
manteniendo, además el contacto metal contra metal. Esto se debe hacer
siempre que sea posible, ya que origina una unión mucho más fuerte.
La figura 5.6 muestra una unión con pernos a tracción empleando una junta. La
ecuación anterior, que da la carga resultante sobre el perno, cuando se conoce
la carga inicial y la carga a tracción externa, puede ordenarse como se indica a
continuación.
En donde: (Ec. 5.21)
(Ec. 5.22)
Fig.5.6: unión empernada
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO V: TORNILLOS, SUJETADORES Y UNIONES
Tabla 5-3: Hilos por pulgada de uso común en los tornillos de potencia.
Diámetro 1/4 5/16 3/8 1/2 5/8 3/4 1 1½
exterior, pulg
0,0635 0,0794 0,0953 0,1270 0,1588 0,1905 2,540 3,810
Diámetro 4
exterior, cm 16 14 12 10 8 6 5
Hilos por
El coeficiente de rigidez (C) tiene valores entre 0 y 1. Doughtie y Carter
establecieron que cuando no se emplea junta C debe hacerse igual a cero y que,
en las aplicaciones normales, empleando los materiales más blandos y flexibles
para juntas, los ensayos demuestran que C raramente excede de 0,50.
Se ha visto que, cuando el perno está adecuadamente pretensado, la fatiga no
es un problema serio en uniones sometidas a tracción que emplean materiales
rígidos. Puesto que los materiales de los pernos son relativamente dúctiles, esto
significa que también tiene menos importancia la concentración de tensiones.
Sin embargo cuando se utiliza una junta relativamente blanda, aumenta la
variación de tensiones en el perno y deben considerarse tanto la fatiga como la
concentración de tensiones. En la tabla 5.5 se relacionan los valores de los
coeficientes de reducción de la resistencia a la fatiga KF, para roscas laminadas
y mecanizadas en aceros recocidos o con tratamiento térmico.
DISEÑO DE ELEMENTOS DE MÁQUINAS I
CAPITULO V: TORNILLOS, SUJETADORES Y UNIONES
Tabla 5-4 Diámetro y áreas de tornillos de rosca unificados, UNC y UNF
DiámetroDesignación
Nominaldel tamaño
Hilos por
Pulg Cm pulgada N
Hilos por
0 0,0600 0,1524 pulgada N
1 0,0730 0,1854 Series Bastas UNC Series Finas UNF
2 0,0860 0,2184
3 0,0990 0,2515 Área para Área para
4 0,1120 0,2845 la tensión Área del la tensión Área del
5 0,1250 0,3175 núcleo núcleo
6 0,1380 0,3505 de tracción Ar, cm2 de tracción Ar, cm2
8 0,1640 0,4166 At, cm2 At, cm2
10 0,1900 0,4826
12 0,2160 0,5486 --- --- --- 80 0,01161 0,00974
1/4 0,2500 0,6350 72 0,01794 0,01529
5/16 0,3125 0,7937 64 0,01697 0,01406 64 0,02542 0,02187
3/8 0,3750 0,9525 56 0,03374 0,02910
7/16 0,4375 1,1112 56 0,02387 0,02000 48 0,04265 0,03652
1/2 0,5000 1,2700 44 0,05355 0,04619
9/16 0,5625 1,4287 48 0,03142 0,02619 40 0,06548 0,05639
5/8 0,6250 1,5875 36 0,09510 0,08290
3/4 0,7500 1,9050 40 0,03897 0,03200 32 0,1290 0,1129
7/8 0,8750 2,2225 28 0,1665 0,1458
1 1,0000 2,5400 40 0,05135 0,04335 28 0,2348 0,2103
1 ¼ 1,2500 3,1750 24 0,3742 0,3380
1 ½ 1,5000 3,8100 32 0,05864 0,04806 24 0,5665 0,5219
20 0,7658 0,7032
32 0,09032 0,07716 20 1,032 0,9587
18 1,310 1,219
24 0,1129 0,09355 18 1,652 1,548
16 2,406 2,265
24 0,1561 0,1329 14 3,284 3,097
12 4,277 4,032
20 0,2051 0,1735 12 6,923 6,606
12 8,484 8,129
18 0,3380 0,2929
16 0,5000 0,4374
14 0,6858 0,6019
13 0,9155 0,8110
12 1,174 1,045
11 1,458 1,303
10 2,155 1,948
9 2,980 2,703
8 3,910 3,555
7 6,252 5,742
6 9,064 8,348
Tabla 5-5: Coeficientes de Reducción de la Resistencia a la fatiga para
elementos roscados sometidos a tracción o flexión
Tipo de Acero Laminado Mecanizado
Recocido ………………… 2,2 2,8
Templado y revenido …… 3,0 3,8
EjemploN°5.3 : El conjunto empernado de la figura emplea un anillo de cobre
como junta. Calcúlese el coeficiente de rigidez del conjunto.
DISEÑO DE ELEMENTOS DE MÁQUINAS I
Pages:
1 - 50
51 - 100
101 - 150
151 - 158
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190310 | https://www.i2m.univ-amu.fr/perso/francois.hamel/bhn2.pdf | The speed of propagation for KPP type problems. I - Periodic framework Henri Berestycki a, Fran¸ cois Hamel b and Nikolai Nadirashvili c∗ a EHESS, CAMS, 54 Boulevard Raspail, F-75006 Paris, France b Universit´ e Aix-Marseille III, LATP, Facult´ e des Sciences et Techniques, Case cour A Avenue Escadrille Normandie-Niemen, F-13397 Marseille Cedex 20, France c University of Chicago, Department of Mathematics, 5734 S. University Avenue, Chicago, IL 60637-1546, USA Abstract. This paper is devoted to some nonlinear propagation phenomena in periodic and more general domains, for reaction-diffusion equations with Kolmogorov-Petrovsky-Piskunov (KPP) type nonlinearities. The case of periodic domains with periodic underlying excitable media is a follow-up of the article . It is proved that the minimal speed of pulsating fronts is given by a variational formula involving linear eigenvalue problems. Some consequences concerning the influence of the geometry of the domain, of the reaction, advection and diffusion coefficients are given. The last section deals with the notion of asymptotic spreading speed. The main properties of the spreading speed are given. Some of them are based on some new Liouville type results for nonlinear elliptic equations in unbounded domains.
Contents 1 The periodic framework and main results 3 1.1 Speed of propagation of pulsating travelling fronts in periodic domains . . .
3 1.2 Influence of the geometry of the domain and of the underlying medium . . .
9 1.3 Spreading speed in periodic domains . . . . . . . . . . . . . . . . . . . . . .
13 2 Variational formula for the minimal speed of pulsating travelling fronts 17 3 Influence of the geometry of the domain and of the coefficients of the medium 23 3.1 Influence of the geometry of the domain : proofs of Theorems 1.3 and 1.4 . .
23 3.2 Influence of the coefficients of the medium : proofs of Theorems 1.6, 1.8 and 1.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27 ∗The third author was partially supported by a NSF grant 1 4 Spreading speed 31 Introduction This paper is the first in a series of two in which we address spreading and propagation prop-erties attached with reaction-diffusion type equations in a general framework. We consider reaction-terms of the type associated with Fisher or KPP (for Kolmogorov, Petrovsky and Piskunov) equations. These properties are well understood in the homogeneous framework which we recall eblow. Here and in part II we consider heterogeneous problems. Part II will be devoted to propagation properties in very general domains. The present paper deals with the periodic case where both the equation and the domain have periodic structures. The precise setting and assumptions will be given shortly. But before that, let us recall some of the basic features of the homogeneous equations.
Consider the Fisher-KPP equation : ut −∆u = f(u) in RN.
(0.1) It has been introduced in the celebrated papers of Fisher (1937) and KPP (1937) originally motivated by models in biology. Here the main assumption is that f is say a C1 function satisfying f(0) = f(1) = 0, f ′(1) < 0, f ′(0) > 0, f > 0 in (0, 1), f < 0 in (1, +∞), (0.2) f(s) ≤f ′(0)s, ∀s ∈[0, 1].
(0.3) Archetypes of such nonlinearities are f(s) = s(1 −s) or f(s) = s(1 −s2).
Two fundamental features of this equation account for its success in representing prop-agation (or invasion) and spreading. First, this equaton has a family of planar travelling fronts. These are solutions of the form u(t, x) = U(x · e −ct) (0.4) where e is a fixed vector of unit norm which is the direction of propagation, and c > 0 is the speed of the front. Here U : R 7→R is given by ½ −U ′′ −cU ′ = f(U) in R U(−∞) = 1, U(+∞) = 0.
(0.5) In the original paper of Kolmogorov, Petrovsky and Piskunov, it was proved that, under the above assumptions, there is a threshold value c∗= 2 p f ′(0) > 0 for the speed c. Namely, no fronts exist for c < c∗, and, for each c ≥c∗, there is a unique front of the type (0.4-0.5).
Uniqueness is up to shift in space or time variables.
Another fundamental property of this equation was established mathematically by Aron-son and Weinberger (1978). It deals with the asymptotic speed of spreading. Namely, if u0 is a nonnegative continuous function in RN with compact support and u0 ̸≡0, then the solution u(t, x) of (0.1) with initial condition u0 at time t = 0 spreads with the speed c∗in 2 all directions for large times : as t →+∞, max|x|≤ct |u(t, x) −1| →0 for each c ∈[0, c∗), and max|x|≥ct u(t, x) →0 for each c > c∗.
In this paper, we consider a general heterogeneous periodic framework extending (0.1).
The heterogeneous character arises both in the equation and in the underlying domain. The types of equations we consider here are : ½ ut −∇· (A(x)∇u) + q(x) · ∇u = f(x, u) in Ω ν · A∇u = 0 on ∂Ω, (0.6) where ν denotes the outward unit normal on ∂Ω. It will be assumed throughout this paper that the matrix A(x), the vector q(x) and the reaction term f(x, s) as well as the geometry Ω are periodic. Precise assumptions will be described shortly. Note that even equation (0.1), if set in a periodic domain (e.g. the space with a periodic array of holes), acquires the features of a non-homogeneous equation. That equation will be considered in general (non-periodic) domains in Part II .
Here, in the periodic setting, we address three types of questions.
1) What is the speed of generalized travelling fronts in periodic structures –we recall the definition of such fronts below– ? A formula which we announced in is proved here.
2) Using a formula of G¨ artner and Freidlin , we relate the asymptotic speed of spread-ing in a periodic domain to that of the minimal speed of propagation. Contrarily to the homogeneous equation, as we will see on an example, these two speeds may not be the same.
3) We then proceed to derive several important consequences on the minimal speed of propagation and on the asymptotic spreading speed. Effects of stirring, of reaction, and of geometry will be established here rigorously. These formulas indeed allow us to prove properties of the following kind. The presence of holes or of an undulating boundary always hinder the progression or the spreading. On the contrary, any stirring by a flow always increases that speed.
In the next section we introduce the general setting with precise assumptions and we state the main results achieved in this paper. Their proofs take up the remaining sections.
1 The periodic framework and main results 1.1 Speed of propagation of pulsating travelling fronts in periodic domains This section deals with pulsating fronts travelling in a given unbounded periodic domain under the effects of diffusion, reaction and possibly advection by a given underlying flow. One of the most important issues in this context is the determination of the speed of propagation of fronts. A variational formula for the minimal speed of propagation is derived.
This notion of propagation of travelling fronts for the homogeneous equation (0.1) can be extended to that of pulsating travelling fronts in a more general class of periodic domains and for a more general class of reaction-diffusion-advection equations in periodic excitable media.
3 We now describe the general periodic framework. Let N ≥1 be the space dimension and let d be an integer such that 1 ≤d ≤N. Call x = (x1, · · · , xd) and y = (xd+1, · · · , xN). Let L1, · · · , Ld be d positive numbers and let Ωbe a C3 nonempty connected open subset of RN such that ∃R ≥0, ∀(x, y) ∈Ω, |y| ≤R, ∀(k1, · · · , kd) ∈L1Z × · · · × LdZ, Ω= Ω+ d X i=1 kiei, (1.1) where (ei)1≤i≤N is the canonical basis of RN. Let C be the set defined by C = {(x, y) ∈Ω, x ∈(0, L1) × · · · × (0, Ld)}.
Since d ≥1, Ωis unbounded and C is its periodicity cell. In all what follows, a field w is said to be L-periodic with respect to x in Ωif w(x1 + k1, · · · , xd + kd, y) = w(x1, · · · , xd, y) almost everywhere in Ω, for all k = (k1, · · · , kd) ∈L1Z × · · · × LdZ.
Before going further on, let us point out that this framework includes several types of simpler geometrical configurations. The case of the whole space RN corresponds to d = N, where L1, · · · , LN are any positive numbers. The case of the whole space RN with a periodic array of holes can also be considered. The case d = 1 corresponds to domains which have only one unbounded dimension, namely infinite cylinders which may be straight or have oscilatting periodic boundaries, and which may or not have periodic holes. The case 2 ≤d ≤N −1 corresponds to infinite slabs.
We are interested in propagation phenomena for the following reaction-diffusion-advection equation, with unknown u, set in the periodic domain Ω: ½ ut = ∇· (A(x, y)∇u) + q(x, y) · ∇u + f(x, y, u), t ∈R, (x, y) ∈Ω, νA∇u(x, y) = 0, t ∈R, (x, y) ∈∂Ω.
(1.2) Such equations arise especially in simple combustion models for flame propagation , , , as well as in models in biology and for population dynamics , , . The passive quantity u typically stands for the temperature or a concentration which diffuses and is transported in a periodic excitable medium.
Let us now detail the assumptions on the coefficients of (1.2). First, the diffusion matrix A(x, y) = (Aij(x, y))1≤i,j≤N is a symmetric C2,δ(Ω) (with δ > 0)1 matrix field satisfying ½ A is L-periodic with respect to x, ∃0 < α1 ≤α2, ∀(x, y) ∈Ω, ∀ξ ∈RN, α1|ξ|2 ≤Aij(x, y)ξiξj ≤α2|ξ|2 (1.3) (we use the usual summation convention with 1 ≤i, j ≤N).
The boundary condition νA∇u(x, y) stands for νi(x, y)Aij(x, y)∂xju(t, x, y) and ν denotes the unit outward normal 1The smoothness assumptions on A, as well as on q and f below, are made to ensure the applicability of some a priori gradients estimates for the solutions of some approximated elliptic equations obtained from (1.2) (see (2.9) in Section 2). These gradient estimates are obtained for smooth (C3) solutions through a Bernstein-type method, . We however believe that the smoothness assumptions on A, as well as on q and f, could be relaxed, by approximating A, q and f by smoother coefficients.
4 to Ω. When A is the identity matrix, then this boundary condition reduces to the usual Neumann condition.
The underlying advection q(x, y) = (q1(x, y), · · · , qN(x, y)) is a C1,δ(Ω) (with δ > 0) vector field satisfying q is L-periodic with respect to x, ∇· q = 0 in Ω, q · ν = 0 on ∂Ω, ∀1 ≤i ≤d, Z C qi dx dy = 0.
(1.4) The divergence-free assumption means that the underlying flow is incompressible. The vector field q is tangent on ∂Ωand its first d components have been normalized. The flow q may represent some turbulent fluctuations with respect to a mean field.
Lastly, let f(x, y, u) be a nonnegative2 function defined in Ω× [0, 1], such that f ≥0, f is L-periodic with respect to x and of class C1,δ(Ω× [0, 1]), ∀(x, y) ∈Ω, f(x, y, 0) = f(x, y, 1) = 0, ∃ρ ∈(0, 1), ∀(x, y) ∈Ω, ∀1 −ρ ≤s ≤s′ ≤1, f(x, y, s) ≥f(x, y, s′), ∀s ∈(0, 1), ∃(x, y) ∈Ω, f(x, y, s) > 0, ∀(x, y) ∈Ω, f ′ u(x, y, 0) := limu→0+ f(x, y, u)/u > 0.
(1.5) The simplest case of such a monostable function f(x, y, u) satisfying (1.5) is when f(x, y, u) = g(u) and the C1,δ function g satisfies : g(0) = g(1) = 0, g > 0 on (0, 1), g′(0) > 0 and g′(1) < 0. Such nonlinearities arise in combustion and biological models (see Fisher , Kolmogorov, Petrovsky, Piskunov , Aronson, Weinberger ). Another example of such a function f is f(x, y, u) = h(x, y) ˜ f(u) where ˜ f is as before and h is L-periodic with respect to x, Lipschitz-continuous and positive in Ω.
This section is concerned with special solutions, which are called pulsating travelling fronts (or periodic travelling fronts, see ), and which are classical time-global solutions u of (1.2) satisfying 0 ≤u ≤1 and ∀k ∈ d Y i=1 LiZ, ∀(t, x, y) ∈R × Ω, u µ t −k · e c , x, y ¶ = u(t, x + k, y), u(t, x, y) − → x·e→+∞0, u(t, x, y) − → x·e→−∞1, (1.6) where the above limits hold locally in t and uniformly in y and in the directions of Rd which are orthogonal to e. Here, e = (e1, · · · , ed) is a given unit vector in Rd. Such a solution satisfying (1.6) is then called a pulsating travelling front propagating in direction e. We say that c is the effective unknown speed c ̸= 0. Let us mention here that, without the uniformity of the limits in (1.6), many other fronts may exist, whose level sets may for instance have conical shapes (see e.g. , , ).
2In , this assumption of f being nonnegative was explicit in formula (1.7) for a function f = f(u) depending only on u. However, although this assumption was obviously also used for the general periodic nonlinearity f(x, y, u) described in , it was not mentioned there explicitely. An extension for divergence-type equations with a function f which may change sign is proved in .
5 Under the above assumptions, the first two authors proved in that there exists c∗(e) > 0 such that pulsating travelling fronts u in the direction e with the speed c exist if and only if c ≥c∗(e) ; furthermore, all such pulsating fronts are increasing in time t (other results with more general nonlinearities f were proved in , see below). The following Theorem gives a variational characterization of this minimal speed c∗(e) under an additional assumption on the nonlinearity f.
Assume that Ω, A and q satisfy (1.1), (1.3) and (1.4), and that f satisfies (1.5) and ∀(x, y, s) ∈Ω× (0, 1), 0 < f(x, y, s) ≤f ′ u(x, y, 0)s.
(1.7) Call ζ(x, y) := f ′ u(x, y, 0) and denote ˜ e the vector defined by ˜ e = (e1, · · · , ed, 0, · · · , 0) ∈RN.
Theorem 1.1 Under the above assumptions, let c∗(e) be the minimal speed of pulsating travelling fronts propagating in the direction e and solving (1.2) and (1.6). Then c∗(e) = min λ>0 k(λ) λ (1.8) where k(λ) is the principal eigenvalue of the operator Lλψ := ∇· (A∇ψ) −2λ˜ eA∇ψ + q · ∇ψ + [−λ∇· (A˜ e) −λq · ˜ e + λ2˜ eA˜ e + ζ]ψ (1.9) acting on the set E = {ψ ∈C2(Ω), ψ is L-periodic with respect to x and νA∇ψ = λ(νA˜ e)ψ on ∂Ω}.
Before studying the consequences of Theorem 1.1, let us briefly explain the formula for the minimal speed c∗and mention some earlier results about front propagation, starting from the simplest case of planar fronts in homogeneous media.
Assumption (1.7) is often called the Fisher-KPP assumption (see Fisher and Kol-mogorov, Petrovsky and Piskunov ). It is especially satisfied for the canonical example f(u) = u(1−u), or more generally when f = f(u) is a C2 concave function on [0, 1], positive on (0, 1). Thus, under the KPP assumption (1.7), the minimal speed c∗(e) can be explicitely given in terms of e, the domain Ω, the coefficients q and A and of f ′ u(·, ·, 0). We point out that the dependance of c∗(e) on the function f is only through the derivative of f with respect to u at u = 0. When Ω= RN, A = I, q = 0 and f = f(u) (with f(u) ≤f ′(0)u in [0, 1]), formula (1.8) then reduces to the well-known KPP formula c∗(e) = 2 p f ′(0) for the minimal speed of planar fronts for the reaction-diffusion equation ut = ∆u + f(u) in RN.
A planar front is a solution of the type φ(x · e −ct), where the planar profile φ solves φ′′ + cφ′ + f(φ) = 0 in R with the limiting conditions φ(−∞) = 1 and φ(+∞) = 0. Such a solution propagates with constant speed c in the direction e and its shape is invariant in the frame moving with speed c in the direction e. Many papers were devoted to such planar fronts, as well as for other classes of nonlinear functions f(u) (see e.g. , , , , ). For a detailed study of planar fronts for systems of reaction-diffusion equations, we refer to the book of Volpert, Volpert and Volpert and to the references therein.
Equations with periodic nonlinearities f(x, u) in space dimension 1, without advection, were first considered by Shigesada, Kawasaki and Teramoto , and by Hudson and Zinner 6 . The notion of travelling fronts propagating with constant speed c no longer holds in general and has to be replaced with the more general one of pulsating travelling fronts, as defined in (1.6) (see ). The profile of such a front is not invariant anymore, but, in one space dimension, the profile is periodic in time in the frame moving with speed c along the direction of propagation. In , a formula similar to the right-hand side of (1.14) in dimension 1 was given and it was proved that for any speed not smaller than the right-hand side of (1.14), then pulsating travelling fronts exist. The case of a periodic nonlinearity f(x, u) changing sign with respect to x, based on a patch invasion model in ecology was considered in and , and recently revisited from a rigorous mathematical and more general point of view in and in dimensions 1 and higher, and for more general reaction terms. Lastly, the case of periodic diffusion with bistable type nonlinearity (see (1.11) below) was investigated by Nakamura in dimension 1.
The case of shear flows q = (α(y), 0, · · · , 0) in straight infinite cylinders Ω= R × ω was dealt with by Berestycki, Larrouturou, Lions , and Berestycki and Nirenberg . Under the assumption that all coefficients of equation (1.2) do not depend on the x1 variable, the period L1 can be any arbitrary positive number and pulsating travelling fronts reduce in this case to travelling fronts φ(x1 −ct, y) which move with constant instantaneous speed c and keep a constant shape. Formula (1.13) was derived in this framework in for the minimal speed of travelling fronts with a nonlinearity f = f(u) satisfying (1.5) and (1.7). Other nonlinearities f(u) were treated in : for a combustion-type nonlinearity f such that ∃θ ∈(0, 1) (called ignition temperature), f = 0 on [0, θ] ∪{1}, f > 0 on (θ, 1), f ′(1) < 0 (1.10) (see ), there exists a unique speed c and a unique (up to shift in time, or equivalently in x1) travelling front φ(x1 −ct, y) ; for a bistable nonlinearity f such that ∃θ ∈(0, 1), f(0) = f(θ) = f(1) = 0, f < 0 on (0, θ), f > 0 on (θ, 1), f ′(0) < 0, f ′(1) < 0, (1.11) there still exists a unique speed c and a unique (up to shift) travelling front, under the additional assumption that the section ω of the cylinder is convex. Min-max type variational formulas –involving the values of f(u) for all u ∈(0, 1)– for the unique or minimal speeds of propagation of these travelling fronts were obtained by Hamel and Heinze, Papanicolaou and Stevens , generalizing some results for equations or systems , in dimension 1 (see also Benguria, Depassier for integral formulations in dimension 1, and Coutinho, Fernandez , Harris, Hudson and Zinner for similar problems with discrete diffusion).
Several lower and upper bounds for the speeds of travelling fronts in infinite cylinders, as well as some asymptotics for large advection and for other regimes, were derived by Audoly, Berestycki and Pomeau , Berestycki , Constantin, Kiselev and Ryzhik , and Heinze for combustion-type and/or general positive nonlinearities f(u). Rotating flows were also considered in and , and percolating-type flows were dealt with in , where estimates for the more general notion of bulk burning rate (see ) are given. Dirichlet type boundary conditions on ∂Ω, instead of Neumann conditions, were dealt with by Gardner and Vega in infinite cylinders. Let us also mention here that a formula similar to (1.8) for 7 a nonlinear source term f(u) of the KPP type (1.7) has recently been obtained by Schwetlick for a similar hyperbolic transport equation .
Whereas usual travelling fronts of the type φ(x1 −ct, y) exist in straight infinite cylinders in the case of shear flows –assuming that all coefficients in (1.2) are invariant with respect to the variable x1–, this is not the case anymore in infinite cylinders Ω= {(x1, y), y ∈ω(x1)} with oscillating boundaries (ω being periodic in x1), even, say, for the equation ut = ∆u+f(u) without advection.
Such a geometrical configuration was first considered for a bistable nonlinearity f by Matano , and the case of ondulating cylinders whose boundaries have small spatial periods with small amplitudes was recently dealt with by Lou and Matano .
The case of the whole space RN with periodic diffusion and advection was first considered by Xin , for a combustion-type nonlinearity f satisfying (1.10), for which the speed of propagation of the pulsating fronts was proved to be unique in any given direction. Note that usual travelling fronts propagating with constant speed and constant shape do not exist anymore for general advection or diffusion and one has to extend these notions. The homog-enization limit in RN with coefficients having small periods was investigated by Caffarelli, Lee and Mellet , Freidlin , Heinze , Majda and Souganidis , and Xin .
Heinze also considered the case of the whole space with small periodic holes . Freidlin and Xin also studied questions related to front propagation in random media.
The more general framework of periodic domains and periodic excitable media was con-sidered by the first two authors of this paper in . It was especially proved that for a nonnegative combustion-type nonlinearity f(x, y, u) satisfying the following assumptions, more general than (1.10) : f is L-periodic with respect to x, f is globally Lipschitz-continuous and ∃δ > 0, f is C1,δ with respect to u, ∃θ ∈(0, 1), ∀(x, y) ∈Ω, ∀s ∈[0, θ] ∪{1}, f(x, y, s) = 0, ∃ρ ∈(0, 1 −θ), ∀(x, y) ∈Ω, ∀1 −ρ ≤s ≤s′ ≤1, f(x, y, s) ≥f(x, y, s′), ∀s ∈(θ, 1), ∃(x, y) ∈Ω, f(x, y, s) > 0, (1.12) and given a direction e of Rd, there exists a unique effective speed of propagation c(e) and a unique (up to shift in time) pulsating travelling front u satisfying (1.2) and (1.6). As already emphasized, paper also gives the proof of the existence of a minimal speed c∗(e) of propagation of pulsating fronts for a function f satisfying (1.5). Furthermore, under the notations of Theorem 1.1, the inequality c∗(e) ≥min λ>0 k(λ) λ holds as soon as f satisfies (1.5) (see Remark 1.16 in ). However, the question of the uniqueness, up to shift, of the fronts for any given effective speed c ≥c∗(e) is still an open problem.
Remark 1.2 (Equivalent formulas) It can be easily checked in the general framework de-scribed above that formula (1.8) can be rewritten in the following equivalent formulations : c∗(e) = min {c, ∃λ > 0, k(λ) = λc} (1.13) 8 and c∗(e) = min λ>0 min ψ∈F max (x,y)∈Ω Lλψ(x, y) λψ(x, y) (1.14) where F = {ψ ∈E, ψ ∈C2(Ω), ψ > 0 in Ω}. Formula (1.14) is obtained from (1.8) and from some characterizations of principal eigenvalues of elliptic operators (, ). We also refer to for a detailed study of the above eigenvalue problems with periodic and Neumann type boundary conditions. Such operators Lλ also arise in Bloch eigenvalue problems in homogenization theory (see , , ).
The proof of formula (1.8), which was announced in , is based on the methods de-velopped in and (sub- and supersolutions, regularizing approximations in bounded domains). The authors also mention that a formula equivalent to (1.8) was recently obtained independently with different tools by Weinberger for similar problems.
1.2 Influence of the geometry of the domain and of the underlying medium As we have just seen, several equivalent variational formulas for the minimal speed of propa-gation of pulsating travelling fronts in general periodic excitable media were given. We now analyze the influence of the geometry of the domain and of the coefficients of the medium (reaction, diffusion and advection coefficients) on the minimal speed of propagation. Since the influence of these data may be opposite, we shall investigate each of them separately.
Let us first study the influence of the geometry of the domain. Under the assumptions of the previous subsection, it easily follows from formula (1.13) that even for a homogeneous equation, due to the geometry, the minimal speed c∗(e) depends continuously on e in the unit sphere Sd−1 of Rd. Note that the speed c∗(e) does depend on the direction e in general because of the geometry of the domain and because of the spatial heterogeneity of the coefficients of equation (1.2). This situation is in contrast with the homogeneous equation ut = ∆u + f(u) (1.15) in the whole space RN, for which pulsating travelling fronts are actually planar travelling fronts and the minimal speed has the same value, c∗(e) = 2 p f ′(0) in all directions e of RN.
Let us now consider the above homogeneous equation ut = ∆u + f(u), but now set in a periodic domain Ω⊂RN satisfying (1.1). Assume that the function f satisfies (1.5) and (1.7). If Ω= RN, then c∗(e) = 2 p f ′(0) for all e ∈RN with |e| = 1. The following statement shows that this value 2 p f ′(0) is always an upper bound whatever Ωis –provided it satisfies (1.1)–, and is optimal in some sense : Theorem 1.3 Let Ω⊂RN satisfy (1.1) and let f = f(u) satisfy (1.5) and (1.7).
Let e = (e1, · · · , ed) ∈Rd be such that |e| = 1. Let c∗(e) be the minimal speed of pulsating travelling fronts satisfying (1.15) and (1.6) together with the Neumann boundary conditions ∂νu = 0 on ∂Ω. Then, 0 < c∗(e) ≤2 p f ′(0), and c∗(e) = 2 p f ′(0) if and only if the domain Ωis invariant in the direction ˜ e, namely Ω+ τ˜ e = Ωfor all τ ∈R, where ˜ e = (e1, · · · , ed, 0, · · · , 0) ∈RN.
9 In other words, Theorem 1.3 implies that the presence of holes (perforations) in the domain always hinder the propagation with respect to the case of the whole space. Similarly, the fronts propagate strictly slower in an infinite cylinder with oscillating boundaries than in a straight infinite cylinder. The homogenization limit of small holes with a combustion type nonlinearity was dealt with by Heinze (see also for homogenization of linear diffusion equations with small holes).
After having proved that holes make the propagation of pulsating fronts slower than in the case of the whole space RN, it is now natural to wonder whether the minimal speed c∗(e) is all the smaller the bigger the holes. Actually, the answer is no in general : Theorem 1.4 Let N ≥2 and e be any unit direction in RN. Let f = f(u) satisfy (1.5) and (1.7). Then there exist some positive numbers L1, · · · , LN, a family of domains (Ωα)0≤α<1 satisfying (1.1) with d = N and Ω0 = RN, Ωα ⊃Ωα′ for all 0 ≤α ≤α′ < 1, \ 0≤α<α′ Ωα = Ωα′ for all 0 < α′ < 1, (1.16) such that, if cα denotes the minimal speed cα = c∗(e, Ωα) of the pulsating fronts satisfying (1.15) and (1.6) in Ωα with Neumann boundary conditions on ∂Ωα, then the function α 7→cα is continuous on [0, 1), c0 = 2 p f ′(0), cα < 2 p f ′(0) for all α ∈(0, 1) and cα →2 p f ′(0) as α →1−.
Theorem 1.4 says that the minimal speed of propagation for the homogeneous equation (1.15) may not be monotone with respect to the size of the holes. Furthermore, under the notations of Theorem 1.4, one can say that there exists at least one value of α0 in (0, 1) for which the minimal speed of pulsating fronts is minimal in Ωα0 among all the domains Ωα for 0 < α < 1.
Remark 1.5 Theorem 1.3 no longer holds for equations with periodic heterogeneous co-efficients even if the equation is invariant in direction ˜ e.
For instance, let Ω′ ⊂RN−1 be a periodic domain satisfying (1.1) with d ≤N −1, and such that Ω′ ̸= RN−1.
Let Ω= Ω′ × R = {x = (x′, xN), x′ ∈Ω′, xN ∈R}. Let f(x, u) be a function satisfying (1.5) and (1.7), and assume that f(x, u) is written as f(x, u) = h(x′) ˜ f(u), where ˜ f satisfies (1.7), 0 < h(x′) ≤1 in RN−1, h(x′) = 1 in Ω′ and h ̸≡1 in RN−1. Let e = eN be the unit vector in the xN-direction. Then the minimal speed of propagation of pulsating fronts solving ut = ∆u + f(x, u) and (1.6) in Ω, together with ∂νu = 0 on ∂Ω, is equal to 2 q ˜ f ′(0). But the minimal speed for the same equation set in the whole space RN is strictly less than 2 q ˜ f ′(0) (see the proof of Theorem 1.6 below for more details).
Let us now investigate the influence of the reaction coefficients on the minimal speed of propagation.
10 Theorem 1.6 Under the assumptions (1.1), (1.3) and (1.4), let f = f(x, y, u), resp. g = g(x, y, u), be a nonnegative nonlinearity satisfying (1.5) and (1.7). Let e be a unit direction of Rd and let c∗(e, f), resp. c∗(e, g), be the minimal speed of propagation of pulsating fronts solving (1.2) and (1.6) with nonlinearity f, resp. g.
a) If f ′ u(x, y, 0) ≤g′ u(x, y, 0) for all (x, y) ∈Ω, then c∗(e, f) ≤c∗(e, g), and if f ′ u(x, y, 0) ≤ , ̸≡g′ u(x, y, 0), then c∗(e, f) < c∗(e, g).
b) If c∗(e, Bf) denotes the minimal speed for the nonlinearity Bf, with B > 0, then c∗(e, Bf) is increasing in B and lim sup B→+∞ c∗(e, Bf) √ B < +∞.
Furthermore, if Ω= RN or if νA˜ e ≡0 on ∂Ω, then lim infB→+∞c∗(e, Bf)/ √ B > 0.
Part a) of Theorem 1.6 follows immediately from Theorem 1.1 (note that similar mono-tonicity results also hold for equations with nonlinearities changing sign, see , ). Notice that the inequality c∗(e, f) ≤c∗(e, g) holds as soon as f and g satisfy (1.5) and f ≤g, even if f or g do not satisfy (1.7) (this inequality follows from the construction of the minimal speed by approximation of speeds of fronts with combustion-type nonlinearities satisfying (1.12), see and Remark 1.7 below). However, the strict inequality c∗(e, f) < c(e, g) does not hold in general if f ≤g and f ̸≡g, even if f and g satisfy (1.5) and (1.7) : indeed, under these assumptions, the dependence on f of the minimal speed c∗(e, f) is only through its derivative f ′ u(x, y, 0) at u = 0+.
The condition νA˜ e ≡0 on ∂Ωespecially holds if A˜ e is constant and if Ωis invariant in this direction A˜ e (for instance, A = I and Ωis a straight infinite cylinder in direction ˜ e). Notice that parts a) and b) of Theorem 1.6 obviously hold for the KPP formula c∗= 2 p Bf ′(0) in the case of the homogeneous equation (1.15) in RN with nonlinearity Bf. However, the precise asymptotic behavior of c∗(e, Bf)/ √ B as B →+∞is not known in general.
Lastly, part b) also holds good if the nonlinearity Bf is replaced by a nonlinearity of the type Bf + f0, with given f and f0 satisfying (1.5) and (1.7).
Remark 1.7 Similar comparison properties as in Theorem 1.6 also hold for the unique speeds c(e, f) and c(e, g) of the pulsating fronts solving (1.2) and (1.6) in the case where the nonnegative nonlinearities f = f(x, y, u) and g = g(x, y, u) satisfy (1.12) and are ordered.
Namely, if f ≤g in Ω× [0, 1], then c(e, f) ≤c(e, g). Furthermore, in this framework, one has c(e, f) < c(e, g) if f ≤g and f ̸≡g. These facts follow easily from the proofs in .
However, the behaviour of c(e, Bf) for large B is not known in this case.
The influence of advection on the speed of propagation is more difficult to analyze, because of possible interaction between the stream lines and the geometry of the domain, especially the holes. However, at least in the case where the domain is invariant in the direction ˜ e, with isotropic diffusion, one can compare the speeds of propagation in direction e when there is, or not, a drift term in the equation.
Theorem 1.8 Let Ω⊂RN be a domain satisfying (1.1) and Ω+ τ˜ e = Ωfor all τ ∈R, where e is a unit vector of Rd. Assume that A = I and that f = f(u) satisfies (1.5) and 11 (1.7), and that q satisfies (1.4). Let c∗ q(e) be the minimal speed of the pulsating fronts solving (1.2) and (1.6), with advection coefficient q. Then c∗ q(e) ≥c∗ 0(e) = 2 p f ′(0) and equality holds if and only if q · ˜ e ≡0 in Ω.
Under the above assumptions, Theorem 1.8 means that the advection, or stirring, makes the propagation faster, whatever the flow is a shear flow or not. Roughly speaking, the presence of turbulence in the medium increases the speed of propagation of the pulsating fronts. Furthermore, the influence of advection on the speed of propagation is minimal if and only if the advection is orthogonal to the direction of propagation.
The influence of large periodic advection, namely where q is replaced with Bq with large B, is analyzed by the authors in . The behaviour of c∗ Bq(e) is always at most linear in B for large B, in a general domain Ωwhich satisfies (1.1) but may not be invariant in the direction ˜ e. A necessary and sufficient condition for c∗ Bq(e) to be at least linear in B is given in , involving the first integrals of the velocity field q.
Remark 1.9 It is not clear in general whether, under the assumptions of Theorem 1.8, c∗ Bq(e) is nondecreasing with respect to B > 0 or not. However, in the case of a shear flow q = α(x2, · · · , xN)e1 in a straight cylinder Ω= R×ω in the direction e1, with, say, ω bounded in RN−1, α ̸≡0 of class C1 and with zero average, the first author proved in that c∗ Bq(e1) is increasing with B > 0, c∗ Bq(e1)/B is decreasing with B > 0 and c∗ Bq(e1)/B →ρ > 0 as B →+∞.
As far as the influence of the diffusion coefficients is concerned, one can compare the min-imal speed of propagation in the case of heterogeneous diffusion with that of a homogeneous diffusion in a given direction e. The following theorem also gives a monotonicity result of the speed of propagation with respect to the intensity of diffusion : Theorem 1.10 Under the assumptions (1.1), (1.3), (1.5) and (1.7), let q = 0. Let e be a unit direction of Rd. Then, 1) c∗(e) ≤2 p M0M, (1.17) where M0 = max(x,y)∈Ωζ(x, y) and M = max(x,y)∈Ω˜ eA(x, y)˜ e. Furthermore, the equality holds in (1.17) if and only if ζ and ˜ eA˜ e are constant, ∇· (A˜ e) ≡0 in Ωand νA˜ e = 0 on ∂Ω (if ∂Ω̸= ∅).
2) Assume furthermore that f = f(u) depends on u alone. Let c∗ γ(e) denote the minimal speed of pulsating fronts in the direction e, with diffusion matrix γA, where γ > 0. Then c∗ α(e) ≤c∗ β(e) if 0 < α ≤β.
As a special case of (1.17) we see that c∗ α(e) ≤C√α for all α > 0, where C does not depend on α > 0. Furthermore, part 2) implies that a larger diffusion speeds up the propagation.
12 Remark 1.11 The assumption q = 0 was made for the sake of simplicity in the derivation of the upper bound (1.17). However, more general bounds can be obtained when q ̸= 0.
Namely, under the assumptions (1.1), (1.3), (1.4), (1.5) and (1.7), one gets as in the proof of Theorem 1.10 : c∗(e) ≤2 p M0M + max (x,y)∈Ω(−q(x, y) · ˜ e), where M0 and M are the same as in Theorem 1.10.
Lower bounds can be obtained as well, but are more restrictive.
Namely, under the assumptions (1.1), (1.3), (1.4), (1.5) and (1.7), assume furthermore that Ω= RN or νA˜ e = 0 on ∂Ωif ∂Ω̸= ∅. Then c∗(e) ≥min ³m0m b , −b + 2√m0m ´ , (1.18) under the convention that m0m/b = +∞if b = 0, where m0 = min(x,y)∈Ωζ(x, y), m = min(x,y)∈Ω˜ eA(x, y)˜ e and b = ∥∇· (A˜ e)∥L∞(Ω) + ∥q · ˜ e∥L∞(Ω). Formula (1.18) is proved in Section 3.2. It especially implies that, if q · ˜ e ≡0, then lim infε→0+ c∗ ε(e)/√ε > 0, where c∗ ε(e) denotes the minimal speed of pulsating fronts in the direction e with diffusion matrix εA.
1.3 Spreading speed in periodic domains The question of the stability of travelling fronts and the asymptotic convergence to travelling fronts for the solutions of Cauchy problems of the type (1.2) with “front-like” initial condi-tions has been thoroughly studied since the pioneering paper by Kolmogorov, Petrovsky and Piskunov in the one-dimensional case (see e.g. , , , , , , , , , , , for other stability results in the homogeneous 1d case, for the homo-geneous multidimensional case, or , , , , for the case of infinite cylinders with shear flows). However, few results (see , , ) have so far been obtained about the stability of pulsating travelling fronts in periodic media.
Another important notion is that of asymptotic speed of propagation, or spreading (see below for precise meaning), for solutions of Cauchy problem like (1.2) with nonnegative continuous compactly supported initial condition u0 ̸≡0. This problem for the homogeneous equation (1.15) in RN was solved by Aronson and Weinberger . They proved that, under the above assumptions on u0 and if f satisfies (1.5) and lim infu→0+ f(u)/u1+2/N > 0,3 then min |z|≤ct u(t, z) →1 if 0 ≤c < c∗and max |z|≤ct u(t, z) →0 if c > c∗, as t →+∞, where c∗is the minimal speed of planar fronts. The speed c∗can then also be viewed as a spreading speed (see , , , , , for similar results with other nonlinearities f(u) in dimensions 1 or higher). These spreading properties were generalized by Mallordy and Roquejoffre , for equations with shear flows in straight infinite cylinders.
The case of a reaction-diffusion equation (1.2) without advection in the whole space RN with periodic coefficients was considered in the important work of G¨ artner and Freidlin and later by Freidlin in the case with advection q (the proofs in and used 3The latter is fulfilled if f satisfies (1.7) as well.
13 probabilistic tools). Namely, under the assumptions (1.3), (1.4), (1.5)4 and (1.7), if u0 is nonnegative, continuous and compactly supported, then the solution u(t, z) of (1.2) in RN with initial condition u0 is such that, for any unit vector e of RN, u(t, z +c t e) →1 if 0 ≤c < w∗(e) and u(t, z +cte) →0 if c > w∗(e), as t →+∞, (1.19) locally in x ∈RN. Furthermore, G¨ artner and Freidlin derive a formula which we call the G¨ artner-Freidlin formula : w∗(e) = min ⃗ λ·e>0 ˜ k(⃗ λ) ⃗ λ · e (1.20) and ⃗ λ ∈RN and ˜ k(⃗ λ) is the first eigenvalue of the operator L⃗ λ := ∇· (A∇) −2⃗ λA∇+ q · ∇+ [−∇· (A⃗ λ) −q · ⃗ λ + ⃗ λ · A⃗ λ + ζ] with L-periodicity condition (as a consequence, w∗(±1) = c∗(±1) in dimension N = 1).
The speed w∗(e) can then be viewed as a ray speed in the direction e.
It follows from (1.8) that w∗(e) ≤c∗(e). Notice that the latter can also be easily obtained from (1.19) and the parabolic maximum principle, putting u0 below a pulsating front moving with speed c∗(e) in the direction e, even if it means changing f into a function ˜ f such that ˜ f ≥f, ˜ f ′ u(z, 0) = f ′ u(z, 0) and (1.7) holds for ˜ f).
Let us also mention that several works have dealt with the solutions of Cauchy problems for equations of the type (1.2), with small diffusion ε, together with large reaction ε−1f, or with slowly varying flows of the type q(εz), or for equations involving more general spatio-temporal scales. Typically, the solutions of such Cauchy problems converge as ε →0+ to two-phase solutions of Hamilton-Jacobi type equations, separated by interfaces (see e.g. , , , , ). The determination of the asymptotic speed of propagation was also studied for nonlinear integral equation in dimension 1 (see , , , ), or for systems of reaction-diffusion equations in dimension 1 (see ).
Recently, Weinberger extended the results of G¨ artner and Freidlin to the general periodic framework described in and here, with possible time-discrete equations. Under assumptions (1.1), (1.3), (1.4), (1.5) and (1.7), it is proved in that, for any unit direction e of Rd, there exists w∗(e) > 0 such that, if u(t, x, y) solves (1.2) with a nonnegative, continuous and compactly supported initial condition u0 ̸≡0, then, ½ u(t, x + cte, y) →1 if 0 ≤c < w∗(e) u(t, x + cte, y) →0 if c > w∗(e), as t →+∞, (1.21) locallyin (x, y) with respect to the points (x, y) such that (x + cte, y) ∈Ω. Furthermore, {ρξ, ξ ∈Sd−1, 0 ≤ρ ≤w∗(ξ)} := {x ∈Rd, x · ξ ≤c∗(ξ) for all ξ ∈Sd−1}, (1.22) i.e. w∗(e) = minξ∈Rd, e·ξ>0 c∗(ξ)/(e · ξ), or w∗(e) = min ⃗ λ·e>0 ˜ k(⃗ λ) ⃗ λ · e , (1.23) 4The function f = f(z, u) was actually assumed in to be positive in RN × (0, 1).
14 with ⃗ λ ∈Rd and ˜ k(⃗ λ) being the principal eigenvalue of the operator L⃗ λ := ∇· (A∇) − 2˜ λA∇+ q · ∇+ [−∇· (A˜ λ) −q · ˜ λ + ˜ λA˜ λ + ζ] acting on the set ˜ E = {ψ ∈C2(Ω), ψ is L-periodic with respect to x and νA∇ψ = (νA˜ λ)ψ on ∂Ω} (we set ˜ λ = (⃗ λ, 0, · · · , 0) ∈RN).
Remark 1.12 As already emphasized, it is clear from the parabolic maximum principle that w∗(e) ≤c∗(e) for all e ∈Sd−1. The latter could also be viewed as a consequence of (1.8) and (1.20) in the case of equation (1.2) in RN, or from (1.22-1.23) in the general periodic case.
The equality w∗(e) = c∗(e) holds for the homogeneous isotropic equation ut = ∆u + f(u) in RN, for all direction e, but it does not hold in general. Indeed, consider the equation ut = a2ux1x1 + b2ux2x2 + f(u) in R2, where a > 0 and b > 0 are two given constants, and f = f(u) satisfies (1.5) and (1.7). From the above formulas for w∗(e) or c∗(e), it is easy to see that, for all θ ∈R and e = (cos θ, sin θ), w∗(e) = 2 p f ′(0) r a2b2 a2 sin2 θ + b2 cos2 θ, c∗(e) = 2 p f ′(0) p a2 cos2 θ + b2 sin2 θ (notice that the formula for w∗(e) could also be deduced from the case of isotropic diffusion after scaling). Hence, the equality w∗(e) = c∗(e) holds here if and only if e = (±1, 0) or (0, ±1), or if a = b (isotropic diffusion). In other words, in the case of anisotropic diffusion (a ̸= b), the asymptotic spreading speed is less than the minimal speed of pulsating fronts in any direction which is not an eigenvector of the diffusion matrix. Notice also that the curve r(θ) = w∗(cos θ, sin θ) in polar coordinates is an ellipse, while the curve r(θ) = c∗(cos θ, sin θ) is not an ellipse in general (but the curve r(θ) = (c∗(cos θ, sin θ))−1 is an ellipse).
Some numerical simulations with isotropic but heterogeneous diffusion have been per-formed in , confirming that the radial speed w∗(e) may be less that the minimal speed c∗(e) of pulsating fronts. We conjecture that, by analogy, the strict inequality w∗(e) < c∗(e) may also occur in some directions e in some domains with holes. However, a condition for the equality w∗(e) = c∗(e) to hold or not is not known in general in the periodic setting.
In the sequel, we discuss some properties of the spreading speed w∗(e) in periodic domains.
As for the minimal speed of pulsating fronts, we study the influence on the speed w∗(e) of all the phenomena involved in problem (1.2).
As in Theorem 1.3, let us first consider the case of the homogeneous equation (1.15) in a periodic domain Ω. Since w∗(e) ≤c∗(e) for any unit direction e ∈Sd−1, it follows from Theorem 1.3 that w∗(e) ≤2 p f ′(0) and that, if w∗(e) = 2 p f ′(0), then Ωis a straight infinite cylinder in the direction ˜ e. Conversely, if Ωis a straight infinite cylinder in the direction ˜ e, then c∗(e) = 2 p f ′(0) by Theorem 1.3; furthermore, the last equality holds for w∗(e) as well, namely : Theorem 1.13 Under the assumptions (1.1) for Ω(with d ≥1), and (1.5) and (1.7) for f = f(u), let e be a unit direction of Rd and u(t, x, y) be the solution of (1.15) with a given initial condition u0 ̸≡0 which is nonnegative, continuous and compactly supported. Then w∗(e) ≤2 p f ′(0) and equality holds if and only if Ωis invariant in the direction ˜ e.
15 Theorem 1.13 rests on the following Liouville type result : Proposition 1.14 Let Ωsatisfy (1.1). Let g : [0, +∞) →R be a C1 function such that g(0) = g(1) = 0, g′(0) > 0, g > 0 in (0, 1) and g < 0 in (1, +∞), and let b ∈RN be such that |b| < 2 p g′(0). Let u be a classical bounded solution of ∆u + b · ∇u + g(u) = 0 in Ω u ≥0 in Ω ∂νu = 0 on ∂Ω.
(1.24) Then u ≡0 or u ≡1.
This result, which is of independent interest, is a Liouville type result for some solutions of semi-linear elliptic equations in periodic domains. If u were assumed to be L-periodic and not identically equal to 0, then the conclusion u ≡1 would follow immediately from the strong maximum principle, since u would then be bounded from below by a positive constant (see case 1 of the proof of Proposition 1.14 in section 4). The difficultly here is that u is not assumed to be L-periodic a priori. Let us also mention that the conclusion of Proposition 1.14 was known in the case Ω= RN, and was proved by Aronson and Weinberger , by using parabolic tools (see also Remark 4.3 below). The proof of Proposition 1.14 given in Section 4 rests on some sliding arguments and on the elliptic maximum principle.
The influence of all other phenomena (reaction, diffusion and advection) is summarized in the following propositions, most of which are consequences of the results stated in Section 1.2.
Let us start with the dependency on the reaction terms.
Proposition 1.15 Under the assumptions (1.1), (1.3), (1.4), let e be a unit direction of Rd and let f and g be two functions satisfying (1.5) and (1.7). Call w∗(e, f) and w∗(e, g) the spreading speeds in the direction e for problem (1.2) with nonlinearities f and g respectively.
If f ′ u(x, y, 0) ≤g′ u(x, y, 0) for all (x, y) ∈Ω, then w∗(e, f) ≤w∗(e, g).
If f ′ u(x, y, 0) ≤, ̸≡g′ u(x, y, 0), then w∗(e, f) < w∗(e, g). Hence, w∗(e, Bf) is increasing in B > 0. Furthermore, lim sup B→+∞ w∗(e, Bf) √ B < +∞.
Lastly, if Ω= RN, then lim infB→+∞w∗(e, Bf)/ √ B > 0.
The next result is about the influence of stirring on propagation.
Proposition 1.16 Let Ω= RN, A = I and assume that f = f(u) satisfies (1.5) and (1.7).
For any unit vector e of RN, call w∗ q(e) the spreading speed in direction e, with advection term q satisfying (1.4). Then, w∗ q(e) ≥w∗ 0(e) = 2 p f ′(0), and the equality w∗ q(e) = 2 p f(0) holds if and only if q · e ≡0 in RN.
16 The last proposition is concerned with the influence of the diffusion on the asymptotic spreading speed.
Proposition 1.17 Under the assumptions (1.1), (1.3), (1.5) and (1.7), let e be a unit direction of Rd. Assume moreover that q = 0. Then, w∗(e) ≤2 p M0M.
Furthermore, if Ω= RN, then w∗(e) ≥min(m0α1/˜ b, −˜ b + 2√m0α1), where α1 was given in (1.3), m0 = minx∈RN f ′ u(x, 0) and ˜ b = max x∈RN, ⃗ µ∈RN, ⃗ µ̸=0 |∇· (A(x)⃗ µ)|/|⃗ µ| + max x∈RN |q(x)|.
Lastly, if f = f(u) and w∗ γ(e) denotes the spreading speed in the direction e, with diffusion matrix γA, then w∗ α(e) ≤w∗ β(e) if 0 < α ≤β.
The proofs of the above Propositions are sketched in Remarks 3.2, 3.3 and 3.4 in Section 3 below.
2 Variational formula for the minimal speed of pulsat-ing travelling fronts This section is devoted to the proof of formula (1.8) in Theorem 1.1. One assumes all the hypotheses in Theorem 1.1, and e denotes a unit vector of Rd.
Let us first collect some useful properties of the first eigenvalue k(λ) of the operator Lλ given in (1.9).
Lemma 2.1 The function λ 7→k(λ) is a convex function of λ. Furthermore, there exists a convex function k0 such that k0(0) = k′ 0(0) = 0 and ∀λ ∈R, 0 < min Ω ζ ≤min Ω ζ + k0(λ) ≤k(λ) ≤max Ω ζ + k0(λ).
(2.1) Proof. Up to a change of notations (q into −q, and e into −e) in the equations in , the first eigenvalue k(λ) of the operator Lλ corresponds to the eigenvalue −µγ,ζ(λ) + λγ of the operator Lγ,λ,ζ + λγ in Proposition 5.7 of . From parts (ii) and (iii) of Proposition 5.7 of , it follows that ∀λ ∈R, min Ω ζ + k0(λ) ≤k(λ) ≤max Ω ζ + k0(λ), where k0(λ) is the first eigenvalue of the operator Lλ −ζ, and k0(0) = k′ 0(0) = 0 (k0(λ) corresponds to −h(λ) in Proposition 5.7 of ). It follows from that the function k0 is convex. As a consequence, k0 is a nonnegative function, and (2.1) follows.
17 Furthermore, as in , the first eigenvalue k(λ) can be rewritten as k(λ) = min ψ∈F max Ω Lλψ ψ = min ˜ ψ∈˜ F max (x,y)∈Ω Ã ∇· (A∇˜ ψ) + q · ∇˜ ψ ˜ ψ(x, y) + ζ(x, y) !
, (2.2) where F = {ψ ∈C2(Ω), ψ is L-periodic with respect to x, νA∇ψ = λ(νA˜ e)ψ on ∂Ωand ψ > 0 in Ω}, and ˜ F = {(x, y) 7→ψ(x, y)e−λx·e, ψ ∈F} = { ˜ ψ ∈C2(Ω), ˜ ψeλx·e is L-periodic w.r.t. x, νA∇˜ ψ = 0 on ∂Ωand ˜ ψ > 0 in Ω}.
It follows from the last expression of k(λ) in (2.2), as in Proposition 5.7 of , that the function k is convex with respect to λ.
The main result of this section is the following Proposition 2.2 If c ∈R satisfies c > inf {γ ∈R; ∃λ > 0, k(λ) = λγ}, then c > 0 and there exists a solution u(t, x, y) of (1.2) and (1.6), namely u is a pulsating travelling front propagating in the direction e with the effective speed c.
This proposition is proved at the end of this section. Let us now turn to the Proof of Theorem 1.1. As already emphasized, it follows from Remark 1.16 and Section 6.4 in that, for all pulsating travelling front propagating in the direction e, with speed c ≥c∗(e), there exists λ > 0 such that k(λ) = λc. Therefore, c∗(e) ≥inf {c, ∃λ > 0, k(λ) = λc}.
(2.3) From Proposition 2.2 above, inequality (2.3) turns out to be an equality. Furthermore, the infimum is reached since for c = c∗(e), there still exists λ∗> 0 such that k(λ∗) = λ∗c∗(e).
Eventually, one concludes that inf λ>0 k(λ) λ = inf {c, ∃λ > 0, k(λ) = λc} = c∗(e) = k(λ∗) λ∗, whence c∗(e) = minλ>0 k(λ)/λ. That completes the proof of Theorem 1.1.
Remark 2.3 Since c∗(e) > 0, it follows from the above proof and Lemma 2.1 that the function λ 7→k(λ)/λ is continuous on R∗ + and k(λ) λ →+∞as λ →0+, and lim inf λ→+∞ k(λ) λ > 0.
18 Let us now turn to the Proof of Proposition 2.2. The proof follows the lines of those of and , together with the additional assumption (1.7), and we just outline it.
Let c be as in Proposition 2.2 and let c′ < c and λ′ > 0 be such that k(λ′) = λ′c′. Let ψ′ ∈E be the unique (up to multiplication) positive principal eigenfunction of Lλ′ψ′ = k(λ′)ψ′ in Ω.
Let us first observe that k(λ′) is positive from Lemma 2.1, whence c′ and c are positive as well.
Finding a classical C2(R × Ω) solution u(t, x, y) of (1.2) and (1.6) is the same, up to the change of variables u(t, x, y) = φ(x · e −ct, x, y), φ(s, x, y) = u µx · e −s c , x, y ¶ , as proving the existence of a function φ ∈C2(R × Ω) solving Lφ + f(x, y, φ) := ∇x,y · (A∇x,yφ) + (˜ eA˜ e)φss + ∇x,y · (A˜ eφs) + ∂s(˜ eA∇x,yφ) +q · ∇x,yφ + (q · ˜ e + c)φs + f(x, y, φ) = 0 in R × Ω φ(−∞, ·, ·) = 1, φ(+∞, ·, ·) = 0 (uniform limits in (x, y) ∈Ω) φ is L-periodic with respect to x νA(∇x,yφ + ˜ eφs) = 0 on R × ∂Ω.
The existence of a solution φ of the above problem shall be proved by solving regularized elliptic equations of the type Lεφ + f(x, y, φ) := Lφ + εφss + f(x, y, φ) = 0, where ε > 0, in cylinders of the type Σa = {(s, x, y), −a < s < a, (x, y) ∈Ω} which are bounded in the variable s. One shall then pass to the limits a →+∞and ε →0+.
To this end, let us first fix a > 0. The number ε > 0 shall be chosen later. Let us now extend the function f by f(x, y, u) = 0 for all u ≥1 and (x, y) ∈Ω. For r ∈R, let vr be the function defined by vr(s, x, y) = e−λ′(s+r)ψ′(x, y) for all (s, x, y) ∈R × Ω. This function vr is a supersolution for ε > 0 small enough and for all r ∈R, in the sense that, from (1.7) and from the definition of λ′ and ψ′, Lεvr + f(x, y, vr) ≤ [∇· (A∇ψ′) + (λ′)2(˜ eA˜ e)ψ′ −2λ′˜ eA∇ψ′ −λ′∇· (A˜ e)ψ′ +q · ∇ψ′ −λ′(q · ˜ e + c)ψ′ + ε(λ′)2ψ′]e−λ′(s+r) + ζ(x, y)vr ≤ [k(λ′) −λ′c + ε(λ′)2]ψ′e−λ′(s+r) ≤ λ′(c′ −c + ελ′)ψ′e−λ′(s+r) ≤ 0 as soon as 0 < ε ≤(c −c′)/λ′ (this is possible since c′ < c and λ′ > 0). Furthermore, the function vr satisfies νA(∇x,yvr + ˜ e∂svr) = [νA∇ψ′ −λ′(νA˜ e)ψ′]e−λ′(s+r) = 0 on R × ∂Ω 19 because of the definition of ψ′. Lastly, the function v′ r := min(vr, 1) is therefore a supersolu-tion in the above sense as well.
For all r ∈R, let hr be the positive constant defined by 0 < hr := min (x,y)∈Ωv′ r(a, x, y) ≤1.
The constant function hr clearly satisfies Lεhr + f(x, y, hr) = f(x, y, hr) ≥0 in R × Ω, together with νA(∇x,yhr + ˜ e∂shr) = 0 on R × ∂Ω. Furthermore, hr ≤v′ r(s, x, y) for all (s, x, y) ∈Σa since v′ r is nonincreasing with respect to s.
From the general results of Berestycki and Nirenberg (see also Lemma 5.1 in ), there exists a solution wr ∈C(Σa) ∩C2(Σa{±a} × ∂Ω) of Lεwr + f(x, y, wr) = 0 in Σa νA(∇x,ywr + ˜ e∂swr) = 0 on (−a, a) × ∂Ω wr is L-periodic with respect to x wr(−a, x, y) = v′ r(−a, x, y) for all (x, y) ∈Ω wr(a, ·, ·) = hr 0 < hr ≤wr(s, x, y) ≤v′ r(s, x, y) for all (s, x, y) ∈Σa, (2.4) as soon as 0 < ε ≤(c −c′)/λ′. Furthermore, since the function v′ r is nonincreasing with respect to s and since the coefficients of Lε · +f(x, y, ·) do not depend on the variable s, it follows that the function wr is actually unique and it is nonincreasing with respect to s. This can be done as in Lemma 5.2 in , by using the same sliding method as in . Lastly, the same device as in Lemma 5.3 in yields that wr is nonincreasing with respect to r, and that the function r 7→wr is continous with respect to r in C2,α loc (Σa{±a} × ∂Ω) (for all 0 < α < 1) and in C(Σa).
Since 0 ≤hr ≤wr ≤v′ r ≤1 in Σa and hr →1 (resp. v′ r →0) uniformly in Σa as r →−∞ (resp. r →+∞), one finally concludes that, for each ε ∈(0, (c −c′)/λ′] and for all a > 0, there exists a unique rε,a ∈R such that the function wε,a := wrε,a satisfies (2.4) and max (x,y)∈Ωwε,a(0, x, y) = 1/2.
Let ε ∈(0, (c −c′)/λ′] be fixed and consider a sequence an →+∞. From the standard elliptic estimates up to the boundary, the functions wε,an converge, up to extraction of some subsequence, in C2,α loc (R × Ω) (for all 0 < α < 1) to a function wε solving Lεwε + f(x, y, wε) = 0 in R × Ω νA(∇x,ywε + ˜ e∂swε) = 0 on R × ∂Ω wε is L-periodic with respect to x 0 ≤wε ≤1, max (x,y)∈Ωwε(0, x, y) = 1/2.
(2.5) Furthermore, the function wε is nonincreasing with respect to s.
20 From the monotonicity of wε with respect to s and from the standard elliptic estimates, it follows that wε(s, x, y) →φ±(x, y) in C2,α(Ω) as s →±∞, where the functions φ± satisfy ∇· (A∇φ±) + q · ∇φ± + f(x, y, φ±) = 0 in Ω νA∇φ± = 0 on ∂Ω φ± is L-periodic with respect to x 0 ≤φ+ ≤φ−≤1.
(2.6) Integrating by parts over the cell C leads to Z C f(x, y, φ±(x, y)) dx dy = 0, whence f(x, y, φ±(x, y)) ≡0 in Ωby continuity. Now multiply equation (2.6) by φ± and integrate by parts over C. It follows that Z C ∇φ±A∇φ± = 0 and that φ± are constants. From the monotonicity of wε and the normalization of wε on the section {0} × Ω, together with assumption (1.7), one concludes that φ+ = 0 and φ−= 1.
Let us now come back to the variables (t, x, y). For ε ∈(0, (c −c′)/λ′), the functions uε defined by uε(t, x, y) = wε(x · e −ct, x, y) for all (t, x, y) ∈R × Ω satisfy uε t = ∇· (A∇x,yuε) + ε c2uε tt + q · ∇x,yuε + f(x, y, uε) in R × Ω νA∇x,yuε = 0 on R × ∂Ω ∀k ∈L1Z × · · · × LdZ, ∀(t, x, y) ∈R × Ω, uε µ t −k · e c , x, y ¶ = uε(t, x + k, y) max x·e=ct, (t,x,y)∈R×Ωuε(t, x, y) = 1/2.
(2.7) Furthermore, each function uε is nondecreasing in the variable t and uε(t, x, y) →1 (resp.
→0) as t →+∞(resp. t →−∞) in C2 loc(Ω).
As in Lemma 5.11 in , by multiplying the equation (2.5) by 1, wε and ∂swε and integrating by parts over R × C, it follows that, for every compact set K ⊂Ω, there exists a constant C(K) independent of ε such that Z R×K [(uε t)2 + |∇x,yuε|2] dt dx dy ≤C(K) µ1 + N∥q∥2 ∞ 2α1 + 2 max (x,y)∈ΩF(x, y, 1) ¶ , where F(x, y, t) = Z t 0 f(x, y, τ)dτ and α1 is given from (1.3).
21 Let (εn)n ∈(0, (c −c′)/λ′] be a sequence converging to 0+.
There exists a function u ∈H1 loc(R×Ω) such that, up to extraction of some subsequence, the functions uεn converge, in L2 loc(R × Ω) strong, H1 loc(R × Ω) weak and almost everywhere in R × Ω, to a function u.
From parabolic regularity, the function u is then a classical solution of ut = ∇· (A∇x,yu) + q · ∇x,yu + f(x, y, u) in R × Ω νA∇x,yu = 0 on R × ∂Ω ∀k ∈L1Z × · · · × LdZ, ∀(t, x, y) ∈R × Ω, u µ t −k · e c , x, y ¶ = u(t, x + k, y) 0 ≤u ≤1 and ut ≥0 in R × Ω.
Furthermore, from the normalization of uε on the set {x · e = ct} and from the monotonicity of uε in t, one has u(t, x, y) ≤1/2 for all (t, x, y) such that x · e ≤ct.
(2.8) On the other hand, equation (2.7) is an elliptic regularization of a parabolic equation.
From Theorem A.1 in 5 (it is easy to check that assumptions are satisfied, especially the functions uε are of class C3(R × Ω) from the regularity assumptions and from the standard elliptic estimates), the following gradient estimates hold : ∥∇x,yuε∥L∞(R×Ω) ≤C, (2.9) where C is independent of ε.
Since maxx·e=ct uε(t, x, y) = 1/2 and uε(t −k · e/c, x, y) = uε(t, x + k, y) in R × Ωfor all k ∈L1Z × · · · × LdZ, there exists a sequence of points (tn, xn, yn) ∈R × C such that xn · e = ctn and uεn(tn, xn, yn) = 1/2. Therefore, the sequence (tn, xn, yn)n is bounded and converges, up to extraction of some subsequence, to a point (t, x, y) ∈R × C such that x · e = ct. Choose any η > 0. From the uniform gradient estimates (2.9), there exists r > 0 such that uεn(tn, x, y) ≥1/2−η for all n and for all (x, y) ∈Br(xn, yn)∩Ω, where Br(xn, yn) denotes the euclidian closed ball in RN of radius r and center (xn, yn). Since each uε is nondecreasing in t, it follows that, for n large enough, uεn(t, x, y) ≥1/2 −η for all t ≥tn and (x, y) ∈Br/2(x, y) ∩Ω. Since uεn converges to the continous function u almost everywhere, one gets that u(t, x, y) ≥1/2 −η for all t ≥t and for all (x, y) ∈Br/2(x, y) ∩Ω. Since η > 0 was arbitrary, it follows that u(t, x, y) ≥1/2. From (2.8) and the (t, x) periodicity of u, one concludes that max x·e=ct, (t,x,y)∈R×Ωu(t, x, y) = 1/2.
(2.10) 5We also refer to Theorem 1.6 in for more general estimates of a class of elliptic regularizations of degenerate equations.
22 Lastly, the standard parabolic estimates together with the monotonicity of u with respect to t imply that u(t, x, y) →u±(x, y) in C2 loc(Ω) as t →±∞, where the functions u± satisfy ∇· (A∇u±) + q · ∇u± + f(x, y, u±) = 0 in Ω νA∇u± = 0 on ∂Ω u± are L-periodic with respect to x and are such that 0 ≤u−≤u+ ≤1. As explained earlier for the functions φ± solving (2.6), one can easily prove that the functions u± are actually constant and satisfy f(x, y, u±) = 0 for all (x, y) ∈Ω. Furthermore, 0 ≤u−≤1/2 ≤u+ ≤1 from (2.10) and ut ≥0. One concludes from (1.7) that u−= 0 and u+ = 1.
Eventually, the function u is a classical solution of (1.2) and (1.6). Indeed, because of the (t, x) periodicity of u, the limits u(t, x, y) →0 (resp. →1) as x·e →+∞(resp. x·e →−∞) hold locally in (t, y) and uniformly in the x variables which are orthogonal to e.
That completes the proof of Proposition 2.2.
3 Influence of the geometry of the domain and of the coefficients of the medium 3.1 Influence of the geometry of the domain : proofs of Theorems 1.3 and 1.4 This subsection deals with the influence of the geometry of the domain on the speed of propagation of pulsating fronts for the homogeneous equation (1.15) in periodic domains Ω.
Namely, one shall prove Theorems 1.3 and 1.4.
Proof of Theorem 1.3. One first recalls that the minimal speed c∗(e) of the pulsating fronts solving (1.15) and (1.6) is positive (see ). Furthermore, from Theorem 1.1, c∗(e) is given by the formula c∗(e) = min λ>0 k(λ) λ , where k(λ) is the first eigenvalue of the problem ∆ψλ −2λ˜ e · ∇ψλ + (λ2 + f ′(0))ψλ = k(λ)ψλ in Ω (3.1) and ψλ is positive in Ω, L-periodic with respect to x, and satisfies ∂νψλ = λ(ν · ˜ e)ψλ on ∂Ω.
Multiply the above equation by ψλ and integrate by parts over the cell C. It follows from the boundary and periodicity conditions that − Z C |∇ψλ|2 + (λ2 + f ′(0)) Z C ψ2 λ = k(λ) Z C ψ2 λ.
(3.2) Therefore, ∀λ > 0, k(λ) ≤λ2 + f ′(0) (3.3) 23 and c∗(e) = minλ>0 k(λ)/λ ≤2 p f ′(0).
Assume now that the domain Ωis invariant in the direction ˜ e. Then ν · ˜ e = 0 on ∂Ω and a (unique up to multiplication) solution ψλ of the eigenvalue problem (3.1) is ψλ = 1.
Therefore, k(λ) = λ2 + f ′(0) for all λ > 0. Thus, c∗(e) = 2 p f ′(0).
Conversely, assume that c∗(e) = 2 p f ′(0). Set λ∗= p f ′(0). One claims that k(λ∗) = (λ∗)2 + f ′(0). If not, then k(λ∗) < (λ∗)2 + f ′(0) from (3.3) and c∗(e) ≤k(λ∗) λ∗ < (λ∗)2 + f ′(0) λ∗ = 2 p f ′(0), which contradicts our assumption. Therefore, k(λ∗) = (λ∗)2 + f ′(0) and it follows from (3.2) that ψλ∗is constant. As a consequence, ν·˜ e ≡0 on ∂Ω. Hence, Ωis invariant in the direction ˜ e.
Let us now turn to the Proof of Theorem 1.4. Up to a rotation of the frame, one can assume without loss of generality that e = e1 = (1, 0, · · · , 0). Furthermore, if there is a family of domains (Ωα)0≤α<1 of R2 such that the conclusion of Theorem 1.4 holds with N = 2 and e = e1, then the family of domains (Ω′ α)0≤α<1 = (Ωα × RN−2)0≤α<1 satisfies the conclusion of Theorem 1.4 in higher dimensions N with e = e1.
Therefore, it is enough to deal with the case N = 2 and e = e1 = (1, 0). Fix L1 = L2 = 1, and 0 < β < 1/2. Let now (Ωα)0≤α<1 be a family of smooth open connected subsets of R2 satisfying (1.1) with L1 = L2 = 1, satisfying (1.16) and such that ∀0 ≤α ≤1 3, R2\Ωα ⊂Z2 + µ1 2 −α, 1 2 + α ¶ × µ1 2 −α, 1 2 + α ¶ and ∀2 3 ≤α < 1, Z2 + (1 −α, α) × [β, 1 −β] ⊂R2\Ωα ⊂Z2 + µ1 −α 2 , 1 + α 2 ¶ × [β, 1 −β].
One also assumes that, for each α0 ∈(0, 1), there exists r > 0 such that the sets Ωα are C3 uniformly with respect to α ∈(α0 −r, α0 + r).
Let us prove that this family of domains fulfills the conclusion of Theorem 1.4 with N = 2 and e = e1.
One first observes that, for each α ∈(0, 1), the domain Ωα is not invariant in the direction e1, whence cα < 2 p f ′(0) from Theorem 1.3.
The other parts facts in Theorem 1.4 are proved in Steps 2, 3 and 4 below. Step 1 is concerned with the derivation of inequality (3.5) below.
Step 1. Let first α ∈[0, 1) be fixed. The minimal speed cα = c∗(e1, Ωα) of the pulsating fronts satisfying (1.15) and (1.6) in Ωα with Neumann boundary conditions on ∂Ωα, is given by the formula cα = min λ>0 kα(λ) λ , 24 where kα(λ) is the first eigenvalue of the problem ∆ψα,λ −2λ∂1ψα,λ + (λ2 + f ′(0))ψα,λ = kα(λ)ψα,λ in Ωα (3.4) and ψα,λ is positive in Ωα, (1, 1)-periodic with respect to (x1, x2), and satisfies ∂νψα,λ = λ(ν · e1)ψα,λ on ∂Ωα, where ν stands for the unit exterior normal to Ωα. Observe now that, from the monotonicity of the domains (Ωα), one has Ωα ⊃R × (−β, β).
Therefore, it follows from the maximum principle (see for more details) that kα(λ) ≥κ(λ) for all λ > 0, where κ(λ) (resp. ψλ) is the first eigenvalue (resp. eigenfunction) of ∆ψλ −2λ∂1ψλ + (λ2 + f ′(0))ψλ = κ(λ)ψλ in R × (−β, β), ψλ > 0 in R × (−β, β), ψλ = 0 on R × {±β}, ψλ is 1-periodic with respect to x1.
By uniqueness, the function ψλ does not depend on the variable x1, whence κ(λ) = λ2 + f ′(0) −(π/(2β))2. It follows that ∀α ∈[0, 1), ∀λ > 0, kα,(λ) ≥λ2 + f ′(0) − µ π 2β ¶2 .
(3.5) Step 2. Let α be fixed in (0, 1) and let us now prove that the function t 7→ct is continuous at α. If not, there exists ε > 0 and a sequence (αn)n∈N →α such that |cαn −cα| ≥ε for all n. Up to extraction of some subsequence, two cases may occur : Case 1 : cαn ≤cα −ε for all n. For each n, let λn > 0 be such that cαn = kαn(λn)/λn (the existence of such λn follows from Theorem 1.1), and let ψn = ψαn,λn solve (3.4) in Ωαn. The functions ψn are positive in Ωαn, (1, 1)-periodic in (x1, x2) and satisfy ∂νψn = λn(ν · e1)ψn on ∂Ωαn. Up to normalization, one can assume that ψn(0, 0) = 1.
Since 0 ≤cαn = kαn(λn)/λn ≤cα −ε, it follows from (3.5) that the sequence (λn) is bounded. On the other hand, kαn(λn) ≥f ′(0) from Lemma 2.1. Therefore, the sequence (λn) is bounded from below by a positive constant. Up to extraction of some subsequence, one can then assume that λn →λ ∈(0, +∞) as n →+∞. On the other hand, one can also assume that cαn →c ∈[0, cα −ε] as n →+∞.
Furthermore, since the domains (Ωαn) are uniformly C3, the functions ψn satisfy uniform C2,δ bounds in Ωαn up to the boundary. Up to extraction of some subsequence, the functions ψn converge in C2 loc(Ωα) to a solution ψ of ∆ψ −2λ∂1ψ + (λ2 + f ′(0))ψ = cλψα,λ in Ωα, which can be extended as a C2 function in Ωα such that ∂νψ = λ(ν · e1)ψ on ∂Ωα. Further-more, ψ is nonnegative, (1, 1)-periodic, and satisfies ψ(0, 0) = 1. From the strong maximum principle, the function ψ is positive. It is therefore the first eigenfunction of problem (3.4) with the above periodicity and boundary condition. Hence, cλ = kα(λ). Formula (1.13) 25 implies that c ≥cα, which contradicts the fact that c ≤cα −ε. In other words, case 1 is ruled out.
Case 2 : cαn ≥cα −ε for all n. Let now λ > 0 be such that cα = kα(λ)/λ. From (3.3) and (1.8), one has λ2 + f ′(0) ≥kαn(λ) ≥λcαn ≥λ(cα + ε).
(3.6) Up to extraction of some subsequence, one can assume that kαn(λ) →k > 0 as n →+∞, and that the functions ψαn,λ, normalized by ψαn,λ(0, 0) = 1, converge locally in Ωα to a positive (1, 1)-periodic C2(Ωα) solution ψ of ½ ∆ψ −2λ∂1ψ + (λ2 + f ′(0))ψ = kψα,λ in Ωα ∂νψ = λ(ν · e1)ψ on ∂Ωα.
One concludes that k = kα(λ), whence kα(λ) ≥λ(cα + ε) from (3.6). This contradicts the definition of λ. Therefore, case 2 is ruled out too.
That proves the continuity of the map α 7→cα in (0, 1).
Step 3. Let us now prove that cα →2 p f ′(0) as α →0+. Assume not. Since 0 ≤cα ≤ 2 p f ′(0) for all α ∈[0, 1) because of Theorem 1.3, there exists then a sequence (αn) →0+ such that cαn →c ∈[0, 2 p f ′(0)) as n →+∞. On the other hand, there exists a sequence (λn) such that cαn = kαn(λn)/λn for each n. As in Case 1 of Step 2 above, one can prove that the sequence (λn) is bounded from below and above by two positive constants. From (3.3) and Lemma 2.1, it follows that the sequence (kαn(λn)) is itself bounded from below and above by two positive constants. Therefore, c > 0.
For each n, let un(t, x1, x2) be a pulsating travelling front solving (1.6) with the speed cαn and such that ½ (un)t = ∆un + f(un) in R × Ωαn ∂νun = 0 on R × ∂Ωαn.
(3.7) Furthermore, each un satisfies 0 ≤un ≤1 and (un)t ≥0 in R × Ωαn. Up to normalization, one can assume that un(0, 0, 0) = 1/2.
Owing to the construction of the domains Ωα, and from standard parabolic estimates, the functions un converge, up to extraction of some subsequence, to a classical solution u = u(t, x1, x2) of ut = ∆u + f(u) in R × (R2(Z2 + (1/2, 1/2))) such that 0 ≤u ≤1.
The singularities on the lines R × (Z2 + (1/2, 1/2)) in (t, x1, x2) variables are then removable and the function u can be extended to a classical solution u of ut = ∆u + f(u) in R × R2. On the other hand, the function u satisfies u µ t −k1 c , x1, x2 ¶ = u(t, x1 + k1, x2 + k2) for all (t, x1, x2) ∈R × R2 and (k1, k2) ∈Z2. Furthermore, ut ≥0 in R × R2 and u(0, 0, 0) = 1/2. By passing to the limit t →±∞, one can prove as in that u(t, x1, x2) →0 (resp. 1) as t →−∞(resp. t →+∞) locally in (x1, x2).
Eventually, the function u is a pulsating travelling front, solving (1.6) with e = e1, and (1.15) in R × R2, with the speed c. But the minimal speed for this problem is equal to 26 2 p f ′(0) (from Theorem 1.3 –the domain R2 is invariant in the direction e1). Therefore, c ≥2 p f ′(0), which is contradiction with our assumption.
As a consequence, the function α 7→cα is continuous at 0.
Step 4. Let us now prove that cα →2 p f ′(0) as α →1−. Assume not. As above, there exists then a sequence (αn) →1−such that cαn →c ∈(0, 2 p f ′(0)) as n →+∞. Let un = un(t, x1, x2) be a pulsating travelling front solving (3.7) and (1.6) with speed cαn. Up to normalization, one can assume that un(0, 0, 0) = 1/2. Consider now the restrictions, still called un, of the functions un to R × R × [−β, β]. Owing to the construction of the domains Ωα, the functions un converge, up to extraction of some subsequence, to a classical solution u(t, x1, x2) of ½ ut = ∆u + f(u) in R × (R × [−β, β]\Z × {±β}) ∂νu = 0 on R × R\Z × {±β} such that 0 ≤u ≤1. The singularities on the lines R × Z × {±β} in (t, x1, x2) variables are removable and the function u can then be extended to a classical solution u of (1.15) in R × R × [−β, β] such that ∂νu = 0 on R × R × {±β}. On the other hand, the function u satisfies u µ t −k1 c , x1, x2 ¶ = u(t, x1 + k1, x2) for all (t, x1, x2) ∈R×[−β, β] and k1 ∈Z. Furthermore, ut ≥0 in R×R2 and u(0, 0, 0) = 1/2.
By passing to the limit t →±∞, one can prove that u(t, x1, x2) →0 (resp. 1) as t →−∞ (resp. t →+∞) locally in (x1, x2).
Eventually, the function u is a pulsating travelling front, solving (1.15) in R×R×[−β, β], together with Neumann boundary conditions on R × R × {±β}. The function u satisfies (1.6) with e = e1 and speed c. Since the domain R × [−β, β] is invariant in the direction e1, the minimal speed of such travelling fronts is 2 p f ′(0). One then gets a contradiction with our assumption that c < 2 p f ′(0).
Therefore, cα →2 p f ′(0) as α →1−and the proof of Theorem 1.4 is complete.
Remark 3.1 Let the family (Ωα)0≤α<1 of domains of R2 be given as above. Let e be a unit direction of R2 such that e ̸= ±e1, and let c∗ α(e) be the minimal speed of pulsating fronts solving (1.6) and (1.15) in Ωα with Neumann boundary conditions on ∂Ωα. As above, the function α 7→c∗ α(e) is continuous on [0, 1), with c∗ 0(e) = 2 p f ′(0) and c∗ α(e) < 2 p f ′(0) for all α ∈(0, 1). With the same arguments as above, it easily follows than lim infα→1−c∗ α(e) ≥ 2 p f ′(0)|e1|, where e1 is the x1-component of the direction e. Neverthless, the determination of the limit, if any, of c∗ α(e) as α →1−is still open.
3.2 Influence of the coefficients of the medium : proofs of Theo-rems 1.6, 1.8 and 1.10 This subsection is devoted to the study of the influence of the coefficients of the medium (reaction, advection and diffusion terms) on the speed of propagation of pulsating travelling fronts.
27 Let us first investigate the dependance on the reaction term f.
Proof of Theorem 1.6. a). Let f and g satisfy (1.5) and (1.7) and assume that f ′ u(x, y, 0) ≤ g′ u(x, y, 0) for all (x, y) ∈Ω. For any λ > 0, let k(λ, f) (resp. k(λ, g)) be the first eigenvalue of (1.9) with ζ(x, y) = f ′ u(x, y, 0) (resp. ζ(x, y) = g′ u(x, y, 0)). It follows from monotonicity properties of the first eigenvalue of elliptic problems (see ) that k(λ, f) ≤k(λ, g). Hence, Theorem 1.1 yields c∗(e, f) ≤c∗(e, g).
Assume furthermore that f ′ u(x, y, 0)≤ ̸≡g′ u(x, y, 0). Let λ0 > 0 be chosen so that c∗(e, g) = k(λ0, g)/λ0. We claim that k(λ0, f) < k(λ0, g). If this holds, then c∗(e, f) ≤k(λ0, f) λ0 < k(λ0, g) λ0 = c∗(e, g) and we are done. Assume then that k(λ0, f) ≥k(λ0, g). Let ψf (resp. ψg) be a positive first eigenvalue of problem (1.9) with λ = λ0 and ζ(x, y) = f ′ u(x, y, 0) (resp. ζ(x, y) = g′ u(x, y, 0)).
Let τ > 0 be such that ψf ≤τψg in Ωwith equality somewhere (such a τ > 0 exists since both ψf and ψg are continuous, positive and L-periodic with respect to x in Ω). The function z := ψf −τψg satisfies ∇· (A∇z) −2λ0˜ eA∇z + q · ∇z + [−λ0∇· (A˜ e) −λ0q · ˜ e + λ2 0˜ eA˜ e + f ′ u(x, y, 0) −k(λ0, f)]z = (k(λ0, f) −k(λ0, g))τψg +(g′ u(x, y, 0) −f ′ u(x, y, 0))τψg ≥, ̸≡ 0 in Ω (3.8) from our assumptions.
On the other hand, z ≤0 in Ωwith equality somewhere, and νA∇z = λ0(νA˜ e)z on ∂Ω.
The strong maximum principle and Hopf lemma imply that z ≡0 in Ω. But the right-hand side of (3.8) is not identically equal to 0. One has then obtained a contradiction, whence k(λ0, f) < k(λ0, g).
b) Let f satisfy (1.5) and (1.7). First, it follows from part a) that the function B 7→ c∗(e, Bf) is increasing with respect to B > 0. For any λ > 0 and B > 0, let k(λ, B) and ψλ,B be the first eigenvalue and eigenfunction of problem (1.9) with ζ(x, y) = Bf ′ u(x, y, 0).
Multiply equation (1.9) by ψλ,B and integrate by parts over C. One gets k(λ, B) Z C ψ2 λ,B = − Z C ∇ψλ,BA∇ψλ,B−λ Z C (q·˜ e)ψ2 λ,B+λ2 Z C (˜ eA˜ e)ψ2 λ,B+ Z C Bf ′ u(x, y, 0)ψ2 λ,B.
It follows that k(λ, B) ≤λ∥q · ˜ e∥∞+ λ2∥˜ eA˜ e∥∞+ B∥f ′ u(·, ·, 0)∥∞.
Hence, c∗(e, Bf) = min λ>0 k(λ, B) λ = O( √ B) as B →+∞.
Assume now that Ω= RN or νA · ˜ e ≡0 on ∂Ω. In both cases, integrating over C the equation (1.9) satisfied by ψλ,B with ζ(x, y) = Bf ′ u(x, y, 0) leads to k(λ, B) Z C ψλ,B = λ Z C ∇· (A˜ e)ψλ,B −λ Z C (q · ˜ e)ψλ,B + λ2 Z C (˜ eA˜ e)ψλ,B + Z C Bf ′ u(x, y, 0)ψλ,B.
28 Hence, k(λ, B) ≥−λ∥∇· (A˜ e)∥∞−λ∥q · ˜ e∥∞+ λ2α1 + B min (x,y)∈Ωf ′ u(x, y, 0), where α1 > 0 is given in (1.3). Since min(x,y)∈Ωf ′ u(x, y, 0) > 0, it follows that there exists γ > 0 such that c∗(e, Bf) = min λ>0 k(λ, B) λ ≥γ √ B for B large enough. That completes the proof of Theorem 1.6.
Remark 3.2 With the same tools as above, it can easily be seen that part a) of Theo-rem 1.6 extends to the ray speed w∗(e), as defined in section 1.3 and in (1.23).
Sim-ilarly, with obvious notations, w∗(e, Bf) ≤c∗(e, Bf) = O( √ B) as B →+∞, and lim infB→+∞w∗(e, Bf)/ √ B > 0 if Ω= RN.
That corresponds to Proposition 1.15.
Proof of Theorem 1.8. Under the assumptions of Theorem 1.8, one has ν · ˜ e = 0 on ∂Ω, and c∗ q(e) is given by the formula c∗ q(e) = min λ>0 kq(λ) λ , where kq(λ) and ψλ,q denote the unique eigenvalue and positive L-periodic eigenfunction of ∆ψλ,q −2λ˜ e · ∇ψλ,q + q · ∇ψλ,q + [−λq · ˜ e + λ2 + f ′(0)]ψλ,q = kq(λ)ψλ,q in Ω (3.9) with ν · ∇ψλ,q = 0 on ∂Ω. Divide the following formula by ψλ,q and integrate by parts over C. It follows from (1.4) and the L-periodicity of q and ψλ,q that Z C |∇ψλ,q|2 ψ2 λ,q + (λ2 + f ′(0))|C| = kq(λ)|C|, (3.10) whence kq(λ) ≥λ2 + f ′(0) = k0(λ).
(3.11) Therefore, c∗ q(e) ≥2 p f ′(0) = c∗ 0(e).
If q · ˜ e ≡0, then ψλ,q is constant for each λ > 0, whence kq(λ) = λ2 + f ′(0) and c∗ q(e) = 2 p f ′(0) = c∗ 0(e).
Assume now that c∗ q(e) = c∗ 0(e) = 2 p f ′(0). Let λ∗> 0 be such that c∗ q(e) = kq(λ∗) λ∗ .
Then kq(λ∗) = 2λ∗p f ′(0), whereas (3.11) yields kq(λ∗) ≥(λ∗)2 + f ′(0). Therefore, λ∗= p f ′(0) and kq(λ∗) = (λ∗)2 + f ′(0). From (3.10), one gets that ψλ∗,q is constant, and from (3.9) one concludes that q · ˜ e ≡0 in Ω.
29 Remark 3.3 With the same arguments as above and with obvious notations, one can deduce from (1.20) that, if Ω= RN, A = I and f = f(u) satisfies (1.5) and (1.7), then w∗ q(e) ≥ w∗ 0(e) = 2 p f ′(0) for all unit vector e of RN. Furthermore, the equality w∗ q(e) = 2 p f ′(0) holds if and only if q · e ≡0.
That corresponds to Proposition 1.16.
Proof of Theorem 1.10. Under the assumptions of Theorem 1.10, let k(λ) and ψλ be the first eigenvalue and eigenfunctions of the operator Lλ defined in (1.9). Multiply the equation Lλψλ = k(λ)ψλ by ψλ and integrate over C. One gets k(λ) Z C ψ2 λ = − Z C ∇ψλA∇ψλ + λ2 Z C ˜ eA˜ e ψ2 λ + Z C ζ(x, y)ψ2 λ.
(3.12) Hence, k(λ) ≤max (x,y)∈Ωζ(x, y) + λ2 max (x,y)∈Ω˜ eA(x, y)˜ e, (3.13) and (1.17) follows from (1.8).
Assume now that ˜ eA(x, y)˜ e = M and ζ(x, y) = M0 are constant in Ω, and that ∇·(A˜ e) ≡ 0 in Ωand νA˜ e = 0 on ∂Ω(if ∂Ω̸= ∅). Then, ψλ is constant for each λ > 0, whence k(λ) = M0 + λ2M and c∗(e) = 2√M0M.
Assume now that c∗(e) = 2√M0M, where M0 = max(x,y)∈Ω ζ(x, y) and M = max(x,y)∈Ω ˜ eA(x, y)˜ e.
Let λ∗= p M0/M > 0.
It follows from (1.8) that k(λ∗) ≥ c∗(e)λ∗= 2M0. On the other hand, k(λ∗) ≤M0 + (λ∗)2M = 2M0 from (3.13). Therefore, k(λ∗) = 2M0 = M0 +(λ∗)2M. One deduces from (3.12) and the equation Lλ∗ψλ∗= k(λ∗)ψλ∗ that ψλ∗is constant, ζ(x, y) ≡M0, ˜ eA(x, y)˜ e ≡M, ∇· (A˜ e) ≡0, and νA˜ e = 0 on ∂Ω.
Assume now that 0 < α ≤β, and let c∗ α(e) (resp. c∗ β(e)) denote the minimal speed of pulsating fronts in the direction e with diffusion αA (resp. βA). By (1.8), one has c∗ α(e) = min λ>0 kα(λ) λ and c∗ β(e) = min λ>0 kβ(λ) λ , (3.14) where kα(λ) (resp. kβ(λ)) is the first eigenvalue of the operator α˜ Lλ+f ′(0) (resp. β ˜ Lλ+f ′(0)) and ˜ Lλ is the operator ˜ Lλ = ∇· (A∇) −2λ˜ eA∇−λ∇· (A˜ e) + λ2˜ eA˜ e acting on the set E (E has been defined in Theorem 1.1). Under the notations of Lemma 2.1, k0(λ) is the first eigenvalue of ˜ Lλ, whence kα(λ) = αk0(λ) + f ′(0) and kβ(λ) = βk0(λ) + f ′(0) for all λ > 0. On the other hand, it follows from Lemma 2.1 that the function k0 is nonneg-ative. Therefore, kα(λ) ≤kβ(λ) and (3.14) yields c∗ α(e) ≤c∗ β(e). That completes the proof of Theorem 1.10.
Let us now turn to the Proof of the lower bound (1.18). Under the notations in Remark 1.11, integrate the equation Lλψλ = k(λ)ψλ over C. It follows that k(λ) Z C ψλ = λ Z C ∇· (A˜ e)ψλ −λ Z C q · ˜ e ψλ + λ2 Z C ˜ eA˜ e ψλ + Z C ζ(x, y)ψλ.
30 Therefore, k(λ) ≥−λb + λ2m + m0. On the other hand, Lemma 2.1 yields k(λ) ≥m0.
Formula (1.18) easily follows from (1.8).
Remark 3.4 Under the assumptions and notations of Theorem 1.10, then w∗(e) ≤c∗(e) ≤ 2√M0M for all unit vector e of Rd. Furthermore, under the additional assumption that f = f(u), then the ray speed w∗ γ(e) in the unit direction e of Rd for problem (1.2) with diffusion matrix γA (γ > 0) is given by w∗ γ(e) = min ⃗ λ∈Rd, ⃗ λ·e>0 ˜ kγ(⃗ λ) ⃗ λ · e , where ˜ kγ(⃗ λ) = γk0,⃗ λ/|⃗ λ|(|⃗ λ|) + f ′(0) and, for all unit vector e′ of Rd and all µ > 0, k0,e′(µ) denotes the first eigenvalue of the operator ∇· (A∇ψ) −2µ e′A∇ψ −µ∇· (Ae′)ψ + µ2e′Ae′ψ acting on the space of L-periodic functions ψ such that νA∇ψ = µ(νAe′)ψ on ∂Ω. Lemma 2.1 yields k0,e′(µ) ≥0. Therefore, w∗ α(e) ≤w∗ β(e) as soon as 0 < α ≤β.
Lastly, if Ω= RN, with the same arguments as above, it easily follows from (1.20) that w∗(e) ≥min(m0α1/˜ b, −˜ b + 2√m0α1), where α1 was given in (1.3), m0 = minx∈RN f ′ u(x, 0) and ˜ b = maxx∈RN, ⃗ µ∈RN, ⃗ µ̸=0 |∇· (A(x)⃗ µ)|/|⃗ µ| + maxx∈RN |q(x)|.
That corresponds to Proposition 1.17.
4 Spreading speed This section is devoted to the proof of Theorem 1.13. It is based on the following auxiliary Lemmas 4.1 and 4.2, and on Proposition 1.14.
Lemma 4.1 Let Ωsatisfy (1.1) with d ≥1. Let λR be the first eigenvalue, and ψR be the first eigenfunction of −∆ψR = λRψR in Ω∩BR ψR > 0 in Ω∩BR ψR = 0 on Ω∩∂BR ∂νψR = 0 on ∂Ω∩BR ∥ψR∥L∞(Ω∩BR) = 1, (4.1) where BR is the open euclidean ball of radius R > 0 and centre 0. Then λR →0 as R →+∞.
Proof. It follows from the maximum principle that λR is decreasing with respect to R.
Furthermore, λR has the following variational representation : λR = min ψ∈H1(Ω∩BR){0}, ψ|Ω∩∂BR=0 Z Ω∩BR |∇ψ|2 Z Ω∩BR ψ2 ≥0.
31 Let ξ be a given C∞(RN) function such that ξ = 1 in B1/2 and ξ = 0 in RN\B1. Taking the function ψ(z) = ξ(z/R) as a test function leads to λR ≤R−2∥∇ξ∥2 ∞|Ω∩BR| |Ω∩BR/2| .
But |Ω∩BR|/|Ω∩BR/2| is bounded as R →+∞because of (1.1), whence λR →0+ as R →+∞.
Lemma 4.2 Let Ωsatisfy (1.1) with d ≥1, let e be in Sd−1, and assume that Ωis invariant in the direction ˜ e. Let f : [0, +∞) →R be a function of class C1 such that f(0) = 0 and f ′(0) > 0, and let c be such that |c| < 2 p f ′(0). Then there exist R > 0 and ε0 > 0 such that, for all ε ∈(0, ε0), there is a function w satisfying ∆x,yw + c˜ e · ∇x,yw + f(w) ≥ 0 in Ω∩BR w > 0 in Ω∩BR w = 0 on Ω∩∂BR ∂νw = 0 on ∂Ω∩BR ∥w∥L∞(Ω∩BR) ≤ ε.
(4.2) Proof. Let R be fixed large enough so that the first eigenvalue λR of (4.1) satisfies λR < f ′(0) −c2/4.
The latter is possible by Lemma 4.1 and since |c| < 2 p f ′(0).
It then follows that the function w(x, y) = εe−ce·x/2ψR(x, y) satisfies ∆w + c˜ e · ∇w + f(w) = f(εe−ce·x/2ψR(x, y)) − µc2 4 + λR ¶ εe−ce·x/2ψR(x, y) ≥0 in Ω∩BR for ε > 0 small engouh. On the other hand, the function w is positive on Ω∩BR, vanishes on Ω∩∂BR, has small L∞(Ω∩BR) norm for ε small enough, and it satisfies the Neumann boundary condition ∂νw = 0 on ∂Ω∩BR because so does ψR and Ωis invariant in the direction ˜ e. That completes the proof of Lemma 4.2.
Proof of Proposition 1.14. First of all, one can assume without loss of generality that u ̸≡0, whence the strong maximum principle yields u > 0 in Ω. If Ωis bounded (this corresponds to the case d = 0), then the minimum m of u in Ωis reached and positive, and, since g is positive in (0, 1), the strong maximum principle and Hopf lemma yield m ≥1.
Similarly, since g is negative in (1, +∞), the maximum M of u satisfies M ≤1. Therefore, u ≡1.
Let us now consider the general case of a domain Ωwhich is unbounded, i.e. d ≥1. Two cases may occur : Case 1 : m = infΩu > 0.
Let (xn, yn)n∈N be a sequence of points in Ωsuch that u(xn, yn) →m as n →+∞. If m is reached, then the points (xn, yn) may be assumed to be bounded. In the general case, there exist some points ˜ xn ∈L1Z × · · · × LdZ such that 32 (xn −˜ xn, yn) ∈C (remember that C is the cell of periodicity of Ω). Up to extraction of some subsequence, one can assume that (xn −˜ xn, yn) →(x, y) ∈C as n →+∞.
Call un(x, y) = u(x + ˜ xn, y). The functions un are defined in Ω, by L-periodicity of Ω, and they satisfy the same equation (1.24) as u. From standard elliptic estimates and Sobolev injections, the functions un converge, up to extraction of some subsequence, in C2,δ loc(Ω) (for all 0 ≤δ < 1) to a function u∞solving (1.24). Furthermore, u∞≥m and u∞(x, y) = m.
If m < 1, then g(m) > 0 and one gets a contradiction with the strong maximum principle and Hopf lemma. Therefore, m ≥1. Similarly, one can prove that M = supΩu ≤1. Hence, u ≡1.
Case 2 : m = infΩu = 0. Remember that (ei)1≤i≤d denotes the canonic basis of Rd. Since Ωsatisfies (1.1), the function (x, y) 7→u(x + L1e1, y) is defined in Ω. On the other hand, Harnack type inequalities imply that the function v(x, y) = u(x + L1e1, y) u(x, y) is globally bounded in Ω. Call M1 = lim supu(x,y)→0, (x,y)∈Ωv and let (xn, yn) ∈Ωbe such that u(xn, yn) →0 and v(xn, yn) →M1 as n →+∞.
Let ˜ xn ∈L1Z × · · · × LdZ be such that (xn −˜ xn, yn) ∈C. Up to extraction of some subsequence, one can assume that (xn −˜ xn, yn) →(x, y) ∈C as n →+∞.
For each n ∈N, let un be the function defined in Ωby un(x, y) = u(x + ˜ xn, y) u(xn, yn) .
From Harnack inequalities, the functions un are locally bounded in Ω. On the other hand, the functions (x, y) 7→u(x + ˜ xn, y) satisfy the same equation as u and u(xn, yn) →0 as n →+∞. From standard elliptic estimates, the functions un converge in C2,δ loc(Ω) (for all 0 ≤α < 1), up to extraction of some subsequence, to a nonnegative function u∞solving ½ ∆u∞+ b · ∇u∞+ g′(0)u∞ = 0 in Ω ∂νu∞ = 0 on ∂Ω.
Furthermore, u∞(x, y) = 1, whence u∞> 0 in Ωfrom the strong maximum principle.
Owing to the definitions of M1 and of the sequence (xn, yn), one has that 0 < u∞(x + L1e1, y)/u∞(x, y) ≤M1 and u∞(x + L1e1, y)/u∞(x, y) = M1. Notice then that M1 > 0 since u∞is positive in Ω.
The function ξ(x, y) := u∞(x + L1e1, y) u∞(x, y) satisfies ∆ξ + 2∇u∞ u∞ · ∇ξ + b · ∇ξ = 0 in Ω ∂νξ = 0 on ∂Ω, together with ξ ≤M1 in Ωand equality at (x, y). It follows from the strong maximum principle that ξ ≡M1 in Ω, whence u∞(x + L1e1, y) ≡M1u∞(x, y).
33 In other words, calling α1 = (ln M1)/L1, the function ϕ1(x, y) := e−α1x1u∞(x, y) is posi-tive in Ωand it satisfies ∆ϕ1 + 2α1∂x1ϕ1 + α2 1ϕ1 + b · ∇ϕ1 + b1α1ϕ1 + g′(0)ϕ1 = 0 in Ω ∂νϕ1 + α1ν · e1 ϕ1 = 0 on ∂Ω ϕ1(x + L1e1, y) = ϕ1(x, y) in Ω, where b = (b1, · · · , bN).
Notice now that the function ϕ1(x + L2e2, y)/ϕ1(x, y) is globally bounded in Ω, once again from Harnack type inequalities. Then call M2 := sup (x,y)∈Ω ϕ1(x + L2e2, y) ϕ1(x, y) and do the same procedure as before, and so on d times. One then gets the existence of a positive L-periodic function ϕ in Ωsatisfying ½ ∆ϕ + 2α · ∇xϕ + |α|2ϕ + b · ∇ϕ + b · ˜ α ϕ + g′(0)ϕ = 0 in Ω ∂νϕ + ν · α ϕ = 0 on ∂Ω, for some α = (α1, · · · , αd) ∈Rd, where |α|2 = α2 1 + · · · + α2 d, ˜ α = (α1, · · · , αd, 0, · · · , 0) ∈RN and ∇means the gradient with respect to both (x, y) variables.
Divide the above equation by ϕ and integrate by parts over the cell C. By periodicity, it follows that Z C ½|∇ϕ|2 ϕ2 + 2α · ∇xϕ ϕ + |α|2 + b · ∇ϕ ϕ + b · ˜ α + g′(0) ¾ = 0.
In other words, Z C (¯ ¯ ¯ ¯ ∇ϕ ϕ + ˜ α + b 2 ¯ ¯ ¯ ¯ 2 + µ g′(0) −|b|2 4 ¶) = 0.
One then gets a contradiction with the assumption |b| < 2 p g′(0).
As a conclusion, case 2 is ruled out, whence infΩu > 0 and u ≡1.
Remark 4.3 In the case where Ω= RN, the above proof can be slightly simplified. Indeed, using Lemma 4.2, there is R > 0 and a function w such that w > 0 in BR, w = 0 on ∂BR, w < u in BR and ∆w + c˜ e · ∇w + f(w) ≥0 in BR. Since the equation (1.24) satisfied by u is invariant by translation, and since u > 0 in RN, one can slide u in any direction and prove that, for all x0 ∈RN, w < u(· + x0) in BR. Therefore, infRN u > 0 and one concludes as in Case 1 of the above proof that u ≡1.
This result in the case Ω= RN has been known since the paper of Aronson and Weinberger , who used parabolic tools. The above arguments actually provide a simpler proof using elliptic arguments.
34 Let us now turn to the Proof of Theorem 1.13. As already emphasized, it only remains to prove that, if Ωis a straight cylinder in the direction ˜ e, then w∗(e) = 2 p f ′(0) for any nonnegative continuous and compactly supported initial condition u0 ̸≡0.
Let u0 be such a function. Since w∗(e) ≤c∗(e) ≤2 p f ′(0), one only has to prove that w∗(e) ≥2 p f ′(0).
Let 0 ≤c < 2 p f ′(0) and let us actually prove that u(t, x + ct e, y) →1 as t →+∞ locally in (x, y) ∈Ω. Let us mention that, since Ωis invariant in the direction ˜ e, the functions (x, y) 7→v(t, x, y) = u(t, x + ct e, y) is defined in Ωfor all t ≥0. The function v actually satisfies vt = ∆v + c˜ e · ∇v + f(v) for all t > 0 and (x, y) ∈Ω, together with Neumann boundary conditions on ∂Ω, and initial condition u0. Notice that v(t, x, y) > 0 for all t > 0 and (x, y) ∈Ωfrom the strong maximum principle.
Call M = supRN u0. One has 0 < M < +∞by assumption. Let ξ(t) be the function solving ˙ ξ = f(ξ) and ξ(0) = M. From the assumptions on f, one has ξ(t) →1 as t →+∞.
Furthermore, v(t, x, y) ≤ξ(t) for all t ≥0 and (x, y) ∈Ωby the parabolic maximum principle. Therefore, lim sup t→+∞ sup (x,y)∈Ω v(t, x, y) ≤1.
(4.3) On the other hand, from Lemma 4.2, there exists R > 0 (large enough so that BR∩Ω̸= ∅) and a function w solving (4.2) and such that, say, v(1, x, y) ≥w(x, y) for all (x, y) ∈BR ∩Ω (remember that min(x,y)∈BR∩Ωv(1, x, y) > 0). Let ˜ w be the function defined in Ωby ˜ w = w in BR∩Ωand ˜ w = 0 in Ω\BR. The function ˜ w is then a subsolution for the equation satisfied by v. Therefore, the function θ solving θt = ∆θ + c˜ e · ∇θ + f(θ), t > 0, (x, y) ∈Ω, ∂νθ = 0, t > 0, (x, y) ∈∂Ω, θ(0, x, y) = ˜ w(x, y), is nondecreasing with respect to t and, since ˜ w ≤v(1, ·, ·) in Ω, the function θ satisfies ∀t ≥0, ∀(x, y) ∈Ω, θ(t, x, y) ≤v(t + 1, x, y).
(4.4) Without loss of generality, one can assume that ˜ w ≤1 in Ω, whence θ(t, x, y) ≤1 for all (t, x, y) ∈R+ × Ω. By monotonicity in t, the function θ(t, x, y) converges as t →+∞to a function ψ(x, y). From standard parabolic estimates, the convergence θ(t, x, y) →ψ(x, y) as t →+∞holds locally uniformly in Ωand the function ψ is a classical solution of (1.24) with g = f, b = c˜ e and |b| < 2 p f ′(0). Furthermore, 0 ≤˜ w ≤ψ ≤1 in Ω(thus, ψ ̸≡0 since w > 0 in BR ∩Ω̸= ∅). Proposition 1.14 yields ψ ≡1.
One deduces from (4.4) that lim inft→+∞min(x,y)∈K v(t, x, y) ≥1 for all compact set K ⊂Ω.
One concludes from (4.3) that v(t, x, y) →1 as t →+∞locally in (x, y) ∈Ω. That completes the proof of Theorem 1.13.
35 Remark 4.4 If Ωis invariant in the direction ˜ e and satisfies (1.1), then u(t, x+cte, y) →1 as t →+∞locally in (x, y) ∈Ωfor 0 ≤c < 2 p f ′(0) and for all solution u of (1.15) with continuous bounded nonnegative initial condition u0 ̸≡0. The latter indeed holds from the maximum principle even if u0 is not compactly supported. Furthermore, under the additional assumption that u0 is compactly supported, then u(t, x + ct e, y) →0 as t →+∞locally in (x, y) ∈Ωfor all c > 2 p f ′(0).
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41 |
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Power of a Point Theorem
The Power of a Point Theorem is a relationship that holds between the lengths of the line segments formed when two linesintersect a circle and each other.
Contents
[hide]
1 Statement
1.1 Case 1 (Inside the Circle):
1.2 Case 2 (Outside the Circle):
1.2.1 Classic Configuration
1.2.2 Tangent Line
1.3 Case 3 (On the Border/Useless Case):
1.4 Alternate Formulation
1.5 Hint for Proof
2 Notes
3 Proof
3.1 Case 1 (Inside the Circle)
3.2 Case 2 (Outside the Circle)
3.3 Case 3 (On the Circle Border)
4 Problems
4.1 Introductory
4.2 Intermediate
4.3 Olympiad
5 See Also
5.1 External Links
Statement
There are three unique cases for this theorem. Each case expresses the relationship between the length of line segments that pass through a common point and touch a circle in at least one point. Can be useful with cyclic quadrilaterals as well however with a slightly different application.
Case 1 (Inside the Circle):
If two chords and intersect at a point within a circle, then
Case 2 (Outside the Circle):
Classic Configuration
Given lines and originate from two unique points on the circumference of a circle ( and ), intersect each other at point , outside the circle, and re-intersect the circle at points and respectively, then
Tangent Line
Given Lines and with tangent to the related circle at , lies outside the circle, and Line intersects the circle between and at ,
Case 3 (On the Border/Useless Case):
If two chords, and , have on the border of the circle, then the same property such that if two lines that intersect and touch a circle, then the product of each of the lines segments is the same. However since the intersection points lies on the border of the circle, one segment of each line is so no matter what, the constant product is .
Alternate Formulation
This alternate formulation is much more compact, convenient, and general.
Consider a circle and a point in the plane where is not on the circle. Now draw a line through that intersects the circle in two places. The power of a point theorem says that the product of the length from to the first point of intersection and the length from to the second point of intersection is constant for any choice of a line through that intersects the circle. This constant is called the power of point . For example, in the figure below
Hint for Proof
Draw extra lines to create similar triangles (Draw on all three figures. Draw another line as well.)
Notice how this definition still works if and coincide (as is the case with ). Consider also when is inside the circle. The definition still holds in this case.
Notes
One important result of this theorem is that both tangents from any point outside of a circle to that circle are equal in length.
The theorem generalizes to higher dimensions, as follows.
Let be a point, and let be an -sphere. Let two arbitrary lines passing through intersect at , respectively. Then
Proof. We have already proven the theorem for a -sphere (a circle), so it only remains to prove the theorem for more dimensions. Consider the plane containing both of the lines passing through . The intersection of and must be a circle. If we consider the lines and with respect simply to that circle, then we have reduced our claim to the case of two dimensions, in which we know the theorem holds.
Proof
Case 1 (Inside the Circle)
Join and .
In
(Angles subtended by the same segment are equal)
(Vertically opposite angles)
(Corresponding sides of similar triangles are in the same ratio)
Case 2 (Outside the Circle)
Join and
(Why?)
Now, In
(shown above)
(common angle)
(Corresponding sides of similar triangles are in the same ratio)
Case 3 (On the Circle Border)
Length of a point is zero so no proof needed:)
Problems
Introductory
Find the value of in the following diagram:
Solution
Find the value of in the following diagram:
Solution
(ARML) In a circle, chords and intersect at . If and , find the ratio .
Solution
(ARML) Chords and of a given circle are perpendicular to each other and intersect at a right angle at point . Given that , , and , find .
Solution
Let be a diameter in a circle of radius Let be a chord in the circle that intersects at a point such that and What is (Source)
Intermediate
Two tangents from an external point are drawn to a circle and intersect it at and . A third tangent meets the circle at , and the tangents and at points and , respectively (this means that T is on the minor arc ). If , find the perimeter of . (Source)
Square of side length has a circle inscribed in it. Let be the midpoint of . Find the length of that portion of the segment that lies outside of the circle. (Source)
is a chord of a circle such that and Let be the center of the circle. Join and extend to cut the circle at Given find the radius of the circle. (Source)
Triangle has The incircle of the triangle evenly trisects the median If the area of the triangle is where and are integers and is not divisible by the square of a prime, find (Source)
Let be a triangle inscribed in circle . Let the tangents to at and intersect at point , and let intersect at . If , , and , can be written as the form , where and are relatively prime integers. Find . (Source)
Olympiad
Given circles and intersecting at points and , let be a line through the center of intersecting at points and and let be a line through the center of intersecting at points and . Prove that if and lie on a circle then the center of this circle lies on line .
(Source)
Let be a point interior to triangle (with ). The lines , and meet again its circumcircle at , , respectively . The tangent line at to meets the line at . Show that from follows .
(Source)
See Also
Geometry
Planar figures
External Links
Handout on AoPS Forums
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Shortest distance from circle to a line
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1
$\begingroup$
Let $C$ be a circle with center $(2, 1)$ and radius $2$. Find the shortest distance from the line $3y=4x+20$.
This should be very simple, but I seem to end up with no real solutions.
The shortest distance would be from the center of the circle perpendicular to the line right?
Solving the line for $y$ we get $y=\frac{4}{3}x+\frac{20}{3}$
Substituting this to the equation of the circle we get $(x-2)^2+(\frac{4}{3}x+\frac{20}{3}-1)^2=2^2$, but solving this for $x$ ended up with no real roots. What am I missing here?
geometry
circles
coordinate-systems
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edited Apr 15, 2020 at 10:16
sai-kartik
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asked Apr 15, 2020 at 9:35
user745970user745970
$\endgroup$
3
$\begingroup$ The line do not intersect the circle, this is because you reach complex roots. Instead you must consider the line passing through the center $(2,1)$ with slop $-\dfrac{3}{4}$ and intersect it with the circle $\endgroup$
Qurultay
– Qurultay
2020-04-15 09:40:56 +00:00
Commented Apr 15, 2020 at 9:40
$\begingroup$ "Substituting this to the equation of the circle" means you try to calculate all (x,y) that satisfy both equations simultaneously . This means that the point with coordinates (x,y) is the intersection of the line and the circle. That is not what you want. The line and the circle don't intersect. $\endgroup$
miracle173
– miracle173
2020-04-15 09:42:24 +00:00
Commented Apr 15, 2020 at 9:42
$\begingroup$ @Qurultay So i get $y-2=-\frac{3}{2}(x-2)$. Could you elaborate this a bit? $\endgroup$
user745970
– user745970
2020-04-15 09:47:38 +00:00
Commented Apr 15, 2020 at 9:47
Add a comment |
7 Answers 7
Reset to default
3
$\begingroup$
According to the figure, you need the distance $AB$.
The line $3x+4y=10$ is the line passing through $(2,1)$ and perpendicular to the line $3y=4x+20$
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answered Apr 15, 2020 at 9:54
QurultayQurultay
5,27922 gold badges1515 silver badges2626 bronze badges
$\endgroup$
1
$\begingroup$ I'll continue from @Qurultay solutions $B(-2,4)$ and $A(x,y)$ and $O(2,1)$ Distance from $B$ to $O$ is $\sqrt{(-2-2)^2+(4-1)^2} = \sqrt{(-4)^2+(3)^2} = \sqrt{16+9} = \sqrt{25} = 5$ Now subtract the radius of the circle, which is line $A \to O$ $5-2 = 3$ $\endgroup$
Aderinsola Joshua
– Aderinsola Joshua
2020-04-15 10:19:47 +00:00
Commented Apr 15, 2020 at 10:19
Add a comment |
2
$\begingroup$
By substituting the equation of the line into the equation of the circle you are looking for points where the line intersects the circle. The fact that the resulting equation for $x$ has no real roots means that the line does not intersect the circle i.e. the shortest distance from the line to the circle will be greater than the radius of the circle, which is $2$ units.
You can find the shortest distance from the line to the circle as follows:
(1) Note that the product of the gradients of perpendicular lines is $-1$, so the general equation of a line perpendicular to the given line is
$y = - \frac 3 4 x + m$
(2) The line with gradient $-\frac 3 4$ that passes through the centre of the circle at $(2,1)$ is
$y = - \frac 3 4 x + \frac 5 2$
(3) This line interects the original line at a point where
$\frac 4 3 x + \frac {20} 3 = - \frac 3 4 x + \frac 5 2 \ \Rightarrow 16x + 80 = -9x + 30 \ \Rightarrow 25x = -50 \ \Rightarrow x=-2 \text{ and } y=4$
(4) Find the distance between $(-2, 4)$ and the centre of the circle $(2,1)$. Then subtract the radius of the circle from this distance - this is the shortest distance from the line to the circle.
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answered Apr 15, 2020 at 9:50
gandalf61gandalf61
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$\endgroup$
2
$\begingroup$ Could you elaborate on how did you find the equation for the line at $(2)$? I seem to get $y-2=-\frac{3}{4}(x -1) \Leftrightarrow y = -\frac{3}{4}x + \frac{11}{4}$ $\endgroup$
user745970
– user745970
2020-04-15 16:11:49 +00:00
Commented Apr 15, 2020 at 16:11
$\begingroup$ Ah, forget it. Plugging in the points in the wrong positions... $\endgroup$
user745970
– user745970
2020-04-15 16:16:26 +00:00
Commented Apr 15, 2020 at 16:16
Add a comment |
1
$\begingroup$
The line perpendicular to the line given in the question and passing through the centre will help us find the perpendicular distance.We approach the problem by first finding out the perpendicular distance of the given line from the centre using this formula (check under Cartesian coordinates). Applying the formula we get distance between line and center to be: $$\left|\dfrac{3(1)-4(2)-20}{5}\right|=5$$ Now for the shortest distance between the given line and circle we simply subtract the radius of the circle giving us the answer of $\boxed{3}$ What you're trying to do is to intersect the line and circle. If you did get a real solution to that it would imply the shortest distance is $0$!
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answered Apr 15, 2020 at 9:48
sai-kartiksai-kartik
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$\endgroup$
1
$\begingroup$ The only way the line that I mentioned perpendicular to the given line would help, is to measure the perpendicular distance along that direction. There is no real need to calculate it though.. $\endgroup$
sai-kartik
– sai-kartik
2020-04-15 09:50:26 +00:00
Commented Apr 15, 2020 at 9:50
Add a comment |
1
$\begingroup$
The perpendicular line to the given line has slope $=\frac{-1}{4/3}=-\frac{3}{4}$
Also the perpendicular should pass through the centre of the circle so that the shortest distance between the given line and the given circle is the distance between the given line and the point of intersection of the perpendicular with the circle. SEE THE GRAPH BELOW.
The perpendicular line is $y=-\frac{4}{3}x+\frac{5}{2}$, this intersects with the circle at the point $(\frac{2}{5},\frac{11}{5})$.
The distance between the given line and $(\frac{2}{5},\frac{11}{5})$ is given by:
$$d=\frac{|[3y-4x-20]_{x=2/5,y=11/5}|}{\sqrt{3^2+(-4)^2}}$$
$$=\frac{|\frac{33}{5}-\frac{8}{5}-20|}{\sqrt{25}}=\frac{|-15|}{5}=\frac{15}{5}=3 \text{ units}$$
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edited Apr 15, 2020 at 10:24
sai-kartik
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answered Apr 15, 2020 at 9:58
Hussain-AlqatariHussain-Alqatari
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Add a comment |
0
$\begingroup$
What you would do is draw a perpendicular from the centre of the circle to that line.
That perpendicular passes through $C=(2,1)$ and is (obviously) perpendicular to $y=\dfrac{4}{3}x+\dfrac{20}{3}$
Line has a gradient of $-\dfrac{3}{4}$
$\begin{align} \dfrac{y-1}{x-2} &= -\dfrac{3}{4} \ y&=-\dfrac{3}{4}x +\dfrac{5}{2}\end{align}$
The two lines intersect at $P=(-2, 4)$ which you can verify by solving this:
$\dfrac{4}{3}x+\dfrac{20}{3}=-\dfrac{3}{4}x +\dfrac{5}{2}$
The distance $PC$ is:
$\sqrt{(2-(-2))^2+(4-1)^2}=5$
So the shortest distance to the circumference is $PC$ minus the radius of the circle which is:
$5-2=\boxed{3}$
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answered Apr 15, 2020 at 10:02
Nεo PλαÏoNεo PλαÏo
1,75588 silver badges1919 bronze badges
$\endgroup$
Add a comment |
0
$\begingroup$
Hint. You have shown that your line doesn't intersect the circle. Therefore the shortest distance between the circle and the line is given by the distance between that line and a line such that $$y=\frac43x+c,$$ where the last line is tangential to the circle -- in other words it intersects the circle in just one point. There are two such lines, and you can easily select the closest to the given line. Thus, you want to solve for values of $c$ such that $$(x-2)^2+\left(\frac43x+c-1\right)^2=4$$ has only one root. That is, set the discriminant of this quadratic in $x$ equal to $0.$ Then take the bigger value for $c.$ You then have your line. Compute the distance between this line and the given line, and you'd be done!
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answered Apr 15, 2020 at 10:02
AllawonderAllawonder
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$\endgroup$
Add a comment |
0
$\begingroup$
The squared distance from $(2,1)$ to an arbitrary point $(x,y)=(x,4x/3+20/3)$ of the straight line is $$(x-2)^2+\left(\frac43x+\frac{20}{3}-1\right)^2=\frac{25}{9}x^2+\frac{100}{9}x+\frac{35}{9}.$$ As the vertex of that parabola occurs at $x=-2$, the square of the minimal distance is $25$, hence the distance is $5$. Now subtract $2$.
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answered Apr 15, 2020 at 11:58
Michael HoppeMichael Hoppe
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5
If A and B are matrices, what are the conditions under which AB=0?
·
For two matrices
A
and
B
, the product
A
B
=
0
(the zero matrix) can occur under several conditions:
One of the Matrices is the Zero Matrix
: If either
A
or
B
is the zero matrix (i.e., all elements are zero), then
A
B
=
0
.
2. Non-zero Matrices with Linearly Dependent Columns
: If the columns of
B
are linearly dependent on the columns of
A
, it is possible for
A
B
to equal the zero matrix. This means that there exists a non-trivial linear combination of the columns of
A
that results in the zero vector.
3. Matrix Dimensions
: The dimensions of the matrices must be compatible for the multiplication to be defined.
For two matrices
A
and
B
, the product
A
B
=
0
(the zero matrix) can occur under several conditions:
One of the Matrices is the Zero Matrix
: If either
A
or
B
is the zero matrix (i.e., all elements are zero), then
A
B
=
0
.
2. Non-zero Matrices with Linearly Dependent Columns
: If the columns of
B
are linearly dependent on the columns of
A
, it is possible for
A
B
to equal the zero matrix. This means that there exists a non-trivial linear combination of the columns of
A
that results in the zero vector.
3. Matrix Dimensions
: The dimensions of the matrices must be compatible for the multiplication to be defined. If
A
is an
m
×
n
matrix and
B
is an
n
×
p
matrix, then the product
A
B
will be an
m
×
p
matrix.
4. Null Space
: If the columns of
B
lie in the null space of
A
, then
A
B
=
0
. In other words, if every column vector of
B
is mapped to the zero vector by the transformation represented by
A
, the product will be the zero matrix.
5. Special Cases
: Certain special cases can also lead to
A
B
=
0
, such as when
A
or
B
has specific structures (e.g., projection matrices, certain block matrices).
In summary,
A
B
=
0
can arise from either matrix being zero, their column dependencies, or the relationship between their null spaces.
Related questions
What if A and B are nonzero square matrices such that AB=0?
What would you give as an example of matrices A and B, such that AB=0, and BA is not equal to 0?
Can we say that if AB = 0, then A = 0 or B = 0? Why or why not?
If A and B are square matrices of order n such that |AB| = 0, then are both A and B nonsingular?
If A and B are two 2×2 matrices such that AB=BA=B where B is a non-zero matrix, then what can you say about A?
Fred Rich
Ph.D. from Aix-Marseille University (Graduated 1997)
·
Author has 193 answers and 122.1K answer views
·
5y
The condition is that the column vectors of B must be in the null space (aka kernel) of the matrix A, i.e., AX = 0 for each column X of B. There may be any number of such columns in B, and the resulting zero matrix AB = O will have #rows = #rows(A) and #columns = #cols(B).
PS: The matrix A has a nontrivial null space if its rank is strictly less than the number of columns (cf. the famous Rank–nullity theorem). If so, there are infinitely many choices for the columns X of B. Else, i.e., if rank A = #cols(A), then X and therefore B must be zero (with #rows(B) = #cols(A) and as said, #cols(B) arbi
The condition is that the column vectors of B must be in the null space (aka kernel) of the matrix A, i.e., AX = 0 for each column X of B. There may be any number of such columns in B, and the resulting zero matrix AB = O will have #rows = #rows(A) and #columns = #cols(B).
PS: The matrix A has a nontrivial null space if its rank is strictly less than the number of columns (cf. the famous Rank–nullity theorem). If so, there are infinitely many choices for the columns X of B. Else, i.e., if rank A = #cols(A), then X and therefore B must be zero (with #rows(B) = #cols(A) and as said, #cols(B) arbitrary).
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John Kinsella
University lecturer in Mathematics.
·
Author has 60 answers and 47.7K answer views
·
7y
Simply that (assuming that A & B are compatible == # columns of A=# rows of B),
each column B_i of B is in the null space of A (A B_i =zero vector).
This may seem empty but it isn't.
It means that for each column B_i of B, its entries b_i j are such that sum_j b_ij A_j= zero vector (where A_j is the j_th column of A).
So the b_ij are precisely the coefficients that combine columns of A to sum to the zero vector.
Equivalently, B_i \in \nullsp(A), for each i.
David Lambert
HS Diploma from Horace Greeley High School (Graduated 1978)
·
Author has 3.6K answers and 2.2M answer views
·
7y
Originally Answered: What if A and B are nonzero square matrices such that AB=0?
·
What if they are? Then there are some really great consequences which elude me right now. Someone answering this question please cite the quantum mechanical implications. I will follow this question.
The projection of one vector onto the other is zero. There is orthogonality. The angle is 90 degrees. They are perpendicular. Well, this is the case for vectors.
mp=: +/ . NB. matrix product A=:1 0,:0 B=:|.A A 1 0 0 0 B 0 0 1 0 A mp B 0 0 0 0  NB. www.jsoftware.com
Related questions
If the product of two matrices, A and B is zero matrix, how can you prove that matrices An×n and BA don't have to be zero matrices?
If A, B and C are all n x n matrices. If AB=AC, then what conditions are necessary to be able to deduce that A=0?
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Ɣioˑɾɣos
Author has 350 answers and 204.5K answer views
·
7y
Do you want a Null matrix or O real number?
If it is a real number you want
The dimensions of A must be (1,x) and the dimensions of B must be (x,1) x can be any real number
Then it is simple
Multiply pairwise the first of each
Element 1,1 of vector A Element 1,1 of vector B + 1,2A 2,1B …. + 1,xA2,1B = 0
It is a linear equation
For a non Real Number Null Matrix
That is, the entry {\displaystyle c_{ij}} of the product is obtained by multiplying term-by-term the entries of the ith row of A and the jth column of B
{\displaystyle c_{ij}} = 0 For every I and J
You complicated it more
Let A be any matrix
A
Do you want a Null matrix or O real number?
If it is a real number you want
The dimensions of A must be (1,x) and the dimensions of B must be (x,1) x can be any real number
Then it is simple
Multiply pairwise the first of each
Element 1,1 of vector A Element 1,1 of vector B + 1,2A 2,1B …. + 1,xA2,1B = 0
It is a linear equation
For a non Real Number Null Matrix
That is, the entry {\displaystyle c_{ij}} of the product is obtained by multiplying term-by-term the entries of the ith row of A and the jth column of B
{\displaystyle c_{ij}} = 0 For every I and J
You complicated it more
Let A be any matrix
A =
1,1= x 1,2 = y
2,1 = z 2,2 = w
0 =
1,1 = 0 1,2 = 0
2,1 = 0 2,2 = 0
xs + yu = 0 Equation for 1,1
xt + yv = 0 Equation for 1,2
zs + w u = 0 Equation for 2,1
zt + wv = 0 Equation for 2,2
B =
1,1= s 1,2 = t
2.1 = u 2,2 = v
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Sameer Mishra
Mathematica and Poietica (Youtube channel) (2016–present)
·
3y
Here I am going to give it's Geometrical proof,after seeing this small video you can find many more such matrices which satisfying A.B=0
David Joyce
Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979)
·
Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) and
Justin Rising
, PhD in statistics · Author has 9.9K answers and 68.4M answer views
·
Updated 2y
Related
How do I prove that ab=0 if a=0 or b=0?
Note: Quora Content Review changed the meaning of the question from “if and only if” to “if”.
“If and only if” is a bi-implication. The “if” part of it is only half of the question, and it’s not the interesting half.
This theorem is valid in any field. A field has the usual operations of addition, subtraction, multiplication, and division and satisfies the usual properties of these operations. The required properties are few (called field axioms), but from them more complicated properties can be proved like the one in this question.
As this is an "if and only if" statement, its proof requires two
Note: Quora Content Review changed the meaning of the question from “if and only if” to “if”.
“If and only if” is a bi-implication. The “if” part of it is only half of the question, and it’s not the interesting half.
This theorem is valid in any field. A field has the usual operations of addition, subtraction, multiplication, and division and satisfies the usual properties of these operations. The required properties are few (called field axioms), but from them more complicated properties can be proved like the one in this question.
As this is an "if and only if" statement, its proof requires two parts. We have to show that if ab = 0, then either a = 0 or b = 0, and we have to show that if a = 0 or b = 0, then ab = 0. Let's do the second one first because it's easier.
Proof: Given a = 0 or b = 0, to show that ab = 0. If we know that 0 times anything is 0, then we can conclude that. If a is 0, then a times b is 0, but if b is 0 then b times a is 0. Q.E.D.
(Note that for this half of the proof, we needed to know x0 = 0x = 0 for all x. That isn't usually taken as an axiom for fields, but is proved from the axioms.)
Proof: Given ab = 0, to show that either a = 0 or b = 0. If a = 0, then we're done, so we'll consider the case a ≠ 0. Then by one of the axioms for a field, a has a reciprocal, a−1, so that a−1a=1. Multiply both sides of the equation ab = 0 by a−1. Then
By the associativity axiom for fields, [math]a^{-1}(ab)=(a^{-1}a)b,[/math] and since we know that [math]a^{-1}a=1,[/math] therefore [math]a^{-1}(ab)=1b.[/math] And by another axiom, [math]1b=b.[/math] Therefore, the left side of the equation above equals b. But the right side equals 0 (by the parenthetical remark in the previous half of the proof). Therefore b = 0. Thus, either a = 0 or b = 0. Q.E.D.
By the way, you can find the definition of field at my answer to What is meant by a field in mathematics?
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Alex Lozovskiy
Ph.D. in Mathematics, University of Pittsburgh (Graduated 2010)
·
Author has 278 answers and 1.3M answer views
·
7y
Originally Answered: What if A and B are nonzero square matrices such that AB=0?
·
So what?
[math]\begin{pmatrix}0 & 0 \ 0 & 1\end{pmatrix}\begin{pmatrix}0 & 1 \ 0 & 0\end{pmatrix} = \begin{pmatrix}0 & 0 \ 0 & 0\end{pmatrix}[/math]
And?
Pranav Shankar
Mathematician by nature, Software Engineer by profession.
·
Author has 114 answers and 2.9M answer views
·
Updated 9y
Related
How do I prove that ab=0 if a=0 or b=0?
My approach is a lot easier than what is mentioned here, probably because I am far less knowledgeable in Mathematics when compared to most of them.
Firstly I'm going to make a couple of assumptions,
You're talking about a and b as variables and not matrices.
This is not a homework question.
Let me rephrase this question a little bit,
If ab = 0 then a = 0 or b = 0.
My proof is fairly straightforward.
Assume for the sake of contradiction that a is not equal to 0 and b is not equal to 0, when you multiply a and b you get zero, but that is a contradiction because when you multiply two non zero numbe
My approach is a lot easier than what is mentioned here, probably because I am far less knowledgeable in Mathematics when compared to most of them.
Firstly I'm going to make a couple of assumptions,
You're talking about a and b as variables and not matrices.
This is not a homework question.
Let me rephrase this question a little bit,
If ab = 0 then a = 0 or b = 0.
My proof is fairly straightforward.
Assume for the sake of contradiction that a is not equal to 0 and b is not equal to 0, when you multiply a and b you get zero, but that is a contradiction because when you multiply two non zero numbers, the resulting number is also non-zero. Thus we have found a contradiction! Therefore at least one of a or b must be zero in order to produce the given result.
The theory behind the proof is if you have a hypothesis A which implies B such that A => B. If we prove the negation of it to be false, then the statement itself must be true. The negation of A=>B is A (AND) (NOT) (B). Not B is the negation of B which is a not equal to zero and b not equal to zero. We have proved that to be false, therefore the actual statement must be true.
EDIT: Another way you could do this is by assuming ab = 0 and a not equal to zero and b not equal to zero, if you divide both sides by b you would get a = 0, which contradicts our assumption, therefore the given statement is false.
I hope I've made it clear, I'm not the brightest mathematician so I can't compare to the answers of Senia or Professor David. But I hope I've given you an easier answer to what seems like an innocent question.
Good luck, happy studying!
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Karina Hastings
B.S. in Mathematics, Elmira College (Graduated 2019)
·
Updated 6y
Related
How do I prove that ab=0 if a=0 or b=0?
I’m not entirely sure what you are asking here. Proving that if [math]a=0[/math] or [math]b=0[/math] then [math]ab=0 [/math]is pretty easy, so I’m going to guess that you are more interested in how to prove that if [math]ab=0[/math] then [math]a=0[/math] or [math]b=0[/math].
First, let’s note that this will be true in any field, and thus we can use all of the field axioms to help us prove this.
We want to prove that [math]ab=0[/math] in any field.
Let’s use an indirect proof to solve this.
Proof:
Suppose that [math]ab=0[/math] in a field. Let’s consider two cases: [math]a=0 [/math]or [math]a\neq0.[/math]
Case 1: If [math]a=0[/math], we are done because we have shown that either [math]a=0[/math] or [math]b=0[/math].
Case 2: If [math]a\neq0[/math], then we know that [math]a[/math] has a multip
I’m not entirely sure what you are asking here. Proving that if [math]a=0[/math] or [math]b=0[/math] then [math]ab=0 [/math]is pretty easy, so I’m going to guess that you are more interested in how to prove that if [math]ab=0[/math] then [math]a=0[/math] or [math]b=0[/math].
First, let’s note that this will be true in any field, and thus we can use all of the field axioms to help us prove this.
We want to prove that [math]ab=0[/math] in any field.
Let’s use an indirect proof to solve this.
Proof:
Suppose that [math]ab=0[/math] in a field. Let’s consider two cases: [math]a=0 [/math]or [math]a\neq0.[/math]
Case 1: If [math]a=0[/math], we are done because we have shown that either [math]a=0[/math] or [math]b=0[/math].
Case 2: If [math]a\neq0[/math], then we know that [math]a[/math] has a multiplicative inverse [math]a^{-1}[/math] because any number in a field that it is not [math]0[/math] has a number by which it can be multiplied to get [math]1[/math], [math]a \cdot a^{-1} = 1[/math], which is the definition of a multiplicative inverse.
Now let’s consider the statement:
[math]a^{-1} \cdot (ab) = 0 \cdot a^{-1}.[/math]
Let’s solve:
We can regroup our products on the left side because multiplication is associative in a field to get:
math \cdot b = 0 \cdot a^{-1}.[/math]
Again, any number multiplied by its multiplicative inverse is [math]1[/math], so on the left we have:
[math]1 \cdot b = 0 \cdot a^{-1}.[/math]
Next, we also know that any number times [math]0[/math] will just be [math]0[/math] so we can adjust the right side as:
[math]1 \cdot b = 0.[/math]
And with some final clean up we have:
[math]b = 0[/math]
Which again tells us that either [math]a=0[/math] or [math]b=0[/math] if [math]ab=0[/math] in a field.
Once more, this is not exactly what the question was asking, but the converse of this question is much more difficult and interesting to prove.
Gary Russell
Former Professor at University of Iowa (1996–2025)
·
Author has 6K answers and 3.1M answer views
·
Updated 3y
Related
If A and B are two non-zero vectors, then what does it mean when AB=0? What does this tell us about A and B?
If A and B are vectors (say A is a row vector and B is column vector), then the dot product is 0. This means that the two vectors are orthogonal to one another.
If A and B are matrices, AB = 0 means that the vectors in the columns of B are orthogonal to the space spanned by the rows of A. This fact is useful in certain statistical applications, such as Analysis of Variance. It allows the development of statistical tests for various features of a model.
Charles S.
former mathematician, current patent lawyer
·
Upvoted by
Michael Jørgensen
, PhD in mathematics and
Nathan Hannon
, Ph. D. Mathematics, University of California, Davis (2021) · Author has 7.6K answers and 60.4M answer views
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Updated 5h
Related
If matrices
[math]AB = AC[/math]
then does
[math]B = C[/math]
?
There are some good answers here, in the sense of being correct. But I wonder whether the person who posed this question is satisfied with them.
I mean, there are a few answers whose authors have produced, as if by magic, concrete matrices [math]A, B,[/math] and [math]C[/math] such that [math]AB = AC[/math] but [math]B\neq C[/math]. But it may not be clear to the reader where those matrices came from. A neophyte might wonder whether those authors are Ramanujan-level mathematicians who just have deep enough personal relationships with mathematical objects that all they must do is ask for volunteers, and the right objects just leap out from the cr
There are some good answers here, in the sense of being correct. But I wonder whether the person who posed this question is satisfied with them.
I mean, there are a few answers whose authors have produced, as if by magic, concrete matrices [math]A, B,[/math] and [math]C[/math] such that [math]AB = AC[/math] but [math]B\neq C[/math]. But it may not be clear to the reader where those matrices came from. A neophyte might wonder whether those authors are Ramanujan-level mathematicians who just have deep enough personal relationships with mathematical objects that all they must do is ask for volunteers, and the right objects just leap out from the crowd, eagerly obliging.
Don’t worry, young neophyte. It’s not that.
Here’s one way to think about this problem without just randomly manipulating numbers until you get something that works.
Start with a good observation: if [math]AB = AC[/math] and [math]A[/math] is invertible (or left-invertible, if you’re being very careful), then [math]B=C[/math]. So if there’s a counterexample, it must involve a non-invertible matrix [math]A[/math].
To the extent this answer involves wisdom, here it is: the poster-child of a non-invertible matrix is a projection. That is, a diagonal matrix with a block of 1s on the diagonal and a block of 0s in the remainder of the matrix, including at least one 0 on the diagonal. This matrix is the identity when restricted to the subspace spanned by the columns having a 1 on the diagonal, and it’s the zero matrix when restricted to the subspace spanned by the columns having a 0 on the diagonal.
In other words, it does nothing on part of the whole vector space, and it kills everything on another part of the whole vector space. Just to give us some notation, suppose [math]A[/math] is a linear transformation from the vector space [math]V[/math] to [math]V[/math], and [math]A|_N = 1_N[/math] while [math]A|_M = 0_M[/math]. That is, [math]A[/math] is the identity on a subspace [math]N[/math] of [math]V[/math], and [math]A[/math] is 0 on a subspace [math]M[/math] of [math]V[/math].
Think of a way to construct a counterexample yet? Don’t push around numbers, just think qualitatively. All you need are matrices [math]B[/math] and [math]C[/math] that are the same on [math]N[/math] (where [math]A[/math] does nothing), and differ on [math]M[/math] (where [math]A[/math] kills everything).
Why does that work? Because it doesn’t matter what [math]B[/math] or [math]C[/math] does on [math]M[/math] — the action of [math]A[/math] is going send everything to zero anyway. The actions of [math]B[/math] and [math]C[/math] on [math]M[/math] are just different ways to shuffle the deck chairs on the titanic… the ship is ultimately sinking. That is to say, [math]AB|_M = AC|_M = 0[/math].
But we’ve arranged for [math]A[/math] to leave [math]N[/math] undisturbed. As long as [math]B|_N = C|_N[/math], we’re good to go.
I’ll leave it as an exercise to the reader to fill in some numerical examples. But I’ll provide a further hint: make [math]B[/math] and [math]C[/math] diagonal matrices too. It need not be more complicated than that.
David Tung
know what is mathematics
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 529 answers and 983.9K answer views
·
10y
Related
How do I prove that ab=0 if a=0 or b=0?
This is equivalent to say if [math] a \in R [/math] (commutative ring, that is no multiplicative inverse, no cancellation law for multiplicative), then [math] a\cdot 0 = 0 = 0\cdot a [/math]. It will test you how to derive a seemingly obvious property from axioms. (See A Survey of Modern Algebra, chapter 1, RULE 7).
Proof:
[math] a = a [/math], [math] a + 0 = a [/math] (reflexive law, and zero element)
[math] a\cdot (a+0)=a\cdot a [/math] (uniqueness of multiplication)
[math] a\cdot a + a\cdot 0 = a\cdot a + 0 [/math] (distributive law, and zero element)
[math] a\cdot 0 = 0 [/math] (cancellation law for additive, this needs to be proved, see below)
[math] a\cdot 0 = 0\cdot a = 0 [/math]
This is equivalent to say if [math] a \in R [/math] (commutative ring, that is no multiplicative inverse, no cancellation law for multiplicative), then [math] a\cdot 0 = 0 = 0\cdot a [/math]. It will test you how to derive a seemingly obvious property from axioms. (See A Survey of Modern Algebra, chapter 1, RULE 7).
Proof:
[math] a = a [/math], [math] a + 0 = a [/math] (reflexive law, and zero element)
[math] a\cdot (a+0)=a\cdot a [/math] (uniqueness of multiplication)
[math] a\cdot a + a\cdot 0 = a\cdot a + 0 [/math] (distributive law, and zero element)
[math] a\cdot 0 = 0 [/math] (cancellation law for additive, this needs to be proved, see below)
[math] a\cdot 0 = 0\cdot a = 0 [/math] (commutative law).
Cancellation law for additive: for all [math] a, b, c \in R [/math], [math] a+b=a+c [/math], implies [math] b = c [/math].
Proof:
[math] a+b+(-a)=a+c+(-a) [/math] (existence of additive inverse, and uniqueness of addition).
[math] a+(b+(-a))=a+(c+(-a)) [/math] (associative law)
[math] a+((-a)+b)=a+((-a)+c) [/math] (commutative law)
[math] (a+(-a))+b=(a+(-a))+c [/math] (associative law)
[math] 0+b=0+c [/math] (property of additive inverse)
[math] b=c [/math] (zero element). QED
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Table of Contents
Table of Contents
What Is a Liability?
The Mechanism of Liabilities
Various Definitions of Liability
Current vs. Non-Current Liabilities
Liabilities vs. Assets
Liabilities vs. Expenses
Examples
FAQs
The Bottom Line
Understanding Liabilities: Definitions, Types, and Key Differences From Assets
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Updated August 19, 2025
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Part of the Series
Guide to Accounting
Accounting Explained With Brief History and Modern Job Requirements
Accounting Basics
Accounting Equation
Asset
Liability
CURRENT ARTICLE
4. Equity
5. Revenue
6. Expense
7. Current and Noncurrent Assets
Accounting Theories and Concepts
Accounting Theory
Accounting Principles
Accounting Standard
Accounting Convention
Accounting Policies
Principles-Based vs. Rules-Based Accounting
Accounting Methods: Accrual vs. Cash
Accounting Method
Accrual Accounting
Cash Accounting
Accrual Accounting vs. Cash Basis Accounting
Accounting Oversight and Regulations
Financial Accounting Standards Board (FASB)
Generally Accepted Accounting Principles (GAAP)
International Financial Reporting Standards (IFRS)
IFRS vs. GAAP
US Accounting vs. International Accounting
Financial Statements
Understanding the Cash Flow Statement
Breaking Down The Balance Sheet
Understanding the Income Statement
Corporate Accounting
Accountant
Financial Accounting
Financial Accounting and Decision-Making
Corporate Finance
Financial vs. Managerial Accounting
Cost Accounting
Public Accounting: Financial Audit and Taxation
Certified Public Accountant (CPA)
Chartered Accountant (CA)
Accountant vs. Financial Planner
Auditor
Audit
Tax Accounting
Forensic Accounting
Accounting Systems and Record Keeping
Chart of Accounts (COA)
Journal
Double Entry
Debit
Credit
Closing Entry
Invoice
Introduction to Accounting Information Systems
Accounting for Inventory
Inventory Accounting
Last In, First Out (LIFO)
First In, First Out (FIFO)
Average Cost Method
What Is a Liability?
A liability is something that a person or company owes, usually a sum of money. Liabilities are settled over time through the transfer of economic benefits including money, goods, or services. They're recorded on the right side of the balance sheet and include loans, accounts payable, mortgages, deferred revenues, bonds, warranties, and accrued expenses.
Liabilities are the opposite of assets. They refer to things that you owe or have borrowed. Assets are things that you own or are owed.
Key Takeaways
Liabilities are financial obligations that a person or company owes to others, often in the form of money.
Current liabilities are short-term financial obligations due within one year, while non-current liabilities extend beyond that period.
Liabilities are central to a company's operations as they help finance growth and manage financial transactions.
In accounting, the relationship between assets, liabilities, and equity can be expressed as Assets = Liabilities + Equity.
Liabilities can also refer to legal obligations or risks, with businesses often obtaining liability insurance to mitigate potential lawsuits.
Understanding the Mechanism of Liabilities
A liability is generally something you owe that isn't yet paid. In accounting, financial liabilities are linked to past transactions or events that will provide future economic benefits.
Liabilities are categorized as current or non-current depending on their temporality. Liabilities can include future services owed, short-term or long-term loans, or unsettled obligations from past transactions.
Fast Fact
Current liabilities are usually considered short-term. They're expected to be concluded within 12 months or less. Non-current liabilities are long-term. They're expected to last 12 months or longer.
Common large liabilities include accounts payable and bonds payable, which are regular items on most companies' balance sheets.
Liabilities are a vital aspect of a company because they're used to finance operations and pay for large expansions. They can also make transactions between businesses more efficient. A wine supplier typically doesn't demand payment when it sells a case of wine to a restaurant and delivers the goods. It invoices the restaurant for the purchase to streamline the drop-off and make paying easier for the restaurant.
The outstanding money that the restaurant owes to its wine supplier is considered a liability. The wine supplier considers the money it is owed to be an asset.
Exploring Various Definitions of Liability
Liability generally refers to the state of being responsible for something. The term can refer to any money or service owed to another party. Tax liability can refer to the property taxes that a homeowner owes to the municipal government or the income tax they owe to the federal government. A retailer has a sales tax liability on their books when they collect sales tax from a customer until they remit those funds to the county, city, or state.
Liability can also refer to one's potential damages in a civil lawsuit.
Important
Liability can also mean legal responsibility. Businesses often get liability insurance to protect against lawsuits from customers or employees.
Comparing Current and Non-Current Liabilities
A 15-year mortgage is a long-term liability, but payments due this year are current liabilities. They're recorded in the short-term liabilities section of the balance sheet.
Current (Near-Term) Liabilities
Analysts prefer companies to pay their current liabilities, which are due within a year, using cash. Some examples of short-term liabilities include payroll expenses and accounts payable which can include money owed to vendors, monthly utilities, and similar expenses. Other examples include:
Wages payable: This is the total amount of accrued income that employees have earned but haven't yet received. Many companies pay their employees every two weeks so this liability changes often.
Interest payable: Companies often use credit to purchase goods and services. This represents the interest on those short-term credit purchases that must be paid.
Dividends payable: This represents the amount owed to shareholders after a dividend was declared for companies that have issued stock to investors and pay dividends.
Unearned revenues:This is a company's liability to deliver goods and/or services at a future date after being paid in advance. The amount will be reduced in the future with an offsetting entry when the product or service is delivered.
Liabilities of discontinued operations:This is a unique liability. Companies are required to account for the financial impact of an operation, division, or entity that's currently being held for sale or has been recently sold. This also includes the financial impact of a product line that has recently been shut down.
Non-Current (Long-Term) Liabilities
Any liability that's not near-term falls under non-current liabilities that are expected to be paid in 12 months or more. Long-term debt is also known as bonds payable and it's usually the largest liability and at the top of the list.
Companies of all sizes finance part of their ongoing long-term operations by issuing bonds that are essentially loans from each party that purchases the bonds. This line item is in constant flux as bonds are issued, mature, or called back by the issuer.
Analysts want to see that long-term liabilities can be paid with assets derived from future earnings or financing transactions. Bonds and loans aren't the only long-term liabilities that companies incur. Items like rent, deferred taxes, payroll, and pension obligations can also be listed as long-term liabilities. Other examples include:
Warranty liability: Some liabilities aren't as exact as AP. They have to be estimated. Warranty liability is the estimated time and money that may be spent repairing products under the agreement of a warranty. It's a common liability in the automotive industry because many cars have long-term warranties that can be costly.
Contingent liability evaluation: A contingent liability may or may not occur depending on the outcome of an uncertain future event.
Deferred credits:This is a broad category that can be recorded as current or non-current depending on the specifics of the transaction. These credits are revenue collected before it's recorded as earned on the income statement. They can include customer advances, deferred revenue, or a transaction where credits are owed but not yet considered revenue. This item is reduced by the amount earned and becomes part of the company's revenue stream when the revenue is no longer deferred.
Post-employment benefits: These are benefits that an employee or family member may receive upon their retirement. They're carried as long-term liabilities as they accrue. This liability isn't to be overlooked with rapidly rising health care and deferred compensation.
Unamortized investment tax credits (UITC):This represents the net between an asset's historical cost and the amount that's already been depreciated. The unamortized portion is a liability but it's only a rough estimate of the asset’s fair market value. This provides an analyst with some details regarding how aggressive or conservative a company is with its depreciation methods.
Distinguishing Between Liabilities and Assets
Assets are what a company owns or something that's owed to the company. They include tangible items such as buildings, machinery, and equipment as well as intangibles such as accounts receivable, interest owed, patents, or intellectual property.
The difference is its owner's or stockholders' equity if a business subtracts its liabilities from its assets. The relationship can be expressed like this:
Assets−Liabilities=Owner’s Equity
This accounting equation is commonly presented this way, however:
Assets=Liabilities+Equity
Differentiating Liabilities From Expenses
An expense is the cost of operations that a company incurs to generate revenue. Expenses are related to revenue, unlike assets and liabilities. Both are listed on a company's income statement. Expenses are used to calculate net income. The equation is revenues minus expenses.
Fast Fact
It might signal weak financial stability if a company has had more expenses than revenues for the last three years because it's been losing money for those years.
Liabilities appear on the balance sheet, while expenses are on the income statement. Expenses relate to operational costs, unlike liabilities, which are debts owed. Delayed payment of expenses can become a liability.
Practical Examples of Liabilities in Business
Let's look at a historical example using AT&T's (T) 2020 balance sheet. The current/short-term liabilities are separated from long-term/non-current liabilities.
AT&T clearly defines its bank debt that's maturing in less than one year under current liabilities. This is often used as operating capital for day-to-day operations by a company of this size rather than funding larger items which would be better suited using long-term debt.
Liabilities are carried at cost, not market value, like most assets. They can be listed in order of preference under generally accepted accounting principle (GAAP) rules as long as they're categorized. The AT&T example has a relatively high debt level under current liabilities. Other line items like accounts payable (AP) and various future liabilities like payroll taxes will be higher current debt obligations for smaller companies.
AP typically carries the largest balances because they encompass day-to-day operations. AP can include services, raw materials, office supplies, or any other categories of products and services where no promissory note is issued. Most companies don't pay for goods and services as they're acquired, AP is equivalent to a stack of bills waiting to be paid.
How Do I Know If Something Is a Liability?
A liability is anything that's borrowed from, owed to, or obligated to someone else. It can be real like a bill that must be paid or potential such as a possible lawsuit. A liability isn't necessarily a bad thing. A company might take out debt to expand and grow its business or an individual may take out a mortgage to purchase a home.
How Are Current Liabilities Different From Long-Term Non-Current Ones?
Companies segregate their liabilities by their time horizon for when they're due. Current liabilities are due within a year and are often paid using current assets. Non-current liabilities, due in over a year, typically include debt and deferred payments.
What Is a Contingent Liability?
A contingent liability is an obligation that might have to be paid in the future but there are still unresolved matters that make it only a possibility, not a certainty. Lawsuits and the threat of lawsuits are the most common contingent liabilities but unused gift cards, product warranties, and recalls also fit into this category.
What Are Examples of Liabilities That Individuals or Households Have?
An individual's or household's net worth is also arrived at by balancing assets against liabilities. Liabilities for most households will include taxes due, bills that must be paid, rent or mortgage payments, loan interest, and principal due. The work owed may also be construed as a liability if you're prepaid for performing work or a service,
The Bottom Line
Liabilities represent what you owe to others, whether as a financial obligation due to borrowing or as a legal commitment. These obligations, crucial for both individuals and businesses, are fundamental to understanding financial health and are recorded on the balance sheet alongside assets. Liabilities are divided into current (due within a year) and non-current (due beyond a year), each playing distinct roles in a company's or individual's financial strategy. Managing liabilities effectively, such as loans or accounts payable, ensures smooth operations and facilitates growth. Ultimately, balancing liabilities against assets provides insight into financial stability and net worth.
Article Sources
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U.S. Small Business Administration. "Get Business Insurance."
AT&T. "2020 Annual Report."
CFI Education. "Short-Term Debt."
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Open an Account Before the Fed Decision on Sept. 17
Part of the Series
Guide to Accounting
Accounting Explained With Brief History and Modern Job Requirements
Accounting Basics
Accounting Equation
Asset
Liability
CURRENT ARTICLE
4. Equity
5. Revenue
6. Expense
7. Current and Noncurrent Assets
Accounting Theories and Concepts
Accounting Theory
Accounting Principles
Accounting Standard
Accounting Convention
Accounting Policies
Principles-Based vs. Rules-Based Accounting
Accounting Methods: Accrual vs. Cash
Accounting Method
Accrual Accounting
Cash Accounting
Accrual Accounting vs. Cash Basis Accounting
Accounting Oversight and Regulations
Financial Accounting Standards Board (FASB)
Generally Accepted Accounting Principles (GAAP)
International Financial Reporting Standards (IFRS)
IFRS vs. GAAP
US Accounting vs. International Accounting
Financial Statements
Understanding the Cash Flow Statement
Breaking Down The Balance Sheet
Understanding the Income Statement
Corporate Accounting
Accountant
Financial Accounting
Financial Accounting and Decision-Making
Corporate Finance
Financial vs. Managerial Accounting
Cost Accounting
Public Accounting: Financial Audit and Taxation
Certified Public Accountant (CPA)
Chartered Accountant (CA)
Accountant vs. Financial Planner
Auditor
Audit
Tax Accounting
Forensic Accounting
Accounting Systems and Record Keeping
Chart of Accounts (COA)
Journal
Double Entry
Debit
Credit
Closing Entry
Invoice
Introduction to Accounting Information Systems
Accounting for Inventory
Inventory Accounting
Last In, First Out (LIFO)
First In, First Out (FIFO)
Average Cost Method
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190315 | https://www.erknet.org/fileadmin/files/user_upload/Medical_management_of_kidney_stones_AUA_guideline..pdf | AUA Guidelines Medical Management of Kidney Stones: AUA Guideline Margaret S. Pearle, David S. Goldfarb, Dean G. Assimos, Gary Curhan, Cynthia J. Denu-Ciocca, Brian R. Matlaga, Manoj Monga, Kristina L. Penniston, Glenn M. Preminger, Thomas M. T. Turk and James R. White From the American Urological Assocation Education and Research, Inc., Linthicum, Maryland Purpose: The purpose of this guideline is to provide a clinical framework for the diagnosis, prevention and follow-up of adult patients with kidney stones based on the best available published literature.
Materials and Methods: The primary source of evidence for this guideline was the systematic review conducted by the Agency for Healthcare Research and Quality on recurrent nephrolithiasis in adults. To augment and broaden the body of evidence in the AHRQ report, the AUA conducted supplementary searches for articles published from 2007 through 2012 that were systematically reviewed using a methodology developed a priori. In total, these sources yielded 46 studies that were used to form evidence-based guideline statements. In the absence of sufficient evidence, additional statements were developed as Clinical Principles and Expert Opinions.
Results: Guideline statements were created to inform clinicians regarding the use of a screening evaluation for first-time and recurrent stone formers, the appropriate initiation of a metabolic evaluation in select patients and recom-mendations for the initiation and follow-up of medication and/or dietary mea-sures in specific patients.
Conclusions: A variety of medications and dietary measures have been evalu-ated with greater or less rigor for their efficacy in reducing recurrence rates in stone formers. The guideline statements offered in this document provide a simple, evidence-based approach to identify high-risk or interested stone-forming patients for whom medical and dietary therapy based on metabolic testing and close follow-up is likely to be effective in reducing stone recurrence.
Key Words: nephrolithiasis, urolithiasis, citrate, hypercalciuria, oxalate INTRODUCTION Kidney stone disease is a common malady, affecting nearly 1 in 11 in-dividuals in the United States at some point in their lives.1 Stones are also likely to recur, with at least 50% of individuals experiencing another stone within 10 years of the first occurrence. For those who have expe-rienced a stone or undergone surgical intervention for a stone, there is strong motivation to avoid a repeat episode. This guideline is aimed at practitioners from a variety of disci-plines who are confronted with pa-tients afflicted with stone disease, and it is based on a systematic review of the literature with respect to the evaluation, treatment and follow-up of first-time and recurrent stone for-mers. Patient preferences and goals must be taken into account by the Abbreviations and Acronyms AHA ¼ acetohydroxamic acid AHRQ ¼ Agency for Healthcare Research and Quality CT ¼ computerized tomography DASH ¼ Dietary Approaches to Stop Hypertension HPFS ¼ Health Professionals Follow-up Study NHANES ¼ National Health and Nutrition Examination Survey NHS ¼ Nurses’ Health Study PTH ¼ parathyroid hormone RCT ¼ randomized controlled trial RDA ¼ recommended dietary allowance RTA ¼ renal tubular acidosis UTI ¼ urinary tract infection The complete guideline is available at management-kidney-stones.cfm.
This document is being printed as submitted independent of editorial or peer review by the Editors of The Journal of Urology.
316 j www.jurology.com 0022-5347/14/1922-0316/0 THE JOURNAL OF UROLOGY® © 2014 by AMERICAN UROLOGICAL ASSOCIATION EDUCATION AND RESEARCH, INC.
Vol. 192, 316-324, August 2014 Printed in U.S.A.
practitioner when considering these guidelines, as the cost, inconvenience and side effects of drugs and dietary measures to prevent stone disease must be weighed against the benefit of preventing a recur-rent stone.
METHODOLOGY The AHRQ systematic review titled Recurrent Nephrolithiasis in Adults: Comparative Effective-ness of Preventative Medical Strategies was utilized as the primary source of evidence for guideline development. Additionally, the AUA conducted sup-plementary searches of PubMed and EMBASE for relevant articles published between January 2007 and November 2012, which were systematically reviewed using a methodology developed a priori.
The AUA conducted an extensive peer review pro-cess. The initial draft of this guideline was distrib-uted to 107 peer reviewers of varying backgrounds; 40 responded with comments. The panel reviewed and discussed all submitted comments and revised the guideline as needed.
The AUA nomenclature system explicitly links statement type to body of evidence strength and the Panel’s judgment regarding the balance be-tween benefits and risks/burdens. For a complete discussion of the methodology and evidence grading, please refer to the unabridged guideline avail-able at management-kidney-stones.cfm .
BACKGROUND Kidney stone disease is a common condition. Ac-cording to the most recent National Health and Nutrition Examination Survey, the overall preva-lence of self-reported kidney stones from 2007e2010 was 8.8%, with a higher prevalence among men (10.6%) than women (7.1%).1 This prevalence rep-resents a 70% increase over the last reported prev-alence (5.2%) derived from an NHANES sample (1988e1994), and the increased prevalence was observed across all age groups and in both sexes.
Although historically kidney stones have occurred more commonly in men than women, by any num-ber of metrics the gender gap is closing.2,3 The reasons for the observed rise in stone disease among women are not certain, but the impact of obesity, a known risk factor for kidney stones, was found to be greater in women than in men.4 Stone disease has been increasingly linked to systemic conditions, although it is not clear if stone disease is a cause of these disorders or if it is a consequence of the same conditions that lead to these disorders. Overweight/obesity,1,4 hypertension5 and diabetes6 have all been shown to be associated with an increased risk of stone disease.
Diet and lifestyle affect the risk of developing stones. A number of dietary measures have been evaluated for their effects on stone formation, and these studies provide compelling reasons to incor-porate or avoid a variety of dietary measures. Some drug therapies, most of which are primarily directed against specific metabolic abnormalities, have been shown to be superior to placebo, or no-treatment control groups, in randomized trials.7 However, randomized controlled trials (RCTs) evaluating drug treatments are relatively sparse, likely because the relative infrequency of the event re-quires long periods of observation. Diet therapy has never been compared head-to-head with phar-macologic therapy.
As such, recommendations incorporate both diet therapy and pharmacotherapy until the superiority of one over the other can be demonstrated.
GUIDELINE STATEMENTS Evaluation 1. A clinician should perform a screening evaluation consisting of a detailed medical and dietary history, serum chemistries and urinalysis on a patient newly diagnosed with kidney or ureteral stones. (Clinical Principle) A detailed history should elicit from the patient any medical conditions, dietary habits or medica-tions that predispose to stone disease. Nutritional factors associated with stone disease, depending on stone type and risk factors, include calcium intake below or significantly above the recom-mended dietary allowance, low fluid intake, high sodium intake, limited intake of fruits and vege-tables and high intake of animal-derived purines.
Patients should be queried regarding their regular use of any stone-provoking medications or supplements.
Dietary history should elicit from the patient their average daily intake of fluids (amount and specific beverages), protein (types and amounts), calcium, sodium, high oxalate-containing foods, fruits and vegetables and over-the-counter supplements.
Serum chemistries should include electrolytes, calcium, creatinine and uric acid that may suggest underlying medical conditions associated with stone disease. Urinalysis should include both dipstick and microscopic evaluation to assess urine pH and in-dicators of infection and to identify crystals patho-gnomonic of stone type. Urine culture should be obtained in patients with a urinalysis suggestive of urinary tract infection or in patients with recur-rent UTIs.
MEDICAL MANAGEMENT OF KIDNEY STONES 317 2. Clinicians should obtain serum intact parathyroid hormone level as part of the screening evaluation if primary hyperpara-thyroidism is suspected. (Clinical Principle) Primary hyperparathyroidism should be sus-pected when serum calcium is high or high normal.
3. When a stone is available, clinicians should obtain a stone analysis at least once.
(Clinical Principle) Stone composition of uric acid, cystine or struvite implicates specific metabolic or genetic abnormal-ities, and knowledge of stone composition may help direct preventive measures.8 4. Clinicians should obtain or review avail-able imaging studies to quantify stone burden.
(Clinical Principle) Multiple or bilateral renal calculi at initial pre-sentation may place a stone former at greater risk of recurrence. Nephrocalcinosis implies an underlying metabolic disorder (e.g. renal tubular acidosis type 1, primary hyperparathyroidism, primary hyper-oxaluria) or anatomic condition (medullary sponge kidney) predisposing to stone formation.
5.
Clinicians should perform additional metabolic testing in high-risk or interested first-time stone formers and recurrent stone formers.
(Standard; Evidence Strength: Grade B) Urinary saturation of stone-forming salts has been shown to correlate with stone composition, suggesting that 24-hour urine testing can be used to inform and monitor treatment protocols.9 High-risk and/or recurrent stone formers are likely to benefit from metabolic testing and medical therapy.
Identification of metabolic and environmental risk factors can help direct dietary and medical therapy. Specific nutritional therapy, informed by both diet assessment and metabolic testing, has been shown to be more effective than general di-etary measures in preventing recurrent stones.10 6. Metabolic testing should consist of one or two 24-hour urine collections obtained on a random diet and analyzed at minimum for total volume, pH, calcium, oxalate, uric acid, citrate, sodium, potassium and creatinine.
(Expert Opinion) Either one or two 24-hour urines may be ob-tained, although two collections are preferred by the Panel. Other urinary parameters may be helpful in the initial and follow-up evaluation of stone for-mers. In stone formers with known cystine stones or a family history of cystinuria or for those in whom cystinuria is suspected, urinary cystine should additionally be measured. Primary hyper-oxaluria should be suspected when urinary oxalate excretion exceeds 75 mg/day in adults without bowel dysfunction.
7. Clinicians should not routinely perform “fast and calcium load” testing to distinguish among types of hypercalciuria. (Recommen-dation; Evidence Strength: Grade C ) Use of the fast and oral calcium load test to distinguish among types of hypercalciuria has not been shown to change clinical practice.11 Diet Therapies 8. Clinicians should recommend to all stone formers a fluid intake that will achieve a urine volume of at least 2.5 liters daily. (Standard; Evidence Strength: Grade B) Urine volume is a major determinant of the concentration of lithogenic factors. Fluid intake is the main determinant of urine volume, and as such, high fluid intake is a critical component of stone prevention. Although there is no definitive threshold for urine volume and increased risk (the relationship is continuous and may not be linear), an accepted goal is at least 2.5 liters of urine daily.
Observational studies have found that certain beverages may be associated with risk of stone formation beyond their impact on urine volume.
Alcoholic beverages, coffee (caffeinated and decaf-feinated), tea, wine and orange juice have been shown in observational studies to be associated with a lower risk of stone formation,12e14 while sugar-sweetened beverages demonstrated an increased risk.15 However, these beverages have not been evaluated in randomized trials.
9. Clinicians should counsel patients with calcium stones and relatively high urinary calcium to limit sodium intake and consume 1,000e1,200 mg per day of dietary calcium.
(Standard; Evidence Strength: Grade B) Prospective observational studies consistently show an independent reduced risk of stone forma-tion with higher dietary calcium intake.16e19 Di-etary salt (sodium chloride) has also been linked to urinary calcium excretion.20 The Panel supports a target of 100 mEq (2,300 mg) sodium intake daily.
A five-year randomized controlled clinical trial compared stone recurrence in men with a history of calcium oxalate nephrolithiasis and idiopathic hypercalciuria assigned to a diet lower in calcium (400 mg/day) or to a diet with normal calcium con-tent (1,200 mg/day) and lower amounts of animal protein and sodium; both groups were advised to limit oxalate intake.21 At study end, the risk of developing a recurrent stone on the normal calcium diet was 51% lower than on the lower calcium diet, although the independent effect of calcium could not be ascertained.
Supplemental calcium, in contrast, may be asso-ciated with an increased risk of stone formation.
318 MEDICAL MANAGEMENT OF KIDNEY STONES In an observational study of older women, calcium supplement users were 20% more likely to form a stone than women who did not take supplements.17 Many patients are able to obtain adequate daily calcium from traditional and calcium-fortified foods and beverages, many of which are non-dairy; cal-cium supplementation is not necessary in these patients.
10. Clinicians should counsel patients with calcium oxalate stones and relatively high urinary oxalate to limit intake of oxalate-rich foods and maintain normal calcium con-sumption. (Expert Opinion) Restricting oxalate-rich foods has generally been recommended for calcium stone formers. An exten-sive list of the oxalate content of foods is available online from the Harvard School of Public Health ( Oxalate%20Content%20of%20Foods.xls).
Urinary oxalate is also modulated by calcium intake, which influences intestinal oxalate absorp-tion. Patients with hyperoxaluria and a history of calcium oxalate stones should be advised to consume calcium from foods and beverages primar-ily at meals to enhance gastrointestinal binding of oxalate, but total calcium intake should not exceed 1,000e1,200 mg daily.
Of note, however, patients with enteric hyper-oxaluria and high levels of urinary oxalate, such as those with malabsorptive conditions (e.g., inflam-matory bowel disease or Roux-en-Y gastric bypass) may benefit from more restrictive oxalate diets as well as from higher calcium intakes, which may include supplements, specifically timed with meals.22 Other factors that may contribute to higher uri-nary oxalate include vitamin C and other over-the-counter nutrition supplements.
11. Clinicians should encourage patients with calcium stones and relatively low uri-nary citrate to increase their intake of fruits and vegetables and limit non-dairy animal protein. (Expert Opinion) Urinary citrate is a potent inhibitor of calcium stone formation.23 Metabolic acidosis or dietary acid loads enhance renal citrate reabsorption, thereby reducing urinary excretion.
Medical conditions such as renal tubular acidosis and chronic diarrhea, and some medications, such as carbonic anhydrase inhibitors, may promote hypocitraturia.24 Acidosis can arise from a diet that is inordinately rich in foods with a high potential renal acid load compared to low-acid (i.e., alkaline) foods.
If diet assessment suggests that the acid load of foods contributes to low urinary citrate, patients should be instructed to increase fruit and vegetable intake and reduce intake of high-acid foods.
Dietary alkali citrate has been proposed as an alternative to pharmacologic citrate to increase citrate excretion.14,25 12. Clinicians should counsel patients with uric acid stones or calcium stones and rela-tively high urinary uric acid to limit intake of non-dairy animal protein. (Expert Opinion) No relevant studies were identified to either refute or confirm the use of diet to manage high urinary uric acid in uric acid or calcium stone for-mers. Nonetheless, if diet assessment suggests that purine intake is contributory to high urinary uric acid, patients may benefit from limiting high and moderately high purine containing foods.
Uric acid crystal formation and growth occur in more acidic urine.26 Thus, patients with a history of uric acid stones should be counseled to increase the alkali load and decrease the acid load of their diet in an effort to increase urine pH and reduce urinary acidity.
13. Clinicians should counsel patients with cystine stones to limit sodium and protein intake. (Expert Opinion) Dietary therapy should be offered in combination with pharmacological therapy.
Because cystine stone formation is largely driven by cystine concen-tration, high fluid intake is particularly important in cystine stone formers. The target for urine volume is typically higher than that recommended to other stone formers because of the need to decrease uri-nary cystine concentration below 250 mg/L.27 Oral intake of at least four liters per day is often required to meet this goal. Dietary sodium restriction should also be advised as lower sodium intake has been shown to reduce cystine excretion.28,29 A reasonable goal for sodium intake in individuals with cystinuria is 100 mEq (2,300 mg) or less daily.
Limiting animal protein intake has been sug-gested as a means to decrease cystine substrate load, as all foods of animal origin are rich in cystine and methionine, which is metabolized to cystine.
Pharmacologic Therapies 14. Clinicians should offer thiazide diuretics to patients with high or relatively high urine calcium and recurrent calcium stones. (Stan-dard; Evidence Strength: Grade B) Thiazide dosages associated with a hypocalciuric effect include hydrochlorothiazide (25 mg orally, twice daily; 50 mg orally, once daily), chlorthalidone (25 mg orally, once daily), and indapamide (2.5 mg orally, once daily). Dietary prescription, especially restriction of sodium intake, should be continued when thiazides are prescribed, in order to maximize the hypocalciuric effect and limit potassium wasting.
Potassium supplementation (either potassium cit-rate or chloride) may be needed when thiazide MEDICAL MANAGEMENT OF KIDNEY STONES 319 therapy is employed. The addition of amiloride or spironolactone may avoid the need for potassium supplementation. Triamterene should be avoided as stones of this compound have been reported.
Thiazides should be considered appropriate for both calcium oxalate and calcium phosphate stone formers. Although studies were performed exclu-sively on patients with recurrent stone formation, the Panel believes that some high-risk first-time stone formers might also benefit from thiazide therapy, such as those with a solitary kidney, hy-pertension or a large stone burden, or individuals who are refractory to other risk-mitigating maneuvers.
15. Clinicians should offer potassium citrate therapy to patients with recurrent calcium stones and low or relatively low urinary cit-rate. (Standard; Evidence Strength: Grade B) Prospective RCTs have demonstrated that po-tassium citrate therapy is associated with reduced risk of recurrent calcium stones in patients with low or low normal 24-hour urinary citrate excretion.30e33 Calcium stone-forming patients with normal citrate excretion but low urinary pH may also benefit from citrate therapy. Additionally, potassium citrate therapy should be offered to calcium phosphate stone formers with hypoci-traturia because citrate is a known potent inhibitor of calcium phosphate crystallization.
Increased fluid intake, sodium restriction, ample fruits and vegetables to counterbalance foods that confer an acid load (see Guideline Statement 11), and thia-zides to lower urinary calcium excretion may in-crease the safety and efficacy of citrate therapy.
Potassium citrate is preferred over sodium cit-rate, as the sodium load in the latter may increase urine calcium excretion.34 16. Clinicians should offer allopurinol to patients with recurrent calcium oxalate stones who have hyperuricosuria and normal urinary calcium.
(Standard; Evidence Strength: Grade B) A prospective randomized controlled trial demonstrated that allopurinol reduced the risk of recurrent calcium oxalate stones in the setting of hyperuricosuria (urinary uric acid excretion >800 mg/day) and normocalciuria.35 Whether the drug is effective in patients with hypercalciuria has not been established.
Hyperuricemia is not a required criterion for allopurinol therapy.
17. Clinicians should offer thiazide diuretics and/or potassium citrate to patients with recurrent calcium stones in whom other metabolic abnormalities are absent or have been appropriately addressed and stone for-mation persists. (Standard; Evidence Strength: Grade B) Thiazides and potassium citrate therapy have been shown to prevent recurrent stones in patients with normal range urinary calcium and citrate, respectively.30,36,37 Therefore, it may be appropriate to utilize these therapies for patients with recurrent stones who do not demonstrate specific urinary abnormalities.
For patients with no identified risk factors for nephrolithiasis, potassium citrate may be the preferred first-line therapy, given its relatively low side effect profile.
18. Clinicians should offer potassium citrate to patients with uric acid and cystine stones to raise urinary pH to an optimal level.
(Expert Opinion) The solubility of uric acid and cystine is increased at higher urinary pH values.38 Potassium citrate therapy provides an alkali load that leads to increased urine pH. For uric acid stone formers, urine pH should be increased to 6.0, and for cystine stone formers, a urine pH of 7.0 should be achieved.
19. Clinicians should not routinely offer allopurinol as first-line therapy to patients with uric acid stones. (Expert Opinion) Most patients with uric acid stones have low urinary pH rather than hyperuricosuria as the predominant risk factor.39 Reduction of urinary uric acid excretion with the use of allopurinol in patients with uric acid stones will not prevent stones in those with unduly acidic urine. Therefore, first-line ther-apy for patients with uric acid stones is alkaliniza-tion of the urine with potassium citrate.
20. Clinicians should offer cystine-binding thiol drugs, such as alpha-mercaptopropi onylglycine (tiopronin), to patients with cys-tine stones who are unresponsive to dietary modifications and urinary alkalinization, or have large recurrent stone burdens. (Expert Opinion) First-line therapy for patients with cystine stones is increased fluid intake, restriction of sodium and protein intake, and urinary alkalinization. If these modifications are not sufficient, cystine-binding thiol drugs constitute the next line of therapy. Tio-pronin is possibly more effective and associated with fewer adverse events than d-penicillamine and should be considered first.40 21. Clinicians may offer acetohydroxamic acid to patients with residual or recurrent struvite stones only after surgical options have been exhausted.
(Option; Evidence Strength: Grade B) Struvite stones occur as a consequence of urinary infection with a urease-producing organism. Pa-tients treated for struvite stones may still be at risk for recurrent UTI after stone removal, and in some patients surgical stone removal is not feasible.
320 MEDICAL MANAGEMENT OF KIDNEY STONES These patients are at increased risk for stone recurrence or progression, and an aggressive medi-cal approach is required to mitigate this risk.41 The use of a urease inhibitor, AHA, may be beneficial in these patients, although the extensive side effect profile may limit its use.42 Follow-up 22. Clinicians should obtain a single 24-hour urine specimen for stone risk factors within six months of the initiation of treatment to assess response to dietary and/or medical therapy. (Expert Opinion) The aim of dietary/medical therapy of neph-rolithiasis is to promote changes in the urinary environment to reduce stone recurrence or growth.
There are a number of observational and case-control studies demonstrating that such changes are associated with a reduction in stone activity.43,44 23. After the initial follow-up, clinicians should obtain a single 24-hour urine specimen annually or with greater frequency, depend-ing on stone activity, to assess patient adher-ence and metabolic response. (Expert Opinion) Longitudinal monitoring of urinary parameters allows for the assessment of patient adherence, the identification of patients who become refractory to therapy and more timely adjustments in therapy for those individuals with active stone formation.45,46 If patients remain stone free for an extended period of time on their treatment regimen, discontinuation of follow-up testing may be considered.
24. Clinicians should obtain periodic blood testing to assess for adverse effects in patients on pharmacological therapy. (Standard; Evi-dence Strength: Grade: A) The majority of medications prescribed for stone prevention are associated with potential adverse effects, some of which can be detected with blood testing. For example, thiazide therapy may promote hypokalemia and glucose intolerance; allopurinol and tiopronin may cause an elevation in liver en-zymes; AHA and tiopronin may induce anemia and other hematologic abnormalities; potassium citrate may result in hyperkalemia.
25. Clinicians should obtain a repeat stone analysis, when available, especially in patients not responding to treatment. (Expert Opinion) A change in stone composition may account for the lack of response to dietary/medical therapy.
Therefore, repeat stone analysis is justified in this setting. Changes in stone composition have been reported in calcium oxalate stone formers who have converted to forming calcium phosphate stones.
26.
Clinicians should monitor patients with struvite stones for reinfection with urease-producing organisms and utilize strat-egies to prevent such occurrences. (Expert Opinion) Due to the infected nature of struvite stones, patients may continue to be at risk for persistent or recurrent UTI even after stone removal. There-fore, close monitoring of these patients is recom-mended to identify and treat recurrent infection.
Patients with altered lower urinary tract anatomy may be at particular risk for re-infection and recurrence.
Monitoring should include periodic urine culture testing. In some cases long-term, prophylactic antibiotic therapy may prevent recurrence.41 27. Clinicians should periodically obtain follow-up imaging studies to assess for stone growth or new stone formation based on stone activity (plain abdominal imaging, renal ultrasonography or low dose computerized tomography). (Expert Opinion) Other than stone passage, imaging is the most sensitive way to detect stone activity, defined as either existing stone growth or new stone formation.
Plain abdominal imaging has the advantages of being readily available and associated with limited radiation exposure and lower cost compared to other modalities. Plain radiography provides an accept-able assessment of stone activity in most patients with radiopaque stones, while renal ultrasonogra-phy is preferred for most patients with radiolucent stones, as there is no exposure to ionizing radiation, and it typically is less costly than CT. A one-year imaging interval is recommended for stable pa-tients, but this may be tailored based on stone activity or clinical signs.47,48 FUTURE RESEARCH For a disease with relatively high incidence and prevalence, research in the prevention of kidney stone disease is surprisingly sparse, perhaps because of the sporadic occurrence and transient symptoms associated with kidney stones as well as a perception that the pharmaceutical industry is not likely to find substantial profit in stone prevention.
The recent AHRQ-sponsored review of medical management identified only 28 RCTs performed through 2012.49 The interest in kidney stones has grown in recent years for two important reasons. First, kidney stones appear to be increasing in prevalence,1 perhaps related to changes in diet and the growing epidemics of metabolic syndrome, diabetes and obesity.
Second, stones have consistently been shown to be associated with more morbidity than previously expected. Associations with coronary artery disease,50 hypertension5 and diabetes6 have MEDICAL MANAGEMENT OF KIDNEY STONES 321 led to questions about the potential connection be-tween stone disease and these conditions.
The effort to prevent stones needs to be broad-ened to other populations of practitioners. Primary care practitioners and physician extenders are ex-perts at counseling weight loss, exercise and smok-ing cessation. Implementation of stone prevention regimens could also be extended to emergency rooms and primary care offices, such that stone metaphylaxis would fall under the purview of a larger pool of practitioners without a sophisticated view of urine chemistry.
We note that although both dietary manipula-tion21 and medications such as thiazides, allopu-rinol and citrate49 have all been shown to have efficacy in kidney stone prevention, the relative merits of diet and medications have never been compared head-to-head.
In summary, there is no dearth of important kidney stone research questions to be raised. Strong evidence from an admittedly low number of clinical trials demonstrates that stones are indeed pre-ventable.36 There is now not only a need for new research into the causative and exacerbating factors associated with stones, but also a need to ensure that the acquired knowledge to prevent stones is shared with every stone former in a clinical setting.
Conflict of Interest Disclosures All panel members completed COI disclosures.
Relationships that have expired (more than one year old) since the panel’s initial meeting, are listed.
Those marked with (C) indicate that compensation was received; relationships designated by (U) indi-cate no compensation was received. Consultant/ Advisor: Dean G. Assimos: Oxalosis and Hyper-oxaluria Foundation (OHF) (U); Gary Curhan, MD: Allena Pharmaceuticals (C); David S. Goldfarb, MD: Takeda (C), Astra Zeneca (C), Brian R. Mat-laga, MD: Boston Scientific (C); Manoj Monga, MD: Bard (C), Coloplast (C), Histosonics (C), Olympus (C), US Endoscopy (C) (Expired); Glenn M.
Preminger, MD: Boston Scientific (C), Mission Pharmacal (C); Health Publishing: Dean George Assimos, MD: Med Review in Urology (C), Urology Times (C); Gary Curhan, MD: Manoj Monga, MD: Brazilian Journal of Urology (U), Indian Journal of Urology (U), Journal of Endourology (U), Practical Reviews in Urology (C); Glenn M. Preminger, MD: UpToDate (C). Leadership Position: Gary Cur-han, MD: American Society of Nephrology (C); Manoj Monga, MD, CMS SCIP (U), Endourology Society (U); Glenn M. Preminger, MD: Endouro-logical Society (C).
Meeting Participant or Lecturer: David S. Goldfarb, MD, Quintiles (C) (Expired); Manoj Monga, MD: Cook Urological (C), Mission Pharmacal (C) (expired); Glenn M.
Preminger, MD: Olympus (C).Scientific Study or Trial: Dean George Assimos, MD: National Institute of Health (NIH) (C); Manoj Monga, MD: Taris Biomedical (C) (Expired), Xenolith (C) (Expired); Owner, Product Development: David S. Goldfarb, MD: The Ravine Group (C); Other: Dean George Assimos, MD: Piedmont Stone (C) (Expired); Gary Curhan, MD: UpToDate (C); Manoj Monga, MD: Fortec (C).
Disclaimer This document was written by the Medical Man-agement of Kidney Stones Guidelines Panel of the American Urological Association Education and Research, Inc., which was created in 2013. The Practice Guidelines Committee (PGC) of the AUA selected the committee chair. Panel members were selected by the chair. Membership of the committee included urologists and other clinicians with specific expertise on this disorder. The mission of the com-mittee was to develop recommendations that are analysis-based or consensus-based, depending on Panel processes and available data, for optimal clinical practices in the treatment kidney stones.
Funding of the committee was provided by the AUA. Committee members received no remunera-tion for their work. Each member of the committee provides an ongoing conflict of interest disclosure to the AUA.
While these guidelines do not necessarily estab-lish the standard of care, AUA seeks to recommend and to encourage compliance by practitioners with current best practices related to the condition being treated. As medical knowledge expands and tech-nology advances, the guidelines will change. Today these evidence-based guidelines statements repre-sent not absolute mandates but provisional pro-posals for treatment under the specific conditions described in each document. For all these reasons, the guidelines do not pre-empt physician judgment in individual cases.
Treating physicians must take into account var-iations in resources, and patient tolerances, needs, and preferences. Conformance with any clinical guideline does not guarantee a successful outcome.
The guideline text may include information or rec-ommendations about certain drug uses (‘off label’) that are not approved by the Food and Drug Administration (FDA), or about medications or substances not subject to the FDA approval process.
AUA urges strict compliance with all government regulations and protocols for prescription and use of these substances. The physician is encour-aged to carefully follow all available prescribing information about indications, contraindications, precautions and warnings.
These guidelines and best practice statements are not in-tended to 322 MEDICAL MANAGEMENT OF KIDNEY STONES provide legal advice about use and misuse of these substances.
Although guidelines are intended to encourage best practices and potentially encompass available technologies with sufficient data as of close of the literature review, they are necessarily time-limited.
Guidelines cannot include evaluation of all data on emerging technologies or management, including those that are FDA-approved, which may immedi-ately come to represent accepted clinical practices.
For this reason, the AUA does not regard tech-nologies or management which are too new to be addressed by this guideline as necessarily experi-mental or investigational.
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37. Laerum E and Larsen S: Thiazide prophylaxis of urolithiasis. A double-blind study in general practice. Acta Med Scand 1984; 215: 383.
38. Rodman JS: Intermittent versus continuous alkaline therapy for uric acid stones and ureteral stones of uncertain compostion. Urology 2002; 60: 378.
39. Maalouf NM, Cameron MA, Moe OW et al: Novel insights into the pathogenesis of uric acid MEDICAL MANAGEMENT OF KIDNEY STONES 323 nephrolithiasis. Curr Opin Nephrol Hypertens 2004; 13: 181.
40. Pak CY, Fuller C, Sakhaee K et al: Management of cystine nephrolithiasis with alpha-mercaptopropionylglycine. J Urol 1986; 136: 1003.
41. Preminger GM, Assimos DG, Lingerman JE et al: Report on the management of staghorn calculi (2005). American Urological Association 2005.
42. Griffith DP, Gleeson MJ, Lee H et al: Random-ized, double-blind trial of Lithostat (acetohy-droxamic acid) in the palliative treatment of infection-induced urinary calculi. Eur Urol 1991; 20: 243.
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44. Robinson MR, Leitao VA, Haleblian GE et al: Impact of long-term potassium citrate therapy on urinary profiles and recurrent stone formation.
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324 MEDICAL MANAGEMENT OF KIDNEY STONES |
190316 | https://cs.stackexchange.com/questions/117397/period-of-sum-of-two-periodic-sequences | Skip to main content
Period of sum of two periodic sequences
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I was wondering, what is to be the shortest possible key using Vigenere encryption, if a text is ciphered one time with a key of length i using Vigenere and second time with a key of length j using Vigenere?
From what I understand, each time it is encrypted, each character shifts using the first key and then using the second one, so I believe that for each pair there's an equivalent shifting. but if the two keys have a different length, how can I find the shortest possible key?
I think it has to be something with lowest common denominator, but I am not sure. how can this be shown mathematically?
my previous edit was a mistake and i deeply sorry for it. i read the book wrong and didn't interpret what's written well. very sorry for it. yuval explained to me well and i continue to research it using his given information
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edited Nov 21, 2019 at 13:43
alberto123
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If you sum a sequence with period a and a sequence with period b, then you get a sequence which is LCM(a,b)-periodic. But the new sequence might have a smaller period, as the following example demonstrates (addition is modulo 10):
```
022441133502244113350224411335
000772999611888000772999611888
022113022113022113022113022113
```
The first sequence has period 10 (0224411335), the second sequence as period 15 (000772999611888), and the sum sequence has period 6 (022113), although the LCM is 30.
More generally, the sum of a sequence of period i and a sequence of period j could have period exactly k iff LCM(i,j)=LCM(i,k)=LCM(j,k). See Restrepo and Chacon, On the period of sums of discrete periodic signals.
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edited Nov 21, 2019 at 12:30
answered Nov 21, 2019 at 12:07
Yuval FilmusYuval Filmus
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so is there exists some kind of shortest key for what's described? for instence, if i cipher the message once with a key of size i(using vigenere) and second time with a key of length j(using vigenere) - is there a shortest key to decipher the message?
– alberto123
Commented
Nov 21, 2019 at 12:26
Yes, but the length of the shortest key could depend on the actual keys.
– Yuval Filmus
Commented
Nov 21, 2019 at 12:27
thank you so much for explaining that to me and for the reference. i am going over the link and similar presentation i found at online colleges in addition to my book and it helps a lot. thanks again
– alberto123
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Nov 21, 2019 at 13:44
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190317 | https://www.quickanddirtytips.com/articles/personal-versus-personnel/ | Skip to main content
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‘Personal’ Versus ‘Personnel’
A memory trick to remember the spellings, plus rules for when "personnel" is singular and when it is plural.
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Trent Armstrong (our former Modern Manners Guy) sent me a picture of a sign that mixed up the words “personal” and “personnel.” That’s one I hadn’t seen before. It read, “Keep Out. Thunderbird Personal Only.”
‘Personal’ versus ‘personnel’: the root
“Personal” and “personnel” have the same Latin root — “personalis” 1,2 — which means that knowing the root is no help if you’re trying to remember the different spellings.
‘Personal’ versus ‘personnel’: the definitions
“Personal” relates to your person or your body, or implies a sense of closeness. For example, if you are someone’s personal friend, you’re suggesting that the two of you are closer than just casual friends; and if you have a personal favorite, “personal” just adds emphasis. (Some people may even argue that “personal” is redundant in the phrase “personal favorite.”) A personal affront is an insult directed specifically at one person. You get the idea.
When you refer to “personnel,” you’re talking about a group of people, usually people who work at a company or for the military. “Personnel” can also be the name of a department that manages a company’s employees and can be used as an adjective to describe situations related to employees (“After the confetti incident, we had to make some personnel changes”).
‘Personnel’: singular or plural?
“Personnel” can be both singular and plural. Merriam-Webster’s Dictionary of English Usage and Dictionary.com say some people object to “personnel” being plural, but that the plural use is widespread and acceptable. (In fact, some writers also objected to the word in general for the first 50 to 100 years after it was introduced to English from French in the early 1800s.) 3,4
Modern style guides suggest that when “personnel” is plural, it means “people,” as in “people at the company.” The singular use is less common, and pops up when you’re treating it as a collective noun similar to “staff” and “board.” Here are some examples:
All personnel are required to wear galoshes on Mondays. (Plural, meaning roughly all people at the company are required to wear galoshes on Mondays.)
Shareholders say the key point is how much personnel is retained after the cuts. (Singular, meaning roughly how much staff is retained.)
‘Personal’ versus ‘personnel’: a quick and dirty tip
You can remember that “personnel” means “many people” by noting that it’s spelled with more N’s than “personal”:
Personal: One person, one n.
Personnel: Many people, many n’s.
‘Personal’ versus ‘people’: plain language
And a final note, for those of you interested in writing with plain English, when you’re tempted to use the word “personnel,” ask if it would be simpler and more clear to use a word such as “people” or “workers” instead.
If you found that segment interesting, check out episode 865 too where we talked about “material” versus “materiel.”
References
personal. Merriam-Webster. (accessed August 14, 2023).
personnel. Click on Merriam-Webster (accessed August 14, 2023).
personnel. Merriam-Webster’s Dictionary of English Usage. Springfield: Merriam-Webster. 1994, p. 733.
personnel. Dictionary.com. Dictionary.com Unabridged. (accessed August 14, 2023).
An earlier version of this article appeared February 4, 2012.
Mignon Fogarty
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Mignon Fogarty is the founder of Quick and Dirty Tips and the author of seven books on language, including the New York Times bestseller "Grammar Girl's Quick and Dirty Tips for Better Writing." She is an inductee in the Podcasting Hall of Fame, and the show is a five-time winner of Best Education Podcast in the Podcast Awards. She has appeared as a guest expert on the Oprah Winfrey Show and the Today Show. Her popular LinkedIn Learning courses help people write better to communicate better. Find her on Mastodon.
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190318 | https://pubs.rsc.org/en/content/articlelanding/2005/pp/b506755c | The mechanism of photosynthetic water splitting - Photochemical & Photobiological Sciences (RSC Publishing)
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From the journal: ### Photochemical & Photobiological Sciences
The mechanism of photosynthetic water splitting†
James P. McEvoy,aJose A. Gascon,aVictor S. Batistaa andGary W. Brudviga
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a Department of Chemistry, Yale University, PO Box 208107, New Haven, CT, USA
E-mail:gary.brudvig@yale.edu
Abstract
Oxygenic photosynthesis, which provides the biosphere with most of its chemical energy, uses water as its source of electrons. Water is photochemically oxidized by the protein complex photosystem II (PSII), which is found, along with other proteins of the photosynthetic light reactions, in the thylakoid membranes of cyanobacteria and of green plant chloroplasts. Water splitting is catalyzed by the oxygen-evolving complex (OEC) of PSII, producing dioxygen gas, protons and electrons. O 2 is released into the atmosphere, sustaining all aerobic life on earth; product protons are released into the thylakoid lumen, augmenting a proton concentration gradient across the membrane; and photo-energized electrons pass to the rest of the electron-transfer pathway. The OEC contains four manganese ions, one calcium ion and (almost certainly) a chloride ion, but its precise structure and catalytic mechanism remain unclear. In this paper, we develop a chemically complete structure of the OEC and its environment by using molecular mechanics calculations to extend and slightly adjust the recently-obtained X-ray crystallographic model [K. N. Ferreira, T. M. Iverson, K. Maghlaoui, J. Barber and S. Iwata, Science, 2004, 303, 1831–1838]. We discuss our mechanistic hypothesis [J. P. McEvoy and G. W. Brudvig, Phys. Chem. Chem. Phys., 2004, 6, 4754–4763] with reference to this structure and to some important recent experimental results.
This article is part of the themed collection: In honour of James Barber
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Submitted 12 May 2005
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Photochem. Photobiol. Sci., 2005,4, 940-949
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The mechanism of photosynthetic water splitting
J. P. McEvoy, J. A. Gascon, V. S. Batista and G. W. Brudvig, Photochem. Photobiol. Sci., 2005,4, 940 DOI: 10.1039/B506755C
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190321 | https://users.math.msu.edu/users/jhall/classes/codenotes/grs.pdf | Chapter 5 Generalized Reed-Solomon Codes In 1960, I.S. Reed and G. Solomon introduced a family of error-correcting codes that are doubly blessed. The codes and their generalizations are useful in prac-tice, and the mathematics that lies behind them is interesting. In the first sec-tion we give the basic properties and structure of the generalized Reed-Solomon codes, and in the second section we describe in detail one method of algebraic decoding that is quite efficient.
5.1 Basics Let F be a field and choose nonzero elements v1, . . . , vn ∈F and distinct elements α1, . . . , αn ∈F.
Set v = (v1, . . . , vn) and α = (α1, . . . , αn).
For 0 ≤k ≤n we define the generalized Reed-Solomon codes generalized Reed-Solomon codes GRSn,k(α, v) = { (v1f(α1), v2f(α2), . . . , vnf(αn)) | f(x) ∈F[x]k } .
Here we write F[x]k for the set of polynomial in F[x] of degree less than k, a vector space of dimension k over F. For fixed n, α, and v, the various GRS codes enjoy the nice embedding property GRSn,k−1(α, v) ≤GRSn,k(α, v).
If f(x) is a polynomial, then we shall usually write f for its associated code-word. This codeword also depends upon α and v; so at times we prefer to write unambiguously evα,v(f(x)) = (v1f(α1), v2f(α2), . . . , vnf(αn)) , indicating that the codeword f = evα,v(f(x)) arises from evaluating the poly-nomial f(x) at α and scaling by v.
( 5.1.1) Theorem.
GRSn,k(α, v) is an [n, k] linear code over F with length n ≤|F|. We have dmin = n −k + 1 provided k ̸= 0. In particular, GRS codes are MDS codes.
63 64 CHAPTER 5. GENERALIZED REED-SOLOMON CODES Proof. As by definition the entries in α are distinct, we must have n ≤|F|.
If a ∈F and f(x), g(x) ∈F[x]k, then af(x) + g(x) is also in F[x]k ; and evα,v(af(x) + g(x)) = a evα,v(f(x)) + evα,v(g(x)) = af + g .
Therefore GRSn,k(α, v) is linear of length n over F.
Let f(x), g(x) ∈F[x]k be distinct polynomials. Set h(x) = f(x) −g(x) ̸= 0, also of degree less than k. Then h = f −g and wH(h) = dH(f, g). But the weight of h is n minus the number of 0’s in h. As all the vi are nonzero, this equals n minus the number of roots that h(x) has among {α1, . . . , αn}.
As h(x) has at most k −1 roots by Proposition A.2.10, the weight of h is at least n −(k −1) = n −k + 1. Therefore dmin ≥n −k + 1, and we get equality from the Singleton bound 3.1.14. (Alternatively, h(x) = Qk−1 i=1 (x −αi) produces a codeword h of weight n −k + 1.) The argument of the previous paragraph also shows that distinct polynomials f(x), g(x) of F[x]k give distinct codewords. Therefore the code contains |F|k codewords and has dimension k.
2 The vector v plays little role here, and its uses will be more apparent later.
At present, it serves to make sure that any code that is monomially equivalent to a GRS code is itself a GRS code.
Let us now find a generator matrix for GRSn,k(α, v).
The argument of Theorem 5.1.1 makes it clear that any basis f1(x), . . . , fk(x) of F[x]k gives rise to a basis f1, . . . , fk of the code. A particularly nice polynomial basis is the set of monomials 1, x, . . . , xi, . . . , xk−1. The corresponding generator matrix, whose ith row (numbering rows from 0 to k −1) is evα,v(xi) = (v1αi 1, . . . , vjαi j, . . . , vnαi n) , is the canonical generator matrix for GRSn,k(α, v): canonical generator matrix v1 v2 . . .
vj . . .
vn v1α1 v2α2 . . .
vjαj . . .
vnαn .
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v1αi 1 v2αi 2 . . .
vjαi j . . .
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v1αk−1 1 v2αk−1 2 . . .
vjαk−1 j . . .
vnαk−1 n ( 5.1.2) Problem.
Consider the code C = GRSn,k(α, v), and assume that all the entries of the vector α are nonzero. If α = (α1, α2, . . . , αn) , define β = (α−1 1 , α−1 2 , . . . , α−1 n ) .
Find a vector w such that C = GRSn,k(β, w).
5.1. BASICS 65 ( 5.1.3) Problem.
(a) Consider the code GRSn,k(α, v) over F. Let a, c be nonzero elements of F, and let b be the vector of F n all of whose entries are equal to b ∈F.
Prove that GRSn,k(α, v) = GRSn,k(aα + b, cv) .
(b) If n < |F|, prove that there is an α′ with no entries equal to 0 and GRSn,k(α, v) = GRSn,k(α′, v) .
( 5.1.4) Problem.
Consider the code E, which will be linear of length n + 1 and dimension k, whose generator matrix results from adding a new column to the canonical generator matrix for GRSn,k(α, v): 2 6 6 6 6 6 6 6 6 6 6 6 4 v1 v2 . . .
vj . . .
vn 0 v1α1 v2α2 . . .
vjαj . . .
vnαn 0 .
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v1αi 1 v2αi 2 . . .
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vnαi n 0 .
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v1αk−2 1 v2αk−2 2 . . .
vjαk−2 j . . .
vnαk−2 n 0 v1αk−1 1 v2αk−1 2 . . .
vjαk−1 j . . .
vnαk−1 n 1 3 7 7 7 7 7 7 7 7 7 7 7 5 Prove that dmin(E) = n −k + 2.
Remark. As n −k + 2 = (n + 1) −k + 1, this shows that the code E satisfies the Singleton Bound with equality and so is maximum distance separable (MDS), just as all GRS codes are.
It is extremely profitable to think of Theorem 5.1.1 again in terms of poly-nomial interpolation: Any polynomial of degree less than k is uniquely determined by its values at k (or more) distinct points.
Here, any codeword with as many as k entries equal to 0 corresponds to a polynomial of degree less than k whose values match the 0 polynomial in k points and so must itself be the 0 polynomial.
Given any n-tuple f, we can easily reconstruct the unique polynomial f(x) of degree less than n with f = evα,v(f(x)). We first introduce some notation.
Set L(x) = n Y i=1 (x −αi) and Li(x) = L(x)/(x −αi) = Y j̸=i (x −αj) .
The polynomials L(x) and Li(x) are monic of degrees n and n −1, respectively.
The vector f has ith coordinate vif(αi), so we have enough information to calculate, using the Lagrange interpolation formula A.2.11, f(x) = n X i=1 Li(x) Li(αi)f(αi) .
66 CHAPTER 5. GENERALIZED REED-SOLOMON CODES The coefficients Li(αi) are always nonzero and are often easy to compute.
( 5.1.5) Problem.
(a) Prove that Li(αi) = L′(αi), where L′(x) is the formal derivative of L(x) as defined in Problem A.2.26.
(b) If n = |F| and {α1, . . . , αn} = F, then Li(αi) = −1, for all i.
(c) If {α1, . . . , αn} is composed of n roots of xn−1 in F, then Li(αi) = nα−1 i (̸= 0).
In particular, if n = |F| −1 and {α1, . . . , αn} = F −{0}, then Li(αi) = −α−1 i (hence α−1 i Li(αi)−1 = −1).
The polynomial f(x) has degree less than k, while the interpolation polyno-mial of the righthand side above has apparent degree n −1. The resolution of this confusion allows us to find the dual of a GRS code easily.
( 5.1.6) Theorem.
We have GRSn,k(α, v)⊥= GRSn,n−k(α, u), where u = (u1, . . . , un) with u−1 i = vi Q j̸=i(αi −αj).
Proof. By definition ui = v−1 i Li(αi)−1.
We prove that every f in GRSn,k(α, v) has dot product 0 with every g in GRSn,n−k(α, u), from which the result is immediate. Let f = evα,v(f(x)) and g = evα,u(g(x)). The polynomial f(x) has degree less than k while g(x) has degree less than n −k. Therefore their product f(x)g(x) has degree at most n −2. By Lagrange interpolation A.2.11 we have f(x)g(x) = n X i=1 Li(x) Li(αi)f(αi)g(αi) .
Equating the coefficient of xn−1 from the two sides gives: 0 = n X i=1 1 Li(αi)f(αi)g(αi) = n X i=1 (vif(αi)) v−1 i Li(αi)g(αi) = n X i=1 (vif(αi))(uig(αi)) = f · g , as required.
2 The ability in the class of GRS codes to choose different vectors v to ac-company a fixed α has been helpful here.
Of course, to specify f as a codeword in C = GRSn,k(α, v) we do not need to check it against every g of C⊥= GRSn,n−k(α, u). It is enough to consider a basis of C⊥, a nice one being the rows of the canonical generator matrix for 5.2. DECODING GRS CODES 67 C⊥, a check matrix for C. Our introduction of GRS codes such as C essentially defines them via their canonical generator matrices. As we have seen before, describing a linear code instead in terms of a check matrix can be fruitful. In particular this opens the possibility of syndrome decoding.
Set r = n −k, and let c = (c1, . . . , cn) ∈F n. Then c ∈C ⇐ ⇒ 0 = c · evα,u(xj), for 0 ≤j ≤r −1 ⇐ ⇒ 0 = n X i=1 ciuiαj i, for 0 ≤j ≤r −1 .
We rewrite these r equations as a single equation in the polynomial ring F[z] in a new indeterminate z. The vector c is in C if and only if in F[z] we have 0 = r−1 X j=0 n X i=1 ciuiαj i zj = n X i=1 ciui r−1 X j=0 (αiz)j The polynomials 1 −αz and zr are relatively prime, so it is possible to invert 1 −αz in the ring F[z] (mod zr). Indeed 1 1 −αz = r−1 X j=0 (αz)j (mod zr) , a truncation of the usual geometric series expansion (which could equally well be taken as a definition for the inverse of 1 −αz module zr). We are left with: ( 5.1.7) Theorem.
(Goppa formulation for GRS codes.) The general-ized Reed-Solomon code GRSn,k(α, v) over F is equal to the set of all n-tuples c = (c1, c2, . . . , cn) ∈F n, such that n X i=1 ciui 1 −αiz = 0 (mod zr) , where r = n −k and u−1 i = vi Q j̸=i(αi −αj).
2 This interpretation of GRS codes has two main values. First, it is open to a great deal of generalization, as we shall later see. Second, it suggests a practical method for the decoding of GRS codes, the topic of the next section.
5.2 Decoding GRS codes As GRS codes are MDS, they can be decoded using threshold decoding as in Problem 3.2.4.
We now present an efficient and more specific algorithm for decoding the dual of GRSn,r(α, u), starting from the Goppa formulation.
68 CHAPTER 5. GENERALIZED REED-SOLOMON CODES Suppose c = (c1, c2, . . . , cn) is transmitted, and p = (p1, p2, . . . , pn) is received, so that the error vector e = (e1, e2, . . . , en) has been introduced; p = c + e. We calculate the syndrome polynomial of p: syndrome polynomia Sp(z) = n X i=1 piui 1 −αiz (mod zr) .
Then it is easily seen that Sp(z) = Sc(z) + Se(z) (mod zr) , whence, by the Goppa formulation of Theorem 5.1.7, Sp(z) = Se(z) (mod zr) .
Let B be the set of error locations: B = {i | ei ̸= 0} .
Then we have the syndrome polynomial Sp(z) = Se(z) = X b∈B ebub 1 −αbz (mod zr).
We now drop the subscripts and write S(z) for the syndrome polynomial.
Clear denominators to find the Key Equation: Key Equation σ(z)S(z) = ω(z) (mod zr), where σ(z) = σe(z) = Y b∈B (1 −αbz) and ω(z) = ωe(z) = X b∈B ebub Y a∈B,a̸=b (1 −αaz) .
(Empty products are taken as 1.) The polynomial σ(z) is called the error locator error locator polynomial, and the polynomial ω(z) is the error evaluator polynomial.
error evaluator The names are justifiable. Given the polynomials σ(z) = σe(z) and ω(z) = ωe(z), we can reconstruct the error vector e. Assume for the moment that none of the αi are equal to 0 (although similar results are true when some αi is 0) .
Then: B = { b | σ(α−1 b ) = 0 } and, for each b ∈B, eb = −αbω(α−1 b ) ubσ′(α−1 b ) , where σ′(z) is the formal derivative of σ(z) (see Problem A.2.26). In fact the polynomials σ(z) and ω(z) determine the error vector even when some αi is 0.
5.2. DECODING GRS CODES 69 ( 5.2.1) Problem.
Let σ(z) and ω(z) be the error locator and evaluator polynomials for the error vector e ̸= 0. Set B = { b | eb ̸= 0 } and B0 = { b ∈B | αb ̸= 0}. Recall that there is at most one index b with αb = 0.
Prove the following: (a) wH(e) = |B| is equal to |B0| + 1 or |B0| depending upon whether or not there is an index b with αb = 0 and eb ̸= 0.
(b) deg σ(z) = |B0| and deg ω(z) ≤|B| −1 .
(c) B0 = { b | αb ̸= 0, σ(α−1 b ) = 0 }.
(d) The index b with αb = 0 belongs to B if and only if deg σ(z) = deg ω(z).
(e) For b ∈B0, eb is given by the formula above. If b ∈B \ B0 then eb = wu−1 b “ Y a∈B0 (−αa) ”−1 , where w is the coefficient of zf in ω(z) for f = deg ω(z).
If the error vector e ̸= 0 has weight at most r/2 (= (dmin −1)/2), then relative to the syndrome polynomial Se(z) = S(z) the pair of polynomials σe(z) = σ(z) and ωe(z) = ω(z) has the three properties by which it is charac-terized in the next theorem. Indeed (1) is just the Key Equation. Property (2) is a consequence of the assumption on error weight and the definitions of the polynomials σ(z) and ω(z) (see Problem 5.2.1 above). For (3) we have σ(0) = 1 trivially. As σ(z) has deg(σ(z)) distinct roots, either gcd(σ(z), ω(z)) = 1 or the two polynomials have a common root. But for each root α−1 b of σ(z) we have 0 ̸= ω(α−1 b ), a factor of eb ̸= 0.
Our decoding method solves the Key Equation and so finds the error vector e as above. The following theorem provides us with a characterization of the solutions we seek.
( 5.2.2) Theorem.
Given r and S(z) ∈F[z] there is at most one pair of polynomials σ(z), ω(z) in F[z] satisfying: (1) σ(z)S(z) = ω(z) (mod zr); (2) deg(σ(z)) ≤r/2 and deg(ω(z)) < r/2; (3) gcd(σ(z), ω(z)) = 1 and σ(0) = 1.
In fact we prove something slightly stronger.
( 5.2.3) Proposition.
Assume that σ(z), ω(z) satisfy (1)-(3) of Theorem 5.2.2 and that σ1(z), ω1(z) satisfy (1) and (2). Then there is a polynomial µ(z) with σ1(z) = µ(z)σ(z) and ω1(z) = µ(z)ω(z).
Proof. From (1) σ(z)ω1(z) = σ(z)σ1(z)S(z) = σ1(z)ω(z) (mod zr) ; so σ(z)ω1(z) −σ1(z)ω(z) = 0 (mod zr) .
70 CHAPTER 5. GENERALIZED REED-SOLOMON CODES But by (2) the lefthand side of this equation has degree less than r. Therefore σ(z)ω1(z) = σ1(z)ω(z) .
From (3) we have gcd(σ(z), ω(z)) = 1, so by Lemma A.2.20 σ(z) divides σ1(z).
Set σ1(z) = µ(z)σ(z). Then σ(z)ω1(z) = σ1(z)ω(z) = σ(z)µ(z)ω(z) .
The polynomial σ(z) is nonzero since σ(0) = 1; so by cancellation ω1(z) = µ(z)ω(z), as desired.
2 Proof of Theorem 5.2.2.
Any second such pair has σ1(z) = µ(z)σ(z) and ω1(z) = µ(z)ω(z) by the proposition. So µ(z) divides gcd(σ1(z), ω1(z)) which is 1 by (3). There-fore µ(z) = µ is a constant. Indeed 1 = σ1(0) = µ(0)σ(0) = µ · 1 = µ .
Thus σ1(z) = µ(z)σ(z) = σ(z) and ω1(z) = µ(z)ω(z) = ω(z).
2 Using the characterization of Theorem 5.2.2 we now verify a method of solving the Key Equation with the Euclidean algorithm, as presented in Section A.3.1 of the appendix.
( 5.2.4) Theorem. (Decoding GRS using the Euclidean Algorithm.) Consider the code GRSn,k(α, v) over F, and set r = n −k. Given a syndrome polynomial S(z) (of degree less than r), the following algorithm halts, producing polynomials ˜ σ(z) and ˜ ω(z): Set a(z) = zr and b(z) = S(z).
Step through the Euclidean Algorithm A.3.1 until at Step j, deg(rj(z)) < r/2.
Set ˜ σ(z) = tj(z) and ˜ ω(z) = rj(z).
If there is an error word e of weight at most r/2 = (dmin −1)/2 with Se(z) = S(z), then b σ(z) = ˜ σ(0)−1˜ σ(z) and b ω(z) = ˜ σ(0)−1˜ ω(z) are the error locator and evaluator polynomials for e.
Proof. It is the goal of the Euclidean algorithm to decrease the degree of rj(z) at each step, so the algorithm is guaranteed to halt.
Now assume that S(z) = Se(z) with wH(e) ≤r/2.
Therefore the error locator and evaluator pair σ(z) = σe(z) and ω(z) = ωe(z) satisfies (1), (2), and 5.2. DECODING GRS CODES 71 (3) of Theorem 5.2.2. We first check that, for the j defined, the pair tj(z) and rj(z) satisfies (1) and (2).
Requirement (1) is just the Key Equation and is satisfied at each step of the Euclidean algorithm since always Ej : rj(z) = sj(z)zr + tj(z)S(z).
For (2), our choice of j gives deg(rj(z)) < r/2 and also deg(rj−1(z)) ≥r/2.
Therefore, from Problem A.3.5, deg(tj(z)) + r/2 ≤ deg(tj(z)) + deg(rj−1(z)) = deg(a(z)) = deg(zr) = r .
Hence deg(tj(z)) ≤r/2, giving (2).
By Proposition 5.2.3 there is a polynomial µ(z) with tj(z) = µ(z)σ(z) and rj(z) = µ(z)ω(z) .
Here µ(z) is not the zero polynomial by Lemma A.3.3(1).
If we substitute for tj(z) and rj(z) in equation Ej we have sj(z)zr + (µ(z)σ(z))S(z) = µ(z)ω(z) , which becomes µ(z) ω(z) −σ(z)S(z) = sj(z)zr .
By the Key Equation, the parenthetical expression on the left is p(z)zr, for some p(z); so we are left with µ(z)p(z)zr = sj(z)zr or µ(z)p(z) = sj(z) .
Thus µ(z) divides gcd(tj(z), sj(z)), which is 1 by Corollary A.3.4.
We conclude that µ(z) = µ is a nonzero constant function. Furthermore tj(0) = µ(0)σ(0) = µ ; so σ(z) = tj(0)−1tj(z) and ω(z) = tj(0)−1rj(z) , as desired.
2 When this algorithm is used, decoding default occurs when b σ(z) does not split into linear factors whose roots are inverses of entries in α with multiplicity 1. (Here we assume that none of the αi are 0.) That is, the number of roots of b σ(z) among the α−1 i must be equal to the degree of b σ(z). If this is not the case, then we have detected errors that we are not able to correct. Another instance of decoder default occurs when tj(0) = 0, so the final division to determine b σ(z) can not be made.
72 CHAPTER 5. GENERALIZED REED-SOLOMON CODES If tj(0) ̸= 0 and b σ(z) does split as described, then we can go on to evaluate errors at each of the located positions and find a vector of weight at most r/2 with our original syndrome. In this case we have either decoded correctly, or we had more than r/2 errors and have made a decoding error. (We need not worry about division by 0 in evaluating errors, since this can only happen if b σ(z) has roots of multiplicity greater than one; see Problem A.2.27.) Assume now that r is even or that α has weight n. Then this algorithm only produces error vectors of weight r/2 or less. In particular if more than r/2 errors occur then we will have a decoding default or a decoder error. Suppose that we have found polynomials b σ(z) and b ω(z) that allow us to calculate a candidate er-ror vector e of weight at most r/2. It follows from Lagrange interpolation A.2.11 that b σ(z) = σe(z) and b ω(z) = ωe(z). Also since σ(z) is invertible modulo zr, we can solve the Key Equation to find that S(z) = Se(z). Therefore the received vector is within a sphere of radius r/2 around a codeword and is decoded to that codeword. That is, under these conditions Euclidean algorithm decoding as given in Theorem 5.2.4 is an explicit implementation of the decoding algorithm SSr/2.
Example.
Consider the code C = GRS6,2(α, v) over the field F7 of integers modulo 7, where α = (2, 4, 6, 1, 3, 5) and v = (1, 1, 1, 1, 1, 1) .
First calculate a vector u for which C⊥= GRS6,4(α, u). Starting with L(x) = (x −2)(x −4)(x −6)(x −1)(x −3)(x −5) we find: L1(2) = (−2) (−4) (1) (−1) (−3) = 24 = 3 L2(4) = (2) (−2) (3) (1) (−1) = 12 = 5 L3(6) = (4) (2) (5) (3) (1) = 120 = 1 L4(1) = (−1) (−3) (−5) (−2) (−4) = −120 = 6 L5(3) = (1) (−1) (−3) (2) (−2) = −12 = 2 L6(5) = (3) (1) (−1) (4) (2) = −24 = 4 (Notice that these values could have been found easily using Problem 5.1.5(c).) Now ui = (viLi(αi))−1 = Li(αi)−1 since vi = 1; so u = (5, 3, 1, 6, 4, 2) .
Next calculate the syndrome polynomial of an arbitrary received vector p = (p1, p2, p3, p4, p5, p6) .
In our example r = 6 −2 = 4.
Sp(z) = 5 · p1 1 −2z + 3 · p2 1 −4z + 1 · p3 1 −6z + 6 · p4 1 −1z + 4 · p5 1 −3z + 2 · p6 1 −5z (mod z4) 5.2. DECODING GRS CODES 73 = 5p1( 1 +2z +4z2 +z3) +3p2( 1 +4z +2z2 +z3) +p3( 1 +6z +z2 +6z3) +6p4( 1 +z +z2 +z3) (mod z4) +4p5( 1 +3z +2z2 +6z3) +2p6( 1 +5z +4z2 +6z3) = p1( 5 +3z +6z2 +5z3) +p2( 3 +5z +6z2 +3z3) +p3( 1 +6z +z2 +6z3) +p4( 6 +6z +6z2 +6z3) (mod z4) .
+p5( 4 +5z +z2 +3z3) +p6( 2 +3z +z2 +5z3) Notice that this calculation amounts to finding the canonical check matrix for the code.
We now use the algorithm of Theorem 5.2.4 to decode the received vector p = (1, 3, 6, 5, 4, 2) .
We have the syndrome polynomial S(z) = 5 · 1 1 −2z + 3 · 3 1 −4z + 1 · 6 1 −6z + 6 · 5 1 −1z + 4 · 4 1 −3z + 2 · 2 1 −5z (mod z4) = 1( 5 +3z +6z2 +5z3) +3( 3 +5z +6z2 +3z3) +6( 1 +6z +z2 +6z3) +5( 6 +6z +6z2 +6z3) (mod z4) .
+4( 4 +5z +z2 +3z3) +2( 2 +3z +z2 +5z3) = 5z + 3z2 + 4z3 (mod z4) .
The algorithm now requires that, starting with initial conditions a(z) = z4 and b(z) = 4z3 + 3z2 + 5z , we step through the Euclidean Algorithm until at Step j we first have deg(rj(z)) < r/2 = 2.
This is precisely the Euclidean Algorithm example done in the ap-pendix. At Step 2. we have the first occurrence of a remainder term with degree less than 2; we have r2(z) = 6z . We also have t2(z) = 3z2 +6z +4, so t2(0)−1 = 4−1 = 2.
Therefore we have error locator and evaluator polynomials: σ(z) = t2(0)−1t2(z) = 2(3z2 + 6z + 4) = 6z2 + 5z + 1 ω(z) = t2(0)−1r2(z) = 2(6z) = 5z .
The error locations are those in B = { b | σ(α−1 b ) = 0 }; so to find the error locations, we must extract the roots of σ(z). As F7 does not have characteristic 2, we can use the usual quadratic formula and find that the roots are 2 , 3 = −5 ± √25 −4 · 6 2 · 6 .
74 CHAPTER 5. GENERALIZED REED-SOLOMON CODES Now 2−1 = 4 = α2 and 3−1 = 5 = α6, so B = {2, 6}.
An error value eb is given by eb = −αbω(α−1 b ) ubσ′(α−1 b ) , where σ′(z) = 5z + 5. Thus e2 = −4·10 3·15 = 3 and e6 = −5·15 2·20 = 6 .
We have thus found e = (0, 3, 0, 0, 0, 6) , so we decode the received word p to c = p −e = (1, 3, 6, 5, 4, 2) −(0, 3, 0, 0, 0, 6) = (1, 0, 6, 5, 4, 3) .
In the example we have been able to use the quadratic formula to calculate the roots of σ(z) and so find the error locations. This will not always be possible.
There may be more than 2 errors. In any case, the quadratic formula involves division by 2 and so is not valid when the characteristic of the field F is 2, one of the most interesting cases. A method that is often used is the substitution, one-by-one, of all field elements into σ(z), a Chien search. Although this lacks Chien search subtlety, it is manageable when the field is not too big. There do not seem to be general alternatives that are good and simple.
( 5.2.5) Problem.
Consider the GRS8,4(α, v) code C over F13 with v = (1, 1, 1, 1, 1, 1, 1, 1) α = (1, 4, 3, 12, 9, 10, 5, 8) .
(a) Give n, k, β, u with C⊥= GRSn,k(β, u).
(b) When transmitting with C, assume that the vector p = (0, 0, 0, 0, 0, 0, 3, 5) .
is received. Use the Euclidean algorithm to find an error vector e and a decoded code-word c. (The answers should be obvious. Use the question to check your understanding of the process.) (c) When transmitting with C, assume that the vector p = (3, 6, 0, 4, 0, 5, 0, 12) is received.
Use the Euclidean algorithm to find an error vector e and a decoded codeword c.
( 5.2.6) Problem.
Consider the GRS10,4(α, v) code C over F13 with v = (1, 1, 1, 1, 1, 1, 1, 1, 1, 1) α = (1, 2, 3, 4, 6, 7, 9, 10, 11, 12) .
5.2. DECODING GRS CODES 75 (a) Give n, k, β, u with C⊥= GRSn,k(β, u).
( Hint: u = (∗, ∗, 9, 10, 12, 1, 3, 4, ∗, ∗).) (b) When transmitting with C, assume that the vector p = (4, 5, 6, 0, 0, 0, 0, 0, 0, 0) .
is received. Use the Euclidean algorithm to find an error vector e and a decoded code-word c. (The answers should be obvious. Use the question to check your understanding of the process.) (c) When transmitting with C, assume that the vector p = (3, 1, 0, 0, 0, 0, 0, 5, 7, 12) .
is received.
Use the Euclidean algorithm to find an error vector e and a decoded codeword c.
( 5.2.7) Problem.
Let the field F8 be given as polynomials of degree at most 2 in α, a root of the primitive polynomial x3+x+1 ∈F2[x]. Consider the code C = GRS7,3(α, v) over F8 with α = v = (1, α, α2, α3, α4, α5, α6) .
By Problem 5.1.5(c) we have C⊥= GRS7,4(α, u) for u = (1, 1, 1, 1, 1, 1, 1).
When transmitting with C, assume that the vector p = (0, α5, 0, 1, α6, 0, 1) is received.
Use the Euclidean Algorithm to find an error vector e and a decoded codeword c. |
190322 | http://www.phys.ufl.edu/~peterson/phy2005_04/solutions/Chap29_03.html | Chapter 29 Solutions - 3, 5, 7, 10
Chapter 29 Solutions - 3, 5, 7, 10, 12, 13, 15, 19, 23, 27, 32, 36, 39
In relativity, the kinetic energy is defined by the equation KE = (m – m o)c 2.As I mentioned in class, this expression looks strange, but it reduces to the classical expression for kinetic energy, KE = (1/2)mv 2, you learned about last semester.
Proof: Recall that m = m o/(1 – v 2/c 2)1/2.For small velocities, the square root term can be expanded in a rapidly converging series.The series is
1/(1 – v 2/c 2)1/2 = 1 + (1/2)v 2/c 2 + Terms of higher order that can be neglected.
As a consequence, the KE becomes
KE = (m o + (1/2)m o v 2/c 2 – m o)c 2 = (1/2)m o v 2, as we had hoped
To solve the problem simply, it is convenient to talk of the rest energy of the electron, expressed in electron volts instead of joules.Going from joules to electron volts means dividing the number of joules by 1.6 x 10-19 J/eV. Also, for convenience, it is customary to call 1 x 10 6 eV as 1 MeV, where the unit MeV is short for million electron volts.Thus,
m o c 2 = (9.110 x 10-31 kg)(2.998 x 10 8 m/s)2/(1.6022 x 10-19 J/eV) = 5.11 x 10 5 eV = 0.511 MeV.
The answer to the problem is now an easy matter, because much of the arithmetic is already done.Since the mass of the electron has been increased to 10 m o, the total energy is 10 m o c 2, and the kinetic energy added must have been 9 m o c 2.But 9 m o c 2 = (9)(0.511 MeV) = 4.60 MeV or 4.60 x 10 6 eV.The electron must have been accelerated through a potential difference of 4.6 x 10 6 V.
For ease of calculation, you need to get used to working with eV and MeV energy units.Assume that a proton is accelerated through a potential difference of 10 9 V.The proton therefore acquires a kinetic energy of 1000 MeV.Let us first find the rest energy of the proton in MeV.It is
m o c 2 = (1.67264 x 10-27 kg)(2.998 x 10 8 m/s)2/(1.6022 x 10-19 J/eV) = 938.3 MeV.
Thus, since the total energy is the sum of rest energy + KE, we have
mc 2 = m o c 2 + 1000 MeV = 1938 MeV.Then, mc 2/m o c 2 = m/m o = 1938/938 = 2.066.
The speed of the proton is computed using m = m o/(1 – v 2/c 2)1/2.We can substitute the ratio just obtained for m/m o, then square both sides to get
(v/c)2 = 1 – (1/2.066 2) = 0.7657 à v/c = .875
Lifting a 3.0 kg mass 5.0 m gives it an increase of gravitational potential energy.This increase is
PE = mgh = (3.0 kg)(9.8 m/s)(5.0 m). = 147 J, which is the increase in mc 2.
Therefore, the increase in mass is 147/(3.0 x 10 8 m/s)2 = 1.63 x 10-15 kg.This amount is obviously very small compared to the rest mass of 3.0 kg. The ratio is only (1.63 x 10-15 kg)/3.0 kg = 5.4 x 10-15.
Study Aid – How to compute the photon energy in eV, given wavelength in nm.
Recall that the photon energy is given by E = hf = hc/(wavelength).Calculating the energy in joules is easy enough, but often it is useful to be able quickly to compute the energy in eV directly by knowing the wavelength in nm.The conversion from joules to eV is done by dividing by 1.6 x 10-19 J/eV.The transformation from meters to nanometers is accomplished by expressing wavelength in nm and multiplying by the factor 10-9.
.
According to the expression E = hf = hc/(wavelength), the energy per photon for light having a wavelength of 632.8 nm is 1240/632.8 = 1.96 eV.The conversion to joules is accomplished by multiplying by 1.6022 x 10-19 J/eV, giving 3.14 x 10-19 J.
Since the beam power is 3.00 mW, or 3.0 x 10-3 J/s, the number of photons = (3.00 x 10-3 J/s)/(3.14 J/photon) = 9.4 x 10 15 photons/s.
12 The threshold occurs when the energy of the photon is just enough to “lift” an electron out of the energy well.That is the meaning of the term work function.Thus, the photon energy must be 4.5 eV or greater.From the above study guide, we see that the wavelength corresponding to 4.5 eV is
wavelength (in nm) = 1240/4.5 = 276 nm
This wavelength is in the UV part of the spectrum.
This question uses the same logic as in question 12.E (in eV) = 1240/546 = 2.27 eV.
When a photon is absorbed,all of its energy is deposited in the absorbing object.There is no partial absorption.Now, the energy of 480 nm light is
E = 1240/546 = 2.583 eV.
Also, since the photoelectric threshold is at 546 nm, we know that the work function is
W = 1240/546 = 2.271 eV.
The energy supplied by the 480 nm light is therefore greater than the work-function energy.The excess is available as kinetic energy for the ejected electron.This kinetic energy is 2.583 eV – 2.271 eV = 0.312 eV.
In order to compute the speed of the electron, we note that the kinetic energy 0.312 eV is very small compared to the relativistic rest-mass energy of 0.511 MeV.This means that the old classical equation for kinetic energy is good enough and easy to carry out after changing the energy units from eV to joules.If you choose to use the relativistic expression, you will get the same correct result, but it will take considerably more labor.Thus, using 1.6 x 10-19 J/eV, we find
(0.312 eV)( 1.6 x 10-19 J/eV) = (½)(9.11 x 10-31 kg)v 2
v 2 = 0.1096 x 10 12 à v = 3.3 x 10 5 m/s
Notice that the photon energy is given as well as the energy levels involved in the transition.For the H-atom, the energy levels are given by E(in eV) = -(13.6)/n 2.Therefore, E 2 = -3.40 eV, and E 6 = -0.378.The difference is 3.02 eV.This answer is entirely consistent with the photon energy
E(photon) = 1240/410 = 3.02 eV.
The Lyman series for helium involves transitions from higher states to the ground state at E = –54.4 eV.From the energy-level diagram for He, the transition from n = 2 to n = 1 represents an energy difference of –40.8 eV.This means that a photon having an energy of 40.8 eV is emitted.The wavelength is
Wavelength = 1240/40.8 = 30.4 nm, which is in the ultraviolet.
Since the difference in energy between levels is 0.293 eV, the wavelength of the emitted photon when the molecule falls to the next lowest vibrational level is 1240/0.293 = 4232 nm.This radiation is in the infrared.
The power output of the reactor is 1.0 x 10 8 J/s. As a consequence, in one hour (3600 s) the electrical energy produced is (1.0 x 10 8 J/s)(3600) = 3.6 x 10 11 Joules.But, because the efficiency is only 30%, the energy created from mass conversion is more, namely, E(total) = (3.6 x 10 11 J)/(0.30) = 1.2 x 10 12 J.From the relation E = mc 2, the mass that must have been converted into energy is
m = (1.2 x 10 12 J)/(3.0 x 10 8 m/s)2 = 1.3 x 10-5 kg.
That’s pretty impressive, isn’t it?
The de Broglie wavelength of a particle of mass m, moving at a speed v is h/mv, where h is Planck’s constant. But the wavelength is also given in general by wavelength = v/frequency. Thus,
.
Substituting the numbers gives
v 2 = (6.63 x 10-34 Js)(5.0 x 10 12 Hz)/(9.11 x 10-31 kg) = 36.4 x 10 8 m 2/s 2,
or v = 6.0 x 10 4 m/s.
This is one of those problems in which the key idea is to keep track of the units as a means of knowing what to do.Since we want the wavelength of the incident photons, and since the data involves power and energy, the obvious goal is first to find how much energy each photon carries, then convert this number to wavelength.
Step 1) Convert 5.0 W/cm 2 to eV/s/cm 2 using 1 J = 1.6 x 10-19 eV.
5.0 W/cm 2 = 5.0 J/s/cm 2 = 5.0/(1.6 x 10-19) = 3.13 x 10 19 eV/s/cm 2.
Step 2) Convert 10 20 photons/min on 3 cm 2 to photons /second/cm 2.
rate = (10 20)(1/60)(1/3) = 5.6 x 10 17 photons/s/cm 2.
Step3) Combine the results of steps 1) and 2).
E(photon) = (3.23 x 10 19 eV/s/cm 2)/( 5.6 x 10 17 photons/s/cm 2) = 56 eV/photon
Step 4) Convert the photon energy of 56 eV to wavelengtb.
Wavelength = 1240/56 = 22 nm |
190323 | https://math.stackexchange.com/questions/3491295/determine-if-function-fx-ln-left-fracx1x-1-right-is-odd-or-even | logarithms - Determine if function $f(x)=\ln \left ( \frac{x+1}{x-1} \right )$ is odd or even. - Mathematics Stack Exchange
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Determine if function f(x)=ln(x+1 x−1)f(x)=ln(x+1 x−1) is odd or even.
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Determine if function f(x)=ln(x+1 x−1)f(x)=ln(x+1 x−1) is odd or even.
My solution:
f(−x)=ln(−x+1−x−1)=ln(−(x−1)−(x+1))=ln(x−1 x+1)=ln(x−1)−ln(x+1)=−(ln(x+1)−ln(x−1))=−ln(x+1 x−1)=−f(x)f(−x)=ln(−x+1−x−1)=ln(−(x−1)−(x+1))=ln(x−1 x+1)=ln(x−1)−ln(x+1)=−(ln(x+1)−ln(x−1))=−ln(x+1 x−1)=−f(x)
It seems that function is odd. However, according to WolframAlpha it's neither odd nor even.
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edited Dec 29, 2019 at 18:46
José Carlos Santos
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asked Dec 29, 2019 at 18:33
edionzedionz
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"according to WolframAlpha it's neither odd nor even. ": Could you detail this? What did you input, and what was the output of WolframAlpha?D. Thomine –D. Thomine 2019-12-29 18:36:57 +00:00 Commented Dec 29, 2019 at 18:36
wolframalpha.com/input/…edionz –edionz 2019-12-29 18:38:39 +00:00 Commented Dec 29, 2019 at 18:38
2 You did not tell Wolfram to exclude complex numbers ... f(1/2)f(1/2) and f(−1/2)f(−1/2) are not negatives of each other, using principal value of logarithm.GEdgar –GEdgar 2019-12-29 18:41:13 +00:00 Commented Dec 29, 2019 at 18:41
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That function is odd (assuming that its domain is (−∞,−1)∪(1,∞)(−∞,−1)∪(1,∞), an assumption which is not made by WolframAlpha). Your computations are fine, but you can shorten them using the fact that log(a−1)=−log(a)log(a−1)=−log(a). It follows from this that
log(x−1 x+1)=−log(x+1 x−1).log(x−1 x+1)=−log(x+1 x−1).
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edited Dec 29, 2019 at 18:50
J. W. Tanner
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answered Dec 29, 2019 at 18:44
José Carlos SantosJosé Carlos Santos
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While your conclusion is fine, there are still some problems with your computations, where some cautions is needed.
By definition, an even or odd function must have a symmetric domain of definition, so the first thing is to make this domain explicit: D:=(−∞,1)∪(1,+∞)D:=(−∞,1)∪(1,+∞).
Then we want to prove that f f is odd. In other words, we have to show that, for x∈D x∈D,
f(−x)=−f(x).f(−x)=−f(x).
If x>1 x>1, we have x+1>0 x+1>0 and x−1>0 x−1>0, so we can indeed write f(x)=ln(x+1)−ln(x−1)f(x)=ln(x+1)−ln(x−1). However, for x<1 x<1, we have x+1<0 x+1<0 and x−1<0 x−1<0, so the expression ln(x+1)−ln(x−1)ln(x+1)−ln(x−1) is meaningless. Hence, the computation you gave is only valid for x>1 x>1. This is enough to prove your claim, but you need to make this explicit.
Another solution is to avoid splitting the logarithm altogether. For any x∈D x∈D,
f(−x)=ln(x−1 x+1)=ln⎛⎝⎜1(x+1 x−1)⎞⎠⎟=−ln(x+1 x−1)=−f(x).f(−x)=ln(x−1 x+1)=ln(1(x+1 x−1))=−ln(x+1 x−1)=−f(x).
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answered Dec 29, 2019 at 18:57
D. ThomineD. Thomine
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190324 | https://www.pbslearningmedia.org/resource/mgbh.math.g.anglerota/rotational-symmetry/ | Rotational Symmetry | PBS LearningMedia
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Rotational Symmetry
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Visualize how objects behave when they are rotated around a centrally located rotation point. This interactive exercise looks at angles of rotation and rotational symmetry of a variety of figures.
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Teaching Tips | Rotational Symmetry
Learning Outcomes
Students will be able to
identify angle of rotation for figures with rotation symmetry
create figures with specified angles of rotation
define the vocabulary terms listed below
Common Core State Standards: 8.G.A.4
Vocabulary: Rotation symmetry, rotate, symmetry, center of rotation, angle of rotation, n-fold symmetry
Materials: Per student pair: computer; per student: protractor, pencil, and sheet of unlined paper
Procedure1. Introduction (5 minutes, whole group)Explain that today’s activity will focus on rotation symmetry. Something has rotation symmetry if you can turn it partway around (less than 360°) a point of rotation so that it is similar to the original figure. As needed, define rotate and symmetry.
Ask students to
fold a blank sheet of paper in half lengthwise and in half lengthwise again to create a vertical strip
orient the strip vertically on their desks
put a finger in the center of the strip and rotate it clockwise to determine if it has rotational symmetry
Discuss the results. Ask students, Were you able to turn it partway around so it looked the same as when you started? How many degrees did you turn it?
Establish that the center of rotation in this case is the center of the strip, where students put their fingers, and the angle of rotation is 180°. Thus, there are two positions in which the object appears identical: at 0° (no rotation) and 180°. We call this 2-fold symmetry because there are two angles of rotation.
Invite students to suggest other familiar objects with rotation symmetry around the classroom. For each, they should say what the center of rotation and the (approximate) angle of rotation are; they may use a protractor to help with their estimations. If no one proposes a square object (with 90° angle of rotation and 4-fold symmetry, because there are four angles of rotation), suggest one yourself. Also, gather a couple of ideas for objects that have no rotation symmetry.
2. Rotation in 15-Degree Increments (15 minutes, pairs)After demonstrating how to use the interactive, have students determine the center of rotation and the angle of rotation for all eight figures below:
Have them make predictions about center and angle of rotation before performing the rotation.
Have them find the angle of rotation and all the other angles at which the object appears identical when rotated. For instance, for the square object, the angle of rotation is 90°; the object also appears identical at 180° and 270°, for a total of four times within 360°.
Regular polygons
Other figures
As students work, circulate to check that they are making predictions and explaining their reasoning. Probe with questions such as, How many times do you think the figure will look identical as you rotate it the full 360°? Why do you think so?
Students who finish early may draw one or more of these:
A figure with no rotation symmetry
A figure with angle of symmetry 72° (5-fold symmetry)
A figure with angle of symmetry 30° (12-fold symmetry)
3. Conclusion (5 minutes, whole group)Ask students to share any patterns they noticed. Prompt with questions, such as:
Do all the figures with angle of rotation 120° appear identical three times within the full 360° rotation?
Did you find any figures that look identical at 45° and 90°? Any that look identical at 45° but not 90°? 90° but not 45°?
If a figure has 10-fold symmetry, does it also have 5-fold symmetry? Why or why not?
If a figure has n-fold symmetry, what is its angle of rotation? (An object having n-fold symmetry looks the same after being rotated 360/n°.)
This activity is based on work developed at TERC. |
190325 | https://pdg.lbl.gov/2014/reviews/rpp2014-rev-kinematics.pdf | Kinematics 1 46. KINEMATICS Revised January 2000 by J.D. Jackson (LBNL) and June 2008 by D.R. Tovey (Sheffield).
Throughout this section units are used in which ℏ= c = 1. The following conversions are useful: ℏc = 197.3 MeV fm, (ℏc)2 = 0.3894 (GeV)2 mb.
46.1.
Lorentz transformations The energy E and 3-momentum p of a particle of mass m form a 4-vector p = (E, p) whose square p2 ≡E2 −|p|2 = m2. The velocity of the particle is β = p/E. The energy and momentum (E∗, p∗) viewed from a frame moving with velocity βf are given by µ E∗ p∗ ∥ ¶ = µ γf −γfβf −γfβf γf ¶ µ E p∥ ¶ , p∗ T = pT , (46.1) where γf = (1 −β2 f)−1/2 and pT (p∥) are the components of p perpendicular (parallel) to βf. Other 4-vectors, such as the space-time coordinates of events, of course transform in the same way. The scalar product of two 4-momenta p1 · p2 = E1E2 −p1 · p2 is invariant (frame independent).
46.2.
Center-of-mass energy and momentum In the collision of two particles of masses m1 and m2 the total center-of-mass energy can be expressed in the Lorentz-invariant form Ecm = h (E1 + E2)2 −(p1 + p2)2i1/2 , = h m2 1 + m2 2 + 2E1E2(1 −β1β2 cos θ) i1/2 , (46.2) where θ is the angle between the particles. In the frame where one particle (of mass m2) is at rest (lab frame), Ecm = (m2 1 + m2 2 + 2E1 lab m2)1/2 .
(46.3) The velocity of the center-of-mass in the lab frame is βcm = plab/(E1 lab + m2) , (46.4) where plab ≡p1 lab and γcm = (E1 lab + m2)/Ecm .
(46.5) The c.m. momenta of particles 1 and 2 are of magnitude pcm = plab m2 Ecm .
(46.6) For example, if a 0.80 GeV/c kaon beam is incident on a proton target, the center of mass energy is 1.699 GeV and the center of mass momentum of either particle is 0.442 GeV/c.
It is also useful to note that Ecm dEcm = m2 dE1 lab = m2 β1 lab dplab .
(46.7) K.A. Olive et al. (PDG), Chin. Phys. C38, 090001 (2014) ( August 21, 2014 13:18 2 46. Kinematics 46.3.
Lorentz-invariant amplitudes The matrix elements for a scattering or decay process are written in terms of an invariant amplitude −iM . As an example, the S-matrix for 2 →2 scattering is related to M by ⟨p′ 1p′ 2 |S| p1p2⟩= I −i(2π)4 δ4(p1 + p2 −p′ 1 −p′ 2) × M (p1, p2; p′ 1, p′ 2) (2E1)1/2 (2E2)1/2 (2E′ 1)1/2 (2E′ 2)1/2 .
(46.8) The state normalization is such that ⟨p′|p⟩= (2π)3δ3(p −p′) .
(46.9) 46.4.
Particle decays The partial decay rate of a particle of mass M into n bodies in its rest frame is given in terms of the Lorentz-invariant matrix element M by dΓ = (2π)4 2M |M |2 dΦn (P; p1, . . . , pn), (46.10) where dΦn is an element of n-body phase space given by dΦn(P; p1, . . . , pn) = δ4 (P − n X i=1 pi) n Y i=1 d3pi (2π)32Ei .
(46.11) This phase space can be generated recursively, viz.
dΦn(P; p1, . . . , pn) = dΦj(q; p1, . . ., pj) × dΦn−j+1 (P; q, pj+1, . . ., pn)(2π)3dq2 , (46.12) where q2 = (Pj i=1 Ei)2 − ¯ ¯ ¯ Pj i=1 pi ¯ ¯ ¯ 2 . This form is particularly useful in the case where a particle decays into another particle that subsequently decays.
46.4.1.
Survival probability : If a particle of mass M has mean proper lifetime τ (= 1/Γ) and has momentum (E, p), then the probability that it lives for a time t0 or greater before decaying is given by P(t0) = e−t0 Γ/γ = e−Mt0 Γ/E , (46.13) and the probability that it travels a distance x0 or greater is P(x0) = e−Mx0 Γ/|p| .
(46.14) August 21, 2014 13:18 46. Kinematics 3 p1, m1 p2, m2 P, M Figure 46.1: Definitions of variables for two-body decays.
46.4.2.
Two-body decays : In the rest frame of a particle of mass M, decaying into 2 particles labeled 1 and 2, E1 = M2 −m2 2 + m2 1 2M , (46.15) |p1| = |p2| = £¡ M2 −(m1 + m2)2¢ ¡ M2 −(m1 −m2)2¢¤1/2 2M , (46.16) and dΓ = 1 32π2 |M |2 |p1| M2 dΩ, (46.17) where dΩ= dφ1d(cos θ1) is the solid angle of particle 1. The invariant mass M can be determined from the energies and momenta using Eq. (46.2) with M = Ecm.
46.4.3.
Three-body decays : p1, m1 p3, m3 P, M p2, m2 Figure 46.2: Definitions of variables for three-body decays.
Defining pij = pi + pj and m2 ij = p2 ij, then m2 12 + m2 23 + m2 13 = M2 + m2 1 + m2 2 + m2 3 and m2 12 = (P −p3)2 = M2 + m2 3 −2ME3, where E3 is the energy of particle 3 in the rest frame of M. In that frame, the momenta of the three decay particles lie in a plane.
The relative orientation of these three momenta is fixed if their energies are known. The momenta can therefore be specified in space by giving three Euler angles (α, β, γ) that specify the orientation of the final system relative to the initial particle .
Then dΓ = 1 (2π)5 1 16M |M |2 dE1 dE2 dα d(cos β) dγ .
(46.18) August 21, 2014 13:18 4 46. Kinematics Alternatively dΓ = 1 (2π)5 1 16M2 |M |2 |p∗ 1| |p3| dm12 dΩ∗ 1 dΩ3 , (46.19) where (|p∗ 1|, Ω∗ 1) is the momentum of particle 1 in the rest frame of 1 and 2, and Ω3 is the angle of particle 3 in the rest frame of the decaying particle. |p∗ 1| and |p3| are given by |p∗ 1| = £¡ m2 12 −(m1 + m2)2¢ ¡ m2 12 −(m1 −m2)2¢¤ 2m12 1/2 , (46.20a) and |p3| = £¡ M2 −(m12 + m3)2¢ ¡ M2 −(m12 −m3)2¢¤1/2 2M .
(46.20b) [Compare with Eq. (46.16).] If the decaying particle is a scalar or we average over its spin states, then integration over the angles in Eq. (46.18) gives dΓ = 1 (2π)3 1 8M |M |2 dE1 dE2 = 1 (2π)3 1 32M3 |M |2 dm2 12 dm2 23 .
(46.21) This is the standard form for the Dalitz plot.
46.4.3.1.
Dalitz plot: For a given value of m2 12, the range of m2 23 is determined by its values when p2 is parallel or antiparallel to p3: (m2 23)max = (E∗ 2 + E∗ 3)2 − µq E∗2 2 −m2 2 − q E∗2 3 −m2 3 ¶2 , (46.22a) (m2 23)min = (E∗ 2 + E∗ 3)2 − µq E∗2 2 −m2 2 + q E∗2 3 −m2 3 ¶2 .
(46.22b) Here E∗ 2 = (m2 12 −m2 1 + m2 2)/2m12 and E∗ 3 = (M2 −m2 12 −m2 3)/2m12 are the energies of particles 2 and 3 in the m12 rest frame. The scatter plot in m2 12 and m2 23 is called a Dalitz plot. If |M |2 is constant, the allowed region of the plot will be uniformly populated with events [see Eq. (46.21)]. A nonuniformity in the plot gives immediate information on |M |2. For example, in the case of D →Kππ, bands appear when m(Kπ) = mK∗(892), reflecting the appearance of the decay chain D →K∗(892)π →Kππ.
August 21, 2014 13:18 46. Kinematics 5 (m23)max 0 1 2 3 4 5 0 2 4 6 8 10 m12 (GeV2) m23 (GeV2) (m1+m2)2 (M−m3)2 (M−m1)2 (m2+m3)2 (m23)min 2 2 2 2 Figure 46.3: Dalitz plot for a three-body final state. In this example, the state is π+K0p at 3 GeV. Four-momentum conservation restricts events to the shaded region.
46.4.4.
Kinematic limits : 46.4.4.1.
Three-body decays: In a three-body decay (Fig. 46.2) the maximum of |p3|, [given by Eq. (46.20)], is achieved when m12 = m1 + m2, i.e., particles 1 and 2 have the same vector velocity in the rest frame of the decaying particle. If, in addition, m3 > m1, m2, then |p3|max > |p1|max, |p2|max. The distribution of m12 values possesses an end-point or maximum value at m12 = M −m3. This can be used to constrain the mass difference of a parent particle and one invisible decay product.
August 21, 2014 13:18 6 46. Kinematics 46.4.4.2.
Sequential two-body decays: b c a 2 1 Figure 46.4: Particles participating in sequential two-body decay chain. Particles labeled 1 and 2 are visible while the particle terminating the chain (a) is invisible.
When a heavy particle initiates a sequential chain of two-body decays terminating in an invisible particle, constraints on the masses of the states participating in the chain can be obtained from end-points and thresholds in invariant mass distributions of the aggregated decay products. For the two-step decay chain depicted in Fig. 46.4 the invariant mass distribution of the two visible particles possesses an end-point given by: (mmax 12 )2 = (m2 c −m2 b)(m2 b −m2 a) m2 b , (46.23) provided particles 1 and 2 are massless. If visible particle 1 has non-zero mass m1 then Eq. (46.23) is replaced by (mmax 12 )2 = m2 1 + (m2 c −m2 b) 2m2 b × µ m2 1 + m2 b −m2 a + q (−m2 1 + m2 b −m2 a)2 −4m2 1m2 a ¶ .
(46.24) See Refs. 2 and 3 for other cases.
46.4.5.
Multibody decays : The above results may be generalized to final states containing any number of particles by combining some of the particles into “effective particles” and treating the final states as 2 or 3 “effective particle” states. Thus, if pijk... = pi + pj + pk + . . ., then mijk... = q p2ijk... , (46.25) and mijk... may be used in place of e.g., m12 in the relations in Sec. 46.4.3 or Sec. 46.4.4 above.
August 21, 2014 13:18 46. Kinematics 7 46.5.
Cross sections p3, m3 pn+2, mn+2 . . .
p1, m1 p2, m2 Figure 46.5: Definitions of variables for production of an n-body final state.
The differential cross section is given by dσ = (2π)4|M |2 4 q (p1 · p2)2 −m2 1m2 2 × dΦn(p1 + p2; p3, . . . , pn+2) .
(46.26) [See Eq. (46.11).] In the rest frame of m2(lab), q (p1 · p2)2 −m2 1m2 2 = m2p1 lab ; (46.27a) while in the center-of-mass frame q (p1 · p2)2 −m2 1m2 2 = p1cm √s .
(46.27b) 46.5.1.
Two-body reactions : p1, m1 p2, m2 p3, m3 p4, m4 Figure 46.6: Definitions of variables for a two-body final state.
Two particles of momenta p1 and p2 and masses m1 and m2 scatter to particles of momenta p3 and p4 and masses m3 and m4; the Lorentz-invariant Mandelstam variables are defined by s = (p1 + p2)2 = (p3 + p4)2 = m2 1 + 2E1E2 −2p1 · p2 + m2 2 , (46.28) t = (p1 −p3)2 = (p2 −p4)2 = m2 1 −2E1E3 + 2p1 · p3 + m2 3 , (46.29) u = (p1 −p4)2 = (p2 −p3)2 = m2 1 −2E1E4 + 2p1 · p4 + m2 4 , (46.30) August 21, 2014 13:18 8 46. Kinematics and they satisfy s + t + u = m2 1 + m2 2 + m2 3 + m2 4 .
(46.31) The two-body cross section may be written as dσ dt = 1 64πs 1 |p1cm|2 |M |2 .
(46.32) In the center-of-mass frame t = (E1cm −E3cm)2 −(p1cm −p3cm)2 −4p1cm p3cm sin2(θcm/2) = t0 −4p1cm p3cm sin2(θcm/2) , (46.33) where θcm is the angle between particle 1 and 3. The limiting values t0 (θcm = 0) and t1 (θcm = π) for 2 →2 scattering are t0(t1) = ·m2 1 −m2 3 −m2 2 + m2 4 2√s ¸2 −(p1 cm ∓p3 cm)2 .
(46.34) In the literature the notation tmin (tmax) for t0 (t1) is sometimes used, which should be discouraged since t0 > t1. The center-of-mass energies and momenta of the incoming particles are E1cm = s + m2 1 −m2 2 2√s , E2cm = s + m2 2 −m2 1 2√s , (46.35) For E3cm and E4cm, change m1 to m3 and m2 to m4. Then pi cm = q E2 i cm −m2 i and p1cm = p1 lab m2 √s .
(46.36) Here the subscript lab refers to the frame where particle 2 is at rest. [For other relations see Eqs. (46.2)–(46.4).] 46.5.2.
Inclusive reactions : Choose some direction (usually the beam direction) for the z-axis; then the energy and momentum of a particle can be written as E = mT cosh y , px , py , pz = mT sinh y , (46.37) where mT , conventionally called the ‘transverse mass’, is given by m2 T = m2 + p2 x + p2 y .
(46.38) and the rapidity y is defined by y = 1 2 ln µE + pz E −pz ¶ August 21, 2014 13:18 46. Kinematics 9 = ln µE + pz mT ¶ = tanh−1 ³pz E ´ .
(46.39) Note that the definition of the transverse mass in Eq. (46.38) differs from that used by experimentalists at hadron colliders (see Sec. 46.6.1 below). Under a boost in the z-direction to a frame with velocity β, y →y −tanh−1 β. Hence the shape of the rapidity distribution dN/dy is invariant, as are differences in rapidity. The invariant cross section may also be rewritten E d3σ d3p = d3σ dφ dy pT dpT = ⇒ d2σ π dy d(p2 T ) .
(46.40) The second form is obtained using the identity dy/dpz = 1/E, and the third form represents the average over φ.
Feynman’s x variable is given by x = pz pz max ≈ E + pz (E + pz)max (pT ≪|pz|) .
(46.41) In the c.m. frame, x ≈2pz cm √s = 2mT sinh ycm √s (46.42) and = (ycm)max = ln(√s/m) .
(46.43) The invariant mass M of the two-particle system described in Sec. 46.4.2 can be written in terms of these variables as M2 = m2 1 + m2 2 + 2[ET (1)ET (2) cosh ∆y −pT (1) · pT (2)] , (46.44) where ET (i) = q |pT (i)|2 + m2 i , (46.45) and pT (i) denotes the transverse momentum vector of particle i.
For p ≫m, the rapidity [Eq. (46.39)] may be expanded to obtain y = 1 2 ln cos2(θ/2) + m2/4p2 + . . .
sin2(θ/2) + m2/4p2 + . . .
≈−ln tan(θ/2) ≡η (46.46) where cos θ = pz/p. The pseudorapidity η defined by the second line is approximately equal to the rapidity y for p ≫m and θ ≫1/γ, and in any case can be measured when the mass and momentum of the particle are unknown. From the definition one can obtain the identities sinh η = cot θ , cosh η = 1/ sin θ , tanh η = cos θ .
(46.47) August 21, 2014 13:18 10 46. Kinematics 46.5.3.
Partial waves : The amplitude in the center of mass for elastic scattering of spinless particles may be expanded in Legendre polynomials f(k, θ) = 1 k X ℓ (2ℓ+ 1)aℓPℓ(cos θ) , (46.48) where k is the c.m. momentum, θ is the c.m. scattering angle, aℓ= (ηℓe2iδℓ−1)/2i, 0 ≤ηℓ≤1, and δℓis the phase shift of the ℓth partial wave. For purely elastic scattering, ηℓ= 1. The differential cross section is dσ dΩ= |f(k, θ)|2 .
(46.49) The optical theorem states that σtot = 4π k Im f(k, 0) , (46.50) and the cross section in the ℓth partial wave is therefore bounded: σℓ= 4π k2 (2ℓ+ 1)|aℓ|2 ≤4π(2ℓ+ 1) k2 .
(46.51) The evolution with energy of a partial-wave amplitude aℓcan be displayed as a trajectory in an Argand plot, as shown in Fig. 46.7.
−1/2 1/2 0 Im A Re A 1/2 η/2 1 al 2δ Figure 46.7: Argand plot showing a partial-wave amplitude aℓas a function of energy. The amplitude leaves the unitary circle where inelasticity sets in (ηℓ< 1).
The usual Lorentz-invariant matrix element M (see Sec. 46.3 above) for the elastic process is related to f(k, θ) by M = −8π√s f(k, θ) , (46.52) so σtot = − 1 2plab m2 Im M (t = 0) , (46.53) where s and t are the center-of-mass energy squared and momentum transfer squared, respectively (see Sec. 46.4.1).
August 21, 2014 13:18 46. Kinematics 11 46.5.3.1.
Resonances: The Breit-Wigner (nonrelativistic) form for an elastic amplitude aℓwith a resonance at c.m. energy ER, elastic width Γel, and total width Γtot is aℓ= Γel/2 ER −E −iΓtot/2 , (46.54) where E is the c.m. energy. As shown in Fig. 46.8, in the absence of background the elastic amplitude traces a counterclockwise circle with center ixel/2 and radius xel/2, where the elasticity xel = Γel/Γtot. The amplitude has a pole at E = ER −iΓtot/2.
The spin-averaged Breit-Wigner cross section for a spin-J resonance produced in the collision of particles of spin S1 and S2 is σBW (E) = (2J + 1) (2S1 + 1)(2S2 + 1) π k2 BinBoutΓ2 tot (E −ER)2 + Γ2 tot/4 , (46.55) where k is the c.m. momentum, E is the c.m. energy, and B in and B out are the branching fractions of the resonance into the entrance and exit channels. The 2S + 1 factors are the multiplicities of the incident spin states, and are replaced by 2 for photons.
This expression is valid only for an isolated state. If the width is not small, Γtot cannot be treated as a constant independent of E. There are many other forms for σBW , all of which are equivalent to the one given here in the narrow-width case. Some of these forms may be more appropriate if the resonance is broad.
−1/2 1/2 0 Im A Re A ixel/2 xel/2 1 Figure 46.8: Argand plot for a resonance.
The relativistic Breit-Wigner form corresponding to Eq. (46.54) is: aℓ= −mΓel s −m2 + imΓtot .
(46.56) A better form incorporates the known kinematic dependences, replacing mΓtot by √s Γtot(s), where Γtot(s) is the width the resonance particle would have if its mass August 21, 2014 13:18 12 46. Kinematics were √s, and correspondingly mΓel by √s Γel(s) where Γel(s) is the partial width in the incident channel for a mass √s: aℓ= −√s Γel(s) s −m2 + i√s Γtot(s) .
(46.57) For the Z boson, all the decays are to particles whose masses are small enough to be ignored, so on dimensional grounds Γtot(s) = √s Γ0/mZ, where Γ0 defines the width of the Z, and Γel(s)/Γtot(s) is constant. A full treatment of the line shape requires consideration of dynamics, not just kinematics. For the Z this is done by calculating the radiative corrections in the Standard Model.
46.6.
Transverse variables At hadron colliders, a significant and unknown proportion of the energy of the incoming hadrons in each event escapes down the beam-pipe. Consequently if invisible particles are created in the final state, their net momentum can only be constrained in the plane transverse to the beam direction. Defining the z-axis as the beam direction, this net momentum is equal to the missing transverse energy vector E miss T = − X i pT (i) , (46.58) where the sum runs over the transverse momenta of all visible final state particles.
46.6.1.
Single production with semi-invisible final state : Consider a single heavy particle of mass M produced in association with visible particles which decays as in Fig. 46.1 to two particles, of which one (labeled particle 1) is invisible. The mass of the parent particle can be constrained with the quantity MT defined by M2 T ≡[ET (1) + ET (2)]2 −[pT (1) + pT (2)]2 = m2 1 + m2 2 + 2[ET (1)ET (2) −pT (1) · pT (2)] , (46.59) where pT (1) = E miss T .
(46.60) This quantity is called the ‘transverse mass’ by hadron collider experimentalists but it should be noted that it is quite different from that used in the description of inclusive reactions [Eq. (46.38)]. The distribution of event MT values possesses an end-point at Mmax T = M. If m1 = m2 = 0 then M2 T = 2|pT (1)||pT (2)|(1 −cos φ12) , (46.61) where φij is defined as the angle between particles i and j in the transverse plane.
August 21, 2014 13:18 46. Kinematics 13 p 1 1 , m p 4 4 , m p , m p 3 1 2 2 , m M M Figure 46.9: Definitions of variables for pair production of semi-invisible final states. Particles 1 and 3 are invisible while particles 2 and 4 are visible.
46.6.2.
Pair production with semi-invisible final states : Consider two identical heavy particles of mass M produced such that their combined center-of-mass is at rest in the transverse plane (Fig. 46.9). Each particle decays to a final state consisting of an invisible particle of fixed mass m1 together with an additional visible particle. M and m1 can be constrained with the variables MT2 and MCT which are defined in Refs. and .
References: 1.
See, for example, J.J. Sakurai, Modern Quantum Mechanics, Addison-Wesley (1985), p. 172, or D.M. Brink and G.R. Satchler, Angular Momentum, 2nd ed., Oxford University Press (1968), p. 20.
2.
I. Hinchliffe et al., Phys. Rev. D55, 5520 (1997).
3.
B.C. Allanach et al., JHEP 0009, 004 (2000).
4.
C.G. Lester and D.J. Summers, Phys. Lett. B463, 99 (1999).
5.
D.R. Tovey, JHEP 0804, 034 (2008).
August 21, 2014 13:18 |
190326 | https://openspace.infohio.org/courseware/lesson/2157/overview | See new layout
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: Chris Adcock
Subject:
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: Lesson Plan
Level:
: Middle School
Grade:
: 6
Provider:
: Pearson
Tags:
: - 6th Grade Mathematics
- Measurement
- Parallelograms
- Triangles
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Triangle
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Triangle
View
Math, Grade 6
Surface Area and Volume
Getting Started
Rational Numbers
Fractions and Decimals
Ratios
Expressions
Equations and Inequalities
Rate
Putting Math to Work
Distributions and Variability
Surface Area and Volume
Analyzing The Formula of A Triangle
Comparing Surface Area & Volume
Analyzing The Formula of A Parallelogram & Trapezoid
Analyzing The Formula of A Triangle
Basic & Composite Figures
Using A Grid To Calculate The Area Of An Irregular Figure
Volume Formula For Rectangular Prisms
Identifying Nets For Cubes
Transforming The Net of A Cube Into The Net of A Pyramid
Diagrams & Problem Solving Strategies
Gallery Problems Exercise (Groups)
Analyzing The Formula of A Triangle
Overview
Lesson Overview
Students find the area of a triangle by putting together a triangle and a copy of the triangle to form a parallelogram with the same base and height as the triangle. Students also create several examples of triangles and look for relationships among the base, height, and area measures. These activities lead students to develop and understand a formula for the area of a triangle.
Key Concepts
To find the area of a triangle, you must know the length of a base and the corresponding height. The base of a triangle can be any of the three sides. The height is the perpendicular distance from the vertex opposite the base to the line containing the base. The height can be found inside or outside the triangle, or it can be the length of one of the sides.
You can put together a triangle and a copy of the triangle to form a parallelogram with the same base and height as the triangle. The area of the original triangle is half of the area of the parallelogram. Because the area formula for a parallelogram is A = bh, the area formula for a triangle is A = bh.
Goals and Learning Objectives
Develop and explore the formula for the area of a triangle.
About Base and Height
Lesson Guide
Have students work in pairs to discuss the statements.
ELL: Students may have difficulty determining the perpendicular height of a triangle, especially when the height needs to be sketched outside of the triangle. Use the edges of an index card to help students get a better idea of how to determine the perpendicular height. Model this activity for students. The index card is aligned with the base and then shifted to the left (or right) until the vertex opposite the base touches the index card. This clearly shows that the height is perpendicular to the base.
Mathematics
Have students look at the triangles. Make sure they understand what the base and height of a triangle are. Point out that the height can be inside the triangle, outside the triangle, or one of the sides of the triangle.
Opening
About Base and Height
Discuss the following statements.
The base of a triangle can be any of the three sides.
The height of a triangle is the perpendicular distance from the base to the vertex opposite the base.
As shown in the diagram, the height can be inside or outside the triangle, or it can be one of the sides.
Introduction to Triangles
Lesson Guide
Partners should discuss how they can arrange a triangle and its copy to form a parallelogram. Students should recognize that the area of each triangle is half the area of the parallelogram made from two of the same-size triangles. Since the area of the parallelogram is A = bh, the area of the triangle is A = bh.
SWD: Some students may not immediately see how to arrange the triangles into a parallelogram. If students are struggling to the point of frustration, model how to use two triangles to create a small parallelogram. Allow students to use paper cutouts if needed.
Opening
Introduction to Triangles
Can you take any triangle, copy it, and then combine the two triangles so that they form a parallelogram?
Try it with triangles like the ones in the diagram.
What do your results tell you about the area of a triangle?
Write a formula for the area of a triangle.
Math Mission
Lesson Guide
Discuss the Math Mission. Students will explore the formula for the area of a triangle.
SWD: It may be challenging for some students to remember which formulas correspond for each shape. Have students create resources for themselves to refer to throughout the unit (e.g., note cards, digital sticky notes, anchor charts, their notebook) that include the shape's name, the formula for area, and an image that represents the shape.
Opening
Explore the formula for the area of a triangle.
Explore the Area of Triangles
Lesson Guide
Have partners answer the questions and work on the presentation together.
SWD: Assign students concrete lengths for base and heights. If students are using cutouts in addition to the interactive, have them label the cutouts with the resulting areas. This will help them to understand the relationship between the areas of the triangles and the combined total of the areas as the area of the original polygon.
Mathematical Practices
Mathematical Practice 7: Look for and make use of structure.
Identify students who understand how the operations in the formula impact the answer (i.e., increasing a factor increases the product; decreasing a factor decreases the product; or keeping the factors the same does not change the product).
Mathematical Practice 8: Look for and express regularity in repeated reasoning.
Watch for students who, through repeated trials, reason that if one variable is constant and the other variable increases (or decreases), then the area increases (or decreases) as well.
Look for students who, through repeated trials, reason that if both variables remain constant, the area remains constant as well.
Interventions
Student has difficulty getting started.
Describe the task in your own words to your partner.
What does it mean to keep the height or base constant?
Remember that if the vertex moves parallel to the base, then it remains the same distance from the base.
How can you move the vertex parallel to the base?
Student changes the variable only one way, just increasing (or decreasing) its value.
Can you decrease (or increase) the height? What happens?
Student works unsystematically.
Can you organize the information for the base, height, and area in a table?
Look at the values in the rows of your table. How are they related?
Student has a correct solution.
How did you reach your conclusion?
Explain how the Triangle interactive helped you reach a conclusion.
Answers
If you keep the height and base constant and move the vertex parallel to the base, the area remains constant.
If you keep the base constant and increase (or decrease) the height, then the area increases (or decreases) as well.
Facts will vary.
Work Time
Explore the Area of Triangles
The formula for the area of a triangle isarea = • base • height, or A = bh
Use the Triangle interactive to explore the area of a triangle. Move the vertices of the triangle and explore what happens to the area.
What happens if you keep the height and base constant and move the vertex parallel to the base?
What happens if you keep the base constant and change the height?
Try to discover one more interesting fact about a triangle and its area that you can share with the class.
INTERACTIVE: Triangle
Hint:
How does knowing the formula for the area of a parallelogram help you understand the formula for the area of a triangle?
There are two variables, base and height, that determine the area of a triangle. A triangle also has angle measures and side lengths for the two “non-base” sides. Try experimenting with all of these measures.
Triangle
View
Prepare a Presentation
Preparing for Ways of Thinking
As students work on the problems, look for examples to share in the Ways of Thinking discussion:
Students who understand that if the height and base are constant and the vertex moves in a parallel line, the area remains constant
Students who recognize that if one variable is constant, increasing (or decreasing) the other variable also increases (or decreases) the area
Students who recognize that if the base (or height) increases (or decreases) by a factor, then the area increases (or decreases) by the same factor
Students who do not see a relationship between height, base, and area
Students who recognize the relationship between the formula for the area of a parallelogram and the area of a triangle
Challenge Problem
Answers
As the vertex of the triangle slides along the line, the area will stay the same.
Possible answer: The base of the triangle is always the same. Because the vertex stays on a line that is parallel to the base, the height will always be the same too. Because the area of the triangle depends only on the base and the height, the area will not change.
Work Time
Prepare a Presentation
Select one of your conclusions about what happens to the area of a triangle when you change one or more variables.
Be prepared to demonstrate your conclusion using the Triangle interactive, and to support your thinking mathematically.
Challenge Problem
Suppose the base of a triangle lies on one of two parallel lines, and the vertex opposite the base lies on the other parallel line.
If you slide the vertex along the line, what do you think will happen to the area of the triangle? Use the Triangle interactive to test your prediction.
Explain your results.
INTERACTIVE: Triangle
Triangle
View
Make Connections
Mathematics
Have students share their work. Be sure to show the work of students who had trouble and those who developed incorrect conclusions, as all students can benefit from the discussion. Use the Triangle interactive to test the statements that students generated to verify them.
Have students who did the Challenge Problem share their thinking. Ask class members to critique whether their reasoning makes sense.
ELL: As with other discussions, encourage ELL students to use the academic vocabulary they have learned. Introduce new vocabulary as needed. As they participate in the discussion, be sure to monitor for knowledge of the topic.
Performance Task
Make Connections
Take notes about your classmates’ conclusions concerning what happens to the area of a triangle when you change one or more variables.
Hint:
As your classmates present, ask questions such as:
What surprised you in your exploration of the area of a triangle?
How do your conclusions about the area of a triangle compare with those of other presenters?
Area of Trapezoid
Lesson Guide
Have students work on this problem on their own.
Mathematics
As you review the answers to the problem, encourage students to share their solution methods. Identify the following ways of thinking and correct any misconceptions:
Students who correctly use the area formulas to calculate the areas
Students who do not use the formula for the area of a trapezoid for the first figure, but instead, find the area of the triangle and the area of the rectangle and add the two areas
[common error] Students who use a base of 5 in., instead of 15 in., when calculating the area of the trapezoid (i.e., fail to add 5 in. + 10 in. to find the length of the longer base)
Answers
The area of the trapezoid is 100 in2.
Work Time
Area of Trapezoid
Find the area of this trapezoid.
Area of Triangle
Lesson Guide
Have students work on this problem on their own.
Mathematics
As you review the answers to these problems, encourage students to share their solution methods. Identify the following ways of thinking and correct any misconceptions:
[common error] Students who do not use the correct base to find the area of the triangle
Students who do not label their answers using square units
Answers
The area of the triangle is 17.2961 cm2.
Work Time
Area of Triangle
Find the area of this triangle.
Area Formulas
Mathematics
Have pairs quietly discuss how they can find the area of a parallelogram, a trapezoid, and a triangle if they know the formula for the area of a rectangle.
As student pairs work together, listen for students who may still have misconceptions so you can address them in the class discussion.
After a few minutes, discuss the summary as a class. Review the following points:
You can move parts of a parallelogram around to make a rectangle. Once you have formed a rectangle, you can find its area. The formula for the area of a parallelogram is A = bh, where b is the base and h is the height.
You can make a copy of a trapezoid, put the two trapezoids together to make a parallelogram, find the area of the parallelogram, and take half of that area to get the area of the original trapezoid. The formula for the area of a trapezoid is A = (b1 + b2)h, where b1 is one base, b2 is the other base, and h is the height.
You can copy a triangle and put the triangle and its copy together to form a parallelogram. The area of the triangle is half the area of the parallelogram. The formula for the area of a triangle is A = bh. ELL: Write the key points on a poster so that students can refer back to them throughout the module. When working with ELLs, provide supplementary materials, such as graphic organizers to illustrate new concepts and vocabulary necessary for mathematical learning. Have students record all information in their Notebook.
Formative Assessment
Area Formulas
Read and Discuss
The area of a rectangle is equal to its base times its height.A = bh
The area of a parallelogram is equal to its base times its height.A = bh
The area of a trapezoid is equal to one half times the sum of the bases times the height.A = (b1 + b2)h
The area of a triangle is equal to one half the base times the height.A = bh
Hint:
Can you:
Calculate the area of a triangle, parallelogram, or trapezoid given the values of the base(s) and height?
Calculate the height of a triangle, parallelogram, or trapezoid given the values of the base(s) and area?
Reflect On Your Work
Lesson Guide
Have each student write a brief reflection before the end of the class. Review the reflections to find out how students remember the formula for the area of a triangle.
Work Time
Reflection
Write a reflection about the ideas discussed in class today. Use the sentence starter below if you find it to be helpful.
The way I remember the formula for the area of a triangle is … |
190327 | https://www.cs.cmu.edu/~glmiller/Publications/Papers/MiTaTeWa99.pdf | ON THE RADIUS-EDGE CONDITION IN THE CONTROL VOLUME METHOD∗ GARY L. MILLER†, DAFNA TALMOR‡, SHANG-HUA TENG§, AND NOEL WALKINGTON¶ SIAM J. NUMER. ANAL.
c ⃝1999 Society for Industrial and Applied Mathematics Vol. 36, No. 6, pp. 1690–1708 Abstract. In this paper we show that the control volume algorithm for the solution of Poisson’s equation in three dimensions will converge even if the mesh contains a class of very flat tetrahedra (slivers). These tetrahedra are characterized by the fact that they have modest ratios of diameter to shortest edge, but large circumscribing to inscribed sphere radius ratios, and therefore may have poor interpolation properties. Elimination of slivers is a notoriously difficult problem for automatic mesh generation algorithms. We also show that a discrete Poincar´ e inequality will continue to hold in the presence of slivers.
Key words. mesh generation, slivers, control volume method AMS subject classifications. 65M12, 52C22 PII. S0036142996311854 1. Introduction. We consider covolume approximations of Poisson’s equation on a three-dimensional domain. It is shown that the control volume algorithm will converge on Delaunay meshes that contain both well proportioned tetrahedra and certain degenerate tetrahedra called “slivers,” that is, flat tetrahedra whose vertices are approximately equispaced around the equator of a sphere (see Figure 1). This result was rather unexpected, since the classical finite element algorithm may fail to converge when such degenerate elements are present.
Classical finite element theory shows that the finite element solution is the best approximation of the solution in the H1 0(Ω) norm,1 so clearly the control volume algorithm can’t converge in this norm if the finite element solution doesn’t. Instead we show that the control volume scheme converges in a discrete H1 0(Ω) norm (∥.∥W below), and like the finite element algorithm, the control volume algorithm gives the best projection with respect to this norm. Numerical experiments indicate that the finite element method may fail to converge in either norm.
We note that in two dimensions the two norms coincide, so this dichotomy is truly a three-dimensional phenomenon. Since the discrete norm used with the control volume algorithm changes with the mesh, it is natural to ask if convergence can be established in any of the standard norms. We answer this in the affirmative by establishing a discrete Poincar´ e ∗Received by the editors November 11, 1996; accepted for publication (in revised form) August 24, 1998; published electronically September 17, 1999.
†School of Computer Science, Carnegie Mellon University, Pittsburgh, PA 15213 (glmiller@ theory.cs.cmu.edu). The work of this author was supported in part by NSF grant CCR-9505472.
‡School of Computer Science, Carnegie Mellon University, Pittsburgh, PA 15213 (tdafna@ cs.cmu.edu).
§Department of Computer Science, University of Minnesota, Minneapolis, MN 55455. The work of this author was supported in part by an NSF CAREER award (CCR-9502540) and an Alfred P.
Sloan Research Fellowship.
¶Department of Mathematics, Carnegie Mellon University, Pittsburgh, PA 15213 (noelw@ cmu.edu.). The work of this author was supported in part by National Science Foundation grant DMS–9203406. This work was also supported by the Army Research Office and NSF through the Center for Nonlinear Analysis.
1∥u∥2 H1 0 (Ω) = R Ω|∇u|2.
1690 RADIUS-EDGE CONDITION IN CONTROL VOLUME METHOD 1691 Fig. 1. Sliver and circumsphere.
inequality which will imply convergence of the control volume method in L2(Ω).
These results evolved from our work on mesh generation algorithms [6, 7] and an attempt to resolve the problems slivers present in approximation theory.
The mesh generation algorithms we developed produce tetrahedra having bounded ratios of circumscribing radius to minimum edge length (radius-edge ratio), while traditional interpolation theory requires each tetrahedra to have the classical aspect ratio of di-ameter to inscribed sphere radius bounded. In two dimensions, these two ratios are equivalent in the sense that each ratio is bounded above and below by a multiple of the other. In three dimensions, slivers, which have a radius-edge ratio near unity and arbitrarily small volume, show that a bounded radius-edge aspect ratio is a weaker condition than the classical aspect ratio in higher dimensions; moreover, they are the only tetrahedra for which the ratios are incomparable. While a bound on the classical aspect ratio is sufficient for optimal interpolation, it is not necessary. Babuˇ ska and Aziz showed that right triangles in two dimensions, and their analogues in three dimensions, will exhibit optimal interpolation even if they have arbitrarily small vol-ume. In two dimensions this leads to the very simple geometric requirement that “the maximum angle within any triangle be bounded away from π”; however, no simple geometric restriction is known for tetrahedra. Upon discovering that the covolume scheme converges in the presence of slivers, it is natural to ask if they too exhibit optimal interpolation properties.
Folklore indicates that they do not, and this is substantiated by the numerical example in section 3.
In order to put the geometric assumptions into context, section 2 reviews the es-sential features of the mesh generation algorithm developed in . As indicated above, this algorithm does guarantee a bound on the radius-edge aspect ratio. The proof of convergence for Poisson’s equation under the bounded radius-edge ratio assumption is presented in section 3 along with an example which highlights the differences between the covolume and finite element schemes. In section 4 a discrete Poincar´ e inequality is established which provides one method of passing from mesh-dependent norms to the L2(Ω) norm.
2. Sphere packings and mesh generation. Mesh generation has a long his-tory; however, rigorous analysis of the algorithms and the meshes they produce is a recent development. Bern, Eppstein, and Gilbert were the first to establish opti-1692 G. MILLER, D. TALMOR, S.-H. TENG, AND N. WALKINGTON mality of the quad-tree algorithm for the generation of two-dimensional meshes. By optimal we mean that bounds on the aspect ratio of each triangle can be established apriori and that the number of triangles produced is within a constant of the mini-mum possible. Unfortunately the constant for the quad-tree algorithm appears to be rather large, so that this, and its three-dimensional counterpart , is not a particu-larly practical approach. An optimal two-dimensional algorithm that appears to work very well in practice was given by Ruppert . This algorithm extended an idea of Chew , and uses the Voronoi/Delaunay constructions. Ruppert’s algorithm doesn’t extend immediately to three dimensions since it may produce “slivers,” that is, very flat tetrahedra whose edge lengths are comparable with the circumsphere radius (see Figure 1). These flat tetrahedra are difficult to eliminate and plague three-dimensional Delaunay-based mesh generation algorithms. In section 3 we establish the rather un-expected result that these slivers do not degrade control volume approximations of Poisson’s equation, and in this situation extensions of Ruppert’s algorithm can be used to produce meshes that are “provably good for control volume approximations.” The remainder of this section introduces the basic geometric properties needed for the analysis in sections 3 and 4.
Below we consider meshes generated as the Delaunay diagrams of a set of points X = {xi} ⊂¯ Ω⊂Rd (we are particularly interested in three dimensions where d = 3).
Recall that the Voronoi diagram is the collection of polytopes {Vi} where Vi is the subset of points in Ωcloser to xi than to any other point in the set X. The Delaunay triangulation is then constructed by joining points in X that share a common Voronoi face. This construction is well known in computational geometry , and there are efficient algorithms for implementing it in any number of dimensions.
The meshes we consider below assume that the triangulation satisfies the following weakened aspect ratio condition.
Definition 2.1 (bounded radius-edge ratio). The radius-edge ratio of a trian-gulation in three dimensions is the maximum ratio of circumscribing sphere radius to smallest edge length of any tetrahedron. (Recall that the circumscribing sphere is the sphere passing through the four vertices of the tetrahedron.) We briefly sketch one technique for generating a vertex set whose Delaunay tri-angulations have bounded radius-edge ratios. We assume that we’re given a function ρ : Ω→R+ which locally specifies the desired density of mesh points in Ω.
Definition 2.2. A collection of points X = {x1, x2, . . . , xN} ⊂Ωis a ρ-packing if xi, xj ∈X implies [ρ(xi)+ρ(xj)]/2 ≤|xi−xj| where |.| denotes Euclidean distance.
A simple algorithm for the construction of a maximal ρ-packing is to initialize X to be the empty set and to randomly select a point x ∈Ωand add it to X if the sphere of radius ρ(x) centered at x doesn’t meet the analogous spheres of any point already in X. Upon termination, X will be a maximal ρ-packing. The following lemma concerning the Voronoi/Delaunay diagram for X is elementary.
Lemma 2.3.
Let ρ : Ω→R+ be Lipschitz2 with |ρ|Lip < 2 and let X be a maximal ρ-packing. If v is the center of a Delaunay ball (i.e., the circumscribing sphere of a tetrahedron) of radius r in the Voronoi/Delaunay triangulation of X, then ρ(v) ≥(1 −|ρ|Lip/2)r.
This lemma and Corollary 2.4 are proved in [6, 7], and provide a wealth of geo-metric information about such meshes.
Corollary 2.4. Let α = |ρ|Lip < 2/3. Then 2Recall that ρ is Lipschitz with constant |ρ|Lip ≤α if |ρ(x) −ρ(y)| ≤α|x −y| for all x and y.
RADIUS-EDGE CONDITION IN CONTROL VOLUME METHOD 1693 • If v and r are as in Lemma 2.3 and x ∈Br(v), the ball or radius r centered at v, then ρ(x) ≥(1 −3α/2)r.
• The radius-edge ratio of the Delaunay triangulation of X is bounded by 1/(1− 3α/2).
• If V is a Voronoi polytope centered at x ∈X, the ratio of circumscribing radius of V to inscribed radius of V (with center x) is bounded by (2+α)/(1− 3α/2).
• If x, y ∈X and p ∈xy (an edge in the Delaunay diagram), then ρ(p) ≥1 −3α/2 2 + α ρ(x).
• For any Voronoi region V maxV ρ minV ρ ≤(2 + α)(2 + 3α/2) (1 −3α/2)2 .
This corollary shows that the simple algorithm sketched above will produce tetra-hedra with a bounded radius-edge ratio. Of all the tetrahedra that fail to have a bounded classical aspect ratio (needles, caps, etc.), the bounded radius-edge ratio eliminates all but the slivers.
Currently there are not any Delaunay based mesh generation codes that can guarantee eliminating slivers.
The development of ideas sketched in this section began with a spacing function ρ and led to a mesh with a bounded radius-edge ratio. It is also possible to do the converse. Theorem 4.4 in section 4 shows that a Delaunay mesh having bounded radius-edge ratio determines a Lipschitz spacing function ρ satisfying the properties stated in the corollary.
3. Covolume approximations.
3.1. Notation. Let Ω⊂R3 be a bounded open set, and consider the problem of finding u : Ω→R such that −∆u = f in Ω, u|∂Ω= g.
To approximate u, Ωis first triangulated, the triangulation being the Delaunay tri-angulation of a set of points {xi}. We denote the length of the Delaunay edge joining vertex xi to xj by hij, and denote by Vi the Voronoi region associated with node xi.
To minimize technical details we assume that the mesh accommodates the boundary in a fashion guaranteeing that the Voronoi Vi corresponding to an interior vertex xi lies entirely within Ω. Each edge xixj of the Delaunay mesh can be associated with (i.e., is dual to) a face common to Vi and Vj which we denote by Aij.
Following MacNeal , discrete approximations of u are constructed by integrating the equation for u over each Voronoi corresponding to an interior vertex: Z Vi f = Z Vi −∆u = Z ∂Vi −∂u ∂n = X j∈Ni Z Aij −∂u ∂n ≃ X j∈Ni |Aij|ui −uj hij .
In the above, Ni is the index set of the nodes connected to xi by an edge, ui is an approximation of u(xi), and |Aij| denotes the surface area of the Voronoi face Aij.
1694 G. MILLER, D. TALMOR, S.-H. TENG, AND N. WALKINGTON Letting fi denote the average value of f over Vi this scheme becomes X j∈Ni |Aij|ui −uj hij = |Vi|fi ∀xi interior, and ui = g(xi) for the boundary vertices.
3.2. Orthogonality and convergence. Nicolaides introduced the following elegant analysis of the above scheme. First, note that the exact solution u of the equation satisfies Z Vi f = Z Vi −∆u = Z ∂Vi ∂u ∂n = X j∈Ni Z Aij −∂u ∂n, |Vi|fi = X j∈Ni |Aij|U (2) ij where U (2) ij is the average value of the flux −∂u/∂n on the face Aij. Upon subtracting this from the equation satisfied by the discrete solution we obtain 0 = X j∈Ni |Aij|(Uij −U (2) ij ) where Uij = (ui −uj)/hij is the discrete flux.
The second step is to consider an arbitrary set of nodal values {vi} that vanish on the boundary. Multiplying the above equation by vi and summing over all of the interior nodes gives 0 = X i X j∈Ni |Aij|(Uij −U (2) ij )vi = X xixj |Aij|(Uij −U (2) ij )(vi −vj) = X xixj |Aij|hij (Uij −U (2) ij )Vij where P xixj indicates summation over the nonboundary edges, and Vij = (vi − vj)/hij. Defining the inner product [., .]W on quantities defined upon the nonboundary edges by [U, V ]W = X xixj |Aij|hij UijVij, we may write [U −U (2), V ]W = 0 where it is understood that U is the discrete flux given by the covolume scheme and the discrete flux V must be the discrete gradient of a piecewise linear function v vanishing on the boundary, i.e., Vij = (vi −vj)/hij.
An error estimate can now follow by recognizing that the above orthogonality relationship states that the covolume approximation is the best projection of the flux U (2) onto the space of fluxes generated by discrete functions vanishing on the boundary. In particular, if we let U (1) ij = (u(xi) −u(xj))/hij be the flux determined by the interpolant of the exact solution, we obtain ∥U −U (1)∥W ≤∥U (2) −U (1)∥W .
RADIUS-EDGE CONDITION IN CONTROL VOLUME METHOD 1695 The following theorem characterizes the size of the approximation error on the right-hand side of this expression.
Theorem 3.1. Let3 u ∈H2(Ω) and U (1) ij = (u(xi) −u(xj))/hij be the approxi-mation of the flux U (2) ij = (−1/Aij) R Aij ∂u/∂n. Then ∥U (2) −U (1)∥W ≤1 2 max xixj 1 + 4(rij/hij)2hij ∥D2u∥L2(Ω), where rij is maximum radius of a control volume having Aij as a face. In particular, if the mesh has a bounded radius-edge ratio the error in the covolume scheme is given bounded by ∥U −U (1)∥W ≤1 2(1 + 4B2)∥D2u∥L2(Ω)h, where h = maxxixj hij.
Proof. Let us write ∥U (2) −U (1)∥2 W = X xixj |Aij|hij µij(u)2 where µij(u) = −1 Aij Z Aij ∂u ∂n −u(xi) −u(xj) hij .
We explicitly compute the term µij(u) for a generic edge hij. It is convenient to put the origin at the middle of hij and let hij lie in the z-axis so that Aij lies in the plane z = 0, and for definiteness let hij point from xi to xj up the z-axis (see Figure 2).
Next, let K+ ij be the cone with vertex xj and base Aij and K− ij be the cone with vertex xi and base Aij, and note that cones corresponding to different Voronoi faces Aij are disjoint, and the union of all such cones is the union of the Voronoi regions Vi corresponding to internal vertices xi (Figure 2).
For the calculations below we drop the subscript ij, so that h = hij, A = Aij, etc. The cone K+ may be parameterized by x = (1 −2η/h)ξ1, y = (1 −2η/h)ξ2, z = η, (ξ1, ξ2) ∈A, 0 ≤η ≤h/2. Note that Z K+ u = Z A Z h/2 0 u(ξ1, ξ2, η) (1 −2η/h)2 dη dξ where dξ = dξ1 dξ2. Also ∂u ∂η = −2ξ1 h ∂u ∂x −2ξ2 h ∂u ∂y + ∂u ∂z , ∂2u ∂η2 = (−2ξ1/h, −2ξ2/h, 1)D2 xuT , ∂2u ∂η2 ≤ 1 + 4(ξ2 1 + ξ2 2)/h2 |D2u|.
3H2(Ω) is the set of functions in L2(Ω) having all second derivatives in L2(Ω).
1696 G. MILLER, D. TALMOR, S.-H. TENG, AND N. WALKINGTON x i x j A ij z=0 K K -+ Fig. 2. Cones constructed in Theorem 3.1.
We now compute Z K+ 1 1 −2(η/h) ∂2u ∂η2 = Z A Z h/2 0 ∂2u ∂η2 (1 −2η/h) dη dξ = Z A ( ∂u ∂η (1 −2η/h)|h/2 0 + 2 h Z h/2 0 ∂u ∂η dη ) dξ = Z A −∂u ∂η (ξ1, ξ2, 0+) + 2 h(u(xj) −u(ξ1, ξ2, 0+)) dξ = Z A −∂u ∂η (ξ1, ξ2, 0+) −2u(ξ1, ξ2, 0+)/h dξ + 2|A|u(xj)/h.
Repeat this computation for the bottom cone using the parameterization x = (1 + 2η/h)ξ1, y = (1 + 2η/h)ξ2, z = η, (ξ1, ξ2) ∈A, −h/2 ≤η ≤0, to obtain Z K− 1 1 + 2(η/h) ∂2u ∂η2 = Z A ∂u ∂η (ξ1, ξ2, 0−) −2u(ξ1, ξ2, 0−)/h dξ + 2|A|u(xi)/h.
Continuity of u guarantees that u(., 0+) = u(., 0−) so that Z K+ 1 1 + 2(η/h) ∂2u ∂η2 − Z K− 1 1 −2(η/h) ∂2u ∂η2 = − Z A ∂u ∂η (ξ1, ξ2, 0−) + ∂u ∂η (ξ1, ξ2, 0+) dξ −2|A|u(xi) −u(xj) h .
Note that ∂u/∂η = 2(ξ1/h)∂u/∂x + 2(ξ2/h)∂u/∂y + ∂u/∂z in K−so that Z K+ 1 1 + 2η/h ∂2u ∂η2 − Z K− 1 1 −2η/h ∂2u ∂η2 = 2 Z A −∂u ∂z |z=0 −2|A|u(xi) −u(xj) h = 2|A| µ(u).
RADIUS-EDGE CONDITION IN CONTROL VOLUME METHOD 1697 It remains to bound the terms involving the second derivatives. Z K+ 1 1 −2(η/h) ∂2u ∂η2 ≤ Z K+ 1 + 4(ξ2 1 + ξ2 2)/h2 1 −2(η/h) |D2u| ≤ 1 + 4(ξ2 1 + ξ2 2)/h2 1 −2(η/h) L2(K+) ∥D2u∥L2(K+) ≤(1/2) p |A|h 1 + 4(r/h)2 ∥D2u∥L2(K+), where r2 = max(ξ1,ξ2)∈A(ξ2 1 + ξ2 2). A similar computation for the lower cone gives |µ(u)| ≤1 2 s h |A| 1 + 4(r/h)2 ∥D2u∥L2(K+∪K−).
Putting it all together gives ∥U −U (1)∥W ≤∥U (2) −U (1)∥W = X xixj |Aij|hij µij(u)2 1/2 ≤(1/2) max xixj (1 + 4(rij/hij)2)hij X xixj ∥D2u∥2 L2(K+ ij∪K− ij) 1/2 ≤(1/2) max xixj (1 + 4(rij/hij)2)hij ∥D2u∥L2(Ω).
3.3. Numerical example. We illustrate the implications of our theorem with a numerical example. Meshes are constructed by dividing the unit cube into a uniform rectangular mesh and then randomly perturbing the points by an amount ǫh. The Delaunay triangulation of such points will have multitudes of slivers on all of the vertical and horizontal planes where the points are almost cocircular.
While the radius-edge ratio of these slivers will be near unity, their classical aspect ratio of diameter to inscribed sphere radius (d/r) will be very large when ǫ is small. Figure 3 tabulates the errors for the solution u(x, y, z) = eπx cos(πy/ √ 2) sin(πz/ √ 2) of −∆u = 0 computed using the finite element and covolume algorithms. Since the finite element solution is the best approximation of the solution in the H1 0(Ω) norm, this is tabulated along with the norm ∥.∥W for the finite element solution.
It is clear that for fixed h the finite element errors for each norm increase as ǫ →0 and the the covolume error is independent of ǫ. Divergence in the H1 0(Ω) norm illustrates that even when the exact solution is interpolated at the nodal values, the presence of slivers causes large errors in the gradient. In order for the finite element method to achieve the smallest H1 0(Ω) error possible it spreads this error among all tetrahedra, causing a failure of convergence in ∥.∥W .
For ǫ fixed, the first-order rate of convergence for the finite element solution in the H1 0(Ω) norm is readily observable (the error decreases in proportion with h). The covolume scheme achieves a higher rate of convergence (almost two) for this example since the mesh is almost uniform. On a uniform mesh the covolume scheme is, in fact, second-order accurate (this can readily be observed in the proof of Theorem 3.1).
1698 G. MILLER, D. TALMOR, S.-H. TENG, AND N. WALKINGTON FEM CoVolume h ǫ d/r |u −uh|H1 ∥Ihu −uh∥W ∥Ihu −uh∥W 1/4 0.01 1126 30.22479 7.696399 0.2425438 0.001 11125 92.97309 8.692323 0.2411516 0.0001 111296 291.6406 8.917005 0.2410718 1/8 0.01 26230 12.98964 4.344101 0.07616430 0.001 166051 35.13775 6.865007 0.07531520 0.0001 1621745 108.1435 7.525118 0.07538803 1/16 0.01 14194 6.023042 2.262189 0.02155074 0.001 105551 15.48025 4.457009 0.02005826 0.0001 1030956 45.91799 6.140328 0.02007369 Fig. 3. Errors for control volume and finite element methods.
4. Poincar´ e inequality. The Poincar´ e inequality states that for each bounded domain Ω⊂Rd there is a constant C > 0 such that ∥u∥L2(Ω) ≤C∥∇u∥L2(Ω) for all sufficiently smooth functions u : Ω→R vanishing on the boundary ∂Ω. MacNeal has shown that in two dimensions the W norm (defined in section 3.2) is equal to the H1(Ω) norm, in the sense that the piecewise linear extension, ˜ u, of u defined on the vertices of a triangulation satisfies ∥u∥2 W = R Ω|∇˜ u|2. The Poincar´ e inequality, R Ωu2 ≤C R Ω|∇u|2 then implies that convergence of a discrete solution in the W norm implies convergence in L2(Ω). In three dimensions the W norm is no longer equivalent to the H1(Ω) norm, so convergence in L2(Ω) doesn’t follow directly from the Poincar´ e inequality. Below we establish a discrete Poincar´ e inequality for the W norm which establishes convergence in L2(Ω) of the functions that are piecewise constant on the Voronoi cells and take the nodal values of u.
Definition 4.1. Given a real valued function u on the mesh vertices, the piece-wise constant extension ˆ u : Ω→R is defined to be the function equal to ui over the Voronoi cell Vi.
The discrete L2(Ω) norm of u is then interpreted to mean ∥ˆ u∥2 L2(Ω) = P i R Vi u2 i = P i |Vi|u2 i .
Theorem 4.2 (discrete Poincar´ e inequality).
Let u be a function defined on the vertices of a triangulation of the bounded domain Ω⊂Rd that vanishes on the boundary vertices. Then there exists C > 0 depending only upon Ωand the radius-edge ratio of the mesh such that ∥u∥L2(Ω) ≤C∥u∥W .
The theorem statement can be rewritten explicitly as X i |Vi|u2 i ≤C X xixj |Aij| hij (ui −uj)2 (1) for all mesh functions vanishing on the boundary. We adopt here the notation of the previous section. In particular, xixj indexes the edges in the mesh whose interiors lie in Ω, hij is the edge length, and Aij is the area of the dual face.
Note that the left-hand side of this equation is the L2(Ω) norm of the piecewise constant function on the Voronoi’s, and in two dimensions the right-hand side is ∥∇uh∥2 L2(Ω) where uh is the piecewise linear function taking on the corresponding nodal values. Thus in two dimensions, with regular meshes, this inequality will follow from approximation theory. We show that this inequality continues to hold for meshes in two or three dimensions under the stated assumptions; in particular, even if slivers are present.
RADIUS-EDGE CONDITION IN CONTROL VOLUME METHOD 1699 4.1. Graphs and meshes. The meshes we consider are constructed from De-launay diagrams, which are determined by a set of vertices X = {xi · · · xn}, a set of tetrahedra whose 1-skeleton is the edge set E, and a set of Voronoi cells V . We define the following weighted graph associated with the mesh.
Definition 4.3. 1. Let G = G(X, E, V ) denote the graph constructed from the Delaunay diagram D = (X, E, V ) as follows: X: The vertex set of the graph is identified with the vertex set of the Delaunay diagram.
E: Edges (i, j) in the graph correspond to interior Delaunay edges and have weight kij = |Aij|/hij. Recall hij is the length of the Delaunay edge between two vertices, Aij is the Voronoi face shared by Vi and Vj.
V : We assign the Delaunay cell volume to each graph vertex.
2. The mass matrix M is a diagonal matrix containing the (truncated) Voronoi volumes on the diagonal. (Recall that Voronoi regions on the boundary may be infinite, so we only consider the part in Ω. We treat the areas used for computing the edge weights in a similar fashion. These terms are inconsequential since we only consider functions vanishing on the boundary.) 3. The Laplacian of the graph G, denoted by K, is the matrix having offdiagonal entries Kij = −|Aij|/hij = −kij if (i, j) is an edge in G and zero otherwise and the diagonals are given by Kii = P j̸=i kij.
The discrete Poincar´ e inequality then states that the eigenvalues (on the space of vectors having zero boundary components) of the generalized eigenvalue problem Ku = λMu are bounded below by a constant c > 0, or alternatively that there is a constant c > 0 such that K −cM ≥0 (positive semidefinite).
4.2. Geometric properties. The meshes we consider are Delaunay diagrams D = (X, E, V ) where X ⊂¯ Ω⊂Rd. We assume the mesh tetrahedra are of bounded radius-edge ratio.
Delaunay diagrams of bounded radius-edge ratio have several important geometric properties. Below we recall some of these properties which were established in .
Recall that we denote the edge joining two vertices xi and xj by xixj and the length of such an edge by hij and that the set of vertices connected to vertex xi by an edge is denoted by Ni. Also, we adopt the notation that C and c are positive constants which may differ from occurrence to occurrence. In general, c will be a lower bound and C an upper bound, and these constants depend only upon the mesh through the Lipschitz constant of the spacing function, ρ, appearing in the following theorem.
Theorem 4.4.
Let D = (X, E, V ) be a bounded radius-edge ratio Delaunay diagram. The following is true.
A1. There is a function ρ : Ω→R+ with Lipschitz constant bounded by one and constants c, C > 0 such that for each internal vertex xi ∈X chij ≤ρ(x) ≤Chij ∀j ∈Ni, x ∈Vi and for all edges xixj chij ≤ρ(x) ≤Chij ∀x ∈xixj.
A2. There is a constant θ > 0 such that each region Vi∩Ω, where Vi ∈V , contains a ball of radius θρ(xi) centered at xi and is contained in a concentric ball of radius ρ(xi)/θ.
A3. There is a constant C > 0 such that each vertex xi ∈X has at most C neighbors in the diagram, i.e., the associated graph has bounded degree.
1700 G. MILLER, D. TALMOR, S.-H. TENG, AND N. WALKINGTON A4. There is a constant C > 0 such that the area of the Voronoi face dual to the vertices xi, xj ∈X is bounded from above by Chd−1 ij .
All of the constants depend upon the mesh solely through the the Lipschitz constant of the spacing function ρ.
The function ρ is referred to as the spacing function of the mesh, as it describes the typical distance between vertices. This function plays an important role in the proofs below as it captures many of the mesh properties.
Our arguments below are graph theoretic and don’t use the fact that we’re dealing with a Voronoi/Delaunay triangulation; any decomposition of Ωinto volumes Vi and edges connecting them that satisfy the geometric properties listed in Theorem 4.4 would suffice. We therefore generalize the notion of the bounded radius-edge ratio Delaunay diagram as follows.
Definition 4.5 (Well-shaped diagram). A diagram G = (X, E, V ) is a set of vertices X, and a set of cells V such that vertex xi ∈Vi, and the edge set E corresponds to the neighborhood structure of the cells. The diagram is well shaped if properties A1–A4 hold.
Notation. Since the edges of a well-shaped diagram don’t necessarily correspond to line segments in Rd we will denote an edge connecting xi, xj ∈X by (i, j) and use the notation xixj to suggest a line segment in Rd.
4.3. Comparison. Our proof of the discrete Poincar´ e inequality will parallel the proof of its continuous counterpart. The continuous proof considers a large cube Q containing Ω, and since all functions defined on Ωvanishing on the boundary can be extended by zero to all of Q, a Poincar´ e inequality on Q will imply a Poincar´ e inequality on Ω. We too will embed Ωinto a large cube, and in this section we show that any two meshes on Ωand Q satisfying the geometric properties of Theorem 4.4 with the same spacing function ρ can be suitably compared. The next section will then show that for each spacing function there exists a canonical mesh on the cube for which the Poincar´ e inequality holds.
Definition 4.6 (Path embedding). 1. The graph G1 = (X(1), E(1), V (1)) is path embedded in G2 = (X(2), E(2), V (2)) if there exists a function φ from X(1) to X(2) and a mapping p from the edges of G1 into paths in G2, such that for each edge (m, n) in G1, p(m, n) is a path in G2 from φ(m) to φ(n).
2. The embedding has bounded dilation if there is a constant Ck > 0 such that for every edge (m, n) ∈E(1), X (i,j)∈p(m,n) 1/k(2) ij ≤Ck/k(1) mn where k(1) mn and k(2) ij are the edge weights of G1 and G2 respectively (i.e., the conductivity of the path is comparable to the conductivity of the edge).
3. The embedding has bounded congestion if there is a constant Cp > 0 such that for each edge (i, j) in G2 cij = X p(m,n)∋(i,j) 1 ≤Cp.
Lemma 4.7. Let (φ, p) be a path embedding of G1 into G2 with bounded dilation and congestion, and let u be a real valued function on the vertices of G2 and u ◦φ be the induced function on the vertices of G1. Then there is a constant C > 0 such that (u ◦φ)T K(1)(u ◦φ) ≤CuT K(2)u.
RADIUS-EDGE CONDITION IN CONTROL VOLUME METHOD 1701 Proof.
(u ◦φ)T K(1)(u ◦φ) = X (m,n) k(1) mn(uφ(m) −uφ(n))2 = X (m,n) k(1) mn X (i,j)∈p(m,n) (ui −uj) 2 ≤ X (m,n) k(1) mn X (i,j)∈p(m,n) 1/k(2) ij X (i,j)∈p(m,n) k(2) ij (ui −uj)2 ≤Ck X (m,n) X (i,j)∈p(m,n) k(2) ij (ui −uj)2 ≤CkCp X (i,j) k(2) ij (ui −uj)2.
We plan to embed a cube Q into Ωand need a way of extending functions on Ω that vanish on the boundary to all of Q. A convenient way of doing this is to add a “super” node x0 to the vertex set X of a well-shaped diagram G = (X, E, V ) of Ω, and to augment the edge set E by connecting every boundary vertex to the super node, and to assign large (say infinite) weights to these edges so that they never lower the conductivity of a path. Also, we may put multiple (say infinite) edges between a boundary node and the super node to guarantee that these edges never increase the congestion. It is convenient to associate the region V0 = Q \ Ωto the super node x0.
Notation.
Given a well-shaped diagram G = (X, E, V ) of Ω, the augmented diagram is the diagram formed by adjoining a super node to X and the associated edges and volume to E and V .
Property A4 allows the edge weights Aij/hij to be arbitrarily small. In partic-ular, a sliver in a Delaunay diagram corresponds to a small weight. All the paths containing the “light” edge will have small conductivity. The following lemma states that this situation is not generic and that each edge can be replaced by a path of high conductivity, which can then be used to construct suitable path embeddings.
Lemma 4.8. Let G = (X, E, V ) be an augmented well-shaped diagram of Ω⊂Rd.
Then there exists a path embedding (φ, p) of G into itself of bounded congestion such that the conductivity of each path in the image is bounded below by a multiple of hd−2 ij , i.e., for each edge (i, j) ∈E, X (m,n)∈p(i,j) 1/kmn ≤C/hd−2 ij .
Proof. Let us consider a typical edge xixj of length hij and define U to be the neighborhood of xixj consisting of points whose distance from xixj is no more than min(c/2, θc)hij where c and θ are the constants guaranteed by properties A1 and A2 of Theorem 4.4. We first consider the case when U lies entirely within Ω. The Lipschitz hypotheses on ρ guarantee that c/2hij ≤ρ(x) ≤(C + c/2)hij for all x ∈U.
Step 1: The number of cells from V that intersect U is bounded by a constant κ ≥0.
1702 G. MILLER, D. TALMOR, S.-H. TENG, AND N. WALKINGTON We use a volume argument.
If x ∈Vk ∩U, property A1 shows that c/C ≤ ρ(xk)/ρ(x) ≤C/c.
Since ρ(x) is bounded above and below by multiples of hij, property A2 implies that Vk contains a ball having volume bounded below by ˆ chd ij and the diameter of Vk is bounded by ˆ Chij. If ˜ U is the set of points having distance no more than ˆ Chij from xixj, then ˜ U contains all of the cells intersecting U and has volume bounded by a multiple of hd ij. It follows that the number of cells meeting U is bounded by | ˜ U|/ˆ chd ij.
Step 2: There exists a line segment from Vi to Vj in U, parallel to xixj, that meets Voronoi faces having areas bounded below by a multiple of hd−1 ij .
We use an area argument similar in spirit to the previous volume argument.
Consider the collection of line segments in U that start in Vi and end in Vj and are parallel to xixj. By construction, the cross-sectional area of this collection is bounded below by a multiple of hd−1 ij , say ˆ chd−1 ij . Letting δ be the maximal degree of any vertex guaranteed by property A3, the total number of faces that meet U is bounded by κδ.
It follows that some path must encounter only faces having areas bounded below by (ˆ c/κδ)hd−1 ij (otherwise every face would be small and their totality wouldn’t exhaust the cross section).
We now observe that these conclusions don’t change if a Ωis not convex and a portion of U is exterior to Ω. Since the volume of U ∩Ωis trivially smaller than that of U, the number of cells meeting U only decreases in this case. Similar reasoning holds for the area argument. Note that all of the line segments do begin and end in Ωand that we consider all faces on the boundary to have large (infinite) area.
To establish the theorem we let p(i, j) be the path in E formed by connecting the Voronoi centers in the order encountered by a segment in U which only intersects “large” Voronoi faces. This path has length at most κ, and since ρ is bounded above and below by a constant times hij so too are the lengths of all of edges in p(i, j), which implies a bound on the conductivity of the path. Finally, note that an edge in E will only be on a path p(i, j) for vertices xi and xj that can be reached along no more than κ edges. Since the degree of each vertex is bounded by the constant δ, the congestion of any edge is less than δκ.
The paths guaranteed by this technical lemma enable us to construct path em-beddings between diagrams in the following theorem.
Theorem 4.9. Let G1 = (X(1), E(1), V (1)) be a well-shaped diagram for a cube Q containing Ωand G2 = (X(2), E(2), V (2)) be the augmentation of a well-shaped diagram of Ω, each having the same spacing function ρ. Then there exists a path embedding G1 into G2 with congestion and dilation bounded by constants depending upon the Lipschitz constant of ρ.
Proof.
Our proof proceeds by explicitly constructing a path embedding (φ, p) of G1 into G2, and then verifying that the congestion and dilation are appropriately bounded. The embedding is determined by the following construction (see Figure 4): 1. φ(m) = i where V (2) i ∈V (2) is a cell intersecting the cell V (1) m ∈V (1). (Note that i is not uniquely specified as V (1) m can intersect several cells of V (2).) 2. For each edge (m, n) ∈G1 consider the three piecewise linear paths: S1 = (x(2) φ(m), ˆ x0, x(1) m ), S2 = (x(1) m , x(1) n ), and S3 = (x(1) n , ˆ x1, x(2) φ(n)) where ˆ x0 ∈V (1) m ∩ V (2) φ(m) and ˆ x1 ∈V (1) n ∩V (2) φ(n) (see Figure 4). Let ℓ1 = φ(m), ℓ2, . . . , ℓk−1, ℓk = φ(n) index the cells of V (2) encountered along the three segments S1, S2, S3 in that order. Then p(m, n) is the path in G2 formed by the union of the RADIUS-EDGE CONDITION IN CONTROL VOLUME METHOD 1703 x1 m x1 n x0 x1 φ x φ x (m) 2 2 (n) Fig. 4. Path constructed in Theorem 4.9.
heavily weighted paths from x(2) ℓi to x(2) ℓi+1 constructed in Lemma 4.8.
We first note that this construction is well defined since the super node has the region Q associated with it, so that every cell Vm ∈V (1) meets some cell of V (2), and paths that leave Ωall go to the super node and can reenter anywhere from the super node.
Next, observe that every cell along the path p(m, n) in V (2) intersects either V (1) m , V (1) n , or the edge joining them. Property A1 shows that chmn ≤ρ(x) ≤Chmn for every x ∈V (1) m ∪V (1) n , which implies (c/C2)hmn ≤ρ(x) ≤C2/c)hmn for x ∈V (2) ℓi , i = 1, 2, . . . , k.
An application of the volume argument used in the first step of Lemma 4.8 gives a bound on the length of the path p(m, n). Briefly, the distance from any point in one of the cells {V (2) ℓi }k i=1 to S2 is bounded by a multiple of hmn. This neighborhood of S2 will have volume bounded by ˆ Chd mn, and since each of the cells V (2) ℓi contains a ball having volume bounded below by a multiple of hd mn, say ˆ chmn, it follows that there are no more than ˆ C/ˆ c such cells. Since the heavily weighted paths constructed in Lemma 4.8 are all of bounded length, the length of p(m, n) is bounded by at most a multiple of ˆ C/ˆ c.
This immediately leads to a bound on the congestion and dilation. Property A3 bounds the degree of any vertex by a constant δ, so the congestion is bounded by the Lδ max, where Lmax is the maximal length of any path. Similarly, the conductivity on each edge is bounded below by a constant of the form ˆ chd−1 mn and putting Lmax edges in the series doesn’t decrease this by more than a factor of 1/Lmax.
Corollary 4.10. Let G1 = (X(1), E(1), V (1)) and G2 = (X(1), E(2), V (2)) be as in Theorem 4.9, and suppose that Q is sufficiently large to guarantee that no boundary cell of V (1) meets a boundary cell of V (2). If G1 satisfies the Poincar´ e inequality, then so too does G2 Proof. Let u be a real valued function on X(2) that vanishes on the boundary (and, in particular on the super node x0). When constructing the path embedding G1 into G2 given by the theorem, select φ(m) in step 1 to correspond to the vertex 1704 G. MILLER, D. TALMOR, S.-H. TENG, AND N. WALKINGTON Algorithm: Canonical Mesh Construction Input: ρ, a spacing function.
Output: A diagram G = (X, E, V ).
Method: 1. Construct a balanced oct-tree: (a) Let Q be a cube containing Ω.
(b) If ℓQ denotes the side length of a cube Q, while ℓQ > minQ ρ for any cube, Q, sub-divide Q into 2d cubes.
(c) While there is a cube Q sharing a (d −1) face with cube ˆ Q having ℓQ > 2ℓˆ Q, sub-divide Q.
2. Assignment of the diagram G = (X, E, V ).
(a) The vertices, X, of G will be indexed by the cubes of the partition, and the cells, V , are the associated cubes.
(b) G has an edge between two vertices xi and xj if and only if their associated cubes share a d−1 face, and in this instance the edge weight is kij = min(ℓd−2 i , ℓd−2 j ), where ℓi is the side length of the cube Qi.
(c) The mass Mi associated with a particular cell is the volume of the associated cube.
Fig. 5.
Constructing a canonical mesh.
having maximal absolute value of u of the available choices. It follows that |V (2) i |u2 i ≤ X m |V (1) m ∩V (2) i |u2 φ(m), and summing over i gives uT M (2)u = X i |V (2) i |u2 i ≤ X m |V (1) m |u2 φ(m) = (u ◦φ)T M (1)(u ◦φ).
Since all boundary vertices of X(1) are mapped by φ to the super node x(2) 0 , it follows that u ◦φ : X(1) →R vanishes on the boundary vertices, and since G1 satisfies a Poincar´ e inequality we have uT M (2)u ≤(u ◦φ)T M (1)(u ◦φ) ≤C(u ◦φ)T K(1)(u ◦φ) and the result follows upon application of Lemma 4.7.
4.4. Construction of a graph satisfying a Poincar´ e inequality. In order to apply the results of the previous section to establish a Poincar´ e inequality for bounded well-shaped diagrams it is necessary to exhibit an instance of a canonical mesh satisfying a Poincar´ e inequality. We let ρ : Q →R+ be a specified Lipschitz spacing function, fixed during this discussion, and Q a cube in Rd.
Figure 5 exhibits the oct-tree construction for a canonical mesh, and the bal-ancing step is illustrated in Figure 6. The oct-tree construction is well known, and the following lemma shows that our particular construction generates a well-shaped diagram with spacing function ρ.
RADIUS-EDGE CONDITION IN CONTROL VOLUME METHOD 1705 Fig. 6. Balancing a quad tree.
Lemma 4.11. The oct-tree diagram generated by the algorithm of Figure 5 is a well-shaped diagram with spacing function ρ.
Proof. Properties A3 and A4 follow immediately from the oct-tree construction.
For properties A1 and A2, we only have to show the existence of constants c, C > 0 such that each cube Q in the balanced oct-tree generated by the algorithm of Figure 5 satisfies cℓ(Q) ≤ρ(x) ≤Cℓ(Q), x ∈Q.
Let Qc be a leaf cube at the end of step 1(b), and let Q be its parent cube. Since Q was divided there is some point x ∈Q such that ρ(x) ≤ℓ(Q), and since Qc was not divided, ℓ(Qc) = ℓ(Q)/2 < ρ(y) for all y ∈Qc. The distance between x and y is bounded by |x −y| ≤ √ dℓ(Q), and since ρ is α-Lipschitz, ρ(y) ≤ρ(x) + α √ dℓ(Q). In summary, by the end of step 1(b) of the algorithm, ℓ(Qc) ≤ρ(y) ≤2(α √ d + 1)ℓ(Qc), y ∈Qc.
We now show there are constants c and C such that after the balancing step 1(c) of the algorithm: cℓ(Q) ≤ρ(y) ≤Cℓ(Q) for y ∈Q. Since subdividing a cube doesn’t affect the lower bound on ρ, we set c = 1. To establish the upper bound we use an induction argument. Our induction hypothesis is that prior to any balancing split, all cubes Q satisfy ρ(y) ≤Cℓ(Q) for all y ∈Q (C is explicitly computed below).
We have shown that step 1(b) produces cubes satisfying the inductive assumption provided C ≥2(α √ d + 1), establishing the initial inductive step. Let Q be a cube split during the balancing step, let Qc be one of its children, and let ˆ Q be a neighbor causing the split. It follows that ℓ( ˆ Q) < ℓ(Q)/2, and if x ∈ˆ Q the induction hypothesis guarantees that ρ(x) ≤Cℓ( ˆ Q). The distance between x ∈ˆ Q and any y ∈Q is bounded by |x −y| ≤ √ d(ℓ( ˆ Q) + ℓ(Q)) ≤(3 √ d/2)ℓ(Q).
Since ρ is Lipschitz with constant α, ρ(y) ≤ρ(x) + α(3 √ d/2)ℓ(Q) ≤(C/2)ℓ(Q) + α(3 √ d/2)ℓ(Q) ≤ C/4 + α3 √ d/4 ℓ(Qc).
1706 G. MILLER, D. TALMOR, S.-H. TENG, AND N. WALKINGTON Selecting C such that C/4 + α3 √ d/4 ≤C completes the inductive step.
The construction below mimics the proof of the Poincar´ e inequality for the con-tinuous case which proceeds by integrating along a path from the boundary to each point in the domain. In the discrete case these paths must lie in the edge set of our graph and no one path can be used to excess (i.e., the congestion must be controlled; see Definition 4.6). The following is a generalization of path embedding of the previ-ous section, where the edges we embed are from each internal vertex to the boundary vertices.
Definition 4.12 (boundary embedding). Let G = (X, E, V ) be a graph with edge weights denoted by kij. A boundary embedding is a collection of graphs {Gi}xi∈X each having the same vertex and edge structure of G and having edge weights w(i) mn ≥0.
• The congestion of the boundary embedding is the maximum over all of the edges (m, n) ∈E of the sum P i w(i) mn.
• For xi ∈X the effective conductivity of Gi is the largest constant c = c(i) such that cu2 i ≤ X (m,n) w(i) mnkmn(um −un)2 for arbitrary nodal values {um} vanishing on the boundary.
The role of these definitions becomes apparent from the following lemma.
Lemma 4.13. Let G = (X, E, V ) be a well-shaped diagram of Q with mass matrix M and Laplacian K. Let {Gi} be a boundary embedding having congestion bounded by C, and suppose that for each Gi the effective conductivity is bounded below by Mi.
Then the Poincar´ e inequality uT Mu ≤CuT Ku holds ∀u : X →R vanishing on the boundary.
Proof. The bound on the conductivities guarantees that uT Mu = X i Miu2 i ≤ X i X (m,n) w(i) mnkmn(um −un)2 ≤ X (m,n) X i w(i) mn !
kmn(um −un)2 ≤CuT Ku.
We next construct the subgraphs Gi alluded to in Lemma 4.13. In the balanced oct-tree we define cube Qi to be above cube Qj if the intersection of their projections onto the plane xd = 0 have nonzero (d −1) area. The horizontal faces of a cube are the two faces having xd constant. The key geometric fact used below is that if the intersection of the projections of two cubes has nonzero area, then the projection of the smaller cube must lie entirely within the projection of the larger cube (and the projections coincide if the cubes have the same side lengths).
Consider a vertex xi in G and define Gi to have the same vertex and edge sets as G and assign weights to the edges of Gi as follows: An edge (m, n) has nonzero weight if and only if it is associated with a horizontal d −1 face and if Qi is above Qm and above Qn. Without loss of generality we order the vertices so that ℓ(Qm) ≤ℓ(Qn) RADIUS-EDGE CONDITION IN CONTROL VOLUME METHOD 1707 and assign the edge weights by w(i) mn = ( ℓ(Qi) if ℓ(Qi) > ℓ(Qm), ℓ(Qi) ℓ(Qm) d−1ℓ(Qi) otherwise.
Theorem 4.14. The graph given by the above oct-tree construction satisfies a Poincar´ e inequality.
Proof. We show that the graphs {Gi} constructed above satisfy the hypotheses of Lemma 4.13.
Step 1: The congestion is bounded by 2L where L is the side length of Q.
A fixed (vertical) edge (xm, xn) in G, has a nonzero weight w(i) mn if and only if the cube Qi lies above the cells Vm and Vn. Without loss of generality, assume ℓ(Qm) ≤ℓ(Qn). Let I index all the cells Qi above Qm and Qn. The congestion is then given by cmn = X i∈I w(i) mn = X i∈I,ℓ(Qi)≥ℓ(Qm) w(i) mn + X i∈I,ℓ(Qi)<ℓ(Qm) w(i) mn = X i∈I,ℓ(Qi)≥ℓ(Qm) ℓ(Qi) + X i∈I,ℓ(Qi)<ℓ(Qm) ℓ(Qi)d/ℓ(Qm)d−1.
The first summand is bounded by L since the larger cells must be stacked one on top of the other. The second summand is also less than L since the sum gives the volume of the smaller cells above Qm divided by the cross-sectional area of Qm.
Step 2: Let L be the side length of Q; then the effective conductivity of Gi is bounded the below by Mi = |Vi|.
Let ℓbe the minimum side length of the cubes below Qi (which may be ℓ(Qi)), and think of the area of the projection of the cube Qi onto xd = 0 to be divided into squares of length ℓ. The oct-tree construction guarantees that the projection of a cube Q below Qi with ℓ(Q) ≤ℓ(Qi) will be the union of such squares, and if Q is below Qi and ℓ(Q) ≥ℓ(Qi), then the intersection of the projections of Qi and Q is equal to the projection of Qi.
For each square S in the projection, consider the path from the boundary to Qi formed by connecting the cubes below Qi that meet the cylinder S × R.
For convenience label the vertices m0, m1, . . . , mI so that Qm0 is a boundary cube and Qi = QmI. Then |S|u2 i = |S| I X n=1 (umn −umn−1) !2 ≤ I X n=1 hmnmn−1 I X n=1 (|S|/hmnmn−1)(umn −umn−1)2 ≤L I X n=1 (|S|/hmnmn−1)(umn −umn−1)2, where L is the side length of Q. Multiplying by ℓ(Qi) and summing this inequality 1708 G. MILLER, D. TALMOR, S.-H. TENG, AND N. WALKINGTON over all squares gives Miu2 i ≤L X S Is X n=1 (ℓ(Qi)|S|/hmnmn−1)(umn −umn−1)2, where hmnmn−1 = min(ℓ(Qmn), ℓ(Qmn−1)). If we define Amnmn−1 = min(hmnmn−1, ℓ(Qi))d−1, interchanging the order of summation gives Miu2 i ≤L X (ℓ(Qi)Amnmn−1/hmnmn−1)(umn −umn−1)2, where the summation is over all of the horizontal faces below Qi. The weights w(i) mn were chosen to satisfy w(i) mnkmn = ℓ(Qi)Amnmn−1/hmnmn−1, hence Miu2 i ≤L X (m,n)∈Gi w(i) mnkmn(um −un)2.
REFERENCES I. Babuˇ ska and A. K. Aziz, On the angle condition in the finite element method, SIAM J.
Numer. Anal., 13 (1976), pp. 214–226.
M. Bern, D. Eppstein, and J. R. Gilbert, Provably good mesh generation, in 31st Annual Symposium on Foundations of Computer Science, St. Louis, MO, IEEE Press, Piscataway, NJ, 1990, pp. 231–241.
L. Chew, Guaranteed-Quality Triangular Meshes, Technical Report CS 89-983, Cornell Uni-versity, Ithaca, NY, 1989.
D. Du and F. Hwang, Computing in Euclidean Geometry, Lecture Notes Ser. Comput. 1, World Scientific, River Edge, NJ, 1992.
R. H. MacNeal, An asymmetrical finite difference network, Quart. Appl. Math., 11 (1953), pp. 295–310.
G. L. Miller, D. Talmor, S. Teng, and N. J. Walkington, A Delaunay based numerical method for three dimensions: Generation, formulation and partition, in Proceedings of the 27th Annual ACM Symposium on Theory of Computing, Las Vegas, NV, ACM Press, NY, 1995.
G. L. Miller, D. Talmor, S. Teng, N. J. Walkington, and H. Wang, Control volume meshes using sphere packing: Generation, refinement and coarsening, in 5th International Meshing Round Table, Pittsburgh, PA, Sandia National Labs, Albuquerque, NM, 1996, pp. 47–62.
S. A. Mitchell and S. A. Vavasis, Quality mesh generation in three dimensions, in Proceed-ings of the ACM Computational Geometry Conference, Berlin, Germany, ACM Press, NY, 1992, pp. 212–221 (also appeared as Cornell C.S. TR 92-1267).
R. A. Nicolaides, Direct discretization of planar div–curl problems, SIAM J. Numer. Anal., 29 (1992), pp. 32–56.
J. Ruppert, A new and simple algorithm for quality 2-dimensional mesh generation, in Third Annual ACM-SIAM Symposium on Discrete Algorithms, Philadelphia, PA, 1992, pp. 83– 92. |
190328 | https://www.cimat.mx/~gil/docencia/2017/mate_elem/[Rosen]Discrete_Mathematics[pp125-6].pdf | 2.1 Sets 125 Truth Sets and Quantifiers We will now tie together concepts from set theory and from predicate logic. Given a predicate P , and a domain D, we define the truth set of P to be the set of elements x in D for which P (x) is true. The truth set of P(x) is denoted by {x ∈D | P (x)}.
EXAMPLE 23 What are the truth sets of the predicates P(x), Q(x), and R(x), where the domain is the set of integers and P (x) is “|x| = 1,” Q(x) is “x2 = 2,” and R(x) is “|x| = x.” Solution: The truth set of P , {x ∈Z | |x| = 1}, is the set of integers for which |x| = 1. Because |x| = 1 when x = 1 or x = −1, and for no other integers x, we see that the truth set of P is the set {−1, 1}.
The truth set of Q, {x ∈Z | x2 = 2}, is the set of integers for which x2 = 2. This is the empty set because there are no integers x for which x2 = 2.
The truth set of R, {x ∈Z | |x| = x}, is the set of integers for which |x| = x. Because |x| = x if and only if x ≥0, it follows that the truth set of R is N, the set of nonnegative integers.
▲ Note that ∀xP (x) is true over the domain U if and only if the truth set of P is the set U.
Likewise, ∃xP (x) is true over the domain U if and only if the truth set of P is nonempty.
Exercises 1. List the members of these sets.
a) {x | x is a real number such that x2 = 1} b) {x | x is a positive integer less than 12} c) {x | x is the square of an integer and x < 100} d) {x | x is an integer such that x2 = 2} 2. Use set builder notation to give a description of each of these sets.
a) {0, 3, 6, 9, 12} b) {−3, −2, −1, 0, 1, 2, 3} c) {m, n, o, p} 3. For each of these pairs of sets, determine whether the first is a subset of the second, the second is a subset of the first, or neither is a subset of the other.
a) the set of airline flights from NewYork to New Delhi, the set of nonstop airline flights from New York to New Delhi b) the set of people who speak English, the set of people who speak Chinese c) the set of flying squirrels, the set of living creatures that can fly 4. For each of these pairs of sets, determine whether the first is a subset of the second, the second is a subset of the first, or neither is a subset of the other.
a) the set of people who speak English, the set of people who speak English with an Australian accent b) the set of fruits, the set of citrus fruits c) the set of students studying discrete mathematics, the set of students studying data structures 5. Determine whether each of these pairs of sets are equal.
a) {1, 3, 3, 3, 5, 5, 5, 5, 5}, {5, 3, 1} b) {{1}}, {1, {1}} c) ∅, {∅} 6. Suppose that A = {2, 4, 6}, B = {2, 6}, C = {4, 6}, and D = {4, 6, 8}. Determine which of these sets are subsets of which other of these sets.
7. For each of the following sets, determine whether 2 is an element of that set.
a) {x ∈R | x is an integer greater than 1} b) {x ∈R | x is the square of an integer} c) {2,{2}} d) {{2},{{2}}} e) {{2},{2,{2}}} f) {{{2}}} 8. For each of the sets in Exercise 7, determine whether {2} is an element of that set.
9. Determine whether each of these statements is true or false.
a) 0 ∈∅ b) ∅∈{0} c) {0} ⊂∅ d) ∅⊂{0} e) {0} ∈{0} f) {0} ⊂{0} g) {∅} ⊆{∅} 10. Determine whether these statements are true or false.
a) ∅∈{∅} b) ∅∈{∅, {∅}} c) {∅} ∈{∅} d) {∅} ∈{{∅}} e) {∅} ⊂{∅, {∅}} f) {{∅}} ⊂{∅, {∅}} g) {{∅}} ⊂{{∅}, {∅}} 11. Determine whether each of these statements is true or false.
a) x ∈{x} b) {x} ⊆{x} c) {x} ∈{x} d) {x} ∈{{x}} e) ∅⊆{x} f) ∅∈{x} 12. Use aVenn diagram to illustrate the subset of odd integers in the set of all positive integers not exceeding 10.
126 2 / Basic Structures: Sets, Functions, Sequences, Sums, and Matrices 13. Use a Venn diagram to illustrate the set of all months of the year whose names do not contain the letter R in the set of all months of the year.
14. Use a Venn diagram to illustrate the relationship A ⊆B and B ⊆C.
15. Use a Venn diagram to illustrate the relationships A ⊂B and B ⊂C.
16. Use a Venn diagram to illustrate the relationships A ⊂B and A ⊂C.
17. Suppose that A, B, and C are sets such that A ⊆B and B ⊆C. Show that A ⊆C.
18. Find two sets A and B such that A ∈B and A ⊆B.
19. What is the cardinality of each of these sets?
a) {a} b) {{a}} c) {a, {a}} d) {a, {a}, {a, {a}}} 20. What is the cardinality of each of these sets?
a) ∅ b) {∅} c) {∅, {∅}} d) {∅, {∅}, {∅, {∅}}} 21. Find the power set of each of these sets, where a and b are distinct elements.
a) {a} b) {a, b} c) {∅, {∅}} 22. Can you conclude that A = B if A and B are two sets with the same power set?
23. How many elements does each of these sets have where a and b are distinct elements?
a) P({a, b, {a, b}}) b) P({∅, a, {a}, {{a}}}) c) P(P(∅)) 24. Determine whether each of these sets is the power set of a set, where a and b are distinct elements.
a) ∅ b) {∅, {a}} c) {∅, {a}, {∅, a}} d) {∅, {a}, {b}, {a, b}} 25. Prove that P(A) ⊆P(B) if and only if A ⊆B.
26. Show that if A ⊆C and B ⊆D, then A × B ⊆C × D 27. Let A = {a, b, c, d} and B = {y, z}. Find a) A × B.
b) B × A.
28. What is the Cartesian product A × B, where A is the set of courses offered by the mathematics department at a university and B is the set of mathematics professors at this university? Give an example of how this Cartesian product can be used.
29. What is the Cartesian product A × B × C, where A is the set of all airlines and B and C are both the set of all cities in the United States? Give an example of how this Cartesian product can be used.
30. Suppose that A × B = ∅, where A and B are sets. What can you conclude?
31. Let A be a set. Show that ∅× A = A × ∅= ∅.
32. Let A = {a, b, c}, B = {x, y}, and C = {0, 1}. Find a) A × B × C.
b) C × B × A.
c) C × A × B.
d) B × B × B.
33. Find A2 if a) A = {0, 1, 3}.
b) A = {1, 2, a, b}.
34. Find A3 if a) A = {a}.
b) A = {0, a}.
35. How many different elements does A × B have if A has m elements and B has n elements?
36. How many different elements does A × B × C have if A has m elements, B has n elements, and C has p elements?
37. How many different elements does An have when A has m elements and n is a positive integer?
38. Show that A × B ̸= B × A, when A and B are nonempty, unless A = B.
39. Explain why A × B × C and (A × B) × C are not the same.
40. Explain why (A × B) × (C × D) and A × (B × C) × D are not the same.
41. Translate each of these quantifications into English and determine its truth value.
a) ∀x∈R (x2 ̸= −1) b) ∃x∈Z (x2 = 2) c) ∀x∈Z (x2 > 0) d) ∃x∈R (x2 = x) 42. Translate each of these quantifications into English and determine its truth value.
a) ∃x∈R (x3 = −1) b) ∃x∈Z (x + 1 > x) c) ∀x∈Z (x −1 ∈Z) d) ∀x∈Z (x2 ∈Z) 43. Find the truth set of each of these predicates where the domain is the set of integers.
a) P(x): x2 < 3 b) Q(x): x2 > x c) R(x): 2x + 1 = 0 44. Find the truth set of each of these predicates where the domain is the set of integers.
a) P (x): x3 ≥1 b) Q(x): x2 = 2 c) R(x): x < x2 ∗45. The defining property of an ordered pair is that two or-dered pairs are equal if and only if their first elements are equal and their second elements are equal. Surpris-ingly, instead of taking the ordered pair as a primitive con-cept, we can construct ordered pairs using basic notions from set theory. Show that if we define the ordered pair (a, b) to be {{a}, {a, b}}, then (a, b) = (c, d) if and only if a = c and b = d. [Hint: First show that {{a}, {a, b}} = {{c}, {c, d}} if and only if a = c and b = d.] ∗46. This exercise presents Russell’s paradox. Let S be the set that contains a set x if the set x does not belong to itself, so that S = {x | x / ∈x}.
a) Show the assumption that S is a member of S leads to a contradiction.
b) Show the assumption that S is not a member of S leads to a contradiction.
By parts (a) and (b) it follows that the set S cannot be de-fined as it was. This paradox can be avoided by restricting the types of elements that sets can have.
∗47. Describe a procedure for listing all the subsets of a finite set. |
190329 | https://www.khanacademy.org/science/in-in-class9th-physics-india/in-in-work-energy/in-in-work/e/work-equation-ap-physics-1 | Calculating work done by a force (practice) | Khan Academy
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Class 9 Physics (India)
Course: Class 9 Physics (India)>Unit 4
Lesson 1: Work
Intro to work
Positive & negative work
Work done on lifting/falling things - Solved numerical
Calculating work done by a force
Science>
Class 9 Physics (India)>
Work & Energy>
Work
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Calculating work done by a force
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A box moves 5 m horizontally when force F=20 N is applied at angle θ=135°.
What is the work done on the box by F during the displacement?
Choose 1 answer:
Choose 1 answer:
(Choice A) 70.7 J A 70.7 J
(Choice B) −70.7 J B −70.7 J
(Choice C) 100 J C 100 J
(Choice D) −100 J D −100 J
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5: Polynomial and Polynomial Functions
Intermediate Algebra 1e (OpenStax)
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5.3: Properties of Exponents and Scientific Notation
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5147
Mary Cameron
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5.3: Properties of Exponents and Scientific Notation
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5.2E: Exercises
5.3E: Exercises
Page ID 5147
OpenStax
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Learning Objectives
2. Note
3. Simplify Expressions Using the Properties for Exponents
1. DEFINITION: EXPONENTIAL NOTATION
2. DEFINITION: PRODUCT PROPERTY FOR EXPONENTS
3. Example 5.3.1
4. Example 5.3.2
5. Example 5.3.3
6. Definition: QUOTIENT PROPERTY FOR EXPONENTS
7. Example 5.3.4
8. Example 5.3.5
9. Example 5.3.6
10. DEFINITION: ZERO EXPONENT PROPERTY
11. ExAMPLE 5.3.7
12. Example 5.3.8
13. Example 5.3.9
Use the Definition of a Negative Exponent
DEFINITION: PROPERTIES OF NEGATIVE EXPONENTS
Example 5.3.10
Example 5.3.11
Example 5.3.12
QUOTIENT TO A NEGATIVE POWER PROPERTY
Example 5.3.13
Example 5.3.14
Example 5.3.15
Example 5.3.16
Example 5.3.17
Example 5.3.18
DEFINITION: POWER PROPERTY FOR EXPONENTS
Example 5.3.19
Example 5.3.20
Example 5.3.21
DEFINITION: PRODUCT TO A POWER PROPERTY FOR EXPONENTS
Example 5.3.22
Example 5.3.23
Example 5.3.24
DEFINITION: QUOTIENT TO A POWER PROPERTY FOR EXPONENTS
Example 5.3.25
Example 5.3.26
Example 5.3.27
DEFINITION: SUMMARY OF EXPONENT PROPERTIES
Example 5.3.28
Example 5.3.29
Example 5.3.30
Use Scientific Notation
DEFINITION: SCIENTIFIC NOTATION
DEFINITION: TO CONVERT A DECIMAL TO SCIENTIFIC NOTATION.
EXAMPLE 5.3.31
EXAMPLE 5.3.32
EXAMPLE 5.3.33
DEFINITION: CONVERT SCIENTIFIC NOTATION TO DECIMAL FORM.
EXAMPLE 5.3.34
EXAMPLE 5.3.35
EXAMPLE 5.3.36
EXAMPLE 5.3.37
EXAMPLE 5.3.38
EXAMPLE 5.3.39
Key Concepts
Glossary
Learning Objectives
By the end of this section, you will be able to:
Simplify expressions using the properties for exponents
Use the definition of a negative exponent
Use scientific notation
Note
Before you get started, take this readiness quiz.
Simplify: (−2)(−2)(−2).
If you missed this problem, review [link].
2. Simplify: 8x 24y.
If you missed this problem, review [link].
3. Name the decimal (−2.6)(4.21).
If you missed this problem, review [link].
Simplify Expressions Using the Properties for Exponents
Remember that an exponent indicates repeated multiplication of the same quantity. For example, in the expression a m, the exponent m tells us how many times we use the base a as a factor.
a m=a⋅a⋅a⋅…⋅a⏟m factors
For example
(−9)5=(−9)⋅(−9)⋅(−9)⋅(−9)⋅(−9)⏟5 factors
Let’s review the vocabulary for expressions with exponents.
DEFINITION: EXPONENTIAL NOTATION
This is read a to the m th power.
In the expression a m, the exponent m tells us how many times we use the base a as a factor.
When we combine like terms by adding and subtracting, we need to have the same base with the same exponent. But when you multiply and divide, the exponents may be different, and sometimes the bases may be different, too.
First, we will look at an example that leads to the Product Property.
x 2⋅x 3
What does this mean?
x⋅x⏟2factors⋅x⋅x⋅x⏟3factors
x 5
Notice that 5 is the sum of the exponents, 2 and 3. We see x 2⋅x 3 is x 2+3 or x 5.
The base stayed the same and we added the exponents. This leads to the Product Property for Exponents.
DEFINITION: PRODUCT PROPERTY FOR EXPONENTS
If a is a real number and m and n are integers, then
a m·a n=a m+n
To multiply with like bases, add the exponents.
Example 5.3.1
Simplify each expression:
y 5·y 6
2 x·2 3x
2a 7·3a.
Answer
ⓐ
Use the Product Property, a m·a n=a m+n.
Simplify.
ⓑ
Use the Product Property, a m·a n=a m+n.
Simplify.
ⓒ
Rewrite, a=a 1.
Use the Commutative Property and
use the Product Property, a m·a n=a m+n.
Simplify.
ⓓ
Add the exponents, since bases are the same.
Simplify.
Example 5.3.2
Simplify each expression:
b 9·b 8
4 2x·4 x
3p 5·4p
x 6·x 4·x 8.
Answer
ⓐ b 17 ⓑ 4 3x ⓒ 12p 6
ⓓ x 18
Example 5.3.3
Simplify each expression:
x 12·x4
10·10 x
2z·6z 7
b 5·b 9·b 5.
Answer a
x 16
Answer b
10 x+1
Answer c
12z 8
Answer d
b 19
Now we will look at an exponent property for division. As before, we’ll try to discover a property by looking at some examples.
Consider x 5 x 2 and x 2 x 3
What do they mean?x·x·x·x·x x·x x·x x·x·x
Use the Equivalent Fractions Property.x·x·x·x·x x·x x·x·1 x·x·x
Simplify.x 3 1 x
Notice, in each case the bases were the same and we subtracted exponents. We see x 5 x 2 is x 5−2 or x 3. We see x 2 x 3 is or 1 x. When the larger exponent was in the numerator, we were left with factors in the numerator. When the larger exponent was in the denominator, we were left with factors in the denominator--notice the numerator of 1. When all the factors in the numerator have been removed, remember this is really dividing the factors to one, and so we need a 1 in the numerator. x x=1. This leads to the Quotient Property for Exponents.
Definition: QUOTIENT PROPERTY FOR EXPONENTS
If a is a real number, a≠0, and m and n are integers, then
a m a n=a m−n,m>n and a m a n=1 a n−m,n>m
Example 5.3.4
Simplify each expression:
x 9 x 7
3 10 3 2
b 8 b 12
7 3 7 5.
Answer
To simplify an expression with a quotient, we need to first compare the exponents in the numerator and denominator.
ⓐ
Since 9>7, there are more factors of x in the numerator.
Use Quotient Property, a m a n=a m−n.
Simplify.
ⓑ
Since 10>2, there are more factors of 3 in the numerator.
Use Quotient Property, a m a n=a m−n.
Simplify.
Notice that when the larger exponent is in the numerator, we are left with factors in the numerator.
ⓒ
Since 12>8, there are more factors of bb in the denominator.
Use Quotient Property, a m a n=a m−n.
Simplify.
ⓓ
Since 5>3, there are more factors of 3 in the denominator.
Use Quotient Property, a m a n=a m−n.
Simplify.
Simplify.
Notice that when the larger exponent is in the denominator, we are left with factors in the denominator.
Example 5.3.5
Simplify each expression:
x 15 x 10
6 14 6 5
x 18 x 22
12 15 12 30.
Answer
ⓐ x 5
ⓑ 6 9
ⓒ 1 x 4
ⓓ 1 12 15
Example 5.3.6
Simplify each expression:
y 43 y 37
10 15 10 7
m 7 m 15
9 8 9 19.
Answer
ⓐ y 6
ⓑ 108
ⓒ 1m8
ⓓ 1 9 11
A special case of the Quotient Property is when the exponents of the numerator and denominator are equal, such as an expression like a m a m. We know, x x=1, for any x(x≠0) since any number divided by itself is 1.
The Quotient Property for Exponents shows us how to simplify a m a m. when m>n and when n<mn<m by subtracting exponents. What if m=n? We will simplify a m a m in two ways to lead us to the definition of the Zero Exponent Property. In general, for a≠0:
We see a m a m simplifies to a 0 and to 1. So a 0=1. Any non-zero base raised to the power of zero equals 1.
DEFINITION: ZERO EXPONENT PROPERTY
If a is a non-zero number, then a 0=1.
If a is a non-zero number, then a to the power of zero equals 1.
Any non-zero number raised to the zero power is 1.
In this text, we assume any variable that we raise to the zero power is not zero.
ExAMPLE 5.3.7
Simplify each expression: ⓐ 9 0 ⓑ n 0.
Answer
The definition says any non-zero number raised to the zero power is 1.
ⓐ Use the definition of the zero exponent. 9 0=1
ⓑ Use the definition of the zero exponent. n 0=1
To simplify the expression n raised to the zero power we just use the definition of the zero exponent. The result is 1.
Example 5.3.8
Simplify each expression: ⓐ 11 0 ⓑ q 0.
Answer
ⓐ 1
ⓑ 1
Example 5.3.9
Simplify each expression: ⓐ 23 0 ⓑ r 0.
Answer
ⓐ 1
ⓑ 1
Use the Definition of a Negative Exponent
We saw that the Quotient Property for Exponents has two forms depending on whether the exponent is larger in the numerator or the denominator. What if we just subtract exponents regardless of which is larger?
Let’s consider x 2 x 5. We subtract the exponent in the denominator from the exponent in the numerator. We see x 2 x 5 is x 2−5 or x−3.
We can also simplify x 2 x 5 by dividing out common factors:
This implies that x−3=1 x 3 and it leads us to the definition of a negative exponent. If n is an integer and a≠0, then a−n=1 a n.
Let’s now look at what happens to a fraction whose numerator is one and whose denominator is an integer raised to a negative exponent.
1 a−n Use the definition of a negative exponent,a−n=1 a n 1 1 a n Simplify the complex fraction.1·a n 1 Multiply.a n
This implies 1 a−n=a n and is another form of the definition of Properties of Negative Exponents.
DEFINITION: PROPERTIES OF NEGATIVE EXPONENTS
If n is an integer and a≠0, then a−n=1 a n or 1 a−n=a n.
The negative exponent tells us we can rewrite 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 x−3, we will take one more step and write 1 x 3. The answer is considered to be in simplest form when it has only positive exponents.
Example 5.3.10
Simplify each expression: ⓐ x−5 ⓑ 10−3 ⓒ 1 y−4 ⓓ 13−2.
Answer
ⓐ
x−5 Use the definition of a negative exponent,a−n=1 a n.1 x 5
ⓑ
10−3 Use the definition of a negative exponent,a−n=1 a n.1 10 3 Simplify.1 1000
ⓒ
1 y−4 Use the property of a negative exponent,1 a−n=a n.y 4
ⓓ
1 3−2 Use the property of a negative exponent,1 a−n=a n.3 2 Simplify.9
Example 5.3.11
Simplify each expression: ⓐ z−3 ⓑ 10−7 ⓒ 1 p−8 ⓓ 1 4−3.
Answer
ⓐ 1 z 3 ⓑ 1 10 7 ⓒ p 8 ⓓ 64
Example 5.3.12
Simplify each expression: ⓐ n−2 ⓑ 10−4 ⓒ 1 q−7 ⓓ 1 2−4.
Answer
ⓐ 1 n 2 ⓑ 1 10,000 ⓒ q 7
ⓓ 16
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.
(3 4)−2 Use the definition of a negative exponent,a−n=1 a n.1(3 4)2 Simplify the denominator.1 9 16 Simplify the complex fraction.16 9 But we know that 16 9 is(4 3)2.This tells us that(3 4)−2=(4 3)2
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 PowerProperty.
QUOTIENT TO A NEGATIVE POWER PROPERTY
If a and b are real numbers, a≠0, b≠0 and n is an integer, then
(a b)−n=(b a)n.
Example 5.3.13
Simplify each expression: ⓐ (5 7)−2 ⓑ (−x y)−3.
Answer
ⓐ
(5 7)−2 Use the Quotient to a Negative Exponent Property,(a b)−n=(b a)n.Take the reciprocal of the fraction and change the sign of the exponent.(7 5)2 Simplify.49 25
ⓑ
(−x y)−3 Use the Quotient to a Negative Exponent Property,(a b)−n=(b a)n.Take the reciprocal of the fraction and change the sign of the exponent.(−y x)3 Simplify.−y 3 x 3
Example 5.3.14
Simplify each expression: ⓐ (2 3)−4 ⓑ (−m n)−2.
Answer
ⓐ 81 16 ⓑ n 2 m 2
Example 5.3.15
Simplify each expression: ⓐ (3 5)−3 ⓑ (−a b)−4.
Answer
ⓐ 125 27 ⓑ b 4 a 4
Now that we have negative exponents, we will use the Product Property with expressions that have negative exponents.
Example 5.3.16
Simplify each expression: ⓐ z−5·z−3 ⓑ (m 4n−3)(m−5n−2) ⓒ (2x−6y 8)(−5x 5y−3).
Answer
ⓐ
z−5·z−3 Add the exponents, since the bases are the same.z−5−3 Simplify.z−8 Use the definition of a negative exponent.1 z 8
ⓑ
(m 4n−3)(m−5n−2)Use the Commutative Property to get like bases together.m 4m−5·n−2n−3 Add the exponents for each base.m−1·n−5 Take reciprocals and change the signs of the exponents.1 m 1·1 n 5 Simplify.1 mn 5
ⓒ
(2x−6y 8)(−5x 5y−3)Rewrite with the like bases together.2(−5)·(x−6x 5)·(y 8y−3)Multiply the coefficients and add the exponents of each variable.−10·x−1·y5 Use the definition of a negative exponent,a−n=1 a n.−10·1 x·y 5 Simplify.−10y 5x
Example 5.3.17
Simplify each expression:
ⓐ z−4·z−5 ⓑ (p 6q−2)(p−9q−1) ⓒ (3u−5v 7)(−4u 4v−2).
Answer
ⓐ 1 z 9 ⓑ 1 p 3q 3 ⓒ −12v 5 u
Example 5.3.18
Simplify each expression:
ⓐ c−8·c−7 ⓑ (r 5s−3)(r−7s−5) ⓒ (−6c−6d 4)(−5c−2d−1).
Answer
ⓐ 1 c 15 ⓑ 1 r 2s 8 ⓒ 30d 3 c 8
Now let’s look at an exponential expression that contains a power raised to a power. See if you can discover a general property.
(x 2)3 What does this mean?x 2·x 2·x 2
How many factors altogether?
So we have
Notice the 6 is the product of the exponents, 2 and 3. We see that (x 2)3 is x 2·3 or x 6.
We multiplied the exponents. This leads to the Power Propertyfor Exponents.
DEFINITION: POWER PROPERTY FOR EXPONENTS
If a is a real number and m and n are integers, then
(a m)n=a m·n
To raise a power to a power, multiply the exponents.
Example 5.3.19
Simplify each expression: ⓐ (y 5)9 ⓑ (4 4)7 ⓒ (y 3)6(y 5)4.
Answer
ⓐ
Use the Power Property, (a m)n=a m·n.
Simplify.
ⓑ
Use the Power Property.
Simplify.
ⓒ
(y 3)6(y 5)4 Use the Power Property.y 18·y 20 Add the exponents.y 38
Example 5.3.20
Simplify each expression: ⓐ (b 7)5 ⓑ (5 4)3 ⓒ (a 4)5(a 7)4.
Answer
ⓐ b 35 ⓑ 5 12 ⓒ a 48
Example 5.3.21
Simplify each expression: ⓐ (z 6)9 ⓑ (3 7)7 ⓒ (q 4)5(q 3)3.
Answer
ⓐ z 54 ⓑ 3 49 ⓒ q 29
We will now look at an expression containing a product that is raised to a power. Can you find this pattern?
(2x)3 What does this mean?2x·2x·2x We group the like factors together.2·2·2·x·x·x How many factors of 2 and of x 2 3·x 3
Notice that each factor was raised to the power and (2x)3 is 2 3·x 3.
The exponent applies to each of the factors! This leads to the Product to a PowerProperty for Exponents.
DEFINITION: PRODUCT TO A POWER PROPERTY FOR EXPONENTS
If a and b are real numbers and m is a whole number, then
(ab)m=a mb m
To raise a product to a power, raise each factor to that power.
Example 5.3.22
Simplify each expression: ⓐ (−3mn)3 ⓑ (−4a 2b)0 ⓒ (6k 3)−2 ⓓ (5x−3)2.
Answer
ⓐ
Use Power of a Product Property, (ab)m=a mb m.
Simplify.
ⓑ
(−4a 2b)0 Use Power of a Product Property,(ab)m=a mb m.(−4)0(a 2)0(b)0 Simplify.1·1·1 Multiply.1
ⓒ
(6k 3)−2 Use Power of a Product Property,(ab)m=a mb m.(6)−2(k 3)−2 Use the Power Property,(a m)n=a m·n.6−2k−6 Use the Definition of a negative exponent,a−n=1 a n.1 6 2·1 k 6 Simplify.1 36k 6
ⓓ
(5x−3)2 Use Power of a Product Property,(ab)m=a mb m.5 2(x−3)2 Simplify.25·x−6 Rewrite x−6 using,a−n=1a n.25·1 x 6 Simplify.25 x 6
Example 5.3.23
Simplify each expression: ⓐ (2wx)5 ⓑ (−11pq3)0 ⓒ (2b 3)−4 ⓓ (8a−4)2.
Answer
ⓐ 32w 5x 5 ⓑ 1 ⓒ 1 16b 12
ⓓ 64 a 8
Example 5.3.24
Simplify each expression: ⓐ (−3y)3 ⓑ (−8m 2n 3)0 ⓒ (−4x 4)−2 ⓓ (2c−4)3.
Answer
ⓐ −27y 3 ⓑ 1 ⓒ 1 16x 8
ⓓ 8c 12
Now we will look at an example that will lead us to the Quotient to a Power Property.
(x y)3 This means x y·x y·x y Multiply the fractions.x·x·x y·y·y Write with exponents.x 3 y 3
Notice that the exponent applies to both the numerator and the denominator.
We see that (x y)3 is x 3 y 3.
This leads to the Quotient to a Power Propertyfor Exponents.
DEFINITION: QUOTIENT TO A POWER PROPERTY FOR EXPONENTS
If a and b are real numbers, b≠0, and m is an integer, then
(a b)m=a m b m
To raise a fraction to a power, raise the numerator and denominator to that power.
Example 5.3.25
Simplify each expression:
ⓐ (b 3)4 ⓑ (k j)−3 ⓒ (2xy 2 z)3 ⓓ (4p−3 q 2)2.
Answer
ⓐ
Use Quotient to a Power Property, (ab)m=a mb m.
Simplify.
ⓑ
Raise the numerator and denominator to the power.
Use the definition of negative exponent.
Multiply.
ⓒ
(2xy 2 z)3 Use Quotient to a Power Property,(a b)m=a m b m.(2xy 2)3 z 3 Use the Product to a Power Property,(ab)m=a mb m.8x 3y 6 z 3
ⓓ
(4p−3 q 2)2 Use Quotient to a Power Property,(a b)m=a m b m.(4p−3)2(q 2)2 Use the Product to a Power Property,(ab)m=a mb m.4 2(p−3)2(q 2)2 Simplify using the Power Property,(a m)n=a m·n.16p−6 q 4 Use the definition of negative exponent.16 q 4·1 p 6 Simplify.16 p 6q 4
Example 5.3.26
Simplify each expression:
ⓐ (p 10)4 ⓑ (m n)−7 ⓒ(3ab 3 c 2)4 ⓓ (3x−2 y 3)3.
Answer
ⓐ p 4 10000 ⓑ n 7 m 7
ⓒ 81a 4b 12 c 8 ⓓ 27 x 6y 9
Example 5.3.27
Simplify each expression:
ⓐ (−2 q)3 ⓑ (w x)−4 ⓒ (xy 3 3z 2)2 ⓓ (2m−2 n−2)3.
Answer
ⓐ −8 q 3 ⓑ x 4 w 4 ⓒ x 2y 6 9z 4
ⓓ 8n 6 m 6
We now have several properties for exponents. Let’s summarize them and then we’ll do some more examples that use more than one of the properties.
DEFINITION: SUMMARY OF EXPONENT PROPERTIES
If a and b are real numbers, and m and n are integers, then
| Property | Description |
--- |
| Product Property | a m·a n=a m+n |
| Power Property | (a m)n=a m·n |
| Product to a Power | (ab)n=a nb n |
| Quotient Property | a m a n=a m−n,a≠0 |
| Zero Exponent Property | a 0=1,a≠0 |
| Quotient to a Power Property | (a b)m=a m b m,b≠0 |
| Properties of Negative Exponents | a−n=1 a n and 1 a−n=a n |
| Quotient to a Negative Exponent | (a b)−n=(b a)n |
Example 5.3.28
Simplify each expression by applying several properties:
ⓐ (3x 2y)4(2xy 2)3 ⓑ (x 3)4(x−2)5(x 6)5 ⓒ (2xy 2 x 3y−2)2(12xy 3 x 3y−1)−1.
Answer
ⓐ
(3x 2y)4(2xy 2)3 Use the Product to a Power Property,(ab)m=a mb m.(3 4x 8y 4)(2 3x 3y 6)Simplify.(81x 8y 4)(8x 3y 6)Use the Commutative Property.81·8·x 8·x 3·y 4·y 6 Multiply the constants and add the exponents.648x 11y 10
ⓑ
(x 3)4(x−2)5(x 6)5 Use the Power Property,(a m)n=a m·n.(x 12)(x−10)(x 30)Add the exponents in the numerator.x 2 x 30 Use the Quotient Property,a m a n=1 a n−m.1 x 28
ⓒ
(2xy 2 x 3y−2)2(12xy 3 x 3y−1)−1 Simplify inside the parentheses first.(2y 4 x 2)2(12y 4 x 2)−1 Use the Quotient to a Power Property,(a b)m=a m b m.(2y 4)2(x 2)2(12y 4)−1(x 2)−1 Use the Product to a Power Property,(ab)m=a mb m.4y 8 x 4·12−1y−4 x−2 Simplify.4y 4 12x 2 Simplify.y 4 3x 2
Example 5.3.29
Simplify each expression:
ⓐ (c 4d 2)5(3cd 5)4 ⓑ (a−2)3(a 2)4(a 4)5 ⓒ (3xy 2 x 2y−3)2
Answer
ⓐ 81c 24d 30 ⓑ 1 a 18
ⓒ 9y 10 x 2
Example 5.3.30
Simplify each expression:
ⓐ (a 3b 2)6(4ab 3)4 ⓑ (p−3)4(p 5)3(p 7)6 ⓒ (4x 3y 2 x 2y−1)2(8xy−3 x 2y)−1.
Answer
ⓐ 256a 22b 24 ⓑ 1 p 39
ⓒ 2x 3y 10
Use Scientific Notation
Working with very large or very small numbers can be awkward. Since our number system is base ten we can use powers of ten to rewrite very large or very small numbers to make them easier to work with. Consider the numbers 4,000 and 0.004.
Using place value, we can rewrite the numbers 4,000 and 0.004. We know that 4,000 means 4×1,000 and 0.004 means 4×1 1,000.
If we write the 1,000 as a power of ten in exponential form, we can rewrite these numbers in this way:
4,000 4×1,000 4×103
0.004 4×1 1,000 4×1 103 4×10−3
When a number is written as a product of two numbers, where the first factor is a number greater than or equal to one but less than ten, and the second factor is a power of 10 written in exponential form, it is said to be in scientific notation.
DEFINITION: SCIENTIFIC NOTATION
A number is expressed in scientific notation when it is of the form
a×10 n where 1≤a<10 and n is an integer.
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 first factor between 1 and 10.
The power of 10 is positive when the number is larger than 1:4,000=4×10 3
The power of 10 is negative when the number is between 0 and 1: 0.004=4×10−3
DEFINITION: TO CONVERT A DECIMAL TO SCIENTIFIC NOTATION.
Move the decimal point so that the first factor 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 10 n.
between 0 and 1, the power of 10 will be 10−n.
Check.
EXAMPLE 5.3.31
Write in scientific notation: ⓐ 37,000 ⓑ 0.0052.
Answer
ⓐ
The original number, 37,000, is greater than 1
so we will have a positive power of 10.37,000
Move the decimal point to get 3.7, a number
between 1 and 10.
Count the number of decimal places the point
was moved.
Write as a product with a power of 10.
3.7×10 4 Check:3.7×10,000 37,000
ⓑ
The original number, 0.0052, is between 0
and 1 so we will have a negative power of 10.0.0052
Move the decimal point to get 5.2, a number
between 1 and 10.
Count the number of decimal places the point
was moved.
Write as a product with a power of 10.
Check:5.2×10−3 5.2×1 10 3 5.2×1 1000 5.2×0.001 0.0052
EXAMPLE 5.3.32
Write in scientific notation: ⓐ 96,000 ⓑ 0.0078.
Answer
ⓐ 9.6×10 4 ⓑ 7.8×10−3
EXAMPLE 5.3.33
Write in scientific notation: ⓐ 48,300 ⓑ 0.0129.
Answer
ⓐ 4.83×10 4
ⓑ 1.29×10−2
How can we convert from scientific notation to decimal form? Let’s look at two numbers written in scientific notation and see.
9.12×10 4 9.12×10−4 9.12×10,000 9.12×0.0001 91,200 0.000912
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.
DEFINITION: CONVERT SCIENTIFIC NOTATION TO DECIMAL FORM.
Determine the exponent, n, on the factor 10.
Move the decimal n places, adding zeros if needed.
If the exponent is positive, move the decimal point n places to the right.
If the exponent is negative, move the decimal point |n| places to the left.
Check.
EXAMPLE 5.3.34
Convert to decimal form: ⓐ 6.2×10 3 ⓑ −8.9×10−2.
Answer
ⓐ
Determine the exponent, n, on the factor 10.
The exponent is 3.
Since the exponent is positive, move the
decimal point 3 places to the right.
Add zeros as needed for placeholders.
ⓑ
Determine the exponent, n, on the factor 10.The exponent is −2.−2.
Since the exponent is negative, move the
decimal point 2 places to the left.
Add zeros as needed for placeholders.
EXAMPLE 5.3.35
Convert to decimal form: ⓐ 1.3×10 3 ⓑ −1.2×10−4.
Answer
ⓐ 1,300 ⓑ −0.00012
EXAMPLE 5.3.36
Convert to decimal form: ⓐ −9.5×10 4 ⓑ 7.5×10−2.
Answer
ⓐ −950,000 ⓑ 0.075
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 5.3.37
Multiply or divide as indicated. Write answers in decimal form: ⓐ (−4×10 5)(2×10−7) ⓑ 9×10 3 3×10−2.
Answer
ⓐ
(−4×10 5)(2×10−7)Use the Commutative Property to rearrange the factors.−4·2·10 5·10−7 Multiply.−8×10−2 Change to decimal form by moving the decimal two places left.−0.08
ⓑ
9×10 3 9×10−2 Separate the factors, rewriting as the product of two fractions.9 3×10 3 10−2 Divide.3×10 5 Change to decimal form by moving the decimal five places right.300,000
EXAMPLE 5.3.38
Multiply or divide as indicated. Write answers in decimal form:
ⓐ (−3×10 5)(2×10−8) ⓑ 8×10 2 4×10−2.
Answer
ⓐ −0.006 ⓑ 20,000
EXAMPLE 5.3.39
Multiply or divide as indicated. Write answers in decimal form:
ⓐ (−3×10−2)(3×10−1) ⓑ 8×10 4 2×10−1.
Answer
ⓐ −0.009 ⓑ 400,000
Access these online resources for additional instruction and practice with using multiplication properties of exponents.
Properties of Exponents
Negative exponents
Scientific Notation
Key Concepts
Exponential Notation
This is read a to the m th power.
In the expression a m, the exponent m tells us how many times we use the base a as a factor.
Product Property for Exponents
If a is a real number and m and n are integers, then
a m·a n=a m+n
To multiply with like bases, add the exponents.
Quotient Property for Exponents
If a is a real number, a≠0, and m and n are integers, then
a m a n=a m−n,m>n and a m a n=1 a n−m,n>m
Zero Exponent
If a is a non-zero number, then a 0=1.
If a is a non-zero number, then a to the power of zero equals 1.
Any non-zero number raised to the zero power is 1.
Negative Exponent
If n is an integer and a≠0, then a−n=1 a n or 1 a−n=a n.
Quotient to a Negative Exponent Property
If a and b are real numbers, a≠0, b≠0 and n is an integer, then
(ab)−n=(ba)n
Power Property for Exponents
If a is a real number and m and n are integers, then
(a m)n=a m·n
To raise a power to a power, multiply the exponents.
Product to a Power Property for Exponents
If a and b are real numbers and m is a whole number, then
(ab)m=a mb m
To raise a product to a power, raise each factor to that power.
Quotient to a Power Property for Exponents
If a and b are real numbers, b≠0, and m is an integer, then
(a b)m=a m b m
To raise a fraction to a power, raise the numerator and denominator to that power.
Summary of Exponent Properties
If a and b are real numbers, and m and n are integers, then
| Property | Description |
--- |
| Product Property | a m·a n=a m+n |
| Power Property | (a m)n=a m·n |
| Product to a Power | (ab)n=a nb n |
| Quotient Property | a m a n=a m−n,a≠0 |
| Zero Exponent Property | a 0=1,a≠0 |
| Quotient to a Power Property: | (a b)m=a m b m,b≠0 |
| Properties of Negative Exponents | a−n=1 a n and 1 a−n=a n |
| Quotient to a Negative Exponent | (a b)−n=(b a)n |
Scientific Notation
A number is expressed in scientific notation when it is of the form
a×10 n where 1≤a<10 and n is an integer.
How to convert a decimal to scientific notation.
Move the decimal point so that the first factor 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 10 n.
between 0 and 1, the power of 10 will be 10−n.
Check.
How to convert scientific notation to decimal form.
Determine the exponent, n, on the factor 10.
Move the decimal n places, adding zeros if needed.
If the exponent is positive, move the decimal point n places to the right.
If the exponent is negative, move the decimal point |n| places to the left.
Check.
Glossary
Product PropertyAccording to the Product Property, a to the m times a to the a equals a to the m plus n.Power PropertyAccording to the Power Property, a to the m to the n equals a to the m times n.Product to a PowerAccording to the Product to a Power Property, a times b in parentheses to the m equals a to the m times b to the m.Quotient PropertyAccording to the Quotient Property, a to the m divided by a to the n equals a to the m minus n as long as a is not zero.Zero Exponent PropertyAccording to the Zero Exponent Property, a to the zero is 1 as long as a is not zero.Quotient to a Power PropertyAccording to the Quotient to a Power Property, a divided by b in parentheses to the power of m is equal to a to the m divided by b to the m as long as b is not zero.Properties of Negative ExponentsAccording to the Properties of Negative Exponents, a to the negative n equals 1 divided by a to the n and 1 divided by a to the negative n equals a to the n.Quotient to a Negative ExponentRaising a quotient to a negative exponent occurs when a divided by b in parentheses to the power of negative n equals b divided by a in parentheses to the power of n.
This page titled 5.3: Properties of Exponents and Scientific Notation is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.
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190331 | https://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl?ele=Fe&ascii=html&isotype=all | Atomic Weights and Isotopic Compositions for Iron
Atomic Weights and Isotopic Compositions for Iron
| Isotope | | Relative Atomic Mass | | Isotopic Composition | Standard Atomic Weight | Notes |
--- --- ---
| |
| |
| | | | | |
| 26 | Fe | 45 | 45.014 42(43#) | |
| 46 | 46.000 63(54#) | |
| 47 | 46.991 85(54#) | |
| 48 | 47.980 23(43#) | |
| 49 | 48.973 429(26) | |
| 50 | 49.962 975(64) | |
| 51 | 50.956 8410(96) | |
| 52 | 51.948 1131(70) | |
| 53 | 52.945 3064(18) | |
| 54 | 53.939 608 99(53) | 0.058 45(35) | 55.845(2) | |
| 55 | 54.938 291 99(51) | |
| 56 | 55.934 936 33(49) | 0.917 54(36) |
| 57 | 56.935 392 84(49) | 0.021 19(10) |
| 58 | 57.933 274 43(53) | 0.002 82(4) |
| 59 | 58.934 874 34(54) | |
| 60 | 59.934 0711(37) | |
| 61 | 60.936 7462(28) | |
| 62 | 61.936 7918(30) | |
| 63 | 62.940 2727(46) | |
| 64 | 63.940 9878(54) | |
| 65 | 64.945 0115(73) | |
| 66 | 65.946 2500(44) | |
| 67 | 66.950 54(23) | |
| 68 | 67.952 95(39) | |
| 69 | 68.958 07(43#) | |
| 70 | 69.961 02(54#) | |
| 71 | 70.966 72(64#) | |
| 72 | 71.969 83(75#) | |
| 73 | 72.975 72(75#) | |
| 74 | 73.979 35(86#) | |
| |
| | |
190332 | https://brainly.com/question/47702889 | [FREE] If the sum of three consecutive terms of an arithmetic progression is 30 and their product is 910, find - brainly.com
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If the sum of three consecutive terms of an arithmetic progression is 30 and their product is 910, find these three terms.
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To find the three consecutive terms of an AP that sum to 30 and multiply to 910, we represent them as B-d, B, B+d, solve for B as the middle term, and deduce the terms to be 9, 10, and 11.
Explanation
Finding three consecutive terms of an arithmetic progression (AP) that sum to 30 and have a product of 910 involves both algebraic and arithmetic reasoning. Let’s denote these terms as A, B, and C, with B being the middle term. Since they are consecutive terms in an AP, we can represent them as B-1, B, and B+1.
The sum of these terms is 30:
(B - d) + B + (B + d) = 30, simplifying to 3B = 30, giving B = 10.
Given the product is 910, our equation becomes (B - d) B (B + d) = 910. Substituting B = 10, we have (10 - d) 10 (10 + d) = 910. Solving this for d will provide us the common difference necessary to find the actual terms.
Upon calculating, it is evident that the common difference, d, is not explicitly needed for this problem because the symmetry in the expression allows simplification to easily find B, which is the central term. Knowing B = 10 helps us deduce the three terms directly: 9, 10, and 11, since the sum is 30 and confirms their product as 910 through calculation: 9 10 11 = 910.
This method illustrates the importance of recognizing and utilizing the properties of arithmetic progressions and symmetry in algebraic expressions to simplify and efficiently solve for unknowns.
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The three consecutive terms of the arithmetic progression that sum to 30 and have a product of 910 are 7, 10, and 13. This was determined by setting up equations based on the properties of arithmetic progressions. The solution confirmed both the sum and product conditions successfully.
Explanation
To solve the problem of finding three consecutive terms in an arithmetic progression (AP) that sum to 30 and have a product of 910, we can represent these terms mathematically.
Let's denote the three consecutive terms of the AP as:
First term: A = B - d
Second term: B (the middle term)
Third term: C = B + d
Here, B is the middle term, and d is the common difference.
Step 1: Set Up the Equations
We are given two conditions:
The sum of the three terms:
(B−d)+B+(B+d)=30
This simplifies to:
3 B=30
So, we find:
B=10
The product of the three terms:
(B−d)×B×(B+d)=910
Substituting B=10:
(10−d)×10×(10+d)=910
This simplifies to:
10(100−d 2)=910
Then, dividing both sides by 10 yields:
100−d 2=91
Therefore:
d 2=9
Hence:
d=3 or d=−3
Step 2: Find the Terms
Now, we can find the three terms:
If d=3:
First term: B−d=10−3=7
Second term: B=10
Third term: B+d=10+3=13
The terms are 7, 10, and 13.
If d=−3:
First term: B−d=10−(−3)=13
Second term: B=10
Third term: B+d=10+(−3)=7
The terms are again 7, 10, and 13.
Conclusion
Thus the three consecutive terms of the arithmetic progression that satisfy the conditions are 7, 10, and 13.
To verify:
Sum: 7+10+13=30
Product: 7×10×13=910
Both conditions hold true.
Examples & Evidence
For example, if we take consecutive terms such as 5, 6, and 7 in an AP, their sum would be 18 and product would be 210, which doesn't satisfy the original problem. However, the terms 7, 10, and 13 satisfy both conditions perfectly as shown in the detailed explanation.
The calculations show that the sum of the terms 7, 10, and 13 equals 30, and their product equals 910, thereby proving that these terms fulfill the requirements of the problem.
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### Free Mathematics solutions and answers
Community Answer the sum of three consecutive terms of an arithmetic sequence is 78 and their product is 17 472. find the three numbers.
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190333 | https://www.gauthmath.com/solution/1812710808438790/-The-area-enclosed-by-the-curve-y-10x-x1-and-the-curve-y-x2-3x-4-lies-partially- | Solved: ⑰ The area enclosed by the curve y=10x-x^1 and the curve y=x^2+3x-4 lies partially above a [Calculus]
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⑰ The area enclosed by the curve y=10x-x^1 and the curve y=x^2+3x-4 lies partially above and partially below the x-axis. Find the ratio of the area above the x-axis to the area below the x-axis.
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Answer
The ratio of the area above the x-axis to the area below the x-axis is determined through the above steps. (Exact numerical values require further computation based on the definite integrals calculated in Steps 4-6).
Explanation
Find the points of intersection of the curves $$y = 10x - x^{1}$$y=10 x−x 1 and $$y = x^{2} + 3x - 4$$y=x 2+3 x−4. Set them equal to each other:
$$10x - x = x^2 + 3x - 4$$10 x−x=x 2+3 x−4
This simplifies to:
$$9x = x^2 + 3x - 4$$9 x=x 2+3 x−4
Rearranging gives:
$$x^2 - 6x - 4 = 0.$$x 2−6 x−4=0.
Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$x=2 a−b±b 2−4 a c where $$a = 1, b = -6, c = -4$$a=1,b=−6,c=−4:
$$x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(-4)}}{2(1)} = \frac{6 \pm \sqrt{36 + 16}}{2} = \frac{6 \pm \sqrt{52}}{2} = \frac{6 \pm 2\sqrt{13}}{2} = 3 \pm \sqrt{13}.$$x=2(1)6±(−6)2−4(1)(−4)=2 6±36+16=2 6±52=2 6±2 13=3±13.
Thus, the points of intersection are $$x_{1} = 3 - \sqrt{13}$$x 1=3−13 and $$x_{2} = 3 + \sqrt{13}$$x 2=3+13
Calculate the area between the curves from $$x_{1}$$x 1 to $$x_{2}$$x 2:
The area $$A$$A is given by:
$$A = \int_{x_1}^{x_2} \left( (10x - x) - (x^2 + 3x - 4) \right) dx.$$A=∫x 1x 2((10 x−x)−(x 2+3 x−4))d x.
This simplifies to:
$$A = \int_{x_1}^{x_2} (7x + 4 - x^2) dx.$$A=∫x 1x 2(7 x+4−x 2)d x.
Compute the integral:
$$A = \left[ \frac{7x^2}{2} + 4x - \frac{x^3}{3} \right]_{x_1}^{x_2}.$$A=[2 7 x 2+4 x−3 x 3]x 1x 2.
Calculate the definite integral at $$x_{1}$$x 1 and $$x_{2}$$x 2 to find the total area.
Determine the areas above and below the x-axis by evaluating the integral separately for the regions where the curves are above and below the x-axis.
Find the ratio of the area above the x-axis to the area below the x-axis.
Final calculations will yield the ratio.
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Visual math talk prompts used to help emerge the visualization and understanding of measures of central tendency: mean, median and mode.
In This Set of Math Visual Prompts…
Students will determine a single value to represent a data set.
Intentionality…
This set of visual math talk prompts is taken from the Math Talk section of Day 2 in the Make Math Moments Problem Based Unit called Olympics. The purpose of the Math Talk is to reinforce key concepts and big ideas from this problem based math unit including:
There are different types of data;
Data is often characterized as continuous or discrete;
Discrete data is often countable and can only take on certain values;
A measure of central tendency for a data set summarizes all of its values with a single number;
Mode is the number that appears most often in a data set;
Mean is the great equalizer; it can be determined by redistributing the data evenly or through partitive division.
Want to take a deeper dive into these concepts? Consider starting from the beginning of this 5-day problem based unit.
Preparing to Facilitate
Present the following data set. Ask students to identify a single number to summarize all the values. Name the measure of central tendency used each time and ask students to justify their answer.
12, 13, 18, 15, 17
Two (2) additional visual math prompts are shared in the Teacher Guide from Day 2 of the Olympics problem based math unit.
The following Visual Math Talk Prompt intends to Spark Curiosity and lower the floor so all students can enter into each problem.
Visual Math Talk Prompt #1
Begin playing the video to share the first visual math talk prompt and be ready to pause the video to allow for think time.
In the video, students will see the number of points that a basketball player scored for the past 5 games this season:
12, 13, 18, 15, 17
Students are then prompted to:
Determine the mean, median and mode for this data set.
Which measure of central tendency do you think best represents the number of points scored by this player over the last 5 games?
After pausing the Visual Math Talk Prompt video to give students an opportunity to think and work through this problem, students can then share their thinking including their models and strategies used to determine the mean, median and mode.
While students who have experience calculating the mean by simply adding the values together and dividing by the number of values in the data set, others might choose to create a bar graph and “redistribute” the extra points scored in game 3 and 5 to games 1 and 2 to average out the number of points.
This approach is worth highlighting because it truly demonstrates the big idea of the mean as the “great equalizer”.
For the median, rearranging the bars from least to greatest can allow us to easily find the median (or middle) value in the data set.
Finally, with the data set organized from least to greatest, we can easily identify the mode: the value occurring most often. In this case, since the number of points scored in each game was unique, there is no mode.
Want to Explore These Concepts & Skills Further?
Two (2) additional math talk prompts are available in Day 2 of the Olympics problem based math unit that you can dive into now.
Why not start from the beginning of this contextual 5-day unit of real world lessons from the Make Math Moments Problem Based Units page.
Did you use this in your classroom or at home? How’d it go? Post in the comments!
Math IS Visual. Let’s teach it that way.
Grade 4 Grade 5 Grade 6 Grade 7 Grade 8 Grade 9
kylepearce3
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Related
Visualizing the Mean (Average) of a Dataset
Visualizing the Mean (Average) of a Dataset With Large Values
Mean, Median and Mode
Visualizing Measures of Central Tendency On The Number Line |
190335 | http://www.kutasoftware.com/FreeWorksheets/PreAlgWorksheets/Proportion%20Word%20Problems.pdf | ©7 m2k0n1Q2E yKeu6t6aM 5SQoJfztkwKaArjeu VLsL5CU.2 h TABltl1 MrziRguhktOsn PrDe0sAesrRvoekdK.q C mMYawdEel PwKiXtFhy OIgnefKi4nUigtbed RPHrDer-MAWlggXewbur6aE.c Worksheet by Kuta Software LLC Kuta Software - Infinite Pre-Algebra Name_____ Period_ Date___ Proportion Word Problems Answer each question and round your answer to the nearest whole number.
1) If you can buy one can of pineapple chunks for $2 then how many can you buy with $10?
2) One jar of crushed ginger costs $2. How many jars can you buy for $4?
3) One cantaloupe costs $2. How many cantaloupes can you buy for $6?
4) One package of blueberries costs $3. How many packages of blueberries can you buy for $9?
5) Shawna reduced the size of a rectangle to a height of 2 in. What is the new width if it was originally 24 in wide and 12 in tall?
6) Ming was planning a trip to Western Samoa. Before going, she did some research and learned that the exchange rate is 6 Tala for $2. How many Tala would she get if she exchanged $6?
7) Jasmine bought 32 kiwi fruit for $16. How many kiwi can Lisa buy if she has $4?
8) If you can buy four bulbs of elephant garlic for $8 then how many can you buy with $32?
9) One bunch of seedlees black grapes costs $2. How many bunches can you buy for $20?
10) The money used in Jordan is called the Dinar. The exchange rate is $3 to 2 Dinars. Find how many dollars you would receive if you exchanged 22 Dinars.
-1-©Z d2H0Z1Z2I 1KGu7teaR NS8oafgtewAavrveb 6LiLmC4.J v JAJlhlz yrfi9gXhJtWs9 1rieBsue8rNv0eQdq.q w CM2aVdxeL qwiiQt5hg yIrnxffirnri1tiei bPqr1eW-cA1lDg9esbqrza7.7 Worksheet by Kuta Software LLC 11) Gabriella bought three cantaloupes for $7. How many cantaloupes can Shayna buy if she has $21?
12) Jenny was planning a trip to the United Arab Emirates. Before going, she did some research and learned that the exchange rate is 4 Dirhams for every $1. How many Dirhams would she get if she exchanged $5?
13) Castel bought four bunches of fennel for $9. How many bunches of fennel can Mofor buy if he has $18?
14) If you can buy one fruit basket for $30 then how many can you buy with $60?
Answer each question. Round your answer to the nearest tenth. Round dollar amounts to the nearest cent.
15) Asanji took a trip to Mexico. Upon leaving he decided to convert all of his Pesos back into dollars. How many dollars did he receive if he exchanged 42.7 Pesos at a rate of $5.30 = 11.1 Pesos?
16) The currency in Argentina is the Peso. The exchange rate is approximately $3 = 1 Peso. At this rate, how many Pesos would you get if you exchanged $121.10?
17) Mary reduced the size of a painting to a width of 3.3 in. What is the new height if it was originally 32.5 in tall and 42.9 in wide?
18) Molly bought two heads of cabbage for $1.80. How many heads of cabbage can Willie buy if he has $28.80?
-2-©l u2y0u1K2p 7K2uStbaC LSvoxfGtnwma2rMe5 kLKLnCP.D 8 KAQlNlr YrRiTgDhFtcsw YrJe4s2ewrQv3eNdf.D B 8MVaodIeh Ew0iotRhc BI8nffIiynzivtkeR 9PWrceB-IAHl2gpelbqrVa7.G Worksheet by Kuta Software LLC Kuta Software - Infinite Pre-Algebra Name_____ Period Date__ Proportion Word Problems Answer each question and round your answer to the nearest whole number.
1) If you can buy one can of pineapple chunks for $2 then how many can you buy with $10?
5 2) One jar of crushed ginger costs $2. How many jars can you buy for $4?
2 3) One cantaloupe costs $2. How many cantaloupes can you buy for $6?
3 4) One package of blueberries costs $3. How many packages of blueberries can you buy for $9?
3 5) Shawna reduced the size of a rectangle to a height of 2 in. What is the new width if it was originally 24 in wide and 12 in tall?
4 in 6) Ming was planning a trip to Western Samoa. Before going, she did some research and learned that the exchange rate is 6 Tala for $2. How many Tala would she get if she exchanged $6?
18 Tala 7) Jasmine bought 32 kiwi fruit for $16. How many kiwi can Lisa buy if she has $4?
8 8) If you can buy four bulbs of elephant garlic for $8 then how many can you buy with $32?
16 9) One bunch of seedlees black grapes costs $2. How many bunches can you buy for $20?
10 10) The money used in Jordan is called the Dinar. The exchange rate is $3 to 2 Dinars. Find how many dollars you would receive if you exchanged 22 Dinars.
$33 -1-©q R2y0R152V hKduQtUa5 WS0onfwtswBaurVeX KLZLCCk.A b iAmljl0 zr1iqgQh1tZsU CraewsWeVrMvDeGdg.0 I HMFaYdqeA 3wOiEtAhn IIznefIimnAi0tUeK MPQr8e5-zA6lwgxeebJr1ab.1 Worksheet by Kuta Software LLC 11) Gabriella bought three cantaloupes for $7. How many cantaloupes can Shayna buy if she has $21?
9 12) Jenny was planning a trip to the United Arab Emirates. Before going, she did some research and learned that the exchange rate is 4 Dirhams for every $1. How many Dirhams would she get if she exchanged $5?
20 Dirhams 13) Castel bought four bunches of fennel for $9. How many bunches of fennel can Mofor buy if he has $18?
8 14) If you can buy one fruit basket for $30 then how many can you buy with $60?
2 Answer each question. Round your answer to the nearest tenth. Round dollar amounts to the nearest cent.
15) Asanji took a trip to Mexico. Upon leaving he decided to convert all of his Pesos back into dollars. How many dollars did he receive if he exchanged 42.7 Pesos at a rate of $5.30 = 11.1 Pesos?
$20.39 16) The currency in Argentina is the Peso. The exchange rate is approximately $3 = 1 Peso. At this rate, how many Pesos would you get if you exchanged $121.10?
40.4 Pesos 17) Mary reduced the size of a painting to a width of 3.3 in. What is the new height if it was originally 32.5 in tall and 42.9 in wide?
2.5 in 18) Molly bought two heads of cabbage for $1.80. How many heads of cabbage can Willie buy if he has $28.80?
32 -2-Create your own worksheets like this one with Infinite Pre-Algebra. Free trial available at KutaSoftware.com |
190336 | https://artofproblemsolving.com/wiki/index.php/Sum_and_difference_of_powers?srsltid=AfmBOoqX8jGXUkwiJGcKTQ0ncFCF-Tbx2lKeth8qbLsy7ByDQzBMIEh_ | Art of Problem Solving
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Sum and difference of powers
The sum and difference of powers are powerful factoring techniques that, respectively, factor a sum or a difference of certain powers.
Contents
[hide]
1 Sums of Odd Powers
2 Differences of Powers
3 Sum of Cubes
4 Factorizations of Sums of Powers
5 See Also
Sums of Odd Powers
Differences of Powers
If is a positive integer and and are real numbers,
For example:
Note that the number of terms in the second factor is equal to the exponent in the expression being factored.
An amazing thing happens when and differ by , say, . Then and
.
For example:
If we also know that then:
Sum of Cubes
Factorizations of Sums of Powers
Note that all these sums of powers can be factorized as follows:
If we have a difference of powers of degree , then
Note, of course, that some of the signs simply change when we have sum of powers instead of difference. Regardless, observe that the "longer" polynomial in the factorization is simply a binomial theorem expansion of the binomial , except for the fact that the coefficient on each of the terms is . This can be quite useful in problems that might have a sum of powers expression as well as an application of the binomial theorem.
icecreamrolls8
See Also
Factoring
Difference of squares, an extremely common specific case of this.
Binomial Theorem
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190337 | https://go.drugbank.com/drugs/DB09154 | The Next Evolution of DrugBank is Here
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Sodium citrate
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Data Packages
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Overview
Description
: A substance that causes the blood to clot during a medical procedure where the donor's whole blood is separated.
Structure
Description
: A substance that causes the blood to clot during a medical procedure where the donor's whole blood is separated.
DrugBank ID
: DB09154
Modality
: Small Molecule
US Approved
: YES
Other Approved
: YES
Patents
: 0
Indicated Conditions
: 13
Clinical Trials
: Phase 0
: 1
Phase 1
: 5
Phase 2
: 8
Phase 3
: 7
Phase 4
: 13
Therapeutic Categories
: Antacids and Adsorbents
Identification
Summary
: Sodium citrate is an ingredient used for the anticoagulation of whole blood as part of automated apheresis procedures.
Brand Names
: As 3, Bss, Bss Ophthalmic Solution, Cpda-1 Blood Collection System, Dalmacol, EnLyte, Fluad Tetra, Intersol, Leukotrap, Nauzene, Oracit, Tricitrasol
Generic Name
: Sodium citrate
DrugBank Accession Number
: DB09154
Background
: Sodium citrate is the sodium salt of citric acid. It is white, crystalline powder or white, granular crystals, slightly deliquescent in moist air, freely soluble in water,
practically insoluble in alcohol. Like citric acid, it has a sour taste.
From the medical point of view, it is used as alkalinizing agent. It works by neutralizing excess acid in the blood and urine. It has been indicated for the treatment of metabolic acidosis.
Modality
: Small Molecule
Groups
: Approved, Investigational
Structure
: MOLSDFPDBSMILESInChI
Similar Structures
#### Structure for Sodium citrate (DB09154)
Weight
: Average: 258.068
Monoisotopic: 257.97283534
Chemical Formula
: C6H5Na3O7
Synonyms
: Anhydrous sodium citrate
Anhydrous trisodium citrate
Citric acid, trisodium salt
Natrii citras
Natrocitral
Sodium citrate anhydrous
Sodium citrate, anhydrous
Sodium citrate,anhydrous
trisodium citrate anhydrous
Trisodium citrate concentration
Trisodium citrate, anhydrous
Trisodium-citrate
External IDs
: E-331(III)
FEMA NO. 3026, ANHYDROUS-
INS NO.331(III)
INS-331(III)
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Pharmacology
Indication
: Used as an anticoagulant during plasmophoresis as well as a neutralizing agent in the treatment of upset stomach and acidic urine Label 6 7.
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Associated Conditions
: | Indication Type | Indication | Combined Product Details | Approval Level | Age Group | Patient Characteristics | Dose Form |
--- --- ---
| Used in combination to treat | Acidosis | Combination Product in combination with: Citric acid (DB04272), Sodium chloride (DB09153), Potassium citrate (DB09125) | •••••••••••• Create Account | | | •••••••• |
| Used in combination to treat | Allergic cough | Combination Product in combination with: Diphenhydramine (DB01075), Dextromethorphan (DB00514) | •••••••••••• Create Account | | | ••••• |
| Used in combination to treat | Allergies | Combination Product in combination with: Dextromethorphan (DB00514), Diphenhydramine (DB01075) | •••••••••••• Create Account | | | ••••• |
| Used in combination to treat | Asthma | Combination Product in combination with: Phenylephrine (DB00388), Chlorpheniramine (DB01114), Ammonium chloride (DB06767) | •••••••••••• Create Account | | | ••••• |
| Used in combination to treat | Asthma chronic, cough | Combination Product in combination with: Dextromethorphan (DB00514), Diphenhydramine (DB01075) | •••••••••••• Create Account | | | ••••• |
Create Account
Associated Therapies
: Airway secretion clearance therapy
Oral rehydration therapy
Plasmapheresis
Urine alkalinization therapy
Fluid and electrolyte maintenance therapy
Irrigation during surgical procedures
Irrigation of the ocular surface therapy
Contraindications & Blackbox Warnings
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Pharmacodynamics
: Citrate prevents activation of the clotting cascade by chelating calcium ions. Citrate neutralizes acid in the stomach and urine, raising the pH 8.
Mechanism of action
: Citrate chelates free calcium ions preventing them from forming a complex with tissue factor and coagulation factor VIIa to promote the activation of coagulation factor X 1 2. This inhibits the extrinsic initiation of the coagulation cascade. Citrate may also exert an anticoagulant effect via a so far unknown mechanism as restoration of calcium concentration does not fully reverse the effect of citrate 1. Citrate is a weak base and so reacts with hydrochloric acid in the stomach to raise the pH. It it further metabolized to bicarbonate which then acts as a systemic alkalizing agent, raising the pH of the blood and urine 8. It also acts as a diuretic and increases the urinary excretion of calcium.
| Target | Actions | Organism |
---
| UCarbonic anhydrase 4 | inhibitor | Humans |
Absorption
: Tmax of 98-130min 3.
Volume of distribution
: 19-39L 3.
Protein binding
: Not Available
Metabolism
: Citrate is metabolized to bicarbonate in the liver and plays a role as an intermediate in the citric acid cycle 5 9.
Route of elimination
: Largely eliminated through hepatic metabolism with very little cleared by the kidneys 3 4.
Half-life
: 18-54 min 3
Clearance
: Total clearance of 313-1107mL/min 3.
Adverse Effects
: Improve decision support & research outcomes
With structured adverse effects data, including: blackbox warnings, adverse reactions, warning & precautions, & incidence rates. View sample adverse effects data in our new Data Library!
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Toxicity
: Overdose toxicity is mainly due to alkalosis as well as tetany or depressed heart function due to lack of free calcium 8.
Pathways
: Not Available
Pharmacogenomic Effects/ADRs
: Not Available
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Interactions
Drug Interactions
: This information should not be interpreted without the help of a healthcare provider. If you believe you are experiencing an interaction, contact a healthcare provider immediately. The absence of an interaction does not necessarily mean no interactions exist.
Approved
Vet approved
Nutraceutical
Illicit
Withdrawn
Investigational
Experimental
All Drugs
| Drug | Interaction |
--- |
| Integrate drug-drug interactions in your software | |
| Abciximab | The risk or severity of bleeding can be increased when Abciximab is combined with Sodium citrate. |
| Aceclofenac | The risk or severity of bleeding and hemorrhage can be increased when Aceclofenac is combined with Sodium citrate. |
| Acemetacin | The risk or severity of bleeding and hemorrhage can be increased when Sodium citrate is combined with Acemetacin. |
| Acenocoumarol | The risk or severity of bleeding can be increased when Acenocoumarol is combined with Sodium citrate. |
| Acetophenazine | Acetophenazine may increase the neurotoxic activities of Sodium citrate. |
Food Interactions
: Take with or without food. Recommendations vary from product to product - consult individual product monographs for additional information.
Products
: Drug product information from 10+ global regions
Our datasets provide approved product information including:
dosage, form, labeller, route of administration, and marketing period.
Access nowAccess drug product information from over 10 global regions.
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Product Ingredients
: | Ingredient | UNII | CAS | InChI Key |
--- --- |
| Sodium citrate dihydrate | B22547B95K | 6132-04-3 | NLJMYIDDQXHKNR-UHFFFAOYSA-K |
Active Moieties
: | Name | Kind | UNII | CAS | InChI Key |
--- ---
| Citric acid | unknown | XF417D3PSL | 77-92-9 | KRKNYBCHXYNGOX-UHFFFAOYSA-N |
| Sodium cation | ionic | LYR4M0NH37 | 17341-25-2 | FKNQFGJONOIPTF-UHFFFAOYSA-N |
Brand Name Prescription Products
: | Name | Dosage | Strength | Route | Labeller | Marketing Start | Marketing End | Region | Image |
--- --- --- ---
| Anticoagulant Sod Citrate Soltn 4gm/100ml | Solution | 4 g | Extracorporeal | Baxter Laboratories | 1990-12-31 | Not applicable | Canada | |
| Anticoagulant Sodium Citrate | Solution | 40 mg/1mL | Extracorporeal | Baxter Healthcare Corporation | 2024-04-02 | Not applicable | US | |
| Anticoagulant Sodium Citrate | Solution | 10 g/250mL | Intravenous | Medsep Corporation | 2011-04-15 | Not applicable | US | |
| Anticoagulant Sodium Citrate 4% W/V Solution, USP | Solution | 4 g / 100 mL | Unknown | Baxter Laboratories | 2012-10-01 | Not applicable | Canada | |
| Anticoagulant Sodium Citrate Sol 4% USP | Liquid | 4 % | Intravenous | Cutter Med & Biol, Division Of Miles Canada Ltd. | 1988-12-31 | 1996-09-09 | Canada | |
Generic Prescription Products
: | Name | Dosage | Strength | Route | Labeller | Marketing Start | Marketing End | Region | Image |
--- --- --- ---
| Anticoagulant Sodium Citrate | Solution | 40 mg/1mL | Extracorporeal | Laboratorios Grifols s.a. | 2019-10-25 | Not applicable | US | |
| Anticoagulant Sodium Citrate | Solution | 40 mg/1mL | Intravenous | CSL Plasma Inc. | 2022-07-01 | Not applicable | US | |
| Sodium Citrate 4% w/v Anticoagulant | Injection, solution | 4 g/100mL | Intravenous | Terumo Bct, Ltd | 2018-06-26 | Not applicable | US | |
| Sodium Citrate w/v Anticoagulant | Injection, solution | 4 g/100mL | Intravenous | Terumo Bct, Ltd | 2018-06-26 | Not applicable | US | |
Over the Counter Products
: | Name | Dosage | Strength | Route | Labeller | Marketing Start | Marketing End | Region | Image |
--- --- --- ---
| Bromo Seltzer | Powder | 3.5 g / 5.83 g | Oral | Warner Lambert Canada Inc. | 1976-12-31 | 1999-08-13 | Canada | |
| CVS Health Nausea Relief | Tablet, chewable | 230 mg/1 | Oral | CVS PHARMACY | 2019-05-22 | Not applicable | US | |
| Emetrol Chewables | Tablet, chewable | 230 mg/1 | Oral | WellSpring Pharmaceutical Corporation | 2022-03-15 | Not applicable | US | |
| Emetrol Chewables Mixed Berry | Tablet, chewable | 230 mg/1 | Oral | WellSpring Pharmaceutical Corporation | 2023-07-01 | Not applicable | US | |
| Emetrol Chewables Orange | Tablet, chewable | 230 mg/1 | Oral | WellSpring Pharmaceutical Corporation | 2023-07-01 | Not applicable | US | |
Mixture Products
: | Name | Ingredients | Dosage | Route | Labeller | Marketing Start | Marketing End | Region | Image |
--- --- --- ---
| A-C-D Solution | Sodium citrate (250 mg / 10 mL) + Citric acid (80 mg / 10 mL) + Dextrose, unspecified form (132 mg / 10 mL) | Liquid | Intravenous | Draximage A Division Of Draxis Specialty Pharmaceuticals Inc | 1959-12-31 | 2003-07-08 | Canada | |
| Acd A | Sodium citrate dihydrate (2.2 g/100mL) + Citric acid monohydrate (0.8 g/100mL) + D-glucose monohydrate (2.45 g/100mL) | Injection, solution | Intravenous | Terumo Bct, Ltd | 2002-02-25 | Not applicable | US | |
| Acd A | Sodium citrate dihydrate (2.2 g/100mL) + Citric acid monohydrate (0.8 g/100mL) + D-glucose monohydrate (2.45 g/100mL) | Injection, solution | Intravenous | Terumo Bct, Ltd | 2002-02-25 | Not applicable | US | |
| Acd A | Sodium citrate (2.2 g/100mL) + Citric acid monohydrate (0.80 g/100mL) + D-glucose monohydrate (2.45 g/100mL) | Injection, solution | Intravenous | Terumo Bct, Ltd | 2002-02-25 | Not applicable | US | |
| ACD Blood-Pack Units (PL 146 Plastic) | Sodium citrate dihydrate (1.48 g/67.5mL) + Citric acid (493 mg/67.5mL) + D-glucose monohydrate (1.65 g/67.5mL) | Solution | Extracorporeal | Fenwal, Inc. | 2007-03-01 | Not applicable | US | |
Unapproved/Other Products
: | Name | Ingredients | Dosage | Route | Labeller | Marketing Start | Marketing End | Region | Image |
--- --- --- ---
| Anticoagulant Citrate Dextrose Solution-A | Sodium citrate (2.20 g/100mL) + Citric acid monohydrate (0.73 g/100mL) + Dextrose, unspecified form (2.45 g/100mL) | Solution | Extracorporeal | Biomet Biologics | 2010-01-01 | 2010-09-06 | US | |
| CVS Health Nausea Relief | Sodium citrate dihydrate (230 mg/1) | Tablet, chewable | Oral | CVS PHARMACY | 2019-05-22 | Not applicable | US | |
| Cytosol Ophthalmics - Balanced Salt Solution | Sodium citrate dihydrate (1.70 mg/1mL) + Calcium chloride dihydrate (48 mg/1mL) + Magnesium chloride hexahydrate (30 mg/1mL) + Potassium chloride (75 mg/1mL) + Sodium acetate trihydrate (3.90 mg/1mL) + Sodium chloride (6.40 mg/1mL) | Solution | Irrigation | Biomet Biologics | 2010-01-01 | 2010-09-04 | US | |
| Cytra 3 | Sodium citrate (500 mg/5mL) + Citric acid monohydrate (334 mg/5mL) + Potassium citrate monohydrate (550 mg/5mL) | Syrup | Oral | Pegasus Laboratories | 2006-05-05 | 2009-10-17 | US | |
| Cytra 3 | Sodium citrate (500 mg/5mL) + Citric acid monohydrate (334 mg/5mL) + Potassium citrate monohydrate (550 mg/5mL) | Syrup | Oral | Cypress Pharmaceuticals, Inc. | 2008-07-29 | 2014-10-11 | US | |
Categories
ATC Codes
: B05CB02 — Sodium citrate
B05CB — Salt solutions
B05C — IRRIGATING SOLUTIONS
B05 — BLOOD SUBSTITUTES AND PERFUSION SOLUTIONS
B — BLOOD AND BLOOD FORMING ORGANS
Drug Categories
: Acids, Acyclic
Antacids and Adsorbents
Anticoagulants
Blood and Blood Forming Organs
Blood Substitutes and Perfusion Solutions
Buffers
Citrates
Citric Acid Derivatives
Compounds used in a research, industrial, or household setting
Diet, Food, and Nutrition
Food
Food Additives
Food Ingredients
Food Preservatives
Hematologic Agents
Irrigating Solutions
Laboratory Chemicals
Miscellaneous Anticoagulants
Miscellaneous GI Drugs
Neurotoxic agents
Physiological Phenomena
Salt Solutions
Tricarboxylic Acids
Urinary Alkalinisers
Chemical TaxonomyProvided by Classyfire
: Description
: This compound belongs to the class of organic compounds known as tricarboxylic acids and derivatives. These are carboxylic acids containing exactly three carboxyl groups.
Kingdom
: Organic compounds
Super Class
: Organic acids and derivatives
Class
: Carboxylic acids and derivatives
Sub Class
: Tricarboxylic acids and derivatives
Direct Parent
: Tricarboxylic acids and derivatives
Alternative Parents
: Tertiary alcohols / Carboxylic acid salts / Carboxylic acids / Organic sodium salts / Organic oxides / Hydrocarbon derivatives / Carbonyl compounds
Substituents
: Alcohol / Aliphatic acyclic compound / Carbonyl group / Carboxylic acid / Carboxylic acid salt / Hydrocarbon derivative / Organic alkali metal salt / Organic oxide / Organic oxygen compound / Organic salt
Molecular Framework
: Aliphatic acyclic compounds
External Descriptors
: organic sodium salt (CHEBI:53258)
Affected organisms
: Not Available
Chemical Identifiers
UNII
: RS7A450LGA
CAS number
: 68-04-2
InChI Key
: HRXKRNGNAMMEHJ-UHFFFAOYSA-K
InChI
: InChI=1S/C6H8O7.3Na/c7-3(8)1-6(13,5(11)12)2-4(9)10;;;/h13H,1-2H2,(H,7,8)(H,9,10)(H,11,12);;;/q;3+1/p-3
IUPAC Name
: trisodium 2-hydroxypropane-1,2,3-tricarboxylate
SMILES
: [Na+].[Na+].[Na+].OC(CC([O-])=O)(CC([O-])=O)C([O-])=O
References
General References
: 1. Mann KG, Whelihan MF, Butenas S, Orfeo T: Citrate anticoagulation and the dynamics of thrombin generation. J Thromb Haemost. 2007 Oct;5(10):2055-61. [Article]
2. Palta S, Saroa R, Palta A: Overview of the coagulation system. Indian J Anaesth. 2014 Sep;58(5):515-23. doi: 10.4103/0019-5049.144643. [Article]
3. Kramer L, Bauer E, Joukhadar C, Strobl W, Gendo A, Madl C, Gangl A: Citrate pharmacokinetics and metabolism in cirrhotic and noncirrhotic critically ill patients. Crit Care Med. 2003 Oct;31(10):2450-5. [Article]
4. Zheng Y, Xu Z, Zhu Q, Liu J, Qian J, You H, Gu Y, Hao C, Jiao Z, Ding F: Citrate Pharmacokinetics in Critically Ill Patients with Acute Kidney Injury. PLoS One. 2013 Jun 18;8(6):e65992. doi: 10.1371/journal.pone.0065992. Print 2013. [Article]
5. Li K, Xu Y: Citrate metabolism in blood transfusions and its relationship due to metabolic alkalosis and respiratory acidosis. Int J Clin Exp Med. 2015 Apr 15;8(4):6578-84. eCollection 2015. [Article]
6. FDA Monograph: Sodium Citrate Tablet [Link]
7. FDA Monograph: Sodium Citrate Liquid [Link]
8. TOXNET: Trisodium Citrate [Link]
9. SMPDB: Citric Acid Cycle [Link]
External Links
: KEGG Drug
: D05855
PubChem Compound
: 6224
PubChem Substance
: 310265067
ChemSpider
: 5989
RxNav
: 56466
ChEBI
: 53258
ChEMBL
: CHEMBL1355
Wikipedia
: Sodium\_citrate
FDA label
: Download (124 KB)
MSDS
: Download (77.2 KB)
Clinical Trials
Clinical Trials ###### Clinical Trial & Rare Diseases Add-on Data Package Explore 4,000+ rare diseases, orphan drugs & condition pairs, clinical trial why stopped data, & more. Preview package
: | Phase | Status | Purpose | Conditions | Count | Start Date | Why Stopped | 100+ additional columns |
--- --- --- --- |
| Unlock 175K+ rows when you subscribe.View sample data | | | | | | | |
| Not Available | Completed | Not Available | Conotruncal Cardiac Defects / Heart Defects, Congenital / Hypoparathyroidism / Pulmonary Atresia / Pulmonary Valve Stenosis / Tetralogy of Fallot (TOF) | 1 | somestatus Unlock 75,000+ rows when you subscribe Explore data packages curated & structured to speed up your pharmaceutical research View Sample Data | stop reason | just information to hide |
| Not Available | Completed | Basic Science | Dental Caries Extending Into Dentine | 1 | somestatus | stop reason | just information to hide |
| Not Available | Completed | Prevention | Other Complication of Vascular Dialysis Catheter | 1 | somestatus | stop reason | just information to hide |
| Not Available | Recruiting | Not Available | Impaired Renal Function | 1 | somestatus | stop reason | just information to hide |
| Not Available | Unknown Status | Treatment | Anticoagulation / End Stage Renal Disease (ESRD) / Hemodialysis Treatment | 1 | somestatus | stop reason | just information to hide |
Unlock 75,000+ rows when you subscribe
Explore data packages curated & structured to speed up your pharmaceutical research
View Sample Data
Pharmacoeconomics
Manufacturers
: Not Available
Packagers
: Not Available
Dosage Forms
: | Form | Route | Strength |
---
| Liquid | Intravenous | |
| Kit; solution | Intravenous | |
| Solution | Irrigation; Parenteral | |
| Solution | Intraocular | |
| Injection, solution | Extracorporeal | |
| Solution | Unknown | |
| Solution | Extracorporeal | |
| Solution | Extracorporeal | 4 g |
| Solution | Extracorporeal | 40 mg/1mL |
| Solution | Intravenous | 10 g/250mL |
| Solution | Intravenous | 40 mg/1mL |
| Solution | Unknown | 4 g / 100 mL |
| Liquid | Intravenous | 4 % |
| Solution | Extracorporeal | 0.8 g/100ml |
| Solution | Other | |
| Solution | Ophthalmic | |
| Injection, powder, for solution | Intramuscular | |
| Injection, solution | Intramuscular | |
| Suppository | Rectal | |
| Tablet, effervescent | Oral | |
| Powder | Oral | 3.5 g / 5.83 g |
| Solution | Parenteral | 136 mmol/l |
| Powder, for solution | Oral | |
| Syrup | Oral | 12.5 mg/5mL |
| Kit | Intravenous | |
| Solution | Rectal | 0.63 g |
| Solution | Rectal | 630 mg |
| Tablet, chewable | Oral | 230 mg/1 |
| Solution | Irrigation | |
| Liquid | Oral | |
| Solution | Oral | |
| Solution | Oral | 334 mg/5ml |
| Solution | Oral | 62.5 mg/5ml |
| Solution | Intravenous | 4.21 g/1000ml |
| Solution | Oral | 0.2600 g |
| Powder, for solution | Oral | 460 mg/1 |
| Capsule, liquid filled | Oral | |
| Capsule, delayed release pellets | Oral | |
| Injection, suspension | Intramuscular | |
| Syrup | Oral | 135 mg/5ml |
| Syrup | Oral | 50 mg/5mL |
| Liquid | Intraocular | |
| Syrup | | |
| Powder | Oral | 2.65 g / pck |
| Enema | Rectal | |
| Enema; liquid | Rectal | |
| Liquid | Rectal | |
| Enema | Rectal | 12.5 % w/w |
| Syrup | Oral | 5 mg/5ml |
| Syrup | Oral | |
| Syrup | Oral | 0.921 g/15mL |
| Emulsion | Rectal | |
| Solution, concentrate | Oral | |
| Solution | Intraocular; Ophthalmic | |
| Powder | Oral | 2.645 g / pck |
| Capsule | Oral | |
| Solution | Oral | 1.188 g |
| Powder | Oral | |
| Kit | Oral | |
| Tablet, chewable | Oral | |
| Solution | Intravenous | 5.03 g/l |
| Enema | Rectal | 0.35 g |
| Syrup | Oral | 62.5 mg/5mL |
| Syrup | Oral | 100 mg/5ml |
| Syrup | Oral | 14 mg |
| Injection, solution | Extracorporeal | |
| Solution | Extracorporeal | |
| Solution | Intravenous | 10.0 g/250mL |
| Injection, solution | Intravenous | 4 g/100mL |
| Solution | Intravenous | 4 g/100mL |
| Solution | Intravenous | 2 g/50mL |
| Powder, for solution | Oral | 2.9 g |
| Solution | Intraocular; Irrigation | |
| Syrup | Oral | 102 mg/5ml |
| Solution | Intravenous | |
| Injection, solution | Intravenous | |
| Granule | Oral | |
| Suspension | Oral | 62.5 mg/5ml |
| Solution | Extracorporeal | 14 g/30mL |
| Granule, effervescent | Oral | 0.72 g |
| Granule, for solution | Oral | |
| Powder, for solution | Oral | |
| Granule, effervescent | Oral | |
| Suspension | Oral | 135 mg/5ml |
| Liquid | Ophthalmic | |
| Tablet | | |
Prices
: Not Available
Patents
: Not Available
Properties
State
: Solid
Experimental Properties
: | Property | Value | Source |
---
| melting point (°C) | 300 | MSDS |
| water solubility | 29.4g/L | MSDS |
Predicted Properties
: | Property | Value | Source |
---
| Water Solubility | 73.7 mg/mL | ALOGPS |
| logP | -0.55 | ALOGPS |
| logP | -1.3 | Chemaxon |
| logS | -0.54 | ALOGPS |
| pKa (Strongest Acidic) | 3.05 | Chemaxon |
| pKa (Strongest Basic) | -4.2 | Chemaxon |
| Physiological Charge | -3 | Chemaxon |
| Hydrogen Acceptor Count | 7 | Chemaxon |
| Hydrogen Donor Count | 1 | Chemaxon |
| Polar Surface Area | 140.62 Å2 | Chemaxon |
| Rotatable Bond Count | 5 | Chemaxon |
| Refractivity | 68.14 m3·mol-1 | Chemaxon |
| Polarizability | 14.23 Å3 | Chemaxon |
| Number of Rings | 0 | Chemaxon |
| Bioavailability | 1 | Chemaxon |
| Rule of Five | Yes | Chemaxon |
| Ghose Filter | No | Chemaxon |
| Veber's Rule | No | Chemaxon |
| MDDR-like Rule | No | Chemaxon |
Predicted ADMET Features
: Not Available
Spectra
Mass Spec (NIST)
: Not Available
Spectra
: | Spectrum | Spectrum Type | Splash Key |
---
| Predicted GC-MS Spectrum - GC-MS | Predicted GC-MS | splash10-0a4i-0090000000-159e2fe9d3ea097b6f77 |
Chromatographic Properties
: ###### Collision Cross Sections (CCS)
| Adduct | CCS Value (Å2) | Source type | Source |
--- --- |
| [M-H]- | 125.66848 | predicted | DeepCCS 1.0 (2019) |
| [M+H]+ | 127.655396 | predicted | DeepCCS 1.0 (2019) |
| [M+Na]+ | 133.56429 | predicted | DeepCCS 1.0 (2019) |
Targets
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Details1. Carbonic anhydrase 4
Kind
: Protein
Organism
: Humans
Pharmacological action
: Unknown
Actions
: Inhibitor
General Function
: Catalyzes the reversible hydration of carbon dioxide into bicarbonate and protons and thus is essential to maintaining intracellular and extracellular pH (PubMed:15563508, PubMed:16686544, PubMed:16807956, PubMed:17127057, PubMed:17314045, PubMed:17652713, PubMed:17705204, PubMed:18618712, PubMed:19186056, PubMed:19206230, PubMed:7625839). May stimulate the sodium/bicarbonate transporter activity of SLC4A4 that acts in pH homeostasis (PubMed:15563508). It is essential for acid overload removal from the retina and retina epithelium, and acid release in the choriocapillaris in the choroid (PubMed:15563508)
Specific Function
: carbonate dehydratase activity
Gene Name
: CA4
Uniprot ID
: P22748
Uniprot Name
: Carbonic anhydrase 4
Molecular Weight
: 35032.075 Da
References
Zhou Y, Zhang Y, Zhao D, Yu X, Shen X, Zhou Y, Wang S, Qiu Y, Chen Y, Zhu F: TTD: Therapeutic Target Database describing target druggability information. Nucleic Acids Res. 2024 Jan 5;52(D1):D1465-D1477. doi: 10.1093/nar/gkad751. [Article]
Transporters
Details1. Mitochondrial 2-oxodicarboxylate carrier
Kind
: Protein
Organism
: Humans
Pharmacological action
: No
Actions
: Substrate
General Function
: Transports dicarboxylates across the inner membranes of mitochondria by a counter-exchange mechanism (PubMed:11083877). Can transport 2-oxoadipate (2-oxohexanedioate), 2-oxoglutarate, adipate (hexanedioate), glutarate, and to a lesser extent, pimelate (heptanedioate), 2-oxopimelate (2-oxoheptanedioate), 2-aminoadipate (2-aminohexanedioate), oxaloacetate, and citrate (PubMed:11083877). Plays a central role in catabolism of lysine, hydroxylysine, and tryptophan, by transporting common metabolite intermediates (such as 2-oxoadipate) into the mitochondria, where it is converted into acetyl-CoA and can enter the citric acid (TCA) cycle (Probable)
Specific Function
: alpha-ketoglutarate transmembrane transporter activity
Gene Name
: SLC25A21
Uniprot ID
: Q9BQT8
Uniprot Name
: Mitochondrial 2-oxodicarboxylate carrier
Molecular Weight
: 33302.88 Da
References
UniProt: Mitochondrial 2-oxodicarboxylate carrier [Link]
Details2. Solute carrier family 13 member 2
Kind
: Protein
Organism
: Humans
Pharmacological action
: No
Actions
: Substrate
General Function
: Low-affinity sodium-dicarboxylate cotransporter, that mediates the entry of citric acid cycle intermediates, such as succinate, citrate, fumarate and alpha-ketoglutarate (2-oxoglutarate) into the small intestine and renal proximal tubule (PubMed:10894787, PubMed:8967342, PubMed:9668069). Transports the dicarboxylate into the cell with a probable stoichiometry of 3 Na(+) for 1 divalent dicarboxylate, rendering the process electrogenic (PubMed:10894787, PubMed:8967342, PubMed:9668069). Citrate is transported in protonated form as a divalent anion, rather than the trivalent form which is normally found in blood (PubMed:10894787). Has a critical role in renal dicarboxylate transport (By similarity)
Specific Function
: alpha-ketoglutarate transmembrane transporter activity
Gene Name
: SLC13A2
Uniprot ID
: Q13183
Uniprot Name
: Solute carrier family 13 member 2
Molecular Weight
: 64409.495 Da
References
UniProt: Solute carrier family 13 member 2 [Link]
Details3. Na(+)/citrate cotransporter
Kind
: Protein
Organism
: Humans
Pharmacological action
: No
Actions
: Substrate
General Function
: High-affinity sodium/citrate cotransporter that mediates the entry of citrate into cells, which is a critical participant of biochemical pathways (PubMed:12445824, PubMed:12826022, PubMed:26324167, PubMed:26384929, PubMed:30054523, PubMed:33597751). May function in various metabolic processes in which citrate has a critical role such as energy production (Krebs cycle), fatty acid synthesis, cholesterol synthesis, glycolysis, and gluconeogenesis (PubMed:12826022). Transports citrate into the cell in a Na(+)-dependent manner, recognizing the trivalent form of citrate (physiological pH) rather than the divalent form (PubMed:12445824, PubMed:12826022, PubMed:26324167, PubMed:26384929, PubMed:30054523, PubMed:33597751). Can recognize succinate as a substrate, but its affinity for succinate is several fold lower than for citrate (PubMed:26324167). The stoichiometry is probably 4 Na(+) for each carboxylate, irrespective of whether the translocated substrate is divalent or trivalent, rendering the process electrogenic (PubMed:12445824, PubMed:12826022). Involved in the regulation of citrate levels in the brain (By similarity)
Specific Function
: citrate transmembrane transporter activity
Gene Name
: SLC13A5
Uniprot ID
: Q86YT5
Uniprot Name
: Na(+)/citrate cotransporter
Molecular Weight
: 63061.605 Da
References
Inoue K, Zhuang L, Maddox DM, Smith SB, Ganapathy V: Structure, function, and expression pattern of a novel sodium-coupled citrate transporter (NaCT) cloned from mammalian brain. J Biol Chem. 2002 Oct 18;277(42):39469-76. Epub 2002 Aug 11. [Article]
Kind
: Protein
Organism
: Humans
Pharmacological action
: No
Actions
: Substrate
General Function
: Mitochondrial electroneutral antiporter that exports citrate from the mitochondria into the cytosol in exchange for malate (PubMed:26870663, PubMed:29031613, PubMed:29238895, PubMed:39881208). Also able to mediate the exchange of citrate for isocitrate, phosphoenolpyruvate, cis-aconitate and to a lesser extent trans-aconitate, maleate and succinate (PubMed:29031613). In the cytoplasm, citrate plays important roles in fatty acid and sterol synthesis, regulation of glycolysis, protein acetylation, and other physiopathological processes (PubMed:29031613, PubMed:29238895, PubMed:39881208)
Specific Function
: antiporter activity
Gene Name
: SLC25A1
Uniprot ID
: P53007
Uniprot Name
: Tricarboxylate transport protein, mitochondrial
Molecular Weight
: 34012.46 Da
References
Sun J, Aluvila S, Kotaria R, Mayor JA, Walters DE, Kaplan RS: Mitochondrial and Plasma Membrane Citrate Transporters: Discovery of Selective Inhibitors and Application to Structure/Function Analysis. Mol Cell Pharmacol. 2010;2(3):101-110. [Article] |
190338 | https://www.youtube.com/watch?v=QyUqGYPFDGQ | The structures of simple solids | Chapter 3 - Shriver & Atkins’ Inorganic Chemistry (5th Edition)
Last Minute Lecture
2 subscribers
Description
31 views
Posted: 4 Sep 2025
Chapter 3 of Shriver & Atkins’ Inorganic Chemistry (Fifth Edition) examines the structures of simple solids, beginning with the description of crystal structures in terms of unit cells, lattice points, and the seven crystal systems. Concepts such as primitive, body-centered, and face-centered lattices are introduced, along with the use of fractional coordinates and close-packing arrangements of spheres. The discussion covers hexagonal close packing (hcp) and cubic close packing (ccp or fcc), highlighting coordination numbers and the role of octahedral and tetrahedral holes. The structures of metals and alloys are analyzed, including non-close-packed arrangements, polymorphism, polytypism, and the trends in atomic radii across the metallic elements. The chapter then explores the structures of ionic solids, focusing on characteristic lattice types such as rock salt, cesium chloride, zinc blende, wurtzite, rutile, fluorite, and perovskite, with rationalization of structural choices through ionic sizes and coordination preferences. Energetics of ionic bonding are detailed through lattice enthalpy, the Born–Haber cycle, and approaches to calculate or estimate lattice energies, including the Born–Mayer and Kapustinskii equations. The consequences of lattice enthalpy for stability, solubility, and melting points are emphasized. Defects in crystals are also treated, from Schottky and Frenkel defects to nonstoichiometry and solid solutions, showing how these imperfections affect conductivity and material properties. Finally, the electronic structures of solids are explained, distinguishing conductors, semiconductors, and insulators through band theory, orbital overlap, and band gaps. By integrating structural description, energetics, and electronic properties, this chapter provides the foundation for understanding metals, alloys, ionic compounds, and the role of crystal structure in dictating material behavior.
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Shriver and Atkins Inorganic Chemistry Chapter 3 summary, structures of simple solids explained, unit cells and crystal systems chemistry, primitive body-centered face-centered lattices, cubic hexagonal crystal packing, close packing hcp fcc ccp structures, coordination numbers octahedral and tetrahedral holes, structures of metals alloys polymorphism, zinc blende rock salt cesium chloride fluorite rutile perovskite, rationalization of ionic crystal structures, lattice enthalpy Born Haber cycle, Born Mayer and Kapustinskii equation, ionic bonding energetics, lattice stability solubility melting point, Schottky Frenkel defects nonstoichiometry, solid solutions and crystal defects, conductors semiconductors insulators band theory, electronic structures of solids chemistry
Transcript:
Welcome to the deep dive, your guide to understanding the essentials. And today, uh, we're diving head first into solid state chemistry, which sounds maybe a bit abstract, right? But think about it. The sturdy alloys in your car, maybe the vibrant pigments and paint, even, you know, cutting edge stuff like nanomaterials and high temperature superconductors. It all comes down to how atoms arrange themselves in solids. It's fundamental. Exactly. And that's our mission for this deep dive. breaking down a key chapter from uh Shrivever and Atkins in organic chemistry. Yep. Making sense of structures, energetics, electronic properties, step by step, really so you can follow along even without the textbook diagrams in front of you. That's the plan. And this chapter, it's great because it builds things up logically. We start, you know, really simple with models like atoms being just hard spheres, like tiny billiard balls packing together. Precisely. And then we layer on the complexities, the energy involved, the quantum electronic theories. It really helps you connect the microscopic arrangement to the macroscopic properties we actually see and use. Okay, let's frame this. The core questions are uh why do atoms and ions line up in these specific ordered ways in solids? Right? Why this structure and not another? And then following from that, what properties actually emerge because of that specific arrangement? That's the heart of it. We'll look at the main bonding types, metallic, ionic, and the uh the interesting interplay between them sometimes. So, the journey covers how we describe these crystal structures first. Yeah. Then the efficiency of packing, this idea of close packing, then the energy tied up in ionic bonds. Okay. And imperfections, you mentioned those defects. Absolutely crucial. They're unavoidable and actually enable many properties. And finally, we'll get into the electronic structure. Why some things conduct electricity and others don't. It sounds like a foundational piece of the puzzle for understanding well a huge chunk of inorganic materials. It really is. It unlocks a lot. All right, let's start at the beginning then describing the structures. You mentioned Legos earlier. If a crystal is a giant Lego structure, what's the smallest repeating brick? That's basically our unit cell. It's a great analogy. A crystal is just a repeating pattern, right, of atoms or ions or molecules. This pattern has a name. We call the underlying grid of points the crystal lattice. And the unit cell is the smallest kind of imaginary box you can define. Like a 3D box. Exactly. A parallel-sided box that if you just copy and shift it over and over in all directions without rotating it, right? Just translation. It perfectly reproduces the entire crystal structure. And we typically pick the smallest one that shows the most symmetry. Okay? And there are different fundamental shapes for these boxes. There are seven crystal systems based on the lengths of the sides and the angles between them. But honestly, for a lot of the simpler inorganic solids we'll talk about, we mostly run into cubic and hexagonal systems, they're very common. And inside these unit cells, the atoms aren't always in the same spots, right? How many atoms effectively belong to one cell? Good question. It depends on the type of cell. The simplest is the primitive cell labeled P. It only has lattice points at the corners. And since each corner is shared by what, eight cells? Exactly. Each corner only counts as 18th for a given cell. So eight corners 18th gives you just one effective lattice point or atom per p cell. Okay. What else is there? Then you have body centered or eye cells. Same corner points but there's one extra point right in the dead center of the box. So that's one from the corners plus the one inside. Two total. Yep. Two lattice points per eye is eye cell. And the last main type is face centered. F. Again points at the corners but also one in the center of each of the six faces. Okay. A face is shared by two cells. So each face center counts as half, right? So you get one from the corners, 818, plus half from each of the six faces, 612. That adds up to four lattice points per F cell. It's a neat way to count how densely packed things are. And how do we specify exactly where an atom is inside that unit cell box? We use fractional coordinates. Think of it like a map reference within the cell using fractions of the cell's side lengths. So like XYZ, where X, Y, and Z are numbers between 0 and one. Exactly. 0 0 0 is one corner, 111 is the opposite corner, and 0.5 would be right smack in the middle of the cell. That's a universal coordinate system. Useful for complex structures, I bet, and visualizing them. Yeah, 3D structures can get messy. So often we simplify by drawing them in projection, like looking down one axis of the unit cell. You draw the base and note the fractional height next to each atom symbol. Okay, that makes sense. Now, you mentioned packing spheres like oranges. If you just have identical spheres, how do they naturally pack together most tightly? Ah, the close packing of spheres. This is a really powerful concept because many metals and even parts of ionic structures behave as if they're just hard spheres trying to get as close as possible. Trying to maximize connections pretty much maximizing attractions if there's no specific directional bonding involved. This leads to close pack structures. In these every single sphere is touching uh 12 other spheres. 12 neighbors. That's the maximum possible. That's the highest coordination number geometry allows for identical spheres. So how do you build this structure layer by layer? Exactly. Start with one flat layer. Call it layer A. Each sphere has six neighbors in a hexagon around it. Okay. Now the next layer, layer B, doesn't sit directly on top. The spheres nestle into the dips or hollows of layer A. Makes sense. More stable. Right now for the third layer, you have a choice. Where do its spheres sit? This choice leads to two main polytype structures that look the same in 2D layers but differ in the 3D stacking. What are the choices? Choice one. Place the third layer directly above the spheres in layer A. So the sequence goes ABA. This gives you a hexagonal close packed HCP structure. It has a hexagonal unit cell. Okay, ABP is HCP. What's the other option? Choice two. Place the third layer over the other set of hollows in layer B. The ones that are not directly above layer A. Let's call this layer C. So A then B then C. Exactly. The sequence goes ABC ABC. This results in a cubic close packed CCP structure. And interestingly, this ABC stacking actually generates a face centered cubic unit cell. So CCP and FCC are essentially the same packing arrangement. Fundamentally, yes. The underlying lattice is FCC. Both HCP and CCP are incredibly efficient ways to pack spheres, but not perfectly efficient. You mentioned some empty space, right? Even in these densest packings, about 26% of the total volume is still unoccupied space. You can actually calculate that for CCP by looking at the volume of the four spheres inside the cubic unit cell versus the volume of the cube itself. 26% seems like quite a bit. Is that space just wasted? Oh, not at all. That unoccupied space is actually structured. It forms specific geometric gaps or holes between the packed spheres. And these holes are absolutely critical for understanding more complex structures, especially ionic ones. Okay, tell me about these holes. What are they like? There are two main types of these interstitial sites or holes in close pack structures. First, you have ocaedral holes. Why ocaedral? Because each one is surrounded by six of the pack spheres arranged at the corners of an ocahedron. Imagine a sphere from the layer above and one below capping a triangle of spheres in each layer. Okay, six neighbors. How many are there? If you have n spheres making the close packed structure, you get exactly n ocaedral holes and they can fit a smaller sphere inside up to a radius of about 414 times the radius of the main packing spheres. Got it. N spheres and ocahedral holes. What's the other type? The other type is tetrahedral holes. These are smaller. They're formed by a triangle of three spheres in one layer capped by a single sphere in the layer directly above or below it. So surrounded by four spheres like a tetrahedra. Exactly. Four nearest neighbors define a tetrahedral hole. And for n packing spheres, you actually get twice as many tetrahedral holes. Two n of them. Half point up, half point down. Twice as many. And they're smaller. Yes. They can only accommodate a sphere with a maximum radius of about 225 times the packing sphere radius. So let me summarize the whole situation. Yeah, in any closed packed structure for every sphere there's one octaedral hole and two tetrahedral holes. That's the crucial ratio. Spheres ocaedral holes. Tetrahedral holes is 1.1.2. Remember that it's fundamental for predicting the formulas and structures of many ionic compounds where you have larger annions forming the packing and smaller cations fitting into these holes. Right? That makes perfect sense. Okay, so we've got the theory of packing in holes. How does this play out with real metals? Do they actually pack this way? Many of them do. Uh, a lot of metallic elements adopt either HCP or CCP structures. Think copper, gold, silver, aluminum, there's CCP, magnesium, zinc, titanium, often HCP. Why clues packed? Well, metallic bonding is generally nondirectional. The atoms are like positively charged cores in a sea of electrons. So, there's no strong preference for specific bond angles, allowing them to just pack together as densely as possible to maximize stability. This density is why metals feel heavy osmium. The densest element is close packed. Can we calculate that density from the structure? Absolutely. If you know the structure, say gold is CCP, which is FCC. You know there are four gold atoms per unit cell. Measure the unit cellside length using X-ray defraction. Use the molar mass of gold and boom, you can calculate its density very accurately. Cool. You mentioned HCP, AB, and CCP. ABC. Are those the only stacking sequences metals use? They're the simplest and most common for close packing, but no, more complex sequences or poly types can occur. Cobalt, for example, is CCP at high temperatures, but below about 500° C, it can form structures with more random like stacking, maybe AB by BC. The only rule is adjacent layers can't be identical. No AA or BB stacking. Are all metals close pack though? No, definitely not. Some adopt less dense structures. A very important one is body centered cubic BCC. We saw that unit cell earlier. two atoms per cell, less than the four in FCCCCP. Right? Each atom in BCC only has eight nearest neighbors, not 12. It fills space less efficiently, about 32% empty space compared to 26% in close packed. But it does have six slightly further second nearest neighbors. Many alkali metals and metals like iron and tungsten use BCC. Any others? Primitive cubic primitive cubic cubic P with only six nearest neighbors is extremely rare for elements. Alpha palonium is pretty much the only example under normal conditions. So metals can switch between these structures. You mentioned cobalt and temperature. Yes, that's polymorphism. Many metals can exist in different crystal structures or polymorphs depending on temperature and pressure. Generally lower temperatures and higher pressures favor denser, more close-packed structures. Like iron, you mentioned that changes structure with temperature. Iron is the classic example. At room temp, it's alpha, which is BCC. Heated above 96° C, it flips to gamma fee, which is CCP. heat it even more above 1401 degrees, it actually goes back to a BCC structure delta FE before it melts. Why would it go back to BCC at high temperature? Isn't that less packed? It seems counterintuitive, but at very high temperatures, the atoms are vibrating much more vigorously. This increased vibration can sometimes favor the slightly more open BCC structure entropically. Understanding these phase changes is absolutely critical in metallurgy like for steel production. Okay. Now when we talk about the size of metal atoms, their radi does the packing structure affect the measurement? It does. Atomic radius is usually defined as half the distance between the centers of adjacent atoms. But an atom in a 12 coordinate environment like CCPHCP will appear slightly larger than the same atom in an 8 coordinate environment like BCC. So comparing raw radi between different structures can be misleading. Exactly. To compare the intrinsic sizes properly, we often use a goldmid correction. This adjusts the measured radius to what it would be in a standard 12 coordinate setup. Ah, so it levels the playing field for comparison. Precisely. Using these corrected radi lets us see the periodic trends more clearly. Radi getting bigger down a group, smaller across a period and explains things like the lanthnide contraction where electrons healed poorly. Makes sense. Now what happens when we deliberately mix metals together? Alloys, right? Alloys are blends of metals. They can be quite simple or complex. One type is a solid solution where the atoms are mixed randomly like dissolving one metal and another kind of in a substitutional solid solution atoms of one metal replace atoms of the host metal on the lice sites. Does that work for any pair of metals? No. There are rules of thumb the humory rules. It works best if the two metals have one similar atomic radi usually within about 15% of each other. Okay. to the same crystal structure in their pure form and three similar chemical character meaning similar electro negativity. Do you have an example? Sure. Copper and nickel both are CCP radi are very close. 128 p.m. versus 125 p.m. similar electro negativity. They can mix in any proportion to form a continuous solid solution. What about metals that don't meet those criteria? They might only mix partially. Copper and zinc, for example. Zinc prefers HCP. Copper is CCP. So they have limited solubility in each other. Is substitution the only way to make an alloy solution? No. There's also interstitial solid solutions. This happens when you have very small atoms. Think boron, carbon, nitrogen that can fit into the holes within the host metals lattice like those ocaedral or tetral holes we talked about. Exactly. The classic example is steel. Carbon atoms which are tiny compared to iron atoms slip into the ocahedral holes in the iron lattice and that changes the properties dramatically. Even a small amount of interstitial carbon makes iron much harder and stronger. That's how we get different grades of steel. Are there alloys that aren't just random mixtures like actual compounds? Yes, those are called inner metallic compounds. They form between two metals but have a distinct crystal structure different from either parent metal and a definite chemical formula a fixed stoic geometry. Examples might be kusen beta brass or mgzn2. Some called zintal phases like kg form between very electropositive and less electropositive metals and show properties somewhere between ionic and metallic. Okay, that transitions us nicely. Let's move away from the metallicy of electrons and into the world of ionic solids like table salt, necl. Right? Ionic solids are typically formed between elements with a large difference in electro negativity like a metal and a non-metal. Think sodium and chlorine. So instead of a sea of electrons, we have distinct positive and negative ions. Exactly. You have positive ions and annions negative ions held together primarily by strong nondirectional electrostatic forces. the attraction between opposite charges. We often model them as hard charged spheres. What are their typical properties? Because the electrostatic forces are strong, they tend to have high melting points. They're often hard but brittle. They shatter rather than bend, and many dissolve in polar solvents like water. Though there are important exceptions like calcium fluoride, which is quite insoluble, and their structures. Does the close packing idea still apply? Very much so, but with a twist. Often you can view ionic structures as one type of ion, usually the larger one, often the anion, forming a close-acked array, CCP or HCP, like the onions or the oranges in the box. Exactly. And then the smaller counter ions, usually the cations fit into some or all of the octahedral or tetrahedral holes within that array. The annion packing might expand a bit to make room, but the basic principle holds. So understanding those holes, octaedral, tetrahedral, is key here, too. absolutely fundamental. Let's look at some classic structure types based on this whole filling idea. Okay, let's start with a simple 1.1 ratio AX compounds like NaCCl itself. Perfect example, the rock salt NaCCl structure. You can describe it as a cubic closepacked CCP or FCC array of the larger chloride annions. Okay, Annions make the CCP lattice. Where are the sodium cations? The smaller sodium cations occupy all of the octahedral holes within that chloride array. We learned there's one ocahedral hole per sphere in CCP. So one N+ for every Cl. Exactly. That gives the 1.1 stoometry AX. And if you look closely, each sodium ion is surrounded by six chloride ions and each chloride ion is surrounded by six sodium ions. We call this 6.6 coordination. And this structure is common. Incredibly common. Many alkali hallides, alkaline earth oxides like MGO and even some more complex materials adopt this basic structure. What's another common 1.1 structure? The cesium chloride cscl structure. This one's a bit different. It's based on a primitive cubic unit cell. Imagine chloride ions at the eight corners of a cube. Okay, that's the P lattice. And then the cesium sits right in the center of that cube in what's called a cubic hole. So each ion is surrounded by eight of the other type, right? It has eight eight coordination. This structure is typically adopted when the cation and annion are closer in size like C's plus and CL or C's plus and BR. So coordination number depends on relative ion sizes too. Yes, that's a major factor. Larger cations relative to onions tend to favor higher coordination numbers like eight while smaller cations fit better in six coordinate octahedral or even four coordinate tetrahedral sites. Let's talk four coordination. Zinc sulfide ZNS has two forms, right? It does. Both are AX structures with 4.4 coordination. The first is spherate also called zinc blend. Here you have sulfide annions forming a CCP FCC array. Okay. Annions are CCP. Where are the zinc cations? The ZN2 plus cations occupy half of the available tetrahedral holes. Half? Why not all? Remember the ratio one sphere two tetrahedral holes. To get a 1.1 ZNS stoic geometry, the can only fill half the tetrahedral sites. All right, makes sense. What's the other ZNS structure? That's wartsite. It's very similar, but the sulfide annions form an htp array instead of CCP. So ABA stacking of annions instead of ABC. Exactly. But again, the ZN2 plus cations fill half the tetrahedral holes leading to the same AX stoometry and 4.4 coordination just a different 3D arrangement. Okay. What about AX2 compounds like K2? Calcium fluoride has the fluoride structure here. Think of the C2 plus forming an expanded CCP array. ation is doing the packing this time. Yes. And then the smaller fluoride annions F occupy all of the tetrahedral holes within that calcium array. Let's check the stochometry. Right. One CA2 plus makes the CCP site. Two tetraal holes per C2 plus. If F fills all of them, you get KF2, a 1.2 ratio AX2 and the coordination is 8 four. Each calcium is surrounded by eight fluorides in a cube and each theorite is surrounded by four calciums in a tetrahedrin. Is there an inverse structure A2X? Yes, the anti fluoride structure. Lithium oxide I2O is an example. Here the annions O2 form a CCP array and the small cations all A2 plus fill all the tetrahedral holes. So A2X stoometry and 4PN8 coordination. It's like flipping the roles of and in fluoride. Exactly. Another important AX2 structure is routile one form of TiO2. This is based on a distorted HCP array of oxide annions. Okay. HCPensions. Where are the titanium cations? The TY4 plus cations occupy half of the ocaedral holes. Half the ocaedral holes. Why half? Remember one sphere one ocahedral hole in the packing. To get ti stoic geometry only half the octahedral sites can be filled by T4 plus val. This gives six ring coordination. Titanium is six coordinate octaedral. Oxygen is three coordinate trigonal planer. These simple packing and hole filling ideas explain a lot of structures. What about more complex ones like with three different elements abx3? That brings us to structures like perovskate kio3. This is a hugely important structure type especially in material science. Many oxides with the ABX3 formula adopt this arrangement. What does it look like ideally? In the ideal cubic corate structure you have the large aation like K2 plus or B2 plus at the corners of the cube. The X annions usually O2 are at the center of each face and the smaller Bation like TY4 plus sits right in the center of the cube. Let me picture that. A at corners O on faces B in the middle. What are the coordinations? The Aation is coordinated to 12 oxygen annions. The Bation in the center is coordinated to the six face centered oxygens forming an ocahedron BO6. And the oxygens link these ocahedra at the corners. That sounds like a very connected network of octahedra. It is. And this structure or variations of it is famous for exhibiting properties like ferro electricity, po electricity and even high temperature superconductivity in related cuprit materials. It's a really versatile structural framework. Incredible how geometry translates to function. Okay, we've seen how things are arranged, but why does a specific compound pick, say rock salt over cesium chloride or spherite? You said it comes down to energy. Exactly. Nature is wazy in a thermodynamic sense. Solids adopt the structure that has the lowest Gibbs energy under given conditions. At reasonably low temperatures, that usually means the structure with the lowest enalpy, the one that releases the most energy when it forms. And how do we quantify the energy holding an ionic crystal together? The key concept is lattice enalpy, often symbolized as ahl degrees. It's defined as the energy required to break one mole of the ionic solid completely apart into its separate gaseous ions. So anaccl solid go nty clock plus cl gas. Precisely. It's always an endothermic process. So lice enalpy is always positive. A larger positive value means it takes more energy to break the lattice apart, indicating a more stable solid with stronger ionic bonding. How do we measure this experimentally? We can't just pull crystals apart easily. We can't measure it directly, but we can calculate it indirectly using a Borne Habber cycle. Ah, the cycle. I remember that from thermo. It's a brilliant application of Hessa's law. You construct a thermodynamic cycle that includes the formation of the solid from its elements which we can measure the standard enalpy of formation along with all the steps needed to turn those elements into gaseous ions like atomizing the solid metal ionizing the metal atoms breaking the non-metal molecules apart like Cl2 to CL atoms adding electrons to the non-metal atoms electron gain enalpy and then the final unknown step is the lattice enalpy forming the solid from the gaseous ions. Exactly. Since the total enthalpy change around the cycle must be zero. If you know all the other enthalpy changes, you can calculate the lattice enthalpy. It gives us a solid experimental value. Okay, that's the experimental route. Can we predict lattice enalpy theoretically just based on the ions in the structure? Yes, we can estimate it using theoretical models. The most common is the Bourne mayor equation. What does that take into account? It primarily considers two things. First, the strong electrostatic attractions between oppositeely charged ions and repulsions between like charged ions throughout the entire crystal. That sounds complicated to sum up for a whole crystal. It is, but it's captured mathematically by the mealin constant. This constant a is specific to each crystal structure type like rock salt or cesium chloride and reflects the precise geometric arrangement of all the ions. So, the meline constant bundles up all the plus minus attractions and repulsions for that geometry essentially. Yes, higher coordination numbers generally lead to slightly larger meatline constants. What's the second factor in the Bourne mayor equation? It accounts for the short range repulsion that happens when the electron clouds of adjacent ions get too close and start to overlap. This repulsion stops the crystal from collapsing in on itself. The equation balances these attractions and repulsions. So what does lattice enthalpy depend on most strongly according to this equation? Two main things. the charges on the ions za and zaba b and the distance between them b the sum of their radi. Latis enthalpy scales with the product of the charges zaca b and inversely with the distance 1d meaning higher charges lead to much stronger lises. Absolutely. Compare na charges plus1 na1a1 with mg charges plus 2 na2. The charge product is four times larger for MGO. Even though the ions are similar in size, MGO's lattice enalpy is vastly higher, making it much harder and higher melting. And smaller ions mean stronger lises, too. Yes, because the distance D is smaller, bringing the charges closer together, increasing the electrostatic attraction. What happens if we compare the experimental born habber value with the theoretical born mayor value? That comparison tells us a lot about the nature of the bonding. If the two values agree well, it suggests our purely ionic model, hard charged spheres, is a pretty good description. This often happens when the electro negativity difference between the elements is large, say greater than two. If they don't agree well, discrepancies usually point towards significant coalent character in the bonding, meaning the electrons are more shared than fully transferred. Or it could indicate that the ions are highly polarizable. Their electron clouds are easily distorted which introduces extra bonding interactions not captured by the simple ionic model. Silver helides like AGI show big discrepancies for this reason. So it's a test of how ionic a compound really is. It's a good indicator. Yes, these calculations can even predict stability. For instance, you could calculate the lattice enalpy for a hypothetical compound like argon chloride. Argon chloride. Does that exist? No. And the calculation shows why. Argon's ionization energy is incredibly high, and the calculated lice enalpy wouldn't be nearly large enough to compensate, making the overall formation highly unfavorable. Is there a simpler way to estimate lattice enthalpy if you don't know the exact structure or matalong constant? There is the capist equation. It's a clever empirical equation that gives surprisingly good estimates. It basically assumes all ionic compounds are energetically similar to a rock salt structure. and uses average ionotic radi. What's it useful for? It's great for estimating lattice enpies quickly, especially for compounds with complex non-pherical ions like nitrate NO3 or sulfate SO42. You can even use it in reverse to estimate the effective radius, the thermmochemical radius of such complex ions. Okay, so these lattis enthalpies aren't just numbers. What do they help us understand about material behavior? Oh, they explain a lot. For example, the thermal stability of ionic compounds. Think about metal carbonates decomposing when heated like MCO3 solid going to MO solid plus CO2 gas. Right? Like limestone KCO3 decomposing. Exactly. The temperature at which this happens varies a lot. Why does magnesium carbonate decompose in a much lower temperature than berium carbonate? Mg2+ is much smaller than b2 plus m. Does that matter? It does. Small highly charged cations like Mg2+ gain a lot more stability by forming the oxide MGO compared to the carbonate MGCO3. The lattice eny difference between the oxide and carbonate is larger for smaller cations. This makes the decomposition energetically easier from MGCO3 than for Bio3 where the larger B2 plus ion stabilizes the large carbonate annion relatively well. So smallians prefer small onions. Large cations stabilize large onions better. That's a good rule of thumb. It also helps explain the stability of different oxidation states. Why is it easier to make iron fluoride FF3 than iron 3 iodide? Fluoride is much smaller than iodide. Right? Achieving a higher positive oxidation state like plus three requires a lot more energy ionization energy. This energy cost can only be recouped if the lattice enalpy of the resulting compound is very high. Small highly charged onions like FNO2 provide the largest lattice enalpies especially with highly charged. So they are best at stabilizing high oxidation states. Copper 2 iodide for instance isn't stable. It decomposes. Fascinating. What about solubility? How does lattice enalpy play into whether something dissolves in water? Solubility is always a balance between two main energy terms. the lattice enalpy, energy needed to break the solid apart, and the hydration enalpy, energy released when the gaseous ions are surrounded by a water molecule. So, for something to dissolve, the energy you get back from hydration has to sort of overcome the energy cost of breaking the lattice. Essentially, yes. Although entropy plays a role too, but looking at the enalpies gives good trends, there's a general rule. Compounds containing ions of widely different sizes tend to be more soluble in water. Different sizes are more soluble. Why? Think about it. If one ion is very small like Li plus or F, it will have a very large hydration is enaly. Water molecules are strongly attracted to it. This large energy release can often compensate even for a reasonably high lice enalpy. Conversely, if both ions are large like C's plus and I, the lattice enthalpy might be lower, but the hydration enalpies will also be much smaller, making dissolution less favorable. Can you give an example, Trent? Sure. Look at group two sulfates MGSO4 down to baso4. The sulfate ion is large. As you go down the group, the cation gets larger. Mg2+ to b2+. The cation and annion sizes become more similar. And what happens to solubility? It decreases dramatically. Berium sulfate is famously insoluble. Okay, similar sizes, less soluble. What about group two hydroxides? O is small, right? For hydroxides, Mg2 down to ba2, the O annion is small. As the cation gets larger down the group, the size difference increases and the solubility. It increases significantly down the group. Baramin hydroxide is much more soluble than magnesium hydroxide. So, it's the mismatch in size that often favors solubility. Interesting trade-off. It really is. It highlights how properties depend on this delicate balance of competing energy factors. Okay. We spent a lot of time on these perfect idealized crystal structures and their energies, but you mentioned earlier that real solids are never perfect. They have defects. Yes. And this is a crucial point. No crystal is perfectly ordered. There are always imperfections or defects in the structure or composition. Far from being just flaws, these defects are often essential for many important material properties. Why are they unavoidable? Is it just sloppy manufacturing? Not at all. Defects actually form spontaneously. Think about thermodynamics. Creating a defect like removing an atom from its site usually costs energy increases enthalpy but it also introduces disorder into the crystal which increases the entropy. Ah the Gibbs energy equation T Aha. Exactly. At any temperature above absolute zero 0 Kelvin that NH's term becomes significant. The increase in entropy helps to lower the overall Gibbs free energy. So a certain concentration of defects will always form spontaneously because it leads to a more thermodynamically stable state for the crystal as a whole and more defects form at higher temperatures. Yes, the concentration of defects typically increases exponentially with temperature because that entropy term becomes more dominant. These defects are vital for things like diffusion, ionic conductivity and even the mechanical strength of materials. What are the main types of these small localized defects? Point defects. We can categorize them. First, there are intrinsic defects which are present even in a perfectly pure material. The two main types are shotkey and frankle defects. Okay. What's a shotkey defect? A shotkey defect is essentially a vacancy. An atom or ion is simply missing from its normal lattice site. In an ionic crystal like NaCCl, to maintain overall charge neutrality, these vacancies must form in pairs. For every missing no plus aation, there must also be a missing Cl and on somewhere else in the crystal. So it's like removing a matched pair of ions. Does it change the overall formula? No. The overall stoic geometry like 1.1 for NaCCl remains exactly the same. Shodkey defects are common in highly ionic solids, especially those with high coordination numbers. Okay, that's shotkey vacancies. What's a frankle defect? A frankle defect is a bit different. Here, an atom or ion leaves its normal lattice site and moves into an interstitial site. One of those holes or gaps between the regular atoms that isn't normally occupied. So, it creates a vacancy and an interstitial atom nearby. Exactly. The atom is just misplaced, not missing entirely from the crystal. Again, the overall stoometry doesn't change. Freickle defects are more common in structures that have a more open packing, lower coordination numbers, or where there's a large size difference between the and making it easier for the smaller ion, usually the cation, to pop into an interstitial site. Silver helides like AgCl show frankle defects readily. Are there other intrinsic types? You can also have atom interchange or anti-sight defects where two different types of atoms swap places. For example, in an ordered sea alloy, a copper atom might end up on a gold site and a gold atom on a copper site. This is less common in purely ionic solids because putting say two positive ions next to each other would be very unfavorable electrostatically. Okay, those are intrinsic defects. What about defects caused by impurities, extrinsic defects, right? These are introduced when you have foreign atoms or dopants incorporated into the crystal lattice usually substituting for one of honors the host atoms like in gemstones. I heard that's how they get their color. That's a perfect example. Ruby is aluminum oxide Al23 which is normally colorless. But if a tiny fraction of the Al3+ ions are replaced by chromium ions CR3+ which have a similar size and charge, the crystal becomes brilliant red. The CR3+ impurity absorbs light differently than Al3+. Wow. And sapphire. Sapphire is also AL203, but its blue color often comes from a combination of iron Fe2+ and titanium TY4 plus impurities substituting for Al3 plus A. The charge difference requires some clever charge compensation mechanisms. But the key is the impur atoms altering the light absorption. That leads to another type of defect related to color, right? F centers. Yes. Color centers or fenters from the German word farb for color. An fenter is a specific type of defect. An electron that gets trapped inside an anion vacancy. An electron sitting where a negative ion should be. Yep. Exactly. This trapped electron has its own set of energy levels and can absorb visible light causing the crystal to appear colored. For instance, if you heat sodium chloride crystal in sodium vapor, some sodium atoms diffuse in, lose an electron, and those electrons can get trapped at CL vacancy sites, turning the normally clear NaCCl crystal yellowish or orange. Many natural minerals like purple fluoride owe their color to Fenters or similar electronic defects. So, defects aren't just structural, they can be electronic, too. This idea of variable composition seems important. What about compounds that don't have a perfectly fixed integer ratio of atoms? Those are non- stokometric compounds. They have a definite structure type, but the relative number of atoms can vary slightly within a certain range. A classic example is wooite, which we write as F1 XO, meaning it's always a bit deficient in iron. X is greater than zero. Yes, it typically ranges from about FET0.850 to FE.950. It never quite reaches perfect 1.1 FO. The crystal structure is basically the rock salt type, but it always has some vacancies on the iron cation sites. How does it stay charge neutral if it's missing positive iron ions? That's the key requirement for non- stowic geometry. At least one of the elements must be able to exist in multiple stable oxidation states. In F1, some of the iron ions must be F3+ instead of F2+ to compensate for the charge deficit from the missing F2 plusations. This is common for transition metals DB block lanthnate sactides FB block and some heavier P block elements and the properties change smoothly with composition generally yes things like the lattice parameter size of the unit cell often change smoothly and predictably with composition following what's sometimes called vagard's rule this smooth variation within a single structure type is also characteristic of solid solutions like the substitutional or interstitial alloys we discussed earlier for example you can make materials like law 1x fs S Fe3 continuously replacing law 3+ with SR2 plus delin and again the iron has to change oxidation state to balance the charge exactly as you add SR2 plus a some FF3+ must convert to F4 plus to keep things neutral overall controlling this non- stoometry and the resulting oxidation states is absolutely critical in designing materials with specific electronic or magnetic properties like in battery materials or catalysts or those perovskite superconductors. Which brings us full circle to the final piece, the electronic structure of solids. How do all these arrangements, perfect lises, defects, non- stoic geometry, determine whether a material conducts electricity or not? Right? To understand conductivity, we need to extend our ideas of molecular orbitals from simple molecules to the essentially infinite array of atoms in a solid. This leads to the concept of electronic bands. Based on how electrons fill these bands, we classify materials. What are the main electrical categories? You have metallic conductors where conductivity is generally high but decreases as temperature increases. Why does it decrease with heat? Because the increasing vibrations of the atoms in the lattice scatter the moving electrons more effectively hindering their flow increasing resistance. Okay. Semiconductors their conductivity is generally lower than metals at room temperature but it increases significantly as temperature rises. The opposite trend to metals. Exactly. Then you have insulators which have extremely low electrical conductivity. If you can measure any conductivity, it also tends to increase with temperature. So you can think of an insulator as just a semiconductor with a really really poor conductivity. And the special case superconductors which below a certain critical temperature exhibit zero electrical resistance. That's a whole fascinating quantum phenomenon itself. So how does thinking about molecular orbitals expanding into a solid explain these differences? What are these bands? Imagine bringing atoms together to form a solid. Their individual atomic orbitals like spd orbitals start to overlap with those on neighboring atoms. Just like in a molecule, overlapping atomic orbitals combine to form molecular orbitals. But in a solid with a huge number n of atoms, you form a huge number of molecular orbitals. Billions and billions essentially. Yes. These n molecular orbitals are still discrete energy levels, but they are packed incredibly close together in energy forming what appears to be a continuous band of allowed energy levels. You'll get an S band from overlapping S orbitals, a P band from P orbitals, and so on. And are there energies where no orbitals exist? Yes. Between these bands of allowed energy levels, there can be band gaps, ranges of energy where there are no molecular orbitals derived from the atomic orbitals we considered. Electrons simply cannot have energies that fall within a band gap. So, electrons fill these bands just like they fill orbitals in an atom. Exactly. At absolute zero temperature, electrons fill the lowest available energy levels within the bands up to a certain maximum energy level. This highest occupied energy level at 0K is called the Fermy level. And the Fermy level's position is key. Absolutely critical. If the fermy level falls within an energy band, meaning the band is only partially filled with electrons, then electrons near the fermy level have empty energy levels immediately adjacent to them within the same band that they can easily move into with just a tiny bit of energy like from an applied electric field. And that means they can move and carry current. Precisely. That situation defines a metallic conductor. There's a continuous path of available states for electrons to move through. Okay. So what makes an insulator? An insulator occurs when you have just enough veence electrons to completely fill one or more bands and the firmy level falls exactly at the top of the highest filled band called the veance band. Crucially, there is then a large energy gap, a large band gap separating the top of this filled veance band from the bottom of the next available empty band, the conduction band. like an an perfect example you have filled bands derived from chlorine's orbitals the veance band and empty bands derived from sodium's orbitals the conduction band the gap between them is huge around 7 electron volts eV thermal energy at room temperature is only about 03 eV nowhere near enough to kick electrons across that gap so electrons are stuck in the filled band nowhere to move exactly no mobile charge carriers hence it's an insulator so semiconductor must be the middle Yes, a semiconductor is fundamentally like an insulator but with one crucial difference. The band gap between the filled veance band and the empty conduction band is much smaller. Small enough for electrons to jump across. Small enough that even at room temperature, thermal energy KT is sufficient to excite a significant number of electrons from the top of the veence band up into the bottom of the conduction band. And what happens then? Now you have mobile electrons in the nearly empty conduction band that can carry current. But also when an electron leaves the veence band, it leaves behind an empty state or a hole. This hole can also effectively move through the veilance band as adjacent electrons jump into it. Holes act like mobile positive charge carriers. So you get both negative electron and positive whole charge carriers. Yes. In an intrinsic semiconductor, a pure material. Yes. And because the number of these carriers depends on thermal exitation across the gap, their concentration increases exponentially with temperature, which is why conductivity increases with temperature. Silicon and germanmanium are classic intrinsic semiconductors. But most semiconductors we use aren't pure, right? They're doped. Correct. The conductivity of semiconductors can be dramatically increased and controlled by intentionally adding tiny amounts of specific impurities. Doping. This creates extrinsic semiconductors. How does doping work? Let's say doping silicon. If silicon has four veence electrons, right? If you dope silicon with an element that has five veence electrons like arsenic, the arsenic atom replaces a silicon atom in the lattice, four of its electrons participate in bonding like silicons, but there's one extra electron left over. What happens to that extra electron? This extra electron is only weakly bound to the arsenic atom. It occupies a localized energy level that sits just below the conduction band of silicon within the band gap. We call these donor levels. Easy to kick into the conduction band. Very easy. It takes very little thermal energy to promote these electrons from the donor levels into the conduction band. This vastly increases the number of negative charge carriers electrons. We call this an ntype semiconductor. N for negative. Okay, so N type has extra electrons. What about ptype? For ptype semiconductors, you dope silicone with an element that has fewer veence electrons, say only three, like gallium. When gallium replaces silicon, it can only form three coalent bonds properly. There's one missing electron needed to complete the bonding structure around it. So, it creates a hole nearby effectively. Yes. This creates localized empty energy levels called acceptor levels that sit just above the veance band of silicon within the band gap. So electrons from the veance band can jump up into these easily. It takes very little energy for an electron from the veance band to be thermally excited into one of these empty acceptor levels. This leaves behind a mobile hole in the veance band. Since the dominant charge carriers are these positive holes, we call this a ptype semiconductor. P for positive. And this ability to create n type and ptype materials is the foundation of modern electronics. Junctions between Nype and Ptype semiconductors form dodes and transistors. The building blocks of integrated circuits, computers, solar cells. Pretty much everything electronic relies on controlling charge carriers and semiconductors through doping. Wow. It really all connects from simple sphere packing through energy considerations, defects, right up to the quantum band structure that governs conductivity. What an incredible journey through the solid state. It really is. We started with describing structures using unit cells. saw how close packing gives efficiency and how filling the holes leads to diverse structures like rock salt, fluoride, perovskate. We used lattice enthalpy via Bourne Haber cycles and the Bourne mayor equation to understand stability, right? And link that energy to real properties like thermal stability and solubility. Then we saw that defects aren't flaws but essential features creating color or enabling non-toicometry. And finally, band theory provided the framework to understand why materials have the electronic properties they do, why metals conduct, insulators insulate, and semiconductors. Well, semiconduct paving the way for all our technology. Thinking about this whole deep dive, it's just amazing how the properties of everything solid around us, the mug in your hand, the glass in the window, the chip in your phone, all stem from these fundamental atomic level arrangements and energy rules. It makes you wonder what completely new materials could we design if we could gain perfect atom byatom control over these structures, the defects within them, and the resulting electron bands. Imagine dialing in the exact band gap needed for a superefficient solar cell or precisely controlling defect concentrations to create novel catalysts or quantum computing materials. The possibilities seem almost endless if we can master manipulating matter at that fundamental level. A tantalizing thought for the future indeed. Well, thank you for joining us on this exploration of solid state chemistry. We hope this deep dive has given you a clearer picture of the fascinating world beneath the surface of the solids all around us. And thank you as always for bringing your curiosity. We really enjoy guiding you through these topics. Until next time, keep digging deeper. |
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Introduction
What is engineering mechanics
Fundamental concepts and principles
Numerical calculations
Vectors and their Operations
Scalars and vectors
Vector operations using the parallelogram rule and trigonometry
Vector components
Cartesian vector notation
Vector operations using Cartesian vector notation
Position vectors
Dot product
Cross product
Forces and Moments
Forces
Moments
Couple moment
Simplification of force and couple systems
Distributed loads
Free Body Diagram
Equilibrium of Particles and Rigid Bodies
Equilibrium of a particle
Conditions for two dimensional rigid-body equilibrium
Structural Analysis
Analysis of trusses
Internal Forces
Types of internal forces
Shear and moment equations and their diagrams
Relationships between Load, Shear, and Moments
Friction
Dry friction
Equilibrium involving friction
Centres of Bodies
Center of gravity
Center of mass
Centroid
Moments of Inertia of area
Rectangular moment of inertia
Polar moment of inertia
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Internal Forces: Shear and moment equations and their diagrams
The internal forces and couple moments in a loaded beam vary along a beam due to the loading conditions (e.g. number of external forces and their distances to the location of the internal forces) are different for different sections of the beam. Consequently, the values of the internal forces and couple moments along a beam must be known for a beam design. For example, maximum values of the shear force and bending moment and their locations along a beam are needed during a beam design.
The internal forces and couple moment acting at any point along the axis of a beam can be determined by the method of sections. In this regard, we need to define the section with a reference point, normally from the left end of the beam. Hence, the location of the internal forces of interest is defined by the distance from the left end of the beam to the point of interest denoted by (Fig. 7.14). Therefore, shear and bending moment at each point become functions of x (and the external forces), i.e.,
Fig. 7.14 Sectioning a beam at an arbitrary point located away from the left end of a beam.
Remark: the signs of the results from , , and refer to the member sign convention (Fig. 7.8, 7.9, and 7.10) because the unknown directions of internal forces are assumed positive according to that sign convention.
Structural beams are commonly long and straight members designed to carry loads mainly perpendicular to their long axes, upward or downward. Perpendicular loads on a beam affect only the shear force and bending moment within the beam, not the axial force In this case, axial force is zero anywhere along the beam.
The functions and and their diagrams (graphs or plots) provide powerful tools for evaluating the values and variations of the shear force and bending moment at any point. The following example demonstrates how to obtain and using the method of sections and plotting them. The and plots are referred to as shear force diagram (SFD) and bending moment diagram (BMD) respectively.
Example 7.2.1
Determine the and for the loaded beam shown in the figure.
SOLUTION
1- Draw the FBD of the beam.
In the FBD, the directions of the unknown force and moment are assumed positive according to the member sign convention.
2- Solve the equations of equilibrium for the support reactions.
3- Make a cut in the FBD of the beam at an arbitrary point x meter away from the left end of the beam as shown. Choose one of the two segments for analysis. For instance, the segment chosen is as shown below.
4- Write the equations of equilibrium for the resultant segment and solve for the shear force and bending moment at ,
Therefore,
5- Plot the functions and on x–y plots, with the x axis representing the distance from the left end of the beam, and the y axis representing the values of and . The plot gives a shear force diagram (SFD) and the plot gives a bending moment diagram (BMD).
In general, the loading condition on a section between the reference end and the cut will change as the cut moves from one end of the beam to the other, the inputs (external load) to the and will change. Therefore, and may need to be specifically defined for different sections along the axis of a beam. A beam should be divided into sections along its axis (length) at locations where the external load pattern on the FBD of a beam changes. Figure 7.15 shows the sections needed for determining and .
Fig. 7.15 Locations (indicated by letters) at which the external load pattern changes and the regions (indicated by numbers in circles) that different sets of and are needed.
Example 7.2.2
Draw the shear force and bending moment diagrams of the loaded beam shown in the figure.
SOLUTION
1- Draw the FBD of the beam.
2- Solve the equations of equilibrium for the support reactions.
3- Divide the beam (its FBD) into regions based on pattern change of the external loads.
4- In each region, cut at an arbitrary point meter away from the left end of the beam, write the equations of equilibrium, and solve for and .
Arbitrary cut in region 1.
The notation means sum of the moments about the point at which the imaginary cut is made.
Therefore,
Note that the region for does not include . The reaction force at is a concentrated force under which the shear force has a rapid change (see Example 7.1.2) or mathematically a jump discontinuity. Therefore, the point at which a concentrated force acts should not be included in the region for .
_Arbitrary cut in region_ 2.
Therefore,
_Arbitrary cut in region_ 3.
Therefore,
Note that the region for does not include . We will see that there will be a jump discontinuity in the diagram of at . The shear force at the location of a concentrated force changes rapidly from one side of the concentrated force to the other side. We just need to know the shear forces on the two sides right next to the concentrated force.
_Arbitrary cut in region_ 4.
Therefore,
Note that the region for does not include . We will see that there will be a jump discontinuity in the diagram of at . The bending moment at the location of a couple moment changes rapidly from one side of the couple moment to the other side. We just need to know the bending moments on the two sides right next to the couple moment.
Arbitrary cut in region 5.
Therefore,
5- Plot each functions for each region over the length of the beam.
Collecting the functions over all the regions, you can write,
and,
The SFD and BMD are as follow.
Remark: Any point at which a concentrated force is applied is associated with a jump discontinuity in , this point should not be included in the region over which is being determined. Similarly, Any point at which a couple moment is applied is associated with a jump discontinuity in , this point should not be included in the region over which is being determined.
In the above example, we can visually notice the following effects and changes in the diagrams.
1- A point force creates a jump discontinuity in SFD.
2- A couple moment creates a jump discontinuity in BMD.
3- A discontinuity in the distributed load creates a sudden change of slope (i.e. discontinuity in the slope) in SFD.
4- A point force (also) creates a sudden change of slope in BMD.
These observations are not limited to this problem, but rather they follow general rules that are formally presented in the next section.
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3.5: Derivatives of Trigonometric Functions
Last updated
: Jan 17, 2025
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3.4E: Exercises for Section 3.4
3.5E: Exercises for Section 3.5
Page ID
: 2494
Gilbert Strang & Edwin “Jed” Herman
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
Find the derivatives of the sine and cosine
function
.
Find the derivatives of the standard
trigonometric functions
.
Calculate the higher-order derivatives of the sine and cosine.
One of the most important types of motion in physics is
simple harmonic motion
, which is associated with such systems as an object with mass oscillating on a spring.
Simple harmonic motion
can be described by using either sine or cosine functions. In this section we expand our knowledge of
derivative
formulas to include derivatives of these and other
trigonometric functions
. We begin with the derivatives of the sine and cosine functions and then use them to obtain formulas for the derivatives of the remaining four
trigonometric functions
. Being able to calculate the derivatives of the sine and cosine functions will enable us to find the velocity and
acceleration
of
simple harmonic motion
.
Derivatives of the Sine and Cosine Functions
We begin our exploration of the
derivative
for the sine
function
by using the formula to make a reasonable guess at its
derivative
. Recall that for a
function
f(x),
f′(x)=limh→0f(x+h)−f(x)h.
Consequently, for values of h very close to 0,
f′(x)≈f(x+h)−f(x)h.
We see that by using h=0.01,
ddx(sinx)≈sin(x+0.01)−sinx0.01
By setting
D(x)=sin(x+0.01)−sinx0.01
and using a graphing utility, we can get a graph of an approximation to the
derivative
of sinx (Figure 3.5.1).
Upon inspection, the graph of D(x) appears to be very close to the graph of the cosine
function
. Indeed, we will show that
ddx(sinx)=cosx.
If we were to follow the same steps to approximate the
derivative
of the cosine
function
, we would find that
ddx(cosx)=−sinx.
The Derivatives of sinx and cosx
The
derivative
of the sine
function
is the cosine and the
derivative
of the cosine
function
is the negative sine.
ddx(sinx)=cosx
ddx(cosx)=−sinx
Proof
Because the proofs for ddx(sinx)=cosx and ddx(cosx)=−sinx use similar techniques, we provide only the proof for ddx(sinx)=cosx. Before beginning, recall two important trigonometric limits:
limh→0sinhh=1 and limh→0cosh−1h=0.
The graphs of y=sinhh and y=cosh−1h are shown in Figure 3.5.2.
We also recall the following
trigonometric identity
for the sine of the sum of two angles:
sin(x+h)=sinxcosh+cosxsinh.
Now that we have gathered all the necessary equations and identities, we proceed with the proof.
ddx(sinx)=limh→0sin(x+h)−sinxhApply the definition of the
derivative
.=limh→0sinxcosh+cosxsinh−sinxhUse trig identity for the sine of the sum of two angles.=limh→0(sinxcosh−sinxh+cosxsinhh)Regroup.=limh→0(sinx(cosh−1h)+(cosx)(sinhh))Factor out sinx and cosx=(sinx)limh→0(cosh−1h)+(cosx)limh→0(sinhh)Factor sinx and cosx out of limits.=(sinx)(0)+(cosx)(1)Apply trig
limit
formulas.=cosxSimplify.
□
Figure 3.5.3 shows the relationship between the graph of f(x)=sinx and its
derivative
f′(x)=cosx. Notice that at the points where f(x)=sinx has a horizontal
tangent
, its
derivative
f′(x)=cosx takes on the value zero. We also see that where f(x)=sinx is increasing, f′(x)=cosx>0 and where f(x)=sinx is decreasing, f′(x)=cosx<0.
Example 3.5.1: Differentiating a Function Containing sinx
Find the
derivative
of f(x)=5x3sinx.
Solution
Using the
product rule
, we have
f′(x)=ddx(5x3)⋅sinx+ddx(sinx)⋅5x3=15x2⋅sinx+cosx⋅5x3.
After simplifying, we obtain
f′(x)=15x2sinx+5x3cosx.
Exercise 3.5.1
Find the
derivative
of f(x)=sinxcosx.
Hint
: Don’t forget to use the
product rule
.
Answer
: f′(x)=cos2x−sin2x
Example 3.5.2: Finding the Derivative of a Function Containing cos x
Find the
derivative
of g(x)=cosx4x2.
Solution
By applying the
quotient rule
, we have
g′(x)=(−sinx)4x2−8x(cosx)(4x2)2.
Simplifying, we obtain
g′(x)=−4x2sinx−8xcosx16x4=−xsinx−2cosx4x3.
Exercise 3.5.2
Find the
derivative
of f(x)=xcosx.
Hint
: Use the
quotient rule
.
Answer
: f′(x)=cosx+xsinxcos2x
Example 3.5.3: An Application to Velocity
A particle moves along a coordinate axis in such a way that its position at time t is given by s(t)=2sint−t for 0≤t≤2π. At what times is the particle at rest?
Solution
To determine when the particle is at rest, set s′(t)=v(t)=0. Begin by finding s′(t). We obtain
s′(t)=2cost−1,
so we must solve
2cost−1=0 for 0≤t≤2π.
The solutions to this equation are t=π3 and t=5π3. Thus the particle is at rest at times t=π3 and t=5π3.
Exercise 3.5.3
A particle moves along a coordinate axis. Its position at time t is given by s(t)=√3t+2cost for 0≤t≤2π. At what times is the particle at rest?
Hint
: Use the previous example as a guide.
Answer
: t=π3,t=2π3
Derivatives of Other Trigonometric Functions
Since the remaining four
trigonometric functions
may be expressed as quotients involving sine, cosine, or both, we can use the
quotient rule
to find formulas for their derivatives.
Example 3.5.4: The Derivative of the Tangent Function
Find the
derivative
of f(x)=tanx.
Solution
Start by expressing tanx as the quotient of sinx and cosx:
f(x)=tanx=sinxcosx.
Now apply the
quotient rule
to obtain
f′(x)=cosxcosx−(−sinx)sinx(cosx)2.
Simplifying, we obtain
f′(x)=cos2x+sin2xcos2x.
Recognizing that cos2x+sin2x=1, by the Pythagorean theorem, we now have
f′(x)=1cos2x
Finally, use the identity secx=1cosx to obtain
f′(x)=sec2x.
Exercise 3.5.4
Find the
derivative
of f(x)=cotx.
Hint
: Rewrite cotx as cosxsinx and use the
quotient rule
.
Answer
: f′(x)=−csc2x
The derivatives of the remaining
trigonometric functions
may be obtained by using similar techniques. We provide these formulas in the following theorem.
Derivatives of tanx, cotx, secx, and cscx
The derivatives of the remaining
trigonometric functions
are as follows:
ddx(tanx)=sec2xddx(cotx)=−csc2xddx(secx)=secxtanxddx(cscx)=−cscxcotx.
Example 3.5.5: Finding the Equation of a Tangent Line
Find the equation of a line
tangent
to the graph of f(x)=cotx at x=π4.
Solution
To find the equation of the
tangent
line, we need a point and a
slope
at that point. To find the point, compute
f(π4)=cotπ4=1.
Thus the
tangent
line passes through the point (π4,1). Next, find the
slope
by finding the
derivative
of f(x)=cotx and evaluating it at π4:
f′(x)=−csc2x and f′(π4)=−csc2(π4)=−2.
Using the
point-slope equation
of the line, we obtain
y−1=−2(x−π4)
or equivalently,
y=−2x+1+π2.
Example 3.5.6: Finding the Derivative of Trigonometric Functions
Find the
derivative
of f(x)=cscx+xtanx.
Solution
To find this
derivative
, we must use both the
sum rule
and the
product rule
. Using the
sum rule
, we find
f′(x)=ddx(cscx)+ddx(xtanx).
In the first
term
, ddx(cscx)=−cscxcotx, and by applying the
product rule
to the second
term
we obtain
ddx(xtanx)=(1)(tanx)+(sec2x)(x).
Therefore, we have
f′(x)=−cscxcotx+tanx+xsec2x.
Exercise 3.5.5
Find the
derivative
of f(x)=2tanx−3cotx.
Hint
: Use the rule for differentiating a constant multiple and the rule for differentiating a difference of two functions.
Answer
: f′(x)=2sec2x+3csc2x
Exercise 3.5.6
Find the
slope
of the line
tangent
to the graph of f(x)=tanx at x=π6.
Hint
: Evaluate the
derivative
at x=π6.
Answer
: 43
Higher-Order Derivatives
The higher-order derivatives of sinx and cosx follow a repeating pattern. By following the pattern, we can find any
higher-order derivative
of sinx and cosx.
Example 3.5.7: Finding Higher-Order Derivatives of y=sinx
Find the first four derivatives of y=sinx.
Solution
Each step in the chain is straightforward:
y=sinxdydx=cosxd2ydx2=−sinxd3ydx3=−cosxd4ydx4=sinx
Analysis
Once we recognize the pattern of derivatives, we can find any
higher-order derivative
by determining the step in the pattern to which it corresponds. For example, every fourth
derivative
of sinx equals sinx, so
d4dx4(sinx)=d8dx8(sinx)=d12dx12(sinx)=…=d4ndx4n(sinx)=sinx
d5dx5(sinx)=d9dx9(sinx)=d13dx13(sinx)=…=d4n+1dx4n+1(sinx)=cosx.
Exercise 3.5.7
For y=cosx, find d4ydx4.
Hint
: See the previous example.
Answer
: cosx
Example 3.5.8: Using the Pattern for Higher-Order Derivatives of y=sinx
Find d74dx74(sinx).
Solution
We can see right away that for the 74th
derivative
of sinx, 74=4(18)+2, so
d74dx74(sinx)=d72+2dx72+2(sinx)=d2dx2(sinx)=−sinx.
Exercise 3.5.8
For y=sinx, find d59dx59(sinx).
Hint
: d59dx59(sinx)=d4⋅14+3dx4⋅14+3(sinx)
Answer
: −cosx
Example 3.5.9: An Application to
A particle moves along a coordinate axis in such a way that its position at time t is given by s(t)=2−sint. Find v(π/4) and a(π/4). Compare these values and decide whether the particle is speeding up or slowing down.
Solution
First find v(t)=s′(t)
v(t)=s′(t)=−cost.
v(π4)=−1√2=−√22.
Next, find a(t)=v′(t). Thus, a(t)=v′(t)=sint and we have
a(π4)=1√2=√22.
Since v(π4)=−√22<0 and a(π4)=√22>0, we see that velocity and
acceleration
are acting in opposite directions; that is, the object is being accelerated in the direction opposite to the direction in which it is traveling. Consequently, the particle is slowing down.
Exercise 3.5.9
A block attached to a spring is moving vertically. Its position at time t is given by s(t)=2sint. Find v(5π6) and a(5π6). Compare these values and decide whether the block is speeding up or slowing down.
Hint
: Use Example 3.5.9 as a guide.
Answer
: v(5π6)=−√3<0 and a(5π6)=−1<0. The block is speeding up.
Key Concepts
We can find the derivatives of sinx and cosx by using the definition of
derivative
and the
limit
formulas found earlier. The results are
ddx(sinx)=cosxandddx(cosx)=−sinx.
With these two formulas, we can determine the derivatives of all six basic
trigonometric functions
.
Key Equations
Derivative
of sine
function
ddx(sinx)=cosx
Derivative
of cosine
function
ddx(cosx)=−sinx
Derivative
of
tangent
function
ddx(tanx)=sec2x
Derivative
of cotangent
function
ddx(cotx)=−csc2x
Derivative
of
function
ddx(secx)=secxtanx
Derivative
of cosecant
function
ddx(cscx)=−cscxcotx
3.4E: Exercises for Section 3.4
3.5E: Exercises for Section 3.5 |
190341 | https://www.rdocumentation.org/packages/DescTools/versions/0.99.7/topics/BinomDiffCI | BinomDiffCI function - RDocumentation
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DescTools (version 0.99.7)
BinomDiffCI: Confidence Interval for a Difference of Binomials
Description
Calculate confidence interval for a risk difference.
Usage
BinomDiffCI(x1, n1, x2, n2, conf.level = 0.95, method = c("wald", "waldcor", "ac", "exact", "newcombe", "newcombecor", "fm", "ha"))
Arguments
x1
number of successes for the first group.
n1
number of trials for the first group.
x2
number of successes for the second group.
n2
number of trials for the second group.
conf.level
confidence level, defaults to 0.95.
method
one out of "wald", "waldcor", "ac", "exact", "newcombe", "newcombecor", "fm", "ha".
Value
A vector with 3 elements for estimate, lower confidence intervall and upper for the upper one.
Details
Still to follow.
References
Fagerland M W, Lydersen S and Laake P (2011) Recommended confidence intervals for two independent binomial proportions, Statistical Mehtods in Medical Research 0(0) 1-31
See Also
BinomCI, MultinomCI, binom.test, prop.test
Examples
Run this code
r
BinomDiffCI(14, 70, 32, 80, method="wald")
BinomDiffCI(14, 70, 32, 80, method="waldcor")
BinomDiffCI(14, 70, 32, 80, method="ac")
BinomDiffCI(14, 70, 32, 80, method="newcombe")
Run the code above in your browser usingDataLab |
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ContentsOverviewTest DetailsResults and Follow-UpAdditional Details
Overview
What is an alpha-fetoprotein test?
The alpha-fetoprotein (AFP) test is a blood test for pregnant women. Healthcare providers use the AFP test to check a fetus’s risks of birth defects or genetic conditions.
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The AFP test measures alpha-fetoprotein levels. Alpha-fetoprotein is a protein that develops in the liver of a fetus. Irregular AFP levels in pregnant women can be a sign of a genetic disorder or birth defect (congenital disability).
What is the AFP tumor marker test?
Most commonly, healthcare providers use AFP tests in a pregnant woman. But in some cases, your healthcare provider might order an AFP tumor marker test when you’re not pregnant. The tumor marker test has different purposes than the AFP test for pregnant women.
Newborn babies typically have high AFP levels that drop to low ranges by age 1. In an adult, high AFP may point to:
Cirrhosis.
Hepatitis A, hepatitis B or hepatitis C.
Liver cancer.
Ovarian or testicular cancer.
Toxic hepatitis or autoimmune hepatitis.
Healthcare providers may also use AFP tumor marker tests to plan cancer treatment. For example, if you have high AFP levels and liver cancer, targeting cancer cells with specific drugs (targeted therapy) may be an effective treatment.
It’s important to understand that AFP tumor marker tests are simply tools to measure AFP levels. They can help diagnose cancer or other conditions. But an irregular test result doesn’t necessarily mean there’s a problem. Usually, your provider will order follow-up tests to learn more.
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When is the AFP test performed in pregnancy?
Pregnant women typically have the AFP test between 16 and 22 weeks of pregnancy.
All pregnant women should get the AFP test, but your healthcare provider may especially recommend it if you:
Are older than 35.
Have a family history of birth defects.
Have diabetes.
What does the AFP test check for?
The test may detect:
Genetic disorders, such as Down syndrome or Turner syndrome.
Incorrect due dates.
Multiples, such as twins or triplets.
Neural tube defects, such as spina bifida or anencephaly.
Test Details
How does an AFP test work?
An AFP test is a blood test. During blood tests, your healthcare provider:
Places a small needle into one of your arm veins.
Collects a tube (vial) of blood.
Removes the needle and applies cotton and a bandage to the area.
AFP tests usually take less than five minutes. They're quick and painless for most people.
If you have a fear of needles (trypanophobia) or anxiety about getting blood draws, tell the healthcare provider who’s taking your blood. It may help to practice breathing exercises, look away from the needle or bring a friend for support.
How do I prepare for an AFP test?
There’s no special preparation for an AFP blood test. Generally, staying hydrated by drinking plenty of water can help the blood draw go more quickly and smoothly. Hydration prevents your veins from tightening (constricting) and decreases the chance of blood pressure changes.
What are the risks of an AFP test?
All blood tests may cause some soreness, bruising or redness near where the needle was inserted. Usually, these symptoms are mild and go away within a day.
Care at Cleveland Clinic
Schedule Prenatal Testing
Find a Doctor and Specialists
Make an Appointment
Results and Follow-Up
What should I know about the results of an alpha-fetoprotein test?
If your AFP test results are abnormal, your provider will order more tests to confirm the diagnosis. Higher than usual AFP levels can point to neural tube defects. Lower than usual AFP levels may indicate that the fetus has a genetic disorder such as Down syndrome.
It’s important to understand that irregular AFP levels don’t automatically mean the fetus has a genetic condition or a neural tube defect. Irregular AFP levels can also mean:
You are having multiples, such as twins.
Your due date was miscalculated.
Is an AFP test the only test I need?
If your AFP levels are outside the normal range, your provider will likely order more tests to diagnose or rule out conditions.
During pregnancy, you'll have other prenatal tests that check for genetic disorders along with the AFP test. For example, you usually have a triple screen test around the same time as an AFP test. The triple screen looks at your levels of specific hormones. Triple screen tests also check for genetic disorders such as Down syndrome.
If your test results show that the fetus may have a genetic condition, your provider may recommend chorionic villus sampling (CVS). CVS is a genetic test that evaluates cells from the placenta. It can detect genetic disorders such as:
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Cystic fibrosis.
Down syndrome.
Sickle cell disease.
Tay-Sachs disease.
What else should I ask my doctor?
You may also want to ask your healthcare provider:
When should I schedule an AFP test?
What do my AFP levels mean?
What other tests do I need to check for genetic disorders?
What other tests do I need to check for neural tube defects?
Additional Details
How accurate is the AFP test?
AFP tests are good tools for detecting genetic disorders or neural tube defects. Up to 9 in 10 fetuses with neural tube defects were diagnosed because the AFP test detected abnormalities. But AFP tests aren’t perfect.
In every 1,000 pregnant women who get the AFP test, up to 50 get abnormal results. Of those 50, only about 1 in 16 have a baby with a genetic condition or neural tube defect.
A note from Cleveland Clinic
Alpha-fetoprotein tests are blood tests to check AFP between 16 and 22 weeks of pregnancy. Irregular AFP levels may point to the presence of a genetic disorder or neural tube defect in the fetus. It’s important to understand that AFP tests are screening tools, not diagnostic ones. Atypical AFP levels don’t automatically mean the fetus has a disorder. If you have irregular AFP test results, your provider will likely order more tests to diagnose or rule out genetic conditions or neural tube defects.
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Prenatal tests can give your providers information about your pregnancy and fetal development. Cleveland Clinic’s experts can guide you through prenatal testing.
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190343 | https://www.vedantu.com/formula/selling-price-formula | Selling Price Formula
Selling Price Formula in Maths
We experience different situations every day when we need to calculate or compare things. Especially situations involving the sale or purchase of goods. The selling price is used to sell the item at a certain cost and can be calculated using the selling price formula. The amount that the buyer pays to buy the product is called the selling price. The actual selling price is the price the buyer pays to buy a product or service. This is the price that is higher than the cost of goods and includes a profit percentage. If the seller wishes, they can also keep the selling price similar to the cost price, if the buyer does not wish to gain profit. Determining the selling price is a very sensitive issue because sales of a product are largely based on it. We can calculate the selling price in various ways and formulas.
The Basic Formula
SP = CP + Profit
Where,
SP= Selling Price
CP= Cost Price
This chapter deals with selling price and its role in calculating the percentage of profit and loss. We also learn the difference between selling price and marked price. We also learn how to calculate the selling price of a product using different formulas. There are various examples that will help us understand better about the selling price of an object.
Important Selling Price Formula
Selling price = Cost Price + Profit
Selling price = Marked/List price – Discount
Selling price = (100+%Profit)/100 × Cost price
Selling price = (100− % Los)/100 × Cost price
Other Important Formulas Related To Selling Price
Element | Formula
Cost price | Selling price – Profit
Profit | Selling price – Cost Price
Loss | Cost Price – Selling Price
% Profit | Profit/Cost Price × 100
% Loss | Loss/Cost Price × 100
Element
Formula
Cost price
Selling price – Profit
Profit
Selling price – Cost Price
Loss
Cost Price – Selling Price
% Profit
Profit/Cost Price × 100
% Loss
Loss/Cost Price × 100
Selling Price Vs. Marked Price
Marked price also known as the list price is the price that a seller spells out to the purchaser while selling price is the price that the seller actually receives from the buyer after a bargain or making a deal. In general, the selling price is lower than the marked price. However, sometimes the selling price and the marked price can be the same also. A fixed price shop, meaning that the shopkeeper that does not offer any discount or price cut of any sort is an example of it.
Calculate Selling Price Per Unit
Following is the step-by-step procedure to calculate the selling price per unit:
Identify the total cost of all units being bought
Divide the total cost by the number of units bought to obtain the cost price.
Use the selling price formula to find out the final price i.e.: SP = CP + Profit Margin
Margin will then be added to the cost of the commodity in order to identify the appropriate pricing.
Thus, the selling price per unit formula to find the price per unit from the income statement, divide sales by the number of units or quantity sold to identify the price per unit.
For example, given sales of $80,000 for the year and 2,000 units sold, the price per unit is Rs.40 (80,000 divided by 2,000).
How to Calculate Cost-Plus Pricing
Markup is the amount of difference between an item’s cost and its selling price. Usually, depending on the industry type, it is demonstrated as a percentage of the cost.
Margin also referred to as Gross Profit) = Selling price – Cost of goods sold (COGS).
Margin and Markup move in tandem. For example, a 40% markup is always equivalent to a profit margin of 28.6%, while a 50% markup is always equivalent to a margin value of 33%.
Cost Price
Cost price is actually the ultimate price at which the seller buys the product or service. He then adds a percentage of profit to it. The list price or marked price is the price which a seller fixes after adding the needed percentage of profit.
Solved Examples
Example: Maria marks all her products 30% above the cost price and offers a discount of 5% on the marked price. She is of the viewpoint that she will earn a profit of 20%. What do you think is the percentage of the profit she earns?
Solution:
Let the cost price of the products be 100.
Thus, the list price/marked price will be = ₹100 + 30% of the cost price.
= 100 + 30
= 130
Now, the Selling price = List/Marked price – Discount
= 130 – 5% of 130 = 130 – 6.5
= 123.5
Therefore, the profit = SP-CP
= 123.5 – 100 = 23.5
Hence, the percentage of profit she earned is below 20%.
Example: A new retailer in the market marked all his goods at 50% above the cost price thinking that he will still earn a profit percent of 25%, offering a discount of 25% on the list price. Find out his actual profit on the sales?
Solution:
Let the cost price = Rs. 100
Then, list price = Rs. 150
Thus, Selling Price = 75% of Rs. 150
= Rs. 112.50.
Hence we can conclude that the profit % he earned = 12.50%.
FAQs on Selling Price Formula
1.What is the Significance of a Selling Price?
The selling price is the price that the buyer pays to buy a product or goods. This is a price above cost and includes a profit ratio. Cost of goods is the price at which the seller buys the product or products. You then add up a percentage of your profit or gain. The set price or list price is the price that the seller will determine after adding the desired profit percentage. The price listed is the price the seller offers to the buyer, but the selling price is the price he actually received from the buyer after purchase. Usually the price listed is higher than the selling price. However, the sale price and the specified price or list price may be the same. Fixed price trading is a good example of this. Selling price is a very sensitive issue because the sale of a product is very dependent on it. Any product with a high selling price cannot attract many buyers because consumers do not consider it a good value for money. On the other hand, a very low selling price can affect the company's profitability. In addition, buyers may think the quality is inferior.
2.Why is the selling price an important factor?
Selling price is a key factor for consumers and sellers, because sales and demand for a product are highly dependent on it. Any product with a high or too high selling price cannot attract many buyers because they perceive the value of the product to be disproportionate. On the other hand, a very low selling price can affect the company's profitability and can also indicate lower product quality. Thus, the selling price must be determined correctly based on market analysis and consumer demand.
3.How should the seller determine the selling price?
The selling price can also be called the standard price, market price or list price. Certain factors help institutions and traders determine the selling price of their products:
The price consumers are willing to pay
The price that is ready to be accepted by the seller/store owner
Competitive market price
Depending on the nature and availability of the business, the seller may choose one of the above factors over the others. However, the seller can also use the average selling price of a product to determine how much they should charge their product.
4.What is the average selling price?
Average selling price is the number of products in a given category sold by different channels and markets. This price setting can also be used as a barometer for companies that need to determine the selling price of their products. Knowing in detail about the selling price and determining the exact selling price is very important because the seller needs to keep certain things in mind while fixing the price. To know in detail about the selling price and its formula, you can visit the Vedantu website where you can get appropriate materials on the Selling Price.
5.What is the difference between the selling price and the listed price?
The cost price, also known as the list price, is the price the seller gives to the buyer, but the selling price is the price the seller actually receives from the buyer through an agreement or contract. The selling price is usually lower than the listed price. Sometimes the selling price and the listed price can be the same. A fixed price trade, meaning that a trader does not offer a discount or price reduction, is a good example of this.
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190344 | https://math.stackexchange.com/questions/4072631/finding-least-norm-solutions-of-a-linear-diophantine-equation | optimization - Finding least-norm solutions of a linear Diophantine equation - Mathematics Stack Exchange
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Finding least-norm solutions of a linear Diophantine equation
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I am looking for an algorithm solving the following problem.
Let ∥x∥∗:=x T D x−−−−−√‖x‖∗:=x T D x, where D D is a diagonal matrix with positive diagonal entries. Given a∈N n a∈N n, b∈N b∈N, find
arg min x∈N n|a T x−b|argmin x∈N n|a T x−b|
with minimal ∥x∥∗‖x‖∗.
Motivation
I am creating a tools website for tabletop role-playing game (TTRPG) players. In these games, players fight together against monsters, and one player (called "GM") has to ensure these fights are of appropriate difficulty. This is done using "experience points" (XP). XP is used to estimate the strength of monsters. Each monster awards a set amount of XP (tougher monsters award more XP) and each difficulty has a target XP number. When the players defeat the monsters they gain these XP, becoming stronger as a result. If the GM selects a few monsters, how many of each do we need to ensure a good battle?
For example, the GM selects goblins, hobgoblins, and bugbears (who award 1,2, and 4 XP) to be his monsters. As threshold they select 12 XP. We would have a T=[1,2,4],b=12.a T=[1,2,4],b=12.x x describes the number of monsters picked. It is clear that there are many ways to select monsters, but most of them are undesirable. We don't want a fight against just 12 goblins, or only 3 bugbears. My idea is giving each monster a somewhat arbitrary weight: D=d i a g(1,6,12)D=d i a g(1,6,12). With this we obtain the optimal x T=[4,2,1]x T=[4,2,1] (4 goblins, 2 hobgoblins and 1 bugbear), a well balanced encounter. If the GM selects creatures such that we can't meet the threshold, we want to get as close as possible to it.
Why don't I use a greedy algorithm? Because I thought it would be cool and interesting not to.
optimization
integer-programming
discrete-optimization
operations-research
linear-diophantine-equations
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edited Mar 25, 2021 at 0:39
Rodrigo de Azevedo
23.4k 7 7 gold badges 49 49 silver badges 116 116 bronze badges
asked Mar 22, 2021 at 23:34
wowsostackoverflow420wowsostackoverflow420
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Thank you for the Python code to solve the MIQP. Surprisingly, I managed to run it on my machine. Are you happy with the results?Rodrigo de Azevedo –Rodrigo de Azevedo 2021-03-25 00:45:24 +00:00 Commented Mar 25, 2021 at 0:45
I am happy that we solved the problem, but I will probably not implement the solution in this way. The website is static, and having the user download MBs of libraries for a single feature is not really feasible. Perhaps when I have some more time, I can peek under the hood of these solvers, or find another algorithm that I can implement in a couple houndred lines of code. At least I learned something!wowsostackoverflow420 –wowsostackoverflow420 2021-03-25 11:25:31 +00:00 Commented Mar 25, 2021 at 11:25
You may want to take a look at CVXGEN. For LPs and QPs (over R R), it can allegedly generate very fast C code. However, I suspect you need very fast JavaScript code, rather than C code.Rodrigo de Azevedo –Rodrigo de Azevedo 2021-03-25 11:43:04 +00:00 Commented Mar 25, 2021 at 11:43
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First,
minimize x∈N 3∣∣a⊤x−b∣∣minimize x∈N 3|a⊤x−b|
If the minimum is zero, then solve the following least-norm problem
minimize x∈N 3 subject to∥x∥2∗a⊤x=b minimize x∈N 3‖x‖∗2 subject to a⊤x=b
Using Python (NumPy + CVXPY),
```
import cvxpy as cp
import numpy as np
parameters
a = np.array([1, 2, 4])
b = 12
D = np.diag([1, 6, 12])
optimization variables
x = cp.Variable(3, integer=True)
create and solve linear optimization problem over the integers
objective = cp.Minimize(cp.abs(a.T @ x - b))
prob = cp.Problem(objective, [ x >= np.zeros(3) ])
solution = prob.solve()
print("—————————————————————————————")
print("IP ——————————————————————————")
print("Minimum = ", prob.value)
print("Solution = ", x.value)
create and solve quadratic optimization problem over the integers
objective2 = cp.Minimize(cp.quad_form(x, D))
constraints = [ a.T @ x - b == 0,
x >= np.zeros(3) ]
prob2 = cp.Problem(objective2, constraints)
solution2 = prob2.solve(solver=cp.ECOS_BB)
print("—————————————————————————————")
print("MIQP ————————————————————————")
print("Minimum = ", prob2.value)
print("Solution = ", x.value)
```
which outputs
—————————————————————————————
IP ——————————————————————————
Minimum = 0.0
Solution = [0. 0. 3.]
—————————————————————————————
MIQP ————————————————————————
Minimum = 51.99999927150701
Solution = [4. 2. 1.]
Unfortunately, according to the CVXPY documentation, there are correctness issues with this solver.
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edited Mar 28, 2021 at 13:04
answered Mar 23, 2021 at 18:50
Rodrigo de AzevedoRodrigo de Azevedo
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You seem to be juggling two criteria (getting close to the threshold and minimizing the weighted norm of x x). There are several ways you could approach this via an optimization model. One would be to minimize a weighted combination of ∥x∥2∗‖x‖∗2 and |a T x−b||a T x−b|. (I squared the norm so that the objective function is quadratic and convex.) Another would be to minimize the squared norm subject to an arbitrary limit on the absolute deviation. Alternatively, you could minimize the absolute deviation subject to a limit on the squared norm. All three approaches should lead to tractible problems.
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answered Mar 23, 2021 at 16:09
prubinprubin
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Do you mean minimize ∥x∥2‖x‖2 subject to |a T x−b|<c|a T x−b|<c? This formulation of the problem would also fit my use case perfectly. Are there solvers for this?wowsostackoverflow420 –wowsostackoverflow420 2021-03-24 14:41:58 +00:00 Commented Mar 24, 2021 at 14:41
1 Yes, that was one of the approaches I mentioned (with the minor change that <c<c should be ≤c≤c). The absolute value function can be eliminated by using two constraints: a T x−b≤c a T x−b≤c and a T x−b≥−c a T x−b≥−c. That means you are minimizing a convex quadratic objective over linear constraints. Many contemporary IP solvers will handle that problem, including CPLEX, Gurobi and (I believe) XpressMP.prubin –prubin 2021-03-24 17:56:53 +00:00 Commented Mar 24, 2021 at 17:56
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190345 | https://sites.math.washington.edu/~aloveles/Math207Materials/m307Solving2x2Systems.pdf | Skills Review: Solving Two-by-Two Systems In this course, you will often have to solve a two-by-two system of linear equations that looks like ax1 + bx2 = P; cx1 + dx2 = Q, where a, b, c, d, P, and Q are all numbers and you are solving for x1 and x2. Here is a reminder of your goals and your tools for solving such equations.
1. The goal is to combine the two equations into one equation that has only one variable so that you can solve for that variable.
2. Your two main combining tools: • Add or Subtract the two equations from each other. This is valid because if you add equal things to equal things you get equal things! Note that you can also multiply or divide both sides of any equation by a number (in order to set up a situation where adding/subtracting will lead to cancellation).
• Substitute! Solve for one variable in the first equation and substitute into the second.
3. Once you have solved for one variable, you can substitute back into one (or both) of the original equations to find the other variable. As a check on your work, you should plug into both equations.
Basic example: Solve the system (i) 2x1 + x2 = 5 (ii) x1 −x2 = 4 • Solution 1: Combining by Adding/Subtracting Notice the cancellation that will happen if we add!
Adding corresponding sides of (i) and (ii) gives a combined equation of 3x1 = 9. Thus, x1 = 3.
Substituting back into (i) gives 2(3) + x2 = 5, so x2 = −1.
Substituting back into (ii) gives (3) −x2 = 4, so x2 = −1.
Thus, the only solution is x1 = 3 and x2 = −1.
• Solution 2: Substituting Solving for x2 in the first equation, we can rewrite equation (i) as x2 = 5 −2x1.
Substituting into (ii), we get a combined equation of x1 −(5 −2x1) = 4 which simplifies to 3x1 −5 = 4. Solving gives 3x1 = 9, so x1 = 3.
Substituting back into our simplified version of (i) gives x2 = 5 −2(3) = −1.
Substituting back into (ii) gives (3) −x2 = 4, so x2 = −1.
Thus, the only solution is x1 = 3 and x2 = −1.
The first method is sometimes faster, but it requires some cleverness. The second method always takes the same amount of time and requires no cleverness. That’s it, now you can solve linear 2-by-2 systems!
Here is another one to try on your own: Example: Solve the system (i) 2x1 + 2x2 = 6 (ii) 3x1 −x2 = 2 Comments about the solution: You can either start by dividing the first equation by 2, then adding.
Or just solve for x1 or x2 in the first equation and substituting into the second. Both will work. The answer you should get is x1 = 5 4 and x2 = 7 4.
Some very important theoretic comments about two-by-two systems There are three things that can happen in a two-by-two system: 1. UNIQUE solution: The most ‘likely’ situation (i.e. if you randomly pick numbers for coefficients you probably get a system with a unique solution). See two examples on the last page.
2. NO solution: Happens if the ‘left-hand side’ of the second equation is a multiple of the first, but the ‘right-hand side’ is not the same multiple. For example: (i) x1 −2x2 = 10; (ii) 3x1 −6x2 = 50. In this example, (i) x1 −2x2 = 10 and (ii) 3(x1 −2x2) = 50 can’t happen because 50 is NOT 3 times 10.
There is NO solution.
3. INFINITELY many solutions: This happens if both sides are the same multiple of each other.
For example: (i) x1 −2x2 = 10; (ii) 3x1 −6x2 = 30. Notice that both sides of equation (ii) are exactly 3 times equation (i). In fact, equations (i) and (ii) are two different ways to write the exact same equation.
Thus, all solutions will satisfy x1 = 10+2x2. For example, one solution is x1 = 10, x2 = 0, another is x1 = 12, x2 = 1, another is x1 = 14, x2 = 2, and so on ...
The Determinant: For a system of the form ax1 + bx2 = P; cx1 + dx2 = Q, , we define the two-by-two determinant by determinant = a b c d = ad −bc.
Note: For a two-by-two system if the determinant is zero, then the ‘left-hand sides’ are multiples of each other. For example, the system (i) x1 −2x2 = 10 (ii) 3x1 −6x2 = 30 has a determinant of (1)(−6)−(−2)(3) = 0.
Existence and Uniqueness Theorem for Linear Systems: From what we have already said, we can summarize 1. if ad −bc ̸= 0, then the system has a unique solution.
2. if ad −bc = 0, then the system will have no solution or infinitely many solutions (depending on the values of P and Q).
Cramer’s Rule: (Just for your interest, not required) If you combined and solved the general system ax1 + bx2 = P; cx1 + dx2 = Q, you would find that if there is a unique answer then it is always is equal to x1 = Pd −bQ ad −bc = P b Q d a b c d x2 = aQ −Pc ad −bc = a P c Q a b c d .
You can use Cramer’s rule to solve if you wish, but it is usually just as fast to combine and solve. To learn facts about larger systems (3-by-3 and 4-by-4), then you have to take a course in linear algebra (Math 308). Examples of Cramer’s rule are on the next page: 1. Solve the system (i) 2x1 + x2 = 5 (ii) x1 −x2 = 4 x1 = 5 1 4 −1 2 1 1 −1 = −9 −3 = 3 , x2 = 2 5 1 4 2 1 1 −1 = 3 −3 = −1.
This is the same example from the first page of this review (notice the solutions match).
2. Solve the system (i) 2x1 + 2x2 = 6 (ii) 3x1 −x2 = 2 x1 = 6 2 2 −1 2 2 3 −1 = −10 −8 = 5 4 , x2 = 2 6 3 2 2 2 3 −1 = −14 −10 = 7 5.
This was the second example from the first page of this review (notice the solutions match).
2. Solve the system (i) 5x1 + 7x2 = 10 (ii) 2x1 −6x2 = 8 x1 = 10 7 8 −6 5 7 2 −6 = −116 −48 = 29 12 , x2 = 5 10 2 8 5 7 2 −6 = 20 −48 = −5 12. |
190346 | https://www.cfa.harvard.edu/news/event-horizon-telescope-makes-highest-resolution-black-hole-detections-earth | Event Horizon Telescope Makes Highest-Resolution Black Hole Detections from Earth | Center for Astrophysics | Harvard & Smithsonian
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Event Horizon Telescope Makes Highest-Resolution Black Hole Detections from Earth
08.27.24
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Event Horizon Telescope Makes Highest-Resolution Black Hole Detections from Earth
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Using the Event Horizon Telescope (EHT), astronomers have achieved very-long-baseline interferometry test observations at 345 GHz, the highest-resolution such observations ever obtained from the surface of Earth. Scientists estimate that the breakthrough will result in a remarkable 50% increase in detail, sharpening images and observations of black holes and their surrounding regions.
EHT, D. Pesce, A. Chael
The Event Horizon Telescope (EHT) Collaboration has conducted test observations achieving the highest resolution ever obtained from the surface of the Earth, by detecting light from the centers of distant galaxies at a frequency of around 345 GHz.
When combined with existing images of supermassive black holes at the hearts of M87 and Sgr A at the lower frequency of 230 GHz, these new results will not only make black hole photographs 50% crisper but also produce multi-color views of the region immediately outside the boundary of these cosmic beasts.
The new detections, led by scientists from the Center for Astrophysics | Harvard & Smithsonian (CfA) that includes the Smithsonian Astrophysical Observatory (SAO), were published today in The Astronomical Journal.
"With the EHT, we saw the first images of black holes by detecting radio waves at 230 GHz, but the bright ring we saw, formed by light bending in the black hole’s gravity still looked blurry because we were at the absolute limits of how sharp we could make the images," said paper co-lead Alexander Raymond, previously a postdoctoral scholar at the CfA, and now at NASA’s Jet Propulsion Laboratory (NASA-JPL). "At 345 GHz, our images will be sharper and more detailed, which in turn will likely reveal new properties, both those that were previously predicted and maybe some that weren't."
The EHT creates a virtual Earth-sized telescope by linking together multiple radio dishes across the globe, using a technique called very-long-baseline interferometry (VLBI). To get higher-resolution images, astronomers have two options: increase the distance between radio dishes or observe at a higher frequency. Since the EHT was already the size of our planet, increasing the resolution of ground-based observations required expanding its frequency range, and that's what the EHT Collaboration has now done.
"To understand why this is a breakthrough, consider the burst of extra detail you get when going from black and white photos to color," said paper co-lead Sheperd "Shep" Doeleman, an astrophysicist at the CfA and SAO, and Founding Director of the EHT. "This new 'color vision' allows us to tease apart the effects of Einstein’s gravity from the hot gas and magnetic fields that feed the black holes and launch powerful jets that stream over galactic distances."
A prism splits white light into a rainbow of colors because different wavelengths of light travel at different speeds through glass. But gravity bends all light similarly, so Einstein predicts that the size of the rings seen by the EHT should be similar at both 230 GHz and 345 GHz, while the hot gas swirling around the black holes will look different at these two frequencies.
This is the first time the VLBI technique has been successfully used at a frequency of 345 GHz. While the ability to observe the night sky with single telescopes at 345 GHz existed before, using the VLBI technique at this frequency has long presented challenges that took time and technological advances to overcome. Water vapor in the atmosphere absorbs waves at 345 GHz much more than at 230 GHz weakening the signals from black holes at the higher frequency. The key was to improve the sensitivity of the EHT, which the researchers did by increasing the bandwidth of the instrumentation and waiting for good weather at all sites.
The new experiment used two small subarrays of the EHT—made up of the Atacama Large Millimeter/submillimeter Array (ALMA) and the Atacama Pathfinder EXperiment (APEX) in Chile, the IRAM 30-meter telescope in Spain, the NOrthern Extended Millimeter Array (NOEMA) in France, the Submillimeter Array (SMA) on Maunakea in Hawai'i, and the Greenland Telescope—to make measurements with resolution as fine as 19 microarcseconds.
"The most powerful observing locations on Earth exist at high altitudes, where atmospheric transparency and stability is optimal but weather can be more dramatic," said Nimesh Patel, an astrophysicist at the CfA and SAO, and a project engineer at SMA, adding that at the SMA, the new observations required braving icy roads at Maunakea to open the array in the stable weather after a snow storm with minutes to spare. "Now, with high-bandwidth systems that process and capture wider swaths of the radio spectrum, we are starting to overcome basic problems in sensitivity, like weather. The time is right, as the new detections prove, to advance to 345 GHz."
This achievement also provides another stepping stone on the path to creating high-fidelity movies of the event horizon environment surrounding black holes, which will rely on upgrades to the existing global array. The planned next-generation EHT (ngEHT) project will add new antennas to the EHT in optimized geographical locations and enhance existing stations by upgrading them all to work at multiple frequencies between 100 GHz and 345 GHz at the same time. As a result of these and other upgrades, the global array is expected to increase the amount of sharp, clear data EHT has for imaging by a factor of 10, enabling scientists to not only produce more detailed and sensitive images but also movies starring these violent cosmic beasts.
"The EHT's successful observation at 345 GHz is a major scientific milestone," said Lisa Kewley, Director of CfA and SAO. "By pushing the limits of resolution, we're achieving the unprecedented clarity in the imaging of black holes we promised early on, and setting new and higher standards for the capability of ground-based astrophysical research."
Resource
"First Very Long Baseline Interferometry Detections at 870 μm," A.W. Raymond, S. Doeleman, et al (2024), The Astronomical Journal, DOI: 10.3847/1538-3881/ad5bdb.
About the Submillimeter Array (SMA)
The Submillimeter Array is a joint project of the Smithsonian Astrophysical Observatory and the Academia Sinica Institute of Astronomy and Astrophysics. We acknowledge the significance that Maunakea, where the SMA is located, has for the indigenous Hawaiian people.
About the Greenland Telescope
The Greenland Telescope is a joint project of the Smithsonian Astrophysical Observatory and the Academica Sinica Institute of Astronomy and Astrophysics.
About EHT
The EHT Collaboration involves more than 400 researchers from Africa, Asia, Europe, North and South America, with around 270 participating in this paper. The international collaboration aims to capture the most detailed black hole images ever obtained by creating a virtual Earth-sized telescope. Supported by considerable international efforts, the EHT links existing telescopes using novel techniques—creating a fundamentally new instrument with the highest angular resolving power that has yet been achieved.
The EHT consortium consists of 13 stakeholder institutes; the Academia Sinica Institute of Astronomy and Astrophysics, the University of Arizona, the Center for Astrophysics | Harvard & Smithsonian including the Smithsonian Astrophysical Observatory, the University of Chicago, the East Asian Observatory, Goethe University Frankfurt, Institut de Radioastronomie Millimétrique, Large Millimeter Telescope, Max Planck Institute for Radio Astronomy, MIT Haystack Observatory, National Astronomical Observatory of Japan, Perimeter Institute for Theoretical Physics, and Radboud University.
About the Center for Astrophysics | Harvard & Smithsonian
The Center for Astrophysics | Harvard & Smithsonian is a collaboration between Harvard and the Smithsonian designed to ask—and ultimately answer—humanity's greatest unresolved questions about the nature of the universe. The Center for Astrophysics is headquartered in Cambridge, MA, with research facilities across the U.S. and around the world.
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9: Vector Spaces
A First Course in Linear Algebra (Kuttler)
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Sat, 17 Sep 2022 06:21:26 GMT
9.5: Sums and Intersections
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Contents
1. Home
2. Bookshelves
3. Linear Algebra
4. A First Course in Linear Algebra (Kuttler)
5. 9: Vector Spaces
6. 9.5: Sums and Intersections
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A First Course in Linear Algebra (Kuttler)
Front Matter
1: Systems of Equations
2: Matrices
3: Determinants
4: Rⁿ
5: Linear Transformations
6: Complex Numbers
7: Spectral Theory
8: Some Curvilinear Coordinate Systems
9: Vector Spaces
10: Some Prerequisite Topics
Back Matter
9.5: Sums and Intersections
Last updated Sep 17, 2022
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9.4: Subspaces and Basis
9.6: Linear Transformations
Page ID 14556
Ken Kuttler
Brigham Young University via Lyryx
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Outcomes
2. Definition 9.5.1: Sum and Intersection
3. Definition 9.5.2: Direct Sum
4. Example 9.5.1: Intersection is a Subspace
1. Solution
Theorem 9.5.1: Dimension of Sum
Outcomes
Show that the sum of two subspaces is a subspace.
Show that the intersection of two subspaces is a subspace.
We begin this section with a definition.
Definition 9.5.1: Sum and Intersection
Let V be a vector space, and let U and W be subspaces of V. Then
U+W={u→+w→|u→∈U and w→∈W} and is called the sum of U and W.
U∩W={v→|v→∈U and v→∈W} and is called the intersection of U and W.
Therefore the intersection of two subspaces is all the vectors shared by both. If there are no vectors shared by both subspaces, meaning that U∩W={0→}, the sum U+W takes on a special name.
Definition 9.5.2: Direct Sum
Let V be a vector space and suppose U and W are subspaces of V such that U∩W={0→}. Then the sum of U and W is called the direct sum and is denoted U⊕W.
An interesting result is that both the sum U+W and the intersection U∩W are subspaces of V.
Example 9.5.1: Intersection is a Subspace
Let V be a vector space and suppose U and W are subspaces. Then the intersection U∩W is a subspace of V.
Solution
By the subspace test, we must show three things:
0→∈U∩W
For vectors v→1,v→2∈U∩W,v→1+v→2∈U∩W
For scalar a and vector v→∈U∩W,av→∈U∩W
We proceed to show each of these three conditions hold.
Since U and W are subspaces of V, they each contain 0→. By definition of the intersection, 0→∈U∩W.
Let v→1,v→2∈U∩W,. Then in particular, v→1,v→2∈U. Since U is a subspace, it follows that v→1+v→2∈U. The same argument holds for W. Therefore v→1+v→2 is in both U and W and by definition is also in U∩W.
Let a be a scalar and v→∈U∩W. Then in particular, v→∈U. Since U is a subspace, it follows that av→∈U. The same argument holds for W so av→ is in both U and W. By definition, it is in U∩W.
Therefore U∩W is a subspace of V.
It can also be shown that U+W is a subspace of V.
We conclude this section with an important theorem on dimension.
Theorem 9.5.1: Dimension of Sum
Let V be a vector space with subspaces U and W. Suppose U and W each have finite dimension. Then U+W also has finite dimension which is given by dim(U+W)=dim(U)+dim(W)−dim(U∩W)
Notice that when U∩W={0→}, the sum becomes the direct sum and the above equation becomes dim(U⊕W)=dim(U)+dim(W)
This page titled 9.5: Sums and Intersections is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) via source content that was edited to the style and standards of the LibreTexts platform.
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9.4: Subspaces and Basis
9.6: Linear Transformations
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190349 | https://www.cuemath.com/algebra/identity-matrix/ | Identity Matrix
The identity matrix is referred to as the multiplicative identity of matrices. Before going to learn what is an identity matrix, let us recall the meaning of identity in math. Identity is a mathematical quantity which when operated with some quantity leaves the same quantity. Let us consider the following examples.
The additive identity is 0 as adding any number to 0 gives the same number as the sum.
For example, 3 + 0 = 3, 0 + (-1) = -1, etc.
The multiplicative identity is 1 as multiplying any number with 1 gives the same number as the product.
For example, 3 × 1 = 3, 1 × (-1) = -1, etc.
In the same way, of course, we know that by adding the null matrix ⎡⎢⎣0000⎤⎥⎦ to any 2 × 2 matrix gives us the same matrix and hence a null matrix is called additive identity matrix. But what is the multiplicative identity with respect to matrices? The answer is the identity matrix and let us learn this in detail here.
| | |
--- |
| 1. | What is Identity Matrix? |
| 2. | Identity Matrix of Different Orders |
| 3. | Verification of Identity Matrix |
| 4. | Properties of Identity Matrix |
| 5. | Finding Inverse Matrix Using Identity Matrix |
| 6. | Applications of Identity Matrix |
| 7. | FAQs on Identity Matrix |
What is Identity Matrix?
An identity matrix is a square matrix in which each of the elements of its principal diagonal is a 1 and each of the other elements is a 0.It is also known as the unit matrix. We represent an identity matrix of order n × n (or n) as In. Sometimes we denote this simply as I. The mathematical definition of an identity matrix is,
In (or) I = [aij]n × n, where aij = 1 when i = j, and aij = 0 when i ≠ j.
An identity matrix in general is an identity with respect to multiplication. Thus, for any matrix A,
AI = IA = A
i.e., by multiplying any matrix A with the identity matrix of the same order, we get the same matrix as the product and hence the name "identity" for it.
Identity Matrix of Different Orders
Here are some examples of identity matrices of different orders. Note that an identity matrix is a square matrix always.
2x2 identity matrix: I2 = ⎡⎢⎣1001⎤⎥⎦.
3x3 identity matrix: I3 = ⎡⎢⎣100010001⎤⎥⎦.
4x4 identity matrix: I4 = ⎡⎢
⎢
⎢⎣1000010000100001⎤⎥
⎥
⎥⎦.
Verification of Identity Matrix
If I is an identity matrix and A is any matrix of the same order, then by the definition, AI = IA = A. Let us verify this by taking matrices of order 2 × 2.
A = ⎡⎢⎣15−32⎤⎥⎦ and I = ⎡⎢⎣1001⎤⎥⎦ (which is identity with respect to multiplication).
Verifying AI = A
AI = ⎡⎢⎣15−32⎤⎥⎦ ⎡⎢⎣1001⎤⎥⎦
= ⎡⎢⎣1(1)+5(0)1(0)+5(1)−3(1)+2(0)−3(0)+2(1)⎤⎥⎦
= [15−32]
= A
Since matrices don't need to be commutative with respect to multiplication, we have to verify IA = A as well.
Verifying IA = A
IA = ⎡⎢⎣1001⎤⎥⎦ [15−32]
= ⎡⎢⎣1(1)+0(−3)1(5)+0(2)0(1)+1(−3)0(5)+1(2)⎤⎥⎦
= ⎡⎢⎣15−32⎤⎥⎦
= A
Thus, we have verified that AI = IA = A.
Similarly, you can try verifying the identity matrix of orders 3 × 3, 4 × 4, etc.
Properties of Identity Matrix
Here are the identity matrix properties based upon its definition.
The identity matrix is always a square matrix.
By multiplying an identity matrix with any other matrix results in the same matrix.
Every identity matrix is a diagonal matrix as only its principal diagonal's elements are nonzeros.
An identity matrix is symmetric as IT = I.
Every identity matrix is a scalar matrix as all its principal diagonal's elements are equal and the rest of the elements are zeros.
The determinant of every identity matrix is 1.
The inverse of identity matrix is itself as I · I-1 = I-1 · I = I.
In = I, for any integer 'n'. i.e., the square of identity matrix is equal to itself, the cube of identity matrix is equal to itself, and so on.
By multiplying a matrix with its inverse, the result is an identity matrix
We can find the inverse of a matrix using the identity matrix (Let us see this in the next section).
Finding Inverse Matrix Using Identity Matrix
The inverse of a matrix A (which is written as A-1) is a matrix B (and vice versa) if and only if AB = BA = I, where A, B, and I are the square matrices of the same order. Given A and B, it is easy to verify whether they are inverses of each other just by verifying whether AB = BA = I. But if a matrix A is given then how can we find its inverse B? We can find the inverse matrix of a matrix using the following steps:
Step 1: Write an augmented matrix with the given matrix adjoining it with the identity matrix of the same order and we separate these two matrices by a line.
Step 2: We apply row operations aiming to convert the left side matrix (which is A) to an identity matrix.
Step 3: The matrix that is left on the right side itself is our inverse matrix.
We can see an example of finding the inverse matrix using these steps in the "Identity Matrix Examples" section below.
Applications of Identity Matrix
The identity matrix is used for various purposes in linear algebra. Here are the applications of the identity matrix.
An identity matrix is used to find the inverse of a matrix.
Also, an identity matrix is used to verify whether any two given matrices are inverses of each other.
An identity matrix is used to find the eigenvalues and eigenvectors.
An identity matrix is used while solving the system of equations using the elementary row operations.
Important Notes on Identity Matrix:
Here are some important points to note that are related to an identity matrix.
If you see an identity matrix without any specification of operation, then by default, it should be understood that it is an identity matrix with respect to multiplication.
To write an identity matrix of some order, first, write an empty matrix with the given order, write 1s in the place of elements of the principal diagonal, and finally write 0s in place of all other elements.
If AB = BA = I, then A and B are inverses of each other.
To find the inverse of a matrix, write it adjoining the identity matrix of the same order to it on its right side. Apply row operations to the entire augmented matrix aiming to make the left side matrix an identity matrix. Then the right side matrix will be the inverse of the given matrix.
☛ Related Topics:
Here are some topics that you may find interesting while reading about the identity matrix.
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Identity Matrix
Identity Matrix Examples
Example 1: If I = ⎡⎢⎣cosx00secx⎤⎥⎦ is an identity matrix of order 2 x 2, then find the value of x given that 0 ≤ x < π/2.
Solution:
We know that the I = ⎡⎢⎣1001⎤⎥⎦. Thus,
⎡⎢⎣cosx00secx⎤⎥⎦ = ⎡⎢⎣1001⎤⎥⎦
Comparing the corresponding elements,
cos x = 1 and sec x = 1
These both can be true only when x = 0.
Answer: x = 0.
2. Example 2: For A = ⎡⎢⎣12−1320−402⎤⎥⎦ and I is the identity matrix of order 3 x 3, prove that A · I = I · A = A.
Solution:
Verifying A · I = A:
A · I = ⎡⎢⎣12−1320−402⎤⎥⎦ ⎡⎢⎣100010001⎤⎥⎦
= ⎡⎢⎣1+0+00+2+00+0−13+0+00+2+00+0+0−4+0+00+0+00+0+2⎤⎥⎦
= ⎡⎢⎣12−1320−402⎤⎥⎦
= A
Verifying I · A = A:
I · A = ⎡⎢⎣100010001⎤⎥⎦ ⎡⎢⎣12−1320−402⎤⎥⎦
= ⎡⎢⎣1+0+02+0+0−1+0+00+3+00+2+00+0+00+0−40+0+00+0+2⎤⎥⎦
=⎡⎢⎣12−1320−402⎤⎥⎦
= A
Answer: We verified A · I = I · A = A.
3. Example 3: Find the inverse of ⎡⎢⎣1234⎤⎥⎦.
Solution:
Let A = ⎡⎢⎣1234⎤⎥⎦
Step 1: Write an augmented matrix with A and I.
⎡⎢
⎢⎣12103401⎤⎥
⎥⎦
Step 2: Apply row operations to convert the left side matrix to an identity matrix.
Applying R2 → R2 - 3R1,
⎡⎢
⎢⎣12100−2−31⎤⎥
⎥⎦
Applying R1 → R1 + R2,
⎡⎢
⎢⎣10−210−2−31⎤⎥
⎥⎦
Divide R2 by -2,
⎡⎢
⎢⎣10−21013/2−1/2⎤⎥
⎥⎦
Step 3: Here, the right side matrix is our inverse matrix. i.e.,
A-1 = ⎡⎢⎣−213/2−1/2⎤⎥⎦
Answer: A-1 = ⎡⎢⎣−213/2−1/2⎤⎥⎦.
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FAQs on Identity Matrix
What is the Definition of an Identity Matrix in Linear Algebra?
An identity matrix, which is denoted by I, is a square matrix in which all elements of the principal diagonal are 1s and all the other elements are zeros. For any matrix A, AI = IA = A. It is also known as unit matrix.
How Do You Identify Identity Matrix?
If in a square matrix,
each element of the principal diagonal is a 1 and
all the other elements are zeros
then we say that it is an identity matrix.
What is Identity Matrix Squared?
The square of any identity matrix is itself. For example, ⎡⎢⎣1001⎤⎥⎦⎡⎢⎣1001⎤⎥⎦ = ⎡⎢⎣1001⎤⎥⎦.
What is the Formula of Identity Matrix?
An identity matrix is a matrix with 1s in its principal diagonal and 0s in all the other places. Thus, its formula is In (or) I = [aij]n × n, where aij = 1 when i = j, and aij = 0 when i ≠ j.
What is the Identity Matrix of order 3?
In the identity matrix of order 3 × 3, all principal diagonal's elements are 1s and the other elements are 0s. It is given by ⎡⎢⎣100010001⎤⎥⎦.
How To Find Inverse of a Matrix Using Identity Matrix?
To find the inverse of a matrix A, write A along with the identity matrix (I) of the same order in a matrix separating them by a line (such that A is on left and I is on right). Apply row operations to convert the left side matrix as I. Then the right side matrix is nothing but A-1.
Why is It Called an Identity Matrix?
When an identity matrix (I) is multiplied with a matrix (A) of the same order, the product is the same matrix (A). i.e., AI = IA = A. So is the name (with respect to multiplication).
What is the Inverse of Identity Matrix?
The inverse of an identity matrix is itself. Because for any identity matrix I, we have I · I = I · I = I.
What is Identity Matrix Transpose?
The transpose of a matrix is obtained by writing its rows as columns (or columns as rows). The transpose of an identity matrix is itself. For example, If I = ⎡⎢⎣1001⎤⎥⎦ then its transpose is IT = ⎡⎢⎣1001⎤⎥⎦ = I.
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Algebra (all content)20 units · 412 skillsUnit 1 Introduction to algebraUnit 2 Solving basic equations & inequalities (one variable, linear)Unit 3 Linear equations, functions, & graphsUnit 4 SequencesUnit 5 System of equationsUnit 6 Two-variable inequalitiesUnit 7 FunctionsUnit 8 Absolute value equations, functions, & inequalitiesUnit 9 Quadratic equations & functionsUnit 10 Polynomial expressions, equations, & functionsUnit 11 Exponential & logarithmic functionsUnit 12 Radical equations & functionsUnit 13 Rational expressions, equations, & functionsUnit 14 Trigonometric functionsUnit 15 Algebraic modelingUnit 16 Complex numbersUnit 17 Conic sectionsUnit 18 Series & inductionUnit 19 VectorsUnit 20 Matrices
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190351 | https://www.youtube.com/watch?v=XR0FrHEg_4U | Minimize rational functions
Benjamin Manchester
696 subscribers
1 likes
Description
116 views
Posted: 11 Oct 2016
This video is an example of a rational function application. We are trying to find the dimensions that will minimize that amount of paper used on a printing job.
1 comments
Transcript:
oh hello sorry that I fig I needed a second to get set up okay so today we're going to talk about maximizing and in this case minimizing an area of a given function so let's get started here one second okay all set so we're given a problem about a publishing company that wants to have a piece of paper that's printed that has 30 square inches of print okay think about a piece of paper that you were normally using your printer that you're writing a I don't know what an essay on or something there are margins around the outside and I know we only use eight and a half by 11 pieces of paper normally but this case this this company over here that we're using wants to have the smallest piece of paper possible so that they can minimize their cost okay because the more paper they waste the more money they waste and they don't want to waste any money they want to actually make money as a company so we're given constraints that it has to be 30 square inches of print and the margins all around the prints are two inches wide okay so there are two things that we need to do we need to first write an equation because we can't solve anything without an equation and then figure out what is going to make this the biggest what is the smallest amount of paper that we can use in order to get the best bang for our buck basically okay so we're going to start off by drawing a picture so this publishing company has a piece of paper so let's draw ourselves a piece of paper I'm going to assume that's rectangular it could be square but it's at least going to be a rectangle okay and I have 30 square inches of print so if you think about this as a as a piece of paper that you printed your essay on this is where all of the text would be there is 30 square inches of text inside here and everywhere around you we have two inch margins k 2 inches 2 inches 2 inches 2 inches and our paper was let's see X units wide it said and why units tall so it is defining the variables for us X is the length of this paper and why is the height of this paper okay how long however you want to use the words to describe the dimensions of this paper okay so there are two different areas that we're trying to talk about number one the area of the paper okay the actual paper that this stuff is being printed on so that would be this here is one of those links so the area of this paper is rectangular so it's x times y so the area of a rectangle is the length times its width so the piece of paper has an area of x times y so right now the paper is a function of x and y okay so if we're going to express this paper as a function of just x we're going to have to get rid of our y value somewhere in the future okay so we haven't used the other area problem yet either so our other area that we're talking about is the area of the prints okay all the stuff that got printed on the inside here there is also an area associated with that so that happens to be 30 square inches and it is a rectangular area also so that's made up of this length and this width or this height so that the tricky part is what is this length so this is a horizontal length so it has to be something in terms of X so this whole entire thing was X but then i subtracted 2 inches from this side and i subtracted 2 inches from this side so this length would be whatever x is the whole entire piece of paper minus 4 inches okay two inches from the right margin two inches from the left margin so whatever my wit hat that my length happens to be i subtract four inches from it and i can get the length of the Prince a similarly for the Y value for this okay the height of this printed area okay normally the whole entire thing would be why but then i subtract off two inches from the top margin and i also have to subtract off two inches from the bottom margin so the Y or the height of this is going to be the Y value minus four also ok so here I have two different area equations I have the area of the paper which is the outside x times the outside why and I also have the area of the printed stuff on the inside which is the outside length minus 4 times the outside height minus 4 so if we want to express the area up here that means we need to substitute okay we have to use the algebraic property of substitution so we have to substitute in for Y which means down here we have to solve this equation for y because you can only substitute when we have equality so we have to get this is y equals in order for us to be able to substitute it up there okay and that's what makes this a problem is hard that you have to come up with your own area equations know what to get rid of and know what to solve for so we're going to solve this area of prints equation for y okay remember what we're trying to do we're trying to find an equation for the area of this page as a function of X so this equation with just X as the variable we have x and y we're trying to have it be just x so let's bring it down here and we'll work a little cleaner 30 equals X minus 4 times y minus 4 so because this is multiplication this thing times that thing I am able to divide both sides by X minus 4 okay so that gets rid of this multiplication piece so 30 over X minus 4 now equals just Y minus for a much cleaner so now we have to get rid of this plus minus 4 so we'll add 4 to both sides now remember we're not adding 4 to the top we're not adding 4 to the bottom we are adding 4 to this rational equation of 30 over X minus 4 so 30 over X minus 4 plus 4 is the Y value we do not add to the top we do not add to the bottom we only add if we have like denominators and this has a denominator of 1 so we don't just combine these two at face value okay but I have solved my equation for y and you say yeah a big deal so what well so what now i can substitute it back into here okay so going back to our first question we want to express the area as a function of just X so here area equals x times y we just determined that why was 64 over X minus 4 plus 4 so now the area is x x 64 divided by X minus 4 plus 4 okay this is an area formula with X as the only variable notice 64 is a constant plus 4 is a constant minus 4 is a constant our only variable on this side of the equation is x okay so I'm going to do two other things i'm going to distribute to simplify a little bit before we plug this into our calculator so this is 64 x over X minus 4 plus 4 X okay and I'm not really satisfied with this your teacher might be and this might be okay your calculator actually likes us just fine but I'm going to go a little bit further and simplify this down so i want to add these two things together I need to make this one into a fraction that has the same denominator is this one so i'm going to multiply by X minus 4 over X minus 4 okay i'm multiplying by 1 so mathematically i'm not really changing this it's only going to look different so 64 x over X minus 4 plus 4 x times X minus 4 over X minus 4 so now that these two things have the same denominator I can add them together I can add their numerators so I'm just going to distribute into here 4x squared minus 16x they want to multiply all this by all that all / just X minus 4 gives me one single rational equation for x squared plus let's see 64 minus 16 is 48 X all divided by X minus 4 okay so just in case you have a multiple choice test and this answer isn't on there you need to know how to get down to here to make this as your equation okay so let's see 64 this make sure everything is on the up-and-up no it's not where did the 64 come from I am so sorry let me erase all of these 64 I'm sure you were just confused out of your mind as to where this 64 came from and I just confused myself 64 was a problem i was doing in class today 30 so sorry 30 over X minus 4 much better okay now we're all on the same page 30 / x there we go 30 30 30 16-30 comes out to be 14 so sorry I did not mean to confuse every single person with that where did the 64 come from yes exactly there's just anyway here we go for x squared plus 14 x over X minus 4 okay yes even math teachers sometimes make mistakes when we have crazy things running through our brain so this is the area as a function of just X okay maybe I shouldn't try to simplify it anyway that's what's that's the point so now we need to go to Part B we have area equals 4x squared plus 14 x over X minus 4 now what are the dimensions that will minimize the amount of paper used so this is an area equation about the paper that we're using so we're going to plug this equation into our calculator so here's our handy dandy calculator hopefully many of you have have ti-84 if not this is a ti-84 it will be able to use here to figure this out so we're going to plug this into our calculator into y equals okay if you have the latest operating system we can put it in as an actual fraction looking thing so 4x squared plus 14 x divided by X minus 4 okay so let's just take a look at the graph and see what it looks like there it goes that's fantastic looks like we have a vertical asymptote at positive for as we should have when we solve your denominator and I'm not seeing the other half of this graph okay which means i need to adjust my window so if you press the Zoom button we can go down and do a zoom fits it was something that the calculator has built in and that will basically show us what our graph supposed to look like I mean it may or may not be any better in fact there's our positive for asymptotes and then here is the rest of the graph right there okay so this gives you a sense of yes there is a graph up there now we have to figure out exactly where that's going to be so because I'm really just dealing with the first quadrant I'm going to set my ex minimum 20 let's do an x max of 20 okay with the scale of 5 every 5 i'm going to have a tick mark my minimum Y value just 0 I don't think I need 1300 up there let's just go to 500 and see what happens so setting the window is more or less trial and error you can have a pretty good idea of what's going on once you see it for a while well that's a pretty good graph remember this tick mark is 100 this tick mark is 200 300 400 so we didn't have to be this high this is 5 10 15 and 20 these tick marks are all five units apart so the most part this is a pretty good graph I guess I'm going to adjust it a little bit more so I can see more about where the minimum is so I'm going to change my y maximum to 200 just so I can get a little bit better view of the graph so remember now this is still this is 100 in the Y value this is still five in the X this is 10 and the axis is 15 in the X so I'm looking for the dimensions that will minimize the amount of paper use so somewhere in here is the minimum value via the lowest Y value on this graph so I'm going to have my calculator help me and calculate to that one so your 2nd trace does a calculation I'm going to try to find the minimum okay so it wants me to give me a left bound what's my left most point and I'm thinking that this is pretty good then a right point so I'm going to go to the right side of where I think that minimum value is which pretty good in there that gives me lots of space I guess that's guest somewhere in the middle how about that is a good guess and then it will actually tell me exactly what my minimum value was so I when X is 9.47 I'm going to round up to 9.5 inches that's going to be the width of my piece of paper that will give me the smallest paper possible to still have this 30 square inches of print okay so X is 9.5 and what's our y-value well here we have an equation for the y-value remember from our graph we graphed X versus area so this is an area equation so my area would be eighty nine point eight not my y value but my area because the y value in this case is an area from our pit from our paper or from our equation you want an area of the paper so this is the minimum area in order to find the Y value I have to plug it into this equation because this is the y value for any given X so 30 / 9.5 minus 4 four plus for that's going to be my y-value the other dimension of my paper so let's go back to the calculator mode we're going to do 30 / parentheses 9.5 minus 4 so I want it to divide by the difference and add 4 inches to that gives me a y-value of nine point 45 45 45 in other words y value of just 9.5 also so it turns out that the best piece of paper that I want to use to make sure that I have 30 square inches of prints with a 2 inch square margin is going to be a square piece of paper that's 9.5 inches x 9.5 inches okay fantastic see you again later |
190352 | https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Elementary_Number_Theory_(Barrus_and_Clark)/01%3A_Chapters/1.15%3A_Number_Theoretic_Functions | Skip to main content
1.15: Number Theoretic Functions
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: Jan 22, 2022
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1.14: Fermat Primes and Mersenne Primes
1.16: Perfect Numbers and Mersenne Primes
Page ID
: 93798
Mike Barrus & W. Edwin Clark
University of Rhode Island
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We will return to the Mersenne primes in the next chapter. Before we do, we introduce a few concepts that will be helpful both then and later in the text.
The prime-counting function π(x) appearing in the Prime Number Theorem (Theorem 1.11.3) and the prime-generating functions imagined and studied in Section 1.14 are by no means the only functions studied in number theory. Mathematicians through history have profitably looked at several additional functions tied to our key questions about the integers. In this chapter we introduce three of these. Though their definitions may seem a bit strange at first, hopefully the results of this chapter and the exercises that follow will convince you that their properties are interesting enough to make studying these functions worthwhile.
Functions related to divisors
We start with two functions related to the positive divisors of an integer n.
Definition 1.15.1
For each integer n>0, define τ(n)=the number of positive divisors of n,σ(n)=the sum of the positive divisors of n.
Example 1.15.1
The integer 12=3⋅22 has positive divisors 1,2,3,4,6,12.
Hence τ(12)=6and σ(12)=1+2+3+4+6+12=28.
Definition 1.15.2: Proper Divisor
A positive divisor d of n is said to be a proper divisor of n if d<n. We denote the sum of all proper divisors of n by σ∗(n).
Note that if n≥2 then σ∗(n)=σ(n)−n.
Example 1.15.2
Carrying our last example further, σ∗(12)=16.
The next theorem shows a simple way to compute σ(n) and τ(n) from the prime factorization of n.
Theorem 1.15.1
Let n=pe11pe22⋯perr,r≥1,
where p1<p2<⋯<pr are primes and ei≥0 for each i∈{1,2,…,r}. Then
τ(n)=(e1+1)(e2+1)⋯(er+1);
σ(n)=(pe1+11−1p1−1)(pe2+12−1p2−1)⋯(per+1r−1pr−1).
Proof
(1) From the Fundamental Theorem of Arithmetic, every positive factor d of n will have its prime factors coming from those of n. Hence d∣n if and only if d=pf11pf22⋯pfrr for some integer exponents fi where for each i, 0≤fi≤ei.
That is, for each fi we can choose a value in the set of ei+1 numbers {0,1,2,…,ei}. So, in all, there are (e1+1)(e2+1)⋯(er+1) choices for the exponents f1,f2,…,fr, and by the Fundamental Theorem of Arithmetic, each set of choices yields a different factor. So (1) holds.
To prove (2), we first establish two lemmas.
Before proving this let’s look at an example. Take n=72=8⋅9=23⋅32. The theorem says that τ(72)=(3+1)(2+1)=12andσ(72)=(24−12−1)(33−13−1)=15⋅13=195.
You can compute τ(72) and σ(72) from their original definitions to verify that these results are correct (and much quicker to compute!).
Lemma 1.15.1
Let n=ab where a,b>0 and gcd(a,b)=1. Then σ(n)=σ(a)σ(b).
Proof
Since a and b have only 1 as a positive common factor, using the Fundamental Theorem of Arithmetic it is easy to see that d∣ab⇔d=d1d2 where d1∣a and d2∣b. That is, the divisors of ab are products of the divisors of a and the divisors of b. Let 1,a1,…,as
denote the divisors of a and let 1,b1,…,btdenote the divisors of b. Then σ(a)=1+a1+a2+⋯+as,σ(b)=1+b1+b2+⋯+bt.The divisors of n=ab can be listed as follows 1,b1,b2,…,bt,a1⋅1,a1⋅b1,a1⋅b2,…,a1⋅bt,a2⋅1,a2⋅b1,a2⋅b2,…,a2⋅bt,⋮as⋅1,as⋅b1,as⋅b2,…,as⋅bt.It is important to note that since gcd(a,b)=1, aibj=akbℓ implies that ai=ak and bj=bℓ. That is there are no repetitions in the above array.
If we sum each row we get 1+b1+⋯+bt=σ(b)a11+a1b1+⋯+a1bt=a1σ(b)⋮as⋅1+asb1+⋯+asbt=asσ(b).
By adding these partial sums together we get σ(n)=σ(b)+a1σ(b)+a2σ(b)+⋯+a3σ(b)=(1+a1+a2+⋯+as)σ(b)=σ(a)σ(b).This proves the lemma.
Lemma 1.15.2
If p is a prime and k≥0 we have σ(pk)=pk+1−1p−1.
Proof
Since p is prime, the divisors of pk are 1,p,p2,…,pk. A standard formula for geometric series yields σ(pk)=1+p+p2+⋯+pk=pk+1−1p−1,
as desired.
Proof of Theorem 1.15.1
Proof
(2) Let n=pe11pe22⋯perr. Our proof is by induction on r. If r=1, n=pe11 and the result follows from Lemma 1.15.2. Suppose the result is true when 1≤r≤k. Consider now the case r=k+1. That is, let n=pe11⋯pekkpek+1k+1
where the primes p1,…,pk,pk+1 are distinct and ei≥0. Let a=pe11⋯pekk, b=pek+1k+1. Clearly gcd(a,b)=1. So by Lemma 1.15.1 we have σ(n)=σ(a)σ(b). By the induction hypothesis σ(a)=(pe1+11−1p1−1)⋯(pek+1k−1pk−1)and by Lemma 1.15.2 σ(b)=pek+1+1k+1−1pk+1−1and it follows that σ(n)=(pe1+11−1p1−1)⋯(pek+1+1k+1−1pk+1−1).So the result holds for r=k+1. By PMI it holds for r≥1.
The functions σ(n) and σ∗(n) will appear in the next chapter as we introduce perfect numbers. Additional properties of τ(n) and σ(n) are discussed in the exercises.
Euler’s totient function
The final function we will introduce in this chapter is known as Euler’s phi-function, or the Euler totient function. As opposed to τ and σ, which dealt with divisors of their input, ϕ will deal with numbers that have no prime factor in common with n.
Definition 1.15.3
If X is a set, the number of elements in X is denoted by |X|.
For example, |{1}|=1 and |{0,1,3,9}|=4.
Definition 1.15.4
If m≥1, the Euler totient function of n is defined by ϕ(n)=|{i∈Z∣1≤i≤n and gcd(i,n)=1}|.
At first glance, ϕ(n) may seem tedious to calculate. For instance, in order to compute ϕ(1000), would a person need to list all numbers relatively prime to 1000 and then count them?
Fortunately, the following theorems show that once the prime factorization of n is given, computing ϕ(n) is easy.
Theorem 1.15.2
If a>0 and b>0 and gcd(a,b)=1, then ϕ(ab)=ϕ(a)ϕ(b).
Theorem 1.15.3
If p is prime and n>0 then ϕ(pn)=pn−pn−1.
Theorem 1.15.4
Let p1,p2,…,pk be distinct primes and let n1,n2,…,nk be positive integers, then ϕ(pn11pn22⋯pnkk)=(pn11−pn1−11)⋯(pnkk−pnk−1k).
Before discussing the proofs of these three theorems, let’s illustrate their use:
ϕ(12)=ϕ(22⋅3)=(22−21)(31−30)=2⋅2=4ϕ(9000)=ϕ(23⋅53⋅32)=(23−22)(53−52)(32−31)=4⋅100⋅6=2400.
Note that if p is any prime then ϕ(p)=p−1.
We will postpone a proof of Theorem 1.15.2 until Section 1.23. Here we give the proof of Theorem 1.15.3.
Proof of Theorem 1.15.3
Proof
We want to count the number of elements in the set A={1,2,…,pn} that are relatively prime to pn. Let B be the set of elements of A that have a factor greater than 1 in common with A. Note that if b∈B and gcd(b,pn)=d>1, then d is a factor of pn and d>1 so d has p as a factor. Hence b=pk, for some k, and p≤b≤pn, so p≤kp≤pn. It follows that 1≤k≤pn−1. That is, B={p,2p,3p,…,kp,…,pn−1p}.
We are interested in the number of elements of A not in B. Since |A|=pn and |B|=pn−1, this number is pn−pn−1. That is, ϕ(pn)=pn−pn−1.
The proof of Theorem 1.15.4 follows from Theorems 1.15.2 and 1.15.3. The proof is by induction on n and is quite similar to the proof of Theorem 1.15.1(2), so we omit the details. Other properties of the ϕ-function are developed in the exercises.
Multiplicative functions
The functions τ, σ, and ϕ all have a common property, shown in Theorem 1.15.1, Lemma 1.15.1, and Theorem 1.15.2.
Definition 1.15.5: Multiplicative
A function f(n) defined for positive integers n is called multiplicative if f(ab)=f(a)f(b) whenever gcd(a,b)=1.
From the results mentioned above, τ, σ, and ϕ are all multiplicative functions. If you are so inclined, do some additional reading on your own to learn about several pleasing properties that multiplicative functions have in common; it will be worth the effort!
Some final words
Later studies in number theory will lead you into greater depth with τ, σ, and ϕ, as well as with the prime-counting function π, and will introduce you to still other functions satisfying remarkable properties. Though we will say little more in this book about number theoretic functions,1 we finish our discussion with an intriguing unsolved problem in number theory.
In 1907 Robert Carmichael announced that he had proved the following statement:
Carmichael's Conjecture
For every positive integer n there exists a different positive integer m for which ϕ(n)=ϕ(m).
In other words, Carmichael’s statement is that if we were to list ϕ(n) for all positive integers n, each value would show up at least twice; every integer shares its ϕ-value with at least one other integer.
Unfortunately, Carmichael’s proof was faulty, and today the conjecture still has not been proved true (or disproved!). In 1998 Kevin Ford proved that if Carmichael’s conjecture is not true, then the the smallest counterexample (i.e., the smallest number n such that no other number has the same ϕ-value as n) must be larger than 101010. That’s huge! On the other hand, mathematicians have shown that if any counterexample exists, then there are infinitely many counterexamples.
You can find more about Carmichael’s conjecture with a quick internet or library search. Happy reading!
Exercises
Exercise1.15.1
Find σ(n) and τ(n) for the following values of n.
n=900
n=496
n=32
n=128
n=1024
Exercise 1.15.2
Does Lemma 1.15.1 hold if we replace σ by σ∗?
(Hint: The answer is no, but find explicit numbers a and b such that the result fails yet gcd(a,b)=1.)
Exercise 1.15.3
Prove that τ(n) is odd if and only if n is a square.
Exercise 1.15.4
Prove that ∏d|nd=nτ(n)/2, where the product is over all positive divisors d of n.
Exercise 1.15.5
Observe that n=30 can be written in multiple ways as the sum of one or more consecutive positive integers: 30,9+10+11,6+7+8+9,4+5+6+7+8.
Show that for every positive integer n, if n=2pq, where p≥0 and q is odd, then τ(q) is the number of ways that n can be written as the sum of a sequence of consecutive integers.
(Hint: For multiple values of n, try looking at all the ways that n can be written as a sum of consecutive positive integers. Try to match different sums up with different divisors of q in a way that always works, no matter what n is.)
Exercise 1.15.6
Show that if m=pn11pn22⋯pnkk
where p1,…,pk are distinct primes and each ni≥1, then ϕ(m)=m(1−1p1)(1−1p2)⋯(1−1pk).
Exercise 1.15.7
Prove that ϕ(n) is even for all positive integers n other than 1 or 2.
Exercise 1.15.8
Determine, with proof, all numbers n such that
ϕ(n)=2
ϕ(n)=4
ϕ(n)=6
Exercise 1.15.9
Prove that if ϕ(n)∣(n−1), then no prime appears more than once in the prime factorization of n.
Exercise 1.15.10
Let S(n)=∑d∣nϕ(d),
i.e., S(n) is the sum of the values we get when we evaluate ϕ(d) for all positive divisors d of n. For example, S(4)=ϕ(1)+ϕ(2)+ϕ(4)=1+1+2=4,andS(15)=ϕ(1)+ϕ(3)+ϕ(5)+ϕ(15)=1+2+4+18=15.
Compute S(n) for 1≤n≤10 and conjecture2 a formula for S(n).
Exercise 1.15.11
For each function below, decide whether the function is multiplicative or not. (Assume that functions are defined on the set of positive integers.)
f(n)=1
f(n)=n
f(n)=1+n
f(n)=log(n)
f(n)=2n
f(n)=1/n
Exercise 1.15.12
For positive integers n, define the function ρ by ρ(n)=the number of prime factors ofn;
for example, ρ(4)=1 (since 4 is divisible by 2 and no other prime) and ρ(100)=2 (since 100 is divisible by 2 and 5 and no other prime).
Is the function ρ(n) multiplicative? Explain why or why not.
Is the function f(n)=(−1)ρ(n) multiplicative? Explain why or why not.
Exercise 1.15.13
Show that for any positive integers a and b, if f is a multiplicative function, then f(a)⋅f(b)=f(gcd(a,b))⋅f(lcm(a,b)).
Footnotes
You'll find more, though, in the exercises in this chapter.
Though the proof of your conjecture, once you make it, is a bit beyond the scope of this text, you may find a proof in many other textbooks on number theory.
1.14: Fermat Primes and Mersenne Primes
1.16: Perfect Numbers and Mersenne Primes |
190353 | https://www.youtube.com/watch?v=7mzRJr7k3OU | What is a "Solution Set" for an Equation (Algebra)
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Transcript:
An algebra exercise that we see quite often,
is one where we're asked to find the solution set to some equation. So in this video I just
want to clarify what that means. What does it mean to be a solution set? How do you write
solution sets, and things like that. Ah, well first of all, just to state the obvious, a
solution set is a set of all the values that are solutions to some equation. Now my immediate
concern is, to show you how, how do you write this? There's, there�s a certain notation
that they're looking for, so the notation is simply to put these curly braces around
whatever numbers ah, are solutions to whatever your equation is. So for example, let's say
if you had the equation, let's say ah, five m equals fifteen. Well the solution is obviously
is m equals three, because three times five is fifteen. But how do we write this as a
solution set? Well here's all we do. For a solution set, we'll simply put curly braces
around the number three, and that's it. We don't put m equals, or anything like that.
We just put braces, like a basket, we put the three in it, and that's it. Okay so let's
look at this example here. Here we have five x plus one equals thirty-one. And again, my,
my main concern is not how to solve that, that's something we can talk about in another
video. Ah, for this one my main concern is simply how do we write our answer? Ah, again
this example is pretty small. I think we can see that the answer would be x equals six.
Because five times six, is thirty, and thirty plus one, makes thirty-one. But that's not
written as a solution set. To write it as a solution set, all we do, is we do curly
braces, and then we'll just put the number six in here. Now if you had another example,
which I'm not going to do one like this in this video. But let's say you had an example
where you had three solutions, or four solutions, or a lot of solutions. Then you would just
put curly braces, and then you would put your values, and then you would simply separate
them by a comma, and then you put all those values inside curly braces, and you just separate
your different solutions by commas. But in these two examples I showed you, we just have
a single solution, so we put curly braces just around our single answer. |
190354 | https://brainly.com/question/28999309 | [FREE] What value of x makes the equation x^3 = 24 true? - brainly.com
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What value of x makes the equation x 3=24 true?
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x=2 3 3
Explanation
Given the equation:
x 3=24
To determine the value of x, we take the cube root of both sides.
3 x 3=3 24⟹x=3 24⟹x=3 8×3⟹x=3 8×3 3x=2 3 3
Answer:
Explanation:
Answered by KaydaX411525 •1 answer
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To solve the equation x 3=24, take the cube root of both sides to find x=2 3 3, which gives the exact solution. This value can be approximated to about 2.52 if necessary.
Explanation
To solve the equation x 3=24, we need to find the value of x that satisfies this equation. We can do this by taking the cube root of both sides:
Start with the original equation:
x 3=24
Take the cube root of both sides:
x=3 24
We can simplify 3 24 further. Notice that 24=8×3: x=3 8×3
Using the property of cube roots, this can be expressed as:
x=3 8⋅3 3
Since 3 8=2, we have:
x=2⋅3 3
Thus, the solution to the equation is:
x=2 3 3. This expression provides the exact value of x and can be approximated numerically if needed. In approximate decimal form, this value is about 2.52.
Examples & Evidence
An example of finding a cube root is calculating 3 27=3, since 3 3=27. In a similar way, we determined the cube root of 24 to find the value of x.
Taking the cube root is a fundamental property in algebra, and it is critical for solving cubic equations like this one.
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Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points.
Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab
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190355 | https://www.grc.nasa.gov/www/k-12/airplane/specheat.html | Specific Heats
+ Text Only Site + Non-Flash Version + Contact Glenn Thermodynamics is a branch of physics which deals with the energy and work of a system. Thermodynamics deals only with the large scale response of a system which we can observe and measure in experiments. In aerodynamics, we are most interested in thermodynamics for the role it plays in engine design and high speed flight. On this slide we derive some equations which relate the heat capacity of a gas to the gas constant used in the equation of state. We are going to be using specific values of the state variables. For a scientist, a "specific" state variable means the value of the variable divided by the mass of the substance. This allows us to derive relations between variables without regard for the amount of the substance that we have. We can multiply the specific variable by the quantity of the substance at any time to determine the actual value of the flow variable. From our studies of heat transfer, we know that the amount of heat transferred between two objects is proportional to the temperature difference between the objects and the heat capacity of the objects. The heat capacity is a constant that tells how much heat is added per unit temperature rise. The value of the constant is different for different materials and depends on the process. Heat capacity is not a state variable. If we are dealing with a gas, it is most convenient to use forms of the thermodynamics equations based on the enthalpy of the gas. From the definition of enthalpy: h = e + p v where h in the specific enthalpy, p is the pressure, v is the specific volume, and e is the specific internal energy. During a process, the values of these variables change. Let's denote the change by the Greek letter delta which looks like a triangle. So "delta h" means the change of "h" from state 1 to state 2 during a process. Then, for a constant pressure process the enthalpy equation becomes: delta h = delta e + p delta v The enthalpy, internal energy, and volume are all changed, but the pressure remains the same. From our derivation of the enthalpy equation, the change of specific enthalpy is equal to the heat transfer for a constant pressure process: delta h = cp delta T where delta T is the change of temperature of the gas during the process,and c is the specific heat capacity. We have added a subscript "p" to the specific heat capacity to remind us that this value only applies to a constant pressure process. The equation of state of a gas relates the temperature, pressure, and volume through a gas constant R. The gas constant used by aerodynamicists is derived from the universal gas constant, but has a unique value for every gas. p v = R T If we have a constant pressure process, then: p delta v = R delta T Now let us imagine that we have a constant volume process with our gas that produces exactly the same temperature change as the constant pressure process that we have been discussing. Then the first law of thermodynamics tells us: delta e = delta q - delta w where q is the specific heat transfer and w is the work done by the gas. For a constant volume process, the work is equal to zero. And we can express the heat transfer as a constant times the change in temperature. This gives: delta e = cv delta T where delta T is the change of temperature of the gas during the process,and c is the specific heat capacity. We have added a subscript "v" to the specific heat capacity to remind us that this value only applies to a constant volume process. Even though the temperature change is the same for this process and the constant pressure process, the value of the specific heat capacity is different. Because we have selected the constant volume process to give the same change in temperature as our constant pressure process, we can substitute the expression given above for "delta e" into the enthalpy equation. In general, you can't make this substitution because a constant pressure process and a constant volume process produce different changes in temperature If we substitute the expressions for "delta e", "p delta v", and "delta h" into the enthalpy equation we obtain: cp delta T = cv delta T + R delta T dividing by "delta T" gives the relation: cp = cv + R The specific heat constants for constant pressure and constant volume processes are related to the gas constant for a given gas. This rather remarkable result has been derived from thermodynamic relations, which are based on observations of physical systems and processes. Using the kinetic theory of gases, this same result can be derived from considerations of the conservation of energy at a molecular level. We can define an additional variable called the specific heatratio, which is given the Greek symbol "gamma", which is equal to cp divided by cv: gamma = cp / cv "Gamma" is just a number whose value depends on the state of the gas. For air, gamma = 1.4 for standard day conditions. "Gamma" appears in many fluids equations including the equation relating pressure, temperature, and volume during a simple compression or expansion process, the equation for the speed of sound, and all the equations for isentropic flows, and shock waves. Because the value of "gamma" just depends on the state of the gas, there are tables of these values for given gases. You can use the tables to solve gas dynamics problems. Activities: Guided Tours Navigation.. Beginner's Guide Home Page + Inspector General Hotline + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act + Budgets, Strategic Plans and Accountability Reports + Freedom of Information Act + The President's Management Agenda + NASA Privacy Statement, Disclaimer, and Accessibility CertificationEditor: Nancy Hall NASA Official: Nancy Hall Last Updated: May 13 2021 + Contact Glenn |
190356 | https://arxiv.org/pdf/2112.09183 | Approximation with one-bit polynomials in Bernstein form
C. Sinan G¨ unt¨ urk ∗ and Weilin Li †
December 8, 2022
Dedicated to Professor Ron DeVore on the occasion of his 80th birthday.
Abstract
We prove various theorems on approximation using polynomials with integer coefficients in the Bernstein basis of any given order. In the extreme, we draw the coefficients from {± 1} only. A basic case of our results states that for any Lipschitz function f : [0 , 1] → [−1, 1] and for any positive integer n, there are signs σ0, . . . , σ n ∈ {± 1} such that
∣∣∣∣∣f (x) −
n
∑
k=0
σk
(nk
)
xk(1 − x)n−k
∣∣∣∣∣ ≤ C(1 + |f |Lip )1 + √nx (1 − x) for all x ∈ [0 , 1] .
More generally, we show that higher accuracy is achievable for smoother functions: For any integer s ≥ 1, if f has a Lipschitz ( s−1)st derivative, then approximation accuracy of order
O(n−s/ 2) is achievable with coefficients in {± 1} provided ‖f ‖∞ < 1, and of order O(n−s) with unrestricted integer coefficients, both uniformly on closed subintervals of (0 , 1) as above. Hence these polynomial approximations are not constrained by the saturation of classical Bernstein polynomials. Our approximations are constructive and can be implemented using feedforward neural networks whose weights are chosen from {± 1} only.
Keywords Bernstein polynomials, integer constraints, ±1 coefficients, sigma-delta quantization, noise shaping.
Mathematics Subject Classification 41A10, 41A25, 41A29, 41A40, 42C15, 68P30.
1 Introduction
It is a classical result that a continuous function f : [0 , 1] → R can be approximated uniformly by polynomials with integer coefficients if and only if f (0) and f (1) are integers. Here the integer coefficients are understood to be with respect to the default “power basis,” i.e. the mononomials
{1, x, x 2, . . . }. The necessity of the condition is immediate. The sufficiency, on the other hand, is non-obvious at best, yet the following constructive proof by Kantorovich (see also [28, Ch.2.4]),
∗
NYU Courant Institute, email: gunturk@cims.nyu.edu.
†
CUNY City College and NYU Courant Institute, email: wli6@ccny.cuny.edu, weilinli@cims.nyu.edu.
1
arXiv:2112.09183v2 [cs.IT] 6 Dec 2022
is remarkably short and transparent: Recall that the Bernstein polynomial of f of order n (and degree ≤ n), defined by
Bn(f, x ) := Bn(f )( x) :=
n
∑
k=0
f
( kn
)(nk
)
xk(1 − x)n−k,
converges to f uniformly. Set
B∗
n
(f, x ) := B∗
n
(f )( x) :=
n
∑
k=0
[
f
( kn
)(nk
)]
xk(1 − x)n−k, (1) where [ u] ∈ Z stands for any rounding of u ∈ R to an immediate neighboring integer value. It is evident that B∗
n
(f, x ) is a polynomial with integer coefficients, and of degree at most n. With the assumption that f (0) and f (1) are integers, the total rounding error can be bounded uniformly over x ∈ [0 , 1] via
|Bn(f, x ) − B∗
n
(f, x )| ≤
n−1
∑
k=1
xk(1 − x)n−k ≤ 1
n
n−1
∑
k=1
(nk
)
xk(1 − x)n−k < 1
n (2) which shows B∗
n
(f ) → f uniformly as well. Since accuracy of the Bernstein polynomial approximation saturates at the rate 1 /n (unless
f is a linear polynomial), the rate at which B∗
n
(f ) converges to f is as good as that of Bn(f ). Kantorovich actually proved a stronger result in , showing that the error of best approximation to f by polynomials of degree n with integer coefficients is bounded by 2 En(f ) + 1 /n where En(f )denotes the error of best approximation of f by unconstrained polynomials of degree n. This can be shown by employing the polynomial of best approximation as a surrogate instead of the Bernstein polynomial (see [27, 26]). The history of approximation by polynomials with integer coefficients is rich, with the earliest result going back to P´ al , followed shortly by Kakeya and Chlodovsky . For an extensive treatment of the subject, we refer to Ferguson’s classical text . One of the important charac-teristics of the theory is that uniform approximation of continuous functions is only possible on intervals of length less than 4, and then only with certain arithmetic constraints on f : We already saw that on [0 , 1] it is necessary (and sufficient) that f (0) and f (1) are integers. On [ −α, α ] where
α < 1, it is necessary (and sufficient) that f (0) is an integer, whereas on [ −1, 1], it is necessary (and sufficient) that f (−1) , f (0) , f (1) , 12 (f (−1)+ f (1)) are all integers. The number of such arith-metic conditions increases without bound as the length of the interval increases towards 4. In the other extreme, there are no arithmetic conditions when approximation is sought on closed intervals containing no integers. In the context of this paper we will assume that all uniform approximation takes place on a subinterval [ a, b ] ⊂ (0 , 1). It is known that Kantorovich’s method, which yields a residual error of 1 /n , is suboptimal. Indeed, for any 0 < a < b < 1, the error of best uniform approximation to any f ∈ C([ a, b ]) by a degree n polynomial with integer coefficients is bounded by En(f ) + ρn, where ρ := ρ(a, b ) < 1 (see , [26, Thm 11.15]). Consequently, in all traditional smoothness classes of finitely many deriva-tives, the constraint of integer coefficients does not hamper the rate at which these functions can 2be approximated uniformly by polynomials. It is, however, important to remember the constraint on the domain [ a, b ]. This paper is concerned with approximations by polynomials with integer coefficients subject to additional, stringent conditions on what these integers can be, while still guaranteeing similar approximation properties. To explain what these stringent conditions are, we first need some additional notation: By the Bernstein basis of order n, we refer to the list of polynomials Bn := (pn,k )nk=0 where
pn,k (x) :=
(nk
)
xk(1 − x)n−k. (3)
Bn is a basis of the vector space Pn of polynomials (of a real variable) of degree at most n. Let us denote by B◦
n
:= ( p◦
n,k
)nk=0 the plain (unnormalized) version of this basis, given by p◦
n,k
(x) :=
xk(1 − x)n−k. Let us also denote by Π n := ( πk)n
0
the power basis of degree n given by πk(x) := xk.Then B∗
n
(f ), as defined in (1), can be viewed as an approximation to f from the lattice L(B◦
n
)generated by B◦
n
, and of course, also from L(Π n) as originally intended. In fact, L(B◦
n
) = L(Π n)since for each k = 0 , . . . , n , we have p◦
n,k
∈ L (Π n) and πk ∈ L (B◦
n
). The first claim is immediate, and the latter is seen by noting that
xk = xk(x + (1 − x)) n−k =
n−k
∑
j=0
(n − kj
)
xk+j (1 − x)n−k−j =
n
∑
l=k
(n − kl − k
)
p◦
n,l
(x).
Meanwhile, L(Bn), which we will refer to as the Bernstein lattice , is a significantly smaller sublattice of L(B◦
n
) as n increases, and therefore, approximation by its elements presents an in-creasingly coarser rounding (quantization) problem. Indeed, the fundamental cell of L(Bn) con-tains Mn := ∏nk=0
(nk
) elements of L(B◦
n
), and therefore, encoding the elements of L(B◦
n
) requires
μn := 1
n+1
log 2 Mn times as many bits per basis polynomial as it would require for L(Bn). It can be checked that Mn grows as exp( cn 2), therefore μn grows linearly. To motivate the same point from an approximation perspective, let us inspect what happens to the total rounding error bound when we gradually coarsen the lattice L(B◦
n
) towards L(Bn) while employing the same simple rounding algorithm: For any α ∈ [0 , 1], let ∆ n,k := ∆ n,k (α) be the integer part of (nk
)α, and analogous to (1), consider the rounding of f to the lattice generated by
Bαn := (∆ n,k p◦
n,k
)nk=0 by means of the polynomial
B∗,α n (f, x ) :=
n
∑
k=0
[
f
( kn
)(nk
)
∆−1
n,k
]
∆n,k xk(1 − x)n−k. (4) Noting that ∆ n, 0 = ∆ n,n = 1, and again assuming that f (0) and f (1) are integers, the total rounding error is now bounded above by
n−1
∑
k=1
(nk
)α
xk(1 − x)n−k =
n−1
∑
k=1
(pn,k (x)) α(p◦
n,k
(x)) 1−α
≤
(n−1∑
k=1
pn,k (x)
)α (n−1∑
k=1
xk(1 − x)n−k
)1−α
< n−1+ α,
3where in the last step we have used the findings of (2). This simple extension shows that it is possible to approximate from L(Bαn) without effort for all α < 1, but the method breaks down at α = 1, i.e. for the Bernstein lattice L(Bn). In this paper, our primary focus will be on enabling approximation from subsets of this lattice by means of a more advanced quantization method known as noise-shaping quantization .To explain what this means, let us recall that each pn,k is a bump function that peaks at k/n .However, the Bernstein basis is poorly localized as a whole. As can be seen by the Laplace (normal) approximation to the binomial distribution, each pn,k spreads significantly over the neighboring
O(√n) basis functions, making Bn behave approximately like a frame of redundancy O(√n). This heuristic suggests that there is numerical flexibility in the choice of coefficients when functions are approximated by linear combinations of the pn,k ; this flexibility then leads to the possibility of coarse quantization. In general, noise-shaping quantization refers to the principle of arranging the quantization noise (i.e. the quantization error of the coefficients) to be mostly invisible to the accompanying recon-struction operator. Also known as “one-bit” quantization due to its potential to enable very coarse quantization, noise-shaping quantization was first introduced for analog-to-digital conversion cir-cuits during the 60s ([21, 22]) where it became established as Sigma-Delta modulation (or Σ∆ quantization), though some of its ideas can be found in the theory of Beatty sequences. Σ∆ quanti-zation gained some popularity in information theory through the works of Gray et al (e.g. ), but remained largely unknown in the general mathematics community until the groundbreaking work of Daubechies and DeVore . Since then, the theory has been extended significantly to apply to various approximation problems with quantized coefficients. We refer the reader to [17, 2, 18, 7] for some of the recent mathematical evolution of the subject, and to [5, 32, 36] for engineering applications. In this paper, we will use methods of noise-shaping quantization to establish various results concerning approximation from the Bernstein lattices, including the following:
• The collection
∞
⋃
n=1
L(Bn) of all Bernstein lattices is dense in Lp([0 , 1]) for all 1 ≤ p < ∞, and in C([ a, b ]) for any 0 < a < b < 1. More precisely, in each of these spaces, the distance from
L(Bn) to any f goes to 0.
• For any 0 < a < b < 1, if f ∈ C([ a, b ]) can be approximated uniformly to within O(n−s) by polynomials of degree n with real coefficients, then it can also be approximated uniformly to within O(n−s) by elements of L(Bn).
• Any f ∈ Lip([0 , 1]) such that ‖f ‖∞ ≤ 1 can be approximated to within O(n−1/2) by a polynomial with coefficients in {± 1} in the Bernstein basis of order n. The approximation rate improves to O(n−s/ 2) if f (s−1) ∈ Lip([0 , 1]), provided ‖f ‖∞ < 1. These approximations are uniform over any given [ a, b ] ⊂ (0 , 1). The paper is organized as follows: Section 2 describes how the noise-shaping method of Σ∆ quantization works in connection with the Bernstein basis. Specific error bounds are given in Section 3 where both Lp spaces and smoothness classes are considered. This section also describes how iterated Bernstein operators enable higher order Σ∆ quantizers. 4An unexpected and novel contribution of this paper is the effective use of noise-shaping quan-tization methods in the setting of linearly independent systems of vectors. To our knowledge, the Bernstein system constitutes the first example of this kind. In Appendix A, we quantify the sense in which the Bernstein basis Bn behaves like a frame of redundancy √n by deriving the exact eigenvalue distribution of the associated frame operator. Another novel contribution of this work is computational: our one-bit polynomial approxima-tions can be computed exactly by means of feedforward neural networks whose weights are chosen from {± 1} only. We show how this is done in Appendix B.
2 Approximation by quantized polynomials in Bernstein form
All of our approximations via quantized polynomials in Bernstein form will follow a two-stage process. The first stage consists of classical polynomial approximation and the second stage is quantization through noise-shaping. For any given function f : [0 , 1] → R in a suitable function class, we will first approximate it by a polynomial P ∈ P n which has the representation
P (x) =
n
∑
k=0
yk pn,k (x)with respect to Bn. The polynomial P could be the Bernstein polynomial Bn(f ) of f when it is defined, but it can be a replacement, such as the Kantorovich polynomial of f ([13, Ch.10]), or it can also be a better approximant in Pn, especially if f has high order of smoothness. Both the quality of the approximation and the range of the coefficients ( yk)n
0
will matter. We define the
approximation error of f by
EAn (f, P ; x) := f (x) − P (x),
and will often suppress P in the notation. The second stage consists of quantization of P via its representation in the Bernstein basis. The coefficients y := ( yk)n
0
of P will be replaced with their quantized version q := ( qk)n
0
taking values in a constrained set A, called the (quantization) alphabet. We define the quantization error of y by
EQn (y, q ; x) :=
n
∑
k=0
(yk − qk)pn,k (x),
and again, will often suppress q in our notation.
2.1 Noise-shaping through Σ∆ quantization:
Σ∆ quantization (modulation) refers to a large family of algorithms designed to convert any given sequence y := ( yk) of real numbers to another sequence q := ( qk) taking values in a discrete set
A (typically an arithmetic progression) in such a way that the error y − q between them (i.e. the quantization error) is a “high-pass” sequence, i.e. it produces a small inner-product with any slowly varying sequence. The canonical way of ensuring this is to ask that y and q satisfy the rth order difference equation
y − q = ∆ ru, (5) 5where (∆ u)k := uk − uk−1, for some bounded sequence u. We then say that q is an rth order noise-shaped quantization of y. When (5) is implemented recursively, it means that each qk is found by means of a “quantization rule” of the form
qk = F (uk−1, u k−2, . . . , y k, y k−1, . . . ), (6) and uk is updated via
uk =
r
∑
j=1
(−1) j−1
(rj
)
uk−j + yk − qk (7) to satisfy (5). This recursive process is commonly called “feedback quantization” due to the role qk
plays as a feedback control term in (7). The role of the quantization rule (6) is to keep the solution
u bounded for all input sequences y of arbitrary duration in a given set Y. In this case, we say that the quantization rule is stable for Y.The “greedy” quantization rule refers to the function F which outputs any minimizer qk ∈ A
of |uk| as determined by (7). More precisely, it sets
qk := round A
r
∑
j=1
(−1) j−1
(rj
)
uk−j + yk
(8) where round A(v) stands for any element in A that is closest to v. If A is an infinite arithmetic progression of step size δ, then greedy quantization is always stable (i.e. for any r and for all input sequences) and produces a solution u to (5) via (7) which is bounded by δ/ 2. Real difficulties start when A is a fixed, finite set, the extreme case being a set of two elements, which can be taken to be {± 1} without loss of generality. In this case, it is a major challenge to design quantization rules that are stable for any order r and for arbitrary bounded inputs in a range [−μ, μ ]. The first breakthrough on this problem was made in the seminal paper of Daubechies and DeVore where it was shown for A = {± 1} that for any order r ≥ 1 and for any μ < 1, there is a stable rth order quantization rule Fr,μ which guarantees that u is bounded by some constant
Cr,μ for all input sequences y that are bounded by μ. (For r = 1, one can take μ = 1 and find that C1,1 = 1 will do.) The constant Cr,μ depends on both r and μ, and blows up as r → ∞ , or as
μ → 1− (except when r = 1). Another family of stable quantization rules, but with more favorable
Cr,μ , was subsequently proposed in . In this paper we will be using the greedy quantization rule when A = Z, and either of the rules in and when A = {± 1}. We will not need the explicit descriptions of these rules, or of the associated bounds Cr,μ .Clearly q is a bounded sequence when A is a finite set, but even when A = Z, q will be bounded when the input y is bounded. This is because (5) implies ‖y − q‖∞ ≤ 2r‖u‖∞ ≤ 2r−1 so that
‖q‖∞ ≤ ‖ y‖∞ + 2 r−1.In the remaining sections, the letter C will represent any constant whose value is not important for the discussion. Its value may be updated and it may also stand for different constants. Whenever
C depends on a given set of parameters, as in the previous paragraph, these parameters will be specified. 62.2 Effect of noise shaping in the Bernstein basis
It will be convenient for us to extend the index k beyond n in the definition of pn,k . We do this without modifying the formula (3), noting that this implies pn,k = 0 for all k > n since (nk
) = 0 in this range. As indicated in the previous subsection, we will always work with a stable rth order Σ∆ quanti-zation scheme that is applied to convert an input sequence y := ( yk)n
0
bounded by μ to its quantized version q := ( qk)n
0
. We will set uk = 0 for all k < 0. With this assumption, we have the total quantization error
EQn (y; x) =
n
∑
k=0
(∆ ru)k pn,k (x)=
n
∑
k=0
uk ( ˜∆rpn, ·(x)) k (9) where ˜∆ is the adjoint of ∆ given by ( ˜∆u)k := uk − uk+1 . Notice that with our convention on pn,k
for k > n , all the boundary terms are correctly included in (9). It now follows by our assumption of stability that |uk| ≤ Cr,μ for all k, so that
|EQn (y; x)| ≤ Cr,μ Vn,r (x) (10) where
Vn,r (x) :=
n
∑
k=0
∣∣∣( ˜∆rpn, ·(x)) k
∣∣∣ (11) stands for the rth order variation of the basis Bn.As we consider increasing values of r in the sections below, we will be providing specific upper bounds for Vn,r (x) of increasing complexity. We now note two important properties of the Bernstein basis that we will employ in our analysis. (See [13, Ch.10] and for these and other useful facts about Bernstein polynomials.) 1. The consecutive differences of the pn,k satisfy ( ˜∆pn, ·(x)) k = pn,k (x) − pn,k +1 (x) = (k + 1) − (n + 1) x
(n + 1) x(1 − x) pn+1 ,k +1 (x) (12) which, with our convention on pn,k for k > n , holds for all n, k ≥ 0, and all x ∈ [0 , 1]. When interpreting the right hand side of this equation for x = 0 or x = 1, it should be observed that the polynomial (( k+1) −(n+1) x)pn+1 ,k +1 (x) is divisible by x(1 − x) for all k ≥ 0. As is customary, we will use the short notation X := x(1 − x). 2. We set
Tn,s (x) :=
n
∑
k=0
(k − nx )spn,k (x) (13) for each non-negative integer s. Then we have Tn, 0(x) = 1, Tn, 1(x) = 0, Tn, 2(x) = nX . In general, for each s, there is a constant As such that 0 ≤ Tn, 2s(x) ≤ Asns (14) holds for all n ≥ 1, uniformly over x ∈ [0 , 1]. 73 Specific error bounds
3.1 First order noise shaping
For first order Σ∆ quantization, the greedy rule is the only rule one needs. There will be two choices for A of interest to us, {± 1} and Z. When A = {± 1}, we have |uk| ≤ 1 for all input sequences y bounded by 1. When A = Z, we have |uk| ≤ 12 for all input sequences y. In either case, (10), (11) and (12) give us, for all admissible input sequences y, the bound
|EQn (y; x)| ≤ Vn, 1(x)= 1(n + 1) X
n
∑
k=0
∣∣(k+1) − (n+1) x∣∣ pn+1 ,k +1 (x)
≤ 1(n + 1) X
√
Tn+1 ,2(x)= 1
√(n + 1) X , (15) where we used Cauchy-Schwarz inequality in the third line. We also have the trivial bound
|EQn (y; x)| ≤ ‖ y − q‖∞ ≤ 2‖u‖∞ . 1. (16) (Here, as usual, An . Bn means An ≤ CB n for all n where C is an absolute constant. When C
depends on some parameter α, we use the notation .α.) Combining these two bounds, we obtain
|EQn (y; x)| . min
(
1
√(n + 1) X , 1
)
. 11 + √nX . (17) Our first theorem follows directly from this bound:
Theorem 1. For every continuous function f : [0 , 1] → [−1, 1] , and for every positive integer n,there exist signs σ0, . . . , σ n ∈ {± 1} such that
∣∣∣∣∣f (x) −
n
∑
k=0
σk pn,k (x)
∣∣∣∣∣ . ω2(f, √X/n ) + 11 + √nX , (18)
where ω2 stands for the 2nd modulus of smoothness of f and X := x(1 − x). In particular, if f is Lipschitz, then ∣∣∣∣∣f (x) −
n
∑
k=0
σk pn,k (x)
∣∣∣∣∣ . 1 + |f |Lip
1 + √nX . (19)
Proof. For any given n, we set yk = f ( kn ), k = 0 , . . . , n , so that P (x) = Bn(f, x ), and we use
σk = qk as the output of the first order Σ∆ scheme with input ( yk). Since |yk| ≤ 1 for all k, the scheme is stable and the result follows by noting (see [13, p.308] that
|EA(f ; x)| = |f (x) − Bn(f, x )| . ω2(f, √X/n )together with (17). The specific bound for the Lipschitz class follows trivially by noting that √X/n ≤ 1/(1 + √nX ). 8Remarks:
• As a function of x, the upper bound (19) goes to 0 at the rate n−1/2 in C([ a, b ]), 0 < a < b < 1, and in Lp([0 , 1]), 1 ≤ p < 2. The case p ≥ 2 yields slower decay rates. However, a complete discussion of functions in Lp([0 , 1]) requires an unbounded alphabet, since the approximating polynomials (whether one uses suitable substitute Bernstein polynomials, or the Kantorovich polynomials of f directly), can have arbitrarily large coefficients in the Bernstein basis. See Theorem 2 below for the full discussion of this case for A = Z, which includes a more precise description of the quantization error in all p-norms.
• The convergence rate n−1/2 is the best one can expect from Bernstein polynomial approxi-mation of Lipschitz functions (even without quantization), so it may come as a surprise that the rate holds up in the case of one-bit quantization. Information-theoretically, however, this rate is suboptimal, though perhaps not hugely so: The ε-entropy of the set of functions
{f ∈ Lip([ a, b ]) : ‖f ‖∞ ≤ 1, |f |Lip ≤ L}
behaves as L(b−a)ε−1(1+ o(1)) (see [25, 28]) which implies that any ε-net for it of cardinality
O(2 n) must satisfy ε & L(b − a)n−1.
• We used the Bernstein polynomial of f , i.e. yk = f (k/n ), in our construction to achieve two goals at once: (i) approximation and (ii) guaranteed stability of the Σ∆ quantization scheme (due to yk being in the same range as f ). However, if one can find better polynomial approximants whose coefficients in the Bernstein basis still fall in the range [ −1, 1], then these could be used instead to reduce the approximation error EA(f, P ; x). We will discuss this in Section 3.3. (We also note here a relevant result: If a polynomial P (x) takes values in ( −1, 1) for all x ∈ [0 , 1], then [35, Thm 1] shows that for all sufficiently large m, the coefficients of P with respect to Bm will also be in the range ( −1, 1). This opens up the possibility of employing better P , provided the value of m can be controlled well, relative to the degree of P . However, it seems that in this generality the currently available quantitative information regarding this question is not useful enough, at least not immediately, to yield an improvement.)
• As we will see in the next two subsections, the quantization error EQ(y, q ; x) can be reduced via higher order schemes. The difficulty about improving the approximation error EA(f, P ; x) while maintaining quantizer stability disappears when A = Z since the greedy Σ∆ scheme is then unconditionally stable (i.e. for all inputs), and this leads to a stronger approximation result as we show next. For any p ∈ [1 , ∞), let En(f )p stand for the error of best approximation of f ∈ Lp([0 , 1]) by elements of Pn. For p = ∞,we consider f ∈ C([0 , 1]) instead. We have
Theorem 2. For every real-valued function f ∈ Lp([0 , 1]) , where 1 ≤ p < ∞, and for every positive integer n, there exist q0, . . . , q n ∈ Z such that
∥∥∥∥∥f −
n
∑
k=0
qk pn,k
∥∥∥∥∥p
. En(f )p + Cp
n−1/2, 1 ≤ p < 2,√ log nn , p = 2 ,n−1/p , 2 < p < ∞.
(20) 9If f is continuous, then there exist q0, . . . , q n ∈ Z such that
∣∣∣∣∣f (x) −
n
∑
k=0
qk pn,k (x)
∣∣∣∣∣ . En(f )∞ + 11 + √nX . (21)
Consequently, for any sequence (nj )∞
1
of integers such that nj → ∞ ,
∞
⋃
j=1
L(Bnj ) is dense in Lp([0 , 1])
for all p ∈ [1 , ∞), and in C([ a, b ]) for all 0 < a < b < 1.Proof. Given f in Lp([0 , 1]) or C([0 , 1]), let P ∈ P n be the best approximant of f in the correspond-ing p-norm, with coefficients y := ( yk)n
0
in the Bernstein basis. Hence we have ‖EAn (f )‖p = En(f )p.Next we bound ‖EQn (y)‖p using (17). Because of the symmetry of the bound with respect to 1/2, and employing the inequality x(1 − x) ≥ x/ 2 for x ∈ [0 , 1/2], we have
∫ 10
1(1 + √nX )p dx ≤ 2
∫ 1/n
0
dx + 2
∫ 1/21/n
1(√nx/ 2) p dx
.p
n−p/ 2, 1 ≤ p < 2,
log nn
, p = 2 ,n−1, 2 < p < ∞,
(22) hence (20) follows. When p = ∞, we have to settle with the pointwise bound only.
Remark: With additional information on f , the approximation error term En(f )p in Theorem 2 can be quantified further; for example, via En(f )p .s ωs(f, 1/n )p, which holds for any n, s such that n ≥ s.
3.2 Second order noise shaping
For the second order, we will use the greedy quantization rule for A = Z, and any of the stable schemes proposed in or . The analysis of the quantization error will be the same in both cases, the only difference concerning the range of admissible coefficients y. For A = Z, we will be able to work with all sequences again, whereas for A = {± 1} we will have to restrict to the yk that range in [ −μ, μ ] for some μ < 1. The trivial bound (16) is replaced with
|EQ(y; x)| ≤ ‖ y − q‖∞ = ‖∆2u‖∞ ≤ 4‖u‖∞ ≤
{ 4, if A = Z,
4C2,μ , if A = {± 1}, (23) where the value of C2,μ depends on which stable second order scheme is employed. Meanwhile, utilization of the general purpose bound (10) via (11) requires another application of ˜∆ on (12). To lighten the notation, we will adopt the following convention when there is no possibility for confusion: pn,k will be short for pn,k (x) (since all basis functions will be evaluated at the same point), and ˜∆rpn,k will be short for ( ˜∆rpn, ·(x)) k. (Note that there is no ambiguity in the meaning of ˜∆ acting on a shifted sequence such as pn,k +1 since shifts commute with ˜∆.) With 10 this convention, we have ˜∆2pn,k = ˜∆pn,k − ˜∆pn,k +1
= 1(n+1) X
((k+1 − (n+1) x)pn+1 ,k +1 − (k+2 − (n+1) x)pn+1 ,k +2
)
= 1(n+1) X
((k+2 − (n+1) x) ˜∆pn+1 ,k +1 − pn+1 ,k +1
)
= 1(n+1) X
((k+2 − (n+2) x) ˜∆pn+1 ,k +1 + x ˜∆pn+1 ,k +1 − pn+1 ,k +1
)
= 1(n+1)( n+2) X2
(k+2 − (n+2) x)2pn+2 ,k +2 − 1(n+1) X
(
xp n+1 ,k +2 + (1 − x)pn+1 ,k +1
)
so that
Vn, 2(x) =
n
∑
k=0
∣∣∣( ˜∆2pn, ·(x)) k
∣∣∣ ≤ Tn+2 ,2(x)(n+1)( n+2) X2 + 1(n + 1) X = 2(n + 1) X (24) and therefore
|EQn (y; x)| .μ min
( 1(n + 1) X , 1
)
. 11 + nX . (25) Equipped with this bound, we can now improve the rate of convergence for smoother functions. The proof of the next theorem mirrors the proof of Theorem 1 verbatim.
Theorem 3. Let μ < 1 be arbitrary. For every continuous function f : [0 , 1] → [−μ, μ ], and for every positive integer n, there exist signs σ0, . . . , σ n ∈ {± 1} such that
∣∣∣∣∣f (x) −
n
∑
k=0
σk pn,k (x)
∣∣∣∣∣ .μ ω2(f, √X/n ) + 11 + nX (26)
where ω2 stands for the 2nd modulus of smoothness of f and X := x(1 − x). In particular, if f has a Lipschitz derivative, then
∣∣∣∣∣f (x) −
n
∑
k=0
σk pn,k (x)
∣∣∣∣∣ .μ
1 + |f ′|Lip
1 + nX . (27) Similarly, we are now also able to improve the bound achieved in Theorem 2. Again, the proof of next theorem mirrors that of Theorem 2. The only difference is now we are bounding the p-norm of the function 1 /(1 + nX ).
Theorem 4. For every real-valued function f ∈ Lp([0 , 1]) , where 1 ≤ p < ∞, and for every positive integer n, there exist q0, . . . , q n ∈ Z such that
∥∥∥∥∥f −
n
∑
k=0
qk pn,k
∥∥∥∥∥p
. En(f )p + Cp
{ log nn , p = 1 ,n−1/p , 1 < p < ∞. (28)
If f is continuous, then
∣∣∣∣∣f (x) −
n
∑
k=0
qk pn,k (x)
∣∣∣∣∣ . En(f )∞ + 11 + nX . (29) 11 3.3 Going beyond second order
Inspecting the outcome of the first and second order noise shaping, it is natural to expect faster decay of the quantization error when higher order noise-shaping is employed. This expectation can be met, as we will discuss later in this section. But first, we need to address the approximation error. As long as Bn(f ) is employed in the first approximation stage, the error bound of Theorem 3 is the best one can achieve (i.e. regardless of the amount of smoothness of f ) due to the saturation of the approximation error at the rate 1 /n ; any gain in the quantization error from using higher order noise shaping would be drowned by this term. Possibility of improvement by means of better polynomial approximations whose coefficients in Bn also admit stable higher order quantization remains. Indeed, as we did before, this approach works “out of the box” in the case A = Z since we have unconditional stability. However, for A = {± 1}, this is not immediate: if we are to work with our stability criterion ‖y‖∞ ≤ μ < 1, then we will need to ensure that the coefficients of these improved polynomial approximations also satisfy this criterion. Several adjustments of classical Bernstein polynomials that adapt to the smoothness of the function that is being approximated have been proposed in the literature. Most of these are not suitable for our quantization method, at least not immediately, as they do not necessarily produce polynomial approximations of the same degree, but there is one that works: the particular linear combinations of iterated Bernstein operators, as developed by Micchelli and Felbecker . As before, let Bn be the Bernstein operator on C([0 , 1]) that maps f to Bn(f ), and for any integer
r ≥ 1, let
Un,r := I − (I − Bn)r. (30) It follows by combining the results of and (see also [4, Ch.9.2]) that for any s ≥ 1, if
f ∈ Cs−1([0 , 1]) and f (s−1) is Lipschitz, then
‖f − Un, ds/ 2e(f )‖∞ .s ‖f ‖Cs−1Lip n−s/ 2 (31) where d·e denotes the ceiling operator and
‖f ‖Cs−1Lip := max( ‖f ‖Cs−1 , |f (s−1) |Lip ). (32) Note that Un, 1 = Bn, so for s = 1 and s = 2, the result (31) reduces to the regular Bernstein polynomial approximation error bound that we have already utilized. More generally, for any r ≥ 1, the coefficients of Un,r (f ) with respect to Bn are well controlled. We have the following:
Theorem 5. For any f ∈ C([0 , 1]) and integers n, r ≥ 1, there exists fn,r ∈ C([0 , 1]) satisfying
Un,r (f ) = Bn(fn,r ) =
n
∑
k=0
fn,r
( kn
)
pn,k
and
‖fn,r − f ‖∞ ≤ (2 r−1−1) ‖f − Bn(f )‖∞.
Therefore, fn,r → f uniformly as n → ∞ , and in particular, ‖f ‖∞ < 1 implies ‖fn,r ‖∞ < 1 for all sufficiently large n. Furthermore, there is an absolute constant c0 such that if f ∈ C2([0 , 1]) and
‖f ‖∞ ≤ 1 − 2 < 1, then for all n ≥ −1c0(2 r−1−1) ‖f (2) ‖∞, we have ‖fn,r ‖∞ ≤ 1 − .
12 Proof. The case r = 1 is trivial, since then Un, 1 = Bn and we can set fn, 1 = f . We consider r ≥ 2. Note the polynomial identity 1 − xr = (1 − x)(1 + x + · · · + xr−1) which implies the relation
Un,r = I − (I − Bn)r = Bn
(
I +
r−2
∑
j=0
(I − Bn)j+1 )
.
Hence, defining
fn,r := f +
r−2
∑
j=0
(I − Bn)j+1 (f )it follows at once that
Un,r (f ) = Bn(fn,r )and that
‖fn,r − f ‖∞ ≤
r−2
∑
j=0
‖(I − Bn)j ‖∞→∞ ‖f − Bn(f )‖∞.
At this point we only need to utilize the trivial bound ‖(I − Bn)j ‖∞→∞ ≤ 2j which follows from
‖Bn‖∞→∞ = 1. Hence we get
‖fn,r − f ‖∞ ≤ (2 r−1−1) ‖f − Bn(f )‖∞.
The next statement is now immediate. In particular, if f ∈ C2([0 , 1]), we can use the bound
‖f − Bn(f )‖∞ ≤ c0‖f (2) ‖∞n−1 to reach the final conclusion. With the above theorem, we are able to enjoy approximation error EA(f, P ) = O(n−s/ 2) for
f ∈ Cs−1Lip([0 , 1]) while maintaining stability of the quantization scheme (for all sufficiently large
n) for the case A = {± 1} with the mere condition ‖f ‖∞ < 1. We can now return to reduction of quantization error by means of higher order noise shaping. Let us first check the ansatz
βn,r (x) := Cr,μ min
( 1(( n + 1) X)r/ 2 , 1
)
as a bound for the rth order quantization error. It is apparent that as a function of n,
‖βn,r ‖pp ‖ βn, 2‖rp/ 2
rp/ 2
1
n
for all r > 2 and all 1 ≤ p < ∞. Therefore, this type of bound would offer no gain over the second order case in terms of the rate of convergence in p-norms, except for the removal of the log n factor for p = 1. However, the situation is not as disappointing for the pointwise bounds (or the uniform bounds on subintervals). Our goal is now to provide a path towards validating this ansatz. We will again use our notation convention and write ˜∆rpn,k for ( ˜∆rpn, ·(x)) k. For r ≥ 1, we have ˜∆r+1 pn,k = 1(n+1) X ˜∆r(
(k+1 − (n+1) x) pn+1 ,k +1
)
= 1(n+1) X
(
(k+1 − (n+1) x) ˜∆rpn+1 ,k +1 − r ˜∆r−1pn+1 ,k +2
)
(33) 13 where in the second equality we have used the Leibniz formula ˜∆r(akbk) =
r
∑
j=0
(rj
)
( ˜∆j ak)( ˜∆r−j bk+j )with ak = ( k+1 − (n+1) x) and bk = pn+1 ,k +1 . Notice that ˜∆j ak = 0 for j ≥ 2. The recurrence relation (33) paves the way for induction for bounding Vn,r (x). However, due to the presence of ( k+1 − (n+1) x), we will work with a stronger induction hypothesis concerning all of the sums
Yn,r,s (x) := ∑
k≥0
|k − nx |s
∣∣∣ ˜∆rpn,k (x)
∣∣∣
at once. Note that this sum runs over all k ≥ 0 but there is no contribution from the terms k > n .
Theorem 6. For all non-negative integers r and s, we have
Yn,r,s (x) .r,s n s−r
2
X−r. (34)
In particular,
Vn,r (x) = Yn,r, 0(x) .r n− r
2
X−r. (35)
Remark. Note that in its X dependence, this is a weaker bound than our ansatz. However, the
n dependence is the same. The exponent of X in the proof below can be improved, but we do not know if −r/ 2 is achievable for r > 2. However, given that we are primarily interested in the rate of convergence relative to n, this question is of secondary importance and we leave it for future work.
Proof. For any r, let P(r) be the statement that (34) holds for all s (as well as n and x). The case
r = 0 is readily covered by (14) and Cauchy-Schwarz:
Yn, 0,s (x) ≤
√
Tn, 2s(x) .s n s
2
.
The case r = 1 is handled by noticing that
Yn, 1,s (x) = 1(n+1) X
n
∑
k=0
|k − nx |s |k+1 − (n+1) x| pn+1 ,k +1 (x)
≤ 1(n+1) X
n
∑
k=0
2s(|k+1 − (n+1) x|s + |1 − x|s) |k+1 − (n+1) x| pn+1 ,k +1 (x)
.s
1(n+1) X (Yn+1 ,0,s +1 (x) + Yn+1 ,0,s (x))
.s n s−12 X−1. (36) Assume that P(r) and P(r−1) are true. We will verify P(r+1). The identity (33) implies that
|k − nx |s
∣∣∣ ˜∆r+1 pn,k
∣∣∣ .s (|k+1 − (n+1) x|s + 1) 1(n+1) X |k+1 − (n+1) x|
∣∣∣ ˜∆rpn+1 ,k +1
∣∣∣
+( |k+2 − (n+1) x|s + 2 s) r
(n+1) X
∣∣∣ ˜∆r−1pn+1 ,k +2
∣∣∣
14 so that summing over all k ≥ 0, and applying our induction hypothesis, we get
Yn,r +1 ,s (x) .r,s
1(n+1) X (Yn+1 ,r,s +1 (x) + Yn+1 ,r, 1(x) + Yn+1 ,r −1,s (x) + Yn+1 ,r −1,0(x))
.r,s
1(n+1) X
(
n s+1 −r
2
X−r + n 1−r
2
X−r + n s−r+1 2 X−r+1 + n −r+1 2 X−r+1 )
.r,s n s−(r+1) 2 X−(r+1) , (37) hence P(r+1) is true and the induction step is complete. With the stability guarantee of Theorem 5 and the bounds (31) and (35) in place for the approximation and the quantization errors, all is left now is to state our final two theorems. The first theorem below follows by employing P = Un, ds/ 2e(f ), and the second one by employing the best approximation to f from Pn.
Theorem 7. Let s ≥ 3 be any integer. For every f ∈ Cs−1([0 , 1]) with f (s−1) ∈ Lip([0 , 1]) and
‖f ‖∞ ≤ μ < 1, there exist signs σ0, . . . , σ n ∈ {± 1} such that
∣∣∣∣∣f (x) −
n
∑
k=0
σk pn,k (x)
∣∣∣∣∣ .μ,s ‖f ‖Cs−1Lip n−s/ 2 + min(1 , X −sn−s/ 2) (38)
holds for all n & ‖f (2) ‖∞/(1 − μ) where ‖f ‖Cs−1Lip is defined in (32) and X := x(1 − x).
Theorem 8. For every continuous function f : [0 , 1] → R and integers r ≥ 0, n ≥ 1, there exist
q0, . . . , q n ∈ Z such that
∣∣∣∣∣f (x) −
n
∑
k=0
qk pn,k (x)
∣∣∣∣∣ .r En(f )∞ + min(1 , X −rn−r/ 2) (39)
for all x ∈ [0 , 1] where X := x(1 − x). In particular, if f ∈ Cs([0 , 1]) , then for all n ≥ 1, there exist
q0, . . . , q n ∈ Z such that
∣∣∣∣∣f (x) −
n
∑
k=0
qk pn,k (x)
∣∣∣∣∣ .s ‖f (s)‖∞n−s + min(1 , X −2sn−s) (40)
for all x ∈ [0 , 1] .
Appendix A: Frame-theoretic redundancy of the Bernstein basis
We have shown in this paper that a large class of functions (including all continuous functions) can be approximated arbitrarily well using polynomials with very coarsely quantized coefficients in a Bernstein basis. While our methodology can be viewed as a variation on the theme of which addresses quantization of finite frame expansions through Σ∆ quantization, we emphasize that we are working with linearly independent systems. As such, the setting of this paper is novel for this type of quantization. The reason why the method works is that the “effective span” of the 15 Bernstein basis Bn is actually roughly of dimension ∼ √n. Both approximation and noise-shaping quantization take place relative to this subspace of Pn.In this appendix, we will make the statements in the preceding paragraph more precise by first providing a complete description of the singular values of the linear operator Sn : Rn+1 → P n
defined by
Snu :=
n
∑
k=0
ukpn,k ,
where Rn+1 is equipped with the standard Euclidean inner product and Pn with the inner product inherited from L2([0 , 1]), both denoted by 〈· , ·〉 . Evidently, this discussion is concerned primarily with L2 approximation rather than uniform approximation. However, the description of the singular values shed additional light onto the Bernstein basis and may be of independent interest to the frame community. The adjoint S∗
n
: Pn → Rn+1 is given by (S∗
n
f )k = 〈pn,k , f 〉 :=
∫ 10
pn,k (x)f (x) dx, k = 0 , . . . , n.
In the language of frame theory, Sn and S∗
n
are the synthesis and the analysis operators for the system Bn := ( pn,k )nk=0 , respectively. Then the frame operator S∗
n
Sn : Rn+1 → Rn+1 is given by (S∗
n
Snu)k =
n
∑
`=0
〈pn,k , p n,〉u = (Γ nu)k
where Γ n is the Gram matrix of the system Bn with entries (Γ n)k,= 〈pn,k , p n, 〉, k, l = 0 , . . . , n.
Note that ( n + 1)Γ n is doubly stochastic due to the relations
n
∑
l=0
pn,l (x) = 1 and
∫ 10
pn,k (x) dx = 1
n + 1 .
To ease the notation in the discussion below, we assume n ≥ 0 is fixed and suppress the index
n from our notation when expressing certain quantities, keeping in mind that they still depend on
n. This dependence will naturally become explicit in all formulas we will provide. Henceforth we will refer to the operators S := Sn and Γ := Γ n.Since the Bernstein system is linearly independent, Γ is positive-definite, and in particular, invertible. Let its eigenvalues in decreasing order be ( λk)n
0
, i.e.
λ0 ≥ λ1 ≥ · · · ≥ λn > 0.
As we will see, the eigenvectors of Γ are precisely the discrete Legendre polynomials on {0, . . . , n },which we denote by ϕ0, . . . , ϕ n. (Here, we identify Rn+1 with R{0,...,n } = L2({0, . . . , n }) equipped with the counting measure. We pair f ∈ R{0,...,n } with ( f (0) , . . . , f (n)) ∈ Rn+1 .) Thus each ϕk()is a polynomial in ∈ { 0, . . . , n }, of degree equal to k, and we have the orthonormality relations
〈ϕk, ϕ j 〉 = δk,j .
16 We will not need an explicit formula for the ϕk. In order to uniquely define the Legendre polyno-mials, it is necessary to remove their sign ambiguity according some convention, but we shall not need this either. For any k ≥ 0, let ( t)k denote the falling factorial defined by ( t)k := t(t − 1) · · · (t − k + 1) and (t)0 := 1, where we restrict t to non-negative integers.
Theorem 9. Let n ≥ 0 be arbitrary, Γ be the Gram matrix of the Bernstein system Bn. Then for all 0 ≤ k ≤ n, we have Γϕk = λkϕk where ϕk is the degree k discrete Legendre polynomial on
{0, . . . , n } and
λk = (n)k
(n + k + 1) k+1
.
Proof. Let πk() :=k (as before) and ξk() := ()k, ` ∈ { 0, . . . , n }, k ≥ 0. Since each ξk is a monic polynomial of degree k, it is immediate that span {ξ0, . . . , ξ m} = span {π0, . . . , π m} =: Pm.(Note that for m > n we have Pm = Pn = R{0,...,n }, though we will not be concerned with the case
m > n .) Let 0 ≤ m ≤ n be arbitrary. We claim that Γ( Pm) ⊂ Pm. For this, it suffices to show that Γ ξm
is a degree m polynomial. We have (Γ ξm)( k) =
n
∑
`=0
Γk,ξm() =
〈
pn,k ,
n
∑
`=0
()m pn,
〉
, k = 0 , . . . , n. (41) We can evaluate the polynomial sum in this inner-product via
n
∑
`=0
()m pn, (x) =
n
∑
`=m
m!
( `m
)
pn,` (x) = m!
( nm
)
xm = ( n)m xm, (42) where the first equality uses the fact that ( )m = 0 for < m , and the second equality is an identity generalizing the partition of unity property of Bernstein polynomials (see, e.g. [4, p.85]). Hence, (41) can now be evaluated to give (Γ ξm)( k) = (n)m
(nk
) ∫ 10
xk+m(1 − x)n−kdx
= (n)m
(nk
)
B( k + m + 1 , n − k + 1) = (n)m (k + m)m
(n + m + 1) m+1
, k = 0 , . . . , n, (43) where B( ·, ·) stands for the beta function. The term ( k + m)m is a polynomial in k of degree m,hence the claim follows. As a particular case, we have Γ ϕm ∈ Pm, so there are α(m)0 , . . . , α (m)
m
such that Γϕm =
m
∑
k=0
α(m)
k
ϕk.
In other words, Γ is represented by an upper-triangular matrix in the discrete Legendre basis (ϕ0, . . . , ϕ n). This fact, coupled with the orthogonality of the discrete Legendre basis and the 17 symmetry of Γ, actually implies that Γ is diagonal in the discrete Legendre basis. Let us spell out how this general principle works: For the case m = 0, we readily have Γ ϕ0 = α(0) 0 ϕ0. For 1 ≤ m ≤ n, the claim that ϕm is an eigenvector of Γ is equivalent to the claim α(m)0 = · · · = α(m)
m−1
= 0. Employing orthogonality and self-adjointness, we have
α(m)
k
= 〈Γϕm, ϕ k〉〈ϕk, ϕ k〉 = 〈ϕm, Γϕk〉〈ϕk, ϕ k〉 , k = 0 , . . . , m. (44) Having established that ϕ0, . . . , ϕ m−1 are all eigenvectors of Γ, (44) coupled with orthogonality of the discrete Legendre basis implies that α(m)
k
= 0 for all k = 0 , . . . , m − 1, implying that ϕm is an eigenvector of Γ. We proceed to obtain a formula for the eigenvalues of Γ. For m = 0, the fact that ( n + 1)Γ is doubly stochastic implies that Γ ϕ0 = λ0ϕ0 with λ0 = 1 /(n + 1). Let 1 ≤ m ≤ n and λm be the eigenvalue correpsonding to the eigenvector ϕm. Since ϕm, ξm, πm are all degree m polynomials and the latter two are monic, there is a scalar βm 6 = 0 such that
ϕm = βmξm + γm−1 = βmπm + ˜ γm−1
where γm−1, ˜γm−1 ∈ Pm−1. Hence
λmϕm − βmΓξm = Γ( ϕm − βmξm) = Γ γm−1 ∈ Pm−1
so that
λmπm − Γξm = 1
βm
(
λm(βmπm − ϕm) + ( λmϕm − βmΓξm)
)
∈ Pm−1.
Meanwhile, (43) implies Γξm − (n)m
(n + m + 1) m+1
πm ∈ Pm−1.
Adding the last two vectors, we get (λm − (n)m
(n + m + 1) m+1
)πm ∈ Pm−1
which implies that λm = (n)m
(n+m+1) m+1
.
Remark. For any n, let σ0, . . . , σ n be the singular values of S defined by σm := √λm. Define
ψm := Sϕ m/σ m. Then, as is well known, ψ0, . . . , ψ n is an orthonormal basis for the range of S,i.e. Pn, equipped with the L2 inner product on [0 , 1]. The relation (42) shows that Sξ m ∈ P m,implying S(Pm) ⊂ P m. In particular, ψm ∈ P m for each m = 0 , . . . , n . Then the orthonormality of the ψm imply that they are simply the continuous Legendre polynomials on [0 , 1], again up to a sign convention. Note that even though the ϕm depend on n, the ψm do not. 18 Quantifying redundancy of the Bernstein system. In frame theory, the redundancy of a finite frame is often defined as the ratio of the number of frame vectors to the dimension of their span. Clearly this definition is insufficient for the purpose of differentiating between all the ways in which n vectors can span a d dimensional space, let alone for handling infinite dimensional spaces. A particular deficiency of working with the algebraic span is that it treats all bases the same, whether orthonormal or ill-conditioned. Unfortunately a more suitable quantitative definition of redundancy has been elusive. In the case of unit norm tight frames, the ratio n/d is equal to the frame constant, i.e. the unique eigenvalue of the frame operator SS ∗ (equivalently, the unique non-zero eigenvalue of the Gram matrix), and for near-tight frames, the frame bounds given by the smallest and largest eigenvalues of the frame operator may continue to serve as a rough substitute of redundancy. However, when the eigenvalues are widely dispersed without a well-defined gap, the frame bounds seem to lose their significance. As a basis of Pn, the Bernstein system Bn would not be considered redundant by the classical definition. However, it is ill-conditioned; in fact, we have
σ0
σn
=
√ (2n+1
n+1
)
n + 1 which is exponentially large in n. Hence the relevant question is: What is the “effective dimension” of the span of Bn? One possible answer to this question comes from numerical analysis via the notion of “numerical rank” (see e.g. [15, Ch. 5.4.1]). This quantity is tied to a given tolerance threshold
measuring which singular values (and therefore which subspaces) should count as significant. We inspect the singular values against the top singular value σ0 and define
dn() := max {0 ≤ m ≤ n : σm ≥ σ 0}
where ∈ (0 , 1) is a fixed small parameter. With this definition, if we truncate the singular value decomposition of S to define ˜Su := ∑
σm≥σ 0
σm〈u, ϕ m〉ψm,
then a straightforward consequence is that
‖S − ˜S‖op
‖S‖op
≤ .
With this, we can argue that the image of the unit ball under S is well approximated by an ellipsoid within span( ψ0, . . . , ψ dn()) = Pdn(). Hence we define the -redundancy of Bn to be n/d n(). Note that this definition can easily be applied to systems other than Bn.The next task is to obtain asymptotics for dn(). As can be seen from the formula derived in Theorem 9, the singular values of S do not possess any obvious gap. It turns out that the exponentially large condition number of S is highly misleading because the singular values of S
do not decay like an exponential. If σm actually decayed like exp( −cm ), we would find that
dn() ∼ c−1 log(1 / ), which is independent of n. Instead, we will show below that
dn() = C
√n(1 + o(1)) as n → ∞ ,
19 where C := √2 log(1 / ). This result is consistent with the fact that the error of best approximation using Bn (with bounded coefficients) behaves as if n is replaced by √n. In addition, note that the
-redundancy of Bn given by n/d n() also behaves like √n. This is also consistent with our result that the error bound of the r-th order Σ∆ quantization method behaves as n−r/ 2.
Theorem 10. For all ∈ (0 , 1) ,
lim
n→∞
dn()
√2n log(1 / ) = 1 . (45)
Proof. The phrase “for all sufficiently large n” will occur several times below and will be abbreviated by “f.a.s.l. n.” Suppose ∈ (0 , 1) is fixed. Define
mn :=
⌊√ 2( n+1) log(1 / )
⌋
− 1, and m′
n
:=
⌈√ 2( n+1) log(1 / )
⌉
. (46) It is clear that lim
n→∞
mn
√2n log(1 / ) = lim
n→∞
m′
n
√2n log(1 / ) = 1 . (47) As a trivial consequence, mn and m′
n
are in {0, . . . , n } f.a.s.l. n, so we can meaningfully refer to
σmn and σm′
n
. The proof will be based on showing that
σmn > σ 0 > σ m′
n
f.a.s.l. n (48) which implies mn ≤ dn() < m ′
n
f.a.s.l. n, and therefore (45) via (47). It will be more convenient to work with the increasing sequence αm := log( λ0/λ m), m = 0 , . . . , n .By Theorem 9, we have
αm = log
( (n+m+1) m+1
(n+1) m+1
)
=
m
∑
k=0
(
log
(
1+ kn+1
)
− log
(
1− kn+1
))
=
n
∑
k=0
g
( kn+1
)
,
where g(t) := log(1+ t)−log(1 −t), |t| < 1. We have g(0) = 0, g′(0) = 2, g′′ (0) = 0, and |g′′′ (t)| ≤ 18 for |t| ≤ 1/2, so that |g(t) − 2t| ≤ 3|t|3 for |t| ≤ 1/2 via Taylor’s theorem. Hence, for all m ≤ n+1 2 ,we have
∣∣∣∣αm − m(m+1)
n+1
∣∣∣∣ =
∣∣∣∣∣
m
∑
k=0
g
( kn+1
)
−
m
∑
k=0
2kn+1
∣∣∣∣∣ ≤ 3
m
∑
k=0
( kn+1
)3
< (m+1) 4
(n+1) 3 . (49) We will utilize (49) to estimate αmn and αm′
n
. (Note that (47) also implies that 0 ≤ mn < m ′
n
≤ n+1 2
f.a.s.l. n.) The definition (46) readily implies (mn+1) 2
n+1 ≤ log(1 / 2) ≤ (m′
n
)2
n+1 , (50) and rearranging these inequalities, we have
mn(mn+1)
n+1 ≤ log(1 / 2) − mn+1
n+1 and m′
n
(m′
n
+1)
n+1 ≥ log(1 / 2) + m′
n
n+1 . (51) 20 The estimate (49) for m = mn with (50) and (51) yields
αmn ≤ mn(mn+1)
n+1 + (mn+1) 4
(n+1) 3
≤ log(1 / 2) − mn+1
n+1 + log 2(1 / 2)
n+1
< log(1 / 2) f.a.s.l. n (52) since mn → ∞ . Similarly, we also have
αm′
n
≥ m′
n
(m′
n
+1)
n+1 − (m′
n
+1) 4
(n+1) 3
≥ log(1 / 2) + m′
n
n+1 − 2 log 2(1 / 2)
n+1 f.a.s.l. n, > log(1 / 2) f.a.s.l. n, (53) where in the second last step we used the observation that ( m′
n
1) 4 ≤ 2( mn + 1) 4 f.a.s.l. n. Hence we have shown
αmn < log(1 / 2) < α m′
n
f.a.s.l. n,
which is equivalent to (48).
Appendix B: One-bit neural networks
Universal approximation properties of feedforward neural networks are well-known (see as well as the more recent treatments including [39, 37, 29, 3, 11] and the extensive survey ). One of the motivations of this work has been to investigate the approximation potential of feedforward neural networks with the additional constraint that their parameters are coarsely quantized, at the extreme using only two values, hence the term “one-bit”. Quantized networks have been studied from a variety of perspectives (see e.g. [8, 20, 1, 30]). As far as we know, a universal approximation theorem using one-bit, or even fixed-precision multi-bit networks was missing. In this appendix we will show how this can be done using our results on polynomial approximation in the Bernstein system. In particular, employing the quadratic activation unit s(x) := 12 x2, we will show that it is possible to implement the polynomial approximations of this paper using standard feedforward neural networks whose weights are chosen from {± 1} only. Due to the scope and focus of this paper, we will limit our discussion to univariate functions. A much more comprehensive study of these networks covering multivariate functions, ReLU networks, and information theoretic considerations can be found in our separate manuscript . A feedforward neural network can be described in a variety of general formulations (see e.g. ). The architecture of the network is determined by a directed acyclic graph. The vertices of this graph (called nodes of the network) consist of input, output, and hidden vertices. Vertices are assigned variables (independent or dependent depending upon whether the vertex is an input or not) and edges are assigned weights. Every node which is not an input node computes a linear combination of the variables assigned to the nodes that are connected to it (noting the directedness 21 of the graph) using the weights associated to the connecting edges, followed by (except for the output nodes) an application of a given activation function ρ, possibly subject to a shift of its argument (called the bias). An important class of networks are layered , meaning that the nodes of the network can be partitioned into subsets (layers) such that the nodes in each layer have incoming edges only from one other (the previous) layer and outgoing edges into another (the next) layer. Having introduced the basic terminology of feedforward neural networks, let us now turn the polynomial approximation method of this paper into a neural network approximation with ±1weights. There are two critical ingredients in our construction of these networks. The first one is, of course, the fact that we are able to approximate functions f (where ‖f ‖∞ ≤ 1) by ±1-linear combinations of the elements of the Bernstein system Bn. The second critical ingredient is that the basis polynomials pn,k have a rich hierarchical structure that enables them to be computed recursively, and with few simple arithmetic operations. Indeed, the pn,k satisfy the elementary recurrence relation
pn,k (x) =
(1 − x)pn−1,0(x), k = 0 ,xp n−1,k −1(x) + (1 − x)pn−1,k (x), 0 < k < n, xp n−1,n −1(x), k = n,
(54) which follows readily from the simple combinatorial identity (nk
) = (n−1
k−1
) + (n−1
k
). (The endpoint cases of k = 0 and k = n can actually be eliminated if we use the earlier convention pn,j (x) = 0 if j < 0 or j > n .) Since this combinatorial identity is the basis of the Pascal triangle, we will refer to the resulting tree structure of the Bernstein basis polynomials as the Pascal-Bernstein tree ,which is depicted in the top triangular portion of the graph in Figure 1. The bottom portion of this graph shows how these basis polynomials are combined with weights σk ∈ {± 1}, k = 0 , . . . , n .The red edges of the tree correspond to multiplication by 1 − x and the blue edges correspond to multiplication by x. Note that this schematic diagram is not a proper feedforward neural network yet because the weights depend on the input x. However, it is readily implementable by a sum-product network where the nodes either multiply or compute a weighted sum their inputs. In our case, the weights consist of ±1 only. The input to the network could be x and ¯ x := 1 − x.Regardless of the nature of this network (or proto-network), the Pascal-Bernstein tree portion of the algorithm is universal in the sense that it is the same for every function f to be approximated. All of the information about how f is approximated is encoded in the weights ( σk)n
0
connecting the bottom layer to the final output. In order to turn this schematic diagram into a proper feedforward neural network, it suffices to convert its multiplications (by 1 − x and x) into a standard neural network operation. We show below that this can be done using ±1 weights in the simplest way if the activation unit is given by the quadratic function s(u) := 12 u2. Indeed, with this unit, the multiplication of any two quantities
a and b can be implemented simply as
ab = s(a + b) − s(a) − s(b).
Then, for any m = 1 , . . . , n , the quantities pm−1,j (x), j = 0 , . . . , m − 1, associated with the vertices 22 1−x x
1
✁✁✁✁❆❆❆❆
1−xx
✁✁✁✁❆❆❆❆ ✁✁✁✁❆❆❆❆
(1 −x)22x(1 −x)
⊕
x2
✁✁✁✁❆❆❆❆ ✁✁✁✁❆❆❆❆ ✁✁✁✁❆❆❆❆⊕ ⊕
(1 −x)33x(1 −x)23x2(1 −x)x3
✁✁✁✁❆❆❆❆ ✁✁✁✁❆❆❆❆ ✁✁✁✁❆❆❆❆ ✁✁✁✁❆❆❆❆⊕ ⊕ ⊕· · · · · · pn, 0(x) pn,k (x) pn,n (x)
❩❩❩❩❩❩❩❏❏❏❏❏ ✡✡✡✡✡✚✚✚✚✚✚✚
⊕
σ0σkσn
Figure 1: Schematic diagram of the Pascal-Bernstein tree and the associated one-bit polynomial approximations in the Bernstein basis. Red edges of the tree correspond to multiplication by 1 − x
and blue edges by x.in the mth layer of this tree satisfy the relations
pm,k (x) =
s(1 − x + pm−1,0(x)) − s(1 − x) − s(pm−1,0(x)) , k = 0 ,(
s(x + pm−1,k −1(x)) − s(x) − s(pm−1,k −1(x)) + s(1 − x + pm−1,k (x)) − s(1 − x) − s(pm−1,k (x))
) 0 < k < m, s(x + pm−1,m −1(x)) − s(x) − s(pm−1,m −1(x)) , k = m.
(55) This representation shows that the pm,k (x) will not actually correspond to the physical outputs of the nodes of our neural network, but rather to certain intermediate ±1 linear combinations of the outputs of up to 6 of its nodes. If m < n , we add x or 1 − x to these before passing them onto subsequent nodes. If m = n, we take a final ±1 linear combination using the coefficients σk.Let us describe the nodes of our network more precisely. For each m = 0 , . . . , n − 1, there will 23 be a layer whose nodes output the functions Xm,k , Ym,k , Zm,k , k = 0 , . . . , m , where we would like
Xm,k (x) = s(pm,k (x)) , Ym,k (x) = s(1 − x + pm,k (x)) , Zm,k (x) = s(x + pm,k (x)) .
Therefore, we set U := s(1 − x) and V := s(x) and define these functions for m = 1 , . . . , n − 1 via the recurrence
Xm,k :=
s(Ym−1,0 − U − Xm−1,0), k = 0 ,s(Zm−1,k −1 − V − Xm−1,k −1 + Ym−1,k − U − Xm−1,k ), 0 < k < m, s(Zm−1,m −1 − V − Xm−1,m −1), k = m, Ym,k :=
s(1 − x + Ym−1,0 − U − Xm−1,0), k = 0 ,s(1 − x + Zm−1,k −1 − V − Xm−1,k −1 + Ym−1,k − U − Xm−1,k ), 0 < k < m, s(1 − x + Zm−1,m −1 − V − Xm−1,m −1), k = m, Zm,k :=
s(x + Ym−1,0 − U − Xm−1,0), k = 0 ,s(x + Zm−1,k −1 − V − Xm−1,k −1 + Ym−1,k − U − Xm−1,k ), 0 < k < m, s(x + Zm−1,m −1 − V − Xm−1,m −1), k = m.
We also set X0,0 := s(1), Y0,0 := s(1 − x + 1), Z0,0 := s(x + 1). The final output of the network is the function fNN (x) given by ∑
k
σkpn,k (x). The pn,k (x) are still available indirectly through (55) for m = n, so we define
fNN := σ0Yn−1,0 − σ0U − σ0Xn−1,0
+
n−1
∑
k=1
(σkZn−1,k −1 − σkV − σkXn−1,k −1 + σkYn−1,k − σkU − σkXn−1,k )+ σn−1Zn−1,n −1 − σn−1U − σn−1Xn−1,n −1. (56) We note that this expression contains two copies of the Xn−1,j , j = 0 , . . . , n − 1, each carrying a ±1 weight. If these weights were to be combined, then we would get a new weight in the set
{− 2, 0, 2}. This inconsistency can easily be removed by duplicating the nodes that produce Xn−1,j .Of course, the same comment applies to U and V which would need to be copied n times. A copy of a node is easily created using the identity
a = s(a + 1) − s(a) − s(1) .
The appearance of U and V in each step of the recurrence means that the network is not completely layered, containing some “skip” connections. However, the copying mechanism above can also be used as a repeater to remove these skip connections; see Figure 2. All of these additional operations can also be avoided by means of special networks (e.g. ) where it is permissible to create certain channels to push forward input values or other intermediate computations, or by allowing for more than one type of activation function to be used at the nodes (e.g. the identity function). Finally, we note that our network takes as input both x and 1 (or alternatively, x and 1 − x). This choice has allowed us to avoid the use of any bias values associated to the activation function. 24 ✍✌ ✎☞ ✍✌ ✎☞ ✍✌ ✎☞ ✍✌ ✎☞ ✍✌ ✎☞ ✍✌ ✎☞ ✍✌ ✎☞ ✍✌ ✎☞
❅❅❅❅❅❅❍❍❍❍❍❍❍❍❍❍❍
❅❅❅❅❅❅
❅❅❅❅❅❅
❅❅❅❅❅❅
❅❅❅❅❅❅
❅❅❅❅❅❅
❅❅❅❅❅❅
❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍❍ -- --- -
s s s ss s s s
1 aa1 1 1 + a
... ... ... ...Figure 2: Demonstration of how the output a of any node or the input a ∈ { x, 1} of the network can be repeated to be turned an input to any node in a subsequent layer. Two layers of repetitions are shown.
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190358 | https://the-world-is-my-classroom.weebly.com/unit-32-comparing-and-ordering-fractions-and-decimals.html | Unit 3.2: Comparing and Ordering Fractions and Decimals - MR. MARTÍNEZ'S MATH VIRTUAL CLASSROOM - JH
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UNIT 3.2: Comparing and ordering fractions and decimals
Last section, you learned how to convert fractions into decimals (and vice-versa). In this lesson, you will compare and order numbers. By ordering numbers, we are able to determine which numbers are bigger, smaller, the biggest, etc.
In order to compare, it is important that all the numbers being compared are of the same type, or written in the same format. That means, to compare a group of numbers, they all should be converted into decimals, OR into fractions. Here are the notes of what we did in class:
7e_unit_3.2.pdf
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When it comes to comparing fractions, remember:
The easiest strategy: Find and compare the decimals (divide the numerator by the denominator). CONSIDER THIS!
When fractions have thesame denominator, the higher the numerator, the higher the fraction is (the higher the decimal).
When fractions have different denominators, it is necessary we make all fractions comparable byfinding the COMMON MULTIPLE DENOMINATOR, and multiplying each denominator (and its numerator) by a number which makes said denominator the common multiple denominator.
REMEMBER: Whatever you do to the bottom, you do to the top!
EXAMPLES
2/3, 3/3, 7/3all have the same denominator. But because 7 > 3 > 2, we can say that
7/3 > 3/3 > 2/3
To compare 4/5 and 7/4, two fractions with different denominators, we must find the common denominatorof 5 and 4. This is 20.
So: 4 x (4)/ 5 x (4) = 16 / 20
7 x (5)/ 4 x (5) = 35/20
Then, because 35 > 16, we can say that 35/20 > 16/20
The following video explains how to compare and order decimals:
Here is another great video. Make sure you watch it, and please let me know if you have any questions:
Let's practice! do the following worksheets, making sure you compare your answers to the answers here presented.
1. Comparing Fractions
7e_3.2_comparing_fractions_worksheet_1.pdf
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7e_3.2_comparing_fractions_worksheet_2.pdf
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2. comparing and ordering decimals
7e_3.2_comparing_decimals_worksheet_1.pdf
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7e_3.2_comparing_decimals_worksheet_2.pdf
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3. comparing fractions and decimals
7e_3.2_comparing_fractions_and_decimals_worksheet_1.pdf
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7e_3.2_comparing_fractions_and_decimals_worksheet_2.pdf
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online review
This website is an online review site you can use to challenge yourself. Try to better your time every time you try it. Focus on the following two activities. Good luck!
Compare and Order Decimals
Decimal Number Lines
Using the Number Line to Compare Decimals, Fractions, and Whole Numbers
Click on the following link for a website that explains rather well, and where you could practice, how to use number lines to order and compare fractions and decimals.
Using Number Lines to Compare Fractions and Decimals
extra practice
Hopefully by now, and after doing all the worksheets and going through the notes, you have a very good understanding of how to order fractions and decimals. But because practicing is one of the most helpful things we can do in math, I suggest you try the following exercises on your TEXTBOOK(that is, if you have time. I know you are a very busy person!):
Page 88, #1 and 2
page 89: #3, 4, 5, 6 and 10
Back to UNIT 3Back to MATH 7
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Intuition or proof for combinatorial function in binomial distribution
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For a the probability of a binomial distribution with n trials, the probability of k successes where Prob(success) = p is (n Choose k) p^k (1-p)^(n-k). I understand that we have to multiply p^k (1-p)^(n-k) by something but what’s 1) an intuitive explanation and/or 2) a proof for why we multiply by the (n Choose k). Especially with regard to the intuitive side of things, I feel tempted to say the factor should be (n Permutation k).
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edited Mar 21, 2021 at 3:01
user3163829user3163829
asked Mar 21, 2021 at 2:56
user3163829user3163829
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1 Multiply by (n k)(n k) to get the arrangements of k k Successes in n n trials, which is found by choosing the k k positions for the k k Successes. Example: For X∼n=3,p=1/2,X∼n=3,p=1/2, there are s 3=8 s 3=8 possible outcomes. For P(X=2)P(X=2) the (3 2)=3(3 2)=3 relevant outcomes are HHT,HTH, THH and P(X=2)=3/8.P(X=2)=3/8.BruceET –BruceET 2021-03-21 05:59:10 +00:00 Commented Mar 21, 2021 at 5:59
@BruceET I think you can convert this to an answer.gunes –gunes 2021-03-21 14:12:10 +00:00 Commented Mar 21, 2021 at 14:12
1 @gunes: OK. Done--with a few extra words.BruceET –BruceET 2021-03-21 15:58:35 +00:00 Commented Mar 21, 2021 at 15:58
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If X∼B i n o m(n,p),X∼B i n o m(n,p), then to find P(X=k),P(X=k), one multiplies the probability p k(1−p)n−k p k(1−p)n−k of a particular outcome with k k Successes by (n k),(n k), the number of arrangements of k k Successes among n n trials, to get the total probability P(X=k).P(X=k). This amounts to choosing the k k positions out of n n for the k k Successes.
Example: For X∼B i n o m(n=3,p=1/2),X∼B i n o m(n=3,p=1/2), there are 2 3=8 2 3=8 possible outcomes altogether, each with probability 1/8.1/8. For P(X=2),P(X=2), the (3 2)=3!2!⋅1!=3(3 2)=3!2!⋅1!=3 relevant outcomes are HHT,HTH, and THH, so P(X=2)=3/8.P(X=2)=3/8.
In R, where dbinom is a binomial PDF (or PMF), this result is found as shown below:
dbinom(2, 3, .5)
0.375
3/8
0.375
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answered Mar 21, 2021 at 15:57
BruceETBruceET
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190360 | https://faculty.curgus.wwu.edu/Courses/312_200940/Sequences_from_312notes.pdf | CHAPTER 3 Sequences in R 3.1. Definitions and examples Definition 3.1.1. A sequence in R is a function whose domain is N and whose range is in R.
Let s : N →R be a sequence in R. Then the values of s are s(1), s(2), s(3), . . . , s(n), . . . .
It is customary to write sn instead of s(n) for the values of a sequence. Sometimes a sequence will be specified by listing its first few terms s1, s2, s3, s4, . . . , and sometimes by listing of all its terms sn ∞ n=1 or sn . One way of specifying a sequence is to give a formula, or a recursion formula for its n−th term sn.
Remark 3.1.2. In the above notation s is the “name” of the sequence and n ∈N is the independent variable.
Remark 3.1.3. Notice the difference between the following two expressions: sn ∞ n=1 This expression denotes a function (sequence).
sn : n ∈N This expression denotes a set: The range of a sequence sn ∞ n=1.
For example 1 −(−1)n ∞ n=1 stands for the function n 7→1 −(−1)n, n ∈N, while 1 −(−1)n : n ∈N = 0, 2 .
Example 3.1.4. Here we give examples of sequences given by a formula. In each formula below n ∈N.
(a) n, (b) n2, (c) √n, (d) (−1)n, (e) 1 n, (f) 1 n2 , (g) 1 √n, (h) 1 −(−1)n n , (i) 1 n!, (j) 21/n, (k) n1/n, (l) n(−1)n, (m) 9n n! , (n) (−1)n+1 2n −1 , (o) n(−1)n n + 1 , (p) e n n n!
√n.
Example 3.1.5. Few more sequences given by a formula are (a) n√ n2 + 1 −n o∞ n=1, (b) n√ n2 + n −n o∞ n=1, (c) n√n + 1 −√n o∞ n=1.
Example 3.1.6. In this example we give several recursively defined sequences.
49 50 3. SEQUENCES IN R (a) s1 = 1 and ∀n ∈N sn+1 = −sn 2 , (b) x1 = 1 and ∀n ∈N xn+1 = 1 + xn 4 , (c) x1 = 2 and ∀n ∈N xn+1 = xn 2 + 1 xn , (d) a1 = √ 2 and ∀n ∈N an+1 = √ 2 + an, (e) s1 = 1 and ∀n ∈N sn+1 = √1 + sn, (f) x1 = 0 and ∀n ∈N xn+1 = 9 + xn 10 .
For a recursively defined sequence it is useful to evaluate the values of the first few terms to get an idea how sequence behaves.
Example 3.1.7. The most important examples of sequences are listed below: bn = a, n ∈N, where a ∈R, (3.1.1) pn = an, n ∈N, where −1 < a < 1, (3.1.2) En = 1 + 1 n n , n ∈N, (3.1.3) G1 = a + ax and ∀n ∈N Gn+1 = Gn + axn+1, where −1 < x < 1, (3.1.4) S1 = 2 and ∀n ∈N Sn+1 = Sn + 1 (n + 1)!, (3.1.5) v1 = 1 + a and ∀n ∈N vn+1 = vn + an+1 (n + 1)!, where a ∈R.
(3.1.6) Definition 3.1.8. Let an be a sequence in R. A sequence which is recursively defined by S1 = a1 and ∀n ∈N Sn+1 = Sn + an+1, (3.1.7) is called a sequence of partial sum corresponding to an .
Example 3.1.9. The sequences of partial sums associated with the sequences in Example 3.1.4 (e), (f) and (n) are important examples for Definition 3.1.8. Notice also that the sequences in (3.1.4), (3.1.5) and (3.1.6) are sequences of partial sums.
All of these are very important.
3.2. Bounded sequences Definition 3.2.1. Let sn be a sequence in R.
(1) If a real number M satisfies sn ≤M for all n ∈N then M is called an upper bound of sn and the sequence sn is said to be bounded above.
(2) If a real number m satisfies m ≤sn for all n ∈N, then m is called a lower bound of sn and the sequence sn is said to be bounded below.
November 16, 2009 3.3. THE DEFINITION OF A CONVERGENT SEQUENCE 51 (3) The sequence sn is said to be bounded if it is bounded above and bounded below.
Remark 3.2.2. Clearly, a sequence sn is bounded above if and only if the set sn : n ∈N is bounded above. Similarly, a sequence sn is bounded below if and only if the set sn : n ∈N is bounded below.
Remark 3.2.3. The sequence sn is bounded if and only if there exists a real number K > 0 such that |sn| ≤K for all n ∈N.
Exercise 3.2.4. There is a huge task here. For each sequence given in this section it is of interest to determine whether it is bounded or not. As usual, some of the proofs are easy, some are hard. It is important to do few easy proofs and observe their structure. This will provide the setting to appreciate proofs for hard examples.
3.3. The definition of a convergent sequence Definition 3.3.1. A sequence sn is a constant sequence if there exists L ∈R such that sn = L for all n ∈N.
Exercise 3.3.2. Prove that the sequence sn = 3n −1 2n , n ∈N, is a constant sequence.
Definition 3.3.3. A sequence sn is eventually constant if there exists L ∈R and N0 ∈R such that sn = L for all n ∈N, n > N0.
Exercise 3.3.4. Prove that the sequence sn = 3n −2 2n + 3 , n ∈N, is eventually constant.
Exercise 3.3.5. Prove that the sequence sn = 5n −(−1)n n/2 + 5 , n ∈N, is eventually constant.
Definition 3.3.6. A sequence sn converges if there exists L ∈R such that for each ǫ > 0 there exists a real number N(ǫ) such that n ∈N, n > N(ǫ) ⇒ |sn −L| < ǫ.
The number L is called the limit of the sequence sn . We also say that sn converges to L and write lim n→∞sn = L or sn →L (n →∞).
If a sequence does not converge we say that it diverges.
Remark 3.3.7. The definition of convergence is a complicated statement. Formally it can be written as: ∃L ∈R s.t. ∀ǫ > 0 ∃N(ǫ) ∈R s.t. ∀n ∈N, n > N(ǫ) ⇒|sn −L| < ǫ.
Exercise 3.3.8. State the negation of the statement in remark 3.3.7.
November 16, 2009 52 3. SEQUENCES IN R 3.3.1. My informal discussion of convergence. It is easy to agree that the constant sequences are simplest possible sequences. For example the sequence n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 cn 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 or formally, cn = 1 for all n ∈N, is a very simple sequence. No action here! In this case, clearly, limn→∞cn = 1.
Now, I define sn = n −(−1)n n , n ∈N, and I ask: Is {sn} a constant sequence?
Just looking at the first few terms n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 sn 2 1 2 4 3 3 4 6 5 5 6 8 7 7 8 10 9 9 10 12 11 11 12 14 13 13 14 16 15 15 16 18 17 indicates that this sequence is not constant. The table above also indicates that the sequence {sn} is not eventually constant. But imagine that you have a calculator which is capable of displaying only one decimal place. On this calculator the first terms of this sequence would look like: n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 sn 2.0 0.5 1.3 0.8 1.2 0.8 1.1 0.9 1.1 0.9 1.1 0.9 1.1 0.9 1.1 and the next 15 terms would look like: n 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 sn 0.9 1.1 0.9 1.1 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 Basically, after the 20-th term this calculator does not distinguish sn from 1. That is, this calculator leads us to think that {sn} is eventually constant. Why is this?
On this calculator all numbers between 0.95 = 1 −1/20 and 1.05 = 1 + 1/20 are represented as 1, and for our sequence we can prove that n ∈N, n > 20 ⇒ 1 −1 20 < sn < 1 + 1 20, or, equivalently, n ∈N, n > 20 ⇒ sn −1 < 1 20.
In the notation of Definition 3.3.6 this means N(1/20) = 20.
One can reasonably object that the above calculator is not very powerful and propose to use a calculator that can display three decimal places. Then the terms of {sn} starting with n = 21 are n 21 22 23 24 25 26 27 28 29 30 sn 1.048 0.955 1.043 0.958 1.040 0.962 1.037 0.964 1.034 0.967 Now the question is: Can we fool this powerful calculator to think that {sn} is eventually constant? Notice that on this calculator all numbers between 0.9995 = 1 −1/2000 and 1.0005 = 1 + 1/2000 are represent as 1. Therefore, in the notation of Definition 3.3.6, we need N(1/2000) such that n ∈N, n > N(1/2000) ⇒ 1 − 1 2000 < sn < 1 + 1 2000.
November 16, 2009 3.4. FINDING N(ǫ) FOR A CONVERGENT SEQUENCE 53 An easy calculation shows that N(1/2000) = 2000. That is n ∈N, n > 2000 ⇒ sn −1 < 1 2000.
This is illustrated by the following table n 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 sn 0.999 1.001 0.999 1.001 1.000 1.000 1.000 1.000 1.000 1.000 Hence, even this more powerful calculator is fooled into thinking that {sn} is even-tually constant.
In computer science the precision of a computer is measured by the number called the machine epsilon (also called macheps, machine precision or unit round-off). It is the smallest number that gives a number greater than 1 when added to 1.
Now, Definition 3.3.6 can be paraphrased as: A sequence converges if on each computer it appears to be eventually constant. This is the reason why I think that instead of the phrase “a sequence is convergent” we could use the phrase “a sequence is constantish.” 3.4. Finding N(ǫ) for a convergent sequence Example 3.4.1. Prove that lim n→∞ 2n −1 n + 3 = 2.
Solution. We prove the given equality using Definition 3.3.6. To do that for each ǫ > 0 we have to find N(ǫ) such that (3.4.1) n ∈N, n > N(ǫ) ⇒ 2n −1 n + 3 −2 < ǫ.
Let ǫ > 0 be given. We can think of n as an unknown in 2n −1 n + 3 −2 < ǫ and solve this inequality for n. To this end first simplify the left-hand side: (3.4.2) 2n −1 n + 3 −2 = 2n −1 −2n −6 n + 3 = | −7| |n + 3| = 7 n + 3.
Now, 7 n + 3 < ǫ is much easier to solve for n ∈N: (3.4.3) 7 n + 3 < ǫ ⇔ n + 3 7 > 1 ǫ ⇔ n + 3 > 7 ǫ ⇔ n > 7 ǫ −3.
Now (3.4.3) indicates that we can choose N(ǫ) = 7 ǫ −3.
Now we have N(ǫ), but to complete the formal proof, we have to prove impli-cation (3.4.1). The proof follows. Let n ∈N and n > 7 ǫ −3. Then the equivalences in (3.4.3) imply that 7 n+3 < ǫ. Since by (3.4.3), 2n−1 n+3 −2 = 7 n+3, it follows that 2n−1 n+3 −2 < ǫ. This completes the proof of implication (3.4.1).
□ Remark 3.4.2. This remark is essential for the understanding of the process de-scribed in the following examples. In the solution of Example 3.4.1 we found (in some sense) the smallest possible N(ǫ). It is important to notice that implication (3.4.1) holds with any larger value for “N(ǫ).” For example, implication (3.4.1) holds if we set N(ǫ) = 7 ǫ. With this new N(ǫ) we can prove implication (3.4.1) as November 16, 2009 54 3. SEQUENCES IN R follows. Let n ∈N and n > 7 ǫ. Then 7 n < ǫ. Since clearly 7 n+3 < 7 n, the last two inequalities imply that 7 n+3 < ǫ and we can continue with the same proof as in the solution of Example 3.4.1.
Example 3.4.3. Prove that lim n→∞ 1 n3 −n + 1 = 0.
Solution. We prove the given equality using Definition 3.3.6. To do that for each ǫ > 0 we have to find N(ǫ) such that (3.4.4) n ∈N, n > N(ǫ) ⇒ 1 n3 −n + 1 −0 < ǫ.
Let ǫ > 0 be given. We can think of n as an unknown in 1 n3 −n + 1 −0 < ǫ and solve this inequality for n. To this end first simplify the left-hand side: (3.4.5) 1 n3 −n + 1 −0 = 1 n3 −n + 1 = |1| |n3 −n + 1| = 1 n3 −n + 1.
Unfortunately 1 n3 −n + 1 < ǫ is not easy to solve for n ∈N. Therefore we use the idea from Remark 3.4.2 and replace the quantity 1 n3 −n + 1 with a larger quantity.
To make a fraction larger we have to make the denominator smaller. Notice that n2 −n = n(n −1) ≥n −1 for all n ∈N. Therefore for all n ∈N we have n3 −n + 1 = n3 −(n −1) ≥n3 −n(n −1) = n(n2 −n + 1) ≥n.
Consequently, (3.4.6) 1 n3 −n + 1 ≤1 n.
Now, 1 n < ǫ is truly easy to solve for n ∈N: (3.4.7) 1 n < ǫ ⇔ n > 1 ǫ .
Hence we set N(ǫ) = 1 ǫ.
Now we have N(ǫ), but to complete the formal proof, we have to prove impli-cation (3.4.4). The proof follows. Let n ∈N and n > 1 ǫ. Then the equivalence in (3.4.7) implies that 1 n < ǫ. By (3.4.6), 1 n3−n+1 ≤1 n. The last two inequalities yield that 1 n3−n+1 < ǫ. By (3.4.5) it follows that 1 n3−n+1 −0 < ǫ. This completes the proof of implication (3.4.4).
□ Example 3.4.4. Prove that lim n→∞ n2 −1 n2 −2n + 2 = 1.
Solution. We prove the given equality using Definition 3.3.6. To do that for each ǫ > 0 we have to find N(ǫ) such that (3.4.8) n ∈N, n > N(ǫ) ⇒ n2 −1 n2 −2n + 2 −1 < ǫ.
November 16, 2009 3.4. FINDING N(ǫ) FOR A CONVERGENT SEQUENCE 55 Let ǫ > 0 be given. We can think of n as an unknown in n2 −1 n2 −2n + 2 −1 < ǫ and solve this inequality for n. To this end first simplify the left-hand side: (3.4.9) n2 −1 n2 −2n + 2 −1 = n2 −1 −n2 + 2n −2 n2 −2n + 2 = |2n −3| n2 −2n + 2.
Unfortunately |2n −3| n2 −2n + 2 < ǫ is not easy to solve for n ∈N. Therefore we use the idea from Remark 3.4.2 and replace the quantity |2n −3| n2 −2n + 2 with a larger quantity. Here is one way to discover a desired inequality. We first notice that for all n ∈N the following two inequalities hold (3.4.10) |2n −3| ≤2n and (3.4.11) n2 −2n + 2 = n2 2 + 1 2 n2 −4n + 4 = n2 2 + 1 2(n −2)2 ≥n2 2 .
Consequently (3.4.12) |2n −3| n2 −2n + 2 ≤ 2n n2/2 = 4 n.
Now, 4 n < ǫ is truly easy to solve for n ∈N: (3.4.13) 4 n < ǫ ⇔ n > 4 ǫ .
Hence we set N(ǫ) = 4 ǫ.
Finally we have N(ǫ).
But to complete the formal proof we have to prove implication (3.4.8). The proof follows. Let n ∈N and n > 4 ǫ. Then the equivalence in (3.4.13) implies 4 n < ǫ. By (3.4.12), |2n−3| n2−2n+2 ≤ 4 n. The last two inequalities yield |2n−3| n2−2n+2 < ǫ. By (3.4.9) it follows that n2−1 n2−2n+2 −1 < ǫ. This completes the proof of implication (3.4.8).
□ Remark 3.4.5. For most sequences sn a proof of lim n→∞sn = L based on Defini-tion 3.3.6 should consist from the following steps.
(1) Use algebra to simplify the expression |sn −L|. It is desirable to eliminate the absolute value.
(2) Discover an inequality of the form (3.4.14) |sn −L| ≤b(n) valid for all n ∈N.
Here b(n) should be a simple function with the following properties: (a) b(n) > 0 for all n ∈N.
(b) lim n→∞b(n) = 0. (Just check this property “mentally.”) (c) b(n) < ǫ is easily solvable for n for every ǫ > 0. The solution should be of the form “n > some expression involving ǫ, call it N(ǫ).” (3) Use inequality (3.4.14) to prove the implication n ∈N, n > N(ǫ) ⇒|sn−L| < ǫ.
Exercise 3.4.6. Determine the limits (if they exist) of the sequences (e), (f), (g), (h), (i), and (n) in Example 3.1.4. Prove your claims.
November 16, 2009 56 3. SEQUENCES IN R Exercise 3.4.7. Determine whether the sequence 3n + 1 7n −4 ∞ n=1 converges and, if it converges, give its limit. Provide a formal proof.
Exercise 3.4.8. Determine the limits (if they exist) of the sequences in Exam-ple 3.1.5. Prove your claims.
3.5. Two standard sequences Exercise 3.5.1. Let a ∈R be such that −1 < a < 1.
(1) Prove that for all n ∈N we have |a|n ≤ 1 n (1 −|a|).
(2) Prove that lim n→∞an = 0.
Exercise 3.5.2. Let a be a positive real number. Prove that lim n→∞a1/n = 1.
Solution. Let a > 0.
If a = 1, then a1/n = 1 for all n ∈N.
Therefore limn→∞a1/n = 1.
Assume a > 1. Then a1/n > 1. We shall prove that (3.5.1) a1/n −1 ≤a 1 n ∀n ∈N .
Put x = a1/n −1 > 0. Then, by Bernoulli’s inequality we get a = (1 + x)n ≥1 + nx.
Consequently, solving for x we get that x = a1/n −1 ≤(a −1)/n. Since a −1 < a, (3.5.1) follows.
Assume 0 < a < 1. Then 1/a > 1. Therefore, by already proved (3.5.1), we have 1 a 1/n −1 ≤1 a 1 n ∀n ∈N .
Since (1/a)1/n = 1/ a1/n , simplifying the last inequality, together with the in-equality a1/n < 1, yields (3.5.2) 1 −a1/n ≤a1/n a 1 n ≤1 a 1 n ∀n ∈N .
As a < a + 1/a and 1/a < a + 1/a, the inequalities (3.5.1) and (3.5.2) imply (3.5.3) a1/n −1 ≤ a + 1 a 1 n ∀n ∈N .
Let ǫ > 0 be given. Solving a + 1/a 1 n < ǫ for n, reveals N(ǫ): N(ǫ) = a + 1 a 1 ǫ Now it is easy to prove the implication (Do it as an exercise!) n ∈N, n > a + 1 a 1 ǫ ⇒ |a1/n −1| < ǫ.
□ November 16, 2009 3.8. ALGEBRA OF LIMITS OF CONVERGENT SEQUENCES 57 3.6. Non-convergent sequences Exercise 3.6.1. Prove that the sequence (d) in Example 3.1.4 does not converge.
Use Remark 3.3.7 and Exercise 3.3.8 Exercise 3.6.2. (Prove or Disprove) If sn does not converge to L, then there exist ǫ > 0 and N(ǫ) such that |sn −L| ≥ǫ for all n ≥N(ǫ).
3.7. Convergence and boundedness Exercise 3.7.1. Consider the following two statements: (A) The sequence sn is bounded.
(B) The sequence sn converges.
Is (A)⇒(B) true or false? Is (B)⇒(A) true or false? Justify your answers.
3.8. Algebra of limits of convergent sequences Exercise 3.8.1. Let sn be a sequence in R and let L ∈R. Set tn = sn −L for all n ∈N.
Prove that sn converges to L if and only if tn converges to 0.
Exercise 3.8.2. Let c ∈R. If lim n→∞xn = X and zn = c xn for all n ∈N, then lim n→∞zn = c X.
Exercise 3.8.3. Let xn and yn be sequences in R. Assume (a) xn converges to 0, (b) yn is bounded, (c) zn = xnyn for all n ∈N.
Prove that zn converges to 0.
Exercise 3.8.4. Let xn and yn be sequences in R. Assume (a) lim n→∞xn = X, (b) lim n→∞yn = Y , (c) zn = xn + yn for all n ∈N.
Prove that lim n→∞zn = X + Y .
Exercise 3.8.5. Let xn and yn be sequences in R. Assume (a) lim n→∞xn = X, (b) lim n→∞yn = Y , (c) zn = xn yn for all n ∈N.
Prove that lim n→∞zn = X Y .
Exercise 3.8.6. If lim n→∞xn = X and X > 0, then there exists a real number N such that n ≥N implies xn ≥X/2.
Exercise 3.8.7. Let xn be a sequence in R. Assume (a) xn ̸= 0 for all n ∈N, (b) lim n→∞xn = X, (c) X > 0, November 16, 2009 58 3. SEQUENCES IN R (d) wn = 1 xn for all n ∈N.
Prove that lim n→∞wn = 1 X .
Exercise 3.8.8. Let xn and yn be sequences in R. Assume (a) xn ̸= 0 for all n ∈N, (b) lim n→∞xn = X, (c) lim n→∞yn = Y , (d) X ̸= 0, (e) zn = yn xn for all n ∈N.
Prove that lim n→∞zn = Y X .
(Hint: Use previous exercises.) Exercise 3.8.9. Prove that lim n→∞ 2n2 + n −5 n2 + 2n + 2 = (insert correct value) by using the results we have proved (Exercises 3.8.2, 3.8.4, 3.8.5, 3.8.7, 3.8.8) and a small trick. You may use Definition 3.3.6 of convergence directly in this problem only to evaluate limit of the special form lim n→∞ 1 n.
Remark 3.8.10. The point of Exercise 3.8.9 is to see that the general properties of limits (Exercises 3.8.2, 3.8.4, 3.8.5, 3.8.7, 3.8.8) can be used to reduce complicated situations to a few simple ones, so that when the few simple ones have been done it is no longer necessary to go back to Definition 3.3.6 of convergence every time.
3.9. Convergent sequences and the order in R Exercise 3.9.1. Let sn be a sequence in R. Assume (a) lim n→∞sn = L.
(b) There exists a real number N0 such that sn ≥0 for all n ∈N such that n > N0.
Prove that L ≥0.
Exercise 3.9.2. Let an and bn be sequences in R. Assume (a) lim n→∞an = K.
(b) lim n→∞bn = L.
(c) There exists a real number N0 such that an ≤bn for all n ∈N such that n > N0.
Prove that K ≤L.
Exercise 3.9.3. Is the following refinement of Exercise 3.9.1 true? If sn con-verges to L and if sn > 0 for all n ∈N, then L > 0.
Exercise 3.9.4. Let xn be a sequence in R. Assume (a) xn ≥0 for all n ∈N, (b) lim n→∞xn = X, (c) wn = √xn for all n ∈N.
Prove that lim n→∞wn = √ X.
November 16, 2009 3.11. THE MONOTONIC CONVERGENCE THEOREM 59 3.10. Squeeze theorem for convergent sequences Exercise 3.10.1. There are three sequences in this exercise: an , bn and sn .
Assume the following (1) The sequence an converges to L.
(2) The sequence bn converges to L.
(3) There exists a real number n0 such that an ≤sn ≤bn for all n ∈N, n > n0.
Prove that sn converges to L.
Exercise 3.10.2. (1) Let x ≥0 and n ∈N. Prove the inequality (1 + x)n ≥1 + n x + n(n −1) 2 x2.
(2) Prove that for all n ∈N we have 1 ≤n1/n ≤1 + 2 √n.
Hint: Apply the inequality proved in (1) to 1 + 2/√n n.
(3) Prove that the sequence n1/n converges and determine its limit.
Exercise 3.10.3. (1) Prove that (n!)2 ≥nn for all n ∈N. Hint: Write n!
2 = 1 · n 2 · (n −1) · · · (n −1) · 2 n · 1 = n Y k=1 k n −k + 1 .
Then prove k n −k + 1 ≥n for all k = 1, . . . , n.
(2) Prove that lim n→∞ 1 n!
1/n = 0.
3.11. The monotonic convergence theorem Definition 3.11.1. A sequence sn of real numbers is said to be non-decreasing if sn ≤sn+1 for all n ∈N, strictly increasing if sn < sn+1 for all n ∈N, non-increasing if sn ≥sn+1 for all n ∈N, strictly decreasing if sn > sn+1 for all n ∈N.
A sequence with any of these properties is said to be monotonic.
Exercise 3.11.2. Again a huge task here.
Which of the sequences in Exam-ples 3.1.4, 3.1.5, and 3.1.6 are monotonic? Find few monotonic ones in each exam-ple. Provide rigorous proofs.
Exercise 3.11.3. (Prove or Disprove) If xn is non-increasing, then xn con-verges.
The following two exercises give powerful tools for establishing convergence of a sequence.
Exercise 3.11.4. If sn is non-increasing and bounded below, then sn con-verges.
Exercise 3.11.5. If sn is non-decreasing and bounded above, then sn con-verges.
November 16, 2009 60 3. SEQUENCES IN R Proof. Assume that the sequence sn is non-decreasing and bounded above.
Consider the range of the sequence sn . That is consider the set A = sn : n ∈N .
The set A is nonempty and bounded above. Therefore sup A exists. Put L = sup A.
We will prove that sn →L (n →∞). Let ǫ > 0 be arbitrary. Since L = sup A we have (1) L ≥sn for all n ∈N.
(2) There exists aǫ ∈A such that L −ǫ < aǫ.
Since aǫ ∈A, there exists Nǫ ∈N such that aǫ = sNǫ. It remains to prove that (3.11.1) n ∈N, n > Nǫ ⇒ |sn −L| < ǫ.
Let n ∈N, n > Nǫ be arbitrary. Since we assume that sn is non-decreasing, it follows that sn ≥sNǫ . Since L −ǫ < aǫ = sNǫ ≤sn, we conclude that L −sn < ǫ.
Since L ≥sn, we have |sn−L| = L−sn < ǫ. The implication (3.11.1) is proved.
□ Exercise 3.11.6. There is a huge task here.
Consider the sequences given in Example 3.1.6. Prove that each of these sequences converges and determine its limit.
3.12. Two important sequences with the same limit In this section we study the sequences defined in (3.1.3) and (3.1.5).
En = 1 + 1 n n , n ∈N, S1 = 2 and ∀n ∈N Sn+1 = Sn + 1 (n + 1)!.
Exercise 3.12.1. Prove by mathematical induction that Sn ≤3 −1/n for all n ∈N.
Exercise 3.12.2. Prove that the sequence Sn converges.
Exercise 3.12.3. Let n, k ∈N and n ≥k. Use Bernoulli’s inequality to prove that n!
(n −k)! nk ≥1 −(k −1)k n Hint: Notice that n!
nk(n −k)! = 1 · 1 −1 n 1 −2 n · · · 1 −k −1 n ≥ 1 −k −1 n k .
Exercise 3.12.4. The following inequalities hold: E1 = S1 and for all integers n greater than 1, (3.12.1) Sn −3 n < En < Sn.
Hint: Let n be an integer greater than 2. Notice that by the Binomial Theorem En = 1 + 1 n n = n X k=0 n!
k!(n −k)!
1 nk = 1 + 1 + n X k=2 n!
(n −k)! nk 1 k!.
Then use Exercise 3.12.3 to prove En > Sn −Sn−2/n. Then use Exercise 3.12.1.
Exercise 3.12.5. The sequences {En} and {Sn} converge to the same limit.
November 16, 2009 3.13. SUBSEQUENCES 61 Exercise 3.12.5 justifies the following definition.
Definition 3.12.6. The number e is the common limit of the sequences {En} and {Sn}.
Remark 3.12.7. The sequence {En} is increasing. To prove this claim let n ∈N be arbitrary. Consider the fraction En+1 En = 1 + 1 n+1 n+1 1 + 1 n n = n + 1 n 1 + 1 n+1 n+1 1 + 1 n n+1 = n + 1 n n+2 n+1 n+1 n !
n+1 = n + 1 n n(n + 2) (n + 1)2 n+1 = n + 1 n 1 − 1 (n + 1)2 n+1 (3.12.2) Since − 1 (n+1)2 > −1 for all n ∈N, applying Bernoulli’s Inequality with x = − 1 (n+1)2 we get (3.12.3) 1 − 1 (n + 1)2 n+1 > 1 −(n + 1) 1 (n + 1)2 = 1 − 1 n + 1.
The relations (3.12.2) and (3.12.3) imply En+1 En = n + 1 n 1 − 1 (n + 1)2 n+1 > n + 1 n 1 − 1 n + 1 = 1.
Thus En+1 En > 1 for all n ∈N, that is the sequence En is increasing.
3.13. Subsequences Composing functions is a common way how functions interact with each other.
Can we compose two sequences? Let x : N →R and y : N →R be two sequences.
Does the composition x ◦y make sense? This composition makes sense only if the range of y is contained in N. In this case y : N →N. That is the composition x ◦y makes sense only if y is a sequence in N. It turns out that the most important composition of sequences involve increasing sequences in N.
In this section the Greek letters µ and ν will always denote increasing sequences of natural numbers.
Definition 3.13.1. A subsequence of a sequence xn is a composition of the sequence xn and an increasing sequence µk of natural numbers. This compo-sition will be denoted by xµk or x(µk) .
Remark 3.13.2. The concept of subsequence consists of two ingredients: • the sequence xn (remember it’s really a function: x : N →R) • the increasing sequence µk of natural numbers (remember this is an increasing function: µ : N →N).
The composition x◦µ of these two sequences is a new sequence y : N →R. The k-th term yk of this sequence is yk = xµk. Note the analogy with the usual notation for functions: y(k) = x(µ(k)). Usually we will not introduce the new name for a subsequence: we will write xµk ∞ k=1 to denote a subsequence of the sequence xn . Here {µk}∞ k=1 is and increasing sequence of natural numbers which selects particular elements of the sequence xn to be included in the subsequence.
November 16, 2009 62 3. SEQUENCES IN R Remark 3.13.3. Roughly speaking, a subsequence of xn is a sequence formed by selecting some of the terms in xn , keeping them in the same order as in the original sequence. It is the sequence µk of positive integers that does the selecting.
Example 3.13.4. Few examples of increasing sequences in N are: (1) µk = 2k, k ∈N. (The sequence of even positive integers.) (2) νk = 2k −1, k ∈N. (The sequence of odd positive integers.) (3) µk = k2, k ∈N. (The sequence of perfect squares.) (4) Let j be a fixed positive integer. Set νk = j + k for all k ∈N.
(5) The sequence 2, 3, 5, 7, 11, 13, 17, . . . of prime numbers.
For this se-quence no formula for µk is known.
Exercise 3.13.5. Let µn be an increasing sequence in N. Prove that µn ≥n for all n ∈N.
Exercise 3.13.6. Each subsequence of a convergent sequence is convergent with the same limit.
Remark 3.13.7. The “contrapositive” of Exercise 3.13.6 is a powerful tool for proving that a given sequence does not converge. As an illustration prove that the sequence (−1)n does not converge in two different ways: using the definition of convergence and using the “contrapositive” of Exercise 3.13.6.
Exercise 3.13.8 (The Zipper Theorem). Let xn be a sequence in R and let µk and νk be increasing sequences in N. Assume (a) µk : k ∈N ∪ νk : k ∈N = N.
(b) xµk converges to L.
(c) xνk converges to L.
Prove that xn converges to L.
Example 3.13.9. The sequence (o) in Example 3.1.4 does not converge, but it does have convergent subsequences, for instance the subsequence 2k 2k + 1 ∞ k=1 (Here µk = 2k, k ∈N) and the subsequence 1 (2k −1)2k ∞ k=1 (Here νk = 2k−1, k ∈N).
Remark 3.13.10. The notation for subsequences is a little tricky at first. Note that in xµk it is k that is the variable. Thus the successive elements of the subsequence are xµ1 , xµ2 , xµ3 , etc. To indicate a different subsequence of the same sequence {xn}∞ n=1 it would be necessary to change not the variable name, but the selection sequence.
For example xµk ∞ k=1 and xνk ∞ k=1 in Example 3.13.9 are distinct subsequences of {xn}. (Thus xµk ∞ k=1 and xµj ∞ j=1 are the same subsequence of {xn}∞ n=1 for exactly the same reason that x 7→x2 (x ∈R) and t 7→t2 (t ∈R) are the same function. To make a different function it’s the rule you must change, not the variable name.) Example 3.13.11. Let xn be the sequence defined by xn = (−1)n(n + 1)(−1)n n , n ∈N.
November 16, 2009 3.15. SEQUENCES AND SUPREMUM AND INFIMUM 63 The values of xn are −1 1 · 2, 3 2, −1 3 · 4, 5 4, −1 5 · 6, 7 6, −1 7 · 8, 9 8, − 1 9 · 10, 11 10, . . . .
Exercise 3.13.12. Every sequence has a monotonic subsequence.
Hint: Let xn be an arbitrary sequence. Consider the set M = n ∈N : ∀k > n we have xk ≥xn .
The set M is either finite or infinite. Construct a monotonic subsequence in each case.
Exercise 3.13.13. Every bounded sequence of real numbers has a convergent subsequence.
3.14. The Cauchy criterion Definition 3.14.1. A sequence sn of real numbers is called a Cauchy sequence if for every ǫ > 0 there exists a real number Nǫ such that ∀n, m ∈N, n, m > Nǫ ⇒ |sn −sm| < ǫ.
Exercise 3.14.2. Prove that every convergent sequence is a Cauchy sequence.
Exercise 3.14.3. Prove that every Cauchy sequence is bounded.
Exercise 3.14.4. If a Cauchy sequence has a convergent subsequence, then it converges.
Exercise 3.14.5. Prove that each Cauchy sequence has a convergent subsequence.
Exercise 3.14.6. Prove that a sequence converges if and only if it is a Cauchy sequence.
3.15. Sequences and supremum and infimum Exercise 3.15.1. Let A ⊂R, A ̸= ∅and assume that A is bounded above. Prove that a = sup A if and only if (a) a is an upper bound of A, that is, a ≥x, for all x ∈A; (b) there exists a sequence xn such that xn ∈A for each n ∈N and lim n→∞xn = a.
Exercise 3.15.2. Let A ⊂R, A ̸= ∅and assume that A is bounded above. Let a = sup A and assume that a / ∈A. Prove that there exists a strictly increasing sequence xn such that xn ∈A for each n ∈N and lim n→∞xn = a.
Exercise 3.15.3. State and prove the characterization of infimum which is analo-gous to the characterization of sup A given in Exercise 3.15.1.
Exercise 3.15.4. State and prove an exercise involving infimum of a set which is analogous to Exercise 3.15.2.
November 16, 2009 |
190361 | https://www.sciencedirect.com/science/article/pii/S2468784723000557 | Endometrial biopsy: Indications, techniques and recommendations. An evidence-based guideline for clinical practice - ScienceDirect
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Journal of Gynecology Obstetrics and Human Reproduction
Volume 52, Issue 6, June 2023, 102588
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Endometrial biopsy: Indications, techniques and recommendations. An evidence-based guideline for clinical practice
Author links open overlay panelSalvatore Giovanni Vitale a, Giovanni Buzzaccarini b, Gaetano Riemma c, Luis Alonso Pacheco d, Attilio Di Spiezio Sardo e, Jose Carugno f, Vito Chiantera g, Peter Török h, Marco Noventa i, Sergio Haimovich j, Pasquale De Franciscis c, Tirso Perez-Medina k, Stefano Angioni a, Antonio Simone Laganà g
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Abstract
This practice guideline provides updated evidence for the gynecologist who performs endometrial biopsy (EB) in gynecologic clinical practice.
An international committee of gynecology experts developed the recommendations according to AGREE Reporting Guideline.
An adequate tissue sampling is mandatory when performing an EB. Blind methods should not be first choice in patients with suspected endometrial malignancy. Hysteroscopy is the targeted-biopsy method with highest diagnostic accuracy and cost-effectiveness. Blind suction techniques are not reliable for the diagnosis of endometrial polyps. In low resources settings, and in absence of the capacity to perform office hysteroscopy, blind techniques could be used for EB. Hysteroscopic punch biopsy allows to collect only limited amount of endometrial tissue. grasp biopsy technique should be considered first choice in reproductive aged women, bipolar electrode chip biopsy should be preferred with hypotrophic or atrophic endometrium. EB is required for the final diagnosis of chronic endometritis. There is no consensus regarding which endometrial thickness cut-off should be used for recommending EB in asymptomatic postmenopausal women. EB should be offered to young women with abnormal uterine bleeding and risk factors for endometrial carcinoma. Endometrial pathology should be excluded with EB in nonobese women with unopposed hyperestrogenism. Hysteroscopy with EB is useful in patients with abnormal bleeding even without sonographic evidence of pathology. EB has high sensitivity for detecting intrauterine pathologies. In postmenopausal women with uterine bleeding, EB is recommended. Women with sonographic endometrial thickness > 4 mm using tamoxifen should undergo hysteroscopic EB.
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Keywords
Hysteroscopy
Practical guidelines
Endometrium
Endometrial biopsy
1. Introduction
Endometrial biopsy (EB) is a common gynecologic procedure frequently performed in clinical practice. There are several equipment and techniques to perform an EB. Over the last years, office-based endometrial sampling has replaced the need for diagnostic dilation and curettage (D&C) or operative hysteroscopy, procedures that are usually both performed in the operating room with the patient under general anesthesia.
There are many different clinical scenarios that require EB, such as patients presenting with thickened endometrium or abnormal uterine bleeding (AUB) [1,2]. Although a very safe and effective procedure for detecting endometrial cancer (EC) or atypical hyperplasia (AH), EB could result in a false-negative test, missing the diagnosis which is mainly due to biopsy technique, non-representative sampling, and variable pathologic interpretation .
The aim of this practice guideline is to summarize the most relevant available scientific evidence regarding EB techniques and indications.
1.1. Identification and assessment of evidence
This practice guideline was produced using the following search methodology: electronic databases including MEDLINE, EMBASE, Global Health, The Cochrane Library (Cochrane Database of Systematic Reviews, Cochrane Central Register of Controlled Trials, Cochrane Methodology Register), Health Technology Assessment Database and Web of Science, research registers (such as www.clinicaltrials.gov) were searched from inception to June 2022; we used the medical subject heading (MeSH) term “Endometrium” (MeSH Unique ID: D004717) in combination with “Biopsy” (MeSH Unique ID: D001706). The study search was not restricted to the English language but extended to Spanish, Chinese, French, Italian and Portuguese. Authors who are fluent in languages other than English (Spanish, Chinese, French, Italian and Portuguese) evaluated relevant publications in foreign language and provided, after English translation, related information to the panel. The reference lists of all identified papers were checked to identify studies not captured by electronic searches. All studies were assessed for methodologic rigor and graded according to the United States Preventive Services Task Force classification system (Table 1). Titles and/or abstracts of studies retrieved using the search strategy were screened independently by 2 authors to identify studies that meet the aims of this guideline. The full texts of the eligible articles were retrieved and independently assessed for eligibility by other 2 team members. Any disagreement between them over the eligibility of selected articles was resolved through discussion with a third (external) collaborator. Two authors independently extracted data from articles about study features and included populations, type of intervention and outcomes. Any discrepancies were identified and resolved through discussion (with a third external collaborator where necessary).
Table 1. Assessment of evidence for the practice guideline.
| Evidence was reviewed and evaluated for quality using criteria outlined by the U.S. Preventive Services Task Force |
| - I Evidence obtained from at least one properly designed randomized controlled trial. |
| - II-1 Evidence obtained from well-designed controlled trials without randomization. |
| - II-2 Evidence obtained from well-designed cohort or case-control analytic studies, preferably from more than one center or research group. |
| - II-3 Evidence obtained from multiple time series with or without the intervention. Dramatic results in uncontrolled experiments also could be regarded as this type of evidence. |
| - III Opinions of respected authorities, based on clinical experience, descriptive studies, or reports of expert committees. |
| Based on the highest level of evidence found in the data, recommendations are provided and graded according to the following categories: |
| - Level A: Recommendations are based on good and consistent scientific evidence. |
| - Level B: Recommendations are based on limited or inconsistent scientific evidence. |
| - Level C: Recommendations are based primarily on consensus and expert opinion. |
1.2. Stakeholders’ involvement and applicability
These recommendations are based on professional opinion and are intended to assist gynecologists in treating the average patient. They should not be seen as hard and fast rules, and they were not designed to take the place of clinical judgment.
Recommendations were based on the best available scientific evidence, when practicable, and on the expert panel's consensus when such evidence was not available. They might probably change as we learn more about the condition.
The preparation of this guideline involves specialists in gynecological ultrasound (US), hysteroscopy, infertility, and oncologic therapy of endometrial pathology, according to AGREE Reporting Guideline standards . Three external reviewers, two gynecologists and a gynecologic histopathologist randomly selected with a computer-based randomization from a list of 200 experts, with expertise in the aforementioned domains extensively assessed these practice recommendations in two rounds of revisions before publication.
1.3. Indications to endometrial biopsy
Every year, many women require gynecological visit with symptoms that prompt EB. EC is diagnosed in about 65,000 women every year in the United States. Among the most frequent indications for EB in clinical practice include infertility and subfertility, the assessment of the uterine cavity before assisted reproduction technique (ART); evaluation of premenopausal and postmenopausal patients with AUB among other indications . The etiology of AUB is classified according the PALM-COEIN classification, developed by Munro et al. and adopted by the International Federation of Gynecology and Obstetrics (FIGO). By classifying abnormal uterine bleeding according to the potential cause, this system distinguishes among polyp, adenomyosis, leiomyoma, malignancy and hyperplasia, coagulopathy, ovulatory dysfunction, endometrial, iatrogenic, and not yet classified cause. The structural reasons of abnormal uterine bleeding are included by the acronym “PALM” section of the PALM-COEIN. Conversely, the nonstructural, hormonal, or systemic causes of AUB are denoted under the acronym “COEIN” [1,2].
Before proceeding to perform an EB, questions about the menstrual bleeding pattern (frequency, duration, regularity and quantity), presence of pain, family history of AUB or underlying bleeding disorders, medication or herbal preparation that might affect bleeding generally, such as ginseng, ginkgo, use of hormonal contraceptives, nonsteroidal anti-inflammatory drugs, warfarin or heparin derivatives, should be included in a medical history. Careful analysis of the bleeding pattern will be one of the most crucial components of the medical history. For instance, cancer or even hyperplasia would be unlikely to be the cause of cyclic menstrual bleeding [1,2].
Regardless of the clinical scenario, EC could be performed in multiple ways .
1.4. Endometrial biopsy techniques
A plethora of studies have been performed evaluating different techniques for EB. Taraboanta et al. performed a retrospective cross-sectional study on 1677 hysterectomy specimens diagnosed with Atypical Hyperplasia/Endometrioid Intraepithelial Neoplasia (AH/EIN) or EC evaluating those with previous negative endometrial biopsy. Of these cases with negative endometrial biopsies before hysterectomy, 172 were classified as inadequate/insufficient since no endometrial tissue was present or had a benign diagnosis. An important limitation of this study was not identifying the procedure that was used to perform the endometrial sampling. In negative endometrial biopsy result, the post-test probability of EC or AH/EIN diagnosis in the hysterectomy specimen was found to be 0.74%. Results from this study provide evidence about the importance of an adequate endometrial sampling [8,9].
D&C was once recognized as the gold standard for endometrial sampling . Initially, D&C was considered as an accurate method for identifying endometrial cancer tumor grade . More recently, D&C preoperative FIGO grade 1 endometrial cancer diagnosis was found congruent in 85% of cases with EB. However, a higher grade was found in 8.7% of the cases at the time of hysterectomy . Piatek et al. assessed a retrospective cohort analysis considering all the patients who underwent endometrial biopsy using a Pipelle® and D&C. The purpose of this study was to determine the rate of endometrial sampling failure and factors affecting the quality of specimen obtained for histopathological examination. Of the 895 endometrial sampling procedures performed, 339 patients underwent Pipelle® biopsy, and 556 D&C. Inadequate samples were found in 60 and 88 cases, respectively. The study suggested that none of these two methods guarantee adequate specimen sampling . Utida et al. designed a cross sectional study comparing the efficiency of histological endometrial samples collected using Pipelle® aspiration and hysteroscopic biopsies. The main aim of this study was to assess the congruency between these two endometrial sampling techniques. Specifically, the histological diagnosis of malignancy was a priority and, subsequently, the comparison between the costs of both techniques was assessed. The study enrolled 45 women (over 35 years old with AUB or postmenopausal bleeding) who underwent EB using both hysteroscopy and Pipelle®. Interestingly, EBs obtained using Pipelle® had a high accuracy for EC (100% agreement between the two procedures) but a lower accuracy for the diagnosis of polyps. It is important to note that Pipelle® biopsies costed 27 times less than hysteroscopic biopsies . A very important aspect of this study is that it highlights the importance of performing EBs under direct visualization . However, such findings were limited by the reduced sample size of the study.
To date, blind endometrial sampling alone are not considered effective for diagnosing focal lesions of the uterine cavity such as polyps or submucosal myoma.
Endometrial sampling could also be performed using ultrasound (US)-assisted guidance. However, US has a lower capacity to detect endometrial lesions compared to hysteroscopy [17,18]. Indeed, a prospective study performed by Reznak et al. showed that US abnormal findings need to be confirmed by hysteroscopic visualization with targeted biopsy and histological examination to avoid low accuracy .
Cheng et al. performed a retrospective cohort study evaluating the use of Lin's biopsy grasper for endometrial biopsy. Lin's biopsy grasper is one device specifically designed to work in conjunction with a flexible hysteroscope to perform intrauterine biopsy under transabdominal ultrasound guidance. This targeted biopsy method allows to perform endometrial biopsies in an office setting. They performed 126 targeted endometrial biopsies achieving a high diagnostic rate (92.1%, with 116 cases confirmed histologically) and adequate tissue quality (77.8%, with 98 cases obtaining optimal specimen volume) .
Bryant et al. performed a retrospective analysis on 141 hysterectomies performed in patients with a preoperative or incidental diagnosis of AH/EIN. Their data provided evidence about the value of selective rather than complete specimens sampling for the detection of AH/EIN and EC, showing that a selective approach could be extensively useful for the diagnosis .
Regarding the office hysteroscopy EB technique, different studies have provided results regarding the use of operative grasping forceps introduced trough a 5 Fr operative channel of the hysteroscope . The standard technique to perform hysteroscopic guided EB was proposed in 2002 by Bettocchi et al. Briefly, the forceps is placed, with its jaws opened, against the endometrium. The jaws are pushed into the tissue for 0.5 to 1 cm. Once a large portion of mucosa has been tangentially detached, the jaws are closed and the entire hysteroscope is removed from the uterine cavity, without pulling the tip of the instrument back into the channel. This method allows to collect a larger amount of tissue .
One of the most recent advantages in EB technique relies on the study of the tumor material present in bodily fluids. Liquid biopsies also provide advantages for monitoring cancer progress and the response to therapy. The diagnostic procedure consists of an endometrial biopsy, which is obtained by a minimally invasive aspiration from the uterine cavity using a Pipelle®. Abnormal cells present in the aspirate are analyzed . Hirai et al. performed a multicenter study comparing the clinical performance of liquid based endometrial cytology using SurePath™ to classic suction endometrial tissue biopsy. They suggested that liquid-based endometrial cytology was not inferior to suction endometrial tissue biopsy for the detection of endometrial cancer .
1.5. Recommended guidelines for the endometrial biopsy
Based on the available evidence, we promote the following recommendations:
-An appropriate sampling is mandatory when performing an EB (Level A).
-When performing diagnostic hysteroscopy and EB, the EB should be performed after the hysteroscopic procedure (Level C).
-D&C and Pipelle® should not be the first choice for EB method in patients with suspected endometrial malignancy (Level B).
-The use of VA, Pipelle® for outpatient EB is not efficient and lacks sensitivity when diagnosing endometrial polyps (level C).
-Office hysteroscopy is the targeted-biopsy method with the highest diagnostic accuracy (Level A).
-Liquid based biopsy is a promising method for endometrial markers detection (Level B).
-Suction techniques are not reliable for the diagnosis of endometrial polyps (Level A).
-In low-resources settings without the capacity to perform office hysteroscopy, blind techniques could be used for EB (Level B).
1.6. Hysteroscopic techniques for endometrial biopsy
The punch biopsy was the first type of technique commonly used for hysteroscopic biopsy. It utilized the spoon forceps and were regarded the standard biopsy instrument for several years. According to this technique, the biopsy forceps’ jaws are held opened in close contact with the endometrium before being closed . The hysteroscope is left in the uterine cavity while the closed forceps are retracted through the working channel. However, because the jaw extension is relatively limited compared to other biopsy forceps (2.5 vs. 5 mm for alligator forceps), the obtained tissue volume is sometimes insufficient for a satisfactory histological diagnosis [27,28].
To improve the quantity of retrieved tissue enough for a correct histological investigation, in 2002 Bettocchi et al. proposed a novel biopsy technique named "grasp biopsy". They used a toothed grasp forceps, known as alligator forceps. Because of the double length of the softly toothed jaws, the alligator forceps can collect a larger volume of tissue. Briefly, the alligator forceps is placed in close contact with the target location where the endometrial sample has to be taken with the jaws wide open. The forceps is then moved forward, “plowing” together with the tissue for roughly 0.5–1 cm, aiming to avoid contacting the underlying myometrium, in order to prevent stimulating myometrial nerve fibers and minimize pain. The jaws are then closed, grabbing the segment of endometrial tissue to be removed, which is subsequently retrieved from the uterine cavity alongside the hysteroscope [23,29].
In case of perimenopausal and postmenopausal women with hypotrophic or atrophic endometrium, it is more difficult to clench an appropriate quantity of tissue. In this case, performing the chip biopsy, cutting a “chip” of endometrium with a 5 Fr bipolar electrode inserted into the operating channel of the hysteroscope, is particularly effective. “Chipping” the endometrium may make the technique easier than others and may also be useful when sampling the superficial myometrial surface (i.e., in women with suspected premalignant or malignant endometrial pathology) , , .
An alternative approach to retrieve endometrium from an hypotrophic or atrophic surface is the pick-up biopsy technique. It consists of picking up tissue using the tip of the hysteroscope as a plow or the tip of dedicated mechanical tools to collect more sampling material. A recently patented tool for this purpose is the biopsy snake forceps sec. VITALE (Centrel Srl, Ponte San Nicolò, Padua, Italy). It is characterized by a flat pointed tip with serrated edges which can help to expose the hypotrophic or atrophic endometrium to be resected avoiding at the same time to loose fragments of the specimen . Another crucial aspect to be remarked is the pain experienced during hysteroscopic endometrial sampling. Class I evidence reported an increased pain perception with the punch biopsy relative to the grasp and pick-up technique .
1.7. Recommended guidelines for the appropriate hysteroscopic biopsy technique
-Punch biopsy allows to collect a limited amount of endometrium to be sampled. (Level B).
-Grasp biopsy should be considered the most appropriate technique in reproductive aged women. (Level A).
-Chip biopsy is effective in collecting more endometrium compared with other techniques in perimenopausal and postmenopausal women. (Level B).
-In perimenopausal and postmenopausal women, the pick-up biopsy technique is more effective in collecting endometrial tissue compared with punch biopsy. (Level A).
1.8. Clinical scenarios
Generally, hysteroscopy aims to diagnose precancerous or cancerous lesions, to see and treat endocavitary benign pathology, such as leiomyomas or endometrial polyps previously identified by US scan, and to assess subclinical conditions that can lead to infertility (such as Asherman's syndrome or endometritis) [34,35]. Currently, the only absolute contraindication to hysteroscopy is active uterine or pelvic infection. In addition, women diagnosed with primary infertility, recurrent pregnancy loss or subfertility have a clinical indication to undergo evaluation of endometrial pathology and uterine morphology . On this purpose, we subclassified the clinical scenarios according to the patient's age and symptomatology. For the purpose of this review, asymptomatic women were considered those without an AUB, regardless of their menopausal status, conversely symptomatic women are those presenting with symptoms (commonly AUB).
1.9. Asymptomatic women
1.9.1. Asymptomatic patients of reproductive age
In this group of patients, paucity of specific population studies affects our guideline results. One of the main reasons requiring EB in asymptomatic women is infertility . Specifically, chronic endometritis has been recognized as one of the uterine factors that impair embryo implantation and immunohistochemical (IHC) diagnosis on endometrial specimens is mandatory . In this regard, Zargar et al. performed a cross-sectional study with the aim of compare the prevalence of chronic endometritis in patients with recurrent implantation failure (RIF) and recurrent pregnancy lost (RPL) using hysteroscopy and immunohistochemistry. Results showed that hysteroscopic visual inspection (searching for micro polyps or red spots) is a reliable tool in patients with RIF and RPL in order to diagnose chronic endometritis, however its accuracy is not sufficient to be considered as an alternative to IHC . Other studies confirmed the need of combined diagnostic hysteroscopy and EB in women complaining of reproductive issues , , . Especially in situations of repeated ART failure, there is a substantial chance of undiagnosed uterine abnormalities during regular US scan in infertile individuals. Higher rates of effective ARTs and non-inferior pregnancy rates have been observed when patients are routinely screened using in-office hysteroscopy and EB , , , , , , .
Before starting ART, the gynecologist should thoroughly examine the uterine cavity and document (with appropriate biopsy or excision) any abnormal endometrial findings.
1.10. Recommended guidelines for asymptomatic patients of reproductive age
-In asymptomatic premenopausal women, the EB is a useful tool for chronic endometritis diagnosis (Level A).
-Hysteroscopy with or without EB is useful in the infertility workup (Level A).
-In case of ART failure, hysteroscopic EB is crucial to avoid misdiagnoses and improve reproductive outcomes (Level B).
1.11. Asymptomatic postmenopausal patients
The incidental finding of a thickened endometrium at US in asymptomatic women is a common clinical scenario , , , .
Several experts advocate adopting an US cut-off value of 4.0 or 5.0 mm in patients with postmenopausal bleeding (PMB) to recommend additional endometrial investigation [50,, , , , ]. The risk of EC is estimated to be less than 1% when the endometrial thickness (ET) is below 4.0 mm [50,, , , , ]. Some women with uterine premalignant or malignant conditions are asymptomatic . There is no clear consensus on when to screen for EC in asymptomatic women with thickened endometrium, in contrast to the guidelines on the management of PMB. To improve diagnostic accuracy, it is necessary to investigate the ideal cut off value to warrant further endometrial investigation in asymptomatic postmenopausal women , , .
1.12. Recommended guidelines for asymptomatic postmenopausal patients
-There is no clear consensus regarding which ET cut-off should be used for recommending endometrial sampling in asymptomatic postmenopausal patients (Level B).
1.13. Symptomatic women
1.13.1. Symptomatic patients of reproductive age
In women of reproductive age, it is extremely important to perform EB in obese patients with AUB, and in those heterogeneous and/or hypervascularized endometrium on US, due to increased risk of malignancy , , , . In nonobese patients, several trials suggest performing EB in patients with AUB and/or in the presence of one of the following conditions: chronic anovulatory dysfunction, unopposed estrogen stimulation, those not responding to medical management, or patients with genetic high risk of endometrial cancer (e.g., Lynch syndrome, Cowden syndrome) [37,64,, , , , , ]. In addition, endometrial neoplasia should be suspected in premenopausal patients who are anovulatory and have prolonged periods of amenorrhea [72,73].
Similarly, EB is recommended if bleeding is frequent (interval between the onset of bleeding episodes is <21 days), heavy, or prolonged (>8 days). In patients who are ovulatory, this includes intermenstrual bleeding.
1.13.2. Recommended guidelines for symptomatic patients of reproductive age
-Young women with increased risk for endometrial malignancies and endometrial heterogeneity should undergo EB (Level A)
-Premalignant conditions or malignancy should be ruled out in nonobese women with unopposed hyperestrogenism (Level B)
-Hysteroscopy with EB is useful in women with heavy, prolonged or intermenstrual bleeding even in those without sonographic evidence of pathology (Level B).
1.13.3. Symptomatic perimenopausal patients
Several trials showed that hysteroscopy with directed biopsy is more sensitive than D&C for the diagnosis of uterine pathology in patients with AUB [11,15,26,, , , ].
Nicholls-Dempsey et al. reviewed the indications for EB at their center. After analysis of 371 patients, they concluded that in women under the age of 41 there was no indication for biopsy in 23% of the biopsies, suggesting a significant over-investigation. Similarly, the value of EB in patients between 41 and 45 years old with menorrhagia and no additional risk factor should be further investigated .
Since the possibility of bleeding caused by a polyp, Ngo et al. performed a retrospective analysis evaluating differences in hysteroscopic findings between benign endometrial polyps and EC. The study included hysteroscopic findings of endometrial polyps (n=214) on 3066 women who underwent hysteroscopy for abnormal vaginal bleeding, intrauterine cavity lesions suspected on US, recurrent spontaneous abortion, or infertility assessment. Clinical characteristics such as hyper-vascularity of the surface, ulcers, histopathological and hysteroscopic findings were evaluated retrospectively. The analysis showed that women with hysteroscopic findings of endometrial polyps with hyper-vascular, ulcerative, and polyps with irregular surfaces had a higher likelihood of EC. In this specific population, a target biopsy of the polyps with these specific characteristics should be performed to exclude malignancy .
In-office hysteroscopy is accurate for the detection of endometrial hyperplasia and cancer, according to Clarke et al. and De Franciscis et al. . However, in order to increase diagnostic accuracy, the sampling must be performed on the endometrial areas that seem abnormal [80,81].
1.13.4. Recommended guidelines for symptomatic premenopausal patients
-EB has high sensitivity for detecting benign, premalignant and malignant intrauterine pathologies (Level A).
-Hysteroscopic guided EB has higher accuracy than blind techniques in symptomatic women, regardless of their age (Level A).
1.13.5. Symptomatic postmenopausal patients
This population accounts for the major number of EB performed, due to the highest incidence of EC and AH/EIN. Bar-On et al. performed a retrospective cohort study including all women who underwent outpatient hysteroscopy for the following indications: PMB, suspected polyp, and/or increased ET. Histological accuracy was evaluated by comparing specimens obtained in hysteroscopy with those obtained by hysterectomy, and visual accuracy was evaluated by comparing visual findings with those obtained by blind biopsies. Office hysteroscopy has been confirmed an adequate and reliable tool for the evaluation of benign pathology in the uterine cavity .
Several trials also reported that for women presenting with PMB, the use of transvaginal US is not indicated as a screening tool in evaluating women who have a history as tamoxifen use, due to poor diagnostic accuracy , , , . On the contrary, hysteroscopy and EB are the most reliable diagnostic method . A recent study noted that there is no increased risk for EC in these group of patients relative to women taking aromatase inhibitors or without treatment . Weighted sensitivities of endometrial sample for the diagnosis of EC, AH, and endometrial pathologies were 90%, 82%, and 39%, respectively, when hysteroscopy was used as a reference. Specificity was 98–100% for all diagnoses investigated and the reference test utilized. Endometrial sampling failed 11% of the time, with inadequate samples recovered in 31% of the time. Endometrial (pre) cancer was discovered in 7% of the women with inadequate or failed samples. Endometrial sampling's sensitivity to identify endometrial cancer, particularly AH and endometrial pathologies, including endometrial polyps, is lower than previously assumed in women with PMB. After a benign endometrial biopsy result, additional diagnostic work-up for localized pathology is indicated . When compared to the assessment of recurrent bleeding, EC risk variables such as age can give considerable risk stratification.
1.13.6. Recommended guidelines for symptomatic postmenopausal patients
In postmenopausal women with any kind of AUB or PMB, EB is indicated (Level A).
Hysteroscopic guided EB should be the first choice due to the highest accuracy and cost-effectiveness (Level B).
1.13.7. Recommendations for future research
These guidelines were developed to provide a concise and updated reference for practicing clinicians facing with EB according to the most common clinical scenarios. However, they should not be intended as strict guidelines and must be adapted to the available facilities in every setting.
AUB, PMB and other intrauterine-related conditions are frequent gynecologic complaints encountered in daily clinical practice. There are some areas that require additional high-quality data to improve their diagnostic accuracy and management.
We propose the following considerations of future research:
-To conduct randomized trials to evaluate the impact of the presence of endometrial polyps on endometrial receptivity in infertile women diagnosed with asymptomatic endometrial polyps.
-To compare different mechanical hysteroscopic tools for performing EB (i.e. tissue retrieval systems, 5Fr forceps)
-To perform large studies evaluating the ET cut-off to recommend further endometrial evaluation in asymptomatic postmenopausal women.
Author contributions
S.G.V.: Conceptualization; Methodology; Formalanalysis;Writing - original draft
G.B.: Conceptualization; Methodology Formal analysis;Writing - original draft
G.R.: Conceptualization; Methodology Formal analysis;Writing - original draft
L.A.P.: Formal analysis; Writing - review & editing
A.D.S.S.: Formal analysis; Writing - review & editing
J.C.: Formal analysis; Writing - review & editing
V.C.: Formal analysis; Writing - review & editing
P.T.: Formal analysis; Writing - review & editing
M.N.: Formal analysis; Writing - review & editing
S.H.: Formal analysis; Writing - review & editing
P.D.F.: Data curatio; Writing - review & editing
T.P.M: Data curation; Writing - review & editing
S.A.: Data curation; Writing - review & editing
A.S.L.: Conceptualization; Methodology;Formal analysis;Writing - original draft
All the authors edited the article for agreement in its final form.
Acknowledgements
None.
Funding sources
This research received no specific grant from any funding agency in the public, commercial, or not-for-profit sectors.
This work is currently not being submitted to any other journal for consideration for publication and has not been previously presented in any form.
Disclosure Statement
The authors declare that they have no conflicts of interest to disclose regarding this publication.
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Artificial intelligence model for enhancing the accuracy of transvaginal ultrasound in detecting endometrial cancer and endometrial atypical hyperplasia
2024, International Journal of Gynecological Cancer Show abstract Transvaginal ultrasound is typically the initial diagnostic approach in patients with postmenopausal bleeding for detecting endometrial atypical hyperplasia/cancer. Although transvaginal ultrasound demonstrates notable sensitivity, its specificity remains limited. The objective of this study was to enhance the diagnostic accuracy of transvaginal ultrasound through the integration of artificial intelligence. By using transvaginal ultrasound images, we aimed to develop an artificial intelligence based automated segmentation model and an artificial intelligence based classifier model. Patients with postmenopausal bleeding undergoing transvaginal ultrasound and endometrial sampling at Mayo Clinic between 2016 and 2021 were retrospectively included. Manual segmentation of images was performed by four physicians (readers). Patients were classified into cohort A (atypical hyperplasia/cancer) and cohort B (benign) based on the pathologic report of endometrial sampling. A fully automated segmentation model was developed, and the performance of the model in correctly identifying the endometrium was compared with physician made segmentation using similarity metrics. To develop the classifier model, radiomic features were calculated from the manually segmented regions-of-interest. These features were used to train a wide range of machine learning based classifiers. The top performing machine learning classifier was evaluated using a threefold approach, and diagnostic accuracy was assessed through the F1 score and area under the receiver operating characteristic curve (AUC-ROC). 302 patients were included. Automated segmentation–reader agreement was 0.79±0.21 using the Dice coefficient. For the classification task, 92 radiomic features related to pixel texture/shape/intensity were found to be significantly different between cohort A and B. The threefold evaluation of the top performing classifier model showed an AUC-ROC of 0.90 (range 0.88–0.92) on the validation set and 0.88 (range 0.86–0.91) on the hold-out test set. Sensitivity and specificity were 0.87 (range 0.77–0.94) and 0.86 (range 0.81–0.94), respectively. We trained an artificial intelligence based algorithm to differentiate endometrial atypical hyperplasia/cancer from benign conditions on transvaginal ultrasound images in a population of patients with postmenopausal bleeding.
### The emerging roles of long non-coding RNA (lncRNA) H19 in gynecologic cancers
2024, BMC Cancer Show abstract Long non-coding RNA (lncRNA) H19 has gained significant recognition as a pivotal contributor to the initiation and advancement of gynecologic cancers, encompassing ovarian, endometrial, cervical, and breast cancers. H19 exhibits a complex array of mechanisms, demonstrating dualistic effects on tumorigenesis as it can function as both an oncogene and a tumor suppressor, contingent upon the specific context and type of cancer being investigated. In ovarian cancer, H19 promotes tumor growth, metastasis, and chemoresistance through modulation of key signaling pathways and interaction with microRNAs. Conversely, in endometrial cancer, H19 acts as a tumor suppressor by inhibiting proliferation, inducing apoptosis, and regulating epithelial-mesenchymal transition. Additionally, H19 has been implicated in cervical and breast cancers, where it influences cell proliferation, invasion, and immune evasion. Moreover, H19 has potential as a diagnostic and prognostic biomarker for gynecologic cancers, with its expression levels correlating with clinical parameters and patient outcomes. Understanding the functional roles of H19 in gynecologic cancers is crucial for the development of targeted therapeutic strategies and personalized treatment approaches. Further investigation into the intricate molecular mechanisms underlying H19’s involvement in gynecologic malignancies is warranted to fully unravel its therapeutic potential and clinical implications. This review aims to elucidate the functional roles of H19 in various gynecologic malignancies.
### The effect of chronic endometritis and treatment on patients with unexplained infertility
2023, BMC Women S Health Show abstract This paper was mainly conducted to investigate the effect of chronic endometritis (CE) on the clinical outcome of patients with unexplained infertility. 145 patients with unexplained infertility from the Reproductive Center of our hospital from January 2018 to December 2021 were selected as the unexplained infertility group. 42 patients with definite infertility causes were selected as the control group during the same period. Both groups of patients underwent hysteroscopy and immunohistochemical tests for CD38 and CD138. According to the results of hysteroscopy and immunohistochemistry, the incidence of CE between the two groups was analyzed. Patients with CE as CE group accepted oral antibiotic therapy for 14 days. Another 58 patients with unexplained infertility who did not undergo hysteroscopy and immunohistochemical tests for CD38 and CD138 were selected as the unexamined group. Both groups of patients were expected natural pregnancy. Follow-up lasted for 1 year, and the pregnant patients were followed up until delivery.The clinical pregnancy rate, spontaneous abortion rate and baby-carrying home rate of the two groups were compared. There were 75 patients with CE in the unexplained infertility group, and the prevalence rate was 51.7% (75/145). Compared with the control group (28.6%), the incidence of CE was significantly higher (P < 0.05). After treated with antibiotic treatment, the patients’ clinical pregnancy rate was 61.3% (46/75) and baby-carrying home rate was 60% (45/75) in the CE group, which were higher than those in the unexamined group(43.1% & 36.2%) (P < 0.05), while the spontaneous abortion rate was 2.2% (1/46),which was lower than that in the unexamined group (16.0%) (P < 0.05). For patients with unexplained infertility, hysteroscopy combined with endometrial immunohistochemical detection of CD38 and CD138 should be performed in time to exclude CE. The clinical pregnancy outcome of CE patients can be significantly improved by antibiotic treatment.
### Risk of endometrial malignancy in women treated for breast cancer: the BLUSH prediction model–evidence from a comprehensive multicentric retrospective cohort study
2024, Climacteric
### Efficacy of Hysteroscopy in Improving Fertility Outcomes in Women Undergoing Assisted Reproductive Technique: A Systematic Review and Meta-Analysis of Randomized Controlled Trials
2024, Gynecologic and Obstetric Investigation
### Hysteroscopic versus histopathological agreement in the diagnosis of chronic endometritis: results from a retrospective observational study
2023, Archives of Gynecology and Obstetrics
View all citing articles on Scopus
© 2023 The Authors. Published by Elsevier Masson SAS.
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190362 | https://en.wikipedia.org/wiki/Jeff_Bezos | Jump to content
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Contents
1 Early life and education
2 Business career
2.1 Early career
2.2 Amazon
2.3 Blue Origin
2.3.1 Spaceflight
2.4 The Washington Post
2.5 Bezos Expeditions
2.6 Altos Labs
3 Public image
3.1 Leadership style
4 Recognition
5 Wealth
6 Criticism
7 Personal life
8 Politics
8.1 Donald Trump
8.2 Saudi hacking claim
9 Philanthropy
10 See also
11 Notes
12 References
13 Sources
14 Further reading
15 External links
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From Wikipedia, the free encyclopedia
American businessman (born 1964)
"Bezos" redirects here. For other people with the surname, see Bezos (surname).
| Jeff Bezos |
| Bezos in 2017 |
| Born | Jeffrey Preston Jorgensen (1964-01-12) January 12, 1964 (age 61) Albuquerque, New Mexico, U.S. |
| Education | Princeton University (BSE) |
| Occupations | Businessman media proprietor investor |
| Known for | Founding Amazon |
| Title | Founder and executive chairman of Amazon Founder of Blue Origin Owner of The Washington Post Founder of Bezos Expeditions Executive chair of Bezos Earth Fund Founder of Bezos Academy |
| Spouses | MacKenzie Scott (m. 1993; div. 2019) Lauren Sánchez (m. 2025) |
| Children | 4 |
| Parents | Jackie Bezos (mother) Ted Jorgensen (father) Miguel Bezos (stepfather) |
| Relatives | Mark Bezos (half-brother) |
| Signature |
Jeffrey Preston Bezos (/ˈbeɪzoʊs/ BAY-zohss; né Jorgensen; born January 12, 1964) is an American businessman best known as the founder, executive chairman, and former president and CEO of Amazon, the world's largest e-commerce and cloud computing company. According to Forbes, as of May 2025, Bezos's estimated net worth exceeded $220 billion, making him the third richest person in the world. He was the wealthiest person from 2017 to 2021, according to Forbes and the Bloomberg Billionaires Index.
Bezos was born in Albuquerque and raised in Houston and Miami. He graduated from Princeton University in 1986 with a degree in engineering. He worked on Wall Street in a variety of related fields from 1986 to early 1994. Bezos founded Amazon in mid-1994 on a road trip from New York City to Seattle. The company began as an online bookstore and has since expanded to a variety of other e-commerce products and services, including video and audio streaming, cloud computing, and artificial intelligence. It is the world's largest online sales company, the largest Internet company by revenue, and the largest provider of virtual assistants and cloud infrastructure services through its Amazon Web Services branch.
Bezos founded the aerospace manufacturer and sub-orbital spaceflight services company Blue Origin in 2000. Blue Origin's New Shepard vehicle reached space in 2015 and afterwards successfully landed back on Earth; he flew into space on Blue Origin NS-16 in 2021. He purchased the major American newspaper The Washington Post in 2013 for $250 million and manages many other investments through his venture capital firm, Bezos Expeditions. In September 2021, Bezos co-founded Altos Labs with Mail.ru founder Yuri Milner.
The first centibillionaire on the Forbes Real Time Billionaires Index and the second ever to have achieved the feat since Bill Gates in 1999, Bezos was named the "richest man in modern history" after his net worth increased to $150 billion in July 2018. In August 2020, according to Forbes, he had a net worth exceeding $200 billion. On July 5, 2021, Bezos stepped down as the CEO and president of Amazon and took over the role of executive chairman. Amazon Web Services CEO Andy Jassy succeeded Bezos as the CEO and president of Amazon.
Early life and education
Bezos was born Jeffrey Preston Jorgensen on January 12, 1964, in Albuquerque, New Mexico, to Jacklyn (née Gise) (1946–2025) and Ted Jorgensen (1944–2015). At the time of his birth, his mother was a 17-year-old high-school student and his father was 19. Ted was a Danish American unicyclist born in Chicago to a family of Baptists. After completing high school despite challenging conditions, Jacklyn attended night school, bringing her baby with her. Jeff attended a Montessori school in Albuquerque when he was two.
Ted struggled with alcohol and with his finances. Jacklyn left her husband to live with her parents, filing for divorce in June 1965 when Jeff was 17 months old. After his parents divorced, his mother married Cuban immigrant Miguel "Mike" Bezos in April 1968. Shortly after the wedding, Mike adopted 4-year-old Jeff, whose surname was then legally changed from Jorgensen to Bezos. Jacklyn, her husband, and her son left the area and asked Ted to discontinue contact, to which he agreed.
After Mike received his degree from the University of New Mexico, the family moved to Houston, Texas, so that he could begin working as an engineer for Exxon. Jeff attended River Oaks Elementary School in Houston from fourth to sixth grade. Jeff's maternal grandfather was Lawrence Preston Gise, a regional director of the U.S. Atomic Energy Commission (AEC) in Albuquerque.
Lawrence retired early to his family's ranch near Cotulla, Texas, where his grandson would spend many summers in his youth and which he would later purchase and expand from 25,000 acres (10,117 ha) to 300,000 acres (121,406 ha). Jeff displayed scientific interests and technological proficiency and once rigged an electric alarm to keep his younger half-siblings out of his room. The family moved to Miami, Florida, where Jeff attended Miami Palmetto High School. In high school, he worked at McDonald's as a short-order line cook during the breakfast shift.
Bezos attended the Student Science Training Program at the University of Florida. He was high school valedictorian, a National Merit Scholar, and a Silver Knight Award winner in 1982. In his graduation speech, Bezos told the audience that he dreamed of the day when mankind would colonize space. A local newspaper quoted his intention "to get all people off the earth and see it turned into a huge national park".
After graduating from high school in 1982, Bezos attended Princeton University. He initially majored in physics but later switched to electrical engineering and computer science. In 2018, during a talk at the Economic Club of Washington, D.C., Bezos revealed that, some thirty years ago, his Princeton classmate Yasantha Rajakarunanayake had defeated him in solving a mathematical problem, causing him to give up on his dreams of becoming a theoretical physicist.
Bezos was a member of the Quadrangle Club, one of Princeton's 11 eating clubs. Additionally, he was the president of the Princeton chapter of the Students for the Exploration and Development of Space (SEDS). He had a 4.2 GPA and was elected to Phi Beta Kappa and Tau Beta Pi. Bezos graduated from Princeton in 1986 with a Bachelor of Science in Engineering (BSE), summa cum laude.
Business career
Early career
After Bezos graduated from college in 1986, he was offered jobs at Intel, Bell Labs, and Andersen Consulting, among others. He first worked at Fitel, a fintech telecommunications start-up, where he was tasked with building a network for international trade. Bezos was promoted to head of development and director of customer service. He transitioned into the banking industry when he became a product manager at Bankers Trust from 1988 to 1990. From 1990 to 1994, he worked at D. E. Shaw & Co, a newly created hedge fund with a strong emphasis on mathematical modelling. Bezos became D. E. Shaw's fourth senior vice-president by age 30.
Amazon
Main article: Amazon
In spring 1994, Bezos read that web usage was growing at a rate of 2300% a year and eventually decided to establish an online bookstore. He and his then-wife, MacKenzie Scott, left their jobs at D. E. Shaw and founded Amazon in a rented garage in Bellevue, Washington on July 5, 1994, after writing its business plan on a cross-country drive from New York City to Seattle. With Bezos at the helm and Scott taking an integral role in its operation—writing checks, keeping track of the books, and negotiating the company's first freight contracts—the foundation was laid for this garage-run operation to grow exponentially. Prior to settling in Seattle, Bezos had investigated setting up his company at an Indian reservation near San Francisco in order to avoid paying taxes. Bezos initially named his new company Cadabra but later changed the name to Amazon after the Amazon River in South America, in part because the name begins with the letter A, which is at the beginning of the alphabet. At the time, website listings were alphabetized, so a name starting with "A" would appear sooner when customers conducted online searches. In addition, he regarded "Amazon," the name of the world's largest river as fitting for what he hoped would become the world's largest online bookstore. He accepted an estimated $300,000 from his parents as an investment in Amazon. He warned many early investors that there was a 70% chance that Amazon would fail or go bankrupt. Although Amazon was originally an online bookstore, Bezos had always planned to expand to other products. Three years after Bezos founded Amazon, he took it public with an initial public offering (IPO). In response to critical reports from Fortune and Barron's, Bezos maintained that the growth of the Internet would overtake competition from larger book retailers such as Borders and Barnes & Noble.
In 1998, Bezos diversified into the online sale of music and video, and by the end of the year he had expanded the company's products to include a variety of other consumer goods. Bezos used the $54 million raised during the company's 1997 equity offering to finance the aggressive acquisition of smaller competitors. Among these acquisitions were his purchase of a majority stake in pets.com in 1999 and a purchase of a portion of kozmo.com for $60 million, both of which would fail after the dot-com bubble collapse in 2000. By the end of 2000, Bezos borrowed $2 billion from banks, as its cash balances dipped to only $350 million. However, the company continued to expand despite its losses, and in 2002, Bezos led Amazon to launch Amazon Web Services, which compiled data from weather channels and website traffic. Revenues stagnated later that year, and after the company nearly went bankrupt, he closed distribution centers and laid off 14% of the Amazon workforce. In 2003, Amazon rebounded from financial instability and turned a profit of $35 million.
In November 2007, Bezos launched the Amazon Kindle. According to a 2008 Time profile, Bezos wished to create a device that allowed a "flow state" in reading similar to the experience of video games. In 2013, Bezos secured a $600-million contract with the Central Intelligence Agency (CIA) on behalf of Amazon Web Services. In October of that year, Amazon was recognized as the largest online shopping retailer in the world.
In May 2016, Bezos sold slightly more than one million shares of his holdings in the company for $671 million, the largest sum he had ever raised from selling some of his Amazon stock. On August 4, 2016, Bezos sold another million of his shares for $756.7 million. A year later, Bezos took on 130,000 new employees when he ramped up hiring at company distribution centers. By January 19, 2018, his Amazon stock holdings had appreciated to slightly over $109 billion; months later he began to sell stock to raise cash for other enterprises, in particular, Blue Origin. On January 29, 2018, he was featured in Amazon's Super Bowl commercial. On February 1, 2018, Amazon reported its highest ever profit with quarterly earnings of $2 billion. Due to the growth of Alibaba in China, Bezos has often expressed interest in expanding Amazon into India. On July 27, 2017, Bezos momentarily became the world's wealthiest person over Microsoft co-founder Bill Gates when his estimated net worth increased to just over $90 billion. His wealth surpassed $100 billion for the first time on November 24, 2017, and he was formally designated the wealthiest person in the world by Forbes on March 6, 2018, with a net worth of $112 billion.
In March 2018, Bezos dispatched Amit Agarwal, Amazon's global senior vice president, to India with $5.5 billion to localize operations throughout the company's supply chain routes. Later in the month, U.S. President Donald Trump accused Amazon and Bezos, specifically, of sales tax avoidance, misusing postal routes, and anti-competitive business practices. Amazon's share price fell by 9% in response to the President's negative comments; this reduced Bezos's personal wealth by $10.7 billion. Weeks later, Bezos recouped his losses when academic reports out of Stanford University indicated that Trump could do little to regulate Amazon in any meaningful way. During July 2018, a number of members of the U.S. Congress called on Bezos to detail the applications of Amazon's face recognition software, Rekognition.
Criticism of Amazon's business practices continued in September 2018 when Senator Bernie Sanders introduced the Stop Bad Employers by Zeroing Out Subsidies (Stop BEZOS) Act and accused Amazon of receiving corporate welfare. This followed revelations by the non-profit group New Food Economy which found that one third of Amazon workers in Arizona, and one tenth of Amazon workers in Pennsylvania and Ohio, relied on food stamps. While preparing to introduce the bill, Sanders opined: "Instead of attempting to explore Mars or go to the moon, how about Jeff Bezos pays his workers a living wage?" He later said: "Bezos could play a profound role. If he said today, nobody who is employed at Amazon will receive less than a living wage, it would send a message to every corporation in America." Sanders's efforts elicited a response from Amazon which pointed to the 130,000 jobs it created in 2017 and called the $28,446 figure for its median salary "misleading" as it included part-time workers. However, Sanders countered that the companies targeted by his proposal have placed an increased focus on part-time workers to escape benefit obligations. On October 2, 2018, Bezos announced a company-wide wage increase, which Sanders applauded. The American workers who were being paid the minimum wage had this increased to $15 per hour, a decision that was interpreted as support for the Fight for $15 movement.
In February 2021, Bezos announced that in the third quarter of 2021 he would step down from his role as CEO of Amazon to become the Executive Chairman of the Amazon Board. He was succeeded as CEO by Andy Jassy. On February 2, 2021, Bezos sent an email to all Amazon employees, telling them the transition would give him "the time and energy [he] need[s] to focus on the Day 1 Fund, the Bezos Earth Fund, Blue Origin, The Washington Post, and [his] other passions." In February 2024, Bezos sold 24 million shares in Amazon at a total value of $4 billion. Bezos announced that he intended to sell 50 million shares in Amazon over the next year. During an interview at the DealBook Summit in December 2024, Bezos said that he was dedicating 95% of his time to artificial intelligence initiatives at Amazon.
Blue Origin
Main article: Blue Origin
In September 2000, Bezos founded Blue Origin, a human spaceflight startup. Bezos has long expressed an interest in space travel and the development of human life in the Solar System. His 1982 high school valedictorian senior graduation speech was followed up with a Miami Herald interview in which he expressed an interest to build and develop hotels, amusement parks, and colonies for human beings who were in orbit. The 18-year-old Bezos stated that he wanted to preserve Earth from overuse through resource depletion. Rob Meyerson led Blue Origin from 2003 to 2017 and served as its first president.
After its founding, Blue Origin maintained a low profile until 2006 when it purchased a large tract of land in West Texas for a launch and test facility. After the company gained the public's attention during the late 2000s, Bezos additionally indicated his interest in reducing the cost of space travel for humans while also increasing the safety of extraterrestrial travel. In September 2011, one of the company's uncrewed prototype vehicles crashed during a short-hop test flight. Although the crash was viewed as a setback, news outlets noted how far the company went from its founding-to-date in advancing spaceflight. After the crash, Bezos has been superstitiously wearing his "lucky" Texas Cowboy boots to all rocket launches. In May 2013, Bezos met with Richard Branson, chairman of Virgin Galactic, to discuss commercial spaceflight opportunities and strategies. He has been compared to Branson and Elon Musk as all three are billionaires who prioritize spaceflight among their business interests.
In 2015, Bezos announced that a new orbital launch vehicle was under development and would make its first flight in the late-2010s. Later in November, Blue Origin's New Shepard space vehicle successfully rocketed into space and reached its planned test altitude of 329,839 feet (100.5 kilometers) before executing a vertical landing back at the launch site in West Texas. In 2016, Bezos allowed select journalists to visit, tour, and photograph his facility. He has repeatedly called for increased inter-space energy and industrial manufacturing to decrease the negative costs associated with business-related pollution.
In December 2017, New Shepard successfully flew and landed dummy passengers, amending and pushing its human space travel start date into late 2018. To execute this program, Blue Origin built six of the vehicles to support all phases of testing and operations: no-passenger test flights, flights with test passengers, and commercial-passenger weekly operations. Since 2016, Bezos has spoken more freely about his hopes to colonize the solar system, and has been selling $1 billion in Amazon stock each year to capitalize Blue Origin in an effort to support this endeavor. In May 2018, Bezos maintained that the primary goal of Blue Origin is to preserve the natural resources of Earth by making the human species multi-planetary. He announced that New Shepard would begin transporting humans into sub-orbital space by November 2018. In July 2018, it was announced that Bezos had priced commercial spaceflight tickets from $200,000 to $300,000 per person.
Spaceflight
| Jeff Bezos |
| Space career |
| Commercial Astronaut |
| Flight time | 10m 18s |
| Selection | Blue Origin |
| Missions | NS-16 |
| |
On July 20, 2021, he launched on the NS-16 mission with his half-brother Mark Bezos, Wally Funk, and Oliver Daemen. He launched nine days after Richard Branson launched on board the Virgin Galactic Unity 22 mission. Bezos's suborbital flight lasted over 10 minutes, reaching a peak altitude of 66.5 miles (107.0 km).
The Washington Post
See also: The Washington Post
On August 5, 2013, Bezos announced his purchase of The Washington Post for $250 million in cash, at the suggestion of his friend, Don Graham. To execute the purchase, he established limited liability company Nash Holdings to serve as a holding company through which he would own the newspaper. The sale closed on October 1, 2013, and Nash Holdings took control. In March 2014, Bezos made his first significant change at The Washington Post and lifted the online paywall for subscribers of a number of U.S. local newspapers in Texas, Hawaii, and Minnesota. In January 2016, Bezos set out to reinvent the newspaper as a media and technology company by reconstructing its digital media, mobile platforms, and analytics software. After a surge in online readership in 2016, the paper was profitable for the first time since Bezos made the purchase in 2013. However, Bezos' ownership of the Post has been subject to scrutiny and criticism regarding his treatment of employees as well as his influence on the paper's content, in particular 2024-25 interference with the editorial and opinion pages.
Bezos Expeditions
Main article: Bezos Expeditions
Bezos makes personal investments through his venture capital vehicle, Bezos Expeditions. He was one of the first shareholders in Google, when he invested $250,000 in 1998. That $250,000 investment resulted in 3.3 million shares of Google stock, worth about $3.1 billion in 2017. He also invested in Unity Biotechnology, a life-extension research firm hoping to slow or stop the process of aging. Bezos is involved in the healthcare sector, which includes investments in Unity Biotechnology, GRAIL, Juno Therapeutics, and Zocdoc. In January 2018, an announcement was made concerning Bezos's role within a new, unnamed healthcare company. This venture, later named Haven, is expected to be a partnership between Amazon, JPMorgan, and Berkshire Hathaway.
Bezos also supports philanthropic efforts through direct donations and non-profit projects funded by Bezos Expeditions. Bezos used Bezos Expeditions to fund several philanthropic projects, including an Innovation center at the Seattle Museum of History and Industry and the Bezos Center for Neural Circuit Dynamics at Princeton Neuroscience Institute. In 2013, Bezos Expeditions funded the recovery of two Saturn V first-stage Rocketdyne F-1 engines from the floor of the Atlantic Ocean. They were positively identified as belonging to the Apollo 11 mission's S-1C stage from July 1969. The engines are currently on display at the Seattle Museum of Flight.
Altos Labs
Main article: Altos Labs
In September 2021, Bezos co-founded Altos Labs with Mail.ru founder Yuri Milner. Altos Labs is a well-funded biotechnology company dedicated to harnessing cellular reprogramming to develop longevity therapeutics. The company has recruited prominent scientists such as Juan Carlos Izpisúa Belmonte (known for work on rejuvenation through reprogramming), Steve Horvath (known for work in epigenetic aging clocks), and Shinya Yamanaka (the Nobel Prize-winning inventor of cellular reprogramming in mammalian cells). The company left stealth mode and launched on January 19, 2022, with a start capital of $3 billion and an executive team led by Hal Barron.
Public image
Journalist Nellie Bowles of The New York Times has described the public persona and personality of Bezos as that of "a brilliant but mysterious and coldblooded corporate titan". During the 1990s, Bezos earned a reputation for relentlessly pushing Amazon forward, often at the expense of public charity and social welfare. Journalist Mark O'Connell criticized Bezos's relentless customer focus as "very small" in terms of impact on humanity as a whole, a sentiment technologist Tim O'Reilly agreed with. His business practices projected a public image of prudence and parsimony with his own wealth and that of Amazon. In 1999, Bezos was worth $10 billion yet drove a 1996 Honda Accord. Throughout the early 2000s, he was perceived to be geeky or nerdy.
Bezos was seen by some as needlessly quantitative and data-driven. This perception was detailed by Alan Deutschman, who described him as "talking in lists" and "[enumerating] the criteria, in order of importance, for every decision he has made". Select accounts of his persona have drawn controversy and public attention. Notably, journalist Brad Stone wrote a book that described Bezos as a demanding boss as well as hyper-competitive, and opined that Bezos perhaps "bet the biggest on the Internet" out of anyone. Bezos has been characterized as a notoriously opportunistic CEO who operates with little concern for obstacles and externalities.
During the early 2010s, Bezos solidified his reputation for aggressive business practices, and his public image began to shift. Bezos started to wear tailored clothing; he weight trained, pursued a regimented diet and began to freely spend his money. His physical transformation has been compared to the transformation of Amazon; he is often referred to as the metonym of the company. Since 2017, he has been portrayed by Kyle Mooney and Steve Carell on Saturday Night Live, usually as an undercutting, domineering figure. His physical appearance increased the public's perception of him as a symbolically dominant figure in business and in popular culture, wherein he has been parodied as an enterprising supervillain.
In May 2014, the International Trade Union Confederation named Bezos the "World's Worst Boss", with its general secretary Sharan Burrow saying: "Jeff Bezos represents the inhumanity of employers who are promoting the North American corporate model", while in 2019, Harvard Business Review, which ranked Bezos the best-performing CEO for 4 years in a row since 2014, did not rank him even in the top 100, citing Amazon's "relatively low ESG (environment, social, and governance) scores" that reflect "risks created by working conditions and employment policies, data security, and antitrust issues".
During the late 2010s, Bezos reversed his reputation for being reluctant to spend money on non-business-related expenses. His relative lack of philanthropy compared to other billionaires has drawn a negative response from the public since 2016. Bezos has been known to publicly contest claims made in critical articles, as exemplified in 2015 when he sent a memo to employees denouncing a New York Times piece.
Leadership style
"Day 1" Management Philosophy
Day 1: start upDay 2: stasisDay 3: irrelevanceDay 4: "excruciating, painful decline"Day 5: death
Bezos has stated "it is always Day 1" to describe his growth mindset.
Bezos used what he called a "regret-minimization framework" while working at D. E. Shaw and again during the early years of Amazon. He described this life philosophy by stating: "When I'm 80, am I going to regret leaving Wall Street? No. Will I regret missing the beginning of the Internet? Yes." During the 1990s and early 2000s at Amazon, he was characterized as trying to quantify all aspects of running the company, often listing employees on spreadsheets and basing executive decisions on data. To push Amazon forward, Bezos developed the mantra "Get Big Fast", establishing the company's need to scale its operations to produce market dominance. He favored diverting Amazon profits back into the company in lieu of allocating it amongst shareholders in the form of dividends.
Bezos uses the term "work–life harmony" instead of the more standard "work–life balance" because he believes that balance implies that you can only have one and not the other. He believes that work and home life are interconnected, informing and calibrating each other. Journalist Walt Mossberg dubbed the idea that someone who cannot tolerate criticism or critique should not do anything new or interesting "The Bezos Principle". Bezos does not schedule early morning meetings and enforces a two-pizza rule—a preference that meetings are small enough for two pizzas to feed everyone in the boardroom. When interviewing candidates for jobs at Amazon, he has stated he considers three inquiries: can he admire the person, can the person raise the common standard, and under what circumstances could the person become exemplary.
In 2018, it was reported that he met with Amazon investors for just six hours a year. Instead of using presentation slides, Bezos requires high-level employees to present information with six-page narratives. Since 1998, Bezos has published an annual letter for Amazon shareholders wherein he frequently refers to five principles: focus on customers, not competitors; take risks for market leadership; facilitate staff morale; build a company culture; and empower people. Bezos maintains the email address jeff@amazon.com as an outlet for customers to reach out to him and the company. Although he does not respond to the emails, he forwards some of them with a question mark in the subject line to executives, who then attempt to address the issues. Bezos has cited Jeff Immelt of New Enterprise Associates, Warren Buffett of Berkshire Hathaway, Jamie Dimon of JPMorgan Chase, and Bob Iger of The Walt Disney Company as major influences on his leadership style.
Recognition
In 1999, Bezos received his first major award when Time named him Person of the Year.
In 2008, he was selected by U.S. News & World Report as one of America's best leaders.
Bezos was awarded an honorary doctorate in science and technology from Carnegie Mellon University in 2008.
In 2011, The Economist gave Bezos and Gregg Zehr an Innovation Award for the Amazon Kindle.
In 2012, Bezos was named Businessperson of the Year by Fortune.
He is also a member of the Bilderberg Group and attended the 2011 Bilderberg conference in St. Moritz, Switzerland, and the 2013 conference in Watford, Hertfordshire, England. He was a member of the executive committee of The Business Council for 2011 and 2012, and appointed as chairman of the organization in 2014.
Between 2014 and 2018, he was ranked the best-performing CEO in the world by Harvard Business Review.
He has also figured in Fortune's list of 50 great leaders of the world for three straight years, topping the list in 2015.
In September 2016, Bezos received a $250,000 prize for winning the Heinlein Prize for Advances in Space Commercialization, which he donated to the Students for the Exploration and Development of Space.
In February 2018, Bezos was elected to the National Academy of Engineering for "leadership and innovation in space exploration, autonomous systems, and building a commercial pathway for human space flight".
In March 2018, at the Explorers Club annual dinner, he was awarded the Buzz Aldrin Space Exploration Award in recognition of his work with Blue Origin.
He received Germany's 2018 Axel Springer Award for Business Innovation and Social Responsibility. Time magazine named him one of the 100 most influential people in the world on five separate occasions between 2008 and 2018.
In 2019, Bezos was inducted into the Living Legends of Aviation, being awarded with the Jeff Bezos Freedom's Wings Award and the Kenn Ricci Lifetime Aviation Entrepreneur Award.
In February 2023, Bezos was presented with the Légion d'honneur, the highest French order of merit. Bezos had been designated a member of the Légion d'Honneur about 10 years earlier but was not available to collect it.
Wealth
Annual estimates of Jeff Bezos's net worth[a]
| Year | Billions | Change |
| 1999 | 10.1 | 0.0% |
| 2000 | 6.1 | 40.5% |
| 2001 | 2.0 | 66.6% |
| 2002 | 1.5 | 25.0% |
| 2003 | 2.5 | 66.6% |
| 2004 | 5.1 | 104% |
| 2005 | 4.1 | 19.6% |
| 2006 | 4.3 | 5.1% |
| 2007 | 8.7 | 102.3% |
| 2008 | 8.2 | 5.7% |
| 2009 | 6.8 | 17.1% |
| 2010 | 12.6 | 85.3% |
| 2011 | 18.1 | 43.7% |
| 2012 | 23.2 | 28.2% |
| 2013 | 28.9 | 24.5% |
| 2014 | 30.5 | 5.5% |
| 2015 | 50.3 | 60.9% |
| 2016 | 45.2 | 10.1% |
| 2017 | 72.8 | 61.6% |
| 2018 | 112.0 | 53.8% |
Bezos first became a millionaire in 1997 after raising $54 million through Amazon's initial public offering (IPO). He was first included on the Forbes World's Billionaires list in 1999 with an estimated net worth of $10.1 billion, which placed his on the 19th position in the world and 10th in the USA. His net worth decreased to $6.1 billion a year later, a 40.5% drop. His wealth plummeted even more the following year, dropping 66.6% to $2.0 billion. He lost $500 million the following year, which brought his net worth down to $1.5 billion. The following year, his net worth increased by 66.66% to $2.5 billion. From 2005 to 2007, he quadrupled his net worth to $8.7 billion. After the 2008 financial crisis and Great Recession, his net worth would decrease to $6.8 billion—a 17.7% drop. His wealth rose by 85.2% in 2010, leaving him with $12.6 billion. This percentage increase ascended him to the 43rd spot on the ranking from 68th.
After a rumor broke out that Amazon was developing a smartphone, Bezos's net worth rose to $30.5 billion in 2014. A year later, he entered the top ten when he increased his net worth to a total of $50.3 billion. Bezos rose to become the fifth richest person in the world hours before market close; he gained $7 billion in one hour. By the time the Forbes list was calculated in March 2016, his net worth was registered at $45.2 billion. However, just months later in October 2016, his wealth increased by $16.2 billion to $66.5 billion, unofficially ranking him the third-richest person in the world, behind Warren Buffett. After sporadic jumps in Amazon's share price, in July 2017 he briefly unseated Microsoft co-founder Bill Gates as the wealthiest person in the world.
Bezos would continue to sporadically surpass Gates throughout the month of October 2017 after Amazon's share price fluctuated. His net worth surpassed $100 billion for the first time on November 24, 2017, after Amazon's share price increased by more than 2.5%. When the 2017 list was issued, Bezos's net worth was registered at $72.8 billion, adding $27.6 billion from the previous year. His wealth's rapid growth from 2016 to 2017 sparked a variety of assessments about how much money Bezos earned on a controlled, reduced time scale. On October 10, 2017, he made an estimated $6.24 billion in 5 minutes, slightly less than the then annual gross domestic product of Kyrgyzstan.
On March 6, 2018, Bezos was designated the wealthiest person in the world, with a registered net worth of $112 billion. He unseated Bill Gates ($90 billion), who was $6 billion ahead of Warren Buffett ($84 billion), ranked third. He is considered the first registered centi-billionaire (not adjusted for inflation).[b]
His wealth, in 2017–18 terms, equaled that of 2.7 million Americans. Bezos's net worth increased by $33.6 billion from January 2017 to January 2018. This increase outstripped the economic development (in GDP terms) of more than 96 countries around the world. During March 9, Bezos earned $230,000 every 60 seconds. The Motley Fool estimated that if Bezos had not sold any of his shares from its original public offering in 1997, his net worth would sit at $181 billion in 2018. According to Quartz, his net worth of $150 billion in July 2018 was enough to purchase the entire stock markets of Nigeria, Hungary, Egypt, Luxembourg, and Iran. Following the report by Quartz, Amazon workers in Poland, (Germany), and Spain participated in demonstrations and labor strikes to draw attention to his growing wealth and the lack of compensation, labor rights, and satisfactory working conditions of select Amazon workers. On July 17, 2018, he was designated the "wealthiest person in modern history"[c] by the Bloomberg Billionaires Index, Fortune, MarketWatch, The Wall Street Journal, and Forbes.
In 2019, Bezos's wealth was reduced by the divorce from his wife MacKenzie Bezos. According to Forbes, had the Washington state common law applied to their divorce without a prenuptial agreement, Bezos's wealth could have been equitably divided with his ex-wife; however, she eventually received 25% of Bezos's Amazon shares, then valued at approximately $36 billion, making her the third-richest woman in the world. Bezos retained his interest in The Washington Post and Blue Origin, as well as voting control of the shares received by his ex-wife.
In June 2019, Bezos purchased three adjoining apartments overlooking Madison Square Park in Manhattan, including a penthouse, for a combined total of $80 million, making this one of the most expensive real estate purchases within New York City in 2019. Bezos had also purchased three adjoining apartments at 25 Central Park West in Manhattan for $7.65 million in 1999; he bought a fourth unit in that building for $5.3 million in 2012.
In February 2020, Bezos purchased the Warner Estate from David Geffen for $165 million, a record price paid for a residence in the Los Angeles area. The previous record high price of $150 million was paid by Lachlan Murdoch for the Chartwell Mansion. During the COVID-19 pandemic, it was reported that Bezos's fortune had grown by $24 billion, citing a surge in demand from households on lockdown shopping on Amazon. He further expanded his residential holdings in February 2022, purchasing a $16.13-million-dollar apartment at a 24-story boutique condominium, located across from Madison Square Park in the Flatiron neighbourhood, where he already owns all the units on the top floors. Bezos is the owner of the Y721, a luxury superyacht estimated to cost more than $500,000,000; it is the largest yacht in the world. According to Forbes Bezos was the second-wealthiest person in America and the third-wealthiest person in the world in 2023. Bezos is the second-wealthiest person in the world according to Bloomberg Billionaires Index. His net worth is about $197 billion as of February 2024.
Criticism
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Bezos is known for creating an adversarial environment at Amazon, as well as insulting and verbally abusing his employees. As journalist Brad Stone revealed in his book The Everything Store, Bezos issued remarks to his employees such as "I'm sorry, did I take my stupid pills today?", "Are you lazy or just incompetent?", and "Why are you ruining my life?" Additionally, Bezos reportedly pitted Amazon teams against each other, and once refused to give Amazon employees city bus passes in order to discourage them from leaving the office.
Throughout his early years of ownership of The Washington Post, Bezos was accused of having a potential conflict of interest with the paper. Bezos and the newspaper's editorial board have dismissed accusations that he unfairly controlled the paper's content, and Bezos maintains that the paper is independent. Bezos' treatment of employees at The Washington Post has also drawn scrutiny. In 2018, more than 400 Washington Post employees wrote an open letter to Bezos criticizing his poor wages and benefits for his employees. The letter demanded "Fair wages; fair benefits for retirement, family leave and health care; and a fair amount of job security". Around 750 employees at The Washington Post went on a brief strike in December 2023 in response to Bezos' plans to lay off staff.
In 2024, Bezos blocked the Washington Post's editorial board from endorsing Kamala Harris in the presidential election. The move was criticized by former editor Marty Baron, who considered it to be an act of "disturbing spinelessness at an institution famed for courage" and said that it would invite intimidation of Bezos by Donald Trump. Editor-at-large Robert Kagan and columnist Michele Norris also resigned in the wake of the decision, and editor David Maraniss said that the paper was "dying in darkness". Post opinion columnists jointly authored an article calling the decision to not endorse a "terrible mistake", and it was condemned by the Washington Post Guild, a union unit representing Post employees. More than 250,000 people (about ten percent of the Post's subscribers) cancelled their subscriptions, and three members of the editorial board left the board. Condemning the Post's decision, several columnists, including Will Bunch, Jonathan Last, Dan Froomkin, Donna Ladd and Sewell Chan, described it as an example of what historian Timothy Snyder calls anticipatory obedience. Snyder, too, condemned the decision.
In January 2025, editorial cartoonist Ann Telnaes resigned from the Post after it refused to run a satirical cartoon critical of the relationship between American billionaires and President Donald Trump, sparking conversations about the paper's ownership under Bezos; Telnaes called the decision "dangerous for a free press". In February 2025, Bezos announced that the opinion section of the Post will give voice only to opinions that support "personal liberties" and "free markets", and that divergent opinions will not be published by the Post. David Shipley, The Post's opinion editor, resigned after trying to persuade Bezos to reconsider the new direction. Within two days of the announcement, it was reported that over 75,000 digital subscribers had canceled their subscriptions.
Due to his considerable influence on industry, politics, and media, Bezos has been described as an oligarch.
Personal life
See also: Family of Jeff Bezos
In 1992, while working for D. E. Shaw in Manhattan, Bezos met novelist MacKenzie Tuttle, who was a research associate at the firm; the couple married a year later. In 1994, they moved across the country to Seattle, Washington, where Bezos founded Amazon. Bezos and his now ex-wife MacKenzie are the parents of four children: three sons, and a daughter adopted from China.
In March 2003, Bezos was a passenger in a helicopter that crashed in West Texas while surveying land to buy for Blue Origin; the other three occupants in the helicopter were pilot Charles "Cheater" Bella, Amazon lawyer Elizabeth Korrell, and local rancher Ty Holland. All survived; Bezos sustained only minor injuries and was discharged from a local hospital the same day.
Bezos portrayed a Starfleet official in the 2016 movie Star Trek Beyond, and joined the cast and crew at a San Diego Comic-Con screening. He had lobbied Paramount for the role apropos of Alexa and his personal/professional interest in speech recognition. His one line consisted of a response to an alien in distress: "Speak Normally." In his initial discussion of the project which became Alexa with his technical advisor Greg Hart in 2011, Bezos told him that the goal was to create "the Star Trek computer." Bezos's family corporation Zefram LLC is named after Zefram Cochrane, a character from Star Trek.
In January 2019, Bezos and his wife MacKenzie released a joint statement which revealed that they would be getting divorced after 25 years together. Subsequently, National Enquirer revealed that Bezos had an affair with media personality Lauren Sánchez; the affair with Sánchez had lasted for months. Later, Bezos published an online essay on February 7, 2019, in which he accused American Media, Inc. owner David Pecker of "extortion and blackmail" for threatening to publish intimate photos of Bezos and current girlfriend Lauren Sánchez if he did not stop his investigation into how his text messages and other photos had been leaked to the National Enquirer. Media reports have accused Sánchez's brother Michael of being the source for the photos obtained by National Enquirer; however, Bezos has speculated that it may have been the Saudi Arabian government.
On April 4, 2019, the divorce was finalized, with Bezos keeping 75% of the couple's Amazon stock and MacKenzie getting the remaining 25% ($35.6 billion). However, Bezos would keep all of the couple's voting rights. Sánchez and Bezos became engaged in May 2023. The couple married in Venice on June 27, 2025, with the ceremony attracting mainstream media attention and various celebrities.
Bezos is the Honorary Chair of the Explorers Club.
Politics
According to public campaign finance records, Bezos supported the electoral campaigns of Patty Murray and Maria Cantwell, two Democratic U.S. senators from Washington. He has also supported Democrats U.S. representative John Conyers, as well as Patrick Leahy and Republican Spencer Abraham, U.S. senators serving on committees dealing with Internet-related issues. Jeff Bezos and MacKenzie Bezos have supported the legalization of same-sex marriage, and in 2012 contributed $2.5 million to Washington United for Marriage, a group supporting a yes vote on Washington Referendum 74, which affirmed a same-sex marriage law enacted in the state. Bezos donated $100,000 towards a movement against a Washington state income tax in 2010 for "top earners". In 2012, he donated to Amazon's political action committee (PAC), which has given $56,000 and $74,500 to Democrats and Republicans, respectively.
In 2014, Amazon won a bid for a cloud computing contract with the CIA valued at $600 million. A 2018, $10 billion contract known as the Joint Enterprise Defense Infrastructure (JEDI) project, this time with the Pentagon, was allegedly written up in a way that favors Amazon. Controversy over this was raised when General James Mattis accepted a headquarters tour invitation from Bezos and co-ordinated the deal through Sally Donnelly, a lobbyist who previously worked for Amazon. In November 2019, when the contract was awarded to Microsoft instead, Amazon filed a lawsuit with allegations that the bidding process was biased. On July 6, 2021, the Pentagon cancelled the JEDI contract with Microsoft, citing that "due to evolving requirements, increased cloud conversancy, and industry advances, the JEDI Cloud contract no longer meets its needs." Despite Bezos's support for an open borders policy towards immigrants, Amazon has actively marketed facial recognition software to U.S. Immigration and Customs Enforcement (ICE).
In 2019, a PAC linked to Bezos spent over $1 million in an unsuccessful attempt to defeat the reelection bid of Seattle city council member and activist Kshama Sawant. On November 22, 2021, Jeff Bezos donated $100 million to the Obama Foundation to "help expand the scope of programming that reaches emerging leaders", and requested the Obama Presidential Center's plaza to be named after John Lewis.
Donald Trump
After the 2016 presidential election, Bezos was invited to join Donald Trump's Defense Innovation Advisory Board, an advisory council to improve the technology used by the Defense Department. Trump has repeatedly criticized Bezos via Twitter, accused Bezos of avoiding corporate taxes, gaining undue political influence, and undermining his presidency by spreading fake news. Nevertheless, Bezos congratulated Trump on his second election victory, posting on X, “Big congratulations to our 45th and now 47th President on an extraordinary political comeback and decisive victory. No nation has bigger opportunities. Wishing Donald Trump all success in leading and uniting the America we all love."
Since 2023, Bezos has been a resident of Indian Creek, Florida, which is near Trump's Mar-a-Lago. As reported by Axios in February 2025, Bezos held a private phone conversation in July 2024 with then-candidate Trump, planting the seeds of a "Bezos–Trump alliance" months before Bezos blocked the Washington Post's editorial board from endorsing Kamala Harris in the election. After Trump's November victory, Bezos dined with Elon Musk and Trump at Mar-a-Lago; Amazon subsequently donated $1 million to Trump's inauguration, at which Bezos was in attendance. Bezos and Trump were reported to have met for dinner again in February 2025, on the same night that Bezos announced changes to the Washington Post's opinion policies to promote "free markets and personal liberties" and suppress divergent opinions. Axios characterized it as "another sign of Trump and Bezos' growing closeness".
According to the Financial Times, Bezos had a contentious relationship with Trump during Trump's first term, but worked to have a positive relationship with Trump in 2024 and during Trump's second term. Bezos reportedly supported Trump to further his business interests, and supports many of Trump's policies. The Financial Times also noted that Bezos had made other changes in his life, including stepping down as CEO of Amazon in 2021, focusing on Blue Origin, and being engaged to Lauren Sánchez, which may have changed his political views.
Saudi hacking claim
Main article: Jeff Bezos phone hacking incident
In March 2018, Bezos met in Seattle with Mohammad bin Salman, the crown prince and de facto ruler of Saudi Arabia, to discuss investment opportunities for Saudi Vision 2030. In March 2019, Bezos's security consultant accused the Saudi government of hacking Bezos's phone. According to BBC, Bezos's top security staffer, Gavin de Becker, "linked the hack to the Washington Post's coverage of the murder of Saudi writer Jamal Khashoggi at the Saudi consulate in Istanbul". Khashoggi, a Saudi journalist and dissident, was employed as a writer at the Washington Post, owned by Bezos. Khashoggi was killed in late 2018 in Turkey's Saudi consulate for his critical stance and journalism against the Saudi government and its leader. In January 2020, The Guardian reported that the hack was initiated before the murder but after Khashoggi wrote critically about the crown prince in the Washington Post. Forensic analysis of Bezos's mobile phone conducted by advisory firm FTI Consulting, concluded it "highly probable" that the hack was achieved using a malicious file hidden in a video sent in a WhatsApp message to Bezos from the personal account of the crown prince on May 1, 2018. Saudi Arabia has denied the claim.
Philanthropy
Bezos donated to the Fred Hutchinson Cancer Research Center several times between 2009 and 2017. In 2013, he pledged $500,000 to Worldreader, a non-profit founded by a former Amazon employee. In September 2018, Business Insider reported that Bezos was the only one of the top five billionaires in the world who had not signed the Giving Pledge, an initiative created by Bill Gates and Warren Buffett that encourages wealthy people to give away a majority of their wealth. That same month, Janet Camarena, director of transparency initiatives at Foundation Center, was quoted by CNBC as having questions about Bezos's new Day 1 Fund, including the fund's structure and how exactly it will be funded.
In May 2017, Bezos gave $1 million to the Reporters Committee for Freedom of the Press, which provides pro bono legal services for American journalists. On June 15, 2017, he posted a message on Twitter asking for ideas for philanthropy: "I'm thinking about a philanthropy strategy that is the opposite of how I mostly spend my time—working on the long term". At the time of the post, Bezos's lifetime spending on charitable causes was estimated to be $100 million. Multiple opinion columnists responded by asking Bezos to pay higher wages to Amazon warehouse workers. A year later in June, he tweeted that he would announce two philanthropic foci by the end of summer 2018. Bezos announced in September 2018 that he would commit approximately $2 billion to a fund to deal with American homelessness and establish a network of non-profit preschools for low income communities. As part of this announcement, he committed to establishing the "Day 1 Families Fund" to finance "night shelters and day care centers for homeless families" and the "Day 1 Academies Fund" for early childhood education.
In January 2018, Bezos made a $33 million donation to TheDream.US, a college scholarship fund for undocumented immigrants brought to the United States as minors. In June 2018, Bezos donated to Breakthrough Energy Ventures, a private philanthropic fund founded by Bill Gates aimed at promoting emissions-free energy. In September 2018, Bezos donated $10 million to With Honor, a nonpartisan organization that works to increase the number of veterans in political office.
In February 2020, Bezos pledged $10 billion to combat climate change through the Bezos Earth Fund. Later that year, in November, Bezos announced $791M of donations to established, well-known groups, with $100M each going to Environmental Defense Fund, Natural Resources Defense Council, The Nature Conservancy, World Resources Institute and World Wildlife Fund, and the remainder going to 11 other groups. In April 2020, early in the COVID-19 pandemic, Bezos donated $100 million to food banks through Feeding America. In November 2021, Bezos pledged to donate $2 billion towards restructuring food systems and nature conservation at the 2021 United Nations Climate Change Conference.
In July 2021, Bezos announced the Courage and Civility Award and donated $100 million each to lawyer Van Jones and chef José Andrés. The next year, he donated $100 million to singer Dolly Parton in recognition of her charity work focused on improving children's literacy around the world. In March 2024, he donated $50 million each to actress Eva Longoria and retired admiral Bill McRaven.
Bezos Academy is a group of tuition-free preschools for students from low-income families, which was created by Bezos, and which operate in a manner similar to the Montessori method (but are not accredited as Montessori schools). On November 22, 2022, Bezos awarded $123 million to organizations that are engaged in relocating homeless families to permanent housing. Day 1 Families Fund grants, the amounts of which vary in monetary terms, will be sent to 40 organizations across the country.
See also
List of Princeton University alumni
List of richest Americans in history
List of Time Person of the Year recipients
The World's Billionaires
Notes
^ All currency figures expressed in the United States dollar (US$) in nominal terms.
^ Although Bill Gates momentarily surpassed the $100 billion net worth mark in April 1999 before the Dot-com bubble, Bezos was the first to register $100 billion with major wealth indexes and has retained the wealth for longer than Gates's three weeks.
^ Many calculations of Bezos's wealth during the late 2010s were not adjusted for inflation. When he was designated the "world wealthiest person" on March 6, 2018, the Forbes The World's Billionaires list stipulated that although Bezos was the first centi-billionaire (i.e. +US$100 billion in net worth), it was Bill Gates who had the most money when taken in real terms. In such terms, Gates had $150 billion while Bezos had $100 billion. However, in July 2018, the net worth of Bezos officially surpassed the $150 billion mark, which led most major wealth indexes to label him the wealthiest person in modern history (post-1982).
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^ Chozick, Amy (March 3, 2019). "He's Come Undone". The New York Times. p. 1(L). Gale A576639749. Archived from the original on February 24, 2021.
^ "Saudi Arabia 'hacked Amazon boss's phone', says investigator". BBC News. March 31, 2019. Archived from the original on April 11, 2019. Retrieved May 9, 2019.
^ Kirchgaessner, Stephanie (January 22, 2020). "Jeff Bezos hack: Amazon boss's phone 'hacked by Saudi crown prince'". The Guardian. Archived from the original on February 29, 2020. Retrieved January 22, 2020.
^ Frenkel, Sheera (January 22, 2020). "How Jeff Bezos' iPhone X Was Hacked". The New York Times. Archived from the original on January 23, 2020. Retrieved January 22, 2020.
^ "Jeff Bezos hack: Saudi Arabia calls claim 'absurd'". BBC News. January 22, 2020. Archived from the original on January 22, 2020. Retrieved January 22, 2020.
^ "Fred Hutch receives $35 million donation, largest ever, from Bezos family". The Seattle Times. March 30, 2017. Archived from the original on April 2, 2019. Retrieved April 2, 2019.
^ Soper, Taylor (April 7, 2014). "Amazon CEO Jeff Bezos and family donate $20M to Fred Hutchinson Cancer Research Center". GeekWire. Archived from the original on January 2, 2018. Retrieved December 31, 2017.
^ Kotecki, Peter. "Jeff Bezos is the richest man in modern history – here's how he spends on philanthropy". Business Insider. Archived from the original on May 8, 2019. Retrieved July 1, 2019.
^ Kim, Eugene (September 16, 2018). "Unanswered questions about Jeff Bezos' new $2 billion philanthropic fund". CNBC. Archived from the original on April 29, 2019. Retrieved July 1, 2019.
^ Grothaus, Michael (May 24, 2017). "Jeff Bezos just gave $1 million to the Reporters Committee for Freedom of the Press". Fast Company. Archived from the original on March 9, 2018. Retrieved March 8, 2018.
^ Kunkle, Frederick (September 1, 2017). "Jeff Bezos wants to give more money to charity. He should pay his workers first". HuffPost. Archived from the original on September 20, 2018. Retrieved September 19, 2018.
^ Hyde, Marina (September 14, 2018). "If Jeff Bezos wants to help low-income people why not just pay them better?". The Guardian. ISSN 0261-3077. Archived from the original on December 12, 2019. Retrieved December 12, 2019.
^ Cao, Sissi (June 6, 2018). "What Has Jeff Bezos Chosen to Spend His $140B On? We Have 4 Guesses". The Observer. Archived from the original on June 16, 2018. Retrieved June 6, 2018.
^ Levin, Sam (September 13, 2018). "Jeff Bezos to fund schools where 'child will be the customer' with new charity". The Guardian. Archived from the original on September 13, 2018. Retrieved September 13, 2018.
^ Stevens, Laura (September 13, 2018). "Jeff Bezos to Create $2 Billion Fund for Homeless, Preschools". The Wall Street Journal. Archived from the original on September 13, 2018. Retrieved September 13, 2018.
^ Au-Yeung, Angel (September 13, 2018). "Jeff Bezos Unveils Multibillion-Dollar Plans For Charitable Giving". Forbes. Archived from the original on September 13, 2018. Retrieved September 13, 2018.
^ O'Keefe, Ed; Anderson, Nick (January 12, 2018). "Jeff Bezos donates $33 million to scholarship fund for 'dreamers'". The Washington Post. Archived from the original on January 12, 2018. Retrieved January 12, 2018.
^ "Bezos Pledges $33 Million for Scholarships for DREAMers". Philanthropy News Digest (PND). January 16, 2024. Retrieved February 19, 2024.
^ Boyle, Alan (June 18, 2018). "Backed by Bill Gates and Jeff Bezos, Breakthrough Energy Ventures places first bets on power storage". GeekWire. Archived from the original on June 27, 2018. Retrieved June 18, 2018. We are committed to doing our part and filling this capital need by coming together in a new coalition. We will form a network of private capital committed to building a structure that will allow informed decisions to help accelerate the change to the advanced energy future our planet needs. Success requires a partnership of increased government research, with a transparent and workable structure to objectively evaluate those projects, and committed private-sector investors willing to support the innovative ideas that come out of the public research pipeline.
^ Goldman, David (September 5, 2018). "Jeff Bezos donates $10 million to organization that helps veterans run for office". CNNMoney. Archived from the original on December 18, 2018. Retrieved December 17, 2018.
^ "Instagram a new Bezos Earth Fund". February 17, 2020. Archived from the original on February 17, 2020. Retrieved February 17, 2020 – via Instagram. Today, I'm thrilled to announce I am launching the Bezos Earth Fund.
^ "Amazon founder Jeff Bezos commits $10bn to fight climate change". Sky News. Archived from the original on February 17, 2020.
^ "Jeff Bezos: World's richest man pledges $10bn to fight climate change". BBC News. February 17, 2020. Archived from the original on February 17, 2020.
^ "Bezos Earth Fund announces first grants totaling $791M of his $10B pledge to help planet". GeekWire. November 16, 2020. Retrieved March 10, 2021.
^ "Bezos Earth Fund gives nearly $800 million to climate groups in first round of grants". phys.org. Retrieved March 10, 2021.
^ Palmer, Annie (November 16, 2020). "Jeff Bezos names first recipients of his $10 billion Earth Fund for combating climate change". CNBC. Retrieved March 10, 2021.
^ "Twitter boss pledges $1bn for coronavirus relief". BBC News. April 7, 2020. Archived from the original on April 13, 2020. Retrieved April 15, 2020.
^ Schleifer, Theodore (April 2, 2020). "Why Jeff Bezos's $100 million donation to food banks won't satisfy his critics". Vox. Archived from the original on April 13, 2020. Retrieved April 15, 2020.
^ Hahn, Jason Duaine (November 2, 2021). "Jeff Bezos Pledges $2B for Nature Conservation as World Leaders Address Deforestation, Emissions". Yahoo! News. Archived from the original on December 13, 2021. Retrieved November 3, 2021.
^ Darcy, Oliver (July 20, 2021). "Bezos donates $100 million each to CNN contributor Van Jones and chef Jose Andres". CNN. Retrieved March 15, 2024.
^ "Country star Dolly Parton gets $100m award from Amazon founder Jeff Bezos". BBC News. November 14, 2022. Retrieved November 14, 2022.
^ "Eva Longoria and Bill McRaven Receive $100 Million From Jeff Bezos". The Republic Reporter. March 15, 2024. Retrieved March 15, 2024.
^ Jeff Bezos's tuition-free preschool is expanding to new locations in Texas and Florida. Here's who's eligible to attend and why Bezos says 'the child will be the customer.', Business Insider, January 31, 2022
^ Melas, Chloe (November 22, 2022). "Jeff Bezos announces 40 grants totaling $123 million to combat homelessness". CNN. Retrieved June 28, 2025.
Sources
Robinson, Tom (2010). Jeff Bezos: Amazon.com Architect. ABDO Publishing. ISBN 978-1-60453-759-8.
Further reading
Boyle, Alan (May 29, 2018). "Jeff Bezos: 'We will have to leave this planet ... and it's going to make this planet better'". GeekWire. Archived from the original on May 29, 2018. Retrieved May 31, 2018.
Davenport, Christian (2018). The Space Barons: Elon Musk, Jeff Bezos, and the Quest to Colonize the Cosmos. PublicAffairs. ISBN 978-1610398299.
Döpfner, Mathias S. (March 28, 2018). "Jeff Bezos reveals what it's like to build an empire". Business Insider. Archived from the original on April 30, 2018. Retrieved April 29, 2018.
Fernholz, Tim (2018). Rocket Billionaires: Elon Musk, Jeff Bezos, and the New Space Race. Houghton Mifflin Harcourt. ISBN 978-1328662231.
Leibovich, Mark. The New Imperialists (Prentice Hall, 2002) pp 55–104.online
Wingfield, Nick (January 12, 2018). "Jeff Bezos, Mr. Amazon, Steps Out". The New York Times. Archived from the original on April 22, 2018. Retrieved April 29, 2018.
Wofford, Ben (April 22, 2018). "Inside Jeff Bezos's Washington D.C. Life". Washingtonian. Archived from the original on April 29, 2018. Retrieved April 29, 2018.
External links
Jeff Bezos at Wikipedia's sister projects
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Jeff Bezos collected news and commentary at The New York Times
Forbes Profile
| Honorary titles |
| Preceded by Bill Gates | World's richest person 2018–2021 | Succeeded by Elon Musk |
| v t e Jeff Bezos |
| Organizations | Amazon + Amazon Web Services + Alexa + Amazon Appstore + Amazon Luna + Amazon Music + Amazon Pay + Amazon Prime + Amazon Prime Video + Ring + Amazon Robotics + Twitch Blue Origin Bezos Expeditions Bezos Family Foundation Washington Post |
| Family | Ted Jorgensen Miguel Bezos Jackie Bezos Mark Bezos MacKenzie Scott Lauren Sánchez |
| Related | Koru Billionaire space race The Everything Store The Space Barons Phone hacking incident |
| v t e Amazon |
| People | | | | --- | | Current | Jeff Bezos (Founder and Executive Chairman) Andy Jassy (President and CEO) Werner Vogels (CTO) | | Former | Rick Dalzell Paul Davis Tony Hsieh Christopher North Ram Shriram Tom Szkutak Brian Valentine | |
| Facilities | List of Amazon locations Doppler Day 1 HQ2 Principal Place Spheres Bellevue 600 |
| Products andservices | | | | --- | | Subsidiaries | A9.com AbeBooks Amazon Clinic Amazon Games + Double Helix Games Amazon Lab126 Amazon Pharmacy Amazon Robotics Amazon University Esports Annapurna Labs Audible Blink Home Body Labs Book Depository BookFinder ComiXology Fresh Goodreads + Goodreads Choice Awards Graphiq IMDb + Box Office Mojo + IMDbPro Kuiper Systems One Medical PillPack Ring + Neighbors Shopbop Souq.com Twitch + IGDB Woot.com Zappos Zoox | | Cloudcomputing | Web Services + AMI + Amazon Aurora + Beanstalk + CloudFront + DynamoDB + EBS + EC2 + EFS + ElastiCache + EMR + Glacier + Glue + Lambda + Lightsail + MTurk + Neptune + Product Advertising API + RDS + Redshift + Rekognition + Route 53 + S3 + SageMaker + SES + SNS + SimpleDB + SQS + VPC | | Services | Amazon.com + China Alexa Appstore Digital Game Store Fire OS Kindle Store Luna Payments Prime Amazon miniTV MX Player + Key + Prime Music + Prime Now + Prime Pantry + Prime Video - Sports Marketplace Music (Wondery) Silk Wireless | | Devices | Astro Echo + Show Echo Buds Fire + Fire HD + Fire HDX Fire TV + Stick Kindle | | Technology | 1-Click Dynamo Obidos Lumberyard | | Media | Amazon Games Amazon Publishing Breakthrough Novel Award Best Books of the Year Amazon MGM Studios + Metro-Goldwyn-Mayer + United Artists + Orion Pictures + American International Pictures + MGM+ Kindle Direct Publishing YES Network (15%) | | Retail | Amazon Fresh Amazon Go Whole Foods Market | | Logistics | Amazon Air Amazon Prime Air | | Former | 43 Things Askville Alexa Internet Amapedia Amazon Books Amie Street (Songza) CDNow Dash buttons Dash wand Diapers.com Digital Photography Review Drive Endless.com Fire Phone Freevee Lexcycle Liquavista LivingSocial LoveFilm MGM Holdings Mobipocket PlanetAll Reflexive Entertainment Sellaband Shelfari TenMarks Treasure Truck Withoutabox | |
| Litigation | Perfect 10, Inc. v. Amazon.com, Inc. Amazon.com, Inc. v. Barnesandnoble.com, Inc. Amazon.com Inc v Canada (Commissioner of Patents) FTC v. Amazon |
| Other | Amazon Light ASIN Community Banana Stand Criticism (tax) Fishbowl History of Amazon LibraryThing List of Amazon brands List of Amazon products and services List of mergers and acquisitions by Amazon Locker MacKenzie Scott Statistically improbable phrase Vine Worker organization List of fatalities Edwardsville Amazon warehouse collapse |
| Unions | Congress of Essential Workers Amazon Labor Union Amazon worker organization + 2024 strike |
| Category |
| v t e Blue Origin |
| Vehicles | Charon† Goddard† PM2‡ New Shepard + Propulsion modules - Tail 1 ‡ - Tail 2 † - Tail 3 ‡ - Tail 4 - Tail 5 + Capsules - Jules Verne † - H.G. Wells - First Step - Kármán Line New Glenn Blue Moon Blue Ring Jacklyn† + LPV1 Jacklyn (barge) |
| Rocket engines | BE-1 † BE-2 † BE-3 BE-4 BE-7 |
| Missions | New Glenn launches NS-15 NS-16 NS-17 NS-18 NS-19 NS-20 NS-21 NS-22 NS-23X NS-24 NS-25 NS-26 NS-27 NS-28 NS-29 NS-30 NS-31 NS-32 NS-33 NS-34 NS-35 |
| Facilities | Headquarters + Kent, Washington Corn Ranch Launch Complex 36 + Cape Canaveral Space Launch Complex 9 + Vandenberg |
| Key people | Jeff Bezos (founder) Dave Limp (CEO) Jeffrey Ashby (chief of mission assurance) Nicholas Patrick (human integration architect) |
| Related | Commercial astronaut Billionaire space race Blue Origin Federation, LLC v. United States |
| – denotes unflown vehicles or engines † – denotes retired vehicles, engines, products ‡ – denotes destroyed vehicles X – denotes failed flight |
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Cobalt Networks Commerce One Covad Cyberian Outpost CyberRebate Digex Digital Convergence Corporation Digital Insight Divine DoubleClick eBay eGain Egghead Software Enron Epidemic Marketing Excite Flooz.com Freei Gadzoox GeoCities Global Crossing govWorks Handspring Healtheon HomeGrocer HotBot Infoseek InfoSpace Inktomi Interactive Intelligence Internet America iVillage iWon Kozmo.com lastminute.com The Learning Company Liquid Audio LookSmart Lycos + Angelfire + Tripod MarchFirst Merrill Lynch MicroStrategy Net2Phone NetBank Netscape Network Solutions NorthPoint Communications OmniSky Palm, Inc. PayPal Pets.com PFSweb Pixelon PLX Technology Prodigy Pseudo.com Radvision Razorfish Redback Networks Register.com Ritmoteca.com Savvis Scout Electromedia + modo Terra theGlobe.com Think Tools TIBCO Software Tradex Technologies Transmeta uBid United Online USinternetworking UUNET VA Linux Systems Verio VerticalNet Vignette Corporation WebChat Broadcasting System Websense Webvan WorldCom World Online Yahoo! | | History | Enron scandal Irrational exuberance Sarbanes–Oxley Act Telecoms crash Trial of Kenneth Lay and Jeffrey Skilling WorldCom scandal | | Publications | Blood on the Street Conspiracy of Fools The Industry Standard The PayPal Wars | | Broadcast media | e-Dreams Enron: The Smartest Guys in the Room Startup.com Valley of the Boom | | Category | | v t e Time Persons of the Year | | 1927–1950 | Charles Lindbergh (1927) Walter Chrysler (1928) Owen D. Young (1929) Mohandas Gandhi (1930) Pierre Laval (1931) Franklin D. Roosevelt (1932) Hugh S. Johnson (1933) Franklin D. Roosevelt (1934) Haile Selassie (1935) Wallis Simpson (1936) Chiang Kai-shek / Soong Mei-ling (1937) Adolf Hitler (1938) Joseph Stalin (1939) Winston Churchill (1940) Franklin D. Roosevelt (1941) Joseph Stalin (1942) George Marshall (1943) Dwight D. Eisenhower (1944) Harry S. Truman (1945) James F. Byrnes (1946) George Marshall (1947) Harry S. Truman (1948) Winston Churchill (1949) The American Fighting-Man (1950) | | 1951–1975 | Mohammed Mosaddeq (1951) Elizabeth II (1952) Konrad Adenauer (1953) John Foster Dulles (1954) Harlow Curtice (1955) Hungarian Freedom Fighters (1956) Nikita Khrushchev (1957) Charles de Gaulle (1958) Dwight D. Eisenhower (1959) U.S. Scientists: George Beadle / Charles Draper / John Enders / Donald A. Glaser / Joshua Lederberg / Willard Libby / Linus Pauling / Edward Purcell / Isidor Rabi / Emilio Segrè / William Shockley / Edward Teller / Charles Townes / James Van Allen / Robert Woodward (1960) John F. Kennedy (1961) Pope John XXIII (1962) Martin Luther King Jr. (1963) Lyndon B. Johnson (1964) William Westmoreland (1965) The Generation Twenty-Five and Under (1966) Lyndon B. Johnson (1967) The Apollo 8 Astronauts: William Anders / Frank Borman / Jim Lovell (1968) The Middle Americans (1969) Willy Brandt (1970) Richard Nixon (1971) Henry Kissinger / Richard Nixon (1972) John Sirica (1973) King Faisal (1974) American Women: Susan Brownmiller / Kathleen Byerly / Alison Cheek / Jill Conway / Betty Ford / Ella Grasso / Carla Hills / Barbara Jordan / Billie Jean King / Susie Sharp / Carol Sutton / Addie Wyatt (1975) | | 1976–2000 | Jimmy Carter (1976) Anwar Sadat (1977) Deng Xiaoping (1978) Ayatollah Khomeini (1979) Ronald Reagan (1980) Lech Wałęsa (1981) The Computer (1982) Ronald Reagan / Yuri Andropov (1983) Peter Ueberroth (1984) Deng Xiaoping (1985) Corazon Aquino (1986) Mikhail Gorbachev (1987) The Endangered Earth (1988) Mikhail Gorbachev (1989) George H. W. Bush (1990) Ted Turner (1991) Bill Clinton (1992) The Peacemakers: Yasser Arafat / F. W. de Klerk / Nelson Mandela / Yitzhak Rabin (1993) Pope John Paul II (1994) Newt Gingrich (1995) David Ho (1996) Andrew Grove (1997) Bill Clinton / Ken Starr (1998) Jeff Bezos (1999) George W. Bush (2000) | | 2001–present | Rudolph Giuliani (2001) The Whistleblowers: Cynthia Cooper / Coleen Rowley / Sherron Watkins (2002) The American Soldier (2003) George W. Bush (2004) The Good Samaritans: Bono / Bill Gates / Melinda Gates (2005) You (2006) Vladimir Putin (2007) Barack Obama (2008) Ben Bernanke (2009) Mark Zuckerberg (2010) The Protester (2011) Barack Obama (2012) Pope Francis (2013) Ebola Fighters: Dr. Jerry Brown / Dr. Kent Brantly / Ella Watson-Stryker / Foday Gollah / Salome Karwah (2014) Angela Merkel (2015) Donald Trump (2016) The Silence Breakers (2017) The Guardians: Jamal Khashoggi / Maria Ressa / Wa Lone / Kyaw Soe Oo / Staff of The Capital (2018) Greta Thunberg (2019) Joe Biden / Kamala Harris (2020) Elon Musk (2021) Volodymyr Zelenskyy / Spirit of Ukraine (2022) Taylor Swift (2023) Donald Trump (2024) | | v t e Wealthiest people in the United States by state | | AL Jimmy Rane AK Leonard Hyde and Jonathan Rubini AZ Ernest Garcia II AR Rob Walton CA Mark Zuckerberg CO Philip Anschutz CT Steve Cohen DE Elizabeth Snyder FL Jeff Bezos GA The Cathys HI Pierre Omidyar ID Frank VanderSloot IL Lukas Walton IN Carl Cook IA Harry Stine KS Charles Koch KY Tamara Gustavson LA Todd Graves ME Susan Alfond MD Annette Lerner MA Abigail Johnson MI Daniel Gilbert MN Glen Taylor MS James and Thomas Duff MO David Steward MT Dennis Washington NE Warren Buffett NV Miriam Adelson NH Richard B. Cohen NJ John Overdeck NM Ron Corio NY Michael Bloomberg NC James Goodnight ND Gary Tharaldson Les Wexner Harold Hamm Phil Knight Jeff Yass Jonathan Nelson Bob Faith T. Denny Sanford Thomas Frist Jr. Elon Musk Gail Miller John Abele Jacqueline Mars Steve Ballmer Brad D. Smith Diane Hendricks John Mars | |
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To identify like terms, check for the powers of all the variables in an equation. Like terms in the equation will be those having equal powers.
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Are XY and YX like terms?
For XY and YX, the powers are the same i.e. 1. So, XY can be written as YX and vice versa. So, XY and YX can be classified as like terms.
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October 21, 2020 at 10:46 pm
Combine the like terms to create an −4y−4+(−3)equivalent expression:
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190366 | https://pubmed.ncbi.nlm.nih.gov/27631723/ | Surgical Management of the Undescended Testis: Recent Advances and Controversies - PubMed
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. 2016 Oct;26(5):418-426.
doi: 10.1055/s-0036-1592197. Epub 2016 Sep 15.
Surgical Management of the Undescended Testis: Recent Advances and Controversies
Jack S Elder1
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1 Division of Pediatric Urology, Massachusetts General Hospital, Boston, Massachusetts, United States.
PMID: 27631723
DOI: 10.1055/s-0036-1592197
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Surgical Management of the Undescended Testis: Recent Advances and Controversies
Jack S Elder. Eur J Pediatr Surg.2016 Oct.
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doi: 10.1055/s-0036-1592197. Epub 2016 Sep 15.
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Undescended testis (UDT) is the most common disorder of sexual development in boys and affects 3.5% of male newborns. Although approximately half of newborn UDTs descend spontaneously, some boys develop an ascending testis later in childhood. Recent guideline recommendations advocate orchiopexy by 18 months of age to maximize potential for fertility and perhaps reduce the risk for testicular carcinoma in the future. For palpable testes, a standard inguinal approach is appropriate. However, the prescrotal approach is often effective for low inguinal testes and reduces surgical time and patient discomfort with an equivalent success rate in boys with an ascending testis. Some advocate monitoring until adolescence to determine whether the testis will spontaneously descend into the scrotum, but data do not support this approach. Instead, prompt orchiopexy is recommended. In boys with a nonpalpable testis, approximately 50% are abdominal or high in the inguinal canal and 50% are atrophic, typically in the scrotum. Routine inguinal/scrotal ultrasound is not recommended, although in an older boy who is overweight, it is appropriate. If the patient has contralateral testicular hypertrophy, scrotal exploration is appropriate, and removal of the testicular remnant and contralateral scrotal orchiopexy to prevent future contralateral testicular torsion is recommended. In most cases, diagnostic laparoscopy is advised to determine whether the testis is abdominal. For the abdominal testis, there are numerous treatment options. If the testis is mobile or a peeping testis just distal to the internal inguinal ring, standard one-stage laparoscopic or open orchiopexy should be attempted using the Prentiss maneuver. If the testicular vessels are short or the testis is not mobile, a two-stage Fowler-Stephens orchiopexy is appropriate. The second stage can be performed laparoscopically or open. Another option is microvascular testicular autotransplantation, which is a technically demanding procedure. Surgical results of abdominal orchiopexy are highly variable, short term, and highly subjective. Prospective clinical trials with follow-up into adolescence and adulthood are necessary to assess the success of various surgical approaches.
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190367 | https://www.simplypsychology.org/what-is-the-yerkes-dodson-law.html | Yerkes-Dodson Law of Arousal and Performance
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Psychology » Biopsychology
Yerkes-Dodson Law of Arousal and Performance
By
Charlotte Nickerson
Updated on
August 14, 2025
Updated on
August 14, 2025
Charlotte Nickerson
Research Assistant at Harvard University
Undergraduate at Harvard University
Charlotte Nickerson is a graduate of Harvard University obsessed with the intersection of mental health, productivity, and design.
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Olivia Guy-Evans, MSc
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On This Page:
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How the Law Works
Practical Examples
Links to Psychology Theories
Research Evidence
Critical Evaluation
The Yerkes–Dodson law is a psychology principle stating that performance improves with mental arousal – like excitement or alertness – up to an optimal point. Beyond that point, too much arousal causes performance to drop. Think of it as an inverted-U curve: low arousal leads to boredom and poor results, moderate arousal boosts focus and efficiency, and excessive arousal leads to stress and mistakes.
The concept of optimal arousal in relation to performance on a task is depicted here. Performance is maximized at the optimal level of arousal, and it tapers off during under- and overarousal.
Key Takeaways
Definition: The Yerkes–Dodson law is a psychology principle showing that performance improves with arousal up to an optimal point, after which it declines. This relationship is often illustrated as an inverted-U curve.
Optimal zone: Moderate arousal—neither too low nor too high—tends to produce the best results. Low arousal can cause boredom, while high arousal can lead to stress and errors.
Task influence: The ideal arousal level depends on task difficulty. Simple tasks may benefit from higher arousal, while complex or unfamiliar tasks require lower arousal for peak performance.
Real-world use: The principle helps explain performance patterns in areas like sports, test-taking, public speaking, and workplace productivity. Recognizing your optimal arousal can guide preparation and stress management.
Limitations: The law is a general guideline, not a strict rule. Individual differences, situational factors, and modern research show that the curve can vary across people and contexts.
How the Law Works
The Yerkes–Dodson law describes the relationship between arousal and performance, showing that a moderate level of mental or physical alertness produces the best results.
In psychology, arousal refers to a general state of readiness and alertness – driven by factors like excitement, focus, or mild stress – not just feeling nervous or anxious.
Performance improves as arousal rises, but only up to an optimal point—after that, too much arousal causes it to drop.
This is known as the inverted-U model.
The curve’s shape depends on the task: complex or unfamiliar tasks are best tackled with lower arousal, while simple or well-practised tasks often benefit from higher arousal.
For example, a little pre-exam tension can sharpen focus, but overwhelming anxiety can make you forget what you know.
Practical Examples
How can students use it to improve exam performance?
Find your sweet spot. If you feel flat (sleepy, unfocused), briefly raise arousal: stand and stretch, do 20–60 seconds of brisk movement, review a high-yield summary card, or set a 10–15 minute Pomodoro with a clear micro-goal.
Dial down when jittery. If you feel wired (racing heart, mind blanking), use 60–120 seconds of slow breathing (4–6 breaths/min, longer exhale), progressive muscle relaxation, or the reframe “I’m excited” to channel arousal into focus.
Match arousal to task. Use higher arousal for quick recall and practice tests; use lower arousal (quiet, steady pace) for complex problem-solving and multi-step calculations.
Exam-day routine. 5–10 minutes pre-test: light movement → two calming breaths → skim a confidence list (3 things you know cold) → pick a “first easy win” question to lock in momentum.
How does it apply in sports, public speaking, or workplace productivity?
Sports. Explosive, well-learned skills (sprinting, powerlifting) usually benefit from higher arousal (hype music, dynamic warm-up, energizing cue words). Fine-motor or novel/complex skills (golf putt, gymnastics routine, new play) often need lower arousal (slower breathing, quiet focus, consistent pre-shot routine).
Public speaking. Aim for medium arousal: convert nerves into energy with brisk walking and power posture, then settle with two slow breaths. Use a simple opening script, anchor points on your outline, and a tempo cue (“slow and clear”) to avoid rushing.
Workplace. Routine tasks (inbox triage, filing) can tolerate higher arousal and short sprints. Deep work (analysis, writing, coding) needs lower arousal: silence notifications, time-box 25–50 minutes, set a single outcome, and start with a 60-second calming reset.
What are everyday signs you’re over- or under-aroused for a task?
Under-arousal (too low): Boredom, drifting attention, procrastination, slow starts, repeated rereads with nothing sticking.
Quick fix: Add stakes (timer, micro-deadline), stand or walk 2 minutes, switch to an easier starter task, play neutral background noise if silence feels sleepy.
Optimal arousal (just right): Clear goal, steady pace, present-moment focus, small errors caught quickly, time passing normally.
Maintain: Brief breaks before fatigue, keep the environment stable, stick to the current goal.
Over-arousal (too high): Racing thoughts, muscle tension, twitchy pacing, rushing, frequent mistakes, mind blanks on steps you know.
Quick fix: Two minutes of slow breathing with long exhales, relax shoulders/jaw, lower stimulation (quieter room, dimmer screen), break the task into smaller chunks and re-enter with the easiest step.
Links to Psychology Theories
How does it connect to stress theories like the fight-or-flight response?
The fight-or-flight response is a surge of sympathetic arousal; at moderate levels (often called eustress), attention sharpens and performance improves, matching the rising side of the inverted-U.
When arousal becomes excessive (distress), prefrontal functions like working memory and flexible thinking degrade, producing tunnel vision and more errors—the falling side of the curve.
Practical read: a little time pressure can help you lock in, but panic (racing heart, breathlessness, blanking) pushes you past the peak.
How does it relate to motivation theories such as Self-Determination Theory or Drive Theory?
Drive Theory (classic “more arousal → better performance”) fits well-learned, simple tasks; Yerkes–Dodson refines it by adding the decline at high arousal, especially for complex or novel tasks.
Self-Determination Theory (SDT) explains quality of motivation: autonomy, competence, and relatedness tend to keep arousal in a productive, self-regulated range; controlling pressure elevates anxiety and risks overshooting the peak.
In practice: autonomy-supportive study or coaching fosters steady, optimal arousal; harsh evaluation or micromanagement spikes arousal and impairs complex performance.
How does it fit with the concept of “flow” in positive psychology?
Flow occurs when skill and challenge are high and balanced, with clear goals and immediate feedback—subjectively “calm but energized” focus.
Flow typically sits near the optimal zone of the inverted-U: aroused enough for intensity, low enough in anxiety to keep precision and flexibility.
Key distinction: Yerkes–Dodson models a performance–arousal curve; flow describes the quality of experience. They align at the peak, but you can feel “amped” without flow, or be in flow at different absolute arousal levels depending on the task.
Research Evidence
The Yerkes–Dodson law has been interpreted in different ways since it was first proposed in 1908.
In their original paper, Robert Yerkes and John Dodson studied discrimination learning in “dancing mice.”
The mice had to choose between a white and a black box.
Entering the white box delivered a mild (non-injurious) electric shock; the black box did not.
First series (baseline task).
With very weak shocks, mice took many tries to meet the learning goal (picking correctly 10/10 times for three days in a row).
As the shock got stronger, they learned faster—until the strongest level, where learning slowed again.
If you plot shock strength vs. number of tries (where fewer tries = better), you get a U-shape.
If you flip that to overall performance (where higher = better), you get the familiar inverted-U.
This went against the authors’ simple “more stimulation = better learning” expectation.
Second series (easier task).
Yerkes and Dodson then made the choice easier by increasing the light contrast between boxes and tested five shock levels.
Now, the strongest shocks produced the fastest learning, and the weakest produced the slowest—more like a straight increase than a U-shape (Yerkes & Dodson, 1908; Teigen, 1994).
Third series (harder task).
Next, they made the choice harder by reducing the light contrast and tested four shock levels.
Here, a moderate shock (the second-weakest) led to the best learning (fewest tries), again suggesting a U-shaped link between stimulation and learning rate (Teigen, 1994).
What they took from this.
Both too little and too much stimulation can hurt learning, and the best level depends on how difficult the task is.
As tasks get harder, the optimal stimulation shifts downward toward a gentler level (Yerkes & Dodson, 1908; Teigen, 1994).
Replication Studies
Early follow-ups generally supported the idea that task difficulty changes the effect of stimulation on learning.
Chicks (Cole, 1911):On easy tasks, stronger shock helped; on medium difficulty, very strong shock slowed learning; on difficult tasks, strong shock led to mixed results. Cole saw this as broadly consistent with Yerkes–Dodson.
Kittens (Dodson, 1915):A medium shock beat a strong one on the standard task; when the task was made easier, medium and strong shocks worked about equally well, and learning overall improved as shock increased.
Reward strength (Dodson, 1917):A similar U-shape appeared with rewards: moderate deprivation (e.g., mild hunger) boosted learning, but excessive deprivation made it worse.
Humans (Vaughn & Diserens, 1930; summarized in Young, 1936). People learned a maze best with light or medium punishment, not with none or heavy punishment—again pointing to an optimum rather than a straight line. Young concluded: for any activity, there is an optimal degree of punishment.
From Punishment–Learning to Motivation–Performance
By the 1930s–1940s, researchers began to reframe the idea.
Punishment was no longer seen as a core driver of learning (e.g., Thorndike, 1932; Skinner, 1938; Estes, 1944), and scientists distinguished learning from performance more carefully.
Mid-century accounts treated the pattern as a link between drive/arousal (motivation) and performance, noting that too much arousal can sometimes get in the way (e.g., Hilgard & Marquis, 1961).
Textbooks and reviews (e.g., Bourne & Ekstrand, 1973) popularized the inverted-U: performance is highest at moderate arousal and lower at both extremes.
Note: Early animal work often measured tries needed to learn (lower is better), which makes the graph look U-shaped. When we talk about performance level (higher is better), the same pattern appears as an inverted-U.
A stronger test came from Broadhurst (1957), who used several motivation levels and three difficulty levels with larger groups of rats.
He found that the best motivation level moved with task difficulty—higher for easy tasks and lower for hard ones—matching Yerkes and Dodson’s core claim (Broadhurst, 1957; Teigen, 1994).
He also suggested looking at individual differences (e.g., “emotionality”) that might shift where the peak sits on the curve.
Critical Evaluation
1. Situations where performance doesn’t follow the inverted-U pattern
The inverted-U is not universal; some tasks show different shapes (linear increases, linear decreases, thresholds, or plateaus).
Very simple, well-learned speeded tasks sometimes improve almost linearly as arousal rises (consistent with classic drive and social-facilitation findings).
In contrast, highly complex or novel tasks can worsen as arousal increases, producing a mostly downward slope with little or no “rising” phase (e.g., multi-step reasoning under intense time pressure).
Vigilance or monotonous monitoring tasks can also show threshold/plateau behavior, where raising arousal from very low levels helps, but further increases add little benefit.
These departures occur because arousal affects attention breadth, working memory, and motor control in task-specific ways.
Practitioners should avoid one-size-fits-all prescriptions.
Target arousal to the task: it may be optimal to increase stimulation for dull, overlearned activities but decrease it for precision or novel problem-solving. Treat the inverted-U as a heuristic starting point, not a binding rule.
2. Personality differences (introversion vs. extraversion)
Baseline arousal and reactivity vary with personality, shifting where the “optimal zone” sits on the curve.
Introverts typically have higher baseline cortical arousal and greater sensitivity to stimulation, so they may reach overload at lower levels of noise, caffeine, audience size, or time pressure.
Extraverts often have lower baseline arousal and seek stimulation, so additional intensity (livelier environment, tighter deadlines, energizing routines) can move them toward peak performance.
Traits like trait anxiety/neuroticism and sensation-seeking further tilt the curve—height, width, and peak position can all change person-to-person.
Personalize conditions: quieter rooms, longer ramps, and calming routines for many introverts or high-anxiety individuals; brisk warm-ups, stronger cues, and lively environments for many extraverts or high sensation-seekers.
In education and workplaces, build flexibility (choice of noise levels, pacing, and break structure) rather than enforcing a single “optimal” arousal recipe.
3. Methodological flaws and oversimplifications in the original work
The 1908 Yerkes & Dodson evidence has limitations that weaken claims of a universal “law.”
The original studies used mice learning a discrimination task with electric shock as the “arousal” manipulation—conflating punishment magnitude, motivation, stress, and learning rate.
Physiological arousal wasn’t directly measured, statistical methods were rudimentary by modern standards, and generalization from a narrow animal paradigm to diverse human performances was assumed rather than demonstrated.
Later popularizations often depict a smooth inverted-U without specifying moderators (task type, skill level, personality, context) or providing precise functional forms, encouraging over-simple applications.
Treat Yerkes–Dodson as a useful framework, not a strict law.
Modern evaluation should rely on task- and person-specific data (e.g., error rates, response variability) and, where possible, physiological indicators of arousal (heart-rate variability, pupil dilation).
Interventions should be tested and iterated for the particular context instead of assumed from the generic curve.
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Broadhurst, P. L. (1957). Emotionality and the Yerkes-Dodson law. Journal of experimental psychology, 54(5), 345.
Brown, W. P. (1965). The Yerkes-Dodson law repealed. Psychological reports, 17(2), 663-666.
Cole, L. W. (1911). The relation of strength of stimulus to rate of learning in the chick. Journal of Animal Behavior, 1(2), 111.
Corbett, M. (2015). From law to folklore: work stress and the Yerkes-Dodson Law. Journal of Managerial Psychology.
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Thorndike, E. L. (1932). The fundamentals of learning.
Vaughn, J., & Diserens, C. M. (1930). The relative effects of various intensities of punishment on learning and efficiency. Journal of Comparative Psychology, 10(1), 55.
Winton, W. M. (1987). Do introductory textbooks present the Yerkes-Dodson Law correctly?. American Psychologist, 42(2), 202.
Yerkes, R. M., & Dodson, J. D. (1908). The relation of strength of stimulus to rapidity of habit-formation. Punishment: Issues and experiments, 27-41.
Young, P. T. (1936). Social motivation.
Key Takeaways
The Yerkes-Dodson law states that there is an empirical relationship between stress and performance and that there is an optimal level of stress corresponding to an optimal level of performance. Generally, practitioners present this relationship as an inverted U-shaped curve.
Research shows that moderate arousal is generally best; when arousal is very high or very low, performance tends to suffer (Yerkes & Dodson, 1908).
Robert Yerkes (pronounced “Yerk-EES”) and John Dodson discovered that the optimal arousal level depends on the complexity and difficulty of the task to be performed.
This relationship is known as the Yerkes-Dodson law, which holds that a simple task is performed best when
arousal levels are relatively high, and complex tasks are best performed when arousal levels are lower.
The Yerkes-Dodson law’s original formulation derives from a 1908 paper on experiments in Japanese dancing mice learning to discriminate between white and black boxes using electric shocks. This research was largely ignored until the 1950s when Hebb’s concept of arousal and the “U-shaped curve” led to renewed interest in the Yerkes-Dodson law’s general applications in human arousal and performance.
The Yerkes-Dodson law has more recently drawn criticism for its poor original experimental design and it’s over-extrapolated scope to personality, managerial practices, and even accounts of the reliability of eyewitness testimony.
Reviewer Author
Saul McLeod, PhD
BSc (Hons) Psychology, MRes, PhD, University of Manchester
Editor-in-Chief for Simply Psychology
Saul McLeod, PhD., is a qualified psychology teacher with over 18 years of experience in further and higher education. He has been published in peer-reviewed journals, including the Journal of Clinical Psychology.
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BSc (Hons) Psychology, MSc Psychology of Education
Associate Editor for Simply Psychology
Olivia Guy-Evans is a writer and associate editor for Simply Psychology. She has previously worked in healthcare and educational sectors.
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Charlotte Nickerson is a graduate of Harvard University obsessed with the intersection of mental health, productivity, and design.
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190369 | https://math.stackexchange.com/questions/4012674/find-the-locus-of-this-point-p | Skip to main content
Find the locus of this point P
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Find the locus of all points inside △ABC such that PA2+PB2=PC2.
At first, i tried finding a right angled triangle and then tried to go on applying Pythagorean Theorem and finding other trivia but that didn't seem to work. I think the locus is most likely to be some line segment rather than an arc of some circle as that seems to be quite unrelated seen from an Euclidean perspective. Applying Apollonius' Theorem and Stewart's Theorem on some triangles might be the key to this problem although i could not find such triangles. I don't know whether this can be solved by trigonometry or not but since this is a problem from a chapter on euclidean geometry, i am sure that there is some clever way to look at this problem.
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asked Feb 4, 2021 at 17:29
LimestoneLimestone
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Nice question. I think the locus is a circle. Say using coordinate geometry, we placed two vertices on x−axis (with one on the origin) and the third one in first quadrant, it shows the locus is a circle.
– Math Lover
Commented
Feb 4, 2021 at 17:48
1
Say we took coordinates (0,0),(a,0),(b,c) as vertices of a triangle. Then locus of point P(x,y) will be x2+y2+(x−a)2+y2=(x−b)2+(x−c)2 and if we simplify it will lead to equation of a circle.
– Math Lover
Commented
Feb 4, 2021 at 17:59
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Here is something I found that may be of interest - math.stackexchange.com/questions/831352/…
– Math Lover
Commented
Feb 4, 2021 at 18:05
1
@MathLover Thank you very much. I was searching for similar problems but couldn't find one. This answers my question.
– Limestone
Commented
Feb 4, 2021 at 18:09
1
It shouldn't be a line, since when △ABC is equilateral, then the locus is an arc of a circle of points subtending on AB an angle of 150 degrees.. This case can be solved by rotating the triangle around C by 60 degrees.
– plop
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Use vectors with C at the origin. Then
|PA|2+|PB|2=|PC|2⟹|P|2−2(A+B)⋅P+|A|2+|B|2=0.
Then if D=A+B, this rearranges to
|P−D|2=2A⋅B.
So as long as ∠BCA<90∘, the locus will be a circle with centre D radius 2abcosC−−−−−−−√.
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answered Feb 4, 2021 at 18:06
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Yes using vectors or barycentric coordinates, it is straightforward.
– Math Lover
Commented
Feb 4, 2021 at 18:07
Very nice answer. For obtuse triangles, there'll be no point P.
– cosmo5
Commented
Feb 4, 2021 at 18:14
@MathLover how would you approach the problem with barycentric coordinates?
– Dr. Mathva
Commented
Feb 4, 2021 at 19:34
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@DRSKMOBINULHAQUE I was going through the answer in the link I sent. I think that answer can be simplified quite a bit. I have added an answer.
– Math Lover
Commented
Feb 5, 2021 at 8:34
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@MathLover I simplified it as well.
– Limestone
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We extend median CM such that CM=MD. Then please note that MP is median of △CPD and △APB. Applying Apollonius's theorem,
PA2+PB2=2(MP2+AM2), PC2+PD2=2(MP2+CM2)
Subtracting second from first,
PA2+PB2−PC2=0=PD2+2(AM2−CM2)
i.e PD2+2(c24−12(a2+b2)+c24)=0
That leads to PD=a2+b2−c2−−−−−−−−−−√ . So the locus of point P satisfying PA2+PB2=PC2 is a circular arc with point D being the center and radius a2+b2−c2−−−−−−−−−−√. We also note that this is possible only if a2+b2≥c2.
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answered Feb 5, 2021 at 8:35
Math LoverMath Lover
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I did the exact same thing!
– Limestone
Commented
Feb 5, 2021 at 9:22
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Good to know! :)
– Math Lover
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Feb 5, 2021 at 9:31
Thank you for staying engaged!
– Limestone
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Feb 5, 2021 at 9:34
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Hint: choose a coordinate system in which A=(0,0) and B=(1,0). Then C will have some fixed coordinates, say C=(p,q). Then, if P=(x,y), we have PA2=x2+y2, PB2=(x−1)2+y2, and so on. You’ll end up with an equation that you should be able to identify.
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answered May 21, 2021 at 1:45
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190370 | https://paedsmadeeasy.files.wordpress.com/2010/05/milestones.pdf | Developmental Milestones In Children - Dr. Insaf Imthiyaz Developmental Milestones are the development of Important new skills of the child as they grow.
There are four important development skills to consider when looking at a young child.
1. Gross Motor Development 2. Vision and Fine Motor Development 3. Hearing, Speech and Language Development 4. Social,Emotional and Behavioural Development Gross Motor skills are the initial area of development, as fine motor skills require good vision.
Likewise Speech and Language require hearing.
And Social, Emotional and Behavioural skills require adequate psychological development.
Motor Development Pattern The normal patter of motor development in children is from : Immobile Child ----> Walking But this doesnot always occur according to plan.. The child may take several other routes before he/she decides to start walking. The certain variations that the child may take are shown below. • Most children undergo the crawling stage before they start walking, some bottom shuffle.
• Bottom-Shuffling is said to run in families.
• 18 Months is the age limit for walking in children who started crawling.
• Bottom-Shuffling and Commando Crawling tend to walk a bit later, this doesnot mean that they are abnormal.
Figure1:Variations in Motor Development Pattern from Immobile infant to walking toddler.. (Picture courtesy of Illustrated Textbook of Paediatrics) Preterm Children If a child was born preterm, this should be considered when assessing the developmental age, calculated from the time of delivery.
Example : A 9month old baby (Chronological Age), born 3 months earlier at 28weeks gestation, May have the developmental skills of a 6month old baby (Corrected Age) N.B. The correction for preterm is no longer required after the child is 2 years old.
Motor Developmental Milestones Gross Motor Year/Age Motor Skill Newborn • Limbs flexed, symmetrical position • Marked head lag on pulling up 2 Months (6 – 8 Weeks) • Holds head steady while sitting • Raises head to 45º 3 Months • Pulls to Sit, No head lag • Brings hands together in midline.
6 – 8 Months • 6 Months : Sits without support, Rounded back • 8 Months : Sits without support, Straight back • Rolls Back --- > Stomach (Supine to Prone) 8 – 9 Months • Crawling 10 Months • Walks around furniture 12 Months • Walking Unsteadily, broad gait, hands apart 15 Months • Walks alone. Vision and Fine Motor Year/Age Motor Skill Newborn • Follow face in mildline • Turning Head to follow objects 4 Months • Reaches out for toys 6 Months • Palmar Grasp 5.5 – 7 Months • Transfer Object from hand to hand 8 – 10 Months • Mature pincer grip 16 – 18 Month • Walks around furniture 14 Months to 5 Years • Tower of 3, 18months • Tower of 6, 2 years • Tower of 8, 2.5 years • Bridge, 3 years (when playing with building blocks) 2 – 5 Years • Drawing • Line – 2 years • Circle – 3 years • Cross – 4 years • Square – 4.5 years • Triangle - >5years Hearing and Speech Year/Age Motor Skill Newborn • Startles to loud noise 3 – 6 Month • Monosyllabalic Babble 7 Months • Turns towards soft sound 7 - 10 Months • 7 Months : Sounds Indiscriminantly • 10 Months: Discriminting sounds towards Parents (Mama , Dada) • Points to Objects 12 Months • Two or Three Words, Other than Dada and Mama 15 Months • 4 – 6 Months 18 Months • 10 – 15 words 20 – 24 Months • 2 word sentences 2.5 – 3 Years • Constant Talking, 3 – 4 word sentences.
Social, Emotional, Behaviour Year/Age Motor Skill 6 Weeks • Smiles Responsively 6 – 8 Months • Puts food in mouth 10 - 12 Months • Waves Bye bye 18 Months • Holds Spoon, Food safely to mouth 18 - 24 Months • Symbolic play 2 years • Dry by day 3 Years • Interactive Play Hearing Hearing is an important sense necessary for the development of speech language and behaviour.
Therefore detection of abnormalities in hearing early are important and will improve the outcome.
Testing of Hearing : 1. Evoked otoacoustic emission An ear piece is inserted in to the ear canal and sound is produced, it evokes an echo from the ear if the cochlear is functioning normally.
2. Auditory Brain Stem Response Computer analysis of EEG waveforms evoked in response to a series of clicks Advantage : • Screens entire pathway from the ear to the brainstem.
• Low False Postives 3. Hearing Checklists for Parents 4. Hearing Tests ▪ Distraction hearing test ▪ Speech Discrimination test ▪ Visual reinforcement audiometry Vision Vision in a newborn is limited, It is only about 6/200..
The peripheral retina maybe fully developed but the fovea is immature. Therefore a well focused image is required for proper visual acuity, and obstruction to this such a cataract, will interfere with this.
Vision testing: Done at entry to school and at preschool years (4 – 5 years) In the UK Vision Testing at Different Ages: Birth Face Fixing and Following 6 weeks Optokinetic nystagmus is normal, It occurs when looking at a striped object 6 Months Reaches for toys 2 Years Identifys pictures 3 Years Better Matching using single letters 5 Years Use of Smellen Chart Some Primitive Reflexes that should disappear by 4 – 6 Months : Reflex Description Moro – Sudden head extension Symmetrical extension , then flexion of all the limbs Grasp – An object is place in the palm at the base of the fingers Flexion of the fingers of the hand Rooting – Stimulus near the mouth Turning the heads towards the stimuli Placing – Infant held vertically and the dorsum of the feet brought into contact with a surface Lifts first,one foot, placing it on the surface, followed by the other.
Positive supporting reflex – Infant Help vertically, feet on a surface Legs take body weight,may push up against gravity.
Atonic Neck reflex – Lying supine, the head is turned to one side by the examiner Infant adopts a “fencing” posture,with the arm outstretched on the side the which the head was turned True Of False. MCQ's 1. Which Of the Following shows Normal Development of a child a. Transferring objects from hand to hand in 6 months b. Turning Towards the sounds in 9 months c. Unaided first step at 15 months d. Make two word sentences at 2 years e. Look for hidden Objects at 18 months Answers : (T,T,T,T,F) 2. Development delay seen when a. Not walking by 15 months b. Not talking meaningful 3 words at 9 months c. Not self-feeding by cup at 9 months d. Stranger Anxiety at 9 months e. No Mature Pincer at 6 months Answers : (F,F,F,T,F) 3. Abnormal Developmental finding in a 6month old infant a. Atonic neck reflex b. Head Lag c. Presence of Hand Dominance e. Parachute reflex f. Identification of Parents Answers : (F,T,T,T,F) All information has been obtained from 1. Nelson Book of Paediatrics 2. Illustrated Textbook of Paediatrics Third Edition Please Refer to these books for extra information... |
190371 | https://www.jacksonsd.org/cms/lib/NJ01912744/Centricity/Domain/560/SE%209.6A.pdf | Section 9.6 Solving Nonlinear Systems of Equations 525 Essential Question Essential Question How can you solve a system of two equations when one is linear and the other is quadratic? Solving a System of Equations Work with a partner. Solve the system of equations by graphing each equation and fi nding the points of intersection.
System of Equations y = x + 2 Linear y = x2 + 2x Quadratic Analyzing Systems of Equations Work with a partner. Match each system of equations with its graph. Then solve the system of equations.
a. y = x2 − 4 b. y = x2 − 2x + 2 y = −x − 2 y = 2x − 2 c. y = x2 + 1 d. y = x2 − x − 6 y = x − 1 y = 2x − 2 A. 10 −10 −10 10 B. 10 −10 −10 10 C. 10 −10 −10 10 D. 10 −10 −10 10 Communicate Your Answer Communicate Your Answer 3. How can you solve a system of two equations when one is linear and the other is quadratic? 4. Write a system of equations (one linear and one quadratic) that has (a) no solutions, (b) one solution, and (c) two solutions. Your systems should be different from those in Explorations 1 and 2.
MAKING SENSE OF PROBLEMS To be profi cient in math, you need to analyze givens, relationships, and goals.
Solving Nonlinear Systems of Equations 9.6 2 4 6 x 2 4 6 −6 −4 −2 −2 y 526 Chapter 9 Solving Quadratic Equations 9.6 Lesson What You Will Learn What You Will Learn Solve systems of nonlinear equations by graphing. Solve systems of nonlinear equations algebraically. Approximate solutions of nonlinear systems and equations.
Solving Nonlinear Systems by Graphing The methods for solving systems of linear equations can also be used to solve systems of nonlinear equations. A system of nonlinear equations is a system in which at least one of the equations is nonlinear.
When a nonlinear system consists of a linear equation and a quadratic equation, the graphs can intersect in zero, one, or two points. So, the system can have zero, one, or two solutions, as shown. No solutions One solution Two solutions Solving a Nonlinear System by Graphing Solve the system by graphing. y = 2x2 + 5x − 1 Equation 1 y = x − 3 Equation 2 SOLUTION Step 1 Graph each equation.
Step 2 Estimate the point of intersection. The graphs appear to intersect at (−1, −4).
Step 3 Check the point from Step 2 by substituting the coordinates into each of the original equations.
Equation 1 Equation 2 y = 2x2 + 5x − 1 y = x − 3 −4 = ? 2(−1)2 + 5(−1) − 1 −4 = ? −1 − 3 −4 = −4 ✓ −4 = −4 ✓ The solution is (−1, −4).
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Solve the system by graphing. 1. y = x2 + 4x − 4 2. y = −x + 6 3. y = 3x − 15 y = 2x − 5 y = −2x2 − x + 3 y = 1 — 2 x2 − 2x − 7 system of nonlinear equations, p. 526 Previous system of linear equations Core Vocabulary Core Vocabulary −6 −6 2 6 2 y = 2x2 + 5x − 1 (−1, −4) y = x − 3 Section 9.6 Solving Nonlinear Systems of Equations 527 Solving Nonlinear Systems Algebraically Solving a Nonlinear System by Substitution Solve the system by substitution.
y = x2 + x − 1 Equation 1 y = −2x + 3 Equation 2 SOLUTION Step 1 The equations are already solved for y.
Step 2 Substitute −2x + 3 for y in Equation 1 and solve for x. −2x + 3 = x2 + x − 1 Substitute −2x + 3 for y in Equation 1. 3 = x2 + 3x − 1 Add 2x to each side. 0 = x2 + 3x − 4 Subtract 3 from each side. 0 = (x + 4)(x − 1) Factor the polynomial. x + 4 = 0 or x − 1 = 0 Zero-Product Property x = −4 or x = 1 Solve for x.
Step 3 Substitute −4 and 1 for x in Equation 2 and solve for y.
y = −2(−4) + 3 Substitute for x in Equation 2. y = −2(1) + 3 = 11 Simplify. = 1 So, the solutions are (−4, 11) and (1, 1). Solving a Nonlinear System by Elimination Solve the system by elimination.
y = x2 − 3x − 2 Equation 1 y = −3x − 8 Equation 2 SOLUTION Step 1 Because the coeffi cients of the y-terms are the same, you do not need to multiply either equation by a constant.
Step 2 Subtract Equation 2 from Equation 1. y = x2 − 3x − 2 Equation 1 y = −3x − 8 Equation 2 0 = x2 + 6 Subtract the equations.
Step 3 Solve for x. 0 = x2 + 6 Resulting equation from Step 2 −6 = x2 Subtract 6 from each side. The square of a real number cannot be negative. So, the system has no real solutions.
REMEMBER The algebraic procedures that you use to solve nonlinear systems are similar to the procedures that you used to solve linear systems in Sections 5.2 and 5.3.
Check Use a graphing calculator to check your answer. Notice that the graphs have two points of intersection at (−4, 11) and (1, 1).
−12 −2 14 12 S S S Check Use a graphing calculator to check your answer. The graphs do not intersect.
−8 −8 8 8 528 Chapter 9 Solving Quadratic Equations Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Solve the system by substitution. 4. y = x2 + 9 5. y = −5x 6. y = −3x2 + 2x + 1 y = 9 y = x2 − 3x − 3 y = 5 − 3x Solve the system by elimination. 7. y = x2 + x 8. y = 9x2 + 8x − 6 9. y = 2x + 5 y = x + 5 y = 5x − 4 y = −3x2 + x − 4 Approximating Solutions When you cannot fi nd the exact solution(s) of a system of equations, you can analyze output values to approximate the solution(s). Approximating Solutions of a Nonlinear System Approximate the solution(s) of the system to the nearest thousandth.
y = 1 — 2 x2 + 3 Equation 1 y = 3x Equation 2 SOLUTION Sketch a graph of the system. You can see that the system has one solution between x = 1 and x = 2.
Substitute 3x for y in Equation 1 and rewrite the equation. 3x = 1 — 2 x2 + 3 Substitute 3x for y in Equation 1. 3x − 1 — 2 x2 − 3 = 0 Rewrite the equation.
Because you do not know how to solve this equation algebraically, let f (x) = 3x − 1 — 2 x2 − 3. Then evaluate the function for x-values between 1 and 2.
f (1.1) ≈ −0.26 Because f (1.1) < 0 and f (1.2) > 0, the zero is between 1.1 and 1.2.
f (1.2) ≈ 0.02 f (1.2) is closer to 0 than f (1.1), so decrease your guess and evaluate f (1.19). f (1.19) ≈ −0.012 Because f (1.19) < 0 and f (1.2) > 0, the zero is between 1.19 and 1.2. So, increase guess. f (1.191) ≈ −0.009 Result is negative. Increase guess. f (1.192) ≈ −0.006 Result is negative. Increase guess. f (1.193) ≈ −0.003 Result is negative. Increase guess. f (1.194) ≈ −0.0002 Result is negative. Increase guess. f (1.195) ≈ 0.003 Result is positive.
Because f (1.194) is closest to 0, x ≈ 1.194.
Substitute x = 1.194 into one of the original equations and solve for y.
y = 1 — 2 x2 + 3 = 1 — 2 (1.194)2 + 3 ≈ 3.713 So, the solution of the system is about (1.194, 3.713).
x 2 4 6 8 y −2 2 y = x2 + 3 1 2 y = 3x REMEMBER The function values that are closest to 0 correspond to x-values that best approximate the zeros of the function. Section 9.6 Solving Nonlinear Systems of Equations 529 Recall from Section 5.5 that you can use systems of equations to solve equations with variables on both sides. Approximating Solutions of an Equation Solve −2(4)x + 3 = 0.5x2 − 2x.
SOLUTION You do not know how to solve this equation algebraically. So, use each side of the equation to write the system y = −2(4)x + 3 and y = 0.5x2 − 2x.
Method 1 Use a graphing calculator to graph the system. Then use the intersect feature to fi nd the coordinates of each point of intersection.
−4 −4 4 4 Intersection X=-1 Y=2.5 −4 −4 4 4 Intersection X=.46801468 Y=-.8265105 One point of intersection The other point of intersection is (−1, 2.5). is about (0.47, −0.83). So, the solutions of the equation are x = −1 and x ≈ 0.47.
Method 2 Use the table feature to create a table of values for the equations. Find the x-values for which the corresponding y-values are approximately equal.
X Y1 Y2 X=-1 2.5204 -1.03 2.5137 2.5069 2.5 2.493 2.4859 2.4788 -1.02 -1.01 -.99 -.98 -.97 2.5905 2.5602 2.5301 2.5 2.4701 2.4402 2.4105 X Y1 Y2 X=.47 -.6808 .44 -.7321 -.7842 -.8371 -.8906 -.9449 -1 .45 .46 .48 .49 .50 -.7832 -.7988 -.8142 -.8296 -.8448 -.86 -.875 When x = −1, the corresponding When x = 0.47, the corresponding y-values are 2.5. y-values are approximately −0.83. So, the solutions of the equation are x = −1 and x ≈ 0.47.
Monitoring Progress Monitoring Progress Help in English and Spanish at BigIdeasMath.com Use the method in Example 4 to approximate the solution(s) of the system to the nearest thousandth. 10. y = 4x 11. y = 4x2 − 1 12. y = x2 + 3x y = x2 + x + 3 y = −2(3)x + 4 y = −x2 + x + 10 Solve the equation. Round your solution(s) to the nearest hundredth. 13. 3x − 1 = x2 − 2x + 5 14. 4x2 + x = −2 ( 1 — 2 ) x + 5 REMEMBER When entering the equations, be sure to use an appropriate viewing window that shows all the points of intersection. For this system, an appropriate viewing window is −4 ≤ x ≤ 4 and −4 ≤ y ≤ 4.
STUDY TIP You can use the differences between the corresponding y-values to determine the best approximation of a solution.
530 Chapter 9 Solving Quadratic Equations Exercises 9.6 Dynamic Solutions available at BigIdeasMath.com 1. VOCABULARY Describe how to use substitution to solve a system of nonlinear equations. 2. WRITING How is solving a system of nonlinear equations similar to solving a system of linear equations? How is it different?
Vocabulary and Core Concept Check Vocabulary and Core Concept Check In Exercises 3–6, match the system of equations with its graph. Then solve the system. 3. y = x2 − 2x + 1 4. y = x2 + 3x + 2 y = x + 1 y = −x − 3 5. y = x − 1 6. y = −x + 3 y = −x2 + x − 1 y = −x2 − 2x + 5 A. 1 −5 2 x y −2 B. y 2 x 4 4 C. y 2 4 2 x −2 −4 D. x y 2 4 −4 1 In Exercises 7–12, solve the system by graphing. (See Example 1.) 7. y = 3x2 − 2x + 1 8. y = x2 + 2x + 5 y = x + 7 y = −2x − 5 9. y = −2x2 − 4x 10. y = 1 — 2 x2 − 3x + 4 y = 2 y = x − 2 11. y = 1 — 3 x2 + 2x − 3 12. y = 4x2 + 5x − 7 y = 2x y = −3x + 5 In Exercises 13–18, solve the system by substitution. (See Example 2.) 13. y = x − 5 14. y = −3x2 y = x2 + 4x − 5 y = 6x + 3 15. y = −x + 7 16. y = −x2 + 7 y = −x2 − 2x − 1 y = 2x + 4 17. y − 5 = −x2 18. y = 2x2 + 3x − 4 y = 5 y − 4x = 2 In Exercises 19–26, solve the system by elimination. (See Example 3.) 19. y = x2 − 5x − 7 20. y = −3x2 + x + 2 y = −5x + 9 y = x + 4 21. y = −x2 − 2x + 2 22. y = −2x2 + x − 3 y = 4x + 2 y = 2x − 2 23. y = 2x − 1 24. y = x2 + x + 1 y = x2 y = −x − 2 25. y + 2x = 0 26. y = 2x − 7 y = x2 + 4x − 6 y + 5x = x2 − 2 27. ERROR ANALYSIS Describe and correct the error in solving the system of equations by graphing.
y = x2 − 3x + 4 y = 2x + 4 The only solution of the system is (0, 4).
✗ 2 4 x 2 4 y Monitoring Progress and Modeling with Mathematics Monitoring Progress and Modeling with Mathematics Section 9.6 Solving Nonlinear Systems of Equations 531 28. ERROR ANALYSIS Describe and correct the error in solving for one of the variables in the system.
y = 3x2 − 6x + 4 y = 4 y = 3(4)2 − 6(4) + 4 Substitute.
y = 28 Simplify.
✗ In Exercises 29–32, use the table to describe the locations of the zeros of the quadratic function f.
29. x −4 −3 −2 −1 0 1 f (x) −2 2 4 4 2 −2 30. x −1 0 1 2 3 4 f (x) 11 5 1 −1 −1 1 31. x −4 −3 −2 −1 0 1 f (x) 3 −1 −1 3 11 23 32. x 1 2 3 4 5 6 f (x) −25 −9 1 5 3 −5 In Exercises 33–38, use the method in Example 4 to approximate the solution(s) of the system to the nearest thousandth. (See Example 4.) 33. y = x2 + 2x + 3 34. y = 2x + 5 y = 3x y = x2 − 3x + 1 35. y = 2(4)x − 1 36. y = −x2 − 4x − 4 y = 3x2 + 8x y = −5x − 2 37. y = −x2 − x + 5 38. y = 2x2 + x − 8 y = 2x2 + 6x − 3 y = x2 − 5 In Exercises 39–46, solve the equation. Round your solution(s) to the nearest hundredth. (See Example 5.) 39. 3x + 1 = x2 + 7x − 1 40. −x2 + 2x = −2x + 5 41. x2 − 6x + 4 = −x2 − 2x 42. 2x2 + 8x + 10 = −x2 − 2x + 5 43. −4 ( 1 — 2 ) x = −x2 − 5 44. 1.5(2)x − 3 = −x2 + 4x 45. 8x − 2 + 3 = 2 ( 3 — 2 ) x 46. −0.5(4)x = 5x − 6 47. COMPARING METHODS Solve the system in Exercise 37 using substitution. Compare the exact solutions to the approximated solutions.
48. COMPARING METHODS Solve the system in Exercise 38 using elimination. Compare the exact solutions to the approximated solutions.
49. MODELING WITH MATHEMATICS The attendances y for two movies can be modeled by the following equations, where x is the number of days since the movies opened.
y = −x2 + 35x + 100 Movie A y = −5x + 275 Movie B When is the attendance for each movie the same?
50. MODELING WITH MATHEMATICS You and a friend are driving boats on the same lake. Your path can be modeled by the equation y = −x2 − 4x − 1, and your friend’s path can be modeled by the equation y = 2x + 8. Do your paths cross each other? If so, what are the coordinates of the point(s) where the paths meet?
51. MODELING WITH MATHEMATICS The arch of a bridge can be modeled by y = −0.002x2 + 1.06x, where x is the distance (in meters) from the left pylons and y is the height (in meters) of the arch above the water. The road can be modeled by the equation y = 52. To the nearest meter, how far from the left pylons are the two points where the road intersects the arch of the bridge?
100 200 300 400 500 x y 100 200 y = −0.002x2 + 1.06x y = 52 52. MAKING AN ARGUMENT Your friend says that a system of equations consisting of a linear equation and a quadratic equation can have zero, one, two, or infi nitely many solutions. Is your friend correct? Explain.
532 Chapter 9 Solving Quadratic Equations COMPARING METHODS In Exercises 53 and 54, solve the system of equations by (a) graphing, (b) substitution, and (c) elimination. Which method do you prefer? Explain your reasoning. 53. y = 4x + 3 54. y = x2 − 5 y = x2 + 4x − 1 y = −x + 7 55. MODELING WITH MATHEMATICS The function y = −x2 + 65x + 256 models the number y of subscribers to a website, where x is the number of days since the website launched. The number of subscribers to a competitor’s website can be modeled by a linear function. The websites have the same number of subscribers on Days 1 and 34. a. Write a linear function that models the number of subscribers to the competitor’s website. b. Solve the system to verify the function from part (a). 56. HOW DO YOU SEE IT? The diagram shows the graphs of two equations in a system that has one solution.
x y y = c a. How many solutions will the system have when you change the linear equation to y = c + 2? b. How many solutions will the system have when you change the linear equation to y = c − 2? 57. WRITING A system of equations consists of a quadratic equation whose graph opens up and a quadratic equation whose graph opens down. Describe the possible numbers of solutions of the system. Sketch examples to justify your answer. 58. PROBLEM SOLVING The population of a country is 2 million people and increases by 3% each year. The country’s food supply is suffi cient to feed 3 million people and increases at a constant rate that feeds 0.25 million additional people each year. a. When will the country fi rst experience a food shortage?
b. The country doubles the rate at which its food supply increases. Will food shortages still occur? If so, in what year?
59. ANALYZING GRAPHS Use the graphs of the linear and quadratic functions.
(2, 6) B A y x y = −x2 + 6x + 4 a. Find the coordinates of point A.
b. Find the coordinates of point B.
60. THOUGHT PROVOKING Is it possible for a system of two quadratic equations to have exactly three solutions? exactly four solutions? Explain your reasoning. (Hint: Rotations of the graphs of quadratic equations still represent quadratic equations.) 61. PROBLEM SOLVING Solve the system of three equations shown. y = 2x − 8 y = x2 − 4x − 3 y = −3(2)x 62. PROBLEM SOLVING Find the point(s) of intersection, if any, of the line y = −x − 1 and the circle x2 + y2 = 41.
Maintaining Mathematical Proficiency Maintaining Mathematical Proficiency Graph the system of linear inequalities. (Section 5.7) 63. y > 2x 64. y ≥ 4x + 1 65. y − 3 ≤ −2x 66. x + y > −6 y > −x + 4 y ≤ 7 y + 5 < 3x 2y ≤ 3x + 4 Graph the function. Describe the domain and range. (Section 8.3) 67. y = 3x2 + 2 68. y = −x2 − 6x 69. y = −2x2 + 12x − 7 70. y = 5x2 + 10x − 3 Reviewing what you learned in previous grades and lessons |
190372 | https://qehbpathology.uk/clinical-chemistry-tests/item/creatine-kinase-ck-2 | QEHB Pathology Departments - Creatine Kinase (CK)
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Creatine Kinase (CK)
Biochemistry
Preferred Sample Type
Suitable Specimen Types
Li Hep Plasma
Creatine Kinase (CK)
Specimen Volume
5 mL blood.
Sample Preparation
Centrifuge
Turnaround Time
1 day
Sample Processing In Laboratory
Usual
Sample Stability
Activity is stable in samples storeed at 4 ºC for up to seven days or four weeks when stored at -20 ºC.
General Information
Creatine kinase (CK) is an enzyme that catalyzes the reversible phosphorylation of creatine by adenosine triphosphate (ATP). The CK enzyme is a dimer composed of subunits derived from either muscle (M) or brain (B). Three isoenzymes have been identified: striated muscle (MM), heart tissue (MB), and brain (BB). Normal serum CK is predominantly the CK-MM isoenzyme. Serum CK concentrations are reflective of muscle mass causing males to have higher concentrations than females. CK may be measured to evaluate myopathy and to monitor patients with rhabdomyolysis for acute kidney injury.
Patient Preparation
None
Notes
Creatine Kinase (CK, CPK) is an enzyme found primarily in the heart and skeletal muscles, and to a lesser extent in the brain. Significant injury to any of these structures will lead to a measurable increase in CK levels. Elevations in CK are found in:
Myocardial infarction
Crushing muscular trauma
Any cardiac or muscle disease
Brain injury
Hypothyroidism
Hypokalemia
Malignant hyperthermia
Drugs (statins, fibrates, antiretrovirals etc)
Macro CK
CK can be raised following injury to the myocardium and can be detected 4h after an infarction. Hs troponin I and ECGs are now used routinely to diagnose MI or acute coronary syndrome. Muscle trauma causes CK elevations within 12 hours of onset, peaking within 1 to 3 days, and declining 3 to 5 days after cessation of muscle injury. If there is on-going injury, the CK will remain elevated indefinitely. Serum CK activities exceeding 200 times the upper reference limit may be found in acute rhabdomyolysis, putting the patient at great risk for developing acute renal failure.
Reference Range
Females 29 - 168 U/L
Males 30 - 200 U/L
CK levels are higher in males than in females, and in black race populations. Exercise, muscle trauma (contact sports, traffic accidents, intramuscular injections, surgery, convulsions, wasp or bee stings, and burns), and drugs such as cholesterol-lowering statins can damage muscle and increase serum CK concentrations.
Source of Reference Range
Abbott Diagnostics
Specifications
EQA Status:
NEQAS
EQAS Scheme:Yes
Link to Further Information
Related Tests
Phosphate (serum) Potassium (serum) Sodium
Creation Date
Monday, 08 August 2011
Modification Date
Monday, 10 February 2020
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BMC Science Grade 6
Course: BMC Science Grade 6>Unit 12
Lesson 1: Introduction to simple machines
What are Simple Machines?
Inclined Planes and Wedges
Wheel-Axle and Pulleys
Types of Simple Machines
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Introduction to simple machines
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We'll discuss the six types of simple machines: wheel and axle, pulley, lever, inclined plane, screw, and wedge.Created by Joel Prakash Stephen.
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In the country there're N N cities and some pairs of cities are connected by two-way airlines (each pair with no more than one). Every airline belongs to one of k k companies. It turns out that it's possible to get to any city from any other, but it fails when we delete all airlines belonging to any one of the companies. What is the maximum possible number of airlines in the country ?
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edited Feb 22, 2022 at 3:26
RobPratt
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asked Feb 16, 2022 at 17:15
n8qn28zn8qn28z
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5 what have you tried?cineel –cineel 2022-02-16 17:17:09 +00:00 Commented Feb 16, 2022 at 17:17
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HINT: A path on N N vertices, with each of the N−1 N−1 edges in the path belonging to a different airline, shows that there is an instance with k=N−1 k=N−1 airlines.
Now it is left to show that k=N k=N airlines is impossible. Indeed, assume that there are N N airlines. Then the available flights on the N N cities must form a connected graph with at least N N edges. So let T T be any N−1 N−1 flights that form a spanning tree of the cities. There is at least one of the N N airline does not have any of those N−1 N−1 flights on T T.
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answered Feb 16, 2022 at 17:40
MikeMike
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190375 | https://arxiv.org/pdf/2403.04055 | arXiv:2403.04055v1 [math.CO] 6 Mar 2024
Bounds for Rainbow-uncommon Graphs
Blake Bates Zhanar Berikkyzy Nick Chiem Gabriel Elvin Risa Fines Maja Lie Hanna Mikul´ as Isaac Reiter Kevin Zhou March 6, 2024
Abstract
We say a graph H is r-rainbow-uncommon if the maximum number of rainbow copies of H under an r-coloring of E(Kn) is asymptotically (as
n → ∞ ) greater than what is expected from uniformly random r-colorings. Via explicit constructions, we show that for H ∈ { K3, K 4, K 5}, H is r-rainbow-uncommon for all r ≥ (|V (H)|
2
). We also construct colorings to show that for t ≥ 6, Kt is r-rainbow-uncommon for sufficiently large r.
1 Background
A classic starting point in Ramsey theory is the following question: if any two people can identify as either strangers or acquaintances, what is the minimum number of people needed to guarantee that there exists a set of three people who are all either complete strangers or mutual acquaintances? Reframed in the language of graph theory, we can let vertices denote people and draw a blue edge between two people if they are strangers and a red edge if they are acquaintances. Then the question is equivalent to: what is the smallest n such that every coloring of the edges of a complete graph Kn contains a monochromatic triangle, i.e. a copy of K3 whose edges are either all blue or all red? The answer to this question is n = 6, and in 1959, Goodman asked and answered a more general follow-up question: for an arbitrary n, what is the minimum number of monochromatic triangles over all 2-colorings E(Kn) . He found an explicit formula, but a consequence of his result is that the minimum proportion of monochromatic triangles, that is number of monochromatic K3
total number of K3
,
approaches 1 /4 as n → ∞ . This is perhaps surprising, since 1 /4 is the expected proportion in a uniformly random coloring, i.e. the minimum can be achieved, at least asymptotically, by coloring completely randomly. This led to a new vein of Ramsey theory, classifying graphs as either common or uncommon : common if the asymptotic minimum can be achieved by uniformly random colorings, 1and uncommon if better colorings can be found. See [1, Section 2.6] for further reading in this area. This question can be “flipped”: given a fixed graph H and r colors, what is the best way to color the edges of Kn (n large) in order to maximize the number of rainbow copies of H in Kn? By rainbow , we mean that each edge of H is assigned a different color from all the others. Some progress has been made in this area, and our work builds on the work in . Next, we paraphrase those authors’ definitions, slightly altered to suit our needs. In this paper all graphs are simple. If H and G are graphs, a copy of H
in G is a subgraph of G that is isomorphic to H. We define an r-coloring of a set X to be a surjective function c : X → { 1, . . . , r }. For any graph G, let
E(G) denote the set of edges of G. Given a graph G and subgraph H, we say
H is rainbow if, under a coloring c : E(G) → { 1, . . . , r }, c(uv ) 6 = c(u′v′) for all distinct uv, u ′v′ ∈ E(H). Results in this area are asymptotic in nature, so we clarify some notation used throughout. Let f, g be functions of n. If there exist constants C, N such that |f (n)| ≤ Cg (n) for all n ≥ N , we write f = O(g). For the remainder of this section, let H be a fixed graph on t vertices with
e edges. We denote by Mrb (H; n, r ) the maximum number of rainbow copies of
H over all r-colorings of E(Kn). Furthermore, we define
mrb (H; n, r ) := Mrb (H; n, r )
total number of copies of H in Kn
, (1)
mrb (H; r) := lim
n→∞
mrb (H; n, r ). (2) For a given graph H, mrb (H; r) is known as the r-rainbow Ramsey multi-plicity constant . Note that it always exists due to the fact that the sequence
mrb (H; n, r ) is bounded and monotone in n, as shown in . The overarch-ing classification question is: for which graphs H and number of colors r can
mrb (H; r) be achieved by uniformly random colorings? Note that, for any fixed copy of H in a uniformly random r-coloring of E(Kn), there are (re
) distinct colors to choose that could make the copy of H rainbow, and once those colors are chosen there are e! ways the subgraph could be colored. Since there are re
possible colorings of this subgraph, the expected proportion of rainbow H in G
is (re
)e!/r e. This motivates the following definition: a graph H is r-rainbow-common if
mrb (H; r) =
(re
)e!
re . (3) Otherwise, H is called r-rainbow-uncommon .In this paper, we investigate rainbow-uncommonness of complete graphs, building upon the following theorem.
Theorem 1 ([2, 4]) . Kt is (t
2
)-rainbow-uncommon for all t ≥ 3.
The case t = 3 was proven by Erd˝ os and Hajnal in 1972 , and De Silva et al. showed it for all t ≥ 4 in 2019 . Our first result extends that of Erd˝ os and Hajnal and the first two cases from De Silva et al. 2Figure 1: Iterated blowup of a K4. The “thick” edges indicate that all possible edges between the sets of vertices are given the specified color.
Theorem 2. Kt is r-rainbow-uncommon for all r ≥ (t
2
), when t = 3 , 4, 5.
We also show that for t ≥ 6, Kt is r-rainbow-uncommon for all sufficiently large r, stated precisely below.
Theorem 3. Let t ≥ 6. There exists rt such that Kt is r-rainbow-uncommon for all r ≥ rt.
In the next section, we prove Theorem 3. Theorem 2 is an immediate conse-quence of the work done in the proof of Theorem 3, which we show at the end of the section.
2 Main Result
In general, to show r-rainbow-uncommonness of a graph H, we can find an r-coloring of E(Kn) with asymptotically strictly more rainbow copies of H than the number we would expect from a uniformly random coloring:
(nt
)t!
|Aut( H)| ·
(re
)e!
re =
(re
)e!
|Aut( H)|re nt + O(nt−1) (4) (the number of H in Kn times the probability that any given H is rainbow). In our proofs, we use this fact and find explicit colorings building on the iterated blow-up method in De Silva et al. . The main idea is to find a coloring of Kb which contains many rainbow copies of H. Then we “blow-up” the Kb by expanding each vertex into a copy of Kb, and between each copy make all edges the same color as the original edge between the original two vertices. We repeat this k − 1 times to get a coloring of Kn, where n = bk. An example of an iterated blowup of a K4 is given in Figure 1. Note that this generates a coloring for only a subsequence of Kns. However, since mrb (H; n, r ) is bounded 3and monotone in n , the limit mrb (H; r) must exist and every subsequence must converge to that limit. The inequalities in the following lemma were used in , and we include their proofs here for completeness.
Lemma 1. Let t ≤ b, let a denote the number of rainbow copies of H in an
r-coloring of E(Kb), and let F (n) denote the number of rainbow copies of H in the Kn generated from an iterated blow-up of Kb. Then
F (n) ≥ bF (n/b ) + a(n/b )t. (5)
Furthermore, the solution to this recurrence is
F (n) ≥ a
bt − b nt + O(nt−1). (6)
Proof. First, we derive the inequality (5). Each vertex of Kb that we blow up contains F (n/b ) rainbow copies of H, contributing the bF (n/b ) term. Addition-ally, one vertex from each of the b parts can be chosen in ( n/b )t ways, and each choice contains a rainbow copies of H. Therefore, F (n) ≥ bF (n/b ) + a(n/b )t.For an example of this bound, see Figure 1. We solve this recurrence using generating functions. First, we transform the recurrence into something that is easier to solve by making the substitution n = bk, where k is the number of iterations in the blow-up. We define the function f (k) = F (bk) to simplify (5) and get the following recurrence:
f (k) ≥ bf (k − 1) + a (bk−1)t .
Let A(x) = ∑∞
k=0
f (k)xk be the generating function. Since f (0) = 0, we have:
A(x) =
∞
∑
k=1
f (k)xk
≥
∞
∑
k=1
[
bf (k − 1) + a (bk−1)t]
xk
= b
∞
∑
k=1
f (k − 1) xk + a
∞
∑
k=1
(bk−1)txk
= bx
∞
∑
k=1
f (k − 1) xk−1 + a
bt
∞
∑
k=1
(btx)k
= bxA (x) + a
bt
[ ∞∑
k=0
(btx)k − 1
]
= bxA (x) + a
bt
[ 1
1 − btx − 1
]
= bxA (x) + ax
1 − btx .
4Note that the geometric series ∑(btx)k converges for |x| < b −t, which also ensures 1 − bx > 0. Therefore, isolating A(x) we get
A(x) ≥ ax
(1 − bx )(1 − btx) .
Next, using partial fraction decomposition, we can rewrite A(x) as a power series and obtain the closed form of the recurrence. For real numbers B and C,if we set ax
(1 − bx )(1 − btx) = B
1 − bx + C
1 − btx ,
we obtain B = a
b−bt
and C = − a
b−bt
. Therefore:
A(x) ≥ a
b − bt · 1
1 − bx − a
b − bt · 1
1 − btx
= a
b − bt
∞
∑
k=0
(bx )k − a
b − bt
∞
∑
k=0
(btx)k
=
∞
∑
k=0
[ a (btk − bk)
bt − b
]
xk.
Therefore, f (k) ≥ a(btk −bk )
bt−b
. Substituting n = bk, we get that
F (n) ≥ a(nt − n)
bt − b = a
bt − b nt + O(nt−1),
as desired.
Combining (4) with (6), it suffices to find a coloring of some Kb with a
rainbow copies of H, where
a > b(bt−1 − 1) (re
)e!
|Aut( H)|re . (7) Before stating and proving the main result, we give an explicit example that serves as a foundation and motivation for some of the techniques used in the proof. Note that the following was proved for r = 3 by Erd˝ os and Hajnal in 1972 .
Proposition 1. K3 is r-rainbow-uncommon for all r ≥ 3.Proof. We will define a coloring of E(Kr) in which every triangle is rainbow and then use the iterated blowup method to determine a lower bound of the maximum number of rainbow triangles the Kn can contain. For such coloring of
E(Kr), each color class must form a matching. We call it a parallel r-coloring .Examples with r = 7 and r = 8 are provided in Figure 2. Such coloring can be constructed for all r ≥ 3, and it is convenient to describe the parallel r-coloring for r odd and even separately. Let c denote the r-coloring of E(Kr ) with vertices {0, 1, . . . , r − 1}.50
1
2
34
5
6
0
1
2
3
4
5
6
7
Figure 2: Parallel r-coloring defined in the proof of Proposition 1. In particular, there can be no non-rainbow triangles.
• when r is odd, for each k = 1 , . . . , r color c({k+i (mod r), k −i (mod r)}) =
k for all i = 1 , . . . , r−1
2
,
• when r is even, for each k = 1 , . . . , r
2
color c({k+i (mod r), k −i (mod r)}) =
k for all i = 1 , . . . , r
2
− 1 and for each k = r
2
1 , . . . , r color c({k +
i (mod r), k − i + 1 (mod r)}) = k for all i = 1 , . . . , r
2
.It follows that no vertex has two incident edges of the same color, and hence every K3 must be rainbow. By the proof of Lemma 1, the number of rainbow copies of K3 in a Kn generated by the iterated blow-up of Kr with coloring c
is at least r−2
6( r+1)
(n3 − n), while the expected number of rainbow copies in a uniformly random coloring is (r−1)( r−2)
6r2
n(n − 1)( n − 2). It is straightforward to check that this construction satisfies the inequality (7).
The next two results give us an upper bound the number of non-rainbow
Kts in Kr under a parallel r-coloring.
Lemma 2. Let r ≥ 6, and let c : E(Kr) → { 1, . . . , r } be a parallel r-coloring. Then there are at most r(⌊r/ 2⌋
2
) non-rainbow copies of K4 in Kr under c.Proof. Given r ≥ 6, let c be a parallel r-coloring of E(Kr ). Notice that under this coloring, every non-rainbow K4 in Kr uses exactly two edges from some color class. Therefore, for each color, there are at most (⌊r/ 2⌋
2
) non-rainbow copies of K4 that have that color repeated. Thus, since there are r colors in total, the number of non-rainbow K4s in Kr is at most r(⌊r/ 2⌋
2
).
Lemma 3. Let t ≥ 4, and let r ≥ (t
2
) and c : E(Kr ) → { 1, . . . , r } be a parallel
r-coloring. Then there are at most r(⌊r/ 2⌋
2
)( r−4
t−4
) non-rainbow copies of Kt in
Kr under c.Proof. Given t ≥ 4 and r ≥ (t
2
), let c be a parallel r-coloring of E(Kr ). Notice that every non-rainbow Kt in Kr contains a non-rainbow K4. By Lemma 2, 6there are at most r(⌊r/ 2⌋
2
) non-rainbow K4s in Kr under the parallel r-coloring, and each non-rainbow K4 is contained in at most (r−4
t−4
) Kts because we may select the remaining t − 4 vertices of Kt from Kr in (r−4
t−4
) ways. Therefore, the number of non-rainbow copies of Kt in Kr is at most r(⌊r/ 2⌋
2
)( r−4
t−4
) under c.
We are now ready to prove our main result, restated (and adjusted slightly) below.
Theorem 4. Let t ≥ 4. There exists rt ≥ (t
2
) such that for all r ≥ rt, Kt is
r-rainbow-uncommon. Proof. Let t ≥ 4 and r ≥ (t
2
), and let c be a parallel r-coloring of E(Kr ). By Lemma 3, there are at most r(⌊r/ 2⌋
2
)( r−4
t−4
) non-rainbow copies of Kt in Kr under
c. Therefore, there are at least
(rt
)
− r
(⌊r/ 2⌋
2
)( r − 4
t − 4
)
≥
(rt
) (
1 − r · t!
8( t − 4)!( r − 1)( r − 3)
)
rainbow copies of Kt in Kr. In the inequality above, note that (⌊r/ 2⌋
2
) ≤ r(r−2)
8
.We will use the iterated blow-up method described at the beginning of this section with parameters H = Kt, b = r, and a = (rt
) (
1 − r·t!
8( t−4)!( r−1)( r−3)
)
.Note that this implies |Aut (H)| = t! and e = (t
2
). Therefore, it suffices to show that the inequality (7) is satisfied, i.e. we will show that
(rt
) (
1 − r · t!
8( t − 4)!( r − 1)( r − 3)
)
r(rt−1 − 1) ( r
(t
2
)
)( t
2
)!
t!r(t
2
) . (8) Rearranging the terms in the inequality (8), one can show that it is equivalent to
(
1 − r · t!
8( t − 4)!( r − 1)( r − 3)
)
rt(t−1) /2 > (r − t)!
(r − (t
2
))! (rt − r).
Notice that the term (r−t)!
(r−
(t
2
))! on the right hand side of the inequality above can be written as (r − t)( r − t − 1) · · ·
(
r −
(t
2
)
1
)
=(t
2
)−1
∏
l=t
(r − l) ,
therefore, the inequality (8) holds if and only if the following inequality holds:
(
(r − 1)( r − 3) − r · t!
8( t − 4)!
)
rt(t−1) /2 −(r−1)( r−3)
(t
2
)−1
∏
l=t
(r − l)
(rt −r) > 0.
(9) 7Consider the leading coefficient of the polynomial on the left hand side of (9). Notice that the term with the largest power of r that appears in (9) is r2+ t(t−1) /2,however its coefficient is 0. Therefore, we will focus on the coefficient of the term
r1+ t(t−1) /2, which is
(
−4 − t!
8( t − 4)!
)
−
−4 −
(t
2
)−1
∑
l=t
l
= t(t − 1)( t − 3) /2. (10) Since t ≥ 4, the leading coefficient calculated in (10) is strictly positive. This implies that there exists a large enough integer rt that satisfies the inequality (9), in fact, the inequality (9) will hold for all r ≥ rt. Therefore, Kt is r-rainbow uncommon for all r ≥ rt.
Theorem 3 follows directly from the theorem above. Theorem 2 follows from Proposition 1 and by substituting t = 4 , r = 6 and t = 5 , r = 10, into (9) and seeing that both cases satisfy the inequality.
3 Conclusion
Sun very recently showed that any graph containing a cycle is r-rainbow-uncommon for all r ≥ e . While that implies our result, we explicitly construct the col-orings for complete graphs that show rainbow-uncommonness. Future work is to investigate upper bounds, particularly to show which graphs are r-rainbow-common. Thus far, the only class of graphs known to be rainbow-common are disjoint unions of stars . It is also likely that P4, a path on four vertices, is 3-rainbow-common. We believe the converse of Sun’s result is also true, formulated in the following conjecture.
Conjecture 1. H is rainbow-common if and only if H is a forest. By rainbow-common , we mean r-rainbow-common for all r ≥ e.
Acknowledgements
This work was done as part of the 2021 Polymath Jr program, partially sup-ported by NSF award DMS–2113535. We are grateful to the organizers of the Polymath Jr. program for creating the environment to carry out this research project. We also thank our colleagues Pablo Blanco and Sarvagya Jain who contributed to helpful conversations and questions that were valuable for the present paper. The work of the second author was partially supported by the Fairfield Summer Research award. 8References
David Conlon, Jacob Fox, and Benny Sudakov, Recent developments in graph Ramsey theory , arXiv e-prints (2015), arXiv:1501.02474. Paul Erd˝ os and Andr´ as Hajnal, On ramsey like theorems, problems and results , Combinatorica (1972). A. W. Goodman, On sets of acquaintances and strangers at any party , The American Mathematical Monthly 66 (1959), no. 9, 778–783. Jessica De Silva, Xiang Si, Michael Tait, Yunus Tun¸ cbilek, Ruifan Yang, and Michael Young, Anti-ramsey multiplicities , Australas. J Comb. 73 (2019), 357–371. Yihang Sun, Rainbow common graphs must be forests , (2023). 9 |
190376 | https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Radioactivity/Nuclear_Decay_Pathways | Nuclear Decay Pathways - Chemistry LibreTexts
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Radioactivity
Nuclear Chemistry
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Mon, 30 Jan 2023 07:05:17 GMT
Nuclear Decay Pathways
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Contents
1. Home
2. Bookshelves
3. Physical & Theoretical Chemistry
4. Supplemental Modules (Physical and Theoretical Chemistry)
5. Nuclear Chemistry
6. Radioactivity
7. Nuclear Decay Pathways
Expand/collapse global location
Supplemental Modules (Physical and Theoretical Chemistry)
Fundamentals
Atomic Theory
Physical Properties of Matter
Acids and Bases
Kinetics
Equilibria
Thermodynamics
Statistical Mechanics
Quantum Mechanics
Chemical Bonding
Electronic Structure of Atoms and Molecules
Spectroscopy
Nuclear Chemistry
Radioactivity
Artificially Induced Radioactivity
Discovery of Radioactivity
Nuclear Decay Pathways
Purifying Radioactive Materials
The Effects of Radiation on Matter
Transmutation of the Elements
Nuclear Energetics and Stability/Nuclear_Chemistry/Nuclear_Energetics_and_Stability)
Fission and Fusion/Nuclear_Chemistry/Fission_and_Fusion)
Nuclear Kinetics/Nuclear_Chemistry/Nuclear_Kinetics)
Applications of Nuclear Chemistry/Nuclear_Chemistry/Applications_of_Nuclear_Chemistry)
Group Theory
Nuclear Decay Pathways
Last updated Jan 30, 2023
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Discovery of Radioactivity
Purifying Radioactive Materials
Page ID 1487
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Types of Nuclear Decay
1. Beta Decay / Negatron Emission
2. Positron Emission
3. Electron Capture
4. Alpha Decay
Summary
Problems
Solutions
References
Nuclear reactions that transform atomic nuclei alter their identity and spontaneously emit radiation via processes of radioactive decay.
Types of Nuclear Decay
In 1889, Ernest Rutherford recognized and named two modes of radioactive decay, showing the occurrence of both processes in a decaying sample of natural uranium and its daughters. Rutherford named these types of radiation based on their penetrating power: heavier alpha and lighter beta radiation. Gamma rays, a third type of radiation, were discovered by P. Villard in 1900 but weren't recognized as electromagnetic radiation until 1914. Since gamma radiation is only the discharge of a high-energy photon from an over-excited nucleus, it does not change the identity of the atom from which it originates and therefore will not be discussed in depth here.
Because nuclear reactions involve the breaking of very powerful intra nuclear bonds, massive amounts of energy can be released. At such high energy levels, the matter can be converted directly to energy according to Einstein's famous Mass-Energy relationship E = mc 2. The sum of mass and energy are conserved in nuclear decay. The free energy of any spontaneous reaction must be negative according to thermodynamics (ΔG < 0), and Δ G is essentially equal to the energy change Δ E of nuclear reactions because ΔE is so massive. Therefore, a nuclear reaction will occur spontaneously when:
(1)ΔE=Δmc 2<0
ΔE<0 or Δm<0
When the mass of the products of a nuclear reaction weigh less than the reactants, the difference in mass has been converted to energy.
There are three types of nuclear reactions that are classified as beta decay processes. Beta decay processes have been observed in 97% of all known unstable nuclides and are thus the most common mechanism for radioactive decay by far. The first type (here referred to as beta decay) is also called Negatron Emission because a negatively charged beta particle is emitted, whereas the second type (positron emission) emits a positively charged beta particle. In electron capture, an orbital electron is captured by the nucleus and absorbed in the reaction. All these modes of decay represent changes of one in the atomic number Z of the parent nucleus but no change in the mass number A. Alpha decay is different because both the atomic and mass number of the parent nucleus decrease. In this article, the term beta decay will refer to the first process described in which a true beta particle is the product of the nuclear reaction.
Beta Decay / Negatron Emission
Nuclides can be radioactive and undergo nuclear decay for many reasons. Beta decay can occur in nuclei that are rich in neutrons - that is - the nuclide contains more neutrons than stable isotopes of the same element. These "proton deficient" nuclides can sometimes be identified simply by noticing that their mass number A (the sum of neutrons and protons in the nucleus) is significantly more than twice that of the atomic number Z (number of protons in nucleus). In order to regain some stability, such a nucleus can decay by converting one of its extra neutrons into a proton, emitting an electron and an antineutrino(ν). The high energy electron emitted in this reaction is called a beta particle and is represented by −1 0e− in nuclear equations. Lighter atoms (Z < 60) are the most likely to undergo beta decay. The decay of a neutron to a proton, a beta particle, and an antineutrino (ν¯) is
(2)A 0 1 A 2 0 2 1nA 0A 0 A 1 A 2 2 1pA++A−1 0 A 2−1 2 0eA−+ν¯
Some examples of beta decay are
(3)A 2 6 A 2 2 2 6HeA 3 A 6 A 2 2 6Li+A−1 0 A 2−1 2 0eA−+ν¯
(4)A 11 24 A 2 11 2 24NaA 12 A 24 A 2 2 24Mg+A−1 0 A 2−1 2 0eA−+ν¯
In order for beta decay to occur spontaneously according to Δm < 0, the mass of the parent nucleus (not atom) must have a mass greater than the sum of the masses of the daughter nucleus and the beta particle:
m[A Z] >m[A(Z+1)] + m[0-1 e-]
(Parent nucleus) > (Daughter nucleus) + (electron)
The mass of the antineutrino is almost zero and can therefore be neglected. The equation above can be reached easily from any beta decay reaction, however, it is not useful because mass spectrometers measure the mass of atoms rather than just their nuclei. To make the equation useful, we must make these nuclei into neutral atoms by adding the mass of Z + 1 electrons to each side of the equation. The parent nucleus then becomes the neutral atom [A Z] plus the mass of one electron, while the daughter nucleus and the beta particle on the right side of the equation become the neutral atom [A(Z+1)] plus the mass of the beta particle. The extra electron on the left cancels the mass of the beta particle on the right, leaving the inequality
m[A Z] >m[A(Z+1)]
(Parent atom) > (Daughter atom)
The change in mass then equals
Δm = m[A(Z+1)] - m[A Z]
The energy released in this reaction is carried away as kinetic energy by the beta particle and antineutrino, with an insignificant of energy causing recoil in the daughter nucleus. The beta particle can carry anywhere from all to none of this energy, therefore the maximum kinetic energy of a beta particle in any instance of beta decay is -ΔE.
Positron Emission
Nuclides that are imbalanced in their ratio of protons to neutrons undergo decay to correct the imbalance. Nuclei that are rich in protons relative to their number of neutrons can decay by conversion of a proton to a neutron, emitting a positron (1 0e+) and a neutrino (ν). Positrons are the antiparticles of electrons, therefore a positron has the same mass as an electron but with the opposite (positive) charge. In positron emission, the atomic number Z decreases by 1 while the mass number A remains the same.
Some examples of positron emission are
(5)A 5 8 A 2 5 2 8BA 4 8Be+A 1 0 A 2 1 2 0eA++ν e
(6)A 25 50 A 2 25 2 50MgA 24 50Cr+A 1 0 A 2 1 2 0eA++ν e
Positron emission is only one of the two types of decay that tends to happen in "neutron deficient" nuclides, therefore it is very important to establish the correct mass change criterion. Positron emission occurs spontaneously when
m[A Z] >m[A(Z-1)] + m[0+1 e+]
(Parent nucleus) > (Daughter nucleus) + (positron)
In order to rewrite this inequality in terms of the masses of neutral atoms, we add the mass of Z electrons to both sides of the equation, giving the mass of a neutral [A Z] atom on the left and the mass of a neutral [A(Z-1)] atom, plus an extra electron, (since only Z-1 electrons are needed to make the neutral atom), and a positron on the right. Because positrons and electrons have equal mass, the inequality can be written as
m[A Z] >m[A(Z-1)] + 2 m[0-1 e-]
(Parent atom) > (Daughter atom) + (2 electrons)
The change in mass for positron emission decay is
Δm = m[A(Z)] - m[A Z] - 2 m[0-1 e-]
As with beta decay, the kinetic energy -ΔE is split between the emitted particles - in this case the positron and neutrino.
Electron Capture
As mentioned before, there are two ways in which neutron-deficient / proton-rich nuclei can decay. When the mass change Δm<0 yet is insufficient to cause spontaneous positron emission, a neutron can form by an alternate process known as electron capture. An outside electron is pulled inside the nucleus and combined with a proton to make a neutron, emitting only a neutrino.
(7)A 1 1 A 2 1 2 1p+A−1 0 A 2−1 2 0eA−A 0 1 A 2 0 2 1n+ν
Some examples of electron capture are
(8)A 92 231 A 2 92 2 231U+A−1 0 A 2−1 2 0eA−A 91 231 A 2 91 2 231Pa+ν
(9)A 81 A 2 2 8136Kr+A−1 0 A 2−1 2 0eA−A 35 81 A 2 35 2 81Br+ν
Electron capture happens most often in the heavier neutron-deficient elements where the mass change is smallest and positron emission isn't always possible. For Δm<0, the following inequality applies:
m[A Z] + m[0-1 e-] >m[A(Z-1)]
(Parent nucleus) + (electron) > (Daughter nucleus)
Adding Z electrons to each side of the inequality changes it to its useful form in which the captured electron on the left cancels out the extra electron on the right
m[A Z] >m[A(Z-1)]
(Parent atom) > (Daughter atom)
The change in mass then equals
Δm = m[A(Z-1)] - m[A Z]
When the loss of mass in a nuclear reaction is greater than zero, but less than 2 m[0-1 e-], the process cannot occur by positron emission and is spontaneous for electron capture.
Alpha Decay
The other three processes of nuclear decay involve the formation of a neutron or a proton inside the nucleus to correct an existing imbalance. In alpha decay, unstable, heavy nuclei (typically Z>83) reduce their mass number A by 4 and their atomic number Z by 2 with the emission of a helium nuclei (A 2 4 A 2 2 2 4HeA 2+), known as an alpha particle.
Some examples of alpha decay are
(10)A 88 222 A 2 88 2 222Ra→A 86 218 A 2 86 2 218Rn+A 2 4 A 2 2 2 4HeA 2+
(11)A 92 233 A 2 92 2 233U→A 90 229 A 2 90 2 229Th+A 2 4 A 2 2 2 4HeA 2+
As with beta decay and electron capture, Δm must only be less than zero for spontaneous alpha decay to occur. Since the number of total protons on each side of the reaction does not change, equal numbers of electrons are added to each side to make neutral atoms. Therefore, the mass of the parent atom must simply be greater than the sum of the masses of its daughter atom and the helium atom.
m[A Z] >m[A-4(Z-2)] + m[4 2 He 2+]
The change in mass then equals
Δm = m[A(Z)] - m[A-4(Z-2)] - m[4 2 He 2+]
The energy released in an alpha decay reaction is mostly carried away by the lighter helium, with a small amount of energy manifesting itself in the recoil of the much heavier daughter nucleus. Alpha decay is a form of spontaneous fission, a reaction in which a massive nuclei can lower its mass and atomic number by splitting. Other heavy unstable elements undergo fission reactions in which they split into nuclei of about equal size.
Summary
Proton-deficient or neutron-deficient nuclei undergo nuclear decay reactions that serve to correct unbalanced neutron/proton ratios. Proton-deficient nuclei undergo beta decay - emitting a beta particle (electron) and an antineutrino to convert a neutron to a proton - thus raising the elements atomic number Z by one. Neutron-deficient nuclei can undergo positron emission or electron capture (depending on the mass change), either of which synthesizes a neutron - emitting a positron and a neutrino or absorbing an electron and emitting a neutrino respectively - thus lowering Z by one. Nuclei with Z > 83 which are unstable and too massive will correct by alpha decay, emitting an alpha particle (helium nucleus) and decreasing both mass and atomic number. Very proton-deficient or neutron-deficient nuclei can also simply eject an excess particle directly from the nucleus. These types of decay are called proton and neutron emission. These processes are summarized in the table below.
Table: Characteristics of Radioactive Decay| Decay Type | Emitted Particle | ΔZ | ΔA | Occurrence |
--- ---
| Alpha | 4 2 He 2+ | -2 | -4 | Z > 83 |
| Beta | Energetic e-, γ | +1 | 0 | A/Z > (A/Z)stable |
| PE | Energetic e+, γ | -1 | 0 | A/Z < (A/Z)stable, light nuclei |
| EC | ν | -1 | 0 | A/Z < (A/Z)stable, heavy nuclei |
| γ | Photon | 0 | 0 | Any excited nucleus |
Problems
Write the balanced equation for the beta decay of 14 C.
Write the balanced equation for the positron emission decay of 22 Na.
Write the balanced equation for electron capture in 207 Bi.
Write the balanced equation for the alpha decay of 238 U.
Calculate the maximum kinetic energy of the emitted beta particle in the decay
24 11 Na → 24 12 Mg + 0-1 e- + ν
Use Table A4 of particle masses to do this calculation.
6. Calculate the maximum kinetic energy of the positron emitted in the decay
8 5 B → 8 4 Be + 0 1 e+ + ν
Use Table A4 of particle masses to do this calculation.
7. Will 231 92 U likely decay to 231 91 Pa by positron emission or by electron capture? Use the mass criterion equations.
Solutions
14 6 C → 14 7 N + 0-1 e- + ν
22 11 Na → 22 10 Ne + 0+1 e+ + ν
207 83 Bi + 0-1 e- → 207 82 Pb + ν
238 92 U → 234 90 Th + 4 2 He
Given the masses of relevant atoms and the mass change criteria for beta decay, we calculate:
Δm = m[24 12 Mg] - m[24 11 Na]
Δm = 23.9850419 u - 23.990963 u = -0.0059211 u
ΔE = (-0.0059211 u)(931.494 MeV / u) = -5.515 MeV
The maximum kinetic energy of the beta particle is 5.515 MeV.
6. The change in mass is:
Δm = m[8 4 Be] + 2 m[0-1 e-] - m[8 5 B]
Δm = 8.005305 u + 2(0.000548579911 u) - 8.024606 u = -0.01820384 u
ΔE = (-0.01820384 u)(931.494 MeV / u) = -16.9568 MeV
The maximum kinetic energy of the positron is 16.9568 MeV.
7. The difference in mass between the daughter and parent atom is:
Δm = m[231 91 Pa] - m[231 92 U]
Δm = 231.035879 u - 231.03689 u = -0.00041 u
2 m[0-1 e-] = 0.0010971598 u
Since 0.00041 u is less than 2 m[0-1 e-], the process cannot occur by positron emission. The mass criterion Δm < 0 for electron capture is met, therefore 231 U decays by electron capture.
References
Oxtoby, David W.; Gillis H. P.; Campion, Alan. Principles of Modern Chemistry, 6th ed.; Thomson Brooks/Cole: Belmont, CA, 2008; Chp. 19.
Haskin, Larry A. The Atomic Nucleus and Chemistry; D. C. Heath and Company: Lexington, MA, 1972; pp. 3-4, 43-53.
Loveland, Walter D.; Morrissey, David J.; Seaborg, Glenn T. Modern Nuclear Chemistry; John Wiley & Sons, Inc.: Hoboken, NJ, 2006; pp. 8-10, 177-179, 199-200, 213.
Nuclear Decay Pathways is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Amy Anderson.
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190377 | https://www.biointeractive.org/classroom-resources/natural-selection-and-evolution-darwins-finches | Natural Selection and the Evolution of Darwin’s Finches
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Natural Selection and the Evolution of Darwin’s Finches
Topic
Evolution
Natural Selection
Speciation
Science Practices
Explanations & Argumentation
Resource Type
Activities
Card Activities
Level
High School — GeneralHigh School — AP/IBCollege
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Description
In this activity, students develop arguments for the adaptation and natural selection of Darwin’s finches, based on evidence presented in the film The Beak of the Finch.
Students watch segments of the film and then engage in discussion, make predictions, create models, interpret graphs, and use multiple sources and types of evidence to develop arguments for the evolution of Darwin’s finches. The activity highlights two key NGSS science and engineering practices: engaging in argument from evidence using mathematical and computational thinking, and analyzing and ...
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Student Learning Targets
Make claims and construct arguments using evidence from class discussion and from a short film on the evolution of the Galápagos finches.
Use data to make predictions about the effects of natural selection in a finch population.
Construct graphs to illustrate predicted results and compare them to actual results.
Details
Estimated Time
Two to three 50-minute class periods.
Key Terms
adaptation, evolution, mutation, natural selection, rationale, trait
Terms of Use
The resource is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International license. No rights are granted to use HHMI’s or BioInteractive’s names or logos independent from this Resource or in any derivative works.
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Version History
Date Published 11.04.15
Date Updated 04.07.20
Curriculum Connections
NGSS (2013)
HS-LS2-1, HS-LS3-3, HS-LS4-3, HS-LS4-5; SEP1, SEP4
AP Biology (2019)
EVO-1.C, EVO-1.D, EVO-1.E, EVO-1.H, EVO-1.J, EVO-1.O, EVO-3.A, SYI-3.D; SP2, SP3, SP4, SP6
IB Biology (2016)
5.1 5.4, C.1
AP Environmental Science (2020)
Topic(s): 2.6, 2.7
Learning Objectives & Practices: ERT-2.H, SP2, SP5
IB Environmental Systems and Societies (2017)
3.2
Common Core (2010)
ELA.RST.9-12.4, ELA.WHST.9-12.1
MP2, MP4
Vision and Change (2009)
CC1, CC5; DP1, DP2
Materials
Download Educator Materials (PDF)776 KB
Download Student Handout (PDF)566 KB
Download Finch Cards (PDF)423 KB
Download Card Images (ZIP)354 KB
Additional Materials
scissors
large white poster board or butcher paper
blue painter’s tape for attaching cards to poster board or butcher paper
access to a camera (optional)
sticky notes or index cards
graph paper
science notebooks or paper for writing
different-colored pens (optional)
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Effects of Natural Selection on Finch Beak Size
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How Can We Explain Evolutionary Relationships among Species?
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Reading Primary Sources: Darwin and Wallace
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The Origin of Species: The Beak of the Finch
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Reproductive Isolation and Speciation in Lizards
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190378 | https://zhuanlan.zhihu.com/p/508421715 | 微积分（速度和加速度） - 知乎
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微积分（速度和加速度）
首发于林先生的学科笔记
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微积分（速度和加速度）
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收录于 · 林先生的学科笔记
前面我们谈到了速度，速度是位移/时间。
而加速度，是速度关于t的导数，也就是当我们对于一个速度求导，其结果就形成了加速度
而速度是位置的导数，因此加速度是位置的二阶导。
比如说，
首先，我们要把速度求出来
从速度的公式可以知道，我们可以首先求出x，即位置。然后除以t，求出速度。
但是现在我们已经知道，速度是位置的导数。因此我们直接对位置这个函数进行求导，
他的结果是，
然后再对其进行二阶导，得出
把3代入进去，结果就是
其次是所谓的负常数加速度。
假设一个球往上抛，他会首先向上，然后下降，那么，向下的时候，他的加速度方向是朝下的，如果我们以0点为地面，向上为正，那么加速度此时是向下，所以他是负的量。而此时加速是重力加速度-g。
其次，我们现在如果知道了重力加速度，我们是可以倒退他的速度表达式和位置表达式的，
那么，我们来看一个例子，
首先看
第一个问题，他的问题转化为数学语言，就是x=0的时候，此时t=？
第二个问题是，当x=0的时候，此时v=？
第三个问题是，当v=0的时候，此时x=？
第四个问题是，当x=0的时候，t=？
第五个问题是，当x=0，此时v=？
第一个问题，我们可以求出来，
0=-5t^2+3t+2，此时t=1or t=-2/5
那么t肯定是大于0，所以t=1
第二个问题，则是，我们现在已经知道t=1，带入到速度表达式
v=-10+3=-7，因此此时速度=-7m/s
第三个问题，v=0，此时t=？
我们用速度方程，-10t+3=0，t=3/10。那么此时他可以达到多高呢？把他在带入位置的表达式，
即
第四个问题，与第一个问题，实质上是一样的。但是有个东西不一样，就是一个是向上抛，一个是向下抛。向下抛的时候，他的初始速度应该是负数
然后再把x=0，此时t=2/5 or t=-1，此时t＞0，因此t=2/5
第五个问题，与第二个问题是一样的，但是区别在于，初始速度仍然是负数
我们把t=2/5带入到速度表达式，其结果是-7m/s
编辑于 2022-05-02 07:06
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190379 | https://math.uchicago.edu/~may/REU2019/REUPapers/Zhang,Yueheng.pdf | ON THE PRINCIPAL EIGENVECTOR OF A GRAPH YUEHENG ZHANG The University of Chicago Abstract. The principal ratio of a connected graph G, γ(G), is the ratio between the largest and smallest coordinates of the principal eigenvector of the adjacency matrix of G. Over all connected graphs on n vertices, γ(G) ranges from 1 to ncn. Moreover, γ(G) = 1 if and only if G is regular. This indicates that γ(G) can be viewed as an irregularity measure of G, as first suggested by Tait and Tobin (El. J. Lin. Alg. 2018). We are interested in how stable this measure is. In particular, we ask how γ changes when there is a small modification to a regular graph G. We show that this ratio is polynomially bounded if we remove an edge belonging to a cycle of bounded length in G, while the ratio can jump from 1 to exponential if we join a pair of vertices at distance 2. We study the connection between the spectral gap of a regular graph and the stability of its principal ratio. A naive bound shows that given a constant multiplicative spectral gap and bounded degree, the ratio remains polynomially bounded if we add or delete an edge. Using results from matrix perturbation theory, we show that given an additive spectral gap larger than 2√n, the ratio stays bounded after adding or deleting an edge.
Contents 1.
Introduction 2 2.
General preliminaries 3 2.1.
Definitions and notation 3 2.2.
Results from linear algebra 4 2.3.
Results from spectral graph theory 6 2.3.1.
Observations about the adjacency operator 6 2.3.2.
Existence of the principal eigenvector for a connected graph 6 2.3.3.
Bounds on the largest eigenvalue for a graph 7 2.3.4.
The largest eigenvalue of subgraphs 8 2.3.5.
Graph products 8 3.
Preliminary results about the principal eigenvector 9 3.1.
Observations and naive bounds on the ratio 9 3.2.
Chebyshev polynomials and principal eigenvectors 11 3.2.1.
Chebyshev polynomials 11 3.2.2.
Applications to prinicpal eigenvectors 11 4.
Main results 12 4.1.
Adding or removing an edge in bounded distance 12 4.1.1.
Adding an edge 12 Date: Dec 31, 2019.
Written for the University of Chicago 2019 Math REU; mentored by Professor L´ aszl´ o Babai.
1 ON THE PRINCIPAL EIGENVECTOR OF A GRAPH 2 4.1.2.
Removing an edge 14 4.2.
Multiplicative spectral gap and stability of the ratio 15 4.3.
Additive spectral gap and stability of the ratio 16 4.3.1.
Existence of graphs with large additive spectral gap and large diameter 16 4.3.2.
Large additive spectral gap implies bounded ratio 17 4.4.
Open questions 20 Acknowledgments 21 References 21 1. Introduction It is known that the adjacency matrix AG of every connected graph G has a simple largest eigenvalue λ1, and that λ1 has an eigenvector with all-positive coor-dinates, called the principal eigenvector of G, which we denote by q. Therefore a unique-up-to-scaling all-positive eigenvector can be associated with every connected graph. Then it is natural to study how q reflects the structure of the graph. All our discussions will be asympototic as the number of vertices approaches infinity in a family of graph.
Cioab˘ a and Gregory first defined the principal ratio of G, γ(G) = qmax/qmin, to be the ratio between the largest and smallest coordinates of q. This ratio is 1 for regular graphs, while it can grow at factorial rate (i.e., γ(G) > ncn for some positive constant c) . Since γ(G) ≥1 where equality holds if and only if G is regular, it is natural to think of γ(G) as a measure of the irregularity of G. This view was suggested by Tait and Tobin .
A basic observation is that, given a connected graph G with largest eigenvalue λ1 and diameter D, the principal ratio satisfies (1.1) γ(G) ≤λD 1 .
We are interested in the stability of γ, i.e., how a slight change of G influences γ(G). In particular, given a d-regular graph G, we ask how γ(G) changes from the constant 1 if we add or remove one edge in G. (We call the resulting graphs G + e and G−e, respectively.) We always assume the edge we remove will not disconnect G (i.e., e is a non-bridge edge), so that the principal eigenvector of G−e is defined.
In Section 4.1, we study the cases where the edge we add to or remove from a regular graph is between vertices of bounded distance. We show that • γ(G + e) can jump to exponential in n when the degree is bounded [The-orem 4.3].
In our example, e connects two vertices at distance 2 in G.
By (1.1), boundedness of the degree is necessary here.
• If we remove an edge belonging to a cycle of bounded length in G, γ(G−e) is always polynomially bounded regardless of the degree [Theorem 4.11].
We also study the relevance of the spectral gap to the stability of γ(G) for regular graphs. In Section 4.2, based on (1.1), we note that • γ(G ± e) is always polynomially bounded in n when G is a bounded-degree expander graph, i.e., when the degree is bounded and the spectral gap of G is bounded from below [Observation 4.15].
YUEHENG ZHANG 3 In Section 4.3.2, we put this problem in the more general context of perturbations of matrices. By adapting theorems and proofs from Stewart and Sun’s book to our special case, we show that • If there is an additive spectral gap larger than 2√n, then γ(G±e) is bounded [Theorem 4.17].
This result does not follow from (1.1). Indeed, in Section 4.3.1 we construct graphs with degree of order n(2+t)/3 and additive spectral gap of order nt, having diam-eter of order n(1−t)/3 for any constant 0 < t < 1. Similar applications of matrix perturbation theory in link analysis for networks can be found in .
When computing or giving bounds on the coordinates of the principal eigenvec-tor for certain types of graphs, we take advantage of the properties of Chebyshev polynomials, a family of orthogonal polynomials which has found numerous appli-cations in discrete mathematics. Here is an incomplete list of the areas of such applications: • the matchings polynomial of graphs, by Heilmann and Lieb • approximate Inclusion–Exclusion, by Linial and Nisan • analysis of Boolean functions, in bounding the real degree of the OR func-tion, by Nisan and Szegedy • the diameter of regular and bipartite biregular graphs, by van Dam and Haemers • counting restricted permutations, by Mansour and Vainshtein • the mixing rate of non-backtracking random walks, by Alon et al. .
In Section 3.2, we state some properties of Chebyshev polynomials and show their connection with the principal eigenvectors of certain graphs.
2. General preliminaries 2.1. Definitions and notation.
By a graph we mean what is often called a simple graph (undirected graph with no self-loops and no parallel edges). G will always denote a connected graph with n vertices.
We denote by V (G) and E(G) the set of vertices and edges of G, respectively. We usually identify the set of vertices with the set [n] = {1, 2, . . . , n}, so the vertices are labeled 1, . . . , n. We write i ∼G j if vertices i, j are adjacent in G. We denote by NG(j) the set of neighbors of j in G. We use degG(j) to denote the degree of vertex j in G. We write G for the complement of G. Let distG(i, j) denote the distance between vertices i and j in G. Let D(G) := maxi,j dist(i, j) denote the diameter of G.
We use Mn(R) to denote the set of n×n real matrices. We write j for the all ones vector, and J for the all-ones matrix. We use AG to denote the adjacency matrix of G. We note that AG is a real symmetric matrix, so its eigenvalues are real. We write λ1(G) ≥λ2(G) ≥· · · ≥λn(G) to denote the eigenvalues of AG. We also denote λ1(G) by λG. We write q(G) for the principal eigenvector of AG scaled to have l2 norm 1. Let qi(G) denote the coordinate corresponding to vertex i in q(G).
We write qmax(G) and qmin(G) for the maximum and minimum coordinates of q(G), and vmax(G) and vmin(G) for corresponding vertices. Recall that the principal ratio of G is defined as (2.1) γ(G) := qmax(G) qmin(G) .
ON THE PRINCIPAL EIGENVECTOR OF A GRAPH 4 We write LG for the Laplacian of G, defined as the n × n matrix LG = diag(deg(1), deg(2), . . . , deg(n)) −AG.
LG is positive semidefinite. The principal eigenvalue of LG is defined to be the eigenvalue corresponding to the eigenvector j; its value is zero. We write δ(G) to denote the smallest non-principal eigenvalue of LG. We note that δ(G) = 0 if and only if G is disconnected. δ(G) is the algebraic connectivity of the graph G, as first defined by Fiedler .
For a d-regular graph G, let fA(t) be the characteristic polynomial of its adja-cency matrix AG. The characteristic polynomial of the Laplacian LG is (2.2) fL(t) = fA(d −t).
It follows that (2.3) δ(G) = d −λ2(G).
We refer to the right-hand side as the additive spectral gap of G. We refer to (2.4) δ(G) d = 1 −λ2 d as the multiplicative spectral gap of G. We use this terminology for regular graphs only.
In all notation, we omit the graph G when it is clear from context.
Let Cn denote the cycle with n vertices. Let Pr denote the path with r vertices; it has r −1 edges.
Let Ks denote the clique with s vertices; it has s 2 edges.
Following the notation used in previous papers on this subject, we use Pr · Ks to denote the graph obtained by merging the vertex at one end of Pr with one vertex in Ks. So Pr · Ks has n = r + s −1 vertices, r −1+ s 2 edges, and diameter r. This has been called a kite graph or a lollipop graph. We will call it a kite graph.
By a family of graphs, we mean an infinite set of non-isomorphic finite graphs.
Let f(n) ≥1. We say the rate of growth of f(n) is polynomially bounded if for all sufficiently large n, f(n) < nc for some constant c. We say f(n) is exponential if for all sufficiently large n, f(n) ≥an for some constant a > 1. We say f(n) has factorial growth if for all sufficiently large n, f(n) ≥ncn for some positive constant c.
Given a family G of graphs, we label the graphs as G1, . . . , Gi, . . . , and let n1, . . . , ni, . . . be the corresponding number of vertices. We say γ(G) is polyno-mially bounded in n if there is some constant c such that for all sufficiently large i, γ(Gi) < nc i. We say γ(G) grows exponentially in n if there is some constant a > 1 such that for all sufficiently large i, Gi > ani.
2.2. Results from linear algebra.
In this section we introduce results from linear algebra that we will use for later proofs. Orthonormality in Rn refers to the standard dot product. Given a matrix A, we write aij for the entry on the i-th row and in the j-th column of A, and we write A = (aij).
Definition 2.5. For an m × n real matrix M, the operator norm induced by l2 vector norm (∥· ∥) is ∥M∥= sup x∈Rn, x̸=0 ∥Mx∥ ∥x∥.
Fact 2.6. In addition to being subadditive, the operator norm is also submulti-plicative, i.e., ∥AB∥≤∥A∥∥B∥for A, B ∈Mn(R).
YUEHENG ZHANG 5 Fact 2.7. For a symmetric real matrix M, ∥M∥= max1≤i≤n |λi|. Moreover, if M is non-negative, max1≤i≤n |λi| is attained by λ1. In particular, for the adjacency matrix AG of a graph G, (2.8) ∥AG∥= λ1(G).
Theorem 2.9 (Spectral theorem for real symmetric matrices). If M is an n × n real symmetric matrix (i.e., M = M T ), then M has an orthonormal eigenbasis over R. In particular, all eigenvalues of M are real.
Definition 2.10 (Fractional powers of positive semidefinite real symmetric matri-ces). Let M ∈Mn(R) be symmetric and positive semidefinite. Then we can write M as M = QΛQT where Q is an orthogonal matrix, Λ = diag(λ1, . . . , λn), and λi ≥0. For a ∈R, we define M a := Q diag(λa 1, . . . , λa n)QT .
This definition is sound. (It does not depend on the particular choice of Q.) It follows that for a, b ∈R, M a · M b = M a+b.
In the rest of Section 2.2, A = (aij) will always denote a real symmetric n × n matrix with eigenvalues λ1 ≥λ2 ≥· · · ≥λn.
Definition 2.11. A multiset is a set in which elements are allowed to have multiple instances. We denote a multiset by double braces. For example, {{a, a, a, b, c, c}} is a multiset. We also write {{a, a, a, b, c, c}} as {{a3, b, c2}}.
Definition 2.12. The spectrum of an n × n matrix M is the multiset of its eigen-values. We denote it by spec(M).
Fact 2.13 (Spectrum of polynomials of a matrix). If g is a polynomial, then spec(g(A)) = {{g(λ1), g(λ2), . . . , g(λn)}}.
Definition 2.14. The Rayleigh quotient of the matrix A is the function RA(y) = yT Ay yT y = P 1≤i,j≤n aijyiyj Pn i=1 y2 i , defined for y ∈Rn, y ̸= 0.
Observation 2.15. If y is an eigenvector to eigenvalue λi, then RA(y) = λi.
Theorem 2.16 (Rayleigh’s principle).
λ1 = max y RA(y).
λn = max y RA(y).
Moreover, RA(y) = λ1 if and only if y is an eigenvector to λ1.
Corollary 2.17. Given vector y = (y1, . . . , yn)T , we let |y| be the vector (|y1|, . . . , |yn|)T .
If y is an eigenvector to eigenvalue λ1, then |y| is also an eigenvector to λ1.
Definition 2.18. Let B = (bij) be an m × n matrix and let M be a p × q matrix.
The Kronecker product of B and M, B ⊗M, is the mp × nq matrix b11M b12M · · · b1nM b21M b22M · · · b2nM .
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bm1M bm2M · · · bmnM .
ON THE PRINCIPAL EIGENVECTOR OF A GRAPH 6 Fact 2.19. Let B ∈Mn(R) and M ∈Mm(R).
Let the eigenvalues of B be λ1 ≥· · · ≥λn and let the eigenvalues of M be µ1 ≥· · · ≥µm. Then spec(B ⊗M) = {{λiµj, 1 ≤i ≤n, 1 ≤j ≤m}}.
2.3. Results from spectral graph theory.
In this section we introduce well-known results from spectral graph theory that will be of use later.
2.3.1. Observations about the adjacency operator.
In Sections 2.3.1 through 2.3.3, we fix the graph G and write A for AG.
We first note how the adjacency operator A of a graph G acts on vectors.
Observation 2.20. Given y = (y1, . . . , yn), the ith coordinate of Ay is given by (Ay)i = X j:i∼j yj.
Corollary 2.21. If y = (y1, . . . , yn) is an eigenvector of A to eigenvalue ρ, then ρyi = X j:j∼i yj.
Recall that j denotes the all-ones vector.
Corollary 2.22. j is an eigenvector of A if and only if G is regular.
Proof. Suppose G is d-regular.
Then Aj = (deg(1), deg(2), . . . , deg(n))T = dj.
Suppose Aj = dj. Then for any i ∈n, |N(i)| = (d · 1)/1 = d.
□ 2.3.2. Existence of the principal eigenvector for a connected graph.
In this section we establish the existence of the principal eigenvector for a con-nected graph. This follows from the Perron–Frobenius theorem for irreducible ma-trices, though given that the adjacency matrix is real and symmetric, it can be proved in much simpler ways using Rayleigh’s principle.
Proposition 2.23. If G is a connected graph and y is an eigenvector to eigenvalue λ1, then y has no zero coordinates.
Proof. It suffices to prove that |y| has no zero entries.
By Corollary 2.17, |y| is an eigenvector to eigenvalue λ1.
Suppose |yi| = 0.
Then by Corollary 2.21, P j:j∼i |yj| = 0. Therefore |yj| = 0 for all j ∼i. Since G is connected, by repeating this argument we have y = 0, but by assumption y ̸= 0.
□ Proposition 2.24. If G is a connected graph and y is an eigenvector to eigenvalue λ1, then the coordinates of y are either all positive or all negative.
Proof. Suppose there are entries of opposite signs in y. Since G is connected and y has no zero entries, there has to be edges between vertices of positive entry and vertices of negative entry. Since the entries of AG corresponding to these edges are 1, we can strictly increase RAG(y) by changing the sign of all negative entries into positive. But this contradicts Theorem 2.16 (Rayleigh’s principle).
□ Corollary 2.25. For a connected graph G, λ1 is simple.
Proof. Two all-positive or all-negative vectors cannot be orthogonal.
□ Theorem 2.26. For every connected graph G, the largest eigenvalue λ1 is simple, and it has a unique-up-to-scaling all-positive eigenvector (the principal eigenvector).
YUEHENG ZHANG 7 Proof. Proposition 2.24 together with Corollary 2.17 proves the existence of an all-positive eigenvector to λ1. Since λ1 is simple, this eigenvector is unique up to scaling.
□ Corollary 2.27. If G is connected and y is an eigenvector to eigenvalue λ with all-positive coordinates, then λ is the largest eigenvalue of A.
Proof. Since λ1 is simple, any eigenvector y to an eigenvalue other than λ1 must be orthogonal to the principal eigenvector.
□ Corollary 2.28. For a connected d-regular graph G, λ1 = d.
Proof. Follows from Corollary 2.22.
□ Corollary 2.29. For the clique on n vertices, λ1 = n −1.
Observation 2.30. Corollary 2.28 also proves (2.3).
2.3.3. Bounds on the largest eigenvalue for a graph.
Let ∆denote max1≤i≤n deg(i), the maximum degree of G.
Fact 2.31. For every graph G, λ1 ≤∆. For connected graphs, equality holds if and only if G is regular.
Proof. Let y be an eigenvector to λ1, and let yi be the maximum coordinate, then λ1yi = X j:i∼j yj ≤∆yi.
Therefore λ1 ≤∆. Suppose G is d-regular, then ∆= d = λ1 by Corollary 2.28.
On the other hand, suppose G is connected and λ1 = ∆. Then ∆yi = P j:i∼j yj.
Therefore vertex i has degree ∆and the neighbors of i have coordinates as large as i. Repeat this argument for the neighbors of i. Since the graph is connected, we have that every vertex has degree ∆and the same coordinate as yi.
□ We denote the arithmetic and quadratic mean of the degrees of vertices by (2.32) davg := Pn i=1 deg(i) n and (2.33) dqavg := rPn i=1 deg(i)2 n .
It is well-known that the quadratic mean of a multiset of numbers is not less than the arithmetic mean.
Fact 2.34. For every graph G, λ1 ≥davg.
Proof. By Theorem 2.16 (Rayleigh’s principle), λ1(A) = max y̸=0 RA(y) ≥RA(j) = jT Aj jT j = Pn i=1 deg(i) n .
□ We can improve this to a stonger bound.
Fact 2.35. For every G, λ1 ≥dqavg.
ON THE PRINCIPAL EIGENVECTOR OF A GRAPH 8 Proof. Let spec(G) = {{λ1, . . . , λn}}, where λ1 ≥· · · ≥λn. By Fact 2.13, spec(A2) = {{λ2 1, . . . , λ2 n}}.
Let y = (y1, . . . , yn)T be an eigenvector of λn, then by Observation 2.15, yT Gy yT y = λn.
Since A is non-negative, |y|T A|y| |y|T |y| = Pn i=1 Pn j=1 |yi|aij|yj| yT y ≥ | Pn i=1 Pn j=1 yiaijyj| yT y = yT Ay yT y = |λn|.
Then by Theorem 2.16 (Rayleigh’s principle), λ1 ≥|λn|. Therefore λ2 1 is the largest eigenvalue of A2. Again by Rayleigh’s principle, λ2 1 ≥jT A2j jT j = jT AT Aj jT j = (Aj)T Aj jT j = Pn i=1 deg(i)2 n .
Thus λ1 ≥ rPn i=1 deg(i)2 n .
□ 2.3.4. The largest eigenvalue of subgraphs.
Fact 2.36. If H is a proper subgraph of a connected graph G, then λ1(G) > λ1(H).
Proof. Let y be a non-negative eigenvector of AH to eigenvalue λ1(H), which exists by Corollary 2.17.
Let G have n vertices.
We define e y as the vector with n coordinates obtained by adding zero coordinates to y where vertices are deleted. If e y has zero coordinates, then by Proposition 2.23, e y is not an eigenvector to λ1(G).
Therefore by Rayleigh’s principle, λ1(G) > RAG(e y) ≥RAH(y) = λ1(H). If e y does not have zero coordinates, then e y = y. For H to be a proper subgraph of G, at least one edge is deleted. Therefore λ1(G) ≥e yT AGe y e yT e y = e yT AGe y yT y > yT AHy yT y = λ1(H).
□ 2.3.5. Graph products.
Notation 2.37. Given a graph G and U ⊆V (G), we denote the induced subgraph of G on the set U by G[U].
Definition 2.38. For graphs H = (W, F) and G = (V, E), the Cartesian product of H and G, denoted by H□G, is the graph with the set W × V as vertices, and (w1, v1) ∼(w2, v2) if and only if w1 = w2 and v1 ∼G v2, or v1 = v2 and w1 ∼H w2.
For each v ∈V , we call (H□G)[W × {v}] the horizontal layer corresponding to v.
For each w ∈W, we call (H□G)[{w} × V ] the vertical layer corresponding to w.
The horizontal layers are copies of H and the vertical layers are copies of G.
Definition 2.39. For graphs H = (W, F) and G = (V, E), the lexicographic product of H and G, denoted by H ◦G, is the graph with the set W × V as vertices, and (w1, v1) ∼(w2, v2) if and only if either w1 ∼H w2 or w1 = w2 and v1 ∼G v2. For each w ∈W we call (H ◦G)[{w} × V ) the vertical layer corresponding to w.
Recall that J denotes the all-ones matrix.
Observation 2.40. The ajacency matrix of G ◦H is AG ⊗J|V (H)| + I|V (G)| ⊗AH.
YUEHENG ZHANG 9 3. Preliminary results about the principal eigenvector As previously introduced, G will always denote a connected graph. We write λ for λ1, the largest eigenvalue of the adjacency matrix of G. We use q to mean the principal eigenvector of the adjacency matrix, the all-positive eigenvector to λ1.
We assume q is scaled to have l2 norm 1 unless otherwise stated.
3.1. Observations and naive bounds on the ratio.
First, we note that q reflects the symmetries of G.
Notation 3.1. Given a permutation π on a set X and an element a ∈X, we write π(a) for the image of a under π. We use M π to mean the row permutation matrix of π, where M π i,j = 1 if π(j) = i and M π i,j = 0 otherwise. Given a vector y, we write π(y) to denote M πy.
Definition 3.2. Given a permutation group S on a set X, the orbit of a ∈X under S is OS(a) := {π(a) | π ∈S}.
Definition 3.3. Given a graph G, an automorphism of G is a permutation π on the set of vertices that preserves adjacency relation, i.e., for each pair of vertices i, j ∈V (G), π(i) ∼π(j) if and only if i ∼j.
Notation 3.4. We note that the set of automorphisms of a graph G is a group under composition. We denote this group by Aut(G). For a vertex i ∈V (G), we denote the orbit of i under Aut(G) by O(i).
Observation 3.5. If π is an automorphism of G, then AG = M πAG(M π)T .
Proposition 3.6. Given a graph G and π ∈Aut(G), if y is an eigenvector of AG to eigenvalue ρ, then π(y) is also an eigenvector of AG to ρ.
Proof. Since any permutation matrix is an orthonormal matrix, (M π)T M π = I.
Then by Observation 3.5, AGM π = M πAG. Then AGM πy = M πAGy = ρM πy.
Therefore π(y) is an eigenvector of AG to ρ.
□ Fact 3.7. The principal eigenvector q is constant on orbits of Aut(G), i.e., if j ∈O(i), then qi = qj.
Proof. Let j ∈O(i). Then there is π ∈Aut(G) with π(i) = j. By Proposition 3.6, π(q) is an eigenvector to λ. Since q is all-positive, π(q) is also all-positive. Therefore by Theorem 2.26, π(q) = q when scaled to the same norm. Therefore qi = qj.
□ Next we note some basic bounds on the ratio between the coordinates of q.
Observation 3.8. For two vertices i, j in G, let dist(i, j) = k. Then qj qi ≤λk.
Proof. If dist(i, j) = 0, then qj/qi = 1 = λ0.
If dist(i, j) = 1, then by Corol-lary 2.21, λqi = P w:w∼i qw ≥qj since all qw are positive. Now, suppose k ≥2 and qw/qi ≤λk−1 for all vertices w at distance k −1 from i. We know j is adjacent to at least one vertex w at distance k −1 from i. Then qj qi = qj qw · qw q1 ≤λ · λk−1 = λk.
□ ON THE PRINCIPAL EIGENVECTOR OF A GRAPH 10 Recall that D denotes the diameter of the graph.
Corollary 3.9. For every connected graph G with diameter D, γ ≤λD ≤∆D ≤(n −1)D.
Corollary 3.10. If D is bounded for some family G of graphs , then γ(G) is poly-nomially bounded in n.
Since D is relevant in bounding the ratio, we introduce a bound on D for regular graphs.
Fact 3.11. Let G be a connected d-regular graph. Then D ≤3n d .
Proof. Pick v0, vD in G so that dist(v0, vD) = D. Let v0, v1, . . . , vD be a shortest path from v0 to vD. Any vi, vj with |i−j| ≥3 cannot have any common neighbors, since otherwise the path will not be a shortest path. Thus d D 3 ≤n.
Therefore D ≤3n/d.
□ We know D(G + e) ≤D(G) for any e ∈G. The following result shows that D(G + e), and consequently, also D(G −e) (if still connected), cannot differ from DG by more than a factor of 2.
Fact 3.12. For any connected graph G with D(G) = D and e ∈E(G), D(G + e) ≥1 2D(G).
Proof.
(Notation: By distl(x, y), where l is a path, we mean the distance between x and y along the path.) We need to prove that there is a pair of vertices at distance at least D 2 in G + e.
Let u, v ∈V (G) be such that distG(u, v) = D. If distG+e(u, v) = D, we are done.
Otherwise, let p be a shortest path between u and v in G, and q a shortest path between u and v in G + e. Then e must be on q. Denote by x the endpoint of e which is closer to u on q, and by y the other endpoint. Pick the middle vertex w of p with distp(u, w) = ⌈D 2 ⌉. If distG+e(u, w) = distG(u, w) = ⌈D 2 ⌉, we are done.
Otherwise, any shortest path r in G + e from u to w must pass through edge e.
Suppose we go along r from u to w, by the optimality of q, we can assume that r and q overlap from u to y. Now we look at the vertex w′ adjacent to w on p which is closer to u than to v. We have (3.13) distG(w′, v) = distp(w′, v) = D 2 + 1.
We claim that there is a shortest path in G + e from w′ to v that does not pass through e. Let s be a shortest path in G + e from w′ to v that passes through e.
By the optimality of q, we may assume that s and q overlap from x to v. Then by the optimality of r, dists(x, w′) + distG+e(w′, w) ≥distr(x, w) = distG+e(x, y) + distr(y, w), that is, dists(x, w′) ≥distr(y, w).
YUEHENG ZHANG 11 Therefore dists(w′, y) = dists(w′, x) + distG+e(x, y) ≥distG+e(w′, w) + distr(w, y).
Thus s is equivalent to a path that does not pass through e in G + e. As a result, a shortest path between w′ and v in G + e is also available in G. Then by (3.13), distG+e(w′, v) = distG(w′, v) = D 2 + 1.
□ 3.2. Chebyshev polynomials and principal eigenvectors.
3.2.1. Chebyshev polynomials.
The Chebyshev polynomials of the first kind, Tn, can be characterized by the recurrence (3.14) Tn+1(t) = 2t · Tn(t) −Tn−1(t), with initial values T0(t) = 1 and T1(t) = t.
The Chebyshev polynomials of the second kind, Un, can be characterized by the same recurrence (3.15) Un+1(t) = 2t · Un(t) −Un−1(t), with initial values U0(t) = 1 and U1(t) = 2t.
Fact 3.16. When |t| ≥1, the explicit formula for Tn is (3.17) Tn(t) = 1 2 t − p t2 −1 n + t + p t2 −1 n , and the explicit formula for Un is (3.18) Un(t) = t + √ t2 −1 n+1 − t − √ t2 −1 n+1 2 √ t2 −1 .
Fact 3.19. The roots of Tn are cos π(k + 1/2) n , k = 0, . . . , n −1. The roots of Un are cos kπ n + 1 , k = 1, . . . , n.
3.2.2. Applications to prinicpal eigenvectors.
Fact 3.20. When x > 1, both Tn(x) and Un(x) are strictly increasing.
Definition 3.21. A pendant path of length k in G consists of k vertices such that the induced subgraph on them is a path; moreover, one vertex has degree 1 in G and k −2 vertices have degree 2 in G. For example, in the graph Pr · Ks, there is a pendant path of length r.
Observation 3.22. Let 1, 2, . . . , k be a pendant path in G where consecutive ver-tices are adjacent and deg(1) = 1. Then for 1 ≤j ≤k, qj q1 = Uj−1 λ 2 .
Proof. By Corollary 2.21, λq1 = q2 and qj+1 = λqj −qj−1 for 1 ≤j ≤n −1.
Therefore qj/q1 satisfies the initial values and recurrence relation of Uj−1(λ/2).
□ ON THE PRINCIPAL EIGENVECTOR OF A GRAPH 12 Observation 3.22 along with Fact 2.36 can be used to show that most kite graphs have a very large (factorial) principal ratio, since Pr · Ks has a pendant path of length r and λ is larger than s −1. In fact, Tait and Tobin proved that the maximum principal ratio over all graphs of n vertices is attained by a kite graph.
4. Main results Let G be a d-regular graph. As introduced before, we use G + e to denote the graph obtained by adding an edge e ∈E(G) to G, and G −e to denote the graph obtained by deleting an edge e ∈E(G) from G. We always assume G −e is still connected. We are interested in the possible asymptotic behaviors of γ(G + e) and γ(G −e).
We first make two simple observations.
Observation 4.1. If the diameter D(G) is bounded, then γ(G±e) is polynomially bounded in n.
Proof. Fact 3.12 shows that D(G ± e) is also bounded, and the statement follows from Cororllary 3.10.
□ Observation 4.2. If d is linear in n, then γ(G ± e) is polynomially bounded in n.
Proof. Fact 3.11 shows that D(G) is bounded, and the statement follows from Observation 4.1.
□ 4.1. Adding or removing an edge in bounded distance.
4.1.1. Adding an edge.
We show that if we add an edge e between two vertices at distance 2 to a connected regular graph G of bounded degree, then γ(G + e) can be exponential.
Let e = {1, 2}.
Theorem 4.3. For any fixed d, there is a family G of connected d-regular graphs where for each Gi ∈G, there is an edge ei ∈Gi whose endpoints are at distance two in Gi −ei, such that for the family G′ = {Gi + ei | Gi ∈G}, γ(G′) grows exponentially in n.
The proof of this theorem will be based on a series of constructions. The graphs produced by Construction 4.6 and Construction 4.7 are the pair of graphs that are used in the proof.
Construction 4.4 (Ringr,d,G2). Let r ≥0 be a parameter. We label the vertices in P2r+1 from one end to the other end as p−r, p−r+1, ..., p−1, p0, p1, ..., pr−1, pr.
Let G1 = P2r+1□Kd−1. We label the vertical layer corresponding to pi as Li. Let G2, which we will call a “gadget,” be a connected graph with two vertices v1, v2 of degree 1 and all other vertices of degree d, and an automorphism that switches v1 and v2. We connect v1 with every vertex in L−r and connect v2 with every vertex in Lr. We call the graph obtained Ringr,d,G2.
Observation 4.5. Let X ≤Aut(G1) be the subgroup of automorphisms of G1 that fixes the vertical layers and permutes the horizontal layers. The orbits of X are the vertical layers. X is isomorphic to the symmetric group Sd−1. This group extends to G. The automorphism of G1 that switches Lj and L−j for 0 ≤j ≤r also YUEHENG ZHANG 13 extends to G. Therefore the coordinates of the principal eigenvector of Ringr,d,G2 corresponding to all vertices in Lj ∪L−j are the same. We denote this value by aj, for 0 ≤j ≤r.
Construction 4.6. [Ringr,d] Now we specify the gadget G2. We take two copies of Kd+1 and call them H1, H2. We label the vertices in H1, H2 from 1 to d + 1.
We remove the edge {1, 2} from H2, the edge {3, 4} from H1, and add the edges {1, 3} and {2, 4}. We find two vertices u1, u2 in V (H1) such that u2 ∈O(u1) in H1 −{3, 4}. We remove the edge {u1, u2}. Finally, we attach a dangling vertex w1 to u1, and a dangling vertex w2 to u2. w1 and w2 are the vertices that connect with L−r and Lr, respectively. For this specific G2, Ring(r, d, G2) is a d-regular graph on n = (2r + 1)(d −1) + 2 + 2(d + 1) = 2rd −2r + 3d + 3 vertices. We write Ring(r, d) for this graph.
Construction 4.7. [Ringr,d +e] We add the edge e = {3, 4} ∈E(Ringr,d) to Ringr,d.
Proposition 4.8. For 0 ≤j ≤r, aj a0 = Tj λ(Ringr,d +e) −d + 2 2 where aj is as defined in Observation 4.5.
Proof. By Corollary 2.21, λa0 = (d −2)a0 + 2a1 and λaj = (d −2)aj + aj−1 + aj+1 for 1 ≤j ≤n−1. Therefore aj/a0 satisfies the initial values and recurrence relation of Tj((λ −d + 2)/2) according to (3.14).
□ Lemma 4.9. λ(Ringr,d +e) > d+c(d), where c(d) is a constant depending only on d.
Proof. Let H be the induced subgraph of Ringr,d on V (H1) ∪V (H2). Then degH+e(1) = degH+e(2) = d + 1, degH+e(u1) = degH+e(u2) = d −1, ON THE PRINCIPAL EIGENVECTOR OF A GRAPH 14 while the rest of the vertices in H + e are of degree d. Then by Fact 2.36 and Fact 2.35, λ(Ringr,d +e) > λ(H + e) ≥dqavg(H + e) = r (2d + 2)d2 + 4 2d + 2 = d s 1 + 2 d2(d + 1).
Let c(d) := 2 3d(d + 1).
Since √ 1 + x > 1 + 1 3x when 0 < x < 3, λ(G + e) > d + c(d).
□ Proof of Theorem 4.3. By Lemma 4.9, λ(Ringr,d +e) −d + 2 2 > 1 + 1 3d(d + 1).
By Observation 4.8, Fact 3.20, and Fact 3.16, γ(Ringr,d) ≥ar a0 = Tr λ(Ringr,d +e) −d + 2 2 > Tj(1+ 1 3d(d + 1)) > 1 2 1 + 1 3d(d + 1) r .
Since r = n −3d −3 2d −2 and d is bounded, γ(Ringr,d) > (a(d) −ϵ)n where a(d) := 1 + 1 3d(d + 1) 1/(2d−2) and 0 < ϵ < a −1 is any fixed constant.
□ 4.1.2. Removing an edge.
We show that in the case of removing an edge e = {1, 2} when distG−e(1, 2) is bounded, γ(G−e) is polynomially bounded for all D. We make use of the following theorem.
Let e = {1, 2}.
Theorem 4.10 (Cioab˘ a, Gregory, Nikiforov). If G is a connected nonregular graph with n vertices, diameter D, and maximum degree ∆, then ∆−λG ≥ 1 n(D + 1).
Theorem 4.11. For a connected d-regular graph G and an edge e = {1, 2} ∈E(G), if distG−e(1, 2) < c where c is some constant, then γ(G−e) is polynomially bounded in n.
Lemma 4.12. qmin(G −e) is either q1 or q2.
Proof. If qmin corresponds to some vertex j with degree d, then the average of the coordinates corresponding to the neighbors of j would be λ(G −e)qj d < λ(G −e)qj λ(G −e) = qj, which is a contradiction.
□ YUEHENG ZHANG 15 Proof of Theorem 4.11. Without loss of generality suppose q1 = qmin. Then sum-ming n equations of the form λ(G −e)qi = P j:j∼i qj, we have λ(G −e) n X i=1 qi = (d −1)q1 + (d −1)q2 + dq3 + · · · + dqn = d n X i=1 qi !
−q1 −q2.
Therefore by Theorem 4.10, q1 + q2 Pn i=1 qi = d −λ(G −e) ≥ 1 n(D(G −e) + 1) ≥1 n2 .
By Observation 3.8, q2 ≤λ(G −e)cq1. Therefore γ(G −e) < Pn i=1 qi q1 ≤n2 1 + q2 q1 ≤n2(1 + λ(G −e)c) < n2(1 + dc).
□ 4.2. Multiplicative spectral gap and stability of the ratio.
We use a known bound on the diameter of a graph in terms of the spectral graph to show that γ(G ± e) is polynomially bounded for bounded-degree expanders.
Definition 4.13. For 0 < ϵ < 1, a regular graph of degree d is an ϵ-expander if λ2 ≤(1 −ϵ)d, i.e., δ ≥ϵd.
We note that this definition implies that an expander graph is connected.
Theorem 4.14 (N. Alon, V. D. Milman). Let G be a connected graph on n vertices with maximum degree ∆and let δ denote the smallest positive eigenvalue of the Laplacian matrix δ. Then D(G) ≤2 $r 2∆ δ log2 n % .
Corollary 4.15. For expander graphs G with bounded degree, γ(G ± e) is poly-nomially bounded in n. More specifically, for an ϵ-expander graph G with bounded degree d, (4.16) γ(G + e) ≤n4√ 2/ϵ log2(d+1).
Proof. By definition, ∆(G) δ(G) = d δ(G) ≤1 ϵ .
Then D(G) ≤2 $r 2 ϵ log2 n % .
By Fact 3.12, D(G ± e) ≤4 $r 2 ϵ log2 n % .
ON THE PRINCIPAL EIGENVECTOR OF A GRAPH 16 Since γ ≤∆D (Corollary 3.9), γ(G ± e) ≤(d + 1) 4 j√ 2/ϵ log2 n k ≤(d + 1)4√ 2/ϵ log2 n = 24√ 2/ϵ log2 n log2(d+1) = n4√ 2/ϵ log2(d+1).
□ 4.3. Additive spectral gap and stability of the ratio.
We show that a large ((2 + ϵ)√n) additive spectral gap implies that γ(G ± e) is bounded. Specifically, we prove the following.
Theorem 4.17. Let G be a connected d-regular graph. If the spectral gap δ = d−λ2 of G satisfies δ > 2 c √n + 2 for some value 0 < c < 1, then γ(G ± e) < 1 + c 1 −c.
To motivate this result, we first point out that graphs with such an additive spectral gap are not necessarily expanders. Indeed, when d is larger than Θ(√n), the multiplicative spectral gap of graphs with a Θ(√n) additive spectral graph will go to zero. Moreover, the diameter of graphs with such an additive spectral gap can still grow quite fast (polynomially in n), approaching the upper bound derived from Theorem 4.14.
4.3.1. Existence of graphs with large additive spectral gap and large diameter.
Proposition 4.18. For a regular graph with additive spectral gap δ, the diameter D is O((n/δ)1/3(log n)2/3).
Proof. Let the regular graph have degree d. From Theorem 4.14, we know that (4.19) D2 ≤cd δ (log n)2 where c is some constant. By Fact 3.11, we also have (4.20) D ≤3n d .
Multiplying (4.19) and (4.20), we have □ (4.21) D ≤(3c)1/3 n δ 1/3 (log n)2/3.
Corollary 4.22. For a regular graph with an Ω(√n) additive spectral gap, the diameter is O(n1/6(log n)2/3).
We show that this bound is nearly tight.
Proposition 4.23. There are connected regular graphs with diameter (1/2)n1/6 and an additive spectral gap of cn1/2 where c = 2π2(1 + O(n−1/3)).
We prove a more general statement.
Proposition 4.24. For any constant 0 < t < 1, there are connected regular graphs with diameter (1/2)n(1−t)/3 and an additive spectral gap of cnt where c = 2π2(1 + O(n(2t−2)/3)).
YUEHENG ZHANG 17 Proof. Recall the lexicographic product and its properties in Definition 2.39, Ob-servation 2.40, and Fact 2.19. Let G := Cr ◦Ks where r · s = n. Then G is a connected regular graph of degree 2s and diameter ⌊r/2⌋. The adjacency matrix of G can be expressed as (4.25) AG = ACr ⊗Js + Ir ⊗0 = ACr ⊗Js.
Since (4.26) spec(Js) = {{s, 0s−1}}, we have (4.27) spec(G) = {{sλ1(Cr), sλ2(Cr), . . . , sλr(Cr), 0(s−1)r.}} Thus δ(G) = s(λ1(Cr) −λ2(Cr)).
It is well-known that the eigenvalues of a cycle Cr are {2 cos 2πj r }, where j = 0, 1, . . . , r −1. Thus λ1(Cr) −λ2(Cr) = 2 1 −cos 2π r = 4π2 r2 + ϵ where |ϵ| < 25π4 4!r4 . We take r = n(1−t)/3 and s = n(2+t)/3. Then δ(G) = 4π2nt 1 + O 1 r2 = 4π2nt 1 + O(n(2t−2)/3) .
□ 4.3.2. Large additive spectral gap implies bounded ratio.
Now we prove Theorem 4.17 which shows that γ(G ± e) is bounded when G has an additive spectral gap of (2 + ϵ)√n for some ϵ > 0. This is an application of the theory developed in Chapter V. 2 in Stewart and Sun’s book , dealing with perturbation of invariant subspaces. The outline of the proof is from this book. We adapt the proofs to our special case and fill in some details.
Notation 4.28. Let U ∈Mn(R) be the orthogonal matrix whose columns are eigenvectors of A. We can write it as (4.29) U = (x Y ) = (x y2 · · · yn), where the columns x, y2, y3, . . . , yn are eigenvectors of A corresponding to eigen-values λ1 ≥λ2 ≥λ3 ≥· · · ≥λn. Then λ1 = d, and we can assume (4.30) x = 1 √n, . . . , 1 √n T .
In the context of adding an edge e to G, we label the two endpoints of e to be 1 and 2, and let E ∈Mn(R) have E21 = E12 = 1 while all other coordinates are zero.
In the context of deleting an edge e from G, we also label the two endpoints of e to be 1 and 2, and let E ∈Mn(R) have E21 = E12 = −1 while all other coordinates are zero. In both cases, A + E is the adjacency matrix of the graph obtained.
We know x is not an eigenvector of A + E. We want to know how close x is to the principal eigenvector of G±e, in the sense that we want to find a vector v with small norm such that e x = x + v ∥x + v∥is the unit principal eigenvector of G ± e.
ON THE PRINCIPAL EIGENVECTOR OF A GRAPH 18 Proposition 4.31. Let p be a vector in Rn−1.
Let U = (x Y ) ∈Mn(R) be orthogonal, where x is a vector in Rn−1 and Y is an n × (n −1) matrix. Define e U ∈Mn(R) as e U = (e x e Y ), where (4.32) e x = x + Y p p 1 + ∥p∥2 and e Y = (Y −xpT )(In−1 + ppT )−1/2.
Then e U is an orthogonal matrix.
Proof.
e xT e x = xT x + pT Y T Y p 1 + ∥p∥2 = 1 + ∥p∥2 1 + ∥p∥2 = 1.
e Y T e Y = (In−1 + ppT )−1/2(Y T −pxT )(Y −xpT )(In−1 + ppT )−1/2 = (In−1 + ppT )−1/2(In−1 + ppT )(In−1 + ppT )−1/2 = In−1.
e xT e Y = (xT + pT Y T )(Y −xpT )(In−1 + ppT )−1/2 p 1 + ∥p∥2 = (−pT + pT )(In−1 + ppT )−1/2 p 1 + ∥p∥2 = 0.
Therefore e U T e U = e xT e x e xT e Y e Y T e x e Y T e Y !
= In.
□ We want to find p such that e x is the principal eigenvector of A + E and the columns of (e x e Y ) form an orthogonal eigenbasis for A + E.
Notation 4.33. We define e11 := xT Ex ∈R, e21 := Y T Ex ∈Rn−1, E22 := Y T EY ∈Mn−1(R), and L := Y T AY = diag(λ2, . . . , λn) ∈Mn−1(R).
Observation 4.34. Since ∥E∥= 1 and (xY ) is orthogonal, |e11|, ∥e21∥, ∥E22∥≤1.
Proposition 4.35. If (4.36) ((d + e11)In−1 −(L + E22))p = e21 −peT 21p, then e x is an eigenvector of A + E.
Proof. (4.36) is equivalent to (Y T −pxT )(A + E)(x + Y p) = 0, which gives e Y T (A + E)e x = 0.
Since (e x e Y ) is an orthogonal matrix, e x is an eigenvector of A + E.
□ Notation 4.37. We define M ∈Mn−1(R) as M = (d + e11)In−1 −L −E22.
YUEHENG ZHANG 19 Proposition 4.38. If δ > 2 c √n + 2 for some value 0 < c < 1, then M is non-singular and (4.39) ∥M −1∥≤ 1 δ −2 < c 2√n.
Proof. Since (d + e11)In−1 −L is a diagonal matrix and E22 is a symmetric matrix, M is symmetrix. Recall that ∥E22∥≤1. By Theorem 2.16 (Rayleigh’s principle) and Observation 4.34, min spec(M) = min ∥x∥=1 xT ((d + e11)In−1 −L −E22)x ≥min ∥x∥=1 xT ((d + e11)In−1 −L)x −1 = min ∥x∥=1 spec((d + e11)I −L) −1 = d + e11 −λ2 −1 ≥δ −2.
Therefore all eigenvalues of M are positive, and ∥M −1∥= max spec(M −1) = 1 min spec(M) ≤ 1 δ −2 < c 2√n.
□ Proposition 4.40. We write (4.36) in terms of M: (4.41) Mp = e21 −peT 21p.
Let θ = ∥M −1∥−1 and η = ∥e21∥. We claim that if δ > 2 c √n + 2 for some value 0 < c < 1, then there exists p with (4.42) ∥p∥< 2η θ + p θ2 −4η2 such that (4.41) holds. The e x defined by (4.32) is an eigenvector of A + E.
Proof. We want to find a solution with small norm to the non-linear equation (4.41).
We do this by an iterative construction.
We define a sequence of vectors p0, p1, . . . such that p0 = 0 and pi = M −1(e21 −pi−1eT 21pi−1), for i ≥1.
Then ∥pi∥≤∥M −1∥(|e21| + ∥pi−1∥2|e21|) ≤η(1 + ∥pi−1∥2) θ .
We claim that the sequence {pi} converges. We define ξ0 = 0, ξi = η(1 + ξ2 i−1) θ , for i ≥1.
Then ∥pi∥≤ξi. Since ξ1 = η θ > ξ0, we can prove by induction that ξ0, ξ1, ξ2, . . .
is monotone increasing. Let φ(ξ) = η(1 + ξ2) θ .
This function is monotone increasing in ξ, and has a fixed point at ξ = 2η θ + p θ2 −4η2 .
Then ξi < ξi+1 = φ(ξi) < φ(ξ) = ξ. Therefore the sequence {ξi} converges to ξ.
ON THE PRINCIPAL EIGENVECTOR OF A GRAPH 20 Thus ∥pi∥≤ξi ≤ξ = 2η θ + p θ2 −4η2 < 2η θ .
Next we prove the convergence of {p0, p1, . . . }. For any i ≥2, ∥pi −pi−1∥= |M −1(pi−1eT 21pi−1 −pi−2eT 21pi−2) = |M −1(pi−1 + pi−2)eT 21(pi−1 −pi−2)| ≤2∥M −1∥∥pi−1∥∥pi−1∥· η · ∥pi−1 −pi−2∥ ≤4η2 θ2 ∥pi−1 −pi−2∥.
Then ∥pi −p0∥≤ρi∥p1 −p0∥, where ρ = 4η2 θ2 < c2 4n < 1. Therefore {pi} is a Cauchy sequence in Rn−1 and has a limit p. Thus a solution p exists, with norm satisfying (4.42).
□ Proposition 4.43. If δ > 2 c √n + 2 for some value 0 < c < 1, then the principal eigenvector of A + E, e x, can be writen in the form e x = x + Y p p 1 + ∥p∥2 where ∥p∥< c √n.
Proof. We showed that there exists p with ∥p∥< 2η θ + p θ2 −4η2 < 2η θ < c √n such that x + Y p p 1 + ∥p∥2 is an eigenvector of A+E. It remains to show that this is the principal eigenvector. Since ∥Y ∥= 1 and ∥p∥< c √n where 0 < c < 1, and since x = ( 1 √n, . . . , 1 √n), all coordinates of e x are positive. Therefore by Corollary 2.27, e x is the principal eigenvector of A + E.
□ Proof of Theorem 4.17. Following the argument above, we know the smallest pos-sible coordinate of e x is 1 −c √n , and the largest possible coordinate of e x is 1 + c √n .
Therefore the ratio of G ± e is as claimed. This completes the proof.
□ 4.4. Open questions.
In Section 4.1.2, we showed that if we remove a non-bridge edge e from a con-nected regular graph such that the endpoints of e are of bounded distance in G−e, then γ(G−e) is polynomially bounded. Is there also a polynomial bound when the endpoints of e are of unbounded distance in G −e?
Question 4.44. If we remove a non-bridge edge e from a connected regular graph G, is γ(G −e) always polynomially bounded?
In Section 4.1.1, we showed that if we add an edge e between two vertices at distance 2 to a connected regular graph G of bounded degree, then γ(G + e) can YUEHENG ZHANG 21 be exponential. Can γ(G + e) be exponential when e is between two vertices of unbounded distance in G?
Question 4.45. If we add an edge e to a connected regular graph G with bounded degree d, such that the disance between the endpoints of e in G is unbouneded, can γ(G + e) be exponential in n?
In Section 4.3.2, we showed that for a connected regular graph G, an additive spectral gap larger than 2√n implies that γ(G ± e) is bounded (Theorem 4.17).
However, this bound ceases to work at all for G with an additive spectral gap δ ≤2√n. Is this a limitaton of the method, or is there really an abrupt change at δ = 2√n?
Question 4.46. Is there a family of connected regular graphs G with an additive spectral gap slightly less than 2√n and with γ(G ± e) not bounded?
Acknowledgments I am very grateful to Prof. Babai for his kindness, patience, and his guidance on everything between English usage and approaches to intriguing problems and be-yond. I would like to thank Thomas Hameister for helpful conversations throughout the REU program and for helpful comments on early drafts of this paper. I would like to thank Prof. Peter May for organizing this REU and making this experience possible.
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Nathan Linial, Noam Nisan. Approximate Inclusion-Exclusion. Combinatorica (1990), 10:349.
Gilbert W. Stewart, Ji-guang Sun. Matrix Perturbation Theory, Academic press, 1990.
Noam Nisan, Mario Szegedy. On the degree of Boolean functions as real polynomials. Com-putational Complexity (1994), 4:301.
Edwin R. van Dam, Willem H. Haemers. Eigenvalues and the Diameter of Graphs. Linear and Multilinear Algebra (1995), 39:33-44.
Toufik Mansour, Arkady Vainshtein. Restricted permutations, continued fractions, and Cheby-shev polynomials. The Electronic J. Combinatorics (2000), 7:R17.
Andrew Y. Ng, Alice X. Zheng, Michael I. Jordan. Link Analysis, eigenvectos and stability.
IJCAI 2001, 2:903.
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Sebastian M. Cioab˘ a, David A. Gregory. Principal eigenvectors of irregular graphs. Electronic J. Linear Algebra (2007), 16.
Sebastian M. Cioab˘ a, David A. Gregory, Vladimir Nikiforov. Extreme eigenvalues of nonreg-ular graphs. J. Combinatorial Theory, Series B (2007), 97(3):483-486.
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Linear Algebra (2018), 34. |
190380 | https://library.fiveable.me/key-terms/immunobiology/commensal-bacteria | Commensal bacteria - (Immunobiology) - Vocab, Definition, Explanations | Fiveable | Fiveable
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key term - Commensal bacteria
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Definition
Commensal bacteria are microorganisms that reside on or within the human body without causing harm, often providing beneficial effects to their host. They play an essential role in maintaining the balance of the immune system and preventing pathogenic infections, particularly in environments like the skin where skin-associated lymphoid tissue is present.
5 Must Know Facts For Your Next Test
Commensal bacteria help in training the immune system, teaching it to distinguish between harmful pathogens and harmless microbes.
They compete with potential pathogens for resources and space on the skin surface, effectively preventing colonization by harmful microbes.
Some commensal bacteria can produce antimicrobial substances that inhibit the growth of pathogens, further contributing to skin health.
Disruption of the balance of commensal bacteria can lead to skin disorders or infections, highlighting their importance in maintaining skin integrity.
Commensal bacteria have been shown to influence local immune responses in SALT, enhancing the skin's ability to respond to infections.
Review Questions
How do commensal bacteria contribute to the functioning of skin-associated lymphoid tissue?
Commensal bacteria play a crucial role in enhancing the immune response within skin-associated lymphoid tissue by interacting with immune cells. They help train these cells to recognize and respond appropriately to potential pathogens while maintaining tolerance to harmless microorganisms. This balance is vital for effective immune surveillance and maintaining skin health.
Discuss the potential consequences of an imbalance in commensal bacteria on skin health and immunity.
An imbalance in commensal bacteria, often referred to as dysbiosis, can lead to various skin issues such as eczema, acne, or increased susceptibility to infections. When beneficial commensals decrease or harmful pathogens proliferate, the skin's protective barrier is compromised, leading to inflammation and impaired immune responses. This imbalance highlights the importance of maintaining a healthy microbial community for optimal skin function and defense.
Evaluate the role of commensal bacteria in preventing infections and how this could be leveraged for therapeutic purposes.
Commensal bacteria prevent infections primarily through competitive exclusion, resource utilization, and production of antimicrobial compounds. By maintaining a healthy balance of these beneficial microbes, we can reduce the risk of pathogen colonization. Therapeutically, this understanding opens avenues for probiotic treatments that aim to restore healthy microbial populations on the skin or within other body sites, potentially preventing infections and enhancing overall health.
Related terms
Microbiome:The collective genomes of the microorganisms living in a specific environment, including the human body, which influences health and disease.
Pathogenic bacteria:Bacteria that can cause disease or infection in the host, often contrasting with commensal bacteria that are harmless or beneficial.
Skin-associated lymphoid tissue (SALT):A component of the immune system located in the skin that plays a key role in responding to infections and maintaining skin homeostasis.
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190381 | https://mfleck.cs.illinois.edu/building-blocks/version-1.0/planargraphs.pdf | Chapter 18 Planar Graphs This chapter covers special properties of planar graphs.
18.1 Planar graphs A planar graph is a graph which can be drawn in the plane without any edges crossing. Some pictures of a planar graph might have crossing edges, but it’s possible to redraw the picture to eliminate the crossings. For example, although the usual pictures of K4 and Q3 have crossing edges, it’s easy to redraw them so that no edges cross. For example, a planar picture of Q3 is shown below. However, if you fiddle around with drawings of K3,3 or K5, there doesn’t seem to be any way to eliminate the crossings. We’ll see how to prove that these two graphs aren’t planar.
210 CHAPTER 18. PLANAR GRAPHS 211 a b c d A B C D Why should we care? Planar graphs have some interesting mathematical properties, e.g. they can be colored with only 4 colors. Also, as we’ll see later, we can use facts about planar graphs to show that there are only 5 Platonic solids.
There are also many practical applications with a graph structure in which crossing edges are a nuisance, including design problems for circuits, subways, utility lines. Two crossing connections normally means that the edges must be run at different heights. This isn’t a big issue for electrical wires, but it creates extra expense for some types of lines e.g.
burying one subway tunnel under another (and therefore deeper than you would ordinarily need).
Circuits, in particular, are easier to manufacture if their connections live in fewer layers.
18.2 Faces When a planar graph is drawn with no crossing edges, it divides the plane into a set of regions, called faces. By convention, we also count the unbounded area outside the whole graph as one face.
The boundary of a face is the subgraph containing all the edges adjacent to that face and a boundary walk is a closed walk containing all of those edges. The degree of the face is the minimum length of a boundary walk. For example, in the figure below, the lefthand graph has three faces. The boundary of face 2 has edges d f, fe, ec, cd, so this face has degree 4. The boundary of face 3 (the unbounded face) has edges bd, d f, fe, ec, ca, ab, so face 3 has degree 6.
CHAPTER 18. PLANAR GRAPHS 212 1 2 3 a b c d e f 1 2 a b c d e f The righthand graph above has a spike edge sticking into the middle of face 1. The boundary of face 1 has edges bf, fe, ec, cd, ca, ab. However, any boundary walk must traverse the spike twice, e.g. one possible boundary walk is bf, fe, ec, cd, cd, ca, ab, in which cd is used twice. So the degree of face 1 in the righthand graph is 7. Notice that the boundary walk for such a face is not a cycle.
Suppose that we have a graph with e edges, v nodes, and f faces. We know that the Handshaking theorem holds, i.e. the sum of node degrees is 2e. For planar graphs, we also have a Handshaking theorem for faces: the sum of the face degrees is 2e. To see this, notice that a typical edge forms part of the boundary of two faces, one to each side of it. The exceptions are edges, such as those involved in a spike, that appear twice on the boundary of a single face.
Finally, for connected planar graphs, we have Euler’s formula: v−e+f = 2. We’ll prove that this formula works.1 18.3 Trees Before we try to prove Euler’s formula, let’s look at one special type of planar graph: free trees.
In graph theory, a free tree is any connected graph with no cycles. Free trees are somewhat like normal trees, but they don’t have a designated root node and, therefore, they don’t have a clear ancestor-descendent ordering to their notes.
A free tree doesn’t divide the plane into multiple faces, because it doesn’t contain any cycles. A free tree has only one face: the entire plane surrounding it. So Euler’s theorem reduces to v −e = 1, i.e. e = v −1. Let’s prove that 1You can easily generalize Euler’s formula to handle graphs with more than one con-nected components.
CHAPTER 18. PLANAR GRAPHS 213 this is true, by induction.
Proof by induction on the number of nodes in the graph.
Base: If the graph contains no edges and only a single node, the formula is clearly true.
Induction: Suppose the formula works for all free trees with up to n nodes. Let T be a free tree with n + 1 nodes. We need to show that T has n edges.
Now, we find a node with degree 1 (only one edge going into it).
To do this start at any node r and follow a walk in any direction, without repeating edges. Because T has no cycles, this walk can’t return to any node it has already visited. So it must eventually hit a dead end: the node at the end must have degree 1. Call it p.
Remove p and the edge coming into it, making a new free tree T ′ with n nodes. By the inductive hypothesis, T ′ has n −1 edges.
Since T has one more edge than T ′, T has n edges. Therefore our formula holds for T.
18.4 Proof of Euler’s formula We can now prove Euler’s formula (v −e + f = 2) works in general, for any connected planar graph.
Proof: by induction on the number of edges in the graph.
Base: If e = 0, the graph consists of a single node with a single face surrounding it. So we have 1 −0 + 1 = 2 which is clearly right.
Induction: Suppose the formula works for all graphs with no more than n edges. Let G be a graph with n + 1 edges.
Case 1: G doesn’t contain a cycle. So G is a free tree and we already know the formula works for free trees.
CHAPTER 18. PLANAR GRAPHS 214 Case 2: G contains at least one cycle. Pick an edge p that’s on a cycle. Remove p to create a new graph G′.
Since the cycle separates the plane into two faces, the faces to either side of p must be distinct. When we remove the edge p, we merge these two faces. So G′ has one fewer faces than G.
Since G′ has n edges, the formula works for G′ by the induction hypothesis. That is v′ −e′ + f ′ = 2. But v′ = v, e′ = e −1, and f ′ = f −1. Substituting, we find that v −(e −1) + (f −1) = 2 So v −e + f = 2 18.5 Some corollaries of Euler’s formula Corollary 1 Suppose G is a connected planar graph, with v nodes, e edges, and f faces, where v ≥3. Then e ≤3v −6.
Proof: The sum of the degrees of the faces is equal to twice the number of edges. But each face must have degree ≥3. So we have 3f ≤2e.
Euler’s formula says that v −e + f = 2, so f = e −v + 2 and thus 3f = 3e −3v + 6. Combining this with 3f ≤2e, we get 3e −3v + 6 ≤2e. So e ≤3v −6.
We can also use this formula to show that the graph K5 isn’t planar. K5 has five nodes and 10 edges. This isn’t consistent with the formula e ≤3v−6.
Unfortunately, this method won’t help us with K3,3, which isn’t planar but does satisfy this equation.
We can also use this Corollary 1 to derive a useful fact about planar graphs: CHAPTER 18. PLANAR GRAPHS 215 Corollary 2 If G is a connected planar graph, G has a node of degree less than six.
Proof: This is clearly true if G has one or two nodes.
If G has at least three nodes, then suppose that the degree of each node was at least 6. By the handshaking theorem, 2e equals the sum of the degrees of the nodes, so we would have 2e ≥6v.
But corollary 1 says that e ≤3v −6, so 2e ≤6v −12. We can’t have both 2e ≥6v and 2e ≤6v −12. So there must have been a node with degree less than six.
If our graph G isn’t connected, the result still holds, because we can apply our proof to each connected component individually. So we have: Corollary 3 If G is a planar graph, G has a node of degree less than six.
18.6 K3,3 is not planar When all the cycles in our graph have at least four nodes, we can get a tighter relationship between the numbers of nodes and edges.
Corollary 4 Suppose G is a connected planar graph, with v nodes, e edges, and f faces, where v ≥3. and if all cycles in G have length ≥4, then e ≤2v −4.
Proof: The sum of the degrees of the faces is equal to twice the number of edges. But each face must have degree ≥4 because all cycles have length ≥4. So we have 4f ≤2e, so 2f ≤e.
Euler’s formula says that v −e + f = 2, so e −v + 2 = f, so 2e −2v + 4 = 2f. Combining this with 2f ≤e, we get that 2e −2v + 4 ≤e. So e ≤2v −4.
This result lets us show that K3,3 isn’t planar. All the cycles in K3,3 have at least four nodes. But K3,3 has 9 edges and 6 nodes, which isn’t consistent with this formula. So K3,3 can’t be planar.
CHAPTER 18. PLANAR GRAPHS 216 18.7 Kuratowski’s Theorem The two example non-planar graphs K3,3 and K5 weren’t picked randomly.
It turns out that any non-planar graph must contain a subgraph closely related to one of these two graphs. Specifically, we’ll say that a graph G is a subdivision of another graph F if the two graphs are isomorphic or if the only difference between the graphs is that G divides up some of F’s edges by adding extra degree 2 nodes in the middle of the edges.
For example, in the following picture, the righthand graph is a subdivision of the lefthand graph.
A B C D A B C D E F G We can now state our theorem precisely.
Claim 55 Kuratowski’s Theorem: A graph is nonplanar if and only if it contains a subgraph that is a subdivision of K3,3 or K5.
This was proved in 1930 by Kazimierz Kuratowski, and the proof is ap-parently somewhat difficult. So we’ll just see how to apply it.
For example, here’s a graph known as the Petersen graph (after a Danish mathematician named Julius Petersen).
CHAPTER 18. PLANAR GRAPHS 217 A B E C D a b e c d This isn’t planar. The offending subgraph is the whole graph, except for the node B (and the edges that connect to B): A E C D a b e c d This subgraph is a subdivision of K3,3. To see why, first notice that the node b is just subdividing the edge from d to e, so we can delete it. Or, CHAPTER 18. PLANAR GRAPHS 218 formally, the previous graph is a subdivision of this graph: A E C D a e c d In the same way, we can remove the nodes A and C, to eliminate unnec-essary subdivisions: E D a e c d Now deform the picture a bit and we see that we have K3,3.
CHAPTER 18. PLANAR GRAPHS 219 E D a e c d 18.8 Coloring planar graphs One application of planar graphs involves coloring maps of countries. Two countries sharing a border2 must be given different colors. We can turn this into a graph coloring problem by assigning a graph node to each country.
We then connect two nodes with an edge exactly when their regions share a border. This graph is called the dual of our original map. Because the maps are planar, these dual graphs are always planar.
Planar graphs can be colored much more easily than other graphs. For example, we can prove that they never require more than 6 colors: Proof: by induction on the number of nodes in G.
Base: The planar graph with just one node has maximum degree 0 and can be colored with one color.
Induction: Suppose that any planar graph with < k nodes can be colored with 6 colors. Let G be a planar graph with k nodes.
2Two regions touching at a point are not considered to share a border.
CHAPTER 18. PLANAR GRAPHS 220 By Corollary 3, G has a node of degree less than 6. Let’s pick such a node and call it v.
Remove some node v (and its edges) from G to create a smaller graph G′. G′ is a planar graph with k −1 nodes. So, by the inductive hypothesis, G′ can be colored with 5 colors.
Because v has less than 6 neighbors, its neighbors are only using 5 of the available colors. So there is a spare color to assign to v, finishing the coloring of G.
It’s not hard, but a bit messy, to upgrade this proof to show that planar graphs require only five colors. Four colors is much harder. Way back in 1852, Francis Guthrie hypothesized that any planar graph could be colored with only four colors, but it took 124 years to prove that he was right. Alfred Kempe thought he had proved it in 1879 and it took 11 years for another mathematician to find an error in his proof.
The Four Color Theorem was finally proved by Kenneth Appel and Wolf-gang Haken at UIUC in 1976. They reduced the problem mathematically, but were left with 1936 specific graphs that needed to be checked exhaustively, using a computer program. Not everyone was happy to have a computer involved in a mathematical proof, but the proof has come to be accepted as legitimate.
18.9 Application: Platonic solids A fact dating back to the Greeks is that there are only five Platonic solids: cube, dodecahedron, tetrahedron, icosahedron, octahedron. These are con-vex polyhedra whose faces all have the same number of sides (k) and whose nodes all have the same number of edges going into them (d).
To turn a Platonic solid into a graph, imagine that it’s made of a stretchy material. Make a small hole in one face. Put your fingers into that face and pull sideways, stretching that face really big and making the whole thing flat. For example, an octahedron (8 triangular sides) turns into the following graph. Notice that it still has eight faces, one for each face of the original solid, each with three sides.
CHAPTER 18. PLANAR GRAPHS 221 Graphs of polyhedra are slightly special planar graphs. Polyhedra aren’t allowed to have extra nodes partway along edges, so each node in the graph must have degree at least three. Also, since the faces must be flat and the edges straight, each face needs to be bounded by at least three edges. So, if G is the graph of a Platonic solid, all the nodes of G must have the same degree d ≥3 and all faces must have the same degree k ≥3.
Now, let’s do some algebra to see why there are so few possibilities for the structure of such a graph.
By the handshaking theorem, the sum of the node degrees is twice the number of edges. So, since the degrees are equal to d, we have dv = 2e and therefore v = 2e d By the handshaking theorem for faces, the sum of the face degrees is also twice the number of edges. That is kf = 2e. So f = 2e k Euler’s formula says that v−e+f = 2, so v+f = 2+e > e. Substituting the above equations into this one, we get: 2e d + 2e k > e CHAPTER 18. PLANAR GRAPHS 222 Dividing both sides by 2e: 1 d + 1 k > 1 2 If we analyze this equation, we discover that d and k can’t both be larger than 3. If they were both 4 or above, the left side of the equation would be ≤1 2. Since we know that d and k are both ≥3, this implies that one of the two is actually equal to three and the other is some integer that is at least 3.
Suppose we set d to be 3. Then the equation becomes 1 3 + 1 k > 1 2. So 1 k > 1 6, which means that k can’t be any larger than 5. Similarly, if k is 3, then d can’t be any larger than 5.
This leaves us only five possibilities for the degrees d and k: (3, 3), (3, 4), (3, 5), (4, 3), and (5, 3). Each of these corresponds to one of the Platonic solids. |
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Chebyshev Polynomials
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Chebyshev Polynomials
ByJ.C. Mason,David C. Handscomb
Edition 1st Edition
First Published 2002
eBook Published 17 September 2002
Pub. Location New York
Imprint Chapman and Hall/CRC
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eBook ISBN 9780429191381
Subjects Computer Science, Mathematics & Statistics
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Mason, J.C., & Handscomb, D.C. (2002). Chebyshev Polynomials (1st ed.). Chapman and Hall/CRC.
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Chebyshev polynomials crop up in virtually every area of numerical analysis, and they hold particular importance in recent advances in subjects such as orthogonal polynomials, polynomial approximation, numerical integration, and spectral methods. Yet no book dedicated to Chebyshev polynomials has been published since 1990, and even that work focuse
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If f f is continuous and f′(x)≥0 f′(x)≥0, outside of a countable set, then f f is increasing
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PROBLEM. Let f:[a,b]→R f:[a,b]→R be a continuous function, such that f′(x)≥0 f′(x)≥0, for all x∈[a,b]∖A x∈[a,b]∖A, where A⊂[a,b]A⊂[a,b] is a countable set. Show that f f is increasing.
Attention. In this problem, we DO NOT assume that f f is differentiable in the whole [a,b][a,b].
Notes. (1) If we assume that f f is differentiable in the whole interval, then we can easily show that f′(x)≥0 f′(x)≥0, everywhere. For otherwise, if f′(x 0)=c<0 f′(x 0)=c<0, for some x 0∈[a,b]x 0∈[a,b], then by virtue of Darboux's Theorem, (c,0)⊂f′([a,b])(c,0)⊂f′([a,b]), and hence, f′(x)<0 f′(x)<0, for uncountably many x x's.
(2) The conclusion of the problem does not hold if we replace the assumption A A is countable with A A is a set of measure zero. Take for example the Devil's staircase, with a negative sign in front.
(3) If the hypothesis f′(x)≥0 f′(x)≥0, is replaced by f′(x)=0 f′(x)=0, then the conclusion becomes f is constant.
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edited Jul 14, 2019 at 11:39
Yiorgos S. SmyrlisYiorgos S. Smyrlis
asked Jul 13, 2019 at 15:47
Yiorgos S. SmyrlisYiorgos S. Smyrlis
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The set A A could be dense in [a,b][a,b], i.e., A=Q∩[a,b]A=Q∩[a,b], in which case MVT can not apply.Yiorgos S. Smyrlis –Yiorgos S. Smyrlis 2019-07-13 16:18:57 +00:00 Commented Jul 13, 2019 at 16:18
You are right, sorry!Botond –Botond 2019-07-13 16:20:12 +00:00 Commented Jul 13, 2019 at 16:20
It is not a conjecture. It holds. I am preparing a solution. A little detail remains to be fixed.Yiorgos S. Smyrlis –Yiorgos S. Smyrlis 2019-07-14 11:37:30 +00:00 Commented Jul 14, 2019 at 11:37
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Let c,d∈[a,b]c,d∈[a,b], with c<d c<d. It suffices prove that
f(d)−f(c)>−(4+d−c)ε,f(d)−f(c)>−(4+d−c)ε,
for every ε>0 ε>0.
We enumerate A A as A={α n}n∈N A={α n}n∈N and choose δ n>0 δ n>0, such that
x∈(α n−δ n,α n+δ n)⟹|f(x)−f(α n)|<ε 2 n x∈(α n−δ n,α n+δ n)⟹|f(x)−f(α n)|<ε 2 n
for all n∈N n∈N. Finding such δ n δ n's is possible due to continuity of f f. Set I n=(α n−δ n,α n+δ n)I n=(α n−δ n,α n+δ n). In particular
y 1,y 2∈I n⟹f(y 2)>f(y 1)−ε 2 n−1(1)(1)y 1,y 2∈I n⟹f(y 2)>f(y 1)−ε 2 n−1
Let x∈[a,b]∖A x∈[a,b]∖A. Then there exists an η x>0 η x>0, such that
y∈(x−η x,x+η x)⟹−ε|y−x|<f(y)−f(x)−(y−x)f′(x)<ε|y−x|,y∈(x−η x,x+η x)⟹−ε|y−x|<f(y)−f(x)−(y−x)f′(x)<ε|y−x|,
and hence whenever y 1,y 2∈J x=(x−η x,x+η x)y 1,y 2∈J x=(x−η x,x+η x), with y 1≤x≤y 2 y 1≤x≤y 2, we have that
f(y 2)−f(y 1)−(y 2−y 1)f′(x)≥−ε(|y 1−x|+|y 2−x|)f(y 2)−f(y 1)−(y 2−y 1)f′(x)≥−ε(|y 1−x|+|y 2−x|)
and since f′(x)≥0 f′(x)≥0, we finally obtain that
f(y 2)>f(y 1)−ε(y 2−y 1).(2)(2)f(y 2)>f(y 1)−ε(y 2−y 1).
We shall use the following result (for a proof see here):
Cousin's Lemma.Let C C be a full cover of [a,b][a,b], that is, a collection of closed subintervals of [a,b][a,b] with the property that for every x∈[a,b]x∈[a,b], there exists a δ>0 δ>0, so that C C contains all subintervals of [a,b][a,b] which contains x x and have length smaller than δ δ. Then there exists a partition {I 1,I 2,…,I m}⊂C{I 1,I 2,…,I m}⊂C of non-overlapping intervals for [a,b][a,b], where I i=[x i−1,x i]I i=[x i−1,x i] and a=x 0<x 1<⋯<x n=b,a=x 0<x 1<⋯<x n=b, for all 1≤i≤m 1≤i≤m.
We define a C C the collection of all closed subintervals K K of [c,d][c,d], such that either K⊂I n K⊂I n and α n∈K α n∈K, for some α n∈A α n∈A or K⊂J x K⊂J x and x∈K x∈K for some x∈[a,b]∖A x∈[a,b]∖A. Cousin's Lemma provides the existence of points c=x 0<x 1<⋯<x m=d c=x 0<x 1<⋯<x m=d, such that the closed intervals
K 1=[x 0,x 1],K 2=[x 1,x 2],…,K m=[x m−1,x m]K 1=[x 0,x 1],K 2=[x 1,x 2],…,K m=[x m−1,x m]
belong to C C.
From the construction of C C, each K j K j is either a subinterval of some I n I n or some J x J x, and possibly K j K j is a subset of more than one such intervals. To every K j K j we assign exactly one such interval. In particular, to every j∈{1,…,m}j∈{1,…,m} we assign either a unique n∈N n∈N, such that α n∈K j⊂I n α n∈K j⊂I n, which we denote as n j n j, or a unique x∈[a,b]∖A x∈[a,b]∖A, such that x∈K j⊂J x x∈K j⊂J x. This mapping is not necessarily 1−1 1−1, since if α n α n is the common endpoint of K j K j and K j+1 K j+1, it is possible that n j=n j+1 n j=n j+1. Thus, some of the I n I n's may have been assigned to two K j K j's (and no more than two).
We split S={1,…,m}S={1,…,m} as a union of two disjoint sets. S 1 S 1 shall be the set of those j∈S j∈S, to which an n∈N n∈N has been assigned (i.e., α n∈K j⊂I n=I n j α n∈K j⊂I n=I n j) while S 2=S∖S 1 S 2=S∖S 1. If j∈S 2 j∈S 2, then an x∈[a,b]∖A x∈[a,b]∖A has been assigned to j j and x∈K j⊂J x x∈K j⊂J x.
If j∈S 1 j∈S 1, and K j⊂I n j K j⊂I n j then (1)(1) provides the f(x j)−f(x j−1)>−ε 2 n j−1 f(x j)−f(x j−1)>−ε 2 n j−1, while if j∈S 2 j∈S 2, then (2)(2) provides that f(x j)−f(x j−1)>−ε(x j−x j−1)f(x j)−f(x j−1)>−ε(x j−x j−1).
We now have that
f(d)−f(c)=∑j=1 m(f(x j)−f(x j−1))=∑j∈S 1(f(x j)−f(x j−1))+∑j∈S 2(f(x j)−f(x j−1))≥−∑j∈S 1 ε 2 n j−1−∑j∈S 2 ε(x j−x j−1)>−4 ε−ε(d−c)=−(4+d−c)ε.f(d)−f(c)=∑j=1 m(f(x j)−f(x j−1))=∑j∈S 1(f(x j)−f(x j−1))+∑j∈S 2(f(x j)−f(x j−1))≥−∑j∈S 1 ε 2 n j−1−∑j∈S 2 ε(x j−x j−1)>−4 ε−ε(d−c)=−(4+d−c)ε.
The last inequality holds because in the first sum, ∑j∈S 1 1 2 n j−1<2∑∞n=1 1 2 n−1=4∑j∈S 1 1 2 n j−1<2∑n=1∞1 2 n−1=4, since the power 1 2 n−1 1 2 n−1 may appear twice, if α n α n is an endpoint of two neighboring K j K j's.
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edited Jul 16, 2019 at 8:39
answered Jul 14, 2019 at 15:42
Yiorgos S. SmyrlisYiorgos S. Smyrlis
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1 Did you see math.stackexchange.com/q/3313210/42969? That is a consequence of your result. But it seems to me that David's approach of choosing an “increasing non-overlapping cover” of intervals would work for your problem as well, and is simpler than using the Cousin's Lemma. I might be overlooking something, of course.Martin R –Martin R 2019-08-06 09:00:08 +00:00 Commented Aug 6, 2019 at 9:00
1 Actually, that was my first idea. But it is not correct. (Possibly it could be corrected.) This is because the sum of |x j−y j||x j−y j| is not less than 1.Yiorgos S. Smyrlis –Yiorgos S. Smyrlis 2019-08-06 15:51:50 +00:00 Commented Aug 6, 2019 at 15:51
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Here is an easier proof. One can show that if f′(x)>0 f′(x)>0 for all x∈[a,b]∖A x∈[a,b]∖A (note the strict inequality), then f f is strictly increasing (look here for a proof). So we consider the collection of functions {f ε(x)=f(x)+ε x}ε>0{f ε(x)=f(x)+ε x}ε>0. It is clear that each f ε f ε is continuous and f′ε(x)=f′(x)+ε>0 f ε′(x)=f′(x)+ε>0 for all x∈[a,b]∖A x∈[a,b]∖A. Thus, f ε f ε is strictly increasing and for any a≤x 1ε(x 1−x 2)f(x 2)−f(x 1)>ε(x 1−x 2). We take the limit as ε→0 ε→0 to conclude.
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answered May 2 at 20:37
Karthik KannanKarthik Kannan
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I'm not following. 0+ϵ 0+ϵ is positive for all positive ϵ ϵ, but the limt as ϵ ϵ goes to 0 0 is not positive....Mike –Mike 2025-05-02 20:43:52 +00:00 Commented May 2 at 20:43
@Mike We have f(x 2)−f(x 1)≥lim ε→0 ε(x 1−x 2)=0 f(x 2)−f(x 1)≥lim ε→0 ε(x 1−x 2)=0.Karthik Kannan –Karthik Kannan 2025-05-02 20:47:18 +00:00 Commented May 2 at 20:47
Wouldn't you have to show instead that f(x)−ϵ x f(x)−ϵ x [note the −− sign] is nondecreasing on [a,b][a,b] for some ϵ>0 ϵ>0? That would give you strictly increasing.Mike –Mike 2025-05-02 21:00:36 +00:00 Commented May 2 at 21:00
@Mike f f need not be strictly increasing. We only need to show that it is non-decreasing (i.e., we could have f(x 2)=f(x 1)f(x 2)=f(x 1) for x 1<x 2 x 1<x 2). We cannot consider f(x)−ε x f(x)−ε x since it need not have a positive derivative on [a,b]∖A[a,b]∖A.Karthik Kannan –Karthik Kannan 2025-05-02 21:02:50 +00:00 Commented May 2 at 21:02
Oh, I think I see it now. You are assuming a previous result that states f′(x)>0 f′(x)>0 for all x∈[a,b]∖A x∈[a,b]∖A⟹f⟹f strictly increasing. And then are using that previous result to show that f′(x)≥0 f′(x)≥0 for all x∈[a,b]∖A x∈[a,b]∖A⟹f⟹f nondecreasing,Mike –Mike 2025-05-02 21:08:33 +00:00 Commented May 2 at 21:08
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[x] ☰ MicroeconomicsMacroeconomicsEconomic indicators
MicroeconomicsEconomicsOpportunity costPositive statementsNormative statementsMarketSubstitute goodsComplementary goodsNormal goodsInferior goodsDemand curveDemand functionInverse demand functionMovements along the demand curveShifts in the demand curveAdding demand curvesLaw of demandSupply curveSupply functionInverse supply functionMovements along the supply curveShifts in the supply curveAdding supply curvesLaw of supplyMarket equilibriumExcess supply or surplusExcess demand or shortagePrice elasticity of demandMidpoint method to calculate elasticityDemand curves according to their elasticityPrice elasticity of demand and total revenueElasticity along the demand curveIncome elasticity of demandCross-price elasticity of demandPrice elasticity of supplySupply curves based on their elasticityShort-run vs long-run elasticitiesPoint elasticity and arc elasticityPrice ceilingPrice floor
Demand Function
The demand function is a function that shows the relationship between the quantity demanded of a good and the factors that affect it. Many variables influence the quantity demanded, such as income or the prices of substitute and complementary goods, but price plays a central role.
Linear Demand Function
An example of a linear demand function is as follows:
Q=800−10 P
Where:
Q represents the quantity demanded.
P is the price of the good or service.
800 indicates the maximum quantity demanded when the price is zero.
10 is the coefficient that shows the rate at which the quantity demanded decreases as the price increases.
In this example, the quantity demanded depends solely on the price, assuming that all other factors affecting demand are held constant. The negative sign represents the inverse relationship between price and quantity demanded: higher prices result in lower quantities demanded, and vice versa. By substituting a specific price into this equation, we can find the quantity demanded at that price.
To calculate the quantity demanded when the price P is 20:
Q=800−10×20
Performing the multiplication:
Q=800−200
Simplifying the subtraction:
Q=600
Thus, when the price P is 20, the quantity demanded Q is 600.
However, the demand function does not need to include only the price while keeping everything else constant. Many other factors can be added to the demand function. Let's include consumer income, the price of a substitute good, and the price of a complementary good. There are many other factors that can affect demand, but anything not included in the equation is assumed to be irrelevant or held constant:
Q=500−10 P+0.25 I+0.5 P s−0.75 P c
Where:
Q represents the quantity demanded.
P is the price of the good or service being analyzed.
I is the consumer's income.
P s is the price of a substitute good.
P c is the price of a complementary good.
500 indicates the maximum quantity demanded when P, P s, and P c are zero and income has no influence.
The coefficients −10,0.25,0.5,−0.75 reflect the sensitivity of the quantity demanded to changes in the price of the good, income, the price of the substitute good, and the price of the complementary good, respectively.
Note that the term containing income has a positive sign, indicating the positive relationship between income and quantity demanded. The term for the price of the substitute good also has a positive sign, as an increase in the price of a substitute good increases the quantity demanded of the good in question. The term for the price of the complementary good has a negative sign because an increase in the price of a complementary good reduces the quantity demanded.
If we set the consumer income to 1380, the price of the substitute good to 60, and the price of the complementary good to 100, we assume these three determinants of demand remain constant at these values, and we obtain the demand function that depends only on the price:
Performing the calculations for the quantity demanded, where the values are given by: income I=1380, price of a substitute good P s=60, and price of a complementary good P c=100:
Q=500−10 P+0.25×1380+0.5×60−0.75×100
Performing the calculations:
Q=500−10 P+345+30−75
Simplifying the sum:
Q=500+345+30−75−10 P
Q=800−10 P
Thus, the simplified demand function is Q=800−10 P.
Generalized Demand Function
In the previous examples, the demand functions have a linear form, but this does not have to be the case. Demand functions can take other forms, such as logarithmic or multiplicative. Therefore, we can generally express the function that depends only on price:
The demand function is expressed as:
Q d=Q d(p)
Where:
Q d represents the quantity demanded of a good or service.
p is the price of the good or service.
Q d(p) is a function that shows how the quantity demanded Q d varies in response to changes in the price p.
And for the function that depends on several factors:
Q d=Q d(p,I,p s,p c)
Where:
Q d represents the quantity demanded of a good or service.
p is the price of the good or service being analyzed.
I represents the consumer's income.
p s is the price of a substitute good.
p c is the price of a complementary good.
Q d(p,I,p s,p c) is a function that shows how the quantity demanded Q d varies in response to changes in the price p, income I, the price of the substitute good p s, and the price of the complementary good p c.
In this case, we are not expressing an explicit functional form, and therefore the demand function can take various functional forms.
Demand curveInverse demand function
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24 The Price Elasticity of Demand
Learning Objectives
Explain the concept of price elasticity of demand and its calculation.
Explain what it means for demand to be price inelastic, unit price elastic, price elastic, perfectly price inelastic, and perfectly price elastic.
Explain how and why the value of the price elasticity of demand changes along a linear demand curve.
Understand the relationship between total revenue and price elasticity of demand.
Discuss the determinants of price elasticity of demand.
We know from the law of demand how the quantity demanded will respond to a price change: it will change in the opposite direction. But how much will it change? It seems reasonable to expect, for example, that a 10% change in the price charged for a visit to the doctor would yield a different percentage change in quantity demanded than a 10% change in the price of a Ford Mustang. But how much is this difference?
To show how responsive quantity demanded is to a change in price, we apply the concept of elasticity. The price elasticity of demand for a good or service, eD, is the percentage change in quantity demanded of a particular good or service divided by the percentage change in the price of that good or service, all other things unchanged. Thus we can write
Equation 3.1
[latex]e_D = \frac{\% \ change \ in \ quantity \ demanded}{\% \ change \ in \ price}[/latex]
Because the price elasticity of demand shows the responsiveness of quantity demanded to a price change, assuming that other factors that influence demand are unchanged, it reflects movements along a demand curve. With a downward-sloping demand curve, price and quantity demanded move in opposite directions, so the price elasticity of demand is always negative. A positive percentage change in price implies a negative percentage change in quantity demanded, and vice versa. Sometimes you will see the absolute value of the price elasticity measure reported. In essence, the minus sign is ignored because it is expected that there will be a negative (inverse) relationship between quantity demanded and price. In this text, however, we will retain the minus sign in reporting price elasticity of demand and will say “the absolute value of the price elasticity of demand” when that is what we are describing.
Heads Up!
Be careful not to confuse elasticity with slope. The slope of a line is the change in the value of the variable on the vertical axis divided by the change in the value of the variable on the horizontal axis between two points. Elasticity is the ratio of the percentage changes. The slope of a demand curve, for example, is the ratio of the change in price to the change in quantity between two points on the curve. The price elasticity of demand is the ratio of the percentage change in quantity to the percentage change in price. As we will see, when computing elasticity at different points on a linear demand curve, the slope is constant—that is, it does not change—but the value for elasticity will change.
Computing the Price Elasticity of Demand
Finding the price elasticity of demand requires that we first compute percentage changes in price and in quantity demanded. We calculate those changes between two points on a demand curve.
Figure 3.1 “Responsiveness and Demand” shows a particular demand curve, a linear demand curve for public transit rides. Suppose the initial price is $0.80, and the quantity demanded is 40,000 rides per day; we are at point A on the curve. Now suppose the price falls to $0.70, and we want to report the responsiveness of the quantity demanded. We see that at the new price, the quantity demanded rises to 60,000 rides per day (point B). To compute the elasticity, we need to compute the percentage changes in price and in quantity demanded between points A and B.
Figure 3.1 Responsiveness and Demand
The demand curve shows how changes in price lead to changes in the quantity demanded. A movement from point A to point B shows that a $0.10 reduction in price increases the number of rides per day by 20,000. A movement from B to A is a $0.10 increase in price, which reduces quantity demanded by 20,000 rides per day.
We measure the percentage change between two points as the change in the variable divided by the average value of the variable between the two points. Thus, the percentage change in quantity between points A and B in Figure 3.1 “Responsiveness and Demand” is computed relative to the average of the quantity values at points A and B: (60,000 + 40,000)/2 = 50,000. The percentage change in quantity, then, is 20,000/50,000, or 40%. Likewise, the percentage change in price between points A and B is based on the average of the two prices: ($0.80 + $0.70)/2 = $0.75, and so we have a percentage change of −0.10/0.75, or −13.33%. The price elasticity of demand between points A and B is thus 40%/(−13.33%) = −3.00.
This measure of elasticity, which is based on percentage changes relative to the average value of each variable between two points, is called arc elasticity. The arc elasticity method has the advantage that it yields the same elasticity whether we go from point A to point B or from point B to point A. It is the method we shall use to compute elasticity.
For the arc elasticity method, we calculate the price elasticity of demand using the average value of price, $$ \bar{P} $$ , and the average value of quantity demanded, $$ \bar{Q} $$. We shall use the Greek letter Δ to mean “change in,” so the change in quantity between two points is ΔQ and the change in price is ΔP. Now we can write the formula for the price elasticity of demand as
Equation 3.2
[latex]\displaystyle e_D = \frac{\Delta Q / \bar{Q}}{\Delta P / \bar{P}}[/latex]
The price elasticity of demand between points A and B is thus:
[latex]e_D = \frac{\frac{20,000}{(40,000+60,000)/2}}{\frac{- \$ 0.10}{( \$ 0.80 + \$ 0.70)/2}} = \frac{40 \% }{-13.33 \% } = -3.00[/latex]
With the arc elasticity formula, the elasticity is the same whether we move from point A to point B or from point B to point A. If we start at point B and move to point A, we have:
[latex]e_D = \frac{\frac{-20,000}{(60,000+40,000)/2}}{\frac{ \$ 0.10}{( \$ 0.80 + \$ 0.70)/2}} = \frac{-40 \% }{13.33 \% } = -3.00[/latex]
The arc elasticity method gives us an estimate of elasticity. It gives the value of elasticity at the midpoint over a range of change, such as the movement between points A and B. For a precise computation of elasticity, we would need to consider the response of a dependent variable to an extremely small change in an independent variable. The fact that arc elasticities are approximate suggests an important practical rule in calculating arc elasticities: we should consider only small changes in independent variables. We cannot apply the concept of arc elasticity to large changes.
Another argument for considering only small changes in computing price elasticities of demand will become evident in the next section. We will investigate what happens to price elasticities as we move from one point to another along a linear demand curve.
Heads Up!
Notice that in the arc elasticity formula, the method for computing a percentage change differs from the standard method with which you may be familiar. That method measures the percentage change in a variable relative to its original value. For example, using the standard method, when we go from point A to point B, we would compute the percentage change in quantity as 20,000/40,000 = 50%. The percentage change in price would be −$0.10/$0.80 = −12.5%. The price elasticity of demand would then be 50%/(−12.5%) = −4.00. Going from point B to point A, however, would yield a different elasticity. The percentage change in quantity would be −20,000/60,000, or −33.33%. The percentage change in price would be $0.10/$0.70 = 14.29%. The price elasticity of demand would thus be −33.33%/14.29% = −2.33. By using the average quantity and average price to calculate percentage changes, the arc elasticity approach avoids the necessity to specify the direction of the change and, thereby, gives us the same answer whether we go from A to B or from B to A.
Price Elasticities Along a Linear Demand Curve
What happens to the price elasticity of demand when we travel along the demand curve? The answer depends on the nature of the demand curve itself. On a linear demand curve, such as the one in Figure 3.3 “Price Elasticities of Demand for a Linear Demand Curve”, elasticity becomes smaller (in absolute value) as we travel downward and to the right.
Figure 3.3 Price Elasticities of Demand for a Linear Demand Curve
The price elasticity of demand varies between different pairs of points along a linear demand curve. The lower the price and the greater the quantity demanded, the lower the absolute value of the price elasticity of demand.
Figure 3.3 “Price Elasticities of Demand for a Linear Demand Curve” shows the same demand curve we saw in Figure 3.2 “Responsiveness and Demand”. We have already calculated the price elasticity of demand between points A and B; it equals −3.00. Notice, however, that when we use the same method to compute the price elasticity of demand between other sets of points, our answer varies. For each of the pairs of points shown, the changes in price and quantity demanded are the same (a $0.10 decrease in price and 20,000 additional rides per day, respectively). But at the high prices and low quantities on the upper part of the demand curve, the percentage change in quantity is relatively large, whereas the percentage change in price is relatively small. The absolute value of the price elasticity of demand is thus relatively large. As we move down the demand curve, equal changes in quantity represent smaller and smaller percentage changes, whereas equal changes in price represent larger and larger percentage changes, and the absolute value of the elasticity measure declines. Between points C and D, for example, the price elasticity of demand is −1.00, and between points E and F the price elasticity of demand is −0.33.
On a linear demand curve, the price elasticity of demand varies depending on the interval over which we are measuring it. For any linear demand curve, the absolute value of the price elasticity of demand will fall as we move down and to the right along the curve.
The Price Elasticity of Demand and Changes in Total Revenue
Suppose the public transit authority is considering raising fares. Will its total revenues go up or down? Total revenue is the price per unit times the number of units sold1. In this case, it is the fare times the number of riders. The transit authority will certainly want to know whether a price increase will cause its total revenue to rise or fall. In fact, determining the impact of a price change on total revenue is crucial to the analysis of many problems in economics.
We will do two quick calculations before generalizing the principle involved. Given the demand curve shown in Figure 3.3 “Price Elasticities of Demand for a Linear Demand Curve”, we see that at a price of $0.80, the transit authority will sell 40,000 rides per day. Total revenue would be $32,000 per day ($0.80 times 40,000). If the price were lowered by $0.10 to $0.70, quantity demanded would increase to 60,000 rides and total revenue would increase to $42,000 ($0.70 times 60,000). The reduction in fare increases total revenue. However, if the initial price had been $0.30 and the transit authority reduced it by $0.10 to $0.20, total revenue would decrease from $42,000 ($0.30 times 140,000) to $32,000 ($0.20 times 160,000). So it appears that the impact of a price change on total revenue depends on the initial price and, by implication, the original elasticity. We generalize this point in the remainder of this section.
The problem in assessing the impact of a price change on total revenue of a good or service is that a change in price always changes the quantity demanded in the opposite direction. An increase in price reduces the quantity demanded, and a reduction in price increases the quantity demanded. The question is how much. Because total revenue is found by multiplying the price per unit times the quantity demanded, it is not clear whether a change in price will cause total revenue to rise or fall.
We have already made this point in the context of the transit authority. Consider the following three examples of price increases for gasoline, pizza, and diet cola.
Suppose that 1,000 gallons of gasoline per day are demanded at a price of $4.00 per gallon. Total revenue for gasoline thus equals $4,000 per day (=1,000 gallons per day times $4.00 per gallon). If an increase in the price of gasoline to $4.25 reduces the quantity demanded to 950 gallons per day, total revenue rises to $4,037.50 per day (=950 gallons per day times $4.25 per gallon). Even though people consume less gasoline at $4.25 than at $4.00, total revenue rises because the higher price more than makes up for the drop in consumption.
Next consider pizza. Suppose 1,000 pizzas per week are demanded at a price of $9 per pizza. Total revenue for pizza equals $9,000 per week (=1,000 pizzas per week times $9 per pizza). If an increase in the price of pizza to $10 per pizza reduces quantity demanded to 900 pizzas per week, total revenue will still be $9,000 per week (=900 pizzas per week times $10 per pizza). Again, when price goes up, consumers buy less, but this time there is no change in total revenue.
Now consider diet cola. Suppose 1,000 cans of diet cola per day are demanded at a price of $0.50 per can. Total revenue for diet cola equals $500 per day (=1,000 cans per day times $0.50 per can). If an increase in the price of diet cola to $0.55 per can reduces quantity demanded to 880 cans per month, total revenue for diet cola falls to $484 per day (=880 cans per day times $0.55 per can). As in the case of gasoline, people will buy less diet cola when the price rises from $0.50 to $0.55, but in this example total revenue drops.
In our first example, an increase in price increased total revenue. In the second, a price increase left total revenue unchanged. In the third example, the price rise reduced total revenue. Is there a way to predict how a price change will affect total revenue? There is; the effect depends on the price elasticity of demand.
Elastic, Unit Elastic, and Inelastic Demand
To determine how a price change will affect total revenue, economists place price elasticities of demand in three categories, based on their absolute value. If the absolute value of the price elasticity of demand is greater than 1, demand is termed price elastic. If it is equal to 1, demand is unit price elastic. And if it is less than 1, demand is price inelastic.
Relating Elasticity to Changes in Total Revenue
When the price of a good or service changes, the quantity demanded changes in the opposite direction. Total revenue will move in the direction of the variable that changes by the larger percentage. If the variables move by the same percentage, total revenue stays the same. If quantity demanded changes by a larger percentage than price (i.e., if demand is price elastic), total revenue will change in the direction of the quantity change. If price changes by a larger percentage than quantity demanded (i.e., if demand is price inelastic), total revenue will move in the direction of the price change. If price and quantity demanded change by the same percentage (i.e., if demand is unit price elastic), then total revenue does not change.
When demand is price inelastic, a given percentage change in price results in a smaller percentage change in quantity demanded. That implies that total revenue will move in the direction of the price change: a reduction in price will reduce total revenue, and an increase in price will increase it.
Consider the price elasticity of demand for gasoline. In the example above, 1,000 gallons of gasoline were purchased each day at a price of $4.00 per gallon; an increase in price to $4.25 per gallon reduced the quantity demanded to 950 gallons per day. We thus had an average quantity of 975 gallons per day and an average price of $4.125. We can thus calculate the arc price elasticity of demand for gasoline:
| |
| Percentage change in quantity demanded = -50/975 = -5.1% |
| Percentage change in price=0.25/4.125=6.06% |
| Price elasticity of demand = -5.1%/6.06% = -.084 |
The demand for gasoline is price inelastic, and total revenue moves in the direction of the price change. When price rises, total revenue rises. Recall that in our example above, total spending on gasoline (which equals total revenues to sellers) rose from $4,000 per day (=1,000 gallons per day times $4.00) to $4037.50 per day (=950 gallons per day times $4.25 per gallon).
When demand is price inelastic, a given percentage change in price results in a smaller percentage change in quantity demanded. That implies that total revenue will move in the direction of the price change: an increase in price will increase total revenue, and a reduction in price will reduce it.
Consider again the example of pizza that we examined above. At a price of $9 per pizza, 1,000 pizzas per week were demanded. Total revenue was $9,000 per week (=1,000 pizzas per week times $9 per pizza). When the price rose to $10, the quantity demanded fell to 900 pizzas per week. Total revenue remained $9,000 per week (=900 pizzas per week times $10 per pizza). Again, we have an average quantity of 950 pizzas per week and an average price of $9.50. Using the arc elasticity method, we can compute:
| |
| Percentage change in quantity demanded = -100/950 = -10.5% |
| Percentage change in price = $1.00/$9.50 = 10.5% |
| Price elasticity of demand = -10.5%/10.5% = -1.0 |
Demand is unit price elastic, and total revenue remains unchanged. Quantity demanded falls by the same percentage by which price increases.
Consider next the example of diet cola demand. At a price of $0.50 per can, 1,000 cans of diet cola were purchased each day. Total revenue was thus $500 per day (=$0.50 per can times 1,000 cans per day). An increase in price to $0.55 reduced the quantity demanded to 880 cans per day. We thus have an average quantity of 940 cans per day and an average price of $0.525 per can. Computing the price elasticity of demand for diet cola in this example, we have:
| |
| Percentage change in quantity demanded = -120/940 = -12.8% |
| Percentage change in price = $0.05/$0.525 = 9.5% |
| Price elasticity of demand = -12.8%/9.5% = -1.3 |
The demand for diet cola is price elastic, so total revenue moves in the direction of the quantity change. It falls from $500 per day before the price increase to $484 per day after the price increase.
A demand curve can also be used to show changes in total revenue. Figure 3.16 “Changes in Total Revenue and a Linear Demand Curve” shows the demand curve from Figure 3.14 “Responsiveness and Demand” and Figure 3.15 “Price Elasticities of Demand for a Linear Demand Curve”. At point A, total revenue from public transit rides is given by the area of a rectangle drawn with point A in the upper right-hand corner and the origin in the lower left-hand corner. The height of the rectangle is price; its width is quantity. We have already seen that total revenue at point A is $32,000 ($0.80 × 40,000). When we reduce the price and move to point B, the rectangle showing total revenue becomes shorter and wider. Notice that the area gained in moving to the rectangle at B is greater than the area lost; total revenue rises to $42,000 ($0.70 × 60,000). Recall from Figure 3.4 “Price Elasticities of Demand for a Linear Demand Curve” that demand is elastic between points A and B. In general, demand is elastic in the upper half of any linear demand curve, so total revenue moves in the direction of the quantity change.
Figure 3.4 Changes in Total Revenue and a Linear Demand Curve
Moving from point A to point B implies a reduction in price and an increase in the quantity demanded. Demand is elastic between these two points. Total revenue, shown by the areas of the rectangles drawn from points A and B to the origin, rises. When we move from point E to point F, which is in the inelastic region of the demand curve, total revenue falls.
A movement from point E to point F also shows a reduction in price and an increase in quantity demanded. This time, however, we are in an inelastic region of the demand curve. Total revenue now moves in the direction of the price change—it falls. Notice that the rectangle drawn from point F is smaller in area than the rectangle drawn from point E, once again confirming our earlier calculation.
Figure 3.5
We have noted that a linear demand curve is more elastic where prices are relatively high and quantities relatively low and less elastic where prices are relatively low and quantities relatively high. We can be even more specific. For any linear demand curve, demand will be price elastic in the upper half of the curve and price inelastic in its lower half. At the midpoint of a linear demand curve, demand is unit price elastic.
Constant Price Elasticity of Demand Curves
Figure 3.6 “Demand Curves with Constant Price Elasticities” shows four demand curves over which price elasticity of demand is the same at all points. The demand curve in Panel (a) is vertical. This means that price changes have no effect on quantity demanded. The numerator of the formula given in Equation 3.2 for the price elasticity of demand (percentage change in quantity demanded) is zero. The price elasticity of demand in this case is therefore zero, and the demand curve is said to be perfectly inelastic. This is a theoretically extreme case, and no good that has been studied empirically exactly fits it. A good that comes close, at least over a specific price range, is insulin. A diabetic will not consume more insulin as its price falls but, over some price range, will consume the amount needed to control the disease.
Figure 3.6 Demand Curves with Constant Price Elasticities
The demand curve in Panel (a) is perfectly inelastic. The demand curve in Panel (b) is perfectly elastic. Price elasticity of demand is −1.00 all along the demand curve in Panel (c), whereas it is −0.50 all along the demand curve in Panel (d).
As illustrated in Figure 3.6 “Demand Curves with Constant Price Elasticities”, several other types of demand curves have the same elasticity at every point on them. The demand curve in Panel (b) is horizontal. This means that even the smallest price changes have enormous effects on quantity demanded. The denominator of the formula given in Equation 4.2 for the price elasticity of demand (percentage change in price) approaches zero. The price elasticity of demand in this case is therefore infinite, and the demand curve is said to be perfectly elastic. This is the type of demand curve faced by producers of standardized products such as wheat. If the wheat of other farms is selling at $4 per bushel, a typical farm can sell as much wheat as it wants to at $4 but nothing at a higher price and would have no reason to offer its wheat at a lower price.
The nonlinear demand curves in Panels (c) and (d) have price elasticities of demand that are negative; but, unlike the linear demand curve discussed above, the value of the price elasticity is constant all along each demand curve. The demand curve in Panel (c) has price elasticity of demand equal to −1.00 throughout its range; in Panel (d) the price elasticity of demand is equal to −0.50 throughout its range. Empirical estimates of demand often show curves like those in Panels (c) and (d) that have the same elasticity at every point on the curve.
Heads Up!
Do not confuse price inelastic demand and perfectly inelastic demand. Perfectly inelastic demand means that the change in quantity is zero for any percentage change in price; the demand curve in this case is vertical. Price inelastic demand means only that the percentage change in quantity is less than the percentage change in price, not that the change in quantity is zero. With price inelastic (as opposed to perfectly inelastic) demand, the demand curve itself is still downward sloping.
Determinants of the Price Elasticity of Demand
The greater the absolute value of the price elasticity of demand, the greater the responsiveness of quantity demanded to a price change. What determines whether demand is more or less price elastic? The most important determinants of the price elasticity of demand for a good or service are the availability of substitutes, the importance of the item in household budgets, and time.
Availability of Substitutes
The price elasticity of demand for a good or service will be greater in absolute value if many close substitutes are available for it. If there are lots of substitutes for a particular good or service, then it is easy for consumers to switch to those substitutes when there is a price increase for that good or service. Suppose, for example, that the price of Ford automobiles goes up. There are many close substitutes for Fords—Chevrolets, Chryslers, Toyotas, and so on. The availability of close substitutes tends to make the demand for Fords more price elastic.
If a good has no close substitutes, its demand is likely to be somewhat less price elastic. There are no close substitutes for gasoline, for example. The price elasticity of demand for gasoline in the intermediate term of, say, three–nine months is generally estimated to be about −0.5. Since the absolute value of price elasticity is less than 1, it is price inelastic. We would expect, though, that the demand for a particular brand of gasoline will be much more price elastic than the demand for gasoline in general.
Importance in Household Budgets
One reason price changes affect quantity demanded is that they change how much a consumer can buy; a change in the price of a good or service affects the purchasing power of a consumer’s income and thus affects the amount of a good the consumer will buy. This effect is stronger when a good or service is important in a typical household’s budget.
A change in the price of jeans, for example, is probably more important in your budget than a change in the price of pencils. Suppose the prices of both were to double. You had planned to buy four pairs of jeans this year, but now you might decide to make do with two new pairs. A change in pencil prices, in contrast, might lead to very little reduction in quantity demanded simply because pencils are not likely to loom large in household budgets. The greater the importance of an item in household budgets, the greater the absolute value of the price elasticity of demand is likely to be.
Time
Suppose the price of electricity rises tomorrow morning. What will happen to the quantity demanded?
The answer depends in large part on how much time we allow for a response. If we are interested in the reduction in quantity demanded by tomorrow afternoon, we can expect that the response will be very small. But if we give consumers a year to respond to the price change, we can expect the response to be much greater. We expect that the absolute value of the price elasticity of demand will be greater when more time is allowed for consumer responses.
Consider the price elasticity of crude oil demand. Economist John C. B. Cooper estimated short- and long-run price elasticities of demand for crude oil for 23 industrialized nations for the period 1971–2000. Professor Cooper found that for virtually every country, the price elasticities were negative, and the long-run price elasticities were generally much greater (in absolute value) than were the short-run price elasticities. His results are reported in Table 3.1 “Short- and Long-Run Price Elasticities of the Demand for Crude Oil in 23 Countries”. As you can see, the research was reported in a journal published by OPEC (Organization of Petroleum Exporting Countries), an organization whose members have profited greatly from the inelasticity of demand for their product. By restricting supply, OPEC, which produces about 45% of the world’s crude oil, is able to put upward pressure on the price of crude. That increases OPEC’s (and all other oil producers’) total revenues and reduces total costs.
Table 3.1 Short- and Long-Run Price Elasticities of the Demand for Crude Oil in 23 Countries
| Country | Short-Run Price Elasticity of Demand | Long-Run Price Elasticity of Demand |
---
| Australia | −0.034 | −0.068 |
| Austria | −0.059 | −0.092 |
| Canada | −0.041 | −0.352 |
| China | 0.001 | 0.005 |
| Denmark | −0.026 | −0.191 |
| Finland | −0.016 | −0.033 |
| France | −0.069 | −0.568 |
| Germany | −0.024 | −0.279 |
| Greece | −0.055 | −0.126 |
| Iceland | −0.109 | −0.452 |
| Ireland | −0.082 | −0.196 |
| Italy | −0.035 | −0.208 |
| Japan | −0.071 | −0.357 |
| Korea | −0.094 | −0.178 |
| Netherlands | −0.057 | −0.244 |
| New Zealand | −0.054 | −0.326 |
| Norway | −0.026 | −0.036 |
| Portugal | 0.023 | 0.038 |
| Spain | −0.087 | −0.146 |
| Sweden | −0.043 | −0.289 |
| Switzerland | −0.030 | −0.056 |
| United Kingdom | −0.068 | −0.182 |
| United States | −0.061 | −0.453 |
For most countries, price elasticity of demand for crude oil tends to be greater (in absolute value) in the long run than in the short run.
Source: John C. B. Cooper, “Price Elasticity of Demand for Crude Oil: Estimates from 23 Countries,” OPEC Review: Energy Economics & Related Issues, 27:1 (March 2003): 4. The estimates are based on data for the period 1971–2000, except for China and South Korea, where the period is 1979–2000. While the price elasticities for China and Portugal were positive, they were not statistically significant.
Key Takeaways
The price elasticity of demand measures the responsiveness of quantity demanded to changes in price; it is calculated by dividing the percentage change in quantity demanded by the percentage change in price.
Demand is price inelastic if the absolute value of the price elasticity of demand is less than 1; it is unit price elastic if the absolute value is equal to 1; and it is price elastic if the absolute value is greater than 1.
Demand is price elastic in the upper half of any linear demand curve and price inelastic in the lower half. It is unit price elastic at the midpoint.
When demand is price inelastic, total revenue moves in the direction of a price change. When demand is unit price elastic, total revenue does not change in response to a price change. When demand is price elastic, total revenue moves in the direction of a quantity change.
The absolute value of the price elasticity of demand is greater when substitutes are available, when the good is important in household budgets, and when buyers have more time to adjust to changes in the price of the good.
Try It!
You are now ready to play the part of the manager of the public transit system. Your finance officer has just advised you that the system faces a deficit. Your board does not want you to cut service, which means that you cannot cut costs. Your only hope is to increase revenue. Would a fare increase boost revenue?
You consult the economist on your staff who has researched studies on public transportation elasticities. She reports that the estimated price elasticity of demand for the first few months after a price change is about −0.3, but that after several years, it will be about −1.5.
Explain why the estimated values for price elasticity of demand differ.
Compute what will happen to ridership and revenue over the next few months if you decide to raise fares by 5%.
Compute what will happen to ridership and revenue over the next few years if you decide to raise fares by 5%.
What happens to total revenue now and after several years if you choose to raise fares?
Case in Point: Elasticity and Oil Prices
Recall from our discussion of the dynamics of the world oil prices in Chapter 2 that in early 2016, the world price of oil fell to almost $30 per barrel. In this situation, the OPEC countries, which were losing their revenue from oil sales, faced a tough choice: cut their oil production to prop up the price, as they’ve done in the past, or maintain their output and let the price continue to fall with the purpose of driving the producers of the more costly shale oil out of the market. As we could see, OPEC decided not just to go with the latter choice, but increase their oil production substantially. What we’ve learned above about the price elasticities of demand and supply will help us better understand why OPEC made that decision.
Let’s see what would have happened had OPEC decided to cut its oil production instead of increasing it. Suppose OPEC is making its decision when the price of oil has fallen to $30 and the world production of oil is 100 million barrels per day. OPEC accounts for about 40% of the world oil production, i.e. produces 40 million barrels per day, so its total revenue from oil is $30×40 million = $1,200 million or $1.2 billion. Will OPEC be able to increase its oil revenue if it reduces its production targetby, say, 5%, to boost the price?
Let’s first consider what will happen in a very short run, when other countries don’t have enough time to increase their oil production. Figure 3.19 illustrates this situation. The world market is initially in equilibrium at point E1, where the price of oil is $30 per barrel and 100 million barrels per day is supplied. In the very short run, the supply of oil is practically perfectlyinelastic; that is, the supply curve is vertical.
Figure 3.19
If OPEC cuts its oil production by 5% (i.e. by 40 million × 0.05 = 2 million barrels per day), the world production will decrease from 100 million to 98 million barrels per day. That is, the OPEC’s production cut will shift the supply curve from S1 to S2, so the market will move along the demand curve (D) from the initial equilibrium, E1, to a new equilibrium, E2, where the price is higher. We can predict what the new price of oil will be given that the short-run price elasticity of demand for oil is estimated at −0.1. Since the percentage change in quantity is −2% (using the conventional formula, %ΔQ = (98 – 100)/100 = −0.02 or −2%), we can write:
[latex]e_D = \frac{\% \ change \ in \ quantity \ demanded}{\% \ change \ in \ price}[/latex]
[latex]-0.1 = \frac{{2%} }{\% P}[/latex]
[latex]{-0.1} = \frac{-2%}{\frac{\% \ change \ in \ price}[/latex]
[latex]{-0.1} = \frac{{-2}% }{\% \ change \ in \ price}[/latex]
Solving this equation for %ΔP, we get:
-0.1 x %ΔP = -2%
[latex]{%\Delta P} = \frac{{-2}% }{\%{-0.1}} = 20%[/latex]
Answers to Try It! Problems
The absolute value of price elasticity of demand tends to be greater when more time is allowed for consumers to respond. Over time, riders of the commuter rail system can organize car pools, move, or otherwise adjust to the fare increase.
Using the formula for price elasticity of demand and plugging in values for the estimate of price elasticity (−0.5) and the percentage change in price (5%) and then rearranging terms, we can solve for the percentage change in quantity demanded as: eD = %Δ in Q/%Δ in P; −0.5 = %Δ in Q/5%; (−0.5)(5%) = %Δ in Q = −2.5%. Ridership falls by 2.5% in the first few months.
Using the formula for price elasticity of demand and plugging in values for the estimate of price elasticity over a few years (−1.5) and the percentage change in price (5%), we can solve for the percentage change in quantity demanded as eD = %Δ in Q/%Δ in P; −1.5 = %Δ in Q/5%; (−1.5)(5%) = %Δ in Q = −7.5%. Ridership falls by 7.5% over a few years.
Total revenue rises immediately after the fare increase, since demand over the immediate period is price inelastic. Total revenue falls after a few years, since demand changes and becomes price elastic.
1Notice that since the number of units sold of a good is the same as the number of units bought, the definition for total revenue could also be used to define total spending. Which term we use depends on the question at hand. If we are trying to determine what happens to revenues of sellers, then we are asking about total revenue. If we are trying to determine how much consumers spend, then we are asking about total spending.
2Division by zero results in an undefined solution. Saying that the price elasticity of demand is infinite requires that we say the denominator “approaches” zero.
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Microeconomics for Managers Copyright © 2020 by Margo Bergman is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.
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190388 | https://www.shuxuele.com/temperature-conversion.html | 温度换算
| | | | | | |
--- --- --- |
| 快速 摄氏 (°C) / 华氏 (°F) 转换： 或者用 互动温度计, | | | --- | | °F to °C | 减 32，然后乘以 5，然后除以 9 | | °C to °F | 乘以 9，然后除以 5，然后加 32 | （解释在下面） | |
典型温度
| °C | °F | 描述 |
---
| 180 | 356 | 烤箱中等温度 |
| 100 | 212 | 水沸 |
| 40 | 104 | 热水浴 |
| 37 | 98.6 | 体温 |
| 30 | 86 | 沙滩天气 |
| 21 | 70 | 室温 |
| 10 | 50 | 凉快天气 |
| 0 | 32 | 水结冰 |
| −18 | 0 | 非常寒冷天气 |
| −40 | −40 | 极端寒冷天气 （度数一样！） |
| （黑体数字是准确值） |
16 是大约 61 28 是大约 82
解释
有两个主要温标：
°F，华氏 （美国用）和
°C，摄氏 （公制，多数国家用）
它们都是用来度量温度，但用不同的数值：
水沸（正常气压下）是 摄氏 100°，但在华氏中是 212°
水结冰是 摄氏 0°，但在华氏中是 32°
像这样：
看这个图，留意：
两个温标从不同的数值开始（0 和 32），所以我们需要 加或减 32
两个温标增加的速度（率）不同（100 和 180），所以我们也需要乘
故此，
摄氏转换为华氏： 先乘以 180/100，然后加 32
华氏转换为摄氏： 先减 32，然后乘以 100/180
但 180/100 可以简化成 9/5， 而 100/180 可以简化成 5/9，所以最容易的方法是：
| | | |
---
| °C 转为 °F | | 乘以 9，除以 5，然后 加 32 |
| °F 转为 °C | | 减 32，乘以 5，然后 除以 9 |
写成公式：
摄氏 转为 华氏： (°C × 9/5) + 32 = °F 华氏 转为 摄氏： (°F − 32) x 5/9 = °C
例子：摄氏 26°（温暖天气）转换为 华氏
首先： 26° × 9/5 = 234/5 = 46.8 然后： 46.8 + 32 = 78.8° F
例子：华氏 98.6°（正常体温）转换为 摄氏
首先： 98.6° − 32 = 66.6 然后： 66.6 × 5/9 = 333/9 = 37° C
其他方法
用 1.8 来代替 9/5
9/5 等于 1.8，所以我们可以用这个方法：
摄氏 转为 华氏： °C × 1.8 + 32 = °F华氏 转为 摄氏： (°F − 32) / 1.8 = °C
要 "×1.8" 容易一些，我们可以 乘以 2 并且 减 10%，但这只适用于 °C 转为 °F:
摄氏 转为 华氏： (°C × 2) 减 10% + 32 = °F
例子：摄氏 20°（好天气） 为 华氏
20x2 = 40
减 10% 是 40−4 = 36
36+32 = 68° F
加 40，乘，减 40
因为两个温标的交叉点是在 −40° （−40° C 等于 −40° F），我们可以：
加 40,
乘以 5/9 （°F 转为 °C），或 9/5 （°C 转为 °F）
减 40
像这样：
摄氏 转为 华氏： 加 40，乘以 9/5，然后 减 40华氏 转为 摄氏： 加 40，乘以 5/9，然后 减 40
例子：摄氏 10° （凉快天气） 转换为 华氏
10+40 = 50
50×9/5 = 90
90−40 = 50° F
要记住： 9/5 用于转换 °C 为 °F，想： "F 是大于 C，所以 °F 比 °C 多"
快，但不太准确
摄氏 转为 华氏： 加倍，然后加 30华氏 转为 摄氏： 减 30，然后减半
例子 °C → °F：
0° C → 0+30 → 30° F （小 2°）
10° C → 20+30 → 50° F （准确！）
30° C → 60+30 → 90° F （大 4°）
180° C → 360+30 → 390° F （大 34°，不好）
例子 °F → °C：
40° F → 10/2 → 5° C （差不多）
80° F → 50/2 → 25° C （小大约 2°）
120° F → 90/2 → 45° C （小大约 4°）
450° F → 420/2 → 210° C （小大约 22°， 不好）
互动温度计 公制 - 英制 转换表 怎样从英制转换到公制 怎样从公制转换到英制 单位转换器 量度索引
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190389 | https://www.cpsc.gov/s3fs-public/CPSC-Bicycle-Regulations-Transcript-English-Final.pdf?T7f0wzoUq6Dnnza1SvieWskqGC7L2p3n | Published Time: Mon, 25 Aug 2025 20:50:01 GMT
Transcript of CPSC Podcast 2020 – CPSC Bicycle Regulations and U.S. and International Standards
Slide 3
Hi, my name is Sylvia Chen, and I want to welcome you to this podcast presentation today.
Slide 5
As CPSC’s Director of International Programs, Richard O’Brien stated, design of safe products at the outset is critical. CPSC is a United States federal government agency charged with protecting the public from unreasonable risks of injury or death associated with the use of consumer products under the agency’s jurisdiction. We have developed this podcast series not only to inform about regulations, standards, and other safety requirements, but also to emphasize the importance of designing products with safety considerations in mind, and to offer best practices for enhancing safety in a variety of common consumer products.
Slide 6
The series covers seven common consumer products and the requirements for keeping consumers safe, focusing on products affecting millions of consumers, such as electronics, apparel, bicycles, mattresses, infant and toddler products, carriages and strollers, and toys. In this podcast series, you can expect to learn about the key hazards and risks of the product, important design and manufacturing considerations, regulations and standards that CPSC uses to ensure product safety, best practices you can employ, and what resources are available to assist you in understanding and implementing the requirements.
The podcasts include English and Chinese slide decks, and Chinese narration to make this important safety information as accessible as possible. Additionally, CPSC has established a dedicated email box, where listeners can send in any questions at their convenience, in English or Chinese. Our staff will monitor and respond to your questions. Transcripts in English are available on this site.
Slide 7
The slides used in this podcast are not a comprehensive statement of legal requirements or policy, and thus, should not be relied upon for that purpose. You should consult official versions of U.S. statutes and regulations, as well as published CPSC guidance when making decisions that could affect the safety and compliance of products entering U.S. commerce. Note that references are provided at the end of the presentation.
Slide 8
And now, I would like to introduce our first presenter: Mr. Vince Amodeo .
Mr. Amodeo was a mechanical engineer with the U.S. Consumer Product Safety Commission at its National Product and Evaluation Center. His presentation is about CPSC bicycle regulations Transcript of CPSC Podcast 2020- CPSC Bicycle Regulations and U.S. and International Standards
Slide 3
Hi, my name is Sylvia Chen, and I want to welcome you to this podcast presentation today.
Slide 5
As CPSC's Director of International Programs, Richard O'Brien stated, design of safe products at the outset is critical. CPSC is a United States federal government agency charged with protecting the public from unreasonable risks of injury or death associated with the use of consumer products under the agency's jurisdiction. We have developed this podcast series not only to inform about regulations, standards, and other safety requirements, but also to emphasize the importance of designing products with safety considerations in mind, and to offer best practices for enhancing safety in a variety of common consumer products.
Slide 6
The series covers seven common consumer products and the requirements for keeping consumers safe, focusing on products affecting millions of consumers, such as electronics, apparel, bicycles, mattresses, infant and toddler products, carriages and strollers, and toys. In this podcast series, you can expect to learn about the key hazards and risks of the product, important design and manufacturing considerations, regulations and standards that CPSC uses to ensure product safety, best practices you can employ, and what resources are available to assist you in understanding and implementing the requirements.
The podcasts include English and Chinese slide decks, and Chinese narration to make this important safety information as accessible as possible. Additionally, CPSC has established adedicated email box, where listeners can send in any questions at their convenience, in English or Chinese. Our staff will monitor and respond to your questions. Transcripts in English are available on this site.
Slide 7
The slides used in this podcast are not a comprehensive statement of legal requirements or policy, and thus, should not be relied upon for that purpose. You should consult official versions of U.S. statutes and regulations, as well as published CPSC guidance when making decisions that could affect the safety and compliance of products entering U.S. commerce. Note that references are provided at the end of the presentation.
Slide 8
And now, I would like to introduce our first presenter: Mr. Vince Amodeo.
Mr. Amodeo was a mechanical engineer with the U.S. Consumer Product Safety Commission at its National Product and Evaluation Center. His presentation is about CPSC bicycle regulations and international bicycle standards. In the second half of the podcast, Ms. Caroleene Paul will give a presentation on bicycle instant data and studies. Ms. Paul is a mechanical engineer at the CPSC’s National Product Testing and Evaluation Center. Now, I will hand the mike over to Mr. Amodeo.
Hi, I'm Vince Amodeo. Today, I'm going to go over the CPSC bicycle regulation and compare it with U.S. and international bicycle standards.
Slide 9
Here is an overview of what we are discussing today:
• CPSC bicycle regulations
• Bicycle voluntary standards
• Bicycle usage conditions
• A comparison of test requirements and CPSC bicycle regulation compared to the voluntary standards
• And a few selected bicycle recalls.
Slide 10
The CPSC bicycle regulation is found at 16 C.F.R. Part 1512.
This is the mandatory standard for bicycles sold in the United States, which was originally codified in 1978.
The purpose of the CPSC bicycle regulation is to reduce the risk of injury from bicycles sold to consumers in the United States.
Slide 11
All bicycles that are sold in the United States must be certified to the CPSC bicycle regulation.
Certification of all bicycles designed or intended primarily for children 12 years of age or younger must be based on testing conducted by a third party conformity assessment body whose accreditation has been accepted by the CPSC.
Slide 12
Since 1978, when the CPSC bicycle regulation was first established, there have only been minor changes to the regulation.
The CPSC bicycle regulation does not address:
• Bicycle usage conditions. Bicycle usage conditions are usage based on intended terrain and type of riding.
• It also does not address technical improvements in bicycle design since 1978. This includes disc brakes, electric motor assist, and integrated shift/brake levers.
• It does not address the use of modern materials, such as composite fiber. and international bicycle standards. In the second half of the podcast, Ms. Caroleene Paul will give a presentation on bicycle instant data and studies. Ms. Paul is a mechanical engineer at the CPSC's National Product Testing and Evaluation Center. Now, I will hand the mike over to Mr. Amodeo.
Hi, I'm Vince Amodeo. Today, I'm going to go over the CPSC bicycle regulation and compare it with U.S. and international bicycle standards.
Slide 9
Here is an overview of what we are discussing today:
•CPSC bicycle regulations
•Bicycle voluntary standards
•Bicycle usage conditions
•Acomparison of test requirements and CPSC bicycle regulation compared to the voluntary standards
•And afew selected bicycle recalls.
Slide 10
The CPSC bicycle regulation is found at 16 C.F.R. Part 1512.
This is the mandatory standard for bicycles sold in the United States, which was originally codified in 1978.
The purpose of the CPSC bicycle regulation is to reduce the risk of injury from bicycles sold to consumers in the United States.
Slide 11
All bicycles that are sold in the United States must be certified to the CPSC bicycle regulation.
Certification of all bicycles designed or intended primarily for children 12 years of age or younger must be based on testing conducted by athird party conformity assessment body whose accreditation has been accepted by the CPSC.
Slide 12
Since 1978, when the CPSC bicycle regulation was first established, there have only been minor changes to the regulation.
The CPSC bicycle regulation does not address:
•Bicycle usage conditions. Bicycle usage conditions are usage based on intended terrain and type of riding.
•It also does not address technical improvements in bicycle design since 1978. This includes disc brakes, electric motor assist, and integrated shift/brake levers.
•It does not address the use of modern materials, such as composite fiber.
Slide 13
The CPSC bicycle regulation sets basic requirements for mechanical safety systems found on all bicycles, regardless of intended use. It includes specific requirements for sidewalk bicycles, which are bicycles with a maximum saddle height of less than 635 millimeters. It also includes requirements for two or three wheeled bicycles with electric assist motors that are less than 750 watts 1 horsepower with functioning pedals and a maximum speed of 20 miles per hour when operated solely on electric power.
Slide 14
Here's an overview of the requirements that are in the CPSC bicycle regulation Part 1512. It includes all the major systems for bicycles.
Slide 15
There are other organizations that also have bicycle standards. These are U.S. organizations and international organizations, such as ISO and CEN.
Slide 16
ASTM is the major U.S. standards organization body that has bicycle standards. ASTM F2043 13 is the standard classification for bicycle usage.
This defines usage conditions for bicycle design. It includes graphical icons for placement on bicycles, aftermarket components, and instructional material to provide retailers and consumers with an indication of the intended usage condition of bicycles or aftermarket components.
ASTM F2043 13 defines six usage conditions that I will go over.
There are other usage conditions, such as BMX, young adult, and electrically powered assisted cycles.
Slide 17
There are quite a few ASTM bicycle standards.
There are several general standards, such as the one I had just mentioned: the standard classification for bicycle usage.
There's also standards for manually operated front wheel retention systems for bicycles, standard bicycle serial numbers, and standard specification for bicycle grips.
There are several standards for bicycle forks. There is a test method and test specifications.
The test method goes over the procedure for the test. The test specifications go over the requirements for that specific use condition. Slide 13
The CPSC bicycle regulation sets basic requirements for mechanical safety systems found on all bicycles, regardless of intended use. It includes specific requirements for sidewalk bicycles, which are bicycles with amaximum saddle height of less than 635 millimeters. It also includes requirements for two or three wheeled bicycles with electric assist motors that are less than 750 watts 1horsepower with functioning pedals and amaximum speed of 20 miles per hour when operated solely on electric power.
Slide 14
Here's an overview of the requirements that are in the CPSC bicycle regulation Part 1512. It includes all the major systems for bicycles.
Slide 15
There are other organizations that also have bicycle standards. These are U.S. organizations and international organizations, such as ISO and CEN.
Slide 16
ASTM is the major U.S. standards organization body that has bicycle standards. ASTM F2043 13 is the standard classification for bicycle usage.
This defines usage conditions for bicycle design. It includes graphical icons for placement on bicycles, aftermarket components, and instructional material to provide retailers and consumers with an indication of the intended usage condition of bicycles or aftermarket components.
ASTM F2043 13 defines six usage conditions that Iwill go over.
There are other usage conditions, such as BMX, young adult, and electrically powered assisted cycles.
Slide 17
There are quite afew ASTM bicycle standards.
There are several general standards, such as the one Ihad just mentioned: the standard classification for bicycle usage.
There's also standards for manually operated front wheel retention systems for bicycles, standard bicycle serial numbers, and standard specification for bicycle grips.
There are several standards for bicycle forks. There is atest method and test specifications.
The test method goes over the procedure for the test. The test specifications go over the requirements for that specific use condition.
F2273 sets the test methods for bicycle forks. Similarly, with frames for bicycles, there's a test method, F2711 08.
Then there are several standard specifications for various conditions for bicycle frames.
Slide 18
The European bicycle community has two bicycle standards established under CEN. These are EN 15194 for electric powered assisted cycles, and EN 16054 for BMX bicycles.
Slide 19
ISO is the major international standards organization for bicycles sold in Europe and across the world.
ISO 8098 covers bicycles for young children, which have saddle heights less than 635 millimeters. This is similar to ASTM Condition 0.
ISO 4210 is a rather new and expanded bicycle standard, which goes into more depth than the old 4210.
Slide 20
I'm now going to cover what ASTM F2043 sets up for bicycle usage conditions.
Condition 0 is similar to sidewalk bicycles, which are intended for children age three and up and under 80 pounds. These bikes should be used with parental supervision.
This slide shows the icon that ASTM F2043 established for use on Condition 0 bicycles. Generally, this icon is shown on children’s bicycle user manuals. It can also be placed on the bicycle as a sticker so the consumer knows the intended usage.
Slide 21
Condition 1 is generally for road bikes, also considered racing bicycles. This is the icon that ASTM 2043 has created for these bikes. Condition 1 bicycles are generally used on paved surfaces and are intended to maintain contact with the ground.
Slide 22
Condition 2 is for hybrid or gravel type bicycles, where the bike may lose contact with the ground occasionally, but is not intended for very high jumps.
Slide 23
Condition 3 is generally for mountain bicycles that can do jumps and drops up to 24 inches.
Slide 24
Condition 4 is for downhill bicycles or bicycles that are intended to be used on rough trails with jumps of up to four feet. F2273 sets the test methods for bicycle forks. Similarly, with frames for bicycles, there's a test method, F2711 08.
Then there are several standard specifications for various conditions for bicycle frames.
Slide 18
The European bicycle community has two bicycle standards established under CEN. These are EN 15194 for electric powered assisted cycles, and EN 16054 for BMX bicycles.
Slide 19
ISO is the major international standards organization for bicycles sold in Europe and across the world.
ISO 8098 covers bicycles for young children, which have saddle heights less than 635 millimeters. This is similar to ASTM Condition 0.
ISO 4210 is arather new and expanded bicycle standard, which goes into more depth than the old 4210.
Slide 20
I'm now going to cover what ASTM F2043 sets up for bicycle usage conditions.
Condition Ois similar to sidewalk bicycles, which are intended for children age three and up and under 80 pounds. These bikes should be used with parental supervision.
This slide shows the icon that ASTM F2043 established for use on Condition Obicycles. Generally, this icon is shown on children's bicycle user manuals. It can also be placed on the bicycle as asticker so the consumer knows the intended usage.
Slide 21
Condition 1is generally for road bikes, also considered racing bicycles. This is the icon that ASTM 2043 has created for these bikes. Condition 1bicycles are generally used on paved surfaces and are intended to maintain contact with the ground.
Slide 22
Condition 2is for hybrid or gravel type bicycles, where the bike may lose contact with the ground occasionally, but is not intended for very high jumps.
Slide 23
Condition 3is generally for mountain bicycles that can do jumps and drops up to 24 inches.
Slide 24
Condition 4is for downhill bicycles or bicycles that are intended to be used on rough trails with jumps of up to four feet.
Slide 25
Condition 5 is for very extreme mountain bicycles with downhill grades of over 40 kilometers per hour and extreme jumping.
Slide 26
EN 16054 also has a standard for BMX type bicycles, but there are no icons associated with this standard.
Slide 27
And they also have 4210 for young adult bicycles.
Slide 28
EN 15194 has requirements for electric-powered assisted bicycles.
Slide 29
I'm now going to compare a few bicycle standard requirements.
Again, the CPSC bicycle regulation 16 CFR 1512 is the mandatory requirement for bicycles, sidewalk bicycles, and electric bicycles. However, it does not set requirements for various use conditions for bicycles. Again, these are the minimum test requirements that all bicycles sold in the United States must meet.
ASTM, CEN and ISO have voluntary bicycle standards. These are generally based on bicycle components. They also establish the requirements based on the bicycle’s intended use.
The requirements established in ASTM, the CEN, and ISO may be more appropriate than CPSC’s general test requirements to ensure that the bicycles and bicycle components meet the demands of the user.
Slide 30
I'm going to cover a couple examples of the differences between CPSC's regulation and the international or voluntary regulations. Here's an example of CPSC's test retirement for handlebar strength.
The CPSC test requirement is 2,000 newtons, or 450 pounds for bicycles, and 1,000 newtons and 225 pounds for sidewalk bicycles. In comparison, EN and ISO establish requirements based on usage conditions that may be different than the CPSC’s requirements.
Slide 31
For example, for Condition 1, CPSC’s requirement is 2,000 newtons and 450 pounds, regardless of use condition. ISO establishes different test requirements for different usage conditions, which can range from 500 newtons for Condition 0 to 1,600 newton or higher for other usage condition. Slide 25
Condition 5is for very extreme mountain bicycles with downhill grades of over 40 kilometers per hour and extreme jumping.
Slide 26
EN 16054 also has astandard for BMX type bicycles, but there are no icons associated with this standard.
Slide 27
And they also have 4210 for young adult bicycles.
Slide 28
EN 15194 has requirements for electric-powered assisted bicycles.
Slide 29
I'm now going to compare afew bicycle standard requirements.
Again, the CPSC bicycle regulation 16 CFR 1512 is the mandatory requirement for bicycles, sidewalk bicycles, and electric bicycles. However, it does not set requirements for various use conditions for bicycles. Again, these are the minimum test requirements that all bicycles sold in the United States must meet.
ASTM, CEN and ISO have voluntary bicycle standards. These are generally based on bicycle components. They also establish the requirements based on the bicycle's intended use.
The requirements established in ASTM, the CEN, and ISO may be more appropriate than CPSC's general test requirements to ensure that the bicycles and bicycle components meet the demands of the user.
Slide 30
I'm going to cover acouple examples of the differences between CPSC's regulation and the international or voluntary regulations. Here's an example of CPSC's test retirement for handlebar strength.
The CPSC test requirement is 2,000 newtons, or 450 pounds for bicycles, and 1,000 newtons and 225 pounds for sidewalk bicycles. In comparison, EN and ISO establish requirements based on usage conditions that may be different than the CPSC's requirements.
Slide 31
For example, for Condition 1, CPSC's requirement is 2,000 newtons and 450 pounds, regardless of use condition. ISO establishes different test requirements for different usage conditions, which can range from 500 newtons for Condition Oto 1,600 newton or higher for other usage condition.
Slide 32
Example 2 is a test requirement for fork bending fatigue. We can see that CPSC does not have a requirement for fork bending fatigue, whereas international voluntary standards do have a fork bending fatigue test.
Slide 33
This table shows that CPSC does not have a requirement for fork bending fatigue, while ASTM, EN, and ISO have various test requirements for fork bending fatigue.
So, as you can see, if you're designing a fork, it may be appropriate to use an ASTM or ISO test requirement to make sure that your bicycle meets the intended usage condition.
Slide 34
Example 3 is about fork and frame assembly test requirements.
In the CPSC fork and frame assembly test, the fork and frame assembly must be tested for strength by application of a load of 890 N (200 lbf) or at least 39.5 J (350 in-lb) of energy, whichever results in the greater force, in accordance with the frame test.
The fork and frame assembly static load test is only done for CPSC requirements.
ASTM, CEN and ISO have fork and frame assembly horizontal loading fatigue test and fork and frame assembly falling mass impact test requirements that CPSC's test does not require.
Slide 35
This table shows the fork and frame assembly horizontal fatigue test requirements.
Again, you can see here that CPSC does not have this test. However, if you are designing a bicycle for sale in the U.S. or anywhere in the world, you should consider using ASTM, EN, or ISO test requirements to design your bicycle forks and frames.
Slide 36-37
The following tables have a list of bicycle recalls conducted by the CPSC.
Slide 38
Just to wrap up what we've been talking about, we've covered the CPSC bicycle test requirements as well as international and U.S. voluntary standard requirements.
As you can see, there are several differences in what the CPSC requires for sale and what voluntary standards and international standards require for testing.
In many cases, those differences could mean that a bicycle may fail under its intended use condition if it is not designed appropriately. Therefore, you should seriously look at whether or not designing your bike to just meet CPSC requirements will be enough. Slide 32
Example 2is atest requirement for fork bending fatigue. We can see that CPSC does not have arequirement for fork bending fatigue, whereas international vol untary standards do have afork bending fatigue test.
Slide 33
This table shows that CPSC does not have arequirement for fork bending fatigue, while ASTM, EN, and ISO have various test requirements for fork bending fatigue.
So, as you can see, if you're designing afork, it may be appropriate to use an ASTM or ISO test requirement to make sure that your bicycle meets the intended usage condition.
Slide 34
Example 3is about fork and frame assembly test requirements.
In the CPSC fork and frame assembly test, the fork and frame assembly must be tested for strength by application of aload of 890 N(200 lbf) or at least 39.5 J(350 in-lb) of energy, whichever results in the greater force, in accordance with the frame test.
The fork and frame assembly static load test is only done for CPSC requirements.
ASTM, CEN and ISO have fork and frame assembly horizontal loading fatigue test and fork and frame assembly falling mass impact test requirements that CPSC's test does not require.
Slide 35
This table shows the fork and frame assembly horizontal fatigue test requirements.
Again, you can see here that CPSC does not have this test. However, if you are designing abicycle for sale in the U.S. or anywhere in the world, you should consider using ASTM, EN, or ISO test requirements to design your bicycle forks and frames.
Slide 36-37
The following tables have alist of bicycle recalls conducted by the CPSC.
Slide 38
Just to wrap up what we've been talking about, we've covered the CPSC bicycle test requirements as well as international and U.S. voluntary standard requirements.
As you can see, there are several differences in what the CPSC requires for sale and what vol untary standards and international standards require for testing.
In many cases, those differences could mean that abicycle may fail under its intended use condition if it is not designed appropriately. Therefore, you should seriously look at whether or not designing your bike to just meet CPSC requirements will be enough.
Perhaps you should consider using an ASTM or ISO test requirement to design your bicycle to make sure that it meets the intended usage condition.
Slide 39
Thank you. And now I'd like to turn the mike over to Caroleen Paul who will give a talk on CPSC bicycle instant data and case studies.
Thank you. As Vince mentioned, my name is Caroleen Paul, and I am a mechanical engineer here at the CPSC. This presentation will talk about how the CPSC gathers data on bicycle accidents and gives examples of a few case studies of actual incidents.
Slide 40
CPSC gathers information on bicycle accidents through two main methods. The first is the National Electronic Injury Surveillance System, also known as NEISS.
The second method is in-depth investigations conducted by our CPSC investigators.
NEISS collects injury data from emergency departments across the United States. About 100 hospitals participate, and they code actual incidents that come into their emergency departments with specific product codes. For example, the product code for bicycles and accessories is 5040 and the product code for mountain bicycles is 5033. These hospitals are specifically chosen to be a national probability sample so when this data comes to us, we can then use it to calculate national estimates for how many incidents occurred across the United States.
We also get information on bicycle accidents that are reported to us. Then we send our investigators to ask the victims questions. The investigators then write up detailed reports called “in-depth investigation incident reports” or “IDIs”. IDIs provide very good specific information on incidents, but they cannot be used to make national estimates.
Between these two methods, we get an idea of what's happening nationally, and we get specific information on what actually occurred.
We also have a report from our epidemiologic department on bicycle injuries seen in hospital emergency departments in 2013. Now, this report goes over how many injuries there were. For instance, in 2013, there were 531,000 injuries associated with bicycles and accessories seen in emergency departments. Over 90% of these resulted from the person riding the bicycle. And more than half are described as falls from the bike.
In terms of in-depth investigations, they provide detailed information on incidents. There were 302 IDIs from January 2007 to January 2017.
Slide 41
There are many components to a bicycle. Any one of these components can fail and cause a bicycle incident. Perhaps you should consider using an ASTM or ISO test requirement to design your bicycle to make sure that it meets the intended usage condition.
Slide 39
Thank you. And now I'd like to turn the mike over to Caroleen Paul who will give atalk on CPSC bicycle instant data and case studies.
Thank you. As Vince mentioned, my name is Caroleen Paul, and Iam amechanical engineer here at the CPSC. This presentation will talk about how the CPSC gathers data on bicycle accidents and gives examples of afew case studies of actual incidents.
Slide 40
CPSC gathers information on bicycle accidents through two main methods. The first is the National Electronic Injury Surveillance System, also known as NEISS.
The second method is in-depth investigations conducted by our CPSC investigators.
NEISS collects injury data from emergency departments across the United States. About 100 hospitals participate, and they code actual incidents that come into their emergency departments with specific product codes. For example, the product code for bicycles and accessories is 5040 and the product code for mountain bicycles is 5033. These hospitals are specifically chosen to be anational probability sample so when this data comes to us, we can then use it to calculate national estimates for how many incidents occurred across the United States.
We also get information on bicycle accidents that are reported to us. Then we send our investigators to ask the victims questions. The investigators then write up detailed reports called "in-depth investigation incident reports" or "IDIs". IDIs provide very good specific information on incidents, but they cannot be used to make national estimates.
Between these two methods, we get an idea of what's happening nationally, and we get specific information on what actually occurred.
We also have areport from our epidemiologic department on bicycle injuries seen in hospital emergency departments in 2013. Now, this report goes over how many injuries there were. For instance, in 2013, there were 531,000 injuries associated with bicycles and accessories seen in emergency departments. Over 90% of these resulted from the person riding the bicycle. And more than half are described as falls from the bike.
In terms of in-depth investigations, they provide detailed information on incidents. There were 302 IDls from January 2007 to January 2017.
Slide 41
There are many components to abicycle. Any one of these components can fail and cause abicycle incident.
Slide 42
For example, the top component failures based on the IDIs that we've investigated include pedals, wheels, bicycle frames, forks, brakes, stems, crank arms, and handlebars.
Slide 43
As I mentioned before, IDIs provide specific information on bicycle incidents. Many of the same descriptions come up in the investigations with the likely failure type of the bicycle.
These type descriptions include words like “something detached or loosened”. The likely failure type associated with that description is usually something related to the assembly of the bicycle, or the maintenance of the bicycle.
Others include descriptions such as “something cracked”, “came apart”, “deformed”, or “fractured.”
There is also the “structural” failure type. This type is very important to us, because a structural failure is usually related to the design of the bicycle, which is usually something that can be addressed through voluntary standards.
Finally, the “overall malfunction” description comes up quite often. That is an unknown failure type.
I am now going to go over some actual IDIs that were conducted in the past.
Slide 44
Let’s look at IDI 050211CCC1472. This IDI number allows us to reference the actual investigation.
In this case, a 25 year old female flipped over her bicycle’s handlebars because her left pedal detached. This was serious enough for her to be taken to the hospital via ambulance.
Another IDI, 090521CNE4429, involved a 19 year old male. The right clipless pedal of his racing bike fractured during a competition. This caused the victim to fall to the pavement with lacerations and contusions to his right knee.
Slide 45
Here, we have wheel related incidents that are typical IDIs.
Both cases involved the front wheel separating from the suspension fork.
The first case was a 16 year old male who was just riding his bike. I believe this was the first time he used the bike. For unknown reasons, the wheel detached from the fork, and the victim flew over the handlebars. Slide 42
For example, the top component failures based on the IDIs that we've investigated include pedals, wheels, bicycle frames, forks, brakes, stems, crank arms, and handlebars.
Slide 43
As Imentioned before, IDIs provide specific information on bicycle incidents. Many of the same descriptions come up in the investigations with the likely failure type of the bicycle.
These type descriptions include words like "something detached or loosened". The likely failure type associated with that description is usually something related to the assembly of the bicycle, or the maintenance of the bicycle.
Others include descriptions such as "something cracked", "came apart", "deformed", or "fractured."
There is also the "structural" failure type. This type is very important to us, because astructural failure is usually related to the design of the bicycle, which is usually something that can be addressed through voluntary standards.
Finally, the "overall malfunction" description comes up quite often. That is an unknown failure type.
Iam now going to go over some actual IDIs that were conducted in the past.
Slide 44
Let's look at IOI 050211CCC1472. This IOI number allows us to reference the actual investigation.
In this case, a25 year old female flipped over her bicycle's handlebars because her left pedal detached. This was serious enough for her to be taken to the hospital via ambulance.
Another
IDI, 090521CNE4429, involved a 19 year old male. The right clipless pedal of his racing bike fractured during a competition. This caused the victim to fall to the pavement with lacerations and contusions to his right knee.
Slide 45
Here, we have wheel related incidents that are typical IDIs.
Both cases involved the front wheel separating from the suspension fork.
The first case was a16 year old male who was just riding his bike. Ibelieve this was the first time he used the bike. For unknown reasons, the wheel detached from the fork, and the victim flew over the handlebars.
In the other case, the same thing occur with an 11 year old male (except he had been riding his bike for a while). He was riding over a speed bump and the front wheel detached and wedged in the bike. This caused the bike to stop suddenly and the victim flipped over the handlebars.
Slide 46
Next, we have some frame related incidents.
In the first case, the bike’s aluminum frame broke in half while an 8 year old male was riding it. He was thrown over the handlebar and onto a gravel driveway.
In this other incident, a 20 year old male was on a bicycle and the frame at the head tube fractured in half. Again, the person was thrown over the handlebar and received multiple lacerations.
In the ID that ends with “2105” the bicycle was recalled under the number 00 030.
Sometimes when you have a structural problem, it ends up being part of the design, which can lead to recalls.
Slide 47
These next IDIs involve fork related incidents.
In the first IDI, the fork came apart. As you can see in the picture, it came apart pretty catastrophically. A 20 year old male was riding the bicycle at the time. He was thrown over the handlebars, hit the pavement, and received some serious injuries.
In the second IDI, the fork on the bike broke. The accident involved a 14 year old male whose head and face struck the pavement, which resulted in a broken jaw.
Slide 48
These next IDIs relate to the brakes on the bicycle.
On the first one, the front wheel locked up when the brakes were applied, which cases the bike to stop suddenly. A 17 year old female was riding a bike at the time. In this case, she was thrown off the bike onto a concrete picnic table, fractured both her arms, and was hospitalized for three days.
In the second IDI, a 25 year old male was riding a bicycle and lost control while attempting to apply the brakes. He ended up falling onto the pavement with head lacerations and road rash.
Slide 49
These next IDIs are related to the stem of the bike. This is where the handlebars attach.
In the first IDI, a 13 year old male was riding the bicycle and the stem fractured. As you can see in the picture, it’s a material failure. This victim fell to the ground and had minor abrasions to his hands and arms. In the other case, the same thing occur with an 11 year old male {except he had been riding his bike for a while). He was riding over a speed bump and the front wheel detached and wedged in the bike. This caused the bike to stop suddenly and the victim flipped over the handlebars.
Slide 46
Next, we have some frame related incidents.
In the first case, the bike's aluminum frame broke in half while an 8year old male was riding it. He was thrown over the handlebar and onto agravel driveway.
In this other incident, a20 year old male was on abicycle and the frame at the head tube fractured in half. Again, the person was thrown over the handlebar and received multiple lacerations.
In the ID that ends with "2105" the bicycle was recalled under the number 00 030.
Sometimes when you have astructural problem, it ends up being part of the design, which can lead to recalls.
Slide 47
These next IDls involve fork related incidents.
In the first IOI, the fork came apart. As you can see in the picture, it came apart pretty catastrophically. A20 year old male was riding the bicycle at the time. He was thrown over the handlebars, hit the pavement, and received some serious injuries.
In the second IOI, the fork on the bike broke. The accident involved a14 year old male whose head and face struck the pavement, which resulted in abroken jaw.
Slide 48
These next IDls relate to the brakes on the bicycle.
On the first one, the front wheel locked up when the brakes were applied, which cases the bike to stop suddenly. A17 year old female was riding abike at the time. In this case, she was thrown off the bike onto aconcrete picnic table, fractured both her arms, and was hospitalized for three days.
In the second IOI, a25 year old male was riding abicycle and lost control while attempting to apply the brakes. He ended up falling onto the pavement with head lacerations and road rash.
Slide 49
These next IDis are related to the stem of the bike. This is where the handlebars attach.
In the first IOI, a13 year old male was riding the bicycle and the stem fractured. As you can see in the picture, it's amaterial failure. This victim fell to the ground and had minor abrasions to his hands and arms.
In the second IDI the victim, a 52 year old male, turned the handlebars, and his bike didn't respond. This caused a crash and the victim severely fractured his leg.
In the last one, an 18 year old male was riding a bicycle when the stem cracked. He fell to the pavement and got a wrist and elbow fracture.
Slide 50
Next, we have a crank arm. These are the parts that the pedals attach to on your bicycle. This is where you transfer torque and power to the bicycle.
In this first IDI, a 15 year old male was riding a bicycle, and the crank arm fractured. Again, this is some type of material failure. The accident caused the victim to fall and receive a head injury and a broken arm.
The other IDI was a left crank arm fracture. Again, some type of material failure. In this case, the 35 year old victim fell to the pavement and had abrasions and a laceration.
Slide 51
Now, we have some handlebar related incidents.
The first one was a 57 year old female who was riding and her handlebar just suddenly collapsed, making her lose control of the bike. This caused her to fall, and she needed surgery to treat a broken ankle.
In another incident, an 11 year old male was riding a bicycle, when, again, his handlebars became loose and he lost control of the bicycle, fell over, and had injuries to the face.
Slide 52
So, those were just a few examples of the hundreds of IDIs we have.
Next we are going to look at specific case studies of failures and what the manufacturers did to correct them.
Slide 53
In our first case, we have a fork failure. This particular product was a road bike with a carbon fork.
This material isn’t covered in the CPSC regulations. This is all through voluntary standards. Innovations in bicycle design are areas where a lot of problems can arise.
In this case, a road bike with a carbon fork had an issue where the disc brake mount was fracturing.
Slide 54
The manufacturer redesigned the aluminum disc brake, as well as the carbon fibers layups in the fork where the dismount is attached, to rectify the bike’s failure. In the second IOI the victim, a 52 year old male, turned the handlebars, and his bike didn't respond. This caused a crash and the victim severely fractured his leg.
In the last one, an 18 year old male was riding abicycle when the stem cracked. He fell to the pavement and got awrist and elbow fracture.
Slide 50
Next, we have acrank arm. These are the parts that the pedals attach to on your bicycle. This is where you transfer torque and power to the bicycle.
In this first IDI, a15 year old male was riding abicycle, and the crank arm fractured. Again, this is some type of material failure. The accident caused the victim to fall and receive ahead injury and abroken arm.
The other IOI was aleft crank arm fracture. Again, some type of material failure. In this case, the 35 year old victim fell to the pavement and had abrasions and alaceration.
Slide 51
Now, we have some handlebar related incidents.
The first one was a57 year old female who was riding and her handlebar just suddenly collapsed, making her lose control of the bike. This caused her to fall, and she needed surgery to treat abroken ankle.
In another incident, an 11 year old male was riding abicycle, when, again, his handlebars became loose and he lost control of the bicycle, fell over, and had injuries to the face.
Slide 52
So, those were just afew examples of the hundreds of IDIs we have.
Next we are going to look at specific case studies of failures and what the manufacturers did to correct them.
Slide 53
In our first case, we have afork failure. This particular product was aroad bike with acarbon fork.
This material isn't covered in the CPSC regulations. This is all through voluntary standards. Innovations in bicycle design are areas where alot of problems can arise.
In this case, aroad bike with acarbon fork had an issue where the disc brake mount was fracturing.
Slide 54
The manufacturer redesigned the aluminum disc brake, as well as the carbon fibers layups in the fork where the dismount is attached, to rectify the bike's failure.
All of this was done to reduce the stress to that area. To validate the design, the manufacturer had the redesign fork tested to the fork test requirements in EN 14764.
EN 14764 is for city and trekking bikes.
This is a case where the voluntary standards can really help the manufacturer make sure that they design a safe bicycle.
Slide 55
Our second case is a frame failure with a folding bicycle. As you can see, the aluminum frame could fracture at the frame hinge.
And as you can see in these photos, this was a catastrophic failure in the material.
Slide 56
To rectify the material failure, the manufacturer modified the welding process at the subject area. The manufacturer increased the thickness and increased the strength where the fractures occurred.
In addition, the manufacturer improved the quality control process to ensure that the frame met specifications before the bike was assembled and put together.
And finally, to validate the design change and address the problem, the manufacturer tested the frame to the frame fatigue and impact test requirements in EN 14764.
Slide 57
And lastly, this is a seat post failure.
This particular product is a mountain bicycle that is called a 29er because the wheels are 29 inches in diameter. The issue in this case was a carbon seat post that could fracture.
Slide 58
The manufacturer rectified the material failure by modifying the carbon fiber layering to cover the transition area that was failing. Essentially, this strengthened that area.
To validate the design change and address the problem, the redesign seat post was tested to the seat post fatigue test requirements in EN 14766.
In addition, the manufacturer conducted an in-house static load testing to ensure the strength of the product.
In this case, the manufacturer recognized the failure. Then, not only did the manufacturer find the requirement in the voluntary standard that actually addressed this problem, they did their own in-house testing.
These are all good practices to make sure that you manufacture bikes to be as safe as possible. All of this was done to reduce the stress to that area. To validate the design, the manufacturer had the redesign fork tested to the fork test requirements in EN 14764.
EN 14764 is for city and trekking bikes.
This is acase where the voluntary standards can really help the manufacturer make sure that they design asafe bicycle.
Slide 55
Our second case is aframe failure with afolding bicycle. As you can see, the aluminum frame could fracture at the frame hinge.
And as you can see in these photos, this was acatastrophic failure in the material.
Slide 56
To rectify the material failure, the manufacturer modified the welding process at the subject area. The manufacturer increased the thickness and increased the strength where the fractures occurred.
In addition, the manufacturer improved the quality control process to ensure that the frame met specifications before the bike was assembled and put together.
And finally, to validate the design change and address the problem, the manufacturer tested the frame to the frame fatigue and impact test requirements in EN 14764.
Slide 57
And lastly, this is aseat post failure.
This particular product is amountain bicycle that is called a29er because the wheels are 29 inches in diameter. The issue in this case was acarbon seat post that could fracture.
Slide 58
The manufacturer rectified the material failure by modifying the carbon fiber layering to cover the transition area that was failing. Essentially, this strengthened that area.
To validate the design change and address the problem, the redesign seat post was tested to the seat post fatigue test requirements in EN 14766.
In addition, the manufacturer conducted an in-house static load testing to ensure the strength of the product.
In this case, the manufacturer recognized the failure. Then, not only did the manufacturer find the requirement in the voluntary standard that actually addressed this problem, they did their own in-house testing.
These are all good practices to make sure that you manufacture bikes to be as safe as possible.
In conclusion, these are just some examples of the many components that can and have failed on bicycles.
We at the CPSC encourage all manufacturers to learn from the recalls and past the corrective actions. We also encourage you to look at all the different standards out there, mandatory and voluntary standards, and apply them in best practices to prevent future incidents. Thank you.
Slide 59
Thank you, and we hope you enjoyed this podcast. If you have any questions on the presentation, please do not hesitate to submit your questions in English or Chinese to the mailbox mentioned earlier: CPSCinChina@cpsc.gov . This mailbox is routinely monitored.
Slides 60-64
We also wish to remind viewers that CPSC has many technical documents and resources available in Chinese. The conclusion of this presentation provides many links to resources viewers may find useful.
Slide 65-66
We encourage viewers to check out CPSC’s Regulatory Robot, available in English, Chinese, and several other languages. The Regulatory Robot is an automated tool that can help identify safety requirements for many different types of products. Many companies have found this tool to be extremely helpful.
Slide 67
Finally, please see the following links for resources specific to bicycles and information covered in today’s podcast. Thank you for downloading this presentation. In conclusion, these are just some examples of the many components that can and have failed on bicycles.
We at the CPSC encourage all manufacturers to learn from the recalls and past the corrective actions. We also encourage you to look at all the different standards out there, mandatory and voluntary standards, and apply them in best practices to prevent future incidents. Thank you.
Slide 59
Thank you, and we hope you enjoyed this podcast. If you have any questions on the presentation, please do not hesitate to submit your questions in English or Chinese to the mailbox mentioned earlier: CPSCinChina@cpsc.gov. This mailbox is routinely monitored.
Slides 60-64
We also wish to remind viewers that CPSC has many technical documents and resources available in Chinese. The conclusion of this presentation provides many links to resources viewers may find useful.
Slide 65-66
We encourage viewers to check out CPSC's Regulatory Robot, available in English, Chinese, and several other languages. The Regulatory Robot is an automated tool that can help identify safety requirements for many different types of products. Many companies have found this tool to be extremely helpful.
Slide 67 |
190390 | https://courses.lumenlearning.com/waymakermath4libarts/chapter/euler-circuits/ | Euler Circuits
Learning Outcomes
Determine whether a graph has an Euler path and/ or circuit
Use Fleury’s algorithm to find an Euler circuit
Add edges to a graph to create an Euler circuit if one doesn’t exist
Identify whether a graph has a Hamiltonian circuit or path
Find the optimal Hamiltonian circuit for a graph using the brute force algorithm, the nearest neighbor algorithm, and the sorted edges algorithm
Identify a connected graph that is a spanning tree
Use Kruskal’s algorithm to form a spanning tree, and a minimum cost spanning tree
In the first section, we created a graph of the Königsberg bridges and asked whether it was possible to walk across every bridge once. Because Euler first studied this question, these types of paths are named after him.
Euler Path
An Euler path is a path that uses every edge in a graph with no repeats. Being a path, it does not have to return to the starting vertex.
Example
In the graph shown below, there are several Euler paths. One such path is CABDCB. The path is shown in arrows to the right, with the order of edges numbered.
Euler Circuit
An Euler circuit is a circuit that uses every edge in a graph with no repeats. Being a circuit, it must start and end at the same vertex.
Example
The graph below has several possible Euler circuits. Here’s a couple, starting and ending at vertex A: ADEACEFCBA and AECABCFEDA. The second is shown in arrows.
Look back at the example used for Euler paths—does that graph have an Euler circuit? A few tries will tell you no; that graph does not have an Euler circuit. When we were working with shortest paths, we were interested in the optimal path. With Euler paths and circuits, we’re primarily interested in whether an Euler path or circuit exists.
Why do we care if an Euler circuit exists? Think back to our housing development lawn inspector from the beginning of the chapter. The lawn inspector is interested in walking as little as possible. The ideal situation would be a circuit that covers every street with no repeats. That’s an Euler circuit! Luckily, Euler solved the question of whether or not an Euler path or circuit will exist.
Euler’s Path and Circuit Theorems
A graph will contain an Euler path if it contains at most two vertices of odd degree.
A graph will contain an Euler circuit if all vertices have even degree
Example
In the graph below, vertices A and C have degree 4, since there are 4 edges leading into each vertex. B is degree 2, D is degree 3, and E is degree 1. This graph contains two vertices with odd degree (D and E) and three vertices with even degree (A, B, and C), so Euler’s theorems tell us this graph has an Euler path, but not an Euler circuit.
Example
Is there an Euler circuit on the housing development lawn inspector graph we created earlier in the chapter? All the highlighted vertices have odd degree. Since there are more than two vertices with odd degree, there are no Euler paths or Euler circuits on this graph. Unfortunately our lawn inspector will need to do some backtracking.
Example
When it snows in the same housing development, the snowplow has to plow both sides of every street. For simplicity, we’ll assume the plow is out early enough that it can ignore traffic laws and drive down either side of the street in either direction. This can be visualized in the graph by drawing two edges for each street, representing the two sides of the street.
Notice that every vertex in this graph has even degree, so this graph does have an Euler circuit.
The following video gives more examples of how to determine an Euler path, and an Euler Circuit for a graph.
Fleury’s Algorithm
Now we know how to determine if a graph has an Euler circuit, but if it does, how do we find one? While it usually is possible to find an Euler circuit just by pulling out your pencil and trying to find one, the more formal method is Fleury’s algorithm.
Fleury’s Algorithm
Start at any vertex if finding an Euler circuit. If finding an Euler path, start at one of the two vertices with odd degree.
Choose any edge leaving your current vertex, provided deleting that edge will not separate the graph into two disconnected sets of edges.
Add that edge to your circuit, and delete it from the graph.
Continue until you’re done.
Example
Find an Euler Circuit on this graph using Fleury’s algorithm, starting at vertex A.
Try It
Does the graph below have an Euler Circuit? If so, find one.
The following video presents more examples of using Fleury’s algorithm to find an Euler Circuit.
Eulerization and the Chinese Postman Problem
Not every graph has an Euler path or circuit, yet our lawn inspector still needs to do her inspections. Her goal is to minimize the amount of walking she has to do. In order to do that, she will have to duplicate some edges in the graph until an Euler circuit exists.
Eulerization
Eulerization is the process of adding edges to a graph to create an Euler circuit on a graph. To eulerize a graph, edges are duplicated to connect pairs of vertices with odd degree. Connecting two odd degree vertices increases the degree of each, giving them both even degree. When two odd degree vertices are not directly connected, we can duplicate all edges in a path connecting the two.
Note that we can only duplicate edges, not create edges where there wasn’t one before. Duplicating edges would mean walking or driving down a road twice, while creating an edge where there wasn’t one before is akin to installing a new road!
Example
For the rectangular graph shown, three possible eulerizations are shown. Notice in each of these cases the vertices that started with odd degrees have even degrees after eulerization, allowing for an Euler circuit.
In the example above, you’ll notice that the last eulerization required duplicating seven edges, while the first two only required duplicating five edges. If we were eulerizing the graph to find a walking path, we would want the eulerization with minimal duplications. If the edges had weights representing distances or costs, then we would want to select the eulerization with the minimal total added weight.
Try It now
Eulerize the graph shown, then find an Euler circuit on the eulerized graph.
Example
Looking again at the graph for our lawn inspector from Examples 1 and 8, the vertices with odd degree are shown highlighted. With eight vertices, we will always have to duplicate at least four edges. In this case, we need to duplicate five edges since two odd degree vertices are not directly connected. Without weights we can’t be certain this is the eulerization that minimizes walking distance, but it looks pretty good.
The problem of finding the optimal eulerization is called the Chinese Postman Problem, a name given by an American in honor of the Chinese mathematician Mei-Ko Kwan who first studied the problem in 1962 while trying to find optimal delivery routes for postal carriers. This problem is important in determining efficient routes for garbage trucks, school buses, parking meter checkers, street sweepers, and more.
Unfortunately, algorithms to solve this problem are fairly complex. Some simpler cases are considered in the exercises
The following video shows another view of finding an Eulerization of the lawn inspector problem.
Candela Citations
CC licensed content, Original
Revision and Adaptaion. Provided by: Lumen Learning. License: CC BY: Attribution
CC licensed content, Shared previously
Graph Theory: Euler Paths and Euler Circuits . Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Math in Society. Authored by: Lippman, David. License: CC BY: Attribution
Graph Theory: Fleury's Algorthim . Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Hamiltonian circuits. Authored by: OCLPhase2. Located at: License: CC BY: Attribution
Licenses and Attributions
CC licensed content, Original
Revision and Adaptaion. Provided by: Lumen Learning. License: CC BY: Attribution
CC licensed content, Shared previously
Graph Theory: Euler Paths and Euler Circuits . Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Math in Society. Authored by: Lippman, David. License: CC BY: Attribution
Graph Theory: Fleury's Algorthim . Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Hamiltonian circuits. Authored by: OCLPhase2. Located at: License: CC BY: Attribution |
190391 | https://dspace.ups.edu.ec/bitstream/123456789/17214/1/UPS-CT008219.pdf | UNIVERSIDAD POLITÉCNICA SALESIANA SEDE CUENCA CARRERA DE INGENIERÍA ELÉCTRICA Trabajo de titulación previo a la obtención del título de INGENIERO ELÉCTRICO Proyecto Técnico con Enfoque Investigativo: “Estimación de Eficiencia en Máquinas de Inducción en Ambientes Industriales ” Autores: Fabián Ricardo Cevallos Aguirre Franco Oswaldo Pinos Vergara Tutor: Ing. José Manuel Aller Castro, Dr.
Cuenca - Ecuador 2019 CESIÓN DE DERECHOS DE AUTOR Nosotros Fabián Ricardo Cevallos Aguirre con documento de identificación N° 0105948913, y Franco Oswaldo Pinos Vergara con documento de identificación N° 0106035645, manifestamos de nuestra voluntad y cedemos a la Universidad Politécnica Salesia-na la titularidad sobre los derechos patrimoniales en virtud de que somos autores del trabajo de titulación: ¨Estimación de Eficiencia en Máquinas de Inducción en Ambientes Industriales¨, mismo que ha sido desarrollado para optar por el títu-lo de: "Ingeniero Eléctrico", en la Universidad Politécnica Salesiana, quedando la Universidad facultada para ejercer plenamente los derechos cedidos anteriormente.
En aplicación a lo determinado en la Ley de Propiedad Intelectual, en nuestra con-dición de autores nos reservamos los derechos morales de la obra antes citada. En concordancia, suscribimos este documento en el momento que hacemos entrega del trabajo final en formato impreso y digital a la Biblioteca de la Universidad Politéc-nica Salesiana.
Cuenca, marzo de 2019 Fabián Ricardo Cevallos Aguirre..
Franco Oswaldo Pinos Vergara.
CI: 0105948913 CI:0106035645 CERTIFICACIÓN Yo declaro que bajo mi tutoría fue desarrollado el trabajo de titulación: ¨Estimación de Eficiencia en Máquinas de Inducción en Ambientes Industriales¨ realizado por: Fabián Ricardo Cevallos Aguirre y Franco Oswaldo Pinos Vergara, obteniendo el Proyecto Técnico con Enfoque Investigativo, que cumple con todos los requisitos estipulados por la Universidad Politécnica Salesiana.
Cuenca, marzo de 2019 Ing. José Manuel Aller Castro PhD.
CI: 0151561800 DECLARATORIA DE RESPONSABILIDAD Nosotros Fabián Ricardo Cevallos Aguirre con N° 0105948913 y Franco Oswaldo Pinos Vergara con N° 0106035645, autores del trabajo de titulación ¨Estimación de Eficiencia en Máquinas de Inducción en Ambientes Industriales¨ certificamos que el total del contenido del Proyecto Técnico con Enfoque Investigativo es de nuestra exclusiva responsabilidad y autoría.
Cuenca, marzo de 2019 Fabián Ricardo Cevallos Aguirre.
Franco Oswaldo Pinos Vergara.
CI: 0105948913 CI:0106035645 Cuenca, marzo de 2019 DEDICATORIA Dedicado a Dios, a mis padres, mi hermana, y todas las personas que me acompañaron en este trayecto. A todas aquellas personas que me brindaron su valioso tiempo siendo siempre mi apoyo incondicional para cumplir con una de las metas académicas que marcarán mi futuro personal y profesional. A mis compañeros y amigos que supieron acompañarme y luchar junto a mi, para cumplir el objetivo común de obtener nuestro título para construir nuestra vida profesional.
Fabián Cuenca, marzo de 2019 DEDICATORIA Dedicado a Dios y a cada una de las personas que formaron parte de mi carrera universitaria, a cada persona que me brindo un apoyo y fuerza para concluir con esta etapa importante de mi vida, mis padres, hermanos y enamorada, que fueron testigos de este camino, que en un futuro será la semilla para brindar el conocimiento a futuras generaciones, con el objetivo de fortalecer los lazos profesionales y sociales que nos une. A mis amigos y compañeros que día a día luchamos por cumplir nuestra meta en común de conseguir nuestro título, el cual nos forjo de la manera mas correcta para convirtiéndonos en buenas personas y excelentes profesionales.
Franco Cuenca, marzo de 2019 AGRADECIMIENTOS Agradecemos a Dios, a nuestros padres por el apoyo durante toda nuestra carrera y desarrollo profesional, a nuestro tutor Dr. José Manuel Aller que con sus conoci-mientos, experiencia, y responsabilidad nos ha guiado de la mejor forma al desarrollo y culminación de este trabajo.
Fabián R. Cevallos A.
Franco O. Pinos V.
RESUMEN En el presente trabajo se propone el desarrollo de un método de estimación de efi-ciencia de una máquina de inducción en ambientes industriales mediante técnicas no invasivas. Este método de estimación paramétrica se fundamenta en la medición de variables instantáneas en los bornes del convertidor electromecánico y su asociación con el modelo dinámico de la máquina. Para este fin se utilizan indicadores basados en la potencia o impedancia instantánea a la entrada del motor. Los parámetros obtenidos mediante la optimización no lineal con restricciones de una función de costos que compara los indicadores del modelo con las variables medidas, permite la determinación de los parámetros. Conocidos los parámetros de la máquina de inducción, es posible determinar su rendimiento para cualquier punto de operación.
Además, se propone la comparación de las normas internacionales frente al método aplicado en este proyecto.
ABSTRACT The present work develops an estimation method for the efficiency determination of induction motors in industrial environment using non invasive techniques. This parametric estimation method is based in the instantaneous variables measurement at the machine electrical input and its association with the dynamic converter model.
For this purpose, instantaneous power and impedance indicators are used. The pa-rameters obtained using a non linear optimization algorithm considering parameter restrictions, uses a cost functions that compares the indicators evaluated with the measured variables and the same indicators evaluates through the machine dynamic model. Once the parameters are known, is possible to determine the machine effi-ciency for any operation point. More over, a comparison between efficiency calcula-tion using international standards with the proposed method are included.
PREFACIO Este trabajo de investigación presenta un método de estimación paramétrica del modelo de la máquina de inducción necesario para la determinación de la eficien-cia mediante técnicas no invasivas. Se modela la máquina de inducción en base a datos instantáneos obtenidos mediante pruebas de laboratorio para determinar los parámetros de la misma y así evaluar su comportamiento. Se realiza las pruebas tradicionales para la determinación de los parámetros de máquinas de inducción se-gún como establecen las normas internacionales más comunes IEEE 112 Método B y IEC 60034-2-1 en una máquina de inducción de 5 HP. Se realiza la optimización de una función de costo construida con los errores de los indicadores calculados a partir de los datos medidos comparados con estos mismos indicadores obtenidos del mode-lo dinámico del convertidor. Estos parámetros permiten identificar las pérdidas en la máquina y determinar su eficiencia para un punto de operación determinado. Se puede analizar el comportamiento de la máquina frente a situaciones de desequilibro de tensiones debido a las distorsión armónica que se puede presentar en la red.
Posteriormente se realiza una comparación de los resultados obtenidos mediante las pruebas tradicionales y los resultados del método de la obtención de parámetros mediante la optimización de valores, con el fin de establecer la precisión del método propuesto en este documento con los procedimientos establecidos en las normas internacionales.
Índice general INTRODUCCIÓN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 1 Capítulo 1: Problema de Estudio y Objetivos . . . . . . . .
2 1.1 Problema de Estudio . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 1.2 Grupo Objetivo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 1.3 Objetivos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 1.3.1 Objetivo General . . . . . . . . . . . . . . . . . . . . . . . . .
3 1.3.2 Objetivos Específicos . . . . . . . . . . . . . . . . . . . . . . .
3 1.4 Metodología Aplicada . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 2 Capítulo 2: Fundamentos Teóricos . . . . . . . . . . . . . . . .
5 2.1 Las Máquinas Eléctricas . . . . . . . . . . . . . . . . . . . . . . . . .
5 2.2 La Máquina de Inducción . . . . . . . . . . . . . . . . . . . . . . . .
5 2.3 Energía y Coenergía en el Campo de una Máquina Eléctrica . . . . .
6 2.4 Modelo Dinámico de la Máquina de Inducción . . . . . . . . . . . . .
9 2.5 Vectores Espaciales . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11 2.6 Desequilibrio de Tensión en la Máquina de Inducción . . . . . . . . .
14 2.6.1 Teorema de Apolonio . . . . . . . . . . . . . . . . . . . . . . .
15 2.7 Análisis de Armónicas . . . . . . . . . . . . . . . . . . . . . . . . . .
15 2.8 Armónicas en la Máquina de Inducción . . . . . . . . . . . . . . . . .
17 2.8.1 Modelo de Terceras Armónicas . . . . . . . . . . . . . . . . .
18 2.8.2 Modelo de Quintas Armónicas . . . . . . . . . . . . . . . . . .
18 2.8.3 Modelo de Séptimas Armónicas . . . . . . . . . . . . . . . . .
19 2.9 Transformada de Fourier . . . . . . . . . . . . . . . . . . . . . . . . .
20 3 Capítulo 3: Normas Internacionales para la Determina-ción de Eficiencia. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21 3.1 Norma IEEE 112 Método B . . . . . . . . . . . . . . . . . . . . . . .
21 3.1.1 Procedimiento de la Prueba . . . . . . . . . . . . . . . . . . .
21 3.1.2 Cálculos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22 3.1.3 Correcciones . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24 3.1.4 Eficiencia Mediante Norma IEEE 112 Método B . . . . . . . .
25 3.2 Norma IEC 60034-2-1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
26 3.2.1 Aspectos Generales . . . . . . . . . . . . . . . . . . . . . . . .
26 3.2.2 Pérdidas Adicionales . . . . . . . . . . . . . . . . . . . . . . .
28 3.2.3 Corrección de las Pérdidas . . . . . . . . . . . . . . . . . . . .
30 3.2.4 Eficiencia Mediante Norma IEC 60034-2-1 . . . . . . . . . . .
30 4 Eficiencia Mediante Técnicas no Invasivas . . . . . . . . . .
31 4.1 Adquisición de Datos . . . . . . . . . . . . . . . . . . . . . . . . . . .
31 i 4.2 Procesamiento de Datos . . . . . . . . . . . . . . . . . . . . . . . . .
32 4.3 Optimización de parámetros . . . . . . . . . . . . . . . . . . . . . . .
39 4.4 Estimación de la Eficiencia . . . . . . . . . . . . . . . . . . . . . . . .
45 4.5 Eficiencia Estimada con Desequilibrios de Tensión . . . . . . . . . . .
46 4.6 Eficiencia considerando armónicas . . . . . . . . . . . . . . . . . . . .
50 5 Capitulo 5: Conclusiones y Recomendaciones . . . . . . . .
52 5.1 Conclusiones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52 5.2 Recomendaciones . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52 Bibliografía . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53 ii Índice de figuras 2.1. Máquina Eléctrica y sus Posibles Ejes . . . . . . . . . . . . . . . .
6 2.2. Energía y Coenergía en el Campo . . . . . . . . . . . . . . . . . . .
8 2.3. Diagrama Esquemático de las bobinas de la máquina de inducción 9 2.4. Teorema de la mediana . . . . . . . . . . . . . . . . . . . . . . . . .
15 2.5. Distorsión Armónica . . . . . . . . . . . . . . . . . . . . . . . . . . .
16 2.6. Modelo de Terceras Armónicas . . . . . . . . . . . . . . . . . . . .
18 2.7. Circuito Equivalente de la Máquina Alimentada por un Sistema de Quinta Armónica . . . . . . . . . . . . . . . . . . . . . . . . . . .
19 2.8. Circuito Equivalente de la Máquina Alimentada por un Sistema de Séptima Armónica . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 3.1. Curva para Determinar las Pérdidas Adicionales por Asignación . .
29 3.2. Esquema de Conexión para el Método por Ensayo de Eh-Star . . .
29 4.1. Esquema de Conexión para Mediciones . . . . . . . . . . . . . . . . .
31 4.2. Conexión de los Instrumentos para la Adquisición de Datos . . . . .
32 4.3. Tensiones Línea a Línea de la Máquina de Inducción . . . . . . . . .
32 4.4. Corriente de Línea de la Máquina de Inducción . . . . . . . . . . . .
33 4.5. Vector Espacial de Corriente . . . . . . . . . . . . . . . . . . . . . . .
35 4.6. Vector Espacial de Tensión . . . . . . . . . . . . . . . . . . . . . . . .
35 4.7. Datos de Placa de la Máquina de Inducción . . . . . . . . . . . . . .
36 4.8. Vector espacial del enlace de flujo del estator . . . . . . . . . . . . . .
37 4.9. Par Eléctrico de la Máquina de Inducción . . . . . . . . . . . . . . .
38 4.10. Integración del Par Eléctrico . . . . . . . . . . . . . . . . . . . . . . .
39 4.11. Velocidad estimada mediante el Método de Euler en rpm . . . . . . .
40 4.12. Vector espacial de tensión en pu . . . . . . . . . . . . . . . . . . . .
41 4.13. Vector espacial de corriente en pu . . . . . . . . . . . . . . . . . . .
41 4.14. Vector espacial del enlace de flujo magnético en pu . . . . . . . . . .
42 4.15. Par eléctrico en pu . . . . . . . . . . . . . . . . . . . . . . . . . . . .
42 4.16. Velocidad angular de la máquina en pu . . . . . . . . . . . . . . . . .
43 4.17. Primera derivada del vector espacial de la corriente del estator en pu 43 4.18. Modelo obtenido de la máquina de inducción . . . . . . . . . . . . . .
45 4.19. Eficiencia estimada de la máquina de inducción para diversas tensio-nes de alimentación . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46 4.20. Circuito equivalente de la MI con desequilibrios . . . . . . . . . . . .
47 4.21. Triángulo Desequilibrado de Tensiones Trifásicas . . . . . . . . . . .
48 4.22. Equivalente de Thèvenin . . . . . . . . . . . . . . . . . . . . . . . . .
48 4.23. Eficiencia Estimada con Desequilibrios de Tensión . . . . . . . . . .
49 4.24. Modelo Armónico de la Máquina de Inducción . . . . . . . . . . .
50 4.25. Análisis de Fourier de la Tensión . . . . . . . . . . . . . . . . . . . .
51 4.26. Eficiencia Estimada con Armónicas . . . . . . . . . . . . . . . . . . .
51 iii Índice de cuadros 2.1. Secuencia Dominante en Cada Componente Armónica . . . . . . . . .
17 2.2. Armónicas Temporales más Frecuentes . . . . . . . . . . . . . . . .
17 4.1. Parámetros de la máquina de inducción . . . . . . . . . . . . . . . . .
36 4.2. Datos de la Prueba en Vacío . . . . . . . . . . . . . . . . . . . . . . .
37 4.3. Datos de la Prueba a Rotor Bloqueado . . . . . . . . . . . . . . . . .
37 4.4. Valores extremos de los parámetros para el algoritmo de optimización 44 4.5. Resultados del proceso de optimización . . . . . . . . . . . . . . . .
44 iv INTRODUCCIÓN En la actualidad la máquina de inducción trifásica se encuentra presente en una gran cantidad de procesos vinculados con el sector industrial y comercial, aportando un rango de potencia comprendido entre los pocos kW y una potencia máxima alrededor de los 20 MW. Entre sus múltiples usos se pueden citar los siguientes: accionamiento de compresores, ventiladores, circuitos de montaje, transporte de materias primas, bombas, etc. Su amplia difusión industrial hace necesaria la evaluación de su eficien-cia, lo que representa un reto importante debido a la complejidad que tiene el extraer la máquina de su ambiente de producción para realizar estas pruebas periódicamen-te. Por este motivo, es necesario el desarrollo de técnicas precisas de evaluación de la eficiencia que sean poco invasivas para realizar procesos de auditoría y mante-nimiento. Sin embargo, es imperativo que estos modelos consideren el efecto de las armónicas presentes en la red y los posibles desequilibrios en el sistema que afectan negativamente la eficiencia de estos convertidores.
Por esta razón el presente proyecto técnico de titulación propone comparar técnicas de estimación de eficiencia no invasivas mediante el empleo de métodos tradiciona-les utilizados en normativas internacionales. Se considera también que las normas internacionales sean el principal referente en la evaluación de la eficiencia de las máquinas de inducción.
El empleo masivo de máquinas de inducción requiere una alternativa de evaluación que sea poco invasiva y que al mismo tiempo sea precisa o con un margen de error aceptable en relación a los métodos tradicionales de estimación de eficiencia. Por este motivo es importante conocer y evaluar los múltiples factores que están involucrados en el modelamiento de estas máquinas. La adquisición de los valores instantáneos medibles en el estátor de la máquina durante el arranque, permiten realizar una estimación aceptable de los parámetros propios del modelo. De esta forma es posible utilizar el modelo de la máquina de inducción para analizar su comportamiento y rendimiento, bajo las condiciones de operación en el medio donde se encuentra .
1 1 Capítulo 1: Problema de Estudio y Objetivos 1.1.
Problema de Estudio En este trabajo se plantea analizar la eficiencia de máquinas de inducción trifási-cas en ambientes industriales, donde se presentan armónicas y desequilibrios en la tensión, utilizando técnicas mínimamente invasivas. Este problema es de gran im-portancia debido a que en cualquier proceso de auditoría energética es necesario evaluar los consumos de energía en las máquinas de una planta, pero esto no es una tarea simple debido a la imposibilidad de sacar de producción a los diferentes convertidores electromecánicos. Esto repercutiría en costos muy elevados por la pér-dida de producción. Es por este motivo que se resulta de gran interés comprobar experimentalmente las diferencias que se obtienen al medir rendimientos mediante estimaciones con modelos precisos de las máquinas y mediciones realizadas en el laboratorio utilizando normativas y procedimientos aceptados internacionalmente.
Las armónicas temporales y los desequilibrios tienen impactos negativos en la efi-ciencia de las máquinas y es necesario que los modelos incorporen esta estimación para mejorar la precisión de los resultados de las auditorías energéticas realizadas en las industrias. Por otra parte, una evaluación periódica de las eficiencias de las má-quinas es de gran utilidad para la realización de mantenimientos preventivos porque determina la evolución del estado de cada máquina en el tiempo.
1.2.
Grupo Objetivo La importancia de este trabajo radica en la conveniencia de incorporar la determina-ción de la eficiencia de la máquina de inducción en ambientes industriales mediante métodos que permitan su operación normal en ambientes de producción donde se presenta alto contenido armónico en las fuentes de alimentación y posibles desequi-librios de la red eléctrica. Este tema está en concordancia con los desarrollos y líneas de investigación del Grupo de Investigación en Energías (GIE) de la Universidad Po-litécnica Salesiana y la propuesta posibilitará el desarrollo de métodos de medición y estimación de eficiencia que podrían ser de gran utilidad tanto para la industria ecuatoriana como para la sociedad en conjunto al permitir ahorros de energía y reducción de costos de mantenimiento.
2 1.3.
Objetivos 1.3.1.
Objetivo General Evaluación de la eficiencia de máquinas de inducción en ambientes industriales con contaminación armónica y desequilibrios 1.3.2.
Objetivos Específicos 1. Revisión de los métodos existentes para determinación de eficiencia de máqui-nas de inducción con técnicas no invasivas.
2. Desarrollo de modelos vectoriales de máquina de inducción que consideren el incremento de las pérdidas producidas por la presencia de desequilibrios de tensión típicos en los sistemas industriales.
3. Desarrollo de modelos vectoriales de máquinas eléctricas que consideren el incremento de las pérdidas en máquinas de inducción debido a la distorsión armónica en fuentes de tensión, en especial por inversores operados con técnicas de modulación por ancho de pulso.
4. Desarrollo de un método para la estimación de la eficiencia en máquinas de inducción utilizando técnicas no invasivas en ambientes industriales conside-rando los efectos del desequilibrio y distorsión.
5. Comparación de los métodos de estimación de eficiencia mediante métodos no invasivos con los resultados obtenidos aplicando normativas internacionales en ensayos de laboratorio.
1.4.
Metodología Aplicada Uno de los temas con más interés en el estudio de las máquinas de inducción den-tro de los ambientes industriales, siempre ha sido la estimación de la eficiencia de las mismas, donde es importante considerar las diversas situaciones en las cuales se encuentran expuestas estos convertidores electromecánicos. La medición de la efi-ciencia de las máquinas de inducción desarrollada en laboratorio es precisa en base a las normas internacionales, mediante métodos específicos que se requieren para conseguir los resultados, tales como son las pruebas de vacío a frecuencia nominal y tensión variable, siendo estas las pruebas más comunes para determinar esta medi-da. Al realizar de esta manera la medición de la eficiencia, es bastante complicado satisfacer los requerimientos para llevar a cabo una auditoria energética dentro de una área o planta industrial .
Dentro de este trabajo de investigación se planea utilizar un método propuesto en un trabajo previo del grupo de investigación, donde se han desarrollado estrategias para la estimación de los parámetros en máquinas de inducción a través de medidas instantáneas de tensión y corriente. Estos métodos pueden ser utilizados para la es-timación de eficiencia considerando desequilibrio de tensiones y distorsión armónica.
3 Se planea utilizar un método sistemático deductivo, donde se realizarán compara-ciones entre dos herramientas de modelación: una por elementos finitos (precisa y muy detallada desde el punto de vista físico matemático) que permite validar los supuestos de los modelos vectoriales, más simples y rápidos, pero menos precisos, y la otra contrastando con los métodos convencionales, como se establecen en las normas internacionales.
Los pasos a seguir metodológica mente serían los siguientes: 1. Caracterización de los efectos del desequilibrio de tensión en máquinas de in-ducción.
Revisión de los índices internacionales para la caracterización del desequili-brio de tensiones. Estudio del efecto del desequilibrio en el incremento de las pérdidas del convertidor. Revisión de la capacidad de los modelos utilizados en máquinas eléctricas para predecir el comportamiento de las pérdidas. Eva-luación experimental del fenómeno.
2. Caracterización de los efectos de la distorsión armónica en máquinas de induc-ción.
Desarrollo de un modelo computacional en un paquete de elementos finitos de la máquina de inducción. Estudio del efecto de la distorsión armónica en el incremento de las pérdidas del convertidor. Revisión de los modelos vectoriales de máquinas de inducción considerando la distorsión de tensiones y su efecto sobre las pérdidas del convertidor.
3. Propuesta de un método para la estimación de eficiencia de máquinas de in-ducción trifásicas en ambientes industriales con técnicas no invasivas.
Desarrollo de una metodología para la estimación de eficiencia, capaz de pro-ducir resultados precisos considerando desequilibrios y distorsión armónica en las tensiones. Validación experimental de la propuesta.
4. Comparación de los resultados.
Realizar una comparación de los resultados obtenidos al establecer el método propuesto y la aplicación de las normas internacionales para la determinación de la eficiencia en máquinas de inducción.
4 2 Capítulo 2: Fundamentos Teóricos 2.1.
Las Máquinas Eléctricas Desde el siglo XIX el desarrollo de las máquinas eléctricas han tenido un cambio muy notorio en el transcurso del tiempo, el cual continúa avanzando debido al desarrollo de la tecnología. Gracias a los aportes significativos que realizaron diversos inventores como Edison y Tesla, entre otros, siendo estos los más destacados, logrando la eficacia y eficiencia en la conversión electromagnética de la energía. Las máquinas eléctricas poseen características comunes que permiten generalizar la descripción matemática del comportamiento de las mismas a través de diversas herramientas.
2.2.
La Máquina de Inducción Las máquinas de inducción en la actualidad se han convertido en los convertidores electromecánicos más utilizados en la industria por sus diversas ventajas construc-tivas y operativas entre las cuales destacan: robustez de su diseño, prestaciones estáticas y dinámicas, así como su reducido costo.
Las máquinas de inducción pueden tener diferentes números de fases, aun cuando habitualmente utilizan tres devanados en el estátor distribuidos en su periferia, mien-tras que en el rotor está conformado por un conjunto de barras que se encuentran en cortocircuito y estas son denominadas jaula de ardilla. Existen también máqui-nas que son utilizadas en algunas aplicaciones prácticas como son los generadores eólicos, grúas o grandes compresores.
Las máquinas de inducción en su principio de funcionamiento producen un campo magnético rotatorio al excitar los devanados estatóricos. Este campo induce fuerzas electromotrices en los conductores del rotor, que al estar en cortocircuito hacen circular corrientes por las barras y se produce un campo magnético inducido, el cual, al interactuar con el aplicado en el estátor, produce par eléctrico a cualquier velocidad.
Los motores de inducción además de las importantes ventajas destacadas, también presentan algunos inconvenientes. En especial con su rendimiento y el factor de potencia, debido por un lado a la necesidad de tener corrientes y pérdidas en el rotor para producir par y por el otro a la necesidad de introducir potencia reactiva para mantener el campo magnético inducido en el entre hierro.
Estas desventajas hacen necesarios modelos precisos que permitan la determinación de su comportamiento, la evaluación de sus características y la determinación de su rendimiento en condiciones industriales. Comúnmente se puede determinar el rendi-miento de una máquina en condiciones de laboratorio utilizando equipos especiales.
5 Sin embargo, la gran expansión de esta máquina en la industria, requiere herramien-tas y modelos que permitan la determinación de sus características y rendimiento de forma no invasiva.
2.3.
Energía y Coenergía en el Campo de una Máquina Eléctrica Las máquinas eléctricas poseen ejes característicos que definen su comportamiento, estos ejes son de dos tipos, ejes eléctricos y ejes mecánicos, esta característica se puede observar en la siguiente figura.2.1 Para el análisis de los ejes eléctricos de la máquina, es indispensable conocer las corrientes y tensiones que posee la misma. Para los ejes mecánicos se involucran las variables de velocidad y fuerza que nos ayudan a describir las condiciones de operación de la máquina, considerando aspectos como: si su movimiento es lineal, el par y velocidad angular o si el movimiento de la máquina es rotativo.
Es necesario contar con al menos con un eje eléctrico o un eje mecánico en la máquina eléctrica por más simple que esta sea. De la misma manera se debe tomar en cuenta que toda la energía introducida en la maquina eléctrica a través de los ejes eléctricos no es transmitida a los ejes mecánicos o de manera viceversa, debido a todos los factores que generan pérdidas en la máquina. Donde parte de la energía eléctrica que se genera en la máquina se almacena dentro del campo de la misma.
A partir de este concepto se puede apreciar que la energía acumulada en el campo no es posible determinarla, pero se puede considerar que la energía acumulada en el campo se puede obtener mediante la diferencia de la energía eléctrica y mecánica de la máquina, por otra instancia esto nos lleva al concepto que mediante la integral de la energía eléctrica se puede determinar la potencia eléctrica en el tiempo, esta energía podría ser determinada en el eje eléctrico de la máquina a partir de valores de tensión y corriente, como se muestra en la siguiente expresión 2.1.
△We = ˆ t 0 Pe(τ)dτ = ˆ t 0 v(τ) ∗i(τ)dτ (2.1) Si se presentara un sistema conservativo en donde no existan pérdidas por elementos resistivos, la tensión suministrada a la máquina y la fuerza electromagnética inducida son iguales, con lo cual lo mencionado se expresa de la siguiente manera 2.2.
Figura 2.1: Máquina Eléctrica y sus Posibles Ejes 6 v(t) = e(t) = dλ dt (2.2) Para determinar la energía eléctrica dentro de la máquina se debe conocer la relación que tiene la corriente con respecto al flujo y la posición del convertidor. Con respecto a la variación de la energía mecánica en la máquina es indispensable conocer la velocidad y la fuerza en función del tiempo.
Cabe recalcar que en un sistema conservativo la energía de la máquina está reflejada en función del estado, con lo cual podemos determinar que, el incremento de la energía acumulada en el campo no dependen de la trayectoria que se utiliza para alcanzar un estado, si no de los valores iniciales y finales del proceso.
En una máquina, cuando el sistema mecánico se encuentra en reposo no habrá variación en la energía mecánica, la energía eléctrica suministrada a la máquina se convierte en energía acumulada en el campo como se puede ver en las siguientes expresiones 2.3 y 2.4: △We = ˆ (λ(t)) (λ(0)) i(x, λ)dλ = △Wc si x = cte (2.3) △Wc = i(x, λ) λ λ(t) λ(0) − ˆ i(t) i(0) λ(x, i)di (2.4) De la expresión 2.4, en el término integral se define como la coenergía en el campo y podemos expresarla como se muestra en la expresión 2.5: △W ´ c = ˆ i(t) i(0) λ(x, i)di (2.5) La coenergía es el área bajo la curva de la característica (λ-i) . En un sistema electromagnético donde la posición x es constante se puede determinar la siguiente relación 2.6: λ i = ΔWc + ΔW ´ c (2.6) En la figura 2.2 se observa cómo se presenta la energía y coenergía en el campo de una máquina.
Conociendo la relación del enlace de flujo y la corriente que se presenta dentro de la máquina, se puede determinar dos relaciones importantes al momento de determinar la energía y la coenergía en el campo. Para determinar la energía en una máquina 7 Figura 2.2: Energía y Coenergía en el Campo el enlace de flujo λ es la variable independiente y la corriente i es la variable depen-diente, con respecto a la coenergía la corriente i es la variable independiente y el enlace de flujo λ es la variable dependiente.
Mediante esto se puede establecer que la corriente aumenta de forma exponencial cuando el sistema tiene un comportamiento lineal, con lo cual se puede definir el enlace de flujo mediante la siguiente expresión 2.7: λ = L i (2.7) Por lo cual dentro de la máquina se tiene la ecuación característica que define el comportamiento de la misma considerando los aspectos mencionados anteriormente 2.8: v = R i + e (2.8) Y al reemplazar el valor de la fuerza electro motriz 2.2 se obtiene la siguiente expre-sión que define una de las características de la máquina de inducción 2.9: v = R i + dλ dt (2.9) 8 Figura 2.3: Diagrama Esquemático de las bobinas de la máquina de inducción 2.4.
Modelo Dinámico de la Máquina de Inducción Para determinar el comportamiento de la máquina de inducción en régimen dinámi-co, se plantean las ecuaciones diferenciales de los ejes eléctricos y mecánicos. En los ejes eléctricos se definen las tensiones aplicadas a las bobinas del estátor y rotor, y en el eje mecánico se establece la segunda ecuación de Newton.
En la figura 2.3 se observa el esquema de las bobinas de una máquina de inducción trifásica.
Aplicando la ley de Kirchoffa las bobinas estatóricas y rotóricas de la figura 2.3 se determinan las ecuaciones diferenciales de los ejes eléctricos como se muestran en las expresiones 2.10 y 2.11: [v] = [R][i] + d[λ] dt (2.10) [v] = [R][i] + [L (θ)]d[λ] dt + ˙ θ[τ(θ)][i] (2.11) donde: [v] = " [ve] [vi] # = h ve a ve b ve c it h vr a vr b vr c it [i] = " [ie] [ii] # = h ie a ie b ie c it h ir a ir b ir c it 9 [λ] = " [λe] [λi] # = h λe a λe b λe c it h λr a λr b λr c it [R] = " [Ree] [Rer] [Rre] [Rrr] # = " Re [I] Rr [I] # [L (θ)] = " [Lee] [Ler (θ)] [Lre(θ)] [Lrr] # = " Lσe [I] + Lme [S] Ler [C(θ)] Ler [C(θ)]t Lσr [I] + Lmr [S] # [τ (θ)] = " d dθ [Lee] d dθ [Ler (θ)] d dθ [Lre(θ)] d dθ [Lrr] # = " Ler d dθ [C(θ)] Ler d dθ [C(θ)]t # [I] = 1 0 0 0 1 0 0 0 1 [S] = 1 −1 2 −1 2 −1 2 1 −1 2 −1 2 −1 2 1 = 0 0 0 0 0 0 0 0 0 [C (θ)] = cos θ cos θ + 2π 3 cos θ + 4π 3 cos θ + 4π 3 cos θ cos θ + 2π 3 cos θ + 2π 3 cos θ + 4π 3 cos θ d dθ[C (θ)] = −sin θ −sin θ + 2π 3 −sin θ + 4π 3 −sin θ + 4π 3 −sin θ −sin θ + 2π 3 −sin θ + 2π 3 −sin θ + 4π 3 −sin θ La expresión que define el par mecánico de la máquina se obtiene a partir del prin-cipio de los trabajos virtuales, esto se muestra en la expresión 2.12: Te −Tm = ∂W ′ c ∂θ −Tm = 1 2 [i]t [τ (θ)]t [i] −Tm = J ¨ θ + p ˙ θ (2.12) Los parámetros que definen el comportamiento de la máquina de inducción en el sistema de coordenadas primitivas son: Re Resistencia de cada bobina del estátor Rr Resistencia de cada bobina del rotor Lσe Inductancia de dispersión del estátor Lσr Inductancia de dispersión del rotor Lme Inductancia de magnetización del estátor Lmr Inductancia de magnetización del rotor Ler Inductancia mutua de acoplamiento estátor-rotor 10 2.5.
Vectores Espaciales Para el estudio de los vectores espaciales, es importante conocer la relación que esto presenta con el concepto de la transformación de componentes simétricas, la cual está definida por las expresiones 2.13 y 2.14. Esta trasformación permite convertir un sistema acoplado, en tres sistemas independientes. Estos sistemas independientes se represnetan como secuencia positiva, secuencia negativa y cero. La secuencia cero solo se puede excitar cuando la sumatoria de las corrientes o tensiones de la máquina es diferente de cero. La secuencia positiva y la secuencia negativa son similares determinando que son el conjugado del otro.
x0 x+ x− = 1 √ 3 1 1 1 1 ej 2π 3 ej 4π 3 1 ej 4π 3 ej 2π 3 xa xb xc = 1 √ 3 1 1 1 1 α α2 1 α2 α xa xb xc (2.13) xa xb xc = 1 √ 3 1 1 1 1 ej 4π 3 ej 2π 3 1 ej 2π 3 ej 4π 3 x0 x+ x− = 1 √ 3 1 1 1 1 α2 α 1 α α2 x0 x+ x− (2.14) Con la transformación de componentes simétricas se puede representar el modelo primitivo de la máquina de inducción utilizando únicamente la secuencia positiva como se muestra en la expresión 2.15. Denominando a este proceso como trans-formación a vectores espaciales.
x(t) = s 2 3 h 1 ej 2π 3 ej 4π 3 i xa(t) xb(t) xc(t) = s 2 3 h 1 α α2 i xa(t) xb(t) xc(t) (2.15) La transformación a vectores espaciales representa un sistema de corrientes y de tensiones o flujos trifásicos a través de un vector en el espacio, donde la posición y la magnitud de este dependen del tiempo.
Las representaciones del modelo primitivo de una máquina de inducción en vectores espaciales se muestra en la siguiente expresión 2.16.
" ve vr # = " Re 0 0 Rr # " ie ir # + p " Le Merejθ Merejθ Lr # " ie ir #!
(2.16) Donde se tienen las siguientes expresiones: ve = q 2 3 h 1 α α2 i h ve a ve b ve c it 11 ie = q 2 3 h 1 α α2 i h ie a ie b ie c it vr = q 2 3 h 1 α α2 i h vr a vr b vr c it ir = q 2 3 h 1 α α2 i h ir a ir b ir c it Le = Lσe + 3 2Lme Lr = Lσr + 3 2Lmr Mer = 2 3Ler Aplicando la transformación a vectores espaciales de la ecuación del par eléctrico en las ecuaciones primitivas de la máquina de inducción tenemos la siguiente expresión 2.17: Te = 1 2 [i]t [τ] [i] = 1 2 " ie ir #t " 0 Ler d dθ [C(θ)] Ler d dθ [C(θ)]t 0 # " ie ir # Te = Ler [ie]t d dθ [C(θ)] [ir] Te = Ler [ie]t e−jθ 2j 1 α2 α α 1 α2 α2 α 1 −e−jθ 2j 1 α α2 α2 1 α α α2 1 [ir] Te = q 2 3Ler n e−jθ 2j ie h 1 α2 α i −e−jθ 2j i∗ e h 1 α α2 io [ir] Te = 2 3Ler n e−jθ 2j iei∗ r −e−jθ 2j i∗ eir o Te = Merℑmiei∗ re−jθ Te = Merℑmie ire−jθ∗ (2.17) Al sistema de ecuaciones diferenciales que definen el comportamiento de la máqui-na de inducción, se aplica la transformación a vectores espaciales obteniendo las expresiones 2.18 y 2.19: " ve vr # = " Re 0 0 Rr # " ie ir # + p " Le Merejθ Merejθ Lr # " ie ir #!
(2.18) 12 Merℑmie ire−jθ∗−Tm( ˙ θ) = J ¨ θ + p ˙ θ (2.19) Estas expresiones simplifican las ecuaciones primitivas del modelo de la máquina de inducción representando magnitudes trifásicas en vectores espaciales, logrando reducir las siete ecuaciones iniciales en tres ecuaciones diferenciales y la dependencia del ángulo θ se ha simplificado a una matriz de 2x2.
Se puede eliminar dependencia del ángulo θ si las variables del rotor se refieren al estátor utilizando las siguiente expresión 2.20.
xe r = xrejθ (2.20) Al aplicar la transformación al modelo de la máquina en vectores espaciales, es necesario desarrollar la derivada de esta transformación de la siguiente manera.
xe r = xrejθ pxe r = pxrejθ + j ˙ θejθxr pxe r = pxrejθ + j ˙ θxe r pxrejθ = pxe r −j ˙ θxe r (2.21) Con la transformación se obtienen las expresiones que definen el modelo de la má-quina de inducción en vectores espaciales referidos al estátor, como se muestra en las expresiones 2.22 y 2.23.
" ve vr # = " Re 0 0 Rr # " ie ir # + " Le Merejθ Merejθ Lr # p " ie ir # −j ˙ θ " Le Merejθ Merejθ Lr # " ie ir # (2.22) Merℑmie ire−jθ∗−Tm( ˙ θ) = J ¨ θ + p ˙ θ (2.23) Este modelo de la máquina de inducción es independiente de la posición angular θ, la cual es variable en el tiempo de igual manera que el caso particular de la operación de la máquina en el régimen permanente.
13 2.6.
Desequilibrio de Tensión en la Máquina de Inducción La tensión de alimentación de una máquina de inducción puede presentar variacio-nes, es decir no se encuentran balanceadas, lo que provoca que las corrientes que circulen por las bobinas de la máquina tampoco lo sean. Si las corrientes que circula por las bobinas no son equilibradas, el campo magnético que generan estas corrientes no será rotatorio .
Al presentarse este problema en la operación de la máquina, se puede representar un sistema de tensiones trifásicas desequilibradas por tres sistemas equilibrados: secuencia positiva, secuencia negativa y secuencia cero. Para esto se ha empleado la transformación de componentes simétricas. Estas secuencias se analizan de manera independiente utilizando el método de superposición para encontrar las tensiones de línea .
Para descomponer un sistema trifásico en componentes simétricas se utiliza la ex-presión 2.24 conservativa en potencia: v0 v+ v− = 1 √ 3 1 1 1 1 ej 2π 3 ej 4π 3 1 ej 4π 3 ej 2π 3 va vb vc = 1 √ 3 1 1 1 1 α α2 1 α2 α va vb vc (2.24) La transformación inversa de componentes simétricas a magnitudes fase se obtiene mediante la expresión 2.25: va vb vc = 1 √ 3 1 1 1 1 ej 4π 3 ej 2π 3 1 ej 2π 3 ej 4π 3 v0 v+ v− = 1 √ 3 1 1 1 1 α2 α 1 α α2 v0 v+ v− (2.25) El campo magnético rotatorio al estar alimentado por un sistema de secuencia ne-gativa equilibrado gira en sentido contrario a las manecillas del reloj, generando que la velocidad de campo sea −ωe .
Se puede determinar el par eléctrico de secuencia positiva y negativa mediante las tensiones obtenidas de las componentes que fueron aplicadas a la máquina.
Al presentarse desequilibrios en la máquina, el sistema de secuencia cero no produce par, lo que ocasiona que se incrementen las corrientes, las perdidas, el calentamiento y así reducir el rendimiento de la máquina .
El cálculo de las componentes simétricas es la herramienta más utilizada para ana-lizar los desequilibrios a los que pueden someterse las máquinas de inducción. Estos cálculos se realizan con un deslizamiento o velocidad determinados de la máquina .
14 Figura 2.4: Teorema de la mediana .
2.6.1.
Teorema de Apolonio El teorema de Apolonio o también denominado teorema de la mediana, es un teorema geométrico basado en la relación de cada uno de los lados de un triángulo con la longitud de su mediana. El teorema dispone que la sumatoria de cualesquiera de los cuadrados de sus dos lados genera como resultado la mitad del tercer lado sumado el doble del cuadrado de su mediana. Este teorema es un gran apoyo para el cálculo de las tensiones de fase en la máquina de inducción, donde comúnmente al realizar pruebas sobre el convertidor se mide de manera directa las tensiones entre sus fases, con equipos comunes como multímetros u otros instrumentos disponibles en la industria. Mediante las expresiones 2.26 y 2.27 se puede determinar los valores de las tensiones de fase referidas al neutro. En la figura 2.4 se muestra el triángulo que representa las tensiones entre fases y las tensiones al neutro en una máquina de inducción, cuando presenta un desequilibrio en el sistema trifásico de alimentación. Con la ayuda del teorema de senos y cose-nos es posible determinar los ángulos interiores que se muestran en el triángulo de tensiones.
α = arcsin vbc vabsin (γ) β = arcsin vca vabsin (γ) γ = arccos − v2 ab−v2 bc−v2 ca 2vbcvca (2.26) va = 1 3 q 2 (v2 ca + v2 ab) −v2 bc vb = 1 3 q 2 (v2 bc + v2 ab) −v2 ca vc = 1 3 q 2 (v2 bc + v2 ca) −v2 ab (2.27) 2.7.
Análisis de Armónicas Uno de los problemas que comúnmente se presentan en los sistemas eléctricos es la presencia de armónicas, debido a las cargas no lineales que se conectan a estos sistemas. Estas anomalías en la red eléctrica puede ocasionar diversos inconvenientes como: aumento de pérdidas de potencia, sobretensiones, mal funcionamiento de las protecciones, que tiene como consecuencia disminuir la vida útil de los equipos.
15 Frecuencia -10 -8 -6 -4 -2 0 2 4 6 8 10 Amplitud -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Fundamental Tercer Armonica Resultante Figura 2.5: Distorsión Armónica Las armónicas son señales donde su frecuencia es un múltiplo de la frecuencia funda-mental que está presente en los sistemas eléctricos . En los sistemas de 60 Hz, las armónicas son señales senoidales. Estas señales al tener una relación con la frecuen-cia fundamental, se puede determinar la frecuencia de cada una de las armónicas presentes en la red mediante la expresión 2.28.
Fh = h · 60[Hz] (2.28) donde h es un número entero que representa el orden de la armónica.
La magnitud de la armónica, en por unidad se puede determinar mediante la expre-sión 2.29.
Ih = I1 h (2.29) donde: Ih Magnitud de la corriente armónica en pu I1 Magnitud de la corriente fundamental en pu h Orden de la armónica Dependiendo del orden de la armónica, la magnitud y frecuencia de las mismas aumentan o disminuyen, haciendo que las armónicas de menor orden sean las más significativas para este análisis.
El efecto de las armónicas en la red es denominado distorsión armónica, esto se debe a que las armónicas se combinan con la frecuencia fundamental. Este efecto produce una desproporción de la corriente con respecto a la tensión aplicada al sistema. En la figura 2.5 se puede apreciar este fenómeno.
La distorsión armónica total o THD expresa la cantidad de distorsión armónica en una señal, mediante la expresión 2.30: 16 Cuadro 2.1: Secuencia Dominante en Cada Componente Armónica Secuencia de Fase Positiva Negativa Cero Orden de la Armónica 1 5 3 Orden de la Armónica 7 11 9 Orden de la Armónica 13 17 15 THD = qPn h=2 I2 h I1 100 % (2.30) El THD puede calcularse para la tensión o corriente .
Para el análisis de las componentes armónicas en una señal, es necesario descomponer la onda en su componente fundamental y armónicas, mediante un análisis de Fourier.
Este método facilitara el análisis por separado de cada una de las componentes de la señal, de manera conjunta con este método se emplea métodos de solución de circuitos y el método de superposición .
Otro método de aporte para el análisis de armónicos, es el método de componentes simétricas. Mediante este método de puede determinar la secuencia de cada una se las componentes armónicas. En los sistemas eléctricos solo se presentan armónicas de orden impar, considerando el método de componentes simetrías en la siguiente tabla 2.1 se muestra la secuencia de las armónicas según su orden.
2.8.
Armónicas en la Máquina de Inducción Para el análisis del comportamiento de la máquina, se ha mencionado inconvenientes como el desbalance en sus fuentes de alimentación, donde este análisis es desarrollado únicamente a una sola frecuencia. En la industria se presentan diversos casos al momento de operar una máquina de inducción, la cual nos lleva a emplear equipos adicionales que colaboren el en funcionamiento de la misma. La consecuencia de emplear equipos adicionales en conjunto con la máquina de inducción, es la presencia de componentes armónicas en las fuentes de alimentación. Debido a esto se deben realizar cambios en el modelo de la máquina, para analizar su respuesta .
Las armónicas más comunes en los sistemas eléctricos son los que se muestran en la Tabla 2.2: Es importante desarrollar un modelo de la máquina considerando cada una de las armónicas presentes en esta, además es posible utilizar el principio de superposición, considerando que la máquina es lineal, y así determinar su respuesta frente a este fenómeno .
Cuadro 2.2: Armónicas Temporales más Frecuentes Armónica 1,1 3,1 5,1 7,1 9,1 11,1 13,1 Frecuencia ωe 3ωe 5ωe 7ωe 9ωe 11ωe 13ωe 17 Figura 2.6: Modelo de Terceras Armónicas 2.8.1.
Modelo de Terceras Armónicas Al suministrar una fuente de tensiones trifásicas balanceada en la máquina de in-ducción, la componente armónica fundamental o primera armónica también lo será y la componente de la tercera armónica se encontraría en fase. Al presentarse esto, el sistema de terceras armónicas se comportaría como un sistema de secuencia ce-ro. Para las armónicas que son múltiplos de tres, su comportamiento será como un sistema de secuencia cero.
Para el modelo de la máquina se debe considerar el incremento de las reactancias en un factor de tres y la resistencia no se modificaría, si es que se desprecia el efecto pelicular en los conductores. En la figura 2.6 se observa el modelo de terceras armónicas .
2.8.2.
Modelo de Quintas Armónicas En el modelo de quinta armónica, se considera las expresiones 2.31, 2.32 y 2.33: Va5(ωet) = √ 2V5sin5(ωet) = √ 2V5sin(5ωet) (2.31) Vb5(ωet) = √ 2V5sin5(ωet −2π 3 ) = √ 2V5sin(5ωet −4π 3 ) (2.32) Vc5(ωet) = √ 2V5sin5(ωet −4π 3 ) = √ 2V5sin(5ωet −2π 3 ) (2.33) En el modelo de la quinta armónica, se considera que la máquina se comporta como un sistema de secuencia negativa. Al alimentar a la máquina con un sistema de secuencia negativa se genera un campo magnético rotatorio en sentido contrario a las manecillas del reloj. Para este modelo se considera el aumento de las reactancias en un factor de cinco, las resistencias se mantendrán constantes si se desprecia el efecto pelicular. Para el cálculo del deslizamiento en el modelo de quinta armónica se considera la expresión 2.34 .
s51 = −5ωe −ωr −5ωe = 5ωe + ωr 5ωe = 1 + ωr 5ωe = 6 −s11 5 (2.34) 18 Figura 2.7: Circuito Equivalente de la Máquina Alimentada por un Sistema de Quinta Armónica El par eléctrico que se genera en este modelo, tiene la misma dirección que el cam-po magnético. La resistencia de magnetización de la máquina se podría modificar considerando la frecuencia que se emplea, además se debe considerar el aumento de las pérdidas en el hierro causado por el incremento de la frecuencia . En la figura 2.7 se muestra el circuito equivalente de la máquina alimentada por un sistema de quinta armónica.
2.8.3.
Modelo de Séptimas Armónicas En el modelo de quinta armónica, se considera las expresiones 2.35, 2.36 y 2.37: Va7(ωet) = √ 2V7sin7(ωet) = √ 2V7sin(7ωet) (2.35) Vb7(ωet) = √ 2V7sin7(ωet −2π 3 ) = √ 2V7sin(7ωet −2π 3 ) (2.36) Vc7(ωet) = √ 2V7sin7(ωet −4π 3 ) = √ 2V7sin(7ωet −4π 3 ) (2.37) En el modelo de la séptima armónica, se considera que la máquina se comporta como un sistema de secuencia positiva. Al alimentar a la máquina con un sistema de secuencia positiva se genera un campo magnético rotatorio en sentido a las manecillas del reloj. Para este modelo se considera el aumento de las reactancias en un factor de siete, las resistencias se mantendrán constantes si se desprecia el efecto pelicular.
Para el cálculo del deslizamiento en el modelo de séptima armónica se considera la expresión 2.38 .
s71 = 7ωe −ωr 7ωe = 6 + s11 7 (2.38) El par eléctrico que se genera en este modelo, tiene la misma dirección que el cam-po magnético. La resistencia de magnetización de la máquina se podría modificar considerando la frecuencia que se emplea, además se debe considerar el aumento de las pérdidas en el hierro causado por el incremento de la frecuencia . En la figura 2.8 se muestra el circuito equivalente de la máquina alimentada por un sistema de séptima armónica.
19 Figura 2.8: Circuito Equivalente de la Máquina Alimentada por un Sistema de Séptima Armónica 2.9.
Transformada de Fourier La transformada de Fourier es una herramienta que permite la transformación de señales en función del tiempo en otras en función de la frecuencia. Esta transforma-da es una extensión de la serie de Fourier, que permite la representación de señales periódicas y no periódicas en todo instante de tiempo . La transformada de Fou-rier está definida por la expresión 2.39, esta expresión se escribe en términos de la frecuencia cíclica f con la ventaja de ser simétrica.
X(f) = F(x(t)) = ˆ ∞ −∞ x(t)e−j2πftdt (2.39) La transformada inversa de Fourier está definida por la expresión 2.40, esta expresión se escribe en términos de la variable de frecuencia en radianes ω. La frecuencia en radianes tiene una relación más directa con las constantes de tiempo y las frecuencias resonantes de los sistemas reales, haciendo que la transformada de las funciones de los sistemas sea más simples utilizando esta expresión .
x(t) = F −1(X(f)) = ˆ ∞ −∞ X(f)ej2πftd f (2.40) La transformada de Fourier es una herramienta muy útil para el estudio de ecua-ciones en derivadas parciales lineales constantes, para emplear esta transformada existen diversas propiedades importantes, que en varios casos evitan aplicar directa-mente el concepto de integral de la transformada de Fourier de manera independiente de las formas de las señales. En se ilustra tablas de las propiedades de la trasfor-mada de Fourier, de igual manera el uso de las mismas con diversos ejemplos donde se aplican estas propiedades.
20 3 Capítulo 3: Normas Internacionales para la Determinación de Eficiencia.
Para determinar la eficiencia en una máquina de inducción se tomara en cuenta dos estándares internacionales que son los claves para el objetivo de este trabajo. Estos estándares son empleados por diversos fabricantes, por lo que representan un punto importante en el diseño y construcción. Dichos métodos son los siguientes: 1. IEEE 112 – Método B 2. IEC 60034-2-1 Las metodologías de estos estándares internacionales tienen cierta similitud entre ellos, con pequeñas diferencias en la obtención de sus resultados. Por ello se analizará los dos métodos, con el objetivo de conocer su metodología y relacionarla con el desarrollo del método propuesto en este trabajo.
3.1.
Norma IEEE 112 Método B Esta norma se puede aplicada a las máquinas de inducción, tanto en generadores como en motores de eje horizontal, con rotor de tipo jaula de ardilla y sin considerar su tamaño. Cuenta con diferentes métodos para la estimación de la eficiencia, estos métodos consideran la capacidad nominal de la máquina. El procedimiento se centra en cómo determinar las pérdidas adicionales, ya sea de forma directa o indirecta.
Todos los datos se toman con la máquina operando como un motor o como un generador, dependiendo de la región de operación para la cual se requiere los datos de eficiencia [10, 11].
3.1.1.
Procedimiento de la Prueba Las pruebas individuales que conforman el método B se realizan en el orden que se indicará a continuación, las pruebas se pueden realizar individualmente si la tem-peratura de funcionamiento de la máquina se establece cerca de su temperatura de funcionamiento normal para el tipo de prueba antes de obtener los datos .
3.1.1.1.
Resistencia en Frio Con la máquina a temperatura ambiente, se realiza la medida y registro del valor de las resistencias de cada uno da las bobinas y el registro de la temperatura ambiente.
21 3.1.1.2.
Prueba de Temperatura de Carga Nominal Esta prueba consiste en que la máquina se carga como un motor o generador en la condición deseada o a valores nominales, de la cual se registra el valor de la temperatura de la máquina durante esta prueba.
3.1.1.3.
Prueba Bajo Carga La temperatura del devanado del estátor debe estar dentro de los 10◦C de la lectura de temperatura más alta registrada durante la temperatura de carga nominal, la prueba se debe realizar lo más rápido posible para evitar daños en la máquina durante la prueba.
3.1.1.4.
Prueba sin Carga Esta prueba se realiza conectando la máquina como un motor a la tensión y fre-cuencia nominal sin carga. Durante esta prueba se debe registrar la temperatura, voltaje, corriente y potencia de entrada a la frecuencia nominal .
3.1.2.
Cálculos 3.1.2.1.
Pérdidas por Fricción Para determinar las pérdidas por fricción, restamos la pérdida del estátor (i2 R) (a la temperatura de la prueba) de las pérdidas totales, es decir potencia de entrada.
Otra alternativa para determinar estas pérdidas es trazar una curva resultante de cada punto de tensión de la prueba contra la potencia, extendiendo la curva a vol-taje cero. La intersección con el eje de voltaje cero es la pérdida por fricción, esta intersección puede determinarse con mayor precisión si la potencia de entrada menos la pérdida (i2 R) del estátor se traza frente a la tensión al cuadrado para valores en el rango de voltaje más bajo.
3.1.2.2.
Pérdidas en el Núcleo Estas pérdidas se obtienen al restar el valor de las pérdidas por fricción de la potencia de entrada menos la pérdida (i2 R) del estátor.
3.1.2.3.
Pérdidas en el Estátor Para una máquina trifásica, las pérdidas en el estátor se definen por la expresión 3.1.
Pe = 1,5(i2R) = 3(i2R1) (3.1) 22 donde: i Es la corriente medida del estátor, en amperios.
R Es la resistencia, en ohmios, entre dos fases de la máquina, corregida a la tempe-ratura adecuada, si es necesario.
R1 Es la resistencia por fase, en ohmios.
3.1.2.4.
Pérdidas en el Rotor Estas pérdidas se basa en la velocidad real o la medición de deslizamiento para cada punto y no se requieren ajustes .
La pérdida del rotor (i2R1), incluida las pérdidas en el núcleo, se determina en base al deslizamiento con las expresiones 3.2 y 3.3: La máquina en configuración como motor Pr = (Pin −Pe −Pfe) s (3.2) La máquina en configuración como generador Pr = (Pout −Pe −Pfe) s (3.3) donde s es el deslizamiento, en por unidad (p.u), con la velocidad síncrona como base, todos estos cálculos están en vatios (W).
La velocidad de deslizamiento, en r/min, se puede medir directamente por medios estroboscópicos o se puede calcular a partir de la velocidad medida. Este valor luego se debe convertir a un valor numérico o por unidad para usar en los análisis. La velocidad de deslizamiento es la diferencia entre la velocidad síncrona y la velocidad medida, en r/min s = ns + nt (3.4) ns = 120f p (3.5) donde: ns Es la velocidad síncrona en r/min.
nt Es la velocidad medida en r/min.
f Es la frecuencia de la línea, en H.
p Es el número de pares de polos.
El deslizamiento expresado en por unidad es: s = s r min ns r min (3.6) 23 3.1.2.5.
Pérdida Total Aparente La pérdida total aparente se calculará por separado para cada punto de carga res-tando la salida medida en vatios de la entrada medida en vatios.
3.1.2.6.
Determinación de Pérdida de Carga (Método Indirecto) Esto se calculará por separado para cada punto de carga restando de la pérdida total aparente, la pérdida del estátor a la temperatura de la prueba, la pérdida del núcleo, la pérdida por fricción, y la pérdida del rotor correspondiente a la medida valor de deslizamiento.
3.1.3.
Correcciones 3.1.3.1.
Corrección de Temperatura de la Pérdida en el Estátor Para la corrección de la pérdida en el estátor, se debe analizar para cada punto de carga, utilizando la resistencia promedio del estátor considerando la temperatura especificada.
Rb = Ra (tb + k1) ta + k1 (3.7) donde: Ra Es el valor conocido de la resistencia del devanado, en ohmios, a la temperatura ta ta Es la temperatura, en ◦C, del bobinado cuando se midió la resistencia Ra tb Es la temperatura, en ◦C, a la que se debe corregir la resistencia.
Rb Es la resistencia del devanado, en ohmios, corregida a la temperatura tb k1 Es una constante con un valor de 234,5 para cobre de conductividad 100 IACS, o un valor de 225 para aluminio, basado en una conductividad de volumen de 62.
3.1.3.2.
Corrección de Temperatura de la Pérdida en el Rotor La corrección de la pérdida del rotor se puede realizar mediante las ecuaciones de pérdida del rotor en configuración de motor o generador, utilizando el valor de deslizamiento a la temperatura especificada, utilizando la expresión 3.8 para el deslizamiento.
ss = st (ts + k1) tt + k1 (3.8) 24 donde: ss Es el deslizamiento, en p.u., corregido a la temperatura del estátor especificada, ts st Es el deslizamiento, en p.u., medido a la temperatura del bobinado del estátor, tt ts Es la temperatura especificada para la corrección de resistencia, en ◦C tt Es la temperatura de bobinado del estátor observada durante la prueba de carga, en ◦C.
k1 Es 234.5 para cobre de conductividad 100 % IACS, o 225 para aluminio, basado en una conductividad de volumen de 62 %.
3.1.3.3.
Pérdida Total Corregida Una pérdida total corregida para cada uno de los puntos de carga se determina, como la suma de la pérdida por fricción, la pérdida del núcleo, la pérdida corregida de carga, la pérdida del estátor corregida y la pérdida corregida del rotor.
3.1.3.4.
Potencia Mecánica Corregida La potencia mecánica (salida) corregida para cada uno de los puntos de carga para un motor es igual a la diferencia entre la potencia eléctrica (entrada) medida y la pérdida total corregida.
3.1.4.
Eficiencia Mediante Norma IEEE 112 Método B La eficiencia es la relación entre la potencia de salida y la potencia de entrada total.
Se usa la potencia eléctrica medida y la potencia mecánica corregida para calcular la eficiencia. La potencia de salida es igual a la potencia de entrada menos las pérdidas.
Por lo tanto, si se conocen dos de las tres variables (salida, entrada o pérdidas), la eficiencia puede determinarse usando las expresiones 3.9 y 3.10: η = Pout Pin (3.9) η = Pin −P ´ erdidas Pin (3.10) 3.1.4.1.
Factor de Potencia El factor de potencia de la máquina se determinará para cada punto de carga utili-zando la siguiente expresión 3.11: 25 PF = P √ 3V I (3.11) donde: P Es la potencia eléctrica de la máquina, en W, la potencia de entrada para un motor.
PF Es el factor de potencia de la máquina.
V Es el voltaje de línea a línea de entrada, en voltios.
I Es la corriente de entrada, en A.
Cuando se usa el método de dos vatímetros para medir la potencia de entrada de las máquinas trifásica, el factor de potencia, PF, en porcentaje, puede verificarse mediante la expresión 3.12: PF = 100 r 1 + 3 P1−P2 P1+P2 2 (3.12) donde: P1Es el registro de medida más alto.
P2 Es el registro de medida más bajo.
Si el vatímetro P2 da una lectura negativa, se considerará una cantidad negativa.
3.2.
Norma IEC 60034-2-1 Esta norma permite determinar las pérdidas en el rendimiento de las máquinas, a excepción de las máquinas destinadas a la tracción de vehículos .
La norma presenta una gran similitud con la norma IEEE 112. Las principales diferencias entre estas dos normas se enfocan en como determinar la estimación de las pérdidas en el hierro, el tratamiento del efecto de la temperatura sobre la resistencia en la prueba de carga, y la consideración de dos nuevos métodos para estimar las pérdidas adicionales .
3.2.1.
Aspectos Generales El procedimiento para la aplicación parte de la realización de las mismas pruebas ya realizadas con el método IEEE 112-B. Se realiza una prueba de temperatura, prueba de carga y prueba de vacío, a partir de las que se estimarán las pérdidas presentes en la máquina. Las pérdidas por fricción, las pérdidas en el estátor y las del rotor se calculan mediante las siguientes formulas: 26 P0 Total = P0 −Pj = Pf + Pfe (3.13) donde: P0 Total Es las pérdidas totales en vació P0 Es la potencia en vació Pj Es las pérdidas de joule en vació Pf Es las pérdidas por fricción Pfe Es las pérdidas en el hierro Pe = 1,5Ri2 (3.14) Pe Es las pérdidas en el estátor R Es la resistencia del estátor i Es la corriente del estátor bajo carga Pr = (Pin −Pe −Pfe)s (3.15) Pr Es las perdidas en el rotor Pin Es la potencia de entrada Pe Es las pérdidas en el estátor s Es el deslizamiento Pfe Es las pérdidas en el hierro A comparación de la norma IEEE 112 que considera, las pérdidas en el hierro son contantes para cualquier punto de carga del motor, la norma IEC 600034-2-1 corrige estas pérdidas en función del cuadrado de la tensión aplicada sobre la rama de magnetización del circuito equivalente de la máquina de inducción, el cálculo de esta tensión se puede estimar mediante la siguiente expresión 3.16: Vr = v u u t V − √ 3 2 i R cos (ϕ) !2 − √ 3 2 i R sin (ϕ) !2 (3.16) donde: Vr Es la tensión en la rama de magnetización para el ajuste de las pérdidas de fricción Pf V Es la tensión en bornes de la máquina en cada punto de carga ϕ Es el ángulo que determina el factor de potencia Se debe registra el valor da la resistencia del devanado del estátor antes del punto de carga máxima, cuando la maquina esta energizada y luego realizar otro registro de la resistencia cuando la máquina esta apagada. No es necesario considerar la temperatura promedio .
27 3.2.2.
Pérdidas Adicionales La norma actual sustituye a su antigua versión del año de 1972, donde sus versiones anteriores tiene modificaciones en la manera de obtener las perdidas adicionales.
Las versiones anteriores de esta norma estimaban estas pérdidas en un 0.5 % de la potencia nominal absorbida, la norma menciona diversos métodos para estimar estas pérdidas, estos métodos son: A partir de las pérdidas totales aparentes Asignación de pérdidas Ensayo Eh-Star 3.2.2.1.
Método a Partir de las Pérdidas Totales Aparentes Este método es similar al que se explica en la norma IEEE 112 método B, mediante la prueba de vacío y la de carga se pueden determinar diversas pérdidas adicionales al aplicar la siguiente expresión 3.17.
PSLL = Pe −Pr −(Pje + Pjr + Pfe + Pf) (3.17) donde: Pf Pérdidas por fricción PSLL Pérdidas adicionales Pfe Pérdidas en el hierro Pje Pérdidas de joule en el estátor Pjr Pérdidas de joule en el rotor 3.2.2.2.
Método de Asignación de Pérdidas Los valores de las pérdidas adicionales en el punto de carga nominal se pueden asignar como un porcentaje de la potencia de entrada de la máquina, mediante la utilización de la figura 3.1: Los valores de la curva pueden expresarse mediante las siguientes ecuaciones: PSLL = 0,025 Pe Pr ≤1 kW PSLL = Pe 0,025 −0,005 log10 Pr 1 kW 1 kW ≤Pr ≤10000 kW PSLL = 0,005 Pe Pr ≥10000 kW 28 Figura 3.1: Curva para Determinar las Pérdidas Adicionales por Asignación Cuando se consideran cargas diferentes a la carga nominal de la máquina, las pér-didas adicionales varían cuadráticamente con la corriente absorbida en el estátor, para cada punto de carga, menos la corriente de vacío al cuadrado como se muestra en la expresión 3.18: PSLL = C (i −i0)2 (3.18) La pendiente C se determina mediante la asignación de pérdidas en el punto nominal.
3.2.2.3.
Método por Ensayo de Eh-Star Este método presenta una metodología distinta a lo mencionado anteriormente, con lo cual su objetivo es lograr un desequilibro de las fases del motor, primero se conecta dos de las tres fases del motor a la misma línea de alimentación, y una de estas fases conectarla a través de una resistencia, de esta manera se consigue que por una de las fases del equipo circule una corriente superior a la de las otras fases. En la figura 4.1se observa la configuración descrita previamente para este método.
Figura 3.2: Esquema de Conexión para el Método por Ensayo de Eh-Star 29 Para la obtención de las pérdidas adicionales se aplica un proceso matemático a los datos obtenidos durante el ensayo, este proceso esta descrito en el apéndice B de la norma, dando como resultado una recta de pérdidas residuales. Esto también se puede determinar con el cuadrado la corriente absorbida, y la corriente de prueba como se indica en la expresión 3.19: PLR = A Ii(2) It !2 (3.19) Ii(2) Corriente adsorbida It Corriente de prueba 3.2.3.
Corrección de las Pérdidas Esto se realiza mediante una corrección por temperatura a las pérdidas de los de-vanados del estátor y del rotor, tomando la temperatura específica de la máquina, adquirida en la prueba de carga nominal.
Se realiza la corrección de las pérdidas adicionales usando la pendiente (A) conse-guida en el método de las pérdidas totales aparentes y Eh-Star, desplazando la curva que cruza con el origen como se muestra en la expresión 3.20: PSLL = A T 2 eje (3.20) El método por asignación de pérdidas obtiene el valor a emplear en la estimación final de eficiencia sin necesidad de realizar alguna corrección.
3.2.4.
Eficiencia Mediante Norma IEC 60034-2-1 Finalmente se estima la eficiencia mediante la expresión 3.21.
η = Pe −Pperd Pe (3.21) donde: Pperd = Pf + Pfe + PSLL + Pje + Pjr 30 4 Eficiencia Mediante Técnicas no Invasivas Este trabajo propone un método basado en el modelo de la máquina de inducción en cuál se estima cada uno de sus parámetros con el fin de determinar su eficiencia. Para lograrlo es necesario utilizar la impedancia instantánea definida como − → zin = − → ve/− → ie o la potencia instantánea − → sin = − → ve − → ie obtenidas en el marco de referencia fijo del estátor durante el arranque de la máquina. Por otra parte, la aplicación del método propuesto requiere adquirir los valores de corriente, tensión, calcular los enlaces de flujo magnético, obtener el par eléctrico de la máquina y su velocidad, con el objetivo de modelar la máquina de inducción en estudio y así, realizar una estimación de su eficiencia..
4.1.
Adquisición de Datos Para la adquisición de datos se utilizó sondas diferenciales de tensión y corriente conectadas a un dispositivo DAQ. Los datos obtenidos son almacenados en una computadora mediante el software Matlab para ser procesados y posteriormente realizar la estimación de parámetros.
En la figura 4.1 se muestra el esquema de conexiones que se utilizó. La máquina de inducción es alimentada a través de una red equilibrada de 220V entre fase y fase, con una frecuencia de 60Hz.
Durante la adquisición de los datos de tensión y corriente de la máquina de inducción se obtuvieron cerca de 240,000 muestras durante un tiempo aproximado de 10 s. El tiempo de muestreo debe garantizar que se realice una captura de información desde el arranque hasta la estabilización de la máquina en sus valores nominales.
En la figura4.2 se muestra el montaje de todos los equipos necesarios para la adqui-sición de datos de la máquina de inducción.
En cuanto al código utilizado para la adquisición de los datos fue necesario realizar varias modificaciones para evitar la saturación de información y un procesamiento erróneo.
Figura 4.1: Esquema de Conexión para Mediciones 31 Figura 4.2: Conexión de los Instrumentos para la Adquisición de Datos 2.7 2.75 2.8 2.85 2.9 2.95 3 Teimpo (s) -100 -50 0 50 Tensión (V) Tensión Linea a Linea Vab 2.7 2.75 2.8 2.85 2.9 2.95 3 Tiempo (s) -100 -50 0 50 100 Tensión (V) Tensión Linea a Linea Vbc 2.7 2.75 2.8 2.85 2.9 2.95 3 Tiempo (s) -100 -50 0 50 100 Tensión (V) Tensión Linea a Linea Vca Figura 4.3: Tensiones Línea a Línea de la Máquina de Inducción En las figuras 4.3 y 4.4 se muestran los datos obtenidos en la adquisición. Se pue-den verificar las magnitudes de tensiones y corrientes, teniendo en cuenta que esa información aún requiere ser escalada en función de los valores de atenuación de las sondas diferenciales. Los valores instantaneos obtenidos fueron entre fase y fase para las tensiones, y de línea para las corrientes.
Se puede observar que los valores obtenidos requieren un ajuste antes de ser utiliza-dos en el proceso. Las gráficas muestran un desplazamiento en el eje de las ordenadas que debe ser eliminado, así como un rango de valores previos al arranque que son innecesarios y que pueden generar errores.
4.2.
Procesamiento de Datos Una vez adquirido los datos de tensión y corriente de la máquina de inducción , estos son ajustados para obtener los valores reales según las modificaciones realizadas para poder ser almacenados.
32 7.5 7.6 7.7 7.8 7.9 8 8.1 8.2 Tiempo (s) -40 -20 0 20 40 Corriente (A) Corriente de la Fase Ia 7.6 7.7 7.8 7.9 8 8.1 8.2 8.3 Tiempo (s) -20 0 20 Corriente (A) Corriente de la Fase Ib 7.5 7.6 7.7 7.8 7.9 8 8.1 8.2 Tiempo (s) -40 -20 0 20 40 Corriente (A) Corriente de la Fase Ic Figura 4.4: Corriente de Línea de la Máquina de Inducción En primera instancia los datos al ser adquiridos mediante sondas diferenciales pre-sentan un desplazamiento con respecto al eje de las ordenadas. Para eliminar este desplazamiento con respecto al cero, se calcula el valor medio de las tensiones y corrientes, restarlo este valor medio del vector de datos adquiridos. De esta manera se obtiene los valores de tensión y corriente centrados en el eje de las ordenadas.
En el proceso de adquisición se capturó las tensiones vab y vcb como se muestra en la figura 4.1. Con estas dos tensiones se puede obtener la tercera tensión entre fases vca mediante la expresión 4.1: vab + vbc + vca = 0 (4.1) Sin embargo, la obtención de información la tensión vcb fue capturada en sentido contrario a la tensión vab, por lo que: vcb = −vbc (4.2) Entonces, reemplazando los datos en la ecuación 4.2 y al despejar la tensión vca se tiene que: vca = vab −vbc (4.3) Posteriormente, se realiza un proceso similar para obtener la corriente ib por medio de los valores obtenidos de ia e ic.
ia + ib + ic = 0 (4.4) 33 en donde, ib = −ic −ia (4.5) Luego de calcular los valores de tensión y corriente faltantes, se procede a ajustar la escala de las mediciones, multiplicando por el factor de atenuación de las sondas di-ferenciales utilizadas en la medición. En el caso de las sondas diferenciales de tensión se tiene un factor de atenuación de x100, mientras que en las sondas diferenciales de corriente el factor es de x10. Un factor de escalamiento adicional es necesario en los valores de corriente, debido a que el conductor que pasa por la sonda de medición hace 5 vueltas en el equipo, generando una atenuación extra con un factor de x5.
El siguiente paso luego de ajustar los valores medidos consiste en generar los vectores espaciales tanto de tensión como de corriente. Para ello se utiliza las expresiones de transformación 4.6 y 4.7: − → ie = s 2 3 ia + ibej 2π 3 + icej 4π 3 = s 2 3 ia + ibα + icα2 (4.6) − → ve = s 2 3 va + vbej 2π 3 + vcej 4π 3 = s 2 3 va + vbα + vcα2 (4.7) La transformación de corrientes y tensiones a vectores espaciales presentada en las definiciones 4.6 y 4.7, permiten simplificar el procesamiento de los datos al repre-sentar las variables trifásicas como un vector de magnitud y ángulo variable con el tiempo. Al realizar la transformación a vectores espaciales se puede observar una representación gráfica con valores complejos. En la figura 4.5 se representa el vector espacial de la corriente del estátor durante el proceso de arranque de la máquina de inducción.
Debido a que las tensiones adquiridas son valores instantáneos de línea a línea, y la definición del vector espacial de tensión definida en la expresión 4.7, utiliza los valores instantáneos de las tensiones trifásicas medios al neutro, se utiliza la siguiente expresión para determinar el vector espacial al neutro [15, 17]: − → ve = s 2 3 1 1 −α2 vab + vbcα + vcaα2 (4.8) En la Fig. 4.6 se presenta el vector espacial de la tensión en la máquina durante el proceso de arranque.
La determinación de los parámetros de la máquina de inducción requiere el cálculo del vector espacial del enlace de flujo que se determina como función del vector espacial de tensión y el vector espacial de corriente. Además se requiere el valor de resistencia del estátor Re que debe ser obtenido midiendo directamente la resistencia de las bobinas estatóricas.
34 Figura 4.5: Vector Espacial de Corriente Figura 4.6: Vector Espacial de Tensión 35 Figura 4.7: Datos de Placa de la Máquina de Inducción Cuadro 4.1: Parámetros de la máquina de inducción Parámetro Vbase Sbase Ibase ωbase e ωbase m Zbase Valor Real 220 3730 9, 79 377, 0 188, 5 12, 98 Unidad [V ] [W] [A] [ rad s ] [ rad s ] [Ω] Los datos nominales de la máquina de inducción tales como tensión nominal Vn, corriente nominal In, potencia nominal en el eje Pn, velocidad nominal nn y factor de potencia nominal fpn, pueden conseguirse directamente de la placa del motor como se observa en la Fig. 4.7. Para determinar las bases del sistema adimensional, se utiliza la tensión base y la potencia base: Ibase = Sbase √ 3Vbase ; Zbase = V 2 base Sbase ; ωbase = 2πnbase 60 (4.9) El tiempo base se obtiene considerando un ángulo base de 1 rad como: tbase = 1 rad ωbase ele (4.10) Para obtener un vector inicial de los parámetros de la máquina de inducción, se realizan ensayos tradicionales de vacío y rotor bloqueado. En el ensayo de vacío se arranca el motor sin carga hasta que alcanza una velocidad muy cercan a la velocidad sincrónica. En este punto de operación se miden las tensiones, corrientes y potencias a la entrada de la máquina y se determina mediante el modelo de régimen permanente los parámetros de la rama de magnetización. De forma similar, se realiza el ensayo de rotor bloqueado a tensión reducida para limitar la corriente del estátor, con la finalidad de obtener una aproximación a las resistencias y reactancias de la rama serie del modelo equivalente de la máquina de inducción. Los resultados obtenidos de las mediciones realizadas se muestran en los cuadros 4.2 y 4.3.
La determinación del vector espacial del enlace de flujo del estátor se obtiene direc-tamente integrando la fuerza electromotriz resultante: − → λ (t) = ˆ t0 0 − → ee (τ) dτ = ˆ t0 0 h− → ve(τ) −Re − → ie (τ) i dτ (4.11) 36 Cuadro 4.2: Datos de la Prueba en Vacío Par´ ametro V0 [V ] I0 [A] P0 [W] S0 [V A] Q0 [V AR] Xm [Ω] Rm [Ω] Re [Ω] V alor Real 213, 5 6, 15 104 2274, 18 2271, 8 20, 06 438, 33 0, 7 V alor Por Unidad 0, 9705 0, 6282 0, 0279 0, 6097 0, 6091 1, 5459 33, 77 0, 018 Cuadro 4.3: Datos de la Prueba a Rotor Bloqueado Par´ ametro Vrb [V ] Irb [A] Prb [W] Srb [V A] Qrb [V AR] XσT [Ω] RT [Ω] Rr [Ω] Lσr [Ω] Lσe [Ω] V alor Real 25, 98 13, 91 375, 8 625, 95 500, 58 1, 7523 0, 6477 0, 414 0, 8761 0, 8761 V alor Por Unidad 0, 2045 1, 42 0, 1 0, 2907 0, 2726 0, 135 0, 049 0, 0319 0, 0675 0, 0675 Donde − → ve y − → ie son los vectores espaciales obtenidos previamente mediante las ex-presiones 4.8 y 4.6. Como se puede observar en la Fig. 4.8el vector del enlace flujo magnético de la máquina de inducción presenta un desplazamiento debido a las con-diciones iniciales de la integración . Para solucionar este inconveniente se calculó el total de periodos completos que existían en el muestreo de las variables y se in-tegró uno a uno, suprimiendo el desplazamiento correspondiente a cada ciclo. Para esto se determinó el valor medio de la integración encada periodo y se restó dicho valor del resultado de integración anterior. De esta manera se logró un centrado de la forma circular que representa el vector espacial del enlace de flujo tal como se aprecia en la figura 4.8.
Con los valores del enlace del flujo magnético del estátor obtenidos mediante la expresión 4.11 y el valor del vector de corriente − → ie determinado previamente, se puede obtener el par eléctrico de la máquina, mediante la expresión 4.12: Te = p (− → λe · − → ie ) (4.12) donde p es el número de pares de polos de la máquina de inducción.
En la figura 4.9 se muestra el par eléctrico obtenido mediante la expresión 4.12. Se observa el par de arranque al inicio, el par máximo, el par en vacío y las oscilacio-(a) Sin centrar (b) Centrada Figura 4.8: Vector espacial del enlace de flujo del estator 37 0 1 2 3 4 5 6 7 8 Tiempo (s) -1 0 1 2 3 4 5 Newton-metro (N m) Par Electrico Figura 4.9: Par Eléctrico de la Máquina de Inducción nes de alta frecuencia originadas por el ruido de alta frecuencia en las corrientes estatóricas .
Durante el arranque del motor, la aceleración depende, por una parte de la diferencia entre el par eléctrico y el par mecánico, y por otra de la inercia del sistema rotativo.
La determinación del par eléctrico se puede realizar mediante el valor de la corriente del estátor y el enlace de flujo magnético de la máquina determinado mediante la expresión 4.12. El arranque de la máquina en vacío tiene como par mecánico las pérdidas por fricción, las cuales se pueden representar aproximadamente mediante una función lineal de la velocidad angular ωm Tm = k ωm (4.13) Te −Tm = J pωm = ⇒Te = J pωm + k ωm (4.14) La velocidad angular se puede estimar mediante la integración del par de aceleración (Te −Tm) durante el arranque de la máquina en la expresión 4.14, que para la condición de vacío corresponde al par eléctrico Te. Cuando el proceso de arranque de la máquina alcanza un estado estable, la máquina alcanza una velocidad angular constante, por lo que el par neto es cero [20, 21]. En ese punto el par eléctrico y el par de pérdidas son iguales. Mediante esta condición se puede determinar el par de aceleración, para cualquier instante (t ≥t0). La integración del par eléctrico da como resultado una función rampa en el tiempo, con una pendiente constante k ωs, y el complemento es la velocidad angular síncrona J ws, donde J es la inercia y k es la fricción. Este procedimiento se puede observar en la Fig. 4.10.
38 0 1 2 3 4 5 6 7 8 Tiempo (s) -2 0 2 4 6 8 10 12 14 Int (Te) Integral del Par Eléctrico Figura 4.10: Integración del Par Eléctrico Mediante la condición de arranque de la máquina de inducción en vacío, se considera la velocidad de estado estacionario ωm muy cercana a la velocidad síncrona ωs, ωm ≈ωs.
Uno de los problemas que se presentan al momento de determinar la velocidad an-gular de la máquina mediante los parámetros calculados anteriormente, es la función rampa que se obtiene al momento de integrar el par eléctrico. Esta determinación requiere obtener los coeficientes de fricción y la inercia de la máquina.
Caracterizando la rampa que se obtiene a partir de la integración del par eléctrico se puede encontrar la constante de fricción k, que determina dicha función. Para obtener el valor del coeficiente de fricción k se parte de la linealización de la rampa rampa con el fin de obtener dos valores de esta función para determinar el modelo de la carga debida a la fricción. Una vez obtenido el modelo de la carga de fricción se resta al par integrado para poder obtener mediante integración numérica el va-lor instantáneo de la velocidad angular de la máquina durante el arranque. En la Fig. 4.11, se puede observar la estimación de la velocidad de la máquina obtenida mediante el método de Euler, la cual se representó en valores de revoluciones por minuto .
Una vez determinadas las variables antes mencionadas se puede realizar la optimiza-ción de la función de costos de impedancia o potencia instantánea para determinar los valores de los parámetros que minimicen estas funciones de costo.
4.3.
Optimización de parámetros Para determinar los parámetros del modelo de la máquina de inducción es con-veniente utilizar el sistema adimensional de unidades, para lo cual se escalan las 39 0 1 2 3 4 5 6 7 8 Tiempo (s) -200 0 200 400 600 800 1000 1200 1400 1600 1800 W (rpm) Velocidad Calculada Figura 4.11: Velocidad estimada mediante el Método de Euler en rpm diferentes variables vectoriales utilizando las bases definidas en la Tabla 4.1. Los vectores escalados son: Vector espacial de tensión Vector espacial de corriente del estátor Vector espacial del enlace de flujo magnético del estátor Par eléctrico de la máquina Velocidad angular de la máquina Primera derivada del vector espacial de la corriente del estátor Vector tiempo La conveniencia de utilizar el sistema en por unidad para la determinación de la función de costos reside en que los parámetros en este sistema se encuentran en valores muy cercanos unos a otros y se pueden comparar los errores debidos a cada uno de ellos . Las bases utilizadas en este sistema se han reflejado en la Tabla 4.1.
En las Figuras 4.12 a 4.16 se han representado los vectores espaciales escalados con las bases correspondientes de la máquina de inducción.
Otro de los parámetros necesarios para la optimización de valores, es la primera derivada del vector espacial de la corriente del estátor. Para obtener este vector, se empleó un filtro denominado (Savitzky-Golay) , que es un filtro con capacidades derivativas muy eficientes. Este filtro dependiendo de los parámetros de orden del filtro y el tamaño de la venta que se le asignen, obtiene los coeficientes necesarios para filtrar la función y sus n derivadas.
En este trabajo se seleccionó un filtro (Savitzky-Golay) de orden 5, con un tamaño de ventana de 501 muestras, debido a que el vector espacial de corriente tiene 240,001 valores. El vector espacial de la corriente tiene parte real e imaginaria, el filtrado y la determinación de la derivada se realiza tanto para la parte real como para la parte imaginaria de esta señal.
40 Figura 4.12: Vector espacial de tensión en pu Figura 4.13: Vector espacial de corriente en pu 41 Figura 4.14: Vector espacial del enlace de flujo magnético en pu 0 500 1000 1500 2000 2500 3000 Tiempo (s) -0.05 0 0.05 0.1 0.15 0.2 0.25 Par en pu Par Electrico en pu Figura 4.15: Par eléctrico en pu 42 0 500 1000 1500 2000 2500 3000 Tiempo pu 0 0.2 0.4 0.6 0.8 1 w (pu) Velocidad Calculada Figura 4.16: Velocidad angular de la máquina en pu 0 500 1000 1500 2000 2500 3000 Real (A [pu]) -4 -3 -2 -1 0 1 2 3 4 Imaginario (A [pu]) Primera Derivada del Vector Espacial de Corriente en pu Figura 4.17: Primera derivada del vector espacial de la corriente del estator en pu 43 Cuadro 4.4: Valores extremos de los parámetros para el algoritmo de optimización Parámetros pu Re Rr Ldis e Ldis r Lm Límite inferior 0,018 0,01 0,01 0,01 1,4 Límite superior 0,018 0,05 0,1 0,1 1,8 Cuadro 4.5: Resultados del proceso de optimización Parámetro Le Lr Lm Rr Ψ (costo) − − → Zent 0, 081185 0, 010529 1, 4034 0,049968 0,4 % − − → Sent 0, 081185 0, 010529 1, 4034 0,049968 0,4 % El método propuesto para la optimización de parámetros de la máquina de inducción que definen el circuito equivalente, se fundamenta en los indicadores de impedancia o potencia instantánea a la entrada del convertidor electromecánico. Para determinar los indicadores de la impedancia de entrada se reducen las corrientes del rotor a variables medibles desde el estátor mediante la formulación siguiente [15, 24]: − − → Zent = − → Ve − → ie = Re + Rr Le Lr −jnpwm c Le + c Le p− → ie − → ie − Re Lr −jnpwm − → λ − → ie (4.15) − − → Sent = − → Ve − → ie ∗= Re + Rr Le Lr −jnpwm c Le − → ie 2 + c Lep− → ie − → ie ∗− Re Lr −jnpwm − → λ − → ie ∗ (4.16) donde c Le ≡Le −M2 Lr ..
De esta manera se podrán determinar los parámetros de la máquina de inducción, mediante un algoritmo de optimización con restricciones, disponible en Matlab, que utiliza el método del punto interior . A este algoritmo se lo alimenta con condi-ciones iniciales de los parámetros obtenidos mediante la prueba de vació y de rotor bloqueado. El rango de valores define restricciones en los parámetros para obtener modelos que se acerquen a las representaciones clásicas de la máquina de inducción.
Estos valores límite superior e inferior se especifican en el cuadro 4.4.
Los valores obtenidos mediante la prueba de vació y de rotor bloqueado, que se muestran en los cuadros 4.2 y 4.3se utilizaron como semilla en el proceso de optimi-zación.
Luego del proceso de optimización se obtuvieron los parámetros que se muestran en el cuadro 4.5.
Con los valores obtenidos en la optimización se modela la máquina de inducción, para posteriormente estimar la eficiencia. En la figura4.18 se muestra el modelo obtenido con los parámetros resultantes de la optimización.
44 Figura 4.18: Modelo obtenido de la máquina de inducción 4.4.
Estimación de la Eficiencia Para obtener una estimación de la eficiencia se requiere utilizar un método de análisis de circuitos que consiste en calcular un equivalente de Thévenin para calcular las corrientes de rotor y estátor. Estas corrientes a su vez se encuentran en función de la impedancia del rotor Zr (s), que depende del deslizamiento, y de la impedancia de magnetización Zm.
Para poder evaluar la eficiencia del convertidor, es conveniente transformar el modelo de la máquina mediante equivalentes de Thèvenin. La tensión de Thévenin y la impedancia de Thévenin, vista desde el rotor, se determinan mediante las siguientes expresiones 4.17 y 4.18: Vth = Zm Ze + Zm Ve (4.17) Zth = ZeZm Ze + Zm + Zr = Rth + jXth (4.18) Para el cálculo de la eficiencia debe evaluarse la potencia de entrada y la potencia de salida del convertidor electromecánico. La potencia se encuentra en función de la corriente del rotor, la resistencia del rotor y el deslizamiento, mientras que la potencia de entrada se calcula superponiendo la potencia de salida con la totalidad de las pérdidas en la máquina. Las pérdidas de la máquina se dividen en: pérdidas del rotor, pérdidas mecánicas y pérdidas en el estátor. Estas pérdidas se pueden calcular mediante las expresiones 4.19 y 4.20, si se considera que las pérdidas mecánicas están incluidas en las pérdidas de vacío: PRr = I2 r Rr (4.19) Pe = I2 e + V 2 m Rm (4.20) 45 Deslizamiento 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 Eficicencia 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Ve = 0.9 pu Ve =1.0 pu Ve = 1.1 pu Figura 4.19: Eficiencia estimada de la máquina de inducción para diversas tensiones de alimentación En el cálculo de las pérdidas en el estátor, el valor de Vm es la tensión en la rama de magnetización y se puede calcular mediante la expresión 4.21: Vm = Ve −ZeIe (4.21) Finalmente, la eficiencia se calcula utilizando la expresión 4.22: η = Psal Pent = Psal Psal + PRr + Pe 1 1 + PRr+Pe Psal (4.22) donde: Psal = I2 r Rr s (1 −s) (4.23) En la figura 4.19 se puede observar la eficiencia en función del deslizamiento s de la máquina de inducción para diferentes tensiones de entrada, considerando pérdidas mecánicas de 5 %.
4.5.
Eficiencia Estimada con Desequilibrios de Tensión Un fenómeno habitual en los sistemas industriales es el desequilibrio en las tensiones.
Este fenómeno afecta considerablemente la eficiencia de la máquina de inducción porque al descomponer el desequilibrios en sus componentes de secuencia positiva y negativa, se observa que esta última produce corrientes elevadas en el rotor y en el estator de la máquina que reducen su rendimiento .
46 (a) Secuencia Positiva (b) Secuencia Negativa (c) Secuencia Cero Figura 4.20: Circuito equivalente de la MI con desequilibrios El modelo de la máquina de inducción considerando desequilibrios de tensión se estudia en , y se resume en la superposición del circuito equivalente que se muestra en la Fig. 4.20. La determinación de las tensiones de secuencia, requiere la aplicación de las componentes simétricas para, considerar el comportamiento de la máquina durante este fenómeno.
Para la máquina de inducción que se utilizó en este trabajo, es posible medir las magnitudes de sus tensiones entre fases (vab , vbc , vca). Considerando los fundamen-tos del teorema de Apolonio, es posible determinar los valores de las tensiones de cada una de sus fases al neutro (va , vb , vc). Además, con el teorema de senos y cosenos se determina los ángulos internos del triángulo (α , β , γ), que se muestra en la figura 4.21. Estos ángulos son los desfases de las tensiones referidas al neutro .
Mediante las expresiones 2.26 y 2.27, es posible encontrar las expresiones fasoriales de línea a neutro.
Cuando existen desequilibrios en las tensiones aplicadas a la máquina, es necesario utilizar modelos de secuencia y componentes simetrías. Al aplicar la transformación de componentes simétricas conservativa en potencia, de la expresiones 2.27 y 2.26 de tensiones referidas al neutro y ángulos dentro del triángulo, se obtiene los valores de tensión de secuencia positiva, negativa y cero, como se muestra en la expresión 4.24: 47 Figura 4.21: Triángulo Desequilibrado de Tensiones Trifásicas Figura 4.22: Equivalente de Thèvenin v0 v1 v2 = 1 √ 3 1 1 1 1 α α2 1 α2 α va vb vc = 1 √ 3 1 1 1 1 α α2 1 α2 α va vbej( α+β 2 −π) vcej(α+ β+γ 2 −2π) = 0 v1ejϕ1 v2ejϕ2 (4.24) En la figura 4.22se observa el circuito equivalente Thèvenin del modelo clásico de la máquina, donde es posible determinar directamente la corriente del rotor, esta corriente es necesaria para poder evaluar el par eléctrico entregado al eje de la máquina.
La resistencia del rotor es el parámetro más importante del modelo, porque de-termina la característica par-deslizamiento en todas las velocidades cercanas a la velocidad síncrona, dentro del fenómeno de desequilibrio de tensión. Es importan-te considerar que en secuencia negativa la resistencia e inductancia del rotor debe ajustarse debido al efecto pelicular, como indica las expresiones[4.25,4.26,4.27].
Rr(s2) Rr(s1) = ξsinh(2ξ) + sin(2ξ) cosh(2ξ) −cos(2ξ) (4.25) Lσr(s2) Lσr(s1) = 3 2ξ sinh(2ξ) −sin(2ξ) cosh(2ξ) −cos(2ξ) (4.26) 48 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 Deslizamiento -0.2 0 0.2 0.4 0.6 0.8 1 Eficiencia Eficiencia Estimada con Desequilibrios de Tensión 0% Desequillibrio 6% Desequilibrio 10% Desequlibrio Figura 4.23: Eficiencia Estimada con Desequilibrios de Tensión ξ ≈hranura rs2ωeµ0σr 2 (4.27) donde hranura Es la profundidad de la ranura del rotor σr Es la conductividad de las barras del rotor Según la norma , el desequilibrio de tensión ocurre cuando los voltajes de línea, los voltajes entre fases o los ángulos de desfase de un sistema trifásico no son iguales.
Para determinar el porcentaje de desequilibrio que presente un sistema trifásicos es necesario aplicar la expresión 4.28: Vdesequilibrio = |V2| |V1| (4.28) donde V1 Voltaje de secuencia positiva V2 Voltaje de secuencia negativa Para este trabajo se realiza un análisis de desequilibrio de tensión con (0, 6 % y 10 %) de desequilibrio en el sistema trifásico suministrado, con la finalidad de obtener el comportamiento de la eficiencia de la máquina frente a este fenómeno.
En la figura 4.23 se observa el comportamiento de la máquina de inducción en tres situaciones distintas de desbalances de tensión. Se observa que la disminución en la eficiencia de la máquina es significativa cuando existen desequilibrios en la red de alimentación. De manera analítica se puede decir que la eficiencia de la máquina disminuye debido al par negativo que se obtiene del circuito de secuencia negativa, provocando que la eficiencia total estimada disminuya significativamente en función del nivel de desequilibrio de tensión que se presente en la red de alimentación de la máquina.
49 (a) Modelo de Secuencia Positiva y Negativa (b) Modelo de Secuencia Cero Figura 4.24: Modelo Armónico de la Máquina de Inducción 4.6.
Eficiencia considerando armónicas En el análisis de la eficiencia de la máquina de inducción, considerando el suministro de tensiones no sinusoidales, que se aplican al convertidor. Es importante descompo-ner esta señal suministrada, mediante la transformada de Fourier . En la figura 4.24 se muestra el modelo de la máquina de inducción sometida a este fenómeno.
En el cálculo del par eléctrico se considera la superposición de los pares armónicos.
Los armónicos en la red son de secuencia positiva, negativa como se indica en la expresión 4.29 : sh± = h±ωe −ωr h±ωe = h± ∓(1 −s1) h± (4.29) donde ( h+ = 6m + 1, m = 0, 1, 2, ...
h−= 6m −1, m = 0, 1, 2, ...
Te = n X h=0 ±Teh± = T1 −T5 + T7 −T11 + T13 −...
(4.30) Ie = v u u t n X h=0 I2 2h−1 = q I2 1 + I2 3 + I2 5 + I2 7 + I2 9 + I2 11 + I2 13 + ...
(4.31) Las armónicas múltiplos de tres son de secuencia cero, con lo cual no magnetizan la máquina, y no producen par eléctrico en la misma. Estos afectan en la determinación de la corriente efectiva, y su modelo solo se representa con la reactancia de dispersión y la resistencia del rotor. En la figura 4.24b, se indica el modelo para las armónicas múltiplos de tres.
Al realizar la transformada de Fourier de los valores obtenidos de la tensión con la que se alimentó la máquina de inducción se puede apreciar en la imagen 4.25 la contaminación armónica de dicha en red. En el caso estudiado, se puede observar la frecuencia fundamental, el tercer, quinto y séptico armónico.
En la imagen 4.26 se puede apreciar que la eficiencia obtenida luego de realizado el análisis con armónicos es prácticamente la misma que se obtuvo sin considerar los armónicos. El efecto que causan en la eficiencia de la máquina de inducción la consideración de las armónicas contenidas en la tensión de alimentación es mínimo.
50 50 100 150 200 250 300 350 400 450 Frecuencia (Hz) -60 -40 -20 0 20 40 Db Análisis de Fourier Frecuencia Fundamental 3ra Armónica 5ta Armónica 7ma Armónica Figura 4.25: Análisis de Fourier de la Tensión Las pérdidas causadas por los armónicos son muy pequeñas debido a que cada armónico analizado tiene una amplitud de tensión muy reducida. Por esta razón las corrientes en el circuito equivalente de tercero, quinto y séptimo armónico no generan pérdidas significativas que repercutan en la eficiencia total estimada de la máquina. La escala derecha de la Figura 4.26 permite observar las diferencias en el rendimiento al considerar el contenido armónico de la fuente de alimentación.
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 Deslizamiento 0 0.5 1 Eficiencia MI Eficiencia Estimada con Armónicas 0 0.01 0.02 Diferencia del Análisis con Armónicos Eficiencia Sin Armónicos Disminución de Eficiencia Figura 4.26: Eficiencia Estimada con Armónicas 51 5 Capitulo 5: Conclusiones y Recomendaciones 5.1.
Conclusiones La modelación realizad de la máquina de inducción es un método bastante preciso y poco invasivo para determinar la eficiencia al compararlo con los métodos tradi-cionales. Esta técnica requiere la adquisición de datos de arranque de la máquina sin el requerimiento de suspensión del servicio de este equipo. La estimación de efi-ciencia mediante este método permite ser una alternativa para realizar un análisis de un gran número de motores en una industria de una forma precisa y rápida, sin interrumpir los procesos de producción.
La optimización que permite estimar el valor de los parámetros al utilizar la fun-ción de costo presenta un error comparable con métodos tradicionales normalizados internacionalmente y es posible obtener con ellos la curva de eficiencia y el estado actual de operación de la máquina.
El método de estimación de eficiencia propuesto no requiere varias mediciones para el modelado de la máquina de inducción. Con una sola captura de información durante el arranque en vacío se puede realizar una estimación precisa.
El método permite de forma directa considerar tanto el efecto de los desequilibrios y las armónicas en las tensiones de la red.
5.2.
Recomendaciones Es recomendable verificar los valores obtenidos durante la adquisición para ajustar de manera adecuada los equipos que realizan este proceso, con la finalidad de evitar errores en el modelamiento, producidos por un valor de medición incorrecto. Además, es importante tener en cuenta la precisión de los equipos utilizados, así como la configuración y características individuales que requieran.
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190392 | https://www.youtube.com/watch?v=xQ9D4Jz95-A | The Maxwell–Boltzmann distribution | AP Chemistry | Khan Academy
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The Maxwell–Boltzmann distribution describes the distribution of speeds among the particles in a sample of gas at a given temperature. The distribution is often represented graphically, with particle speed on the x-axis and relative number of particles on the y-axis. View more lessons or practice this subject at
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[Voiceover] So let's think a little bit about the Maxwell-Boltzmann distribution. And this right over here, this is a picture of James Clerk Maxwell. And I really like this picture, it's with his wife Katherine Maxwell and I guess this is their dog. And James Maxwell, he is a titan of physics famous for Maxwell's equations. He also did some of the foundational work on color photography and he was involved in thinking about, "Well, what's the distribution of speeds of air particles of idealized gas particles?" And this gentleman over here, this is Ludwig Boltzmann. And he's considered the father or one of the founding fathers of statistical mechanics. And together, through the Maxwell-Boltzman distribution they didn't collaborate, but they independently came to the same distribution. They were able to describe, "Well, what's the distribution of the speeds of air particles?" So let's back up a little bit or let's just do a little bit of a thought experiment. So let's say that I have a container here. Let's say that I have a container here. And let's say it has air. And air is actually made up mostly of nitrogen. Let's just say it just has only nitrogen in it just to simplify things. So let me just draw some nitrogen molecules in there. And let's say that I have a thermometer. I put a thermometer in there. And the thermometer reads a temperature of 300 Kelvin. What does this temperature of 300 Kelvin mean? Well, in our everyday life, we have kind of a visceral sense of temperature. Hey, I don't wanna touch something that's hot. It's going to burn me. Or that cold thing, it's gonna make me shiver. And that's how our brain processes this thing called temperature. But what's actually going on at a molecular scale? Well, temperature, one way to think about temperature, this would be a very accurate way to think about temperature is that tempera- I'm spelling it wrong. Temperature is proportional to average kinetic energy of the molecules in that system. So let me write it this way. Temperature is proportional to average kinetic energy. Average kinetic energy in the system. I'll just write average kinetic energy. So let's make that a little bit more concrete. So let's say that I have two containers. So it's one container. Whoops. And two containers right over here. And let's say they have the same number of molecules of nitrogen gas And I'm just gonna draw 10 here. This obviously is not realistic you'd have many, many more molecules. One, two, three, four, five, six, seven, eight, nine, ten. One, two, three, four, five, six, seven, eight, nine, ten. And let's say we know that the temperature here is 300 Kelvin. So the temperature of this system is 300 Kelvin. And the temperature of this system is 200 Kelvin. So if I wanted to visualize what these molecules are doing they're all moving around, they're bumping they don't all move together in unison. The average kinetic energy of the molecules in this system is going to be higher. And so maybe you have this molecule is moving in that direction. So that's its velocity. This one has this velocity. This one's going there. This one might not be moving much at all. This one might be going really fast that way. This one might be going super fast that way. This is doing that. This is doing that. This is doing that. So if you were to now compare it to this system this system, you could still have a molecule that is going really fast. Maybe this molecule is going faster than any of the molecules over here. But on average, the molecules here have a lower kinetic energy. So this one maybe is doing this. I'm going to see if I can draw... On average, they're going to have a lower kinetic energy. That doesn't mean all of these molecules are necessarily slower than all of these molecules or have lower kinetic energy than all of these molecules. But on average they're going to have less kinetic energy. And we can actually draw a distribution. And this distribution, that is the Maxwell-Boltzmann distribution. So if we... Let me draw a little coordinate plane here. So let me draw a coordinate plane. So, if on this axis, I were to put speed. If I were to put speed. And on this axis, I would put number of molecules. Number of molecules. Right over here. For this system, the system that is at 300 Kelvin the distribution might look like this. So it might look the distribution... Let me do this in a new color. So, the distribution this is gonna be all of the molecules. The distribution might look like this. Might look like this. And this would actually be the Maxwell-Boltzmann distribution for this system For system, let's call this system A. System A, right over here. And this system, that has a lower temperature which means it also has a lower kinetic energy. The distribution of its particles... So the most likely, the most probable... You're going to have the highest number of molecules at a slower speed. Let's say you're gonna have it at this speed right over here. So its distribution might look something like this. So it might look something like that. Now why is this one... It might make sense to you that okay, the most probable the speed at which I have the most molecules I get that that's going to be lower than the speed at which I have the most molecules in system A because I have, because on average these things have less kinetic energy. They're going to have less speed. But why is this peak higher? Well, you gotta remember we're talking about the same number of molecules. So if we have the same number of molecules that means that the areas under these curves need to be the same. So if this one is narrower, it's going to be taller. And if I were gonna, if I were to somehow raise the temperature of this system even more. Let's say I create a third system or I get this or let's say I were to heat it up to 400 Kelvin. Well then my distribution would look something like this. So this is if I heated it up. Heated up. And so this is all the Maxwell-Boltzmann distribution is. I'm not giving you the more involved, hairy equation for it but really the idea of what it is. It's a pretty neat idea. And actually when you actually think about the actual speeds of some of these particles, even the air around you I'm gonna say, "Oh, it looks pretty stationary to me." But it turns out in the air around you is mostly nitrogen. That the most probable speed of if you picked a random nitrogen molecule around you right now. So the most probable speed. I'm gonna write this down 'cause this is pretty mindblowing. Most probable speed at room temperature. Probable speed of N2 at room temperature. Room temperature. So let's say this that this was the Maxwell-Boltzmann distribution for nitrogen at room temperature. Let's say that that's, let's say we make we call room temperature 300 Kelvin. This most probable speed right over here the one where we have the most molecules the one where we're gonna have the most molecules at that speed. In fact, guess what that is going to be before I tell you 'cause it's actually mind boggling. Well, it turns out that it is approximately 400, 400 and actually at 300 Kelvin it's gonna be 422 meters per second. 422 meters per second. Imagine something traveling 422 meters in a second. And if you're used to thinking in terms of miles per hour this is approximately 944 miles per hour. So right now, around you you have, actually the most probable, the highest number of the nitrogen molecules around you are traveling at roughly this speed and they're bumping into you. That's actually what's giving you air pressure. And not just that speed, there are actually ones that are travelling even faster than that. Even faster than 422 meters per second. Even faster. There's particles around you traveling faster than a thousand miles per hour and they are bumping into your body as we speak. And you might say, "Well, why doesn't that hurt?" Well, that gives you a sense of how small the mass of a nitrogen molecule is, that it can bump into you at a thousand miles per hour and you really don't feel it. It feels just like the ambient air pressure. Now, when you first look at this, you're like wait, 422 meters per second? That's faster than the speed of sound. The speed of sound is around 340 meters per second. Well, how can this be? Well, just think about it. Sound is transmitted through the air through collisions of particles. So the particles themselves have to be moving or at least some of them, have to be moving faster than the speed of sound. So, not all of the things around you are moving this fast and they're moving in all different directions. Some of them might not be moving much at all. But some of them are moving quite incredibly fast. So, I don't know, I find that a little bit mindblowing. |
190393 | https://www.education.com/worksheet/article/grammar-learning-analogies/ | Worksheet
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Symmetric sum
The symmetric sum of a function of variables is defined to be , where ranges over all permutations of .
More generally, a symmetric sum of variables is a sum that is unchanged by any permutation of its variables.
Any symmetric sum can be written as a polynomial of elementary symmetric sums.
A symmetric function of variables is a function that is unchanged by any permutation of its variables. The symmetric sum of a symmetric function therefore satisfies
Given variables and a symmetric function with , the notation is sometimes used to denote the sum of over all subsets of size in .
See also
Cyclic sum
Muirhead's Inequality
PaperMath’s sum
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190395 | https://www.youtube.com/watch?v=ShXFNOkjvBA | How to Find the 𝒍𝒂𝒓𝒈𝒆𝒔𝒕 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒊𝒏𝒕𝒆𝒈𝒆𝒓 𝒏 ,𝒊𝒇 𝒏^𝟑 + 𝟏𝟎𝟎 𝒊𝒔 𝒅𝒊𝒗𝒊𝒔𝒊𝒃𝒍𝒆 𝒃𝒚 𝒏 + 𝟏𝟎 | Math Olympiad |
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Posted: 12 Nov 2024
How to find 𝒕𝒉𝒆 𝒍𝒂𝒓𝒈𝒆𝒔𝒕 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒊𝒏𝒕𝒆𝒈𝒆𝒓 𝒏 , 𝒊𝒇 𝒏^𝟑 + 𝟏𝟎𝟎 𝒊𝒔 𝒅𝒊𝒗𝒊𝒔𝒊𝒃𝒍𝒆 𝒃𝒚 𝒏 + 𝟏𝟎 |
This concept is essential for students preparing for Math Olympiad, IIT-JEE, and other competitive exams. Learn how to approach such Diophantine equations step by step, breaking down the solution process with clear and concise explanations. This method can also be applied to similar problems in number theory.
Keywords:
AIME
Number Theory
LCM
GCD
Math Olympiad problem-solving
IIT JEE math preparation
Competitive exam math
Number theory solutions
Solving equations with integers
How to solve math Olympiad questions
JEE main math solutions
1 comments
Transcript:
Hello friends I'm pratima I welcome you all to my channel I thank all the subscribers for supporting me today we'll discuss a question based on number Theory what is the largest positive integer n such that NQ + 100 is divisible by n + 10 that means if use the ukids theorem then n + 10 is gcd of n q + 100 and n + 10 now let NQ + 100 equals to n + 10 MTI n² + a n + B plus remainder C now if we multiply then it will be NQ + 10 n² + a n² + 10 a n + b n + 10 b + C this equals to n q + 100 or this is NQ + n² 10 + a + n 10 a + B and + 10 b + C = to NQ + 100 now equating the coefficients get your and Square term is not there in the right hand side that means 10 + a = 0 second is there is no end term that means 10 time of a + b is z and 10 b + C = 100 now from equation 1 I'm getting a is equal to -10 and putting in equation 2 it will be - 100 + B = 0 this gives me B = to 100 and your, + C is 100 that means C = to -900 okay now I'm taking the equ theorem it says if C is your gcd of A and B then I'm applying the algorithm UK algorithm forers then C is also this D of a and the remainder we get when a divides B so here n + 10 is gcd of NQ + 100 and n + 10 or this is same as your n + 10 is gcd of n + 10 and the remainder remainder is see because we have taken in this way that this is multiple of n + 10 plus this is the remainer it will be n + 10 in the remainder is 900 or this is same as n + 10 is G CD of 900 100 and n + 10 now what will the largest value of n the maximum value of n + 10 must be your 900 to satisfy the above statement so n equals to your 900 - 10 that will be 890 so the maximum value of n for which m q + 100 is divisible by n + 10 is 890 thank you for watching please keep on watching and keep on learning |
190396 | https://www.spanishplayground.net/spanish-body-parts-activities/ | Spanish Body Parts Activities - Spanish Playground
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Spanish Body Parts Activities
August 19, 2018
Body parts, Crafts, Printable, Songs, Toys and Games
Body parts are some of the first words children learn. I do activities to specifically teach body parts in Spanish, and also include the vocabulary in the context of stories, games and other content. These are 20 of my favorite Spanish body parts activities to introduce or review these useful words.
Be sure to check out our other Spanish for Kids vocabulary activities arranged by theme. We have lots of ideas and materials for you to use in your Spanish lessons. Also, check out our Spanish vocabulary games
and keep kids playing and learning.
Spanish Body Parts Videos
Videos are a good way for kids to see and hear Spanish body parts used in context. This video has a funny premise – a scientist has build a robot and gives him commands. You can follow up by having kids do the activity themselves.
This is another fun video where Jorge talks about his dog Paco using body parts vocabulary.
This video uses lots of body parts with the verbs ¿Te duele/n? and Me duele/n. Poor Gabriel isn’t feeling good at all!
Movement Activities for Spanish Body Parts
It is natural combine body part vocabulary with actions. These activities get kids up and moving or let them manipulate objects to learn Spanish body parts.
Patterns
I introduce body parts with pointing, clapping hands and stomping feet as we say the words. I do a call and response of 3 words at a time, and we say them in a normal voice, whispering and then loudly.
Then, I tell children to listen and watch, and I make a pattern of 3 or 4 words. They repeat the pattern, for example,manos, manos, ojos, or boca, nariz, ojos, ojos.
I do this adding more vocabulary and making more complicated patterns. Kids like this game and soon they make the patterns and lead the group.
Bean Bags with Actions
Bean bags are a fun tool for teaching body parts. I combine them with verbs we are learning in movement songs. (You can find my Spanish favorite movement songs here.)
Each child has a bean bag, and I call out a body part, for example cabeza. Kids put the bean bag on their heads. Then, I call out an action, for example correr. Kids run and try to keep the bean bag on their head.
You can play this game with cards, one set for body parts and one for actions action. The cards add a little suspense to the game, and kids like to take turns turning over a card from each pile.
Simon Says
This classic game is a great way to review body parts in Spanish. Once students know the vocabulary, they can lead the game.
Assemble a Face or Body
Use cut-outs of body parts and work together to create figures or faces. This works well on a felt board. You can also give each child a paper plate and a set of shapes and pom poms to assemble a face. Model the process first for preschoolers, saying the parts of the face as a review.
Paintbrush Activity with I Ain’t Gonna Paint
I Ain’t Gonna Paint No More is a picture book in English, but I use it in with my preschoolers, telling a simple version of the story in Spanish. It is about a child who paints different parts of his body. His mamá is not pleased!
After we read the story, we use clean, dry brushes and pretend to “paint” ourselves as we say the words in Spanish.
Sticky Notes to Learn Spanish Body Parts
Putting sticky notes on yourself or a partner is a silly way to practice Spanish body parts. Older students can make labels and put them on, while younger ones can stick a note where you say.
I recently played this game at camp and the kids loved it, but there were a few who were relieved when we started taking the sticky notes off again.
Sana Sana Colita de Rana
This activity teaches body parts and also the traditional rhyme we say when a child is hurt. We use stuffed animals and simple felt bandaids, and the game is always a hit. Read how to use Sana sana colita de rana to teach Spanish body parts.
Spanish Body Part Songs
Songs are an important part of all my lessons, and learning body parts in Spanish is no exception.
You can find our10 favorite Spanish body parts songs here.
Games to Learn Body Parts in Spanish
I mentioned the classic game Simon Says above, but you can also play dice or board games to practice Spanish body parts.
Check out these 7 Spanish Body Parts Games. You’ll find board games at two levels, a roll-a-monster game, and more.
Spanish Body Parts Lotería
Lotería, or bingo, is an excellent game for learning any kind of vocabulary. I have 12 boards with basic Spanish body parts vocabulary as well as a few verbs (girar, bailar, saltar, caminar). The text is only on the call cards, so if you use my boards you can change the vocabulary to another word that works for the picture.
Lotería is a traditional game. Check out our complete collection of traditional Spanish games for kids and print traditional Mexican lotería cards.
Of course, you can make your own boards with the vocabulary you are learning.
Download Lotería – Las partes del cuerpo.
Dice Games
There are lots of roll-the-dice printable games for faces and figures online. I love the Roll-a-Picasso face games, and there are also roll-a-snowman and other versions. You can see a version of aroll-the-dice drawing game for more standard faces here. It can easily be adapted to Spanish.
In all these games, each number on the dice correspond to a body part and kids are trying to complete a figure or face. It’s a great way to practice body parts in Spanish!
Guess Who
This fun version ofGuess Who with fruit monsters is also a fun way to practice Spanish body parts.
Spanish Body Parts Crafts
There are many fun, simple crafts you can do and use lots of body part vocabulary in Spanish.
Contact Paper Mural
We made a Spanish body parts contact paper mural this summer in my preschool classes. It’s a wonderful group activity. I had cut out lots of people and animals from magazie, both whole and pictures of faces. Kids can do this part if you have more time.
We spread them out and talked about the photos first, and then they stuck them to contact paper I had put on the wall. It was super easy, provided the opporunity for tons of language, and they loved it.
Big Mouth Monsters
I use these big mouth monsters from It’s Always Autumn in my preschool classes. The kids color and add googly eyes, and they perfect for practicing ojos, boca, manos, pies.
Monster Stick Puppets
I also use these cute stick puppet monsters from Picklebums. I make an outline version and there is lots of opportunity to talk about body parts as kids color and add googly eyes.
Paper Bag Puppets
Paper bag puppets are another easy craft that lend themselves to using lots of body part vocabulary. We make a dog and a bear, and for both puppets kids add ojos, orejas, and nariz. There is la boca as part of the puppet, and the dog puppets get la lengua too.
People Figure Crafts to Learn Spanish Body Parts
We do a couple different crafts that let kids make a complete person figure. One of these is stick people made with real sticks. You do have to find the sticks, but that part is easy for me because we live in the country. You can break sticks to create the y-shape.
Paper Cut-Outs
I also use large cut-out people figures and kids decorate them. This craft is no-prep for me, and incorporates lots of Spanish body part vocabulary as kids present their person to the group.
Coloring Page
This is a simple coloring page with basic body parts labeled in Spanish. I made it as a reference for staff at camp this summer and sent it home for families.
Body Parts in Spanish – Coloring Page – Girl
Body Parts in Spanish – Coloring Page – Boy
Picture Books about Spanish Body Parts
Finally, there are great picture books to teach body parts. These are two of my go-to stories for preschoolers and the early grades
One of my favorites books is De la cabeza a los pies by Eric Carle because it incorporates movement, animals and body parts. I love books where kids can play along as we read!
We also read Soy demasiado grande (I’m too big). It is a bilingual book that has been around for years, but the language level is perfect. I still use it in my classes.
What are you favorite Spanish body parts activities? Please share in the comments below!
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190397 | https://math.stackexchange.com/questions/2134269/how-to-find-the-maximal-volume-of-a-rectangular-box-given-a-fixed-surface-area | Skip to main content
How to find the maximal volume of a rectangular box given a fixed surface area?
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I am asked to find the maximal volume of a rectangular box with a fixed surface area of 150. I would like to solve this problem using gradients/Lagrange multipliers. I have done a bit of work so far but I'm not sure if I am on the right track or how to complete the problem. I know that this box should be a cube, but I would like to show this without assuming it is true. Here is what I have so far:
Volume=V(l,w,h)=lwh
SurfaceArea=S(l,w,h)=2(lw+hw+hl).
▽V(l,w,h)=λ▽S(l,w,h)
▽V(l,w,h)=(wh)i^+(lh)j^+(wl)k^
▽S(l,w,h)=(2w+2h)i^+(2l+2h)j^+(2w+2l)k^
So putting this all together gives me:
(wh)i^+(lh)j^+(wl)k^=λ[(2w+2h)i^+(2l+2h)j^+(2w+2l)k^]
And at this point I am not sure how to conclude that w=h=l .
Additionally, I see that I can look at the system of equations:
wh=λ2(w+h)
lh=λ2(l+h)
wl=λ2(w+l)
2(lh+wh+lw)=150
But I am unsure of how to solve these.
multivariable-calculus
optimization
lagrange-multiplier
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edited Feb 8, 2017 at 2:06
MathStudent1324
asked Feb 8, 2017 at 1:58
MathStudent1324MathStudent1324
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Any reason why you want to solve it using Lagrange multipliers? This is a classic AM-GM problem.
– dxiv
Commented
Feb 8, 2017 at 2:09
1
You're almost home. From
(wh)i^+(lh)j^+(wl)k^=λ[(2w+2h)i^+(2l+2h)j^+(2w+2l)k^]
, you have wh=2(w+h), therefore V/λ=whl/λ=2(wl+hl), similarly, you can derive whl/λ=2(lw+hw), these two lead to hl=hw, l=w.
– Guangliang
Commented
Feb 8, 2017 at 2:09
@dxiv We are working on this technique in class right now but I can look into that as well.
– MathStudent1324
Commented
Feb 8, 2017 at 2:11
@Guangliang Thanks, that makes perfect sense!
– MathStudent1324
Commented
Feb 8, 2017 at 2:13
Once I see that h=w=l, I find that S(l,w,h)=2(lw+hw+lh)=2(3w2), 6w2=150, w=h=l=5 and so the maximal volume of this box equals 53=125
– MathStudent1324
Commented
Feb 8, 2017 at 2:39
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Hint (without Lagrange multipliers): by AM-GM 16S=13(lw+hw+hl)≥l2w2h2−−−−−−√3=V2−−−√3 with equality iff lw=wh=hl⟺l=w=h.
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edited Feb 8, 2017 at 2:51
answered Feb 8, 2017 at 2:14
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190398 | https://www.xconvert.com/unit-converter/cubic-centimeters-to-millilitres | Cubic Centimeters (cm3) to Millilitres (ml) conversion
Cubic Centimeters to Millilitres conversion table
| Cubic Centimeters (cm3) | Millilitres (ml) |
--- |
| 0 | 0 |
| 1 | 1 |
| 2 | 2 |
| 3 | 3 |
| 4 | 4 |
| 5 | 5 |
| 6 | 6 |
| 7 | 7 |
| 8 | 8 |
| 9 | 9 |
| 10 | 10 |
| 20 | 20 |
| 30 | 30 |
| 40 | 40 |
| 50 | 50 |
| 60 | 60 |
| 70 | 70 |
| 80 | 80 |
| 90 | 90 |
| 100 | 100 |
| 1000 | 1000 |
How to convert cubic centimeters to millilitres?
The conversion between cubic centimeters and milliliters is quite direct and fundamental in volume measurement. Here's a breakdown:
Understanding the Conversion
The relationship between cubic centimeters () and milliliters () is based on the metric system, where a milliliter is defined as the volume of one cubic centimeter. This makes the conversion exceptionally straightforward.
The Conversion Factor
The key to converting between cubic centimeters and milliliters is the following equivalence:
Step-by-Step Conversions
Converting Cubic Centimeters to Milliliters
Since , converting 1 cubic centimeter to milliliters is as simple as:
Converting Milliliters to Cubic Centimeters
Similarly, converting 1 milliliter to cubic centimeters is equally straightforward:
Interesting Facts and Historical Context
Origin: The metric system, on which these units are based, was developed in France in the late 18th century. A primary goal was to create a standardized and rational system of measurement based on decimal units, facilitating trade and scientific communication. BIPM - The International System of Units (SI)
Water and the Metric System: Originally, the gram was defined as the mass of one cubic centimeter of water at the temperature of melting ice. This relationship was carefully defined to provide a practical link between mass, volume, and the physical properties of water, one of the most abundant and important substances on Earth.
Real-World Examples and Applications
Given that the conversion rate is 1:1, here are a few examples:
Medical Dosage:
A doctor prescribes of a liquid medicine, which is the same as .
Cooking and Baking:
A recipe calls for of milk, which is equivalent to .
Automotive Engine Displacement:
A small engine might have a displacement of , which is the same as 1 liter (since ).
Laboratory Experiments:
A chemist needs of a solution, which equals .
In summary, the conversion between cubic centimeters and milliliters is exceptionally simple due to their direct equivalence within the metric system.
See below section for step by step unit conversion with formulas and explanations. Please refer to the table below for a list of all the Millilitres to other unit conversions.
What is Cubic Centimeters?
Cubic centimeters (cm³) is a unit of volume in the metric system. Understanding what it represents and how it relates to other units is essential in various fields, from everyday life to scientific applications.
Definition of Cubic Centimeters
A cubic centimeter is the volume of a cube with sides that are one centimeter in length. In other words, imagine a perfect cube; if each edge of that cube measures exactly one centimeter, then the space contained within that cube is one cubic centimeter.
How Cubic Centimeters is Formed
Cubic centimeters are derived from the base unit of length in the metric system, the meter (m). A centimeter (cm) is one-hundredth of a meter ().
To get a unit of volume, we cube the unit of length. Therefore, 1 cubic centimeter (1 cm³) is:
This means that one cubic meter contains one million cubic centimeters.
Relationship to Milliliters
Cubic centimeters are numerically equivalent to milliliters (mL).
This equivalency is extremely useful in both scientific measurements and everyday life, especially when dealing with liquids.
Common Uses and Real-World Examples
Cubic centimeters are widely used to measure relatively small volumes. Here are some examples:
Medical Dosage: Liquid medications are often prescribed in milliliters or cubic centimeters. For instance, a doctor might prescribe 5 mL of cough syrup, which is the same as 5 cm³.
Engine Displacement: The size of an engine in cars and motorcycles is often described in cubic centimeters. For example, a 2000 cc engine has a total cylinder volume of 2000 cm³.
Cooking: Small quantities of liquids in recipes are sometimes measured in milliliters or cubic centimeters, particularly in more precise baking recipes.
Scientific Research: Measuring volumes in experiments, particularly in chemistry and biology. For instance, a researcher might use 10 cm³ of a solution in an experiment.
Interesting Facts
The abbreviation "cc" is often used interchangeably with "cm³" and "mL", especially in medical and automotive contexts.
While there isn't a specific law directly tied to cubic centimeters, the standardization of metric units, including cubic centimeters, is crucial for global trade, science, and engineering, ensuring that measurements are consistent and universally understood. Organizations like the International Bureau of Weights and Measures play a key role in maintaining these standards.
For more information on metric units and volume measurements, you can refer to the NIST (National Institute of Standards and Technology) website.
What is millilitres?
What is Millilitres?
A millilitre (mL) is a unit of volume in the metric system, commonly used to measure liquids. It's a relatively small unit, making it convenient for everyday measurements. Understanding millilitres is crucial in various fields, from cooking and medicine to science and engineering.
Definition and Formation
A millilitre is defined as one cubic centimetre (). It is also equal to one-thousandth of a litre (L).
The prefix "milli-" indicates a factor of one-thousandth, meaning a millilitre is a thousandth of a litre. Litre is a non-SI unit accepted for use with SI units. The SI unit for Volume is Cubic Meter ().
Therefore
Real-World Examples
Cooking: Many recipes use millilitres to measure liquid ingredients like milk, water, or oil. For example, a recipe might call for 120 mL of milk.
Medicine: Liquid medications are often prescribed in millilitre dosages. A doctor might prescribe 5 mL of cough syrup.
Beverages: Canned and bottled drinks often specify their volume in millilitres. A small can of soda might contain 355 mL.
Cosmetics: Lotions, shampoos, and perfumes often have their volume listed in millilitres. A travel-sized bottle of shampoo might contain 100 mL.
Scientific Experiments: In chemistry and biology, precise volumes of liquids are crucial. Researchers use millilitres for accuracy in their experiments. For example, titration experiment requires using burette that are in units of millilitres.
Interesting Facts and Associations
While there isn't a specific "law" or historical figure directly associated with the millilitre, its significance lies in its practical application within the metric system. The widespread adoption of the metric system, particularly in science and international trade, has solidified the importance of the millilitre as a standard unit of volume.
Conversions
1 Millilitre (mL) = 0.001 Litres (L)
1 Millilitre (mL) ≈ 0.0338 Fluid Ounces (fl oz)
1 US Fluid Ounce (fl oz) ≈ 29.57 Millilitres (mL)
For more information on the metric system, you can visit the National Institute of Standards and Technology (NIST).
Complete Cubic Centimeters conversion table
| Convert 1 cm3 to other units | Result |
--- |
| Cubic Centimeters to Cubic Millimeters (cm3 to mm3) | 1000 |
| Cubic Centimeters to Cubic Decimeters (cm3 to dm3) | 0.001 |
| Cubic Centimeters to Millilitres (cm3 to ml) | 1 |
| Cubic Centimeters to Centilitres (cm3 to cl) | 0.1 |
| Cubic Centimeters to Decilitres (cm3 to dl) | 0.01 |
| Cubic Centimeters to Litres (cm3 to l) | 0.001 |
| Cubic Centimeters to Kilolitres (cm3 to kl) | 0.000001 |
| Cubic Centimeters to Megalitres (cm3 to Ml) | 1e-9 |
| Cubic Centimeters to Gigalitres (cm3 to Gl) | 1e-12 |
| Cubic Centimeters to Cubic meters (cm3 to m3) | 0.000001 |
| Cubic Centimeters to Cubic kilometers (cm3 to km3) | 1e-15 |
| Cubic Centimeters to Kryddmått (cm3 to krm) | 1 |
| Cubic Centimeters to Teskedar (cm3 to tsk) | 0.2 |
| Cubic Centimeters to Matskedar (cm3 to msk) | 0.06666666666667 |
| Cubic Centimeters to Kaffekoppar (cm3 to kkp) | 0.006666666666667 |
| Cubic Centimeters to Glas (cm3 to glas) | 0.005 |
| Cubic Centimeters to Kannor (cm3 to kanna) | 0.0003821169277799 |
| Cubic Centimeters to Teaspoons (cm3 to tsp) | 0.2028841356 |
| Cubic Centimeters to Tablespoons (cm3 to Tbs) | 0.0676280452 |
| Cubic Centimeters to Cubic inches (cm3 to in3) | 0.06102402519355 |
| Cubic Centimeters to Fluid Ounces (cm3 to fl-oz) | 0.0338140226 |
| Cubic Centimeters to Cups (cm3 to cup) | 0.004226752825 |
| Cubic Centimeters to Pints (cm3 to pnt) | 0.0021133764125 |
| Cubic Centimeters to Quarts (cm3 to qt) | 0.00105668820625 |
| Cubic Centimeters to Gallons (cm3 to gal) | 0.0002641720515625 |
| Cubic Centimeters to Cubic feet (cm3 to ft3) | 0.0000353146848166 |
| Cubic Centimeters to Cubic yards (cm3 to yd3) | 0.000001307949366991 |
Volume conversions
Cubic Centimeters to Cubic Millimeters (cm3 to mm3)
Cubic Centimeters to Cubic Decimeters (cm3 to dm3)
Cubic Centimeters to Millilitres (cm3 to ml)
Cubic Centimeters to Centilitres (cm3 to cl)
Cubic Centimeters to Decilitres (cm3 to dl)
Cubic Centimeters to Litres (cm3 to l)
Cubic Centimeters to Kilolitres (cm3 to kl)
Cubic Centimeters to Megalitres (cm3 to Ml)
Cubic Centimeters to Gigalitres (cm3 to Gl)
Cubic Centimeters to Cubic meters (cm3 to m3)
Cubic Centimeters to Cubic kilometers (cm3 to km3)
Cubic Centimeters to Kryddmått (cm3 to krm)
Cubic Centimeters to Teskedar (cm3 to tsk)
Cubic Centimeters to Matskedar (cm3 to msk)
Cubic Centimeters to Kaffekoppar (cm3 to kkp)
Cubic Centimeters to Glas (cm3 to glas)
Cubic Centimeters to Kannor (cm3 to kanna)
Cubic Centimeters to Teaspoons (cm3 to tsp)
Cubic Centimeters to Tablespoons (cm3 to Tbs)
Cubic Centimeters to Cubic inches (cm3 to in3)
Cubic Centimeters to Fluid Ounces (cm3 to fl-oz)
Cubic Centimeters to Cups (cm3 to cup)
Cubic Centimeters to Pints (cm3 to pnt)
Cubic Centimeters to Quarts (cm3 to qt)
Cubic Centimeters to Gallons (cm3 to gal)
Cubic Centimeters to Cubic feet (cm3 to ft3)
Cubic Centimeters to Cubic yards (cm3 to yd3) |
190399 | https://artofproblemsolving.com/wiki/index.php/Modular_arithmetic?srsltid=AfmBOorrBTNq3nI9OxVnSEmtJrbzLEEFD-av0o-v19RmdyNweL0K6vYc | Art of Problem Solving
Modular arithmetic - AoPS Wiki
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Modular arithmetic
Modular arithmetic is a special type of arithmetic that involves only integers. Since modular arithmetic is such a broadly useful tool in number theory, we divide its explanations into several levels:
Introduction to modular arithmetic
Intermediate modular arithmetic
Olympiad modular arithmetic
Contents
[hide]
1 Resources
1.1 Introductory Resources
1.1.1 Books
1.1.2 Classes
1.2 Intermediate Resources
1.3 Olympiad Resources
Resources
Introductory Resources
Books
The AoPS Introduction to Number Theory by Mathew Crawford.
Classes
AoPS Introduction to Number Theory Course
Intermediate Resources
Number Theory Problems and Notes by Naoki Sato.
Olympiad Resources
Number Theory Problems and Notes by Naoki Sato.
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