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187900 | https://projecteuclid.org/journals/electronic-journal-of-probability/volume-13/issue-none/Near-critical-percolation-in-two-dimensions/10.1214/EJP.v13-565.pdf | E l e c t r o n i c J o u r n a l o f P r o b a b i l i t y Vol. 13 (2008), Paper no. 55, pages 1562–1623.
Journal URL Near-critical percolation in two dimensions Pierre Nolin∗ École Normale Supérieure†and Université Paris-Sud‡ Email: pierre.nolin@ens.fr Abstract We give a self-contained and detailed presentation of Kesten’s results that allow to relate critical and near-critical percolation on the triangular lattice. They constitute an important step in the derivation of the exponents describing the near-critical behavior of this model. For future use and reference, we also show how these results can be obtained in more general situations, and we state some new consequences.
Key words: Near-critical percolation, arm events, critical exponents.
AMS 2000 Subject Classification: Primary 60K35; Secondary: 82B43, 82B27.
Submitted to EJP on January 11, 2008, final version accepted July 17, 2008.
∗Research supported by the Agence Nationale pour la Recherche under the grant ANR-06-BLAN-0058 †Département de Mathématiques et Applications, École Normale Supérieure, 45 rue d’Ulm, 75230 Paris Cedex 05, France ‡Laboratoire de Mathématiques, Bâtiment 425, Université Paris-Sud 11, 91405 Orsay Cedex, France 1562 DOI: 10.1214/EJP.v13-565 1 1 Introduction Since 2000, substantial progress has been made on the mathematical understanding of percolation on the triangular lattice. In fact, it is fair to say that it is now well-understood. Recall that performing a percolation with parameter p on a lattice means that each site is chosen independently to be black with probability p and white with probability 1 −p. We then look at the connectivity properties of the set of black sites (or the set of white ones). It is well-known that on the regular triangular lattice, when p ≤1/2 there is almost surely no infinite black connected component, whereas when p > 1/2 there is almost surely one infinite black connected component. Its mean density can then be measured via the probability θ(p) that a given site belongs to this infinite black component.
Thanks to Smirnov’s proof of conformal invariance of the percolation model at p = 1/2 , allow-ing to prove that critical percolation interfaces converge toward SLE6 as the mesh size goes to zero, and to the derivation of the SLE6 critical exponents [34; 35] by Lawler, Schramm and Werner, it is possible to prove results concerning the behavior of the model when p is exactly equal to 1/2, that had been conjectured in the physics literature, such as the values of the arm exponents [47; 36].
See e.g. for a survey and references.
More than ten years before the above-mentioned papers, Kesten had shown in his 1987 paper Scaling relations for 2D-percolation that the behavior of percolation at criticality (ie when p = 1/2) and near criticality (ie when p is close to 1/2) are very closely related. In particular, the exponents that describe the behavior of quantities such as θ(p) when p →1/2+ and the arm exponents for percolation at p = 1/2 are related via relations known as scaling (or hyperscaling) relations. At that time, it was not proved that any of the exponents existed (not to mention their actual value) and Kesten’s paper explains how the knowledge of the existence and the values of some arm exponents allows to deduce the existence and the value of the exponents that describe “near-critical” behavior.
Therefore, by combining this with the derivation of the arm exponents, we can for instance conclude that θ(p) = (p −1/2)5/36+o(1) as p →1/2+.
The first goal of the present paper is to give a complete self-contained proof of Kesten’s results that are used to describe near-critical percolation. Some parts of the proofs are simplified by using the so-called 5-arm exponent and Reimer’s inequality. We hope that this will be useful and help a wider community to have a clear and complete picture of this model.
It is also worth emphasizing that the proofs contain techniques (such as separation lemmas for arms) that are interesting in themselves and that can be applied to other situations. The second main purpose of the present paper is to state results in a more general setting than in , for possible future use. In particular, we will see that the “uniform estimates below the characteristic length” hold for an arbitrary number of arms and non-homogeneous percolation (see Theorem 11 on separation of arms, and Theorem 27 on arm events near criticality). Some technical difficulties arise due to these generalizations, but these new statements turn out to be useful. They are for instance instrumental in our study of gradient percolation in . Other new statements in the present paper concern arms “with defects” or the fact that the finite-size scaling characteristic length Lε(p) remains of the same order of magnitude when ε varies in (0,1/2) (Corollary 37) – and not only for ε small enough. This last fact is used in to study the “off-critical” regime for percolation.
1563 Figure 1: Percolation on the triangular lattice can be viewed as a random coloring of the dual hexagonal lattice.
2 Percolation background Before turning to near-critical percolation in the next section, we review some general notations and properties concerning percolation. We also sketch the proof of some of them, for which small generalizations will be needed. We assume the reader is familiar with the standard fare associated with percolation, and we refer to the classic references [27; 23] for more details.
2.1 Notations Setting Unless otherwise stated, we will focus on site percolation in two dimensions, on the triangular lattice. This lattice will be denoted by T = (VT,ET), where VT is the set of vertices (or “sites”), and ET is the set of edges (or “bonds”), connecting adjacent sites. We restrict ourselves to this lattice because it is at present the only one for which the critical regime has been proved to be conformal invariant in the scaling limit.
The usual (homogeneous) site percolation process of parameter p can be defined by taking the different sites to be black (or occupied) with probability p, and white (vacant) with probability 1 −p, independently of each other. This gives rise to a product probability measure on the set of configurations, which is referred to as Pp, the corresponding expectation being Ep. We usually represent it as a random (black or white) coloring of the faces of the dual hexagonal lattice (see Figure 1).
More generally, we can associate to each family of parameters ˆ p = (ˆ pv)v a product measure ˆ P for which each site v is black with probability ˆ pv and white with probability 1−ˆ pv, independently of all other sites.
1564 0 Sn Figure 2: We refer to oblique coordinates, and we denote by Sn the “box of size n”.
Coordinate system We sometimes use complex numbers to position points in the plane, but we mostly use oblique coordinates, with the origin in 0 and the basis given by 1 and eiπ/3, ie we take the x–axis and its image under rotation of angle π/3 (see Figure 2). For a1 ≤a2 and b1 ≤b2, the parallelogram R of corners aj + bkeiπ/3 (j, k = 1,2) is thus denoted by [a1, a2] × [b1, b2], its interior being � R := ]a1, a2[×]b1, b2[= [a1 +1, a2 −1]×[b1 +1, b2 −1] and its boundary ∂R := R\� R the concatenation of the four boundary segments {ai} × [b1, b2] and [a1, a2] × {bi}.
We denote by ∥z∥∞the infinity norm of a vertex z as measured with respect to these two axes, and by d the associated distance. For this norm, the set of points at distance at most N from a site z forms a rhombus SN(z) centered on this site and whose sides line up with the basis axes. Its boundary, the set of points at distance exactly N, is denoted by ∂SN(z), and its interior, the set of points at distance strictly less that N, by � SN(z). To describe the percolation process, we often use SN := SN(0) and call it the “box of size N”. Note that it can also be written as SN = [−N, N] × [−N, N]. It will sometimes reveal useful to have noted that |SN(z)| ≤C0N 2 (2.1) for some universal constant C0. For any two positive integers n ≤N, we also consider the annulus Sn,N(z) := SN(z) \� Sn(z), with the natural notation Sn,N := Sn,N(0).
Connectivity properties Two sites x and y are said to be connected if there exists a black path, ie a path consisting only of black sites, from x to y. We denote it by x ⇝y. Similarly, if there exists a white path from x to y, these two sites are said to be ∗–connected, which we denote by x ⇝∗y.
For notational convenience, we allow y to be “∞”: we say that x is connected to infinity (x ⇝∞) 1565 if there exists an infinite, self-avoiding and black path starting from x. We denote by θ(p) := Pp 0 ⇝∞ (2.2) the probability for 0 (or any other site by translation invariance) to be connected to ∞.
To study the connectivity properties of a percolation realization, we often are interested in the connected components of black or white sites: the set of black sites connected to a site x (empty if x is white) is called the cluster of x, denoted by C(x). We can define similarly C∗(x) the white cluster of x. Note that x ⇝∞is equivalent to the fact that |C(x)| = ∞.
If A and B are two sets of vertices, the notation A ⇝B is used to denote the event that some site in A is connected to some site in B. If the connection is required to take place using exclusively the sites in some other set C, we write A C ⇝B.
Crossings A left-right (or horizontal) crossing of the parallelogram [a1, a2] × [b1, b2] is simply a black path connecting its left side to its right side. However, this definition implies that the existence of a crossing in two boxes sharing just a side are not completely independent: it will actually be more convenient to relax (by convention) the condition on its extremities. In other words, we request such a crossing path to be composed only of sites in ]a1, a2[×]b1, b2[ which are black, with the exception of its two extremities on the sides of the parallelogram, which can be either black or white. The existence of such a horizontal crossing is denoted by CH([a1, a2] × [b1, b2]). We define likewise top-bottom (or vertical) crossings and denote their existence by CV([a1, a2] × [b1, b2]), and also white crossings, the existence of which we denote by C ∗ H and C ∗ V.
More generally, the same definition applies for crossings of annuli Sn,N(z), from the internal bound-ary ∂Sn(z) to the external one ∂SN(z), or even in more general domains D, from one part of the boundary to another part.
Asymptotic behavior We use the standard notations to express that two quantities are asymptotically equivalent. For two positive functions f and g, the notation f ≍g means that f and g remain of the same order of magnitude, in other words that there exist two positive and finite constants C1 and C2 such that C1g ≤f ≤C2g (so that the ratio between f and g is bounded away from 0 and +∞), while f ≈g means that log f /log g →1 (“logarithmic equivalence”) – either when p →1/2 or when n →∞, which will be clear from the context. This weaker equivalence is generally the one obtained for quantities behaving like power laws.
2.2 General properties On the triangular lattice, it is known since that percolation features a phase transition at p = 1/2, called the critical point: this means that • When p < 1/2, there is (a.s.) no infinite cluster (sub-critical regime), or equivalently θ(p) = 0.
1566 • When p > 1/2, there is (a.s.) an infinite cluster (super-critical regime), or equivalently θ(p) > 0. Furthermore, the infinite cluster turns out to be unique in this case.
In sub- and super-critical regimes, “correlations” decay very fast, this is the so-called exponential decay property: • For any p < 1/2, ∃C1, C2(p) > 0, Pp(0 ⇝∂Sn) ≤C1e−C2(p)n.
• We can deduce from it that for any p > 1/2, Pp(0 ⇝∂Sn,|C(0)| < ∞) ≤Pp(∃white circuit surrounding 0 and a site on ∂Sn) ≤C′ 1e−C′ 2(p)n for some C′ 1(p), C′ 2(p) > 0.
We would like to stress the fact that the speed at which these correlations vanish is governed by a constant C2 which depends on p – it becomes slower and slower as p approaches 1/2. To study what happens near the critical point, we need to control this speed for different values of p: we will derive in Section 7.4 an exponential decay property that is uniform in p.
The intermediate regime at p = 1/2 is called the critical regime. It is known for the triangular lattice that there is no infinite cluster at criticality: θ(1/2) = 0. Hence to summarize, θ(p) > 0 iff p > 1/2.
Correlations no longer decay exponentially fast in this critical regime, but (as we will see) just like power laws. For instance, non trivial random scaling limits – with fractal structures – arise. This particular regime has some very strong symmetry property (conformal invariance) which allows to describe it very precisely.
2.3 Some technical tools Monotone events We use the standard terminology associated with events: an event A is increasing if “it still holds when we add black sites”, and decreasing if it satisfies the same property, but when we add white sites.
Recall also the usual coupling of the percolation processes for different values of p: we associate to the different sites x i.i.d. random variables Ux uniform on [0,1], and for any p, we obtain the measure Pp by declaring each site x to be black if Ux ≤p, and white otherwise. This coupling shows for instance that p 7→Pp(A) is a non-decreasing function of p when A is an increasing event. More generally, we can represent in this way any product measure ˆ P.
1567 Correlation inequalities The two most common inequalities for percolation concern monotone events: if A and B are increas-ing events, we have ([5; 23]) 1. the Harris-FKG inequality: P(A∩B) ≥P(A)P(B).
2. the BK inequality: P(A◦B) ≤P(A)P(B) if A and B depend only on sites in a finite set, A ◦B meaning as usual that A and B occur “disjointly”.
In the paper where they proved the BK inequality, Van den Berg and Kesten also conjectured that this inequality holds in a more general fashion, for any pair of events A and B (depending on a finite number of sites): if we define A□B the disjoint occurrence of A and B in this situation, we have P(A□B) ≤P(A)P(B).
(2.3) This was later proved by Reimer , and it is now known as Reimer’s inequality. We will also use the following inequality: P1/2(A◦B) ≤P1/2(A∩˜ B), (2.4) where ˜ B is the event obtained by “flipping” the configurations in B. This inequality is an intermediate step in Reimer’s proof. On this subject, the reader can consult the nice review .
Russo’s formula Russo’s formula allows to study how probabilities of events vary when the percolation parameter p varies. Recall that for an increasing event A, the event “v is pivotal for A” is composed of the configurations ω such that if we make v black, A occurs, and if we make it white, A does not occur.
Note that by definition, this event is independent of the particular state of v. An analog definition applies for decreasing events.
Theorem 1 (Russo’s formula). Let A be an increasing event, depending only on the sites contained in some finite set S. Then d dpPp(A) = X v∈S Pp(v is pivotal for A).
(2.5) We now quickly remind the reader how to prove this formula, since we will later (in Section 6) generalize it a little.
Proof. We allow the parameters ˆ pv (v ∈S) to vary independently, which amounts to consider the more general function P : ˆ p = (ˆ pv)v∈S 7→ˆ P(A). This is clearly a smooth function (it is polynomial), and Pp(A) = P (p,..., p). Now using the standard coupling, we see that for any site w, for a small variation ε > 0 in w, ˆ P+ε(A) −ˆ P(A) = ε × ˆ P(w is pivotal for A), (2.6) 1568 so that ∂ ∂ˆ pw ˆ P(A) = ˆ P(w is pivotal for A).
Russo’s formula now follows readily by expressing d dpPp(A) in terms of the previous partial deriva-tives: d dpPp(A) = X v∈S ∂ ∂ˆ pv ˆ P(A) ˆ p=(p,...,p) = X v∈S Pp(v is pivotal for A).
Russo-Seymour-Welsh theory For symmetry reasons, we have: ∀n, P1/2(CH([0, n] × [0, n])) = 1/2.
(2.7) In other words, the probability of crossing an n× n box is the same on every scale. In particular, this probability is bounded from below: this is the starting point of the so-called Russo-Seymour-Welsh theory (see [23; 27]), that provides lower bounds for crossings in parallelograms of fixed aspect ratio τ × 1 (τ ≥1) in the “hard direction”.
Theorem 2 (Russo-Seymour-Welsh). There exist universal non-decreasing functions fk(.) (k ≥2), that stay positive on (0,1) and verify: if for some parameter p the probability of crossing an n × n box is at least δ1, then the probability of crossing a kn × n parallelogram is at least fk(δ1). Moreover, these functions can be chosen satisfying the additional property: fk(δ) →1 as δ →1, with fk(1 −ε) = 1 −Ckεαk + o(εαk) for some Ck,αk > 0.
If for instance p > 1/2, we know that when n gets very large, the probability δ1 of crossing an n× n rhombus becomes closer and closer to 1, and the additional property tells that the probability of crossing a kn × n parallelogram tends to 1 as a function of δ1.
Combined with Eq.(2.7), the theorem implies: Corollary 3. For each k ≥1, there exists some δk > 0 such that ∀n, P1/2(CH([0, kn] × [0, n])) ≥δk.
(2.8) Note that only going from rhombi to parallelograms of aspect ratio slightly larger than 1 is difficult.
For instance, once the result is known for 2n × n parallelograms, the construction of Figure 3 shows that we can take fk(δ) = δk−2(f2(δ))k−1.
(2.9) 1569 Figure 3: This construction shows that we can take fk(δ) = δk−2(f2(δ))k−1.
c1 c2 c3 r2 t l1 l2 b r1 H H′ 1.
2.
3.
Figure 4: Proof of the RSW theorem on the triangular lattice.
1570 Proof. The proof goes along the same lines as the one given by Grimmett for the square lattice.
We briefly review it to indicate the small adaptations to be made on the triangular lattice. We hope Figure 4 will make things clear. For an account on RSW theory in a general setting, the reader should consult Chapter 6 of .
We work with hexagons, since they exhibit more symmetries. Note that a crossing of an N × N rhombus induces a left-right crossing of a hexagon of side length N/2 (see Figure 4.1). We then apply the “square-root trick” – used recurrently during the proof – to the four events {li ⇝rj}: one of them occurs with probability at least 1 −(1 −δ)1/4. This implies that P(l1 ⇝r1) = P(l2 ⇝r2) ≥(1 −(1 −δ)1/4)2 =: τ(δ).
(2.10) (if P(l1 ⇝r1) = P(l2 ⇝r2) ≥1−(1−δ)1/4 we are OK, otherwise we just combine a crossing l1 ⇝r2 and a crossing l2 ⇝r1).
Now take two hexagons H, H′ as on Figure 4.2 (with obvious notation for their sides). With proba-bility at least 1−(1−δ)1/2 there exists a left-right crossing in H whose last intersection with l′ 1∪l′ 2 is on l′ 2. Assume this is the case, and condition on the lowest left-right crossing in H: with probability at least 1 −(1 −τ(δ))1/2 it is connected to t′ in H′. We then use a crossing from l′ 1 to r′ 1 ∪r′ 2, occurring with probability at least 1 −(1 −δ)1/2, to obtain P(l1 ∪l2 ⇝r′ 1 ∪r′ 2) ≥(1 −(1 −δ)1/2) × (1 −(1 −τ(δ))1/2) × (1 −(1 −δ)1/2).
The hard part is done: it now suffices to use t successive “longer hexagons”, and t −1 top-bottom crossings of regular hexagons, for t large enough (see Figure 4.3). We construct in such a way a left right-crossing of a 2N × N parallelogram, with probability at least f2(δ) := (1 −(1 −δ)1/2)2t(1 −(1 −τ(δ))1/2)2t−1.
(2.11) When δ tends to 1, τ(δ), and thus f2(δ), also tend to 1. Moreover, it is not hard to convince oneself that f2 admits near δ = 1 an asymptotic development of the form f2(1 −ε) = 1 −Cε1/8 + o(ε1/8).
(2.12) Eq.(2.9) then provides the desired conclusion for any k ≥2.
3 Near-critical percolation overview 3.1 Characteristic length We will use throughout the paper a certain “characteristic length” L(p) defined in terms of crossing probabilities, or “sponge-crossing probabilities”. This length is often convenient to work with, and it has been used in many papers concerning finite-size scaling, e.g. [15; 16; 6; 7].
Consider the rhombi [0, n]×[0, n]. At p = 1/2, Pp(CH([0, n]×[0, n])) = 1/2. When p < 1/2 (sub-critical regime), this probability tends to 0 when n goes to infinity, and it tends to 1 when p > 1/2 (super-critical regime). We introduce a quantity that measures the scale up to which these crossing probabilities remain bounded away from 0 and 1: for each ε ∈(0,1/2), we define Lε(p) = ( min{n s.t. Pp(CH([0, n] × [0, n])) ≤ε} when p < 1/2, min{n s.t. Pp(C ∗ H([0, n] × [0, n])) ≤ε} when p > 1/2.
(3.1) 1571 Hence by definition, Pp(CH([0, Lε(p) −1] × [0, Lε(p) −1])) ≥ε (3.2) and Pp(CH([0, Lε(p)] × [0, Lε(p)])) ≤ε (3.3) if p < 1/2, and the same with ∗’s if p > 1/2.
Note that by symmetry, we also have directly Lε(p) = Lε(1 −p). Since P1/2(CH([0, n] × [0, n])) is equal to 1/2 on every scale, we will take the convention Lε(1/2) = +∞, so that in the following, the expression “for any n ≤Lε(p)” must be interpreted as “for any n” when p = 1/2. This convention is also consistent with the following property.
Proposition 4. For any fixed ε ∈(0,1/2), Lε(p) →+∞when p →1/2.
Proof. Was it not the case, we could find an integer N and a sequence pk →1/2, say pk < 1/2, such that for each k, Lε(pk) = N, which would imply Ppk(CH([0, N] × [0, N])) ≤ε.
This contradicts the fact that Ppk(CH([0, N] × [0, N])) →1/2, the function p 7→Pp(CH([0, N] × [0, N])) being continuous (it is polynomial in p).
3.2 Russo-Seymour-Welsh type estimates When studying near-critical percolation, we will have to consider product measures ˆ P more general than simply the measures Pp (p ∈[0,1]), with associated parameters ˆ pv which are allowed to depend on the site v: Definition 5. A measure ˆ P on configurations is said to be “between Pp and P1−p” if it is a product measure, and if its parameters ˆ pv are all between p and 1 −p.
The Russo-Seymour-Welsh theory implies that for each k ≥1, there exists some δk = δk(ε) > 0 (depending only on ε) such that for all p, ˆ P between Pp and P1−p, ∀n ≤Lε(p), ˆ P(CH([0, kn] × [0, n])) ≥δk, (3.4) and for symmetry reasons this bound is also valid for horizontal white crossings.
These estimates for crossing probabilities will be the basic building blocks on which most further considerations are built. They imply that when n is not larger than Lε(p), things can still be com-pared to critical percolation: roughly speaking, Lε(p) is the scale up to which percolation can be considered as “almost critical”.
In the other direction, we will see in Section 7.4 that Lε(p) is also the scale at which percolation starts to look sub- or super-critical. Assume for instance that p > 1/2, we know that Pp(CH([0, Lε(p)] × [0, Lε(p)])) ≥1 −ε.
Then using RSW (Theorem 2), we get that Pp(CH([0,2Lε(p)] × [0, Lε(p)])) ≥1 −˜ ε, where 1−˜ ε = f2(1−ε) can be made arbitrarily close to 1 by taking ε sufficiently small. This will be useful in the proof of Lemma 39 (but actually only for it).
1572 3.3 Outline of the paper In the following, we fix some value of ε in (0,1/2). For notational convenience, we forget about the dependence on ε. We will see later (Corollary 37) that the particular choice of ε is actually not relevant, in the sense that for any two ε, ε′, the corresponding lengths are of the same order of magnitude.
In Section 4 we define the so-called arm events. On a scale L(p), the RSW property, which we know remains true, allows to derive separation results for these arms. Section 5 is devoted to critical percolation, in particular how arm exponents – describing the asymptotic behavior of arm events – can be computed. In Section 6 we study how arm events are affected when we make vary the parameter p: if we stay on a scale L(p), the picture does not change too much. It can be used to describe the characteristic functions, which we do in Section 7. Finally, Section 8 concludes the paper with some remarks and possible applications.
With the exception of this last section, the organization follows the implication between the different results: each section depends on the previous ones. A limited number of results can however be obtained directly, we will indicate it clearly when this is the case.
4 Arm separation We will see that when studying critical and near-critical percolation, certain exceptional events play a central role: the arm events, referring to the existence of some number of crossings (“arms”) of the annuli Sn,N (n < N), the color of each crossing (black or white) being prescribed. These events are useful because they can be combined together, and they will prove to be instrumental for studying more complex events. Their asymptotic behavior can be described precisely using SLE6 (see next section), allowing to derive further estimates, especially on the characteristic functions.
4.1 Arm events Let us consider an integer j ≥1. A color sequence σ is a sequence (σ1,...,σj) of “black” and “white” of length j. We use the letters “W” and “B” to encode the colors: the sequence (black,white,black) is thus denoted by “BW B”.
Only the cyclic order of the arms is relevant, and we identify two sequences if they are the same up to a cyclic permutation: for instance, the two sequences “BW BW” and “W BW B” are the same, but they are different from “BBWW”. The resulting set is denoted by ˜ Sj. For any color sequence σ, we also introduce ˜ σ = (˜ σ1,..., ˜ σj) the inverted sequence, where each color is replaced by its opposite.
For any two positive integers n ≤N, we define the event Aj,σ(n, N) := {∂Sn ⇝j,σ ∂SN} (4.1) that there exist j disjoint monochromatic arms in the annulus Sn,N, whose colors are those prescribed by σ, when taken in counterclockwise order (see Figure 5). We denote such an ordered set of crossings by C = {ci}1≤i≤j, and we say it to be “σ-colored”. Recall that by convention, we have relaxed the color prescription for the extremities of the ci’s. Hence for j = 1 and σ = B, Aj,σ(0, N) just denotes the existence of a black path 0 ⇝∂SN.
1573 ∂SN ∂Sn Figure 5: The event A6,σ(n, N), with σ = BBBW BW.
Note that a combinatorial objection due to discreteness can arise: if j is too large compared to n, the event Aj,σ(n, N) can be void, just because the arms do not have enough space on ∂Sn to arrive all together. For instance Aj,σ(0, N) = ∅if j ≥7. In fact, we just have to check that n is large enough so that the number of sites touching the exterior of |∂Sn| (ie |∂Sn+1| with the acute corners removed) is at least j: if this is true, we can then draw straight lines heading toward the exterior. For each positive integer j, we thus introduce n0(j) the least such nonnegative integer, and we have ∀N ≥n0(j), Aj,σ(n0(j), N) ̸= ∅.
Note that n0(j) = 0 for j = 1,...,6, and that n0(j) ≤j. For asymptotics, the exact choice of n is not relevant since anyway, for any fixed n1, n2 ≥n0(j), ˆ P(Aj,σ(n1, N)) ≍ˆ P(Aj,σ(n2, N)).
Remark 6. Note that Reimer’s inequality implies that for any two integers j, j′, and two color sequences σ, σ′ of these lengths, we have: ˆ P(Aj+j′,σσ′(n, N)) ≤ˆ P(Aj,σ(n, N))ˆ P(Aj′,σ′(n, N)) (4.2) for any ˆ P, n ≤N (denoting by σσ′ the concatenation of σ and σ′).
4.2 Well-separateness We now impose some restrictions on the events Aj,σ(n, N). Our main goal is to prove that we can separate macroscopically (the extremities of) any sequence of arms: with this additional condition, the probability of Aj,σ(n, N) does not decrease from more than a (universal) constant factor. This result is not really surprising, but we will need it recurrently for technical purposes.
Let us now give a precise meaning to the property of being “separated” for sets of crossings. In the following, we will actually consider crossings in different domain shapes. We first state the definition for a parallelogram of fixed (1 × τ) aspect ratio, and explain how to adapt it in other cases.
1574 Z+ Z− √ηM 2ηM zi ci ri τM M Figure 6: Well-separateness for a set of crossings C = {ci}.
We first require that the extremities of these crossings are distant from each other. We also need to add a condition ensuring that the crossings can easily be extended: we impose the existence of “free spaces” at their extremities, which will allow then to construct longer extensions. This leads to the following definition, similar to Kesten’s “fences” (see Figure 6).
Definition 7. Consider some M × τM parallelogram R = [a1, a1 + M] × [b1, b1 + τM], and C = {ci}1≤i≤j a (σ-colored) set of j disjoint left-right crossings. Introduce zi the extremity of ci on the right side of the parallelogram, and for some η ∈(0,1], the parallelogram ri = zi+[0,pηM]×[−ηM,ηM], attached to R on its right side.
We say that C is well-separated at scale η (on the right side) if the two following conditions are satisfied: 1. The extremity zi of each crossing is not too close from the other ones: ∀i ̸= j, dist(zi,zj) ≥2pηM, (4.3) nor from the top and bottom right corners Z+, Z−of R: ∀i, dist(zi, Z±) ≥2pηM.
(4.4) 2. Each ri is crossed vertically by some crossing ˜ ci of the same color as ci, and ci ⇝˜ ci in � SpηM(zi).
(4.5) For the second condition, we of course require the path connecting ci and ˜ ci to be of the same color as these two crossings. The crossing ˜ ci is thus some small extension of ci on the right side of R. The free spaces ri will allow us to use locally an FKG-type inequality to further extend the ci’s on the right.
1575 Definition 8. We say that a set C = {ci}1≤i≤j of j disjoint left-right crossings of R can be made well-separated on the right side if there exists another set C ′ = {c′ i}1≤i≤j of j disjoint crossings that is well-separated on the right side, such that c′ i has the same color as ci, and the same extremity on the left side.
The same definitions apply for well-separateness on the left side, and also for top-bottom crossings.
Consider now a set of crossings of an annulus Sn,N. We can divide this set into four subsets, according to the side of ∂SN on which they arrive. Take for instance the set of crossings arriving on the right side: we say it to be well-separated if, as before, the extremities of these crossings on ∂SN are distant from each other and from the top-right and bottom-right corners, and if there exist free spaces ri that satisfy condition 2 of Definition 7. Then, we say that a set of crossings of Sn,N is well-separated on the external boundary if each of the four previous sets is itself well-separated. Note that requiring the extremities to be not too close from the corners ensures that they are not too close from the extremities of the crossings arriving on the other sides either. We take the same definition for the internal boundary ∂Sn: in this case, taking the extremities away from the corners also ensures that the free spaces are included in Sn and do not intersect each other.
We are in a position to define our first sub-event of Aj,σ(n, N): for any η,η′ ∈(0,1), ˜ Aη/η′ j,σ (n, N) := {∂Sn ⇝η/η′ j,σ ∂SN} (4.6) denotes the event Aj,σ(n, N) with the additional condition that the set of j arms is well-separated at scale η on ∂Sn, and at scale η′ on ∂SN.
We can even prescribe the “landing areas” of the different arms, ie the position of their extremities.
We introduce for that some last definition: Definition 9. Consider ∂SN for some integer N: a landing sequence {Ii}1≤i≤j on ∂SN is a sequence of disjoint sub-intervals I1,..., I j on ∂SN in counterclockwise order. It is said to be η-separated if1, 1. dist(Ii, Ii+1) ≥2pηN for each i, 2. dist(Ii, Z) ≥2pηN for each i and each corner Z of ∂SN.
It is called a landing sequence of size η if the additional property 3. length(Ii) ≥ηN for each i is also satisfied.
We identify two landing sequences on ∂SN and ∂SN′ if they are identical up to a dilation. This leads to the following sub-event of ˜ Aη/η′ j,σ (n, N): for two landing sequences I = {Ii}1≤i≤j and I′ = {I′ i}1≤i′≤j, ˜ ˜ Aη,I/η′,I′ j,σ (n, N) := {∂Sn ⇝η,I/η′,I′ j,σ ∂SN} (4.7) denotes the event ˜ Aη/η′ j,σ (n, N), with the additional requirement on the set of crossings {ci}1≤i≤j that for each i, the extremities zi and z′ i of ci on (respectively) ∂Sn and ∂SN satisfy zi ∈Ii and z′ i ∈I′ i.
1As usual, we consider cyclic indices, so that here for instance I j+1 = I1.
1576 We will also have use for another intermediate event between A and ˜ ˜ A: ¯ AI/I′ j,σ (n, N), for which we only impose the landing areas I/I′ of the j arms. We do not ask a priori the sub-intervals to be η-separated either, just to be disjoint.
Note however that if they are η/η′-separated, then the extremities of the different crossings will be η/η′-separated too.
To summarize: Aj,σ(n, N) = {j arms ∂Sn ⇝∂SN, color σ} separated at scale η/η′ + small extensions hhhhhhh thhhhhhhh landing areas I/I′ U U U U U U U U U U U U U U U ˜ Aη/η′ j,σ (n, N) landing areas I/I′ U U U U U U U U U U U U U U ¯ AI/I′ j,σ (n, N) small extensions (if I/I′ are η/η′-separated) iiiiiiii tiiiiii ˜ ˜ Aη,I/η′,I′ j,σ (n, N) Remark 10. If we take for instance alternating colors (¯ σ = BW BW), and as landing areas ¯ I1,..., ¯ I4 the (resp.) right, top, left and bottom sides of ∂SN, the 4-arm event ¯ A./¯ I 4,¯ σ(0, N) (the “.” meaning that we do not put any condition on the internal boundary) is then the event that 0 is pivotal for the existence of a left-right crossing of SN.
4.3 Statement of the results Main result Our main separation result is the following: Theorem 11. Fix an integer j ≥1, some color sequence σ ∈˜ Sj and η0,η′ 0 ∈(0,1). Then we have ˆ P ˜ ˜ Aη,I/η′,I′ j,σ (n, N) ≍ˆ P Aj,σ(n, N) (4.8) uniformly in all landing sequences I/I′ of size η/η′, with η ≥η0 and η′ ≥η′ 0, p, ˆ P between Pp and P1−p, n ≤N ≤L(p).
First relations Before turning to the proof of this theorem, we list some direct consequences of the RSW estimates that will be needed.
Proposition 12. Fix j ≥1, σ ∈˜ Sj and η0,η′ 0 ∈(0,1).
1. “Extendability”: We have ˆ P ˜ ˜ Aη,I/ ˜ η′,˜ I′ j,σ (n,2N), ˆ P ˜ ˜ A˜ η,˜ I/η′,I′ j,σ (n/2, N) ≍ˆ P ˜ ˜ Aη,I/η′,I′ j,σ (n, N) uniformly in p, ˆ P between Pp and P1−p, n ≤N ≤L(p), and all landing sequences I/I′ (resp.
˜ I/˜ I′) of size η/η′ (resp. ˜ η/ ˜ η′) larger than η0/η′ 0. In other words: “once well-separated, the arms can easily be extended”.
1577 2. “Quasi-multiplicativity”: We have for some C = C(η0,η′ 0) > 0 ˆ P(Aj,σ(n1, n3)) ≥C ˆ P(˜ ˜ A ./η,Iη j,σ (n1, n2/4))ˆ P(˜ ˜ A η′,Iη′/.
j,σ (n2, n3)) uniformly in p, ˆ P between Pp and P1−p, n0(j) ≤n1 < n2 < n3 ≤L(p) with n2 ≥4n1, and all landing sequences I/I′ of size η/η′ larger than η0/η′ 0.
3. For any η,η′ > 0, there exists a constant C = C(η,η′) > 0 with the following property: for any p, ˆ P between Pp and P1−p, n ≤N ≤L(p), there exist two landing sequences I and I′ of size η and η′ (that may depend on all the parameters mentioned) such that ˆ P ˜ ˜ Aη,I/η′,I′ j,σ (n, N) ≥C ˆ P ˜ Aη,η′ j,σ (n, N).
Proof. The proof relies on gluing arguments based on RSW constructions. However, the events considered are not monotone when σ is non-constant (there is at least one black arm and one white arm). We will thus need a slight generalization of the FKG inequality for events “locally monotone”.
Lemma 13. Consider A+, ˜ A+ two increasing events, and A−, ˜ A−two decreasing events. Assume that there exist three disjoint finite sets of vertices A , A + and A −such that A+, A−, ˜ A+ and ˜ A−depend only on the sites in, respectively, A ∪A +, A ∪A −, A + and A −. Then we have ˆ P(˜ A+ ∩˜ A−|A+ ∩A−) ≥ˆ P(˜ A+)ˆ P(˜ A−) (4.9) for any product measure ˆ P.
Proof. Conditionally on the configuration ωA in A , the events A+∩˜ A+ and A−∩˜ A−are independent, so that ˆ P(A+ ∩˜ A+ ∩A−∩˜ A−|ωA ) = ˆ P(A+ ∩˜ A+|ωA )ˆ P(A−∩˜ A−|ωA ).
The FKG inequality implies that ˆ P(A+ ∩˜ A+|ωA ) ≥ˆ P(A+|ωA )ˆ P(˜ A+|ωA ) = ˆ P(A+|ωA )ˆ P(˜ A+) and similarly with A−and ˜ A−. Hence, ˆ P(A+ ∩˜ A+ ∩A−∩˜ A−|ωA ) ≥ˆ P(A+|ωA )ˆ P(˜ A+)ˆ P(A−|ωA )ˆ P(˜ A−) = ˆ P(A+ ∩A−|ωA )ˆ P(˜ A+)ˆ P(˜ A−).
The conclusion follows by summing over all configurations ωA .
Once this lemma at our disposal, items 1. and 2. are straightforward. For item 3., we consider a covering of ∂Sn (resp. ∂SN) with at most 8η−1 (resp. 8η′−1) intervals {I} of length η (resp. (I′) of length η′). Then for some I, I′, ˆ P ˜ ˜ Aη,I/η′,I′ j,σ (n, N) ≥(8η−1)−1(8η′−1)−1ˆ P ˜ Aη,η′ j,σ (n, N).
1578 We also have the following a-priori bounds for the arm events: Proposition 14. Fix some j ≥1, σ ∈˜ Sj and η0,η′ 0 ∈(0,1).
Then there exist some exponents 0 < αj,α′ < ∞, as well as constants 0 < Cj, C′ < ∞, such that Cj n N αj ≤ˆ P ˜ ˜ Aη,I/η′,I′ j,σ (n, N) ≤C′ n N α′ (4.10) uniformly in p, ˆ P between Pp and P1−p, n ≤N ≤L(p), and all landing sequences I/I′ of size η/η′ larger than η0/η′ 0.
The lower bound comes from iterating item 1. The upper bound can be obtained by using concentric annuli: in each of them, RSW implies that there is a probability bounded away from zero to observe a black circuit, preventing the existence of a white arm (consider a white circuit instead if σ = BB ... B).
4.4 Proof of the main result Assume that Aj,σ(n, N) is satisfied: our goal is to link this event to the event ˜ ˜ A η0,Iη0/η′ 0,Iη′ 0 j,σ (n, N), for some fixed scales η0,η′ 0.
Proof. First note that it suffices to prove the result for n, N which are powers of two: then we would have, if k, K are such that 2k−1 < n ≤2k and 2K ≤n < 2K+1, ˆ P Aj,σ(n, N) ≤ˆ P Aj,σ(2k,2K) ≤C1ˆ P ˜ ˜ Aη,I/η′,I′ j,σ (2k,2K) ≤C2ˆ P ˜ ˜ Aη,I/η′,I′ j,σ (n, N).
We have to deal with the extremities of the j arms on the internal boundary ∂Sn, and on the external boundary ∂SN.
1. External extremities Let us begin with the external boundary. In the course of proof, we will have use for the intermediate event ˜ A./η′ j,σ (n, N) that there exists a set of j arms that is well-separated on the external side ∂SN only, and also the event ˜ ˜ A./η′,I′ j,σ (n, N) associated to some landing sequence I′ on ∂SN. Each of the j arms induces in S2K−1,2K a crossing of one of the four U-shaped regions U1,ext 2K−1 ,..., U4,ext 2K−1 depicted in Figure 7. The “ext” indicates that a crossing of such a region connects the two marked parts of the boundary.
For the internal extremities, we will use the same regions, but we distinguish different parts of the boundary. The key observation is the following.
In a U-shaped region, any set of disjoint crossings can be made well-separated with high probability.
1579 ∂SN ∂S2N U 1,ext N U 2,ext N U 3,ext N U 4,ext N Figure 7: The four U-shaped regions that we use for the external extremities.
More precisely, if we take such an N × 4N domain, the probability that any set of disjoint crossings can be made η-well-separated (on the external boundary) can be made arbitrarily close to 1 by choosing η sufficiently small, uniformly in N. We prove the following lemma, which implies that on every scale, with very high probability the j arms can be made well-separated.
Lemma 15. For any δ > 0, there exists a size η(δ) > 0 such that for any p, any ˆ P between Pp and P1−p and any N ≤L(p): in the domain U1,ext N , ˆ P(Any set of disjoint crossings can be made η-well-separated) ≥1 −δ.
(4.11) Proof. First we note that there cannot be too many disjoint crossings in U1,ext N . Indeed, there is a crossing in this domain (either white or black) with probability less than some 1 −δ′, by RSW: combined with the BK inequality (and also FKG), this implies that the probability of observing at least T crossings is less than (1 −δ′)T.
(4.12) We thus take T such that this quantity is less than δ/4.
Consider for the moment any η ∈(0,1) (we will see during the proof how to choose it).
We note that we can put disjoint annuli around Z−and Z+ to prevent crossings from arriving there.
Consider Z−for instance, and look at the disjoint annuli centered on Z−of the form S2l−1,2l(Z−), with pηN ≤2l−1 < 2l ≤η3/8N (see Figure 8). We can take at least −C4 logη such disjoint annuli for some universal constant C4 > 0, and with probability at least 1−(1−δ′′)−C4 logη there exists a black circuit in one of the annuli. Consider then the annuli S2l−1,2l(Z−), with η3/8N ≤2l−1 < 2l ≤η1/4N: with probability at least 1−(1−δ′′)−C′ 4 logη we observe a white circuit in one of them. If two circuits as described exist, we say that Z−is “protected”. The same reasoning applies for Z+.
1580 Consider now the following construction: take c1 the lowest (ie closest to the bottom side) monochromatic crossing (which can be either black or white), then c2 the lowest monochromatic crossing disjoint from c1, and so on. The process stops after t steps, and we denote by C = {cu}1≤u≤t the set of crossings so-obtained. Of course, C can be void: we set t = 0 in this case. We have P(t ≥T) ≤(1 −δ′)T ≤δ/4 (4.13) by definition of T. We denote by zu the extremity of cu on the right side, and by σu ∈{B,W} its color.
In order to get some independence and be able to apply the previous construction around the ex-tremities of the crossings, we condition on the successive crossings. Consider some u ∈{1,..., T} and some ordered sequence of crossings ˜ c1,˜ c2,...,˜ cu, together with colors ˜ σ1, ˜ σ2,..., ˜ σu. The event Eu := {t ≥u and cv = ˜ cv, σv = ˜ σv for any v ∈{1,...,u}} is independent from the status of the sites above ˜ cu. Hence, if we condition on Eu, percolation there remains unbiased and we can use the RSW theorem.
We now do the same construction as before. Look at the disjoint annuli centered on zu of the form S2l−1,2l(zu), with pηN ≤2l−1 < 2l ≤η3/8N on one hand, and with η3/8N ≤2l−1 < 2l ≤η1/4N on the other hand. Assume for instance that ˜ σu = B. With probability at least 1 −(1 −δ′′)−C′′ 4 logη we observe a white circuit in one of the annuli in the first set, preventing other disjoint black crossings to arrive near zu, and also a black one in the second set, preventing white crossings to arrive. Moreover, by considering a black circuit in the annuli S2l−1,2l(zu) with η3/4N ≤2l−1 < 2l ≤pηN, we can construct a small extension of cu. If the three circuits described exist, cu is said to be “protected from above”. Summing over all possibilities for ˜ ci, ˜ σi (1 ≤i ≤u), we get that for some C′′′ 4 , P(t ≥u and cu is not protected from above) ≤(1 −δ′′)−C′′′ 4 logη.
(4.14) Now for our set of crossings C , P(C is not η-well-separated) ≤P(t ≥T) + T−1 X u=1 P(t ≥u and cu is not protected from above) + P(Z−is not protected) + P(Z+ is not protected).
First, each term in the sum, as well as the last two terms, are less than (1 −δ′′)−C′′′ 4 logη. We also have P(t ≥T) ≤δ/4, so that the right-hand side is at most (T + 1)(1 −δ′′)−C′′′ 4 logη + δ 4 .
(4.15) It is less than δ if we choose η sufficiently small (T is fixed).
We now assume that C is η-well-separated, and prove that any other set C ′ = {c′ u}1≤u≤t′ of t′ (≤t) disjoint crossings (we take it ordered) can also be made η-well-separated. For that purpose, we replace recursively the tip of each c′ u by the tip of one of the cv’s. If we take c′ 1 for instance, it has to cross at least one of the cv’s (by maximality of C ). Let us call cv1 the lowest one: still by maximality, c′ 1 cannot go below it. Take the piece of c′ 1 between its extremity z′ 1 and its last intersection a1 with cv1, and replace it with the corresponding piece of cv1: this gives c′′ 1 . This new crossing has the 1581 Z+ Z− U 1,ext N zu cu √ηN 2ηN Figure 8: We apply RSW in concentric annuli around Z−and Z+, and then around the extremity zu of each crossing cu.
same extremity as cv1 on the right side, and it is not hard to check that it is connected to the small extension ˜ cv1 of cv1 on the external side. Indeed, this extension is connected by a path that touches cv1 in, say, b1: either b1 is between a1 and z1, in which case c′′ 1 is automatically connected to ˜ cv1, otherwise c′ 1 has to cross the connecting path before a1 and c′′ 1 is also connected to ˜ cv1.
Consider then c2, and cv2 the lowest crossing it intersects: necessarily v2 > v1 (since c1 stays above cv1), and the same reasoning applies. The claim follows by continuing this procedure until c′ t′.
The arms are well-separated with positive probability.
The idea is then to “go down” in successive concentric annuli, and to apply the lemma in each of them. We work with two different scales of separation: • a fixed (macroscopic) scale η′ 0 that we will use to extend arms, associated to a constant exten-sion cost.
• another scale η′ which is very small (η′ ≪η′ 0), so that the j arms can be made well-separated at scale η′ with very high probability.
The proof goes as follows. Take some δ > 0 very small (we will see later how small), and some η′ > 0 associated to it by Lemma 15. We start from the scale ∂S2K and look at the crossings induced by the j arms. The previous lemma implies that with very high probability, these j arms can be modified in S2K−1,2K so that they are η′-well-separated. Otherwise, we go down to the next annulus: there still exist j arms, and what happens in S2K−1,2K is independent of what happens in S2K−1. On each scale, we have a very low probability to fail, and once the arms are separated on scale η′, we go backwards by using the scale η′ 0, for which the cost of extension is constant.
1582 More precisely, after one step we get Aj,σ(2k,2K) ⊆˜ A./η′ j,σ (2k,2K) ∪ {One of the four Ui,ext 2K−1 fails} ∩Aj,σ(2k,2K−1) .
Hence, by independence of the two latter events and Lemma 15, ˆ P(Aj,σ(2k,2K)) ≤ˆ P(˜ A./η′ j,σ (2k,2K)) + (4δ)ˆ P(Aj,σ(2k,2K−1)).
We then iterate this argument: after K −k steps, ˆ P(Aj,σ(2k,2K)) ≤ˆ P(˜ A./η′ j,σ (2k,2K)) + (4δ)ˆ P(˜ A./η′ j,σ (2k,2K−1)) + (4δ)2ˆ P(˜ A./η′ j,σ (2k,2K−2)) + ...
+ (4δ)K−k−1ˆ P(˜ A./η′ j,σ (2k,2k+1)) + (4δ)K−k.
We then use the size η′ 0 to go backwards: if the crossings are η′-separated at some scale m, there exists some landing sequence Iη′ of size η′ where the probability of landing is comparable to the probability of just being η′-well-separated, and then we can reach Iη′ 0 of size η′ 0 on the next scale.
More precisely, there exist universal constants C1(η′), C2(η′) depending only on η′ such that for all 1 ≤i′ ≤i, we can choose some Iη′ (which can depend on i′) such that ˆ P(˜ A./η′ j,σ (2k,2K−i′)) ≤C1(η′)ˆ P(˜ ˜ A ./η′,Iη′ j,σ (2k,2K−i′)) (by item 3. of Proposition 12) and then go to Iη′ 0 on the next scale with cost C2(η′): ˆ P(˜ ˜ A ./η′,Iη′ j,σ (2k,2K−i′)) ≤C2(η′)ˆ P(˜ ˜ A ./η′ 0,Iη′ 0 j,σ (2k,2K−i′+1)) (by item 1.). Now for the size η′ 0, going from ∂Sm to ∂S2m has a cost C′ 0 depending only on η′ 0 on each scale m, we thus have ˆ P(˜ A./η′ j,σ (2k,2K−i′)) ≤C1(η′)C2(η′)C i′−1 0 ˆ P(˜ ˜ A ./η′ 0,Iη′ 0 j,σ (2k,2K)).
There remains a problem with the first term ˆ P(˜ A./η′ j,σ (2k,2K)). . . So assume that we have started from 2K−1 instead, so that the annulus S2K−1,2K remains free: ˆ P(Aj,σ(2k,2K)) ≤ˆ P(Aj,σ(2k,2K−1)) ≤ˆ P(˜ A./η′ j,σ (2k,2K−1)) + (4δ)ˆ P(˜ A./η′ j,σ (2k,2K−2)) + (4δ)2ˆ P(˜ A./η′ j,σ (2k,2K−3)) + ...
+ (4δ)K−k−2ˆ P(˜ A./η′ j,σ (2k,2k+1)) + (4δ)K−k−1 ≤C1(η′)C2(η′) 1 + (4δC0) + ... + (4δC0)K−k−1 ˆ P(˜ ˜ A ./η′ 0,Iη′ 0 j,σ (2k,2K)).
Now C0 is fixed as was noticed before, so we may have taken δ such that 4δC0 < 1/2, so that C1(η′)C2(η′) 1 + (4δC0) + ... + (4δC0)K−k−1 ≤C3(η′) for some C3(η′). We have thus reached the desired conclusion for external extremities: ˆ P(Aj,σ(2k,2K)) ≤C3(η′)ˆ P(˜ ˜ A ./η′ 0,Iη′ 0 j,σ (2k,2K)).
1583 ∂SN ∂S2N U 1,int N U 2,int N U 3,int N U 4,int N Figure 9: For the internal extremities, we consider the same domains but we mark different parts of the boundary.
2. Internal extremities The reasoning is the same for internal extremities, except that we work in the other direction, from ∂S2k toward the interior. If we consider the domains Ui,int N having the same shapes as the Ui,ext N domains, but with different parts of the boundary distinguished (see Figure 9), then the lemma remains true. Hence, ˆ P(˜ ˜ A ./η′ 0,Iη′ 0 j,σ (2k,2K)) ≤ˆ P(˜ ˜ A ./η′ 0,Iη′ 0 j,σ (2k+1,2K)) ≤ˆ P(˜ ˜ A η,./η′ 0,Iη′ 0 j,σ (2k+1,2K)) + (4δ)ˆ P(˜ ˜ A η,./η′ 0,Iη′ 0 j,σ (2k+2,2K)) + ...
+ (4δ)K−k−2ˆ P(˜ ˜ A η,./η′ 0,Iη′ 0 j,σ (2K−1,2K)) + (4δ)K−k−1 ≤C1(η)C2(η) 1 + (4δC0) + ... + (4δC0)K−k−1 ˆ P(˜ ˜ A η0,Iη0/η′ 0,Iη′ 0 j,σ (2k,2K)) and the conclusion follows.
4.5 Some consequences We now state some important consequences of the previous theorem.
1584 Extendability Proposition 16. Take j ≥1 and a color sequence σ ∈˜ Sj. Then ˆ P(Aj,σ(n,2N)), ˆ P(Aj,σ(n/2, N)) ≍ˆ P(Aj,σ(n, N)) (4.16) uniformly in p, ˆ P between Pp and P1−p and n0(j) ≤n ≤N ≤L(p).
Proof. This proposition comes directly from combining the arm separation theorem with the extend-ability property of the ˜ ˜ A events (item 1. of Proposition 12).
Quasi-multiplicativity Proposition 17. Take j ≥1 and a color sequence σ ∈˜ Sj. Then ˆ P(Aj,σ(n1, n2))ˆ P(Aj,σ(n2, n3)) ≍ˆ P(Aj,σ(n1, n3)) (4.17) uniformly in p, ˆ P between Pp and P1−p and n0(j) ≤n1 < n2 < n3 ≤L(p).
Proof. On one hand, we have ˆ P(Aj,σ(n1, n3)) ≤ˆ P(Aj,σ(n1, n2) ∩Aj,σ(n2, n3)) = ˆ P(Aj,σ(n1, n2))ˆ P(Aj,σ(n2, n3)) by independence of the events Aj,σ(n1, n2) and Aj,σ(n2, n3).
On the other hand, we may assume that n2 ≥8n1. Then for some η0, Iη0, the previous results (sep-aration and extendability) allow to use the quasi-multiplicativity for ˜ ˜ A events (item 2. of Proposition 12): ˆ P(Aj,σ(n1, n2))ˆ P(Aj,σ(n2, n3)) ≍ˆ P(Aj,σ(n1, n2/4))ˆ P(Aj,σ(n2, n3)) ≍ˆ P(˜ ˜ A ./η0,Iη0 j,σ (n1, n2/4))ˆ P(˜ ˜ A η0,Iη0/.
j,σ (n2, n3)) ≍ˆ P(Aj,σ(n1, n3)).
Arms with defects In some situations, the notion of arms that are completely monochromatic is too restrictive, and the following question arises quite naturally: do the probabilities change if we allow the arms to present some (fixed) number of “defects”, ie sites of the opposite color?
We define A(d) j,σ(n, N) the event that there exist j disjoint arms a1,..., aj from ∂Sn to ∂SN with the property: for any i ∈{1,..., j}, ai contains at most d sites of color ˜ σi. The quasi-multiplicativity property entails the following result, which will be needed for the proof of Theorem 27: Proposition 18. Let j ≥1 and σ ∈˜ Sj. Fix also some number d of defects. Then we have ˆ P A(d) j,σ(n, N) ≍(1 + log(N/n))d ˆ P Aj,σ(n, N) (4.18) uniformly in p, ˆ P between Pp and P1−p and n0(j) ≤n ≤N ≤L(p).
1585 Actually, we will only need the upper bound on ˆ P A(d) j,σ(n, N). For instance, we will see in the next section that the arm events decay like power laws at the critical point. This proposition thus implies, in particular, that the “arm with defects” events are described by the same exponents: allowing defects just adds a logarithmic correction.
Proof. We introduce a logarithmic division of the annulus Sn,N: we take k and K such that 2k−1 < n ≤2k and 2K ≤N < 2K+1. Roughly speaking, we “take away” the annuli where the defects take place, and “glue” the pieces of arms in the remaining annuli by using the quasi-multiplicativity property.
Let us begin with the upper bound: we proceed by induction on d. The property clearly holds for d = 0. Take some d ≥1: by considering the first annuli S2i,2i+1 where a defect occurs, we get ˆ P(A(d) j,σ(n, N)) ≤ K−1 X i=k ˆ P(Aj,σ(2k,2i))ˆ P(A(d−1) j,σ (2i+1,2K)).
(4.19) We have ˆ P(A(d−1) j,σ (2i+1,2K)) ≤Cd−1(1 + log(N/n))d−1ˆ P(Aj,σ(2i+1,2K)) thanks to the induction hy-pothesis, and by quasi-multiplicativity, ˆ P(A(d) j,σ(n, N)) ≤(1 + log(N/n))d−1Cd−1 K−1 X i=k ˆ P(Aj,σ(2k,2i))ˆ P(Aj,σ(2i+1,2K)) ≤Cd−1(1 + log(N/n))d−1 K−1 X i=k C′ˆ P(Aj,σ(2k,2K)) ≤Cd(1 + log(N/n))d−1(K −k)ˆ P(Aj,σ(2k,2K)), which gives the desired upper bound.
For the lower bound, note that for any k ≤i0 < i1 < ... < id < id+1 = K, A(d) j,σ(n, N) ⊇ A(d) j,σ(2k−1,2K+1) ⊇A(d) j,σ(2k−1,2K+1) ∩{Each of the j arms has exactly one defect in each of the annuli S2ir ,2ir +1, 1 ≤r ≤d}, so that for K −k ≥d + 1, ˆ P(A(d) j,σ(n, N)) ≥ X k=i0<i1<i2<...<id 1, we can get results uniform in the usual parameters and in parallelograms r, R such that Sn ⊆r ⊆Sτn and SN/τ ⊆R ⊆SN for some n, N ≤L(p): ˆ P(∂r ⇝j,σ ∂R) ≍ˆ P(∂Sn ⇝j,σ ∂SN), (4.20) and similarly with separateness conditions on the external boundary or on the internal one.
4.6 Arms in the half-plane So far, we have been interested in arm events in the whole plane: we can define in the same way the event Bj,σ(n, N) that there exist j arms that stay in the upper half-plane H, of colors prescribed by σ ∈˜ Sj and connecting ∂S′ n to ∂S′ N, with the notation ∂S′ n = (∂Sn) ∩H. These events appear naturally when we look at arms near a boundary.
For the sake of completeness, let us just mention that all the results stated here remain true for arms in the half-plane. In fact, there is a natural way to order the different arms, which makes this case easier. We will not use these events in the following, and we leave the details to the reader.
5 Description of critical percolation When studying the phase transition of percolation, the critical regime plays a very special role.
It possesses a strong property of conformal invariance in the scaling limit. This particularity, first observed by physicists ([41; 9; 10]), has been proved by Smirnov in , and later extended by Camia and Newman in . It allows to link the critical regime to the SLE processes (with parameter 6 here) introduced by Schramm in , and thus to use computations made for these processes ([34; 35]).
In the next sections, we will see why our description of critical percolation yields in turn a good description of near-critical percolation (which does not feature a priori any sort of conformal invari-ance), in particular how the characteristic functions behave through the phase transition.
5.1 Arm exponents for critical percolation Color switching We focus here on the probabilities of arm events at the critical point. For arms in the half-plane, a nice combinatorial argument (noticed in [3; 47]) shows that once fixed the number j of arms, pre-scribing the color sequence σ does not change the probability. This is the so-called “color exchange trick”: Proposition 19. Let j ≥1 be any fixed integer. If σ,σ′ are two color sequences, then for any n′ 0(j) ≤ n ≤N, P1/2(Bj,σ(n, N)) = P1/2(Bj,σ′(n, N)).
(5.1) Proof. The proof relies on the fact that there is a canonical way to order the arms. If we condition on the i left-most arms, percolation in the remaining domain is unbiased, so that we can “flip” the sites there: for any color sequence σ, if we denote by ˜ σ(i) = (σ1,...,σi, ˜ σi+1,..., ˜ σj) 1587 the sequence with the same i first colors, and the remaining ones flipped, then P1/2(Bj,σ(n, N)) = P1/2(Bj,˜ σ(i)(n, N)).
It is not hard to convince oneself that for any two sequences σ,σ′, we can go from σ to σ′ in a finite number of such operations.
This result is not as direct in the whole plane case, since there is no canonical ordering any more.
However, the argument can be adapted to prove that the probabilities change only by a constant factor, as long as there is an interface, ie as long as σ contains at least one white arm and one black arm.
Proposition 20. Let j ≥1 be any fixed integer. If σ,σ′ ∈˜ Sj are two non-constant color sequences (ie both colors are present), then P1/2(Aj,σ(n, N)) ≍P1/2(Aj,σ′(n, N)) (5.2) uniformly in n0(j) ≤n ≤N.
Proof. Assume that σ1 = B and σ2 = W, and fix some landing sequence I. If we replace the event Aj,σ(n, N) by the strengthened event ¯ AI/.
j,σ(n, N), we are allowed to condition on the black arm arriving on I1 and on the white arm arriving on I2 that are closest to each other: if we choose for instance I such that the point (N,0) is between I1 and I2, these two arms can be determined via an exploration process starting at (N,0). We can then “flip” the remaining region. More generally, we can condition on any set of consecutive arms including these two arms, and the result follows for the same reasons as in the half-plane case.
We would like to stress the fact that for the reasoning, we crucially need two arms of opposite colors.
In fact, the preceding result is expected to be false if σ is constant and σ′ non-constant (the two probabilities not being of the same order of magnitude), which may seem quite surprising at first sight.
Derivation of the exponents The link with SLE6 makes it possible to prove the existence of the (multichromatic) “arm exponents”, and derive their values ([36; 47]), that had been predicted in the physics literature (see e.g. and the references therein).
Theorem 21. Fix some j ≥1. Then for any non-constant color sequence σ ∈˜ Sj, P1/2 Aj,σ(n0(j), N) ≈N −αj (5.3) when N →∞, with • α1 = 5/48, • and for j ≥2, αj = (j2 −1)/12.
1588 Let us sketch very briefly how it is proved. Consider the discrete (radial) exploration process in a unit disc: using the property of conformal invariance in the scaling limit, we can prove that this process converges toward a radial SLE6, for which we can compute disconnection probabilities. It implies that P1/2(Aj,σ(ηn, n)) →g j(η), for some function g j(η) ∼ηαj as η →0. Then, the quasi-multiplicativity property in concentric annuli of fixed modulus provides the desired result.
As mentioned, this theorem is believed to be false for constant σ, ie when the arms are all of the same color. In this case, the probability should be smaller, or equivalently the exponent (assuming its exis-tence) larger. Hence for each j = 2,3,..., there are two different arm exponents, the multichromatic j-arm exponent αj given by the previous formula (most often simply called the j-arm exponent) and the monochromatic j-arm exponent α′ j, for which no closed formula is currently known, nor even predicted. The only result proved so far concerns the case j = 2: as shown in , the monochro-matic 2-arm exponent can be expressed as the leading eigenvalue of some (complicated) differential operator. Numerically, it has been found (see ) to be approximately α′ 2 ≃0.35...
Note also that the derivation using SLE6 only provides a logarithmic equivalence. However, there are reasons to believe that a stronger equivalence holds, a “≍”: for instance we know that this is the case for the “universal exponents” computed in the next sub-section.
We will often relate events to combinations of arm events, that in turn can be linked (see next section) to arm events at the critical point p = 1/2. It will thus be convenient to introduce the following notation, with σj = BW BW ...: for any n0(j) ≤n < N, πj(n, N) := P1/2(Aj,σj(n, N)) (5.4) (≍P1/2(Aj,σ(n, N)) for any non-constant σ), and in particular πj(N) := P1/2(Aj,σj(n0(j), N)) (≈N −αj).
(5.5) Note that with this notation, the a-priori bound and the quasi-multiplicativity property take the aesthetic forms Cj(n/N)αj ≤πj(n, N) ≤C′(n/N)α′, (5.6) and πj(n1, n2)πj(n2, n3) ≍πj(n1, n3).
(5.7) Let us mention that we can derive in the same way exponents for arms in the upper half-plane, the “half-plane exponents”: Theorem 22. Fix some j ≥1. Then for any sequence of colors σ, P1/2 Bj,σ(n′ 0(j), N) ≈N −βj (5.8) when N →∞, with βj = j(j + 1)/6.
Remark 23. As mentioned earlier, the triangular lattice is at present the only lattice for which conformal invariance in the scaling limit has been proved, and as a consequence the only lattice for which the existence and the values of the arm exponents have been established – with the noteworthy exception of the three “universal” exponents that we are going to derive.
1589 Note: fractality of various sets These arm exponents can be used to measure the size (Hausdorff dimension) of various sets describ-ing percolation clusters. In physics literature for instance (see e.g. ), a set S is said to be fractal of dimension DS if the density of points in S within a box of size n decays as n−xS, with xS = 2 −DS (in 2D). The co-dimension xS is related to arm exponents in many cases: • The 1-arm exponent is related to the existence of long connections, from the center of a box to its boundary. It will thus measure the size of “big” clusters, like the incipient infinite cluster (IIC) as defined by Kesten (), which scales as n(2−5/48) = n91/48.
• The monochromatic 2-arm exponent describes the size of the “backbone” of a cluster. The fact that this backbone is much thinner than the cluster itself was used by Kesten to prove that the random walk on the IIC is sub-diffusive (while it has been proved to converge toward a Brownian Motion on a super-critical infinite cluster).
• The multichromatic 2-arm exponent is related to the boundaries (hulls) of big clusters, which are thus of fractal dimension 2 −α2 = 7/4.
• The 3-arm exponent concerns the external (accessible) perimeter of a cluster, which is the accessible part of the boundary: one excludes “fjords” which are connected to the exterior only by 1-site wide passages. The dimension of this frontier is 2 −α3 = 4/3. These two latter exponents can be observed on random interfaces, numerically and in “real-life” experiments as well (see [43; 21] for instance).
• As mentioned earlier, the 4-arm exponent with alternating colors counts the pivotal (singly-connecting) sites (often called “red” sites in physics literature). This set can be viewed as the contact points between two distinct (large) clusters, its dimension is 2 −α4 = 3/4. We will relate this exponent to the characteristic length exponent ν in Section 7.
5.2 Universal exponents We will now examine as a complement some particular exponents, for which heuristic predictions and elementary derivations exist, namely β2 = 1, β3 = 2 and α5 = 2. They are all integers, and they were established before the complete derivation using SLE6 (and actually they provide crucial a-priori estimates to prove the convergence toward SLE6). Moreover, the equivalence that we get is stronger: we can replace the “≈” by a “≍”.
Theorem 24. When N →∞, 1. For any σ ∈S2, P1/2 B2,σ(0, N) ≍N −1.
2. For any σ ∈S3, P1/2 B3,σ(0, N) ≍N −2.
1590 ∂SN I1 I2 I3 I4 I5 v Figure 10: The landing sequence I1,..., I5.
3. For any non-constant σ ∈˜ S5, P1/2 A5,σ(0, N) ≍N −2.
Proof. We give a complete proof only for item 3., since we will not need the two first ones – we will however sketch at the end how to derive them.
Heuristically, we can prove that the 5-arm sites can be seen as particular points on the boundary of two big black clusters, and that consequently their number is of order 1 in SN/2. Then it suffices to use the fact that the different sites in SN/2 produce contributions of the same order. This argument can be made rigorous by proving that the number of “macroscopic” clusters has an exponential tail: we refer to the first exercise sheet in for more details. We propose here a more direct – but less elementary – proof using the separation lemmas (see , Lemma 5).
By color switching, it is sufficient to prove the claim for σ = BW BBW. In light of our previous results, it is clear that P1/2 v ⇝5,σ ∂SN) ≍P1/2 0 ⇝5,σ ∂SN) uniformly in N, v ∈SN/2. It is thus enough to prove that the number of such 5-arm sites in SN/2 is of order 1.
Let us consider the upper bound first. Take the particular landing sequence I1,..., I5 depicted on Figure 10, and consider the event Av := {v ⇝I 5,σ ∂SN} ∩{v is black}.
Note that P1/2(Av) = 1 2P(v ⇝I 5,σ ∂SN) since the existence of the arms is independent of the status of v, so that P1/2(Av) ≍P1/2 0 ⇝5,σ ∂SN). We claim that Av can occur for at most one site v. Indeed, assume that Av and Aw occur, and denote by r1,..., r5 and r′ 1,..., r′ 5 the corresponding arms. Since r1 ∪r4 ∪{v} separates I3 from I5, necessarily w ∈r1 ∪r4 ∪{v}. Similarly, w ∈r2 ∪r4 ∪{v}: since r1 ∩r2 = ∅, we get that w ∈r4 ∪{v}. But only one arm can “go through” r3 ∪r5: the arm r′ 1 ∪{w} from w to I1 has to contain v, and so does r′ 2 ∪{w}. Since r′ 1 ∩r′ 2 = ∅, we get finally v = w.
1591 Consequently, 1 ≥P1/2 ∪v∈SN/2 Av = X v∈SN/2 P1/2(Av) ≍N 2P1/2 0 ⇝5,σ ∂SN), (5.9) which provides the upper bound.
Let us turn to the lower bound. We perform a construction showing that a 5-arm site appears with positive probability, by using multiple applications of RSW . With a probability of at least δ2 16 > 0, there is a black horizontal crossing in the strip [−N, N] × [0, N/8], together with a white one in [−N, N] × [−N/8,0]. Assume this is the case, and condition on the lowest black left-right crossing c. We note that any site on this crossing has already 3 arms, 2 black arms and a white one. On the other hand, percolation in the region above it remains unbiased.
Now, still using RSW , with positive probability c is connected to the top side by a black path included in [−N/8,0] × [−N, N], and another white path included in [0, N/8] × [−N, N]. Let us assume that these paths exist, and denote by v1 and v2 the respective sites on c where they arrive. Let us then follow c from left to right, and consider the last vertex v before v2 that is connected to the top side: it is not hard to see that there is a white arm from v to the top side, and that v ∈SN/2, since v is between v1 and v2. Hence, P1/2 ∪v∈SN/2 {v ⇝5,σ ∂SN} ≥C (5.10) for some universal constant C > 0. Since we also have P1/2 ∪v∈SN/2 {v ⇝5,σ ∂SN} ≤ X v∈SN/2 P1/2 v ⇝5,σ ∂SN) ≤C′N 2P1/2 0 ⇝5,σ ∂SN), the desired lower bound follows.
We now explain briefly how to obtain the two half-plane exponents (items 1. and 2.). We again use the arm separation theorem, but note that contains elementary proofs for them too. For the 2-arm exponent in the half-plane, we take σ = BW and remark that if we fix two landing areas I1 and I2 on ∂S′ N, at most one site on the segment [−N/2, N/2] × {0} is connected by two arms to I1 and I2. On the other hand, a 2-arm site can be constructed by considering a black path from [−N/2,0] × {0} to I1 and a white path from [0, N/2] × {0} to I2. Then the right-most site on [−N/2, N/2] × {0} connected by a black arm to I1 is a 2-arm site. Several applications of RSW allow to conclude.
For the 3-arm exponent, we take three landing areas I1, I2 and I3, and σ = BW B. It is not hard to construct a 3-arm site by taking a black crossing from I1 to I3 and considering the closest to I2.
We can then force it to be in SN/2 ∩H by a RSW construction. For the upper bound, we first notice that if we require the arms to stay strictly positive (except in the sites neighboring the origin), the probability remains of the same order of magnitude. We then use the fact that at most three sites in SN/2 ∩H are connected to the landing areas by three positive arms.
The proofs given here only require RSW-type considerations (including separation of arms). As a consequence, they also apply to near-critical percolation. It is clear for Pp, on scales N ≤L(p), but a priori only for the color sequences we have used in the proofs (resp. σ = BW, BW B and BW BBW – 1592 and of course those we can deduce from them by the symmetry p ↔1−p): it is indeed not obvious that Pp 0 ⇝5,σ ∂SN ≍Pp 0 ⇝5,σ′ ∂SN for two distinct non-constant σ and σ′. This is essentially Theorem 27, its proof occupies a large part of the next section.
For a general measure ˆ P between Pp and P1−p, we similarly have to be careful: we do not know whether ˆ P(v ⇝5,σ ∂SN) remains of the same order of magnitude when v varies. This also comes from Theorem 27, but in the course of its proof we will need an a-priori estimate on the probability of 5 arms, so temporarily we will be content with a weaker statement that does not use its conclusion: Lemma 25. For σ = BW BBW (= σ5), we have uniformly in p, ˆ P between Pp and P1−p and N ≤L(p): X v∈SN/2 ˆ P v ⇝5,σ ∂SN ≍ X v∈SN/8 ˆ P v ⇝5,σ ∂SN ≍1.
(5.11) Remark 26. We would like to mention that these estimates for critical and near-critical percolation remain also valid on other lattices, like the square lattice (see the discussion in the last section) – at least for the color sequences that we have used in the proofs, no analog of the color exchange trick being available (to our knowledge).
6 Arm events near criticality 6.1 Statement of the theorem We would like now to study how the events Aj,σ(n, N) are affected by a variation of the parameter p. We have defined L(p) in terms of crossing events to be the scale on which percolation can be considered as (approximately) critical, we would thus expect the probabilities of these events not to vary too much if n, N remain below L(p). This is what happens: Theorem 27. Let j ≥1, σ ∈˜ Sj be as usual. Then we have ˆ P Aj,σ(n, N) ≍ˆ P′ Aj,σ(n, N) (6.1) uniformly in p, ˆ P and ˆ P′ between Pp and P1−p, and n0(j) ≤n ≤N ≤L(p).
Note that if we take in particular ˆ P′ = P1/2, we get that below the scale L(p), the arm events remain roughly the same as at criticality: ˆ P Aj,σ(n, N) ≍P1/2 Aj,σ(n, N).
This will be important to derive the critical exponents for the characteristic functions from the arm exponents at criticality.
Remark 28. Note that the property of exponential decay with respect to L(p) (Lemma 39), proved in Section 7.4, shows that we cannot hope for a similar result on a much larger range, so that L(p) is the appropriate scale here: consider for instance Pp with p > 1/2, the probability to observe a white arm tends to 0 exponentially fast (and thus much faster than at the critical point), while the probability to observe a certain number of disjoint black arms tends to a positive constant.
1593 6.2 Proof of the theorem We want to compare the value of ˆ P(Aj,σ(n, N)) for different measures ˆ P. A natural way of doing this is to go from one to the other by using Russo’s formula (Theorem 1). But since for j ≥2 and non-constant σ, the event Aj,σ(n, N) is not monotone, we need a slight generalization of this formula, for events that can be expressed as the intersection of two monotone events, one increasing and one decreasing. We also allow the parameters pv to be differentiable functions of t ∈[0,1].
Lemma 29. Let A+ and A−be two monotone events, respectively increasing and decreasing, depending only on the sites contained in some finite set of vertices S. Let (ˆ pv)v∈S be a family of differentiable functions ˆ pv : t ∈[0,1] 7→ˆ pv(t) ∈[0,1], and denote by (ˆ Pt)0≤t≤1 the associated product measures.
Then d dt ˆ Pt(A+ ∩A−) = X v∈S d dt ˆ pv(t) h ˆ Pt(v is pivotal for A+ but not for A−, and A−occurs) −ˆ Pt(v is pivotal for A−but not for A+, and A+ occurs) i .
Proof. We adapt the proof of standard Russo’s formula. We use the same function P of the param-eters (ˆ pv)v∈S, and we note that for a small variation ε > 0 in w, ˆ P+ε(A+ ∩A−) −ˆ P(A+ ∩A−) = ε × ˆ P(w is pivotal for A+ but not for A−, and A−occurs) −ε × ˆ P(w is pivotal for A−but not for A+, and A+ occurs).
Now, it suffices to compute the derivative of the function t 7→ˆ Pt(A+ ∩A−) by writing it as the composition of t 7→(ˆ pv(t)) and (ˆ pv)v∈S 7→ˆ P(A).
Remark 30. Note that if we take A−= Ωin Lemma 29, we get usual Russo’s formula for A+, with parameters that can be functions of t.
Proof of the theorem. We now turn to the proof itself. It is divided into three main steps.
1. First simplifications Note first that by quasi-multiplicativity, we can restrict ourselves to n = n0(j). It also suffices to prove the result for some fixed ˆ P′, with ˆ P varying: we thus assume that p < 1/2, and take ˆ P′ = Pp.
Denoting by ˆ pv the parameters of ˆ P, we have by hypothesis ˆ pv ≥p for each site v. For technical reasons, we suppose that the sizes of annuli are powers of two: take k0, K such that 2k0−1 < n0 ≤2k0 and 2K ≤N < 2K+1, then Pp(Aj,σ(n0, N)) ≍Pp(Aj,σ(2k0,2K)) and the same is true for ˆ P.
To estimate the change in probability when p is replaced by ˆ pv, we will use the observation that the pivotal sites give rise to 4 alternating arms locally (see Figure 11). However, this does not work so 1594 v ∂S2l ∂S2l+3 ∂S2K ∂S2l(v) Figure 11: If v is pivotal, 4 alternating arms arise locally.
nicely for the sites v which are close to ∂S2k0 or ∂S2K, so for the sake of simplicity we treat aside these sites. We perform the change p ⇝ˆ pv in S2k0,2K \S2k0+3,2K−3. Note that the intermediate measure ˜ P so obtained is between Pp and P1−p, and that ˜ P(Aj,σ(2k0+3,2K−3)) = Pp(Aj,σ(2k0+3,2K−3)). We have ˜ P(Aj,σ(2k0,2K)) ≍˜ P(Aj,σ(2k0+3,2K−3)) (6.2) and also Pp(Aj,σ(2k0,2K)) ≍Pp(Aj,σ(2k0+3,2K−3)), (6.3) which shows that it would be enough to prove the result with ˜ P instead of Pp.
2. Make appear the logarithmic derivative of the probability by applying Russo’s formula The event Aj,σ(2k0,2K) cannot be directly written as an intersection like in Russo’s formula, since the order of the different arms is prescribed. To fix this difficulty, we impose the landing areas of the different arms on ∂S2K, ie we fix some landing sequence I′ = I′ 1,..., I′ j and we consider the event ¯ A./I′ j,σ(2k0,2K). Since we know that ˜ P Aj,σ(2k0,2K) ≍˜ P ¯ A./I′ j,σ(2k0,2K), (6.4) and also with ˆ P instead of ˜ P, it is enough to prove the result for this particular landing sequence.
We study successively three cases. We begin with the case of one arm, which is slightly more direct than the two next ones – however, only small adaptations are needed. We then consider the special case where j is even and σ alternating: due to the fact that any arm is surrounded by two arms of opposite color, the local four arms are always long enough. We finally prove the result for any j and any σ: a technical complication arises in this case, for which the notion of “arms with defects” is needed.
Case 1: j = 1 1595 We first consider the case of one arm, and assume for instance σ = B. We introduce the family of measures (˜ Pt)t∈[0,1] with parameters ˜ pv(t) = tˆ pv + (1 −t)p (6.5) in S2k0+3,2K−3, corresponding to a linear interpolation between p and ˆ pv. For future use, note that ˜ Pt is between Pp and P1−p for any t ∈[0,1]. We have d dt ˜ pv(t) = ˆ pv −p if v ∈S2k0+3,2K−3 (and 0 otherwise), generalized Russo’s formula (with just an increasing event – see Remark 30) thus gives: d dt ˜ Pt ¯ A./I′ 1,σ(2k0,2K) = X v∈S2k0+3,2K−3 (ˆ pv −p)˜ Pt(v is pivotal for ¯ A./I′ 1,σ(2k0,2K)).
The key remark is that the summand can be expressed in terms of arm events: for probabilities, being pivotal is approximately the same as having a black arm, and four arms locally around v.
Indeed, v is pivotal for ¯ A./I′ 1,σ(2k0,2K) iff (1) there exists an arm r1 from ∂S2k0 to I′ 1, with v ∈r1; r1 is black, with a possible exception in v (¯ A./I′ 1,σ(2k0,2K) occurs when v is black), (2) there exists a path c1 passing through v and separating ∂S2k0 from I′ 1 (c1 may be either a circuit around ∂S2k0 or a path with extremities on ∂S2K); c1 is white, except possibly in v (there is no black arm from ∂S2k0 to I′ 1 when v is white).
We now put a rhombus R(v) around v: if it does not contain 0, then v is connected to ∂R(v) by 4 arms of alternating colors. Indeed, r1 provides two black arms, and c1 two white arms.
Let us look at the pieces of the black arm outside of R(v): if R(v) is not too large, we can expect them to be sufficiently large to enable us to reconstitute the whole arm. We also would like that the two white arms are a good approximation of the whole circuit. We thus take R(v) of size comparable to the distance d(0, v): if 2l+1 < ∥v∥∞≤2l+2, we take R(v) = S2l(v). It is not hard to check that R(v) ⊆S2l,2l+3 for this particular choice of R(v) (see Figure 11), so that for any t ∈[0,1], ˜ Pt(v is pivotal for ¯ A./I′ 1,σ(2k0,2K)) ≤˜ Pt {∂S2k0 ⇝∂S2l} ∩{∂S2l+3 ⇝∂S2K} ∩{v ⇝4,σ4 ∂S2l(v)} = ˜ Pt ∂S2k0 ⇝∂S2l ˜ Pt ∂S2l+3 ⇝∂S2K˜ Pt v ⇝4,σ4 ∂S2l(v) by independence of the three events, since they are defined in terms of sites in disjoint sets (recall that σ4 = BW BW). We can then make appear the original event by combining the two first terms, using quasi-multiplicativity and extendability2: ˜ Pt ∂S2k0 ⇝∂S2l ˜ Pt ∂S2l+3 ⇝∂S2K ≤C2˜ Pt ¯ A./I′ 1,σ(2k0,2K) (6.6) for some C2 universal. Hence3, ˜ Pt(v is pivotal for ¯ A./I′ 1,σ(2k0,2K)) ≤C2˜ Pt ¯ A./I′ 1,σ(2k0,2K)˜ Pt v ⇝4,σ4 ∂S2l(v).
(6.7) 2Note that in the case of one arm, the extendability property, as well as the quasi-multiplicativity, are direct conse-quences of RSW and do not require the separation lemmas.
3As we will see in the next sub-section (Proposition 32), the converse bound also holds: the estimate obtained gives the exact order of magnitude for the summand.
1596 We thus get d dt ˜ Pt ¯ A./I′ 1,σ(2k0,2K) ≤C2 X v∈S2k0+3,2K−3 (ˆ pv −p)˜ Pt ¯ A./I′ 1,σ(2k0,2K)˜ Pt v ⇝4,σ4 ∂S2l(v).
Now, dividing by ˜ Pt ¯ A./I′ 1,σ(2k0,2K), we make appear its logarithmic derivative in the left-hand side, d dt log˜ Pt ¯ A./I′ 1,σ(2k0,2K) ≤C2 X v∈S2k0+3,2K−3 (ˆ pv −p)˜ Pt v ⇝4,σ4 ∂S2l(v), (6.8) it thus suffices to show that for some C3 universal, Z 1 0 X v∈S2k0+3,2K−3 (ˆ pv −p) ˜ Pt v ⇝4,σ4 ∂S2l(v)dt ≤C3.
(6.9) We will prove this in the next step, but before that, we turn to the two other cases: even if the computations need to be modified, it is still possible to reduce the proof to this inequality.
Case 2: j even and σ alternating In this case, ¯ A./I′ j,σ(2k0,2K) = A+ ∩A− (6.10) with A+ = A+(2k0,2K) = {There exist j/2 disjoint black arms r1 : ∂S2k0 ⇝I′ 1, r3 : ∂S2k0 ⇝I′ 3 ...} and A−= A−(2k0,2K) = {There exist j/2 disjoint white arms r2 : ∂S2k0 ⇝∗I′ 2, r4 : ∂S2k0 ⇝∗I′ 4 ...}.
We then perform the change p ⇝ˆ pv in S2k0+3,2K−3 linearly as before (Eq.(6.5)), which gives rise to the family of measures (˜ Pt)t∈[0,1], and generalized Russo’s formula reads d dt ˜ Pt ¯ A./I′ j,σ(2k0,2K) = X v∈S2k0+3,2K−3 (ˆ pv −p) h ˜ Pt(v is pivotal for A+ but not for A−, and A−occurs) −˜ Pt(v is pivotal for A−but not for A+, and A+ occurs) i .
We note that v is pivotal for A+(2k0,2K) but not for A−(2k0,2K), and A−(2k0,2K) occurs iff for some i′ ∈{1,3..., j −1}, (1) there exist j disjoint monochromatic arms r1,..., rj from ∂S2k0 to I′ 1,..., I′ j, with v ∈ri′; r2, r4,... are white, and r1, r3,... are black, with a possible exception for ri′ in v (the event ¯ A./I′ j,σ(2k0,2K) is satisfied when v is black), (2) there exists a path ci′ separating ∂S2k0 from I′ i′; this path is white, except possibly in v (∂S2k0 and I′ i′ are separated when v is white).
1597 If we take the same rhombus R(v) ⊆S2l,2l+3 around v, then v is still connected to ∂R(v) by 4 arms of alternating colors. Indeed, ri′ provides two black arms, and ci′ (which can contain parts of ri′−1 or ri′+1 – see Figure 11) provides the two white arms.
Hence for any t ∈[0,1], ˜ Pt(v is pivotal for A+ but not for A−, and A−occurs) ≤˜ Pt Aj,σ(2k0,2l) ∩¯ A./I′ j,σ(2l+3,2K) ∩{v ⇝4,σ4 ∂S2l(v)} = ˜ Pt Aj,σ(2k0,2l)˜ Pt ¯ A./I′ j,σ(2l+3,2K)˜ Pt v ⇝4,σ4 ∂S2l(v) by independence of the three events. We then combine the two first terms using extendability and quasi-multiplicativity: ˜ Pt Aj,σ(2k0,2l)˜ Pt ¯ A./I′ j,σ(2l+3,2K) ≤C1˜ Pt ¯ A./I′ j,σ(2k0,2K) (6.11) for some C1 universal. We thus obtain ˜ Pt(v is pivotal for A+ but not for A−, and A−occurs) ≤C1˜ Pt ¯ A./I′ j,σ(2k0,2K)˜ Pt v ⇝4,σ4 ∂S2l(v).
If we then do the same manipulation on the second term of the sum, we get ¯ ¯ ¯ ¯ d dt ˜ Pt ¯ A./I′ j,σ(2k0,2K) ¯ ¯ ¯ ¯ ≤2C1 X v∈S2k0+3,2K−3 (ˆ pv −p)˜ Pt ¯ A./I′ j,σ(2k0,2K)˜ Pt v ⇝4,σ4 ∂S2l(v), and if we divide by ˜ Pt ¯ A./I′ j,σ(2k0,2K), ¯ ¯ ¯ ¯ d dt log˜ Pt ¯ A./I′ j,σ(2k0,2K) ¯ ¯ ¯ ¯ ≤2C1 X v∈S2k0+3,2K−3 (ˆ pv −p)˜ Pt v ⇝4,σ4 ∂S2l(v).
(6.12) As promised, we have thus reduced this case to Eq.(6.9).
Case 3: Any j, σ In the general case, a minor complication may arise, coming from consecutive arms of the same color: indeed, the property of being pivotal for a site v does not always give rise to four arms in R(v), but to some more complex event E(v) (see Figure 12). If v is on ri, and this arm is black for instance, there are still two black arms coming from ri, but the two white arms do not necessarily reach ∂R(v), since they can encounter neighboring black arms.
We first introduce an event for which the property of being pivotal is easier to formulate. We group consecutive arms of the same color in “packs”: if (riq, riq+1,..., riq+lq−1) is such a sequence of arms, say black, we take an interval ˜ Iq covering all the Ii for iq ≤i ≤iq + lq −1 and replace the condition “ri ⇝Ii for all iq ≤i ≤iq + lq −1” by “ri ⇝˜ Iq for all iq ≤i ≤iq + lq −1”. We construct in this way an event ˜ A = ˜ A+ ∩˜ A−: since it is intermediate between ¯ A./I′ j,σ(2k0,2K) and Aj,σ(2k0,2K), we have ˜ Pt(˜ A) ≍˜ Pt(¯ A./I′ j,σ(2k0,2K)).
1598 ∂S2l(v) v ∂S2l′+1(v) ∂S2l′ (v) Figure 12: More complex events may arise when σ is not alternating.
This new definition allows to use Menger’s theorem (see , Theorem 3.3.1): v is pivotal for ˜ A+ but not for ˜ A−, and ˜ A−occurs iff for some arm ri′ in a black pack (riq, riq+1,..., riq+lq−1), (1) there exist j disjoint monochromatic arms r1,..., rj from ∂S2k0 to ˜ Iq (an appropriate number of arms for each of these intervals), with v ∈ri′; all of these arms are of the prescribed color, with a possible exception for ri′ in v (˜ A occurs when v is black), (2) there exists a path ci′ separating ∂S2k0 from ˜ Iq; this path is white, except in (at most) lq −1 sites, and also possibly in v (∂S2k0 and ˜ Iq can be separated by turning white lq −1 sites when v is white).
Now, we take once again the same rhombus R(v) ⊆S2l,2l+3 around v: if there are four arms v ⇝4,σ4 ∂S2l−1(v), we are OK. Otherwise, if l′, 1 ≤l′ ≤l −2, is such that the defect on ci′ closest to v is in S2l′+1(v) \ S2l′(v), then there are 4 alternating arms v ⇝4,σ4 ∂S2l′(v), and also 6 arms ∂S2l′+1(v) ⇝(j) 6,σ′ 6 ∂S2l(v) having at most j defects, with the notation σ′ 6 = BBW BBW. We denote by E(v) the corresponding event: E(v) := {There exists l′ ∈{1,..., l −2} such that v ⇝4,σ4 ∂S2l′(v) and ∂S2l′+1(v) ⇝(j) 6,σ′ 6 ∂S2l(v)} ∪{v ⇝4,σ4 ∂S2l−1(v)}.
For the 6 arms with defects, Proposition 18 applies and the probability remains roughly the same, with just an extra logarithmic correction: ˜ Pt ∂S2l′+1(v) ⇝(j) 6,σ′ 6 ∂S2l(v) ≤C1(l −l′)j˜ Pt ∂S2l′+1(v) ⇝6,σ′ 6 ∂S2l(v) ≤C1(l −l′)j˜ Pt ∂S2l′+1(v) ⇝4,σ4 ∂S2l(v)˜ Pt ∂S2l′+1(v) ⇝2,BB ∂S2l(v) ≤C2(l −l′)j˜ Pt ∂S2l′+1(v) ⇝4,σ4 ∂S2l(v)2−α′(l−l′) using Reimer’s inequality (its consequence Eq.(4.2)) and the a-priori bound for one arm (Eq.(4.10)).
1599 This implies that ˜ Pt E(v) ≤C5˜ Pt v ⇝4,σ4 ∂S2l(v) (6.13) for some universal constant C5: indeed, by quasi-multiplicativity, l−2 X l′=1 ˜ Pt v ⇝4,σ4 ∂S2l′(v)˜ Pt ∂S2l′+1(v) ⇝(j) 6,σ′ 6 ∂S2l(v) ≤C2 l−2 X l′=1 ˜ Pt v ⇝4,σ4 ∂S2l′(v)˜ Pt ∂S2l′+1(v) ⇝4,σ4 ∂S2l(v)(l −l′)j2−α′(l−l′) ≤C3˜ Pt v ⇝4,σ4 ∂S2l(v) l−2 X l′=1 (l −l′)j2−α′(l−l′) ≤C4˜ Pt v ⇝4,σ4 ∂S2l(v), since Pl−2 l′=1(l −l′)j2−α′(l−l′) ≤ P∞ r=1 r j2−α′r < ∞.
The reasoning is then identical: ˜ Pt(v is pivotal for ˜ A+ but not for ˜ A−, and ˜ A−occurs) ≤˜ Pt Aj,σ(2k0,2l)˜ Pt Aj,σ(2l+3,2K)˜ Pt E(v) ≤C6˜ Pt Aj,σ(2k0,2K)˜ Pt v ⇝4,σ4 ∂S2l(v), and using ˜ Pt Aj,σ(2k0,2K) ≤C7˜ Pt(˜ A), we get ¯ ¯ ¯ ¯ d dt log˜ Pt ˜ A ¯ ¯ ¯ ¯ ≤C8 X v∈S2k0+3,2K−3 (ˆ pv −p)˜ Pt v ⇝4,σ4 ∂S2l(v).
(6.14) Once again, Eq.(6.9) would be sufficient.
3. Final summation We now prove Eq.(6.9), ie that for some universal constant C1, Z 1 0 X v∈S2k0+3,2K−3 (ˆ pv −p) ˜ Pt v ⇝4,σ4 ∂S2l(v)dt ≤C1.
(6.15) Recall that Russo’s formula allows to count 4-arm sites: for any N and any measure ¯ P between Pp and P1−p, Z 1 0 X v∈SN (¯ pv −p) ¯ Pt v ⇝./¯ I 4,σ4 ∂SN dt = ¯ P(CH(SN)) −Pp(CH(SN)) ≤1 (6.16) (we remind that ¯ I consists of the different sides of ∂SN). This is essentially the only relation we have at our disposal, the end of the proof consists in using it in a clever way.
1600 Roughly speaking, when applied to N = L(p), this relation gives that (p −1/2)N 2π4(N) ≤1, since all the sites give a contribution of order ¯ Pt 0 ⇝4,σ4 ∂SN/2 ≍π4(N).
(6.17) This corresponds more or less to the sites in the “external annulus” in Eq.(6.15). Now each time we get from an annulus to the next inside it, the probability to have 4 arms is multiplied by 2α4 ≈25/4, while the number of sites is divided by 4, so that things decay exponentially fast, and the sum of Eq.(6.15) is bounded by something like K−4 X j=k0+3 (25/4−2)K−4−j ≤ ∞ X q=0 (2−3/4)q < ∞.
We have to be more cautious, in particular Eq.(6.17) does not trivially hold, since we do not know at this point that the probability of having 4 arms remains of the same order on a scale L(p) (and the estimate for 4 arms only gives a logarithmic equivalence). The a-priori estimate coming from the 5-arm exponent will allow us to circumvent these difficulties. We also need to take care of the boundary effects.
Assume that v ∈S2l+1,2l+2 as before. We subdivide this annulus into 12 sub-boxes of size 2l+1 (see Figure 13) ˜ Ri 2l+1 (i = 1,...,12). At least one of these boxes contains v: we denote it by ˜ R(v). We then associate to each of these boxes a slightly enlarged box ˜ R′i 2l+1 of size 2l+2, and we also use the obvious notation ˜ R′(v). Since {v ⇝4,σ4 ∂S2l+2(v)} ⊆{v ⇝4,σ4 ∂˜ R′(v)} ⊆{v ⇝4,σ4 ∂S2l(v)}, we have ˜ Pt v ⇝4,σ4 ∂S2l(v) ≍˜ Pt v ⇝4,σ4 ∂˜ R′(v).
We thus have to find an upper bound for K−4 X j=k0+3 12 X i=1 Z 1 0 X v∈˜ Ri 2j (ˆ pv −p) ˜ Pt v ⇝4,σ4 ∂˜ R′i 2j dt.
(6.18) For that purpose, we will prove that for i = 1,...,12, and fixed t ∈[0,1], Si,(4) j := X v∈˜ Ri 2j (ˆ pv −p) ˜ Pt v ⇝4,σ4 ∂˜ R′i 2j indeed decays fast when, starting from j = K −4, we make j decrease. For that, we duplicate the parameters in the box ˜ R′i 2j periodically inside S2K−3: this gives rise to a new measure ¯ P inside S2K (to completely define it, simply take ¯ pv = p outside of S2K−3). This measure contains 22(K−j−3) copies of the original box (of size 2j+1), that we denote by (¯ R′ q). We also consider ¯ Rq the box of size 2j centered inside ¯ R′ q. We know that Z 1 0 X v∈S2K−3 (¯ pv −p) ¯ Pt v ⇝4,σ4 ∂S2Kdt ≤1.
(6.19) 1601 ∂S2l+2 ∂S2l+1 v 0 Figure 13: We replace R(v) = S2l(v) by one of the ˜ R′i 2l+1 (i = 1,...,12).
If we restrict the summation to the sites in the union of the ¯ Rq’s, we get that X v∈S2K−3 (¯ pv −p)¯ Pt(v ⇝4,σ4 ∂S2K) ≥ X q X v∈¯ Rq (¯ pv −p)¯ Pt(v ⇝4,σ4 ∂S2K) ≥C1 X q X v∈¯ Rq (¯ pv −p)¯ Pt v ⇝4,σ4 ∂¯ R′ q ¯ Pt ∂¯ R′ q ⇝4,σ4 ∂S2K = C1 X q ¯ Pt ∂¯ R′ q ⇝4,σ4 ∂S2K Si,(4) j .
Hence, using Reimer’s inequality and the a-priori bound for one arm, X v∈S2K−3 (¯ pv −p)¯ Pt(v ⇝4,σ4 ∂S2K) ≥C2 X q ¯ Pt ∂¯ R′ q ⇝5,σ5 ∂S2K¯ Pt ∂¯ R′ q ⇝∂S2K−1 Si,(4) j ≥C32α′(K−j)Si,(4) j X q ¯ Pt ∂¯ R′ q ⇝5,σ5 ∂S2K .
The same type of manipulation for 5 arms gives, introducing ¯ R′′ q the box of size 2j+2 centered on ¯ R′ q, X v∈S2K−3 ¯ Pt(v ⇝5,σ5 ∂S2K) ≤ X q X v∈¯ R′ q ¯ Pt v ⇝5,σ5 ∂¯ R′′ q ¯ Pt ∂¯ R′′ q ⇝5,σ5 ∂S2K ≤C4 X q ¯ Pt ∂¯ R′′ q ⇝5,σ5 ∂S2K , since we know from Lemma 25 that P v∈¯ R′ q ¯ Pt v ⇝5,σ5 ∂¯ R′′ q ≍1.
We also know that P v∈S2K−3 ¯ Pt(v ⇝5,σ5 ∂S2K) ≍1 (still by Lemma 25) and ¯ Pt ∂¯ R′′ q ⇝5,σ5 ∂S2K ≍¯ Pt ∂¯ R′ q ⇝5,σ5 1602 ∂S2K, we thus have X q ¯ Pt ∂¯ R′ q ⇝5,σ5 ∂S2K ≥C5 (6.20) for some C5 > 0. This implies that Si,(4) j ≤C62−α′(K−j) X v∈S2K−3 (¯ pv −p)¯ Pt(v ⇝4,σ4 ∂S2K), and finally, by integrating and using Eq.(6.19), Z 1 0 X v∈˜ Ri 2j (ˆ pv −p) ˜ Pt v ⇝4,σ4 ∂˜ R′i 2j dt ≤C62−α′(K−j).
The sum of Eq.(6.18) is thus less than K−4 X j=k0+3 12C62−α′(K−j) ≤C7 ∞ X r=0 2−α′r < ∞, which completes the proof.
Remark 31. We will use this theorem in the next section to relate the so-called “characteristic functions” to the arm exponents at criticality. We will have use in fact only for the two cases j = 1 and j = 4, σ = σ4: the general case (3rd case in the previous proof) will thus not be needed there. It is however of interest for other applications, for instance to say that for an interface in near-critical percolation, the dimension of the accessible perimeter is the same as at criticality: this requires the case j = 3, σ = σ3.
6.3 Some complements Theorem for more general annuli We will sometimes need a version of Theorem 27 with non concentric rhombi. For instance, for any fixed η > 0, ˆ P(∂Sn(v) ⇝∂SN) ≍P1/2(∂Sn ⇝∂SN) (6.21) uniformly in v ∈S(1−η)N. It results from the remark on more general annuli (Eq.(4.20)) combined with Theorem 27 applied to ˆ Pv, the measure ˆ P translated by v.
A complementary bound Following the same lines as in the previous proof, we can get a bound in the other direction: Proposition 32. There exists some universal constant ˜ C > 1 such that for all p > 1/2, Pp 0 ⇝∂SL(p) ≥˜ C P1/2 0 ⇝∂SL(p) .
(6.22) In other words, the one-arm probability varies of a non-negligible amount, like the crossing proba-bility: there is a macroscopic difference with the critical regime.
1603 Proof. Take K such that 2K ≤L(p) < 2K+1 and (ˆ Pt) the linear interpolation between P1/2 and Pp.
By gluing arguments, for A = {0 ⇝∂SL(p)}, for any v ∈S2K−4,2K−3, ˜ Pt(v is pivotal for A) ≥C1˜ Pt 0 ⇝∂S2K−5˜ Pt ∂S2K−2 ⇝∂SL(p) ˜ Pt v ⇝4,σ4 ∂S2K−5(v) ≥C2˜ Pt 0 ⇝∂S2K˜ Pt v ⇝4,σ4 ∂S2K−5(v), so that d dt log˜ Pt(A) ≥ X v∈S2K−4,2K−3 (p −1/2)ˆ Pt v ⇝4,σ4 ∂S2K−5(v) ≥C3(p −1/2)L(p)2ˆ Pt 0 ⇝4,σ4 ∂SL(p) , since each of the sites v ∈S2K−4,2K−3 produces a contribution of order ˆ Pt 0 ⇝4,σ4 ∂SL(p) . Proposi-tion 34, proved later4, allows to conclude.
7 Consequences for the characteristic functions 7.1 Different characteristic lengths Roughly speaking, a characteristic length is a quantity intended to measure a “typical” scale of the system. There may be several natural definitions of such a length, but we usually expect the different possible definitions to produce lengths that are of the same order of magnitude. For two-dimensional percolation, the three most common definitions are the following: Finite-size scaling The lengths Lε that we have used throughout the paper, introduced in , are known as “finite-size scaling characteristic lengths”: Lε(p) = ( min{n s.t. Pp(CH([0, n] × [0, n])) ≤ε} when p < 1/2, min{n s.t. Pp(C ∗ H([0, n] × [0, n])) ≤ε} when p > 1/2.
(7.1) Mean radius of a finite cluster The (quadratic) mean radius measures the “typical” size of a finite cluster. It can be defined by the formula ξ(p) = 1 Ep |C(0)|;|C(0)| < ∞ X x ∥x∥2 ∞Pp 0 ⇝x,|C(0)| < ∞ 1/2 .
(7.2) 4This does not raise any problem since we have included this complementary bound only for the sake of completeness, and we will not use it later.
1604 Connection probabilities A third possible definition would be via the rate of decay of correlations. Take first p < 1/2 for example. For two sites x and y, we consider the connection probability between them τx,y := Pp x ⇝y, (7.3) and then τn := sup x∈∂Sn τ0,x, (7.4) the maximum connection probability between sites at distance n (using translation invariance). For any n, m ≥0, we have τn+m ≥τnτm, in other words (−logτn)n≥0 is sub-additive, which implies the existence of a constant ˜ ξ(p) such that −logτn n −→ 1 ˜ ξ(p) = inf m −logτm m (7.5) when n →∞. Note the following a-priori bound: Pp 0 ⇝x ≤e−∥x∥∞/˜ ξ(p).
(7.6) For p > 1/2, we simply use the symmetry p ↔1 −p: we consider τ∗ n := sup x∈∂Sn Pp 0 ⇝∗x (7.7) and then ˜ ξ(p) in the same way. We have in this case Pp 0 ⇝∗x ≤e−∥x∥∞/˜ ξ(p).
(7.8) Note that the symmetry p ↔1 −p gives immediately ˜ ξ(p) = ˜ ξ(1 −p).
Relation between the different lengths As expected, these characteristic lengths turn out to be all of the same order of magnitude: we will prove in Section 7.3 that Lε ≍Lε′ for any two ε,ε′ ∈(0,1/2), in Section 7.4 that L ≍˜ ξ, and in Section 7.5 that L ≍ξ.
7.2 Main critical exponents We focus here on three functions commonly used to describe the macroscopic behavior of percola-tion. We have already encountered some of them: (i) ξ(p) = 1 Ep |C(0)|;|C(0)|<∞ P x ∥x∥2 ∞Pp 0 ⇝x,|C(0)| < ∞ 1/2 the mean radius of a finite cluster.
1605 (ii) θ(p) := Pp(0 ⇝∞). This function can be viewed as the density of the infinite cluster C∞, in the following sense: 1 |SN| ¯ ¯SN ∩C∞ ¯ ¯ a.s.
−→θ(p) (7.9) when N →∞.
(iii) χ(p) = Ep |C(0)|;|C(0)| < ∞ the average size of a finite cluster.
Theorem 33 (Critical exponents). The following power-law estimates hold: (i) When p →1/2, ξ(p) ≍˜ ξ(p) ≍L(p) ≈|p −1/2|−4/3.
(7.10) (ii) When p →1/2+, θ(p) ≈(p −1/2)5/36.
(7.11) (iii) When p →1/2, χ(p) ≈|p −1/2|−43/18.
(7.12) The corresponding exponents are usually denoted by (respectively) ν, β and γ. This theorem is proved in the next sub-sections by combining the arm exponents for critical percolation with the estimates established for near-critical percolation.
7.3 Critical exponent for L We derive here5 the exponent for Lε(p) by counting the sites which are pivotal for the existence of a crossing in a box of size Lε(p). These pivotal sites are exactly those for which the 4-arm event ¯ A./¯ I 4,σ4 with alternating colors (σ4 = BW BW) and sides (¯ I = right, top, left and bottom sides): Proposition 34 ([31; 47]). For any fixed ε ∈(0,1/2), the following equivalence holds: |p −1/2| Lε(p)2π4(Lε(p)) ≍1.
(7.13) Recall now the value α4 = 5/4 of the 4-arm exponent, stated in Theorem 21. If we plug it into Eq.(7.13), we get the value of the characteristic length exponent: when p →1/2, 1 ≈|p −1/2| Lε(p)2 Lε(p)−5/4 = |p −1/2| Lε(p)3/4, so that indeed Lε(p) ≈|p −1/2|−4/3.
5In this sub-section and in the next one, we temporarily choose to stress that L depends on ε – in particular we study this dependence.
1606 ηL ηL L 1/2 1/2 ⇝p Figure 14: We restrict to the sites at distance at least ηL from the boundary of [0, L]2: these sites produce contributions of the same order, since the 4 arms stay comparable in size.
Proof. For symmetry reasons, we can assume that p > 1/2. The proof goes as follows. We first apply Russo’s formula to estimate the variation in probability of the event CH([0, Lε(p)] × [0, Lε(p)]) between 1/2 and p, which makes appear the events ¯ A./¯ I 4,σ4. By construction of Lε(p), the variation of the crossing event is of order 1, and the sites that are “not too close to the boundary” (such that none of the 4 arms can become too small – see Figure 14) each produce a contribution of the same order by Theorem 27: proving that they all together produce a non-negligible variation in the crossing probabilities will thus imply the result. For that, we need the following lemma: Lemma 35. For any δ > 0, there exists η0 > 0 such that for all p, ˆ P between Pp and P1−p, we have: for any parallelogram [0, n] × [0, m] with sides n, m ≤L(p) and aspect ratio less than 2 (ie such that 1/2 ≤n/m ≤2), for any η ≤η0, ¯ ¯ˆ P(CH([0, n] × [0, m])) −ˆ P(CH([0,(1 + η)n] × [0, m])) ¯ ¯ ≤δ.
(7.14) Proof of lemma. First, we clearly have ˆ P(CH([0, n] × [0, m])) ≥ˆ P(CH([0,(1 + η)n] × [0, m])).
For the converse bound, we use the same idea as for Lemma 15, we apply RSW in concentric annuli (see Figure 15). By considering (parts of) annuli centered on the top right corner of [0, n] × [0, m], with radii between η3/4n and pηn, we see that the probability for a crossing to arrive at a distance less than η3/4n from this corner is at most δ/100 for η0 small enough. Assume this is not the case, condition on the lowest crossing and apply RSW in annuli between scales ηn and η3/4n: if η0 is sufficiently small, with probability at least 1 −δ/100, this crossing can be extended into a crossing of [0,(1 + η)n] × [0, m].
Let us return to the proof of the proposition. Take η0 associated to δ = ε/100 by the lemma, and assume that instead of performing the change 1/2 ⇝p in the whole box [0, Lε(p)]2, we make it 1607 m n Figure 15: We extend a crossing of [0, n]×[0, m] into a crossing of [0,(1+η)n]×[0, m] by applying RSW in concentric annuli.
only for the sites in the sub-box [ηLε(p),(1 −η)Lε(p)]2, for η = η0/4. It amounts to consider the measure ˆ P(η) with parameters ˆ p(η) v = ¯ ¯ ¯ ¯ ¯ p if v ∈[ηLε(p),(1 −η)Lε(p)]2, 1/2 otherwise.
(7.15) We are going to prove that ˆ P(η)(CH([0, Lε(p)]2)) and Pp(CH([0, Lε(p)]2)) are very close by show-ing that they are both very close to Pp(CH([ηLε(p),(1 −η)Lε(p)]2)) = ˆ P(η)(CH([ηLε(p),(1 − η)Lε(p)]2)). Indeed, for any ˜ P ∈{ˆ P(η),Pp}, we have by the lemma ˜ P(CH([0, Lε(p)]2)) ≤˜ P(CH([ηLε(p),(1 −η)Lε(p)] × [0, Lε(p)])) = 1 −˜ P(C ∗ V([ηLε(p),(1 −η)Lε(p)] × [0, Lε(p)])) ≤1 − ˜ P(C ∗ V([ηLε(p),(1 −η)Lε(p)]2)) −2δ = ˜ P(CH([ηLε(p),(1 −η)Lε(p)]2)) + 2δ, and in the other way, ˜ P(CH([0, Lε(p)]2)) ≥˜ P(CH([ηLε(p),(1 −η)Lε(p)] × [0, Lε(p)])) −2δ = 1 −˜ P(C ∗ V([ηLε(p),(1 −η)Lε(p)] × [0, Lε(p)])) −2δ ≥1 −˜ P(C ∗ V([ηLε(p),(1 −η)Lε(p)]2)) −2δ = ˜ P(CH([ηLε(p),(1 −η)Lε(p)]2)) −2δ.
The claim follows readily, in particular ˆ P(η)(CH([0, Lε(p)]2)) ≥Pp(CH([0, Lε(p)]2)) −4δ, (7.16) which is at least (1/2+ ε) −4δ ≥1/2+ ε/2 by the very definition of Lε(p). It shows as desired that the sites in [ηLε(p),(1 −η)Lε(p)]2 produce all together a non-negligible contribution.
Now, Russo’s formula applied to the interpolating measures (ˆ P(η) t )t∈[0,1] (with parameters ˆ p(η) v (t) = 1608 t × ˆ p(η) v + (1 −t) × 1/2) and the event CH([0, Lε(p)]2) gives Z 1 0 X v∈[ηLε(p),(1−η)Lε(p)]2 p −1/2 ˆ P(η) t v ⇝¯ I 4,σ4 ∂[0, Lε(p)]2dt = ˆ P(η)(CH([0, Lε(p)]2)) −P1/2(CH([0, Lε(p)]2)), and this quantity is at least ε/2, and thus of order 1.
Finally, it is not hard to see that once η fixed, we have (uniformly in p, ˆ P between Pp and P1−p, and v ∈[ηLε(p),(1 −η)Lε(p)]2) ˆ P v ⇝¯ I 4,σ4 ∂[0, Lε(p)]2 ≍ˆ P v ⇝4,σ4 ∂S η 2 Lε(p)(v) ≍P1/2(0 ⇝4,σ4 ∂S η 2 Lε(p)) ≍P1/2(0 ⇝4,σ4 ∂SLε(p)), which yields the desired conclusion.
Remark 36. Note that the intermediate lemma was required for the lower bound only, the upper bound can be obtained directly from Russo’s formula. To get the lower bound, we could also have proved that for n ≤L(p), X x∈Sn ˆ P x ⇝4,σ4 ∂Sn ≍n2π4(n).
(7.17) Basically, it comes from the fact that when we get closer to ∂SN, one of the arms may be shorter, but the remaining arms also have less space – and the 3-arm exponent in the half-plane appears.
All the results we have seen so far hold for any fixed value of ε in (0,1/2), in particular Proposition 34. Combining it with the estimate for 4 arms, we get an important corollary, that the behavior of Lε does not depend on the value of ε.
Corollary 37. For any ε,ε′ ∈(0,1/2), Lε(p) ≍Lε′(p).
(7.18) Proof. To fix ideas, assume that ε ≤ε′, so that Lε(p) ≥Lε′(p), and we need to prove that Lε(p) ≤ C Lε′(p) for some constant C. We know that |p −1/2| Lε(p)2π4(Lε(p)) ≍1 ≍|p −1/2| Lε′(p)2π4(Lε′(p)), hence for some constant C1, Lε(p)2π4(Lε(p)) Lε′(p)2π4(Lε′(p)) ≤C1.
This yields Lε(p) Lε′(p) 2 ≤C1 π4(Lε′(p)) π4(Lε(p)) ≤C2 π4(Lε′(p), Lε(p))−1 by quasi-multiplicativity. Now we use the a-priori bound for 4 arms given by the 5-arm exponent: π4(Lε′(p), Lε(p)) ≥C3 Lε′(p) Lε(p) −α′ π5(Lε′(p), Lε(p)) ≥C4 Lε′(p) Lε(p) 2−α′ .
1609 Together with the previous equation, it implies the result: Lε(p) ≤(C5)1/α′ Lε′(p).
Remark 38. In the other direction, a RSW construction shows that we can increase Lε by any constant factor by choosing ε small enough.
7.4 Uniform exponential decay, critical exponent for θ Up to now, our reasonings (separation of arms, arm events near criticality, critical exponent for L) were based on RSW considerations on scales n ≤L(p), so that critical and near-critical percolation could be handled simultaneously. In the other direction, the definition of L(p) also implies that when n > L(p), the picture starts to look like super/sub-critical percolation, supporting the choice of L(p) as the characteristic scale of the model.
More precisely, we prove a property of exponential decay uniform in p. This property will then be used to link L with the other characteristic functions, and we will derive the following expressions of θ, χ and ξ as functions of L: (i) θ(p) ≍π1(L(p)), (ii) χ(p) ≍L(p)2π2 1(L(p)), (iii) ξ(p) ≍L(p).
The critical exponents for these three functions will follow readily, since we already know the expo-nent for L.
Uniform exponential decay The following lemma shows that correlations decay exponentially fast with respect to L(p). This allows to control the speed for p varying: Lemma 39. For any ε ∈(0,1/2), there exist constants Ci = Ci(ε) > 0 such that for all p < 1/2, all n, Pp(CH([0, n] × [0, n])) ≤C1e−C2n/Lε(p).
(7.19) Proof. We use a block argument: for each integer n, Pp(CH([0,2n] × [0,4n])) ≤C′[Pp(CH([0, n] × [0,2n]))]2, (7.20) with C′ = 102 some universal constant.
It suffices for that (see Figure 16) to divide the parallelogram [0,2n]×[0,4n] into 4 horizontal sub-parallelograms [0,2n]×[in,(i +1)n] (i = 0,...,3) and 6 vertical ones [in,(i +1)n]×[jn,(j +2)n] (i = 0,1, j = 0,1,2). Indeed, consider a horizontal crossing of the big parallelogram: by considering its pieces in the two regions 0 < x < n and n < x < 2n, we can extract from it two sub-paths, each 1610 Figure 16: Two of the small sub-parallelograms are crossed in the “easy” way.
crossing one of the 10 sub-parallelograms “in the easy way”. They are disjoint by construction, so the claim follows by using the BK inequality.
We then obtain by iterating: C′Pp(CH([0,2kLε(p)] × [0,2k+1Lε(p)])) ≤(C′ ˜ ε)2k (7.21) with ˜ ε ≥Pp(CH([0, Lε(p)] × [0,2Lε(p)])).
Recall that by definition, Pp(CH([0, Lε(p)] × [0, Lε(p)])) ≤ε0 if ε ≤ε0. The RSW theory thus implies (Theorem 2) that for all fixed ˜ ε > 0, we can take ε0 sufficiently small to get automatically (and independently of p) that Pp(CH([0, Lε(p)] × [0,2Lε(p)])) ≤˜ ε.
(7.22) We now choose ˜ ε = 1/(e2C′).
For each integer n ≥Lε(p), we can define k = k(n) such that 2k ≤n/Lε(p) < 2k+1, and then, Pp(CH([0, n] × [0, n])) ≤Pp(CH([0,2kLε(p)] × [0,2k+1Lε(p)])) ≤e−2k+1 ≤e × e−n/Lε(p), which is also valid for n < Lε(p), thanks to the extra factor e.
Hence, we have proved the property for any ε below some fixed value ε0 (given by RSW). The result for any ε ∈(0,1/2) follows readily by using the equivalence of lengths for different values of ε (Corollary 37).
We would like to stress the fact that in the proof, we have not used any of the previous results until the last step. This exponential decay property could thus have been derived much earlier – but only for values of ε small enough. It would for instance provide a more direct way to prove that Lε(p) ≍Lε′(p), but still only for ε, ε′ less than some fixed value.
1611 ∂SL(p) 0 Figure 17: We consider overlapping parallelograms, with size doubling at each step.
Remark 40. It will sometimes reveal useful to know this property for crossings of longer parallelograms “in the easy way”: we also have for any k ≥1, Pp(CH([0, n] × [0, kn])) ≤C(k) 1 e−C(k) 2 n/Lε(p) (7.23) for some constants C(k) i (depending on k and ε). This can be proved by combining the previous lemma with the fact that in Theorem 2, we can take fk satisfying fk(1 −ε) = 1 −Ckεαk + o(εαk) for some Ck,αk > 0.
Consequence for θ When p > 1/2, we now show that at a distance L(p) from the origin, we are already “not too far from infinity”: once we have reached this distance, there is a positive probability (bounded away from 0 uniformly in p) to reach infinity.
Corollary 41. We have θ(p) = Pp 0 ⇝∞ ≍Pp 0 ⇝∂SL(p) (7.24) uniformly in p > 1/2.
Proof. It suffices to consider overlapping parallelograms as in Figure 17, each parallelogram twice larger than the previous one, so that the kth of them has a probability at least 1−C1e−C22k to present a crossing in the “hard” direction (thanks to the previous remark). Since Q k(1 −C1e−C22k) > 0, we are done.
Now, combining Eq.(7.24) with Theorem 27 gives, for p > 1/2, θ(p) ≍Pp 0 ⇝∂SL(p) ≍P1/2 0 ⇝∂SL(p) = π1(L(p)).
(7.25) Using the 1-arm exponent α1 = 5/48 stated in Theorem 21, we get θ(p) ≈ L(p)−5/48 (7.26) as p →1/2+. Together with the critical exponent for L derived previously, this provides the critical exponent for θ: θ(p) ≈ (p −1/2)−4/3−5/48 ≈(p −1/2)5/36.
(7.27) 1612 Equivalence of L and ˜ ξ To fix ideas, we assume in this sub-section that p < 1/2. Performing a RSW-type construction as in Figure 3 yields Pp(CH([0, kL(p)] × [0, L(p)])) ≥δk−1 2 δk−2 1 = C1e−C2kL(p)/L(p), (7.28) so that L(p) measures exactly the speed of decaying. Once knowing this, it is easy to compare L and ˜ ξ.
Corollary 42. We have ˜ ξ(p) ≍L(p).
(7.29) Proof. We exploit the previous remark: on one hand L measures the speed of decaying for cross-ings of rhombi, and on the other hand ˜ ξ was defined to give the optimal bound for point-to-point connections.
More precisely, consider any x ∈∂Sn. If for instance x is on the right side of ∂Sn, then 0 ⇝x implies that CH([0, n] × [−n, n]) occurs, so that τ0,x = Pp(0 ⇝x) ≤Pp(CH([0, n] × [0,2n])) ≤C(2) 1 e−C(2) 2 n/L(p).
By definition of τn (Eq.(7.4)), we thus have τn ≤C(2) 1 e−C(2) 2 n/L(p), which gives τkL(p) ≤C(2) 1 e−C(2) 2 k and − logτkL(p) kL(p) ≥− 1 kL(p) log C(2) 1 −C(2) 2 k − − − → k→∞ C(2) 2 L(p).
Hence, 1 ˜ ξ(p) ≥ C(2) 2 L(p) and finally ˜ ξ(p) ≤C L(p).
Conversely, we know that Pp(CH([0, kL(p)] × [0, kL(p)])) ≥˜ C1e−˜ C2k for some ˜ Ci > 0 (Eq.(7.28)).
Consequently, τkL(p) ≥ 1 (kL(p) + 1)2 Pp(CH([0, kL(p)] × [0, kL(p)])) ≥ 1 (kL(p))2 ˜ C1e−˜ C2k, which implies − logτkL(p) kL(p) ≤− 1 kL(p) log ˜ C1 −2log(kL(p)) −˜ C2k − − − → k→∞ ˜ C2 L(p), whence the conclusion: ˜ ξ(p) ≥C′L(p).
1613 7.5 Further estimates, critical exponents for χ and ξ Estimates from critical percolation We start by stating some estimates that we will need. These estimates were originally derived for critical percolation (see e.g. [28; 29]), but for exactly the same reasons they also hold for near-critical percolation on scales n ≤L(p): Lemma 43. Uniformly in p, ˆ P between Pp and P1−p and n ≤L(p), we have 1. ˆ E|x ∈Sn : x ⇝∂Sn| ≍n2π1(n).
2. For any t ≥0, X x∈Sn ∥x∥t ∞ˆ P 0 ⇝x ≍ X x∈Sn ∥x∥t ∞ˆ P 0 Sn ⇝x ≍nt+2π2 1(n).
Note that item 2. implies in particular for t = 0 that ˆ E|x ∈Sn : x ⇝0| ≍ˆ E|x ∈Sn : x Sn ⇝0| ≍n2π2 1(n).
Proof. We will have use for the fact that we can take α1 = 1/2 for j = 1 in Eq.(5.6) (actually any α < 1 would be enough for our purpose): for any integers n < N, π1(n, N) ≥C(n/N)1/2.
(7.30) This can be proved like (3.15) of : just use blocks of size n instead of individual sites to obtain that N n π2 1(n, N) is bounded below by a constant.
Proof of item 1. We will use that ˆ E|x ∈Sn : x ⇝∂Sn| = X x∈Sn ˆ P(x ⇝∂Sn).
(7.31) For the lower bound, it suffices to note that for any x ∈Sn, ˆ P(x ⇝∂Sn) ≥ˆ P(x ⇝∂S2n(x)) = ˆ Px(0 ⇝∂S2n) (7.32) (where ˆ Px is the measure ˆ P translated by x), and that ˆ Px(0 ⇝∂S2n) ≥C1ˆ Px(0 ⇝∂Sn) ≥C2π1(n) (7.33) by extendability and Theorem 27 for one arm.
For the upper bound, we sum over concentric rhombi around 0: X x∈Sn ˆ P(x ⇝∂Sn) ≤ X x∈Sn ˆ P(x ⇝∂Sd(x,∂Sn)(x)) = X x∈Sn ˆ Px(0 ⇝∂Sd(x,∂Sn)) ≤C1n + n X j=1 C1n × C2P1/2(0 ⇝∂Sj) 1614 using that there are at most C1n sites at distance j from ∂Sn, and Theorem 27. This last sum is at most C3n n X j=1 π1(j) ≤C3nπ1(n) n X j=1 π1(j) π1(n) ≤C4nπ1(n) n X j=1 π1(j, n)−1 by quasi-multiplicativity. Now Eq.(7.30) says that π1(j, n) ≥C(j/n)1/2, so that n X j=1 π1(j, n)−1 ≤ n X j=1 (j/n)−1/2 = n1/2 n X j=1 j−1/2 ≤C5n, which gives the desired upper bound.
Proof of item 2.
Since X x∈Sn ∥x∥t ∞ˆ P 0 Sn ⇝x ≤ X x∈Sn ∥x∥t ∞ˆ P 0 ⇝x, it suffices to prove the desired lower bound for the left-hand side, and the upper bound for the right-hand side.
Consider the lower bound first. We note (see Figure 18) that if 0 is connected to ∂Sn and if there exists a black circuit in S2n/3,n (which occurs with probability at least δ4 6 by RSW), then any x ∈ Sn/3,2n/3 connected to ∂S2n(x) will be connected to 0 in Sn. Using the FKG inequality, we thus get for such an x: ˆ P(0 Sn ⇝x) ≥δ4 6ˆ P(0 ⇝∂Sn)ˆ P(x ⇝∂S2n(x)) which is at least (still using extendability and Theorem 27) C1π2 1(n). Consequently, X x∈Sn ∥x∥t ∞ˆ P 0 Sn ⇝x ≥ X x∈Sn/3,2n/3 ∥x∥t ∞C1π2 1(n) ≥C2n2(n/3)tπ2 1(n).
Let us turn to the upper bound. We take a logarithmic division of Sn: define k = k(n) so that 2k < n ≤2k+1, we have X x∈Sn ∥x∥t ∞ˆ P 0 ⇝x ≤C1 + k+1 X j=3 X x∈S2j−1,2j ∥x∥t ∞ˆ P 0 ⇝x.
(7.34) Now for x ∈S2j−1,2j, take the two boxes S2j−2(0) and S2j−2(x): since they are disjoint, ˆ P 0 ⇝x ≤ˆ P 0 ⇝S2j−2(0)ˆ P x ⇝S2j−2(x), (7.35) which is at most C2π2 1(2j−1) using the same arguments as before. Our sum is thus less than (since |S2j−1,2j| ≤C322j) k+1 X j=3 C322j × (2j)t × (C2π2 1(2j−1)) ≤C42(2+t)kπ2 1(2k) × k+1 X j=3 2(2+t)(j−k) π2 1(2j−1) π2 1(2k) .
(7.36) 1615 x 0 ∂Sn ∂S2n(x) Figure 18: With this construction, any site x in Sn/3,2n/3 connected to a site at distance 2n is also connected to 0 in Sn.
Now, 2(2+t)kπ2 1(2k) ≤C5n2+tπ2 1(n), and this yields the desired result, using as previously π1(2j−1) π1(2k) ≤ C6π1(2j−1,2k)−1 ≤C72−(j−k)/2: k+1 X j=3 2(2+t)(j−k) π2 1(2j−1) π2 1(2k) ≤C7 k+1 X j=3 2(2+t)(j−k)2−(j−k) ≤C8 k−3 X l=−1 2−(1+t)l, and this sum is bounded by P∞ l=−1 2−(1+t)l < ∞.
Main estimate The following lemma will allow us to link directly χ and ξ with L. Roughly speaking, it relies on the fact that the sites at a distance much larger than L(p) from the origin have a negligible contribution, due to the exponential decay property, so that the sites in SL(p) produce a positive fraction of the total sum: Lemma 44. For any t ≥0, we have X x ∥x∥t ∞Pp 0 ⇝x,|C(0)| < ∞ ≍L(p)t+2π2 1(L(p)) (7.37) uniformly in p.
Proof. Lower bound. The lower bound is a direct consequence of item 2. above: indeed, X x ∥x∥t ∞Pp 0 ⇝x,|C(0)| < ∞ ≥Pp(∃white circuit in SL,2L) X x∈SL ∥x∥t ∞Pp 0 SL ⇝x ≥δ4 4 X x∈SL ∥x∥t ∞Pp 0 SL ⇝x 1616 ∂SL Figure 19: For the upper bound, we cover the plane with rhombi of size 2L and sum their different contributions.
by RSW , and item 2. gives X x∈SL ∥x∥t ∞Pp 0 SL ⇝x ≥C Lt+2π2 1(L).
Upper bound. To get the upper bound, we cover the plane by translating SL: we consider the family of rhombi SL(2n1L,2n2L), for any two integers n1 and n2 (see Figure 19). By isolating the contribution of SL, we get: X x ∥x∥t ∞Pp 0 ⇝x,|C(0)| < ∞ ≤ X x∈SL ∥x∥t ∞Pp 0 ⇝x,|C(0)| < ∞ + X (n1,n2)̸=(0,0) X x∈SL(2n1L,2n2L) ∥x∥t ∞Pp 0 ⇝x,|C(0)| < ∞.
Using item 2. above, we see that the rhombus SL gives a contribution X x∈SL ∥x∥t ∞Pp 0 ⇝x,|C(0)| < ∞ ≤C Lt+2π2 1(L), which is of the right order of magnitude.
We now prove that each small rhombus outside of SL at a distance kL gives a contribution of order π1(L) × Lt × Ep |x ∈SL : x ⇝∂SL| ≍Lt+2π2 1(L) (using item 1.), multiplied by some quantity which decays exponentially fast in k and will thus produce a series of finite sum. More precisely, if we regroup the rhombi into concentric annuli around SL, we get that the previous summation is at 1617 most ∞ X k=1 X (n1,n2) ∥(n1,n2)∥∞=k X x∈SL(2n1L,2n2L) ∥x∥t ∞Pp 0 ⇝x,|C(0)| < ∞ ≤ ∞ X k=1 X (n1,n2) ∥(n1,n2)∥∞=k X x∈SL(2n1L,2n2L) [(2k + 1)L]tPp 0 ⇝x,|C(0)| < ∞ ≤ ∞ X k=1 X (n1,n2) ∥(n1,n2)∥∞=k C′kt Lt Ep |C(0) ∩SL(2n1L,2n2L)|;|C(0)| < ∞.
Now, we have to distinguish between the sub-critical and the super-critical cases: we are going to prove that in both cases, Ep |C(0) ∩SL(2n1L,2n2L)|;|C(0)| < ∞ ≤C1L2π2 1(L)e−C2k (7.38) for some constants C1, C2 > 0. When p < 1/2, we will use that Pp(∂SL ⇝∂SkL) ≤C3e−C4k, (7.39) which is a direct consequence of the exponential decay property Eq.(7.23) for “longer” parallelo-grams. When p > 1/2, we have an analog result, which can be deduced from the sub-critical case just as in the discrete case (replace sites by translates of SL): Pp(∂SL ⇝∂SkL,|C(0)| < ∞) ≤Pp(∃white circuit surrounding a site on ∂SL and a site on ∂SkL) ≤C5e−C6k.
Assume first that p < 1/2. By independence, we have (∥(n1, n2)∥∞= k) Ep |C(0) ∩SL(2n1L,2n2L)|;|C(0)| < ∞ ≤Pp(0 ⇝∂SL) × Ep |x ∈SL(2n1L,2n2L) : x ⇝∂SL(2n1L,2n2L)| × Pp(∂SL ⇝∂S(2k−1)L) ≤π1(L) × (C′′L2π1(L)) × C′ 3e−C′ 4k.
If p > 1/2, we write similarly (here we use FKG to separate the existence of a white circuit (decreas-ing) from the other terms (increasing), and then independence of the remaining terms) Ep |C(0) ∩SL(2n1L,2n2L)|;|C(0)| < ∞ ≤Pp(0 ⇝∂SL) × Ep |x ∈SL(2n1L,2n2L) : x ⇝∂SL(2n1L,2n2L)| × Pp(∃white circuit surrounding a site on ∂SL and a site on ∂S(2k−1)L) ≤π1(L) × (C′′L2π1(L)) × C′ 5e−C′ 6k.
1618 Since there are at most C(3)k rhombi at a distance k for some constant C(3), the previous summation is in both cases less than ∞ X k=1 C(3)k × C′kt Lt × C1L2π2 1(L)e−C2k ≤C(4) ∞ X k=1 kt+1e−C2k Lt+2π2 1(L), which yields the desired upper bound, as P∞ k=1 kt+1e−C2k < ∞.
Critical exponents for χ and ξ The previous lemma reads for t = 0: Proposition 45. We have χ(p) = Ep |C(0)|;|C(0)| < ∞ ≍L(p)2π2 1(L(p)).
(7.40) In other words, “χ(p) ≍χnear(p)”. It provides the critical exponent for χ: χ(p) ≈L(p)2L(p)−5/482 ≈|p −1/2|−4/386/48 ≈|p −1/2|−43/18.
(7.41) Recall also that ξ was defined via the formula ξ(p) = 1 Ep |C(0)|;|C(0)| < ∞ X x ∥x∥2 ∞Pp 0 ⇝x,|C(0)| < ∞ 1/2 .
Using the last proposition and the lemma for t = 2, we get ξ(p) ≍ L(p)4π2 1(L(p)) L(p)2π2 1(L(p)) 1/2 = L(p).
(7.42) We thus obtain the following proposition, announced in Section 7.1: Proposition 46. We have ξ(p) ≍L(p).
(7.43) This implies in particular that ξ(p) ≈|p −1/2|−4/3.
(7.44) 8 Concluding remarks 8.1 Other lattices Most of the results presented here (the separation of arms, the theorem concerning arm events on a scale L(p), the “universal” arm exponents, the relations between the different characteristic functions, etc.) come from RSW considerations or the exponential decay property, and remain true 1619 on other regular lattices like the square lattice. The triangular lattice has a property of self-duality which makes life easier, in the general case we have to consider the original lattice together with the matching lattice (obtained by “filling” each face with a complete graph): instead of black or white connections, we thus talk about primal and dual connections. We can also handle bond percolation in this way. We refer the reader to the original paper of Kesten for more details, where results are proved in this more general setting. The only obstruction to get the critical exponents is actually the derivation of the arm exponents at the critical point p = pc (and only two exponents are needed, for 1 arm and 4 alternating arms).
Now, consider site percolation on Z2 for instance. We know that the (hypothetical) arm exponents satisfy 0 < αj,α′ j,βj < ∞for any j ≥1. Hence the a-priori estimate Ppc(0 ⇝4,σ4 ∂SN) ≥N −2+α for some α > 0, coming from the 5-arm exponent, remains true: α4 < 2 (and in the same way α6 > 2). Combined with Proposition 34, this leads to the weaker but nonetheless interesting statement L(p) ≤|p −pc|−A (8.1) for some A > 0. Hence ν < ∞, and then γ < ∞(if these exponents exist). Using α1 < ∞, we also get β < ∞.
If we use a RSW construction in a box, we can make appear 3-arm sites on the lowest crossing and deduce that α1 ≤1/3. Here are rigorous bounds for the critical exponents in two dimensions: triangular lattice general rigorous bounds β = 5/36 0 < β < 1 γ = 43/18 8/5 ≤γ < ∞ ν = 4/3 1 < ν < ∞ For more details, the reader can consult and the references therein.
8.2 Some related issues For the sake of completeness, let us mention finally that the way the characteristic length L was defined also allows to use directly the compactness results of . Indeed, the a-priori estimates on arm events coming from RSW considerations are exactly hypothesis (H1) of this paper. This hypothesis implies that percolation interfaces cannot cross too many times any annulus, and thus cannot be too “intricate”: this is Theorem 1, asserting the existence of Hölder parametrizations with high probability.
This regularity property then implies tightness, using (a version of) Arzela-Ascoli’s theorem for continuous functions on a compact subset of the plane. We can thus show in this way the existence of scaling limits for near-critical percolation interfaces.
Let us also mention that the techniques presented here are important to study various models related to the critical regime, for instance Incipient Infinite Clusters [17; 24; 25], Dynamical Percolation , Gradient Percolation . . .
1620 Acknowledgements I would like to thank W . Werner for suggesting me to look at these questions. He had an important role in reading various intermediate versions of this paper. I also enjoyed many stimulating and enlightening discussions with various people, including V . Beffara, J. van den Berg, F . Camia, L.
Chayes, C. Garban, A. Sapozhnikov and O. Schramm.
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236, 311-334 (2003). MR1981994 H. Kesten, The critical probability of bond percolation on the square lattice equals 1/2, Comm. Math. Phys. 74, 41-59 (1980). MR0575895 H. Kesten, Percolation theory for mathematicians, Birkhäuser, Boston (1982). MR0692943 H. Kesten, The incipient infinite cluster in two-dimensional percolation, Probab. Th. Rel.
Fields 73, 369-394 (1986). MR0859839 H. Kesten, A scaling relation at criticality for 2D-percolation, in Percolation theory and er-godic theory of infinite particle systems, IMA Volumes in Mathematics and its Applications 8, 203-212 (1987). MR0894549 H. Kesten, Subdiffusive behavior of random walk on a random cluster, Ann. I.H.P . Sec. B 22, 425-487 (1986). MR0871905 1622 H. Kesten, Scaling relations for 2D-percolation, Comm. Math. Phys. 109, 109-156 (1987).
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What is the order of the bond lengths for NO2+, NO2-, and NO3-?
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1y
To compare the bond lengths of NO+2 NO 2+, NO−2 NO 2−, and NO−3 NO 3−, we can consider their resonance structures and the overall bonding characteristics:
NO+2 NO 2+ (Nitronium ion):
This species has a formal positive charge and is typically represented as having a double bond between nitrogen and one oxygen, and a single bond with the other oxygen. The bond lengths are shorter due to the increased positive charge and the presence of a double bond.
Bond Lengths: Shortest among the three.
NO−2 NO 2− (Nitrite ion):
This ion has a formal negative charge and typically e
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To compare the bond lengths of NO+2 NO 2+, NO−2 NO 2−, and NO−3 NO 3−, we can consider their resonance structures and the overall bonding characteristics:
NO+2 NO 2+ (Nitronium ion):
This species has a formal positive charge and is typically represented as having a double bond between nitrogen and one oxygen, and a single bond with the other oxygen. The bond lengths are shorter due to the increased positive charge and the presence of a double bond.
Bond Lengths: Shortest among the three.
NO−2 NO 2− (Nitrite ion):
This ion has a formal negative charge and typically exhibits resonance between two structures: one with a double bond to one oxygen and a single bond to the other, and vice versa. The bond lengths are longer than those in NO+2 NO 2+ due to the presence of a negative charge and less double bond character overall.
Bond Lengths: Intermediate length.
NO−3 NO 3− (Nitrate ion):
In this ion, nitrogen is bonded to three oxygen atoms, and the negative charge is delocalized over the three bonds. The bond character is effectively equal due to resonance, leading to bond lengths that are longer than those in NO+2 NO 2+ but shorter than typical single bonds.
Bond Lengths: Longest among the three.
Summary of Bond Length Order:
Shortest:NO+2 NO 2+
Intermediate:NO−2 NO 2−
Longest:NO−3 NO 3−
Thus, the order of bond lengths from shortest to longest is:
NO+2<NO−2<NO−3 NO 2+<NO 2−<NO 3−
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What is the bond order for NO2+?
How do you find the bond order of NO2?
What is the order of N-O bond lengths in NO, NO2-, NO3-, and N2O4?
What is the bond angle relation between No2, No2+, and NO2?
Why is the N-O bond in NO2 shorter than the NO bond in NO3?
Anannya
8y
Bond order = (No. of bonding electrons - No. of anitboding electrons ) ÷ 2
BO = (Nb - Na) / 2
Now according to Molecular Orbital Theory….
Let's start with NO first
Follow Hund’s rule and fill the bonding electrons then ( the stared ones {} ) anitboding electrons
Write the equation and fill 2 electrons max in each orbital….if there is 1 electron remaining put it alone in the last orbital as we normally do
The energy level diagram is just for understanding how the electrons occupy the orbitals …as said before the bonding electrons occupy first then the antibonding.
Look at the equation….fill only 2 el
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Bond order = (No. of bonding electrons - No. of anitboding electrons ) ÷ 2
BO = (Nb - Na) / 2
Now according to Molecular Orbital Theory….
Let's start with NO first
Follow Hund’s rule and fill the bonding electrons then ( the stared ones {} ) anitboding electrons
Write the equation and fill 2 electrons max in each orbital….if there is 1 electron remaining put it alone in the last orbital as we normally do
The energy level diagram is just for understanding how the electrons occupy the orbitals …as said before the bonding electrons occupy first then the antibonding.
Look at the equation….fill only 2 electrons max
(The stared ones are anitbonding electrons )
Count all the bonding electrons
Nb = 10
Count all the antibonding electrons () ones
Na = 5
So now put it into the BO equation
BO = (Nb - Na) / 2
BO = ( 10 - 5) / 2 = 5 / 2 = 2.5
BO ( NO ) = 2.5
This is how you do….
Now NO 2+ is 2 electrons less
As total electrons in NO was 7+8 = 15
In NO 2+ it will be 15-2 = 13
Similarly,
NO 2- will have 2 electrons extra…. 15+2 = 17
And NO 3- will have 3 electrons extra…. 15+3 = 18
Now all you have to do is put these electrons in the pair of 2 in the equation and count the bonding electrons and the antibonding electrons….subtract and divide by 2…you get the bond order (BO)!
So…
BO ( NO 2+ ) = 2.5
BO ( NO 2- ) = 1.5
BO ( NO 3- ) = 1
Hope you get it.
PS : Sorry for my messy handwriting…
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Tamoghna Mukhopadhyay
M.Sc. in Chemistry, Ramakrishna Mission Residential College, Narendrapur (Graduated 2019) · Upvoted by
Kartikey Pandey
, M.Sc. Chemistry, Jamia Millia Islamia (2020) · Author has 164 answers and 643.4K answer views
·8y
To answer this question firstly we have to calculate the bond orders. The higher the bond order is, the shorter the bond length.
Bond order= (No. of bonding electrons-No. of antibonding electrons)/2
Now, to calculate these we need to consider the M.O. of a heteronuclear diatomic molecule NO first.
(Note: For the calculation of bond orders,we
will not consider the non bonding electrons)
Now, NO has 6 electrons in the bonding orbitals & 1 in antibonding orbital. Thus it has bond order of (6–1)/2= 2.5
Now, NO2+ has 2 electrons lesser than NO. It has 6 electrons in 4thhe bonding orbitals and no electron
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To answer this question firstly we have to calculate the bond orders. The higher the bond order is, the shorter the bond length.
Bond order= (No. of bonding electrons-No. of antibonding electrons)/2
Now, to calculate these we need to consider the M.O. of a heteronuclear diatomic molecule NO first.
(Note: For the calculation of bond orders,we
will not consider the non bonding electrons)
Now, NO has 6 electrons in the bonding orbitals & 1 in antibonding orbital. Thus it has bond order of (6–1)/2= 2.5
Now, NO2+ has 2 electrons lesser than NO. It has 6 electrons in 4thhe bonding orbitals and no electron in antibonding. Thus it has bond order of (6–0)/2=3
For NO 2- the bond order is (6–3)/2=1.5
For NO 3- the bond order is (6–4)/2=1.
The order of bond order is
NO2+>NO2->NO3-
Thus, the order of bond length is
NO3->NO2->NO2+
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Chandrodai Pratap Singh
Lives in Pondicherry University, Pondicherry (2018–present)
·7y
If the total number of electrons in a diatomic molecule is 14 then the bond order is 3….now either increase or decrease the number of electron bond order decreases by 0.5…This is shortcut
In case of NO2+ total electrons will be = 7 electrons of nitrogen + 8 electrons of oxygen = 15 electrons.
Two positive charge concludes deficiency of two electrons so total electrons will we 13. Hence bond order will be 2.5
Similarly you can calculate for other two atoms. This works in most of the cases.
But the real way is already answerd.
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Related questions
More answers below
What is the N-O bond order in NO3-?
Among NO2+ and NO2- which bond angle is larger?
What is the order of the bond angles for N2O, NO2+, NO2-, and NO3-?
Why is the bond angle of NO2 greater than NO3-?
How can we compare the bond lengths of n20, no, no2, etc?
Himanshu Kumar
7y
NO3–>NO2–>NO2+.
As the distance between atoms increases so does the bond length.
To simplify it , Gaining electrons increases bond length barring exceptions .
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Raj Kaushik
7y
Bond order is inversely proportional of bond length .so first of all find bond onder…then you will be able to solve it
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Waverly Winters
Student
·8y
Bond length depends on number of shared electron pairs . More number of electron pair shared has less bond length.
Also Bond length is inversely depends on bond order so correct order of bond length will be NO2+<NO2-<NO3-
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Anish Koulgi
BE in Computer Science, Pune Institute of Computer Technology (Graduated 2022) · Author has 56 answers and 304K answer views
·7y
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What is the correct order of increasing bond angles in the following triatomic species: NO2+, NO2, and NO2-?
The bond angle can be compared by comparing the lone pair of e−e− in all the 3 molecules. N O+2 N O 2+ has no lone pair of e^- . N O 2 N O 2 has an odd electron(which tends to increase bond angle). N O−2 N O 2− has a lone pair of electrons.
So the lone pair bond pair repulsion is the least in N O+2 N O 2+, and it has sp hybridisation, while rest two have sp2 hybridisation, while it’s the most in N O−2 N O 2−. So in N O−2 N O 2− the bond is contracted due to the repulsion between the bond pair and lone pair e−e−. Here’s the structures of these three molecules.
Source:
So the order of increa
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The bond angle can be compared by comparing the lone pair of e−e− in all the 3 molecules. N O+2 N O 2+ has no lone pair of e^- . N O 2 N O 2 has an odd electron(which tends to increase bond angle). N O−2 N O 2− has a lone pair of electrons.
So the lone pair bond pair repulsion is the least in N O+2 N O 2+, and it has sp hybridisation, while rest two have sp2 hybridisation, while it’s the most in N O−2 N O 2−. So in N O−2 N O 2− the bond is contracted due to the repulsion between the bond pair and lone pair e−e−. Here’s the structures of these three molecules.
Source:
So the order of increasing bond angle is :
N O−2<N O 2<N O+2 N O 2−<N O 2<N O 2+
Hope it helps!
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Balunga Ranjan
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Sagar Jadhav
, M.S. Pharm Chemistry & Physics, National Institute of Pharmaceutical Education and Research, Mohali (2016) and
Jeff Armstrong
, PhD Chemistry, The University of Texas at Austin (1998) · Author has 305 answers and 729.1K answer views
·9y
Related
How do you find the bond order of NO2?
N O 2 N O 2 is ideally represented by the resonance structure on the right. Bond order for the molecule is given by :
(Total number of electron pairs in N-O bonds)/(total number of N-O bonds)
= 3/2 = 1.5
Continue Reading
N O 2 N O 2 is ideally represented by the resonance structure on the right. Bond order for the molecule is given by :
(Total number of electron pairs in N-O bonds)/(total number of N-O bonds)
= 3/2 = 1.5
Upvote ·
99 27
9 1
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Falling and rising factorials
Peter Burbery
Peter Burbery, Marshall University
Posted 3 years ago
CITE THIS NOTEBOOK: Falling and rising factorials by Peter Burbery. Wolfram Community DEC 15 2022. I wrote about this in another essay on the twelve fold way but I want to reorganize my work for factorials into a separate notebook. I think this will be better organization. This document is based on the Wikipedia article for falling and rising factorials and my postA systematic approach to combinatorics word problems. I wanted to add colors to the cells with the new front end toolbar introduced in 13.1. I like light yellow so I made the background light yellow.. In mathematics, the falling factorial (sometimes called the descending factorial, falling sequential product, or lower factorial) is defined as the polynomial (x) n = n x = n f a c t o r s x(x-1)(x-2)...(x-n+1) = n ∏ k=1 (x-k+1)= n-1 ∏ k=0 (x-k) The rising factorial (sometimes called the Pochhammer function, Pochhammer polynomial, ascending factorial, rising sequential product, or upper factorial) is defined as (n) x = n x = n f a c t o r s x(x+1)(x+2)...(x+n-1) = n ∏ k=1 (x+k-1)= n-1 ∏ k=0 (x+k) The value of each is taken to be 1 (an empty product) when n=0 . These symbols are collectively called factorial powers. When x is a positive integer, (x) n gives the number of n-p e r m u t a t i o n s of an x-e l e m e n t set, or equivalently the number of injective functions from a set of size n to a set of size x . Test if the FactorialPower gives the falling factorial: I n[]:= F a c t o r i a l P o w e r[x,n]== n ∏ k=1 (x-k+1)== n-1 ∏ k=0 (x-k)//F u l l S i m p l i f y O u t[]= T r u e I can also use SameQ or CodeEquivalentQ from the paclet Code Equivalence Utilities. The forms are not the same however in Mathematica: I n[]:= S a m e Q@@F a c t o r i a l P o w e r[x,n], n ∏ k=1 (x-k+1), n-1 ∏ k=0 (x-k) O u t[]= F a l s e I n[]:= C o d e E q u i v a l e n t Q@@F a c t o r i a l P o w e r[x,n], n ∏ k=1 (x-k+1), n-1 ∏ k=0 (x-k) O u t[]= F a l s e I n[]:= N e e d s["W o l f r a mC o d e E q u i v a l e n c e U t i l i t i e s"] EquivalenceTestData shows why the forms are not the same: Show the differences between FactorialPower and the first product: I n[]:= E q u i v a l e n c e T e s t D a t aF a c t o r i a l P o w e r[x,n], n ∏ k=1 (x-k+1) O u t[]= T i m i n gS a m e Q0.× -8 1 0 ,E q u a l Q0.× -8 1 0 ,T o C a n o n i c a l F o r m 10.3 5 9 4 1 3 3,T o C a n o n i c a l F o r m 20.6 4 0 6 2 0 5,E v a l u a t e 10.6 4 0 6 2 0 5,E v a l u a t e 20.6 5 6 2 4 6 1,S a n d b o x E q u i v a l e n t Q0.8 7 4 9 9 4 8,S a m e QF a l s e,E q u a l QF a l s e,C a n o n i c a l F o r m s1H o l d C o m p l e t e[F a c t o r i a l P o w e r[ 1 , 2 ]],2H o l d C o m p l e t e 3 ∏ 1 =1 (1- 1 + 2 ),C a n o n i c a l E q u i v a l e n t QF a l s e,S a n d b o x F o r m s1F a c t o r i a l P o w e r[x,n],2 (1-n+x)P o c h h a m m e r[2-n+x,n] 1+x ,S a n d b o x E q u i v a l e n t QF a l s e,E q u i v a l e n t QF a l s e Show the differences between the second and third terms. Notice the form with Pochhammer. I n[]:= E q u i v a l e n c e T e s t D a t a n ∏ k=1 (x-k+1), n-1 ∏ k=0 (x-k) O u t[]= T i m i n gS a m e Q0.× -8 1 0 ,E q u a l Q0.× -8 1 0 ,T o C a n o n i c a l F o r m 10.0 3 1 2 8 4 4,T o C a n o n i c a l F o r m 20.0 6 2 5 0 6 5,E v a l u a t e 10.0 7 8 1 3 9 3,E v a l u a t e 20.0 7 8 1 3 9 3,S a n d b o x E q u i v a l e n t Q0.3 1 2 5 0 0 0,S a m e QF a l s e,E q u a l QF a l s e,C a n o n i c a l F o r m s1H o l d C o m p l e t e 3 ∏ 1 =1 (1- 1 + 2 ),2H o l d C o m p l e t e -1+ 3 ∏ 1 =0 (- 1 + 2 ),C a n o n i c a l E q u i v a l e n t QF a l s e,S a n d b o x F o r m s1 (1-n+x)P o c h h a m m e r[2-n+x,n] 1+x ,2(1-n+x)P o c h h a m m e r[2-n+x,-1+n],S a n d b o x E q u i v a l e n t QF a l s e,E q u i v a l e n t QF a l s e Show the differences between the third product and FactorialPower: I n[]:= E q u i v a l e n c e T e s t D a t a n-1 ∏ k=0 (x-k),F a c t o r i a l P o w e r[x,n] O u t[]= T i m i n gS a m e Q0.× -8 1 0 ,E q u a l Q0.× -8 1 0 ,T o C a n o n i c a l F o r m 10.0 4 6 8 4 3 7,T o C a n o n i c a l F o r m 20.0 9 3 7 1 6 5,E v a l u a t e 10.0 9 3 7 1 6 5,E v a l u a t e 20.1 0 9 3 4 4 8,S a n d b o x E q u i v a l e n t Q0.3 2 8 0 9 0 0,S a m e QF a l s e,E q u a l QF a l s e,C a n o n i c a l F o r m s1H o l d C o m p l e t e -1+ 3 ∏ 1 =0 (- 1 + 2 ),2H o l d C o m p l e t e[F a c t o r i a l P o w e r[ 1 , 2 ]],C a n o n i c a l E q u i v a l e n t QF a l s e,S a n d b o x F o r m s1(1-n+x)P o c h h a m m e r[2-n+x,-1+n],2F a c t o r i a l P o w e r[x,n],S a n d b o x E q u i v a l e n t QF a l s e,E q u i v a l e n t QF a l s e ### Examples and combinatorial interpretation The first few rising factorials are as follows: I n[]:= G r i d[T a b l e[{i,P o c h h a m m e r[x,i],E x p a n d[P o c h h a m m e r[x,i]]},{i,0,9}],F r a m e->A l l]//T r a d i t i o n a l F o r m O u t[]//T r a d i t i o n a l F o r m= 0 1 1 1 x x 2 x(x+1)2 x +x 3 x(x+1)(x+2)3 x +3 2 x +2 x 4 x(x+1)(x+2)(x+3)4 x +6 3 x +1 1 2 x +6 x 5 x(x+1)(x+2)(x+3)(x+4)5 x +1 0 4 x +3 5 3 x +5 0 2 x +2 4 x 6 x(x+1)(x+2)(x+3)(x+4)(x+5)6 x +1 5 5 x +8 5 4 x +2 2 5 3 x +2 7 4 2 x +1 2 0 x 7 x(x+1)(x+2)(x+3)(x+4)(x+5)(x+6)7 x +2 1 6 x +1 7 5 5 x +7 3 5 4 x +1 6 2 4 3 x +1 7 6 4 2 x +7 2 0 x 8 x(x+1)(x+2)(x+3)(x+4)(x+5)(x+6)(x+7)8 x +2 8 7 x +3 2 2 6 x +1 9 6 0 5 x +6 7 6 9 4 x +1 3 1 3 2 3 x +1 3 0 6 8 2 x +5 0 4 0 x 9 x(x+1)(x+2)(x+3)(x+4)(x+5)(x+6)(x+7)(x+8)9 x +3 6 8 x +5 4 6 7 x +4 5 3 6 6 x +2 2 4 4 9 5 x +6 7 2 8 4 4 x +1 1 8 1 2 4 3 x +1 0 9 5 8 4 2 x +4 0 3 2 0 x The first few falling factorials are as follows: I n[]:= G r i d[T a b l e[{i,F u n c t i o n E x p a n d@F a c t o r i a l P o w e r[x,i],E x p a n d[F u n c t i o n E x p a n d@F a c t o r i a l P o w e r[x,i]]},{i,0,9}],F r a m e->A l l]//T r a d i t i o n a l F o r m O u t[]//T r a d i t i o n a l F o r m= 0 1 1 1 x x 2(x-1)x 2 x -x 3(x-2)(x-1)x 3 x -3 2 x +2 x 4(x-3)(x-2)(x-1)x 4 x -6 3 x +1 1 2 x -6 x 5(x-4)(x-3)(x-2)(x-1)x 5 x -1 0 4 x +3 5 3 x -5 0 2 x +2 4 x 6(x-5)(x-4)(x-3)(x-2)(x-1)x 6 x -1 5 5 x +8 5 4 x -2 2 5 3 x +2 7 4 2 x -1 2 0 x 7(x-6)(x-5)(x-4)(x-3)(x-2)(x-1)x 7 x -2 1 6 x +1 7 5 5 x -7 3 5 4 x +1 6 2 4 3 x -1 7 6 4 2 x +7 2 0 x 8(x-7)(x-6)(x-5)(x-4)(x-3)(x-2)(x-1)x 8 x -2 8 7 x +3 2 2 6 x -1 9 6 0 5 x +6 7 6 9 4 x -1 3 1 3 2 3 x +1 3 0 6 8 2 x -5 0 4 0 x 9(x-8)(x-7)(x-6)(x-5)(x-4)(x-3)(x-2)(x-1)x 9 x -3 6 8 x +5 4 6 7 x -4 5 3 6 6 x +2 2 4 4 9 5 x -6 7 2 8 4 4 x +1 1 8 1 2 4 3 x -1 0 9 5 8 4 2 x +4 0 3 2 0 x The coefficients that appear in the expansions are Stirling numbers of the first kind. I'm hoping to write about about Stirling numbers of the first kind later. When the variable x is a positive integer, the number (x) n is equal to the number of n-p e r m u t a t i o n s f r o m a n x-s e t , that is the number of ways of choosing an ordered list of length n consisting of distinct elements drawn from a collection of size x . For example, what is the number of different podiums—assignments of gold, silver, and bronze metals—possible in an eight-person race? The answer is: (8) 3 =8 7 6=3 3 6 Also (x) n is the "number of ways to arrange n flags on x flagpoles", where all flags must be used and each flagpole can have at most one flag. In this context, other notations like x P n , x P n , or P(x,n) are also sometimes used. ### Properties The rising and falling factorials are simply related to one another: (n) m = (m+n-1) n = n (-1) (-m) n (m) n = (n) (m-n+1) = n (-1) (n) (-m) The rising and falling factorials are directly related to the ordinary factorial: n!= (n) 1 = (n) n (m) n = m! (m-n)! (n) m = (m+n-1)! (m-1)! The rising and falling factorials can be used to express a binomial coefficient: (n) x n! = x+n-1 n (x) n n! = x n Thus many identities on binomial coefficients carry over to the falling and rising factorials. I don't really understand what a unital ring is in the next sentence. I get the idea of what a ring is but I don't fully understand what a unital ring is. The rising and falling factorials are well defined in any unital ring, and therefore x can be taken to be, for example, a complex number, including negative integers, or a polynomial with complex coefficients, or any complex-valued function. The rising factorial can be extended to real values of n using the log-convex gamma function provided x and x+n are real numbers that are not negative integers: (n) x = Γ(x+n) Γ(x) and so can the falling factorial: (x) n = Γ(x+1) Γ(x-n+1) Falling factorials appear in multiple differentiation of simple power functions: n n x a x = (a) n a-n x The rising factorial is also integral to the definition of the hypergeometric function: The hypergeometric function is defined for z<1 by the power series 2 F 1 (a,b;c;z)= ∞ ∑ n=0 (n) a (n) b (n) c n z n! provided that c!=0,-1,-2,… . Note, however, that the hypergeometric function literature typically uses the notation (a) n for rising factorials. I n[]:= H y p e r g e o m e t r i c 2 F 1[a,b,c,z]== ∞ ∑ n=0 ( n a n b ) n z n c n! //F u l l S i m p l i f y O u t[]= a b z c H y p e r g e o m e t r i c 2 F 1[a,b,c,z] I n[]:= F u l l S i m p l i f yH y p e r g e o m e t r i c 2 F 1[a,b,c,z]== ∞ ∑ n=0 ( n a n b ) n z n c n! ,A s s u m p t i o n s->{c∉ ≤0 ,A b s[z]<1} O u t[]= a b z c H y p e r g e o m e t r i c 2 F 1[a,b,c,z] I'm not sure why FullSimplify didn't work. I can find an instance of the function however: I n[]:= F i n d I n s t a n c e a b z c H y p e r g e o m e t r i c 2 F 1[a,b,c,z],{a,b,c,z} O u t[]= a- 2 1 2 9 + 2 8 2 5 ,b- 8 7 2 2 - 9 1 3 ,c 2 9 2 3 - 2 5 3 1 ,z0 ### Relation to umbral calculus The falling factorial occurs in a formula which represents polynomials using the forward difference operator Δ and which is formally similar to Taylor's theorem: f(x)= ∞ ∑ n=0 n Δ f(0) n! (x) n I entered the following input with DifferenceDelta f(x)= ∞ ∑ n=0 ( i f) n! (x) n Here is the same cell in StandardForm: f[x]= ∞ ∑ n=0 D i f f e r e n c e D e l t a[f,i] x n n!
POSTED BY: Peter Burbery
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187903 | https://www.proprep.com/questions/define-preimage-math-definition-and-its-role-in-understanding-inverse-mappings-in-linear-algebra | Define preimage math definition and its role in understanding inverse mappings in linear algebra.
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Define preimage math definition and its role in understanding inverse mappings in linear algebra.
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In linear algebra, the concept of a preimage is fundamental to understanding inverse mappings. The preimage of a set under a given function is the collection of all elements in the domain that map to the elements of that set under the function. To elaborate on this definition and its role in inverse mappings, let's proceed step by step.
aaaaaaaa aaaaaaaa sarfrrfrfrr ffrfrf
yyyynejhgfrrrrrrrrrrrrttttttttttttttttttttttttt
bbbb bbb bbbb bbbb bbbb bbbb bbbb
ccc ccc dsdsddsdcccccccccjrsdccccccc ggc.
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Exercise 11 - Solving 44 system of linear equations using inverse matrix method
Exercise 10 - Solving 33 system of linear equations using inverse matrix method
Exercise 4 Part b - Inverse of linear transformation from M2(R) to P3(R)
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00:00 In this exercise, we have here a system of linear equations.
00:05 I guess could have put curly braces here to
00:08 emphasize the system as a whole, 4 equations,
00:11 4 unknowns, x, y, z, and t.
00:16 We want to solve that using the inverse matrix method.
00:20 First of all, let's write it in matrix form.
00:24 We just take the coefficients making sure that the variables are all in order.
00:29 Then any missing variable gets a 0,
00:31 like here we had a missing t and here we have a missing x.
00:38 Then the column matrix which is a vector,
00:43 x, y, z, t,
00:44 the variables and then the constants 1, 0, 1, 0,
00:48 label these matrix A, vector x, vector b.
00:54 The theory says that when you have it in this form,
00:58 that is the form A x equals b,
01:01 then the solution x is given by the inverse of A times b.
01:09 We do actually have the inverse of A because we did this in
01:13 a previous exercise where we found that the inverse value was this,
01:19 just put the labels on these.
01:22 The way we got to this,
01:26 the exercise number at the time that I'm recording this,
01:31 it may have changed since is this.
01:35 So go and check,
01:37 and that's what we got for A inverse.
01:41 Okay. All that remains to do is to multiply out this by this.
01:50 Here's the result of the calculation.
01:53 I'll go through it with you.
01:55 The first entry here should be equal to this row times this column.
02:03 When I say row times a column, you know what I mean.
02:05 Pairwise product and add,
02:07 7 times 1 is 7,
02:11 minus nothing, minus 20 plus nothing,
02:17 7 minus 20 is minus 13.
02:19 That one's okay.
02:22 Next 1, minus 2, 0, 6, 0, minus 2 and 6 is 4.
02:32 That's okay.
02:33 Next 1, 3, 0, minus 8, 0, 3 minus 8, is minus 5.
02:44 The last, minus 1, 0, 3, 0 altogether 2.
02:55 It might be nicer to write explicitly what x is,
02:59 y is and z and t are and put it in a nice box.
03:04 In case you somehow are missing this exercise,
03:09 I'll give you a summary of what it looks like.
03:13 Here we are, here's a summary of these steps to get from A to A inverse.
03:19 Yes, I know it doesn't quite all fit in.
03:21 I'll scroll down in the moment.
03:23 It starts from here and then continues up to here,
03:28 and then from here to here,
03:36 and then down here.
03:37 Now I'm going to just scroll a bit so you can have the last bit.
03:44 As I said, we did do it in a previous exercise.
03:46 This is just in case you missed it. We're done.
This video explains how to solve a system of linear equations using the inverse matrix method. The system is written in matrix form, with the coefficients of the variables and the constants in the column matrix. The solution is given by the inverse of the matrix multiplied by the column matrix. The steps to get from the matrix to its inverse are explained in detail, with a summary of the steps provided in case the exercise is missed.
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What is the difference with variables and parameters
Course Q&AMachine Learning SpecializationSupervised ML: Regression and Classification
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someone555777
Apr 2023
Hi! I found that you use this terms separately. So after a bit of googling, I understood, that variables is something that is passing to function and parameters is near to term coefficient, which only helps to count function?
So, in the first example variable is “x”
изображение1920×956 106 KB
And in second example variables for J are w,b, isn’t it? In the same time, when they were only parameters for f(x)?
изображение1920×954 75.5 KB
I am python developer and a bit not understand your maths syntax
Are f(x) and y from second screen something like global functions in python with bounded context to them w and b? But in the same time we pass this w and b manually to J?
saifkhanengr
DLS Mentor
Apr 2023
Hello @someone555777!
For example, in f(x)f(x), xx is variable (also input to function ff and also called an argument). But in f(x) = wx + bf(x)=wx+b, xx represents input data (single value), and f(x)f(x) represents the output (predicted value) also denoted by \hat{y}^y. In this equation, ww and bb are parameters (or in math, you can say coefficient and intercept). Furthermore, J_{(w,b)}J(w,b) is a cost function at the given value of ww and bb. You may call them variable or any other name but it is not the input argument. One point to note here is that it is J_{(w,b)}J(w,b), NOT J{(w,b)}J(w,b). The second term can confuse someone that ww and bb are input arguments to function JJ but the correct denotation is J_{(w,b)}J(w,b) (subscript) and it means the value of cost function at given value of ww and bb.
So, in Figure one, xx is the input data, and ww and bb are parameters. And in Figure 2, J_{(w,b)}J(w,b) is a cost function at the given value of ww and bb.
Is it clear now?
Best,
Saif.
Christian_Simonis
Leader
Apr 2023
In addition to @saifkhanengr’s great reply:
One more point, @someone555777: you will encounter many times in your future ML journey that a model is fitted (or a function parametrised): that means that the parameters of the model are determined via an optimization process (e.g. w/ gradient descent). Sometimes this process is also called „calibration“.
Anyhow: fitting a model is something fundamental that you will do very often in machine learning and in data science.
Side note: In deep neural networks with modern transformer architectures, we can even have up to billions (175 billion parameters in GPT3) or even one trillion of parameters (like in current state of the art models such as GPT4).
Best regards
Christian
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187905 | https://api.pageplace.de/preview/DT0400.9780134389967_A26625862/preview-9780134389967_A26625862.pdf | This page intentionally left blank C O N C R E T E M A T H E M A T I C S Second Edition Dedicated to Leonhard Euler (1707–1783) A Foundation for Computer Science C O N C R E T E M A T H E M A T I C S Second Edition Ronald L. Graham AT&T Bell Laboratories Donald E. Knuth Stanford University Oren Patashnik Center for Communications Research 6 77 ADDISON–WESLEY Upper Saddle River, NJ • Boston • Indianapolis • San Francisco New York • Toronto • Montréal • London • Munich • Paris • Madrid Capetown • Sydney • Tokyo • Singapore • Mexico City Library of Congress Cataloging-in-Publication Data Graham, Ronald Lewis, 1935-Concrete mathematics : a foundation for computer science / Ronald L. Graham, Donald E. Knuth, Oren Patashnik. — 2nd ed.
xiii,657 p. 24 cm.
Bibliography: p. 604 Includes index.
ISBN 978-0-201-55802-9 1. Mathematics. 2. Computer science—Mathematics. I. Knuth, Donald Ervin, 1938- . II. Patashnik, Oren, 1954- . III. Title.
QA39.2.G733 1994 510—dc20 93-40325 CIP Internet page contains current information about this book and related books.
Reproduced by Addison-Wesley from camera-ready copy supplied by the authors.
Copyright (c) 1994, 1989 by Addison-Wesley Publishing Company, Inc.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechan-ical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America.
ISBN-13 978-0-201-55802-9 ISBN-10 0-201-55802-5 Text printed in the United States at LSC Communications in Crawfordsville, Indiana.
Thirty-Second Printing, April 2018 32 18 Preface THIS BOOK IS BASED on a course of the same name that has been taught “Audience, level, and treatment— a description of such matters is what prefaces are supposed to be about.” —P.R.Halmos annually at Stanford University since 1970. About fifty students have taken it each year — juniors and seniors, but mostly graduate students — and alumni of these classes have begun to spawn similar courses elsewhere. Thus the time seems ripe to present the material to a wider audience (including sophomores).
It was a dark and stormy decade when Concrete Mathematics was born.
Long-held values were constantly being questioned during those turbulent years; college campuses were hotbeds of controversy. The college curriculum itself was challenged, and mathematics did not escape scrutiny. John Ham-mersley had just written a thought-provoking article “On the enfeeblement of mathematical skills by ‘Modern Mathematics’ and by similar soft intellectual trash in schools and universities” ; other worried mathematicians even asked, “Can mathematics be saved?” One of the present authors had “People do acquire a little brief author-ity by equipping themselves with jargon: they can pontificate and air a superficial expertise.
But what we should ask of educated mathematicians is not what they can speechify about, nor even what they know about the existing corpus of mathematical knowledge, but rather what can they now do with their learning and whether they can actually solve math-ematical problems arising in practice.
In short, we look for deeds not words.” —J.Hammersley embarked on a series of books called The Art of Computer Programming, and in writing the first volume he (DEK) had found that there were mathematical tools missing from his repertoire; the mathematics he needed for a thorough, well-grounded understanding of computer programs was quite different from what he’d learned as a mathematics major in college. So he introduced a new course, teaching what he wished somebody had taught him.
The course title “Concrete Mathematics” was originally intended as an antidote to “Abstract Mathematics,” since concrete classical results were rap-idly being swept out of the modern mathematical curriculum by a new wave of abstract ideas popularly called the “New Math.” Abstract mathematics is a wonderful subject, and there’s nothing wrong with it: It’s beautiful, general, and useful. But its adherents had become deluded that the rest of mathemat-ics was inferior and no longer worthy of attention. The goal of generalization had become so fashionable that a generation of mathematicians had become unable to relish beauty in the particular, to enjoy the challenge of solving quantitative problems, or to appreciate the value of technique. Abstract math-ematics was becoming inbred and losing touch with reality; mathematical ed-ucation needed a concrete counterweight in order to restore a healthy balance.
When DEK taught Concrete Mathematics at Stanford for the first time, he explained the somewhat strange title by saying that it was his attempt v vi PREFACE to teach a math course that was hard instead of soft. He announced that, contrary to the expectations of some of his colleagues, he was not going to teach the Theory of Aggregates, nor Stone’s Embedding Theorem, nor even the Stone–Čech compactification. (Several students from the civil engineering “The heart of math-ematics consists of concrete exam-ples and concrete problems.” —P.R.Halmos department got up and quietly left the room.) Although Concrete Mathematics began as a reaction against other trends, the main reasons for its existence were positive instead of negative. And as the course continued its popular place in the curriculum, its subject matter “solidified” and proved to be valuable in a variety of new applications. Mean-while, independent confirmation for the appropriateness of the name came from another direction, when Z. A. Melzak published two volumes entitled “It is downright sinful to teach the abstract before the concrete.” —Z.A.Melzak Companion to Concrete Mathematics .
The material of concrete mathematics may seem at first to be a disparate bag of tricks, but practice makes it into a disciplined set of tools. Indeed, the techniques have an underlying unity and a strong appeal for many people.
When another one of the authors (RLG) first taught the course in 1979, the students had such fun that they decided to hold a class reunion a year later.
But what exactly is Concrete Mathematics? It is a blend of continuous Concrete Mathe-matics is a bridge to abstract mathe-matics.
and discrete mathematics. More concretely, it is the controlled manipulation of mathematical formulas, using a collection of techniques for solving prob-lems. Once you, the reader, have learned the material in this book, all you will need is a cool head, a large sheet of paper, and fairly decent handwriting in order to evaluate horrendous-looking sums, to solve complex recurrence relations, and to discover subtle patterns in data. You will be so fluent in algebraic techniques that you will often find it easier to obtain exact results than to settle for approximate answers that are valid only in a limiting sense.
The major topics treated in this book include sums, recurrences, ele-“The advanced reader who skips parts that appear too elementary may miss more than the less advanced reader who skips parts that appear too complex.” —G. Polya mentary number theory, binomial coefficients, generating functions, discrete probability, and asymptotic methods. The emphasis is on manipulative tech-nique rather than on existence theorems or combinatorial reasoning; the goal is for each reader to become as familiar with discrete operations (like the greatest-integer function and finite summation) as a student of calculus is familiar with continuous operations (like the absolute-value function and in-definite integration).
Notice that this list of topics is quite different from what is usually taught nowadays in undergraduate courses entitled “Discrete Mathematics.” There-fore the subject needs a distinctive name, and “Concrete Mathematics” has proved to be as suitable as any other.
(We’re not bold enough to try Distinuous Math-ematics.) The original textbook for Stanford’s course on concrete mathematics was the “Mathematical Preliminaries” section in The Art of Computer Program-ming . But the presentation in those 110 pages is quite terse, so another author (OP) was inspired to draft a lengthy set of supplementary notes. The PREFACE vii present book is an outgrowth of those notes; it is an expansion of, and a more leisurely introduction to, the material of Mathematical Preliminaries. Some of the more advanced parts have been omitted; on the other hand, several topics not found there have been included here so that the story will be complete.
The authors have enjoyed putting this book together because the subject began to jell and to take on a life of its own before our eyes; this book almost “. . . a concrete life preserver thrown to students sinking in a sea of abstraction.” —W. Gottschalk seemed to write itself. Moreover, the somewhat unconventional approaches we have adopted in several places have seemed to fit together so well, after these years of experience, that we can’t help feeling that this book is a kind of manifesto about our favorite way to do mathematics. So we think the book has turned out to be a tale of mathematical beauty and surprise, and we hope that our readers will share at least ϵ of the pleasure we had while writing it.
Since this book was born in a university setting, we have tried to capture the spirit of a contemporary classroom by adopting an informal style. Some people think that mathematics is a serious business that must always be cold and dry; but we think mathematics is fun, and we aren’t ashamed to admit the fact.
Why should a strict boundary line be drawn between work and play? Concrete mathematics is full of appealing patterns; the manipulations are not always easy, but the answers can be astonishingly attractive. The joys and sorrows of mathematical work are reflected explicitly in this book because they are part of our lives.
Students always know better than their teachers, so we have asked the first students of this material to contribute their frank opinions, as “graffiti” Math graffiti: Kilroy wasn’t Haar.
Free the group.
Nuke the kernel.
Power to the n.
N=1 ⇒P=NP.
in the margins.
Some of these marginal markings are merely corny, some are profound; some of them warn about ambiguities or obscurities, others are typical comments made by wise guys in the back row; some are positive, some are negative, some are zero. But they all are real indications of feelings that should make the text material easier to assimilate. (The inspiration for such marginal notes comes from a student handbook entitled Approaching Stanford, where the official university line is counterbalanced by the remarks of outgoing students. For example, Stanford says, “There are a few things I have only a marginal interest in this subject.
you cannot miss in this amorphous shape which is Stanford”; the margin says, “Amorphous . . . what the h does that mean? Typical of the pseudo-intellectualism around here.” Stanford: “There is no end to the potential of a group of students living together.” Graffito: “Stanford dorms are like zoos without a keeper.”) The margins also include direct quotations from famous mathematicians This was the most enjoyable course I’ve ever had. But it might be nice to summarize the material as you go along.
of past generations, giving the actual words in which they announced some of their fundamental discoveries. Somehow it seems appropriate to mix the words of Leibniz, Euler, Gauss, and others with those of the people who will be continuing the work. Mathematics is an ongoing endeavor for people everywhere; many strands are being woven into one rich fabric.
viii PREFACE This book contains more than 500 exercises, divided into six categories: I see: Concrete mathemat-ics means drilling.
• Warmups are exercises that every reader should try to do when first reading the material.
• Basics are exercises to develop facts that are best learned by trying one’s own derivation rather than by reading somebody else’s.
The homework was tough but I learned a lot. It was worth every hour.
• Homework exercises are problems intended to deepen an understand-ing of material in the current chapter.
• Exam problems typically involve ideas from two or more chapters si-multaneously; they are generally intended for use in take-home exams Take-home exams are vital—keep them.
(not for in-class exams under time pressure).
• Bonus problems go beyond what an average student of concrete math-ematics is expected to handle while taking a course based on this book; Exams were harder than the homework led me to expect.
they extend the text in interesting ways.
• Research problems may or may not be humanly solvable, but the ones presented here seem to be worth a try (without time pressure).
Answers to all the exercises appear in Appendix A, often with additional infor-mation about related results. (Of course, the “answers” to research problems are incomplete; but even in these cases, partial results or hints are given that might prove to be helpful.) Readers are encouraged to look at the answers, especially the answers to the warmup problems, but only after making a serious attempt to solve the problem without peeking.
Cheaters may pass this course by just copying the an-swers, but they’re only cheating themselves.
We have tried in Appendix C to give proper credit to the sources of each exercise, since a great deal of creativity and/or luck often goes into the design of an instructive problem.
Mathematicians have unfortunately developed a tradition of borrowing exercises without any acknowledgment; we believe that the opposite tradition, practiced for example by books and magazines about chess (where names, dates, and locations of original chess problems are routinely specified) is far superior. However, we have not been Difficult exams don’t take into ac-count students who have other classes to prepare for.
able to pin down the sources of many problems that have become part of the folklore. If any reader knows the origin of an exercise for which our citation is missing or inaccurate, we would be glad to learn the details so that we can correct the omission in subsequent editions of this book.
The typeface used for mathematics throughout this book is a new design by Hermann Zapf , commissioned by the American Mathematical Society and developed with the help of a committee that included B. Beeton, R. P.
Boas, L. K. Durst, D. E. Knuth, P. Murdock, R. S. Palais, P. Renz, E. Swanson, S. B. Whidden, and W. B. Woolf. The underlying philosophy of Zapf’s design is to capture the flavor of mathematics as it might be written by a mathemati-cian with excellent handwriting. A handwritten rather than mechanical style is appropriate because people generally create mathematics with pen, pencil, PREFACE ix or chalk. (For example, one of the trademarks of the new design is the symbol for zero, ‘0’, which is slightly pointed at the top because a handwritten zero rarely closes together smoothly when the curve returns to its starting point.) I’m unaccustomed to this face.
The letters are upright, not italic, so that subscripts, superscripts, and ac-cents are more easily fitted with ordinary symbols. This new type family has been named AMS Euler, after the great Swiss mathematician Leonhard Euler (1707–1783) who discovered so much of mathematics as we know it today.
The alphabets include Euler Text (Aa Bb Cc through Xx Yy Zz), Euler Frak-tur (Aa Bb Cc through Xx Yy Zz), and Euler Script Capitals (A B C through X Y Z), as well as Euler Greek (Aα Bβ Γγ through Xχ Ψψ Ωω) and special symbols such as ℘and ℵ. We are especially pleased to be able to inaugurate the Euler family of typefaces in this book, because Leonhard Euler’s spirit truly lives on every page: Concrete mathematics is Eulerian mathematics.
The authors are extremely grateful to Andrei Broder, Ernst Mayr, An-Dear prof: Thanks for (1) the puns, (2) the subject matter.
drew Yao, and Frances Yao, who contributed greatly to this book during the years that they taught Concrete Mathematics at Stanford. Furthermore we offer 1024 thanks to the teaching assistants who creatively transcribed what took place in class each year and who helped to design the examination ques-tions; their names are listed in Appendix C. This book, which is essentially a compendium of sixteen years’ worth of lecture notes, would have been im-possible without their first-rate work.
Many other people have helped to make this book a reality. For example, we wish to commend the students at Brown, Columbia, CUNY, Princeton, I don’t see how what I’ve learned will ever help me.
Rice, and Stanford who contributed the choice graffiti and helped to debug our first drafts.
Our contacts at Addison-Wesley were especially efficient and helpful; in particular, we wish to thank our publisher (Peter Gordon), production supervisor (Bette Aaronson), designer (Roy Brown), and copy ed-itor (Lyn Dupré). The National Science Foundation and the Office of Naval Research have given invaluable support. Cheryl Graham was tremendously helpful as we prepared the index. And above all, we wish to thank our wives (Fan, Jill, and Amy) for their patience, support, encouragement, and ideas.
I had a lot of trou-ble in this class, but I know it sharpened my math skills and my thinking skills.
This second edition features a new Section 5.8, which describes some important ideas that Doron Zeilberger discovered shortly after the first edition went to press. Additional improvements to the first printing can also be found on almost every page.
We have tried to produce a perfect book, but we are imperfect authors.
Therefore we solicit help in correcting any mistakes that we’ve made. A re-ward of $2.56 will gratefully be conveyed to anyone who is the first to report any error, whether it is mathematical, historical, or typographical.
I would advise the casual student to stay away from this course.
Murray Hill, New Jersey — RLG and Stanford, California DEK May 1988 and October 1993 OP A Note on Notation SOME OF THE SYMBOLISM in this book has not (yet?) become standard.
Here is a list of notations that might be unfamiliar to readers who have learned similar material from other books, together with the page numbers where these notations are explained. (See the general index, at the end of the book, for references to more standard notations.) Notation Name Page ln x natural logarithm: loge x 276 lg x binary logarithm: log2 x 70 log x common logarithm: log10 x 449 ⌊x⌋ floor: max{ n | n » x, integer n } 67 ⌈x⌉ ceiling: min{ n | n – x, integer n } 67 x mod y remainder: x −y⌊x/y⌋ 82 {x} fractional part: x mod 1 70 X f(x) δx indefinite summation 48 Xb a f(x) δx definite summation 49 xn falling factorial power: x!/(x −n)!
47, 211 xn rising factorial power: Γ(x + n)/Γ(x) 48, 211 n¡ subfactorial: n!/0! −n!/1! + · · · + (−1)nn!/n!
194 ℜz real part: x, if z = x + iy 64 If you don’t under-stand what the x denotes at the bottom of this page, try asking your Latin professor instead of your math professor.
ℑz imaginary part: y, if z = x + iy 64 Hn harmonic number: 1/1 + · · · + 1/n 29 H(x) n generalized harmonic number: 1/1x + · · · + 1/nx 277 x A NOTE ON NOTATION xi f(m)(z) mth derivative of f at z 470 n m Stirling cycle number (the “first kind”) 259 n m Stirling subset number (the “second kind”) 258 n m Eulerian number 267 n m Second-order Eulerian number 270 Prestressed concrete mathematics is con-crete mathematics that’s preceded by a bewildering list of notations.
(am . . . a0)b radix notation for Pm k=0 akbk 11 K(a1, . . . , an) continuant polynomial 302 F a, b c z hypergeometric function 205 #A cardinality: number of elements in the set A 39 [zn] f(z) coefficient of zn in f(z) 197 [α . . β] closed interval: the set {x | α » x » β} 73 [m = n] 1 if m = n, otherwise 0 24 [m\n] 1 if m divides n, otherwise 0 102 [m\ \n] 1 if m exactly divides n, otherwise 0 146 [m ⊥n] 1 if m is relatively prime to n, otherwise 0 115 In general, if S is any statement that can be true or false, the bracketed notation [S] stands for 1 if S is true, 0 otherwise.
Throughout this text, we use single-quote marks (‘. . . ’) to delimit text as it is written, double-quote marks (“. . . ”) for a phrase as it is spoken. Thus, the string of letters ‘string’ is sometimes called a “string.” Also ‘nonstring’ is a string.
An expression of the form ‘a/bc’ means the same as ‘a/(bc)’. Moreover, log x/log y = (log x)/(log y) and 2n! = 2(n!).
Contents 1 Recurrent Problems 1 1.1 The Tower of Hanoi 1 1.2 Lines in the Plane 4 1.3 The Josephus Problem 8 Exercises 17 2 Sums 21 2.1 Notation 21 2.2 Sums and Recurrences 25 2.3 Manipulation of Sums 30 2.4 Multiple Sums 34 2.5 General Methods 41 2.6 Finite and Infinite Calculus 47 2.7 Infinite Sums 56 Exercises 62 3 Integer Functions 67 3.1 Floors and Ceilings 67 3.2 Floor/Ceiling Applications 70 3.3 Floor/Ceiling Recurrences 78 3.4 ‘mod’: The Binary Operation 81 3.5 Floor/Ceiling Sums 86 Exercises 95 4 Number Theory 102 4.1 Divisibility 102 4.2 Primes 105 4.3 Prime Examples 107 4.4 Factorial Factors 111 4.5 Relative Primality 115 4.6 ‘mod’: The Congruence Relation 123 4.7 Independent Residues 126 4.8 Additional Applications 129 4.9 Phi and Mu 133 Exercises 144 5 Binomial Coefficients 153 5.1 Basic Identities 153 5.2 Basic Practice 172 5.3 Tricks of the Trade 186 xii CONTENTS xiii 5.4 Generating Functions 196 5.5 Hypergeometric Functions 204 5.6 Hypergeometric Transformations 216 5.7 Partial Hypergeometric Sums 223 5.8 Mechanical Summation 229 Exercises 242 6 Special Numbers 257 6.1 Stirling Numbers 257 6.2 Eulerian Numbers 267 6.3 Harmonic Numbers 272 6.4 Harmonic Summation 279 6.5 Bernoulli Numbers 283 6.6 Fibonacci Numbers 290 6.7 Continuants 301 Exercises 309 7 Generating Functions 320 7.1 Domino Theory and Change 320 7.2 Basic Maneuvers 331 7.3 Solving Recurrences 337 7.4 Special Generating Functions 350 7.5 Convolutions 353 7.6 Exponential Generating Functions 364 7.7 Dirichlet Generating Functions 370 Exercises 371 8 Discrete Probability 381 8.1 Definitions 381 8.2 Mean and Variance 387 8.3 Probability Generating Functions 394 8.4 Flipping Coins 401 8.5 Hashing 411 Exercises 427 9 Asymptotics 439 9.1 A Hierarchy 440 9.2 O Notation 443 9.3 O Manipulation 450 9.4 Two Asymptotic Tricks 463 9.5 Euler’s Summation Formula 469 9.6 Final Summations 476 Exercises 489 A Answers to Exercises 497 B Bibliography 604 C Credits for Exercises 632 Index 637 List of Tables 657 This page intentionally left blank 1 Recurrent Problems THIS CHAPTER EXPLORES three sample problems that give a feel for what’s to come. They have two traits in common: They’ve all been investi-gated repeatedly by mathematicians; and their solutions all use the idea of recurrence, in which the solution to each problem depends on the solutions to smaller instances of the same problem.
1.1 THE TOWER OF HANOI Let’s look first at a neat little puzzle called the Tower of Hanoi, invented by the French mathematician Edouard Lucas in 1883. We are given a tower of eight disks, initially stacked in decreasing size on one of three pegs: Raise your hand if you’ve never seen this.
OK, the rest of you can cut to equation (1.1).
The objective is to transfer the entire tower to one of the other pegs, moving only one disk at a time and never moving a larger one onto a smaller.
Lucas furnished his toy with a romantic legend about a much larger Tower of Brahma, which supposedly has 64 disks of pure gold resting on three Gold—wow.
Are our disks made of concrete?
diamond needles. At the beginning of time, he said, God placed these golden disks on the first needle and ordained that a group of priests should transfer them to the third, according to the rules above. The priests reportedly work day and night at their task. When they finish, the Tower will crumble and the world will end.
1 2 RECURRENT PROBLEMS It’s not immediately obvious that the puzzle has a solution, but a little thought (or having seen the problem before) convinces us that it does. Now the question arises: What’s the best we can do? That is, how many moves are necessary and sufficient to perform the task?
The best way to tackle a question like this is to generalize it a bit. The Tower of Brahma has 64 disks and the Tower of Hanoi has 8; let’s consider what happens if there are n disks.
One advantage of this generalization is that we can scale the problem down even more. In fact, we’ll see repeatedly in this book that it’s advanta-geous to look at small cases first. It’s easy to see how to transfer a tower that contains only one or two disks. And a small amount of experimentation shows how to transfer a tower of three.
The next step in solving the problem is to introduce appropriate notation: name and conquer. Let’s say that Tn is the minimum number of moves that will transfer n disks from one peg to another under Lucas’s rules. Then T1 is obviously 1, and T2 = 3.
We can also get another piece of data for free, by considering the smallest case of all: Clearly T0 = 0, because no moves at all are needed to transfer a tower of n = 0 disks! Smart mathematicians are not ashamed to think small, because general patterns are easier to perceive when the extreme cases are well understood (even when they are trivial).
But now let’s change our perspective and try to think big; how can we transfer a large tower? Experiments with three disks show that the winning idea is to transfer the top two disks to the middle peg, then move the third, then bring the other two onto it. This gives us a clue for transferring n disks in general: We first transfer the n −1 smallest to a different peg (requiring Tn−1 moves), then move the largest (requiring one move), and finally transfer the n−1 smallest back onto the largest (requiring another Tn−1 moves). Thus we can transfer n disks (for n > 0) in at most 2Tn−1 + 1 moves: Tn » 2Tn−1 + 1 , for n > 0.
This formula uses ‘ » ’ instead of ‘ = ’ because our construction proves only that 2Tn−1 + 1 moves suffice; we haven’t shown that 2Tn−1 + 1 moves are necessary. A clever person might be able to think of a shortcut.
But is there a better way? Actually no. At some point we must move the Most of the pub-lished “solutions” to Lucas’s problem, like the early one of Allardice and Fraser , fail to ex-plain why Tn must be – 2Tn−1 + 1.
largest disk. When we do, the n −1 smallest must be on a single peg, and it has taken at least Tn−1 moves to put them there. We might move the largest disk more than once, if we’re not too alert. But after moving the largest disk for the last time, we must transfer the n−1 smallest disks (which must again be on a single peg) back onto the largest; this too requires Tn−1 moves. Hence Tn – 2Tn−1 + 1 , for n > 0.
1.1 THE TOWER OF HANOI 3 These two inequalities, together with the trivial solution for n = 0, yield T0 = 0 ; Tn = 2Tn−1 + 1 , for n > 0.
(1.1) (Notice that these formulas are consistent with the known values T1 = 1 and T2 = 3. Our experience with small cases has not only helped us to discover a general formula, it has also provided a convenient way to check that we haven’t made a foolish error. Such checks will be especially valuable when we get into more complicated maneuvers in later chapters.) A set of equalities like (1.1) is called a recurrence (a.k.a. recurrence Yeah, yeah. . .
I seen that word before.
relation or recursion relation). It gives a boundary value and an equation for the general value in terms of earlier ones. Sometimes we refer to the general equation alone as a recurrence, although technically it needs a boundary value to be complete.
The recurrence allows us to compute Tn for any n we like. But nobody really likes to compute from a recurrence, when n is large; it takes too long.
The recurrence only gives indirect, local information.
A solution to the recurrence would make us much happier. That is, we’d like a nice, neat, “closed form” for Tn that lets us compute it quickly, even for large n. With a closed form, we can understand what Tn really is.
So how do we solve a recurrence? One way is to guess the correct solution, then to prove that our guess is correct.
And our best hope for guessing the solution is to look (again) at small cases. So we compute, successively, T3 = 2·3 + 1 = 7; T4 = 2·7 + 1 = 15; T5 = 2·15 + 1 = 31; T6 = 2·31 + 1 = 63.
Aha! It certainly looks as if Tn = 2n −1 , for n – 0.
(1.2) At least this works for n » 6.
Mathematical induction is a general way to prove that some statement about the integer n is true for all n – n0. First we prove the statement when n has its smallest value, n0; this is called the basis. Then we prove the Mathematical in-duction proves that we can climb as high as we like on a ladder, by proving that we can climb onto the bottom rung (the basis) and that from each rung we can climb up to the next one (the induction).
statement for n > n0, assuming that it has already been proved for all values between n0 and n −1, inclusive; this is called the induction. Such a proof gives infinitely many results with only a finite amount of work.
Recurrences are ideally set up for mathematical induction. In our case, for example, (1.2) follows easily from (1.1): The basis is trivial, since T0 = 20 −1 = 0. And the induction follows for n > 0 if we assume that (1.2) holds when n is replaced by n −1: Tn = 2Tn−1 + 1 = 2(2n−1 −1) + 1 = 2n −1 .
Hence (1.2)holds for n as well. Good! Our quest for Tn has ended successfully.
4 RECURRENT PROBLEMS Of course the priests’ task hasn’t ended; they’re still dutifully moving disks, and will be for a while, because for n = 64 there are 264−1 moves (about 18 quintillion). Even at the impossible rate of one move per microsecond, they will need more than 5000 centuries to transfer the Tower of Brahma. Lucas’s original puzzle is a bit more practical. It requires 28 −1 = 255 moves, which takes about four minutes for the quick of hand.
The Tower of Hanoi recurrence is typical of many that arise in applica-tions of all kinds. In finding a closed-form expression for some quantity of interest like Tn we go through three stages: 1 Look at small cases. This gives us insight into the problem and helps us in stages 2 and 3.
2 Find and prove a mathematical expression for the quantity of interest.
What is a proof?
“One half of one percent pure alco-hol.” For the Tower of Hanoi, this is the recurrence (1.1) that allows us, given the inclination, to compute Tn for any n.
3 Find and prove a closed form for our mathematical expression. For the Tower of Hanoi, this is the recurrence solution (1.2).
The third stage is the one we will concentrate on throughout this book. In fact, we’ll frequently skip stages 1 and 2 entirely, because a mathematical expression will be given to us as a starting point. But even then, we’ll be getting into subproblems whose solutions will take us through all three stages.
Our analysis of the Tower of Hanoi led to the correct answer, but it required an “inductive leap”; we relied on a lucky guess about the answer.
One of the main objectives of this book is to explain how a person can solve recurrences without being clairvoyant. For example, we’ll see that recurrence (1.1) can be simplified by adding 1 to both sides of the equations: T0 + 1 = 1 ; Tn + 1 = 2Tn−1 + 2 , for n > 0.
Now if we let Un = Tn + 1, we have Interesting: We get rid of the +1 in (1.1) by adding, not by subtracting.
U0 = 1 ; Un = 2Un−1 , for n > 0.
(1.3) It doesn’t take genius to discover that the solution to this recurrence is just Un = 2n; hence Tn = 2n −1. Even a computer could discover this.
1.2 LINES IN THE PLANE Our second sample problem has a more geometric flavor: How many slices of pizza can a person obtain by making n straight cuts with a pizza knife? Or, more academically: What is the maximum number Ln of regions 1.2 LINES IN THE PLANE 5 defined by n lines in the plane? This problem was first solved in 1826, by the Swiss mathematician Jacob Steiner .
(A pizza with Swiss cheese?) Again we start by looking at small cases, remembering to begin with the smallest of all. The plane with no lines has one region; with one line it has two regions; and with two lines it has four regions: L0 = 1 1 L1 = 2 1 2 L2 = 4 1 2 3 4 (Each line extends infinitely in both directions.) Sure, we think, Ln = 2n; of course! Adding a new line simply doubles the number of regions. Unfortunately this is wrong. We could achieve the doubling if the nth line would split each old region in two; certainly it can split an old region in at most two pieces, since each old region is convex. (A A region is convex if it includes all line segments be-tween any two of its points. (That’s not what my dictionary says, but it’s what mathematicians believe.) straight line can split a convex region into at most two new regions, which will also be convex.) But when we add the third line — the thick one in the diagram below — we soon find that it can split at most three of the old regions, no matter how we’ve placed the first two lines: hhhhhhhhhhhhhh 2 4a 4b 3b 1b 1a 3a Thus L3 = 4 + 3 = 7 is the best we can do.
And after some thought we realize the appropriate generalization. The nth line (for n > 0) increases the number of regions by k if and only if it splits k of the old regions, and it splits k old regions if and only if it hits the previous lines in k −1 different places. Two lines can intersect in at most one point. Therefore the new line can intersect the n−1 old lines in at most n−1 different points, and we must have k » n. We have established the upper bound Ln » Ln−1 + n , for n > 0.
Furthermore it’s easy to show by induction that we can achieve equality in this formula. We simply place the nth line in such a way that it’s not parallel to any of the others (hence it intersects them all), and such that it doesn’t go 6 RECURRENT PROBLEMS through any of the existing intersection points (hence it intersects them all in different places). The recurrence is therefore L0 = 1 ; Ln = Ln−1 + n , for n > 0.
(1.4) The known values of L1, L2, and L3 check perfectly here, so we’ll buy this.
Now we need a closed-form solution. We could play the guessing game again, but 1, 2, 4, 7, 11, 16, . . . doesn’t look familiar; so let’s try another tack. We can often understand a recurrence by “unfolding” or “unwinding” it all the way to the end, as follows: Ln = Ln−1 + n = Ln−2 + (n −1) + n Unfolding?
I’d call this “plugging in.” = Ln−3 + (n −2) + (n −1) + n .
.
.
= L0 + 1 + 2 + · · · + (n −2) + (n −1) + n = 1 + Sn , where Sn = 1 + 2 + 3 + · · · + (n −1) + n.
In other words, Ln is one more than the sum Sn of the first n positive integers.
The quantity Sn pops up now and again, so it’s worth making a table of small values. Then we might recognize such numbers more easily when we see them the next time: n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Sn 1 3 6 10 15 21 28 36 45 55 66 78 91 105 These values are also called the triangular numbers, because Sn is the num-ber of bowling pins in an n-row triangular array.
For example, the usual four-row array q q q q q q q q q q has S4 = 10 pins.
To evaluate Sn we can use a trick that Gauss reportedly came up with in 1786, when he was nine years old (and which had been used already by Archimedes in propositions 10 and 11 of his classic treatise on spirals): It seems a lot of stuff is attributed to Gauss— either he was really smart or he had a great press agent.
Maybe he just had a magnetic personality.
Sn = 1 + 2 + 3 + · · · + (n −1) + n + Sn = n + (n −1) + (n −2) + · · · + 2 + 1 2Sn = (n + 1) + (n + 1) + (n + 1) + · · · + (n + 1) + (n + 1) We merely add Sn to its reversal, so that each of the n columns on the right sums to n + 1. Simplifying, Sn = n(n + 1) 2 , for n – 0.
(1.5) 1.2 LINES IN THE PLANE 7 OK, we have our solution: Actually Gauss is often called the greatest mathe-matician of all time.
So it’s nice to be able to understand at least one of his discoveries.
Ln = n(n + 1) 2 + 1 , for n – 0.
(1.6) As experts, we might be satisfied with this derivation and consider it a proof, even though we waved our hands a bit when doing the unfolding and reflecting. But students of mathematics should be able to meet stricter standards; so it’s a good idea to construct a rigorous proof by induction. The key induction step is Ln = Ln−1 + n = 1 2(n −1)n + 1 + n = 1 2n(n + 1) + 1 .
Now there can be no doubt about the closed form (1.6).
Incidentally we’ve been talking about “closed forms” without explicitly When in doubt, look at the words.
Why is it “closed,” as opposed to “open”? What image does it bring to mind?
Answer: The equa-tion is “closed,” not defined in terms of itself—not leading to recurrence. The case is “closed”—it won’t happen again.
Metaphors are the key.
saying what we mean.
Usually it’s pretty clear.
Recurrences like (1.1) and (1.4) are not in closed form — they express a quantity in terms of it-self; but solutions like (1.2) and (1.6) are. Sums like 1 + 2 + · · · + n are not in closed form — they cheat by using ‘ · · · ’; but expressions like n(n + 1)/2 are.
We could give a rough definition like this: An expression for a quantity f(n) is in closed form if we can compute it using at most a fixed number of “well known” standard operations, independent of n.
For example, 2n −1 and n(n + 1)/2 are closed forms because they involve only addition, subtraction, multiplication, division, and exponentiation, in explicit ways.
The total number of simple closed forms is limited, and there are recur-rences that don’t have simple closed forms. When such recurrences turn out to be important, because they arise repeatedly, we add new operations to our repertoire; this can greatly extend the range of problems solvable in “simple” closed form. For example, the product of the first n integers, n!, has proved to be so important that we now consider it a basic operation. The formula ‘n!’ is therefore in closed form, although its equivalent ‘1·2·. . .·n’ is not.
And now, briefly, a variation of the lines-in-the-plane problem: Suppose that instead of straight lines we use bent lines, each containing one “zig.” What is the maximum number Zn of regions determined by n such bent lines Is “zig” a technical term?
in the plane? We might expect Zn to be about twice as big as Ln, or maybe three times as big. Let’s see: Z1 = 2 PPPPPPP P 1 2 Z2 = 7 B B B B B B B B PPPPPPP P 1 2 3 4 5 6 7 8 RECURRENT PROBLEMS From these small cases, and after a little thought, we realize that a bent . . . and a little afterthought. . .
line is like two straight lines except that regions merge when the “two” lines don’t extend past their intersection point.
PPPPPP p p p p p p p p p p p p p p p p p p p p 1 2 3 4 Regions 2, 3, and 4, which would be distinct with two lines, become a single region when there’s a bent line; we lose two regions. However, if we arrange things properly — the zig point must lie “beyond” the intersections with the other lines — that’s all we lose; that is, we lose only two regions per bent line.
Thus Exercise 18 has the details.
Zn = L2n −2n = 2n(2n + 1)/2 + 1 −2n = 2n2 −n + 1 , for n – 0.
(1.7) Comparing the closed forms (1.6) and (1.7), we find that for large n, Ln ∼ 1 2n2 , Zn ∼2n2 ; so we get about four times as many regions with bent lines as with straight lines. (In later chapters we’ll be discussing how to analyze the approximate behavior of integer functions when n is large. The ‘∼’ symbol is defined in Section 9.1.) 1.3 THE JOSEPHUS PROBLEM Our final introductory example is a variant of an ancient problem (Ahrens [5, vol. 2] and Herstein and Kaplansky discuss the interest-ing history of this problem. Josephus himself is a bit vague.) named for Flavius Josephus, a famous historian of the first century. Legend has it that Josephus wouldn’t have lived to become famous without his math-ematical talents. During the Jewish–Roman war, he was among a band of 41 Jewish rebels trapped in a cave by the Romans. Preferring suicide to capture, the rebels decided to form a circle and, proceeding around it, to kill every third remaining person until no one was left. But Josephus, along with an unindicted co-conspirator, wanted none of this suicide nonsense; so he quickly calculated where he and his friend should stand in the vicious circle.
. . . thereby saving his tale for us to hear.
In our variation, we start with n people numbered 1 to n around a circle, and we eliminate every second remaining person until only one survives. For 1.3 THE JOSEPHUS PROBLEM 9 example, here’s the starting configuration for n = 10: '$ &% 6 1 2 3 4 5 6 7 8 9 10 The elimination order is 2, 4, 6, 8, 10, 3, 7, 1, 9, so 5 survives. The problem: Determine the survivor’s number, J(n).
Here’s a case where n = 0 makes no sense.
We just saw that J(10) = 5. We might conjecture that J(n) = n/2 when n is even; and the case n = 2 supports the conjecture: J(2) = 1. But a few other small cases dissuade us — the conjecture fails for n = 4 and n = 6.
n 1 2 3 4 5 6 J(n) 1 1 3 1 3 5 It’s back to the drawing board; let’s try to make a better guess. Hmmm . . .
Even so, a bad guess isn’t a waste of time, because it gets us involved in the problem.
J(n) always seems to be odd. And in fact, there’s a good reason for this: The first trip around the circle eliminates all the even numbers. Furthermore, if n itself is an even number, we arrive at a situation similar to what we began with, except that there are only half as many people, and their numbers have changed.
So let’s suppose that we have 2n people originally. After the first go-round, we’re left with '$ &% 6 1 3 5 7 . . .
2n −3 2n −1 and 3 will be the next to go. This is just like starting out with n people, except that each person’s number has been doubled and decreased by 1. That is, This is the tricky part: We have J(2n) = newnumber(J(n)), where newnumber(k) = 2k −1.
J(2n) = 2J(n) −1 , for n – 1.
We can now go quickly to large n. For example, we know that J(10) = 5, so J(20) = 2J(10) −1 = 2·5 −1 = 9 .
Similarly J(40) = 17, and we can deduce that J(5·2m) = 2m+1 + 1.
10 RECURRENT PROBLEMS But what about the odd case? With 2n + 1 people, it turns out that Odd case? Hey, leave my brother out of it.
person number 1 is wiped out just after person number 2n, and we’re left with '$ &% 6 3 5 7 9 . . .
2n −1 2n + 1 .
Again we almost have the original situation with n people, but this time their numbers are doubled and increased by 1. Thus J(2n + 1) = 2J(n) + 1 , for n – 1.
Combining these equations with J(1) = 1 gives us a recurrence that defines J in all cases: J(1) = 1 ; J(2n) = 2J(n) −1 , for n – 1; J(2n + 1) = 2J(n) + 1 , for n – 1.
(1.8) Instead of getting J(n) from J(n −1), this recurrence is much more “efficient,” because it reduces n by a factor of 2 or more each time it’s applied. We could compute J(1000000), say, with only 19 applications of (1.8). But still, we seek a closed form, because that will be even quicker and more informative. After all, this is a matter of life or death.
Our recurrence makes it possible to build a table of small values very quickly. Perhaps we’ll be able to spot a pattern and guess the answer.
n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 J(n) 1 1 3 1 3 5 7 1 3 5 7 9 11 13 15 1 Voilà!
It seems we can group by powers of 2 (marked by vertical lines in the table); J(n) is always 1 at the beginning of a group and it increases by 2 within a group. So if we write n in the form n = 2m + l, where 2m is the largest power of 2 not exceeding n and where l is what’s left, the solution to our recurrence seems to be J(2m + l) = 2l + 1 , for m – 0 and 0 » l < 2m.
(1.9) (Notice that if 2m » n < 2m+1, the remainder l = n −2m satisfies 0 » l < 2m+1 −2m = 2m.) We must now prove (1.9). As in the past we use induction, but this time the induction is on m. When m = 0 we must have l = 0; thus the basis of 1.3 THE JOSEPHUS PROBLEM 11 (1.9) reduces to J(1) = 1, which is true. The induction step has two parts, But there’s a sim-pler way! The key fact is that J(2m) = 1 for all m, and this follows immedi-ately from our first equation, J(2n) = 2J(n)−1.
Hence we know that the first person will survive whenever n is a power of 2.
And in the gen-eral case, when n = 2m + l, the number of people is reduced to a power of 2 after there have been l executions.
The first remaining person at this point, the survivor, is number 2l + 1.
depending on whether l is even or odd. If m > 0 and 2m + l is equal to 2k, for some integer k, then l is even and J(2m + l) = 2J(2m−1 + l/2) −1 = 2(2l/2 + 1) −1 = 2l + 1 , by (1.8) and the induction hypothesis; this is exactly what we want. A similar proof works in the odd case, when 2m + l = 2k + 1. We might also note that (1.8) implies the relation J(2n + 1) −J(2n) = 2.
Either way, the induction is complete and (1.9) is established.
To illustrate solution (1.9), let’s compute J(100). In this case we have 100 = 26 + 36, so J(100) = 2·36 + 1 = 73.
Now that we’ve done the hard stuff (solved the problem) we seek the soft: Every solution to a problem can be generalized so that it applies to a wider class of problems. Once we’ve learned a technique, it’s instructive to look at it closely and see how far we can go with it. Hence, for the rest of this section, we will examine the solution (1.9) and explore some generalizations of the recurrence (1.8). These explorations will uncover the structure that underlies all such problems.
Powers of 2 played an important role in our finding the solution, so it’s natural to look at the radix 2 representations of n and J(n). Suppose n’s binary expansion is n = (bm bm−1 . . . b1 b0)2 ; that is, n = bm2m + bm−12m−1 + · · · + b12 + b0 , where each bi is either 0 or 1 and where the leading bit bm is 1. Recalling that n = 2m + l, we have, successively, n = (1 bm−1 bm−2 . . . b1 b0)2 , l = (0 bm−1 bm−2 . . . b1 b0)2 , 2l = (bm−1 bm−2 . . . b1 b0 0)2 , 2l + 1 = (bm−1 bm−2 . . . b1 b0 1)2 , J(n) = (bm−1 bm−2 . . . b1 b0 bm)2 .
(The last step follows because J(n) = 2l + 1 and because bm = 1.) We have proved that J (bm bm−1 . . . b1 b0)2 = (bm−1 . . . b1 b0 bm)2 ; (1.10) 12 RECURRENT PROBLEMS that is, in the lingo of computer programming, we get J(n) from n by doing a one-bit cyclic shift left! Magic. For example, if n = 100 = (1100100)2 then J(n) = J (1100100)2 = (1001001)2, which is 64 + 8 + 1 = 73. If we had been working all along in binary notation, we probably would have spotted this pattern immediately.
If we start with n and iterate the J function m + 1 times, we’re doing (“Iteration” here means applying a function to itself.) m + 1 one-bit cyclic shifts; so, since n is an (m+1)-bit number, we might expect to end up with n again. But this doesn’t quite work. For instance if n = 13 we have J (1101)2 = (1011)2, but then J (1011)2 = (111)2 and the process breaks down; the 0 disappears when it becomes the leading bit.
In fact, J(n) must always be » n by definition, since J(n) is the survivor’s number; hence if J(n) < n we can never get back up to n by continuing to iterate.
Repeated application of J produces a sequence of decreasing values that eventually reach a “fixed point,” where J(n) = n. The cyclic shift property makes it easy to see what that fixed point will be: Iterating the function enough times will always produce a pattern of all 1’s whose value is 2ν(n) −1, where ν(n) is the number of 1 bits in the binary representation of n. Thus, since ν(13) = 3, we have 2 or more J’s z }| { J(J(. . . J(13) . . . )) = 23 −1 = 7 ; similarly Curiously enough, if M is a compact C∞n-manifold (n > 1), there ex-ists a differentiable immersion of M into R2n−ν(n) but not necessarily into R2n−ν(n)−1 .
I wonder if Jose-phus was secretly a topologist?
8 or more z }| { J(J(. . . J((101101101101011)2) . . . )) = 210 −1 = 1023 .
Curious, but true.
Let’s return briefly to our first guess, that J(n) = n/2 when n is even.
This is obviously not true in general, but we can now determine exactly when it is true: J(n) = n/2 , 2l + 1 = (2m + l)/2 , l = 1 3(2m −2) .
If this number l = 1 3(2m −2) is an integer, then n = 2m +l will be a solution, because l will be less than 2m. It’s not hard to verify that 2m −2 is a multiple of 3 when m is odd, but not when m is even. (We will study such things in Chapter 4.) Therefore there are infinitely many solutions to the equation 1.3 THE JOSEPHUS PROBLEM 13 J(n) = n/2, beginning as follows: m l n = 2m + l J(n) = 2l + 1 = n/2 n (binary) 1 0 2 1 10 3 2 10 5 1010 5 10 42 21 101010 7 42 170 85 10101010 Notice the pattern in the rightmost column. These are the binary numbers for which cyclic-shifting one place left produces the same result as ordinary-shifting one place right (halving).
OK, we understand the J function pretty well; the next step is to general-ize it. What would have happened if our problem had produced a recurrence that was something like (1.8), but with different constants? Then we might not have been lucky enough to guess the solution, because the solution might have been really weird. Let’s investigate this by introducing constants α, β, and γ and trying to find a closed form for the more general recurrence Looks like Greek to me.
f(1) = α ; f(2n) = 2f(n) + β , for n – 1; f(2n + 1) = 2f(n) + γ , for n – 1.
(1.11) (Our original recurrence had α = 1, β = −1, and γ = 1.) Starting with f(1) = α and working our way up, we can construct the following general table for small values of n: n f(n) 1 α 2 2α + β 3 2α + γ 4 4α + 3β 5 4α + 2β + γ 6 4α + β + 2γ 7 4α + 3γ 8 8α + 7β 9 8α + 6β + γ (1.12) It seems that α’s coefficient is n’s largest power of 2. Furthermore, between powers of 2, β’s coefficient decreases by 1 down to 0 and γ’s increases by 1 up from 0. Therefore if we express f(n) in the form f(n) = A(n)α + B(n)β + C(n)γ , (1.13) 14 RECURRENT PROBLEMS by separating out its dependence on α, β, and γ, it seems that A(n) = 2m ; B(n) = 2m −1 −l ; (1.14) C(n) = l .
Here, as usual, n = 2m + l and 0 » l < 2m, for n – 1.
It’s not terribly hard to prove (1.13) and (1.14) by induction, but the Hold onto your hats, this next part is new stuff.
calculations are messy and uninformative. Fortunately there’s a better way to proceed, by choosing particular values and then combining them. Let’s illustrate this by considering the special case α = 1, β = γ = 0, when f(n) is supposed to be equal to A(n): Recurrence (1.11) becomes A(1) = 1 ; A(2n) = 2A(n) , for n – 1; A(2n + 1) = 2A(n) , for n – 1.
Sure enough, it’s true (by induction on m) that A(2m + l) = 2m.
Next, let’s use recurrence (1.11) and solution (1.13) in reverse, by start-ing with a simple function f(n) and seeing if there are any constants (α, β, γ) that will define it. Plugging the constant function f(n) = 1 into (1.11) says that A neat idea!
1 = α; 1 = 2·1 + β; 1 = 2·1 + γ; hence the values (α, β, γ) = (1, −1, −1) satisfying these equations will yield A(n) −B(n) −C(n) = f(n) = 1. Similarly, we can plug in f(n) = n: 1 = α; 2n = 2·n + β; 2n + 1 = 2·n + γ.
These equations hold for all n when α = 1, β = 0, and γ = 1, so we don’t need to prove by induction that these parameters will yield f(n) = n. We already know that f(n) = n will be the solution in such a case, because the recurrence (1.11) uniquely defines f(n) for every value of n.
And now we’re essentially done! We have shown that the functions A(n), B(n), and C(n) of (1.13), which solve (1.11) in general, satisfy the equations A(n) = 2m , where n = 2m + l and 0 » l < 2m; A(n) −B(n) −C(n) = 1 ; A(n) + C(n) = n .
1.3 THE JOSEPHUS PROBLEM 15 Our conjectures in (1.14) follow immediately, since we can solve these equa-tions to get C(n) = n −A(n) = l and B(n) = A(n) −1 −C(n) = 2m −1 −l.
This approach illustrates a surprisingly useful repertoire method for solv-Beware: The au-thors are expecting us to figure out the idea of the repertoire method from seat-of-the-pants examples, instead of giving us a top-down presentation. The method works best with recurrences that are “linear,” in the sense that the solutions can be expressed as a sum of arbitrary param-eters multiplied by functions of n, as in (1.13). Equation (1.13) is the key.
ing recurrences. First we find settings of general parameters for which we know the solution; this gives us a repertoire of special cases that we can solve.
Then we obtain the general case by combining the special cases. We need as many independent special solutions as there are independent parameters (in this case three, for α, β, and γ). Exercises 16 and 20 provide further examples of the repertoire approach.
We know that the original J-recurrence has a magical solution, in binary: J (bm bm−1 . . . b1 b0)2 = (bm−1 . . . b1 b0 bm)2 , where bm = 1.
Does the generalized Josephus recurrence admit of such magic?
Sure, why not? We can rewrite the generalized recurrence (1.11) as f(1) = α ; f(2n + j) = 2f(n) + βj , for j = 0, 1 and n – 1, (1.15) if we let β0 = β and β1 = γ. And this recurrence unfolds, binary-wise: f (bm bm−1 . . . b1 b0)2 = 2f (bm bm−1 . . . b1)2 + βb0 = 4f (bm bm−1 . . . b2)2 + 2βb1 + βb0 .
.
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= 2mf (bm)2 +2m−1βbm−1+ · · · +2βb1+βb0 = 2mα + 2m−1βbm−1 + · · · + 2βb1 + βb0 .
Suppose we now relax the radix 2 notation to allow arbitrary digits instead (‘relax’ = ‘destroy’) of just 0 and 1. The derivation above tells us that f (bm bm−1 . . . b1 b0)2 = (α βbm−1 βbm−2 . . . βb1 βb0)2 .
(1.16) Nice. We would have seen this pattern earlier if we had written (1.12) in I think I get it: The binary repre-sentations of A(n), B(n), and C(n) have 1’s in different positions.
another way: n f(n) 1 α 2 2α + β 3 2α + γ 4 4α + 2β + β 5 4α + 2β + γ 6 4α + 2γ + β 7 4α + 2γ + γ 16 RECURRENT PROBLEMS For example, when n = 100 = (1100100)2, our original Josephus values α = 1, β = −1, and γ = 1 yield n = ( 1 1 0 0 1 0 0 )2 = 100 f(n) = ( 1 1 −1 −1 1 −1 −1 )2 = +64 +32 −16 −8 +4 −2 −1 = 73 as before. The cyclic-shift property follows because each block of binary digits (1 0 . . . 0 0)2 in the representation of n is transformed into (1 −1 . . . −1 −1)2 = (0 0 . . . 0 1)2 .
So our change of notation has given us the compact solution (1.16) to the “There are two kinds of general-izations. One is cheap and the other is valuable.
It is easy to gen-eralize by diluting a little idea with a big terminology.
It is much more difficult to pre-pare a refined and condensed extract from several good ingredients.” —G. Polya general recurrence (1.15). If we’re really uninhibited we can now generalize even more. The recurrence f(j) = αj , for 1 » j < d; f(dn + j) = cf(n) + βj , for 0 » j < d and n – 1, (1.17) is the same as the previous one except that we start with numbers in radix d and produce values in radix c. That is, it has the radix-changing solution f (bm bm−1 . . . b1 b0)d = (αbm βbm−1 βbm−2 . . . βb1 βb0)c .
(1.18) For example, suppose that by some stroke of luck we’re given the recurrence f(1) = 34 ; f(2) = 5 ; f(3n) = 10f(n) + 76 , for n – 1; f(3n + 1) = 10f(n) −2 , for n – 1; f(3n + 2) = 10f(n) + 8 , for n – 1; and suppose we want to compute f(19). Here we have d = 3 and c = 10. Now Perhaps this was a stroke of bad luck.
19 = (201)3, and the radix-changing solution tells us to perform a digit-by-digit replacement from radix 3 to radix 10. So the leading 2 becomes a 5, and the 0 and 1 become 76 and −2, giving f(19) = f (201)3 = (5 76 −2)10 = 1258 , which is our answer.
Thus Josephus and the Jewish–Roman war have led us to some interesting But in general I’m against recurrences of war.
general recurrences.
1 EXERCISES 17 Exercises Warmups 1 All horses are the same color; we can prove this by induction on the Please do all the warmups in all the chapters!
—The Mgm’t number of horses in a given set. Here’s how: “If there’s just one horse then it’s the same color as itself, so the basis is trivial. For the induction step, assume that there are n horses numbered 1 to n. By the induc-tion hypothesis, horses 1 through n −1 are the same color, and similarly horses 2 through n are the same color. But the middle horses, 2 through n −1, can’t change color when they’re in different groups; these are horses, not chameleons. So horses 1 and n must be the same color as well, by transitivity. Thus all n horses are the same color; QED.” What, if anything, is wrong with this reasoning?
2 Find the shortest sequence of moves that transfers a tower of n disks from the left peg A to the right peg B, if direct moves between A and B are disallowed. (Each move must be to or from the middle peg. As usual, a larger disk must never appear above a smaller one.) 3 Show that, in the process of transferring a tower under the restrictions of the preceding exercise, we will actually encounter every properly stacked arrangement of n disks on three pegs.
4 Are there any starting and ending configurations of n disks on three pegs that are more than 2n −1 moves apart, under Lucas’s original rules?
5 A “Venn diagram” with three overlapping circles is often used to illustrate the eight possible subsets associated with three given sets: '$ &% '$ &% '$ &% A B C Can the sixteen possibilities that arise with four given sets be illustrated by four overlapping circles?
6 Some of the regions defined by n lines in the plane are infinite, while others are bounded. What’s the maximum possible number of bounded regions?
7 Let H(n) = J(n + 1) −J(n). Equation (1.8) tells us that H(2n) = 2, and H(2n+1) = J(2n+2)−J(2n+1) = 2J(n+1)−1 − 2J(n)+1 = 2H(n)−2, for all n – 1. Therefore it seems possible to prove that H(n) = 2 for all n, by induction on n. What’s wrong here?
18 RECURRENT PROBLEMS Homework exercises 8 Solve the recurrence Q0 = α ; Q1 = β ; Qn = (1 + Qn−1)/Qn−2 , for n > 1.
Assume that Qn ̸= 0 for all n – 0. Hint: Q4 = (1 + α)/β.
9 Sometimes it’s possible to use induction backwards, proving things from . . . now that’s a horse of a different color.
n to n −1 instead of vice versa! For example, consider the statement P(n) : x1 . . . xn » x1 + · · · + xn n n , if x1, . . . , xn – 0.
This is true when n = 2, since (x1 + x2)2 −4x1x2 = (x1 −x2)2 – 0.
a By setting xn = (x1 + · · · + xn−1)/(n −1), prove that P(n) im-plies P(n −1) whenever n > 1.
b Show that P(n) and P(2) imply P(2n).
c Explain why this implies the truth of P(n) for all n.
10 Let Qn be the minimum number of moves needed to transfer a tower of n disks from A to B if all moves must be clockwise — that is, from A to B, or from B to the other peg, or from the other peg to A. Also let Rn be the minimum number of moves needed to go from B back to A under this restriction. Prove that Qn = 0, if n = 0; 2Rn−1 +1, if n > 0; Rn = 0, if n = 0; Qn +Qn−1 +1, if n > 0.
(You need not solve these recurrences; we’ll see how to do that in Chap-ter 7.) 11 A Double Tower of Hanoi contains 2n disks of n different sizes, two of each size.
As usual, we’re required to move only one disk at a time, without putting a larger one over a smaller one.
a How many moves does it take to transfer a double tower from one peg to another, if disks of equal size are indistinguishable from each other?
b What if we are required to reproduce the original top-to-bottom order of all the equal-size disks in the final arrangement?
[Hint: This is difficult—it’s really a “bonus problem.”] 12 Let’s generalize exercise 11a even further, by assuming that there are n different sizes of disks and exactly mk disks of size k.
Determine A(m1, . . . , mn), the minimum number of moves needed to transfer a tower when equal-size disks are considered to be indistinguishable.
1 EXERCISES 19 13 What’s the maximum number of regions definable by n zig-zag lines, XXXXXXXXXXXXXXX QQQQQQ XXXXXXXXXXXXXXX ZZ2 = 12 each of which consists of two parallel infinite half-lines joined by a straight segment?
14 How many pieces of cheese can you obtain from a single thick piece by making five straight slices? (The cheese must stay in its original position while you do all the cutting, and each slice must correspond to a plane Good luck keep-ing the cheese in position.
in 3D.) Find a recurrence relation for Pn, the maximum number of three-dimensional regions that can be defined by n different planes.
15 Josephus had a friend who was saved by getting into the next-to-last position. What is I(n), the number of the penultimate survivor when every second person is executed?
16 Use the repertoire method to solve the general four-parameter recurrence g(1) = α ; g(2n + j) = 3g(n) + γn + βj , for j = 0, 1 and n – 1.
Hint: Try the function g(n) = n.
Exam problems 17 If Wn is the minimum number of moves needed to transfer a tower of n disks from one peg to another when there are four pegs instead of three, show that Wn(n+1)/2 » 2Wn(n−1)/2 + Tn , for n > 0.
(Here Tn = 2n −1 is the ordinary three-peg number.) Use this to find a closed form f(n) such that Wn(n+1)/2 » f(n) for all n – 0.
18 Show that the following set of n bent lines defines Zn regions, where Zn is defined in (1.7): The jth bent line, for 1 » j » n, has its zig at (n2j, 0) and goes up through the points (n2j −nj, 1) and (n2j −nj −n−n, 1).
19 Is it possible to obtain Zn regions with n bent lines when the angle at each zig is 30◦?
20 Use the repertoire method to solve the general five-parameter recurrence Is this like a five-star general recurrence?
h(1) = α ; h(2n + j) = 4h(n) + γjn + βj , for j = 0, 1 and n – 1.
Hint: Try the functions h(n) = n and h(n) = n2.
20 RECURRENT PROBLEMS 21 Suppose there are 2n people in a circle; the first n are “good guys” and the last n are “bad guys.” Show that there is always an integer q (depending on n) such that, if we go around the circle executing every qth person, all the bad guys are first to go. (For example, when n = 3 we can take q = 5; when n = 4 we can take q = 30.) Bonus problems 22 Show that it’s possible to construct a Venn diagram for all 2n possible subsets of n given sets, using n convex polygons that are congruent to each other and rotated about a common center.
23 Suppose that Josephus finds himself in a given position j, but he has a chance to name the elimination parameter q such that every qth person is executed. Can he always save himself?
Research problems 24 Find all recurrence relations of the form Xn = 1 + a1Xn−1 + · · · + akXn−k b1Xn−1 + · · · + bkXn−k whose solution is periodic regardless of the initial values X0, . . . , Xk−1.
25 Solve infinitely many cases of the four-peg Tower of Hanoi problem by proving that equality holds in the relation of exercise 17.
26 Generalizing exercise 23, let’s say that a Josephus subset of {1, 2, . . . , n} is a set of k numbers such that, for some q, the people with the other n−k numbers will be eliminated first. (These are the k positions of the “good guys” Josephus wants to save.) It turns out that when n = 9, three of the 29 possible subsets are non-Josephus, namely {1, 2, 5, 8, 9}, {2, 3, 4, 5, 8}, and {2, 5, 6, 7, 8}. There are 13 non-Josephus sets when n = 12, none for any other values of n » 12. Are non-Josephus subsets rare for large n?
Yes, and well done if you find them.
2 Sums SUMS ARE EVERYWHERE in mathematics, so we need basic tools to handle them. This chapter develops the notation and general techniques that make summation user-friendly.
2.1 NOTATION In Chapter 1 we encountered the sum of the first n integers, which we wrote out as 1 + 2 + 3 + · · · + (n −1) + n. The ‘ · · · ’ in such formulas tells us to complete the pattern established by the surrounding terms. Of course we have to watch out for sums like 1 + 7 + · · · + 41.7, which are meaningless without a mitigating context. On the other hand, the inclusion of terms like 3 and (n −1) was a bit of overkill; the pattern would presumably have been clear if we had written simply 1 + 2 + · · · + n. Sometimes we might even be so bold as to write just 1 + · · · + n.
We’ll be working with sums of the general form a1 + a2 + · · · + an , (2.1) where each ak is a number that has been defined somehow. This notation has the advantage that we can “see” the whole sum, almost as if it were written out in full, if we have a good enough imagination.
Each element ak of a sum is called a term. The terms are often specified A term is how long this course lasts.
implicitly as formulas that follow a readily perceived pattern, and in such cases we must sometimes write them in an expanded form so that the meaning is clear. For example, if 1 + 2 + · · · + 2n−1 is supposed to denote a sum of n terms, not of 2n−1, we should write it more explicitly as 20 + 21 + · · · + 2n−1.
21 22 SUMS The three-dots notation has many uses, but it can be ambiguous and a “Le signe Pi=∞ i=1 in-dique que l’on doit donner au nombre entier i toutes ses valeurs 1, 2, 3, . . . , et prendre la somme des termes.” —J. Fourier bit long-winded. Other alternatives are available, notably the delimited form n X k=1 ak , (2.2) which is called Sigma-notation because it uses the Greek letter Σ (upper-case sigma).
This notation tells us to include in the sum precisely those terms ak whose index k is an integer that lies between the lower and upper limits 1 and n, inclusive. In words, we “sum over k, from 1 to n.” Joseph Fourier introduced this delimited P-notation in 1820, and it soon took the mathematical world by storm.
Incidentally, the quantity after P (here ak) is called the summand.
The index variable k is said to be bound to the P sign in (2.2), because the k in ak is unrelated to appearances of k outside the Sigma-notation. Any other letter could be substituted for k here without changing the meaning of Well, I wouldn’t want to use a or n as the index vari-able instead of k in (2.2); those letters are “free variables” that do have mean-ing outside the P here.
(2.2). The letter i is often used (perhaps because it stands for “index”), but we’ll generally sum on k since it’s wise to keep i for √ −1.
It turns out that a generalized Sigma-notation is even more useful than the delimited form: We simply write one or more conditions under the P, to specify the set of indices over which summation should take place. For example, the sums in (2.1) and (2.2) can also be written as X 1»k»n ak .
(2.3) In this particular example there isn’t much difference between the new form and (2.2), but the general form allows us to take sums over index sets that aren’t restricted to consecutive integers. For example, we can express the sum of the squares of all odd positive integers below 100 as follows: X 1»k<100 k odd k2 .
The delimited equivalent of this sum, 49 X k=0 (2k + 1)2 , is more cumbersome and less clear. Similarly, the sum of reciprocals of all prime numbers between 1 and N is X p»N p prime 1 p ; 2.1 NOTATION 23 the delimited form would require us to write π(N) X k=1 1 pk , where pk denotes the kth prime and π(N) is the number of primes » N.
(Incidentally, this sum gives the approximate average number of distinct prime factors of a random integer near N, since about 1/p of those inte-gers are divisible by p. Its value for large N is approximately ln ln N + M, where M ≈0.2614972128476427837554268386086958590515666 is Mertens’s constant ; ln x stands for the natural logarithm of x, and ln ln x stands for ln(ln x).) The biggest advantage of general Sigma-notation is that we can manip-ulate it more easily than the delimited form. For example, suppose we want The summation symbol looks like a distorted pacman.
to change the index variable k to k + 1. With the general form, we have X 1»k»n ak = X 1»k+1»n ak+1 ; it’s easy to see what’s going on, and we can do the substitution almost without thinking. But with the delimited form, we have n X k=1 ak = n−1 X k=0 ak+1 ; it’s harder to see what’s happened, and we’re more likely to make a mistake.
On the other hand, the delimited form isn’t completely useless.
It’s nice and tidy, and we can write it quickly because (2.2) has seven symbols A tidy sum.
compared with (2.3)’s eight. Therefore we’ll often use P with upper and lower delimiters when we state a problem or present a result, but we’ll prefer to work with relations-under-P when we’re manipulating a sum whose index variables need to be transformed.
The P sign occurs more than 1000 times in this book, so we should be That’s nothing.
You should see how many times Σ ap-pears in The Iliad.
sure that we know exactly what it means. Formally, we write X P(k) ak (2.4) as an abbreviation for the sum of all terms ak such that k is an integer satisfying a given property P(k). (A “property P(k)” is any statement about k that can be either true or false.) For the time being, we’ll assume that only finitely many integers k satisfying P(k) have ak ̸= 0; otherwise infinitely many nonzero numbers are being added together, and things can get a bit 24 SUMS tricky. At the other extreme, if P(k) is false for all integers k, we have an “empty” sum; the value of an empty sum is defined to be zero.
A slightly modified form of (2.4) is used when a sum appears within the text of a paragraph rather than in a displayed equation: We write ‘P P(k) ak’, attaching property P(k) as a subscript of P, so that the formula won’t stick out too much. Similarly, ‘Pn k=1 ak’ is a convenient alternative to (2.2) when we want to confine the notation to a single line.
People are often tempted to write n−1 X k=2 k(k −1)(n −k) instead of n X k=0 k(k −1)(n −k) because the terms for k = 0, 1, and n in this sum are zero. Somehow it seems more efficient to add up n −2 terms instead of n + 1 terms. But such temptations should be resisted; efficiency of computation is not the same as efficiency of understanding! We will find it advantageous to keep upper and lower bounds on an index of summation as simple as possible, because sums can be manipulated much more easily when the bounds are simple. Indeed, the form Pn−1 k=2 can even be dangerously ambiguous, because its meaning is not at all clear when n = 0 or n = 1 (see exercise 1). Zero-valued terms cause no harm, and they often save a lot of trouble.
So far the notations we’ve been discussing are quite standard, but now we are about to make a radical departure from tradition. Kenneth E. Iverson introduced a wonderful idea in his programming language APL [191, page 11; see also 220], and we’ll see that it greatly simplifies many of the things we want to do in this book. The idea is simply to enclose a true-or-false statement in brackets, and to say that the result is 1 if the statement is true, 0 if the Hey: The “Kro-necker delta” that I’ve seen in other books (I mean δkn , which is 1 if k = n, 0 oth-erwise) is just a special case of Iverson’s conven-tion: We can write [ k = n ] instead.
statement is false. For example, [p prime] = 1, if p is a prime number; 0, if p is not a prime number.
Iverson’s convention allows us to express sums with no constraints whatever on the index of summation, because we can rewrite (2.4) in the form X k ak P(k) .
(2.5) If P(k) is false, the term ak P(k) is zero, so we can safely include it among the terms being summed.
This makes it easy to manipulate the index of summation, because we don’t have to fuss with boundary conditions.
“I am often surprised by new, important applications [of this notation].” —B. de Finetti A slight technicality needs to be mentioned: Sometimes ak isn’t defined for all integers k. We get around this difficulty by assuming that P(k) is “very strongly zero” when P(k) is false; it’s so much zero, it makes ak P(k) 2.1 NOTATION 25 equal to zero even when ak is undefined. For example, if we use Iverson’s convention to write the sum of reciprocal primes » N as X p [p prime][p » N]/p , there’s no problem of division by zero when p = 0, because our convention tells us that [0 prime][0 » N]/0 = 0.
Let’s sum up what we’ve discussed so far about sums. There are two good ways to express a sum of terms: One way uses ‘ · · · ’, the other uses ‘ P ’. The three-dots form often suggests useful manipulations, particularly the combination of adjacent terms, since we might be able to spot a simplifying pattern if we let the whole sum hang out before our eyes. But too much detail can also be overwhelming. Sigma-notation is compact, impressive to family and friends, and often suggestive of manipulations that are not obvious in . . . and it’s less likely to lose points on an exam for “lack of rigor.” three-dots form.
When we work with Sigma-notation, zero terms are not generally harmful; in fact, zeros often make P-manipulation easier.
2.2 SUMS AND RECURRENCES OK, we understand now how to express sums with fancy notation.
But how does a person actually go about finding the value of a sum? One way is to observe that there’s an intimate relation between sums and recurrences.
The sum Sn = n X k=0 ak is equivalent to the recurrence (Think of Sn as not just a single number, but as a sequence defined for all n – 0.) S0 = a0 ; Sn = Sn−1 + an , for n > 0.
(2.6) Therefore we can evaluate sums in closed form by using the methods we learned in Chapter 1 to solve recurrences in closed form.
For example, if an is equal to a constant plus a multiple of n, the sum-recurrence (2.6) takes the following general form: R0 = α ; Rn = Rn−1 + β + γn , for n > 0.
(2.7) Proceeding as in Chapter 1, we find R1 = α + β + γ, R2 = α + 2β + 3γ, and so on; in general the solution can be written in the form Rn = A(n)α + B(n)β + C(n)γ , (2.8) 26 SUMS where A(n), B(n), and C(n) are the coefficients of dependence on the general parameters α, β, and γ.
The repertoire method tells us to try plugging in simple functions of n for Rn, hoping to find constant parameters α, β, and γ where the solution is especially simple. Setting Rn = 1 implies α = 1, β = 0, γ = 0; hence A(n) = 1 .
Setting Rn = n implies α = 0, β = 1, γ = 0; hence B(n) = n .
Setting Rn = n2 implies α = 0, β = −1, γ = 2; hence 2C(n) −B(n) = n2 and we have C(n) = (n2 + n)/2. Easy as pie.
Actually easier; π = P n– 0 8 (4n+1)(4n+3).
Therefore if we wish to evaluate n X k=0 (a + bk) , the sum-recurrence (2.6) boils down to (2.7) with α = β = a, γ = b, and the answer is aA(n) + aB(n) + bC(n) = a(n + 1) + b(n + 1)n/2.
Conversely, many recurrences can be reduced to sums; therefore the spe-cial methods for evaluating sums that we’ll be learning later in this chapter will help us solve recurrences that might otherwise be difficult. The Tower of Hanoi recurrence is a case in point: T0 = 0 ; Tn = 2Tn−1 + 1 , for n > 0.
It can be put into the special form (2.6) if we divide both sides by 2n: T0/20 = 0; Tn/2n = Tn−1/2n−1 + 1/2n , for n > 0.
Now we can set Sn = Tn/2n, and we have S0 = 0; Sn = Sn−1 + 2−n , for n > 0.
It follows that Sn = n X k=1 2−k.
2.2 SUMS AND RECURRENCES 27 (Notice that we’ve left the term for k = 0 out of this sum.) The sum of the geometric series 2−1+2−2+· · ·+2−n = ( 1 2)1+( 1 2)2+· · ·+( 1 2)n will be derived later in this chapter; it turns out to be 1 −( 1 2)n. Hence Tn = 2nSn = 2n −1.
We have converted Tn to Sn in this derivation by noticing that the re-currence could be divided by 2n. This trick is a special case of a general technique that can reduce virtually any recurrence of the form anTn = bnTn−1 + cn (2.9) to a sum. The idea is to multiply both sides by a summation factor, sn: snanTn = snbnTn−1 + sncn .
This factor sn is cleverly chosen to make snbn = sn−1an−1 .
Then if we write Sn = snanTn we have a sum-recurrence, Sn = Sn−1 + sncn .
Hence Sn = s0a0T0 + n X k=1 skck = s1b1T0 + n X k=1 skck , and the solution to the original recurrence (2.9) is Tn = 1 snan s1b1T0 + n X k=1 skck .
(2.10) For example, when n = 1 we get T1 = (s1b1T0+s1c1)/s1a1 = (b1T0+c1)/a1.
(The value of s1 cancels out, so it can be anything but zero.) But how can we be clever enough to find the right sn? No problem: The relation sn = sn−1an−1/bn can be unfolded to tell us that the fraction sn = an−1an−2 . . . a1 bnbn−1 . . . b2 , (2.11) or any convenient constant multiple of this value, will be a suitable summation factor. For example, the Tower of Hanoi recurrence has an = 1 and bn = 2; the general method we’ve just derived says that sn = 2−n is a good thing to multiply by, if we want to reduce the recurrence to a sum. We don’t need a brilliant flash of inspiration to discover this multiplier.
We must be careful, as always, not to divide by zero. The summation-factor method works whenever all the a’s and all the b’s are nonzero.
28 SUMS Let’s apply these ideas to a recurrence that arises in the study of “quick-sort,” one of the most important methods for sorting data inside a computer.
(Quicksort was invented by Hoare in 1962 .) The average number of comparison steps made by a typical implementation of quicksort when it is applied to n items in random order satisfies the recurrence C0 = C1 = 0 ; Cn = n + 1 + 2 n n−1 X k=0 Ck , for n > 1.
(2.12) Hmmm. This looks much scarier than the recurrences we’ve seen before; it includes a sum over all previous values, and a division by n. Trying small cases gives us some data (C2 = 3, C3 = 6, C4 = 19 2 ) but doesn’t do anything to quell our fears.
We can, however, reduce the complexity of (2.12) systematically, by first getting rid of the division and then getting rid of the P sign. The idea is to multiply both sides by n, obtaining the relation nCn = n2 + n + 2 n−1 X k=0 Ck , for n > 1; hence, if we replace n by n −1, (n −1)Cn−1 = (n −1)2 + (n −1) + 2 n−2 X k=0 Ck , for n −1 > 1.
We can now subtract this equation from the first, and the P sign disappears: nCn −(n −1)Cn−1 = 2n + 2Cn−1 , for n > 2.
Therefore the original recurrence for Cn reduces to a much simpler one: C0 = C1 = 0 ; C2 = 3 ; nCn = (n + 1)Cn−1 + 2n , for n > 2.
Progress. We’re now in a position to apply a summation factor, since this recurrence has the form of (2.9) with an = n, bn = n + 1, and cn = 2n −2[n = 1] + 2[n = 2] .
The general method described on the preceding page tells us to multiply the recurrence through by some multiple of sn = an−1an−2 . . . a1 bnbn−1 . . . b2 = (n −1) · (n −2) · . . . · 1 (n + 1) · n · . . . · 3 = 2 (n + 1)n .
2.2 SUMS AND RECURRENCES 29 The solution, according to (2.10), is therefore We started with a P in the recur-rence, and worked hard to get rid of it.
But then after ap-plying a summation factor, we came up with another P.
Are sums good, or bad, or what?
Cn = 2(n + 1) n X k=1 1 k + 1 − 2 3(n + 1) , for n > 1.
The sum that remains is very similar to a quantity that arises frequently in applications. It arises so often, in fact, that we give it a special name and a special notation: Hn = 1 + 1 2 + · · · + 1 n = n X k=1 1 k .
(2.13) The letter H stands for “harmonic”; Hn is a harmonic number, so called because the kth harmonic produced by a violin string is the fundamental tone produced by a string that is 1/k times as long.
We can complete our study of the quicksort recurrence (2.12) by putting Cn into closed form; this will be possible if we can express Cn in terms of Hn.
The sum in our formula for Cn is n X k=1 1 k + 1 = X 1»k»n 1 k + 1 .
We can relate this to Hn without much difficulty by changing k to k −1 and revising the boundary conditions: X 1»k»n 1 k + 1 = X 1»k−1»n 1 k = X 2»k»n+1 1 k = X 1»k»n 1 k −1 1 + 1 n + 1 = Hn − n n + 1 .
Alright! We have found the sum needed to complete the solution to (2.12): But your spelling is alwrong.
The average number of comparisons made by quicksort when it is applied to n randomly ordered items of data is Cn = 2(n + 1)Hn −8 3n −2 3 , for n > 1.
(2.14) As usual, we check that small cases are correct: C2 = 3, C3 = 6.
30 SUMS 2.3 MANIPULATION OF SUMS The key to success with sums is an ability to change one P into another that is simpler or closer to some goal. And it’s easy to do this by Not to be confused with finance.
learning a few basic rules of transformation and by practicing their use.
Let K be any finite set of integers. Sums over the elements of K can be transformed by using three simple rules: My other math books have different definitions for these laws.
X k∈K cak = c X k∈K ak ; (distributive law) (2.15) X k∈K (ak + bk) = X k∈K ak + X k∈K bk ; (associative law) (2.16) X k∈K ak = X p(k)∈K ap(k) .
(commutative law) (2.17) The distributive law allows us to move constants in and out of a P. The associative law allows us to break a P into two parts, or to combine two P’s into one. The commutative law says that we can reorder the terms in any way we please; here p(k) is any permutation of the set of all integers. For example, Why not call it permutative instead of commutative?
if K = {−1, 0, +1} and if p(k) = −k, these three laws tell us respectively that ca−1 + ca0 + ca1 = c(a−1 + a0 + a1) ; (distributive law) (a−1 + b−1) + (a0 + b0) + (a1 + b1) = (a−1 + a0 + a1) + (b−1 + b0 + b1) ; (associative law) a−1 + a0 + a1 = a1 + a0 + a−1 .
(commutative law) Gauss’s trick in Chapter 1 can be viewed as an application of these three basic laws. Suppose we want to compute the general sum of an arithmetic progression, S = X 0»k»n (a + bk) .
By the commutative law we can replace k by n −k, obtaining This is something like changing vari-ables inside an integral, but easier.
S = X 0»n−k»n a + b(n −k) = X 0»k»n (a + bn −bk) .
These two equations can be added by using the associative law: 2S = X 0»k»n (a + bk) + (a + bn −bk) = X 0»k»n (2a + bn) .
2.3 MANIPULATION OF SUMS 31 And we can now apply the distributive law and evaluate a trivial sum: “What’s one and one and one and one and one and one and one and one and one and one?” “I don’t know,” said Alice.
“I lost count.” “She can’t do Addition.” —Lewis Carroll 2S = (2a + bn) X 0»k»n 1 = (2a + bn)(n + 1) .
Dividing by 2, we have proved that n X k=0 (a + bk) = (a + 1 2bn)(n + 1) .
(2.18) The right-hand side can be remembered as the average of the first and last terms, namely 1 2 a + (a + bn) , times the number of terms, namely (n + 1).
It’s important to bear in mind that the function p(k) in the general commutative law (2.17) is supposed to be a permutation of all the integers. In other words, for every integer n there should be exactly one integer k such that p(k) = n. Otherwise the commutative law might fail; exercise 3 illustrates this with a vengeance. Transformations like p(k) = k + c or p(k) = c −k, where c is an integer constant, are always permutations, so they always work.
On the other hand, we can relax the permutation restriction a little bit: We need to require only that there be exactly one integer k with p(k) = n when n is an element of the index set K. If n / ∈K (that is, if n is not in K), it doesn’t matter how often p(k) = n occurs, because such k don’t take part in the sum. Thus, for example, we can argue that X k∈K k even ak = X n∈K n even an = X 2k∈K 2k even a2k = X 2k∈K a2k , (2.19) since there’s exactly one k such that 2k = n when n ∈K and n is even.
Iverson’s convention, which allows us to obtain the values 0 or 1 from logical statements in the middle of a formula, can be used together with the distributive, associative, and commutative laws to deduce additional proper-Additional, eh?
ties of sums. For example, here is an important rule for combining different sets of indices: If K and K′ are any sets of integers, then X k∈K ak + X k∈K ′ ak = X k∈K∩K ′ ak + X k∈K∪K ′ ak .
(2.20) This follows from the general formulas X k∈K ak = X k ak [k ∈K] (2.21) and [k ∈K] + [k ∈K′] = [k ∈K ∩K′] + [k ∈K ∪K′] .
(2.22) 32 SUMS Typically we use rule (2.20) either to combine two almost-disjoint index sets, as in m X k=1 ak + n X k=m ak = am + n X k=1 ak , for 1 » m » n; or to split off a single term from a sum, as in (The two sides of (2.20) have been switched here.) X 0»k»n ak = a0 + X 1»k»n ak , for n – 0.
(2.23) This operation of splitting off a term is the basis of a perturbation method that often allows us to evaluate a sum in closed form.
The idea is to start with an unknown sum and call it Sn: Sn = X 0»k»n ak .
(Name and conquer.) Then we rewrite Sn+1 in two ways, by splitting off both its last term and its first term: Sn + an+1 = X 0»k»n+1 ak = a0 + X 1»k»n+1 ak = a0 + X 1»k+1»n+1 ak+1 = a0 + X 0»k»n ak+1 .
(2.24) Now we can work on this last sum and try to express it in terms of Sn. If we succeed, we obtain an equation whose solution is the sum we seek.
For example, let’s use this approach to find the sum of a general geomet-If it’s geometric, there should be a geometric proof.
B B B BB B B B B B B B B B B B B B B B B B B B B B B B BB B B P P P P P P P P P P P P P P ric progression, Sn = X 0»k»n axk .
The general perturbation scheme in (2.24) tells us that Sn + axn+1 = ax0 + X 0»k»n axk+1 , and the sum on the right is x P 0»k»n axk = xSn by the distributive law.
Therefore Sn + axn+1 = a + xSn, and we can solve for Sn to obtain n X k=0 axk = a −axn+1 1 −x , for x ̸= 1.
(2.25) 2.3 MANIPULATION OF SUMS 33 (When x = 1, the sum is of course simply (n + 1)a.) The right-hand side can be remembered as the first term included in the sum minus the first term Ah yes, this formula was drilled into me in high school.
excluded (the term after the last), divided by 1 minus the term ratio.
That was almost too easy.
Let’s try the perturbation technique on a slightly more difficult sum, Sn = X 0»k»n k 2k .
In this case we have S0 = 0, S1 = 2, S2 = 10, S3 = 34, S4 = 98; what is the general formula? According to (2.24) we have Sn + (n + 1)2n+1 = X 0»k»n (k + 1)2k+1 ; so we want to express the right-hand sum in terms of Sn. Well, we can break it into two sums with the help of the associative law, X 0»k»n k 2k+1 + X 0»k»n 2k+1 , and the first of the remaining sums is 2Sn. The other sum is a geometric progression, which equals (2 −2n+2)/(1 −2) = 2n+2 −2 by (2.25). Therefore we have Sn + (n + 1)2n+1 = 2Sn + 2n+2 −2, and algebra yields X 0»k»n k 2k = (n −1)2n+1 + 2 .
Now we understand why S3 = 34: It’s 32 + 2, not 2·17.
A similar derivation with x in place of 2 would have given us the equation Sn + (n + 1)xn+1 = xSn + (x −xn+2)/(1 −x); hence we can deduce that n X k=0 kxk = x −(n + 1)xn+1 + nxn+2 (1 −x)2 , for x ̸= 1.
(2.26) It’s interesting to note that we could have derived this closed form in a completely different way, by using elementary techniques of differential cal-culus. If we start with the equation n X k=0 xk = 1 −xn+1 1 −x and take the derivative of both sides with respect to x, we get n X k=0 kxk−1 = (1−x) −(n+1)xn + 1−xn+1 (1 −x)2 = 1 −(n+1)xn + nxn+1 (1 −x)2 , 34 SUMS because the derivative of a sum is the sum of the derivatives of its terms. We will see many more connections between calculus and discrete mathematics in later chapters.
2.4 MULTIPLE SUMS The terms of a sum might be specified by two or more indices, not just by one. For example, here’s a double sum of nine terms, governed by two Oh no, a nine-term governor.
indices j and k: Notice that this doesn’t mean to sum over all j – 1 and all k » 3.
X 1»j,k»3 ajbk = a1b1 + a1b2 + a1b3 + a2b1 + a2b2 + a2b3 + a3b1 + a3b2 + a3b3 .
We use the same notations and methods for such sums as we do for sums with a single index. Thus, if P(j, k) is a property of j and k, the sum of all terms aj,k such that P(j, k) is true can be written in two ways, one of which uses Iverson’s convention and sums over all pairs of integers j and k: X P(j,k) aj,k = X j,k aj,k P(j, k) .
Only one P sign is needed, although there is more than one index of sum-mation; P denotes a sum over all combinations of indices that apply.
We also have occasion to use two P’s, when we’re talking about a sum of sums. For example, X j X k aj,k P(j, k) is an abbreviation for X j X k aj,k P(j, k) , which is the sum, over all integers j, of P k aj,k P(j, k) , the latter being the Multiple Σ’s are evaluated right to left (inside-out).
sum over all integers k of all terms aj,k for which P(j, k) is true. In such cases we say that the double sum is “summed first on k.” A sum that depends on more than one index can be summed first on any one of its indices.
In this regard we have a basic law called interchanging the order of summation, which generalizes the associative law (2.16) we saw earlier: X j X k aj,k P(j, k) = X P(j,k) aj,k = X k X j aj,k P(j, k) .
(2.27) 2.4 MULTIPLE SUMS 35 The middle term of this law is a sum over two indices. On the left, P j P k stands for summing first on k, then on j. On the right, P k P j stands for summing first on j, then on k. In practice when we want to evaluate a double sum in closed form, it’s usually easier to sum it first on one index rather than on the other; we get to choose whichever is more convenient.
Sums of sums are no reason to panic, but they can appear confusing to Who’s panicking?
I think this rule is fairly obvious compared to some of the stuff in Chapter 1.
a beginner, so let’s do some more examples. The nine-term sum we began with provides a good illustration of the manipulation of double sums, because that sum can actually be simplified, and the simplification process is typical of what we can do with P P’s: X 1»j,k»3 ajbk = X j,k ajbk[1 » j, k » 3] = X j,k ajbk[1 » j » 3][1 » k » 3] = X j X k ajbk[1 » j » 3][1 » k » 3] = X j aj[1 » j » 3] X k bk[1 » k » 3] = X j aj[1 » j » 3] X k bk[1 » k » 3] = X j aj[1 » j » 3] X k bk[1 » k » 3] = 3 X j=1 aj 3 X k=1 bk .
The first line here denotes a sum of nine terms in no particular order. The second line groups them in threes, (a1b1 + a1b2 + a1b3) + (a2b1 + a2b2 + a2b3) + (a3b1 + a3b2 + a3b3). The third line uses the distributive law to factor out the a’s, since aj and [1 » j » 3] do not depend on k; this gives a1(b1 + b2 + b3) + a2(b1 + b2 + b3) + a3(b1 + b2 + b3). The fourth line is the same as the third, but with a redundant pair of parentheses thrown in so that the fifth line won’t look so mysterious. The fifth line factors out the (b1 + b2 + b3) that occurs for each value of j: (a1 + a2 + a3)(b1 + b2 + b3).
The last line is just another way to write the previous line. This method of derivation can be used to prove a general distributive law, X j∈J k∈K ajbk = X j∈J aj X k∈K bk , (2.28) valid for all sets of indices J and K.
The basic law (2.27) for interchanging the order of summation has many variations, which arise when we want to restrict the ranges of the indices 36 SUMS instead of summing over all integers j and k. These variations come in two flavors, vanilla and rocky road. First, the vanilla version: X j∈J X k∈K aj,k = X j∈J k∈K aj,k = X k∈K X j∈J aj,k .
(2.29) This is just another way to write (2.27), since the Iversonian [j ∈J, k ∈K] factors into [j ∈J][k ∈K]. The vanilla-flavored law applies whenever the ranges of j and k are independent of each other.
The rocky-road formula for interchange is a little trickier. It applies when the range of an inner sum depends on the index variable of the outer sum: X j∈J X k∈K(j) aj,k = X k∈K ′ X j∈J ′(k) aj,k .
(2.30) Here the sets J, K(j), K′, and J′(k) must be related in such a way that [j ∈J] k ∈K(j) = [k ∈K′] j ∈J′(k) .
A factorization like this is always possible in principle: We can let J = K′ be the set of all integers and K(j), J′(k) be sets corresponding to the property P(j, k) that governs a double sum. But there are important special cases where the sets J, K(j), K′, and J′(k) have a simple form. Such cases arise frequently in applications. For example, here’s a particularly useful factorization: [1 » j » n][j » k » n] = [1 » j » k » n] = [1 » k » n][1 » j » k] .
(2.31) This Iversonian equation allows us to write n X j=1 n X k=j aj,k = X 1»j»k»n aj,k = n X k=1 k X j=1 aj,k .
(2.32) One of these two sums of sums is usually easier to evaluate than the other; (Now is a good time to do warmup exercises 4 and 6.) (Or to check out the Snickers bar languishing in the freezer.) we can use (2.32) to switch from the hard one to the easy one.
Let’s apply these ideas to a useful example. Consider the array a1a1 a1a2 a1a3 . . .
a1an a2a1 a2a2 a2a3 . . .
a2an a3a1 a3a2 a3a3 . . .
a3an .
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.
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.
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...
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ana1 ana2 ana3 . . .
anan of n2 products ajak. Our goal will be to find a simple formula for S@ = X 1»j»k»n ajak , 2.4 MULTIPLE SUMS 37 the sum of all elements on or above the main diagonal of this array. Because ajak = akaj, the array is symmetrical about its main diagonal; therefore S@ will be approximately half the sum of all the elements (except for a fudge factor that takes account of the main diagonal).
Does rocky road have fudge in it?
Such considerations motivate the following manipulations. We have S@ = X 1»j»k»n ajak = X 1»k»j»n akaj = X 1»k»j»n ajak = S@ , because we can rename (j, k) as (k, j). Furthermore, since [1 » j » k » n] + [1 » k » j » n] = [1 » j, k » n] + [1 » j = k » n] , we have 2S@ = S@ + S@ = X 1»j,k»n ajak + X 1»j=k»n ajak .
The first sum is Pn j=1 aj Pn k=1 ak = Pn k=1 ak 2, by the general distribu-tive law (2.28). The second sum is Pn k=1 a2 k. Therefore we have S@ = X 1»j»k»n ajak = 1 2 n X k=1 ak 2 + n X k=1 a2 k !
, (2.33) an expression for the upper triangular sum in terms of simpler single sums.
Encouraged by such success, let’s look at another double sum: S = X 1»j<k»n (ak −aj)(bk −bj) .
Again we have symmetry when j and k are interchanged: S = X 1»k<j»n (aj −ak)(bj −bk) = X 1»k<j»n (ak −aj)(bk −bj) .
So we can add S to itself, making use of the identity [1 » j < k » n] + [1 » k < j » n] = [1 » j, k » n] −[1 » j = k » n] to conclude that 2S = X 1»j,k»n (aj −ak)(bj −bk) − X 1»j=k»n (aj −ak)(bj −bk) .
38 SUMS The second sum here is zero; what about the first?
It expands into four separate sums, each of which is vanilla flavored: X 1»j,k»n ajbj − X 1»j,k»n ajbk − X 1»j,k»n akbj + X 1»j,k»n akbk = 2 X 1»j,k»n akbk −2 X 1»j,k»n ajbk = 2n X 1»k»n akbk −2 n X k=1 ak n X k=1 bk .
In the last step both sums have been simplified according to the general distributive law (2.28). If the manipulation of the first sum seems mysterious, here it is again in slow motion: 2 X 1»j,k»n akbk = 2 X 1»k»n X 1»j»n akbk = 2 X 1»k»n akbk X 1»j»n 1 = 2 X 1»k»n akbkn = 2n X 1»k»n akbk .
An index variable that doesn’t appear in the summand (here j) can simply be eliminated if we multiply what’s left by the size of that variable’s index set (here n).
Returning to where we left off, we can now divide everything by 2 and rearrange things to obtain an interesting formula: n X k=1 ak n X k=1 bk = n n X k=1 akbk − X 1»j<k»n (ak −aj)(bk −bj) . (2.34) This identity yields Chebyshev’s monotonic inequalities as a special case: (Chebyshev actually proved the analogous result for integrals instead of sums, (Rb a f(x) dx) · (Rb a g(x) dx) » (b −a) · (Rb af(x)g(x) dx), if f(x) and g(x) are monotone nondecreasing functions.) n X k=1 ak n X k=1 bk » n n X k=1 akbk , if a1 » · · · » an and b1 » · · · » bn; n X k=1 ak n X k=1 bk – n n X k=1 akbk , if a1 » · · · » an and b1 – · · · – bn.
(In general, if a1 » · · · » an and if p is a permutation of {1, . . . , n}, it’s not difficult to prove that the largest value of Pn k=1 akbp(k) occurs when bp(1) » · · · » bp(n), and the smallest value occurs when bp(1) – · · · – bp(n).) 2.4 MULTIPLE SUMS 39 Multiple summation has an interesting connection with the general op-eration of changing the index of summation in single sums. We know by the commutative law that X k∈K ak = X p(k)∈K ap(k) , if p(k) is any permutation of the integers. But what happens when we replace k by f(j), where f is an arbitrary function f : J →K that takes an integer j ∈J into an integer f(j) ∈K? The general formula for index replacement is X j∈J af(j) = X k∈K ak #f−(k) , (2.35) where #f−(k) stands for the number of elements in the set f−(k) = j f(j) = k , that is, the number of values of j ∈J such that f(j) equals k.
It’s easy to prove (2.35) by interchanging the order of summation, X j∈J af(j) = X j∈J k∈K ak f(j) = k = X k∈K ak X j∈J f(j) = k , since P j∈J f(j) = k = #f−(k).
In the special case that f is a one-to-one correspondence between J and K, we have #f−(k) = 1 for all k, and the My other math teacher calls this a “bijection”; maybe I’ll learn to love that word some day.
And then again. . .
general formula (2.35) reduces to X j∈J af(j) = X f(j)∈K af(j) = X k∈K ak .
This is the commutative law (2.17) we had before, slightly disguised.
Our examples of multiple sums so far have all involved general terms like ak or bk. But this book is supposed to be concrete, so let’s take a look at a multiple sum that involves actual numbers: Sn = X 1»j<k»n 1 k −j .
For example, S1 = 0; S2 = 1; S3 = 1 2−1 + 1 3−1 + 1 3−2 = 5 2.
Watch out— the authors seem to think that j, k, and n are “actual numbers.” 40 SUMS The normal way to evaluate a double sum is to sum first on j or first on k, so let’s explore both options.
Sn = X 1»k»n X 1»j<k 1 k −j summing first on j = X 1»k»n X 1»k−j<k 1 j replacing j by k −j = X 1»k»n X 0<j»k−1 1 j simplifying the bounds on j = X 1»k»n Hk−1 by (2.13), the definition of Hk−1 = X 1»k+1»n Hk replacing k by k + 1 = X 0»k<n Hk .
simplifying the bounds on k Alas! We don’t know how to get a sum of harmonic numbers into closed form.
Get out the whip.
If we try summing first the other way, we get Sn = X 1»j»n X j<k»n 1 k −j summing first on k = X 1»j»n X j<k+j»n 1 k replacing k by k + j = X 1»j»n X 0<k»n−j 1 k simplifying the bounds on k = X 1»j»n Hn−j by (2.13), the definition of Hn−j = X 1»n−j»n Hj replacing j by n −j = X 0»j<n Hj .
simplifying the bounds on j We’re back at the same impasse.
But there’s another way to proceed, if we replace k by k + j before deciding to reduce Sn to a sum of sums: Sn = X 1»j<k»n 1 k −j recopying the given sum = X 1»j 0, and assume that (2.39) holds when n is replaced by n −1. Since n = n−1 + n2, we have 3 n = (n −1)(n −1 2)(n) + 3n2 = (n3 −3 2n2 + 1 2n) + 3n2 = (n3 + 3 2n2 + 1 2n) = n(n + 1 2)(n + 1) .
Therefore (2.39) indeed holds, beyond a reasonable doubt, for all n – 0.” Judge Wapner, in his infinite wisdom, agrees.
Induction has its place, and it is somewhat more defensible than trying to look up the answer. But it’s still not really what we’re seeking. All of the other sums we have evaluated so far in this chapter have been conquered without induction; we should likewise be able to determine a sum like n from scratch. Flashes of inspiration should not be necessary. We should be able to do sums even on our less creative days.
Method 2: Perturb the sum.
So let’s go back to the perturbation method that worked so well for the geometric progression (2.25). We extract the first and last terms of n+1 in 44 SUMS order to get an equation for n: n + (n + 1)2 = X 0»k»n (k + 1)2 = X 0»k»n (k2 + 2k + 1) = X 0»k»n k2 + 2 X 0»k»n k + X 0»k»n 1 = n + 2 X 0»k»n k + (n + 1) .
Oops — the n’s cancel each other. Occasionally, despite our best efforts, the perturbation method produces something like n = n, so we lose.
Seems more like a draw.
On the other hand, this derivation is not a total loss; it does reveal a way to sum the first few nonnegative integers in closed form, 2 X 0»k»n k = (n + 1)2 −(n + 1) , even though we’d hoped to discover the sum of first integers squared. Could it be that if we start with the sum of the integers cubed, which we might call n, we will get an expression for the integers squared? Let’s try it. n + (n + 1)3 = X 0»k»n (k + 1)3 = X 0»k»n (k3 + 3k2 + 3k + 1) = n + 3 n + 3(n+1)n 2 + (n+1) .
Sure enough, the n’s cancel, and we have enough information to determine Method 2′: Perturb your TA.
n without relying on induction: 3 n = (n + 1)3 −3(n + 1)n/2 −(n + 1) = (n + 1)(n2 + 2n + 1 −3 2n −1) = (n + 1)(n + 1 2)n .
Method 3: Build a repertoire.
A slight generalization of the recurrence (2.7) will also suffice for sum-mands involving n2. The solution to R0 = α ; Rn = Rn−1 + β + γn + δn2 , for n > 0, (2.40) will be of the general form Rn = A(n)α + B(n)β + C(n)γ + D(n)δ ; (2.41) and we have already determined A(n), B(n), and C(n), because (2.40) is the same as (2.7) when δ = 0. If we now plug in Rn = n3, we find that n3 is the 2.5 GENERAL METHODS 45 solution when α = 0, β = 1, γ = −3, δ = 3. Hence 3D(n) −3C(n) + B(n) = n3 ; this determines D(n).
We’re interested in the sum n, which equals n−1 + n2; thus we get n = Rn if we set α = β = γ = 0 and δ = 1 in (2.40) and (2.41). Consequently n = D(n). We needn’t do the algebra to compute D(n) from B(n) and C(n), since we already know what the answer will be; but doubters among us should be reassured to find that 3D(n) = n3 + 3C(n) −B(n) = n3 + 3(n+1)n 2 −n = n(n+ 1 2)(n+1) .
Method 4: Replace sums by integrals.
People who have been raised on calculus instead of discrete mathematics tend to be more familiar with R than with P, so they find it natural to try changing P to R . One of our goals in this book is to become so comfortable with P that we’ll think R is more difficult than P (at least for exact results).
But still, it’s a good idea to explore the relation between P and R , since summation and integration are based on very similar ideas.
In calculus, an integral can be regarded as the area under a curve, and we can approximate this area by adding up the areas of long, skinny rectangles that touch the curve. We can also go the other way if a collection of long, skinny rectangles is given: Since n is the sum of the areas of rectangles whose sizes are 1 × 1, 1 × 4, . . . , 1 × n2, it is approximately equal to the area under the curve f(x) = x2 between 0 and n.
The horizontal scale here is ten times the vertical scale.
6 -f(x) x 1 2 3 . . .
n f(x) = x2 qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q q The area under this curve is Rn 0 x2 dx = n3/3; therefore we know that n is approximately 1 3n3.
46 SUMS One way to use this fact is to examine the error in the approximation, En = n −1 3n3. Since n satisfies the recurrence n = n−1 + n2, we find that En satisfies the simpler recurrence En = n −1 3n3 = n−1 + n2 −1 3n3 = En−1 + 1 3(n−1)3 + n2 −1 3n3 = En−1 + n −1 3 .
Another way to pursue the integral approach is to find a formula for En by summing the areas of the wedge-shaped error terms. We have This is for people addicted to calculus.
n − Z n 0 x2 dx = n X k=1 k2 − Z k k−1 x2 dx = n X k=1 k2 −k3 −(k −1)3 3 = n X k=1 k −1 3 .
Either way, we could find En and then n.
Method 5: Expand and contract.
Yet another way to discover a closed form for n is to replace the orig-inal sum by a seemingly more complicated double sum that can actually be simplified if we massage it properly: n = X 1»k»n k2 = X 1»j»k»n k = X 1»j»n X j»k»n k = X 1»j»n j + n 2 (n −j + 1) = 1 2 X 1»j»n n(n + 1) + j −j2 = 1 2n2(n + 1) + 1 4n(n + 1) −1 2 n = 1 2n(n + 1 2)(n + 1) −1 2 n .
Going from a single sum to a double sum may appear at first to be a backward (The last step here is something like the last step of the perturbation method, because we get an equation with the unknown quantity on both sides.) step, but it’s actually progress, because it produces sums that are easier to work with. We can’t expect to solve every problem by continually simplifying, simplifying, and simplifying: You can’t scale the highest mountain peaks by climbing only uphill.
Method 6: Use finite calculus.
Method 7: Use generating functions.
Stay tuned for still more exciting calculations of n = Pn k=0 k2, as we learn further techniques in the next section and in later chapters.
2.6 FINITE AND INFINITE CALCULUS 47 2.6 FINITE AND INFINITE CALCULUS We’ve learned a variety of ways to deal with sums directly. Now it’s time to acquire a broader perspective, by looking at the problem of summa-tion from a higher level. Mathematicians have developed a “finite calculus,” analogous to the more traditional infinite calculus, by which it’s possible to approach summation in a nice, systematic fashion.
Infinite calculus is based on the properties of the derivative operator D, defined by Df(x) = lim h→0 f(x + h) −f(x) h .
Finite calculus is based on the properties of the difference operator ∆, defined by ∆f(x) = f(x + 1) −f(x) .
(2.42) This is the finite analog of the derivative in which we restrict ourselves to positive integer values of h. Thus, h = 1 is the closest we can get to the “limit” as h →0, and ∆f(x) is the value of f(x + h) −f(x) /h when h = 1.
The symbols D and ∆are called operators because they operate on functions to give new functions; they are functions of functions that produce functions. If f is a suitably smooth function of real numbers to real numbers, then Df is also a function from reals to reals. And if f is any real-to-real As opposed to a cassette function.
function, so is ∆f. The values of the functions Df and ∆f at a point x are given by the definitions above.
Early on in calculus we learn how D operates on the powers f(x) = xm.
In such cases Df(x) = mxm−1. We can write this informally with f omitted, D(xm) = mxm−1 .
It would be nice if the ∆operator would produce an equally elegant result; unfortunately it doesn’t. We have, for example, ∆(x3) = (x + 1)3 −x3 = 3x2 + 3x + 1 .
But there is a type of “mth power” that does transform nicely under ∆, Math power.
and this is what makes finite calculus interesting.
Such newfangled mth powers are defined by the rule xm = m factors z }| { x(x −1) . . . (x −m + 1) , integer m – 0.
(2.43) Notice the little straight line under the m; this implies that the m factors are supposed to go down and down, stepwise. There’s also a corresponding 48 SUMS definition where the factors go up and up: xm = m factors z }| { x(x + 1) . . . (x + m −1) , integer m – 0.
(2.44) When m = 0, we have x0 = x0 = 1, because a product of no factors is conventionally taken to be 1 (just as a sum of no terms is conventionally 0).
The quantity xm is called “x to the m falling,” if we have to read it aloud; similarly, xm is “x to the m rising.” These functions are also called falling factorial powers and rising factorial powers, since they are closely related to the factorial function n! = n(n −1) . . . (1). In fact, n! = nn = 1n.
Several other notations for factorial powers appear in the mathematical literature, notably “Pochhammer’s symbol” (x)m for xm or xm; notations Mathematical terminology is sometimes crazy: Pochhammer actually used the notation (x)m for the binomial coefficient x m , not for factorial powers.
like x(m) or x(m) are also seen for xm. But the underline/overline convention is catching on, because it’s easy to write, easy to remember, and free of redundant parentheses.
Falling powers xm are especially nice with respect to ∆. We have ∆(xm) = (x + 1)m −xm = (x + 1)x . . . (x −m + 2) −x . . . (x −m + 2)(x −m + 1) = m x(x −1) . . . (x −m + 2) , hence the finite calculus has a handy law to match D(xm) = mxm−1: ∆(xm) = mxm−1 .
(2.45) This is the basic factorial fact.
The operator D of infinite calculus has an inverse, the anti-derivative (or integration) operator R . The Fundamental Theorem of Calculus relates D to R : g(x) = Df(x) if and only if Z g(x) dx = f(x) + C .
Here R g(x) dx, the indefinite integral of g(x), is the class of functions whose derivative is g(x). Analogously, ∆has as an inverse, the anti-difference (or “Quemadmodum ad differentiam denotandam usi sumus signo ∆, ita summam indi-cabimus signo Σ.
. . . ex quo æquatio z = ∆y, si inver-tatur, dabit quoque y = Σz + C.” —L. Euler summation) operator P; and there’s another Fundamental Theorem: g(x) = ∆f(x) if and only if X g(x) δx = f(x) + C .
(2.46) Here P g(x) δx, the indefinite sum of g(x), is the class of functions whose difference is g(x). (Notice that the lowercase δ relates to uppercase ∆as d relates to D.) The “C” for indefinite integrals is an arbitrary constant; the “C” for indefinite sums is any function p(x) such that p(x + 1) = p(x). For 2.6 FINITE AND INFINITE CALCULUS 49 example, C might be the periodic function a + b sin 2πx; such functions get washed out when we take differences, just as constants get washed out when we take derivatives. At integer values of x, the function C is constant.
Now we’re almost ready for the punch line. Infinite calculus also has definite integrals: If g(x) = Df(x), then Z b a g(x) dx = f(x) b a = f(b) −f(a) .
Therefore finite calculus — ever mimicking its more famous cousin — has def-inite sums: If g(x) = ∆f(x), then Xb a g(x) δx = f(x) b a = f(b) −f(a) .
(2.47) This formula gives a meaning to the notation Pb a g(x) δx, just as the previous formula defines Rb a g(x) dx.
But what does Pb a g(x) δx really mean, intuitively? We’ve defined it by analogy, not by necessity. We want the analogy to hold, so that we can easily remember the rules of finite calculus; but the notation will be useless if we don’t understand its significance. Let’s try to deduce its meaning by looking first at some special cases, assuming that g(x) = ∆f(x) = f(x + 1) −f(x). If b = a, we have Xa a g(x) δx = f(a) −f(a) = 0 .
Next, if b = a + 1, the result is Xa+1 a g(x) δx = f(a + 1) −f(a) = g(a) .
More generally, if b increases by 1, we have Xb+1 a g(x) δx − Xb a g(x) δx = f(b + 1) −f(a) − f(b) −f(a) = f(b + 1) −f(b) = g(b) .
These observations, and mathematical induction, allow us to deduce exactly what Pb a g(x) δx means in general, when a and b are integers with b – a: Xb a g(x) δx = b−1 X k=a g(k) = X a»k<b g(k) , for integers b – a.
(2.48) In other words, the definite sum is the same as an ordinary sum with limits, You call this a punch line?
but excluding the value at the upper limit.
50 SUMS Let’s try to recap this in a slightly different way. Suppose we’ve been given an unknown sum that’s supposed to be evaluated in closed form, and suppose we can write it in the form P a»k<b g(k) = Pb a g(x) δx. The theory of finite calculus tells us that we can express the answer as f(b) −f(a), if we can only find an indefinite sum or anti-difference function f such that g(x) = f(x + 1) −f(x).
One way to understand this principle is to write P a»k<b g(k) out in full, using the three-dots notation: X a»k<b f(k + 1) −f(k) = f(a+1) −f(a) + f(a+2) −f(a+1) + · · · + f(b−1) −f(b−2) + f(b) −f(b−1) .
Everything on the right-hand side cancels, except f(b) −f(a); so f(b) −f(a) is the value of the sum. (Sums of the form P a»k<b f(k + 1) −f(k) are often called telescoping, by analogy with a collapsed telescope, because the thickness of a collapsed telescope is determined solely by the outer radius of And all this time I thought it was telescoping because it collapsed from a very long expression to a very short one.
the outermost tube and the inner radius of the innermost tube.) But rule (2.48) applies only when b – a; what happens if b < a? Well, (2.47) says that we must have Xb a g(x) δx = f(b) −f(a) = − f(a) −f(b) = − Xa b g(x) δx .
This is analogous to the corresponding equation for definite integration. A similar argument proves Pb a + Pc b = Pc a, the summation analog of the iden-tity Rb a + Rc b = Rc a. In full garb, Xb a g(x) δx + Xc b g(x) δx = Xc a g(x) δx , (2.49) for all integers a, b, and c.
At this point a few of us are probably starting to wonder what all these parallels and analogies buy us. Well for one, definite summation gives us a Others have been wondering this for some time now.
simple way to compute sums of falling powers: The basic laws (2.45), (2.47), and (2.48) imply the general law X 0»k<n km = km+1 m + 1 n 0 = nm+1 m + 1 , for integers m, n – 0.
(2.50) This formula is easy to remember because it’s so much like the familiar Rn 0 xm dx = nm+1/(m + 1).
2.6 FINITE AND INFINITE CALCULUS 51 In particular, when m = 1 we have k1 = k, so the principles of finite calculus give us an easy way to remember the fact that X 0»k<n k = n2 2 = n(n −1)/2 .
The definite-sum method also gives us an inkling that sums over the range 0 » k < n often turn out to be simpler than sums over 1 » k » n; the former are just f(n) −f(0), while the latter must be evaluated as f(n + 1) −f(1).
Ordinary powers can also be summed in this new way, if we first express them in terms of falling powers. For example, k2 = k2 + k1 , hence X 0»k<n k2 = n3 3 + n2 2 = 1 3n(n −1)(n −2 + 3 2) = 1 3n(n −1 2)(n −1) .
Replacing n by n + 1 gives us yet another way to compute the value of our old friend n = P 0»k»n k2 in closed form.
With friends like this. . .
Gee, that was pretty easy. In fact, it was easier than any of the umpteen other ways that beat this formula to death in the previous section. So let’s try to go up a notch, from squares to cubes: A simple calculation shows that k3 = k3 + 3k2 + k1 .
(It’s always possible to convert between ordinary powers and factorial powers by using Stirling numbers, which we will study in Chapter 6.) Thus X a»k<b k3 = k4 4 + k3 + k2 2 b a .
Falling powers are therefore very nice for sums. But do they have any other redeeming features? Must we convert our old friendly ordinary powers to falling powers before summing, but then convert back before we can do anything else? Well, no, it’s often possible to work directly with factorial powers, because they have additional properties.
For example, just as we have (x + y)2 = x2 + 2xy + y2, it turns out that (x + y)2 = x2 + 2x1y1 + y2, and the same analogy holds between (x + y)m and (x + y)m. (This “factorial binomial theorem” is proved in exercise 5.37.) So far we’ve considered only falling powers that have nonnegative expo-nents. To extend the analogies with ordinary powers to negative exponents, 52 SUMS we need an appropriate definition of xm for m < 0. Looking at the sequence x3 = x(x −1)(x −2) , x2 = x(x −1) , x1 = x , x0 = 1 , we notice that to get from x3 to x2 to x1 to x0 we divide by x −2, then by x −1, then by x. It seems reasonable (if not imperative) that we should divide by x + 1 next, to get from x0 to x−1, thereby making x−1 = 1/(x + 1).
Continuing, the first few negative-exponent falling powers are x−1 = 1 x + 1 , x−2 = 1 (x + 1)(x + 2) , x−3 = 1 (x + 1)(x + 2)(x + 3) , and our general definition for negative falling powers is x−m = 1 (x + 1)(x + 2) . . . (x + m) , for m > 0.
(2.51) (It’s also possible to define falling powers for real or even complex m, but we How can a complex number be even?
will defer that until Chapter 5.) With this definition, falling powers have additional nice properties. Per-haps the most important is a general law of exponents, analogous to the law xm+n = xmxn for ordinary powers. The falling-power version is xm+n = xm (x −m)n , integers m and n.
(2.52) For example, x2+3 = x2 (x −2)3; and with a negative n we have x2−3 = x2 (x −2)−3 = x(x −1) 1 (x −1)x(x + 1) = 1 x + 1 = x−1 .
If we had chosen to define x−1 as 1/x instead of as 1/(x + 1), the law of exponents (2.52) would have failed in cases like m = −1 and n = 1. In fact, we could have used (2.52) to tell us exactly how falling powers ought to be defined in the case of negative exponents, by setting m = −n.
When an Laws have their exponents and their detractors.
existing notation is being extended to cover more cases, it’s always best to formulate definitions in such a way that general laws continue to hold. |
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Solving Systems of Equations by Substitution Worksheet
Created by
Robin Cornecki - Round Robin Math
Are you teaching your students how to Solve Systems of Equations by Substitution? Do you need a quiz or a homework assignment on solving systems of equations by substitution? Look no further than this practice worksheet or quiz, with 8 questions, on all things solving systems by substitution. ✅ I used this as a quiz with my 8th graders after I did this activity: Solving Systems of Equations by Substitution Activity. I have a blog post all about how I used the above activity in my classro
8 th - 10 th
Algebra, Algebra 2, Math
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8.EE.C.8a
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Solving Systems of Equations By Substitution Worksheet and Class Follow-along
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David Howard
This is a great worksheet I use during this unit by having my students follow along with me on the first page, and then have them practice by themselves on the other page. They all work out as integer answers.
8 th
Algebra, Algebra 2, Math
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8.EE.C.8b
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Algebra I - Solving Systems of Equations with Substitution Worksheet Free
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This practice worksheet has Algebra I or II students solving systems of equations using substitution. There are 10 practice problems and 1 error analysis problem where students must find and describe an error in solving using substitution. An answer bank is included on the back page. This document includes the blank student copy (pages 1 and 2) and the answer key (pages 3 and 4).
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Systems of Equations (Solve by Substitution) Student Worksheet
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10 Versions of Solving Systems of Equations by Substitutions. Get the complete unit which includes worksheets, notes, quizzes, unit study guide, and unit exam for $15.99.
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Systems of Linear Equations Worksheet, Substitution Method
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Thoughtful Academics
Ten systems of linear equation problems. Good introductory activity: No isolation necessary to perform substitution, and all solutions are integers. Answers included on second page!
8 th - 11 th
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CCSS
HSA-CED.A.3
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Pairs Check Activity - Solving Systems of Equations (Substitution Method)
Created by
The Blakenator
In this partner activity, students practice using the substitution method to solve four challenging systems of equations. Partner 1 will do the problem in the left column while Partner 2 coaches. Then, Partner 2 will do the problem in the right column while partner 1 coaches. Students repeat the process until they reach a teacher check stop point. Easy to read answer key included! Connect with The Blakenator Be sure to follow my TpT store by clicking on the green ‘Follow Me’ next to my Seller
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Solving Systems of Equations by Substitution Activity: Problem Experts
Created by
SR Math
Problem Expert Activities are a wonderful way to make sure each student in your class receives individual attention and feels empowered. It's also a great way to have shy students work with others in a way that is not too overwhelming. To begin, you give each student a problem necklace (or if your students are too cool to wear these necklaces, they can just have the card) and the worksheet to record their answers. Each necklace has a problem on the front and the answer to that problem on the
7 th - 10 th
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Also included in:Solving Systems of Equations Bundle
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Solving Systems of Equations Matching Worksheet
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Shaffer's Secondary Math Resources
This document is a worksheet for solving systems of equations. Students may choose their method (substitution, elimination, or graphing) and are given an answer bank to match the system of equations to the correct answer. Blank graphs are provided if students wish to graph. This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
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Solving Systems of Equations by Substitution Color-by-Number
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Students are able to self-check with a color-by-number! Answer key included. I'm new to selling on TpT so I'd really appreciate the feedback! :)
7 th - 9 th
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Solving Systems of Equations by Substitution Review MAZE
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Students will love practicing Solving Systems by Substitution with this MAZE Activity.In this exciting activity, you will navigate through a maze of equations, using your algebraic skills to solve for the values of two variables. But beware, the maze is tricky, and you will need to be quick on your feet to avoid the pitfalls of incorrect answers! Each equation in the maze will have one variable already isolated and equal to an integer a term or an expression. This means that you will need to
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Systems of Equations by Substitution
Created by
Kenneth deLoera
This is a word document that I created so it can be edited to fit your class. It is a good worksheet that I use with pull out groups to teach kids how to solve system of equation by substitution. It can also be good for supplemental work, it includes some word problems.
9 th - 10 th
Algebra
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Rated 4.81 out of 5, based on 26 reviews
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Solving Systems of Linear Equations Math Bundle worksheets
Created by
nicole berry
Everyone needs to practice! Included are 3 worksheets covering solving systems of equations (one by graphing, one by substitution and one by elimination). Also included are the answer keys to check the answers. ALL worksheets have different equations, so no repeats!
9 th - 12 th, Adult Education, Higher Education
Algebra, Algebra 2, Math
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Solve Systems of Equations by Elimination and Substitution | Systems Activity
Created by
Shawn Henry
Your students will find the punchline to the groan worthy dad joke by practicing solving systems of equations by elimination and substitution. They will solve each question and fill in the blanks at the bottom of the page. Skills reviewed in these 9 problems: Solving systems of equations using EliminationSolving systems of equations using SubstitutionSystems of equations with no solution and infinitely many solutionsAll files come as ready to use PDFs. Questions too hard or too easy? Want to e
7 th - 10 th
Algebra, Math
CCSS
8.EE.C.8
, 8.EE.C.8a
, 8.EE.C.8b
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Also included in:8th Grade Math | Algebra and Pre-Algebra | Lesson, Activity, Assessment Bundle
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Solving Systems of Equations by Elimination Worksheet
Created by
Evan Ohman
Simple Worksheet with work space for students to practice solving systems of equations using elimination or substitution.
8 th - 9 th
Math
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Solving Systems of Equations by Substitution (Barcelona Team PowerPoint Game)
Created by
Math with Beskat
This PowerPoint game is very good resource to practice solving systems of equations by substitution. There are Barcelona team soccer players and coach which represent the problems. Students choose a picture and try to answer it on their recording worksheet. Then, look for the answer on the PowerPoint. Students need to click on “Answer” to see answer of each problem. When you return to the first slide, the problems that have been completed will disappear.
7 th - 12 th
Algebra, Math
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Systems of Equations Graphing, Elimination, Substitution | Murder Mystery Review
Created by
Shawn Henry
In this whodunit murder mystery review activity your students will answer questions about systems of equations involving solving using graphing, elimination, and substitution to help the police find who murdered their math teacher. Students will take the role as a detective to bring down one of their own classmates. In this 9 question review, students will practice the following skills:solve systems by graphing(including parallel and overlapping lines)solve systems of equations by substitutions
7 th - 10 th
Algebra, Math
CCSS
8.EE.C.8
, 8.EE.C.8a
, 8.EE.C.8b
Also included in:8th Grade Math | Algebra and Pre-Algebra | Lesson, Activity, Assessment Bundle
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Solving Systems of Equations by Substitution for 1 Variable Quick Check
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This digital choice board is a great way to give students choice and assess their knowledge on solving a system of linear equations by using substitution. This digital activity is a great warm up or exit ticket for students who are just being introduced to systems or a good scaffold for learners who need a little more support. Students are given the choice of four problems but must complete two in order to complete the assignment. This activity is made up of 4 problems where students must solv
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Systems of Linear-Linear Equations by Substitution
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Students solve Systems of Linear-Linear Equations by substitution. At the end of 10 questions, students will decode a secret message that leads them to a secret mission (reviewing Quadratic Equations before moving on to Quadratic-Linear Systems of Equations the next class). I'll typically give students 30 minutes to complete the worksheet and give two points of extra credit to anyone who finishes the secret mission before time is up. There is also a 1/2 page of guided practice in case students n
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9043 - SYSTEMS of LINEAR EQUATIONS - Substitution
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A lesson on SYSTEMS – Substitution, brought to you by Tom Foolery! This lesson was created for students preparing for the 9th grade NYS Common Core Regents Exam in Algebra. This lesson instructs and assesses students on multiple Common Core State Standards. OBJECTIVE: All students will be able to solve a system of linear equations algebraically using the substitution method. LESSON ELEMENTS: DO NOW: spiraled review designed to build momentum in the opening moments of class MAD MINUTE: time
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Solving Systems of Equations by Substitution
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This product was made for students with learning or cognitive disabilities. These equations require one step or two steps to complete. Most of the equations are already solve for one variable and if the students need to solve for the second variable, the steps are simple.
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Solve Same System of Equations Graphically, Substitution, and Elimination
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Worksheet with System of Equations and directions to solve it graphically, using substitution, and using elimination. Allows students to see the connection between solving systems graphically and algebraically as well as determine the difference between the two algebraic methods. Answer key provided.
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Solving Systems of Linear Equations by Substitution - Free Sample
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Master the method of solving systems of linear equations with this comprehensive worksheet designed for students and educators. This resource provides a step-by-step approach to the substitution method, enabling learners to build confidence and accuracy in solving linear systems. Inside, you'll find: A variety of problems ranging from beginner to advanced levels.Clear, concise instructions on how to solve systems using substitution.Fully worked-out solutions with detailed explanations to enhance
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Solving Systems of Equations using Substitution
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Systems of equations solving practice using the substitution method. This set of 8 problems includes two challenge problems. Use this worksheet as a set of practice problems for all students, a test review, or as a tool for interventions. Cut the problems from this worksheet out to create stations if you want to add some movement to your classroom.
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187908 | https://mathbitsnotebook.com/Algebra1/Quadratics/QDquadFormulaPractice.html | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- | | | | Quadratic Formula Practice MathBitsNotebook.com Topical Outline | Algebra 1 Outline | MathBits' Teacher Resources Terms of Use Contact Person: Donna Roberts | Directions: Solve the following questions, for the indicated variable, using the quadratic formula. It is possible that some of these problems can also be solved by factoring, but for right now, we are practicing the quadratic formula. | | | | | --- --- | | | | | --- | | 1. | Solve using the quadratic formula: x2 - 2x - 24 = 0 | | | | | | | | | --- --- | | | | | --- | | 2. | Solve using the quadratic formula: m2 = 20 - 4m | | | | | | | | | --- --- | | | | | --- | | 3. | Solve using the quadratic formula: | | | | | | | | | --- --- | | | | | --- | | 4. | Given y = 16x2 - 8x. Find all real values of x for which y = -1. | | | | | | | | | --- --- | | | | | --- | | 5. | Solve using the quadratic formula: 3x (3x + 1) = 2 | | | | | | | NOTE: The re-posting of materials (in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". | Topical Outline | Algebra 1 Outline | MathBitsNotebook.com | MathBits' Teacher Resources Terms of Use Contact Person:Donna Roberts | | |
187909 | https://resourcecenter.byupathway.org/math/m04-08 | ### Introduction
Introduction
In this lesson, you will learn how to divide multiple-digit numbers that have a whole number or a decimal as an answer. Unfortunately, answers to division problems are not always whole numbers.
People around the world use different division algorithms or methods. This lesson will teach one method. Feel free to use the method that best works for you.
These videos illustrate the lesson material below. Watching the videos is optional.
### Dividing by a Single Whole Number
Dividing by a Single Whole Number
You will learn how to do division with multiple digits. This process is applicable for numbers that divide evenly, and those with decimal answers. If you are familiar with another algorithm or pattern, you are welcome to use it. The important thing is to understand what you are doing and why so you can apply it to real-world situations.
Steps for division:
The same process of division is used, even with a decimal answer. Remember, you can always add 0’s after the decimal place on a number (for example: (2 = 2.0 = 2.00)). The following instructions will show how to divide with decimals.
Note: This process may be slow at first. Take your time learning to divide accurately. Studying and memorizing your 1-digit multiplication facts will speed up the process and help you as you follow the division steps.
Example 1
(364\div 7)
The goal is to divide 364 into 7 equal pieces. Start by creating a box to help you organize the problem. Put 364 under the box, and 7, the divisor, outside of the box: (7\enclose{longdiv}{364})
Start with the first number on the left. In this case, 3. How many sevens can go into three? The answer is zero, so either put the zero here or just leave it blank. Normally, you leave it blank.
Since seven does not go into three, look at the next number over. Now you’re looking at the number 36. How many times does 7 go into 36, or what multiplied by 7 is close to 36? From your multiplication facts, you know that (7\times5=35). Because 35 is close to 36 while still being less than 36, you can put a 5 in the tens column.
Figure 1
(7\times5=35), so subtract 35 from 36, which leaves 1.
Figure 2
Carry down the number in the next column to the right, which is four. Now you are working with 14.
Figure 3
How many times does 7 go into 14? Your multiplication facts tell you that (2\times7=14), so the answer is two. (14-14=0), and because you have 0, you know that you’re done with the problem. The answer is 52: (364\div7=52).
Figure 4
Consider why this works. When you multiply 5 and 7, it’s actually 50 multiplied by 7 because the 5 is in the tens column. When you subtract, you’re actually subtracting 350 from 364. You couldn’t use a number larger than five here because anything larger than a five in the tens place would give you too large of a number.
You want to know how much is left over, so you subtract 350 from 364. In the first example, you just brought down the four. You can do this because there’s really a zero under the four. The remaining 14 means there’s 14 that still needs to be divided by 7.
Example 2
(8\enclose{longdiv}{984})
(984\div8). Start with the column farthest to the left, which in this case is the hundreds column. 8 goes into 9 one time. (1\times8=8), and (9-8=1).
Figure 5
Bring the remaining one down the next number to the right.
Figure 6
After bringing down the 8, you have 18. How many times does 8 go into 18? (8\times2=16), which is less than 18 but close to it. Put a 2 in the tens place, subtract 16 from 18, and bring the next number down.
Figure 7
When you bring the next number down, you have 24.
Figure 8
(8\times3=24). Subtract 24 from 24, and you get a remainder of zero, meaning the answer is a whole number. (984\div8=123)
Figure 9
### Division Algorithm
Division Algorithm
How can you find the answerto a division problem when the answer is not a whole number? If the answer is not a whole number, it will be a decimal.
Example 3
(17\div5)
From your multiplication facts, you know that (5\times3=15), and (5\times4=20). This means that (17\div5) is going to have an answer somewhere between three and four, but it’s not exactly three or four. The answer will have a decimal in it.
Start by creating a little box: (5\enclose{longdiv}{17})
Use the same division algorithm used in the previous section. Five does not go into one, so move on to the next number: 5 does go into 17. (3\times5=15), which is ideal because it is close to 17 but still smaller than 17.Now subtract: (17-15=2).
Figure 10
The answer means that (17\div5) gives you an answer of three with two pieces remaining.
Now you must figure out how to put the two remaining pieces into decimal form. Put a decimal after 17 because 17 is the same as 17.0.If you put a zero here, you can bring a zero down and continue doing the algorithm as before. Keep in mind that if you add a decimal to the number you are dividing, you must put a decimal in the final answer as well.
Figure 11
(5\times4=20), so put a four in the answer after the decimal place.
Figure 12
You estimated at the beginning that the answer would be somewhere between three and four, and it is.
### Repeating Decimals
Repeating Decimals
There are some division problems where the answer will continue to repeat forever.
Example 4
(1\div3) or (3\enclose{longdiv}{1})
If you solve this equation using the regular division algorithm, you know that three doesn’t go into one, so you put a decimal in the answer and next to the one. Now you pretend like the answer is 10. Three goes into 10 three times.
Figure 13
(3\times3=9), and when you subtract, you get one left over. Bring down another zero. Three goes into 10 three times, and (3\times3=9). When you subtract 9 from 10, you have 1 remaining. If you repeat the process again, you get the same answer, so you find that there’s a pattern going on.
Figure 14
You could continue doing this forever, and you would get a never-ending stream of threes. In this case, the answer is 0.333… The little dots after it mean that the number goes on forever. Another way you can show that the answer repeats is by putting a line over the three: (0.\overline3)
Example 5
(6\enclose{longdiv}{5})
Sometimes it’s not immediately obvious that the answer will repeat forever.
Six doesn’t go into five, so you put a decimal point and a zero: (6\enclose{longdiv}{5.0})
Six can go into 50. (6\times8=48), so there’s two left over from 50.
Figure 15
Add another zero to the decimal point and bring it down.
Figure 16
(3\times6=18), so you know that six goes into 20 three times. (20-18=2), so you’ll bring down another zero.
Figure 17
You didn’t see repetition in the first digit, but you see it in the second and third digits. If you continued on, you’d see the repetition forever, so you can write the answer with a bar over the three to show that the three repeats from here on out: (0.8\overline{3})
Example 6
(1\div7)
A calculator will help demonstrate this example. You get a huge number when you plug this equation into the calculator: 0.1428571428571428571….
Figure 18
It doesn’t look like it’s repeating initially. However, you see that the string of numbers repeats the sequence 142857.
When you run across problems like this, it’s efficient to round to the nearest hundredth. In the example above, because the two in the thousandths position indicates rounding down, the answer would be 0.14.
### Things to Remember
Things to Remember
Practice Problems |
187910 | https://users.math.msu.edu/users/gnagy/teaching/10-fall/mth234/w2-234-h.pdf | Dot product and vector projections (Sect. 12.3) ▶Two definitions for the dot product.
▶Geometric definition of dot product.
▶Orthogonal vectors.
▶Dot product and orthogonal projections.
▶Properties of the dot product.
▶Dot product in vector components.
▶Scalar and vector projection formulas.
There are two main ways to introduce the dot product Geometrical definition → Properties → Expression in components.
Geometrical expression ← Properties ← Definition in components.
We choose the first way, the textbook chooses the second way.
Dot product and vector projections (Sect. 12.3) ▶Two definitions for the dot product.
▶Geometric definition of dot product.
▶Orthogonal vectors.
▶Dot product and orthogonal projections.
▶Properties of the dot product.
▶Dot product in vector components.
▶Scalar and vector projection formulas.
The dot product of two vectors is a scalar Definition Let v , w be vectors in Rn, with n = 2, 3, having length |v | and |w| with angle in between θ, where 0 ≤θ ≤π. The dot product of v and w, denoted by v · w, is given by v · w = |v | |w| cos(θ).
O V W Initial points together.
The dot product of two vectors is a scalar Example Compute v · w knowing that v, w ∈R3, with |v| = 2, w = ⟨1, 2, 3⟩ and the angle in between is θ = π/4.
Solution: We first compute |w|, that is, |w|2 = 12 + 22 + 32 = 14 ⇒ |w| = √ 14.
We now use the definition of dot product: v · w = |v| |w| cos(θ) = (2) √ 14 √ 2 2 ⇒ v · w = 2 √ 7.
◁ ▶The angle between two vectors is a usually not know in applications.
▶It will be convenient to obtain a formula for the dot product involving the vector components.
Dot product and vector projections (Sect. 12.3) ▶Two definitions for the dot product.
▶Geometric definition of dot product.
▶Orthogonal vectors.
▶Dot product and orthogonal projections.
▶Properties of the dot product.
▶Dot product in vector components.
▶Scalar and vector projection formulas.
Perpendicular vectors have zero dot product.
Definition Two vectors are perpendicular, also called orthogonal, iffthe angle in between is θ = π/2.
0 = / 2 V W Theorem The non-zero vectors v and w are perpendicular iffv · w = 0.
Proof.
0 = v · w = |v| |w| cos(θ) |v| ̸= 0, |w| ̸= 0 ) ⇔ ( cos(θ) = 0 0 ⩽θ ⩽π ⇔ θ = π 2 .
The dot product of i, j and k is simple to compute Example Compute all dot products involving the vectors i, j , and k.
Solution: Recall: i = ⟨1, 0, 0⟩, j = ⟨0, 1, 0⟩, k = ⟨0, 0, 1⟩.
y i j k x z i · i = 1, j · j = 1, k · k = 1, i · j = 0, j · i = 0, k · i = 0, i · k = 0, j · k = 0, k · j = 0.
◁ Dot product and vector projections (Sect. 12.3) ▶Two definitions for the dot product.
▶Geometric definition of dot product.
▶Orthogonal vectors.
▶Dot product and orthogonal projections.
▶Properties of the dot product.
▶Dot product in vector components.
▶Scalar and vector projection formulas.
The dot product and orthogonal projections.
The dot product is closely related to orthogonal projections of one vector onto the other. Recall: v · w = |v| |w| cos(θ).
V W = |V| cos(O) O V W |W| |V| O V W V W = |W| cos(O) Dot product and vector projections (Sect. 12.3) ▶Two definitions for the dot product.
▶Geometric definition of dot product.
▶Orthogonal vectors.
▶Dot product and orthogonal projections.
▶Properties of the dot product.
▶Dot product in vector components.
▶Scalar and vector projection formulas.
Properties of the dot product.
Theorem (a) v · w = w · v , (symmetric); (b) v · (aw) = a (v · w), (linear); (c) u · (v + w) = u · v + u · w, (linear); (d) v · v = |v |2 ⩾0, and v · v = 0 ⇔ v = 0, (positive); (e) 0 · v = 0.
Proof.
Properties (a), (b), (d), (e) are simple to obtain from the definition of dot product v · w = |v| |w| cos(θ).
For example, the proof of (b) for a > 0: v · (aw) = |v| |aw| cos(θ) = a |v| |w| cos(θ) = a (v · w).
Properties of the dot product.
(c), u · (v + w) = u · v + u · w, is non-trivial. The proof is: V W w |V+W| cos(0) V+W U 0V 0 0W |W| cos(0 ) |V| cos(0 ) V W |v + w| cos(θ) = u · (v + w) |u| , |w| cos(θw) = u · w |u| , |v| cos(θv) = u · v |u| , ⇒u · (v + w) = u · v + u · w Dot product and vector projections (Sect. 12.3) ▶Two definitions for the dot product.
▶Geometric definition of dot product.
▶Orthogonal vectors.
▶Dot product and orthogonal projections.
▶Properties of the dot product.
▶Dot product in vector components.
▶Scalar and vector projection formulas.
The dot product in vector components (Case R2) Theorem If v = ⟨vx, vy⟩and w = ⟨wx, wy⟩, then v · w is given by v · w = vxwx + vywy.
Proof.
Recall: v = vx i + vy j and w = wx i + wy j . The linear property of the dot product implies v · w = (vx i + vy j ) · (wx i + wy j ) = vxwx i · i + vxwy i · j + vywx j · i + vywy j · j .
Recall: i · i = j · j = 1 and i · j = j · i = 0. We conclude that v · w = vxwx + vywy.
The dot product in vector components (Case R3) Theorem If v = ⟨vx, vy, vz⟩and w = ⟨wx, wy, wz⟩, then v · w is given by v · w = vxwx + vywy + vzwz.
▶The proof is similar to the case in R2.
▶The dot product is simple to compute from the vector component formula v · w = vxwx + vywy + vzwz.
▶The geometrical meaning of the dot product is simple to see from the formula v · w = |v| |w| cos(θ).
Example Find the cosine of the angle between v = ⟨1, 2⟩and w = ⟨2, 1⟩ Solution: v · w = |v| |w| cos(θ) ⇒ cos(θ) = v · w |v| |w|.
Furthermore, v · w = (1)(2) + (2)(1) |v| = p 12 + 22 = √ 5, |w| = p 22 + 12 = √ 5, ⇒ cos(θ) = 4 5.
◁ Dot product and vector projections (Sect. 12.3) ▶Two definitions for the dot product.
▶Geometric definition of dot product.
▶Orthogonal vectors.
▶Dot product and orthogonal projections.
▶Properties of the dot product.
▶Dot product in vector components.
▶Scalar and vector projection formulas.
Scalar and vector projection formulas.
Theorem The scalar projection of vector v along the vector w is the number pw(v) given by pw(v) = v · w |w| .
The vector projection of vector v along the vector w is the vector pw(v) given by pw(v) = v · w |w| w |w|.
P (V) = V W = |V| cos(O) O V W W |W| P (V) = V W W O V W W |W| |W| Example Find the scalar projection of b = ⟨−4, 1⟩onto a = ⟨1, 2⟩.
Solution: The scalar projection of b onto a is the number pa(b) = |b| cos(θ) = b · a |a| = (−4)(1) + (1)(2) √ 12 + 22 .
We therefore obtain pa(b) = −2 √ 5 .
a p (b) a b Example Find the vector projection of b = ⟨−4, 1⟩onto a = ⟨1, 2⟩.
Solution: The vector projection of b onto a is the vector pa(b) = b · a |a| a |a| = −2 √ 5 1 √ 5 ⟨1, 2⟩, we therefore obtain pa(b) = − 2 5, 4 5 .
a p (b) a b Example Find the vector projection of a = ⟨1, 2⟩onto b = ⟨−4, 1⟩.
Solution: The vector projection of a onto b is the vector pb(a) = a · b |b| b |b| = −2 √ 17 1 √ 17 ⟨−4, 1⟩, we therefore obtain pa(b) = 8 17, −2 17 .
b b a p (a) Cross product and determinants (Sect. 12.4) ▶Two definitions for the cross product.
▶Geometric definition of cross product.
▶Parallel vectors.
▶Properties of the cross product.
▶Cross product in vector components.
▶Determinants to compute cross products.
▶Triple product and volumes.
There are two main ways to introduce the cross product Geometrical definition → Properties → Expression in components.
Geometrical expression ← Properties ← Definition in components.
We choose the first way, like the textbook.
Cross product and determinants (Sect. 12.4) ▶Two definitions for the cross product.
▶Geometric definition of cross product.
▶Parallel vectors.
▶Properties of the cross product.
▶Cross product in vector components.
▶Determinants to compute cross products.
▶Triple product and volumes.
The cross product of two vectors is another vector Definition Let v , w be vectors in R3 having length |v | and |w| with angle in between θ, where 0 ≤θ ≤π. The cross product of v and w, denoted as v × w, is a vector perpendicular to both v and w, pointing in the direction given by the right hand rule, with norm |v × w| = |v | |w| sin(θ).
V W W x V V x W O Cross product vectors are perpendicular to the original vectors.
|v × w| is the area of a parallelogram Theorem |v × w| is the area of the parallelogram formed by vectors v and w.
Proof.
V W |V| sin(O) O The area A of the parallelogram formed by v and w is given by A = |w| |v| sin(θ) = |v × w|.
Cross product and determinants (Sect. 12.4) ▶Two definitions for the cross product.
▶Geometric definition of cross product.
▶Parallel vectors.
▶Properties of the cross product.
▶Cross product in vector components.
▶Determinants to compute cross products.
▶Triple product and volumes.
Parallel vectors have zero cross product.
Definition Two vectors are parallel iffthe angle in between them is θ = 0.
v w Theorem The non-zero vectors v and w are parallel iffv × w = 0.
Proof.
Recall: Vector v × w = 0 iffits length |v × w| = 0, then |v| |w| sin(θ) = 0 |v| ̸= 0, |w| ̸= 0 ) ⇔ ( sin(θ) = 0 0 ⩽θ ⩽π ⇔ θ = 0, or θ = π.
Recall: |v × w| is the area of a parallelogram Example The closer the vectors v, w are to be parallel, the smaller is the area of the parallelogram they form, hence the shorter is their cross product vector v × w.
W O V x W V 1 2 V W V x W O ◁ Example Compute all cross products involving the vectors i, j , and k.
Solution: Recall: i = ⟨1, 0, 0⟩, j = ⟨0, 1, 0⟩, k = ⟨0, 0, 1⟩.
y i j k x z i × j = k, j × k = i, k × i = j , i × i = 0, j × j = 0, k × k = 0, i × k = −j , j × i = −k, k × j = −i.
◁ Cross product and determinants (Sect. 12.4) ▶Two definitions for the cross product.
▶Geometric definition of cross product.
▶Parallel vectors.
▶Properties of the cross product.
▶Cross product in vector components.
▶Determinants to compute cross products.
▶Triple product and volumes.
Main properties of the cross product Theorem (a) v × w = −(w × v ), (Skew-symmetric); (b) v × v = 0; (c) (a v ) × w = v × (a w) = a (v × w), (linear); (d) u × (v + w) = u × v + u × w, (linear); (e) u × (v × w) ̸= (u × v ) × w, (not associative).
Proof.
Part (a) results from the right hand rule. Part (b) comes from part (a). Parts (b) and (c) are proven in a similar ways as the linear property of the dot product. Part (d) is proven by giving an example.
The cross product is not associative, that is, u × (v × w) ̸= (u × v) × w.
Example Show that i × (i × k) = −k and (i × i) × k = 0.
Solution: i × (i × k) = i × (−j ) = −(i × j ) = −k ⇒ i × (i × k) = −k, (i × i) × k = 0 × j = 0 ⇒ (i × i) × k = 0.
◁ Recall: The cross product of two vectors vanishes when the vectors are parallel Cross product and determinants (Sect. 12.4) ▶Two definitions for the cross product.
▶Geometric definition of cross product.
▶Parallel vectors.
▶Properties of the cross product.
▶Cross product in vector components.
▶Determinants to compute cross products.
▶Triple product and volumes.
The cross product vector in vector components.
Theorem If the vector components of v and w in a Cartesian coordinate system are v = ⟨v1, v2, v3⟩and w = ⟨w1, w2, w3⟩, then holds v × w = ⟨(v2w3 −v3w2), (v3w1 −v1w3), (v1w2 −v2w1)⟩.
For the proof, recall the non-zero cross products i × j = k, j × k = i, k × i = j , and their skew-symmetric products, while all the other cross products vanish, and then use the properties of the cross product.
Cross product in vector components.
Proof.
Recall: v = v1 i + v2 j + v3 k, w = w1 i + w2 j + w3 k.
Then, it holds v × w = (v1 i + v2 j + v3 k) × (w1 i + w2 j + w3 k).
Use the linearity property. The only non-zero terms are those with products i × j = k and j × k = i and k × i = j . The result is v × w = (v2w3 −v3w2) i + (v3w1 −v1w3) j + (v1w2 −v2w1) k.
Cross product in vector components.
Example Find v × w for v = ⟨1, 2, 0⟩and w = ⟨3, 2, 1⟩, Solution: We use the formula v × w = ⟨(v2w3 −v3w2), (v3w1 −v1w3), (v1w2 −v2w1)⟩ = ⟨[(2)(1) −(0)(2)], [(0)(3) −(1)(1)], [(1)(2) −(2)(3)]⟩ = ⟨(2 −0), (−1), (2 −6)⟩ ⇒ v × w = ⟨2, −1, −4⟩.
◁ Exercise: Find the angle between v and w above, and then check that this angle is correct using the dot product of these vectors.
Cross product and determinants (Sect. 12.4) ▶Two definitions for the cross product.
▶Geometric definition of cross product.
▶Parallel vectors.
▶Properties of the cross product.
▶Cross product in vector components.
▶Determinants to compute cross products.
▶Triple product and volumes.
Determinants help to compute cross products.
We use determinants only as a tool to remember the components of v × w. Let us recall here the definition of determinant of a 2 × 2 matrix: a b c d = ad −bc.
The determinant of a 3 × 3 matrix can be computed using three 2 × 2 determinants: a1 a2 a3 b1 b2 b3 c1 c2 c3 = a1 b2 b3 c2 c3 −a2 b1 b3 c1 c3 + a3 b1 b2 c1 c2 .
Determinants help to compute cross products.
Claim If the vector components of v and w in a Cartesian coordinate system are v = ⟨v1, v2, v3⟩and w = ⟨w1, w2, w3⟩, then holds v × w = i j k v1 v2 v3 w1 w2 w3 A straightforward computation shows that i j k v1 v2 v3 w1 w2 w3 = (v2w3−v3w2) i−(v1w3−v3w1) j +(v1w2−v2w1) k.
Determinants help to compute cross products.
Example Given the vectors v = ⟨1, 2, 3⟩and w = ⟨−2, 3, 1⟩, compute both w × v and v × w.
Solution: We need to compute the following determinant: w × v = i j k w1 w2 w3 v1 v2 v3 = i j k −2 3 1 1 2 3 The result is w×v = (9−2) i−(−6−1) j +(−4−3) k ⇒ w × v = ⟨7, 7, −7⟩.
From the properties of the determinant we know that v × w = −w × v, therefore v × w = ⟨−7, −7, 7⟩.
◁ Cross product and determinants (Sect. 12.4) ▶Two definitions for the cross product.
▶Geometric definition of cross product.
▶Parallel vectors.
▶Properties of the cross product.
▶Cross product in vector components.
▶Determinants to compute cross products.
▶Triple product and volumes.
The triple product of three vectors is a number Definition Given vectors u, v , w, the triple product is the number given by u · (v × w).
The parentheses are important. First do the cross product, and only then dot the resulting vector with the first vector.
Property of the triple product.
Theorem The triple product of vectors u, v , w satisfies u · (v × w) = w · (u × v ) = v · (w × u).
The triple product is related to the volume of the parallelepiped formed by the three vectors Theorem If u, v , w are vectors in R3, then |u · (v × w)| is the volume of the parallelepiped determined by the vectors u, v , w.
W x V U V W The triple product and volumes A W x V 0 U V W h Proof.
Recall the definition of a dot product: x · y = |x| |y| cos(θ). So, |u · (v × w)| = |u| |v × w| cos(θ) = h |v × w|.
|v × w| is the area A of the parallelogram formed by v and w. So, |u · (v × w)| = h A, which is the volume of the parallelepiped formed by u, v, w.
The triple product and volumes Example Compute the volume of the parallelepiped formed by the vectors u = ⟨1, 2, 3⟩, v = ⟨3, 2, 1⟩, w = ⟨1, −2, 1⟩.
Solution: We use the formula V = |u · (v × w)|. We must compute the cross product first: v × w = i j k 3 2 1 1 −2 1 = (2 + 2) i −(3 −1) j + (−6 −2) k, that is, v × w = ⟨4, −2, −8⟩. Now compute the dot product, u · (v × w) = ⟨1, 2, 3⟩· ⟨4, −2, −8⟩= 4 −4 −24, that is, u · (v × w) = −24. We conclude that V = 24.
◁ The triple product is computed with a determinant Theorem The triple product of vectors u = ⟨u1, u2, u3⟩, v = ⟨v1, v2, v3⟩, and w = ⟨w1, w2, w3⟩is given by u · (v × w) = u1 u2 u3 v1 v2 v3 w1 w2 w3 .
Example Compute the volume of the parallelepiped formed by the vectors u = ⟨1, 2, 3⟩, v = ⟨3, 2, 1⟩, w = ⟨1, −2, 1⟩.
Solution: u · (v × w) = 1 2 3 3 2 1 1 −2 1 .
The triple product is computed with a determinant Example Compute the volume of the parallelepiped formed by the vectors u = ⟨1, 2, 3⟩, v = ⟨3, 2, 1⟩, w = ⟨1, −2, 1⟩.
Solution: u · (v × w) = 1 2 3 3 2 1 1 −2 1 .
The result is: u · (v × w) = (1)(2 + 2) −(2)(3 −1) + (3)(−6 −2), = 4 −4 −24, that is, u · (v × w) = −24. We conclude that V = 24.
◁ |
187911 | https://www.swegon.com/na/knowledge-hub/technical-guides/mollier-diagram/ | Mollier diagram | www.swegon.com
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Mollier diagram
Mollier diagram
A Mollier diagram is the name given to a chart that describes the properties of a gas and has enthalpy as one of its scales. Mollier diagrams exist for steam, refrigerants and air for example. Here the focus is on air. Technically, the Mollier diagram for humid air is a psychrometric chart, which is the term generally used for it in English speaking countries.
The diagram is used to illustrate heating, cooling and humidification processes. We can see the effect of the process and we can see how it relates to requirements for the comfort conditions in the room, for example.
History
Richard Mollier was a German professor of Applied Physics and Mechanics and a pioneer of research in thermodynamics, particularly for water, steam and moist air. He created a diagram in 1904, when he plotted the total heat H against entropy S.At the 1923 Thermodynamics Conference it was decided to name any thermodynamic diagram using enthalpy as one of its axes "Mollier diagram", in his honor.
Mollier diagrams are used by engineers in the design work associated with power plants, compressors, steam turbines, refrigeration systems and air conditioning equipment to visualize the working cycles of thermodynamic systems.
Source: Wikipedia
Properties of air
It is important to understand the properties of air and what they are used for.
Dry airDry air is made up of a mixture of gases. 78% is Nitrogen, 21% is Oxygen and the remaining 1% is made up of gases such as carbon dioxide, argon, neon, helium etc. Dry air is mixed with an amount of water vapour up to about 4%.
Humid airThe main physical properties of humid air that needs to be consider are:
Temperature
Pressure
Humidity
Density
Enthalpy
The Mollier diagram describes the enthalpy, humidity and temperature at a fixed density and pressure.Read more about the properties of humid air below.
Composition of dry air
Properties of humid air
Temperature
Temperature is the degree of hotness and is measured with a thermometer. It is a key parameter in thermal comfort. Adding heat to the air will raise the temperature while removing heat will lower the temperature.
Pressure
Pressure is the force exerted over an area given in Newtons per square meter, also known as Pascal or Pa. Pressure varies with the weather and also depends on altitude. The Mollier diagram is defined at the mean air pressure at sea level; which is 101325 Pa.
Humidity
Humidity is the amount of water vapour in the air. Air can hold up to about 4% water at the normal range of outdoor air temperature. The maximum amount of water that the air can carry depends on the temperature and increases rapidly with temperature. Air at minus 20 degrees centigrade can hold less than 1 gram of water per kilogram of air while at plus 20 degrees the air can hold 15 grams.
Air holding the maximum possible amount of water is called saturated. Saturation depends on temperature and the water content but is also affected by pressure.
Saturation line (x-axis shows moisture content, y-axis shows temperature)
There are two important measures of humidity: Relative humidity and specific humidity.
Relative humidity tells us how much water there is in the air compared with the amount of water at saturation at the same temperature. It is expressed as a ratio in percent, e.g. 50% RH.
Specific humidity tells us how much water there is in the air as a ratio of the mass of water to the mass of the air-water mixture in kg/kg or g/kg.
Two other measures of humidity:
Wet bulb temperatureis measured using a thermometer that has a bulb soaked in water. The water evaporates and cools the thermometer. The lower the humidity, the greater the evaporation and the reduction in temperature. The difference between wet bulb and dry bulb temperature is related to humidity.
Dew Point temperature is the dry bulb temperature at which air becomes saturated when cooled.
Density
Density is the mass of air per cubic meter. The normal density of air is 1,2 kg/m3. Density of air reduces with increasing temperature and also with increasing altitude.
Enthalpy
Enthalpy is a measure of the amount of energy in the air. The enthalpy of moist air is the sum of the enthalpy of the dry air and the enthalpy of the water moisture. Enthalpy increases with increasing temperature and with increasing moisture content. Enthalpy is given in kJ/kg.
When air is heated or cooled, the enthalpy changes and the change in enthalpy multiplied by the mass flow of air gives us the power, making it important in air conditioning processes.
The Mollier diagram layout
Relative humidity and enthalpy are plotted in a diagram against temperature and specific humidity to form the Mollier diagram. The curved lines in the diagram are relative humidity while the diagonal lines are enthalpy.
The left hand diagram is the Mollier diagram used in many countries. The right hand diagram is normally called a Psychrometric chart and is used in most English speaking countries.The diagrams are really the same – they give the same results but the temperature and specific humidity scales are reversed.
Mollier diagrams for typical HVAC system processes
The diagram is used to illustrate heating, cooling and humidification processes. The effect of the process is visualised as well as how it relates to requirements for the comfort conditions in the room.
The equations used to plot the Mollier chart are used in selection programs to calculate the results such as the change in temperature and humidity, as well as power consumption etc.
Learn more and see diagrams of typical processes used in HVAC systems
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187912 | http://faculty.up.edu/wootton/Complex/Chapter10.pdf | The Residue Theorem “Integration Methods over Closed Curves for Functions with Singular-ities” We have shown that if f(z) is analytic inside and on a closed curve C, then Z C f(z)dz = 0.
We have also seen examples where f(z) is analytic on the curve C, but not inside the curve C and Z C f(z)dz ̸= 0 (for example f(z) = 1/z over the unit circle centered at 1). In the latter instance however, we had to calculate the integral directly by brute force. In the following few sections, we shall develop methods to integrate functions with singularities over closed curves which avoid direct computation and then we shall use them to solve other related (and seemingly unrelated) problems in math.
1. The Cauchy Residue Theorem Before we develop integration theory for general functions, we observe the following useful fact.
Proposition 1.1. Suppose that f(z) has an isolated singularity at z0 and f(z) = ∞ X k=−∞ ak(z −z0)k is its Laurent expansion in a deleted neighbourhood of z0. Then if C is any circle surrounding z0 and containing no other isolated singularities and it is oriented counterclockwise, then Z C f(z)dz = 2πia−1.
Proof. By our earlier results, in the Laurent expansion for f(z) around z0, for a given k we have ak = 1 2πi Z C f(z) (z −z0)k+1dz, so using k = 1, the result follows.
□ 1 2 The coefficient a−1 will be very important for our uses so we give it its own name.
Definition 1.2. If f(z) = ∞ X k=−∞ akzk in a deleted neighbourhood of z0, then we call the coefficient a−1 the residue of f at z0 and we denote it by Res(f; z0).
Evaluation of residues is fairly straight forward and we do not (always) have to find the Laurent expansion explicitly to find residues. Specifi-cally, if we have a pole, we can use the following results.
Proposition 1.3. Suppose z0 is a pole of f(z) = A(z)/B(z).
(i) If z0 is a simple pole (of order 1) and B(z) has a simple zero at z0, then a−1 = lim z→z0(z −z0)f(z) = A(z) B′(z) (ii) If z0 is a pole of order k, then a−1 = lim z→z0 1 (k −1)!
dk−1 dk (z −z0)kf(z) Proof.
(i) By assumption, we have f(z) = a−1 z −z0 + a0 + a1(z −z0) + . . .
so (z −z0)f(z) = a−1 + a0(z −z0) + a1(z −z0)2 + . . .
giving lim z→z0(z −z0)f(z) = lim z→z0(a−1 + a0(z −z0) + a1(z −z0)2 + . . . ) = a−1.
Alternatively, we have a−1 = lim z→z0(z −z0)f(z) = lim z→z0(z −z0)A(z) B(z) = lim z→z0 A(z) (z −z0) B(z) −B(z0) since B(z0) = 0, so we get a−1 = lim z→z0 A(z) (z −z0) B(z) −B(z0) = lim z→z0 A(z) B(z)−B(z0) z−z0 = A(z0) B′(z0) noting that B′(z0) ̸= 0 since the pole is of order 1 (using the homework problem).
□ 3 This proposition can be used to evaluate the residue for functions with simple poles very easily and can be used to evaluate the residue for functions with poles of fairly low order. However, it becomes increas-ingly difficult the higher the order of the pole, and impossible with essential singularities. In these cases, we have no choice but to return to the Laurent expansion.
Example 1.4. Find the residues of f(z) = sin (z)/z2 and g(z) = e−1/z2 at z = 0 and use it to evaluate Z C f(z)dz and Z C g(z)dz where C is the unit circle centered at the origin..
(i) We could apply the above results, but first we would need to determine what the order of the pole of f(z) at z = 0 is (it looks like a pole of order 2, but recall that sin (z)/z has a removable singularity at z = 0). With this in mind, we instead use the Laurent expansion. We have sin (z) = z −z3 3! + z5 5! −. . .
so sin (z) z2 = 1 z −z 3! + z3 5! −. . .
so the residue is 1 (and in fact z = 0 is a pole of order 1).
Using the earlier proposition, we have Z C f(z)dz = 2πi ∗1 = 2πi.
(ii) We have ez = 1 + z + z2 2! + z3 3! + . . .
so e−1/z2 = 1 −1 z2 + 1 2!z4 − 1 3!z6 + . . .
so the residue is 0. Using the earlier proposition, we have Z C f(z)dz = 2πi ∗0 = 0.
By the first proposition we gave, we can use residues to evaluate inte-grals of functions over circles containing a single. To evaluate general integrals, we need to find a way to generalize to general closed curves which can contain more than one singularity.
First we recall some simple facts and definitions about closed curves.
4 Definition 1.5. A closed curve is called simple if it does not intersect itself (the book calls such curves “regular curves”).
Theorem 1.6. Any closed curve in C can be written as a union of simple closed curves.
Proposition 1.7. If C is a closed curve and C = C1∪· · ·∪Cn (with the relevant imposed orientations) where the Ci are simple closed curves, then Z C f(z)dz = n X i=1 Z Ci f(z)dz Example 1.8. Evaluate Z C sin (z) z2 dz where C is one of the following curves: (i) We can break up C into two curves, one oriented counter-clockwise around the origin - C1, and the other clockwise - C2.
Next note that since there are no singularites contained be-tween C1 and a small circle C3 centered at the origin oriented counterclockwise, and C2 and C3, we have Z C1 f(z)dz = Z C3 f(z)dz and Z C2 f(z)dz = − Z C3 f(z)dz Then using our earlier proposition, we have Z C3 f(z)dz = 2πi It follows that Z C f(z)dz = 2πi −2πi = 0.
5 (ii) We can break up C into two curves, both oriented clockwise around the origin. Using the same argument as above, it fol-lows that Z C f(z)dz = −2πi −2πi = −4πi.
(iii) We can break up C into two curves, both oriented counter-clockwise around the origin.
Using the same argument as above, it follows that Z C f(z)dz = 2πi + 2πi = 4πi.
With these results and observing the the examples above, it suffices to determine a formula to integrate a function f(z) over simple closed de-pending upon its orientation. Before we develop the formula however, we have a couple of necessary definitions.
Definition 1.9. If C is a closed simple curve, we call the compact region bounded by C the “inside” of C.
Definition 1.10. We say a simple closed curve C is oriented coun-terclockwise if as a particle moves around C in the direction of the orientation, the “inside” of C is to the left of the particle.
We are now ready to prove the main result.
Theorem 1.11. (Cauchy’s Residue Theorem) Suppose f(z) is analytic in a simply connected region D except for isolated singularities. Let γ be a simple closed curve in D which does not contain any singularities oriented counterclockwise and suppose the singularities z1, . . . , zn lie in 6 the inside of C. Then Z C f(z)dz = 2πi n X i=1 Res(f, zi).
Proof. In order to prove this result, we shall use a generalization of the previous result we proved stating that the integral over a closed sim-ple curve is equal to the integral over any closed curve inside provided f(z) is analytic in between the two curves. Specifically, if z1, . . ., zn are the singularities inside C, we can draw small circles around them, C1, . . . , Cn which are fully contained in the interior of C and lines con-nected these circles to the boundary, L1, . . ., Ln, see illustration.
Cn C L1 C1 z_1 Ln z_n Starting at the end of L1 on C, we define the curve K by traversing L1 to C1, traversing C1 counterclockwise, traversing L1 back toward Cand then traversing C counterclockwise until the end of L2 and continuing in this fashion until we traverse the whole of C. As with the earlier proposition, the interior of K is simply connected, and since z1, . . ., zn are the only singulairites of f(z), we get Z K f(z)dz = 0.
Next note that since all the Li’s cancel (since we are traversing each curve in both directions), we get Z C∪−C1∪−C2∪···∪−Cn f(z)dz = 0 or Z C f(z)dz = n X i=1 Z Ci f(z)dz.
Applying our first result, we get Z C f(z)dz = 2πi n X i=1 Res(f, zi).
□ 7 Corollary 1.12. Suppose f(z) is analytic in a simply connected region D except for isolated singularities. Let γ be a simple closed curve in D which does not contain any singularities oriented clockwise and suppose the singularities z1, . . . , zn lie in the inside of C. Then Z C f(z)dz = −2πi n X i=1 Res(f, zi).
Proof. This is simply due to the fact that C is oriented in the opposite direction to that given in the previous result.
□ 2. Application of the Residue Theorem We shall see that there are some very useful direct applications of the residue theorem. However, before we do this, in this section we shall show that the residue theorem can be used to prove some important further results in complex analysis. We start with a definition.
Definition 2.1. We say f is meromorphic in a domain D if f is analytic in D except possibly isolated singularities.
Theorem 2.2. Suppose C is a simple closed curve. If f is meromor-phic inside and on C and contains no zeros or poles on C and if Z = number of zeros (counted with multiplicity) and P = number of poles (counted with multiplicity), then 1 2πi Z C f ′(z) f(z) dz = Z −P.
Proof. In order to show this, we shall calculate the residues at each of the poles of f ′/f in D, the interior of C and then apply the Residue Theorem. First note that f ′/f will be analytic at all points except the zeroes and poles of f(z), so we consider these two possibilities.
First, if z0 ∈D is a zero of f(z), of multiplicity k, then f(z) = (z − z0)kg(z) for some function g(z) which is analytic and nonzero at z0.
Then we have f ′(z) = k(z −z0)k−1g(z) + (z −a)kg′(z), so f ′(z) f(z) = k(z −z0)k−1g(z) + (z −a)kg′(z) (z −z0)kg(z) = k z −z0 + g′(z) g(z) near z0. Since g′/g is analytic at z0, it follows that f ′/f will have a simple pole at z0 with residue k.
If (z −z0) is a pole of f(z) of order k, then f(z) = g(z)(z −z0)−k for some function g(z) which is analytic and nonzero at z0. Then we have f ′(z) = −k(z −z0)−k−1g(z) + (z −a)−kg′(z), so f ′(z) f(z) = −k(z −z0)−k−1g(z) + (z −a)−kg′(z) (z −z0)−kg(z) = − k z −z0 + g′(z) g(z) 8 near z0. Since g′/g is analytic at z0, it follows that f ′/f will have a simple pole at z0 with residue −k. Applying the residue theorem, the result follows.
□ Theorem 2.3. (Argument Principle) Suppose f(z) is analytic and nonzero on a closed simple curve C, fix some z0 on C and let ∆CArg(f) denote the change in argument (measured in total radians of change -not restricted to 0 ⩽Arg(z) ⩽2π) from z0 when traversing the curve in a counterclockwise direction. Then 1 2π∆CArg(f) = 1 2πi Z C f ′(z) f(z) dz.
Proof. First, if z(t) parameterizes the curve C with a ⩽t ⩽b, then we have Z C f ′(z) f(z) dz = Z b a f ′(z(t)) f(z(t)) ˙ z(t)dt.
Suppose that f(z) = r(t)eiϑ(t). Next, recall (as shown in Chapter 3) that if w(t) = f(z(t)), then w′(t) = f ′(z(t)) ˙ z(t) (this is simply the chain rule for complex functions depending upon a single parameter).
It follows that along smooth arcs making up the curve C, we have f ′(z(t)) ˙ z(t) = d dtf(z(t)) = d dt(r(t)eiϑ(t)) = r′(t)eiϑ(t) + ir(t)eiϑ(t)ϑ′(t).
Therefore Z C f ′(z) f(z) dz = Z b a f ′(z(t)) f(z(t)) ˙ z(t)dt = Z b a r′(t)eiϑ(t) + ir(t)eiϑ(t)ϑ′(t) r(t)eiϑ dt = Z b a r′(t) r(t) dt + Z b a iϑ′(t)dt = ln (r(t)) b 1 + ϑ(t) b a = 0 + ∆CArg(f) since the first integral is purely real and the second integral is simply the change in argument from z(a) to z(b) (which is measured as the actual change, even though z(a) = z(b)).
□ Corollary 2.4. Suppose C is a simple closed curve.
If f is mero-morphic inside and on C and contains no zeros or poles on C and if Z = number of zeros (counted with multiplicity) and P = number of poles (counted with multiplicity), and ∆CArg(f) denotes the change in argument when traversing once around C, then Z −P = 1 2π∆CArg(f).
The argument principle provides a way to measure the angle change when traversing a closed curve once. Since there are 2π radians in a 9 complete revolution, the value of 1 2πi Z C f ′(z) f(z) dz gives the number of times a curve wraps around the origin (and is consequently sometimes referred to as the winding number of C). Given the corollary, it follows that Z −P gives the winding number of a curve C. A consequence of the argument principle and these observations is the following.
Theorem 2.5. (Rouch´ es Theorem) Suppose that f and g are analytic inside and on a closed regular curve C and that |f(z)| > |g(z)| for all z ∈C. Then Z(f + g) = Z(f) inside C (where Z(f) denotes the number of zeros of f inside C).
Proof. First, on C observe that |f(z) + g(z)| ⩾|f(z)| −|g(z)| > 0.
Therefore, since we are assuming |f(z)| > |g(z)| ⩾0, it follows that f(z) has no zeros on C and neither does f(z) + g(z). Since both f and f + g are analytic, it follows that the number of zeros of f and f + g inside C are (∆CArg(f))/2π and (∆CArg(f + g))/2π respectively.
Next using the basic properties of the argument, we observe that ∆CArg(1 + g f ) = ∆CArg(f + g f ) = ∆CArg(f + g) −∆CArg(f).
Therefore it suffices to show that ∆CArg(1 + g f ) = 0.
Observe however that |f| > |g|, so |g/f| < 1 and consequently 1 + g(z) f(z) −1 < 1.
However, this means that all the values of 1+f/g take place inside the circle of radius 1 centered at z = 1, and consequently there is no way the function 1 + g/f could wrap around the origin. It follows that ∆CArg(1 + g f ) = 0 and hence Z(f + g) = Z(f).
□ We finish with a couple of examples.
Example 2.6.
(i) Show that z6 + 3z4 −2z + 8 has all 6 zeros satisfying 1 < |z| < 2.
We shall attempt to apply Rouch/’es Theorem. First ob-serve that on |z| = 2, if f(z) = z6 and g(z) = 3z4 −2z + 8, then |f(z)| = 64 and |g(z)| ⩽3 · 23 + 2 · 2 + 8 = 60, so 10 |f(z)| > |g(z)|. It follows that f(z) + g(z) has the same num-ber of zeros as f(z) for |z| < 2, so all 6 zeros of z6+3z4−2z+8 occur in |z| < 2.
Next observe that for |z| = 1, if we take f(z) = z6 + 8 and g(z) = 3z4−2z, then |f(z)| ⩾8−1 = 7 and |g(z)| ⩽3+2 = 5, so it follows that f(z) + g(z) has the same number of zeros in |z| < 1 as f(z) = z6 + 8. However, all the zeros of z6 + 8 have modulus 8(1/6) > 1, so none occur in |z| < 1. Thus it follows that no zeros of z6 + 3z4 −2z + 8 occur in |z| < 1. Thus all zeros occur in the annulus 1 < |z| < 2.
(ii) Show that the quartic polynomial p(z) = z4 + z3 + 1 has one zero in each quadrant.
First observe that there is no zero on the real axis (since x4+x3 > −1 for all real x using elementary calculus). Likewise, there is no zero on the imaginary axis since (iy4) + (iy)3 + 1 = y4 −iy3 + 1 = (y4 + 1) −iy3 always has positive real part i.e.
y4 + 1 > 0 for all real y. Thus the zeros of p(z) must occur in the quadrants.
Next note that since p(z) has real coefficients, the zeros of p(z) must come in conjugate pairs. Therefore, it suffices to show that exactly one zero occurs in the first quadrant (since its conjugate will appear in the fourth, and the other pair must appear in the second and third).
We shall use the argument principle to count the number of zeros (since f(z) is analytic, the value of 1 2π∆CArg(f(z)) will count the number of zeros contained in C). Let C be the curve which consists of the quarter circle in the first quadrant centered at z = 0 with radius R and the real and imaginary line segments making this a closed curve (see illustration).
R We now consider the image of C under the map f(z). First, we have f(0) = 1. Next, the line segment 0 < x < R along the real axis (A) maps under f(z) to the curve x4 + x3 + 1 which is a portion of the positive real axis. The line line segment 11 0 < y < R along the imaginary axis (B) maps under f(z) to the curve (y4 + 1) −iy3 which has strictly positive real part and strictly negative imaginary part. Finally, the quarter circle Reiϑ with 0 ⩽ϑ ⩽π/2 (D) maps to R4e4iϑ + R3e3iϑ + 1 = R4e4iϑ 1 + e−iϑ R + e−4iϑ R4 .
It follows that the image of C under f(z) lookes something like the following: f(B) f A R B D f(D) f(A) In particular, observe that R4e4iϑ 1 + e−iϑ R + e−4iϑ R4 has argument close to 4ϑ for large values of R, so the image of D under f(z) will have a change in argument of approximately 2π. Next, since A and B both map to curves with strictly real part, they could not wrap around the origin, so the change in ϑ must be 2π. It follows that the number of zeros contained in C for sufficiently large R will be equal to 1 2π∆CArg(f(z)) = 1 2π2π = 1.
Homework: Questions from pages 126-127; 1,2,5,6,7,8 |
187913 | https://medium.com/@2davidspeakman/why-you-are-feeling-so-lonely-2af32917efa1 | Mr. Speakman Explains: Social Atomism | by David Speakman | Medium
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Mr. Speakman Explains: Social Atomism
David Speakman
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Why Are We Feeling So Lonely?
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I’m a former computer and science high school teacher. I’ve got this crazy idea that I can teach most anything as long as it’s in English.
Today’s Topic Is Social Atomism.
First, I will tell you what social atomism is.
Then, I will tell you how social atomization works.
Next, I’ll explain how lame brain politicians, uncaring economic forces, and insidious cultural trends try to convince you that individualism is good for you and social isolation is just the price of progress, when in reality, it’s making life worse for everyone.
Finally, I’ll show you how capitalism not only fuels social atomization but also actively prevents solutions to it.
1. What Is Social Atomism?
Social atomization is the breakdown of communal and social bonds, leaving individuals increasingly isolated from one another.
A few key aspects of social atomization:
Loss of Community Ties — Fewer connections with neighbors, family, and local groups.
Increased Individualism — Prioritizing personal success over collective well-being.
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187914 | https://e.math.cornell.edu/people/belk/measuretheory/LebesgueMeasure.pdf | Lebesgue Measure The idea of the Lebesgue integral is to first define a measure on subsets of R. That is, we wish to assign a number m(S) to each subset S of R, representing the total length that S takes up on the real number line. For example, the measure m(I) of any interval I ⊆R should be equal to its length ℓ(I).
Measure should also be additive, meaning that the measure of a disjoint union of two sets is the sum of the measures of the sets: m(S ⊎T) = m(S) + m(T).
Indeed, if we want m to be compatible with taking limits, it should be countably additive, meaning that m ] n∈N Sn = X n∈N m(Sn) for any sequence {Sn} of pairwise disjoint subsets of R.
Of course, the measure m(R) of the entire real line should be infinite, as should the measure of any open or closed ray. Thus the measure should be a function m: P(R) →[0, ∞] where P(R) is the power set of R.
Question: Measuring Subsets of R Does there exist a function m: P(R) →[0, ∞] having the following properties?
1. m(I) = ℓ(I) for every interval I ⊆R.
2. For every sequence S1, S2, . . . of pairwise disjoint subsets of R, m ] n∈N Sn = X n∈N m(Sn).
Lebesgue Measure 2 Surprisingly, the answer to this question is no, although it will be a while before we prove this. But it turns out that it is impossible to define a function m: P(R) →[0, ∞] satisfying both of the conditions above.
The reason is that there exist certain subsets of R that really cannot be assigned a measure. In fact, there is a rigorous sense in which most subsets of R cannot be assigned a measure. Interestingly, actual examples of this phenomenon are difficult to construct, with all such constructions requiring the axiom of choice. As a result, such poorly behaved sets are quite rare in practice, and it is possible to define a measure that works well for almost any set that one is likely to encounter.
Thus our plan is to restrict ourselves to a certain collection M of subsets of R, which we will refer to as the Lebesgue measurable sets. We will then define a function m: M →[0, ∞] called the Lebesgue measure, which has all of the desired properties, and can be used to define the Lebesgue integral. The following theorem summarizes what we are planning to prove.
Main Theorem Existence of Lebesgue Measure There exists a collection M of subsets of R (the measurable sets) and a function m: M →[0, ∞] satisfying the following conditions: 1.
Every interval I ⊆R is measurable, with m(I) = ℓ(I).
2.
If E ⊆R is a measurable set, then the complement Ec = R −E is also measurable.
3.
For each sequence {En} of measurable sets in R, the union S n∈N En is also measurable. Moreover, if the sets {En} are pairwise disjoint, then m ] n∈N En = X n∈N m(En).
Lebesgue Outer Measure We begin by defining the Lebesgue outer measure, which assigns to each subset S of R an “outer measure” m∗(S). Thus m∗will be a function m∗: P(R) →[0, ∞] where P(R) denotes the power set of R.
Lebesgue Measure 3 Of course, m∗will not be countably additive. Instead, it will have the weaker property of countable subadditivity, meaning that m∗ [ n∈N Sn ≤ X n∈N m∗(Sn) for any sequence {Sn} of subsets of R.
The outer measure m∗should be thought of as our first draft of Lebesgue measure.
Indeed, once we determine which subsets of R are measurable, we will simply restrict m∗to the collection M of measurable sets to obtain the Lebesgue measure m. Thus, even though m∗is not countably additive in general, it will turn out be countably additive on the collection of measurable sets.
For the following definition, we say that a collection C of subsets of R covers a set S ⊆R if S ⊆S C.
Definition: Lebesgue Outer Measure If S ⊆R, the (Lebesgue) outer measure of S is defined by m∗(S) = inf (X I∈C ℓ(I) C is a collection of open intervals that covers S ) .
It should make intuitive geometric sense that m∗(J) = ℓ(J) for any interval J, though we will put offthe proof of this for a little while. The difficult part is to show that if we cover an interval J with open intervals, then the sum of the lengths of the open intervals is greater than or equal to the length of J.
Note that m∗(S) may be infinite if P I∈C ℓ(I) is infinite for every collection C of open intervals that covers S. For example, it is not hard to see that m∗(R) must be infinite.
Proposition 1 Properties of m∗ Lebesgue outer measure m∗has the following properties: 1.
m∗(∅) = 0.
2.
If S ⊆T ⊆R, then m∗(S) ≤m∗(T).
3.
If {Sn} is a sequence of subsets of R, then m∗ [ n∈N Sn ≤ X n∈N m∗(Sn) Lebesgue Measure 4 PROOF Statement (1) is obvious from the definition. For (2), let S ⊆T ⊆R, and let C be any collection of open intervals that covers T. Then C also covers S, so m∗(S) ≤ X I∈C ℓ(I).
This holds for every cover C of T by open intervals, and therefore m∗(S) ≤m∗(T).
For (3), let {Sn} be a sequence of subsets of R, and let S = S n∈N Sn. If m∗(Sn) is infinite for some n, then by statement (2) it follows that m∗(S) = ∞, and we are done. Suppose then that m∗(Sn) < ∞for all n. For each n, let Cn be a cover of Sn by open intervals so that X I∈Cn ℓ(I) ≤m∗(Sn) + ϵ 2n.
Then C = S n∈N Cn is a cover of S by open intervals, so m∗(S) ≤ X I∈C ℓ(I) ≤ X n∈N X I∈Cn ℓ(I) ≤ X n∈N m∗(Sn) + ϵ 2n = ϵ + X n∈N m∗(Sn).
Since ϵ was arbitrary, statement (3) follows.
■ Lebesgue Measure We are now ready to define the measurable subsets of R. There are many possi-ble equivalent definitions of measurable sets, and the following definition is known as Careth´ eodory’s criterion. It is not very intuitive, and we shall see equivalent definitions of measurability later on that make much more sense. The advantage of Careth´ eodory’s criterion is that it is relatively easy to use from a theoretical perspec-tive, and also it can be generalized to many other settings.
Definition: Lebesgue Measure A subset E of R is said to be (Lebesgue) measurable if m∗(T ∩E) + m∗(T ∩Ec) = m∗(T).
for every subset T of R. In this case, the outer measure m∗(E) of E is called the (Lebesgue) measure of E, and is denoted m(E).
The arbitrary subset T of R that appears in the criterion is known as a test set.
Note that m∗(T ∩E) + m∗(T ∩Ec) ≥m∗(T) Lebesgue Measure 5 automatically since m∗is subadditive. Thus a set E is Lebesgue measurable if and only if m∗(T ∩E) + m∗(T ∩Ec) ≤m∗(T) for every test set T.
Note also that Careth´ eodory’s criterion is symmetric between E and Ec. Thus a set E is measurable if and only if its complement Ec is measurable.
Proposition 2 Union of Two Measurable Sets If E and F are measurable subsets of R, then E ∪F is also measurable.
PROOF Let T ⊆R be a test set. Since E is measurable, we know that m∗(T) = m∗(T ∩E) + m∗(T ∩Ec).
(1) Also, if we use T ∩(E ∪F) as a test set, we find that m∗ T ∩(E ∪F) = m∗(T ∩E) + m∗ T ∩Ec ∩F .
(2) Finally, since F is measurable, we know that m∗(T ∩Ec) = m∗(T ∩Ec ∩F) + m∗(T ∩Ec ∩F c).
(3) Combining equations (1), (2), and (3) together yields m∗(T) = m∗ T ∩(E ∪F) + m∗(T ∩Ec ∩F c).
Since Ec ∩F c = (E ∪F)c, this proves that E ∪F is measurable.
■ Corollary 3 Intersection of Two Measurable Sets If E and F are measurable subsets of R, then E ∩F is also measurable.
PROOF Since E and F are measurable, their complements Ec and F c is also mea-surable. It follows that the union Ec ∪F c is measurable, and the complement of this is E ∩F.
■ Lebesgue Measure 6 Proposition 4 Countable Additivity Let {Ek} be a sequence of pairwise disjoint measurable subsets of R. Then the union U k∈N Ek is measurable, and m ] k∈N Ek = X k∈N m(Ek).
PROOF Let T ⊆R be a test set, and let U = U k∈N Ek. We wish to show that m∗(T) ≥m∗(T ∩U) + m∗(T ∩U c).
For each n ∈N, let Un = Un k=1 Ek. By the Proposition 2, each Un is measurable, so m∗(T) = m∗(T ∩Un) + m∗(T ∩U c n).
But each Un ⊆U, so T ∩U c n ⊇T ∩U c, and hence m∗(T) ≥m∗(T ∩Un) + m∗(T ∩U c).
Thus it suffices to show that m∗(T ∩Un) →m∗(T ∩U) as n →∞.
To prove this claim, observe first that m∗(T ∩Uk) = m∗(T ∩Uk ∩Ek) + m∗(T ∩Uk ∩Ec k) = m∗(T ∩Ek) + m∗(T ∩Uk−1).
for each k. By induction, it follows that m∗(T ∩Un) = n X k=1 m∗(T ∩Ek) for each n. Then n X k=1 m∗(T ∩Ek) = m∗(T ∩Un) ≤m∗(T ∩U) ≤ X k∈N m∗(T ∩Ek), where the last inequality follows from the countable subadditivity of m∗. By the squeeze theorem, we conclude that lim n→∞m∗(T ∩Un) = m∗(T ∩U) = X k∈N m∗(T ∩Ek), which proves that U is measurable. Moreover, in the case where T = R, the last equation gives m(U) = X k∈N m(Ek).
■ Lebesgue Measure 7 Corollary 5 Countable Union of Measurable Sets If {Ek} is any sequence of measurable subsets of R, then the union S k∈N Ek is measurable.
PROOF Let Un = Sn k=1 Ek for each k, and let Fn = Un −Un−1, with F1 = U1.
By Proposition 2, we know that each Un is measurable, and thus Fn = Un ∩U c n−1 is measurable by Corollary 3. But the sets {Fn} are disjoint, and ] n∈N Fn = [ k∈N Ek so S k∈N Ek is measurable.
■ The Geometry of Intervals All that remains in proving the desired properties of Lebesgue measure is to show that intervals in R are measurable, with m(I) = ℓ(I) for any interval I. Unlike all of the work so far, proving this requires exploiting the geometry of intervals in a significant way.
We begin with the following proposition.
Proposition 6 Intervals are Measurable Every interval J in R is Lebesgue measurable.
PROOF Since each interval in R is the intersection of two rays, it suffices to prove that each ray in R is measurable.
Let R be a ray in R, and let T ⊆R be a test set. We wish to prove that m∗(T) ≥m∗(T ∩R) + m∗(T ∩Rc) If m∗(T) = ∞then we are done, so suppose that m∗(T) < ∞. Let ϵ > 0, and let C be a cover of T by open intervals so that X I∈C ℓ(I) ≤m∗(T) + ϵ 2.
Since the sum P I∈C ℓ(I) is finite, C must be countable (see the appendix on sums).
Let {I1, I2, . . .} be an enumeration of the elements of C, where we set In = ∅for Lebesgue Measure 8 n > |C| if C is finite. Then each of the intersections In ∩R and In ∩Rc is an interval, with ℓ(In ∩R) + ℓ(In ∩Rc) = ℓ(In).
For each n, let Jn and Kn be open intervals containing In ∩R and In ∩Rc, respectively, such that ℓ(Jn) ≤ℓ(In ∩R) + ϵ 2n+2 and ℓ(Kn) ≤ℓ(In ∩Rc) + ϵ 2n+2.
Then {Jn}n∈N is a cover of T ∩R by open intervals, and {Kn}n∈N is a cover of T ∩Rc by open intervals, so m∗(T ∩R) + m∗(T ∩Rc) ≤ X n∈N ℓ(Jn) + X n∈N ℓ(Kn) ≤ X n∈N ℓ(In ∩R) + ϵ 2n+2 + X n∈N ℓ(In ∩Rc) + ϵ 2n+2 = ϵ 2 + X n∈N ℓ(In) ≤m∗(T) + ϵ.
Since ϵ was arbitrary, this proves the desired inequality.
■ All that remains is to prove that the measure of any interval is equal to its length.
For this we need the famous Heine-Borel theorem, which we will state and prove next.
Those familiar with point-set topology should recognize this theorem as a special case of the statement that closed intervals in R are compact. In fact, the notion of compactness in point-set topology arose as a generalization of this theorem.
Theorem 7 Heine-Borel Theorem Let [a, b] be a closed interval in R, and let C be a family of open intervals that covers [a, b]. Then there exists a finite subcollection of C that covers [a, b].
PROOF Let S be the set of all points s ∈[a, b] for which the interval [a, s] can be covered by some finite subcollection of C. Note that a ∈S, since the interval [a, a] is just a single point. Our goal is to prove that b ∈S.
Let x = sup(S).
Since S ⊆[a, b], we know that x ∈[a, b].
Therefore, there exists an interval (c, d) ∈C that contains x. Since c < x, there is some point s ∈S that lies between c and x. Let {(c1, d1), . . . , (cn, dn)} be a finite subcollection of C that covers [a, x]. Then the collection {(c1, d1), . . . , (cn, dn), (c, d)} covers [a, x], which proves that x ∈S.
Lebesgue Measure 9 Moreover, if x < b, then there exists an ϵ > 0 such that x + ϵ ∈[a, b] and x + ϵ ∈(c, d). Then the collection {(c1, d1), . . . , (cn, dn), (c, d)} covers [a, x + ϵ], which proves that x + ϵ ∈S, a contradiction since x is the supremum of S. We conclude that x = b, and therefore b ∈S.
■ In addition to the Heine-Borel theorem, the following proof will use the Riemann integral and characteristic functions. If S is any subset of R, the characteristic function (or indicator function) for S is the function χS : R →R defined by χS(x) = ( 1 if x ∈S, 0 if x / ∈S.
Note that if I is an interval then Z ∞ −∞ χI(x) dx = ℓ(I).
Proposition 8 Measure of an Interval If J is any interval in R, then m(J) = ℓ(J).
PROOF Note first that, for every ϵ > 0, there exists an open interval J′ containing J so that ℓ(J′) ≤ℓ(J) + ϵ.
Then the singleton collection {J′} of open intervals covers J, so m(J) ≤ℓ(J′) ≤ℓ(J) + ϵ.
Since ϵ was arbitrary, it follows that m(J) ≤ℓ(J).
Now let C be any collection of open intervals that covers J. Let ϵ > 0, and let K be a closed subinterval of J such that ℓ(K) ≥ℓ(J) −ϵ. By the Heine-Borel theorem, there exists a finite subcollection {I1, . . . , In} of C that covers K. Then χI1 + · · · + χIn ≥χK so X I∈C ℓ(I) ≥ℓ(I1) + · · · + ℓ(In) = Z ∞ −∞ χI1(x) dx + · · · + Z ∞ −∞ χIn(x) dx = Z ∞ −∞ χI1(x) + · · · + χIn(x) dx ≥ Z ∞ −∞ χK(x) dx = ℓ(K) ≥ℓ(J) −ϵ.
Lebesgue Measure 10 Since ϵ was arbitrary, it follows that X I∈C ℓ(I) ≥ℓ(J) which proves that m(J) ≥ℓ(J).
■ Exercises 1. If {En} is a sequence of measurable sets, prove that the intersection T n∈N En is measurable.
2. Prove that if S ⊆R and m∗(S) = 0, then S is measurable.
3.
a) If E ⊆F are measurable sets, prove that F −E is measurable.
b) Prove that if m(E) < ∞then m(F −E) = m(F) −m(E).
4. If E and F are measurable sets with finite measure, prove that m(E ∪F) = m(E) + m(F) −m(E ∩F).
5. Suppose that E ⊆S ⊆F, where E and F are measurable.
Prove that if m(E) = m(F) and this measure is finite, then S is measurable as well.
6. Prove that every countable subset of R is measurable and has measure zero.
7. Given a nested sequence E1 ⊆E2 ⊆· · · of measurable sets, prove that m [ n∈N En = sup n∈N m(En).
8.
a) Let E1 ⊇E2 ⊇· · · be a nested sequence of measurable sets with \ n∈N En = ∅.
Prove that if m(E1) < ∞, then m(En) →0 as n →∞.
b) Let E1 ⊇E2 ⊇· · · be a nested sequence of measurable sets, and suppose that m(E1) < ∞. Prove that m \ n∈N En = inf n∈N m(En).
c) Give an example of a nested sequence E1 ⊇E2 ⊇· · · of measurable sets such that m(En) = ∞for all n but m \ n∈N En < ∞. |
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Intermediate Algebra 2e
12.1 Sequences
Intermediate Algebra 2e12.1 Sequences
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Learning Objectives
By the end of this section, you will be able to:
Write the first few terms of a sequence
Find a formula for the general term (nth term) of a sequence
Use factorial notation
Find the partial sum
Use summation notation to write a sum
Be Prepared 12.1
Before you get started, take this readiness quiz.
Evaluate for the integers 1, 2, 3, and 4. If you missed this problem, review Example 1.6.
Be Prepared 12.2
Evaluate for the integers 1, 2, 3, and 4. If you missed this problem, review Example 1.19.
Be Prepared 12.3
If find If you missed this problem, review Example 3.49.
Write the First Few Terms of a Sequence
Let’s look at the function and evaluate it for just the counting numbers.
| |
| |
| | |
| 1 | 2 |
| 2 | 4 |
| 3 | 6 |
| 4 | 8 |
| 5 | 10 |
| … | … |
If we list the function values in order as 2, 4, 6, 8, and 10, … we have a sequence. A sequence is a function whose domain is the counting numbers.
Sequences
A sequence is a function whose domain is the counting numbers.
A sequence can also be seen as an ordered list of numbers and each number in the list is a term. A sequence may have an infinite number of terms or a finite number of terms. Our sequence has three dots (ellipsis) at the end which indicates the list never ends. If the domain is the set of all counting numbers, then the sequence is an infinite sequence. Its domain is all counting numbers and there is an infinite number of counting numbers.
If we limit the domain to a finite number of counting numbers, then the sequence is a finite sequence. If we use only the first four counting numbers, 1, 2, 3, 4 our sequence would be the finite sequence,
Often when working with sequences we do not want to write out all the terms. We want more compact way to show how each term is defined. When we worked with functions, we wrote and we said the expression was the rule that defined values in the range. While a sequence is a function, we do not use the usual function notation. Instead of writing the function as we would write it as The is the nth term of the sequence, the term in the nth position where n is a value in the domain. The formula for writing the nth term of the sequence is called the general term or formula of the sequence.
General Term of a Sequence
The general term of the sequence is found from the formula for writing the nth term of the sequence. The nth term of the sequence, an, is the term in the nth position where n is a value in the domain.
When we are given the general term of the sequence, we can find the terms by replacing n with the counting numbers in order. For
| | | | | | | |
--- --- ---
| | 1 | 2 | 3 | 4 | 5 | |
| | | | | | | 2n |
To find the values of a sequence, we substitute in the counting numbers in order into the general term of the sequence.
Example 12.1
Write the first five terms of the sequence whose general term is
Solution
We substitute the values 1, 2, 3, 4, and 5 into the formula, in order.
The first five terms of the sequence are 1, 5, 9, 13, and 17.
Try It 12.1
Write the first five terms of the sequence whose general term is
Try It 12.2
Write the first five terms of the sequence whose general term is
For some sequences, the variable is an exponent.
Example 12.2
Write the first five terms of the sequence whose general term is
Solution
We substitute the values 1, 2, 3, 4, and 5 into the formula, in order.
The first five terms of the sequence are 3, 5, 9, 17, and 33.
Try It 12.3
Write the first five terms of the sequence whose general term is
Try It 12.4
Write the first five terms of the sequence whose general term is
It is not uncommon to see the expressions or in the general term for a sequence. If we evaluate each of these expressions for a few values, we see that this expression alternates the sign for the terms.
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--- --- --- |
| | 1 | 2 | 3 | 4 | 5 |
| | | | | | |
| | | | | | |
The terms in the next example will alternate signs as a result of the powers of
Example 12.3
Write the first five terms of the sequence whose general term is
Solution
We substitute the values 1, 2, 3, 4, and 5 into the formula, in order.
The first five terms of the sequence are and
Try It 12.5
Write the first five terms of the sequence whose general term is
Try It 12.6
Write the first five terms of the sequence whose general term is
Find a Formula for the General Term (nth Term) of a Sequence
Sometimes we have a few terms of a sequence and it would be helpful to know the general term or nth term. To find the general term, we look for patterns in the terms. Often the patterns involve multiples or powers. We also look for a pattern in the signs of the terms.
Example 12.4
Find a general term for the sequence whose first five terms are shown.
Solution
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--- |
| | |
| | |
| We look for a pattern in the terms. | |
| The numbers are all multiples of 4. | |
| | The general term of the sequence is |
Try It 12.7
Find a general term for the sequence whose first five terms are shown.
Try It 12.8
Find a general term for the sequence whose first five terms are shown.
Example 12.5
Find a general term for the sequence whose first five terms are shown.
Solution
| | |
--- |
| | |
| | |
| We look for a pattern in the terms. | |
| The numbers are powers of 2. The signs arealternating, with even negative. | |
| | The general term of the sequence is |
Try It 12.9
Find a general term for the sequence whose first five terms are shown.
Try It 12.10
Find a general term for the sequence whose first five terms are shown
Example 12.6
Find a general term for the sequence whose first five terms are shown.
Solution
| | |
--- |
| | |
| | |
| We look for a pattern in the terms. | |
| The numerators are all 1. | |
| The denominators are powers of 3. | The general term of the sequence is |
Try It 12.11
Find a general term for the sequence whose first five terms are shown.
Try It 12.12
Find a general term for the sequence whose first five terms are shown.
Use Factorial Notation
Sequences often have terms that are products of consecutive integers. We indicate these products with a special notation called factorial notation. For example,, read 5 factorial, means The exclamation point is not punctuation here; it indicates the factorial notation.
Factorial Notation
If n is a positive integer, then is
We define as 1, so
The values of for the first 5 positive integers are shown.
Example 12.7
Write the first five terms of the sequence whose general term is .
Solution
We substitute the values 1, 2, 3, 4, 5 into the formula, in order.
The first five terms of the sequence are
Try It 12.13
Write the first five terms of the sequence whose general term is
Try It 12.14
Write the first five terms of the sequence whose general term is
When there is a fraction with factorials in the numerator and denominator, we line up the factors vertically to make our calculations easier.
Example 12.8
Write the first five terms of the sequence whose general term is
Solution
We substitute the values 1, 2, 3, 4, 5 into the formula, in order.
The first five terms of the sequence are 2, 6, 12, 20, and 30.
Try It 12.15
Write the first five terms of the sequence whose general term is
Try It 12.16
Write the first five terms of the sequence whose general term is
Find the Partial Sum
Sometimes in applications, rather than just list the terms, it is important for us to add the terms of a sequence. Rather than just connect the terms with plus signs, we can use summation notation.
For example, can be written as We read this as “the sum of a sub i from i equals one to five.” The symbol means to add and the i is the index of summation. The 1 tells us where to start (initial value) and the 5 tells us where to end (terminal value).
Summation Notation
The sum of the first n terms of a sequence whose nth term is is written in summation notation as:
The i is the index of summation and the 1 tells us where to start and the n tells us where to end.
When we add a finite number of terms, we call the sum a partial sum.
Example 12.9
Expand the partial sum and find its value:
Solution
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--- |
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| We substitute the values 1, 2, 3, 4, 5 in order. | |
| Simplify. | |
| Add. | |
| | |
Try It 12.17
Expand the partial sum and find its value:
Try It 12.18
Expand the partial sum and find its value:
The index does not always have to be i we can use any letter, but i and k are commonly used. The index does not have to start with 1 either—it can start and end with any positive integer.
Example 12.10
Expand the partial sum and find its value:
Solution
| | |
--- |
| | |
| We substitute the values 0, 1, 2, 3, in order. | |
| Evaluate the factorials. | |
| Simplify. | |
| Simplify. | |
| Simplify. | |
| | |
Try It 12.19
Expand the partial sum and find its value:
Try It 12.20
Expand the partial sum and find its value:
Use Summation Notation to Write a Sum
In the last two examples, we went from summation notation to writing out the sum. Now we will start with a sum and change it to summation notation. This is very similar to finding the general term of a sequence. We will need to look at the terms and find a pattern. Often the patterns involve multiples or powers.
Example 12.11
Write the sum using summation notation:
Solution
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--- |
| | |
| | |
| We look for a pattern in the terms. | Terms: |
| The numerators are all one. | Pattern: |
| The denominators are the counting numbersfrom one to five. | The sum written in summation notation is |
| | |
Try It 12.21
Write the sum using summation notation:
Try It 12.22
Write the sum using summation notation:
When the terms of a sum have negative coefficients, we must carefully analyze the pattern of the signs.
Example 12.12
Write the sum using summation notation:
Solution
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--- |
| | |
| | |
| We look for a pattern in the terms. | |
| The signs of the terms alternate,and the odd terms are negative. | |
| The numbers are the cubes of thecounting numbers from one to five. | |
| | |
| | The sum written in summation notation is |
| | |
Try It 12.23
Write each sum using summation notation:
Try It 12.24
Write each sum using summation notation:
Media
Access this online resource for additional instruction and practice with sequences.
Series and Sequences-Finding Patterns
Section 12.1 Exercises
Practice Makes Perfect
Write the First Few Terms of a Sequence
In the following exercises, write the first five terms of the sequence whose general term is given.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
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18.
Find a Formula for the General Term (nth Term) of a Sequence
In the following exercises, find a general term for the sequence whose first five terms are shown.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
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Use Factorial Notation
In the following exercises, using factorial notation, write the first five terms of the sequence whose general term is given.
37.
38.
39.
40.
41.
42.
43.
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48.
Find the Partial Sum
In the following exercises, expand the partial sum and find its value.
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59.
60.
Use Summation Notation to write a Sum
In the following exercises, write each sum using summation notation.
61.
62.
63.
64.
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66.
67.
68.
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72.
Writing Exercises
73.
In your own words, explain how to write the terms of a sequence when you know the formula. Show an example to illustrate your explanation.
Which terms of the sequence are negative when the nth term of the sequence is
75.
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Angina
Christine Hermiz; Yub Raj Sedhai.
Author Information and Affiliations
Authors
Christine Hermiz1; Yub Raj Sedhai2.
Affiliations
1 McLaren Oakland Hospital
2 Virginia Commonwealth Un School of Med
Last Update: June 6, 2023.
Continuing Education Activity
Angina is the most common symptom of ischemic heart disease, which is the major cause of morbidity and mortality worldwide. Approximately 9 million patients in the United States have symptoms of angina, though its treatment is challenging. This activity describes the evaluation and management of angina and reviews the role of the interprofessional team in improving care for patients with this condition.
Objectives:
Outline the pathophysiology of angina.
Review the evaluation of angina.
Summarize the treatment and management options available for angina.
Describe some interprofessional team strategies for improving care coordination and communication to advance treatment for angina and improve outcomes.
Access free multiple choice questions on this topic.
Introduction
Angina, or chest pain, is the most common symptom of ischemic heart disease, a major cause of morbidity and mortality worldwide. Chest pain can be due to non-cardiac and cardiac causes, and thorough history and physical is critical in differentiating these causes and identifying patients experiencing acute coronary syndrome. Angina is one of the signs of acute coronary syndrome (ACS) and can further subdivide into stable and unstable angina. Stable angina defines as the occurrence of symptoms with exertion only. Unstable angina or symptoms occurring at rest requires more prompt evaluation and management. Approximately 9 million patients in the United States have symptoms of angina, and recognizing these symptoms is imperative in improving patient outcomes.
Etiology
Chest pain can be due to non-cardiac causes, non-ischemic cardiac disease, and, finally, ischemic cardiac disease. Non-cardiac causes include gastroesophageal reflux disease, lung disease, musculoskeletal causes, and anxiety/panic attacks. Non-ischemic cardiac causes include pericardial disease. The etiology of chest pain caused by cardiac ischemia is largely accepted to be due to atherosclerosis of coronary arteries and coronary vasospasm. This condition leads to myocardial oxygen supply and demand mismatch. In stable angina, the increased demand only occurs with exertion, but in unstable angina, it also happens at rest. Increased myocardial oxygen demand from exercise is most notably due to increased heart rate, increased blood pressure, and increased myocardial contractility, among other factors. Under normal cardiac physiologic conditions, increased oxygen demand occurring with exertion is followed by coronary vasodilation, but in cases of coronary artery atherosclerosis, this function is impeded, and ischemia and chest pain ensue. Vasospastic angina, otherwise known as variant angina or Prinzmetal angina, like stable angina, also occurs at rest but is unrelated to coronary atherosclerosis.
Epidemiology
Chronic stable angina affects approximately 30000 to 40000 people per million people in Western countries. Prevalence increases with age in both men and women. The estimates are prevalence for men and women 45 to 64 years old are 4 to 7% and 5 to 7%, respectively. In men and women 65 to 84 years old, the estimated prevalence is 14 to 15% and 10 to 12%, respectively.
Modifiable risk factors for angina include hyperlipidemia, hypertension, current or past tobacco use, diabetes mellitus, and obesity/metabolic syndrome. Increasing BMI is an independent risk factor for coronary arterial disease (CAD). Non-modifiable risk factors include increasing age, male sex, family history of CAD, and ethnic origin.
Pathophysiology
The heart is dependent on an adequate oxygen supply for energy production to support contractility. At the cellular level, ischemia causes an increase in anaerobic glycolysis. This increases the levels of hydrogen, potassium, and lactate in the venous return of the ischemic or affected area of the myocardium. The hydrogen ions compete with calcium ions causing hypokinesia/akinesia of the affected area. For stable angina, this change in oxygen supply requires a trigger that would cause metabolic mismatch— exercise, stress, and low temperature.
History and Physical
Patients with ACS most commonly present with angina, which patients usually describe as pain, pressure, tightness, or heaviness in the chest, with potential radiation to the jaw or left arm. It may be accompanied by shortness of breath, diaphoresis, nausea, or any combination of the above. The chest pain may be precipitated by exertion and relieved by rest and/or nitroglycerin in the case of stable angina. In cases of unstable angina or myocardial infarction (MI) [non-ST segment elevation MI (NSTEMI)/ST-segment elevation MI (STEMI)], the chest pain will likely not fully resolve with rest or nitroglycerin. In the case of stable angina, the symptoms may last 5 minutes before resolving after rest or the use of nitroglycerin. Although these are classic signs of ACS, classic angina may not occur in some patients, especially diabetics, so one must have a high index of suspicion in patients with significant cardiac risk factors.
A physical exam may be unremarkable; however, the patient may appear uncomfortable or anxious. He or she may be diaphoretic or clutching his or her chest. Vital signs may be normal or reveal tachycardia and tachypnea most commonly.
Evaluation
In patients presenting with angina, the value of cardiac testing is determined by the patient’s pretest, probability of ACS. Pretest probability is evaluated by considering the patient’s presentation, along with their cardiac risk factors. In patients with very high or very low pretest probability, diagnostic tests are less valuable as they are less likely to change management.
Initial testing includes a 12-lead electrocardiogram (ECG), chest X-ray, and basic lab testing, including complete blood count (CBC), and basic metabolic profile (BMP), along with serial troponin levels if ACS is suspected. ECG may not show any abnormalities in the cases of stable angina, unstable angina, or NSTEMI. ECG findings of myocardial ischemia include T-wave flattening or inversions, or ST-segment depressions. Further testing may include exercise or pharmacologic stress testing with or without nuclear perfusion imaging and diagnostic heart catheterization. ECG changes will appear in STEMIs and prompt an immediate need for coronary revascularization.
Treatment / Management
Treatment of chronic stable angina is aimed at managing symptoms as well as slowing the progression to cardiac events. Management is multifactorial and involves lifestyle modifications, risk factor modification, and medical therapy as essential components of treatment. In cases in which symptoms are refractory to medical therapy, revascularization may be attempted; however, although it may be successful in controlling symptoms, it is not shown to reduce major cardiovascular events compared with medical therapy.
Lifestyle modifications include regular exercise, weight control, and smoking cessation, which should be encouraged. Risk factor modification includes controlling blood pressure, cholesterol, and blood sugar. Medications for risk factor modification and to prevent disease progression include aspirin, statins, angiotensin-converting enzyme inhibitors, or angiotensin receptor blockers.
Medical therapy can be used to control symptoms as well as help mitigate the risk of the progression of atherosclerosis and cardiac events. Antianginal agents can be separated based on the mechanism of symptom relief in angina. In general, symptomatic control is achieved by way of decreasing myocardial oxygen consumption.
As heart rate is the main influencer of oxygen consumption, most anginal events are initiated by an increase in heart rate. Three classes of drugs used for angina reduce symptoms by way of heart rate reduction— beta-blockers, ivabradine, and non-dihydropyridine calcium channel blockers. Calcium channel blockers should be avoided in patients with left ventricular dysfunction and decreased ejection fraction.
Another mechanism by which anginal symptoms can be treated is vascular smooth muscle relaxation. This leads to coronary artery dilatation, thereby increasing perfusion ability. The drugs that work on this mechanism are dihydropyridine calcium channel blockers, nitrates, and nicorandil.
Another drug used for chronic stable angina is ranolazine, which inhibits the late sodium current in ventricular myocardial cells. This reduces diastolic contractile dysfunction.
Treatment for unstable angina is aimed at pain reduction, limiting damage to the myocardium, and decreasing morbidity and mortality.
Nitrates - no mortality benefit, but used for chest pain relief. They cause vasodilation, which decreases preload and left ventricular end-diastolic volume. This reduces myocardial oxygen consumption. They are contraindicated in cases of hypotension and previous use of phosphodiesterase inhibitors within the past 48 hours.
Morphine - no mortality benefit, used for pain relief when pain relief is not fully achieved by nitrates. It causes some vasodilation aside from analgesia.
Beta-blockers - reduce mortality. They cause a decrease in heart rate, contractility, and blood pressure, thereby reducing myocardial oxygen demand.
Antiplatelet agents - dual therapy with aspirin and either clopidogrel, ticagrelor, or prasugrel decrease the risk of cardiovascular events in patients with acute coronary syndromes— acute myocardial infarction, cardiovascular death, and stroke.
Anticoagulants - reduce mortality by decreasing re-infarction rates in combination with antiplatelet agents. Used intravenously for acute treatment.
Anatomic assessment of coronary arteries/consideration of revascularization - high-risk patients should be identified with risk stratification methods and considered for urgent revascularization.
Differential Diagnosis
The differential diagnosis of angina can divide into body systems:
Gastrointestinal: gastroesophageal reflux, hiatal hernia, peptic ulcer disease
Pulmonary: pneumothorax, pneumonia, pulmonary embolism
Musculoskeletal: costochondritis, rib injury, muscle spasm, chest wall injury
Psychiatric: panic attack, generalized anxiety
Cardiac non-ischemic: pericarditis, myocarditis
Vascular: aortic dissection
Prognosis
The prognosis of chronic stable angina progression to cardiac events varies among patients. Factors affecting prognosis include cardiovascular comorbidities as well as compliance with lifestyle modifications and medical treatment plans. Long-term prognosis is also affected by left ventricular systolic function, the degree of exercise the patient can tolerate, and the extent of CAD present.
Risk factors for poorer prognosis are diabetes mellitus, previous MI, hypertension, increasing age, and male sex. The use of nitrates has also been demonstrated to be a negative prognostic indicator of mortality, likely because their use indicates more advanced disease.
Complications
As the initial symptom of CAD in many cases, the main complication of angina is a future cardiac event, such as myocardial infarction. In one study, estimates are that the 10-year risk of MI exceeded 10 percent in women with chronic stable angina starting from when they initially began to use nitrates to treat symptoms.
Aside from this potentially fatal complication, chronic stable angina has implications for society as well as patients. These include lower quality of life due to decreased ability to perform daily activities, as well as an increased burden to society from indirect costs, such as early retirement or disability. Angina treatment should be aimed not only at improving mortality but also at treating symptoms so patients can be more active.
Deterrence and Patient Education
Patients with cardiac risk factors should be monitored for symptoms of angina and receive education about alarming signs, such as symptoms occurring at rest and symptoms no longer relieved by nitrates. It is imperative to stress risk factor modification as an essential component of treatment to slow progression. So, dietary and exercise education, along with smoking cessation counseling, if applicable, is beneficial. It is important to stress compliance with medications, as well.
Enhancing Healthcare Team Outcomes
Recognizing the symptoms of angina is critical to making a diagnosis of CAD, but similar symptoms can present in many other conditions. The interprofessional team must recognize signs of cardiac vs. non-cardiac chest pain and understand that the etiology may not be clear based solely on the quality of pain. In specific patient populations, such as patients with diabetes, the presentation may be more inconspicuous. Understanding the patient's medical history and family history is critical to making the diagnosis. In the presence of cardiac risk factors, the clinician must have a high index of suspicion for a cardiac etiology and conduct further workup, versus when risk factors are absent.
While the cardiologist is the team leader in the treatment and management of these patients, it is the primary care physician who usually completes the initial workup and makes the diagnosis in many cases. It is vital to obtain an accurate history and physical and have an open line of communication between physicians, or else the clinician may miss the diagnosis. A missed diagnosis increases morbidity and mortality for these patients as it may delay their treatment.
As the management of the condition progresses, cardiology nurses play an essential role in monitoring, assessing patient compliance with treatment regimens, and encouraging lifestyle changes. The pharmacist will weigh in on the patient's medication regimen by evaluating the effectiveness of chosen agents, checking for drug-drug interactions, and verifying dosing. With this type of interprofessional collaboration, patients with angina can experience increased quality of life and avoid future morbidity. [Level 5]
Review Questions
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Comment on this article.
Figure
Coronary Artery Changes and Corresponding Types of Angina. This image depicts the pathophysiology of stable angina, unstable angina, and variant (Prinzmetal) angina, as related to varying degrees and patterns of coronary artery stenosis. (more...)
References
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: Hjemdahl P, Eriksson SV, Held C, Forslund L, Näsman P, Rehnqvist N. Favourable long term prognosis in stable angina pectoris: an extended follow up of the angina prognosis study in Stockholm (APSIS). Heart. 2006 Feb;92(2):177-82. [PMC free article: PMC1860751] [PubMed: 15951393]
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: Disclosure: Christine Hermiz declares no relevant financial relationships with ineligible companies.
: Disclosure: Yub Raj Sedhai declares no relevant financial relationships with ineligible companies.
Copyright © 2025, StatPearls Publishing LLC.
This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( ), which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal.
Bookshelf ID: NBK557672PMID: 32491604
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Hermiz C, Sedhai YR. Angina. [Updated 2023 Jun 6]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
In this Page
Continuing Education Activity
Introduction
Etiology
Epidemiology
Pathophysiology
History and Physical
Evaluation
Treatment / Management
Differential Diagnosis
Prognosis
Complications
Deterrence and Patient Education
Enhancing Healthcare Team Outcomes
Review Questions
References
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AlgebraInequality
Caity C.
asked • 02/10/13
x^2 is less than or equal to 25
solve the inequality. write the solution set in interval notation. x^2 is less than or equal to 25
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Barry F.answered • 02/10/13
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X^2 less than or equal to 25, X is less than or equal to +5. (5)^2 = 25, and (-5)^2 = 25, so the solution is any value of x between, and including, 5, -5. If you were to plot this on a number line, you would have a "solid"or "bubbled in"circle on -5 and +5 and a line between them. The interval notation would be [-5,5]... since you are including both 5 and -5 in the solution, you have to use brackets, not parenthesis.
The reason you INCLUDE -5 and 5 is because the equation says"less than OR equal to" the phrase "equal to" indicates that you include both numbers in the solution.
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Nataliya D.answered • 02/10/13
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Let's remember that equation
x2 = a (were a ≥ 0 and "x" is real number) has 2 roots:
√(±x)2 = √a , so ±x = √a , x = ±√a , or lxl = √a
To solve the inequality x2 ≤ 25 (52 = 25 and (-5)2 = 25) we have to examine the inequality on two intervals/segments:
if 0 ≤ x ≤ 5 (x is positive, or equal 0) , then x ≤ 5
if -5 ≤ x < 0 (x is negative) , -x ≤ 5 (when we multiply/divide inequality by negative number, we have to switch sign of inequality to opposite) -x/ (-1) ≥ 5/(-1) ------> x ≥ -5
So the answer is -5 ≤ x ≤ 5 , or any number from the segment [-5,5]
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MCAT Curriculum
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General Chemistry
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Intermolecular Forces
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London Dispersion Forces, Dipole-Dipole Interactions, and Hydrogen Bonds
London Dispersion Forces, Dipole-Dipole Interactions, and Hydrogen Bonds
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General Chemistry
Intermolecular forces are electrostatic interactions between molecules and can be classified into three main types: London dispersion forces, dipole-dipole Interactions, and hydrogen bonds. London dispersion forces are a type of Van der Waals interaction caused by random fluctuations in electron density, creating temporary dipoles. These forces are experienced by all atoms and molecules and are stronger in larger atoms and molecules. However, they are the weakest of the intermolecular forces.
Dipole-dipole Interactions occur between molecules with permanent dipoles, which form when atoms of differing electronegativity cause electrons to spend more time around the electronegative atom. These interactions are stronger than London dispersion forces, but weaker than hydrogen bonds. Hydrogen bonds are the strongest of the intermolecular forces and form between an electronegative atom in one molecule (oxygen, nitrogen or fluorine) and a hydrogen atom covalently bonded to another electronegative atom in a different molecule. Compounds that are capable of forming hydrogen bonds have significantly higher boiling points, melting points, and solubility compared to those that do not. These bonds also contribute to a compound's stability, especially in the tertiary structure of proteins.
Lesson Outline
Introducing Intermolecular Forces
Comparison between intermolecular forces and intramolecular forces
Three important intermolecular forces: London dispersion forces, dipole-dipole interactions, and hydrogen bonds
London Dispersion Forces
Caused by random fluctuations in electron density
Temporary dipoles form due to uneven electron distribution
Stronger in larger atoms and molecules, weaker in smaller atoms and molecules
Experienced by all atoms and molecules, even non-polar ones
Weakest of the intermolecular forces
Dipole-Dipole Interactions
Only occur in polar molecules
Formed due to permanent dipoles in polar molecules
Attractive force between partially positive end of one polar molecule and partially negative end of another polar molecule
Stronger than London dispersion forces, but weaker than hydrogen bonds
Hydrogen Bonds
Special kind of dipole-dipole interaction
Occurs between an electronegative atom (O, N, F) and a hydrogen covalently bonded to an electronegative atom in a different molecule
Hydrogen becomes partially positive, and the electronegative atom becomes partially negative
Compounds with hydrogen bonds have higher boiling and melting points
Increased solubility in water and contribution to stability
Strongest type of intermolecular force
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FAQs
What are the differences between London dispersion forces, Dipole-dipole interactions, and hydrogen bonds?
London dispersion forces are weak intermolecular forces that occur between temporary dipoles due to the random motion of electrons. These forces are present between all types of molecules, polar or nonpolar. Dipole-dipole interactions are stronger intermolecular forces occurring between molecules with permanent dipoles, such as polar molecules. Hydrogen bonds are a specific type of strong dipole-dipole interaction that occurs between a hydrogen atom bonded to a highly electronegative atom (usually nitrogen, oxygen, or fluorine) and another highly electronegative atom in a neighboring molecule.
How do intermolecular forces affect boiling points?
Intermolecular forces, such as London dispersion forces, dipole-dipole Interactions, and hydrogen bonds, impact the boiling points of substances. A higher boiling point is indicative of stronger intermolecular forces present, requiring more energy to break these forces and change the substance from liquid to gas. As such, substances with hydrogen bonding typically have the highest boiling points, followed by those with dipole-dipole interactions and then those with London dispersion forces.
What is the difference between intermolecular and intramolecular forces?
Intermolecular forces are forces that occur between different molecules, such as London dispersion forces, dipole-dipole interactions, and hydrogen bonds. These forces are responsible for the physical properties of a substance, like boiling point and melting point. Intramolecular forces, on the other hand, are forces occurring within a single molecule, such as covalent and ionic bonds. Intramolecular forces are generally much stronger than intermolecular forces and dictate a molecule's structure and chemical properties.
What factors impact the strength of London dispersion forces between molecules?
The strength of London dispersion forces between molecules is influenced by the size of the molecules and their corresponding electron cloud. Larger molecules with more electrons generally experience stronger London dispersion forces as their electron cloud is more easily distorted, leading to larger temporary dipoles. Additionally, molecular shape also affects the strength of these forces. Molecules with an elongated shape have a larger surface area, increasing the chances of experiencing stronger London dispersion forces when compared to more compact or spherical molecules.
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187920 | https://eprints.lancs.ac.uk/id/eprint/75498/1/JNStressFinal.pdf | Submitted manuscript No.
(will be inserted by the editor) Stress matrices and global rigidity of frameworks on surfaces B. Jackson · A. Nixon Received: date / Accepted: date Abstract In 2005, Bob Connelly showed that a generic framework in Rd is globally rigid if it has a stress matrix of maximum possible rank, and that this sufficient condition for generic global rigidity is preserved by the 1-extension operation. His results gave a key step in the characterisation of generic global rigidity in the plane. We extend these results to frameworks on surfaces in R3.
For a framework on a family of concentric cylinders, cones or ellipsoids, we show that there is a natural surface stress matrix arising from assigning edge and vertex weights to the framework, in equilibrium at each vertex. In the case of cylinders and ellipsoids, we show that having a maximum rank stress matrix is sufficient to guarantee generic global rigidity on the surface. We then show that this sufficient condition for generic global rigidity is preserved under 1-extension and use this to make progress on the problem of characterising generic global rigidity on the cylinder.
Keywords rigidity, global rigidity, stress matrix, framework on a surface Mathematics Subject Classification (2000) MSC 52C25 · 05C10 · 53A05 1 Introduction A bar-joint framework (G, p) in d-dimensional Euclidean space Rd is a realisation of a (simple) graph G = (V, E), via a map p : V →Rd, with vertices considered as universal joints and edges as fixed length bars. A framework (G, p) is rigid if the only continuous motions of the vertices in Rd that preserve the edge lengths, arise from isometries of Rd. More strongly, (G, p) is globally rigid in Rd if every realisation (G, q) with the same edge lengths as (G, p) arises from an isometry of Rd. We refer the reader to for more information on rigidity and its applications.
It is NP-hard to determine if an arbitrary framework is rigid or globally rigid . These problems become more tractable if we restrict to generic frameworks, for which rigidity and global rigidity can be determined in polynomial time when d = 1, 2. It remains a difficult open problem to B. Jackson School of Mathematical Sciences, Queen Mary, University of London, E1 4NS, UK, E-mail: b.jackson@qmul.ac.uk A. Nixon Department of Mathematics and Statistics, Lancaster University, LA1 4YF, UK, E-mail: a.nixon@lancaster.ac.uk 2 B. Jackson, A. Nixon characterise, in an efficient combinatorial way, when a 3-dimensional generic framework is rigid or globally rigid.
Results have recently been obtained concerning the rigidity and global rigidity of frameworks in R3 that are constrained to lie on a fixed surface [15,18,19]. In this paper we obtain a natural sufficient condition for such a framework to be globally rigid.
We first recall some fundamental results about the generic (global) rigidity of bar-joint frame-works in Euclidean space. A graph G = (V, E) is (2, k)-sparse if for every subgraph G′ = (V ′, E′), with at least one edge, the inequality |E′| ≤2|V ′|−k holds. Moreover G is (2, k)-tight if |E| = 2|V |−k and G is (2, k)-sparse. While the definitions of (2, k)-sparse and (2, k)-tight make sense for graphs with multiple edges and loops, such edges are of no use when considering rigidity so we will restrict our attention to simple graphs.
Theorem 1 () A generic framework (G, p) in R2 is rigid if and only if G contains a spanning (2, 3)-tight subgraph.
A framework (G, p) is said to be redundantly rigid if (G −e, p) is rigid for all edges e of G.
Theorem 2 ([7,12]) A generic framework (G, p) in R2 is globally rigid if and only if G is a complete graph on at most three vertices or (G, p) is redundantly rigid and G is 3-connected.
Hendrickson had previously shown that (d + 1)-connectivity and redundant rigidity are necessary conditions for generic global rigidity in Rd for all d. Examples constructed by Connelly show that they are not sufficient to imply generic global rigidity when d ≥3. Connelly also obtained a different kind of sufficient condition for generic global rigidity in terms of ‘stress matrices’ (which will be defined in Section 4).
Theorem 3 () Let (G, p) be a generic framework in Rd with n ≥d + 2 vertices. Suppose that (G, p) has an equilibrium stress ω whose associated stress matrix Ωhas rank n −d −1. Then (G, p) is globally rigid in Rd.
Gortler, Healy and Thurston have shown that Connelly’s sufficient condition for generic global rigidity is also a necessary condition. This characterization implies that generic global rigidity depends only on the underlying graph (but does not give rise to a polynomial algorithm for deciding which graphs are generically globally rigid in Rd).
In this paper we develop natural analogues of an equilibrium stress and a stress matrix for frameworks constrained to lie on a surface. Our main result is an analogue of Theorem 3, giving a sufficient condition for generic frameworks on families of concentric cylinders and ellipsoids to be globally rigid.
We conclude the introduction by giving a short outline of what follows. Section 2 recalls basic definitions and results for frameworks on surfaces. We describe the rigidity map and rigidity matrix for surfaces in Section 3. In Section 4 we develop basic properties of stresses, stress matrices and energy functions for frameworks on surfaces. Section 5 contains our main result, Theorem 9, an analogue of Theorem 3 for generic frameworks on cylinders and ellipsoids. We use this result in Section 6 to show that the property of having a maximum rank surface stress matrix is preserved by 1-extensions on these surfaces. We finish by applying our results to make some progress on the problem of characterising generic global rigidity on the cylinder.
Stress matrices and global rigidity of frameworks on surfaces 3 2 Frameworks on Surfaces It was shown in that the rigidity of a generic framework on a surface depends crucially on the number of continuous isometries of R3 admitted by the surface, see Theorems 4, 5 and 7 below.
Since generic rigidity and global rigidity on the plane and sphere , the surfaces with 3-dimensional isometry groups, are now well understood, we consider cylinders, cones and ellipsoids as natural examples of surfaces with 2, 1 and 0-dimensional isometry groups, respectively.
Let r = (r1, r2, . . . , rn) be a vector of (not necessarily distinct) positive real numbers. For 1 ≤ i ≤n, let Yi = {(x, y, z) ∈R3 : x2 + y2 = ri}, Ci = {(x, y, z) ∈R3 : x2 + y2 = riz2} and Ei = {(x, y, z) ∈R3 : x2 +αy2 +βz2 = ri} for some fixed α, β ∈Q with 1 < α < β. Let Y = Sn i=1 Yi, C = Sn i=1 Ci and E = Sn i=1 Ei.
We will use S = Sn i=1 Si to denote one of the three surfaces Y, C, E, and ℓfor the dimension of its space of infinitesimal isometries (so ℓ= 2, 1 or 0 when S = Y, C or E, respectively). We will occasionally use S(r) when we wish to specify a particular vector r and S(1) for the special case when r1 = r2 = · · · = rn (there is no loss in generality in assuming that ri = 1 for all 1 ≤i ≤n when the ri are all equal).
A framework on S is a pair (G, p) where G = (V, E) is a graph with V = {v1, v2, . . . , vn}, and p : V →R3 with p(vi) ∈Si for all 1 ≤i ≤n. Two frameworks (G, p) and (G, q) on S are equivalent if ∥p(vi)−p(vj)∥= ∥q(vi)−q(vj)∥for all vivj ∈E and congruent if ∥p(vi)−p(vj)∥= ∥q(vi)−q(vj)∥ for all vi, vj ∈V . The framework (G, p) is globally rigid on S if every framework (G, q) on S which is equivalent to (G, p) is congruent to (G, p). We say that (G, p) is rigid on S if there exists an ϵ > 0 such that every framework (G, q) on S which is equivalent to (G, p), and has ∥p(vi) −q(vi)∥< ϵ for all 1 ≤i ≤n, is congruent to (G, p). (This is equivalent to saying that every continuous motion of the vertices that stays on S and preserves equivalence also preserves congruence). If (G, p) is not rigid on S then (G, p) is said to be flexible on S. The framework (G, p) is minimally rigid on S if it is rigid and, for every edge e ∈E, (G −e, p) is flexible on S. It is redundantly rigid on S if (G −e, p) is rigid on S for all e ∈E.
An infinitesimal flex s of (G, p) on S is a map s : V →R3 such that s(v) is tangential to S at p(v) for all v ∈V and (p(u) −p(v)) · (s(u) −s(v)) = 0 for all uv ∈E. The framework (G, p) is infinitesimally rigid on S if every infinitesimal flex s of (G, p) satisfies (p(u)−p(v))·(s(u)−s(v)) = 0 for all u, v ∈V .
We consider a framework (G, p) on S = S(r) to be generic if td [Q(r, p) : Q(r)] = 2|V |, where td [Q(r, p) : Q(r)] denotes the transcendence degree of the field extension. Thus (G, p) is generic on S if the coordinates of the vertices of G are as algebraically independent as possible. The following results characterise when a generic framework on Y or C(1) is minimally rigid.
Theorem 4 () Let (G, p) be a generic framework on Y. Then (G, p) is minimally rigid if and only if G is a complete graph on at most three vertices or G is (2, 2)-tight.
Theorem 5 () Let (G, p) be a generic framework on C(1). Then (G, p) is minimally rigid if and only if G is a complete graph on at most four vertices or G is (2, 1)-tight.
It remains an open problem to characterise generic minimally rigid frameworks on E. (The natural analogue of the above theorems is known to be false.) The final result of this section gives necessary conditions for generic global rigidity of frameworks on S which are analogous to Hendrickson’s conditions for Rd.
4 B. Jackson, A. Nixon Theorem 6 () Suppose (G, p) is a generic globally rigid framework on S with n ≥7−ℓvertices.
Then (G, p) is redundantly rigid on S, and G is k-connected, where k = 2 if S ∈{Y, C} and k = 1 if S = E.
We believe that these necessary conditions for generic global rigidity are also sufficient when S ∈{Y, C}, see [15, Conjecture 9.1]. One motivation for the current paper is to try to verify this conjecture by using the same proof technique as Theorem 2. We will return to this in Section 7.
3 The Rigidity Map We assume henceforth that G = (V, E) is a graph with V = {v1, v2, . . . , vn} and E = {e1, e2, . . . , em}.
The rigidity map F G : R3n →Rm is defined by F G(p) = (∥e1∥2, . . . , ∥em∥2) where ∥ei∥2 = ∥p(vj) − p(vk)∥2 when ei = vjvk. Its differential at the point p is the map dF G p : R3n →Rm defined by dF G p (q) = 2R(G, p) · q where R(G, p) is the |E| × 3|V | matrix with rows indexed by E and 3-tuples of columns indexed by V in which, for e = vivj ∈E, the submatrices in row e and columns vi and vj are p(vi)−p(vj) and p(vj)−p(vi), respectively, and all other entries are zero. We refer to R(G, p) as the rigidity matrix for (G, p).
We next define a rigidity map and matrix for a framework (G, p) constrained to lie on our surface S. Let ΘS : R3n →Rn be the map defined by ΘS(p) = (h1(p(v1)), . . . , hn(p(vn))) where, for each 1 ≤i ≤n, hi(x, y, z) = x2 + y2 −ri, if S = Y(r1, r2, . . . , rn); x2 + y2 −riz2, if S = C(r1, r2, . . . , rn); x2 + αy2 + βz2 −ri, if S = E(α, β, r1, r2, . . . , rn).
(3.1) Then the differential of ΘS at the point p is the map dΘS p : R3n →Rn defined by dΘS p (q) = 2S(G, p)·q where S(G, p) = s1 0 . . . 0 0 s2 . . . 0 .
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...
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0 0 . . . sn , si = si(p(vi)) and si(x, y, z) = (x, y, 0), if S = Y; (x, y, −riz), if S = C; (x, αy, βz), if S = E.
(3.2) It follows that rank dΘS p = n if p ∈W = S1 × S2 × . . . × Sn and p(vi) ̸= (0, 0, 0) for all 1 ≤i ≤n.
Hence p ∈W is a regular point of ΘS unless S = C and p(vi) = (0, 0, 0) for some 1 ≤i ≤n.
The S-rigidity map F G,S : R3n →Rm+n is defined by F G,S = (F G, ΘS). The rigidity matrix RS(G, p) = R(G, p) S(G, p) for the framework (G, p) on S is (up to scaling) the Jacobian matrix of F G,S evaluated at the point p. It is shown in that the null space of RS(G, p) is the space of infinitesimal flexes of (G, p) on S. This allows us to characterise infinitesimal rigidity in terms of RS(G, p).
Stress matrices and global rigidity of frameworks on surfaces 5 Theorem 7 () Let (G, p) be a framework on S. Then (G, p) is infinitesimally rigid on S if and only if rank RS(G, p) = 3n −ℓ.
Theorem 7 implies that the (redundant) rigidity of a generic framework (G, p) on S depends only on the graph G. Hence we say that G is (redundantly) rigid on S if some, or equivalently every, generic realisation of G on S is (redundantly) rigid.
We close this section by pointing out that Theorem 4 implies that a graph which is (redundantly) rigid on some family of concentric cylinders, is (redundantly) rigid on all families of concentric cylinders irrespective of their radii. We do not know if analogous results hold for families of concentric cones or ellipsoids.
4 Stresses and stress matrices In this section we develop the notion of an equilibrium stress for a framework on our surface S and show that if (G, p) is ‘fully realised’ on S and has a maximum rank positive semi-definite stress matrix then every equivalent framework on S is an affine image of (G, p).
A stress for a framework (G, p) on S is a pair (ω, λ), where ω : E →R and λ : V →R. A stress (ω, λ) is an equilibrium stress if it belongs to the cokernel of RS(G, p). Thus (ω, λ) is an equilibrium stress for (G, p) on S if and only if n X j=1 ωij(p(vi) −p(vj)) + λisi(p(vi)) = 0 for all 1 ≤i ≤n, (4.1) where si(p(vi)) is as defined in Equation (3.2), ωij is taken to be equal to ωe if e = vivj ∈E and to be equal to 0 if vivj ̸∈E. We can think of ω as a weight function on the edges and λ as a weight function on the vertices. Note that, if the rows of RS(G, p) are linearly independent, then the only equilibrium stress for (G, p) is the all-zero equilibrium stress.
Given a stress (ω, λ) for a framework (G, p) on S we define: Ω= Ω(ω) to be the n×n symmetric matrix with off-diagonal entries −ωij and diagonal entries P j ωij; Λ = Λ(λ) to be the n×n diagonal matrix with diagonal entries λ1, λ2, . . . , λn; and ∆= ∆(λ, r) to be the n × n diagonal matrix with diagonal entries λ1r1, λ2r2, . . . , λnrn. The stress matrix associated to (ω, λ) on S is the 3n × 3n symmetric matrix ΩS(ω, λ) = Ω+ Λ 0 0 0 Γ 0 0 0 Σ where: Γ = Ω+ Λ if S ∈{Y, C} and Γ = Ω+ αΛ if S = E; Σ = Ωif S = Y, Σ = Ω−∆if S = C, and Σ = Ω+ βΛ if S = E. Our next result, which follows immediately from the definition of an equilibrium stress, tells us how we can use ΩS(ω, λ) to determine if (ω, λ) is an equilibrium stress for (G, p) on S.
Lemma 1 Let (G, p) be a framework on S with p(vi) = (xi, yi, zi) and let Π = x1 . . . xn 0 . . . 0 0 . . . 0 0 . . . 0 y1 . . . yn 0 . . . 0 0 . . . 0 0 . . . 0 z1 . . . zn .
Then (ω, λ) is an equilibrium stress for (G, p) on S if and only if Π ΩS = 0.
6 B. Jackson, A. Nixon We next define the configuration matrix CS(G, p) for a framework (G, p) on S by modifying the above matrix Π as follows: CS(G, p) = x1 . . . xn 0 . . . 0 0 . . . 0 0 . . . 0 y1 . . . yn 0 . . . 0 0 . . . 0 0 . . . 0 z1 . . . zn y1 . . . yn 0 . . . 0 0 . . . 0 0 . . . 0 x1 . . . xn 0 . . . 0 0 . . . 0 0 . . . 0 1 . . . 1 if M = Y, CS(G, p) = x1 . . . xn 0 . . . 0 0 . . . 0 0 . . . 0 y1 . . . yn 0 . . . 0 0 . . . 0 0 . . . 0 z1 . . . zn y1 . . . yn 0 . . . 0 0 . . . 0 0 . . . 0 x1 . . . xn 0 . . . 0 if M = C, and CS(G, p) = Π if M = E. We can use the configuration matrix to obtain an upper bound on the rank of a stress matrix.
Lemma 2 Let (ω, λ) be an equilibrium stress for a framework (G, p) on S. Then each row of CS(G, p) belongs to the cokernel of ΩS(ω, λ), rank ΩS(ω, λ) ≤3n −rank CS(G, p) and, if equal-ity holds, then the rows of CS(G, p) span the cokernel of ΩS(ω, λ).
Proof Equation (4.1) and the definitions of ΩS(ω, λ) and CS(G, p) imply that CS(G, p) ΩS(ω, λ) = 0.
Thus each row of CS(G, p) belongs to the cokernel of ΩS(ω, λ). Hence dim coker ΩS(ω, λ) ≥rank CS(G, p) and we have rank ΩS(ω, λ) = 3n−dim coker ΩS(ω, λ) ≤3n−rank CS(G, p). Furthermore, if equality holds, then coker ΩS(ω, λ) is equal to the row space of CS(G, p).
⊓ ⊔ We next use Lemma 2 to show that, if a framework (G, p) on S has an equilibrium stress (ω, λ) whose associated stress matrix has maximum rank, then every framework (G, q) on S which has (ω, λ) as an equilibrium stress can be obtained from (G, p) by an affine transformation.
Lemma 3 Let (G, p) and (G, q) be frameworks on S and let (ω, λ) be an equilibrium stress for both (G, p) and (G, q). Suppose that rank ΩS(ω, λ) = 3n −rank CS(G, p). Then, for some fixed a, b, c, d, e, f ∈R, we have q(vi) = a b 0 c d 0 0 0 e p(vi) + 0 0 f for all 1 ≤i ≤n, (4.2) where f = 0 if S ∈{C, E} and b = c = 0 if S = E.
Proof Lemma 2 implies that the rows of CS(G, p) span the cokernel of ΩS(ω, λ), and that each row of CS(G, q) belongs to the cokernel of ΩS(ω, λ). It follows that each row of CS(G, q) is a linear combination of the rows of CS(G, p). The lemma now follows from the structure of the matrices CS(G, p) and CS(G, q).
⊓ ⊔ Stress matrices and global rigidity of frameworks on surfaces 7 We will say that (G, q) is an S-affine image of (G, p) if it satisfies the conclusion of Lemma 3.
Our next result gives a converse to Lemma 3.
Lemma 4 Let (G, p) and (G, q) be frameworks on S such that (G, q) is an S-affine image of (G, p).
Then every equilibrium stress (ω, λ) for (G, p) is an equilibrium stress for (G, q).
Proof Since (G, q) is an S-affine image of (G, p), we have q(vi) = Ap(vi) + t for some fixed A, t satisfying the conclusion of Lemma 3, and all 1 ≤i ≤n. Hence X j ωij(q(vi) −q(vj)) + λisi(q(vi)) = X j ωijA(p(vi) −p(vj)) + λisi(Ap(vi) + t) = A X j ωij(p(vi) −p(vj)) + λisi(p(vi)) , since si(Ap(vi) + t) = Asi(p(vi)) by the definitions of si, A, t. The lemma now follows by applying Equation (4.1).
⊓ ⊔ A framework (G, p) on S is fully realised on S if the rows of its configuration matrix are linearly independent i.e. we have rank CS(G, p) = µ where µ = 6 if S = Y; 5 if S = C; 3 if S = E.
(4.3) It can be seen that (G, p) is fully realised on S if and only if its points do not all lie on: a plane containing or perpendicular to the z-axis when S = Y; a plane containing the z-axis when S = C; one of the planes x = 0, y = 0 or z = 0 when S = E.
We will next use a similar argument to that used by Connelly in to show that, if (G, p) has a positive semi-definite stress matrix of maximum rank then any equivalent framework is an S-affine image of (G, p).
The energy function associated to a stress (ω, λ) for a framework (G, q) and a family of concentric surfaces S is defined as Eω,λ,S(q) = X 1≤i 0 so that rank ΩSq(ω′′, λ′′) = 3n −µ, and ω′′ f ̸= 0 for all f ∈E with ω′ f ̸= 0. This contradicts the choice of (ω′, λ′).
⊓ ⊔ We can now obtain our result on generic 1-extensions.
Theorem 10 Suppose (G, p) is an infinitesimally rigid framework on S, for some S ∈{Y, E}, and (ω, λ) is an equilibrium stress for (G, p) with rank ΩS(ω, λ) = 3n −µ. Let G′ = (V ′, E′) be a 1-extension of G and q : V ′ →R3 such that q is generic in R3(n+1). Then (G′, q) is infinitesimally rigid on Sq and has an equilibrium stress (ω′, λ′) with rank ΩSq(ω′, λ′) = 3(n + 1) −µ.
Proof We may assume that p is a generic point in R3n and that ωe ̸= 0 for all e ∈E by Lemmas 9 and 10. We can now use Lemma 8 to deduce that there exists a map p∗: V ′ →R3 such that (G, p∗) is infinitesimally rigid on Sp∗with an equilibrium stress (ω∗, λ∗) for (G′, p∗) on Sp∗such that rank ΩSp∗(ω∗, λ∗) = 3(n + 1) + 3. The theorem now follows by another application of Lemmas 9 and 10.
⊓ ⊔ 7 Global rigidity on concentric cylinders In this section we apply our results to make progress on the conjectured characterisation of global rigidity on concentric cylinders given in [15, Conjecture 9.1], see also [17, Conjecture 5.7].
Conjecture 2 Suppose (G, p) is a generic framework on a family of concentric cylinders Y. Then (G, p) is globally rigid if and only if G is a complete graph on at most four vertices, or G is 2-connected and redundantly rigid on Y.
We have seen that the redundant rigidity of G on Y is independent of the radii of the cylinders in Y. Thus Conjecture 2 would imply that the global rigidity of a generic realisation of G on a family of concentric cylinders is also independent of the radii of the cylinders.
Theorem 6 shows that the combinatorial conditions given in Conjecture 2 are necessary for global rigidity. We could try to demonstrate sufficiency using a similar proof technique to that of Theorem 2. This would involve two steps: (i) a graph theoretic step obtaining a recursive construction for Stress matrices and global rigidity of frameworks on surfaces 15 2-connected, redundantly rigid graphs; (ii) a geometric step showing that each operation used in the recursive construction preserves global rigidity. Part (i) would be resolved by the following conjecture (which uses the base graphs K5−e, H1, H2 and the operations of 1-, 2- and 3-join illustrated in Figures 1 and 2).
Conjecture 3 Suppose G is a 2-connected graph which is redundantly rigid on some (or equivalently every) family of concentric cylinders. Then G can be obtained from either K5 −e, H1 or H2 by recursively applying the operations of edge addition, 1-extension, and 1-, 2- and 3-join.
The results of verify the special case of this conjecture when |E| = 2|V | −1 i.e. E is a circuit in the generic rigidity matroid for the cylinder.
v5 v1 v3 v2 v4 (a) v4 v1 v5 v2 v6 v3 (b) v4 v1 v5 v2 v6 v3 v7 (c) Fig. 1 The graphs K5 −e, H1 and H2.
a b b1 a1 b2 a2 d2 c2 a b b1 a1 d1 c1 b2 a2 d2 c2 v1 v2 b1 b2 c1 a1 c2 a2 a2 c2 b2 a1 c1 b1 Fig. 2 The 1-, 2- and 3-join operations. The 1- and 2-join operations form the graphs in the centre by merging a1 and a2 into a, and b1 and b2 into b.
16 B. Jackson, A. Nixon We close by showing that all graphs constructed from our base graphs using the edge addition and 1-extension operations are generically globally rigid on concentric cylinders with algebraically independent radii.
Theorem 11 Suppose G is a graph on n vertices which can be constructed from K5 −e, H1, or H2 by a sequence of 1-extensions and edge additions. Then (G, p) is globally rigid on Yp for all generic p ∈R3n.
Proof We use induction on n to show that (G, p) is infinitesimally rigid on Yp and has an equilibrium stress (ω, λ) with rank ΩYp(ω, λ) = 3n −6. The result will then follow from Theorem 9. The base case of the induction is when G ∈{K5 −e, H1, H2}. We construct a particular realisation (G, p) for each such G which is infinitesimally rigid on Yp and has an equilibrium stress with a full rank stress matrix in Appendix A. We may deduce that the same properties hold for all generic p by applying Lemmas 9 and 10. To complete the induction we need to show that the 1-extension and edge addition operations preserve the properties of infinitesimal rigidity and having a maximum rank stress matrix. This is trivially true for edge addition. It holds for 1-extension by Theorem 10.
⊓ ⊔ We conjecture that Theorem 11 can be strengthened to show that, if G can be constructed as in Theorem 11 and (G, p) is a generic framework on any family of concentric cylinders Y, then (G, p) is globally rigid on Y.
8 Closing Remarks 1. We would like to show that Theorem 9 holds for all of our surfaces S rather than just surfaces induced by generic points in R3. This would follow from Conjecture 1, which would in turn follow from Lemma 3 if we could show that equivalent generic frameworks on S must have the same equilibrium stresses. To date we have only been able to prove the following partial result.
Theorem 12 Let (G, p0) be a generic framework on S and (ω, λ) be an equilibrium stress for (G, p0).
Let (G, q0) be equivalent to (G, p0). Then (ω, λ′) is an equilibrium stress for (G, q0) for some λ′ ∈Rn.
The proof of Theorem 12 is given in Appendix B.
2. It follows from and that a generic framework in Rd with n ≥d+2 vertices is globally rigid if and only if it has a stress matrix of rank n −d −1. It is conceivable that the stress matrix condition given in Theorem 9 provides a necessary, as well as a sufficient, condition for the global rigidity of a generic framework on S whenever the framework has at least 7 −ℓvertices. The following examples indicate why we need this lower bound on n.
The smallest redundantly rigid graph on the cone is K5, but no framework (K5, p) on C can have a stress matrix with the maximum possible rank of 3n −µ = 10. To see this consider a generic p ∈R15. Since every equilibrium stress for (K5, p) in R3 is an equilibrium stress for (K5, p) on Cp, and since the spaces of equilibrium stresses for (K5, p) in R3 and on Cp are both 1-dimensional, these spaces are the same. This implies that every equilibrium stress (ω, λ) for (K5, p) has λ = 0 and rank Ω(ω) ≤3. Hence rank ΩCp(ω, λ) ≤9. On the other hand, (K5, p) is globally rigid on Cp for all p.
Similarly, the smallest redundantly rigid graph on the ellipsoid is K6 −{e, f} for two nonadja-cent edges e, f, but no framework (K6 −{e, f}, p) on E can have a stress matrix with the maximum Stress matrices and global rigidity of frameworks on surfaces 17 possible rank of 3n −µ = 15. (We do not know whether every generic framework (K6 −{e, f}, p) on Ep is globally rigid.) Acknowledgements We would like to thank the School of Mathematics, University of Bristol for providing partial financial support for this research. We would also like to thank Lee Butler for helpful discussions concerning semi-algebraic sets and Bob Connelly for many helpful discussions.
References 1. T. Abbott, Generalizations of Kempe’s universality theorem, Master’s thesis, Massachusetts Institute of Tech-nology (2008).
2. S. Basu, R. Pollack and M.-F. Roy, Algorithms in Real Algebraic Geometry. Springer, 2004.
3. J. Bochnak, M. Coste, and M.-F. Roy, Real Algebraic Geometry. Springer-Verlag, Berlin, 1998.
4. A. Berg and T. Jord´ an, Algorithms for graph rigidity and scene analysis, in: Proceedings of the 11th Annual European Symposium on Algorithms 2003, Springer Lecture Notes in Computer Science, vol. 2832, 2003, 78–89.
5. R. Connelly, Rigidity and energy, Invent. Math. 66 (1982) 11-33.
6. R. Connelly, On generic global rigidity, in Applied Geometry and Discrete Mathematics, DIMACS Ser. Discrete Math, Theoret. Comput. Scie 4 (1991) 147-155.
7. R. Connelly, Generic global rigidity, Discrete Comput. Geom. 33 (2005) 549-563.
8. R. Connelly and W. Whiteley, Global rigidity: the effect of coning, Discrete Comput. Geom. 43 4 (2010) 717-735 9. S. Gortler, A. Healy and D. Thurston, Characterizing generic global rigidity, Amer. J. Math. 132 4 (2010) 897-939.
10. B. Hendrickson, Conditions for Unique Graph Realisations, SIAM J. Comput. 21 1 (1992) 65-84.
11. G. Laman, On Graphs and the Rigidity of Plane Skeletal Structures, J. Engrg. Math. 4 (1970) 331-340.
12. B. Jackson and T. Jord´ an, Connected Rigidity Matroids and Unique Realisations of Graphs, J. Comb. Theory B 94 (2005) 1-29.
13. B. Jackson, T. Jord´ an and Z. Szabadaka, Globally Linked Pairs of Vertices in Equivalent Realizations of Graphs, Disc. Comput. Geom. 35 (2006) 493-512.
14. B. Jackson and P. Keevash, Necessary conditions for global rigidity of direction-length frameworks, Disc.
Comput. Geom., 46 (2011), 72-85.
15. B. Jackson, T. McCourt and A. Nixon, Necessary conditions for the generic global rigidity of frameworks on surfaces, Disc. Comput. Geom., 52:2 (2014), 344-360.
16. A. Lee and I. Streinu, Pebble game algorithms and sparse graphs, Discrete Math., 308 8 (2008) 1425-1437.
17. A. Nixon, A constructive characterisation of circuits in the simple (2, 2)-sparsity matroid, European J. of Combin., 42 (2014) 92-106.
18. A. Nixon, J. Owen and S. Power, Rigidity of frameworks supported on surfaces, SIAM J. Discrete Math. 26 (2012) 1733-1757.
19. A. Nixon, J. Owen and S. Power, A characterization of generically rigid frameworks on surfaces of revolution, SIAM J. Discrete Math. 28:4 (2014) 2008-2028.
20. J. Saxe, Embeddability of weighted graphs in k-space is strongly NP-hard, In Seventeenth Annual Allerton Conference on Communication, Control, and Computing, Proceedings of the Conference held in Monticello, Ill., October 1012, 1979.
21. M. Spivak, A comprehensive introduction to differential geometry: volume 1. Publish or perish, inc., 1999.
22. W. Whiteley, Some matroids from discrete applied geometry, In Matroid theory (Seattle, WA, 1995), Contemp.
Math., 197, Amer. Math. Soc., Providence, RI, 1996.
Appendix A: Base graphs We define a framework (G, p) for G ∈{K5 −e, H1, H2} which is infinitesimally rigid on Yp and has a self-stress (ω, λ) on Yp with maximum rank stress matrix. We will use the labeling of the vertices given in Figure 1 and adopt the convention that ωij is the weight on the edge vivj in ω and λi is the weight on the vertex vi in λ.
18 B. Jackson, A. Nixon Case 1: G = K5 −e Let (G, p) and (ω, λ) be defined by p(v1) = (0, 1, 0), p(v2) = (1, 1, 1), p(v3) = (−1, −2, −1), p(v4) = (2, 3, 4), p(v5) = (5, 1, −1), (ω12, ω13, ω14, ω15, ω23, ω24, ω25, ω35, ω45) = (−369, 192, 153, 51, −96, −279, −138, 32, 45) and (λ1, λ2, λ3, λ4, λ5) = (−270, −270, −192, 54, −6)).
It is straightforward to check that rank RYp(G, p) = 13, that (ω, λ) · RYp(G, p) = 0 and that rank ΩYp(ω, λ) = 9.
Case 2: G = H1 Let (G, p) and (ω, λ) be defined by p(v1) = (0, 1, 0), p(v2) = (3, 1, 0), p(v3) = (1, 4, 1), p(v4) = (1, 2, 2), p(v5) = (2, 2, 3), p(v6) = (6, 0, 2), (ω12, ω13, ω15, ω23, ω24, ω25, ω26, ω35, ω36, ω45, ω56) = (41, −246, 369, −123, 30, 48, 60, 50, −40, 492, 56) and (λ1, λ2, λ3, λ4, λ5, λ6) = (−123, −39, 30, 123, −102, 28).
It is straightforward to check that rank RYp(G, p) = 16, that (ω, λ) · RYp(G, p) = 0 and that rank ΩYp(ω, λ) = 12.
Case 3: G = H2 Let (G, p) and (ω, λ) be defined by p(v1) = (0, 1, 0), p(v2) = (3, 1, 0), p(v3) = (1, 4, 1), p(v4) = (1, 2, 2), p(v5) = (2, 2, 3), p(v6) = (6, 0, 2), p(v7) = (3, 4, 3), (ω12, ω13, ω15, ω23, ω24, ω25, ω35, ω36, ω37, ω45, ω56, ω57, ω67) = (−58, 348, −522, −108, −24, −40, 14, 21, −696, 56, 588, −42) and (λ1, λ2, λ3, λ4, λ5, λ6, λ7) = (−174, −6, 24, 174, 372, −28, −252).
It is straightforward to check that rank RYp(G, p) = 19, that (ω, λ) · RYp(G, p) = 0 and that rank ΩYp(ω, λ) = 15.
Stress matrices and global rigidity of frameworks on surfaces 19 Appendix B: Proof of Theorem 12 First we recall some basic concepts from differential and algebraic geometry, and prove a key technical result, Proposition 2, which extends Proposition 1 to the case when the domain of f is an algebraic set.
Let M be a smooth manifold and f : M →Rm be a smooth map. Then x ∈M is a regular point of f if d f|x has maximum rank, and f(x) is a regular value of f if, for all y ∈f −1(f(x)), y is a regular point of f.
Lemma 11 For i = 1, 2, let Mi be an open subset of Rn, pi ∈Mi, and fi : Mi →Rm be a smooth map with rank d fi|pi = m and f1(p1) = f2(p2). Then there exist open neighbourhoods N1 of p1, N2 of p2, and a diffeomorphism g : N1 →N2 such that f2(g(x)) = f1(x) for all x ∈N1.
Proof We first consider the case when m = n. By the Inverse Function Theorem there exist neigh-bourhoods ˜ Ni ⊆Mi of pi such that fi maps ˜ Ni diffeomorphically onto fi( ˜ Ni) for i = 1, 2. Let W = f1( ˜ N1) ∩f2( ˜ N2) and then let Ni = f −1 i (W) for i = 1, 2. We have f1(N1) = W = f2(N2). Thus we may choose g = f −1 2 ◦f1 and find f2(g(x)) = f2(f −1 2 (f1(x))) = f1(x) for all x ∈N1.
We next consider the case when m < n. Let Fi : Mi →Rm × Rn−m be defined by Fi(x) = (fi(x), xm+1, xm+2, . . . , xn). Then rank dFi|pi = n. By the Inverse Function Theorem there exist neighbourhoods ˜ Ni ⊆Mi of pi such that Fi is a diffeomorphism from ˜ Ni to Fi( ˜ Ni) ⊆Rm × Rn−m.
Let Fi( ˜ Ni) = Ui × Vi where Ui ⊆Rm and Vi ⊆Rn−m. Then Vi is diffeomorphic to Rn−m for i = 1, 2 so we can choose a diffeomorphism h : V1 →V2 such that h(p1) = p2, where pi is the projection of pi onto its last n−m coordinates. Let ι be the identity map on U1 and let H = (ι, h) : U1×V1 →U1×V2.
Let F ′ 1 = H ◦F1. Then we have F ′ 1(p1) = (f1(p1), h(p1)) = (f2(p2), p2) = F2(p2). By the previous paragraph there exist neighbourhoods Ni ⊆˜ Ni of pi and a diffeomorphism g : N1 →N2 ⊆Rn such that F2(g(x)) = F ′ 1(x) for all x ∈N1. Since F ′ 1(x) = (f1(x), h(x)) and F2(g(x)) = (f2(g(x)), g(x)) we have f1(x) = f2(g(x)) for all x ∈N1.
⊓ ⊔ Let K be a field such that Q ⊆K ⊆R. A set W ⊆Rn is an algebraic set defined over K if W = {x ∈Rn : Pi(x) = 0 for all 1 ≤i ≤n} where Pi ∈K[X1, . . . , Xn] for 1 ≤i ≤m. An algebraic set W is irreducible if it cannot be expressed as the union of two algebraic proper subsets defined over R. The dimension of W, dim W, is the largest integer t for which W has an open subset homeomorphic to Rt. A point p ∈W is generic over K if every h ∈K[X] satisfying h(p) = 0 has h(x) = 0 for all x ∈W.
Lemma 12 () Let K be a field with Q ⊆K ⊆R, W ⊆Rn be an algebraic set defined over K and p ∈W. Then dim W ≥td [K(p) : K]. Furthermore, if W is irreducible and dim W = td [K(p) : K], then p is a generic point of W.
Note that, if (G, p) is a generic framework on S, then Lemma 12 implies that p is a generic point of the irreducible algebraic set S1 × S2 × . . . × Sn defined over Q(r) in R3n.
A set A ⊆Rn is a semi-algebraic set defined over K if it can be expressed as a finite union of sets of the form {x ∈Rn : Pi(x) = 0 for 1 ≤i ≤s and Qj(x) > 0 for 1 ≤j ≤t}, where Pi, Qj ∈K[X1, . . . , Xn] for 1 ≤i ≤s and 1 ≤j ≤t. It is easy to see that the family of semi-algebraic sets defined over K is closed under union and intersection. A deeper result is that if 20 B. Jackson, A. Nixon A ⊆Rn is a semi-algebraic set defined over K and f : A →Rm is a map in which each coordinate is a polynomial with coefficients in K, then f(A) is a semi-algebraic set defined over K. Another result we shall need is that a semi-algebraic set A can be partitioned into a finite number of semi-algebraic subsets C1, C2, . . . Ct, called cells, such that, for all 1 ≤i ≤t, Ci is diffeomorphic to Rmi for some integer mi ≥0 (where R0 is taken to be a single point). The dimension of A is the largest integer t for which A has an open subset homeomorphic to Rt. The Zariski closure, A∗, of A is the smallest algebraic set defined over R which contains A. It is known that A∗is an algebraic set defined over L, for some finite field extension L of K, and that dim A = dim A∗. We refer the reader to [2,3] for more information on semi-algebraic sets.
We can now obtain our desired extension of Proposition 1.
Proposition 2 Let K be a field with Q ⊆K ⊆R, W ⊂Ra be an irreducible algebraic set defined over K of dimension n, and f : W →Rb be a function where each coordinate is a polynomial with coefficients in K. Suppose that the maximum rank of the differential of f is m, and that p0 ∈W with td [K(p0) : K] = n. Then rank d f|p0 = m. Furthermore, if q0 ∈W and f(p0) = f(q0), then there exist open neighbourhoods Np0 of p0 and Nq0 of q0 in W and a diffeomorphism g : Nq0 →Np0 such that g(q0) = p0 and, for all q ∈Nq0, f(g(q)) = f(q).
Proof We first show that rank d f|p0 = m = rank d f|q0, and that there exist open neighbourhoods Mp0 of p0 and Mq0 of q0 in W such that f(Mp0) = f(Mq0) and f(Mp0) is diffeomorphic to Rm. We then complete the proof by applying Lemma 11.
By Lemma 12, p0 is a generic point of W. We can now use [9, Lemma 2.7 and Proposition 2.32] to deduce that f(p0) is a regular value of f. In particular, we have rank d f|p0 = m = rank d f|q0. The Constant Rank Theorem (see, for example, [21, Theorem 9]) now implies that we can choose disjoint open balls B(p1, ϵ) and B(q1, δ) in Ra such that: p0 ∈B(p1, ϵ) ∩W; q0 ∈B(q1, δ) ∩W; p1, q1 ∈Qa; ϵ, δ ∈Q; both B(p1, ϵ) ∩W and B(q1, δ) ∩W are diffeomorphic to Rn; both f(B(p1, ϵ) ∩W) and f(B(q1, δ) ∩W) are diffeomorphic to Rm.
Let Up0 = B(p1, ϵ)∩W and Uq0 = B(q1, ϵ)∩W. Since Up0 and Uq0 are both semi-algebraic defined over K, f(Up0) and f(Uq0) are both semi-algebraic defined over K, and hence T = f(Up0) ∩f(Uq0) is semi-algebraic defined over K. The facts that f is a polynomial map, td[K(p0) : K] = n and rank d f|p0 = m ≤n imply that td[K(f(p0)) : K] = m, see for example [13, Lemma 3.1]. Let C1, C2, . . . , Ct be a cell decomposition of T with f(p0) ∈C1, and let C∗ 1 be the Zariski closure of C1. Then C∗ 1 is an algebraic set defined over some finite field extension L of K. Since f(p0) ∈C∗ 1, Lemma 12 gives dim C1 = dim C∗ 1 ≥td[L(f(p0)) : L] = td[K(f(p0)) : K] = m.
Since C1 ⊆f(Up0) and f(Up0) is diffeomorphic to Rm, we must have dim C1 = m. We can now take Mp0 = f −1(C1) ∩Up0 and Mq0 = f −1(C1) ∩Uq0. Then f(Mp0) = C1 = f(Mq0) and C1 is diffeomorphic to Rm.
The proposition now follows from Lemma 11 by choosing M1 = Mp0, M2 = Mq0, and fi = f|Mi for i = 1, 2.
⊓ ⊔ Proof of Theorem 12 Let F = F G,S, W = S1 × S2 × . . . Sn and put f = F|W. By Proposition 2 there exist open neighbourhoods Np0 of p0 and Np0 of q0 in W and a diffeomorphism g : Nq0 →Np0 such that g(q0) = p0 and, for all q ∈Nq0, f(g(q)) = f(q). Taking differentials at q0 we obtain d fq0(q) = d fp0(dgq0(q)) for all q in the tangent space TWq0. Since the Jacobian matrix of F evaluated Stress matrices and global rigidity of frameworks on surfaces 21 at p is 2RS(G, p) and d fp(x) = dFp(x) for all p ∈W and all x ∈TWp, we can rewrite this equation as RS(G, q0) q = RS(G, p0) dgq0(q). Thus (ω, λ) RS(G, q0) q = (ω, λ) RS(G, p0) dgq0(q). Since (ω, λ) is an equilibrium stress for (G, p0) we have (ω, λ) RS(G, q0) q = (ω, λ) RS(G, p0) dgq0(q) = 0 dgq0(q) = 0 for all q ∈TWq0. Hence (ω, λ) RS(G, q0) ∈TW⊥ q0, the orthogonal complement of TWq0 in R3n.
Since a vector x ∈R3n belongs to TW⊥ q0 if and only if x = δS(G, q0) for some δ ∈Rn, we have (ω, λ) RS(G, q0) = (0, δ)RS(G, q0). Therefore (ω, λ′) is an equilibrium stress of (G, q0) for λ′ = λ−δ.
⊓ ⊔ |
187921 | https://www.youtube.com/watch?v=X7PWOyh1ZeQ | How to expand and simplify (a - b)^2
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For the following exercises, expand the binomial.
(12 − 4x)^2
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BinomialExpansion #Polynomials #Math
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Transcript:
what's going on it's Christina from the Glazer Tutoring company and today we are going to expand some binomials so in this case we have to expand the binomial the binomial that is given here is 12 - 4x and that 12 - 4x is being raised to the 2 power okay well what does expand the binomial mean well first off binomial by in medical terms or scientific terms mathematical terms by means two and since we're in the realm of pooms a binomial just means that you have a mathematical expression that just has two different terms in here my two terms are just the 12 right this doesn't have any x value or x squ value so that's one term and then this 4X is the other term so I got two terms and they're being raised to the second power or being squared we must expand this so expansion just means to make it bigger right expand on that we're going to I guess enlarge in it or make it bigger basically we just have to write this the longer way right now this all comes from what the exponent is now in this case it's a squared value and if you have binomials or pols in parentheses and it's being squared or cubed or raised to the 4th that exponent is always going to tell you how many you have of your polom or your binomial so in this case how many 12 - 4 X's do we have yeah it said that it was being raised the second so we have two of them so that's your expansion we must write out how many total we have so let's go for it let's write out our first parentheses and here it is 12 - 4x and now let's just write out the second one so here's my next set of parentheses 12 - 4x and we're done with basically the expansion part but we actually have to do the math here right can't just write it out and not do the math the first thing is what operation is in the middle between these two binomials well I didn't say plus and I didn't say minus right if you're um if you're raising something to the second power or the third power it all comes down to multiplication so basically you're just taking these binomials and you're multiplying by them you know by by themselves so let's do it right you got the foil method you got graphs and charts and all that kind of crazy stuff but I think we can just be fair when I multiply polinomial or binomials I always keep fairness in mind you do all the heavy lifting on your first uh binomial or trinomial or whatever and once you get to the end of your last term all you got to do is then just simplify and call it a day so always start with one term at a time in this case I'm going to focus on the 12 and we must multiply but you will never multiply inside the same parentheses because it literally says subtraction so when we're multiplying we must go across the parentheses across the pond so we have to multiply 12 by 12 but as we're doing that first multiplication the 12 is being multiplied by the 12 this -4 says I want to be multiplied too that's where the fairness comes into play the 12 must be multiplied by the first term the 12 and you got to multiply that by the same term so the 12 has to be multiplied by the -4x let's get that down and then we'll come back and do the second term so 12 12 12 12 is 144 so we'll start it off with 144 now numbers first 12 -4 please take note of your signs because signs here is going to make all the difference so I have a 12 a -4 uh 12 a-4 is what 48 48 so Min - 48 this doesn't have any x value but this one does so I just bring that x value along for the ride and notice how I wrote These two numbers in different colors because just numbers only want to get simplified with just numbers and just X values want to get simplified with just One X value okay now we're basically done with the 12 so that's what the heavy lifting is all about once you focus on your first term you do all the math that one you don't even have to look at anymore and now we move on to the next one keep in mind that it's a negative so this -4x will get multiplied by the 12 as that's happening the other -4x is saying I want to get multiplied to so we got to be fair let's be fair we got to take that arrow and do it to the both of them so let's go for it -4x 12 that's the same as what it was before right 4-4 12 is -48 so - 48 this just has the x value there was no x value on the 12 so - 48x and now -4 a -4 is a positive 16 and now we got X values in both terms right so I'm going to collect One X here right this was x rais to the 1 and I'm going to collect another x value that's x to the 1st how many total X's do you have 1 2 so how many X's do I have I got two x's so x^ SAR now let's just simplify this so we can only simplify the X terms in the middle because those are the ones that are written in blue so 4 144 is just going to stay and now let's do the math here minus so - 48 minus 48 so that's what 96 96 x and then + 16 x^2 now this technically is a final answer but if you want to um put it into um standard notation always list your terms with the highest degree aka the biggest exponent so technically the 16 x^2 should go first then the Min - 96x because that has only one x value and since this 144 was a positive value there was no negative in front of it it would be plus 144 and that would be your standard notation and now we have ourselves an answer 16 x^2 - 96x + 144 and you are done what do you think hopefully this helped thank you so much for tuning in and yeah I look forward to helping you guys out I love helping you guys out I love doing these videos uh my brother and I we really do truly appreciate all you guys that have been with us and you know newcomers old comers whoever ever it is right if you've been with us for a while if you're just tuning in hi and thank you for stopping by check out the channel we got tons of videos we got physics uh chemistry and math videos on the channel and soon we'll have more subject as well so I'm super excited and good luck on those tests and quizzes all right have an awesome day and I'll talk to you soon bye-bye |
187922 | https://en.wiktionary.org/wiki/%E4%BA%94 | 五 - Wiktionary, the free dictionary
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Beginning
1 TranslingualToggle Translingual subsection
1.1 Han character
1.1.1 Derived characters
1.1.2 References
2 ChineseToggle Chinese subsection
2.1 Glyph origin
2.2 Etymology 1
2.2.1 Pronunciation
2.2.2 Definitions
2.2.3 See also
2.2.4 Compounds
2.2.5 Descendants
2.3 Etymology 2
2.3.1 Pronunciation
2.3.2 Definitions
2.3.3 Derived terms
2.4 References
3 JapaneseToggle Japanese subsection
3.1 Kanji
3.1.1 Readings
3.1.2 Compounds
3.2 Etymology 1
3.2.1 Pronunciation
3.2.2 Noun
3.2.2.1 Derived terms
3.3 Etymology 2
3.3.1 Pronunciation
3.3.2 Noun
3.3.2.1 Coordinate terms
3.4 Etymology 3
3.4.1 Pronunciation
3.4.2 Noun
4 KoreanToggle Korean subsection
4.1 Etymology
4.2 Pronunciation
4.3 Hanja
4.3.1 Compounds
4.4 References
5 VietnameseToggle Vietnamese subsection
5.1 Han character
5.1.1 Compounds
5.2 References
6 ZhuangToggle Zhuang subsection
6.1 Numeral
五
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From Wiktionary, the free dictionary
See also:伍and𢀑
五U+4E94, 五 CJK UNIFIED IDEOGRAPH-4E94 ← 亓 [U+4E93]CJK Unified Ideographs井 → [U+4E95]
㈤U+3224, ㈤ PARENTHESIZED IDEOGRAPH FIVE ← ㈣ [U+3223]Enclosed CJK Letters and Months㈥ → [U+3225]
㊄U+3284, ㊄ CIRCLED IDEOGRAPH FIVE ← ㊃ [U+3283]Enclosed CJK Letters and Months㊅ → [U+3285]
Wikimedia Commons has media related to:
五
Translingual
[edit]
| Stroke order |
| |
| Stroke order |
| |
Han character
[edit]
五 (Kangxi radical 7,二+2, 4 strokes, cangjie input 一木一 (MDM), four-corner1010 7, composition⿱一𫝀 or ⿻丅ユ or ⿻工𠃍)
Derived characters
[edit]
伍, 㕶, 鿉, 㐏, 吾, 忢, 𫰌, 𭜐, 𬂡, 𥄬, 𥐳, 𦙗, 𮛉, 𮡨, 𭈔, 𮤴, 𬃂, 𠄼, 𡟿, 𮐌, 𬁶, 𫡲, 𫡵, 𫇩, 𠄶, 𬂽, 𥄪, 𭘣, 𠵥, 𠄲, 𠄻, 𭱷
References
[edit]
Kangxi Dictionary: page 86, character 11
Dai Kanwa Jiten: character 257
Dae Jaweon: page 178, character 4
Hanyu Da Zidian (first edition): volume 1, page 11, character 3
Unihan data for U+4E94
Chinese
[edit]
Glyph origin
[edit]
| more ▼Historical forms of the character 五 |
| Shang | Western Zhou | Warring States | Shuowen Jiezi (compiled in Han) | Liushutong (compiled in Ming) |
| Oracle bone script | Bronze inscriptions | Chu slip and silk script | Small seal script | Transcribed ancient scripts |
| | | | | |
| Oracle bone script j29736 j29737 j29738 j29739 j29740 j29741 j29742 j29743 j29744 j29745 j29746 j29747 j29748 j29749 j29750 j29751 j29752 j29753 j29754 j29755 j29756 j29757 j29758 j29759 j29760 j29761 j29762 j29763 j29764 j29765 j29766 j29767 j29768 j29769 j29770 j29771 j29772 j29773 j29774 j29775 j29776 j29777 j29778 j29779 j29780 j29781 Bronze inscriptions b19855 b19856 b19857 b19858 b19859 b19860 b19861 b19862 b19863 b19864 b19865 b19866 b19867 b19868 b19869 b19870 b19871 b19872 b19873 b19874 b19875 b19876 b19877 b19878 b19879 b19880 b19881 b19882 b19883 Small seal script s10919 Transcribed ancient scripts L34450 L34451 L14821 L14822 L14823 L14824 L14825 L14826 L14827 L14828 |
| ExpandReferences: Mostly from Richard Sears' Chinese Etymology site (authorisation), which in turn draws data from various collections of ancient forms of Chinese characters, including: Shuowen Jiezi (small seal), Jinwen Bian (bronze inscriptions), Liushutong (Liushutong characters) and Yinxu Jiaguwen Bian (oracle bone script). |
show ▼Characters in the same phonetic series (五) (Zhengzhang, 2003)
| | Old Chinese |
--- |
| 吾 | ŋraː, ŋaː |
| 衙 | ŋraː, ŋa, ŋaʔ |
| 鼯 | ŋaː |
| 浯 | ŋaː |
| 珸 | ŋaː |
| 郚 | ŋaː |
| 齬 | ŋaː, ŋa, ŋaʔ |
| 鯃 | ŋaː |
| 娪 | ŋaː |
| 梧 | ŋaː |
| 峿 | ŋaː |
| 五 | ŋaːʔ |
| 伍 | ŋaːʔ |
| 寤 | ŋaːs |
| 啎 | ŋaːs |
| 晤 | ŋaːs |
| 悟 | ŋaːs |
| 逜 | ŋaːs |
| 窹 | ŋaːs |
| 捂 | ŋaːs |
| 鋙 | ŋa, ŋaʔ |
| 語 | ŋaʔ, ŋas |
| 圄 | ŋaʔ |
| 敔 | ŋaʔ |
Originally written 𠄡, consisting of 㐅 with a bar on the top and bottom. One hypothesis explains that this may have meant five because when counting on a single hand, one first counts to five and then crosses back the other way to ten.
In 韋, the upper component (𫝀) is the stylization of a footprint, hence it is similar but unrelated to 五.
Etymology 1
[edit]
| simp. and trad. | 五 |
| alternative forms | 伍financial 㐅 𠄡 𫝀 |
From Proto-Sino-Tibetanl-ŋaʔ(“five”).
Pronunciation
[edit]
Mandarin(Standard)(Pinyin): wǔ (wu 3)(Zhuyin): ㄨˇDuration: 1 second.0:01(Chengdu, Sichuanese Pinyin): wu 3(Xi'an, Guanzhong Pinyin): wù(Nanjing, Nanjing Pinyin): wǔ(Dungan, Cyrillic and Wiktionary): ву (vu, II)
Cantonese(Guangzhou–Hong Kong, Jyutping): ng 5Duration: 1 second.0:01(Dongguan, Jyutping++): m 5(Taishan, Wiktionary): m 2(Yangjiang, Jyutping++): ung 2
Gan(Wiktionary): ng 3 / u 3
Hakka(Sixian, PFS): ńg(Hailu, HRS): ng ˊ(Meixian, Guangdong): n 3
Jin(Wiktionary): vu 2
Northern Min(KCR): ngù
Eastern Min(BUC): ngô / ngū
Puxian Min(Pouseng Ping'ing): ngou 5 / gou 3
Southern Min(Hokkien, POJ): gō͘ / gǒ͘ / ňg / ngó͘ / gó͘ / gú(Teochew, Peng'im): ngou 6 / ngou 2 / u 2(Leizhou, Leizhou Pinyin): ngeu 6 / ngu 2
Southern Pinghua(Nanning, Jyutping++): ngu 5
Wu(Wugniu)
(Northern): 6 ng / 4 n / 4 ng / 6 m / 3 ng / 2 ng / 1 u / 5 u / 3 u / 1 ou / 4 wu / 2 wu
(Jinhua): 3 ng; 3 u
Xiang(Changsha, Wiktionary): u 3(Loudi, Wiktionary): u 3(Hengyang, Wiktionary): u 3 more ▼
Mandarin
(Standard Chinese)+
Hanyu Pinyin: wǔ
Zhuyin: ㄨˇ
Tongyong Pinyin: wǔ
Wade–Giles: wu 3
Yale: wǔ
Gwoyeu Romatzyh: wuu
Palladius: у(u)
Sinological IPA(key): /u²¹⁴/
(Chengdu)
Sichuanese Pinyin: wu 3
Scuanxua Ladinxua Xin Wenz: u
Sinological IPA(key): /vu⁵³/
(Xi'an)
Guanzhong Pinyin: wù
Sinological IPA(key): /u⁵³/
(Nanjing)
Nanjing Pinyin: wǔ
Nanjing Pinyin (numbered): wu 3
Sinological IPA(key): /u¹¹/
(Dungan)
Cyrillic and Wiktionary: ву (vu, II)
Sinological IPA(key): /vou⁵¹/
(Note: Dungan pronunciation is currently experimental and may be inaccurate.)
Cantonese
(Standard Cantonese, Guangzhou–Hong Kong)
Jyutping: ng 5
Yale: ńgh
Cantonese Pinyin: ng 5
Guangdong Romanization: ng 5
Sinological IPA(key): /ŋ̍¹³/
(Dongguan, Guancheng)
Jyutping++: m 5
Sinological IPA(key): /m̩¹³/
(Taishanese, Taicheng)
Wiktionary: m 2
Sinological IPA(key): /m̩⁵⁵/
(Yangjiang Yue, Jiangcheng)
Jyutping++: ung 2
Sinological IPA(key): /ʊŋ²¹/
Gan
(Nanchang)
Wiktionary: ng 3 / u 3
Sinological IPA(key): /ŋ̍²¹³/, /u²¹³/
Note:
ng3 - vernacular;
u3 - literary (e.g. 五香).
Hakka
(Sixian, incl. Miaoli and Neipu)
Pha̍k-fa-sṳ: ńg
Hakka Romanization System: ng ˋ
Hagfa Pinyim: ng 3
Sinological IPA: /ŋ̍³¹/
(Hailu, incl. Zhudong)
Hakka Romanization System: ng ˊ
Sinological IPA: /ŋ²⁴/
(Meixian)
Guangdong: n 3
Sinological IPA: /n̩³¹/
Jin
(Taiyuan)+
Wiktionary: vu 2
Sinological IPA (old-style): /vu⁵³/
Northern Min
(Jian'ou)
Kienning Colloquial Romanized: ngù
Sinological IPA(key): /ŋu⁴²/
Eastern Min
(Fuzhou)
Bàng-uâ-cê: ngô / ngū
Sinological IPA(key): /ŋou²⁴²/, /ŋu³³/
Note:
ngô - vernacular;
ngū - literary.
Puxian Min
(Putian)
Pouseng Ping'ing: ngou 5
Báⁿ-uā-ci̍: ngō
Sinological IPA(key): /ŋɔu¹¹/
(Xianyou)
Pouseng Ping'ing: ngou 5
Sinological IPA(key): /ŋɔu²¹/
(Putian)
Pouseng Ping'ing: gou 3
Báⁿ-uā-ci̍: gô
Sinological IPA(key): /kɔu⁴⁵³/
(Xianyou)
Pouseng Ping'ing: gou 3
Sinological IPA(key): /kɔu³³²/
Note:
ngou5 - vernacular;
gou3 - literary.
Southern Min
(Hokkien: Xiamen, Zhangzhou, Changtai, Zhangpu, Yongchun, Taipei, Kaohsiung, Tainan, Taichung, Hsinchu, Yilan, Sanxia, Magong, Kinmen, Penang, Singapore, Klang)
Pe̍h-ōe-jī: gō͘
Tâi-lô: gōo
Phofsit Daibuun: go
IPA (Zhangpu): /ɡɔu³³/
IPA (Changtai): /ɡeu²²/
IPA (Penang): /ɡɔ²¹/
IPA (Xiamen, Zhangzhou, Yongchun, Kinmen, Singapore): /ɡɔ²²/
IPA (Taipei, Kaohsiung, Tainan, Yilan): /ɡɔ³³/
IPA (Klang): /ɡɔ³¹/
(Hokkien: Quanzhou, Jinjiang, Nan'an, Hui'an, Lukang, Philippines)
Pe̍h-ōe-jī: gǒ͘
Tâi-lô: gǒo
IPA (Jinjiang, Lukang, Philippines): /ɡɔ³³/
IPA (Quanzhou, Nan'an, Hui'an): /ɡɔ²²/
(Hokkien: Longyan)
Pe̍h-ōe-jī: ňg
Tâi-lô: ňg
IPA (Longyan): /ŋ̍⁵³/
(Hokkien: Xiamen, Zhangzhou, Changtai, General Taiwanese, Penang)
Pe̍h-ōe-jī: ngó͘
Tâi-lô: ngóo
Phofsit Daibuun: ngor
IPA (Xiamen, Zhangzhou, Changtai, Taipei): /ŋɔ̃⁵³/
IPA (Kaohsiung): /ŋɔ̃⁴¹/
IPA (Penang): /ŋɔ⁴⁴⁵/
(Hokkien: Quanzhou, Jinjiang, Nan'an, Hui'an, Zhangpu)
Pe̍h-ōe-jī: gó͘
Tâi-lô: góo
Phofsit Daibuun: gor
IPA (Quanzhou, Jinjiang, Nan'an): /ɡɔ⁵⁵⁴/
IPA (Zhangpu): /ɡɔu⁵³/
IPA (Hui'an): /ɡɔ⁵⁴/
(Hokkien: Longyan)
Pe̍h-ōe-jī: gú
Tâi-lô: gú
Phofsit Daibuun: guo
IPA (Longyan): /ɡu²¹/
Note:
gō͘/gǒ͘/ňg - vernacular;
ngó͘/gó͘/gú - literary.
(_Teochew_)
_Peng'im_: ngou 6 / ngou 2 / u 2
_Pe̍h-ōe-jī-like_: ngŏu / ngóu / ú
Sinological IPA(key): /ŋou³⁵/, /ŋou⁵²/, /u⁵²/
Note:
ngou6 - vernacular;
ngou2, u2 - literary.
(_Leizhou_)
_Leizhou Pinyin_: ngeu 6 / ngu 2
Sinological IPA: /ŋɛu³³/, /ŋu³¹/
Note:
ngeu6 - vernacular;
ngu2 - literary.
Southern Pinghua
(Nanning Pinghua, Tingzi)
Jyutping++: ngu 5
Sinological IPA(key): /ŋu²⁴/
Wu
(Northern: Shanghai, Jiading, Suzhou, Hangzhou, Ningbo)
Wugniu: 6 ng
MiniDict: ng 去
Wiktionary Romanisation (Shanghai): 3 hhngg
Sinological IPA (Shanghai): /ŋ̍²³/
Sinological IPA (Jiading): /ŋ̍¹³/
Sinological IPA (Suzhou): /ŋ̍²³¹/
Sinological IPA (Hangzhou): /ŋ̍¹¹³/
Sinological IPA (Ningbo): /ŋ̍¹¹³/
(Northern: Chongming)
Wugniu: 4 n
MiniDict: n 上
Sinological IPA (Chongming): /ɦn̩²⁴²/
(Northern: Jiading, Chuansha, Yixing, Jiaxing, Haining, Haiyan, Deqing, Xiaoshan, Shaoxing, Zhoushan)
Wugniu: 4 ng
MiniDict: ng 上
Sinological IPA (Jiading): /ŋ̍¹³/
Sinological IPA (Chuansha): /ŋ̍²¹³/
Sinological IPA (Yixing): /ŋ̍³³/
Sinological IPA (Jiaxing): /ŋ̍²¹³/
Sinological IPA (Haining): /ŋ̍⁴²/
Sinological IPA (Haiyan): /ŋ̍²³²/
Sinological IPA (Deqing): /ŋ̍³¹/
Sinological IPA (Xiaoshan): /ŋ̍²¹³/
Sinological IPA (Shaoxing): /ŋ̍¹¹³/
Sinological IPA (Zhoushan): /ŋ̍²⁴/
(Northern: Kunshan)
Wugniu: 6 m
Sinological IPA (Kunshan): /m̩²¹³/
(Northern: Changzhou, Tongxiang)
Wugniu: 3 ng
MiniDict: 'ng 上
Sinological IPA (Changzhou): /ŋ̍⁴⁵/
Sinological IPA (Tongxiang): /ŋ̍⁵³/
(Northern: Cixi)
Wugniu: 2 ng
Sinological IPA (Cixi): /ŋ̍¹⁴/
(Northern: Shanghai, Chongming)
Wugniu: 1 u
MiniDict: u 平
Wiktionary Romanisation (Shanghai): 1 u
Sinological IPA (Shanghai): /u⁵³/
Sinological IPA (Chongming): /ʔu⁵⁵/
(Northern: Shanghai, Jiading)
Wugniu: 5 u
MiniDict: u 去
Wiktionary Romanisation (Shanghai): 2 u
Sinological IPA (Shanghai): /u³⁴/
Sinological IPA (Jiading): /u³⁴/
(Northern: Jiading, Changzhou, Jingjiang, Tongxiang, Hangzhou)
Wugniu: 3 u
MiniDict: u 上
Sinological IPA (Jiading): /u³⁴/
Sinological IPA (Changzhou): /u⁴⁵/
Sinological IPA (Jingjiang): /u³⁵/
Sinological IPA (Tongxiang): /u⁵³/
Sinological IPA (Hangzhou): /u⁵³/
(Northern: Suzhou)
Wugniu: 1 ou
MiniDict: ou 平
Sinological IPA (Suzhou): /əu⁴⁴/
(Northern: Jiaxing, Haining, Haiyan, Xiaoshan, Shaoxing, Zhoushan)
Wugniu: 4 wu
MiniDict: wu 上
Sinological IPA (Jiaxing): /ɦu²¹³/
Sinological IPA (Haining): /ɦu⁴²/
Sinological IPA (Haiyan): /ɦu²³²/
Sinological IPA (Xiaoshan): /ɦu²¹³/
Sinological IPA (Shaoxing): /ɦu¹¹³/
Sinological IPA (Zhoushan): /ɦu²⁴/
(Northern: Cixi)
Wugniu: 2 wu
Sinological IPA (Cixi): /ɦu¹⁴/
(Jinhua)
Wugniu: 3 ng; 3 u
Sinological IPA (Jinhua): /ŋ̍⁵³⁵/, /u⁵³⁵/
Note:
ng - vernacular;
(o)u, wu - literary;
6 ng (Hangzhou) - younger speakers only, influenced by nearby lects.
Xiang
(Changsha)
Wiktionary: u 3
Sinological IPA(key): /u⁴¹/
(Loudi)
Wiktionary: u 3
Sinological IPA(key): /u⁴²/
(Hengyang)
Wiktionary: u 3
Sinological IPA(key): /u³³/
Middle Chinese: nguX more ▼
| Rime |
| Character | 五 |
| Reading # | 1/1 |
| Initial (聲) | 疑 (31) |
| Final (韻) | 模 (23) |
| Tone (調) | Rising (X) |
| Openness (開合) | Open |
| Division (等) | I |
| Fanqie | 疑古切 |
| Baxter | nguX |
| Reconstructions |
| Zhengzhang Shangfang | /ŋuo X/ |
| Pan Wuyun | /ŋuo X/ |
| Shao Rongfen | /ŋo X/ |
| Edwin Pulleyblank | /ŋɔ X/ |
| Li Rong | /ŋo X/ |
| Wang Li | /ŋu X/ |
| Bernhard Karlgren | /ŋuo X/ |
| Expected Mandarin Reflex | wǔ |
| Expected Cantonese Reflex | ng 5 |
Old Chinesemore ▼(Baxter–Sagart): /C.ŋˤaʔ/(Zhengzhang): /ŋaːʔ/
| Baxter–Sagart system 1.1 (2014) |
| Character | 五 |
| Reading # | 1/1 |
| Modern Beijing (Pinyin) | wǔ |
| Middle Chinese | ‹ ngu X › |
| Old Chinese | /C.ŋˁaʔ/ |
| English | five |
| Expand _Notes for Old Chinese notations in the Baxter–Sagart system:_ Parentheses "()" indicate uncertain presence; Square brackets "[]" indicate uncertain identity, e.g. [t] as coda may in fact be -t or -p; Angle brackets "<>" indicate infix; Hyphen "-" indicates morpheme boundary; Period "." indicates syllable boundary. |
| Zhengzhang system (2003) |
| Character | 五 |
| Reading # | 1/1 |
| No. | 13147 |
| Phonetic component | 五 |
| Rime group | 魚 |
| Rime subdivision | 0 |
| Corresponding MC rime | 五 |
| Old Chinese | /ŋaːʔ/ |
Definitions
[edit]
五
five
a surname, Wu(Mainland China,Taiwan) or Ng(Hong Kong)
(printing) The size of type between 小五 (xiǎowǔ) (little5) and 小四 (xiǎosì) (little4), standardized as 10½ point.
See also
[edit]
| Collapse Chinese numbers |
| | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 10 2 | 10 3 | 10 4 | 10 6 | 10 8 | 10 12 |
| Normal (小寫/ 小写) | 〇, 零, 空 | 一, 蜀 | 二, 兩/ 两 | 三 | 四 | 五 | 六 | 七 | 八 | 九 | 十 | 百 | 千 | 萬/ 万, 十千(Malaysia, Singapore) | 百萬/ 百万, 桶(Philippines), 面桶(Philippines) | 億/ 亿 | 兆(Taiwan) 萬億/ 万亿(Mainland China) |
| Financial (大寫/ 大写) | 零 | 壹 | 貳/ 贰 | 參/ 叁 | 肆 | 伍 | 陸/ 陆 | 柒 | 捌 | 玖 | 拾 | 佰 | 仟 |
In Min Nan numbers, the vernacular (白) pronunciation is the more common pronunciation, while the literary (文) reading is used for reading numbers out loud, such as in phone numbers. Please note that this usage is similar to the usage of the variant 幺 for the numeral 一 in Mandarin.
Compounds
[edit]
一丈五尺
一丸五色
一五一十(yīwǔyīshí)
一百五
一百五日
一目五行
一花五葉/ 一花五叶
一身五心
七五三
七俠五義/ 七侠五义
七相五公
七策五成
七貴五侯/ 七贵五侯
三三五五
三上五落
三不五時/ 三不五时(sānbùwǔshí)
三五
三五之隆
三五傳柑/ 三五传柑
三五夕
三五夜
三五好友
三五成群(sānwǔ-chéngqún)
下五旗(xiàwǔqí)
三五月
三五蟾光
三五門/ 三五门
三令五申(sānlìngwǔshēn)
三俠五義/ 三侠五义
三四五
三回五次
三回五解
三回五轉/ 三回五转
三墳五典/ 三坟五典
三尸五鬼
三山五嶽/ 三山五岳(sānshānwǔyuè)
三差五錯/ 三差五错
三年五載/ 三年五载(sānniánwǔzǎi)
三方五氏
三智五猜
三朝五日
三江五湖
三湯五割/ 三汤五割
三牲五鼎
三環五扣/ 三环五扣
三番五次(sānfānwǔcì)
三百五篇
三皇五帝(Sānhuáng Wǔdì)
三綱五常/ 三纲五常(sāngāngwǔcháng)
三老五更
三荒五月
三葷五厭/ 三荤五厌
三表五餌/ 三表五饵
三貞五烈/ 三贞五烈
三釀五齊/ 三酿五齐
三陽五會/ 三阳五会
三陽五輸/ 三阳五输
不為五斗米折腰(bù wèi wǔ dǒu mǐ zhéyāo)
中五(zhōngwǔ)
九五(jiǔwǔ)
九五之位
九五之尊(jiǔwǔzhīzūn)
二不挂五
二五
二五八
二五更
二五眼(èrwǔyǎn)
二五耦
二五附稅/ 二五附税
二十五史
二十五弦
二十五絃/ 二十五弦
二十五老
二百五(èrbǎiwǔ)
五七
五丁
五丈
五三
五丈原
五三慘案/ 五三惨案(Wǔ-Sān Cǎn'àn)
五丈旗
五不取
五不娶
五世
五世其昌
五丞
五中
五乘
五九國恥/ 五九国耻
五事
五五
五五憲草/ 五五宪草
五交
五京
五亭
五人冢
五人塚/ 五人冢
五人墓
五代(wǔdài)
五代十國/ 五代十国(Wǔdài Shíguó)
五代史
五任
五位
五作
五伯(wǔbà)
五佐
五位君臣
五位百法
五位缾/ 五位瓶
五伯長/ 五伯长
五使
五供
五例
五供兒/ 五供儿
五供養/ 五供养
五保
五俎
五俊
五侯
五侯七貴/ 五侯七贵
五侯九伯
五保戶/ 五保户
五侯第
五侯蠟燭/ 五侯蜡烛
五侯鯖/ 五侯鲭
五倫/ 五伦(wǔlún)
五倖/ 五幸
五倉/ 五仓
五倍子(wǔbèizǐ)
五倍子蟲/ 五倍子虫
五停心觀/ 五停心观
五傳/ 五传
五傳弟子/ 五传弟子
五儀/ 五仪
五元
五光
五兆
五光十色(wǔguāngshísè)
五內/ 五内(wǔnèi)
五內如焚/ 五内如焚
五兩/ 五两
五公
五六六七
五兵
五兵佩
五兵尚書/ 五兵尚书
五典
五典三墳/ 五典三坟
五冕
五冬六夏
五出
五刃
五分
五分明兒/ 五分明儿
五分錢/ 五分钱
五刑(wǔxíng)
五利
五則/ 五则
五剽
五劇/ 五剧
五劍/ 五剑
五力
五功(Wǔgōng)
五加(wǔjiā)
五加皮(wǔjiāpí)
五動/ 五动
五勝/ 五胜
五勞/ 五劳(wǔláo)
五勞七傷/ 五劳七伤(wǔláoqīshāng)
五勢/ 五势
五十三參/ 五十三参
五十弦
五十肩(wǔshíjiān)
五千文
五千言
五卅慘案/ 五卅惨案(Wǔ-Sà Cǎn'àn)
五南
五博
五卜
五印(Wǔyìn)
五印字
五印度(Wǔyìndù)
五厄
五原(Wǔyuán)
五反
五叔
五口(Wǔkǒu)
五口通商
五司
五古
五吏
五君
五和
五命
五味(wǔwèi)
五味子(wǔwèizǐ)
五味神
五品
五品孫/ 五品孙
五善
五單于
五噎
五噫
五器
五噫歌
五四(Wǔ-Sì)
五四三(wǔsìsān)
五四運動/ 五四运动(Wǔsì Yùndòng)
五國城/ 五国城
五圖/ 五图
五土
五地
五圻
五均
五坊
五均六管
五坊小兒/ 五坊小儿
五城
五城兵馬/ 五城兵马
五堂功課/ 五堂功课
五塗/ 五涂
五塥
五塵/ 五尘
五壏/ 五𰊑
五壤
五壺浮漏/ 五壶浮漏
五夏
五夜
五夜元宵
五大(wǔdà)
五大三粗(wǔdàsāncū)
五大夫
五大夫城
五大建設/ 五大建设
五大洋
五大洲(Wǔdàzhōu)
五大運動/ 五大运动
五天
五天竺(Wǔtiānzhú)
五失本
五奴
五妃廟/ 五妃庙
五姓
五始
五姓蕃
五姦/ 五奸
五威將/ 五威将
五威將帥/ 五威将帅
五威將軍/ 五威将军
五子
五子之歌
五子哭墳/ 五子哭坟
五子棋(wǔzǐqí)
五子登科
五字
五字句
五字城
五字訣/ 五字诀
五字詩/ 五字诗
五孝
五季
五孰
五學/ 五学
五官(wǔguān)
五宗
五官將/ 五官将
五官掾
五官郎
五室
五家
五家七宗
五家洲(Wǔjiāzhōu)
五宿
五寸子
五射
五將/ 五将
五專/ 五专
五小工業/ 五小工业
五尚
五就
五尸
五尺
五尺之僮
五尺之孤
五尺之童
五尺童子
五尺豎子/ 五尺竖子
五局
五屬/ 五属
五屬大夫/ 五属大夫
五山(Wǔshān)
五屼
五岳(wǔyuè)
五岳丈人
五岳四瀆/ 五岳四渎
五峰(wǔfēng)
五峰先生
五嶠/ 五峤
五嶺/ 五岭(Wǔlǐng)
五嶽/ 五岳(wǔyuè)
五嶽圖/ 五岳图
五州
五布
五帝(Wǔdì)
五帝坐
五帝車/ 五帝车
五席
五常(wǔcháng)
五帶/ 五带
五幡
五府
五度
五度圈(wǔdùquān)
五庫/ 五库
五廕/ 五荫
五廟/ 五庙
五弄
五引
五弦(wǔxián)
五弦琴
五形
五彩(wǔcǎi)
五彩旗
五彩繽紛/ 五彩缤纷(wǔcǎibīnfēn)
五律(wǔlǜ)
五御
五徵/ 五征
五德
五德終始/ 五德终始
五心
五心六意
五性
五怖
五恭
五惡/ 五恶
五情
五惡趣/ 五恶趣
五愛/ 五爱
五態/ 五态
五慮/ 五虑
五戊
五戎
五戒(wǔjiè)
五戶絲/ 五户丝
五房
五才
五技
五承
五折
五技鼠
五括
五指(wǔzhǐ)
五指山(Wǔzhǐshān)
五推
五排
五搶六奪/ 五抢六夺
五擾/ 五扰
五政
五放家
五教
五數/ 五数
五斂子/ 五敛子(wǔliǎnzǐ)
五斗
五斗先生
五斗子
五斗折腰
五斗櫃(wǔdǒuguì)
五斗祿
五斗米
五斗米師
五斗米道
五斗粟
五斗解酲
五斤手
五方(wǔfāng)
五方佛(Wǔfāngfó)
五方幢
五方旗(wǔfāngqí)
五方色
五方雜厝/ 五方杂厝
五方雜處/ 五方杂处(wǔfāngzáchǔ)
五旅
五旌
五族
五族共和(wǔzú gònghé)
五旗
五日
五日京兆(wǔrìjīngzhào)
五日子
五日節/ 五日节(Wǔrìjié)
五旦(wǔdàn)
五旦七調/ 五旦七调
五旬宗(wǔxúnzōng)
五旬節/ 五旬节(Wǔxúnjié)
五明
五易
五明囊
五明宮/ 五明宫
五明扇
五明馬/ 五明马
五明驥/ 五明骥
五星(wǔxīng)
五是
五星三
五星上將/ 五星上将(wǔ xīng shàngjiàng)
五星紅旗/ 五星红旗(Wǔxīng Hóngqí)
五星聚
五星聯珠/ 五星联珠
五星連珠/ 五星连珠
五時/ 五时
五時副車/ 五时副车
五時節/ 五时节
五時衣/ 五时衣
五時車/ 五时车
五時雞/ 五时鸡
五暉/ 五晖
五暴
五曉/ 五晓
五曜
五曲
五更(wǔgēng)
五更三點/ 五更三点
五更天
五更調/ 五更调
五更雞/ 五更鸡
五更頭/ 五更头
五曹
五會/ 五会
五會念佛/ 五会念佛
五月先兒/ 五月先儿
五月子
五月披裘
五月節/ 五月节(Wǔyuèjié)
五月花號/ 五月花号
五有
五服(wǔfú)
五木
五木香
五本
五朱
五朵雲/ 五朵云
五材(wǔcái)
五果
五松
五柞
五柳
五柳先生
五柞宮/ 五柞宫
五柳心
五桀
五根
五校
五桂
五株
五案
五梁冠
五棱子
五極/ 五极
五楘
五榮/ 五荣
五樓/ 五楼
五樂/ 五乐
五橫/ 五横
五權/ 五权
五權分立/ 五权分立
五權憲法/ 五权宪法
五欲
五正
五步成詩/ 五步成诗
五步蛇
五殘/ 五残
五殖
五殺/ 五杀
五毒(wǔdú)
五毒草
五毒餅/ 五毒饼
五比丘
五毛(wǔmáo)
五氏
五民
五氣/ 五气
五氣朝元/ 五气朝元
五水
五水蠻/ 五水蛮
五沃
五沙
五汶
五泄
五河(Wǔhé)
五泰
五法
五洲
五洩/ 五泄
五流
五津
五洋
五洲四海
五洋雜貨/ 五洋杂货
五浮
五涂
五淨/ 五净
五液
五涼/ 五凉
五淨德/ 五净德
五湖(Wǔhú)
五湖四海(wǔhúsìhǎi)
五湖心
五溪
五溝/ 五沟
五漿/ 五浆
五漏
五潢
五濁/ 五浊
五濁惡世/ 五浊恶世
五火
五煙/ 五烟
五熏
五熟
五熟行
五熟釜
五燈會元/ 五灯会元
五營/ 五营
五爪金龍/ 五爪金龙
五爵
五父
五牙
五牛
五牛旗
五牢
五物
五牲
五牲看碗
五牸
五狄
五狗
五猖
五猖會/ 五猖会
五猶/ 五犹
五獸/ 五兽
五獻/ 五献
五王
五王帳/ 五王帐
五玉
五瑞
五生
五生盆
五甲
五申三令
五男二女
五畜
五畤
五疾
五疵
五痔
五瘟使
五瘟神(wǔwēnshén)
五白
五百(wǔbǎi)
五百年前
五百灘/ 五百滩
五百羅漢/ 五百罗汉
五皓
五相
五省
五盾
五眼
五眾/ 五众
五眼泉(Wǔyǎnquán)
五眼雞/ 五眼鸡
五知
五短身材
五石
五石六鷁/ 五石六鹢
五石散(wǔshísǎn)
五石瓠
五石銅/ 五石铜
五矹
五示
五祀
五祖
五神
五祖七真
五神通
五禁
五福(wǔfú)
五福紙/ 五福纸
五福臨門/ 五福临门(wǔfúlínmén)
五福餅/ 五福饼
五禪/ 五禅
五禮/ 五礼
五禽
五禽嬉
五禽戲/ 五禽戏
五禽言
五秀
五秉
五種/ 五种
五種性/ 五种性
五穀/ 五谷(wǔgǔ)
五稼
五穀不分/ 五谷不分
五穀不升/ 五谷不升
五穀囊/ 五谷囊
五穀精/ 五谷精
五穀蟲/ 五谷虫
五穀豐熟/ 五谷丰熟
五穀豐登/ 五谷丰登(wǔgǔfēngdēng)
五穀豐稔/ 五谷丰稔
五積六受/ 五积六受
五窮/ 五穷
五竇聯珠/ 五窦联珠
五端
五等
五等爵
五節/ 五节
五管
五箭
五籍
五粒松
五粒風/ 五粒风
五粟
五粱禾
五精
五精舍
五紀/ 五纪
五純/ 五纯
五細/ 五细
五絃/ 五弦(wǔxián)
五紽/ 五𰬉
五結/ 五结(Wǔjié)
五絕/ 五绝(wǔjué)
五絲/ 五丝
五經/ 五经(Wǔjīng)
五經博士/ 五经博士
五經家/ 五经家
五經師/ 五经师
五經庫/ 五经库
五經掃地/ 五经扫地
五經笥/ 五经笥
五經解元/ 五经解元
五經魁/ 五经魁
五經魁首/ 五经魁首
五綵/ 五彩
五綦
五緉/ 五𮉧
五綵戲/ 五彩戏
五綵旗/ 五彩旗
五綵衣/ 五彩衣
五緯/ 五纬
五線譜/ 五线谱(wǔxiànpǔ)
五縣/ 五县
五縗/ 五缞
五繇
五總龜/ 五总龟
五纑/ 五𮉡
五罪
五罰/ 五罚
五羊(Wǔyáng)
五羊城(Wǔyángchéng)
五羊皮
五美
五羖
五羖大夫
五羖皮
五義/ 五义
五考
五老
五老峰
五老會/ 五老会
五老榜
五聖/ 五圣
五聲/ 五声
五聽/ 五听
五育
五股(Wǔgǔ)
五胡(Wǔhú)
五胡亂華/ 五胡乱华(Wǔhúluànhuá)
五脈/ 五脉
五脊六獸/ 五脊六兽
五膿/ 五脓
五臘/ 五腊
五臟/ 五脏(wǔzàng)
五臟六腑/ 五脏六腑(wǔzàng liùfǔ)
五臟廟/ 五脏庙
五臟神/ 五脏神
五臣
五臧
五臭
五至
五臺/ 五台(Wǔtái)
五臺山/ 五台山(Wǔtái Shān)
五色(wǔsè)
五色亂目/ 五色乱目
五色土
五色掛錢/ 五色挂钱
五色旗(Wǔsèqí)
五色書/ 五色书
五色棒
五色毫
五色氣/ 五色气
五色水團/ 五色水团
五色泥
五色無主/ 五色无主
五色瓜
五色石
五色祥雲/ 五色祥云
五色筆/ 五色笔
五色線/ 五色线
五色縷/ 五色缕
五色繽紛/ 五色缤纷
五色羽
五色腸/ 五色肠
五色花子
五色衣
五色詔/ 五色诏
五色診病/ 五色诊病
五色話/ 五色话(wǔsèhuà)
五色陸離/ 五色陆离
五色雀
五色雲/ 五色云
五色雲氣/ 五色云气
五色魚/ 五色鱼
五芝
五花(wǔhuā)
五花儀/ 五花仪
五花八門/ 五花八门(wǔhuābāmén)
五花判事
五花大綁/ 五花大绑(wǔhuādàbǎng)
五花官誥/ 五花官诰
五花度牒
五花殺馬/ 五花杀马
五花爨弄
五花繃裂/ 五花绷裂
五花肉(wǔhuāròu)
五花茶(wǔhuāchá)
五花誥/ 五花诰
五花館/ 五花馆
五花馬/ 五花马(wǔhuāmǎ)
五花驄/ 五花骢
五苦
五英
五范
五茄
五茸
五荒六月
五莖/ 五茎
五華/ 五华(Wǔhuá)
五菜
五葷/ 五荤(wǔhūn)
五葉/ 五叶
五蓋/ 五盖
五蒼/ 五苍
五薀/ 五蕰
五藏
五藏六府
五藏神
五藥/ 五药
五蘊/ 五蕴(wǔyùn)
五蘟/ 五𦻕
五虎
五虎將/ 五虎将
五虐
五彪
五處士/ 五处士
五虛/ 五虚
五號/ 五号
五蛇
五蜀
五蟲/ 五虫
五蠹
五行(wǔxíng)
五行並下/ 五行并下
五行俱下
五行八作
五行四柱
五行大布
五行家
五行生剋/ 五行生克
五行生勝/ 五行生胜
五行相剋/ 五行相克
五行相勝/ 五行相胜
五行相生
五行舞
五行陣/ 五行阵
五衍
五衛/ 五卫
五衢
五衰
五衷
五裂
五裁
五褲手/ 五裤手
五褲歌/ 五裤歌
五褲詠/ 五裤咏
五褲謠/ 五裤谣
五褲謳/ 五裤讴
五見/ 五见
五親六眷/ 五亲六眷
五角六張/ 五角六张
五角大廈/ 五角大厦(Wǔjiǎo Dàshà)
五觳
五言(wǔyán)
五言古
五言古詩/ 五言古诗
五言四句
五言城
五言律
五言律詩/ 五言律诗(wǔyán lǜshī)
五言排律
五言絕/ 五言绝
五言絕句/ 五言绝句(wǔyán juéjù)
五言詩/ 五言诗(wǔyánshī)
五言試帖/ 五言试帖
五言金城
五言長城/ 五言长城
五言長律/ 五言长律
五詞/ 五词
五診/ 五诊
五誡/ 五诫(wǔjiè)
五誥/ 五诰
五調/ 五调
五諫/ 五谏
五諸侯/ 五诸侯
五講四美/ 五讲四美
五識/ 五识
五讓/ 五让
五貢/ 五贡
五費/ 五费
五賊/ 五贼
五賢/ 五贤
五起
五趣
五路
五路總頭/ 五路总头
五路財神/ 五路财神
五路酒
五車/ 五车
五車書/ 五车书
五車腹笥/ 五车腹笥
五軍/ 五军
五輅/ 五辂
五輕/ 五轻
五輪/ 五轮
五辛(wǔxīn)
五辛盤/ 五辛盘
五辛菜
五辟
五辭/ 五辞
五辰
五逆
五逆罪(wǔnìzuì)
五通(Wǔtōng)
五通神
五運/ 五运
五遁
五過/ 五过
五達/ 五达
五道
五運六氣/ 五运六气
五道將軍/ 五道将军
五道神
五達道/ 五达道
五邦
五郊
五郎
五部
五部洲
五都
五酉
五酘
五醜/ 五丑
五采
五里(Wǔlǐ)
五里鋪(Wǔlǐpù)
五里霧
五野
五量
五量店
五金(wǔjīn)
五釜
五銖/ 五铢
五銖服/ 五铢服
五銑衣/ 五铣衣
五銖衣/ 五铢衣
五銖錢/ 五铢钱
五鍾/ 五钟
五鎮/ 五镇
五鐘/ 五钟
五鑿/ 五凿
五長/ 五长
五門/ 五门
五閒/ 五闲
五間/ 五间
五關/ 五关(wǔguān)
五院
五陣/ 五阵
五院制
五陳/ 五陈
五陰/ 五阴(wǔ yīn)
五陵(wǔlíng)
五陰世間/ 五阴世间
五陵兒/ 五陵儿
五陵原
五陵年少(wǔlíngniánshào)
五陵氣/ 五陵气
五陵英少
五陵豪氣/ 五陵豪气
五陽/ 五阳
五際/ 五际
五雀
五雀六燕
五雅
五雉
五雜俎/ 五杂俎
五難/ 五难
五雲/ 五云(wǔyún)
五雲判/ 五云判
五雲城/ 五云城
五雲字/ 五云字
五雲樓/ 五云楼
五雲毫/ 五云毫
五雲漿/ 五云浆
五雲裘/ 五云裘
五雲谿/ 五云谿
五雲車/ 五云车
五雲鄉/ 五云乡
五雲體/ 五云体
五雷
五零二落
五零四散
五雷正法
五雷法
五雷轟頂/ 五雷轰顶
五霸(wǔbà)
五靈/ 五灵
五靈神/ 五灵神
五韙/ 五韪
五音(wǔyīn)
五音不全(wǔyīnbùquán)
五音兒/ 五音儿
五音士
五音戲/ 五音戏
五章
五韺
五頂/ 五顶
五領/ 五领
五題/ 五题
五顔六色/ 五颜六色
五顏六色/ 五颜六色(wǔyánliùsè)
五類雜種/ 五类杂种
五顯公/ 五显公
五顯靈官/ 五显灵官
五風/ 五风
五風十雨/ 五风十雨(wǔfēngshíyǔ)
五餌/ 五饵
五餅二魚/ 五饼二鱼(wǔbǐng'èryú)
五館/ 五馆
五香(wǔxiāng)
五香飲/ 五香饮
五馬/ 五马(Wǔmǎ)
五馬分屍/ 五马分尸(wǔmǎfēnshī)
五馬浮江/ 五马浮江
五馬渡/ 五马渡
五馬渡江/ 五马渡江
五馬貴/ 五马贵
五馭/ 五驭
五體/ 五体
五體投地/ 五体投地(wǔtǐtóudì)
五體投誠/ 五体投诚
五鬼(wǔguǐ)
五鬼術/ 五鬼术
五魁
五鳥花/ 五鸟花
五鳩/ 五鸠
五鳧/ 五凫
五鳳/ 五凤
五鳳城/ 五凤城
五鳳樓/ 五凤楼
五鹽/ 五盐
五鹿(Wǔlù)
五麾
五黃/ 五黄
五黄六月
五黃六月/ 五黄六月
五鼎
五鼎亨
五鼎烹
五鼎萬鐘/ 五鼎万钟
五鼎芝
五鼎食
五鼓
五齊/ 五齐
五齏/ 五齑
五龍/ 五龙
五龍山/ 五龙山(Wǔlóngshān)
五龍車/ 五龙车
亨利五世
修五臟廟/ 修五脏庙
假五百
八瓣兒五/ 八瓣儿五
八門五花/ 八门五花
六五
公沙五龍/ 公沙五龙
初一十五(chūyī shíwǔ)
初五隔開/ 初五隔开
前五代
前五子
化色五倉/ 化色五仓
十五時/ 十五时
十光五色
十十五五
十圍五攻/ 十围五攻
十惡五逆/ 十恶五逆
十漿五饋/ 十浆五馈
十變五化/ 十变五化
十雨五風/ 十雨五风
十風五雨/ 十风五雨
南五祖
博覽五車/ 博览五车
去天尺五
參五/ 参五
吆五喝六(yāowǔhèliù)
呼五白
咸五登三
喝五吆三
喓五喝六
四分五剖
四分五落
四分五裂(sìfēnwǔliè)
四功五法
四山五岳
四山五嶽/ 四山五岳
四律五論/ 四律五论
四捨五入/ 四舍五入(sìshěwǔrù)
四斗五方
四紛五落/ 四纷五落
四通五達/ 四通五达
坐五行三
士五
夏五
夏五郭公
大五金
大烹五鼎
天中五瑞
天尺五
學富五車/ 学富五车
宋五嫂
小五(xiǎowǔ)
小五義/ 小五义
小五金
尺五
尺五天
廢五金/ 废五金
後五代/ 后五代
後五日/ 后五日
忌破五
恨五罵六/ 恨五骂六
惠施五車/ 惠施五车
戲五禽/ 戏五禽
折腰五斗
拔十失五
拔十得五
拿三道五
挨三頂五/ 挨三顶五
捱三頂五/ 捱三顶五
攢三聚五/ 攒三聚五
攢三集五/ 攒三集五
敦煌五龍/ 敦煌五龙
新五代史
日誦五車/ 日诵五车
春秋五霸
書富五車/ 书富五车
書讀五車/ 书读五车
杜五
格五
梁鴻五噫
梧鼠五技
歌五褲/ 歌五裤
正五九
民歌五褲/ 民歌五裤
牽五掛四/ 牵五挂四
猜三划五
猜三劃五/ 猜三划五
猜三喝五
獨腳五通/ 独脚五通
王老五(wánglǎowǔ)
王雲五/ 王云五
百五
百五日
目迷五色
直百五銖/ 直百五铢
破五
破家五鬼
破敗五鬼/ 破败五鬼
祭五臟廟/ 祭五脏庙
視窗九五/ 视窗九五
禪門五宗/ 禅门五宗
種五生/ 种五生
端五(Duānwǔ)
第五(dì-wǔ)
第五縱隊/ 第五纵队(dì-wǔzòngduì)
紅五月/ 红五月
紅五類/ 红五类
腹載五車/ 腹载五车
舊五代史/ 旧五代史
華嚴五祖/ 华严五祖
蜀五
蠻攀五經/ 蛮攀五经
行三坐五
行五
褚五
起五更
趙五娘/ 赵五娘
跑五方
踏五花
連三接五/ 连三接五
連三跨五/ 连三跨五
逗五逗六
遁五
過五關/ 过五关
過破五/ 过破五
郭公夏五
鄭五/ 郑五
鄭五歇後/ 郑五歇后
重五(Chóngwǔ)
銘感五內/ 铭感五内
閻老五/ 阎老五
陳三五娘/ 陈三五娘
陶潛五柳/ 陶潜五柳
陰陽五行/ 阴阳五行
隔三差五(gésānchàwǔ)
音學五書/ 音学五书
飛五/ 飞五
馬中關五/ 马中关五
鬧五魁/ 闹五魁
黑五類/ 黑五类(hēiwǔlèi)
墨分五色
鼫鼠五技
龍飛九五/ 龙飞九五
show more ▼
Descendants
[edit]
Sino-Xenic (五):
→ Japanese:五(ご)(go)
→ Korean: 오(五)(o)
→ Vietnamese: ngũ(五)
Others:
→ Proto-Tai: haːꟲ(“five”)
Northern Tai
Bouyei: hac
Saek: ฮ̂า
Tai Do: ha
Zhuang: haj
Central Tai
Nùng: hà
Tày: hả
Southwestern Tai
Ahom: 𑜑𑜡(hā)
Lao: ຫ້າ(hā)
Lü: ᦠᦱᧉ(ḣaa²)
Northern Thai: ᩉ᩶ᩣ
Shan: ႁႃႈ(hāa)
Tai Dam: ꪬ꫁ꪱ
Tai Nüa: ᥞᥣᥲ(hàa)
Thai: ห้า(hâa)
Etymology 2
[edit]
| simp. and trad. | 五 |
Pronunciation
[edit]
Mandarin(Pinyin): wǔ (wu 3)(Zhuyin): ㄨˇ
Cantonese(Jyutping): wu 1more ▼
Mandarin
(Standard Chinese)+
Hanyu Pinyin: wǔ
Zhuyin: ㄨˇ
Tongyong Pinyin: wǔ
Wade–Giles: wu 3
Yale: wǔ
Gwoyeu Romatzyh: wuu
Palladius: у(u)
Sinological IPA(key): /u²¹⁴/
Cantonese
(Standard Cantonese, Guangzhou–Hong Kong)
Jyutping: wu 1
Yale: w ū
Cantonese Pinyin: wu 1
Guangdong Romanization: wu 1
Sinological IPA(key): /wuː ⁵⁵/
Definitions
[edit]
五
(music)Kunqugongchenotation for the notela (6).
(music)Cantonese operagongchenotation for the notela (6).
Derived terms
[edit]
伍(wǔ)
鿉(wǔ)
References
[edit]
“五”, in 漢語多功能字庫 (Multi-function Chinese Character Database), 香港中文大學 (the Chinese University of Hong Kong), 2014–
莆田市政协文化文史和学习委员会[Culture, History and Learning Committee of Putian CPPCC], editor (2021), “五”, in 莆仙方言大词典 [Comprehensive Dictionary of Puxian Dialect] (overall work in Mandarin and Puxian Min), Xiamen University Press, →ISBN, page 236.
莆田市政协文化文史和学习委员会[Culture, History and Learning Committee of Putian CPPCC], editor (2021), “五”, in 莆仙方言大词典 [Comprehensive Dictionary of Puxian Dialect] (overall work in Mandarin and Puxian Min), Xiamen University Press, →ISBN, page 407.
Japanese
[edit]
Japanese cardinal numbers| <4 | 5 | 6> |
| Cardinal: 五 |
| |
Kanji
[edit]
See also:
Category:Japanese terms spelled with 五
五
(First grade kyōiku kanji)
five
Readings
[edit]
Go-on: ご(go, Jōyō)
Kan-on: ご(go, Jōyō)
Kun: いつ(itsu, 五, Jōyō)、いつつ(itsu tsu, 五つ, Jōyō)
Nanori: い(i)、さ(sa)、さつ(satsu)、ち(chi)、ふ(fu)、み(mi)、め(me)
Compounds
[edit]
show ▼Compounds
五(ご)節(せっ)句(く)(gosekku, “the five seasonal festivals of the Japanese calendar”)
五(ご)悪(あく)(Goaku, “(Buddhism) the Five Evils”)
五(ご)位(い)鷺(さぎ)(goisagi), 五(ゴ)位(イ)鷺(サギ)(goisagi, “night heron”)
五(ご)十(じっ)歩(ぽ)百(ひゃっ)歩(ぽ)(gojippo hyappo), 五(ご)十(じゅっ)歩(ぽ)百(ひゃっ)歩(ぽ)(gojuppo hyappo)
五(ご)音(おん)音(おん)階(かい)(gōn onkai, “pentatonic scale”)
五(ご)戒(かい)(Gokai, “(Buddhism) the Five Precepts”)
五(ご)角(かく)形(けい)(gokakukei), 五(ご)角(かっ)形(けい)(gokakkei, “pentagon”)
五(ご)官(かん)(gokan, “the five sense organs”)
五(ご)感(かん)(gokan, “the five senses”)
五(ご)経(きょう)(Gokyō, “(Confucianism) Four Books and Five Classics”)
五(ご)桁(けた)(goketa)
五(ご)月(がつ)(gogatsu), 五(さつ)月(き)(satsuki), 五(サツ)月(キ)(satsuki, “May”)
五(ご)言(ごん)絶(ぜっ)句(く)(gogonzekku)
五(ご)言(ごん)律(りっ)詩(し)(gogonrisshi)
五(ご)更(こう)(gokō)
五(ご)行(ぎょう)(Gogyō, “Wu Xing”)
五(ご)穀(こく)(gokoku)
五(ご)彩(さい)(gosai)
五(ご)指(し)(goshi, “five fingers”)
五(ご)種(しゅ)競(きょう)技(ぎ)(goshu kyōgi, “pentathlon”)
五(いつ)重(え)(itsue)
五(ご)旬(じゅん)節(せつ)(gojunsetsu)
五(ご)書(しょ)(gosho)
五(ご)女(じょ)(gojo)
五(ご)常(じょう)(Gojō, “the five Confucian virtues”)
五(ご)情(じょう)(Gojō, “the Five Passions”)
五(ご)色(しき)(goshiki)
五(ご)人(にん)組(ぐみ)(Goningumi, “Five Households (Edo era civil defence units)”)
五(ご)人(にん)囃(ばや)子(し)(goninbayashi)
五(ご)寸(すん)釘(くぎ)(gosun kugi, “a long nail, roughly 15 cm in length”)
五(ご)節(せつ)(gosetsu)
五(ご)線(せん)(gosen)
五(ご)臓(ぞう)(gozō)
五(ご)体(たい)(gotai)
五(ご)段(だん)(godan)
五(ご)男(なん)(gonan)
五(ご)斗(と)米(べい)(gotobei)
五(ご)徳(とく)(gotoku)
五倍子(ふし)(fushi)
五(ご)風(ふう)十(じゅう)雨(う)(gofū jūu, “clementweather(literally, “five days and one is windy, ten days and one is rainy”)”)
五(ご)分(ぶ)(gobu), 五(ご)分(ふん)(gofun), 五(ご)分(ぶん)(gobun)
五(ご)辺(へん)形(けい)(gohenkei)
五(ご)面(めん)体(たい)(gomentai, “pentahedron”)
五(ご)目(もく)(gomoku, “mixture”)
五(ご)里(り)霧(む)中(ちゅう)(gori muchū, “totallylost, groping in the dark (literally, “five leagues in the fog”)”)
五(ご)稜(りょう)郭(かく)(Goryōkaku, “Goryōkaku, a star-shaped fort in Hakodate”)
五(ご)稜(りょう)堡(ほ)(goryōho)
五(ご)倫(りん)(Gorin, “Gorin, the five Confucian virtues”)
五(ご)輪(りん)(Gorin, “Olympics”)
五(ご)芒(ぼう)(gobō, “pentagram”)
弦(げん)楽(がく)五(ご)重(じゅう)奏(そう)(gengaku gojūsō, “stringquintet”)
Etymology 1
[edit]
| Kanji in this term |
| 五 |
| いつ Grade: 1 |
| kun'yomi |
From Old Japanese, from Proto-Japonicetu.
Pronunciation
[edit]
(Tokyo)い つ[íꜜtsù] (Atamadaka – )
IPA(key): [it͡sɨ]
(Kyoto,Osaka)いつ[íꜜtsù] (Kōki)
Noun
[edit]
五(いつ)• (itsu)
five
Derived terms
[edit]
五(いつ)つ(itsutsu)
Etymology 2
[edit]
| Kanji in this term |
| 五 |
| ご Grade: 1 |
| on'yomi |
From Middle Chinese五 (MC nguX). Compare modern Hokkien五(gō͘ / gǒ͘).
Pronunciation
[edit]
(Tokyo)ご[góꜜ] (Atamadaka – )
IPA(key): [ɡo̞]
Audio:Duration: 1 second.0:01(file)
(Kyoto,Osaka)ご[gó] (Kōki – )
Noun
[edit]
五(ご)• (go)
five
a name of a hole of a wind instrument
Coordinate terms
[edit]
| v•d•e Japanese numbers |
| Number | Kanji | Kana | Romaji |
| 0 | 零 | れい、ゼロ | rei, zero |
| 1 | 一、壱、弌、壹 | いち | ichi |
| 2 | 二、弐、貳、貮 | に | ni |
| 3 | 三、参 | さん | san |
| 4 | 四、肆 | よん、し | yon, shi |
| 5 | 五、伍 | ご | go |
| 6 | 六、陸 | ろく | roku |
| 7 | 七、漆 | なな、しち | nana, shichi |
| 8 | 八、捌 | はち | hachi |
| 9 | 九、玖 | きゅう、く | kyū, ku |
| 10 | 十、拾 | じゅう | jū |
| 100 | 百、陌、佰 | ひゃく | hyaku |
| 1,000 | 千、阡、仟 | せん | sen |
| 10,000 | 一万、万、一萬、萬 | いちまん、まん | ichiman, man |
| 100,000,000 | 一億、億 | いちおく、おく | ichioku, oku |
show more ▼
Etymology 3
[edit]
| Kanji in this term |
| 五 |
| い Grade: 1 |
| kun'yomi |
Pronunciation
[edit]
IPA(key): [i]
Noun
[edit]
五(い)• (i)
five
Korean
[edit]
Etymology
[edit]
Korean numbers(edit)50
←456→
Native isol.: 다섯(daseot)
Native attr.: 다섯(daseot), (archaic)닷(dat)
Sino-Korean: 오(o)
Hanja: 五
Ordinal: 다섯째(daseotjjae)
From Middle Chinese五 (MC nguX).
| Expand Historical Readings |
| Dongguk Jeongun Reading |
| Dongguk Jeongun, 1448 | ᅌᅩᆼ〯 (Yale: ngwǒ) |
| Middle Korean |
| Text | Eumhun |
| Gloss (hun) | Reading |
| Hunmong Jahoe, 1527 | 다ᄉᆞᆺ〮 (Yale: tàsós) | 오〯 (Yale: wǒ) |
Pronunciation
[edit]
(SK Standard/Seoul) IPA(key): [o̞(ː)]
Phonetic hangul: [오(ː)]
Though still prescribed in Standard Korean, most speakers in both Koreas no longer distinguish vowel length.
Hanja
[edit]
Korean Wikisource has texts containing the hanja:
五
Wikisource
五 (eumhun다섯오(daseot o))
hanja form? of 오(“five”)
Compounds
[edit]
show ▼Compounds
오곡 (五穀, ogok)
오만 (五萬, oman)
오십 (五十, osip)
오온 (五蘊, oon)
오월 (五月, owol)
오행 (五行, ohaeng)
십오 (十五, sibo)
References
[edit]
국제퇴계학회 대구경북지부 (國際退溪學會 大邱慶北支部) (2007). Digital Hanja Dictionary, 전자사전／電子字典.
Vietnamese
[edit]
Han character
[edit]
五: Hán Việt readings: ngũ
五: Nôm readings: ngũ, ngủ, ngỗ
chữ Hán form of ngũ(“five”)
Compounds
[edit]
五代十國 (Ngũ Đại Thập Quốc)
五位尊官 (Ngũ Vị Tôn Quan)
五倫 (ngũ luân)
五味 (ngũ vị)
五味香 (ngũ vị hương)
五官 (ngũ quan)
五戒 (ngũ giới)
五果 (ngũ quả)
五福 (ngũ phúc)
五穀 (ngũ cốc)
五經 (ngũ kinh)
五臟 (ngũ tạng)
五色 (ngũ sắc)
五蘊 (ngũ uẩn)
五行 (ngũ hành)
五言 (ngũ ngôn)
五金 (ngũ kim)
五音 (ngũ âm)
重五 (trùng ngũ)
show more ▼
References
[edit]
↑ Jump up to: 1.01.11.21.3Nguyễn et al. (2009).
^Trần (2004).
↑ Jump up to: 3.03.1Bonet (1899).
^Génibrel (1898).
^Taberd & Pigneau de Béhaine (1838).
Zhuang
[edit]
Numeral
[edit]
五
Sawndip form of haj(“five”)
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五
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187923 | https://files.eric.ed.gov/fulltext/EJ1068215.pdf | ALM International Journal, Volume 5(1), pp. 23-35 Volume 5(1) – June 2010 23 Correcting Students’ Misconceptions about Probability in an Introductory College Statistics Course Leonid Khazanov Borough of Manhattan Community College City University of New York lkhazanov@bmcc.cuny.edu Lucio Prado Borough of Manhattan Community College City University of New York lprado@bmcc.cuny.edu Abstract College students’ misconceptions about probability are common and widespread. These misconceptions impede students' ability to make sound judgments in situations of uncertainty and master fundamental concepts of inferential statistics. In this paper the authors report the results of a study undertaken with the objective of correcting three common stochastic misconceptions within the framework of an introductory statistics course. Six instructors were recruited from an urban community college, along with their students. These instructors implemented a number of activities designed by one of the investigators and aimed at correcting the representativeness bias, the equiprobability misconception, and the outcome orientation bias.
The purpose of the activities was to trigger cognitive conflict thereby leading students to bring out and correct their incomplete or erroneous concepts. Three of the instructors attended a workshop on misconceptions before implementing the activities; the other three implemented the activities without going through training. Instructors in the third (control) group did not implement any of the proposed activities and were not trained. An instrument designed by one of the investigators was used to measure the extent of students’ misconceptions at the end of the semester in both treatment groups and the control group. The results show that trained instructors achieved significantly better outcomes than the control group in correcting two of the misconceptions: the representativeness bias and outcome orientation. By contrast, instructors who implemented the activities without being trained did not post better results than the control group in resolving any of the misconceptions. The results suggest that it is possible to improve students’ conceptual understanding of probability and correct their misconceptions by targeting the misconceptions directly in an introductory college statistics course. They also suggest that training on misconceptions is critical in ensuring instructors’ ability to successfully address students’ erroneous concepts about probability.
Introduction A substantial body of research demonstrates that students often hold non-standard concepts in mathematics and science (Garfield, 2001; Morton, 2008; Ozdemir & Clark, 2007; Vosniadu, 2004; Khazanov, 2008). These non-standard concepts are referred to in the literature by different terms: naive concepts, preconceptions, misperceptions, and misconceptions. The most common, and broadly accepted, term used in the literature is misconception, which is defined as Leonid Khazanov &Lucio Prado: Correcting Students’ Misconceptions about Probability in an Introductory College Statistics Course 24 Adults Learning Mathematics – An International Journal “a student conception that produces a systematic pattern of errors” (Smith, diSessa, & Roschelle, 1993, p. 10). It is a well established fact that a large proportion of college students have misconceptions about probability (Antoine, 2000; Das, 2008; Giuliano, 2006; Hirsch & O’Donnell, 2001; Konold, 1989; delMas & Bart, 1989). These non-standard concepts could have developed as a result of informal encounters students have with uncertain events at home, at the workplace, or while playing games of chance. Some of them might have developed misconceptions at school while learning probability, statistics or other subjects. There are scores of misconceptions about probability described in the literature. The most common include the equiprobability bias, conjunction fallacy, outcome orientation, representativeness, availability bias, and various misconceptions of conditional probability (Shaughnessy,1992; Jones, Langrall, & Mooney, 2007).
There is also strong research evidence that misconceptions about probability do not disappear as a result of traditional instruction in probability comprised, for the most part, of formal definitions, rules, and procedures (delMas & Bart, 1989; Khazanov, 2005, 2008; Konold, 1995, Shaughnessy, 1977, 1981). Misconceptions can “peacefully coexist” with correct concepts and interfere with students’ ability to apply these concepts consistently and with confidence (Khazanov & Gourgey, 2009; Ozdemir & Clark, 2007). Garfield (2001, 2007) points out that, although students may learn probability rules and procedures and may actually calculate correct answers on mathematical tests, these same students frequently misunderstand basic ideas and concepts and often ignore the rules when making their own judgment about uncertain events. Most researchers agree that instructional interventions designed specifically to eliminate students’ misconceptions of probability are necessary if any tangible and stable improvements in students’ concepts were to be expected (Khazanov & Gourgey, 2009; Khazanov, 2008; Konold, 1995; Hirsch & O’Donnell, 2001, Shaughnessy, 1981, 1992).
Need for study Correcting students’ misconceptions about probability has been broadly acknowledged as an important instructional goal for stochastic instruction. Shaughnessy, an influential scholar in the field of probability and statistics education, declared that one of the main goals of stochastic instruction should be to provide students with evidence of “how misconceptions of probability can lead to erroneous decisions” (Shaughnessy 1992, p. 482).
Khazanov & Gourgey (2009) surveyed statistics teachers and found that the majority of their respondents agreed that students’ erroneous concepts and beliefs about probability need to be addressed in an introductory college statistics course. In another study, Khazanov (2005) found that the number of misconceptions held by students was significantly associated with, and predictive of, their achievement in an introductory statistics course.
Shaughnessy (2007) quoted examples from a number of studies in which misconceptions students have about probability impede the development of correct understandings of some important statistical concepts.
Teaching probability for conceptual understanding implies a major shift in emphasis from simply providing formulas, rules, and procedures for calculations to addressing students’ erroneous intuitions and preconceptions (Garfield, 1995; Konold, 1995; Sharma, 2006; Khazanov & Gourgey, 2009). Khazanov (2005) pointed out that while misconceptions about probability have been extensively studied at different levels and for different age groups (see, for example, Fiscbein & Schnarch, 1997; Antoine, 2000) only a few studies reported attempts to correct misconceptions about probability in college instruction, and even fewer in introductory college statistics courses. The present study was designed to address the above deficiency.
ALM International Journal, Volume 5(1), pp. 23-35 Volume 5(1) – June 2010 25 Purpose of study The main purpose of this study was to determine whether targeting three common misconceptions about probability – the equiprobability bias, representativeness, and outcome orientation–directly and systematically, by employing activities designed by the investigators in a college statistics course, would be instrumental in resolving the misconceptions.
The equiprobability bias involves attributing the same probability in a random experiment to different events regardless of their actual chances (Jones, Langrall, & Mooney, 2007; Lecoutre, 1992; Lecoutre & Rezrazi, 1998). Outcome orientation is characterized by treating the probability of an occurrence or non-occurrence of an event as an affirmation of certainty rather than a measure of likelihood (Jones, 2005; Konold, 1989, 1995).
Representativeness bias implies estimating the likelihood of an event based on how well an outcome represents some aspect of the parent population (Hirsch & O’Donnell, 2001; Kahneman, Slovic, & Tversky, 1982; Shaughnessy, 1992). Another important purpose was to determine whether training instructors on misconceptions prior to implementing the activities would have an impact on student outcomes.
Review of the literature on the teaching and learning of probability A number of investigators used computer modeling to teach about probability and address students’ misconceptions. Krishnamachari (1988) explored how the use of computer simulations facilitated students’ comprehension of basic concepts of probability. She used workbooks with problems based on computer simulations. The assessment showed that students understood the concepts of probability explored in the study. Konold (1989) used a computer modeling intervention in an attempt to influence students’ misconceptions. The results were mixed; some students changed their interpretation while others persisted in their erroneous concepts. Garfield and delMas (1989) who used a computer program Coin Toss also obtained mixed results. While some students changed their ideas about variability after using their tutorial, others persisted in their misconceptions about sample size and variation. According to Snee (1993), computer simulations may not be helpful in changing misconceptions about probability in some students.
Nevertheless, other researchers found that teaching with computers can facilitate conceptual understanding of probability by allowing students to explore and represent stochastic models, manipulate the parameters, vary assumptions, and analyze data (Jones et al, 2007).
Keeler and Steinhorst (2001) used an approach in which numerical information was presented in the form of frequencies rather than fractions, decimals or percent. They drew on research by Cosmides and Tooby (1996) who found that students have an affinity for counting things. Keeler and Steinhorst emphasize the use of frequencies within the constructivist framework to develop students’ understanding of concepts such as independence and randomness. They also used simple probability density functions to allow students to focus on learning about probabilities associated with continuous random variables without getting bogged down in the mechanics of standard normal tables and z-scores. Keeler and Seinhorst did not report about any studies that measured the effectiveness of the above approach; they encouraged other researchers to investigate how students’ thinking about probability changes as a result of this approach.
DelMas and Bart (1989) employed an instructional activity that required one group of students to evaluate predictions based on their intuitive understanding while the other group was not required to perform this evaluation. They found that the group required to evaluate predictions did significantly better on a mastery test than the other group. Garfield and Chance Leonid Khazanov &Lucio Prado: Correcting Students’ Misconceptions about Probability in an Introductory College Statistics Course 26 Adults Learning Mathematics – An International Journal (1999) demonstrated significant changes in student performance on items designed to assess their understanding of how sample size is related to the shape and variability of the sampling distribution. The activity asked students to select among a set of graphic displays the one that most likely represented a sampling distribution from a population, given a specified sample size.
After making a prediction students used computer software to simulate the sampling distribution and compare the results to their predictions. While a significant number of students were found to overcome their misconceptions, subsequent studies revealed that for some of them the gains were short lived (delMas, Garfield, & Chance, 2002).
Hirsh & O’Donnell (2001) used the instrument they developed for identifying the representativeness bias to determine the effectiveness of a number of approaches aimed at correcting this misconception. In the instructional intervention they carried out, three methods of triggering and resolving cognitive conflict were employed: direct instruction, individual activities, and small group activities. The effectiveness of the three treatment models was compared to a control group that received instruction not specifically designed to trigger cognitive conflict. Although the researchers documented slightly better outcomes in sections were the cognitive conflict approach was utilized, the improvement did not reach the level of statistical significance.
Fast (2001) successfully used analogies and anchors to facilitate conceptual reconstruction of common misconceptions about probability in high school students. He found that his method led to stable outcomes. Students tested half a year after the treatment still retained the correct concepts and a significant proportion of them did not show evidence of misconceptions.
Lecoutre investigated the various opportunities for resolving the equiprobability bias. This bias involves attributing the same probability to different events in a random experiment regardless of the chances in favor or against them. According to Lecoutre: There exists an intra-subject vicariance of different cognitive models in various structurally isomorphic situations. (Lecoutre, 1992, p. 7) The specific activation of a particular model was found to be linked to the “surface features” of the situation; the chance context of a purely random situation evokes in most subjects an implicit model which is not adequate: random events are thought to be “equiprobable by nature.” Lecoutre succeeded in activating an appropriate combinatorial model by masking the random aspect of the situation in spite of the fact that the chance model is highly resistant to change. Nevertheless such activation was found to be superficial. The transfer of an appropriate model to an isomorphic random situation was less frequent than had been expected. Lecoutre surmised that the little transfer occurred because the subjects were unable to construct an abstract representation of the situation. In a follow-up study (Lecoutre & Rezrazi, 1998) it was found that in situations when subject succeed in constructing an adequate abstract representation for themselves (with situations of cognitive conflict, learning situations with feedback, anchoring, etc.) the result of such cognitive activity was, not infrequently, the abandonment of an inappropriate “chance model” and a more stable transfer to isomorphic situations.
Khazanov (2005) developed instructional materials aimed at addressing three misconceptions: representativeness, outcome orientation, and the equiprobability bias. He also developed a test instrument aimed at identifying the above misconceptions. In his 2005 study, Khazanov trained some instructors and then had them use the instructional materials in a statistics classroom at their own discretion. Instructors could also use their own materials in addition to those suggested by Khazanov. The results were inconclusive. While some instructors managed to achieve a significant reduction in certain misconceptions, the majority did not report any major changes. It was not clear whether these outcomes could be attributed to the tenacity ALM International Journal, Volume 5(1), pp. 23-35 Volume 5(1) – June 2010 27 of misconceptions, instructor inability to use the materials properly, or simply to the insufficient number of activities used to address the misconceptions. The Khazanov 2005 study was broad in scope and exploratory in nature. The study recommended that a more structured approach be utilized to determine whether the instructional materials used in the study were helpful in correcting some common misconceptions about probability. The present study was designed to implement the approach recommended by Khazanov.
Method The experimental phase of the study was conducted at an urban community college with a diverse student population situated in a large metropolitan area in the Eastern United States.
One of the principal missions of that college is to provide access to higher education for non-traditional adult learners who seek professional and economic advancement and personal fulfillment. According to the college’s office of institutional research, the median age of students is 22 years old.
The introductory college statistics course offered in this college satisfies the mathematics requirements for students majoring in such diverse fields as Accounting, Business, Liberal Arts, and Mental Health. The prerequisite for this course is basic algebra. A custom edition of a popular textbook was used that included all the topics represented in the syllabus.
The topics in this course are measures of central tendency, measures of dispersion, graphs, the binomial and normal distributions, sampling distributions of statistics, t-test, chi-square, correlation and regression.
In particular, the probability segment of the course includes fundamentals of probability, addition and multiplication rules, and conditional probability.
Six instructors teaching in aggregate 9 sections of the course were recruited in the spring semester of 2007, along with their students. Three of these instructors were randomly selected to attend the workshop on misconceptions that was designed and conducted by Leonid Khazanov. The other three instructors did not partake in training. All participating instructors were provided with the instructional materials and a list of common misconceptions with explanations and examples. All instructors had advanced degrees (doctoral or masters) in statistics, mathematics, or mathematics education obtained from accredited institutions. It must be noted, that the control group used in this study was drawn from sections of Introduction to Statistics offered at the college in 2005. This control group was used in the Khazanov 2005 study. Although the control group was not drawn from the same population as the treatment group, it is a valid comparison group for the following reasons: Both treatment and control sections were drawn from the same statistics course “Introduction to Statistics” offered at the same urban community college In both cohorts students taking the course majored in the same areas Prerequisite requirements for enrolling students in the course did not change from 2005 to 2007 The same syllabus for the course was used for both cohorts The same textbook was adopted by the department The same test instrument was used to identify the misconceptions Four out of nine instructors from the control group were in the treatment groups in the present study (by random selection, two of them were trained and the other two untrained); the remaining five instructors in the control group were similar to the instructors in the treatment group in terms of education and teaching experience Leonid Khazanov &Lucio Prado: Correcting Students’ Misconceptions about Probability in an Introductory College Statistics Course 28 Adults Learning Mathematics – An International Journal The workshop The workshop was designed to inform statistics teachers about common misconceptions of probability and equip them with effective strategies for resolving the misconceptions. The workshop was conducted by one of the investigators in June, 2006 in two 1.5 hour sessions.
Three instructors randomly selected from a list of six participating instructors attended the workshop which covered the following topics: What are misconceptions of probability; overview of prevalent misconceptions Why is it important to address misconceptions in the first statistics course Methods of assessing students’ misconceptions; the PRQ Methods of addressing misconceptions; the Instructional Materials Implementation. The pitfalls and how to avoid them. Case studies This workshop is described in detail by Khazanov (2008). The workshop was attended by all three of the selected participants. They were actively involved in the training and shared with one another their experiences in dealing with students’ naïve intuitions and erroneous concepts.
Each participant received a copy of the Instructional Materials and guidelines on how to use them in class.
Instructional Materials The Instructional Materials (IM) were prepared by one of the investigators (L. Khazanov). The misconceptions addressed in IM are representativeness, equiprobability bias and outcome orientation. The materials are comprised of hands-on activities and problems for discussions.
Some elements of the materials were designed by the investigator; others were borrowed from published sources and modified to varying degrees (Shaughnessy, 1977; Lecoutre, 1998; Fast, 2001; Konold, 1995). The proposed intervention included two hands-on activities and seven topics for discussions. The activities and problems were chosen so that they address all three misconceptions.
The theoretical underpinning for the instructional intervention based on the IM is the theory of ‘cognitive dissonance’ or ‘cognitive conflict. The theory holds that people prefer their multiple cognitions to be consistent with one another. When their cognitions are inconsistent or ‘dissonant’ people feel uneasy and are motivated to make them consistent (Bernstein, et al., 1994).
In addition to suggested activities and questions for discussions, the instructional materials contain guidelines for their implementation in the classroom. For each activity and question there is an indication of the misconception(s) treated, links to important concepts in the statistics course, suggested placement in the course, estimated time and possible organizational formats. Below is an example of activities utilized in the instructional intervention1.
Tossing a coin Misconceptions treated: representativeness Links to important concepts: sample space, independence, binomial probabilities Placement in the course: introduction to probability Time required: approximately 20 minutes if some work is done at home Materials: fair coins 1Reproduced with the permission of the author from Khazanov, 2008 ALM International Journal, Volume 5(1), pp. 23-35 Volume 5(1) – June 2010 29 Organizational format: instructor may choose to do his/her own demonstrations or have students work individually or in small groups Procedure: this activity begins with the teacher asking students to compare the probability of obtaining the following 3 sequences of heads and tails in tossing a coin four times: TTTT, THHT, and THTT. Say this to your students: When you flip a coin 4 times and record the outcomes, the following sequences may occur, among other sequences: THHT, TTTT, and THTT. Compare the probabilities of these sequences. You may state your position in words, for example, the probability of TTTT is less than the probability of THTT, or the probability of TTTT is equal to the probability of THTT. Alternately, you may estimate and compare the probabilities numerically.
Then the experiment (tossing a coin four times) is performed a large number of times and the results are recorded. It may be a good idea to ask the students to perform the experiment at home. Ask each student to do it 30 times at home and record the results. Make sure your students understand that they need to toss the coin 120 times to obtain 30 foursomes. This way you will have at least 600 sequences. Ask a group of diligent students to compile the data and compute the experimental probabilities for the above three sequences. The experimental probabilities for the aforementioned sequences are calculated using the relative frequency model. After that the students compare their predictions with the experimental results (some might observe discrepancies). The next step involves building a theoretical model for this experiment. Students should be asked (and if necessary helped) to list all possible sequences of length four and compute the probability of each sequence using the multiplication principle. Finally the theoretical probabilities are to be compared with the calculated relative frequencies and guesses. It is likely that students who hold the misconceptions of representativeness would predict a greater probability for THHT, than for TTTT. Both, the result of the experiment and the theoretical model are expected to contribute to the resolution of the misconception.
This activity can be performed in a variety of organizational formats depending on instructor preferences. One possible format includes a teacher demonstration followed by students working in small groups. As was mentioned earlier, you may ask the students do some work at home. If you prefer the small group format, I can provide suggestions as to how students might be grouped to maximize their learning. In addition to addressing the misconceptions of representativeness, this activity can be extended to address the equiprobability bias; students may be guided to observe that different random events may have different probabilities (for example, the probability of getting exactly 2 heads without regard to order is greater than the probability of getting exactly 0 heads).
Enrichment: have students explore various outcomes when a coin is flipped 6 times or 4 coins are tossed simultaneously (same denomination or different denominations). Students may be given a link to the website This website allows students to instantly view the results of tossing a large number of coins.
They can make a variety of observations. For example, break down the long sequence into foursomes and count the number of sequences HTTH and HHHH.
Test Instrument The test instrument entitled The Probability Reasoning Questionnaire (PRQ) was developed by one of the authors (L. Khazanov) and is discussed in detail in Khazanov (2005). The instrument has 16 items. Each test item is comprised of two multiple-choice parts: the principal question and justification. This approach has several advantages over the one-part items. A number of Leonid Khazanov &Lucio Prado: Correcting Students’ Misconceptions about Probability in an Introductory College Statistics Course 30 Adults Learning Mathematics – An International Journal studies report that correct conceptual understanding is significantly overstated if no justification for the chosen answer is required (Konold, 1995; Hirsch & O’Donnell, 2001; Rubel, 2002).
Konold (1995) described how a correct answer to a question that involved comparison of probabilities could be obtained using an erroneous concept. On the other hand, answers to both parts that are consistent with a particular misconception serve as strong evidence that the misconception is indeed part of the students’ conceptualization of certain random processes and events.
The instrument was designed to test students for three highly prevalent misconceptions: representativeness, equiprobability and outcome orientation. To increase the utility of the instrument, an effort was made to create distracters consistent with at least two misconceptions on each item. The careful selection of items and creation of suitable distracters achieved this objective. When creating distracters, priority was given to common explanations consistent with one of the three misconceptions subject to investigation and treatment. These common erroneous explanations were extracted from the literature (Shaughnessy, 1992; Lecoutre, 1992, 1998; Konold, 1995, Konold, et al., 1993), as well as the investigator’s own teaching experience. Distracters that reflect some other common misconceptions and skill deficiencies have also been used, where appropriate. About half of the items on this instrument are original.
The other half are similar to those proposed by Konold (1990), Garfeild (1998), and Hirsch and O’Donnell (2001). However, the majority of the borrowed items were modified. The modification normally included adding or replacing distracters, rephrasing the problem or editing it.
The validity of the instrument was confirmed by three experts who have advanced degrees in mathematics or statistics. The experts were asked to pass judgment on how well the questions reflect the construct domain, and whether they are appropriate at the introductory college level. They were also asked to identify correct responses, as well as responses consistent with the specific misconceptions. This method known as “backward translation” is recommended by Sundre (2003). Items about which the experts disagreed were revised until a consensus was achieved. The experts also established the cutoff scores and estimated the time required to complete the test. The consensus was that a student would be considered to have a specific misconception if he/she scored two or more misconception points for that misconception on the test. The instrument was then subject to a known-group validation procedure, another accepted validation method recommended in the literature (Hirsch & O’Donnell, 2001). An agreement coefficient (Berk, 1984; Subcoviak, 1988; Garfield, 1998) was calculated for each misconception to establish the reliability of the test.
Administration of Test The test instrument was administered in class at the end of the spring semester of 2007 in the nine participating sections of the Introduction to Statistics course. Students were given 40 minutes to complete the test. The test was administered on different days of the week for different sections. To maintain the integrity of the study, the instructors were directed to collect all question sheets after the administration of the test and not to discuss it with colleagues or students. In such a way, the independence of the outcomes was assured.
ALM International Journal, Volume 5(1), pp. 23-35 Volume 5(1) – June 2010 31 Results A summary of results is presented in table 1.
Table 1: Percentages of different student groups by categories of probability misconceptions Probability Misconceptions CGNA (263) TFA (106) UTFA (65) Representativeness 67% (176) 54% (57) 68% (44) Equiprobability 83% (218) 80% (85) 80% (52) Outcome orientation 37% (97) 27% (29) 34% (22) CGNA=Control Group of College Students (No Activities); TFA=Trained Faculty Implemented Activities; UTFA= Untrained Faculty Implemented Activities The first column of table 1 contains data based on a large sample of the college population who took introduction to statistics in an earlier semester. Instructors teaching the students in this sample did not receive any training and did not implement any targeted activities aimed at correcting the misconceptions. The second column contains data aggregated from the five sections whose instructors attended the workshop before implementing the activities in the classroom. The third column includes the data from the four sections where the instructors implemented the activities, but had not attended the workshop. The three categorical variables in this study are specified in the header of table 1. The observed values of these variables for each specific misconception are recorded in the respective columns. For analyzing the data, z -test for comparing two sample proportions from the population was used. The results show that trained instructors achieved significantly better outcomes than the control group in correcting the representativeness bias (p-value 0.0089) and outcome orientation (p-value 0.0404). There was no significant difference between the treatment and control groups on the equiprobability bias (p-value 0.7299). Notably, this study did not find any significant evidence that implementing the activities without training was instrumental in resolving any of the misconceptions that students might have.
Figure 1 illustrates the data presented in Table 1. Each triple of bars corresponds to a specific treatment category. The height of the bars reflects the prevalence of the misconceptions after the intervention.
Figure 1: Misconceptions vs. Treatment Categories Discussion Misconceptions vs. Treatment Categories CGNA (263) TFA (106) UTFA (65) 0 20 40 60 80 100 Representativeness Equiprobability Outcome orientation Percentile Leonid Khazanov &Lucio Prado: Correcting Students’ Misconceptions about Probability in an Introductory College Statistics Course 32 Adults Learning Mathematics – An International Journal In this section we will discuss a number of points in our findings that warrant explanation. First, why did the instructors who had been trained prior to implementing the activities overall achieved better results than those who had not been trained? Second, why was there no significant improvement in the resolution of the equiprobability bias in any of the treatment groups? And third, why a large proportion of students in the treatment sections still did not have their misconceptions corrected?
A possible explanation of better outcomes for students taught by trained instructors is that instructors who belonged to the trained cohort gave more thought to the misconceptions and had a better understanding of their nature and, hence, how to address them in instruction. This might have affected how they led and summarized students' discussions and activities and addressed students' misconceptions when confronted with them. Another possibility is that trained instructors were more alert to misconceptions when implementing other activities in the course; this could have reinforced the impact of the activities and affected students’ long term outcomes. Notably, the test instrument was administered at the end of the semester. Hence, the outcomes might have reflected the aggregate effect of teaching the course while being “misconception vigilant” (Khazanov, 2008), rather than just the net effect of the activities from the Instructional Materials. Although, as far as we know, none of the instructors targeted the misconceptions directly in any other activities beyond those recommended by the investigators, they might have kept an eye on misconceptions better than untrained instructors. We have anecdotal evidence to corroborate this speculation. One instructor shared with us that the idea of being “misconception vigilant” emphasized in the workshop had a deep and lasting effect on his teaching.
Our findings did not show a significant response to the attempt to correct the equiprobability bias. (Notably, this bias has the highest prevalence of the three considered in this study). A possible explanation is that there were not enough activities to address this bias which appears to be more difficult to correct than other biases. To learn their way out of the equiprobability bias, students need, in addition to understanding chance events in general, to develop the ability to quantify random events and compare probabilities numerically (Khazanov, 2005). Based on our findings, it appears that in order to resolve the equiprobability bias students would need more exposure to the unequally likely probability models. (These models are not sufficiently represented in many popular textbooks). Additionally, students must become proficient in the fundamentals of combinatorics.
It must be noted, that in spite of the best effort put forth by our participants to correct their students’ misconceptions, a large proportion of students still retained some of them at the end of the term. When we come to think about the factors that detracted from the effectiveness of the intervention, time constraint appears to be a major factor.
An introductory statistics course is replete with topics that need to be covered. Because of the time constraint, we had to limit the number of activities and topics for discussions included in the intervention.
Another important factor that surfaced in the course of carrying out the intervention was that some instructors were unable to trigger cognitive conflict in all students; some students could not see the discrepancy between their concept and the evidence presented. Triggering cognitive conflict is a powerful approach to resolving misconceptions. Its effectiveness has been confirmed in a large number of studies. However, it does not necessarily work for all students. It is possible that for some students, gradual knowledge refinement and reorganization would be a more productive approach to correcting their probability misconceptions (Roschelle, 1995).
This speculation needs to be tested in future studies. The investigators have already begun to revise the IM to incorporate this approach.
And finally, a plausible reason why so many students still had misconceptions in spite of the intervention is that people’s naïve concepts about probability are difficult to change. A deep ALM International Journal, Volume 5(1), pp. 23-35 Volume 5(1) – June 2010 33 insight into this issue is offered by Steven Pinker in his book “How Mind Works” (1999).
Pinker pointed out that human brain has undergone extensive growth about 100,000 years ago.
At that time our brain has evolved modules instrumental in the processing of important survival information. Some probability misconceptions are perfectly consistent with the kind of probability experiences our ancestors had and which we ourselves still experience in everyday life. For example, the belief in Gamblers Fallacy, which is a manifestation of the representativeness bias, may be explained by the experiences created by observations such as ‘a sunny day is more likely to follow a series of rainy days’ or ‘it is more likely not to see a wild animal after several of them passed by’. Change is a norm in nature, not an exception. Thus believing in change had some survival value and is therefore deeply ingrained in human psyche.
To overcome the misconceptions, students sometimes have to go against their natural instincts, and this proves to be difficult.
In conclusion, we would like to map out a few directions for future research in the area of probability misconceptions. We believe that it is important to conduct a study designed with the objective of getting a significant improvement in the resolution of the equiprobability bias, which proved to be particularly resistant to change. Another possible direction is to conduct a qualitative study that will include classroom observations and interviews with instructors and students in order to further our understanding of factors that affect students’ conceptual understanding of probability. It may also be useful to develop a test instrument to identify other misconceptions, especially those effecting comprehension of inferential statistics.
Acknowledgements This paper is based upon research supported by the Professional Staff Congress and the Research Foundation of the City University of New York under Grant No. 68090-00 37. Any opinions, findings, conclusions or recommendations are those of the authors and do not necessarily reflect the position of the Professional Staff Congress or the CUNY Research Foundation. The authors also wish to thank the anonymous referees of the paper for their valuable comments and suggestions.
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AoPS Wiki:Competition ratings
This page contains an approximate estimation of the difficulty level of various competitions. It is designed with the intention of introducing contests of similar difficulty levels (but possibly different styles of problems) that readers may like to try to gain more experience.
Each entry groups the problems into sets of similar difficulty levels and suggests an approximate difficulty rating, on a scale from 1 to 10 (from easiest to hardest). Note that many of these ratings are not directly comparable, because the actual competitions have many different rules; the ratings are generally synchronized with the amount of available time, etc. Also, due to variances within a contest, ranges shown may overlap. A sample problem is provided with each entry, with a link to a solution.
If you have some experience with mathematical competitions, we hope that you can help us make the difficulty rankings more accurate. Currently, the system is on a scale from 1 to 10 where 1 is the easiest level, e.g. early AMC problems and 10 is hardest level, e.g. China IMO Team Selection Test. When considering problem difficulty, put more emphasis on problem-solving aspects and less so on technical skill requirements.
Note that "easier" contests (e.g. MATHCOUNTS) have shorter time lengths, meaning being successful emphasizes speed.
Contents
1 Scale
2 Competitions
2.1 Introductory Competitions
2.1.1 MOEMS
2.1.2 AMC 8
2.1.3 MATHCOUNTS
2.1.4 AMC 10
2.1.5 CEMC Multiple Choice Tests
2.1.6 CEMC Fryer/Galois/Hypatia
2.2 Problem Solving Books for Introductory Students
2.3 Intermediate Competitions
2.3.1 AMC 12
2.3.2 AIME
2.3.3 ARML
2.3.4 HMMT (November)
2.3.5 CEMC Euclid
2.3.6 COMC
2.3.7 Purple Comet
2.3.8 LMT
2.4 Problem Solving Books for Intermediate Students
2.5 Beginner Olympiad Competitions
2.5.1 USAMTS
2.5.2 Indonesia MO
2.5.3 CentroAmerican Olympiad
2.5.4 JBMO
2.6 Olympiad Competitions
2.6.1 USAJMO
2.6.2 HMMT (February)
2.6.3 Canadian MO
2.6.4 Iberoamerican Math Olympiad
2.6.5 APMO
2.6.6 Balkan MO
2.7 Hard Olympiad Competitions
2.7.1 USAMO
2.7.2 USA TST
2.7.3 Putnam
2.7.4 Korea Final Round
2.7.5 China TST (hardest problems)
2.7.6 IMO
2.7.7 IMO Shortlist
Scale
All levels are estimated and refer to averages. The following is a rough standard based on the USA tier system AMC 8 – AMC 10 – AMC 12 – AIME – USAMO/USAJMO - IMO, representing Middle School – Junior High – High School – Challenging High School – Olympiad levels. Other contests can be interpolated against this.
Notes:
Multiple choice tests like AMC are rated as though they are free-response. Test-takers can use the answer choices as hints, and so correctly answer more AMC questions than MATHCOUNTS or AIME problems of similar difficulty.
Some Olympiads are taken in 2 sessions, with 2 similarly difficult sets of questions, numbered as one set. For these the first half of the test (questions 1-3) is similar difficulty to the second half (questions 4-6).
Scale
1: Problems strictly for beginner, on the easiest elementary school or middle school levels (MOEMS, MATHCOUNTS School, AMC 8 1-10, AMC 10 1-10, easier AMC 12 1-5, and others that involve standard techniques introduced up to the middle school level), most traditional middle/high school word problems.
1.5: Problems for stronger beginner students, on the level of the middling problems in most middle school contests (AMC 8 11-20, harder AMC 10 1-10, AMC 12 1-5, and those others that force students to apply their school-level knowledge to slightly more challenging problems), traditional middle/high school word problems with more complex problem solving.
2: For motivated beginners, harder questions from the previous categories (AMC 8 21-25, MATHCOUNTS Chapter (Sprint 21-30, Target 6-8), MATHCOUNTS States/Nationals, AMC 10 11-15, AMC 12 5-10, easiest AIME 1-3)
2.5: More advanced beginner problems, hardest questions from previous categories (Harder AMC 8 21-25, harder MATHCOUNTS States questions, AMC 10 16-20, AMC 12 11-15, usual AIME 1-3)
3: Early intermediate problems that require more creative thinking (harder MATHCOUNTS National questions, AMC 10 21-25, AMC 12 15-20, hardest AIME 1-3, usual AIME 4-6).
4: Intermediate-level problems (AMC 12 21-25, hardest AIME 4-6, usual AIME 7-10).
5: More difficult AIME problems (11-13), simple proof-based Olympiad-style problems (early JBMO questions, easiest USAJMO 1/4).
6: High-leveled AIME-styled questions (14/15). Introductory-leveled Olympiad-level questions (harder USAJMO 1/4 and easier USAJMO 2/5, easier USAMO and IMO 1/4).
7: Tougher Olympiad-level questions, may require more technical knowledge (harder USAJMO 2/5 and most USAJMO 3/6, extremely hard USAMO and IMO 1/4, easy-medium USAMO and IMO 2/5).
8: High-level Olympiad-level questions (medium-hard USAMO and IMO 2/5, easiest USAMO and IMO 3/6).
9: Expert Olympiad-level questions (average USAMO and IMO 3/6).
9.5: The hardest problems appearing on Olympiads which the strongest students could reasonably solve (hard USAMO and IMO 3/6).
10: Historically hard problems, generally unsuitable for very hard competitions (such as the IMO) due to being exceedingly tedious, long, and difficult (e.g. very few students are capable of solving on a worldwide basis).
Examples
For reference, here are some sample problems from each of the difficulty levels 1-10:
<1: Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum? (2003 AMC 8, Problem 1)
1: How many integer values of satisfy ? (2021 Spring AMC 10B, Problem 1)
1.5: A number is called flippy if its digits alternate between two distinct digits. For example, and are flippy, but and are not. How many five-digit flippy numbers are divisible by (2020 AMC 8, Problem 19)
2: A fair -sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number? (2021 Spring AMC 10B, Problem 18)
2.5:, , are three piles of rocks. The mean weight of the rocks in is pounds, the mean weight of the rocks in is pounds, the mean weight of the rocks in the combined piles and is pounds, and the mean weight of the rocks in the combined piles and is pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles and ? (2013 AMC 12A, Problem 16)
3: Triangle with and has area . Let be the midpoint of , and let be the midpoint of . The angle bisector of intersects and at and , respectively. What is the area of quadrilateral ? (2018 AMC 10A, Problem 24)
3.5: Find the number of integer values of in the closed interval for which the equation has exactly one real solution. (2017 AIME II, Problem 7)
4: Define a sequence recursively by andfor all nonnegative integers Let be the least positive integer such thatIn which of the following intervals does lie?
![Image 57: $\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } 729,\infty)$
(2019 AMC 10B, Problem 24 and 2019 AMC 12B, Problem 22)
4.5: Find, with proof, all positive integers for which is a perfect square. (USAJMO 2011/1)
5: Find all triples of real numbers such that the following system holds: (JBMO 2020/1)
5.5: Triangle has , , , and . Let , , and be the orthocenter, incenter, and circumcenter of , respectively. Assume that the area of pentagon is the maximum possible. What is ? (2011 AMC 12A, Problem 25)
6: Let be an acute triangle with circumcircle and let be the intersection of the altitudes of Suppose the tangent to the circumcircle of at intersects at points and with and The area of can be written in the form where and are positive integers, and is not divisible by the square of any prime. Find (2020 AIME I, Problem 15)
6.5: Rectangles and are erected outside an acute triangle Suppose thatProve that lines and are concurrent. (USAMO 2021/1, USAJMO 2021/2)
7: We say that a finite set in the plane is balanced if, for any two different points , in , there is a point in such that . We say that is centre-free if for any three points , , in , there is no point in such that .
Show that for all integers , there exists a balanced set consisting of points.
Determine all integers for which there exists a balanced centre-free set consisting of points.
(IMO 2015/1)
7.5: Let be the set of integers. Find all functions such thatfor all with . (USAMO 2014/2)
8: For each positive integer , the Bank of Cape Town issues coins of denomination . Given a finite collection of such coins (of not necessarily different denominations) with total value at most most , prove that it is possible to split this collection into or fewer groups, such that each group has total value at most . (IMO 2014/5)
8.5: Let be the incentre of acute triangle with . The incircle of is tangent to sides , and at and , respectively. The line through perpendicular to meets at . Line meets again at . The circumcircles of triangle and meet again at .
Prove that lines and meet on the line through perpendicular to . (IMO 2019/6)
9: Let be a positive integer and let be a finite set of odd prime numbers. Prove that there is at most one way (up to rotation and reflection) to place the elements of around the circle such that the product of any two neighbors is of the form for some positive integer . (IMO 2022/3)
9.5: An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following is an anti-Pascal triangle with four rows which contains every integer from to . Does there exist an anti-Pascal triangle with rows which contains every integer from to ? (IMO 2018/3)
10: Prove that there exists a positive constant such that the following statement is true: Consider an integer , and a set of points in the plane such that the distance between any two different points in is at least 1. It follows that there is a line separating such that the distance from any point of to is at least .
(A line separates a set of points S if some segment joining two points in crosses .) (IMO 2020/6)
Competitions
Introductory Competitions
Most middle school and first-stage high school competitions would fall under this category. Problems in these competitions are usually ranked from 1 to 3. A full list is available.
MOEMS
Division E: 1The whole number is divisible by . leaves a remainder of when divided by or . What is the smallest value that can be? (Solution)
Division M: 1.5The value of a two-digit number is times more than the sum of its digits. The units digit is 1 more than twice the tens digit. Find the two-digit number. (Solution)
AMC 8
Problem 1 - Problem 12: 1-1.5The coordinates of are , , and , with . The area of is 12. What is the value of ? (Solution)
Problem 13 - Problem 25: 1.25-2.5How many four-digit numbers have all three of the following properties?
(I) The tens and ones digit are both 9. (II) The number is 1 less than a perfect square. (III) The number is the product of exactly two prime numbers. (Solution)
MATHCOUNTS
Countdown: 0.5-1.5.
Sprint: 1-1.5 (School) 1-2(Chapter), 1.5-3 (State), 2-3.5 (National)
Target: 1-2 (School/Chapter), 1.5-2 (State), 2-2.5 (National)
Team: 1-1.5 (School)1-2(Chapter), 1-3 (State), 1.5-3.5 (National)
AMC 10
Since ~2020, AMC 10 shares about half its questions with AMC 12, but places them 0-3 spots later in the test.
Problem 1 - 10: 1-1.5A rectangular box has integer side lengths in the ratio . Which of the following could be the volume of the box? (Solution)
Problem 11 - 20: 1.5-2.5For some positive integer , the repeating base- representation of the (base-ten) fraction is . What is ? (Solution)
Problem 21 - 25: 3-4.5The vertices of an equilateral triangle lie on the hyperbola , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle? (Solution)
Isosceles trapezoid ABCD has parallel sides and , with and . There is a point P on the plane such that ,,, and . What is ? (Solution)
CEMC Multiple Choice Tests
This covers the CEMC Gauss, Pascal, Cayley, and Fermat tests.
Part A: 0.5-1.5How many different 3-digit whole numbers can be formed using the digits 4, 7, and 9, assuming that no digit can be repeated in a number? (2015 Gauss 7 Problem 10)
Part B: 1-2Two lines with slopes and intersect at . What is the area of the triangle formed by these two lines and the vertical line ? (2017 Cayley Problem 19)
Part C (Gauss/Pascal): 2-2.5Suppose that , where , , and are positive integers with in lowest terms. What is the sum of the digits of the smallest positive integer for which is a multiple of 1004? (2014 Pascal Problem 25)
Part C (Cayley/Fermat): 2.5-3Wayne has 3 green buckets, 3 red buckets, 3 blue buckets, and 3 yellow buckets. He randomly distributes 4 hockey pucks among the green buckets, with each puck equally likely to be put in each bucket. Similarly, he distributes 3 pucks among the red buckets, 2 pucks among the blue buckets, and 1 puck among the yellow buckets. Once he is finished, what is the probability that a green bucket contains more pucks than each of the other 11 buckets? (2018 Fermat Problem 24)
CEMC Fryer/Galois/Hypatia
Problem 1-2: 1-2
Problem 3-4 (early parts): 1.5-2.5
Problem 3-4 (later parts): 3-5
Problem Solving Books for Introductory Students
Remark: There are many other problem books for Introductory Students that are not published by AoPS. Typically the rating on the left side is equivalent to the difficulty of the easiest review problems and the difficulty on the right side is the difficulty of the hardest challenge problems. The difficulty may vary greatly between sections of a book.
Prealgebra by AoPS0.5-1.5
Introduction to Algebra by AoPS0.5-3
Introduction to Counting and Probability by AoPS0.5-3
Introduction to Number Theory by AoPS0.5-3
Introduction to Geometry by AoPS0.5-4.5
105 Algebra by Awesome Math 1.5-5
106 Geometry by Awesome Math 1-6
112 Combinatorial by Awesome Math 1-5
111 Algebra and Number Theory by Awesome Math 1-6
Intermediate Competitions
This category consists of all the non-proof math competitions for the middle stages of high school. The difficulty range would normally be from 3 to 6. A full list is available [Category:Intermediate mathematic competitions|here].
AMC 12
Problem 1-10: 1.5-2What is the value of (Solution)
Problem 11-20: 2.5-3.5An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point? (Solution)
Problem 21-25 (Easier): 3-4Consider polynomials of degree at most , each of whose coefficients is an element of . How many such polynomials satisfy ? (Solution)
Problem 21-25 (Harder): 4.5-6Semicircle has diameter of length . Circle lies tangent to at a point and intersects at points and . If and then the area of equals where and are relatively prime positive integers, and is a positive integer not divisible by the square of any prime. What is ? (Solution)
AIME
Problem 1 - 5: 3Consider the integer Find the sum of the digits of . (Solution)
Problem 6 - 9: 4An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, the ant does not return to its previous vertex, but chooses to move to one of the other two adjacent vertices. All choices are selected at random so that each of the possible moves is equally likely. The probability that after exactly moves that ant is at a vertex of the top face on the cube is , where and are relatively prime positive integers. Find (Solution)
Problem 10 - 12: 5Triangle has side lengths and Circle passes through and is tangent to line at Circle passes through and is tangent to line at Let be the intersection of circles and not equal to Then where and are relatively prime positive integers. Find (Solution)
Problem 13 - 15: 6Let Let be the distinct zeros of and let for where and and are real numbers. Let where and are integers and is not divisible by the square of any prime. Find . (Solution)
ARML
Individuals, Problem 1: 2
Individuals, Problems 2, 3, 4, 5, 7, and 9: 3
Individuals, Problems 6 and 8: 4
Individuals, Problem 10: 5.5
Team/power, Problem 1-5: 3.5
Team/power, Problem 6-10: 5
HMMT (November)
Individual Round, Problem 6-8: 4
Individual Round, Problem 10: 4.5
Team Round: 4-5
Guts: 3.5-5.25
CEMC Euclid
Problem 1-6: 1-3
Problem 7-10: 3-6
COMC
Part A: 1-2.5
Part B: 2.5-4
Part C: 2-5
Purple Comet
Problems 1-10 (MS): 1.5-3
Problems 11-17 (MS): 3-4.5
Problems 18-20 (MS): 4-4.75
Problems 1-10 (HS): 1.5-3.5
Problems 11-20 (HS): 3.5-4.75
Problems 21-30 (HS): 4.5-6
LMT
Easy Problems: 0.5Let trapezoid be such that . Additionally, , , and . Find .
Medium Problems: 2-4Let have side lengths , , and . Let the angle bisector of meet the circumcircle of at a point . Determine the area of .
Hard Problems: 5-7A magic board can toggle its cells between black and white. Define a pattern to be an assignment of black or white to each of the board’s cells (so there are patterns total). Every day after Day 1, at the beginning of the day, the board gets bored with its black-white pattern and makes a new one. However, the board always wants to be unique and will die if any two of its patterns are less than cells different from each other. Furthermore, the board dies if it becomes all white. If the board begins with all cells black on Day 1, compute the maximum number of days it can stay alive.
Problem Solving Books for Intermediate Students
Remark: As stated above, there are many books for Intermediate students that have not been published by AoPS. Below is a list of intermediate books that AoPS has published and their difficulty. The left-hand number corresponds to the difficulty of the easier review problems, while the right-hand number corresponds to the difficulty of the hardest challenge problems.
Intermediate Algebra by AoPS1-6, may vary across chapters
Intermediate Counting & Probability by AoPS1-6, may vary across chapters
Precalculus by AoPS2-7, may vary across chapters
Calculus by AoPS3-8 (not an olympiad book)
108 Algebra by Awesome Math 2.5-8
107 Geometry by Awesome Math 2-8
Beginner Olympiad Competitions
This category consists of beginning Olympiad math competitions. Most junior and first stage Olympiads fall under this category. The range from the difficulty scale would be around 4 to 6. A full list is available.
USAMTS
USAMTS generally has a different feel to it than olympiads, and is mainly for proofwriting practice instead of olympiad practice depending on how one takes the test. USAMTS allows an entire month to solve problems, with internet resources and books being allowed. However, the ultimate gap is that it permits computer programs to be used, and that Problem 1 is not a proof problem. However, it can still be roughly put to this rating scale:
Problem 1-2: 3-4Find three isosceles triangles, no two of which are congruent, with integer sides, such that each triangle’s area is numerically equal to 6 times its perimeter. (Solution)
Problem 3-5: 4-6Call a positive real number groovy if it can be written in the form for some positive integer . Show that if is groovy, then for any positive integer , the number is groovy as well. (Solution)
Indonesia MO
Problem 1/5: 3.5In a drawer, there are at most balls, some of them are white, the rest are blue, which are randomly distributed. If two balls were taken at the same time, then the probability that the balls are both blue or both white is . Determine the maximum amount of white balls in the drawer, such that the probability statement is true? (Solution)
Problem 2/6: 4.5Find the lowest possible values from the function for any real numbers . (Solution)
Problem 3/7: 5A pair of integers is called good if Given 2 positive integers which are relatively prime, prove that there exists a good pair with and , but and . (Solution)
Problem 4/8: 6Given an acute triangle . The incircle of triangle touches respectively at . The angle bisector of cuts and respectively at and . Suppose is one of the altitudes of triangle , and be the midpoint of .
(a) Prove that and are perpendicular with the angle bisector of . (b) Show that is a cyclic quadrilateral. (Solution)
CentroAmerican Olympiad
Problem 1,4: 3.5Find all three-digit numbers (with ) such that is a divisor of 26. (Solution)
Problem 2,5: 4.5Show that the equation has no integer solutions. (Solution)
Problem 3/6: 6Let be a convex quadrilateral. , and , , and are points on , , and respectively, such that . If , , show that . (Solution)
JBMO
Problem 1: 4Find all real numbers such that
Problem 2: 4.5-5Let be a convex quadrilateral with , and . The diagonals intersect at point . Determine the measure of .
Problem 3: 5Find all prime numbers , such that . (Solution:
-paixiao
Problem 4: 6A table is divided into white unit square cells. Two cells are called neighbors if they share a common side. A move consists in choosing a cell and changing the colors of neighbors from white to black or from black to white. After exactly moves all the cells were black. Find all possible values of .
Olympiad Competitions
This category consists of standard Olympiad competitions, usually ones from national Olympiads. Average difficulty is from 5 to 8. A full list is available.
USAJMO
Problem 1/4: 5There are bowls arranged in a row, numbered through , where and are given positive integers. Initially, each of the first bowls contains an apple, and each of the last bowls contains a pear.
A legal move consists of moving an apple from bowl to bowl and a pear from bowl to bowl , provided that the difference is even. We permit multiple fruits in the same bowl at the same time. The goal is to end up with the first bowls each containing a pear and the last bowls each containing an apple. Show that this is possible if and only if the product is even. (Solution)
Problem 2/5: 6-6.5Let be positive real numbers such that . Prove that (Solution)
Problem 3/6: 7Two rational numbers and are written on a blackboard, where and are relatively prime positive integers. At any point, Evan may pick two of the numbers and written on the board and write either their arithmetic mean or their harmonic mean on the board as well. Find all pairs such that Evan can write on the board in finitely many steps. (Solution)
HMMT (February)
Individual Round, Problem 1-5: 5
Individual Round, Problem 6-10: 5.5-6
Team Round: 7.5
HMIC: 8
Canadian MO
Problem 1: 5.5
Problem 2: 6
Problem 3: 6.5
Problem 4: 7-7.5
Problem 5: 7.5-8
Iberoamerican Math Olympiad
Problem 1/4: 5.5
Problem 2/5: 6.5
Problem 3/6: 7.5
APMO
Problem 1: 6.5-7
Problem 2: 7
Problem 3: 7.5
Problem 4: 7.5-8
Problem 5: 8.5-9
Balkan MO
Problem 1: 5Solve the equation in positive integers.
Problem 2: 6.5Let be a line parallel to the side of a triangle , with on the side and on the side . The lines and meet at point . The circumcircles of triangles and meet at two distinct points and . Prove that .
Problem 3: 7.5A rectangle is partitioned into unit squares. The centers of all the unit squares, except for the four corner squares and eight squares sharing a common side with one of them, are coloured red. Is it possible to label these red centres in such way that the following to conditions are both fulfilled
the distances are all equal to the closed broken line has a centre of symmetry?
Problem 4: 8Denote by the set of all positive integers. Find all functions such that
Hard Olympiad Competitions
This category consists of harder Olympiad contests. Difficulty is usually from 7 to 10. A full list is available.
USAMO
Problem 1/4: 6-7Let be a convex polygon with sides, . Any set of diagonals of that do not intersect in the interior of the polygon determine a triangulation of into triangles. If is regular and there is a triangulation of consisting of only isosceles triangles, find all the possible values of . (Solution)
Problem 2/5: 7-8Three nonnegative real numbers , , are written on a blackboard. These numbers have the property that there exist integers , , , not all zero, satisfying . We are permitted to perform the following operation: find two numbers , on the blackboard with , then erase and write in its place. Prove that after a finite number of such operations, we can end up with at least one on the blackboard. (Solution)
Problem 3/6: 8-9Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree with real coefficients is the average of two monic polynomials of degree with real roots. (Solution)
USA TST
Problem 1/4/7: 6.5-7
Problem 2/5/8: 7.5-8
Problem 3/6/9: 8.5-9
Putnam
Problem A/B,1-2: 7Find the least possible area of a concave set in the 7-D plane that intersects both branches of the hyperparabola and both branches of the hyperbola (A set in the plane is called convex if for any two points in the line segment connecting them is contained in ) (Solution)
Problem A/B,3-4: 8Let be an matrix all of whose entries are and whose rows are mutually orthogonal. Suppose has an submatrix whose entries are all Show that . (Solution)
Problem A/B,5-6: 9For any , define the set . Show that there are no three positive reals such that . (Solution)
Korea Final Round
Problem 1/4: 7
Problem 2/5: 7.5-8
Problem 3/6: 8-9
China TST (hardest problems)
Problem 1/4: 8-8.5Given an integer prove that there exist odd integers and a positive integer such that
Problem 2/5: 9Given a positive integer and real numbers such that prove that for any positive real number
Problem 3/6: 9.5-10Let be an integer and let be non-negative real numbers. Define for . Prove that
IMO
Problem 1/4: 6-7Let be the circumcircle of acute triangle . Points and are on segments and respectively such that . The perpendicular bisectors of and intersect minor arcs and of at points and respectively. Prove that lines and are either parallel or they are the same line. (Solution)
Problem 2/5: 7-8Let be a polynomial of degree with integer coefficients, and let be a positive integer. Consider the polynomial , where occurs times. Prove that there are at most integers such that . (Solution)
Problem 3/6: 9-10
Let be an equilateral triangle. Let be interior points of such that , , , andLet and meet at let and meet at and let and meet at
Prove that if triangle is scalene, then the three circumcircles of triangles and all pass through two common points.
(Note: a scalene triangle is one where no two sides have equal length.)
(..this is one of the hardest problems and we had to define a scalene triangle.)
IMO Shortlist
Problem 1-2: 6-6.5
Problem 3-4: 6.5-7.5
Problem 5-6: 8-9
Problem 7+: 8.5-10
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187925 | https://k12mathworksheets.com/worksheets/volume-and-surface-area-of-rectangular-prisms/ | Elementary School
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Volume and Surface Area of Rectangular Prisms
Volume and Surface Area of Rectangular Prisms
Free and premium, printable “Volume and Surface Area of Rectangular Prisms Worksheets!”
Imagine you want to wrap a present. First, you need to put your gift into a box, right? The amount of fillable space inside the box is called the volume. Now imagine wrapping the box. The amount of space that needs to be covered by paper is called the surface area.
Volume = length • width • height
Locate your three values, multiply them together, and voila. Volume!
Surface area is a little different. Because a rectangular prism has six sides (or faces), you need to find the surface area of each side, and then add those 6 values together.
Free and premium, printable “Volume and Surface Area of Rectangular Prisms Worksheets!”
Imagine you want to wrap a present. First, you need to put your gift into a box, right? The amount of fillable space inside the box is called the volume. Now imagine wrapping the box. The amount of space that needs to be covered by paper is called the surface area.
Volume = length • width • height
Locate your three values, multiply them together, and voila. Volume!
Surface area is a little different. Because a rectangular prism has six sides (or faces), you need to find the surface area of each side, and then add those 6 values together.
Resource Details
Here’s how to calculate the volume and surface area of a rectangular prism:
EXAMPLE: Find the volume and surface area of the rectangular prism below.
VOLUME:
The formula for the volume of a rectangular prism is:
Volume = length • width • height
In our figure, the length is 6 m, the width is 2 m, and the height is 4 m.
All we need to do is substitute the values into our formula.
V= l • w • h
V = (6 m)(2 m)(4 m)
V = (12 m²)(4m)
V = 48 m³
SURFACE AREA:
To find the surface area of the entire figure, find the area of each face.
Area of bottom = 6 m • 2 m = 12 m²
Area of top = 6 m • 2 m = 12 m²
Area of right side = 2 m • 4 m = 8 m²
Area of left side = 2 m • 4 m = 8 m²
Area of front = 6 m • 4 m = 24 m²
Area of back = 6 m • 4 m = 24 m²
Surface area = The sum of the areas of all 6 faces
SA = 12 m² + 12 m² + 8 m² + 8 m² + 24 m² + 24 m²
SA = 24 m² + 16 m² + 48 m²
SA = 88 m²
Volume of Rectangular Prisms (Printable PDFs)
In these worksheets, students will practice calculating the space inside simple boxes. The formula for the volume of a rectangular prism is V = length • width • height. Label all answers with the correct cubic units: cubic inches, cubic feet, cubic centimeters, etc. All worksheets include answer keys.
In these worksheets, students will practice calculating the space inside simple boxes. The formula for the volume of a rectangular prism is V = length • width • height. Label all answers with the correct cubic units: cubic inches, cubic feet, cubic centimeters, etc. All worksheets include answer keys.
Worksheet #1
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Surface Area of Rectangular Prisms (Printable PDFs)
In these worksheets, students will find the total area of all six faces on each shape. You could find the area of all six sides individually or use this formula: SA=2(width • depth) + (perimeter of base • height) You could even try it both ways, but check your answers with our included answer keys. And don’t forget your units!
In these worksheets, students will find the total area of all six faces on each shape. You could find the area of all six sides individually or use this formula: SA=2(width • depth) + (perimeter of base • height) You could even try it both ways, but check your answers with our included answer keys. And don’t forget your units!
Worksheet #1
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Volume and Surface Area of Rectangular Prisms (Printable PDFs)
In our volume and surface area worksheets, students will calculate the volume and surface area of various rectangular prisms. Remember: the volume should always be labeled with the appropriate cubic units, and the area should always be labeled with the appropriate square units. All worksheets include answer keys.
In our volume and surface area worksheets, students will calculate the volume and surface area of various rectangular prisms. Remember: the volume should always be labeled with the appropriate cubic units, and the area should always be labeled with the appropriate square units. All worksheets include answer keys.
Worksheet #1
Get instant access to download and print Volume and Surface Area Worksheets, plus all K12 Math Worksheets!
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187926 | https://www.mathed.page/middle-school/pdf/proportional.pdf | Proportional Relationships: Connections Henri Picciotto A set of activities using multiple mathematical tools (and some “real world” applications) to help middle school students make connections between proportional relationships and other concepts, including area, scaling, slope, functions, and rate of change.
For related activities, teacher notes, and in some cases answer keys, go to www.MathEducationPage.org ..................................................................................................
Investigating Rectangle Areas ! p. 1 This lesson is adapted from Lesson 1.A in Algebra: Themes, Tools, Concepts. The whole book can be found free on my Web site, along with a Teachers’ Edition and much in the way of support materials.
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Scaling on the Geoboard, Similar Rectangles ! p. 3 These labs are excerpted from Geometry Labs. The whole book can be found free on my Web site. They are intended to be used with 11 by 11 geoboards, which can also be used for many other lessons, including work on area and the Pythagorean theorem. One implementation comes with a circle geoboard on the back, which is useful for an introduction to geometry and trigonometry. (The CircleTrig Geoboard is available from Nasco.com. More information is available on my Web site.) In the absence of geoboards, you can do these labs using dot paper, also available on my Web site.
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Discounts ! p. 6 Adapted from Lesson 6.A in Algebra: Themes, Tools, Concepts.
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Guess My Function, Nine Function Diagrams, Sixteen Function Diagrams ! p. 7 These can be found in the Function Diagrams section of my Web site.
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Constant Speed ! p. 11 Adapted from Lesson 4.1 in Algebra: Themes, Tools, Concepts.
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In the Lab ! p. 13 Adapted from Lesson 4.6 in Algebra: Themes, Tools, Concepts.
Adapted from Algebra: Themes, Tools, Concepts © 1994 Anita Wah and Henri Picciotto www.MathEducationPage.org Investigating Rectangle Areas In this lesson, we will investigate rectangle areas by looking for patterns in tables and graphs. These are the questions we will be thinking about: ◊ How does the area of a rectangle change if you vary the length or width and leave the other dimension unchanged? ◊ How does the area of a rectangle change if you vary both the length and the width? 1. What is the area of a rectangle with the following dimensions? a. 1 by 9 b. 2 by 9 c. 3 by 9 d. 9 by 9 2. What is the area of a rectangle with the following dimensions if x = 10? a. 1 by x b. 2 by x c. 3 by x d. x by x 3. Fill out this table: Rectangle area x 1 by x 1 1 2 3 4 5 6 4. Draw axes, with x on the horizontal axis, and area on the vertical axis. Plot the points you obtained in the previous exercise for the area of 1 by x rectangles. For example, (1,1) will be on the graph. 5. Does it make sense to connect the points you plotted? What would be the meaning of points on the line, in between the ones you got from your table? p. 1 Adapted from Algebra: Themes, Tools, Concepts © 1994 Anita Wah and Henri Picciotto www.MathEducationPage.org 6. Now fill out these tables: Rectangle area Rectangle area Square area x 2 by x x 3 by x x x by x 1 1 1 2 4 2 2 3 3 9 3 4 4 4 16 5 5 5 6 6 6 7. On the same axes, graph the data you obtained for 2 by x, 3 by x, and x by x rectangles. For more accuracy on the last one, you may use your calculator to find points for x = .5, 1.5, and so on. 8. Discussion: Answer these questions about the graphs: a. Which ones represent a proportional relationship? b. How do the first three graphs differ from each other? Explain. c. What is special about the fourth graph? d. Do the graphs intersect? What is the meaning of the intersections? e. Where would the graphs meet the y-axis if we extended them? f. Which area grows the fastest? why? 9. I am thinking of a number, which I will call k. If I made a table and a graph for rectangles with dimensions k by x, it would start like this: Rectangle area x k by x 1 5 What is k equal to? Explain. 10. Here is one part of a table: Rectangle area x 9 by x ? 13.5 What is the missing number? Explain. p. 2 Geometry Labs Section 10 Similarity and Scaling 133 LAB 10.1 Name(s) Scaling on the Geoboard Equipment: 11 ! 11 geoboard, dot paper The figure shows four houses. House a is the original, and the others have been copied from it following very exact rules.
1. What is the rule that was used for each copy?
2. Among the three copies, two are distorted, and one is a scaled image of the original.
a. Which one is scaled?
b. How would you describe the distortions?
When two figures are scaled images of each other, they are said to be similar. Similar figures have equal angles and proportional sides.The sides of one figure can be obtained by multiplying the sides of the other by one number called the scaling factor.
3. What is the scaling factor that relates the two similar houses in the figure above?
F G D E © 1999 Henri Picciotto, www.MathEducationPage.org p. 3 Geometry Labs Section 10 Similarity and Scaling 135 LAB 10.2 Name(s) Similar Rectangles Equipment: 11 11 geoboard The figure above shows geoboard rectangles nested inside each other.
1. Explain why the two rectangles on the left are similar but the two rectangles on the right are not.
2. Draw the diagonal from the bottom left to the top right of the larger rectangle in a. Note that it goes through the top right vertex of the smaller rectangle.
Repeat on b. How is it different?
3. The rectangles in a are part of a family of similar geoboard rectangles. If we include only rectangles with a vertex at the origin (the bottom left peg of the geoboard), the family includes ten rectangles (five horizontal ones and five vertical ones—two of the vertical ones are shown in the figure).
a. List the rectangles in that family by listing the coordinates of their top right vertex.
b. What is the slope of the diagonal through the origin for the vertical rectangles?
c. What is the slope of the diagonal for the horizontal rectangles? D E © 1999 Henri Picciotto, www.MathEducationPage.org p. 4 136 Section 10 Similarity and Scaling Geometry Labs 4. Including the one in Problem 3, there are ten families of similar rectangles (including squares) on the geoboard.Working with your neighbors, find them all, and list all the rectangles in each family. (Again, assume a vertex at the origin, and use the coordinates of the top right vertex for the list. Only consider families with more than one rectangle.) 5. Working with your neighbors, find every geoboard slope between 1 and 2.
(Express the slopes both as fractions and as decimals.) Counting 1 and 2, there are seventeen different slopes.
Discussion A. Problem 2 is an example of using the diagonal test for similar rectangles. Explain.
B. In Problems 3 and 4, how does symmetry facilitate the job of listing the rectangles?
C. What are the advantages of fractional versus decimal notation in Problem 5?
D. Explain how to use the answers to Problem 5 to create lists of geoboard slopes in the following ranges.
a. Between 0.5 and 1 b. Between –1 and –2 c. Between –0.5 and –1 LAB 10.2 Name(s) Similar Rectangles (continued) © 1999 Henri Picciotto, www.MathEducationPage.org p. 5 Adapted from Algebra: Themes, Tools, Concepts © 1994 Anita Wah and Henri Picciotto www.MathEducationPage.org Discounts A discount card at a movie theater costs $10. With that card, it only costs $3 to attend a movie, instead of $5. The card is valid for three months. 1. Use the same pair of axes for both of the graphs in this problem. Make a graph of the total cost (including the cost of the discount card if you got one) as a function of the number of movies you see: a. if you have the discount card. b. if you do not have the discount card. 2. What is the total cost of seeing n movies in three months a. with the discount card? b. without the discount card? 3. a. Which of those two choices is a proportional relationship? b. Bonus: Can you think of another situation involving money that is a proportional relationship? 4. a. If you saw 12 movies in three months, how much would you save by buying the discount card? b. If you saw only 2 movies in three months, how much would you save by not buying the discount card? 5. What is the break-even point; that is, how many movies would you have to see in order to spend exactly the same amount with and without the discount card? 6. How would your decision be affected if the cost of the discount card were raised to $12? 7. Challenge: How would your decision be affected if the cost of the discount card were changed to $D? p. 6 Guess My Function, p. 1 © Henri Picciotto www.MathEducationPage.org Guess My Function Here is an example of a function: y = 2x – 3. For this function, If x = 0, y = -3. If x = 1, y = -1. 1. a. If x = 2, y = ? b. If x = 1.5, y = ? c. If x = -1, y = ? x is called the independent variable, or input. For this function, you can choose any number for x. y depends on x, so it is called the dependent variable, or output. You can arrange the information about this function in a table: 2. Fill out the rest of the table. You can also arrange the information in a graph: 3. Label the points with their coordinates. Definition:)A"function"is"a"rule"that"assigns"to"each"input"exactly"one"output." On"this"page,"we"made"a"table"and"a"graph"from"knowing"the"formula"for"the"function." On"the"next"page,"you"will"guess"formulas"for"functions,"knowing"a"table"or"a"graph." " " x y -1 0 -3 1 -1 1.5 2 p. 7 Guess My Function, p. 2 © Henri Picciotto www.MathEducationPage.org 4. Guess a formula of the function for each table. (Hint: what can you do to x to get y?) 5. Guess the formula of the function for each graph. (Hint: label the points with their coordinates.) a. b. 6. Which functions in #4 and #5 are proportional relationships? 7. Guess a formula of the function for each table. These are more challenging. a. x y b. x y c. x y -2 6 -2 -7 -2 3 -1 5 -1 -5 -1 0 0 4 0 -3 0 -1 1 3 1 -1 1 0 Functions appear in all sorts of situations in math. For example, the input could be the side of a square, and the output its area. In that case, the formula would be A=s2. 8. For each function, write a formula, and name the input and the output. Which are proportions? a. The perimeter of a square b. Half of a number c. The area of a circle a. x y b. x y c. x y -2 0 -2 -6 -2 -6 -1 1 -1 -3 -1 -5 0 2 0 0 0 -4 1 3 1 3 1 -3 2 4 2 6 2 -2 p. 8 © Henri Picciotto www.MathEducationPage.org Nine Function Diagrams 1. Three diagrams represent functions of the form y = x + b. Which ones? What is b? 2. Three diagrams represent functions of the form y = x – b. Which ones? What is b? 3. Six diagrams represent functions of the form y = mx. Which ones? What is m? 4. Six diagrams represent functions of the form y = x/m. Which ones? What is m? 5. A function diagram has parallel in-out lines. Write as much as you can about the function. 6. If the in-out lines are parallel, in what case do they go up? Down? Straight across? 7. A diagram for y = mx has in-out lines that move closer to each other. What can you say about m? 8. A diagram for y = mx has in-out lines that move apart from each other. What can you say about m? 9. A function diagram for y = mx has in-out lines that cross each other. What can you say about m? 10. Two diagrams above represent functions of the form y = b – x. Which ones? What are the functions? p. 9 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 10 0 a b c d e f g h i j k l m n o p Sixteen Function Diagrams www.MathEducationPage.org from Algebra: Themes, Tools, Concepts by Anita Wah and Henri Picciotto p. 10 Adapted from Algebra: Themes, Tools, Concepts © Anita Wah and Henri Picciotto www.MathEducationPage.org Constant Speed, p. 1 Constant Speed 1. Ophelia and Xavier are traveling along a road. If you could view the road from above and make a sketch of what you saw every ten minutes, your sketches might look something like the figure. a. Which person (0 or X) is traveling faster? b. If the entire length of the road is six miles, can you figure out approximately how fast each person is traveling? Explain. Bea is participating in a long-distance roller-skating race. Her speed is approximately 10 miles per hour. The graph below shows Bea's progress. It shows that after 5 hours she had traveled 50 miles. ! 2. Bea’s motion: a. Copy the graph onto graph paper. Use a whole piece of graph paper. You will be adding more to this graph. ! b. One of the points on the graph is (5,50). Mark and label three more points on the graph of Bea's progress. 3. The distance traveled and the time elapsed are in a proportional relationship. Explain how we know this. 4. In this lesson we are assuming everyone travels at a constant speed. a. What might make it impossible to travel at a constant speed? Explain. b. Why might it be a good idea to assume constant speed anyway? p. 11 Adapted from Algebra: Themes, Tools, Concepts © Anita Wah and Henri Picciotto www.MathEducationPage.org Constant Speed, p. 2 Abe is walking along the same road. Assume he is moving at an approximately constant speed. The table shows how long it took for Abe to go certain distances. 5. a. Copy and complete the table up to 20 miles. b. Use the same axes you !used for Bea. Plot and label the points !from the table in part (a). c. Connect the points with a straight line. !Then find and label a point that is on the line but not in your table. Interpret the coordinates of the point in terms of this problem. Amazingly, Gabe and Al are on the same road, and started at the same time. Gabe is riding a scooter, going 30 mph. Al is driving a van, going 50 mph. 6. Make tables like the one you made for Abe showing Gabe's progress on his scooter and Al's progress in the van. Make graphs of their progress on the same axes you used to show Abe's and Bea's progress. Label the four different lines. 7. Use your graphs to help you answer these questions. If Bea and Abe start out at the same time,! a. how far apart will they be after !one hour?! b. how far apart will they be after !two hours? 8. Look for a pattern. a. How far apart will Abe and Bea be after H hours? Explain. b. Is this a proportional relationship? Explain. 9. Mrs. Gral was traveling at a constant speed. She started on the same road at the same time as all the others, and was two miles ahead of Abe after one hour. a. Add a graph of Mrs. Gral's progress to your axes. b. How far ahead was Mrs. Gral after two hours? c. Was she ahead of, or behind Bea? After three hours, how far ahead or behind? d. How fast was Mrs. Gral going? !What mode of travel do you think she was using? 10. Summary: a. How does the mode of travel affect the steepness of the line? Explain. b. What is the meaning of points on two of the graphs that have the same x-coordinate but different y-coordinates? c. What is the meaning of the vertical distance between two lines for a given value of x? pattern do you notice in the table? ong would it take for someone ravels at a constant speed of S per hour to cover 100 miles? • 2 3 4 56 7 8x TIME (hours) h shows Bea's progress on the ows that after 5 hours of roller-he had traveled 50 miles. the graph onto graph paper. Use a piece of graph paper. You will be g more to this graph. f the points on the graph is (5, 50). and label three more points on the of Bea's progress. sson we are assuming everyone a constant speed. How valid is mption? For each mode of travel ght make it impossible to travel at nt speed? Explain. e The table shows how long it took for Abe to go certain distances. Abe's Progress Time (hours) Distance (miles) 4 2 8 8. a. Copy and complete the table up to 20 miles. b. For this problem, use the same axes you used for Bea. Plot and label the points from the table in part (a). c. Connect the points with a straight line. Then find and label a point that is on the line but not in your table. Interpret the coordinates of the point in terms of this problem. 9. Make a table like the one you made for Abe showing Gabe's progress on his scooter and Al's progress in the van. Make graphs of their progress on the same axes you used to show Abe's and Bea's progress. Label the four different lines. 10. Use your graphs to help you answer these questions. If Bea and Abe start out at the same time, a. how far apart will they be after one hour? b. how far apart will they be after two hours? 11. Look for a pattern. How far apart will Abe and Bea be after H hours? Explain. p. 12 Adapted from Algebra: Themes, Tools, Concepts © 1994 Anita Wah and Henri Picciotto www.MathEducationPage.org In the Lab, page 1 In the Lab A Mystery Liquid Reg, Bea and Gabe were doing an experiment in science class. They had an unknown liquid whose volume they measured in a graduated cylinder. A graduated cylinder is a tall, narrow container that is used for accurately measuring liquid volume. They used one that weighed 50 grams, and measured volume in milliliters. They used a balance to find the mass of the liquid to the nearest gram. Reg's data volume mass 10 ml 16 g 20 ml 32 g 50 ml 80 g 80 ml 128 g 1. a. Plot Reg's data, with volume on the horizontal axis and mass on the vertical axis. b. Does it make sense to connect the points on your graph? Explain. 2. Find an equation relating mass to volume. 3. Estimate the mass of a. 60 ml of liquid b. 1 ml of liquid 4. If you add 30 ml to the volume, how much are you adding to the mass? See if you get the same answer in two different cases. 5. If you double the volume, do you double the mass? Bea's data mass volume 16 g 10 ml 32 g 20 ml 48 g 30 ml 64 g 40 ml 6. Plot Bea’s data on the same axes, using a different symbol or color for those points. Be careful! Because the mass depends on the volume, make sure the volume is on the horizontal axis, and the mass is on the vertical axis, again. 7. Connect the points on your graph with a line and write an equation for the line. 8. Estimate the volume of: a. 100 g of liquid b. 1 g of liquid 9. Compare Bea's graph with Reg's graph. Explain the similarities and differences. 10. If you add 10 ml to the volume, how much are you adding to the mass? See if you get the same answer in three different cases. Is the answer consistent with what you found in Reg’s data? Definition: Density equals mass per unit of volume. This means that to find the density of the mystery liquid, you would find the mass of 1 ml of the liquid. 11. Find the density of the mystery liquid, using three different pairs of mass / volume values from Reg’s and Bea’s data. Do all your answers agree? Explain. 12. Is the relationship between volume and mass a proportional relationship? Explain. p. 13 Adapted from Algebra: Themes, Tools, Concepts © 1994 Anita Wah and Henri Picciotto www.MathEducationPage.org In the Lab, page 2 The Mystery Grows Gabe's data: volume mass 10 ml 66 g 20 ml 82 g 40 ml 114 g 60 ml 146 g 13. a. On the same axes, plot Gabe's data. b. In Gabe’s table, if you double the volume, does the mass double? Check this in two cases. 14. If you add 20 ml, how much mass are you adding? Is this consistent with what you learned from Reg’s and Bea’s data? 15. a. According to Gabe's graph, what is the mass of 0 ml of the liquid? Does this make sense? b. What might be the real meaning of the y-intercept on Gabe's graph? Did Gabe make a mistake? Explain. 16. Divide mass by volume for three different pairs of values from Gabe’s data. Do all your answers agree? Will that method work to find the density of the liquid?. 17. Are the numbers in Gabe’s table examples of a proportional relationship? Explain. 18. Write an equation that expresses mass as a function of volume for Gabe’s data. 19. Compare Gabe’s graph to Reg’s and Bea’s. How are they the same, and how are they different? 20. There are number patterns and graph patterns in all the data in this lesson. a. What pattern is there in all of Reg’s, Bea’s, and Gabe’s data? b. What patterns are only true of Reg’s and Bea’s data? Other Substances 21. The graph shows the relationship between mass in grams, and volume in milliliters for some familiar substances. The substances are aluminum, cork, gold, ice, iron and oak. Which substance do you think is represented by each line? Explain. p. 14 |
187927 | https://www.youtube.com/watch?v=z25imPmrxR0 | Sum of the k^th Powers of the First n Positive Integers
Existsforall Academy
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Many people have seen formulas for the sum of the first n positive integers, or the sum of their squares or cubes. But what about finding a formula for any fixed power of these integers? We show a recursive method of finding such polynomial formulas that is due to Pascal.
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[Music] hi everyone today we're going to be finding the sum of the kth powers of the first and positive integers we already have seen one most basic version of this which is that by gauss's trick one plus two plus three all the way through plus n is equal to n times n plus 1 over 2. the next question is what about 1 to the k plus 2 to the k plus 3 to the k all the way through to n to the k we'll call it s k n for uh we'll have n as a positive integer and k is also a positive integer so we're going to be developing a technique to find a formula for this although some of the formulas will be pretty complicated and what we want to know is is this a polynomial in the variable n for fixed k so we're going to fix k as a positive integer we want to find out whether in the variable n which represents the first n positive integers whether this sum here is a polynomial and it'll turn out yes we can write it as a polynomial we're going to be using an idea of pascal i believe he came up with this method and what he said is use the binomial theorem by the binomial theorem what we get is that m plus one this is sort of like a lemma so we're using a different variable to the k plus one is equal to the sum from t equals zero to k plus one and the expansion has coefficients k plus one choose t times m to the t and what we're gonna do is we're gonna siphon off the top the top degree term so we get m plus 1 to the k plus 1 minus m to the k plus 1 is equal to the sum which is truncated now from k plus 1 to k and we get k plus 1 choose t times m to the t and if you take a look at this over here this thing is just ripe for telescoping so we're going to take the sum of a whole bunch of these so let's do that and see where that leads us we're going to take the sum of m equals 1 through n and this is just an exploration like the idea of exploring this is natural and it happens to lead to a solution to our problem so we have the sum from m equals one through n and we're putting this in here so we have m plus one k to the k plus one minus m to the k plus one and when we telescope we get m plus one to the k plus one actually that that's uh we get because m is an index so we so we actually get n plus one to the k plus one minus one to the k plus one which is just which is just uh one so we don't need to put that in there so we've telescoped and that's what we got and now we're going to use the binomial theorem result over here so we get the sum from m equals 1 through n and we have an inner sum now which has t t equals zero through k of k plus one choose t to the uh times m to the t power and now what we're going to do is you know we're going to apply the discrete fubini principle which sounds fancy but it just means we're going to swap these two which we can because they form full matrices so t equals zero through k with the inner sum m equals one through n of k plus one choose t m to the t and now we're gonna sort of evaluate the inner sum not really evaluate but sort of get rid of this summation so we get t equals zero through k and in here we have k plus one choose t and what we're we're multiplying that by is 1 to the t plus 2 to the t plus 3 to the t all the way through to n to the t and you can see here that this is just t equals 0 through k of k plus 1 choose t times s t n and so now let's let's write down the left side here as well because uh it's way up there it's n plus one to the k plus one minus one and so we can just substitute in successes values of k and find out what this is equal to let's let's let's do an example so that it's clear what's going on let's do uh k equals to 2 because we already know the answers for k equals to zero and one um one for zero it's just n and for one is n times n plus one over two by gauss's formula but we don't know k equals to two yet for so the sum of the first and squares so let's figure that out what we get is n plus one cubed minus one is equal to so for t equals zero we have k plus one choose zero and s zero n plus k plus two sorry k plus one choose actually the k we can just write it as a 2 here so we don't need to put in k so we get a 2 plus 1 2 0 then we get 2 plus 1 choose 1 times s one n and then we get plus two plus one choose two times s two of n and what that equals is [Music] uh this one is just equal to n because it's just 1 to the 0 plus 2 to the 0 all the way through to n to the zero so we get n plus over here we get three choose one which is three times n times n plus one over two and the last term is what we wanna figure out so we get three choose two which is three times s2 of n so you see we can we have this on the left side and we have this on the right side and now we can isolate s2 of n so let's figure that out s2 of n which is equal to 1 square plus 2 square plus 3 square all the way through to n square is simply equal to n plus 1 cubed minus one minus n minus three n times n plus one over two and we're going to divide the whole thing by three now at this point we found our polynomial but let's just go one step further and simplify it so we have n cubed plus three n squared plus three n plus one minus one minus n minus three n squared plus three n over two and we're gonna divide the whole thing again by one by three and so we have the ones cancel out here the n takes away one from this so we have two n cubed plus six n squared plus four n minus three n squared minus three n divided by two and times one third again so we're going to be able to cancel this and put a 3 here and cancel this and put a 1 here so that's just equal to 2 n cubed plus 3 n squared plus n divided by six and if you factor that you get n times n plus one times two n plus one over six and that is the classic formula so it does work out and that should give tell you in general how to find skn okay thank you for watching and i'll see you next time you |
187928 | https://quizlet.com/explanations/questions/a-fluid-is-defined-as-a-material-that-continually-deforms-under-the-action-of-normal-stress-5093f299-5a787803-12c9-41d4-8549-c14b591e44c5 | A fluid is defined as a material that continually deforms un | Quizlet
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The hydrostatic equation is p/γ\gamma γ+z=C, where p is pressure, γ\gamma γ is specific weight, z is elevation, and C is a constant. Prove that the hydrostatic equation is dimensionally homogeneous.
In the following list, identify which parameters are dimensions and which paramenters are units: slug, mass, kg, energy/time, meters, horsepower, pressure, and pascals.
In the given table, determine the hydrostatic equation. For each form of the equation that appears, list the name, symbol, and primary dimensions of each variable.
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A fluid is defined as a material that continually deforms under the action of normal stress.
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A fluid can be defined as a substance that continuously deforms under the action of the s h e a r shear s h e a r stress which is the external force which is not the same stress as the n o r m a l normal n or ma l stress. Therefore we can determine this statement as the false statement.
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1/4
EngineeringList the primary dimensions of each of the units: kWh, poise, slug, cfm, cSt.
EngineeringOf the three lists below, which sets of units are consistent? Select all that apply. a. pounds-mass, pounds-force, feet, and seconds. b. slugs, pounds-force, feet, and seconds c. kilograms, newtons, meters, and seconds.
EngineeringApply the grid method to calculate the cost in U.S. dollars to operate a pump for one year. The pump power is 20 hp. The pump operates for 20 hr/day, and electricity costs $0.10 per kWh.
EngineeringWhen a bicycle rider is traveling at a speed of V=24 mph, the power P she needs to supply is given by P=FV, where F=5 lbf is the force necessary to overcome aerodynamic drag. Apply the grid method to calculate: a. power in watts. b. energy in food calories to ride for 1 hour.
EngineeringApply the grid method to calculate force using F=ma. a. Find force in newtons for m=10 kg and a=10 m/s2. b. Find force in pounds-force for m=10 lbm and a=10 ft/s2. c. Find force in newtons for m=10 slug and a=10 ft/s2.
EngineeringThe pressure rise δ\delta δ p associated with wind hitting a window of a building can be estimated using the formula δ\delta δ p = p(V2/2), where _ is density of air and V is the speed of the wind. Apply the grid method to calculate pressure rise for P = 1.2 kg/m3 and V = 60 mph. a. Express your answer in pascals. b. Express your answer in pounds-force per square inch (psi). c. Express your answer in inches of water column (in H2O).
EngineeringApply the grid method to calculate the density of an ideal gas using the formula p = p/RT. Express your answer in lbm/ft3. Use the following data: absolute pressure is p _ 60 psi, the gas constant is R = 1716 ft -lbf/slug-∘^{\circ}∘R, and the temperature is T = 180∘^{\circ}∘F.
EngineeringSelect an engineered design (e.g., hydroelectric power as in a dam, an artificial heart) that involves fluid mechanics and is also highly motivating to you. Write a one-page essay that addresses the following questions. Why is this application motivating to you? How does the system you selected work? What role did engineers play in the design and development of this system?
EngineeringIf the local atmospheric pressure is 84 k P a\mathrm{84\ kPa}84 kPa, use the grid method to determine the pressure in units of a. psi b. psf c. bar d. atmospheres e. feet of water f. inches of mercury
EngineeringWhich of these is a correct conversion ratio? Select all that apply. a. 1=1 hp/(550 f t−l b f/s)1=1 \operatorname{hp} /(550 \mathrm{ft}-\mathrm{lbf} / \mathrm{s})1=1 hp/(550 ft−lbf/s), b. 1=101.3 k P a\mathrm{kPa}kPa /\left(14.7 l b f\mathrm{lbf}lbf / i n\mathrm{in}in^{2}\right), c. 1=3.785 U.S.\mathrm{U.S.}U.S.g a l\mathrm{gal}gal /(1.0 L\mathrm{L}L).
EngineeringIn your own words, describe what actions need to be taken in each step of the grid method.
EngineeringFor each variable given, list three common units. a. Volume flow rate (Q), mass flow rate, (m ), and pressure, (p) b. Force, energy, power, c. Viscosity.
EngineeringA design team is developing a prototype C O 2\mathrm{CO}_2 CO 2 cartridge for a manufacturer of rubber rafts. This cartridge will allow a user to quickly inflate a raft. A typical raft is shown in the given sketch. Assume a raft inflation pressure of 3 p s i 3\ \mathrm{psi}3 psi (this means that the absolute pressure is 3 p s i 3\ \mathrm{psi}3 psi greater than local atmospheric pressure). Calculate the volume of the raft and the mass of C O 2\mathrm{CO}_2 CO 2 in grams in the prototype cartridge.
1/7
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187929 | https://fiveable.me/key-terms/intro-chem/strong-electrolyte | 💏intro to chemistry review
key term - Strong electrolyte
Citation:
Definition
A strong electrolyte is a substance that completely dissociates into ions when dissolved in water, resulting in high electrical conductivity. Examples include most salts, strong acids, and strong bases.
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A substance that partially dissociates into ions in solution, resulting in lower electrical conductivity compared to strong electrolytes.
Non-Electrolyte: A substance that does not produce ions when dissolved in water, thus it does not conduct electricity.
The process by which molecules split into smaller particles such as atoms, ions, or radicals, usually in a reversible manner.
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187930 | https://www.youtube.com/watch?v=tbXk4kIRnok | Statics Friction Example for a ladder leaning against a wall
Mechanical Engineering with Dr. Sanei
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Posted: 28 Oct 2020
Statics Friction Example for a ladder leaning against a wall
2 comments
Transcript:
hello everyone in this video we are going to solve a friction problem example the uniform 10 kilogram ladder rests against a smooth wall at b so the water smooth means that we don't have any friction at b and at the end a rests on the rough horizontal plane for which the coefficient of a static friction is point tree and because it says rough we have friction and also the friction coefficient is given to us determine the angle of inclination theta of the ladder and the normal reaction at b if the ladder is on the verge of sleeping so this statement tells us that the problem is a type 2 problem so we know the surface of sleeping the first step is to draw the free body diagram to see what known values and unknown values we have so if i draw the free body diagram the ladder is 10 kilogram and it's acting in the middle so 10 is the mass if i multiply it by 9.81 i get the weight i get the fourth at b i only have normal force because i don't have any friction force the problem is telling me that the wall is smooth at a i have both normal force and also the friction force so the ladder tends to move to the left so the friction force would be the opposite direction also i have the angle theta the total length of the ladder is 4 meter that comes into play when i write my moment equations so listing our unknowns what unknowns do we have so mb the normal forces are known here n a theta and f a so i have four unknowns using equilibrium equations i can only find three unknowns but also because the problem is telling me that the latter is on the verge of sleeping i know i can write the friction equation so technically i have three unknowns or you can think of it as four unknowns but also with four equations so let's write our equilibrium equations so i'm gonna start with summation of forces in y equals zero so i have n a and nine point uh my with the weight so that would be 98.1 nit that's the only two forces i have then summation of forces in x direction equal zero i have the friction force at a minus and b equals zero so nb the normal force would be equal to the friction force but what is the friction force at a is simply mu and a so if i write it here mu is 0.3 and a bring it here would be 98.1 therefore mb would be 29.43 mute so i found the normal force at b as well now i need to find theta that's the only remaining unknown i found f a n a m b and only theta is remaining but i have one equilibrium equation left summation of moment i'm gonna write summation of moment about point a so i can get rid of two forces here so summation of moment about point a equals zero counterclockwise positive there are two forces that creating a moment and b that is creating a positive moment because it's rotating my member counter-clockwise about point a also the weight that is creating a negative moment because it's rotating clockwise what is the moment arm for mb the moment arm for mb is this distance which is 4 sine theta and what is the moment arm for my weight that would be this distance and if you think of it this triangle so if the total distance is 4 meter here would be 2 meter and this is theta so this system would be 2 cosine theta so now that i have the moment arm i can write my equation and b again is creating a counter-clockwise moment so it's positive moment arm is 4 sine theta but force yeah i have my weight 98.1 moment arm is 2 cosine theta equals 0. so this problem i have one equation i want to know is theta so if i divide the whole equation by cosine theta then i get rid of cosine theta over there and i can say 4 i have the value for mb here so i can plug in the value for mb i know mb here is 29.43 so i have 4 29.43 if i divide everything by cosine theta i get tangent theta minus 298.1 equals zero so if i rearrange the equation and find tangent inverse i will find theta to be 59.32 degrees and that would be my other unknown so that was a type 2 problem because i knew the surface of sleeping |
187931 | https://arxiv.org/abs/1908.10996 | [1908.10996] The Page curve of Hawking radiation from semiclassical geometry
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arXiv:1908.10996 (hep-th)
[Submitted on 29 Aug 2019 (v1), last revised 4 Nov 2019 (this version, v2)]
Title:The Page curve of Hawking radiation from semiclassical geometry
Authors:Ahmed Almheiri, Raghu Mahajan, Juan Maldacena, Ying Zhao
View a PDF of the paper titled The Page curve of Hawking radiation from semiclassical geometry, by Ahmed Almheiri and 3 other authors
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Abstract:We consider a gravity theory coupled to matter, where the matter has a higher-dimensional holographic dual. In such a theory, finding quantum extremal surfaces becomes equivalent to finding the RT/HRT surfaces in the higher-dimensional theory. Using this we compute the entropy of Hawking radiation and argue that it follows the Page curve, as suggested by recent computations of the entropy and entanglement wedges for old black holes. The higher-dimensional geometry connects the radiation to the black hole interior in the spirit of ER=EPR. The black hole interior then becomes part of the entanglement wedge of the radiation. Inspired by this, we propose a new rule for computing the entropy of quantum systems entangled with gravitational systems which involves searching for "islands" in determining the entanglement wedge.
Comments:21 pages, 12 figures
Subjects:High Energy Physics - Theory (hep-th); General Relativity and Quantum Cosmology (gr-qc)
Cite as:arXiv:1908.10996 [hep-th]
(or arXiv:1908.10996v2 [hep-th] for this version)
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From: Raghu Mahajan [view email]
[v1] Thu, 29 Aug 2019 00:37:46 UTC (4,622 KB)
[v2] Mon, 4 Nov 2019 16:40:19 UTC (4,623 KB)
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187932 | https://www.youtube.com/watch?v=6ezEhxm09N0 | Eng131-What is the difference between "silent" and "quiet" ?
e-genieclimatique
88200 subscribers
Description
155 views
Posted: 20 Jul 2023
Question to chatGPT #learnenglish #shorts
Playlist: English-I'm learn english to become fluent, someday :
1 comments
Transcript:
what's the difference between silent and quiet silent and quiet are both adjectives that refer to the absence or reduction of noise or sound however they differ in the degree of the absence or reduction of sound they describe silent refers to a complete absence of sound or noise no sound at all implies a state of total Stillness or an activity in terms of noise examples the room was completely silent there was a moment of silence in honor of the deceased quiet refers to a low level of noise or sound but not a complete absence implies a relative calmness Tranquility or subdued atmosphere when compared to a noisier environment examples the library was quiet please speak quietly during the presentation in summary silent means there is no sound at all while quiet means there is some sound but it is at a low or subdued level gives four examples for each case sure here are four examples for each case silent after the power outage the once buzzing Factory fell completely silent the audience held their breath during the suspenseful scene making the theater eerily silent as the snow fell gently Outside The Cabin in the Woods remained silent and undisturbed when the teacher asked who broke the vase the students all looked at each other but remained silent quiet the beach was quiet and Serene in the early morning hours before the crowds arrived the baby slept soundly in the quiet Nursery with only the faint sound of a lullaby playing the small town was known for its quiet streets and friendly neighbors offering a peaceful Retreat from the bustling City the quiet corner of the library was a favorite study spot for students seeking a focused atmosphere |
187933 | https://www.youtube.com/watch?v=kDh_Fl0ne60 | Given the Cross Product , Find Angle Between Vectors
Patrick J
1400000 subscribers
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Posted: 23 Sep 2012
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( Thanks to all of you who support me on Patreon. You da real mvps! $1 per month helps!! :) !! Given the Cross Product , Find Angle Between Vectors.
Here we are given the cross product of two unit vectors and we proceed to find the angle between two vectors.
26 comments
Transcript:
all right in this example we have two unknown unit vectors A and B we know that the cross product has components 1/4 comma 1/4 comma 0 we want to find the possible value in degrees for the angle Theta formed by uh formed by vectors A and B and Theta is going to be between 0 and 180° so we've got the nice formula that the magnitude of A cross B that equals the magnitude of a the magnitude of B time sin Theta so again this is going to help relate uh the uh it's going to help us relate the angle between those original vectors well A cross B we can compute that by just taking the magnitude so uh or finding the magnitude so we'll square the components 1/4 SAR plus -14 squared and then we would have 0^ squared all right well again by a assumption uh A and B are unit vectors and if they're unit vectors they both have length of one so that'll make that uh easy for us well let's see underneath the square root we would have 1 over 16+ 1 over 16 + 0 that's going to equal s of theta well that's going to give us the Square < t of 2 over 16 = sin Theta and we can always uh simplify this a little bit more we can make this the < TK of two over the < TK of 16 which would be four and I'm going to simplify this a little bit so just to evaluate it this is definitely you know I don't know an angle off the top of my head uh sign of what angle gives me < tk2 over4 so let's see uh the sare RO T of two uh divid by 4 U I'm getting this to be roughly 0.3 uh five and I don't know we can round it one more 0354 so now I'm just going to use uh the inverse sign of both sides to solve so we'll get theta equals Ark sign of 0.354 so let me plug this into a calculator real quick so Ark sign of 0.354 make sure everything is in degrees I'm getting Theta to be roughly equal to 20 uh 7 we'll just say 20.7 de so that's one of our possible solutions recall though um if we're looking at angles between 0 and 180° so now I'm kind of thinking about unit circle stuff so if we're looking for uh angles between 0 and 180° um we found one of the solutions to be 20 . 7° there would be another solution in the second quadrant and to get that one we'll just take 180 de and we'll subtract away 207° so let's see I guess if we subtract 20 that would give us 160 if we subtract another 7 that should leave us with 159.3max: and then 159.3max: |
187934 | https://www.fishersci.ca/shop/products/sodium-bromide-97-thermo-scientific/p-4407864 | UserName
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Sodium bromide, 97%
CAS: 7647-15-6 | BrNa | 102.89 g/mol
$117.92 - $307.16
Chemical Identifiers
| | |
--- |
| CAS | 7647-15-6 |
| Molecular Formula | BrNa |
| Molecular Weight (g/mol) | 102.89 |
| MDL Number | MFCD00003475 |
| InChI Key | JHJLBTNAGRQEKS-UHFFFAOYSA-M |
| Synonym | sodium bromide, bromide salt of sodium, sodium bromide nabr, sedoneural, sodiumbromide, trisodium tribromide, bromnatrium, nabr, bromnatrium german, caswell no. 750a |
| PubChem CID | 253881 |
| ChEBI | CHEBI:63004 |
| IUPAC Name | sodium bromide |
| SMILES | [Na+].[Br-] |
| Catalog Number | Mfr. No. | Quantity | Price | Quantity | | | | | |
--- --- --- --- --- |
| Catalog Number | Mfr. No. | Quantity | Price | Quantity | | | | | |
--- --- --- --- --- |
|
| |
AA1403736 View Documents View Product Certificates | Thermo Scientific Chemicals 01403736 | 500 g | Each for $117.92 | Only null left | Sign In or Register to check your price and availability. Add to cart | | || | | | | | | | | | |
AA14037A3 View Documents View Product Certificates | Thermo Scientific Chemicals 014037A3 | 2 kg | Each for $307.16 | Only null left | Sign In or Register to check your price and availability. Add to cart | | || | | | | | | | | | |
Description
Sodium bromide is used in the preparation of bromides and as a source of bromide ion.
This Thermo Scientific Chemicals brand product was originally part of the Alfa Aesar product portfolio. Some documentation and label information may refer to the legacy brand. The original Alfa Aesar product / item code or SKU reference has not changed as a part of the brand transition to Thermo Scientific Chemicals.
This Thermo Scientific Chemicals brand product was originally part of the Alfa Aesar product portfolio. Some documentation and label information may refer to the legacy brand. The original Alfa Aesar product / item code or SKU reference has not changed as a part of the brand transition to Thermo Scientific Chemicals.
Chemical Identifiers
| | |
--- |
| CAS | 7647-15-6 |
| Molecular Weight (g/mol) | 102.89 |
| InChI Key | JHJLBTNAGRQEKS-UHFFFAOYSA-M |
| PubChem CID | 253881 |
| IUPAC Name | sodium bromide |
| | |
--- |
| Molecular Formula | BrNa |
| MDL Number | MFCD00003475 |
| Synonym | sodium bromide, bromide salt of sodium, sodium bromide nabr, sedoneural, sodiumbromide, trisodium tribromide, bromnatrium, nabr, bromnatrium german, caswell no. 750a |
| ChEBI | CHEBI:63004 |
| SMILES | [Na+].[Br-] |
Specifications
| | |
--- |
| Boiling Point | 1390°C |
| CAS | 7647-15-6 |
| Quantity | 500 g |
| Molecular Formula | BrNa |
| Synonym | sodium bromide, bromide salt of sodium, sodium bromide nabr, sedoneural, sodiumbromide, trisodium tribromide, bromnatrium, nabr, bromnatrium german, caswell no. 750a |
| InChI Key | JHJLBTNAGRQEKS-UHFFFAOYSA-M |
| IUPAC Name | sodium bromide |
| PubChem CID | 253881 |
| Formula Weight | 102.89 |
| Sensitivity | Hygroscopic |
| Chemical Name or Material | Sodium bromide |
| | |
--- |
| Melting Point | 755°C |
| Physical Form | Granules |
| Assay Percent Range | 97% |
| MDL Number | MFCD00003475 |
| Solubility Information | Soluble in water,alcohol |
| SMILES | [Na+].[Br-] |
| Molecular Weight (g/mol) | 102.89 |
| ChEBI | CHEBI:63004 |
| Refractive Index | 1.6412 |
| Density | 3.203 g/mL |
Safety and Handling
EINECSNumber : 231-599-9
RTECSNumber : VZ3150000
TSCA : Yes
Recommended Storage : Ambient temperatures
missing translation for 'documents'
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187935 | https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Combinatorics_(Morris)/02%3A_Enumeration/09%3A_Some_Important_Recursively-Defined_Sequences/9.01%3A_Derangements | f(i)=n
f(j)=n
j≠i
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9.1: Derangements
Last updated
: Jul 7, 2021
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9: Some Important Recursively-Defined Sequences
9.2: Catalan Numbers
Page ID
: 60115
Joy Morris
University of Lethbridge
( \newcommand{\kernel}{\mathrm{null}\,})
Definition: Derangements
A derangement of a list of objects is a permutation of the objects, in which no object is left in its original position.
A classic example of this is a situation in which you write letters to ten people, address envelopes to each of them, and then put them in the envelopes, but accidentally end up with none of the letters in the correct envelope.
Another example might be a dance class in which five brother-sister pairs are enrolled. The instructor mixes them up so that no one is dancing with a sibling.
Since we’re considering enumeration, it shouldn’t surprise you that the question we want answered is: in how many ways can this happen? That is, given nn objects, how many derangements of the nn objects are there? Let’s use DnDn to denote the number of derangements of nn objects.
We can label the objects with the numbers {1,...,n}{1,...,n}, and think of a derangement as a bijection
f:{1,...,n}→{1,...,n}
f:{1,...,n}→{1,...,n}(9.1.1)
such that ff does not fix any value. There are n−1n−1 choices for f(n)f(n), since the only restriction is f(n)≠nf(n)≠n. Say f(n)=if(n)=i. We consider two possible cases.
Case 1: f(i)=nf(i)=n
Now, on the other n−2n−2 values between 11 and nn that are neither ii nor nn, ff must map {1,...,n−1}∖{i}{1,...,n−1}∖{i} to {1,...,n−1}∖{i}{1,...,n−1}∖{i}, and must be a derangement. So there are Dn−2Dn−2 derangements that have f(n)=if(n)=i and f(i)=nf(i)=n.
Case 2: f(j)=nf(j)=n for some j≠ij≠i
In this case, we define another function
g:{1,...,n−1}→{1,...,n−1}
g:{1,...,n−1}→{1,...,n−1}(9.1.2)
as follows. We set g(j)=ig(j)=i, and for every other value, g(a)=f(a)g(a)=f(a) (that is, for every a∈{1,...,n−1}∖{j}a∈{1,...,n−1}∖{j}). We had f(j)=nf(j)=n andf(n)=if(n)=i, and we are eliminating nn from the derangement while maintaining a bijection, by creating the shortcut ff with g(j)=ig(j)=i but g(a)=f(a)g(a)=f(a) for every other a∈{1,...,n−1}a∈{1,...,n−1}. Since ff is a derangement and j≠ij≠i, we see that gg is also a derangement (this time of n−1n−1 objects). So there are Dn−1Dn−1 possible derangements gg, and for a fixed choice of ii, these are in one-to-one correspondence with derangements ff that have f(j)=nf(j)=n and f(n)=if(n)=i, so there are also Dn−1Dn−1 of these.
We conclude that Dn=(n−1)(Dn−1+Dn−2)Dn=(n−1)(Dn−1+Dn−2).
We also need some initial conditions. We have D1=0D1=0; there is no way of arranging a single object so that it doesn’t end up in the correct place. Also, D2=1D2=1, since there is exactly one way of deranging two objects (by interchanging them).
If we wanted to solve this recursively-defined sequence, we would need to use exponential generating functions, which we’ll introduce in this chapter but won’t really study in this course. Instead, we’ll give the explicit formula for DnDn without proof.
Proposition 9.1.19.1.1
For any n≥1n≥1, the number of derangements of nn objects is
Dn=n!(n∑i=0(−1)ii!)
Exercise 9.1.1
Use induction to prove Proposition 9.1.1.
Which kind of induction did you have to use to prove Proposition 9.1.1?
Calculate D5 using the explicit formula given in Proposition 9.1.1.
Calculate D5 using the recursive relation.
9: Some Important Recursively-Defined Sequences
9.2: Catalan Numbers |
187936 | https://askfilo.com/user-question-answers-mathematics/is-decreasing-for-all-when-a-b-c-d-38353234323631 | {f{{\left({x}\right)}}}={\sin{{x}}}-{k}{x} is decreasing for all {x}\in{R..
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{f{{({x})}}}={{{x}}}-{k}{x} is decreasing for all {x}{R}, when
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f(x)=sin x−k x is decreasing for all x∈R, when a. k<1 b. k≤1 c. k>1 d. k≥1
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C f′(x)=(cos x−h) and therefore,
f(x) is decreasing ⇔f′(x)<0⇒cos x−k<0
⇒cos xcos x⇒k>1
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Question Text f(x)=sin x−k x is decreasing for all x∈R, when a. k<1 b. k≤1 c. k>1 d. k≥1
Updated On Mar 25, 2024
Topic Application of Derivatives
Subject Mathematics
Class Class 12
Answer Type Text solution:1
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187937 | https://www.tiger-algebra.com/zh-CN/%E8%A7%A3%E5%86%B3%E6%96%B9%E6%A1%88/%E7%A7%91%E5%AD%A6%E8%AE%B0%E6%95%B0%E6%B3%95%E8%BD%AC%E6%8D%A2/610000000/ | 科学计数法计算器 610,000,000
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Enter an equation or problem 6 1 0 0 0 0 0 0 0 输入一个方程或问题
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解答 - 科学计数法/标准形式
6.1⋅10 8
6.110^8
查看步骤
其他解决方法
科学计数法/标准形式
逐步解答
1. 将数字写成小数
610000000.0
2. 将其制作成1到10之间的新数字
将小数点移动，使 610000000.0 成为1和10之间的新数字。因为我们的数字大于10，我们将小数点向左移动。删除任何尾随的零，并在第一个非零数字后放置小数点。记住我们移动小数点的次数。
610000000.0 -> 6.1
我们的新数字是 6.1。我们移动了小数点 8 次。
3. 定义10的幂
因为我们的原始数字大于10，所以10的幂是正的。记住，我们移动了小数点 8 次，所以指数是正 8 :
10 8
4. 最终结果
6.1⋅10 8
我们做得怎么样？
给我们反馈
为什么学习这个
科学计数法，或称标准形式，使得处理非常大或非常小的数字变得更容易，这在科学和工程领域中经常出现。例如，在描述天体的质量时就用到了科学计数法：木星的质量是 1.898⋅10 27 公斤，这比写出一个以1,898开头然后还有24个零的数字要清晰很多。科学计数法还使得使用这些很高或很低的数字的问题更容易解决。
术语和主题
科学计数法
相关链接
How to Write in Scientific Notation | Dummies
Concept of Scientific Notation | Brightstorm
Introduction to scientific notation | Khan Academy
最新相关的解决方案
0.000005
189196203210
5.2 x 10 6
4.8 x 10−3
1.5 x 10−5
35000000000000
36 x 10 5
6.2 x 10−4
3 x 10 6
9.9 x 10−4
1.4 x 10 6
−28172372572.772
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How to expend log a(log a x)log a(log ax) for a∈(0;1)∧x∈(0;1)a∈(0;1)∧x∈(0;1)?
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Here are some logarithm rules :
log a y=ln y ln a log ay=lny lna
log a(A⋅B)=log a A+log a B log a(A⋅B)=log aA+log aB
1 ln a=log a e 1 lna=log ae
Hence:
log a(log a x)=log a(1 ln a⋅ln x)=log a(log a e)+log a(ln x)log a(log ax)=log a(1 lna⋅lnx)=log a(log ae)+log a(lnx)
The problem: ln a<0∨log a e<0 lna<0∨log ae<0 to choose- the argument of log a log a is negative. ln x lnx as well.
Origin of the question: log a x=a x log ax=a x
A sample way to solve: log a(log a x)=x log a(log ax)=x
I am not sure if I can say: log a x=|ln x||ln a|log ax=|lnx||lna|
logarithms
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edited Oct 2, 2018 at 14:56
Michael Hardy
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asked Oct 1, 2018 at 14:43
Krzysztof MyśliwiecKrzysztof Myśliwiec
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See my edits to this question and to the answer below for proper MathJax usage.Michael Hardy –Michael Hardy 2018-10-02 15:00:50 +00:00 Commented Oct 2, 2018 at 15:00
@MichaelHardy I see the different (of course it's nice)- I don't know where or how set MathJax 'ON'. I understand to fix my typos (thank you) but the LaTex default front is not my fault Krzysztof Myśliwiec –Krzysztof Myśliwiec 2018-10-02 16:31:56 +00:00 Commented Oct 2, 2018 at 16:31
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Since ln a lna and ln x lnx are negative- we can solve via the complex logarithm:
ln(r⋅e i θ)=ln r+i θ ln(r⋅e i θ)=lnr+i θ
Trick: e i π=−1 e i π=−1
Hence:
ln(−r)=ln r+i π ln(−r)=lnr+i π
For better calculating:
ln(y)→ln(|y|)+i π;y<0 ln(y)→ln(|y|)+i π;y<0
This way:
log a(log a x)=1 ln a⋅ln(ln x ln a)=1 ln a⋅[ln(ln x)−ln(ln a)]log a(log ax)=1 lna⋅ln(lnx lna)=1 lna⋅[ln(lnx)−ln(lna)]
Note that:
ln(ln x)−ln(ln a)=ln(|ln x|)+i π−(ln(|ln a|)+i π)=ln(|ln x|)−ln(|ln a|)ln(lnx)−ln(lna)=ln(|lnx|)+i π−(ln(|lna|)+i π)=ln(|lnx|)−ln(|lna|)
Hence:
log a(log a x)=1 ln a⋅[ln(|ln x|)−ln(|ln a|)]log a(log ax)=1 lna⋅[ln(|lnx|)−ln(|lna|)]
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edited Oct 2, 2018 at 14:58
Michael Hardy
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answered Oct 2, 2018 at 14:51
Krzysztof MyśliwiecKrzysztof Myśliwiec
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187939 | https://pi.math.cornell.edu/~djp282/documents/jmm-2014.pdf | Difference Set Transfers Dylan Peifer1 San Diego State University Summary Difference sets are subsets of groups with combinatorial properties. Fundamental questions involve determining which groups do or do not contain difference sets, finding all difference sets up to a natural definition of equivalence, and developing techniques to construct and classify difference sets. The tools used to approach these questions generally come from algebra, combinatorics, representation theory, algebraic number theory, and computer programming.
In our 8-week program, we addressed the question of the existence of Hadamard difference sets in groups of order 64, discovered and defined the concept of a difference set transfer, and categorized and explained transfers in groups of order 16.
Definition (Difference Set) A (v, k, λ)-difference set is a subset D of a group G such that ▶|G| = v ▶|D| = k ▶each nonidentity element g ∈G, can be represented as a ”difference” g = d1d−1 2 for exactly λ pairs (d1, d2) ∈D2.
Alternatively, we can work in the group ring Z[G] of formal sums n X i=1 ci · gi c1, ..., cn ∈Z, with addition and multiplication defined naturally. Abusing notation, let G := X g∈G g D := X d∈D d D(−1) := X d∈D d−1.
Under this notation, the condition for a group ring element D to be a difference set is having coefficients ci = 0, 1 with k coefficients equal to 1 and satisfying the equation DD(−1) = (k −λ) · 1G + λ · G for some λ.
Example (Difference Set) Consider the group G = C7 = ⟨x | x7 = 1⟩and the subset D = {x, x2, x4}. Organizing all differences d1d−1 2 in a table yields d2 d1 x x2 x4 x 1 x6 x4 x2 x 1 x5 x4 x3 x2 1 Each nonidentity element appears once in the table, so we have that D is a (7,3,1)-difference set.
In the group ring viewpoint we have D = x + x2 + x4, and D(−1) = x6 + x5 + x3. Thus DD(−1) = (x + x2 + x4)(x6 + x5 + x3) = (1 + x6 + x4) + (x + 1 + x5) + (x3 + x2 + 1) = 3 + x + x2 + x3 + x4 + x5 + x6 = 2 + G and again D is a (7,3,1)-difference set.
Definition (Hadamard Difference Set) A Hadamard Difference Set (HDS) is a (v, k, λ)-difference set such that v = 4(k −λ). The name Hadamard refers to the fact that the incidence matrix of the associated block design given by a Hadamard difference set is a regular Hadamard matrix. Hadamard difference sets form the largest category of known examples of difference sets.
Theorem (The Hadamard Parameters) For any (v, k, λ)-HDS, (v, k, λ) = (4m2, 2m2 ± m, m2 ± m) for some m ∈Z>0.
Because (4m2, 2m2 −m, m2 −m)-difference sets and (4m2, 2m2 + m, m2 + m)-difference sets are complementary, we may consider all Hadamard difference sets as having parameters (v, k, λ) = (4m2, 2m2 −m, m2 −m).
Motivation - A Strange Result in Groups of Order 64 Our first task involved finding Hadamard difference sets in groups of order 64, which we accomplished using previous results and the computer algebra system GAP . GAP is specifically designed to do computational group theory, and we used it to automate algorithms for finding and constructing difference sets. GAP also has a catalog of all groups of small order. It is convenient to store difference sets in “GAP notation” by specifying the group number in GAP’s catalog and the elements that form a difference set from GAP’s ordered list of group elements. For example, SmallGroup(64, 12) has the difference set [ 1, 2, 3, 4, 5, 6, 8, 10, 11, 12, 14, 17, 18, 20, 26, 27, 31, 32, 33, 34, 35, 38, 39, 44, 50, 56, 60, 63 ] .
In our work, we noticed that many lists of GAP indices forming a difference set in one group would also form a difference set in another group. For example, the above difference set in SmallGroup(64,12) is a difference set in 7 groups of order 64. This is very surprising, as the chance of randomly finding a difference set in a group of order 64 is vanishingly small.
Transfers in Groups of Order 16 Using GAP Work in order 64 is computationally difficult and complicated by the large number of groups, so we decided to study this strange result in order 16. Our goal was to categorize and explain all cases of shared GAP indices (referred to as transfers) in order 16. Hopefully this would lead to theories and generalizations that could be applied to new existence proofs and construction techniques for other orders.
The following table shows all the places transfers occur in groups of order 16. Each row and column is labeled by GAP’s category number for a group of order 16. The number in each entry indicates how many difference sets in the associated row and column groups are the same when expressed as indices in GAP .
For comparison, note that Hadamard difference sets in groups of order 16 are subsets of 6 elements, and there are 16 6 = 8008 such subsets in each group.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 192 192 192 64 64 0 64 128 192 192 192 128 192 3 0 192 192 192 64 64 0 64 128 192 192 192 128 192 4 0 192 192 192 64 64 0 64 128 192 192 192 128 192 5 0 64 64 64 192 64 0 0 64 192 64 192 64 192 6 0 64 64 64 64 64 0 0 64 64 64 64 64 64 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 0 64 64 64 0 0 0 128 128 64 64 64 64 64 9 0 128 128 128 64 64 0 128 256 128 128 128 128 128 10 0 192 192 192 192 64 0 64 128 448 192 448 192 448 11 0 192 192 192 64 64 0 64 128 192 192 192 128 192 12 0 192 192 192 192 64 0 64 128 448 192 704 256 448 13 0 128 128 128 64 64 0 64 128 192 128 256 320 192 14 0 192 192 192 192 64 0 64 128 448 192 448 192 448 We can focus our attention by reducing the above table to the following chart, which indicates when all difference sets in one group correspond to difference sets in another group by way of GAP indices. In the chart, groups are represented by category number in GAP . Groups are placed in the same box if all their difference sets are the same in GAP indices. At the bottom right of each box is the total number of difference sets found in the group or groups, and directed arrows indicate all difference sets in the starting group are also difference sets in the ending group when viewed as GAP indices.
Definition (Power-Commutator Presentation) Given a group G of order pn, a power-commutator presentation of G consists of a set of generators {f1, f2, . . . , fn} with defining relations fp i = Qn k=i+1 fβ(i,k) k and [fj, fi] = Qn k=j+1 fβ(i,j,k) k , where the value of β is in {1, . . . , p −1} and 1 ⩽i < j ⩽n.
GAP expresses p-groups in power-commutator presentations for efficient computation purposes, and lists their elements in lexicographical order as words in standard form on the generators. This means that the transfers we noticed are just equivalent sets of words.
Definition (Difference Set Transfer) Given two groups G, H of order 2n and their power commutator presentations on generators {g1, . . . gn}, {h1, . . . , hn}, we say a difference set transfer exists between G and H when a difference set in G can be mapped to a difference set in H by mapping words on the generators of G to words on the corresponding (same index) generators of H.
Proving Transfers in Order 16 Using the Spread Construction The spread construction is a standard technique used to build difference sets. It works on groups of order 22s+2 with a normal elementary abelian subgroup of order 2s+1. The construction spreads subgroups of the normal subgroup into cosets of the normal subgroup to form a difference set.
G = C4 × C4 = ⟨x, y | x4 = y4 = [x, y] = 1⟩ E = C2 × C2 = ⟨x2, y2⟩= {1, y2, x2, x2y2} D = {1, y2, x, x3, y, x2y3} 1 y y2 y3 x xy xy2 xy3 x2 x2y x2y2 x2y3 x3 x3y x3y2 x3y3 With some supporting lemmas and observations, the following three theorems prove the results we see in the chart for GAP’s groups 2, 3, 4, 5, 6, 11, and their attached arrows.
Theorem 1 If |G| = 16, we can build 192 difference sets over a normal subgroup E C2 × C2 using the spread construction. If this E is in Z(G), the center of G, these are all difference sets.
Theorem 2 Given |G| = 16, if for E ◁G, E C2 × C2, but E 1 Z(G) then a spread construction over E generates at least 64 difference sets.
Theorem 3 Let G be a group of order 16 that does not contain a subgroup isomorphic to the quaternion group. If the socle of G has order 4, then every difference set in G can be generated via a spread construction over soc(G).
Other Results The chart results for GAP’s groups 10 and 14 follow from basic algebra and casework.
Groups 8 and 9 can be seen to have a quaternion subgroup on the same labeled generators, and their transfers can be proven by examining the interaction of difference sets with this subgroup.
While most of our theorems do not generalize to higher orders, there are still relations to prove and discover. We are convinced by the pervasiveness of difference set transfers in 2-groups that general results exist and can shed light on the study of difference sets.
References Chirashree Bhattacharya and Ken W. Smith. “Factoring (16,6,2) Hadamard difference sets”. In: Electron. J. Combin. 15 (2008).
James A. Davis and Jonathan Jedwab. “A survey of Hadamard difference sets”. In: Groups, difference sets, and the Monster (Columbus, OH, 1993). Vol. 4. Ohio State Univ.
Math. Res. Inst. Publ. Berlin: de Gruyter, 1996, pp. 145–156.
J. F. Dillon. “Variations on a scheme of McFarland for noncyclic difference sets”. In: J.
Combin. Theory Ser. A 40 (1985), pp. 9–21.
Robert E. Kibler. “A summary of noncyclic difference sets, k < 20”. In: J. Combinatorial Theory Ser. A 25 (1978), pp. 62–67.
E. A. O’Brien. “The p-group generation algorithm”. In: J. Symbolic Comput. 9 (1990), pp. 677–698.
1Carleton College Funded by NSF Grant 1061366 |
187940 | https://www.newscientist.com/article/dn9020-oldest-snake-fossil-shows-a-bit-of-leg/ | Advertisement
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Life
Oldest snake fossil shows a bit of leg
By Kimm Groshong
19 April 2006
A legged snake fossil found in 2003 in Patagonia, Argentina, may be the most primitive snake ever found – the sacral region is shown here
(Image: H Zaher)
An articulated portion of the Najash fossil
(Image: H Zaher)
A legged snake fossil found in 2003 in Patagonia, Argentina, may be the most primitive snake ever found – the sacral region is shown here
(Image: H Zaher)
An articulated portion of the Najash fossil
(Image: H Zaher)
Scientists have found fossils of a legged snake with “hips” – a specimen that could be the most primitive snake ever unearthed. The find suggests early snakes were not creatures of the sea and has reignited the debate over how snakes evolved.
Sebastián Apesteguía at the Argentine Museum of Natural History and his team found the snake fossil in a terrestrial deposit in the Río Negro province of north Patagonia, Argentina, in 2003. Unlike a handful of legged fossils found in marine deposits and identified as snakes over the past decade, the new fossil, named Najash rionegrina, has a well-defined sacrum supporting a pelvis and functional hind legs outside of its ribcage.
The creature’s skeletal structure suggests it was evolutionarily closer to its four-legged ancestor than previous fossils. And since the scientists found it in a terrestrial deposit, it is near certain that the animal lived on land.
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“This snake is an important addition because it is the first snake with a sacrum. This represents an intermediate morphology that has never before been seen,” says Hussam Zaher, curator of herpetology at the University of São Paulo in Brazil, and part of the research team.
The burrowers
The fossil was found in a deposit from the late Cretaceous period and Zaher says the snake is at least 90 million years old. “This fills an important morphological gap of information regarding the early evolution of snakes,” he says.
Zaher and Apesteguía argue that the Najash fossil supports the hypothesis that snakes evolved on land, eventually losing their limbs as they became soil burrowers.
That idea was popular for most of the 20th century, but when legged fossils found in marine sediments in and around Israel were identified as snakes around the turn of the millennium, a group of scientists resurrected an older, alternate theory. They say snakes lost their limbs in the oceans and seas rather than on land, and that they evolved from now extinct marine lizards called mosasaurs.
Fins and paddles
Zaher told New Scientist: “We can now reject the hypothesis of marine origin. This Najash snake fossil suggests that mosasaur lizards were not the most closely related group of lizards to snakes.” He says the marine legged snakes are of a more recent lineage and probably represent the first invasion of the sea by snakes.
Blair Hedges, an evolutionary biologist at Pennsylvania State University, US, says: “In one fell swoop, this new fossil kind of casts doubt on the aquatic hypothesis.”
His DNA sequencing studies suggest a terrestrial origin for snakes. And he says that, looking at evolutionary history, it is difficult to find examples of limb loss in an aquatic environment. “We see many cases where animals that walked on land eventually evolved lineages that invaded the oceans. Almost all of them kept their limbs and turned them into fins or paddles,” he says.
Missing ancestor
Zaher admits that even if the new fossil does prove snakes did not lose their legs in the seas, there are many questions about snake evolution left unsolved. At the top of that list is the question of what lizard group is most closely related to snakes. “We do not have an undisputed hypothesis on that question,” he admits.
Michael Caldwell, from the University of Alberta, Canada, and one of the researchers who reintroduced the marine hypothesis, told New Scientist: “These specimens provide important new information on the anatomy of Cretaceous snakes.” But he is also critical of the interpretations of the new fossil’s anatomy.
He argues that without identifying a closest ancestor, there is no robust way of gaining insight into the origin of snakes and says the new study’s assessment of the Najash snake only takes into account snakes rather than all the snakes and lizards in the squamate order, which includes mosasaurs.
Furthermore, he argues, calling the specimen the most primitive may be incorrect because the fossils identified in marine environments are at least 8 million years older.
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187941 | https://en.wikipedia.org/wiki/Pole_position | Jump to content
Pole position
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From Wikipedia, the free encyclopedia
First position on a motor-racing starting grid
For other uses, see Pole position (disambiguation).
"Polesitter" redirects here. For the early Christian ascetics, see Stylite.
| | |
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| | This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. Find sources: "Pole position" – news · newspapers · books · scholar · JSTOR (March 2008) (Learn how and when to remove this message) |
In a motorsports race, pole position is the best position on the track at the start and thus, by definition, the participant in pole position is starting it from first place.
As having pole position is extremely advantageous in most forms of motorsport, the pole position is usually earned by a driver via another competitive method. Most often, this is by having the best qualifying times in timed trials before the race that determine the order of the starting grid, though it can also be awarded in other circumstances (such as current position in the series' championship, the results of a previous race, the drawing of lots, or a reversed grid). The driver in pole position is referred to as the pole-sitter.
Origin
[edit]
The term has its origins in horse racing, in which the fastest qualifying horse would be placed on the inside part of the course, next to the pole marking the start line for the race.
Determination methods
[edit]
Grid position is typically determined by a qualifying session before the race, where race participants compete to ascend to the number 1 grid slot, the driver, pilot, or rider having recorded fastest qualification time awarded the advantage of the number 1 grid slot (i.e., the pole-position) ahead of all other vehicles for the start of the race. Historically, the fastest qualifier was not necessarily the designated pole-sitter. Different sanctioning bodies in motor sport employ different qualifying formats in designating who starts from pole position. Often, a starting grid is derived either by current rank in the championship, or based on finishing position of a previous race. In particularly important events where multiple qualification attempts spanned several days, the qualification result was segmented or staggered, by which session a driver qualified, or by which particular day a driver set his qualification time, only drivers having qualified on the initial day were eligible for pole position.
Some race promoters or sanctioning bodies invert the starting grid, or part of the starting grid, for the purpose of entertainment value (e.g., pack racing; to artificially stimulate passing).[failed verification] An example of a series that does this is the British Touring Car Championship, which reverses the grid for each third race based on a lottery determining how many positions are reversed.
Formula One
[edit]
Main articles: List of Formula One polesitters and Formula One qualifying
Originally in Grand Prix racing, grid positions, including pole, were determined by lottery among the drivers. Before the inception of the Formula One World Championship, the first instance of grid positions being determined by qualifying times was at the 1933 Monaco Grand Prix. Since then, the FIA have introduced many different qualifying systems to Formula One. From the long-standing system of one session on each of Friday and Saturday, to the current knockout-style qualifying leaving 10 out of 20 drivers to battle for pole, there have been many changes to qualifying systems. Between 1996 and 2006, the FIA made 6 significant changes to the qualifying procedure, each with the intention of making the battle for pole more interesting to viewers at home.
Traditionally, pole was always occupied by the fastest driver due to low-fuel qualifying. The race-fuel qualifying era between 2003 and 2009 briefly changed this. Despite the changing formats, drivers attempting pole were required between 2003 and 2009 to do qualifying laps with the fuel they would use to start the race the next day. An underfuelled slower car and driver would therefore be able to take pole ahead of a better but heavier-fueled car. In this situation, pole was not always advantageous to have in the race as the under-fueled driver would have to pit for more fuel before their rivals. With the race refueling ban introduced, low-fuel qualifying returned and these strategy decisions are no longer in play.
Also, when Formula One enforced the 107% rule between 1996 and 2002, a driver's pole time might affect slower cars also posting times for qualifying, as cars that could not get within 107% of the pole time were not allowed start the race unless the stewards decided otherwise. Since the reintroduction of the rule in 2011, this only applies to the quickest time in the first session of qualifying (Q1) and not the pole time.
Top ten most Formula One pole positions
[edit]
: As of the 2025 Hungarian Grand Prix
| | |
--- |
| Bold | Driver has competed in the 2025 season |
| | Driver | Poles |
---
| 1 | Lewis Hamilton | 104 |
| 2 | Michael Schumacher | 68 |
| 3 | Ayrton Senna | 65 |
| 4 | Sebastian Vettel | 57 |
| 5 | Max Verstappen | 44 |
| 6 | Jim Clark | 33 |
| Alain Prost | 33 |
| 8 | Nigel Mansell | 32 |
| 9 | Nico Rosberg | 30 |
| 10 | Juan Manuel Fangio | 29 |
| Source: | | |
Pole position trophy
[edit]
See also: List of Formula One polesitters § Most pole positions per season
From 2014, the FIA awarded a trophy to the driver who won the most pole positions in a Formula One season. In 2018, the FIA Pole Trophy was discontinued and replaced with the Pirelli Pole Position Award, where the polesitter at each race as awarded a Pirelli wind tunnel tyre with the name of the polesitter and their time. A whole-season trophy was awarded in 2018 and 2019, but discontinued from 2020 onwards.
| Year | Winner | Team | Chassis | Pole positions |
--- ---
| 2014 | Nico Rosberg | Mercedes | F1 W05 Hybrid | 11 |
| 2015 | Lewis Hamilton (WC) | Mercedes | F1 W06 Hybrid | 11 |
| 2016 | Lewis Hamilton | Mercedes | F1 W07 Hybrid | 12 |
| 2017 | Lewis Hamilton (WC) | Mercedes | F1 W08 EQ Power+ | 11 |
| 2018 | Lewis Hamilton (WC) | Mercedes | F1 W09 EQ Power+ | 11 |
| 2019 | Charles Leclerc | Ferrari | SF90 | 7 |
(WC) indicates that the driver won the World Championship in the same season.
IndyCar
[edit]
IndyCar uses four formats for qualifying: one for most oval tracks, one for Iowa Speedway, one for the Indianapolis 500, and another for road and street circuits. Oval qualifying is almost like the Indianapolis 500, with two laps, instead of four, averaged together with one attempt, although with just one session.
At Iowa, each car takes one qualifying lap, and the top six cars advance to the feature race for the pole position. Positions from 7th onward are assigned to their races, based on time, with cars in the odd-numbered finishing order starting in one race, and cars in the even-numbered finishing order starting in the second race. The finishing order for the odd-numbered race starts on the inside, starting in Row 6 (11th), and even-numbered race on the outside based on finishing position, again from Row 6 (12th), except for the top two in each race, which start in the inside and outside, respectively (Row 4 and 5) of the race for the pole position. The result of the feature race determines positions 1–10. All three races are 50 laps.
On road and street courses, cars are drawn randomly into two qualifying groups. After each group has one twenty-minute session, the top six cars from each group qualify for a second session. The cars that finished seventh or worse are lined up by their times, with the best of these times starting 13th. The twelve remaining cars run a 15-minute session, after which the top six cars move on to a final 10-minute session to determine positions one through six on the grid.
The Iowa format was instituted in 2012 with major modifications (times set based on open qualifying session in second practice, positions 11th and back in odd positions raced in the inside heat, positions 12th and back in even positions raced in the outside heat, and positions 1–10 raced for the pole, each heat 30 laps), and non-Iowa oval format in August 2010, while the Indianapolis format was in 2010. The road course format was installed for 2008. In prior seasons, oval qualifying ran for four laps, Indianapolis-style, from 2008, and previously two laps with the best lap used for qualification. Street and road circuits used a two-phase format similar to oval qualifying except that cars took one qualifying lap, then the top six advanced to the ten-minute session for the pole.
Indianapolis 500
[edit]
Main article: List of Indianapolis 500 pole-sitters
The pole position for the Indianapolis 500 is determined on the first day (or first full round) of time trials. Cars run four consecutive laps (10 miles), and the total elapsed time on the four laps determines the positioning. The fastest car on the first day of time trials wins the pole position. Times recorded in earlier days (rounds) start ahead of subsequent days (rounds). A driver could record a time faster than that of the pole winner on a subsequent day; however, he will be required to line up behind the previous day(s)' qualifiers.
Starting in 2010, the first day is split into Q1 and Q2. At the end of Q1, positions 10–24 are set. The top nine cars will then have their times wiped out and advance to Q2 where cars will have 90 minutes to run for pole. If inclement weather causes officials to cancel Q2, positions 1–24 are set. If inclement weather in Q1 is early where Q2 is late (past 6 PM usually), drivers will have only one attempt in Q2.
Grand Prix motorcycle racing
[edit]
Main article: List of 500cc/MotoGP polesitters
Since 2006, there has been one hour-long session on Saturday where the riders have an unlimited number of laps to record a fast lap time. Simply, the rider with the fastest lap gains pole position for the race.
In 2013 a new format was introduced whereby qualifying is conducted over two 15-minute sessions labelled Q1 and Q2. The fastest 10 riders over combined practice times advance automatically to Q2, while the rest of the field competes in Q1. At the conclusion of Q1 the fastest 2 riders progress to Q2 with a chance to further improve their grid position.
In 2023 a new format was introduced where the results of qualifying set the grid for a Saturday Sprint Race as well as the Sunday Grand Prix Race.
Top ten riders in Grand Prix motorcycle racing with most pole positions
[edit]
: As of 24 August 2025
| | |
--- |
| Bold | Rider still competing in Grand Prix motorcycle racing as of the 2025 season |
| | Rider | Poles |
---
| 1 | Marc Márquez | 102 |
| 2 | Jorge Lorenzo | 69 |
| 3 | Valentino Rossi | 65 |
| 4 | Mick Doohan | 58 |
| 5 | Max Biaggi | 56 |
| 6 | Dani Pedrosa | 49 |
| 7 | Casey Stoner | 43 |
| 8 | Loris Capirossi | 41 |
| Jorge Martín | 41 |
| 10 | Freddie Spencer | 33 |
| Source: | | |
NASCAR
[edit]
Main article: NASCAR qualifying procedure
See also: Coors Light Pole Award and List of Daytona 500 pole position winners
Before 2001, NASCAR used a two-day qualifying format in its national series. Before 2002 only one lap was run on oval tracks except short tracks and restrictor plate tracks. Until 2014, the pole position has been determined by a two-lap time trial (one lap on road courses) with the faster lap time used as the driver's qualifying speed. In 2014, NASCAR used a knockout qualifying format for all races except the Daytona 500, non-points races, and the Truck Series' Eldora Dirt Derby: after a 25-minute session (on tracks longer than 1.25 miles (2.01 km); tracks shorter than 1.25 miles have a 30-minute session), the 24 fastest cars advance to a ten-minute session, with the top 12 advancing to a final five-minute session. Starting in 2003, if a driver's team changed their car's engine after the qualifying segment was over, the car would be relegated to the rear of the 43-car field. In the case of multiple teams changing engines on the same weekend after a qualifying segment (although this is a rare occurrence), qualifying times from that segment are used to determine the starting order for those cars.
In the Eldora Dirt Derby, practice runs are held, which determine the starting grids for five heat races of eight laps each. The top five fastest qualifiers started on pole for each heat, and the winner of the first heat is awarded the pole for the feature race.
Top ten most Cup Series pole positions
[edit]
: As of 26 October 2024
| | |
--- |
| Bold | Driver still competing in the Cup Series as of the 2024 season |
| | Driver | Poles |
---
| 1 | Richard Petty | 123 |
| 2 | David Pearson | 113 |
| 3 | Jeff Gordon | 81 |
| 4 | Cale Yarborough | 69 |
| 5 | Bobby Allison | 59 |
| 5 | Darrell Waltrip | 59 |
| 7 | Mark Martin | 56 |
| 8 | Bill Elliott | 55 |
| 9 | Ryan Newman | 51 |
| 10 | Bobby Isaac | 48 |
| Source: | | |
Top ten most Xfinity Series pole positions
[edit]
: As of 26 October 2024
| | |
--- |
| Bold | Driver still competing in the Xfinity Series as of the 2024 season |
| | Driver | Poles |
---
| 1 | Kyle Busch | 70 |
| 2 | Joey Logano | 36 |
| 3 | Mark Martin | 30 |
| 4 | Tommy Ellis | 28 |
| 5 | Carl Edwards | 27 |
| 6 | Kevin Harvick | 25 |
| 7 | Sam Ard | 24 |
| 8 | Jeff Green | 23 |
| 9 | David Green | 22 |
| 9 | Brad Keselowski | 22 |
Top ten most Truck Series pole positions
[edit]
: As of 26 October 2024
| | |
--- |
| Bold | Driver still competing in the Truck Series as of the 2024 season |
| | Driver | Poles |
---
| 1 | Mike Skinner | 50 |
| 2 | Jack Sprague | 32 |
| 3 | Ron Hornaday Jr. | 27 |
| 4 | Kyle Busch | 22 |
| 5 | Mike Bliss | 18 |
| 6 | Joe Ruttman | 17 |
| 7 | Matt Crafton | 16 |
| 8 | Austin Dillon | 13 |
| 9 | Greg Biffle | 12 |
| 9 | Ted Musgrave | 12 |
Superbike
[edit]
Superpole for Superbike is a timed event to establish starting positions for motorcycle racers in World Superbike races.
For 2023 a World Superbike weekend typically consists of:
Friday – Two Free Practice sessions.
Saturday – Free Practice, Superpole, WorldSBK Race 1, WorldSSP300 Last Chance Race.
Sunday – Warm-up, WorldSBK Superpole Race, World SSP Race, WorldSBK Race 2, WorldSSP300 Race
: : 1. The final results of the Superpole decide the grid for WorldSBK Race One and Sunday's (sprint) Superpole Race.
: 2. The top six finishers of the "last chance race" take the final six spots of the WorldSSP300 race starting grid.
: 3. The grid for WorldSBK Race 2 will be determined from the first nine positions in the Superpole Race, and the grid from 10th onwards will be the positions from Saturday's Superpole.
The format of Superpole depends on weather conditions:
Dry Superpole: The race director declares a 'dry' Superpole (referring to the weather conditions) then Superpole will consist of 3 laps of the circuit. Riders start one by one in reverse qualifying order. Grid position for the races will be determined by each rider's fastest single lap time.
Wet Superpole: If Superpole is declared 'wet', Superpole will consist of 50 minutes of timed laps, for all 15 riders together, during which a rider may complete up to 12 laps (including in and out laps). Grid position for the races will be determined by each rider's fastest single lap time. For each lap over 12 laps completed, the rider's best lap time will be cancelled.
To qualify for the race, riders must record a lap time no longer than 107% of the time recorded by the pole-position rider. Qualifying tires may be used.
Radio-controlled racing
[edit]
In radio-controlled car racing, the term Top Qualifier (TQ) is used to determine the fastest qualifying driver, usually over a two-day, five/six rounds qualifying sessions, depending on the overall duration of the event. The result is determined by the best half of the driver's performance. As the event bring in over 100 entrants, the fastest driver is guaranteed directly a place in front of the A-main final, the group that carries a chance of being the overall winner. The slower drivers are allocated a spot to compete in their groups to determine the overall positions.
References
[edit]
^ "How Often Does The Pole-Sitter Win In F1?". F1 Chronicle. 2022-07-01. Retrieved 2022-10-29.
^ Jump up to: a b "Pole Position". Lexico Dictionaries | English. Archived from the original on April 11, 2021. Retrieved 2020-02-09.
^ "Grand Prix History – The Circuits". Archived from the original on 27 May 2012. Retrieved 27 April 2011.
^ Jump up to: a b "Which qualifying system is best? (Poll)". F1 Fanatic. 4 December 2008. Retrieved 28 May 2013.
^ Jump up to: a b "Official: Real qualifying returns in 2010". F1 Fanatic. 30 April 2009. Retrieved 28 May 2013.
^ Diepraam, Mattijs (1 December 2019). "Pole positions in World Championship events". 6th Gear. Forix. Archived from the original on 7 July 2018. Retrieved 15 April 2019.
^ "The quickest tyre of all". Racingspot.pirelli.com. Retrieved 25 March 2018.
^ "Leclerc clinches 2019 pole position prize – despite being fourth fastest". www.formula1.com. 16 November 2019. Archived from the original on 17 November 2019. Retrieved 20 January 2021.
^ "MotoGP sprint races: Everything you need to know". Autosport. 21 March 2023. Retrieved 14 April 2023.
^ "Statistics". MotoGP.com. Retrieved 10 October 2022. Select Riders' best result, poles, all seasons, all classes, all countries, all tracks.
^ James, Brant (22 January 2014). "NASCAR sets new qualifying format". ESPN. Retrieved 18 July 2014.
^ Newton, David (April 9, 2013). "NASCAR unveils Eldora qualifying". NASCAR. Retrieved September 6, 2013.
^ "Sprint Cup Series All-Time Pole Winners & Records". Jayski's Silly Season Site. Retrieved 11 February 2015.
^ "SUPERBIKE WORLD CHAMPIONSHIP 2023: RACE SCHEDULE, CALENDAR, HOW TO WATCH, IS ALVARO BAUTISTA THE RIDER TO BEAT?". Eurosport. 28 February 2023. Retrieved 14 April 2023.
^ "ROAR v13.0 Rules" (PDF). roarracing.com. Retrieved 12 August 2022.
^ "Archived copy" (PDF). Archived from the original (PDF) on 2014-11-01. Retrieved 2014-12-01.{{cite web}}: CS1 maint: archived copy as title (link)
^ "Archived copy" (PDF). Archived from the original (PDF) on 2015-07-01. Retrieved 2015-01-02.{{cite web}}: CS1 maint: archived copy as title (link)
External links
[edit]
Media related to Pole position at Wikimedia Commons
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Calculating distances on maps is a crucial skill for travelers, hikers, and anyone looking to navigate effectively. However, it’s easy to make mistakes that can lead to misunderstandings about how far you need to travel. In this article, we will explore some common pitfalls people encounter when calculating distances on maps and provide practical tips for avoiding them.
Mistake #1: Ignoring Map Scale
One of the most frequent errors made when calculating distances is not paying attention to the map scale. The scale indicates the relationship between distance on the map and actual distance on the ground. For example, if a scale shows that 1 inch equals 5 miles, measuring a 2-inch distance means you’ve covered 10 miles in reality. Always check the scale before making your calculations to ensure accuracy.
Mistake #2: Misunderstanding Types of Distance Measurements
Maps can show different types of distances—straight-line (as-the-crow-flies) versus travel distance (the route you actually take). Many people mistakenly assume that straight-line measurements reflect how long it will take them to travel from one point to another. However, roads often wind and curves can add extra distance. Be sure you’re using the right type of measurement for your needs.
Mistake #3: Not Using Technology Effectively
In today’s digital age, there are various tools available that simplify distance calculations on maps such as Google Maps or GPS devices. Some users still rely solely on paper maps or manual calculations without taking advantage of these technological advancements. Familiarize yourself with mapping software features like route planning or measuring tools that can help provide accurate distances efficiently.
Mistake #4: Underestimating Elevation Changes
When calculating distances, many overlook elevation changes which can significantly affect travel time and effort required. If you’re hiking or driving in hilly areas, be aware that steep climbs may slow you down compared to flat terrain even if the horizontal distance appears short on a map. Account for elevation by considering both vertical gain and loss in your planning.
Mistake #5: Forgetting Local Conditions
Local conditions such as traffic patterns, road closures, construction zones or weather conditions also impact how long it takes to cover a certain distance—even if you have accurately calculated your route’s length. Always factor in these variables by checking local news updates or using real-time navigation apps before heading out.
By being aware of these common mistakes when calculating distances on maps—and knowing how to avoid them—you’ll enhance your navigation skills significantly. Whether you’re planning a road trip or just trying to get around town more efficiently, taking care with your calculations ensures you’ll arrive at your destination smoothly.
This text was generated using a large language model, and select text has been reviewed and moderated for purposes such as readability.
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187943 | https://en.wiktionary.org/wiki/percent | percent - Wiktionary, the free dictionary
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Beginning
1 EnglishToggle English subsection
1.1 Alternative forms
1.2 Etymology
1.3 Pronunciation
1.4 Adverb
1.4.1 Usage notes
1.4.2 Derived terms
1.5 Noun
1.5.1 Derived terms
1.5.2 Translations
1.6 Prepositional phrase
1.6.1 Usage notes
1.7 See also
1.8 References
1.9 Further reading
1.10 Anagrams
2 ChineseToggle Chinese subsection
2.1 Etymology
2.2 Pronunciation
2.3 Noun
2.4 References
3 FrenchToggle French subsection
3.1 Pronunciation
3.2 Verb
4 HungarianToggle Hungarian subsection
4.1 Etymology
4.2 Pronunciation
4.3 Verb
percent
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From Wiktionary, the free dictionary
See also:per centandper cent.
English
[edit]
English Wikipedia has an article on:
percent
Wikipedia
Alternative forms
[edit]
%, pc, pct, p.c., p. c.(abbreviation)
per cent(chiefly Commonwealth)
per centum(law, US, dated)
per cent.(archaic)
Etymology
[edit]
From New Latinpercentum(“by the hundred”).
Pronunciation
[edit]
(UK)IPA(key): /pə(ɹ)ˈsɛnt/
Audio (US):Duration: 2 seconds.0:02(file)
(Indic)IPA(key): /ˈpɜː(ɾ).seɳʈ/, /pa(ɾ)ˈseɳʈ/
Rhymes: -ɛnt
Adverb
[edit]
percent (not comparable)
For every hundred (used with preceding numeral to form a noun phrase expressing a proportion). [from 16th c.]quotations▼
2002 May 8, Leon Jaroff, Time:Diane Watson has had a distinguished career in education and politics, and last year was elected to the House of Representatives, winning 75 percent of the vote in her Congressional district.
2011, Frank Dikötter, Mao's Great Famine, Bloomsbury, →ISBN, →OCLC, →OL, pages 297–298:In Sichuan the rates were much higher. In Kaixian county, a close examination by a team sent by the provincial party committee at the time concluded that in Fengle commune, where 17 per cent of the population had perished in less than a year, up to 65 per cent of the victims had died because they were beaten, punished with food deprivation or forced into committing suicide.
2016 July 7, Arthur Neslen, The Guardian:Twelve percent of the world’s population now relies directly or indirectly on the fisheries industry.
Usage notes
[edit]
A percentage is often denoted by the character %. 50% denotes 50 percent.
The difference of two percentages is measured by percentage point, not by percent.
In the UK, per cent is the normal form.
Derived terms
[edit]
100 percent
51 percent
99 percent
a hundred and ten percent
a hundred percent
cent percent
give a hundred percent
not be 100 percent on something
one hundred percent
one hundred percent American
one-percent milk
percentage
percentual
percentwise
two-percent milk
weight/volume percent
show more ▼
Noun
[edit]
percent (pluralpercentorpercents)
A percentage, a proportion (especially per hundred). only a small percent attain the top ranks
One part perhundred; one percent, hundredth. [from 19th c.]quotations▼
2008, Niall Ferguson, The Ascent of Money, Penguin, page 254:And from 1966, under Regulation Q, there was a ceiling of 5.5 per cent on their deposit rates, a quarter of a per cent more than banks were allowed to pay.
The percent sign, %.
An annuity or security with a certain fixed and guaranteed annual percentage rate of return or percentage dividend. quotations▼
1848, Edward W. Cox, Esq., of the Middle Temple, Barrister at Law, editor, Reports of Cases in Criminal Law Argued and Determined in All the Courts of England and Ireland: Volume II 1868 to 1848, page 371:[…]Several stocks in the Three Per Cents and Three Per Cents Reduced to be transferred into the name and to the credit of the prosecutor, without any authority to him (the traverser) to sell, negotiate, transfer or pledge the said 9000l. Three-and-a-Quarter per Cent. Annuities.
1886, Jerome K. Jerome, “On Being Hard Up”, in Idle Thoughts of an Idle Fellow:Why, from the pleasant and businesslike manner in which the transaction is carried out, it might be a large purchase in the three per cents. Yet what a piece of work a man makes of his first "pop."
2018, Nancy Henry, Women, Literature and Finance in Victorian Britain: Cultures of Investment, page 33:picking up on a phrase that was used as early as 1752, Benjamin Disraeli famously referred to the “sweet simplicity of the three percents in his novel Endymion (1880) because of the reliable dividend this form of investment provided.
Derived terms
[edit]
percent-encode, percent-encoding
Translations
[edit]
show ▼±a part or other object per hundred
[Select preferred languages] [Clear all]
Afrikaans: persent(af)
Albanian: përqind
Arabic: فِي الْمِئَة(fī l-miʔa), بِالْمِئَة(bi-l-miʔa)
Aramaic: Assyrian Neo-Aramaic: ܓܵܘ ܐܸܡܵܐ(go imma), ܒܐܸܡܵܐ(bimma)
Armenian: տոկոս(hy)(tokos)
Asturian: por cientu
Azerbaijani: faiz(az), yüzdə
Bashkir: процент(protsent)
Basque: ehuneko
Belarusian: працэ́нтm(pracént), адсо́такm(adsótak)
Bengali: শতাংশ(bn)(śotaṅśo)
Breton: dre gant
Bulgarian: проце́нт(bg)(procént)
Burmese: ရာခိုင်နှုန်း(my)(rahkuinghnun:)
Catalan: per cent
Chinese: Cantonese: 百分之……(baak 3 fan 6 zi 1...)(the number follows it, e.g. 30%: 百分之三十 baak3 fan6 zi1 saam1 sap6), 巴仙(baa 1 sin 1)(unit follows the number with qualifier 個/个)Hakka: 趴先多(phà-sén-tó)(unit follows the number)Hokkien: 百分之……(pah-hun-chi...)(the number follows it, e.g. 30%: 百分之三十 pah-hun-chi saⁿ-cha̍p)Mandarin: 百分之……(zh)(bǎi fēn zhī...)(the number follows it, e.g. 30%: 百分之三十 bǎi fēn zhī sānshí), 巴仙(zh)(bāxiān)(unit follows the number with qualifier 個/个)
Czech: procento(cs)n
Danish: procent(da)c
Dutch: procent(nl)n
Erzya: сядопелькс(śadopeľks)
Esperanto: procento, pocento, elcento, centono
Estonian: protsent
Faroese: prosentn
Fijian: pasede
Finnish: prosentti(fi), sadasosa(fi), (colloquial)prossa(fi), (rare)sadannes(fi)
French: pour cent(fr), pourcent(fr)
Georgian: პროცენტი(ṗrocenṭi)
German: vom Hundert, Prozent(de)n, Perzentn
Greek: τοις εκατό(tois ekató)
Hebrew: אָחוּז(he)(akhúz)
Hindi: प्रतिशतm(pratiśat), फ़ीसदm(fīsad)
Hungarian: százalék(hu)
Icelandic: hundraðshluti(is)m, prósentaf, hundraðstalaf
Ido: po cent, centimi(io)pl
Indonesian: persen(id)
Italian: per cento(it)
Japanese: パーセント(ja)(pāsento)
Kazakh: пайыз(kk)(paiyz), процент(prosent)
Khmer: ភាគរយ(km)(phiək rɔɔy)
Korean: 퍼센트(ko)(peosenteu), 프로(ko)(peuro)
Kurdish: Northern Kurdish: ji sedî
Kyrgyz: процент(protsent), пайыз(payız)
Lao: ເປີເຊັນ(pœ̄ sen)
Latin: per centum, pro centum
Latvian: procentsm
Lithuanian: nuošimtis, procentasm
Macedonian: насто(nasto), процент(procent)
Malay: peratus
Maltese: fil-mija
Maori: ōrau
Mongolian: Cyrillic: процент(procent), хувь(mn)(xuvʹ)
Norwegian: Bokmål: prosent(no)m
Pashto: فيصد(ps)(fisad)
Persian: Iranian Persian: دَرْصَد(darsad), فی صَد(fi sad)
Polish: procent(pl)m inan
Portuguese: por cento
Punjabi: ਪ੍ਰਤਿਸ਼ਤ(prtiśat)
Romanian: procent(ro)n, la sută
Russian: проце́нт(ru)m(procént)
Samoan: pasene
Scottish Gaelic: às a' cheud
Serbo-Croatian: Cyrillic: по̏сто̄, о̏дсто̄, по̀стотакm, проценатRoman: pȍstō(sh), ȍdstō, pòstotak(sh)m, procenat(sh)
Slovak: percenton
Slovene: odstotekm
Spanish: por ciento
Swahili: asilimia(sw)
Swedish: procent(sv)c
Tagalog: bahagdan(tl)
Tajik: дарсад(darsad), фоиз(tg)(foyiz), дар сад(dar sad)
Tamil: விழுக்காடு(ta)(viḻukkāṭu), சதவீதம்(ta)(catavītam)
Telugu: శాతం(śātaṁ)
Thai: เปอร์เซ็นต์(th)(bpəə-sen), ร้อยละ(th)(rɔ́ɔi-lá)
Turkish: yüzde(tr)
Turkmen: protsent, prosent
Ukrainian: відсо́токm(vidsótok), проце́нт(uk)m(procént)
Urdu: فِیصَدm(fīsad), فی صَد(fī sad)
Uyghur: پىرسەنت(pirsent)
Uzbek: protsent(uz), foiz(uz)
Vietnamese: phần trăm(vi)
Walloon: porcint(wa)m
Welsh: y cant
Yiddish: פּראָצענטm(protsent)
Add translation: More
[x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective
Noun class: Plural class:
Transliteration: (e.g. zìmǔ for 字母)
Literal translation:
Raw page name: (e.g. 疲れる for 疲れた)
Qualifier: (e.g. literally, formally, slang)
Script code: (e.g. Cyrl for Cyrillic, Latn for Latin)
Nesting: (e.g. Serbo-Croatian/Cyrillic)
Prepositional phrase
[edit]
percent
Per hundred. quotations▼
2014, Alan Tussy, Diane Koenig, Basic Mathematics for College Students with Early Integers, →ISBN, page 637:By how many percent did the cancer survival rate for breast cancer increase by 2008?
Usage notes
[edit]
Percent/per cent originated as a shortening of the Latin phrase per centum, “per hundred”, and historically the use of the word as a noun (as in “half a percent” or “percents”) was regarded as an error, though such use has now become so common that it is recognized by all other major dictionaries, and a few treat the word as being only a noun. Of those which recognize non-nounal uses, most label it an adverb and many also label it an adjective though it does not meet tests of adjectivity.
See also
[edit]
per mille, permille, ‰
per myriad, ‱
ppm, ppb, ppt, ppq
Typography
ampersand ( & )
asterisk ( ) and asterism ( ⁂ )
at sign ( @ )
backslash ( )
bullet ( • )
dagger ( †‡ )
degree symbol ( ° )
number sign ( # )
prime ( ′ )
tilde ( ~ )
underscore ( _ )
vertical bar/pipe ( | )
References
[edit]
^Various older pedagogic works, e.g. Charles Harvey Raymond's Essentials of English composition (1923), page 461, prescribe: "Per cent is an adverb meaning in the hundred. [...] Percentage is a noun meaning rate per cent."
↑ Jump up to: 2.02.12.2“percent”, in Merriam-Webster Online Dictionary, Springfield, Mass.: Merriam-Webster, 1996–present.
↑ Jump up to: 3.03.13.2“percent”, in The American Heritage Dictionary of the English Language, 5th edition, Boston, Mass.: Houghton Mifflin Harcourt, 2016, →ISBN.
↑ Jump up to: 4.04.14.2“percent”, in Collins English Dictionary.
↑ Jump up to: 5.05.1“percent”, in Cambridge English Dictionary, Cambridge, Cambridgeshire: Cambridge University Press, 1999–present.
↑ Jump up to: 6.06.1“percent”, in Lexico, Dictionary.com; Oxford University Press, 2019–2022.
↑ Jump up to: 7.07.1“percent”, in Dictionary.com Unabridged, Dictionary.com, LLC, 1995–present.
↑ Jump up to: 8.08.1“percent” (US) / “percent” (UK) in Macmillan English Dictionary.
Further reading
[edit]
“percent”, in OneLook Dictionary Search.
William Dwight Whitney, Benjamin E[li] Smith, editors (1911), “percent”, in The Century Dictionary[…], New York, N.Y.: The Century Co., →OCLC.
Anagrams
[edit]
-nercept, precent
Chinese
[edit]
Etymology
[edit]
From Englishpercent. Doublet of 巴仙(bāxiān).
Pronunciation
[edit]
Cantonese(Jyutping): poe 6 sen 1 more ▼
Cantonese
(Standard Cantonese, Guangzhou–Hong Kong)+
Jyutping: poe 6 sen 1
Cantonese Pinyin: poe 6 sen 1
Sinological IPA(key): /pʰœː²² sɛːn ⁵⁵/
Noun
[edit]
percent
(Hong Kong Cantonese)percent(Classifier: 個／个c)
References
[edit]
English Loanwords in Hong Kong Cantonese
French
[edit]
Pronunciation
[edit]
Audio:Duration: 2 seconds.0:02(file)
Verb
[edit]
percent
third-personpluralpresentindicative/subjunctive of percer
Hungarian
[edit]
Etymology
[edit]
percen + -t
Pronunciation
[edit]
IPA(key): [ˈpɛrt͡sɛnt]
Hyphenation: per‧cent
Verb
[edit]
percent
third-personsingularindicativepastindefinite of percen
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This page was last edited on 22 March 2025, at 22:07.
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Radial Nerve Palsy following Frx of the Humerus
see also (Radial Nerve Discussion; Tendon Transfers for Radial Nerve Palsy) Discussion up to 18% of humeral shaft frx have an associated radial nerve palsy; it occurs most commonly with middle third humeral fractures & in Holstein Lewis fractures; types of nerve lesions; laceration of nerve, most common in spiral frx of the distal 1/3 … Read more
Categories Nerves MenuTags fractures of the humerus, Nerves, Radial Nerve Palsy
Anatomy of the Posterior Cruciate Ligament
– Anatomy: – PCL is approximately twice as strong and twice as thick as the normal ACL (therefore less commonly injured); – size: 13 mm, its length, 38 mm, (approximates that of ACL); – origin: – PCL originates from the antero-lateral aspect of … Read more
Categories Muscles TendonsTags Knee, Muscles Tendons, Nerves, Tibia and Fibula
Amyotrophic Lateral Sclerosis
– See:– El Escorial Revisited: Revised Criteria for the Diagnosis of Amyotrophic Lateral Sclerosis – Disorders of Nerves: – Anterior Horn Cell Disorders: – ALS Society Homepage – Discussion: – amyotrophic lateral sclerosis is a syndrome characterized by both upper and lower motor neuron disease; – destruction of anterior horn cells & descending corticospinal tracts; … Read more
Categories Muscles TendonsTags Muscles Tendons, Nerves, Spine
Technique of Total Shoulder Arthroplasty
– Position: – place patient in a beach chair position w/ torso flexed 45 deg and the knees flexed to 30 deg; – move patient to the edge of the table (operative side), and use a McConnel positioner to prevent the patient from failing off the table; –patient must lie … Read more
Categories Joints, ShoulderTags Hand, Hip, Humerus, Knee, Muscles Tendons, Nerves, Shoulder, Spine
Tibio-Talar-Calcaneus Arthrodesis
– Discussion: – surgical indications include diabetic neuropathy, osteoarthritis, posttraumatic injury,talar AVN,and RA involving ankle and subtalar joints; – position for pantalar arthrodesis: – 0-5 deg of calcaneus (not equinus) and 5 deg of valgus; – in the report by Chou, et al, the authors … Read more
Categories OrthopaedicsTags Arthritis, Knee, Muscles Tendons, Nerves, Trauma Fractures
TKR: Soft Tissue Release of Valgus Deformity
– Sequential Order of Release: – PCL: (posterior stabilized prosthesis is generally required); – LCL – in general, sepquential elevation of the femoral origin of the LCL is performed prior to release of other structures because … Read more
Categories Muscles TendonsTags Femur, Knee, Muscles Tendons, Nerves, Tibia and Fibula
TKR: Valgus Deformity
– See: – TKR Menu: – Collateral ligaments: – Medial Collateral Ligament – Lateral Collateral Ligament – Discussion: – more common in females; – in valgus knee, ligament balancing is more difficult to fix; – the anatomic deformity is usually larger in valgus than in varus, and it is on the femoral side; – femur … Read more
Categories Joints, KneeTags Femur, Hip, Knee, Muscles Tendons, Nerves, Tibia and Fibula
Tumors of Hand and Distal radius
– Adamantinoma – Basal Cell Carcinoma: – Calcinosis – Chondrosarcoma: – the most common bony malignancy in the hand; – CMC Boss: bony growth at bases of second and third metacarpals; – special radiographs can demonstrate the osteophytes, which may be resected if symptoms warrent; – associated with ganglia at least 30% of the time, … Read more
Categories OrthopaedicsTags Arthroscopy, Hand, Muscles Tendons, Nerves, Orthobiologics, Tibia and Fibula, Trauma Fractures, Wrist
Tunnel of Guyon
– Discussion: – Guyon (French Urologist) described this space in 1861; – Guyon’s canal is approximately 4 cm long beginning at the proximal extent of transverse carpal ligament and ends at the aponeurotic arch of hypothenar muscles; – depression between pisiform & hook of hamate is converted into fibrosseous tunnel, the tunnel of Guyon, by pisohamate … Read more
Categories OrthopaedicsTags Hand, Muscles Tendons, Nerves, Wrist
Total Knee Replacement – Patient Saftey Packet
How We Are Committed to Protecting Your Safety: How We Protect You From Infection How We Protect You From Blood Clots: What you should know before your surgery? your laboratory results your hematocrit should be between 30 and 45% – if your hematocrit drops below 25-26% during your surgery,we may recommend a … Read more
Categories Joints, KneeTags Knee, Nerves
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187945 | https://it.wikipedia.org/wiki/Battimento_(fisica) | Indice
Battimento (fisica)
Il battimento è, nella teoria musicale, in fisica e particolarmente in acustica, la frequenza risultante dalla sovrapposizione di grandezze periodiche, in genere oscillazioni sinusoidali di diversa e vicina frequenza. Si basa sulle proprietà del principio di sovrapposizione. Oltre ai campi citati, tutti i fenomeni fisici che prevedono onde risentono del fenomeno del battimento, onde meccaniche e onde elettromagnetiche comprese; battimenti si verificano, tra l'altro, in materia di elaborazione dei segnali, quando due frequenze di segnale si trovano vicine le une alle altre.
Descrizione
Entrando più nel dettaglio del fenomeno acustico, al battimento risulta un effetto vibratorio particolare, caratterizzato da rapide ondulazioni acustiche. L'effetto è un rafforzamento seguito da un indebolimento del suono a seconda che le frequenze siano in concordanza o in discordanza di fase. I battimenti si distinguono con difficoltà negli strumenti a corde percosse come il pianoforte, a causa della breve durata dei suoni. Si rileva con minore difficoltà negli strumenti a vento e a serbatoio d'aria, come ad esempio l'organo, in quanto hanno una sonorità più ampia. Questo effetto è anche facilmente riscontrabile nel cosiddetto vibrato degli archi. Infatti, siccome il vibrato si ottiene spostando leggermente il dito sulla corda del violino, esso causa dei suoni leggermente diversi l'uno dall'altro determinando, con la sovrapposizione delle vibrazioni, i battimenti.
Approccio fisico
Supponiamo di avere due corpi che vibrano simultaneamente, i cui suoni si possano rappresentare con onde sinusoidali con la stessa frequenza e la stessa ampiezza. Queste due onde possono sovrapporsi in diverse maniere: in fase (interferenza costruttiva), in opposizione di fase (interferenza distruttiva), o in una via di mezzo. Essendo il suono risultante la somma dei due suoni, nel primo caso questo sarà identico ai primi due, ma di ampiezza doppia (le creste si sommano e le valli si sommano); nel secondo caso non si avrà alcun suono risultante (le creste e le valli si compensano in ogni punto annullandosi tra di loro); nel terzo si avrà un suono di intensità intermedia, a seconda di quanto è lo sfasamento tra i due suoni iniziali. Naturalmente, avendo i due suoni la stessa frequenza, lo sfasamento sarà costante nel tempo: se ad esempio la prima cresta del primo suono è perfettamente sovrapposta alla prima cresta del secondo, lo stesso avverrà per le seconde creste, per le terze e così via (analogamente nel caso di sfasamento arbitrario).
Supponendo ora che le due frequenze non siano proprio identiche, ma che ci sia una piccola differenza tra di esse, lo sfasamento questa volta non sarà più costante, ma varierà nel tempo: se ad esempio le prime creste dei due suoni coincidevano perfettamente (l'intensità totale quindi era il doppio), le seconde non saranno perfettamente sovrapposte, perché una arriverà un po' prima dell'altra; per le terze creste questa differenza di fase sarà ancora più marcata e così via, fino a quando la cresta del primo suono non sarà sovrapposta a una valle del secondo: i due suoni sono passati in opposizione di fase e l'intensità totale è zero. Procedendo ancora in maniera analoga, dopo un certo numero di periodi (dipendente dalla differenza relativa tra le due frequenze iniziali) i due suoni ritorneranno in fase. In altri termini si hanno battimenti quando lo sfasamento (e quindi il tipo di interferenza) tra due suoni di frequenze simili varia nel tempo. Questo meccanismo si vede chiaramente nell'immagine.
Un'elegante spiegazione matematica del fenomeno si dà tramite le formule di prostaferesi: se rappresentiamo i due suoni con due onde sinusoidali di ampiezza unitaria (per semplicità), possiamo applicare le formule al suono risultante:
sin
(
ω
1
t
)
+
sin
(
ω
2
t
)
=
2
cos
(
ω
1
−
ω
2
2
t
)
sin
(
ω
1
+
ω
2
2
t
)
=
2
cos
(
Ω
t
)
⋅
sin
(
ω
t
)
{\displaystyle \sin(\omega _{1}t)+\sin(\omega _{2}t)=2\cos \left({\frac {\omega _{1}-\omega _{2}}{2}}t\right)\sin \left({\frac {\omega _{1}+\omega _{2}}{2}}t\right)=2\cos \left(\Omega t\right)\cdot \sin \left(\omega t\right)}
Ove si è posto
ω
ω
1
+
ω
2
2
{\displaystyle \omega ={\frac {\omega _{1}+\omega _{2}}{2}}}
,
Ω
ω
1
−
ω
2
2
{\displaystyle \Omega ={\frac {\omega _{1}-\omega _{2}}{2}}}
.
Se
Ω
≪
ω
{\displaystyle \Omega \ll \omega }
, (cioè se
ω
1
{\displaystyle \omega _{1}}
e
ω
2
{\displaystyle \omega _{2}}
sono vicine), si può esprimere la somma dei due suoni come un suono di frequenza intermedia, pari a
ω
{\displaystyle \omega }
, la cui ampiezza sia modulata alla frequenza molto più bassa
Ω
{\displaystyle \Omega }
.
Esempi pratici di battimenti
Il fenomeno dei battimenti è facilmente riscontrabile se facciamo vibrare contemporaneamente due corpi che hanno fra loro una leggera differenza di frequenza (per esempio di una sola vibrazione al secondo), nel primo istante i due moti arriveranno all'orecchio nella stessa fase di vibrazione; ma dopo mezzo secondo la prima origine sonora avrà compiuto mezza vibrazione in più della seconda e i due moti saranno in fase opposta. Nel successivo mezzo secondo le vibrazioni si rimetteranno gradatamente in fase e l'orecchio riceverà nuovamente due moti concordi.
L'intensità del suono quindi, nell'alternarsi delle diverse fasi oscillerà continuamente, sì che nel miscuglio dei due suoni, leggermente disuguali d'altezza, si avrà, a intervalli uguali, un susseguirsi di periodici rinforzamenti e di periodici indebolimenti che sono chiamati battimenti.
Vi sono degli strumenti che producono quasi sempre dei battimenti: così sono le campane che, presentando diversità di spessore in diversi punti, producono battimenti assai intensi che conferiscono loro la caratteristica sonorità ondulante.
Spesso i battimenti sono appositamente impiegati per conseguire effetti speciali; nell'esempio dell'organo, il registro della voce umana è formato da due canne non perfettamente intonate, allo scopo di ottenere una specie di tremolio che imita la voce dei cantanti. Anche la voce celeste è ottenuta da due canne intonate in modo da ottenere questo effetto.
Suoni di differenza, addizione
Suonando due note contemporaneamente, l'orecchio percepisce note aggiuntive di varie frequenze pari a opportune somme e differenze delle due note emesse: si parla in questi casi di suoni di combinazione. Fra questi il più importante da un punto di vista pratico è il cosiddetto terzo suono di Tartini, scoperto appunto dal Tartini nel Settecento. Il celebre violinista constatò infatti che suonando un bicordo a un intervallo di 5ª (ovvero con rapporto di frequenze 3:2) si sentiva al basso un'altra nota la cui frequenza corrispondeva a un numero di vibrazioni pari alla differenza fra quelle dei due suoni originari. Così, ad esempio, se un suono aveva 900 vibrazioni e l'altro 600, il suono ulteriore che si sentiva aveva 300 vibrazioni al secondo ed era, quindi, di un'ottava più grave di quest'ultimo.
Da un punto di vista fisico il fenomeno risulta particolarmente evidente suonando due note a un intervallo di 5ª poiché i prodotti di intermodulazione (vedi nel seguito) del second'ordine f2−f1 e del terz'ordine 2f1−f2, che sono normalmente disgiunti, in questo caso coincidono esattamente sommandosi.
Il fenomeno dei suoni di combinazione è ormai noto da oltre mezzo secolo nell'elettronica applicata alle telecomunicazioni dove questi vengono denominati "prodotti di intermodulazione": si generano in ogni amplificatore non lineare, ovvero che produce una distorsione sui segnali in ingresso, in particolare quindi anche all'interno del nostro orecchio quando questo percepisce due suoni da sorgenti distinte.
Due suoni di frequenza
f
1
{\displaystyle f_{1}}
e
f
2
{\displaystyle f_{2}}
sommati in un amplificatore non lineare come il nostro orecchio, producono infatti i prodotti di intermodulazione del second'ordine:
f
1
+
f
2
,
f
2
−
f
1
{\displaystyle f_{1}+f_{2},\,f_{2}-f_{1}}
; del terz'ordine:
2
f
1
+
f
2
,
2
f
1
−
f
2
,
2
f
2
+
f
1
,
2
f
2
−
f
1
{\displaystyle 2f_{1}+f_{2},\,2f_{1}-f_{2},\,2f_{2}+f_{1},2f_{2}-f_{1}}
e degli ordini successivi; oltre alle armoniche
2
f
1
,
2
f
2
,
3
f
1
,
3
f
2
{\displaystyle 2f_{1},\,2f_{2},\,3f_{1},\,3f_{2}}
... multiple delle frequenze fondamentali.
Sono tali frequenze generate all'interno dell'orecchio a produrre i suoni differenza e addizione, i primi a lungo confusi con inesistenti "armonici inferiori" o "ipotoni".
Termini come "ipotoni", "suoni di moltiplicazione", "subarmonici", che si trovano sovente in letteratura non hanno alcun significato in fisica. Il fenomeno dei cosiddetti subarmonici, ad esempio, deriva non tanto da un fenomeno fisico reale, quanto da un errore indotto dall'orecchio quando percepisce due suoni da sorgenti distinte producendo al proprio interno i prodotti di intermodulazione sopra citati.
Applicazioni pratiche
Il fenomeno del "terzo suono" trova una sua applicazione pratica nella costruzione degli organi: talvolta, invece di costruire canne enormi per frequenze molto basse si creano registri in cui due canne a distanza di quinta suonano contemporaneamente creando l'illusione di un terzo suono più profondo; tali registri sono spesso riconoscibili per il loro nome, solitamente Acustico, Risultante o Gravissima. Anche il theremin sfrutta il battimento tra due frequenze non udibili (nel campo degli ultrasuoni) per ottenere un suono udibile e modulabile cambiando la frequenza di una delle due onde.
I registri di Voce umana, Voce celeste, Unda maris, Voce eterea, Timballi degli organi e molti registri delle fisarmoniche (il registro "musette" tipico della musica francese e del liscio) sfruttano il fenomeno dei battimenti per ottenere un suono più caldo ed espressivo. Questi registri fanno suonare contemporaneamente due canne (o ance): una intonata correttamente e una leggermente calante o crescente, in modo da ottenere un certo numero di oscillazioni di intensità al secondo.
Interessante segnalare che nella pratica operativa dell'accordatura di strumenti musicali gli accordatori non sfruttano i battimenti che si verificano tra i fondamentali ma quelli tra alcuni loro armonici facilmente percepibili. Così ad esempio se vengono suonati in concomitanza un Do3 e un Sol3, il terzo suono armonico del Do3 che è un Sol4 "batte" col secondo armonico del Sol3 che è appunto lo stesso Sol4. In base alla frequenza dei battimenti di questi suoni armonici comuni coincidenti si sono realizzate e si realizzano le accordature dette impropriamente "a orecchio" ma che hanno un profondo connotato scientifico basato sulle leggi della spettroscopia acustica. Le accordature storiche quali la pitagorica e mesotonica, ad esempio, potevano essere effettuate sfruttando le caratteristiche timbriche degli strumenti.
Un pianoforte ad esempio non potrà mai essere accordato per terze, proprio perché è impossibile percepire i battimenti tra il quinto armonico del fondamentale (es. Do3) con il quarto armonico della terza (Mi3), cosa invece possibile per un organo o un clavicembalo, tentativi sperimentali nell'accordatura di un pianoforte col sistema mesotonico dove si utilizzano ben otto intervalli di terza maggiore (5/4) portano a innumerevoli imprecisioni nella realizzazione degli otto intervalli stessi e nella conseguente realizzazione della quinta del lupo che costituisce il processo operativo finale dell'accordatura stessa. In sostanza, storicamente è stato possibile ideare e conseguentemente realizzare certe accordature perché le caratteristiche timbriche (percezione dei suoni armonici) dei vari strumenti consentivano agli accordatori di poterle realizzare con una certa facilità.
Intorno al 1965 Pietro Grossi, pioniere della computer music registra un repertorio di eventi sonori nello Studio di Fonologia Musicale di Firenze S 2F M, tra cui una serie di Battimenti. Lo scopo era creare del materiale sonoro da utilizzare per altre composizioni.
I Battimenti però, possono essere ascoltati anche come opera compiuta, tanto che alcuni, come Albert Mayr uno dei maggiori collaboratori di Grossi all'epoca dello studio S 2F M di Firenze, si sbilanciano definendoli come una delle opere più affascinanti del secolo scorso,per la massima espressività pur nell'essenzialità del suono.
Nel 2019 le frequenze dei Battimenti di Grossi vengono riprese dal musicista Sergio Maltagliati e dal pittore astrattista Romano Rizzato. Maltagliati usa 11 frequenze di suoni, da 395 a 405 Hz (le stesse registrate da Grossi), combinate in base a quanto suggerisce la pittura di Rizzato.
Toni binaurali
I toni binaurali sono dei battimenti che vengono generati direttamente dal cervello quando le due onde sonore vengono ascoltate separatamente tramite degli auricolari (quindi non vi è sovrapposizione fisica delle due onde sonore).
Note
Bibliografia
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| Controllo di autorità | GND (DE) 4267871-7 | |
187946 | https://en.wikipedia.org/wiki/Semigroup | Jump to content
Semigroup
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From Wikipedia, the free encyclopedia
Algebraic structure consisting of a set with an associative binary operation
Not to be confused with quasigroups.
In mathematics, a semigroup is an algebraic structure consisting of a set together with an associative internal binary operation on it.
The binary operation of a semigroup is most often denoted multiplicatively (just notation, not necessarily the elementary arithmetic multiplication): x ⋅ y, or simply xy, denotes the result of applying the semigroup operation to the ordered pair (x, y). Associativity is formally expressed as that (x ⋅ y) ⋅ z = x ⋅ (y ⋅ z) for all x, y and z in the semigroup.
Semigroups may be considered a special case of magmas, where the operation is associative, or as a generalization of groups, without requiring the existence of an identity element or inverses.[a] As in the case of groups or magmas, the semigroup operation need not be commutative, so x ⋅ y is not necessarily equal to y ⋅ x; a well-known example of an operation that is associative but non-commutative is matrix multiplication. If the semigroup operation is commutative, then the semigroup is called a commutative semigroup or (less often than in the analogous case of groups) it may be called an abelian semigroup.
A monoid is an algebraic structure intermediate between semigroups and groups, and is a semigroup having an identity element, thus obeying all but one of the axioms of a group: existence of inverses is not required of a monoid. A natural example is strings with concatenation as the binary operation, and the empty string as the identity element. Restricting to non-empty strings gives an example of a semigroup that is not a monoid. Positive integers with addition form a commutative semigroup that is not a monoid, whereas the non-negative integers do form a monoid. A semigroup without an identity element can be easily turned into a monoid by just adding an identity element. Consequently, monoids are studied in the theory of semigroups rather than in group theory. Semigroups should not be confused with quasigroups, which are generalization of groups in a different direction; the operation in a quasigroup need not be associative but quasigroups preserve from groups the notion of division. Division in semigroups (or in monoids) is not possible in general.
The formal study of semigroups began in the early 20th century. Early results include a Cayley theorem for semigroups realizing any semigroup as a transformation semigroup, in which arbitrary functions replace the role of bijections in group theory. A deep result in the classification of finite semigroups is Krohn–Rhodes theory, analogous to the Jordan–Hölder decomposition for finite groups. Some other techniques for studying semigroups, like Green's relations, do not resemble anything in group theory.
The theory of finite semigroups has been of particular importance in theoretical computer science since the 1950s because of the natural link between finite semigroups and finite automata via the syntactic monoid. In probability theory, semigroups are associated with Markov processes. In other areas of applied mathematics, semigroups are fundamental models for linear time-invariant systems. In partial differential equations, a semigroup is associated to any equation whose spatial evolution is independent of time.
There are numerous special classes of semigroups, semigroups with additional properties, which appear in particular applications. Some of these classes are even closer to groups by exhibiting some additional but not all properties of a group. Of these we mention: regular semigroups, orthodox semigroups, semigroups with involution, inverse semigroups and cancellative semigroups. There are also interesting classes of semigroups that do not contain any groups except the trivial group; examples of the latter kind are bands and their commutative subclass – semilattices, which are also ordered algebraic structures.
| Algebraic structures |
| Group-like Group Semigroup / Monoid Rack and quandle Quasigroup and loop Abelian group Magma Lie group Group theory |
| Ring-like Ring Rng Semiring Near-ring Commutative ring Domain Integral domain Field Division ring Lie ring Ring theory |
| Lattice-like Lattice Semilattice Complemented lattice Total order Heyting algebra Boolean algebra Map of lattices Lattice theory |
| Module-like Module Group with operators Vector space Linear algebra |
| Algebra-like Algebra Associative Non-associative Composition algebra Lie algebra Graded Bialgebra Hopf algebra |
| v t e |
Definition
[edit]
A semigroup is a set S together with a binary operation ⋅ (that is, a function ⋅ : S × S → S) that satisfies the associative property:
: For all a, b, c ∈ S, the equation (a ⋅ b) ⋅ c = a ⋅ (b ⋅ c) holds.
More succinctly, a semigroup is an associative magma.
Examples of semigroups
[edit]
Empty semigroup: the empty set forms a semigroup with the empty function as the binary operation.
Semigroup with one element: there is essentially only one (specifically, only one up to isomorphism), the singleton {a} with operation a · a = a.
Semigroup with two elements: there are five that are essentially different.
A null semigroup on any nonempty set with a chosen zero, or a left/right zero semigroup on any set.
The "flip-flop" monoid: a semigroup with three elements representing the three operations on a switch – set, reset, and do nothing. This monoid plays a central role in Krohn-Rhodes theory.
The set of positive integers with addition. (With 0 included, this becomes a monoid.)
The set of integers with minimum or maximum. (With positive/negative infinity included, this becomes a monoid.)
Square nonnegative matrices of a given size with matrix multiplication.
Any ideal of a ring with the multiplication of the ring.
The set of all finite strings over a fixed alphabet Σ with concatenation of strings as the semigroup operation – the so-called "free semigroup over Σ". With the empty string included, this semigroup becomes the free monoid over Σ.
A probability distribution F together with all convolution powers of F, with convolution as the operation. This is called a convolution semigroup.
Transformation semigroups and monoids.
The set of continuous functions from a topological space to itself with composition of functions forms a monoid with the identity function acting as the identity. More generally, the endomorphisms of any object of a category form a monoid under composition.
The product of faces of an arrangement of hyperplanes.
Basic concepts
[edit]
Identity and zero
[edit]
A left identity of a semigroup S (or more generally, magma) is an element e such that for all x in S, e ⋅ x = x. Similarly, a right identity is an element f such that for all x in S, x ⋅ f = x. Left and right identities are both called one-sided identities. A semigroup may have one or more left identities but no right identity, and vice versa.
A two-sided identity (or just identity) is an element that is both a left and right identity. Semigroups with a two-sided identity are called monoids. A semigroup may have at most one two-sided identity. If a semigroup has a two-sided identity, then the two-sided identity is the only one-sided identity in the semigroup. If a semigroup has both a left identity and a right identity, then it has a two-sided identity (which is therefore the unique one-sided identity).
A semigroup S without identity may be embedded in a monoid formed by adjoining an element e ∉ S to S and defining e ⋅ s = s ⋅ e = s for all s ∈ S ∪ {e}. The notation S1 denotes a monoid obtained from S by adjoining an identity if necessary (S1 = S for a monoid).
Similarly, every magma has at most one absorbing element, which in semigroup theory is called a zero. Analogous to the above construction, for every semigroup S, one can define S0, a semigroup with 0 that embeds S.
Subsemigroups and ideals
[edit]
The semigroup operation induces an operation on the collection of its subsets: given subsets A and B of a semigroup S, their product A · B, written commonly as AB, is the set { ab | a ∈ A and b ∈ B }. (This notion is defined identically as it is for groups.) In terms of this operation, a subset A is called
a subsemigroup if AA is a subset of A,
a right ideal if AS is a subset of A, and
a left ideal if SA is a subset of A.
If A is both a left ideal and a right ideal then it is called an ideal (or a two-sided ideal).
If S is a semigroup, then the intersection of any collection of subsemigroups of S is also a subsemigroup of S.
So the subsemigroups of S form a complete lattice.
An example of a semigroup with no minimal ideal is the set of positive integers under addition. The minimal ideal of a commutative semigroup, when it exists, is a group.
Green's relations, a set of five equivalence relations that characterise the elements in terms of the principal ideals they generate, are important tools for analysing the ideals of a semigroup and related notions of structure.
The subset with the property that every element commutes with any other element of the semigroup is called the center of the semigroup. The center of a semigroup is actually a subsemigroup.
Homomorphisms and congruences
[edit]
A semigroup homomorphism is a function that preserves semigroup structure. A function f : S → T between two semigroups is a homomorphism if the equation
: f(ab) = f(a)f(b).
holds for all elements a, b in S, i.e. the result is the same when performing the semigroup operation after or before applying the map f.
A semigroup homomorphism between monoids preserves identity if it is a monoid homomorphism. But there are semigroup homomorphisms that are not monoid homomorphisms, e.g. the canonical embedding of a semigroup S without identity into S1. Conditions characterizing monoid homomorphisms are discussed further. Let f : S0 → S1 be a semigroup homomorphism. The image of f is also a semigroup. If S0 is a monoid with an identity element e0, then f(e0) is the identity element in the image of f. If S1 is also a monoid with an identity element e1 and e1 belongs to the image of f, then f(e0) = e1, i.e. f is a monoid homomorphism. Particularly, if f is surjective, then it is a monoid homomorphism.
Two semigroups S and T are said to be isomorphic if there exists a bijective semigroup homomorphism f : S → T. Isomorphic semigroups have the same structure.
A semigroup congruence ~ is an equivalence relation that is compatible with the semigroup operation. That is, a subset ~ ⊆ S × S that is an equivalence relation and x ~ y and u ~ v implies xu ~ yv for every x, y, u, v in S. Like any equivalence relation, a semigroup congruence ~ induces congruence classes
: [a]~ = {x ∈ S | x ~ a}
and the semigroup operation induces a binary operation ∘ on the congruence classes:
: [u]~ ∘ [v]~ = [uv]~
Because ~ is a congruence, the set of all congruence classes of ~ forms a semigroup with ∘, called the quotient semigroup or factor semigroup, and denoted S / ~. The mapping x ↦ [x]~ is a semigroup homomorphism, called the quotient map, canonical surjection or projection; if S is a monoid then quotient semigroup is a monoid with identity ~. Conversely, the kernel of any semigroup homomorphism is a semigroup congruence. These results are nothing more than a particularization of the first isomorphism theorem in universal algebra. Congruence classes and factor monoids are the objects of study in string rewriting systems.
A nuclear congruence on S is one that is the kernel of an endomorphism of S.
A semigroup S satisfies the maximal condition on congruences if any family of congruences on S, ordered by inclusion, has a maximal element. By Zorn's lemma, this is equivalent to saying that the ascending chain condition holds: there is no infinite strictly ascending chain of congruences on S.
Every ideal I of a semigroup induces a factor semigroup, the Rees factor semigroup, via the congruence ρ defined by x ρ y if either x = y, or both x and y are in I.
Quotients and divisions
[edit]
The following notions introduce the idea that a semigroup is contained in another one.
A semigroup T is a quotient of a semigroup S if there is a surjective semigroup morphism from S to T. For example, (Z/2Z, +) is a quotient of (Z/4Z, +), using the morphism consisting of taking the remainder modulo 2 of an integer.
A semigroup T divides a semigroup S, denoted T ≼ S if T is a quotient of a subsemigroup S. In particular, subsemigroups of S divides T, while it is not necessarily the case that there are a quotient of S.
Both of those relations are transitive.
Structure of semigroups
[edit]
For any subset A of S there is a smallest subsemigroup T of S that contains A, and we say that A generates T. A single element x of S generates the subsemigroup { xn | n ∈ Z+ }. If this is finite, then x is said to be of finite order, otherwise it is of infinite order.
A semigroup is said to be periodic if all of its elements are of finite order.
A semigroup generated by a single element is said to be monogenic (or cyclic). If a monogenic semigroup is infinite then it is isomorphic to the semigroup of positive integers with the operation of addition.
If it is finite and nonempty, then it must contain at least one idempotent.
It follows that every nonempty periodic semigroup has at least one idempotent.
A subsemigroup that is also a group is called a subgroup. There is a close relationship between the subgroups of a semigroup and its idempotents. Each subgroup contains exactly one idempotent, namely the identity element of the subgroup. For each idempotent e of the semigroup there is a unique maximal subgroup containing e. Each maximal subgroup arises in this way, so there is a one-to-one correspondence between idempotents and maximal subgroups. Here the term maximal subgroup differs from its standard use in group theory.
More can often be said when the order is finite. For example, every nonempty finite semigroup is periodic, and has a minimal ideal and at least one idempotent. The number of finite semigroups of a given size (greater than 1) is (obviously) larger than the number of groups of the same size. For example, of the sixteen possible "multiplication tables" for a set of two elements {a, b}, eight form semigroups[b] whereas only four of these are monoids and only two form groups. For more on the structure of finite semigroups, see Krohn–Rhodes theory.
Special classes of semigroups
[edit]
Main article: Special classes of semigroups
A monoid is a semigroup with an identity element.
A group is a monoid in which every element has an inverse element.
A subsemigroup is a subset of a semigroup that is closed under the semigroup operation.
A cancellative semigroup is one having the cancellation property: a · b = a · c implies b = c and similarly for b · a = c · a. Every group is a cancellative semigroup, and every finite cancellative semigroup is a group.
A band is a semigroup whose operation is idempotent.
A semilattice is a semigroup whose operation is idempotent and commutative.
0-simple semigroups.
Transformation semigroups: any finite semigroup S can be represented by transformations of a (state-) set Q of at most |S| + 1 states. Each element x of S then maps Q into itself x : Q → Q and sequence xy is defined by q(xy) = (qx)y for each q in Q. Sequencing clearly is an associative operation, here equivalent to function composition. This representation is basic for any automaton or finite-state machine (FSM).
The bicyclic semigroup is in fact a monoid, which can be described as the free semigroup on two generators p and q, under the relation pq = 1.
C0-semigroups.
Regular semigroups. Every element x has at least one inverse y that satisfies xyx = x and yxy = y; the elements x and y are sometimes called "mutually inverse".
Inverse semigroups are regular semigroups where every element has exactly one inverse. Alternatively, a regular semigroup is inverse if and only if any two idempotents commute.
Affine semigroup: a semigroup that is isomorphic to a finitely-generated subsemigroup of Zd. These semigroups have applications to commutative algebra.
Structure theorem for commutative semigroups
[edit]
There is a structure theorem for commutative semigroups in terms of semilattices. A semilattice (or more precisely a meet-semilattice) (L, ≤) is a partially ordered set where every pair of elements a, b ∈ L has a greatest lower bound, denoted a ∧ b. The operation ∧ makes L into a semigroup that satisfies the additional idempotence law a ∧ a = a.
Given a homomorphism f : S → L from an arbitrary semigroup to a semilattice, each inverse image Sa = f−1{a} is a (possibly empty) semigroup. Moreover, S becomes graded by L, in the sense that SaSb ⊆ Sa∧b.
If f is onto, the semilattice L is isomorphic to the quotient of S by the equivalence relation ~ such that x ~ y if and only if f(x) = f(y). This equivalence relation is a semigroup congruence, as defined above.
Whenever we take the quotient of a commutative semigroup by a congruence, we get another commutative semigroup. The structure theorem says that for any commutative semigroup S, there is a finest congruence ~ such that the quotient of S by this equivalence relation is a semilattice. Denoting this semilattice by L, we get a homomorphism f from S onto L. As mentioned, S becomes graded by this semilattice.
Furthermore, the components Sa are all Archimedean semigroups. An Archimedean semigroup is one where given any pair of elements x, y , there exists an element z and n > 0 such that xn = yz.
The Archimedean property follows immediately from the ordering in the semilattice L, since with this ordering we have f(x) ≤ f(y) if and only if xn = yz for some z and n > 0.
Group of fractions
[edit]
The group of fractions or group completion of a semigroup S is the group G = G(S) generated by the elements of S as generators and all equations xy = z that hold true in S as relations. There is an obvious semigroup homomorphism j : S → G(S) that sends each element of S to the corresponding generator. This has a universal property for morphisms from S to a group: given any group H and any semigroup homomorphism k : S → H, there exists a unique group homomorphism f : G → H with k = fj. We may think of G as the "most general" group that contains a homomorphic image of S.
An important question is to characterize those semigroups for which this map is an embedding. This need not always be the case: for example, take S to be the semigroup of subsets of some set X with set-theoretic intersection as the binary operation (this is an example of a semilattice). Since A.A = A holds for all elements of S, this must be true for all generators of G(S) as well, which is therefore the trivial group. It is clearly necessary for embeddability that S have the cancellation property. When S is commutative this condition is also sufficient and the Grothendieck group of the semigroup provides a construction of the group of fractions. The problem for non-commutative semigroups can be traced to the first substantial paper on semigroups. Anatoly Maltsev gave necessary and sufficient conditions for embeddability in 1937.
Semigroup methods in partial differential equations
[edit]
Further information: C0-semigroup
Semigroup theory can be used to study some problems in the field of partial differential equations. Roughly speaking, the semigroup approach is to regard a time-dependent partial differential equation as an ordinary differential equation on a function space. For example, consider the following initial/boundary value problem for the heat equation on the spatial interval (0, 1) ⊂ R and times t ≥ 0:
Let X = L2((0, 1) R) be the Lp space of square-integrable real-valued functions with domain the interval (0, 1) and let A be the second-derivative operator with domain
where is a Sobolev space. Then the above initial/boundary value problem can be interpreted as an initial value problem for an ordinary differential equation on the space X:
On an heuristic level, the solution to this problem "ought" to be
However, for a rigorous treatment, a meaning must be given to the exponential of tA. As a function of t, exp(tA) is a semigroup of operators from X to itself, taking the initial state u0 at time t = 0 to the state u(t) = exp(tA)u0 at time t. The operator A is said to be the infinitesimal generator of the semigroup.
History
[edit]
The study of semigroups trailed behind that of other algebraic structures with more complex axioms such as groups or rings. A number of sources attribute the first use of the term (in French) to J.-A. de Séguier in Élements de la Théorie des Groupes Abstraits (Elements of the Theory of Abstract Groups) in 1904. The term is used in English in 1908 in Harold Hinton's Theory of Groups of Finite Order.
Anton Sushkevich obtained the first non-trivial results about semigroups. His 1928 paper "Über die endlichen Gruppen ohne das Gesetz der eindeutigen Umkehrbarkeit" ("On finite groups without the rule of unique invertibility") determined the structure of finite simple semigroups and showed that the minimal ideal (or Green's relations J-class) of a finite semigroup is simple. From that point on, the foundations of semigroup theory were further laid by David Rees, James Alexander Green, Evgenii Sergeevich Lyapin [fr], Alfred H. Clifford and Gordon Preston. The latter two published a two-volume monograph on semigroup theory in 1961 and 1967 respectively. In 1970, a new periodical called Semigroup Forum (currently published by Springer Verlag) became one of the few mathematical journals devoted entirely to semigroup theory.
The representation theory of semigroups was developed in 1963 by Boris Schein using binary relations on a set A and composition of relations for the semigroup product. At an algebraic conference in 1972 Schein surveyed the literature on BA, the semigroup of relations on A. In 1997 Schein and Ralph McKenzie proved that every semigroup is isomorphic to a transitive semigroup of binary relations.
In recent years researchers in the field have become more specialized with dedicated monographs appearing on important classes of semigroups, like inverse semigroups, as well as monographs focusing on applications in algebraic automata theory, particularly for finite automata, and also in functional analysis.
Generalizations
[edit]
Group-like structures
| | Total | Associative | Identity | Divisible | Commutative |
| Partial magma | Unneeded | Unneeded | Unneeded | Unneeded | Unneeded |
| Semigroupoid | Unneeded | Required | Unneeded | Unneeded | Unneeded |
| Small category | Unneeded | Required | Required | Unneeded | Unneeded |
| Groupoid | Unneeded | Required | Required | Required | Unneeded |
| Magma | Required | Unneeded | Unneeded | Unneeded | Unneeded |
| Quasigroup | Required | Unneeded | Unneeded | Required | Unneeded |
| Unital magma | Required | Unneeded | Required | Unneeded | Unneeded |
| Loop | Required | Unneeded | Required | Required | Unneeded |
| Semigroup | Required | Required | Unneeded | Unneeded | Unneeded |
| Monoid | Required | Required | Required | Unneeded | Unneeded |
| Group | Required | Required | Required | Required | Unneeded |
| Abelian group | Required | Required | Required | Required | Required |
If the associativity axiom of a semigroup is dropped, the result is a magma, which is nothing more than a set M equipped with a binary operation that is closed M × M → M.
Generalizing in a different direction, an n-ary semigroup (also n-semigroup, polyadic semigroup or multiary semigroup) is a generalization of a semigroup to a set G with a n-ary operation instead of a binary operation. The associative law is generalized as follows: ternary associativity is (abc)de = a(bcd)e = ab(cde), i.e. the string abcde with any three adjacent elements bracketed. n-ary associativity is a string of length n + (n − 1) with any n adjacent elements bracketed. A 2-ary semigroup is just a semigroup. Further axioms lead to an n-ary group.
A third generalization is the semigroupoid, in which the requirement that the binary relation be total is lifted. As categories generalize monoids in the same way, a semigroupoid behaves much like a category but lacks identities.
Infinitary generalizations of commutative semigroups have sometimes been considered by various authors.[c]
See also
[edit]
Absorbing element
Biordered set
Compact semigroup
Empty semigroup
Generalized inverse
Identity element
Light's associativity test
Principal factor
Quantum dynamical semigroup
Semigroup ring
Weak inverse
Notes
[edit]
^ The closure axiom is implied by the definition of a binary operation on a set. Some authors thus omit it and specify three axioms for a group and only one axiom (associativity) for a semigroup.
^ Namely: the trivial semigroup in which (for all x and y) xy = a and its counterpart in which xy = b, the semigroups based on multiplication modulo 2 (choosing a or b as the identity element 1), the groups equivalent to addition modulo 2 (choosing a or b to be the identity element 0), and the semigroups in which the elements are either both left identities or both right identities.
^ See references in Udo Hebisch and Hanns Joachim Weinert, Semirings and Semifields, in particular, Section 10, Semirings with infinite sums, in M. Hazewinkel, Handbook of Algebra, Vol. 1, Elsevier, 1996. Notice that in this context the authors use the term semimodule in place of semigroup.
Citations
[edit]
^ Feller 1971
^ Jacobson 2009, p. 30, ex. 5
^ Jump up to: a b Lawson 1998, p. 20
^ Kilp, Mati; Knauer, U.; Mikhalev, Aleksandr V. (2000). Monoids, Acts, and Categories: With Applications to Wreath Products and Graphs : a Handbook for Students and Researchers. Walter de Gruyter. p. 25. ISBN 978-3-11-015248-7. Zbl 0945.20036.
^ Li͡apin, E. S. (1968). Semigroups. American Mathematical Soc. p. 96. ISBN 978-0-8218-8641-0.
^ Lothaire 2011, p. 463
^ Lothaire 2011, p. 465
^ Pin, Jean-Éric (November 30, 2016). Mathematical Foundations of Automata Theory (PDF). p. 19.
^ Clifford & Preston 2010, p. 3
^ Grillet 2001
^ Farb, B. (2006). Problems on mapping class groups and related topics. Amer. Math. Soc. p. 357. ISBN 978-0-8218-3838-9.
^ Auslander, M.; Buchsbaum, D. A. (1974). Groups, rings, modules. Harper & Row. p. 50. ISBN 978-0-06-040387-4.
^ Clifford & Preston 1961, p. 34
^ Suschkewitsch 1928
^ Preston, G. B. (1990). Personal reminiscences of the early history of semigroups. Archived from the original on 2009-01-09. Retrieved 2009-05-12.
^ Maltsev, A. (1937). "On the immersion of an algebraic ring into a field". Math. Annalen. 113: 686–691. doi:10.1007/BF01571659. S2CID 122295935.
^ "Earliest Known Uses of Some of the Words of Mathematics".
^ Jump up to: a b "An account of Suschkewitsch's paper by Christopher Hollings" (PDF). Archived from the original (PDF) on 2009-10-25.
^ B. M. Schein (1963) "Representations of semigroups by means of binary relations" (Russian), Matematicheskii Sbornik 60: 292–303 MR 0153760
^ B. M. Schein (1972) Miniconference on semigroup Theory, MR 0401970
^ B. M. Schein & R. McKenzie (1997) "Every semigroup is isomorphic to a transitive semigroup of binary relations", Transactions of the American Mathematical Society 349(1): 271–85 MR 1370647
^ Dudek, W.A. (2001). "On some old problems in n-ary groups". Quasigroups and Related Systems. 8: 15–36. Archived from the original on 2009-07-14.
References
[edit]
General references
[edit]
Howie, John M. (1995). Fundamentals of Semigroup Theory. Clarendon Press. ISBN 978-0-19-851194-6. Zbl 0835.20077.
Clifford, Alfred Hoblitzelle; Preston, Gordon Bamford (1961). The Algebraic Theory of Semigroups. Vol. 1. American Mathematical Society. ISBN 978-0-8218-0271-7. Zbl 0111.03403. {{cite book}}: ISBN / Date incompatibility (help)
Clifford, Alfred Hoblitzelle; Preston, Gordon Bamford (2010) . The algebraic theory of semigroups. Vol. 2. American Mathematical Society. ISBN 978-0-8218-0272-4.
Grillet, Pierre Antoine (1995). Semigroups: An Introduction to the Structure Theory. Marcel Dekker. ISBN 978-0-8247-9662-4. Zbl 0830.20079.
Grillet, Pierre Antoine (2001). Commutative Semigroups. Springer Verlag. ISBN 978-0-7923-7067-3. Zbl 1040.20048.
Hollings, Christopher (2009). "The Early Development of the Algebraic Theory of Semigroups". Archive for History of Exact Sciences. 63 (5): 497–536. doi:10.1007/s00407-009-0044-3. S2CID 123422715.
Hollings, Christopher (2014). Mathematics across the Iron Curtain: A History of the Algebraic Theory of Semigroups. American Mathematical Society. ISBN 978-1-4704-1493-1. Zbl 1317.20001.
Petrich, Mario (1973). Introduction to Semigroups. Charles E. Merrill. ISBN 978-0-675-09062-9. Zbl 0321.20037.
Specific references
[edit]
Feller, William (1971). An introduction to probability theory and its applications. Vol. II (2nd ed.). Wiley. MR 0270403.
Hille, Einar; Phillips, Ralph S. (1974). Functional analysis and semi-groups. American Mathematical Society. ISBN 978-0821874646. MR 0423094.
Suschkewitsch, Anton (1928). "Über die endlichen Gruppen ohne das Gesetz der eindeutigen Umkehrbarkeit". Mathematische Annalen. 99 (1): 30–50. doi:10.1007/BF01459084. hdl:10338.dmlcz/100078. ISSN 0025-5831. MR 1512437. S2CID 121081075.
Kantorovitz, Shmuel (2009). Topics in Operator Semigroups. Springer. ISBN 978-0-8176-4932-6. Zbl 1187.47003.
Jacobson, Nathan (2009). Basic algebra. Vol. 1 (2nd ed.). Dover. ISBN 978-0-486-47189-1.
Lawson, Mark V. (1998). Inverse semigroups: the theory of partial symmetries. World Scientific. ISBN 978-981-02-3316-7. Zbl 1079.20505.
Lothaire, M. (2011) . Algebraic combinatorics on words. Encyclopedia of Mathematics and Its Applications. Vol. 90. Cambridge University Press. ISBN 978-0-521-18071-9. Zbl 1221.68183.
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Volume 132, Issue 43 pp. 19026-19044
Kurzaufsatz
Open Access
Electrode Materials in Modern Organic Electrochemistry
Dr. David M. Heard,
Dr. David M. Heard
orcid.org/0000-0003-3005-8222
University of Bristol, School of Chemistry, Cantocks Close, Bristol, Avon, BS8 1TS UK
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Dr. Alastair J. J. Lennox,
Corresponding Author
Dr. Alastair J. J. Lennox
a.lennox@bristol.ac.uk
orcid.org/0000-0003-2019-7421
University of Bristol, School of Chemistry, Cantocks Close, Bristol, Avon, BS8 1TS UK
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Dr. David M. Heard,
Dr. David M. Heard
orcid.org/0000-0003-3005-8222
University of Bristol, School of Chemistry, Cantocks Close, Bristol, Avon, BS8 1TS UK
Search for more papers by this author
Dr. Alastair J. J. Lennox,
Corresponding Author
Dr. Alastair J. J. Lennox
a.lennox@bristol.ac.uk
orcid.org/0000-0003-2019-7421
University of Bristol, School of Chemistry, Cantocks Close, Bristol, Avon, BS8 1TS UK
Search for more papers by this author
First published: 06 July 2020
Citations: 64
Abstract
The choice of electrode material is critical for achieving optimal yields and selectivity in synthetic organic electrochemistry. The material imparts significant influence on the kinetics and thermodynamics of electron transfer, and frequently defines the success or failure of a transformation. Electrode processes are complex and so the choice of a material is often empirical and the underlying mechanisms and rationale for success are unknown. In this review, we aim to highlight recent instances of electrode choice where rationale is offered, which should aid future reaction development.
Graphical Abstract
The choice of electrode material is critical for achieving optimal yields and selectivity in synthetic organic electrochemistry. The material imparts significant influence on the kinetics and thermodynamics of electron transfer. However, electrode processes are complex and so their optimisation is often empirical. Highlighted herein are recent instances of electrode choice where rationale is offered.
1 Introduction
With high control of the rate, location and driving force of electron-transfer processes, electrochemistry is uniquely positioned to provide selectivity and sustainability benefits for the preparation of organic compounds. As these opportunities are being realised by academic and industrial research groups worldwide, the field of synthetic organic electrochemistry has received renewed interest over the last 5–10 years.1-5 New synthetic methodology and reactivity has been developed, including processes that are more inexpensive, safer and produce less waste than “classical” approaches.6-12 In addition, the relative ease with which the technique can be scaled is demonstrated by the fact that several industrial organic electrochemical processes have been developed.13-19
As the electron transfer between electrode and solution-phase electrolyte is heterogenous, the development of synthetic organic electrochemical reactions requires close attention to parameters that are not traditionally encountered by organic chemists. As well as optimising the applied current density or potential difference across a cell, electrochemical reactions can be performed in either batch or flow cells, and divided or undivided cells. However, it is the electrodes that constitute the most important difference, as the success or selectivity of a particular transformation is highly dependent on the material. Not only does the electrode material itself determine the mechanism of electron transfer, but the electrode separation distance, shape and size determine the submerged surface area, the field homogeneity and the resulting current density; all of which can affect the reaction outcome. While the electrode material is an additional parameter that requires optimisation, it can be exploited to control and change the selectivity of a reaction, and provides opportunities to vary reactivity through electrode-catalysis, (electrocatalysis), mediator-modified or chemically-modified electrocatalysis.
The ability of specific materials to give unique outcomes and determine the selectivity for synthetic electrochemical reactions has long been recognised.20, 21 Classical examples include the anodic oxidation of acetic acid in aqueous solutions (Figure 1 A), in which the identity of the products and their distribution varies with different anode materials,22-25 and the reduction of acrylonitrile in which the reaction products strongly depend on the cathode material (Figure 1 B).26-28 In this reaction, the formation of adiponitrile (1) with cadmium and steel electrodes is a mega-tonne per annum industrial process15 that is used in the production of nylon-6,6, thus exemplifying the importance of the control that the electrode material imparts, and the possible ramifications of its variation. The choice of electrode material can impart a more binary outcome by switching reactivity on or off. Classic examples of this include the cathodic hydrodimerization of formaldehyde to ethylene glycol, wherein product is only obtained with the use of mercury or carbon cathodes and no product is observed with lead or cadmium (Figure 1 C).29 In addition, memory of chirality (enantiomeric excess) was only observed with the use of graphite anodes in a decarboxylative etherification reaction (Figure 1 D).30 In a more recent example, Xu reports a drastic change in yield when exploring the electrode material in an aromatic C−H functionalisation reaction with electrochemically-generated amidinyl radicals (Figure 1 E).31 Varying the electrolyte or the applied current had a relatively minor effect on the yield of the reaction, but replacing the reticulated vitreous carbon (RVC) anode with Pt completely shut down reactivity, whereas graphite gave an intermediate outcome.
While the differences in the outcomes of these reactions with different electrode materials are stark, the high complexity of electrode processes commonly renders the generation of conclusive explanations very difficult. Indeed, it has been noted elsewhere that it is “impossible to select the optimum electrode for a given process on a theoretical basis. Instead, an empirical approach must be used”.21 While it is true that an empirical approach is currently the most efficient strategy to optimise a reaction, an appreciation of the influence of electrode materials and a greater understanding of electrode processes should lead to a more informed approach and new opportunities in the field. In addition to this, poor reproducibility is a major challenge that accompanies the use of electrochemistry in organic synthesis, and differences in the electrode material, grade and source all contribute to this problem. Thus, an appreciation of the important factors associated with the electrode should facilitate an improved rationalisation of the differences between reported and achieved yields or selectivity.
In this review, we initially summarise the most important practical and reactivity considerations for electrode materials in organic electrochemistry. Then, our goal is to highlight examples in which the performance of a particular electrode material is found to be unique and important. We place an emphasis on contributions to the literature from the last decade, while focussing on synthetic organic transformations and practical considerations on regular laboratory scales. It should be noted that although an explanation on electrode choice is given in an increasing number of cases, of the protocols that we have surveyed from the last decade, only a small percentage (ca. <5 %) provide some supporting insight. Reaction selectivity and yields can also very much depend on other reaction parameters, and so it is not always clear if the electrode material itself exclusively defines the observed difference. As this point adds to the ambiguity, the examples have been selected as carefully as possible, in preference to providing an exhaustive coverage. Thus, the reader is referred to a number of earlier review articles that are also relevant to these themes.21, 32-40 Beyond the scope of this review are photoelectrodes,41-47 and other practical aspects of electrochemistry, which have been addressed in recent tutorial reviews.48-54
2 Electrode Selection
2.1 Practical Aspects
The primary judgement of candidate materials will be based on their performance in the reaction, i.e., yields and selectivity, but current efficiencies, obvious signs of corrosion, cost, availability and machinability are other critical factors, the relative importance of which will vary according to the specific process. In other applications of electrochemistry-such as those focussed on energy or bulk scale commodity production-small, single digit differences of efficiency gains can be extremely critical, for example from the use of a precise grade of graphite. However, in organic synthesis where the scales are comparatively smaller, larger gains in yield or complete switches in selectivity become more important. This is because the cost of the electrode material and the man-hours that are required to optimise a process must be balanced against the costs of the reagents and the value of the product. For example, as the price of electricity is typically low compared to the contents of a reaction mixture, achieving small gains in current efficiency is not the highest priority for reaction optimisation. This will only become a consideration if the scale is increased and the value of product is lowered.
Whilst electrodes can in principal be made from any conductive material, in order to make an appropriate choice there are a number of mechanical and electrochemical properties to consider. An idealised electrode material should be inexpensive, non-toxic, stable to a wide range of temperatures, pressures and solvents, yet able to be manipulated into forms for electrode construction, such as rods, wires, plates, foams and meshes. Most electrodes consist of a single material, but a support material combined with an electroactive coating, such as Pt, can also be used.20 In organic solvents, which are more resistive than aqueous systems, the use of 3-dimensional, high surface area electrodes is advantageous, as they impart decreased current density and cell potential. Thus, the use of RVC or carbon felt can increase productivity as higher currents can be applied.37, 55 Between electrode materials, surface area can vary dramatically, for example, the surface area of a “smooth electrode” can be up to 3 orders of magnitude lower than a porous surface, such as platinised platinum.56
An electrode should be stable and resist corrosion. An exception to this is when the electrode is sacrificial, for example when metal ionisation is intended as a counter electrode process, or when the metal ions are used to stabilise a product, such as in a carboxylation reaction.57 Degradation of electrodes by mechanical action can occur as a consequence of convection forces in the reaction vessel, such as the release of graphite particulates, which requires separation via filtration. In addition, fragile materials, such as low pore density RVC, can lead to difficulties in physical handling and manipulation. Swelling of the electrode can also be problematic with certain electrode material/electrolyte combinations.
The use of electrodes with high resistivity leads to an ohmic (IR) drop, which creates the requirement for a higher cell potential. This excess energy input is likely lost as heat into the reaction medium, which is inefficient and may be deleterious to the reaction outcome.58 On an industrial scale, this can limit the choice of materials to only those that are highly conductive, or require special electrode architectures.21 Once a material is formed into an electrode, a low ohmic resistance connection should also be made.
2.2 Reactivity Aspects
The mechanism for electron transfer at an electrode occurs between two limiting scenarios. In the first limiting case (Figure 2 A), the electrode surface is intimately involved in the mechanism of electron transfer and acts as a catalyst in the reaction; i.e., electrocatalysis.52 The products, mechanism and kinetics of electrode reaction in this case are highly dependent on the composition of the electrode material, meaning that small differences may be extremely significant in determining the outcome of the reaction. Conversely, in the second limiting case (Figure 2 B), the electrode is completely inert and provides a source or sink of electrons that are transferred in an outer-sphere manner between the substrate and electrode. The identity of the products formed, and the mechanism and kinetics of their formation should be independent of the material.
The potential required beyond that necessitated by thermodynamics to drive a reaction at a practical rate is referred to as the overpotential (η).59 The observed overpotential in a particular system is a sum of the individual overpotentials for each step in the process, such as adsorption, charge-transfer, desorption and mass-transport (diffusion, convection and migration) overpotentials. As the electrode material dictates the mechanism of electron transfer, it is the biggest contributing factor to the overall overpotential of a process. This important factor will be responsible for outcome variations observed during reaction optimisation.
For many reactions, such as the Hydrogen Evolution Reaction (HER) or Oxygen Evolution Reaction (OER), the decrease in overpotential through new electrode materials is the subject of intense investigation.60-69 Small efficiency gains will translate into large cost savings when these processes are conducted on scale. However, of greater importance to synthetic organic electrochemistry is the selectivity changes or suppression of side reactions that are enabled by the different overpotentials for each process on different electrode materials. An example of this control in a substrate-reduction reaction is to suppress competing proton reduction (HER) by choosing a cathode material that has a high overpotential for this process. Indeed, the overpotentials on common electrode materials varies considerably for the HER and OER, Table 1 and Figure 3. A low overpotential for the desired redox reaction will not only ensure the reaction can be driven more efficiently but will improve selectivity over competing processes. The overpotential for solvent oxidation or reduction can also vary significantly on different electrode materials.70 This variation has implications for the width of the potential solvent window that is available to a reaction and therefore to the extent of redox chemistry that can be performed, Figure 4.
Table 1.
Overpotentials (HER and OER) and conductivities for various electrode materials.
| |
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Overpotential is also influenced by supporting electrolyte,93 solvent,99, 111-113 current density,95, 96 concentration,114 temperature,115 and any additives,21, 116-119, which will all affect the direct application of these figures. a Hydrogen evolution (HER) overpotential recorded at 1 mA cm−2, 25 °C, 1 m HCl (or H2SO4) in the solvent specified, unless otherwise indicated. HER data for various alloys (not listed) also available.120 b recorded at 2×10−4 A cm−2. c Oxygen evolution (OER) overpotential recorded at 1 mA cm−2, 25 °C, 1M KOH in water, unless otherwise indicated. d Recorded at 273 K (except Hg, glassy carbon, BDD that were recorded at 298 K), data taken from ref. 121, 122 except where specifically given. e 1 m KOH. f 0.5 m H2SO4. g Highly dependent on doping and treatment. h pH 3.4.
The stability of an electrode is important for ensuring longevity of use. However, the stability of the substrate or the intermediates produced on the electrode is also important for ensuring high yields of product. A compound can irreversibly bind and decompose on the surface, leading to a decreased mass balance and yield of product (Figure 5 A). Strategies for grafting organic compounds onto electrode surfaces for intentional surface modification have also been reported.71-74 However, unintentional grafting can vary in degree depending on the specific redox event, electrolyte and electrode material. The result is a passivated electrode with decreased electrode activity due to the formation of an electrically insulating layer. Electrode passivation can be detected by cycling a cyclic voltammetry (CV) experiment and observing the current decrease with each cycle,75-77 with the current not being restored to its original value until the electrode is cleaned. Examples of electrode passivation are tight oxide films on metals that are formed at high anodic potentials,78, 79 insoluble oxidation products, polymer deposits generated by anodic oxidation of olefinic, aromatic or phenolic compounds,80-82 or solutions of HF or ionic liquids.83, 84 Optimisation of the electrode material is a key task in remedying this effect (Figure 5 B). Other methods that have been shown to be effective include pulse electrolysis,85 sonication,86 alternating the polarity of the electrodes (which can also effect the reaction selectivity or yield),87-89 and the use of mediators to shuttle redox equivalents from the electrode to the substrate in the bulk phase.75, 90 Alternatively, the addition of additives can increase the solubility of the insulating polymer in the electrolyte or protect the electrode surface, which has been shown to be highly effective in a recent electrochemical Birch reduction.91
2.3 Trends
The factors that contribute to the choice of electrode material vary and can be very specific. The number of electrode materials available has increased over time and trends of use have changed and evolved. For example, lead and mercury were previously preferred due to their high hydrogen overpotential (ηH) and stability to acidic media. With mercury being in the liquid state, the surface is constantly renewed and can remain clean and free of impurities. However, concern over the high toxicity of these metals has limited more recent wide-spread use and hence other materials have attracted greater attention. Modern organic electrochemical methodology relies more heavily on platinum, which is robust, easy to clean and redox stable, as well as carbon-based electrodes that are more inexpensive and thus appropriate when the scale of a reaction renders the cost of platinum prohibitive (Figure 6).130, 131 Glassy carbon is the most commonly used carbon material, which is the fullerene allotrope of carbon,132 and includes the high-surface area foam form, RVC.34 Graphite is also a commonly used form of carbon electrode, which is less chemically inert than glassy carbon but more conductive133 and is less expensive. The diamond allotrope of carbon can also be used, Boron Doped Diamond (BDD) has emerged as a unique material and is becoming increasingly popular.134-137 There has also been evidence for the emergence of new materials, metals or alloys used as electrodes in organic synthesis, such as leaded bronze, tantalum, niobium or molybdenum.138-141 No doubt this trend will continue as the electrode processes with each material become better understood, wider range of materials are adopted, and the further development of idealised materials.
3 Electrocatalysis: Specific Adsorption and Surface Interactions
At the extremity of the first limiting case (Figure 2), the electrode surface is explicitly involved in the reaction mechanism through specific adsorption and surface interactions. As well as providing the required redox equivalents, the electrode surface serves to catalyse the reaction, and is thus known as electrocatalysis. Savéant defined “Electrocatalysis” as the term to “name catalysis of electrochemical reactions by surface states of the electrode”,52 which is distinct from mediated electrolysis that employs molecularly defined catalysts. The precise nature of these interactions varies, depends on each specific reaction and can often be the subject of much debate. Nevertheless, theoretical models are improving and can now describe catalytic reactions in great detail.142 The strength of interaction (adsorption vs. desorption), the timing and order of electron transfers and the concertedness of steps are all relevant when considering the mechanism. Adsorption describes a variety of more specific interactions of a substrate with the electrode, such as strong electrostatic interactions, π-interactions and chemical bonds. As well as the smooth material itself, the sites of binding and catalysis may be impurities, edge or end atoms, deposited nanoparticles, thin films or single atoms of a secondary or different material to the bulk material.143-145
The strength of interaction between substrate and electrode should be strong enough to trigger a reaction, but not too strong that the product fails to dissociate and desorb. This balance is known as the Sabatier principle146 and has been shown to contribute to the bell curves observed for rates of electrocatalytic HER.147, 148 In organic electrochemistry, it is common for products to avoid dissociation from the electrode (Figure 5 A), which can lead to decomposition and a low mass balance at the end of reaction.
3.1 Working Electrode Material
In a classical example, the extent of electrocatalysis in the oxidative decarboxylative Kolbe and Hofer–Moest reactions has been the subject of much debate in the literature over the years.149-152 In one study, the product distributions from the use of a platinum and carbon anodes were compared. It was found that the ratio of 1-electron vs. 2-electron oxidation products (i.e., ratio of products-from-radicals over products-from-cations) was much greater with platinum anodes than with graphite anodes (Figure 7).153 Carbon anodes are more efficient than platinum anodes at removing a second electron, to form a cation (with a proton loss).145 The difference was proposed to be due to a greater tendency of radicals to adsorb onto carbon because of the presence of paramagnetic centres in the material. Thus, the adsorbed radicals on carbon undergo further oxidation to form a carbocation that is then electrostatically repelled and primed to react with nucleophiles. However, the radicals produced on a platinum surface are largely desorbed and so participate in radical reactions. This effect has also been recorded in other transformations, such as in the electrochemical cyclisation of dienes, in which Moeller observes a difference in the efficiency of 1- vs. 2-electron pathways when using platinum and carbon anodes.154
More recently, electrocatalysis has been especially noted for cathodic processes; it has even been remarked that “it seems uncertain that totally inert electrodes exist […] within the cathodic range”.155 In particular, the dehalogenation of aryl and alkyl-halides with different cathodic materials has been the subject of significant investigation.21, 155-157 The over-potential, concertedness and degree of interaction varies with different electrode materials and can lead to the formation of different product distributions. For example, the use of silver cathodes significantly decreases the over-potential necessary to cleave a C−X bond,157 compared to mercury or glassy carbon cathodes. In the reduction of linear alkyliodides on smooth silver cathodes, there is evidence for the transient formation of [Ag+−R] I− species on the interface.155 The formation of such species will stabilise the radical and ensure a lower reduction overpotential. Compared with glassy carbon electrodes, copper has also been found to show exceptional electrocatalytic properties, either as a smooth metal or when deposited onto a conducting surface.158
The electrocatalytic dehalogenation of aryl-halides can occur via a stepwise electron transfer-cleavage mechanism, or a concerted process. Recent analysis of an extensive range of cathode materials revealed a strong dependency of the mechanism of debromination on the electrode material (Figure 8).159 Electron transfer coefficients (α) give an indication of how reactant or product-like the transition state is in terms of its electrical behaviour. These were extracted from CVs by analysis of the difference between the peak potential and half-wave potential, and used as an indication for the mechanism. A value of α significantly lower than 0.5 indicates a concerted mechanism, whereas a stepwise mechanism will either have an α significantly higher than 0.5 if cleavage is the RDS or only slightly lower than 0.5 if ET is the RDS. Thus, reduction potentials and electron transfer coefficients were measured by cyclic voltammetry for the reduction of different aromatic bromides on different electrode materials. Only 4 of the 11 materials showed reduction features in the CV. Ag and glassy carbon were found to both follow a concerted mechanism: Ag exhibited a remarkable electrocatalytic activity with a 0.9 V lower overpotential than glassy carbon. This overpotential difference is a considerable thermodynamic improvement and demonstrates the significant effect that materials can have on the overpotential of an electrochemical redox event. Cu and Zn electrodes were found to give step-wise mechanisms, with rate determining ET and cleavage steps, respectively.
The cathode material can determine the distribution of products from a reaction. The reduction products of alkylhalides have been reported by Peters to vary according to the cathode material.160, 161 On vitreous carbon cathodes, n-decane and 1-decene are yielded from the reduction of iododecane, whilst on a silver cathode, a dimeric product was formed. When testing secondary alkyl halides, it was interesting to note that the product distribution switched, such that dimers were predominately formed at carbon-based cathodes.
Avoiding the adsorption of reagents and the subsequent electrocatalysis of competing side-reactions can be critical to the success of a desired reaction. This can be achieved using an electrode material with a high overpotential for the competing processes. For example, in a reduction reaction, competing proton reduction can be avoided through the use of a cathode with a high overpotential for that process, such as glassy carbon, mercury or lead. Lead has found use for this reason in the deoxygenation of amides and sulfoxides (Figure 9 A).162 Amides are thermodynamically difficult to reduce and the presence of acid is necessary to provide the equivalents of protons. Therefore, it was important to use a material that preferentially reduces amides over protons, and lead was found to be superior for that.
In the reduction of menthone oximes to menthylamines (Figure 9 B), Waldvogel screened cathode materials to avoid any competing proton reduction.119, 163 Those with only a moderate hydrogen overpotential, such as titanium, iron, copper, zinc, indium, tin and bismuth, all failed to produce the desired product, and so it was necessary to use a high overpotential material, such as lead or mercury. The electrode material also influenced the selectivity of the reaction: whereas, mercury and cadmium cathodes led to pronounced diastereoselectivity, lead or copper/lead gave either no or little selectivity. The authors proposed that good diastereoselectivity was due to stabilisation of the reactive intermediate by stronger binding to the electrode surface and slowing down conformational switching.
The efficiency of the electrochemical reductive carboxylation of imines to yield N-bromoamino acids also depends on the cathode material (Figure 10 A).164 In this case, the yield was proposed to correlate with the strength of substrate adsorption onto the electrode surface. Silver was noted to exhibit pronounced specific adsorption of imines, which leads to a higher concentration of imines on the electrode surface, and therefore more facile decomposition and accelerated imine dimerisation. Highest yields were reported using nickel cathodes. When adapting the reaction into flow, Atobe considered cathode materials with a high overpotential for carbon dioxide reduction (Figure 10 B).165 In this example, the overpotentials (determined by linear sweep voltammetry) correlated strongly with yield. Glassy carbon gave the highest efficiency, followed by graphite, then platinum and lastly silver. Reduction of the imine to a radical anion is necessary for a productive reaction to take place. Any competing direct CO2 reduction decreases charge efficiency and reaction yields.
The electrochemical hydrodechlorination of 3,5,6-trichloropicolinic acid (3,5,6-T) is an example of a reaction in which the cathode material proved absolutely crucial to reaction success.166 Using copper, glassy carbon or nickel cathodes, the reaction was completely unsuccessful (Figure 11 A). Only the use of silver produced any desired 3,5-dichloropicolinic acid (3) product, a compound with significant pharmaceutical relevance. It was proposed that the overpotential for proton reduction was lower with the use of Cu and Ni, making substrate reduction more difficult. The productivity of silver cathodes compared to glassy carbon (GC) cathodes was ascribed to an electrocatalytic effect of silver that was not possible on carbon (Figure 11 B). Interestingly, the selectivity of hydrodechlorination was found to be dependent on the pH as well. Electrostatic forces engendered high selectivity for the 3,5-substituted isomer (3) at pH 3, and the 3,6-substituted isomer (4) at pH 13 (97 %) (Figure 11 B).
3.2 Counter Electrode Material: Electrogenerated Base
As well as tuning the overpotential on the working electrode, the overpotential on the counter electrode reaction is also important to consider and has often been shown to be key to the success of a reaction. Of particularly frequent consequence is the reduction of protons to evolve hydrogen (HER) at the cathode to form a base. The concept of forming electrogenerated bases (EGBs, Figure 12 A)167, 168 in situ from a pro-base for utilisation in a synthetic transformation was first reported in 1967.169 Using electrochemistry allows the concentration and basicity of the reaction to be carefully controlled,170 and changing the counterion influences both the stability and reactivity of the EGB.171
To generate bases electrochemically (EGB), the reduction potential of the pro-base must be less negative than any other species in the reaction (including the product), which renders the overpotential for proton reduction vital for success to be achieved. By promoting hydrogen evolution over other potential reductive processes, the choice of cathode material influences the outcome of anodic transformations.138, 172, 173 An example of this is in the synthesis of (E)-vinylsulfones from cinnamic acids, in which Wang found a significant dependence of the reaction on the counter electrode material (Figure 12 B).174 Whilst the reaction is oxidative with respect to the substrates, the cathodic generation of base is required for the deprotonation of the carboxylic acid. Platinum was the best performing cathode material, which has the lowest overpotential for proton reduction, whilst glassy carbon, which has a higher overpotential, resulted in a decreased yield. Materials with a medium ηH performed in-between these two cases. Thus, a lower potential difference is necessary with Pt under constant current electrolysis conditions, which limits competing reduction processes. Similarly, Zhang switched from carbon to platinum counter electrodes in an electrochemical Hofmann rearrangement, to more readily form methoxide on the cathode and found that yields improved (Figure 12 C).175
The electrochemical oxidative formation of N-centred radicals and their intramolecular cyclisation onto alkenes has been developed by Moeller and Xu.176-182 Electrode materials were thoroughly analysed by Wirth when the process was transferred to a flow cell set up.183 Interestingly, it was found that the choice of counter electrode material had a stronger influence on yield than the choice of the working electrode material. A yield of over 90 % was achieved with a platinum cathode, around 60 % with nickel, and no reaction occurred with a graphite cathode (Figure 12 D). These results directly correlate with the over-potentials for proton reduction (Pt=−0.09 V; Ni=−0.32 V; graphite=−0.47 V, Table 1). Yields were consistent (ca. <5 %) between the anodic materials tested with a Pt cathode. Correct choice of cathode material therefore meant the anode material could be chosen by cost, rather than due to its influence on the reaction.
In an elegant example of the importance of counter electrode material, Xu recently reported a complete selectivity switch when the cathode material was switched from platinum to lead.184 The reaction is an oxidative TEMPO catalyzed intramolecular arene amination from oxime 5 (Figure 13). When platinum was used as the counter electrode, the low overpotential for proton reduction resulted in the N-oxide product 6 remaining intact. However, when lead was employed, the deoxygenated heterocycle 7 was returned. This is because lead has a much higher hydrogen overpotential for proton reduction, meaning proton reduction is more difficult and so N-oxide 6 now preferentially reduced on the counter electrode.
3.3 Modified Electrode Surfaces
The modification of electrode surfaces to aid catalysis of a reaction and decrease the overpotential for electron transfer is a technique that is well established, especially for energy-related applications, such as the HER, OER and CO2 reduction.60-68 For synthetic applications, the immobilisation or tethering of electron transfer mediators onto electrode surfaces, either covalently185 or non-covalently186 has been shown to improve the efficiency of reactions. Carbon electrodes are especially effective supports for catalysts as they can be readily functionalised.74 For example, oxidation produces a high density of surface carboxyl groups with which amide bonds can be formed, or the single electron reduction of diazonium cations reveals arene radicals that readily combine with graphitic electrodes.187, 188 Further details of the mediator-immobilisation approach are, however, beyond the scope of this review, and the interested reader is directed to other relevant reviews.74, 189
The bulk surface modification of electrodes through, for example, polymer coating or nanoparticle deposition is comparatively less well exploited for organic synthesis compared to energy applications.190 A recent example demonstrated that the in situ generation of an active MoV layer on the surface of a Mo anode was responsible for greatly enhanced yields in the dehydrogenative coupling of arenes (Figure 14).138 Whilst this arene coupling reaction could be performed with BDD, Pt, Au, V, Cr or W electrodes, the efficiency was less optimal than with the use of Mo. Only very low levels of molybdenum were detected by mass spectrometry in the electrolyte solution, which is evidence that the active MoV species is only present on the surface and not released into solution.
4 Double Layer Control
Upon application of a potential to an electrode in solution, an ordering process occurs to form a structure of oppositely charged ions and solvent molecules at the surface, commonly known as the Helmholtz double layer. There have been several theoretical models proposed for this interfacial region (Helmholtz, Gouy–Chapman, Stern) but the precise behaviour depends on the nature of electrode (material and surface properties), as well as the composition of the electrolyte (supporting electrolyte, solvent).21 Unlike under aqueous conditions, the structure and thickness of the double layer in organic solvents is not well-understood. Nevertheless, the structure determines the potential distribution close to the surface and the uniformity of current. The double layer thus influences the local driving force for electron transfer, which determines the kinetics of electron transfer.
Waldvogel manipulated the interfacial region in the reduction of menthone oximes through the addition of quaternary ammonium salts (Figure 9 B).119 These small, hard cationic species form a compact and robust layer on the cathode surface (Figure 15 A). It was found that di- or poly-ammonium salts separated by an alkyl chain also gave superior reaction outcomes, possibly due to an entropic effect. The hard, lipophilic layer was able to exclude both solvent and protons from the surface, and decrease side-reactions. The ammonium salts serve to attenuate corrosion of the lead cathode, suppressing the formation of lead sulfate deposits to keep a shiny intact surface. The adsorption of ammonium cations also serves to increase the hydrogen overpotential of the cathode by reducing the rate of the HER. This effect was further studied in an amide deoxygenation reaction.118 It was proposed that the cationic layer still allows the tunnelling of electrons to reduce substrate, but protons are repelled due to coulombic repulsion. By avoiding competing proton reduction, the double layer protects the electrode from corrosion, and leads to an improved performance. Another example of the use of quaternary ammonium salts to suppress hydrogen evolution was demonstrated by Bhanage in the reduction of N-alkoxyamides to esters.191
Reactive species, such as cations or radicals, can readily react with solvent molecules to form undesired products. However, when formed at an electrode within the double layer, solvent can be excluded, which can aid reactivity and enhance selectivity. An elegant example of this is from Xu, who reported an alkynyl-hydroxylation reaction of alkenes, which is highly regio- and chemo-selective and proposed to only be successful because the selectivity-determining alkyne addition step occurs within the polarised double layer (Figure 15 B).192 The negatively charged alkynyltrifluoroborate nucleophile is attracted to the region and creates a localised high concentration, while competing neutral nucleophiles, such as water, are excluded. Oxidation at an electrode was found to be essential for these reasons, as photochemical or chemical oxidation conditions facilitated direct reaction with water in preference to alkyne.
Moeller also demonstrated that an ordered double layer can improve selectivity in an intramolecular cyclisation reaction by promoting cyclisation over solvent trapping (Figure 15 C).193-195 The anodic oxidation and the ensuing cyclisation occur within the ordered environment of the double layer at the anode surface, which slows diffusion and excludes the methanol from the electrode surface that could otherwise interfere with cyclisation. Silyl protection of the internal alcohol moiety of the substrate was still necessary to prevent the intramolecular trapping by this competing nucleophile.
5 Inert Electrodes
At the other end of the scale to the significant involvement of the electrode and high levels of electrocatalysis is electron transfer from an inert electrode that does not participate in the mechanism and has little substrate or intermediate adsorption. An outer-sphere-type electron transfer mechanism is more dominant, which results in high overpotentials for specific reactions. The best-known inert material is boron doped diamond (BDD), the use of which in organic electrosynthesis has primarily been driven by Waldvogel.36, 196-200 Although, the level of interaction of an electrode in a reaction is very difficult to determine, BDD has the highest known overpotential for the oxygen and hydrogen evolution reactions, which indicates very low levels of interaction. Because of this, BDD also offers a very high potential window and is highly chemically inert. However, it has been shown that the level of boron doping can actually affect selectivity,201 and sp2 non-diamond carbon impurities alters the potential window,144 implying that the material is not perfectly inert. BDD has been reported to yield differences in selectivity to other materials in various reactions, such as CO2 reduction.202-204 However, herein, we describe several synthetic organic examples that have required the use of a more inert electrode, which BDD has fulfilled. More specific features of BDD and its general use has been well reviewed elsewhere.137, 205-207
The electrochemical C−H amination of arenes via Zincke intermediates (8) was reported by Yoshida using a carbon felt anode and platinum plate cathode, Figure 16.208 However, the scope was limited to electron rich rings containing methoxy groups. In an effort to widen the scope toward arenes with less electron density, Waldvogel explored the use of different anode materials in the reaction.200 While retaining the Pt counter cathode in a divided cell set up, the use of carbon felt or fleece anodes were confirmed to give moderate or poor results for the amination of alkylated arenes. These porous carbon materials have high surface area, which causes diffusion of the radical cation away from the electrode to be more difficult, as it is liable to adsorb and oxidise further. Platinum, glassy carbon and isostatic graphite anode materials all returned very poor yields, with electrode fouling observed with the former two and corrosion with the latter. However, switching to a BDD anode resulted in a significant boost in yield, up to 60 %. CV studies were conducted to gain greater insight into the enhanced performance of BDD compared to platinum and glassy carbon. CV traces of xylene (red) and xylene with pyridine (blue) were recorded and compared (Figure 16). With the use of Pt and glassy carbon anodes, the oxidation feature of xylene disappeared upon addition of pyridine. However, with a BBD anode the CV trace was unaffected, and the oxidation of xylene was retained. This trend correlates with the outcome of the amination reaction at each anode material.
An electrochemical dimethoxylation was a key step in Nishiyama and Einaga's synthesis of (±)-parasitenone.209, 210 The use of BDD and Pt for the oxidation of 9 gave excellent yields of the desired product 10. However, glassy carbon or the use of chemical oxidants returned a different, aldehyde product 11 (Figure 17 A). Anodic oxidation of 9 leads to the radical cation 12, from which 11 is formed from methoxide deprotonating the benzylic position (via route c). Although methoxide attack of the ring leads to the product 10 (via route a), the anode material dependency on the reaction selectivity suggests an alternative electrode material-dependent mechanism. ESR studies revealed the formation of methoxyl radicals, most efficiency with a BDD anode, to a lesser extent with Pt, but not at all with GC (Figure 17 B). The authors proposed that these data signal that methoxyl radical attack onto the radical cation is the leading pathway to product 10 (via route b) and were used to explain the selectivity differences observed with each anode material. As methoxyl radicals are highly reactive, an inert electrode proved to be essential for this transformation.
The use of a BDD anode led to the highest yields in the challenging oxidation of isoeugenol (13) to (±)-Licarin A (14) (Figure 17 C).210 This was similarly proposed to be due to the more efficient formation of methoxyl radicals on BDD. Lower yields of the desired product 14 and overall mass balance were observed with Pt and lower still with glassy carbon. The other side-products that also required the formation of highly reactive radicals were also formed in greater quantities with BDD. Interestingly, the oxidation of isoeugenol on BDD in hexafluoro isopropanol (HFIP) give the homo-coupled product, diisoeugenol.211
Waldvogel tested the influence of anode materials in the cross-coupling of phenols and arenes. Preliminary studies revealed that the use of carbon electrodes gave only homo-coupled adducts.196 Platinum plates improved the yield and selectivity of cross to homo-coupled ratio to 1:1, but a switch to BDD gave a further enhancement in the selectivity (1.5:1). Further optimisation led to a set of improved conditions that contained methanol or water as an additive (Figure 18).212 When the electrode material was varied again, BDD, Pt and GC all gave excellent selectivity, but BDD proved to be the superior anode material with respect to yield. These observations led the authors to propose the formation of alkoxyl radicals, stabilised by the HFIP environment and which mediated the generation of the phenoxyl radical intermediate. Further studies revealed the alcohol additive beneficially altered the oxidation potentials of the substrates.213 Nonetheless, the use of BDD as an inert electrode had a positive effect on the outcome of the reaction.
6 Sacrificial Electrode
The use of a metal anode with a very low oxidation potential will leech metal ions into the solution upon its oxidation. In this case, it is termed a sacrificial electrode, as it is being consumed stoichiometrically. This approach is practical, facile and so frequently applied as a counter electrode process in electrochemical reduction reactions. Common choices for a sacrificial anode include zinc, magnesium or aluminium. In many cases, the choice is dictated by price or toxicity concerns, with little effect on the reaction observed. However, commonly, the liberated ions play a role in the reaction by coordinating reactants or products, and maintaining high conductivity. Care should be taken to avoid reduction of the liberated ions on the cathode to avoid short circuiting the system, hence the use of highly reducing metals that thermodynamically disfavour this process.
The reduction of organohalides is a reaction in which a sacrificial anode is frequently used,214, 215 in particular for carboxylation reactions by coupling with CO2 as an electrophile.57, 139, 216-220 The metal ions liberated from the anode stabilise the carboxylate product, which also helps to prevent anodic Kolbe-type reactions of the carboxylate.57, 165, 220 A more recent example is the use of either Mg or Al anodes by Baran in an electrochemical Birch reduction (Figure 19 B). The reaction was shown to be highly scalable and remarkably tolerant to a very wide range of substrates.91 In an earlier study of the same reaction (Figure 19 A),86 Mg anodes were also found to give the highest yields, which was proposed to be due to Mg2+ ions acting as electron transfer catalysts or stabilising anionic intermediates and promoting their reduction. Mechanistic studies were undertaken by Baran to elucidate if liberated Mg2+ ions played a beneficial role in the reaction. The addition of MgBr2 as an additive only served to decrease the product yield, although it seems likely that its reduction may compete with substrate reduction. In the absence of stirring, diminished yields could also be correlated to smaller electrode separation distances. These data suggest the diffusion of metal ions to the anode is deleterious to the reaction and so the liberated ions are not mechanistically relevant.
Rather than a sacrificial anode producing only waste products, it can be used to liberate reagents into solution in a controlled manner that matches reaction progress, often yielding results that are not possible by other means. In an elegant recent example of this, Sevov reported the use of an aluminum anode in the deoxygenation of phosphine oxides (Figure 20).221 The sacrificial electrode oxidises to liberate aluminum ions into solution that combine with chloride ions to form an amine-stabilised AlCl3 complex. This in situ generated Lewis acid activates the phosphine oxide, producing a less negative potential for its reduction and subsequent deoxygenation. As 2 electrons are required for the phosphine oxide reduction, and 3 are removed from Al0 to give Al3+, an additional quantity of added AlCl3 is required to balance the stoichiometry and ensure high reaction efficiency, without which, a lower performance was observed.
Metal anodes have been employed as sacrificial electrodes by Willans to provide metal ions into solution, at a controlled rate and with control of the oxidation state, in order to generate organometallic complexes that are not otherwise obtainable (Figure 21 A). Examples of this approach include the use of copper, iron and manganese anodes in the presence of NHC and salen ligands, to form the corresponding complexes.222-224 The reaction is a paired process (Figure 21 B), that is, the ligand precursor is reduced on the cathode and the metal ions are produced on the anode. The procedures are remarkably straightforward and yield high purity complexes without the necessity for column chromatography. Similarly, Mellah demonstrated SmII complexes could be efficiently prepared through the use of a sacrificial samarium anode, and are important catalysts.225-227
7 Summary and Outlook
While a rational choice of electrode material for use in organic electrochemical transformations cannot yet be made readily and reliably, herein, we have highlighted where efforts have moved beyond screening and empirical investigations. In many cases, efforts to understand the influence of electrode material through analytical electrochemistry and a physical organic chemistry approach have led to the elucidation of trends. Such trends and insight may be applied more broadly, which will lead to enhanced efficiencies and new opportunities. Due to the complexities and variation of electrode-substrate interactions in organic transformations, it is likely that experimentation will remain necessary, even when the choice is guided by principles. To aid exploration of the breadth of materials available, two tables are appended summarising key materials, their properties, and applications in electrosynthesis (Tables 2 and 3).
Table 2.
A summary of the key properties, forms, sources and example uses of common electrode materials used in organic electrosynthesis. Pricing information obtained January 2020.
| |
| |
Table 3.
A summary of the key properties and example uses of electrode materials for organic electrosynthesis.
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| |
The criteria for an ideal electrode material is that it is inexpensive, non-toxic, stable, manipulatable, resist corrosion and, most importantly, provide high yields and exquisite selectivity. While a number of materials perform extremely well and fit many of these criteria, it is clear that there is currently no material that meets all of them. These criteria are also reaction-specific, as cost, selectivity and yield have to be balanced against the cost of product and ease of access to it by other means. The development of new materials and the design of robust electrocatalysts for organic synthesis also still lags many other applications of electrochemistry, yet may provide new opportunities for the field. While the electrode material remains a key optimisation parameter, it holds great opportunity to impart new reactivity and greater reaction efficiency.
Acknowledgements
We would like to thank the Royal Society (University Research Fellowship to A.J.J.L.) and the EPSRC (EP/T001631/1) for funding.
Conflict of interest
The authors declare no conflict of interest.
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Biographical Information
David Heard studied at the University of Sheffield (MChem, 2014) then joined the Bristol Chemical Synthesis CDT at the University of Bristol. He conducted his PhD research into the structural elucidation and total synthesis of maleidride natural products under the supervision of Prof. Chris Willis. David has been a member of the Lennox group since 2019, and is investigating new methods for electrochemical reaction optimisation.
Biographical Information
Alastair Lennox is a graduate of the University of Manchester and completed his PhD in 2012 under the supervision of Professor Guy Lloyd-Jones. Following postdoctoral research with Prof. Matthias Beller as an Alexander von Humboldt Fellow, and with Prof. Shannon Stahl at the University of Wisconsin-Madison, he joined the University of Bristol as a Royal Society University Research Fellow. His research group's interests include the development of new selective and sustainable electrochemical methodologies for pharmaceutical and agrochemical use.
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Keywords
electrocatalysis
electrochemistry
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organic synthesis
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What is the locus of points in a space that are equidistant from 2 parallel lines?
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What is the locus of points in a space that are equidistant from 2 parallel lines?
My instructor says it's a parallel line but I think that's incorrect.
geometry
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edited May 30, 2017 at 19:54
DMcMor
10.1k55 gold badges2828 silver badges4242 bronze badges
asked May 30, 2017 at 19:27
Bran JohnsonBran Johnson
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7
Why do you think this is false? What would be your guess?
– C. Falcon
Commented
May 30, 2017 at 19:28
are you defining parallel to mean non-intersecting or some different way?
– gt6989b
Commented
May 30, 2017 at 19:29
1
Because its in a 3 dimensional space wouldnt that mean that the answer is a parallel plane?
– Bran Johnson
Commented
May 30, 2017 at 19:31
4
in 3+D, yes but in 2D he is right
– gt6989b
Commented
May 30, 2017 at 19:32
The question says "in a space" and not "in a plane" the exact answer to this question is important to me because it costed me points on a test.
– Bran Johnson
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May 30, 2017 at 19:38
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If the two lines are in Rn the points equidistant from them form an n−1 dimensional hyperplane. If n=2 they form a line halfway between the two original lines. If n=3 they form a plane that is perpendicular to the plane the lines are in going through the line between them.
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answered May 30, 2017 at 19:43
Ross MillikanRoss Millikan
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How do you formally prove this in R3?
– smth
Commented
May 8, 2022 at 19:45
@smth: If you know about affine transformations you can find one that takes the two parallel lines to x=1,y=0 and x=−1,y=0. Then observe that any point with x=0 is equidistant from them. Doing it without the transformation is messier but basically the same.
– Ross Millikan
Commented
May 8, 2022 at 20:16
@smth: You can also construct the plane perpendicular to the lines and through a point, then work in that plane to show the set of points equidistant to the two points where the lines cut the plane is a line between them. Project back to 3D and you have a plane.
– Ross Millikan
Commented
May 8, 2022 at 20:21
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187949 | https://libraryguides.centennialcollege.ca/c.php?g=645085&p=5254626 | Vector Subtraction - Math help from the Learning Centre - Library Guides at Centennial College
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Vector Subtraction
Vector subtraction is similar to addition except it is often used to find one of the original vectors from the resultant vector.
For example, if we have v→1+v→2=v→r but we're only given v→1 and v→r, we can find v→2 by solving our previous equation for v→2 like this: v→2=v→r−v→1. To subtract v→1 from v→r, we need to add the negative of v→1 like this:
v→2=v→r−v→1
v→2=v→r+(−v→1)
This is useful for when we want to remove the effects of vectors from others. For example, if a boat is travelling in the same direction as the wind and we want to know how fast the boat would be moving without the wind, we can subtract the wind from the boat's speed, or, as we learned before, we can add the negative of the wind.
But what is the negative of a vector?
The length of the vector will always be a positive quantity, so the negative of a vector will only change the direction. Specifically, it will point in exactly the opposite direction of the original vector.
For example, if we have a vector in rectangular coordinates like v→=(3,−4), we can get it to point in the opposite direction by changing the sign of both the x and y components. If we do that, we would have −v→=(−3,4). As you can see, the vector is reflected through the x and y axes, so the direction is completely flipped, but the length is still the same.
Plane Example
A good example of vector subtraction and vector relativity is with airplanes. Because airplanes generate their lift based on how fast the air moves over the wing, planes need to measure their speed relative to the ground and relative to the air, like we can see in the following problem:
Example: A plane is moving north at 200km/h (relative to the ground) and the wind is moving north at 30km/h (relative to the ground). How fast is the plane moving relative to the air?
Solution:
In this problem, we can find how fast the plane is moving relative to the air by subtracting the wind's speed from the plane's speed. By doing this, we are subtracting the effect of the wind from the system so the wind is no longer moving. This means we can treat the wind as our reference like the ground was before.
First, we can label the plane speed as v→p l a n e and the wind speed as v→w i n d. Next, we can plot the vectors to get an idea of what they look like and what our plane speed relative to the wind, v→r, will look like. If we use north as the +y direction (like on a compass), we will have: v→p l a n e=(0,200) and v→w i n d=(0,30) which can be graphed like this:
Note that v→w i n d is shifted slightly to the right to make it easier to see. It is actually on top of v→p l a n e.
Next, we need to subtract the wind velocity from the plane velocity to find the velocity of the plane relative to the wind, v→r, so that we can use the wind as our reference, as mentioned before. We perform this subtraction like so:
v→r=v→p l a n e−v→w i n d
v→r=v→p l a n e+ (−v→w i n d)
To get −v→w i n d we need to change the sign of the x and y coordinates of v→w i n d. This would give us −v→w i n d=(0,−30).
Now, as we did with vector addition, we need to add v→p l a n e and−v→w i n d tip to tail, which would look like this:
Note that −v→w i n d and v→r are shifted slightly to the right to make them easier to see. They are actually on top of v→p l a n e.
Now we need to add the x and y components of v→p l a n e and −v→w i n d to get the x and y components of v→r which we'll label v r x and v r y.
So for v r x we would have:
v r x=v p l a n e x+ ( −v w i n d x )
v r x=0+0
v r x=0
Similarly, for v r y:
v r y=v p l a n e y+ ( −v w i n d y )
v r y=200+(−30)
v r y=170
So the velocity of the plane relative to the wind is v→r=(v r x, v r y)=(0,170).
If the plane being slower relative to the wind than the ground is unintuitive to you, you can think about it like when you pass by a car on the highway. That car appears to be moving slowly relative to you because your car and the other car are both traveling in the same direction. In this case, both the plane and the wind are traveling in the same direction.
Subtraction Summary
In summary, to subtract vectors in rectangular coordinates we need to:
Graph the original vectors and figure out which one is being subtracted from which
Find the negative of the vector being subtracted by changing the sign on the x and y components
Add the x components of each vector together to get the x component of your resulting vector
Add the y components of each vector together to get the y component of your resulting vector
Write your x and y results in rectangular form as a new vector
Try this interactive tool!
Adjust the start and end points of vectors u→, v→, and the resultant vector u→−v→. Use the checkboxes to toggle the visibility of each vector. You can also choose to use the positive vector v→, or the negative vector −v→.
Practice
To practice, try subtracting the following vectors:
1) v→1=(5,−2) and v→2=(−1,−1). Subtract v→2 from v→1.
2) v→1=(−7,4) and v→2=(3,−3). Subtract v→1 from v→2.
3) v→1=(13,3) and v→2=(5,7). Subtract v→2 from v→1.
Answers:
1) v→r e s u l t=(6,−1)
2) v→r e s u l t=(10,−7)
3) v→r e s u l t=(8,−4)
Another Example
If you would like another example, take a look at the video below:
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187950 | https://en.wikipedia.org/wiki/Mean_corpuscular_volume | Jump to content
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Contents
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1 Calculation
2 Interpretation
2.1 High
2.2 Low
3 Worked example
4 Derivation
5 References
Mean corpuscular volume
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From Wikipedia, the free encyclopedia
Average volume of a red blood cell, which sometimes helps in diagnosis
Medical diagnostic method
| Mean corpuscular volume |
| Types of anemias by MCV |
| Purpose | MCV measurement that allows classification as either a microcytic anemia, normocytic anemia or macrocytic anemia |
The mean corpuscular volume, or mean cell volume (MCV), is a measure of the average volume of a red blood corpuscle (or red blood cell). The measure is obtained by multiplying a volume of blood by the proportion of blood that is cellular (the hematocrit), and dividing that product by the number of erythrocytes (red blood cells) in that volume. The mean corpuscular volume is a part of a standard complete blood count.
In patients with anemia, it is the MCV measurement that allows classification as either a microcytic anemia (MCV below normal range), normocytic anemia (MCV within normal range) or macrocytic anemia (MCV above normal range). Normocytic anemia is usually deemed so because the bone marrow has not yet responded with a change in cell volume. It occurs occasionally in acute conditions, namely blood loss and hemolysis.
If the MCV was determined by automated equipment, the result can be compared to RBC morphology on a peripheral blood smear, where a normal RBC is about the size of a normal lymphocyte nucleus. Any deviation would usually be indicative of either faulty equipment or technician error, although there are some conditions that present with high MCV without megaloblast RBCs.
For further specification, it can be used to calculate red blood cell distribution width (RDW). The RDW is a statistical calculation made by automated analyzers that reflects the variability in size and shape of the RBCs.
Calculation
[edit]
To calculate MCV, the hematocrit (Hct) is divided by the concentration of RBCs ([RBC])
Normally, MCV is expressed in femtoliters (fL, or 10−15 L), and [RBC] in millions per microliter (106 / μL). The normal range for MCV is 80–100 fL.
If the hematocrit is expressed as a percentage, the red blood cell concentration as millions per microliter, and the MCV in femtoliters, the formula becomes
For example, if the Hct = 42.5% and [RBC] = 4.58 million per microliter (4,580,000/μL), then
Using implied units,
The MCV can be determined in a number of ways by automatic analyzers. In volume-sensitive automated blood cell counters, such as the Coulter counter, the red cells pass one-by-one through a small aperture and generate a signal directly proportional to their volume. Other automated counters measure red blood cell volume by means of techniques that measure refracted, diffracted, or scattered light.
Interpretation
[edit]
The normal reference range is typically 80-100 fL.
High
[edit]
In pernicious anemia (macrocytic), MCV can range up to 150 femtolitres. (as are an elevated GGT and an AST/ALT ratio of 2:1). Vitamin B12 and/or folic acid deficiency has also been associated with macrocytic anemia (high MCV numbers).
Low
[edit]
The most common causes of microcytic anemia are iron deficiency (due to inadequate dietary intake, gastrointestinal blood loss, or menstrual blood loss), thalassemia, sideroblastic anemia or chronic disease. In iron deficiency anemia (microcytic anemia), it can be as low as 60 to 70 femtolitres. In some cases of thalassemia, the MCV may be low even though the patient is not iron deficient.[citation needed]
Worked example
[edit]
| Measure | Units | Conventional units | Conversion |
--- --- |
| Hct | 40% | | |
| Hb | 100 grams/liter | 10 grams/deciliter | (deci- is 10−1) |
| RBC | 5E+12 cells/liter | 5E+6 cells/μL | (micro is 10−6) |
| MCV = (Hct/100) / RBC | 8E-14 liters/cell | 80 femtoliters/cell | (femto- is 10−15) |
| MCH = Hb / RBC | 2E-11 grams/cell | 20 picograms/cell | (pico- is 10−12) |
| MCHC = Hb / (Hct/100) | 250 grams/liter | 25 grams/deciliter | (deci is 10−1) |
Derivation
[edit]
The MCV can be conceptualized as the total volume of a group of cells divided by the number of cells. For a real world sized example, imagine you had 10 small jellybeans with a combined volume of 10 μL. The mean volume of a jellybean in this group would be 10 μL / 10 jellybeans = 1 μL / jellybean. A similar calculation works for MCV.[citation needed]
Measure the RBC index in cells/μL. Take the reciprocal (1/RBC index) to convert it to μL/cell.
The 1 μL is only made of a proportion of red cells (e.g. 40%) with the rest of the volume composed of plasma. Multiply by the hematocrit (a unitless quantity) to take this into account.
Finally, convert the units of μL to fL by multiplying by . The result would look like this:
Note: the shortcut proposed above just makes the units work out:
References
[edit]
^ Mondal, Himel; Budh, Deepa P. (2020). "Hematocrit". Hematocrit (HCT). PMID 31194416. {{cite book}}: |journal= ignored (help)
^ Stanley L Schrier, MD; Stephen A Landaw, MD (30 September 2011). "Mean corpuscular volume". uptodate.com.
^ "RBC indices: MedlinePlus Medical Encyclopedia". medlineplus.gov. Retrieved May 13, 2020.
^ Tønnesen H, Hejberg L, Frobenius S, Andersen J (1986). "Erythrocyte Mean Cell Volume-Correlation to Drinking Pattern in Heavy Alcoholics". Acta Med Scand. 219 (5): 515–8. doi:10.1111/j.0954-6820.1986.tb03348.x. PMID 3739755.
| v t e Hematology blood tests |
| Complete blood count | Hemoglobin Hematocrit Red blood cell count + Red blood cell indices - Mean corpuscular hemoglobin - Mean corpuscular hemoglobin concentration - Mean corpuscular volume - Red blood cell distribution width White blood cell count + White blood cell differential + Absolute neutrophil count Platelet count + Mean platelet volume Reticulocyte count + Reticulocyte index |
| Other tests of red blood cells | Blood volume + Total blood volume (TBV) + Plasma volume (PV) + Red cell volume (RCV) Fetal hemoglobin + Apt–Downey test + Kleihauer–Betke test Hemoglobinopathy testing + Hemoglobin electrophoresis + Sickle solubility test + Mentzer index Erythrocyte sedimentation rate Haptoglobin Monocyte distribution width (MDW) Osmotic fragility |
| Coagulation | Clotting factors + Prothrombin time + Partial thromboplastin time + Thrombin time + Activated clotting time + Fibrinogen Bleeding time animal enzyme + Reptilase time + Ecarin clotting time + Dilute Russell's viper venom time Thrombin generation assay + Calibrated automated thrombogram + ST Genesia-based test Thromboelastography + Thrombodynamics test Overall hemostatic potential Coagulation activation markers + Prothrombin fragments 1+2 + Thrombin–antithrombin complex + Fibrinopeptide A + Fibrin monomers + Activated protein C–protein C inhibitor Fibrinolysis + Euglobulin lysis time + D-Dimer + Plasmin-α2-antiplasmin complex Von Willebrand factor Ristocetin-induced platelet aggregation Activated protein C resistance test + aPTT-based activated protein C resistance test + ETP-based activated protein C resistance test |
| Other | Blood film Blood viscosity Nitro blue tetrazolium chloride test Flow cytometry + Immunophenotyping |
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187951 | https://blog.csdn.net/nvd11/article/details/144521937 | Published Time: 2024-12-17T03:39:58+08:00
理解奇函数和偶函数_奇函数偶函数举例-CSDN博客
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理解奇函数和偶函数
最新推荐文章于 2024-12-28 21:41:01 发布
原创 于 2024-12-17 03:39:58 发布·3k 阅读
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奇函数的定义
对于函数f(x)， 如果x是在f(x) 的定义内， 如果任意x ， 都有f(-x) = -f(x) 那么f(x) 是奇函数
数学表达：
f(−x)=−f(x),x∈D(f)f(-x) = - f(x), x \in D(f)f(−x)=−f(x),x∈D(f)
很简单
在函数图像里， 奇函数都是对于原点对称
奇函数的例子
1. 一次线性函数
例如 f(x)=2xf(x) = 2x f(x)=2 x
注意的是， f(x)=2x+n,n≠0f(x) = 2x + n, n \neq 0 f(x)=2 x+n,n=0 这个并不是奇函数
明显 f(−1)=−2+nf(-1) = -2 + n f(−1)=−2+n 和 f(1)=2+nf(1) = 2 + n f(1)=2+n 并不是相反的值对
2. 3次函数 f(x)=x3f(x) = x^3 f(x)=x 3
这个也很明显
3. 反比例函数 f(x)=x−1f(x) = x^{-1}f(x)=x−1
4. 正弦函数 f(x)=x−1f(x) = x^{-1}f(x)=x−1
这个波浪线也是原点对称的！
偶函数的定义
对于函数f(x)， 如果x是在f(x) 的定义内， 如果任意x ， 都有f(-x) = f(x) 那么f(x) 是偶函数
数学表达：
f(−x)=f(x),x∈D(f)f(-x) = f(x), x \in D(f)f(−x)=f(x),x∈D(f)
在函数图像里， 偶函数都是对于y轴对称
偶函数的例子
1. 2次函数 f(x)=x2f(x) = x^2 f(x)=x 2
注意的是 f(x)=x2+n,n≠0f(x) = x^2 + n, n \neq 0 f(x)=x 2+n,n=0 仍然是偶函数， 因为函数图像上下移动一段距离仍然对于y轴对称
但是 f(x)=(x+n)2,n≠0f(x) = (x+n)^2, n \neq 0 f(x)=(x+n)2,n=0 就不是偶函数了，左右移动不行
2. 绝对值函数 f(x)=∣x∣f(x) = |x|f(x)=∣x∣
3. 余弦函数 f(x)=cos(x)f(x) = \cos(x)f(x)=cos(x)
f(x)=cos(x)=sin(x+2π)f(x) = \cos(x) = \sin(x +\frac{2}{\pi})f(x)=cos(x)=sin(x+π 2)
也就将， 正弦函数的图像向左移动半个身位(2π\frac{2}{\pi}π 2) , 就由原点对称变成y轴对称， 由奇函数变成偶函数， 由正弦函数变成余弦函数了。
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目录
奇函数的定义
奇函数的例子
1. 一次线性函数
2. 3次函数 f(x)=x3f(x) = x^3f(x)=x3
3. 反比例函数 f(x)=x−1f(x) = x^{-1}f(x)=x−1
4. 正弦函数 f(x)=x−1f(x) = x^{-1}f(x)=x−1
偶函数的定义
偶函数的例子
1. 2次函数 f(x)=x2f(x) = x^2f(x)=x2
2. 绝对值函数 f(x)=∣x∣f(x) = |x|f(x)=∣x∣
3. 余弦函数 f(x)=cos(x)f(x) = \cos(x)f(x)=cos(x)
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目录
奇函数的定义
奇函数的例子
1. 一次线性函数
2. 3次函数 f(x)=x3f(x) = x^3f(x)=x3
3. 反比例函数 f(x)=x−1f(x) = x^{-1}f(x)=x−1
4. 正弦函数 f(x)=x−1f(x) = x^{-1}f(x)=x−1
偶函数的定义
偶函数的例子
1. 2次函数 f(x)=x2f(x) = x^2f(x)=x2
2. 绝对值函数 f(x)=∣x∣f(x) = |x|f(x)=∣x∣
3. 余弦函数 f(x)=cos(x)f(x) = \cos(x)f(x)=cos(x)
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187953 | https://cdn.ttgtmedia.com/searchEnterpriseLinux/downloads/WGC_Chapter_3.pdf | 3 B I N A R Y A R I T H M E T I C A N D B I T O P E R A T I O N S Understanding how computers represent data in binary is a prerequisite to writing software that works well on those computers. Of equal importance, of course, is under-standing how computers operate on binary data. Exploring arithmetic, logical, and bit operations on binary data is the purpose of this chapter.
3.1 Arithmetic Operations on Binary and Hexadecimal Numbers Because computers use binary representation, programmers who write great code often have to work with binary (and hexadecimal) values. Often, when writing code, you may need to manually operate on two binary values in order to use the result in your source code. Although calculators are available to compute such results, you should be able to perform simple arithmetic operations on binary operands by hand. Write Great Code www.nostarch.com 40 Chapter 3 Hexadecimal arithmetic is sufficiently painful that a hexadecimal calculator belongs on every programmer’s desk (or, at the very least, use a software-based calculator that supports hexadecimal operations, such as the Windows calculator). Arithmetic operations on binary values, however, are actually easier than decimal arithmetic. Knowing how to manually compute binary arithmetic results is essential because several important algorithms use these operations (or variants of them). Therefore, the next several subsections describe how to manually add, subtract, multiply, and divide binary values, and how to perform various logical operations on them.
3.1.1 Adding Binary Values Adding two binary values is easy; there are only eight rules to learn. (If this sounds like a lot, just realize that you had to memorize approximately 200 rules for decimal addition!) Here are the rules for binary addition: 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0 with carry Carry + 0 + 0 = 1 Carry + 0 + 1 = 0 with carry Carry + 1 + 0 = 0 with carry Carry + 1 + 1 = 1 with carry Once you know these eight rules you can add any two binary values together. Here are some complete examples of binary addition: 0101 + 0011 ------Step 1: Add the LO bits (1 + 1 = 0 + carry).
c 0101 + 0011 ------0 Step 2: Add the carry plus the bits in bit position one (carry + 0 + 1 = 0 + carry).
c 0101 + 0011 -------00 Write Great Code www.nostarch.com Binary Arithmetic and Bit Operations 41 Step 3: Add the carry plus the bits in bit position two (carry + 1 + 0 = 0 + carry).
c 0101 + 0011 ------000 Step 4: Add the carry plus the bits in bit position three (carry + 0 + 0 = 1).
0101 + 0011 ------1000 Here are some more examples: 1100_1101 1001_1111 0111_0111 + 0011_1011 + 0001_0001 + 0000_1001 ----------- ----------- -----------1_0000_1000 1011_0000 1000_0000 3.1.2 Subtracting Binary Values Binary subtraction is also easy; like addition, binary subtraction has eight rules: 0 − 0 = 0 0 − 1 = 1 with a borrow 1 − 0 = 1 1 − 1 = 0 0 − 0 − borrow = 1 with a borrow 0 − 1 − borrow = 0 with a borrow 1 − 0 − borrow = 0 1 − 1 − borrow = 1 with a borrow Here are some complete examples of binary subtraction: 0101 − 0011 ------Step 1: Subtract the LO bits (1 − 1 = 0).
0101 − 0011 ------0 Write Great Code www.nostarch.com 42 Chapter 3 Step 2: Subtract the bits in bit position one (0 − 1 = 1 + borrow).
0101 − 0011 b ------10 Step 3: Subtract the borrow and the bits in bit position two (1 − 0 − b = 0).
0101 − 0011 ------010 Step 4: Subtract the bits in bit position three (0 − 0 = 0).
0101 − 0011 ------0010 Here are some more examples: 1100_1101 1001_1111 0111_0111 − 0011_1011 − 0001_0001 − 0000_1001 ----------- ----------- -----------1001_0010 1000_1110 0110_1110 3.1.3 Multiplying Binary Values Multiplication of binary numbers is also very easy. It’s just like decimal multi-plication involving only zeros and ones (which is trivial). Here are the rules you need to know for binary multiplication: 0 × 0 = 0 0 × 1 = 0 1 × 0 = 0 1 × 1 = 1 Using these four rules, multiplication is done the same way you’d do decimal multiplication (in fact, if you just follow the rules for decimal multiplication on your binary values you’ll actually get the correct results, because the rules for decimal multiplication involving the zero and one digits are identical). Here are some examples of binary multiplication: Write Great Code www.nostarch.com Binary Arithmetic and Bit Operations 43 1010 × 0101 -------Step 1: Multiply the LO bit of the multiplier times the multiplicand.
1010 × 0101 -------1010 (1 × 1010) Step 2: Multiply bit one of the multiplier times the multiplicand.
1010 × 0101 -------1010 (1 × 1010) 0000 (0 × 1010) -------01010 (partial sum) Step 3: Multiply bit two of the multiplier times the multiplicand.
1010 × 0101 -------001010 (previous partial sum) 1010 (1 × 1010) -------110010 (partial sum) Step 4: Multiply bit three of the multiplier times the multiplicand.
1010 × 0101 -------110010 (previous partial sum) 0000 (0 × 1010) -------0110010 (product) 3.1.4 Dividing Binary Values Like multiplication of binary numbers, binary division is actually easier than decimal division. You use the same (longhand) division algorithm, but binary division is easier because you can trivially determine whether the divisor goes into the dividend during each step of the longhand division algorithm. Figure 3-1 on the next page shows the steps in a decimal division problem.
Write Great Code www.nostarch.com 44 Chapter 3 Figure 3-1: Decimal division (3456/12) This algorithm is actually easier in binary because at each step you do not have to guess how many times 12 goes into the remainder nor do you have to multiply 12 by your guess to obtain the amount to subtract. At each step in the binary algorithm, the divisor goes into the remainder exactly zero or one times. As an example, consider the division of 27 (11011) by three (11) as shown in Figure 3-2.
12 3456 24 (1) 12 goes into 34 two times.
(3) 12 goes into 105 eight times.
12 3456 24 105 96 28 12 3456 24 105 2 (2) Subtract 24 from 34 and drop down the 105.
12 3456 24 105 96 28 96 (4) Subtract 96 from 105 and drop down the 96.
12 3456 24 105 96 288 96 96 (5) 12 goes into 96 exactly eight times.
12 3456 24 105 96 288 96 96 (6) Therefore, 12 goes into 3456 exactly 288 times.
2 11 11011 11 11 goes into 11 one time.
1 11 11011 11 00 Subtract out the 11 and bring down the zero.
1 11 11011 11 00 00 11 goes into 00 zero times.
10 Write Great Code www.nostarch.com Binary Arithmetic and Bit Operations 45 Figure 3-2: Longhand division in binary 11 11011 11 00 00 01 Subtract out the zero and bring down the one.
10 11 11011 11 00 00 01 00 11 goes into 01 zero times.
100 11 11011 11 00 00 01 00 11 Subtract out the zero and bring down the one.
100 11 11011 11 00 00 01 00 11 11 11 goes into 11 one time.
1001 11 11011 11 00 00 01 00 11 11 00 This produces the final result of 1001.
1001 Write Great Code www.nostarch.com 46 Chapter 3 3.2 Logical Operations on Bits There are four main logical operations we’ll need to perform on hexa-decimal and binary numbers: AND, OR, XOR (exclusive-or), and NOT. Unlike the arithmetic operations, a hexadecimal calculator isn’t necessary to perform these operations.
The logical AND, OR, and XOR operations accept two single-bit operands and compute the following results: AND: 0 and 0 = 0 0 and 1 = 0 1 and 0 = 0 1 and 1 = 1 OR: 0 or 0 = 0 0 or 1 = 1 1 or 0 = 1 1 or 1 = 1 XOR: 0 xor 0 = 0 0 xor 1 = 1 1 xor 0 = 1 1 xor 1 = 0 Table 3-1, Table 3-2, and Table 3-3 show the truth tables for the AND, OR, and XOR operations. A truth table is just like the multiplication tables you encountered in elementary school. The values in the left column correspond to the left operand of the operation. The values in the top row correspond to the right operand of the operation. The value located at the intersection of the row and column (for a particular pair of input values) is the result.
Table 3-1: AND truth table AND 0 1 0 0 0 1 0 1 Table 3-2: OR truth table OR 0 1 0 0 1 1 1 1 Write Great Code www.nostarch.com Binary Arithmetic and Bit Operations 47 In plain English, the logical AND operation translates as, “If the first operand is one and the second operand is one, the result is one; otherwise the result is zero.” We could also state this as “If either or both operands are zero, the result is zero.” The logical AND operation is useful for forcing a zero result. If one of the operands is zero, the result is always zero regardless of the value of the other operand. If one of the operands contains one, then the result is the value of the other operand.
Colloquially, the logical OR operation is, “If the first operand or the second operand (or both) is one, the result is one; otherwise the result is zero.” This is also known as the inclusive-OR operation. If one of the operands to the logical-OR operation is one, the result is always one. If an operand is zero, the result is always the value of the other operand.
In English, the logical XOR operation is, “If the first or second operand, but not both, is one, the result is one; otherwise the result is zero.” If one of the operands is a one, the result is always the inverse of the other operand.
The logical NOT operation is unary (meaning it accepts only one operand). The truth table for the NOT operation appears in Table 3-4. This operator simply inverts (reverses) the value of its operand.
3.3 Logical Operations on Binary Numbers and Bit Strings The logical functions work on single-bit operands. Because most pro-gramming languages manipulate groups of 8, 16, or 32 bits, we need to extend the definition of these logical operations beyond single-bit operands. We can easily extend logical functions to operate on a bit-by-bit (or bitwise) basis. Given two values, a bitwise logical function operates on bit zero of both operands producing bit zero of the result; it operates on bit one of both operands producing bit one of the result, and so on. For example, if you want to compute the bitwise logical AND of two 8-bit numbers, you would logically AND each pair of bits in the two numbers: %1011_0101 %1110_1110 ----------%1010_0100 Table 3-3: XOR truth table XOR 0 1 0 0 1 1 1 0 Table 3-4: NOT truth table NOT 0 1 1 0 Write Great Code www.nostarch.com 48 Chapter 3 This bit-by-bit execution also applies to the other logical operations, as well. The ability to force bits to zero or one using the logical AND and OR operations, and the ability to invert bits using the logical XOR operation, is very important when working with strings of bits (such as binary numbers). These operations let you selectively manipulate certain bits within a value while leaving other bits unaffected. For example, if you have an 8-bit binary value X and you want to guarantee that bits four through seven contain zeros, you could logically AND the value X with the binary value %0000_1111. This bitwise logical AND operation would force the HO four bits of X to zero and leave the LO four bits of X unchanged. Likewise, you could force the LO bit of X to one and invert bit number two of X by logically ORing X with %0000_0001 and then logically exclusive ORing (XORing) X with %0000_0100. Using the logical AND, OR, and XOR operations to manipulate bit strings in this fashion is known as masking bit strings. We use the term masking because we can use certain values (one for AND, zero for OR and XOR) to “mask out” or “mask in” certain bits in an operand while forcing other bits to zero, one, or their inverse. Several languages provide operators that let you compute the bitwise AND, OR, XOR, and NOT of their operands. The C/C++/Java language family uses the ampersand (&) operator for bitwise AND, the pipe (|) oper-ator for bitwise OR, the caret (^) operator for bitwise XOR, and the tilde (~) operator for bitwise NOT. The Visual Basic and Delphi/Kylix languages let you use the and, or, xor, and not operators with integer operands. From 80x86 assembly language, you can use the AND, OR, NOT, and XOR instructions to do these bitwise operations.
// Here's a C/C++ example: i = j & k; // Bitwise AND i = j | k; // Bitwise OR i = j ^ k; // Bitwise XOR i = ~j; // Bitwise NOT 3.4 Useful Bit Operations Although bit operations may seem a bit abstract, they are quite useful for many non-obvious purposes. The following subsections describe some of their useful properties of using the logical operations in various languages.
3.4.1 Testing Bits in a Bit String Using AND You can use the bitwise AND operator to test individual bits in a bit string to see if they are zero or one. If you logically AND a value with a bit string that contains a one in a certain bit position, the result of the logical AND will be zero if the corresponding bit contains a zero, and the result will be nonzero Write Great Code www.nostarch.com Binary Arithmetic and Bit Operations 49 if that bit position contains one. Consider the following C/C++ code that checks an integer value to see if it is odd or even by testing if bit zero of the integer: IsOdd = (ValueToTest & 1) != 0; In binary form, here’s what this bitwise AND operation is doing: xxxx_xxxx_xxxx_xxxx_xxxx_xxxx_xxxx_xxxx // Assuming ValueToTest is 32 bits 0000_0000_0000_0000_0000_0000_0000_0001 // Bitwise AND with the value one ---------------------------------------0000_0000_0000_0000_0000_0000_0000_000x // Result of bitwise AND The result is zero if the LO bit of ValueToTest contains a zero in bit position zero. The result is one if ValueToTest contains a one in bit position one. This calculation ignores all other bits in ValueToTest.
3.4.2 Testing a Set of Bits for Zero/Not Zero Using AND You can also use the bitwise AND operator to check a set of bits to see if they are all zero. For example, one way to check to see if a number is evenly divisible by 16 is to see if the LO four bits of the value are all zeros. The following Delphi/Kylix statement uses the bitwise AND operator to accomplish this: IsDivisibleBy16 := (ValueToTest and $f) = 0; In binary form, here’s what this bitwise AND operation is doing: xxxx_xxxx_xxxx_xxxx_xxxx_xxxx_xxxx_xxxx // Assuming ValueToTest is 32 bits 0000_0000_0000_0000_0000_0000_0000_1111 // Bitwise AND with $F ---------------------------------------0000_0000_0000_0000_0000_0000_0000_xxxx // Result of bitwise AND The result is zero if and only if the LO four bits of ValueToTest are all zero, because ValueToTest is evenly divisible by 16 only if its LO four bits all contain zero.
3.4.3 Comparing a Set of Bits Within a Binary String The AND and OR operations are particularly useful if you need to compare a subset of the bits in a binary value against some other value. For example, you might want to compare two 6-bit values found in bits 0, 1, 10, 16, 24, and 31 of a pair of 32-bit values. The trick is to set all the uninteresting bits to zero and then compare the two results.1 1 It’s also possible to set all the uninteresting bits to ones via the OR operation, but the AND operator is often more convenient.
Write Great Code www.nostarch.com 50 Chapter 3 Consider the following three binary values; the “x” bits denote bits whose values we don’t care about: %1xxxxxx0xxxxxxx1xxxxx0xxxxxxxx10 %1xxxxxx0xxxxxxx1xxxxx0xxxxxxxx10 %1xxxxxx1xxxxxxx1xxxxx1xxxxxxxx11 If we compare the first and second binary values (assuming we’re only interested in bits 31, 16, 10, 1, and 0), we should find that the two values are equal. If we compare either of the first two values against the third value, we’ll find that they are not equal. Furthermore, if we compare either of the first two values against the third, we should discover that the third value is greater than the first two. In C/C++ and assembly, this is how we could compare these values: // C/C++ example if( (value1 & 0x81010403) == (value2 & 0x81010403)) { // Do something if bits 31, 24, 16, 10, 1, and 0 of // value1 and value2 are equal } if( (value1 & 0x81010403) != (value3 & 0x81010403)) { // Do something if bits 31, 24, 16, 10, 1, and 0 of // value1 and value3 are not equal } // HLA/x86 assembly example: mov( value1, eax ); // EAX = value1 and( $8101_0403, eax ); // Mask out unwanted bits in EAX mov( value2, edx ); // EDX = value2 and( $8101_0403, edx ); // Mask out the same set of unwanted bits in EDX if( eax = edx ) then // See if the remaining bits match // Do something if bits 31, 24, 16, 10, 1, and 0 of // value1 and value2 are equal endif; mov( value1, eax ); // EAX = value1 and( $8101_0403, eax ); // Mask out unwanted bits in EAX mov( value3, edx ); // EDX = value2 and( $8101_0403, edx ); // Mask out the same set of unwanted bits in EDX Write Great Code www.nostarch.com Binary Arithmetic and Bit Operations 51 if( eax <> edx ) then // See if the remaining bits do not match // Do something if bits 31, 24, 16, 10, 1, and 0 of // value1 and value3 are not equal endif; 3.4.4 Creating Modulo-n Counters Using AND The AND operation lets you create efficient modulo-n counters. A modulo-n counter counts from zero2 to some maximum value and then resets to zero. Modulo-n counters are great for creating repeating sequences of numbers such as 0, 1, 2, 3, 4, 5, . . . n−1, 0, 1, 2, 3, 4, 5, . . . n−1, 0, 1, . . . . You can use such sequences to create circular queues and other objects that reuse array elements upon encountering the end of the data structure. The normal way to create a modulo-n counter is to add one to the counter, divide the result by n, and then keep the remainder. The following code examples demon-strate the implementation of a modulo-n counter in C/ C++, Pascal, and Visual Basic: cntr = (cntr + 1 ) % n; // C/C++ cntr := (cntr + 1) mod n; // Pascal/Delphi/Kylix cntr = (cntr + 1) Mod n Visual Basic The problem with this particular implementation is that division is an expensive operation, requiring far more time to execute than operations such as addition. In general, you’ll find it more efficient to implement modulo-n counters using a comparison rather than the remainder operator. Here’s a Pascal example: cntr := cntr + 1; // Pascal example if( cntr >= n ) then cntr := 0; For certain special cases, however, you can increment a modulo-n counter more efficiently and conveniently using the AND operation. You can use the AND operator to create a modulo-n counter when n is a power of two. To create such a modulo-n counter, increment your counter and then logically AND it with the value n = 2m−1 (2m−1 contains ones in bit positions 0..m−1 and zeros everywhere else). Because the AND operation is usually much faster than a division, AND-driven modulo-n counters are much more efficient than those using the remainder operator. Indeed, on most CPUs, using the AND operator is quite a bit faster than using an if statement. The following examples show how to implement a modulo-n counter for n = 32 using the AND operation: 2 Actually, they could count down to zero as well, but usually they count up.
Write Great Code www.nostarch.com 52 Chapter 3 //Note: 0x3f = 31 = 25 − 1, so n = 32 and m = 5 cntr = (cntr + 1) & 0x3f; // C/C++ example cntr := (cntr + 1) and $3f; // Pascal/Delphi/Kylix example cntr = (cntr + 1) And &h3f Visual Basic example The assembly language code is especially efficient: inc( eax ); // Compute (eax + 1) mod 32 and( $3f, eax ); 3.5 Shifts and Rotates Another set of logical operations on bit strings are the shift and rotate oper-ations. These functions can be further broken down into shift lefts, rotate lefts, shift rights, and rotate rights. These operations turn out to be very useful in many programs.
The shift left operation moves each bit in a bit string one position to the left, as shown in Figure 3-3. Bit zero moves into bit position one, the previous value in bit position one moves into bit position two, and so on. Figure 3-3: Shift left operation (on a byte) There are two questions that arise: “What goes into bit zero?” and “Where does the HO bit wind up?” We’ll shift a zero into bit zero, and the previous value of the HO bit will be the carry out of this operation.
Several high-level languages (such as C/C++/C#, Java, and Delphi/Kylix) provide a shift left operator. In the C language family, this operator is <<. In Delphi/Kylix, you use the shl operator. Here are some examples: // C: cLang = d << 1; // Assigns d shifted left one position to // variable "cLang" // Delphi: Delphi := d shl 1; // Assigns d shifted left one position to // variable "Delphi" Shifting the binary representation of a number one position to the left is equivalent to multiplying that value by two. Therefore, if you’re using a programming language that doesn’t provide an explicit shift left operator, 7 6 5 4 3 2 1 0 Write Great Code www.nostarch.com Binary Arithmetic and Bit Operations 53 you can usually simulate this by multiplying a binary integer value by two. Although the multiplication operation is usually slower than the shift left operation, most compilers are smart enough to translate a multiplication by a constant power of two into a shift left operation. Therefore, you could write code like the following in Visual Basic to do a shift left: vb = d 2 A shift right operation is similar to a shift left, except we’re moving the data in the opposite direction. Bit seven moves into bit six; bit six moves into bit five; bit five moves into bit four; and so on. During a shift right, we’ll move a zero into bit seven, and bit zero will be the carry out of the operation (see Figure 3-4). C, C++, C#, and Java use the >> operator for a shift right operation. Delphi/Kylix uses the shr operator. Most assembly languages also provide a shift right instruction (shr on the 80x86).
Figure 3-4: The shift right operation (on a byte) Shifting an unsigned binary value right divides that value by two. For example, if you shift the unsigned representation of 254 ($FE) one place to the right, you get 127 ($7F), exactly as you would expect. However, if you shift the 8-bit two’s complement binary representation of −2 ($FE) one position to the right, you get 127 ($7F), which is not correct. To divide a signed number by two using a shift, we must define a third shift operation: arithmetic shift right. An arithmetic shift right operation does not modify the value of the HO bit. Figure 3-5 shows the arithmetic shift right operation for an 8-bit operand.
Figure 3-5: Arithmetic shift right operation (on a byte) This generally produces the result you expect for two’s complement signed operands. For example, if you perform the arithmetic shift right operation on −2 ($FE), you get −1 ($FF). Note, however, that this operation always rounds the numbers to the closest integer that is less than or equal to the actual result. If you arithmetically shift right −1 ($FF), the result is −1, not zero. Because −1 is less than zero, the arithmetic shift right operation rounds towards −1. This is not a “bug” in the arithmetic shift right operation; it just uses a different (though valid) definition of integer division. The bottom line, however, is that you probably won’t be able to use a signed division 0 7 6 5 4 3 2 1 0 7 6 5 4 3 2 1 0 Write Great Code www.nostarch.com 54 Chapter 3 operator as a substitute for arithmetic shift right in languages that don’t support arithmetic shift right, because most integer division operators round towards zero.
One problem with the shift right operation in high-level languages is that it’s rare for a high-level language to support both the logical shift right and the arithmetic shift right. Worse still, the specifications for certain languages leave it up to the compiler’s implementer to decide whether to use an arithmetic shift right or a logical shift right operation. Therefore, it’s only safe to use the shift right operator on values whose HO bit will cause both forms of the shift right operation to produce the same result. If you need to guarantee that a shift right is a logical shift right or an arithmetic shift right operation, then you’ll either have to drop down into assembly language or you’ll have to handle the HO bit manually. Obviously, the high-level code gets ugly really fast, so a quick in-line assembly statement might be a better solution if your program doesn’t need to be portable across different CPUs. The following code demonstrates how to simulate a 32-bit logical shift right and arithmetic shift right in languages that don’t guarantee the type of shift they use: // Written in C/C++, assuming 32-bit integers, logical shift right: // Compute bit 30.
Bit30 = ((ShiftThisValue & 0x800000000) != 0) ? 0x40000000 : 0; // Shifts bits 0..30.
ShiftThisValue = (ShiftThisValue & 0x7fffffff) >> 1; // Merge in Bit #30.
ShiftThisValue = ShiftThisValue | Bit30; // Arithmetic shift right operation Bits3031 = ((ShiftThisValue & 0x800000000) != 0) ? 0xC0000000 : 0; // Shifts bits 0..30.
ShiftThisValue = (ShiftThisValue & 0x7fffffff) >> 1; // Merge bits 30/31.
ShiftThisValue = ShiftThisValue | Bits3031; Many assembly languages also provide various rotate instructions that recir-culate bits through an operand by taking the bits shifted out of one end of the operation and shifting them into the other end of the operand. Few high-level languages provide this operation; fortunately, you won’t need it very often. If you do, you can synthesize this operation using the shift operators available in your high-level language: // Pascal/Delphi/Kylix Rotate Left, 32-bit example: // Puts bit 31 into bit 0, clears other bits.
CarryOut := (ValueToRotate shr 31); ValueToRotate := (ValueToRotate shl 1) or CarryOut; Write Great Code www.nostarch.com Binary Arithmetic and Bit Operations 55 Assembly language programmers typically have access to a wide variety of shift and rotate instructions. For more information on the type of shift and rotate operations that are possible, consult my assembly language programming book, The Art of Assembly Language (No Starch Press).
3.6 Bit Fields and Packed Data CPUs generally operate most efficiently on byte, word, and double-word data types;3 but occasionally you’ll need to work with a data type whose size is something other than 8, 16, or 32 bits. In such cases, you may be able to save some memory by packing different strings of bits together as compactly as possible, without wasting any bits to align a particular data field on a byte or other boundary.
Consider a date of the form “04/02/01.” It takes three numeric values to represent this date: month, day, and year values. Months, of course, use the values 1..12. It will require at least four bits (a maximum of 16 different values) to represent the month. Days use the range 1..31. Therefore, it will take five bits (a maximum of 32 different values) to represent the day entry. The year value, assuming that we’re working with values in the range 0..99, requires seven bits (representing up to 128 different values). Four plus five plus seven is 16 bits, or two bytes. In other words, we can pack our date data into two bytes rather than the three that would be required if we used a separate byte for each of the month, day, and year values. This saves one byte of memory for each date stored, which could be a substantial saving if you need to store many dates. You might arrange the bits as shown in Figure 3-6.
Figure 3-6: Short packed date format (16 bits) MMMM represents the four bits making up the month value, DDDDD represents the five bits making up the day, and YYYYYYY is the seven bits that hold the year. Each collection of bits representing a data item is a bit field. We could represent April 2, 2001, with $4101: 0100 00010 0000001 = %0100_0001_0000_0001 or $4101 04 02 01 Although packed values are space efficient (that is, they use little memory), they are computationally inefficient (slow!). The reason? It takes extra instructions to unpack the data from the various bit fields. These extra instructions take time to execute (and additional bytes to hold the instructions); hence, you must carefully consider whether packed data 3 Some RISC CPUs only operate efficiently on double-word values, so the concept of bit fields and packed data may apply to any object less than 32 bits in size on such CPUs. Y Y Y Y Y Y Y 15 14 13 12 11 10 7 6 5 4 3 2 1 0 9 8 M M M M D D D D D Write Great Code www.nostarch.com 56 Chapter 3 fields will save you anything. The following sample HLA/x86 code demonstrates the effort that must go into packing and unpacking this 16-bit date format.
program dateDemo; #include( "stdlib.hhf" ) static day: uns8; month: uns8; year: uns8; packedDate: word; begin dateDemo; stdout.put( "Enter the current month, day, and year: " ); stdin.get( month, day, year ); // Pack the data into the following bits: // // 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 // m m m m d d d d d y y y y y y y mov( 0, ax ); mov( ax, packedDate ); //Just in case there is an error. if( month > 12 ) then stdout.put( "Month value is too large", nl ); elseif( month = 0 ) then stdout.put( "Month value must be in the range 1..12", nl ); elseif( day > 31 ) then stdout.put( "Day value is too large", nl ); elseif( day = 0 ) then stdout.put( "Day value must be in the range 1..31", nl ); elseif( year > 99 ) then stdout.put( "Year value must be in the range 0..99", nl ); Write Great Code www.nostarch.com Binary Arithmetic and Bit Operations 57 else mov( month, al ); shl( 5, ax ); or( day, al ); shl( 7, ax ); or( year, al ); mov( ax, packedDate ); endif; // Okay, display the packed value: stdout.put( "Packed data = $", packedDate, nl ); // Unpack the date: mov( packedDate, ax ); and( $7f, al ); // Retrieve the year value. mov( al, year ); mov( packedDate, ax ); // Retrieve the day value. shr( 7, ax ); and( %1_1111, al ); mov( al, day ); mov( packedDate, ax ); // Retrieve the month value. rol( 4, ax ); and( %1111, al ); mov( al, month ); stdout.put( "The date is ", month, "/", day, "/", year, nl ); end dateDemo; Keeping in mind the Y2K4 problem, adopting a date format that only supports a two-digit year is rather foolish. So consider a better date format, shown in Figure 3-7.
Figure 3-7: Long packed date format (32 bits) 4 Year 2000, a software engineering disaster that occurred because programmers in the 1900s encoded dates using only two digits and then discovered they couldn’t differentiate 1900 and 2000 when the year 2000 came along.
15 16 31 8 7 0 Month (1-12) Year (0-65535) Day (1-31) Write Great Code www.nostarch.com 58 Chapter 3 Because there are more bits in a 32-bit variable than are needed to hold the date, even accounting for years in the range 0–65,535, this format allots a full byte for the month and day fields. Because these two fields are bytes, an application can easily manipulate them as byte objects, reducing the overhead to pack and unpack these fields on those processors that support byte access. This leaves fewer bits for the year, but 65,536 years is probably sufficient (you can probably assume that your software will not be in use 63,000 years from now).
Of course, you could argue that this is no longer a packed date format. After all, we needed three numeric values, two of which fit just nicely into one byte each and one that should probably have at least two bytes. Because this “packed” date format consumes the same four bytes as the unpacked version, what is so special about this format? Well, in this example packed effectively means packaged or encapsulated. This particular packed format does not use as few bits as possible; by packing the data into a double-word variable the program can treat the date value as a single data value rather than as three separate variables. This generally means that it requires only a single machine instruction to operate on this data rather than three separate instructions.
Another difference you will note between this long packed date format and the short date format appearing in Figure 3-6 is the fact that this long date format rearranges the Year, Month, and Day fields. This is important because it allows you to easily compare two dates using an unsigned integer comparison. Consider the following HLA/assembly code: mov( Date1, eax ); // Assume Date1 and Date2 are double-word variables if( eax > Date2 ) then // using the Long Packed Date format.
<< do something if Date1 > Date2 >> endif; Had you kept the different date fields in separate variables, or organized the fields differently, you would not have been able to compare Date1 and Date2 in such a straightforward fashion. This example demonstrates another reason for packing data, even if you don’t realize any space savings — it can make certain computations more convenient or even more efficient (contrary to what normally happens when you pack data).
Some high-level languages provide built-in support for packed data. For example, in C you can define structures like the following: Write Great Code www.nostarch.com Binary Arithmetic and Bit Operations 59 struct { unsigned bits0_3 :4; unsigned bits4_11 :8; unsigned bits12_15 :4; unsigned bits16_23 :8; unsigned bits24_31 :8; } packedData; This structure specifies that each field is an unsigned object that holds four, eight, four, eight, and eight bits, respectively. The “:n” item appearing after each declaration specifies the minimum number of bits the compiler will allocate for the given field.
Unfortunately, it is not possible to provide a diagram that shows how a C/C++ compiler will allocate the values from a 32-bit double word among the fields. No (single) diagram is possible because C/C++ compiler imple-menters are free to implement these bit fields any way they see fit. The arrangement of the bits within the bit string is arbitrary (for example, the compiler could allocate the bits0_3 field in bits 28..31 of the ultimate object). The compiler can also inject extra bits between fields as it sees fit. The compiler can use a larger number of bits for each field if it so desires (this is actually the same thing as injecting extra padding bits between fields). Most C compilers attempt to minimize the injection of extraneous padding, but different C compilers (especially on different CPUs) do have their differences. Therefore, any use of C/C++ struct bit field declarations is almost guaranteed to be nonportable, and you can’t really count on what the compiler is going to do with those fields.
The advantage of using the compiler’s built-in data-packing capabilities is that the compiler automatically handles packing and unpacking the data for you. For example, you could write the following C/C++ code, and the compiler would automatically emit the necessary machine instructions to store and retrieve the individual bit fields for you: struct { unsigned year :7; unsigned month :4; unsigned day :5; } ShortDate; . . .
ShortDate.day = 28; ShortDate.month = 2; ShortDate.year = 3; // 2003 Write Great Code www.nostarch.com 60 Chapter 3 3.7 Packing and Unpacking Data The advantage of packed data types is efficient memory use. Consider the Social Security identification number in use in the United States. This is a nine-digit code that normally takes the following form (each “X” represents a single decimal digit): XXX–XX–XXXX If we encode a Social Security number using three separate (32-bit) integers, it will take 12 bytes to represent this value. That’s actually more than the 11 bytes needed to represent the number using an array of characters. A better solution is to encode each field using short (16-bit) integers. Now it takes only 6 bytes to represent the Social Security number. Because the middle field in the Social Security number is always between 0 and 99, we can actually shave one more byte off the size of this structure by encoding the middle field with a single byte. Here’s a sample Delphi/Kylix record structure that defines this data structure: SSN :record FirstField: smallint; // smallints are 16 bits in Delphi/Kylix SecondField: byte; ThirdField: smallint; end; If we drop the hyphens in the Social Security number, you’ll notice that the result is a nine-digit number. Because we can exactly represent all values between 0 and 999,999,999 (nine digits) using 30 bits, it should be clear that we could actually encode any legal Social Security number using a 32-bit integer. The problem is that some software that manipulates Social Security numbers may need to operate on the individual fields. This means that you have to use expensive division, modulo, and multiplication operators in order to extract fields from a Social Security number you’ve encoded in a 32-bit integer format. Furthermore, it’s a bit more painful to convert Social Security numbers to and from strings when using the 32-bit format. The advantage of using bit fields to hold a value is that it’s relatively easy to insert and extract individual bit fields using fast machine instructions, and it’s also less work to create a standard string representation (including the hyphens) of one of these fields. Figure 3-8 provides a straightforward implementation of the Social Security number packed data type using a separate string of bits for each field (note that this format uses 31 bits and ignores the HO bit).
Write Great Code www.nostarch.com Binary Arithmetic and Bit Operations 61 Figure 3-8: Social Security number packed fields encoding As you’ll soon see, fields that begin at bit position zero in a packed data object are the ones you can most efficiently access. So it’s generally a good idea to arrange the fields in your packed data type so that the field you access most often begins at bit zero. Of course, you’ll have to determine which field you access most often on an application-by-application basis. If you have no idea which field you’ll access most often, you should try to assign the fields so they begin on a byte boundary. If there are unused bits in your packed type, you should attempt to spread them throughout the structure so that individual fields begin on a byte boundary and have those fields consume multiples of eight bits. We’ve only got one unused bit in the Social Security example shown in Figure 3-8, but it turns out that we can use this extra bit to align two fields on a byte boundary and ensure that one of those fields occupies a bit string whose length is a multiple of eight bits. Consider Figure 3-9, which shows a rearranged version of our Social Security number data type.
Figure 3-9: A (possibly) improved encoding of the Social Security number One problem with the data format shown in Figure 3-9 is that we can’t sort Social Security numbers in an intuitive fashion by simply comparing 32-bit unsigned integers.5 Therefore, if you intend to do a lot of sorting based on the entire Social Security number, the format in Figure 3-8 is probably a better format.
If this type of sorting isn’t important to you, the format in Figure 3-9 has some advantages. This packed type actually uses eight bits (rather than seven) to represent SecondField (along with moving SecondField down to bit position zero); the extra bit will always contain zero. This means that SecondField consumes bits 0..7 (a whole byte) and ThirdField begins on a byte boundary (bit position eight). ThirdField doesn’t consume a multiple of eight bits, and FirstField doesn’t begin on a nice byte boundary, but we’ve done fairly well with this encoding, considering we only had one extra bit to play around with.
5 “Intuitive” meaning that the first field is the most significant portion of the value, the second field is the next most significant, and the third field is the least significant component of the number.
Third field 0000-9999 Second field 00-99 First field 000-999 31 20 0 12 31 21 Third field 0000-9999 Second field 00-99 First field 000-999 0 7 Write Great Code www.nostarch.com 62 Chapter 3 The next question is, “How do we access the fields of this packed type?” There are two separate activities here. We need the ability to retrieve, or extract, the packed fields, and we need to be able to insert data into these fields. The AND, OR, and SHIFT operations provide the tools for this.
When actually operating on these fields, it’s convenient to work with three separate variables rather than working directly with the packed data. For our Social Security number example, we can create the three variables FirstField, SecondField, and ThirdField. We can then extract the actual data from the packed value into these three variables, operate on these variables, and then insert the data from these three variables back into their fields when we’re done.
Extracting the SecondField data from the packed format shown in Figure 3-9 is easy (remember, the field aligned to bit zero in our packed data is the easiest one to access). All you have to do is copy the data from the packed representation to the SecondField variable and then mask out all but the SecondField bits using the AND operation. Because SecondField is a 7-bit value, we can create the mask as an integer containing all one bits in positions zero through six and zeros everywhere else. The following C/C++ code demon-strates how to extract this field into the SecondField variable (assuming packedValue is a variable holding the 32-bit packed Social Security number): SecondField = packedValue & 0x7f; // 0x7f = %0111_1111 Extracting fields that are not aligned at bit zero takes a little more work. Consider the ThirdField entry in Figure 3-9. We can mask out all the bits associated with the first and second fields by logically ANDing the packed value with %_11_1111_1111_1111_0000_0000 ($3F_FF00). However, this leaves the ThirdField value sitting in bits 8 through 21, which is not con-venient for various arithmetic operations. The solution is to shift the masked value down eight bits so that it’s aligned at bit zero in our working variable. The following Pascal/Delphi/Kylix code shows how one might do this: SecondField := (packedValue and $3fff00) shr 8; As it turns out, you can also do the shift first and then do the logical AND operation (though this requires a different mask, $11_1111_1111_1111 or $3FFF). Here’s the C/C++ code that extracts SecondField using that technique: SecondField = (packedValue >> 8) & 0x3FFF; Extracting a field that is aligned against the HO bit, as the first field is in our Social Security packed data type is almost as easy as accessing the data aligned at bit zero. All you have to do is shift the HO field down so that it’s aligned at bit zero. The logical shift right operation automatically fills in the HO bits of the result with zeros, so no explicit masking is necessary. The following Pascal/Delphi code demonstrates this: Write Great Code www.nostarch.com Binary Arithmetic and Bit Operations 63 FirstField := packedValue shr 18; // Delphi's shift right is a logical // shift right.
In HLA/x86 assembly language, it’s actually quite easy to access the second and third fields of the packed data format in Figure 3-9. This is because we can easily access data at any arbitrary byte boundary in memory. That allows us to treat both the second and third fields as though they both are aligned at bit zero in the data structure. In addition, because the SecondField value is an 8-bit value (with the HO bit always containing zero), it only takes a single machine instruction to unpack the data, as shown here: movzx( (type byte packedValue), eax ); This instruction fetches the first byte of packedValue (which is the LO 8 bits of packedValue on the 80x86), and it zero extends this value to 32 bits in EAX (movzx stands for “move with zero extension”). The EAX register, therefore, contains the SecondField value after this instruction executes.
Extracting the ThirdField value from our packed format isn’t quite as easy, because this field isn’t an even multiple of eight bits long. Therefore, we’ll still need a masking operation to clear the unused bits from the 32-bit result we produce. However, because ThirdField is aligned on a byte (8-bit) boundary in our packed structure, we’ll be able to avoid the shift operation that was necessary in the high-level code. Here’s the HLA/x86 assembly code that extracts the third field from our packedValue object: mov( (type word packedValue), ax ); // Extracts bytes 1 & 2 // from packedValue.
and( $3FFF, eax ); // Clears all the undesired bits.
Extracting FirstField from the packedValue object in HLA/x86 assembly code is identical to the high-level code; we’ll simply shift the upper ten bits (which comprise FirstField) down to bit zero: mov( packedValue, eax ); shr( 21, eax ); Inserting a field into a packed structure is only a little more complicated than extracting a field. Assuming the data you want to insert appears in some variable and contains zeros in the unused bits, inserting a field into a packed object requires three operations. First, if necessary, you shift the field’s data to the left so its alignment matches the corresponding field in the packed object. The second step is to clear the corresponding bits in the packed structure. The final step is to logically OR the shifted field into the packed object. Figure 3-10 on the next page displays the details of this operation.
Write Great Code www.nostarch.com 64 Chapter 3 Figure 3-10: Inserting ThirdField into the Social Security packed type Here’s the C/C++ code that accomplishes the operation shown in Figure 3-10: packedValue = (packedValue & 0xFFc000FF) | (ThirdField << 8 ); You’ll note that $FFC000FF is the hexadecimal value that corresponds to all zeros in bit positions 8 through 21 and ones everywhere else.
3.8 For More Information My book, The Art of Assembly Language, provides additional information on bit processing, including several algorithms for counting bits, reversing the bits in an object, merging two bit strings, coalescing sets of bits, and spreading bits out across some value. Please see that text for more details on these low-level bit operations. Donald Knuth’s The Art of Computer Programming, Volume Two: Seminumerical Algorithms provides a discussion of various arithmetic operations (addition, subtraction, multiplication, and division) that you may find of interest.
F F T T T T Step 1: Need to align the bits in the ThirdField variable to bit position eight F F T T T T T T T T T T S S S S S S S S t t t t t t t t t t t t t t F F F F F F Step 2: Need to mask out the corresponding bits in the packed structure t t t t t t t t t t t t t t F T T T T F F T T T T T T T T T T S S S S S S S S F F F F F F F F F F F F F F F F F 0 0 0 0 0 0 0 0 0 0 0 0 0 0 S S S S S S S S t t t t t t t t t t t t t t Step 3: Need to logically OR the two values to produce the final result F F F F F F F F F F S S S S S S S S t t t t t t t t t t t t t t Final Result Write Great Code www.nostarch.com |
187954 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10016947/ | The Spectre of Berkson's Paradox: Collider Bias in Covid-19 Research - PMC
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. 2020 Jul 29;17(4):6–7. doi: 10.1111/1740-9713.01413
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The Spectre of Berkson's Paradox: Collider Bias in Covid-19 Research
Annie Herbert
Annie Herbert
1 Annie Herbert is a Senior research associate at the MRC Integrative Epidemiology Unit, Bristol Medical School, University of Bristol
Find articles by Annie Herbert
1, Gareth Griffith
Gareth Griffith
2 Gareth Griffith is an ESRC postdoctoral fellow at the MRC Integrative Epidemiology Unit, Bristol Medical School, University of Bristol
Find articles by Gareth Griffith
2, Gibran Hemani
Gibran Hemani
3 Gibran Hemani is a Senior research fellow at the MRC Integrative Epidemiology Unit, Bristol Medical School, University of Bristol
Find articles by Gibran Hemani
3, Luisa Zuccolo
Luisa Zuccolo
4 Luisa Zuccolo is a Senior lecturer in epidemiology at the MRC Integrative Epidemiology Unit, Bristol Medical School, University of Bristol
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4
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1 Annie Herbert is a Senior research associate at the MRC Integrative Epidemiology Unit, Bristol Medical School, University of Bristol
2 Gareth Griffith is an ESRC postdoctoral fellow at the MRC Integrative Epidemiology Unit, Bristol Medical School, University of Bristol
3 Gibran Hemani is a Senior research fellow at the MRC Integrative Epidemiology Unit, Bristol Medical School, University of Bristol
4 Luisa Zuccolo is a Senior lecturer in epidemiology at the MRC Integrative Epidemiology Unit, Bristol Medical School, University of Bristol
Collection date 2020 Aug.
© 2020 The Royal Statistical Society
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This article is made available via the PMC Open Access Subset for unrestricted re-use and analyses in any form or by any means with acknowledgement of the original source. These permissions are granted for the duration of the COVID-19 pandemic or until permissions are revoked in writing. Upon expiration of these permissions, PMC is granted a perpetual license to make this article available via PMC and Europe PMC, consistent with existing copyright protections.
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PMCID: PMC10016947 PMID: 37250182
Abstract
When non-random sampling collides with our understanding of Covid-19 risk, we must be careful not to draw incorrect conclusions about cause and effect. By Annie Herbert, Gareth Griffith, Gibran Hemani and Luisa Zuccolo
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There have been many surprising things written and said about the coronavirus pandemic, but perhaps none more so than the claim that smoking might protect against Covid-19 infection (bit.ly/2YLudbR).
The claim originated early in the pandemic and was greeted with disbelief. But as New Scientist explained in a 19 May article (bit.ly/3fMDZQB), “data emerging from the countries first hit by coronavirus gave doctors pause: the proportion of smokers among those being hospitalised for covid-19 was lower than in the general population. In China, for example, about 8 per cent of people in hospital with covid-19 were smokers, while 26 per cent of the general population smoke. The equivalent figures for Italy are 8 and 19 per cent respectively.”
Observational evidence such as this can and must inform the prioritisation of avenues for research. However, while attempting to understand such observations, researchers must be aware of non-causal as well as causal explanations for observed associations. Specifically, a non-causal but credible explanation for the association between smoking and Covid-19 is that it is due to a form of bias known as “Berkson's paradox”, after the US statistician Joseph Berkson (1899–1982). For reasons we explain, it is now often called “collider bias”.
Understanding collider bias
To explain collider bias, consider the following example. In middle age, obesity generally increases with age. However, if we evaluate the cross-sectional relationship between obesity and age among middle-aged individuals tested for Covid-19, we might tempt you to believe that there is a negative correlation: that is, you have a slimmer waistline to look forward to as you get older.
But you would be wrong to conclude this. It is an inferential error arising from the fact that being older or being more obese both influence your likelihood of being tested for Covid-19, because testing in the UK has largely been confined to those presenting at hospital with severe symptoms of the disease. This can be seen in UK Biobank participants (aged 40–69) in Figure 1. In the entire subsample of those tested for Covid-19, those who are more obese will appear to have a lower age, and vice versa, inducing a negative correlation.
FIGURE 1.
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The cross-sectional relationship between z-scores of age and obesity score in the general population (red, r = 0.02, N = 17,613) and in the subsample of individuals tested for Covid-19 (blue, r = –0.12, N = 5,871). Data from the UK Biobank.
In general, if two factors influence being selected into a sample, we say that they “collide on selection” (see Figure 2a). Hence the name “collider bias”. The impact of this could range
FIGURE 2.
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Arrows represent causal effects, and dotted lines represent induced relationships between risk factor (yellow) and outcome (blue) when conditioning on sample selection (red). (a) The association in Figure 1 can be induced if both age and obesity influence the probability of being tested. (b) Sample selection in hospitals is influenced by different factors than those in (a), which means that associations may not be consistent between different study designs.
Like other threats to causal inference, once you become aware of collider bias, you see it lurking everywhere from introducing an association when the two factors were in fact independent, to reducing, exaggerating or even reversing an existing association in ways which are hard to anticipate.
In a recent paper,1 we summarise how collider bias might play a role in Covid-19 studies, and identify hundreds of demographic, genetic and health-related factors all influencing the probability of a person being selected for testing for Covid-19 among UK Biobank participants. But simply having the symptoms of Covid-19 infection also increases the chances of being selected for testing. So all these health-related factors and infection will collide on selection, increasing the risk of associations being distorted by Berkson's paradox.
Like other threats to causal inference (e.g. a third factor explaining a link between the exposure and outcome), once you become aware of the collider bias mechanism, you see it lurking everywhere. This includes studies of hospitalised patients, for example. While hospitalised patients are an obvious source of Covid-19 data, they are not a random sample of the general population, but instead represent individuals who are effectively selected from that population on the grounds of being older, frail with health issues, smokers, and – of course – those suffering from Covid-19.
To see how this could lead to a misleading link between smoking and protection against Covid-19, imagine – for simplicity – that patients are admitted to hospital for just two reasons: smoking-related illness or Covid-19 (Figure 2b). Covid-19 tests on these hospitalised individuals are likely to show lower infection rates among smokers than among non-smokers, because the former are also hospitalised for smoking-related illness, not necessarily Covid-19. This could explain the reports from several studies claiming that smoking might protect against Covid-19 infection.
Protecting against collider bias
To avoid drawing incorrect conclusions from studies that might be affected by collider bias, there are a few things to watch out for.
When reading about the selection or sampling of participants in a study of Covid-19 risk factors, we should look out for the terms “Berkson's”, “collider”, “selection” and “random”. Ideally, investigators would randomly sample from the target population, as this aims to ensure no variables can influence selection and introduce collider bias. Failing this, if the study attempts to recruit individuals who are representative of the target population, or at least uses a sample matching recruitment design, we should not need to worry too much about collider bias. However, selection processes are often not clear from the outset of a study, so even when sampling is representative, it is good practice for studies to report sensitivity analyses exploring selection bias (not only in specific studies, but also in subsequent systematic reviews). This is particularly important where causal inference is the focus of the work.2 Sensitivity analyses may include probability weighting of the samples, or an evaluation of the extent to which collider bias could explain a study's findings.1 But if a scientific paper simply states “collider bias” as a potential limitation, or does not mention it at all, be very cautious.
There is clearly a place for responsive, time-sensitive analyses of convenience samples – like those of people tested for, or hospitalised with, Covid-19. However, we fail to do justice to those data collection efforts if we do not investigate biases that could lead to faulty conclusions.
Contributor Information
Annie Herbert, Annie Herbert is a Senior research associate at the MRC Integrative Epidemiology Unit, Bristol Medical School, University of Bristol.
Gareth Griffith, Gareth Griffith is an ESRC postdoctoral fellow at the MRC Integrative Epidemiology Unit, Bristol Medical School, University of Bristol.
Gibran Hemani, Gibran Hemani is a Senior research fellow at the MRC Integrative Epidemiology Unit, Bristol Medical School, University of Bristol.
Luisa Zuccolo, Luisa Zuccolo is a Senior lecturer in epidemiology at the MRC Integrative Epidemiology Unit, Bristol Medical School, University of Bristol.
References
Griffith, G., Morris, T. M., Tudball, M.. et al. (2020) Collider bias undermines our understanding of COVID-19 disease risk and severity. Epidemiology. doi: 10.1101/2020.05.04.20090506. [DOI] [PMC free article] [PubMed]
Greenland, S. (2003) Quantifying biases in causal models: Classical confounding vs collider-stratification bias. Epidemiology, 14, 300–306. [PubMed] [Google Scholar]
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187955 | https://www.uvm.edu/~cclandry/chem36/Laboratory_files/lab%206.pdf | Synthesis of a Cobalt Complex Lab #6, Chem 36 Spring 2009 -1- Introduction The most extensively studied class of octahedral transition metal compounds are cobalt(III) complexes in which ammonia (or other neutral molecules, closely related to ammonia, called amines) occupy some or all of the six coordination positions. The (III) in the name is a way of indicating the +3 oxidation state of the Co3+ ion. These complexes played a decisive role in early formulations of the structure of transition metal compounds and they continue to be important model systems for contemporary research into the properties of complex ions. The first and simplest cobalt ammine complex ion, [Co(NH3)6]3+, was prepared in 1798. Alfred Werner, a German chemist, studied the cobalt ammines extensively in the late 19th and early 20th centuries. He correctly interpreted his observations as requiring an octahedral geometry of the ligands about the metal. Modern transition metal chemistry has evolved from his work for which he was awarded the Nobel Prize in chemistry in 1913. The intensity of ongoing research interest in cobalt ammine complexes is measured by the fact that a recent Chemical Abstracts cumulative index to the chemical literature has about 5000 entries referring to articles on the subject over a five year period. This laboratory experiment involves the preparation of aquapentaammine-cobalt(III) as a nitrate salt, [Co(NH3)5(H2O)] [NO3]3: Note the spelling of the complex name. There are molecules which, as a class, are called amines, but the ammonia as a ligand is called ammine in the chemical's name. Water as a ligand is called aqua (formerly aquo) in the name. Once a successful synthesis has been carried out, a number of reactions of the complex will be explored that will establish the purity of the product and characterize some of its chemical behavior in weeks two, three, and four. Complexes of amines with cobalt(III) are nearly always prepared from a cobalt(II) salt, the amine, and a reagent which will convert cobalt(II) to cobalt(III). The procedure used here is typical, with hydrogen peroxide serving as the reagent (called an "oxidizing agent" for its ability to remove an electron) and ammonia as the amine. Here is the stoichiometric net reaction for this synthesis: 2 HNO3 + 2 [Co(H2O)6] [NO3]2(s) + H2O2 + 10 NH3 → 2 [Co(NH3)5(H2O)] [NO3]3(s) + 12 H2O The oxidation-reduction half-reactions consist of the oxidation of cobalt (II) to cobalt (III) and the reduction of the hydrogen peroxide: 2 Co+2 → 2 Co+3 + 2 e- 2 H+ + H2O2 + 2 e- → 2 H2O The purpose of each reagent in the mixture is described below. About the Reagents Cobalt Nitrate (Co(H2O)62 or Co(NO3)2•6H2O) is a typical hydrated cobalt(II) salt which consists of octahedral Co(H2O)6 2+ cations and nitrate anions in the solid. It is extremely soluble in water and the solubility increases rapidly with increasing temperature. Ammonia is an aqueous solution of NH3 gas (sometimes laboratory bottles bear the old-fashioned label "ammonium hydroxide" or NH4OH). The solution is basic due to the following equilibrium: NH3(aq) + H2O → NH4 + + OH- The 6 M solutions used here should be treated with respect. Aqueous ammonia is quite volatile, losing ammonia gas. In low concentrations, exposure to the gas causes irritation to the eyes and throat. At higher concentrations it is extremely toxic. Although the 6 M reagent does not present a serious hazard, ammonia solutions should be covered or kept in the hood throughout the procedure. Dilute ammonia solutions are familiar as household cleaning agents. They derive their effectiveness from the degradative action of basic solutions on natural fats. Ammonia provides the ammine ligands for your complex. Ammonium nitrate (NH4NO3) is a common salt that requires no special precautions in standard laboratory applications. It is a major product of the chemical industry, largely because NH3 Co NH3 OH2 H3N NH3 H3N 3 NO3-+3 Synthesis of a Cobalt Complex Lab #6, Chem 36 Spring 2009 -2- of its use as a source of nitrogen in fertilizers. When exposed to extreme heat and/or pressure it is explosive. Several major industrial disasters have occurred when fires have initiated the explosion of large stocks of ammonium nitrate. The truck bomb of the Oklahoma City bombing was filled ammonium nitrate. Ammonium nitrate is added to the reaction mixture to provide nitrate ions and to increase the concentration of NH3 (aq) in solution. According to LeChatelier's Principle, addition of NH4 + ions will push the equilibrium between NH3 and NH4 + to the reactant, or NH3, side of the above equation. Hydrogen peroxide (H2O2) is another ubiquitous chemical. Its commonplace applications include use as a disinfectant and as a bleaching agent. Industrially it is widely used for bleaching, particularly in the processing of paper pulp. Although concentrated solutions, such as 30% by weight, cause severe burns, the 3% solutions used here do not. The concentration of 3% H2O2 is 0.9 M. (% concentrations are almost always expressed as weight percents.) In your synthesis reaction, hydrogen peroxide is the oxidizing agent that oxidizes the cobalt (II) reactant to the cobalt (III) product. Concentrated nitric acid (HNO3, 16 M) is the most hazardous reagent used in this experiment. It is a typical strong acid in its capacity to burn skin and other tissue. It has the added peculiarity of turning exposed skin yellow. Gloves are recommended, although they provide only temporary protection from this strong acid--be careful! Although pure nitric acid solutions are colorless, the laboratory reagent is sometimes significantly discolored due to the presence of NO2 and N2O4, which are decomposition products of nitric acid when exposed to light. Nitric acid provides the H+ that appears in the reduction half-reaction. Ethyl alcohol (ethanol, C2H5OH) should require no introduction. A 95% ethanol-5% water solution is the standard laboratory form. It is that mixture of water and ethanol that has a boiling point, (78.15 °C), which is lower than that of pure ethanol (78.3 °C), pure water (100 °C), or any other mixture. The residual water, therefore, cannot be removed by simple distillation. The solutions are quite volatile and easily ignited. Exposure to open flame should be avoided. Alcohol is included in the reaction mixture to allow the product to precipitate. The product complex is soluble in water, but insoluble in ethanol. Waste Disposal All reaction mixtures and cobalt-containing solutions should be placed in the labelled waste drums. Unused acid and base solutions can be rinsed down the drain with lots of water. Procedure Refer to the stoichiometric net reaction as you first read through this procedure. You will be asked to calculate a final percent of theoretical yield (percent yield, for short) for your product. Thus, you must know the initial numbers of moles of all the reactants involved, to at least two significant figure accuracy. Keep this in mind as you weigh solids or measure solutions. Solutions of reagents will be available in the lab in repipets set to deliver the appropriate amount into separate, clean beakers. Put a watch glass over the top of the beakers and take them to your hood. At the appropriate point in the procedure, add each reagent to your synthesis beaker, in your hood, not at the repipet. Do not remeasure the volume of any solution delivered by repipet, since greater accuracy is not necessary. Rinse the ammonia beaker with water as soon as you add it to your reaction beaker. Keep your reaction beaker in the hood at all times, to minimize fumes. Synthesis Procedure Weigh 3.6 g (0.012 mol) Co(NO3)2•6H2O and 2.5 g (0.031 mol) NH4NO3 on an electronic top-loading balance (not on an analytical balance). Record the weights and place these solids in a 400 mL beaker (or 500 mL Erlenmeyer). Add no more than 10 mL hot water from a hot water bath (dippers made from 10-mL beakers will be provided) and swirl until the solids are dissolved. Place the beaker on a stirring plate in the hood and add 40 mL 6 M ammonia (0.24 mol). Over a period of about 30 min add with stirring 25 mL (0.022 mol) 3% H2O2 in small (0.8 mL/min.) amounts. Best yields will be achieved if the 3% H2O2 is added via an eyedropper while stirring the solution. Addition can be more rapid at first than towards the end. Add a total of 15 mL during the first 15 minute period and a total of 10 mL during the final 15 minutes After addition is complete, allow the solution to sit with occasional stirring for 10 minutes or until bubbling stops, whichever comes first. The slow addition of H2O2 Synthesis of a Cobalt Complex Lab #6, Chem 36 Spring 2009 -3- minimizes the extent of a non-productive side reaction that hydrogen peroxide undergoes by not allowing large concentrations of peroxide to build up. This side reaction causes oxygen gas to bubble out of the solution especially when relatively little unreacted cobalt(II) remains. It is the common mode of peroxide self-decomposition: 2 H2O2 → 2 H2O + O2 (g) Slowly and carefully, add 30 mL (0.48 mol) 16 M HNO3. Let the solution cool for 10 minutes. Add a volume of 95% ethyl alcohol (also called ethanol and abbreviated EtOH) approximately equal to the volume of solution already in the large beaker. The precipitate should become visible at this stage.1 If it does not, notify your TA and review your procedure to this point. Allow the mixture to sit for 10 minutes to allow time for precipitation to proceed and then collect the solid by vacuum filtration on a Büchner funnel, as described in the Techniques section. Rinse out solid remaining in the beaker with ethanol and twice rinse the solid on the filter. Pass air through the solid by suction for a few minutes to remove most of the ethanol. After filtration, the solid should be first weighed and then recrystallized, following the procedure described below. Procedure for the Recrystallization of Aquapentaammine Cobalt(III) Nitrate The main impurity in the initial preparation is the nitrate, Co(NH3)5NO32, rather than the desired aqua complex, Co(NH3)5H2O3. To remove this impurity, the sample is first dissolved in a warm basic solution. The nitrate complex is rapidly converted to the hydroxo complex: [Co(NH3)5NO3]2+ + OH- → [Co(NH3)5OH]2+ + NO3 - Then the solution is made acidic. [Co(NH3)5OH]2+, which is the conjugate base of the weak acid [Co(NH3)5H2O]3+, adds a proton to give the aqua complex which is then precipitated as the nitrate salt: [Co(NH3)5OH]2+ + H+ → [Co(NH3)5H2O]3+ [Co(NH3)5H2O]3+ + 3 NO3 - → Co(NH3)5H2O3 (s) Prepare first a 1 M ammonia solution by diluting the 6 M solution available in the lab. You will need at least 20 mL of this solution per gram of crude product. (If you are uncertain how to make this dilution, check with your TA.) Weigh the sample of crude [Co(NH3)5 (H2O)][NO3]3 (molecular mass = 348.1 g/mol) on an electronic top-loading balance, transfer it to a small (30 mL or 50 mL, if crude weight is >1g) beaker, and dissolve it in 1 M ammonia, using a maximum of 20 mL of ammonia per gram of crude product. Place the solution in a hot water bath and stir with a clean glass rod until all the solid is dissolved. Keep the solution in the hot bath for a full five minutes. (The baths will be adjusted to be close to 80 °C. Note the actual temperature, but do not allow the temperature to rise above 85 °C.) Next cool the solution in an ice bath (take care that your beaker does not tip over as the ice melts!) until the temperature is below 5 °C as measured by your thermometer (with the bulb fully immersed in the solution). When the solution reaches 5 °C, add 16 M nitric acid dropwise while the solution remains in the ice bath, in the hood. A precipitate should form when sufficient HNO3 has been added to neutralize the amount of ammonia used to dissolve the crude product. Add a ~25% excess of nitric acid with an eye dropper, stir, and allow a few minutes for precipitation to proceed in the ice bath. Test a drop of the mixture with pH paper, to confirm a strongly acidic pH. The required amount of nitric acid is easily calculated. Suppose that 3 g of crude product have been dissolved in 60 mL of 1 M NH3. The required volume of 16 M HNO3 is given by the following calculation: V HNO3 = (mmoles NH3 present)/(conc. HNO3) = (60 mL) (1 M)/16 M = 3.8 mL For a 25% excess, (1.25)(3.8 mL) = 4.8 mL. Thus about 5 mL would be required. Precipitation may be aided by scratching the inside of the beaker with a stirring rod or by adding a "seed" crystal of impure material. Collect the solid using vacuum filtration. With the wet solid still on the filter paper in the funnel, rinse twice with 95% ethanol. The ethanol rinses remove residual solution that contains ammonium nitrate from the solid. Ammonium nitrate is reasonably soluble in Synthesis of a Cobalt Complex Lab #6, Chem 36 Spring 2009 -4- ethanol and the complex ion salt is not soluble in ethanol. Ethanol is miscible with water and therefore serves to remove residual aqueous solution. Allow air to pass through the solid on the filter for a few minutes, then carefully transfer the filter and solid, to a watchglass to dry. After you have cleaned up your glassware, return to your product and scrape the solid off the filter paper and onto a piece of folded weighing paper. Transfer the solid from the paper into a clean, tared glass vial and obtain the mass of the product. After Lab The Data and Results section should include a calculation of the theoretical yield of the preparation according to the net reaction stoichiometry and the molar quantities of the different reagents used. The calculation should clearly show how the reagent that limits the overall yield was identified. The yield of product should be calculated as a percentage of the theoretical yield. Relative yields are very low in this experiment, largely because of the substantial solubility of the product. The preparation should be considered a success if 0.5 g or more of good quality recrystallized product is obtained. In this week's data sheet, you will report the crude yield of product, before recrystallization and drying, and the yield of recrystallized product. Include a description of each reagent used in the procedure in the synthesis reaction (acid-base, redox, common ion effect, precipitation, etc.). Relate this chemistry to your observations for the reaction. Include a description of each reagent used in the procedure and its purpose in the reaction. |
187956 | https://www.youtube.com/watch?v=kWQAse8dxog | continuity of piecewise functions (5 examples!)
vinteachesmath
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Posted: 31 Jul 2021
In this video, I go through 5 examples showing how to determine if a piecewise function is continuous. For each of the 5 calculus questions, I show a step by step approach for determining continuity of piecewise functions. The questions used in this video involve limits, piecewise functions, linear functions, trig functions, rational functions and polynomial functions.
If this video was helpful, please LIKE and SUBSCRIBE. If you have any requests, leave the topics you want me to cover in the COMMENT section. Thanks for watching!
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Transcript:
First example what's up i'm vin and today i want to show how to determine continuity of piecewise functions so we're going to go through these five questions here and let's get started so for these questions we have the formal definition of continuity but this is a nice step-by-step approach to answering these questions and what we want to do is determine if f of x is continuous at x equals three now the first thing we should try to do is show that the limit exists so if we're trying to show that the limit as x approaches three of f of x exists we're going to target the top portion of this piecewise function where x is not equal to 3 because when x is not equal to 3 we could approach it from the left and right side now you'll notice that quadratic on top you have x squared minus 4x plus 3. that preemptively i just want to go ahead and factor as x minus 3 times x minus 1. and now this is all being divided by x minus 3. because if we wanted to evaluate this limit in the original form we're going to get 0 over 0 which is indeterminate so we'd have to do this algebra anyway so now we cancel out the matching factors and when we plug in we have x minus one so we're replacing x with three we have three minus one which is equal to two so the limit does exist so this is a good start for us showing this function is continuous at x equals 3. but now we want to show the function is defined at x equals 3 and if we investigate f of 3 this is just the point this is the point 3 comma 2. so the function value of 3 is equal to 2. so this is also checking out here and now the last thing that we have to be able to show is that the limit and the function are equal in this case at x equals 3 you notice the limit was equal to 2 when the function value is equal to 2. so we can say that this function is continuous at x equals three so we went ahead and showed the last step here now just for a little bit more insight as to like what actually happened and if we look at a little sketch of this this will help us understand a little bit more notice all we were left with was x minus one and if we provide a little sketch of this x minus one is a line and it looks rough sketch something like this so this is at the point 1 and then 2 is out here 3 is out here so what we just found here was at x equals 2 this would go up to 1 and at x equals 3 this is going up to 2. so temporarily because x is not equal to 3 this is a line with a hole in it here but notice that the point 3 2 goes exactly where the hole in the line is going ahead and like filling this gap making this a continuous function if this was any other number but 2 then the function would be not continuous at x equals three but this did work out for us so this function is in fact continuous at x equals three Second example so for this next example we'll start the same way we wanna show that the limit exists but this time we wanna see if g of x is continuous at x equals 1. so if we take the limit as x approaches one of g of x we're going to factor this quadratic expression and it would factor as x plus 1 times x minus 1 and we have an x minus 1 on bottom so when these cancel out we're left with x plus 1 so when we plug in x equals 1 we have 1 plus 1 is equal to 2. so the limit does exist but notice this time around if we look at the function value and now this is for g of x the function value at 1 is equal to 4. so right away i would say this is not continuous at x equals 1 because the limit and the function are not equal so this last step is not satisfied if we want to investigate this a little bit more we could just look at a rough sketch of this and see that if we look at the left over here the line x plus one so we'll just get a little sketch here so these would be the intercepts so once again this is just a rough sketch but notice everything that's going on is at x equals one that the limit is equal to two but the function value was equal to four so this is a situation here when we once again have a line with a hole in it but the function value here is up top so this would be a removable discontinuity if this y value here was equal to 2 then this function would be continuous but since they're not equal we can say this function is not continuous at x equals 1. so for this type of function we want to show the same thing that the limit exists the function is defined and show that those two things are equal but this time around we have a slightly different piecewise function to show that the limit exists it's a little bit different now we have to investigate what is the left side limit of h of x and what is the right side limit of h of x so when we do this the thought process that we should be going through is we have to associate left side with less than so what i mean by this is if i have some number line and the target value is x equals two when i think of the phrase two from the left i associate that with x less than two because when i'm on the left side of 2 i'm looking at values less than 2. and when i'm looking at 2 from the right side i associate that with x greater than 2 because values to the right of 2 are greater than 2. right the only thing different here is i would just throw in this less than or equal to to match the actual question but now how does that help us well when we want to find the limit as x approaches 2 from the left of h of x we have to look at the x less than or equal to 2 piece and instead of doing four times x minus two we're going to replace that x with two and we have four times two is eight minus two is equal to six and then when i investigate the limit on the right side of two the right side of two two plus i'm looking at x greater than 2 which is the top part of this piecewise function so that's when i use the 3x plus 1 so i'm replacing that x value with 2 and this is going to give us 7. so notice right away the left and right limits are not equal so since the left and right limits are not equal this tells us the limit doesn't exist and if the limit doesn't exist this tells us the function is not continuous at x equals 2. i should just mention something real quick these three dots mean therefore it's just a nice abbreviation when you're writing explanations in math so for the fourth question we're going Fourth example to start the same way but this one has a little bit of a trap sprinkled into it that when i look at this right away i automatically say that this function is not continuous at x equals one and the fast way i could say it is that this function is not defined at x equals one notice here that p of one is undefined because for this piecewise function we only have these intervals here for x greater than one x less than one but notice none of them include an equal to so since p of one is undefined this tells us we could say and therefore p of x is not continuous at x equals one okay so last question here we're looking at the function q of x the first thing we want to show is that the limit exists so notice this is broken into three pieces we have x less than zero x equals zero and x greater than zero so if we want to show that this limit exists at zero first thing we can look for is what is the limit as x approaches zero from the left of q of x so if i'm looking for the left side limit so we're looking at zero from the left that means i'm using the x less than zero so when i evaluate this this is the limit as x approaches zero from the left and we're using the sine x over x piece now this is a famous limit the limit as x approaches zero of sine x over x is equal to one if you forget this you could always use l'hopital's rule if you've learned that part of derivatives yet but otherwise we could say this limit is equal to one so now we wanna find what is the limit as x approaches 0 from the right side of q of x and for this what we're going to do is we're going to use x greater than 0 for 0 from the right so we're using this piece here so what we could do is we're going to take the limit as x approaches 0 from the right and we're going to take this expression here and factor it on top we have 4x squared plus 2x so we have a greatest common factor of 2x and we would be left with 2x plus 1. and then on bottom here we'd be left with 2x so notice 2x over 2x cancels out and then now when i plug in 0 i have 2 times 0 plus 1 which is equal to 1. so what conclusion we can draw from this is that since the left and right side limits are equal this tells us that the limit as x approaches zero is equal to one so now that we've shown that the limit exists we can investigate the function value so we have the limit as x approaches zero if q of x is equal to one and now we could find q of 0 is equal to 1. so what this tells us we have the limit exists the function is defined and the limit which is equal to 1 is equal to the function value which is equal to 1. so this tells us that q of x is in fact continuous at x equals zero okay well this is going to conclude this video on determining continuity of piecewise functions if you found this video to be helpful please like and subscribe it really helps me grow the channel and if you've got any requests just leave the topics you want me to cover in the comments section below and thanks for watching |
187957 | https://math.stackexchange.com/questions/1620598/projection-of-u-onto-v-and-v-onto-u | Stack Exchange Network
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Projection of u onto v and v onto u
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Given the vector u = $<-2, 6, 4>$ and a vector v such that the vector projection of u onto v is $<2,4,4>$, and the vector projection of v onto u is $<8,24,16>$. What is the vector v?
I tried it like this but can't reach to final answer.
calculus
vectors
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asked Jan 21, 2016 at 4:25
AkiraAkira
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$\begingroup$ You should use the command \langle and rangle to correctly type $\langle, \rangle$, although it seems rare to use these for vectors. $\endgroup$
Zhanxiong
– Zhanxiong
2016-01-21 04:29:21 +00:00
Commented Jan 21, 2016 at 4:29
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The projection of $\def\v#1{{\bf#1}}\v u$ onto $\v v$ is a scalar times $\v v$. So from the given information we have $$\v v=\lambda(2,4,4)\ .$$ Hence $${\rm proj}_{\v u}\v v=\frac{\v v\cdot\v u}{\v u\cdot\v u}\,\v u =\lambda\frac{9}{14}(-2,6,4)\ .$$ It is given that this projection is $(-8,24,16)$, so $\lambda=\frac{56}9$ and hence $$\v v=\tfrac{56}9(2,4,4)\ .$$
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answered Jan 21, 2016 at 4:33
DavidDavid
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Here is a more tangent way to solve this problem:
Notice $v$ and the projection of $u$ onto $v$ must have the same direction, therefore we can assume $$v = \lambda(2, 4, 4),$$ in which $\lambda$ is a constant to be determined. Now use the other condition to establish the equation $$(-8, 24, 16) = \frac{\langle v, u \rangle}{\langle u, u \rangle}u = \frac{36\lambda}{56}(-2, 6, 4).$$ Solve this to get $\lambda = \frac{56}{9}$. Therefore, $$v = \frac{56}{9}(2, 4, 4).$$
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answered Jan 21, 2016 at 4:37
ZhanxiongZhanxiong
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FAQs: Standing Repo Facility
June 25, 2025
The following frequently asked questions (FAQs) provide further information about the Federal Reserve's Standing Repo Facility(SRF)operations.
What are SRF operations?
At its July 2021 meeting, the Federal Open Market Committee (FOMC) established the SRF to serve as a backstop in money markets to support the effective implementation and transmission of monetary policy and smooth market functioning. The FOMC has authorized and directed the New York Fed’s Open Market Trading Desk (the Desk) to conduct overnight repurchase agreement (repo) operations with a specified minimum bid rate and aggregate operation limit. When the Federal Reserve enters into an overnight repo transaction, it buys a security from an eligible counterparty and simultaneously agrees to sell the security back the next day. The difference between the purchase price and the sale price of the securities implies a rate of interest earned by the Federal Reserve on the transaction. The FOMC sets the SRF minimum bid rate, which is the minimum interest rate the Federal Reserve is willing to receive in an SRF operation.
What are the SRF operational parameters?
Standing Repo Facility Parameters
Operation Type:Morning Afternoon
Schedule:Every business day from 8:15 to 8:30 a.m. ET, unless otherwise stated Every business day from 1:30 to 1:45 p.m. ET, unless otherwise stated
Aggregate operation limit:$500 billion$500 billion less the accepted amount from the morning operation
Proposition limit:Two propositions per eligible security type per auction, subject to a $20 billion per proposition limit
Minimum bid rate:4.50 percent
Settlement:Same-day settled. Bank of New York (BNY) tri-party repo conventions apply. Funds will typically be delivered within thirty minutes following the operation close time.Same-day settled. BNY tri-party repo conventions apply.
Eligible counterparties:Primary dealers and SRF counterparties, which include depository institutions
Eligible securities:U.S. Treasuries, agency debt, and agency mortgage-backed securities (for additional details on the security types see Repo Securities Schedule)
How will SRF operations be conducted?
The Desk conducts SRF operations via FedTrade, the Desk's proprietary trading system, using a multiple-price auction format.
How will changes to SRF operations be communicated?
The FOMC directs the Desk to conduct SRF operations as detailed in its Continuing Directive for Domestic Open Market Operations and implementation note. Any changes to the operational parameters not provided by the FOMC’sContinuing Directive for Domestic Open Market Operations and implementation note will be announced on the New York Fed’s website.
How is the minimum bid rate for SRF operations determined?
The minimum bid rate for SRF operations is set by the FOMC and detailed in the most recent implementation note.
Why are there two SRF operations each business day?
Twice daily SRF operations are intended to support the effectiveness of the SRF in meeting its monetary policy implementation and market functioning objectives. The morning SRF operation allows for execution and settlement of SRF trades earlier in the business day, which may better align with the intraday timing of funding needs for some counterparties. The afternoon SRF operation allows counterparties to use the facility later in the day to meet funding needs.
How will the $500 billion SRF aggregate operation limit be allocated between the morning SRF operation and afternoon SRF operation?
The current aggregate SRF operation limit of $500 billion will apply to the combined daily operations. Specifically, the morning SRF operation will have an aggregate operation limit of $500 billion, and the afternoon SRF operation will be sized to $500 billion less the total amount of accepted propositions from the same day’s morning SRF operation. The Desk intends to announce the aggregate limit of each business day’s afternoon SRF operation on the Repo Operations webpage shortly after the morning SRF operation has concluded.
How are propositions submitted in SRF operations?
The minimum proposition size is $1 million, and propositions must be submitted in $1 million increments. Each counterparty is permitted to submit up to two propositions per security type that are at the minimum bid rate or higher and do not exceed the specified proposition limit.
How are awards determined in each SRF operation?
The Desk conducts each SRF operation in a multiple-price auction format, in which allocations are made to participants at the rate they bid in the operation. Bid rates can be at or above the minimum bid rate. Individual awards are made at the rates received by the Desk and awards are allocated in $1 million increments.
If the total amount of propositions received is less than or equal to the aggregate operation limit, all counterparties will be awarded at their submitted rates for their full submitted amounts (up to the per bid maximum amount).
If the total amount of propositions received is greater than the aggregate operation limit, individual propositions reflecting the most competitive rates relative to the benchmark rates set internally for each security type will be awarded at their submitted rates until the aggregate operation limit is reached. After that, individual propositions may either be partially awarded or not awarded based on their proximity to those benchmark rates for each security type.
Whom should SRF counterparties call if they experience difficulties during a FedTrade operation?
Counterparties may call the Desk with submission and verification questions. For system-related problems, primary dealers may call primary dealer support.
How are trades resulting from SRF operations settled?
In the tri-party repo market, trades are settled on the books of a clearing bank.Trades resulting from SRF operations are cleared and settled on the tri-party repo platform operated by BNY. BNY acts as an agent to the Desk and the Desk’s counterparty by taking custody of securities,valuing these securities and ensuring the appropriate margin is applied, and settling the transaction on its books.
Trades resulting from SRF operations settle on a same-day basis. Trades resulting from the daily morning SRF operation will generally be available for counterparties to settle within thirty minutes of the operation close time, as the Desk will make best efforts to position necessary funds on the BNY tri-party repo platform within that time frame. Trades resulting from the daily afternoon SRF auction will settle by the tri-party repo market close, unless otherwise stated by the Desk. Cash settlement amounts arising from SRF trades each day are netted against maturing trades. The repurchase of securities by the Desk on the maturity date will occur following the daily afternoon SRF operation by the tri-party repo market close.
How will the Desk communicate SRF operation results?
After the completion of an SRF operation, the Desk publishes a summary of the results that provides the total amount submitted, total amount accepted, and weighted-average award rate by security type, as well as the stop-out rate, high rate, and low rate of propositions for each security type.
Section 11 of the Federal Reserve Act, as amended by section 1103 of the Dodd-Frank Wall Street Reform and Consumer Protection Act of 2010, requires thatdetailed operational results, including counterparty names, be released two years after each quarterly transaction period.
FAQs: May 28, 2025 »
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Zinc Protoporphyrin
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Also Known As
ZPP
ZP
Free Erythrocyte Protoporphyrin
FEP
Formal Name
Zinc Protoporphyrin
This article was last reviewed on
This article waslast modified on 25 July 2023.
At a Glance
Why Get Tested?
To screen for and monitor chronic exposure to lead in industry; as an aid in the diagnosis of iron deficiency in children
When To Get Tested?
When you have been chronically exposed to lead in your working environment, or when your doctor suspects lead poisoning
Sample Required?
An anti-coagulated blood sample taken from a vein in your arm
Test Preparation Needed?
No test preparation is needed.
On average it takes 7 working days for the blood test results to come back from the hospital, depending on the exact tests requested. Some specialist test results may take longer, if samples have to be sent to a reference (specialist) laboratory. The X-ray & scan results may take longer. If you are registered to use the online services of your local practice, you may be able to access your results online. Your GP practice will be able to provide specific details.
If the doctor wants to see you about the result(s), you will be offered an appointment. If you are concerned about your test results, you will need to arrange an appointment with your doctor so that all relevant information including age, ethnicity, health history, signs and symptoms, laboratory and other procedures (radiology, endoscopy, etc.), can be considered.
Lab Tests Online-UK is an educational website designed to provide patients and carers with information on laboratory tests used in medical care. We are not a laboratory and are unable to comment on an individual's health and treatment.
Reference ranges are dependent on many factors, including patient age, sex, sample population, and test method, and numeric test results can have different meanings in different laboratories.
For these reasons, you will not find reference ranges for the majority of tests described on this web site. The lab report containing your test results should include the relevant reference range for your test(s). Please consult your doctor or the laboratory that performed the test(s) to obtain the reference range if you do not have the lab report.
For more information on reference ranges, please read Reference Ranges and What They Mean.
What is being tested?
The test measures the concentration of zinc protoporphyrin (ZPP) in the blood. ZPP is present in blood in trace amounts and is formed during the production of haem. Concentrations of ZPP are increased in patients with lead poisoning and iron deficiency. Haem is an essential component of haemoglobin, the protein in red blood cells (RBCs) that carries oxygen from the lungs to the body’s tissues and cells. The formation of haem occurs in a series of reactions that conclude with the insertion of an iron into the centre of a molecule called protoporphyrin. If there is not enough iron available, then zinc is incorporated into protoporphyrin form zinc protoporphyrin (ZPP). ZPP serves no useful purpose in the red blood cells since it cannot transport oxygen around the body.
Lead prevents iron (but not zinc) from attaching to protoporphyrin so zinc protoporphyrin will also be elevated in severe cases of lead poisoning and iron deficiency.
ZPP is measured in two ways. The free erythrocyte protoporphyrin (FEP) test measures both ZPP (which accounts for 90% of protoporphyrin in red blood cells) and free protoporphyrin (not bound to zinc). The ZPP/haem ratio gives the proportion of ZPP compared to haem in red blood cells.
How is the sample collected for testing?
To measure FEP, an anticoagulated blood sample is taken by inserting a needle into a vein in your arm. To determine the ZPP/haem ratio, a drop of blood is placed in an instrument called a haematofluorometer. This instrument measures the fluorescence of ZPP and reports the amount of ZPP per number of haem molecules. Since only a single drop of blood is required, this test is well suited for screening children.
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Accordion Title
Common Questions
How is it used?
Zinc protoporphyrin (ZPP) is primarily requested to detect and monitor chronic exposure to lead in adults and to aid in the diagnosis of iron deficiency in children.
ZPP may be requested in addition to lead, to test for chronic lead exposure. Hobbyists who work with products containing lead and people who live in older houses may be at an increased risk of developing lead poisoning. This is because lead is usually ingested or inhaled. Those who inhale dust that contains lead, handle lead directly and then eat, or in the case of children, suck or eat paint that contain lead (common in houses built prior to 1960) can have elevated levels of lead and ZPP in their body.
In the UK, the Health Protection Agency (HPA) offers guidance on safety of chemicals both in the environment and the workplace. Currently, only recommendations based on blood lead concentration are given.
ZPP is not sensitive enough for use as a screening test in children, as values do not rise until lead concentrations exceed the acceptable range. The maximum blood lead concentrations considered safe in children have been set at a very low level to minimise the negative impact of lead exposure on their development. In this age group, blood lead measurements should be done to detect exposure to lead.In children, the ZPP/haem ratio is sometimes requested as an early indicator of iron deficiency. An increase in the ZPP/haem ratio is one of the first signs of insufficient iron stores and will be elevated in most young people before signs or symptoms of anaemia are present. As more specific tests of iron status are required to confirm iron deficiency, this is seldom used but still has a role in developing countries.
When is it requested?
ZPP may be requested alongside lead in adults when chronic exposure to lead is suspected, when an employee is a participant in an occupational lead monitoring programme, or when someone has a hobby, such as stained glass working, that brings them into frequent contact with lead. ZPP may also rarely be requested as a test for iron deficiency in children and adolescents and/or when iron deficiency is suspected.
What does the test result mean?
The ZPP concentration in blood is usually very low. An increase in ZPP indicates a disruption of normal haem production but is not specific as to its cause. The main reasons for increases in ZPP are iron deficiency and lead poisoning. It is important that ZPP concentrations be evaluated in the context of a patient’s history, clinical findings, and the results of other tests such as ferritin, lead, and a full blood count (FBC). It is possible that the patient may have both iron deficiency and lead poisoning.
In cases of chronic lead exposure, ZPP reflects the average lead concentration over the previous 3-4 months. However, the amount of lead currently present in the blood and the burden of lead in the body (the amount in the organs and bones) cannot be determined with a ZPP test. Values for ZPP rise more slowly than the blood lead concentration following exposure, and they take longer to drop after exposure to lead has ceased.
Is there anything else I should know?
An increased ZPP concentration is also seen in erythropoietic porphyrias. In X-linked dominant protoporphyria (XLDPP) there are increases in both free protoporphyrin and ZPP. However, these hereditary diseases are much less common than iron deficiency or lead poisoning.
ZPP may be elevated in inflammatory conditions, anaemia of chronic disease, infections, and several blood-related diseases, but it is not generally used to monitor or diagnose these conditions.
ZPP may also be increased in genetic conditions such as Thalassaemia’s where there are surplus globins which are not used in the formation of haemoglobin.
Depending on the method used to test ZPP, other substances in the blood that fluoresce, such as bilirubin and riboflavin, can produce false positive results. Falsely low values may occur if the sample is not protected from light before testing.
Besides ZPP and lead levels, what other tests might my physician request to monitor exposure to lead?
If you are in an occupational setting where you are frequently exposed to lead, your doctor may use the following tests to evaluate your kidneys and red blood cell production:
Renal Profile
Urinalysis
FBC or haemoglobin and haematocrit
Blood Film
Related Content
On This Site Tests: Serum iron; TIBC; Ferritin; Lead; FBC; Haemoglobin; Haematocrit; Porphyrin tests; Heavy metals Conditions: Lead Poisoning, Anaemia
Elsewhere On The Web Health Protection Agency Patient Health & Safety Executive European Porphyria Initiative
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187960 | https://sites.calvin.edu/scofield/courses/m355/handouts/definiteMatrices.pdf | MATH 355 Supplemental Notes Positive Definite Matrices Positive Definite Matrices A quadratic form qpxq (in the n real variables x “ px1, . . . , xnq P Rn) is said to be • positive definite if qpxq ° 0 for each x , 0 in Rn.
• positive semidefinite if qpxq • 0 for each x , 0 in Rn.
• negative definite if qpxq † 0 for each x , 0 in Rn.
• negative semidefinite if qpxq § 0 for each x , 0 in Rn.
• indefinite if there exist u, v P Rn such that qpuq ° 0 and qpvq † 0.
To any (real) quadratic form q there is an associated real symmetric matrix A for which qpxq “ xx, Axy “ xAx, xy “ xTAx. We apply the same words to characterize this symmetric matrix, calling it positive/negative (semi)definite or indefinite depending on which of the above conditions hold for the quadratic form qpxq “ xx, Axy.
Note that, while it would seem, in classifying a quadratic form q, one must investigate the behavior of qpxq over all x P Rn, it is enough to focus on those x P Rn with unit length }x} “ 1. This is because, for x , 0, qpxq “ xTAx “ }x}2 ˆ x }x} ˙T A ˆ x }x} ˙ “ }x}2qpuq, where u “ x{}x} is a unit vector. This means q is positive definite, if and only if qpuq ° 0 for every unit vector u P Rn, positive semidefinite, if and only if qpuq • 0 for every unit vector u P Rn, and so on.
Now, because the matrix A associated to q is symmetric, the Spectral Theorem says there exists an orthonormal basis tq1, . . . , qnu of Rn consisting of eigenvectors of A. Without loss of generality, we may assume the vectors of this basis have been indexed so that the corresponding (real) eigenvalues are ordered from smallest to largest λ1 § λ2 § ¨ ¨ ¨ § λn.
The orthogonal matrix S whose jth column is qj diagonalizes A: A “ S⇤ST, with ⇤“ » — — — — – λ1 0 ¨ ¨ ¨ 0 0 λ2 ¨ ¨ ¨ 0 .
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...
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.
0 0 ¨ ¨ ¨ λn fi ffi ffi ffi ffi fl .
Thus, for any vector x P Rn with }x} “ 1, qpxq “ xx, Axy “ @ x, S⇤STx D “ @ STx, ⇤STx D “ @ ⇤STx, STx D “ x⇤y, yy “ n ÿ j“1 λjy2 j, 2 MATH 355 Supplemental Notes Positive Definite Matrices where y “ STx, and }y} “ 1, as well (since ST is orthogonal). From this, we deduce several facts: Fact 1: For a real symmetric matrix A with eigenvalues λ1 § λ2 § ¨ ¨ ¨ § λn, (i) λ1 § xx, Axy § λn, for each unit vector x (i.e., }x} “ 1).
(ii) Corresponding to each eigenvalue λj of A, there is a unit eigenvector xj (that is, }xj} “ 1 with Axj “ λjxj), and for this xj, qpxjq “ @ xj, Axj D “ @ λjxj, xj D “ λj @ xj, xj D “ λj.
(iii) A is positive semidefinite if and only if each eigenvalue λj • 0, and positive definite i↵ every eigenvalue is positive. Similarly, A is negative semidefinite i↵every eigenvalue λj § 0, and negative definite i↵they are all negative.
(iv) A is indefinite if and only if λ1 † 0 † λn.
The role of definiteness in optimization Earlier, we said a real-valued function f of multiple real variables x “ px1, . . . , xnq which is smooth about the point a “ pa1, . . . , anq is, for small enough }h}, well approximated by the second-degree Taylor polynomial fpa hq « fpaq r fpaq¨ h B h, 1 2H fpaqh F .
(1) As with the functions of a single variable studied in Calculus I, a necessary condition for di↵eren-tiable f to have an extremum at a is that a be a critical point, a site where the 1st derivative is zero.
A di↵erentiable function fpxq has many first partial derivatives, and for f to have an extremum, they must all be zero simultaneously. This means that a is a critical point when r fpaq “ 0. Thus, in the neighborhood of a critical point a, (1) reduces to fpa hq « fpaq B h, 1 2H fpaqh F (2) for small }h}. We have that @ h, 1 2Hfpaqh D is a quadratic form. If it is the case that @ h, Hfpaqh D is positive definite, then the thing added to fpaq on the right-hand side of (2) is positive for any nonzero vector h representing the magnitude and direction we have ”strayed” from a; this, in turn, means that fpaq is a local minimum of f. By similar reasoning, fpaq is a local maximum if H fpaq is a negative definite matrix. If H fpaq is indefinite, then fpaq is neither a local max nor a local min; in that case, x “ a is called a saddle point. We summarize these findings: 3 MATH 355 Supplemental Notes Positive Definite Matrices Theorem 1 (2nd Derivative Test for Functions of Multiple Variables): Suppose f : Rn Ñ R is a smooth function in an open neighborhood of a P Rn. If a is a critical point, then (i) fpaq is a local minimum if Hfpaq is positive definite.
(ii) fpaq is a local maximum if Hfpaq is negative definite.
(iii) fpaq is a saddle point if H fpaq is indefinite.
Note that we have not stated a conclusion if H fpaq is only positive or negative semidefinite.
We must be careful not to make too much of the approximation (2) to f around a critical point a. With any h , 0, you have strayed from the point a, and the two sides of (2) are (generally) unequal. But for a function f whose 2nd partial derivatives are continuous throughout an open neighborhood containing a, they are enough equal for small }h} that an upswing in the term p1{2q @ h, H fpaqh D (i.e., the e↵ects of H fpaq being positive definite) overrides any other e↵ects in a small neighborhood around a, and makes a the site of a local minimum.
Other tests of definiteness The eigenvalues tell us about the definiteness, or lack thereof, of a symmetric matrix A. It can be time-consuming, however, to compute the eigenvalues of a matrix. We seek another method by which we may determine if a symmetric matrix is (semi)definite. This new method involves the calculation of upper left determinants—i.e., determinants of submatrices emanating from the upper left corner of a matrix. Given an n-by-n (symmetric) matrix A “ paijq, we define the relevant n determinants ∆1, ∆2, . . . , ∆n to be ∆1 “ a11, ∆2 “ !
!
!
!
!
!
a11 a12 a21 a22 !
!
!
!
!
!
∆3 “ !
!
!
!
!
!
!
!
!
a11 a12 a13 a21 a22 a23 a31 a32 a33 !
!
!
!
!
!
!
!
!
, . . . , ∆n “ detpAq.
In the text, Strang notes that a symmetric matrix A is positive definite i↵each ∆j ° 0. This is known as Sylvester’s Criterion. In fact, the same upper left determinants can be used to tell that a matrix is negative definite. We state and prove the theorem below. But first we state and prove a fact relating the determinant of any square matrix to its eigenvalues.
Lemma 1: Suppose λ1, . . . , λn are the eigenvalues of the n-by-n matrix A.
The |A| “ λ1λ2 ¨ ¨ ¨ λn.
4 MATH 355 Supplemental Notes Positive Definite Matrices Proof: Consider the nth degree characteristic polynomial pApλq “ detpA ´ λIq “ p´1qnpλ ´ λ1q ¨ ¨ ¨ pλ ´ λnq, where λ1, . . . , λn are the (perhaps nonreal) eigenvalues of A. Then |A| “ pAp0q “ p´1qnp0 ´ λ1q ¨ ¨ ¨ p0 ´ λnq “ p´1q2n n π j“1 λj “ n π j“1 λj.
⇤ Theorem 2 (Generalized(?) Sylvester’s Criterion): Let qpxq “ xTAx be a quadratic form defined on Rn with symmetric nonsingular matrix A. We may conclude that q is (i) positive definite if and only if ∆k ° 0 for each k “ 1, . . . , n. Here, ∆k is the determinant of the k-by-k submatrix of A comprising entries aij with 1 § i, j § k, (ii) negative definite if and only if p´1qk∆k ° 0 for each k “ 1, . . . , n.
(iii) indefinite if neither of the previous two conditions is satisfied.
The proof, which you may read at your own behest (it continues on through the end of the next page), follows. No doubt it is the most technical proof that has been given in the course.
Proof: Each “if and only if” statement requires a proof of two statements. We begin with the ”i↵” statement in (i), focusing first on the assertion that ∆k ° 0 for each k implies A is positive definite. The proof is by induction on n, the size of the matrix.
When n “ 1 (so the matrix has just one entry, a), then trivially ∆1 “ a ° 0 implies xax “ ax2 ° 0, for all real x , 0.
For our induction hypothesis, assume A is n-by-n for some n ° 1, and that the claim holds for each quadratic form defined on Rn´1. Now, since 0 † ∆n “ λ1λ2 ¨ ¨ ¨ λn, the eigenvalues of A are all nonzero. If they are all positive, then we know qpxq “ xAx, xy is positive definite. So, let us assume that some eigenvalue λi † 0. But for their product to be positive, there must be an even number of negative eigenvalues, so let λj be another negative eigenvalue, and take vi, vj to be eigenvectors of unit length corresponding to λi, λj, respectively. We may assume vi and vj are orthogonal to each other, either because λi , λj, or because Gram-Schmidt may be used to select orthogonal basis vectors in an eigenspace. Let W “ spanptvi, vjuq, a plane in Rn. Given any y P W, with y “ yivi yjvj, we have qpyq “ xAy, yy “ @ λiyivi λjyjvj, yivi yjvj D “ λiy2 i xvi, viy λjyiyj @ vj, vi D λiyiyj @ vi, vj D λjy2 j @ vj, vj D “ λiy2 i λjy2 j † 0.
5 MATH 355 Supplemental Notes Positive Definite Matrices Thus, q is negative for all nonzero y P W. Let q˚ : Rn´1 Ñ R be the restriction of q to Rn´1, so that q˚px1, . . . , xn´1q :“ qpx1, . . . , xn´1, 0q.
Along with that, let A˚ “ paijq1§i,j§n´1. Then q˚px1, . . . , xn´1q “ qpx1, . . . , xn´1, 0q “ ” x1 ¨ ¨ ¨ xn´1 ı A˚ » — — – x1 .
.
.
xn´1 fi ffi ffi fl.
By assumption, ∆1, . . . , ∆n´1 are all positive, so using the induction hypothesis, we get that q˚ is positive definite on Rn´1. But the pn ´ 1q-dimensional hyperplane in Rn consisting of vectors x “ px1, . . . , xn´1, 0q and the plane W have at least a line in common, and we have shown that qpvq for a nonzero vector v from that line is both strictly positive and strictly negative.
Ñ– To prove the converse statement of (i), suppose q is positive definite, and let m ° 0 be the minimum value of q over the set tx P Rn : }x} “ 1u. Similar to above, we take Ak “ paijq1§i,j§k and define a function qk : Rk Ñ R as the ”restriction of q to Rk”, in the sense that qkpx1, . . . , xkq “ qpx1, . . . , xk, 0, . . . , 0q “ ” x1 ¨ ¨ ¨ xk ı Ak » — — – x1 .
.
.
xk fi ffi ffi fl.
Let µ1, . . . , µk be the eigenvalues of Ak. We know there is an x P Rk with }x} “ 1 such that qkpxq “ µi. It follow from qkpxq “ qpx, 0q that each µi • m. Thus, by Lemma 1, ∆k “ |Ak| “ µ1 ¨ ¨ ¨ µk • mk ° 0.
To prove (ii), define q˚pxq “ ´qpxq, so that negative definiteness of q is equivalent to positive definiteness of q˚. By part (i), q˚ is positive definite if and only if each 0 † ∆˚ k “ | ´ Ak| “ p´1qk|Ak| “ p´1qk∆k.
To prove (iii), note first that 0 , ∆n “ λ1 ¨ ¨ ¨ λn means every eigenvalue is nonzero. If all were positive, then part (a) shows each ∆k ° 0; if all were negative, then part (b) implies each p´1qk∆k ° 0. since neither of these conditions holds by supposition, it must be that A has both positive and negative eigenvalues, and is nondefinite.
⇤ Example 1: Classify each of the nonsingular symmetric matrices A and B given below according to its type of definiteness.
A “ » — – ´1 1{2 1 1{2 ´1 1{2 1 1{2 1 fi ffi fl, B “ » — — — — – 3 1 5 3 1 5 2 0 5 2 10 3 3 0 3 14 fi ffi ffi ffi ffi fl .
6 MATH 355 Supplemental Notes Positive Definite Matrices For A, we have ∆1 “ ´1, ∆2 “ !
!
!
!
!
!
´1 1{2 1{2 ´1 !
!
!
!
!
! “ 3 4, ∆3 “ |A| “ 5 2.
Thus, while A is nonsingular, neither of the conditions (i) nor (ii) of Theorem 2 hold, making A indefinite.
For B, we have ∆1 “ 3, ∆2 “ !
!
!
!
!
!
3 1 1 5 !
!
!
!
!
! “ 14, ∆3 “ !
!
!
!
!
!
!
!
!
3 1 5 1 5 2 5 2 10 !
!
!
!
!
!
!
!
!
“ 23, ∆4 “ |B| “ 196.
Since each of these subdeterminants is positive, B is positive definite.
One more result, equivalent to positive definiteness, seems pertinent.
Theorem 3: A symmetric matrix A is positive definite if and only if A “ BTB for some matrix B whose columns are all linearly independent.
Proof: Suppose, first, that A “ BTB, where the columns of B are linearly independent.
Due to the linear independence of its columns, NullpBTBq “ NullpBq “ t0u, showing that A is nonsingular. For x , 0, we have xAx, xy “ @ BTBx, x D “ xBx, Bxy “ }Bx}2 ° 0, since x, being nonzero, is not in NullpBq.
The converse is harder to prove.
Here, I rely on another factorization, called the Cholesky factorization, which exists for any real symmetric matrix. This factorization is A “ RTR, which gives the result with B “ R. We note, in particular, that were the columns of B not linearly independent, then those in A “ RTR would not be either. ⇤ 7 |
187961 | https://www.youtube.com/watch?v=yRSY1ZfNrHA | Vectors Prove that the midpoints of quadrilateral form a paralellogram - EDEXCEL - GCSE
Anil Kumar
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Show that if the midpoints of adjacent sides of any quadrilateral are joined, then the resulting figure is a parallelogram. #edexcel #vectors #MCV4U_Vectors
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Transcript:
welcome or welcome to my series on vectors we'll take a few applications to geometry in this particular series the question here is show that if the midpoints of adjacent sides of any quadrilateral are joined then the resulting figure is a parallelogram so let us construct quadrilaterals so we'll just consider the general correlator not a rectangle or a square right so let's just make a random one so it doesn't look like any regular one now what we are going to do here is join the midpoints so let's say these are the midpoints for our quad little right so we'll just join them we need to prove that they form a parallelogram that is what we need to prove right so I hope the concept is clear so this is the quadrilateral a B C D given to us and the midpoints means these are the points which we are talking about let's say P Q R and s right so these are the midpoints okay these are the midpoints now we need to prove that p q r s is a parallelogram okay so two easy way to probe s that we can begin from any one side so let's talk about Q Bar so what is Q R equals two so Q Bar is equal to Q B plus B R right Q B plus B R so that is Q Bar now Q B could be written as half of a B so Q B is half of a B since we know Q is the midpoint and B R is half of BC right it is half of BC correct now so so it is half off we could also write this as a B plus BC now a B plus BC is what well let's join it a B plus BC we can join this that is what it is right so it is half of A to C right so this is equal to half of AC so we know what Q R is now let us figure out what is PS is it same as half of a see that's what we need to prove right so let's begin with PS now so p2 s is p2 d plus d s right now p2 d is half of 80 since D is the P is the midpoint right okay and D s is half of DC so we could again take 1/2 Commons we get Ad plus DC and ad plus this is AC right so we get this as half of AC so comparing equation one this is our one which is we are adding Q is equal to half of AC and now we got this PS the opposite n is half of AC so both the same right so that shows that these are equal and they are parallel okay so they are opposite sides are parallel correct now you could use the same strategy so I'd like you to use the same strategy and then figure out what PQ is so you can say the same thing so P Q will be equal to P 2 A plus a 2 B half right I should write P 2 a first right ok so P 2 A plus a 2 Q which is half of a B 1/2 of da p 2 a right half of D plus half of a B which is half of D plus a B which is half of so this is half of the other diagonal which is d to be correct so we got PQ now let's see what s RS so same strategy s to R will be SC plus CR which is half of DC plus half of CB that means it is half of DC and CD is half of BT right so you can write this as half of s right DB right half of DB so again we have shown that they are also equal right so we got this PQ equals to SR so they are parallel and they are equal so that is how we can say that p q r s is a parallelogram perfect so that is how you would actually prove it I hope it is straightforward and simple so from here we'll take a few complicated applications of geometry I hope that will help feel free to write your comments and share your views if you have some questions feel free to post them thank you and all the best |
187962 | https://www.3blue1brown.com/lessons/taylor-series | Chapter 14Taylor series
"To many, mathematics is a collection of theorems. For me, mathematics is a collection of examples; a theorem is a statement about a collection of examples and the purpose of proving theorems is to classify and explain the examples."
John B. Conway
Introduction
When I first learned about Taylor series, I definitely didn't appreciate how important they are. But time and time again they come up in math, physics, and many fields of engineering because they're one of the most powerful tools that math has to offer for approximating functions.
One of the first times this clicked for me as a student was not in a calculus class, but in a physics class. We were studying some problem that had to do with the potential energy of a pendulum, and for that you need an expression for how high the weight of the pendulum is above its lowest point, which works out to be proportional to one minus the cosine of the angle between the pendulum and the vertical.
The specifics of the problem we were trying to solve are beyond the point here, but I'll just say that this cosine function made the problem awkward and unwieldy. But by approximating cos(θ) as 1−2θ2, of all things, everything fell into place much more easily. If you've never seen anything like this before, an approximation like that might seem completely out of left field.
If you graph cos(θ) along with this function 1−2θ2, they do seem rather close to each other for small angles near 0, but how would you even think to make this approximation? And how would you find this particular quadratic?
The study of Taylor series is largely about taking non-polynomial functions, and finding polynomials that approximate them near some input. The motive is that polynomials tend to be much easier to deal with than other functions: They're easier to compute, easier to take derivatives, easier to integrate... they're just all around friendly.
Approximating cos(x)
Let's look at the function cos(x), and take a moment to think about how you might find a quadratic approximation near x=0. That is, among all the polynomials that look c0+c1x+c2x2 for some choice of the constants c0, c1 and c2, find the one that most resembles cos(x) near x=0.
First of all, at the input 0 the value of cos(x) is 1, so if our approximation is going to be any good at all, it should also equal 1 when you plug in 0.
cos
(
x
)
cos
(
0
)
c
0
=
c
0
+
c
1
x
+
c
2
x
2
=
c
0
+
c
1
(
0
)
+
c
2
(
0
)
2
=
1
Plugging in 0 just results in whatever c0 is, so we can set that equal to 1.
This leaves us free to choose constant c1 and c2 to make this approximation as good as we can, but nothing we do to them will change the fact that the output of our approximation at 0 is equal to 1.
It would also be good if our approximation had the same tangent slope as cos(x) at x=0. Otherwise, the approximation drifts away from the cosine graph faster than it needs to as x tends away from 0. The derivative of cos(x) is −sin(x), and at x=0 that equals 0, meaning its tangent line is flat.
This is the same as making the derivative of our approximation as close as we can to the derivative of our original function cos(x).
d
x
d
(
cos
(
x
))
−
sin
(
x
)
−
sin
(
0
)
c
1
=
d
x
d
(
c
0
+
c
1
x
+
c
2
x
2
+
)
=
c
1
+
2
c
2
x
=
c
1
+
2
c
2
(
0
)
=
0
Setting c1 equal to 0 ensures that our approximation matches the tangent slope of cos(x) at this point.
This is the same approximation we just had! But we should feel confident that the process is working, because our approximation now equals the value and slope of cos(x) a x=0, leaving us free to change c2.
We can take advantage of the fact that the cosine graph curves downward above x=0, it has a negative second derivative. Or in other words, even though the rate of change is 0 at that point, the rate of change itself is decreasing around that point.
Specifically, since its derivative is −sin(x) its second derivative is −cos(x), so at x=0 its second derivative is −1.
d
x
2
d
2
(
cos
)
(
0
)
=
−
cos
(
0
)
=
−
1
Just as we wanted the derivative of our approximation to match that of cosine, we'll also make sure that their second derivatives match so as to ensure that they curve at the same rate. The slope of our polynomial shouldn't drift away from the slope of cos(x) any more quickly than it needs to.
d
x
2
d
2
(
cos
(
x
))
d
x
d
(
−
sin
(
x
))
−
cos
(
x
)
−
cos
(
0
)
c
2
=
d
x
2
d
2
(
c
0
+
c
1
x
+
c
2
x
2
+
)
=
d
x
d
(
c
1
+
2
c
2
x
)
=
2
c
2
=
2
c
2
=
−
2
1
We can compute that at x=0, the second derivative of our polynomial is 2c2. In fact, that's its second derivative everywhere, it is a constant. To make sure this second derivative matches that of cos(x), we want it to equal −1, which means c2=−21.
This locks in a final value for our approximation:
cos
(
x
)
≈
1
+
0
x
+
(
−
2
1
)
x
2
To get a feel for how good this is, let's try it out for x=0.1
cos
(
0.1
)
P
(
0.1
)
=
1
−
2
1
(
0.1
)
2
=
true value
0.9950042
…
=
approximation
0.995
That's pretty good!
Take a moment to reflect on what just happened. You had three degrees of freedom with a quadratic approximation, the coefficients in the expression c0+c1x+c2x2.
c0 was responsible for making sure that the output of the approximation matches that of cos(x) at x=0.
c1 was in charge of making sure the derivatives match at that point.
c2 was responsible for making sure the second derivatives match up.
This ensures that the way your approximation changes as you move away from x=0, and the way that the rate of change itself changes, is as similar as possible to behavior of cos(x), given the amount of control you have.
Better Approximations
You could give yourself more control by allowing more terms in your polynomial and matching higher-order derivatives of cos(x). For example, let's say we add on the term c3x3 for some constant c3.
cos
(
x
)
≈
1
−
2
1
x
2
+
c
3
x
3
If you take the third derivative of a cubic polynomial, anything quadratic or smaller goes to 0. As for that last term, after three iterations of the power rule it looks like 1⋅2⋅3⋅c3.
d
x
3
d
3
(
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
)
=
d
x
2
d
2
(
c
1
+
2
c
2
x
+
3
c
3
x
2
)
=
d
x
d
(
2
c
2
+
6
c
3
x
)
=
6
c
3
On the other hand, the third derivative of cos(x) is sin(x), which equals 0 at x=0, so to make the third derivatives match, the constant c3 should be 0.
In other words, not only is 1−21x2 the best possible quadratic approximation of cos(x) around x=0, it's also the best possible cubic approximation.
You can actually make an improvement by adding a fourth order term, c4x4. The fourth derivative of cos(x) is also cos(x), which equals 1 at x=0. And what's the fourth derivative of our polynomial with this new term?
When you keep applying the power rule over and over, with those exponents all hopping down to the front, you end up with 1⋅2⋅3⋅4⋅c4, which is 24c4.
d
x
4
d
4
(
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
)
=
d
x
3
d
3
(
c
1
+
2
c
2
x
+
3
c
3
x
2
+
4
c
4
x
3
)
=
d
x
2
d
2
(
2
c
2
+
6
c
3
x
+
12
c
4
x
2
)
=
d
x
d
(
6
c
3
+
24
c
4
x
)
=
24
c
4
So if we want this to match the fourth derivative of cos(x), which at x=0 is 1, then we must set c4=241.
This polynomial 1−21x2+241x4 is a very close approximation for cos(x) around x=0 and for any physics problem involving the cosine of some small angle, for example, predictions would be almost unnoticeably different if you substituted this polynomial for cos(x).
Generalizing
It's worth stepping back to notice a few things about this process. First, factorial terms naturally come up in this process. When you take n derivatives of xn, letting the power rule just keep cascading, what you're left with is 1⋅2⋅3 and on up to n.
d
x
8
d
8
(
c
8
x
8
)
8
!
1
⋅
2
⋅
3
⋅
4
⋅
5
⋅
6
⋅
7
⋅
8
⋅
c
8
So you don't simply set the coefficients of the polynomial equal to whatever derivative value you want, you have to divide by the appropriate factorial to cancel out this effect. For example, in the approximation for cos(x), the x4 coefficient is the fourth derivative of cosine, 1, divided by 4!=24.
The second thing to notice is that adding new terms, like this c4x4, doesn't mess up what old terms should be, and that's important. For example, the second derivative of this polynomial at x=0 is still equal to 2 times the second coefficient, even after introducing higher order terms to the polynomial.
This is because we're plugging in x=0, so the second derivative of any higher order terms, which all include an x, will wash away. The same goes for any other derivative, which is why each derivative of a polynomial at x=0 is controlled by one and only one coefficient.
Comprehension Question
Just like cos(x), the function sin(x) is a situtation where we know its derivative and its value at x=0. Let's apply the same process to find the third degree taylor polynomial of sin(x).
Using what we know about the derivatives of the function sin(x), what are the best possible choices for the coefficients c0, c1, c2 and c3 to approximate the function as a third degree polynomial around the point x=0?
Approximating around other points
If instead, you were approximating near an input other than 0, like x=π, to get the same effect you would have to write your polynomial in terms of powers of (x−π). Or more generally, powers of (x−a) for some constant a.
This makes it look notably more complicated, but it's all to make sure plugging in x=π results in a lot of nice cancelation so that the value of each higher-order derivative is controlled by one and only one coefficient.
Finally, on a more philosophical level, notice how we're taking information about the higher-order derivatives of a function at a single point, and translating it into information (or at least approximate information) about the value of that function near that point. This is the major takeaway for Taylor series: Differential information about a function at one value tells you something about an entire neighborhood around that value.
We can take as many derivatives of cos(x) as we want, it follows this nice cyclic pattern cos(x), −sin(x), −cos(x), sin(x), and repeat.
So the value of these derivative of x=0 have the cyclic pattern 1, 0, −1, 0, and repeat. And knowing the values of all those higher-order derivatives is a lot of information about cos(x), even though it only involved plugging in a single input, x=0.
That information is leveraged to get an approximation around this input by creating a polynomial whose higher order derivatives, match up with those of cos(x), following this same 1, 0, −1, 0 cyclic pattern.
To do that, make each coefficient of this polynomial follow this same pattern, but divide each one by the appropriate factorial which cancels out the cascading effects of many power rule applications. The polynomials you get by stopping this process at any point are called "Taylor polynomials" for cos(x).
Other Functions
More generally, if we were dealing with some function other than cosine, you would compute its derivative, second derivative, and so on, getting as many terms as you'd like, and you'd evaluate each one at x=0.
Then for your polynomial approximation, the coefficient of each xn term should be the value of the n-th derivative of the function at 0, divided by (n!).
P
(
x
)
=
f
(
0
)
+
d
x
df
(
0
)
1
!
x
1
+
d
x
2
d
2
f
(
0
)
2
!
x
2
+
d
x
3
d
3
f
(
0
)
3
!
x
3
+
⋯
When you see this, think to yourself that the constant term ensures that the value of the polynomial matches that of f(x) at x=0, the next term ensures that the slope of the polynomial matches that of the function, the next term ensure the rate at which that slope changes is the same, and so on, depending on how many terms you want.
The more terms you choose, the closer the approximation, but the tradeoff is that your polynomial is more complicated. And if you want to approximate near some input a other than 0, you write the polynomial in terms of (x−a) instead, and evaluate all the derivatives of f at that input a.
P
(
x
)
=
f
(
a
)
+
d
x
df
(
a
)
1
!
(
x
−
a
)
1
+
d
x
2
d
2
f
(
a
)
2
!
(
x
−
a
)
2
+
⋯
This is what Taylor series look like in their fullest generality. Changing the value of a changes where the approximation is hugging the original function; where its higher order derivatives will be equal to those of the original function.
A meaningful example
One of the simplest meaningful examples is to approximate the function f(x)=ex, around the input x=0. Computing its derivatives is nice since the derivative of ex is also ex, so its second derivative is also ex, as is its third, and so on.
So at the point x=0, all of the derivatives are equal to 1. This means our polynomial approximation looks like (1)+(1)x+(1)2!x2+(1)3!x3+(1)4!x4, and so on, depending on how many terms you want. These are the Taylor polynomials for ex.
Infinity and convergence
We could call it an end here, and you'd have you'd have a phenomenally useful tool for approximations with these Taylor polynomials. But if you're thinking like a mathematician, one question you might ask is if it makes sense to never stop, and add up infinitely many terms.
In math, an infinite sum is called a "series", so even though one of the approximations with finitely many terms is called a "Taylor polynomial" for your function, adding all infinitely many terms gives what's called a "Taylor series".
You have to be careful with the idea of an infinite sum because one can never truly add infinitely many things; you can only hit the plus button on the calculator so many times. The more precise way to think about a series is to ask what happens as you add more and more terms. If the partial sums you get by adding more terms one at a time approach some specific value, you say the series converges to that value.
It's a mouthful to always say "The partial sums of the series converge to such and such value", so instead mathematicians often think about it more compactly by extending the definition of equality to include this kind of series convergence. That is, you'd say this infinite sum equals the value its partial sums converge to.
For example, look at the Taylor polynomials for ex, and plug in some input like x=1.
e
x
e
1
→
1
+
1
!
x
1
+
2
!
x
2
+
3
!
x
3
+
4
!
x
4
+
5
!
x
5
+
⋯
→
2.7182818
1
+
1
!
1
1
+
2
!
1
2
+
3
!
1
3
+
4
!
1
4
+
5
!
1
5
+
⋯
As you add more and more polynomial terms, the total sum gets closer and closer to the value e. The precise-but-verbose way to say this is "the partial sums of the series on the right converge to e." More briefly, most people would abbreviate this by simply saying "the series equals e."
In fact, it turns out that if you plug in any other value of x, like x=2, and look at the value of higher and higher order Taylor polynomials at this value, they will converge towards ex, in this case e2.
This is true for any input, no matter how far away from 0 it is, even though these Taylor polynomials are constructed only from derivative information gathered at the input 0.
In a case like this, we say ex equals its Taylor series at all inputs x. This is a somewhat magical fact! It means all of the information about the function is somehow captured purely by higher-order derivatives at a single input, namely x=0.
Limitations for ln(x)
Although this is also true for some other important functions, like sine and cosine, sometimes these series only converge within a certain range around the input whose derivative information you're using. If you work out the Taylor series for ln(x) around the input x=1, which is built from evaluating the higher order derivatives of ln(x) at x=1, this is what it looks like.
When you plug in an input between 0 and 2, adding more and more terms of this series will indeed get you closer and closer to the natural log of that input.
But outside that range, even by just a bit, the series fails to approach anything. As you add more and more terms the sum bounces back and forth wildly. The partial sums do not approach the natural log of that value, even though the ln(x) is perfectly well defined for x>2.
In some sense, the derivative information of ln(x) at x=1 doesn't propagate out that far. In a case like this, where adding more terms of the series doesn't approach anything, you say the series diverges.
That maximum distance between the input you're approximating near, and points where the outputs of these polynomials actually do converge, is called the "radius of convergence" for the Taylor series.
What is the Taylor Series expansion of the function −ln(1−x) around the point x=0?
Summary
There remains more to learn about Taylor series, their many use cases, tactics for placing bounds on the error of these approximations, tests for understanding when these series do and don't converge. For that matter there remains more to learn about calculus as a whole, and the countless topics not touched by this series.
The goal with these videos is to give you the fundamental intuitions that make you feel confident and efficient learning more on your own, and potentially even rediscovering more of the topic for yourself. In the case of Taylor series, the fundamental intuition to keep in mind as you explore more is that they translate derivative information at a single point to approximation information around that point.
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Higher order derivatives
Taylor series (geometric view)
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Table of Contents
Taylor series
Introduction
Approximating cos(x)
Better Approximations
Generalizing
Comprehension Question
Approximating around other points
Other Functions
A meaningful example
Infinity and convergence
Limitations for ln(x)
Summary
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Factor Theorem
Last Updated : 23 Jul, 2025
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Factor theorem is used for finding the roots of the given polynomial. This theorem is very helpful in finding the factors of the polynomial equation without actually solving them.
According to the factor theorem, for any polynomial f(x) of degree n ≥ 1 a linear polynomial (x - a) is the factor of the polynomial if f(a) is zero.
Let's learn about the factor theorem, its proof, and others in detail in this article.
Table of Content
What is the Factor Theorem?
Factor Theorem Statement
Factor Theorem Formula
Zero of a Polynomial
Factor Theorem Proof
How to Use Factor Theorem?
Using the Factor Theorem To Factor a Cubic Polynomial
Factor Theorem and Remainder Theorem
Factor Theorem Examples
What is the Factor Theorem?
A special theorem that links polynomials with their zeros and helps to find the factors of the polynomial is called the Factor Theorem. Factor theorem along with the remainder theorem is very helpful in solving complex polynomial equations. For any polynomial of a higher degree factor, the theorem removes all the known zeros and reduces the polynomial to a lesser degree, and then its factors are easily calculated.
Factor Theorem Statement
According to the factor theorem, any polynomial f(x) of degree greater than or equal to 1 its factor can be easily found if we know the root of the polynomial equation. The root of the polynomial equation is the number that satisfies the polynomial equation. Suppose the root of f(x) is x = a then according to the factor theorem (x- a) is the factor of the polynomial f(x).
The image given below shows the factor theorem statement.
Factor Theorem
Factor Theorem Formula
According to the factor theorem x - a can be considered the factor of the polynomial f(x) if f(a) is zero.Now the Factor Theorem Formula is,
If (x – a) is a factor of f(x) then,
f(a) = 0,
Now f(x) can be expanded as,
f(x) = (x-a)q(x)
where,
a is the root of the equation
q(x) is the quotient when f(x) is divided by q(a)
Also Read:Real Life Applications of Factor Theorem
Zero of a Polynomial
The concept of zero of the polynomialsis very useful for understanding the factor theorem. Zero of any Polynomial is the real number that satisfies the polynomial equation. i.e. For polynomial f(x) 'a' is the zero of f(x) if f(a) is zero.
For example, the zero of the polynomial x2+ 4x - 12 is calculated as,
Solution:
x 2 + 4x - 12
= x 2 + 6x -2x - 12
= x(x + 6) - 2(x + 6)
= (x -2)(x+6)
zeros of the polynomial x 2 + 4x - 12 are,
x - 2 = 0
⇒ x = 2, and
x + 6 = 0
⇒ x = -6
Verification: To verify the zero of the polynomial x2+ 4x - 12
f(2) = 2 2 + 4(2) -12
⇒ f(2) = 4 + 8 -12 = 0
f(-6) = (-6)2 + 4(-6) - 12
⇒ f(-6) = 36 - 24 -12 = 0
Thus, zero of the polynomial x 2 + 4x - 12 is verified.
Factor Theorem Proof
Consider a polynomial p(x) that is being divided by (x – b) only if p(b) = 0.
Given Polynomial can be written as,
Dividend = (Divisor × Quotient) + Remainder
By using the division algorithm,
p(x) = (x – b) q(x) + remainder
Where,
p(x) is the dividend,
(x – b) is the divisor, and
q(x) is the quotient.
From the remainder theorem,
p(x) = (x – b) q(x) + p(b).
Suppose that p(b) =0.
p(x) = (p – b) q(x) + 0
⇒ p(x) = (x – b) q(x)
Thus, we can say that (x – b) is a factor of the polynomial p(x).
Here we can see that the factor theorem is actually a result of the remainder theorem, which states that a polynomial (x) has a factor (x – a), if and only if, a is a root i.e., p(b) = 0.
How to Use Factor Theorem?
Factor Theorem is used widely in solving the polynomial equation.
We can learn about the use of factor theorem by going through the example discussed below,
Example: Find the factor of x2- 4x + 3.
Solution:
Given polynomial,
f(x) = x 2 - 4x + 3
by hit and try method,
taking x = 1
f(1) = 1 2 -4(1) + 3 = 0
Thus, x -1 is the factor of x 2 - 4x + 3
again taking x = 3
f(3) = 3 2 - 4(3) + 3 = 0
Thus, x -3 is the factor of x 2 - 4x + 3
Using the Factor Theorem to Factor a Cubic Polynomial
Factor Theorem To Factor a Third-Degree Polynomial is widely used for solving the three-degree polynomial also called acubic polynomial. Generally, for quadratic polynomial factorization method is used and the factor theorem is used when we have to find the factors of any cubic polynomial.
Use the steps given below to find the Factor of a Cubic Polynomial
Factorize the cubic polynomial f(x) = ax 3 + bx 2 + cx + d.
Step 1: By hit and try method find one of the zeros of the polynomial f(x). Say the zero of the polynomials is "a" such that f(a) is zero.
Step 2: Divide the polynomial f(x) by x-a using synthetic division of the polynomial method.
Step 3: Using the division algorithm, write the given cubic polynomial as the f(x) = (x-a)g(x), where g(x) is a quadratic polynomial.
Step 4: Factorize the cubic polynomial g(x). As, g(x) = (x-b)(x-c) for any real numbers b and c.
Step 5: Express the given cubic polynomial as the product of these factors.
f(x) = (x -a)(x-b)(x-c)
Example Using factor theorem factorize f(x) = x3+ 10x2+ 23x + 14
Solution:
f(x) = x 3 + 10x 2 + 23x + 14
Now applying Factor Theorem.
(x - a) is a factor of f(x) if f(a) = 0.
And g(y) = (y-a)q(a)
Now, by hit and try method we get,
f(-1) = (-1)3 + 10(-1)2 + 23(-1) + 14
⇒ f(-1) = 24 - 24 = 0
Thus, x + 1 is a factor of x 3 + 10x 2 + 23x + 14
Now, by dividing x + 1 by x 3 + 10x 2 + 23x + 14 we get,
f(x) = (x + 1)(x 2 + 9x + 14)
⇒ f(x) = (x + 1)(x 2 + 7x + 2x + 14)
⇒ f(x) = (x + 1){x(x +7) + 2(x + 2)}
⇒ f(x) = (x + 1)(x + 2)(x + 7)
Factor Theorem and Remainder Theorem
Remainder Theoremand Factor Theorem are the basic theorems that help in mathematics and the basic difference between them is stated in the table given below,
Difference Between Factor Theorem and Remainder Theorem
Key difference between factor and remainder theorem are listed as follows:
| Remainder Theorem | Factor Theorem |
--- |
| According to, Remainder Theoremfor any polynomial p(x) when divided by x - a the remainder is p(a). | According to Factor Theorem if (x - a) is a factor of p(x) then this is true only if f(a) = 0. |
| Remainder Theorem helps us to find the remainder of the polynomial without actually dividing it. | Factor theorem helps us to find the factor of the given polynomial. |
For example, for the given polynomial f(x) = x2+ 3x - 4 the remainder when f(x) is divided by x - 2 = 0 is,
Solution:
Given,
x - 2 = 0 ⇒ x = 2
Thus, f(2) = 2 2 + 3(2) - 4 = 6
Hence, the remainder when we divide x 2 + 3x - 4 by x - 2 is 6.
Similarly, for the given polynomial f(x) = x2+ 3x - 4 the remainder when f(x) is divided by x - 2 = 0 is "0" then we can say that x -2 is a factor of x2+ 3x - 4
Given,
x - 1 = 0 ⇒ x = 1
f(2) = 1 2 + 3(1) - 4 = 0
Hence, the remainder when we divide x 2 + 3x - 4 by x - 2 is 0.
Thus, according to the factor theorem x -2 is a factor of x 2 + 3x - 4.
Related Article:
Division
Divisibility Rules
Chinese Remainder Theorem
Quotient Remainder Theorem
Factoring Polynomials
Practice Questions on Remainder theorem
Factor Theorem Examples
Example 1: Find the root of the polynomial f(x) = x2- 5x + 6 using the factor theorem.
Solution:
f(x) = x 2 - 5x + 6.
Now applying factor theorem.
If (x - a) is a factor of f(x) then f(a) = 0
By hit and try method,
f(2) = 4 - 10 + 6= 0
Thus, (x - 2) is a factor of f(x).
Example 2: Suppose a polynomial f(x) = x3+ 3x2- 6x - 18. If x = -3 is a root show that x + 3 is a factor.
Solution:
f(x) = x 3 + 3x 2 - 6x - 18.
Now applying factor theorem.
If x = a is the root of f(x) then x - a is the factor of the f(x)
Now for, f(x) at x = -3
f(-3) = -27 + 27 + 18 - 18= 0
Thus, (x + 3) is a factor of f(x).
Example 3: Using the factor theorem check if y + 2 is a factor of the polynomial 2y4+ y3– 2y2+ 4y -8, or not.
Solution:
Given y + 2
If, y + 2 = 0, then y = -2
Substituting y = -2 in the given polynomial 2y 4 + y 3 – 2y 2 + 4y -8
= 2(-2)4 + (-2)3 - 2(-2)2 + 4(-2) - 8
= 32 - 8 - 8 - 8 - 8
= 0
Thus, we can say that y + 2 is a factor of the polynomial 2y 4 + y 3 – 2y 2 + 4y -8
Example 4: Check if x - 1 is a factor of the polynomial f(x) = 2x4+ 3x2- 5x + 7
Solution:
Given x - 1
If, x - 1 = 0, then x = 1
Substituting x = 1 in the given polynomial 2x 4 + 3x 2 - 5x + 7
= 2(1)4 + 3(1)2 - 5(1) + 7
= 2 + 3 - 5 + 7
= 7
As, f(-1) ≠ 0
We can say that x - 1 is not a factor of the polynomial 2x 4 + 3x 2 - 5x + 7
Example 5: Given a polynomial f(x) = x3- x2- 2x + 1. Factorize and find its roots.
Solution:
f(x) = x 3 - x 2 - x + 1
Now applying Factor Theorem.
(x - a) is a factor of f(x) if f(a) = 0.
And g(y) = (y-a)q(a)
Now, by hit and try method we get,
f(-1) = (-1)3 - (-1)2 -(-1)
⇒ f(-1) = -8 + 4 + 4 = 0
Thus, x + 1 is a factor of x 3 - x 2 - x + 1
Now, by dividing x + 1 byx 3 - x 2 - x + 1 we get,
f(x) = (x + 1)(x 2 - 2x + 2)
⇒ f(x) = (x + 1)(x 2 - x - x -2)
⇒ f(x) = (x + 1){x(x - 1) - 1(x - 1)}
⇒ f(x) = (x + 1)(x - 1)(x - 1)
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187965 | https://www.youtube.com/watch?v=7GAs_R-Rrx0 | PC 20 - Sign Analysis (Mini Lesson)
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foreign analysis is a method of determining solutions to an inequality in one variable we've talked about other methods of doing this before like graphing the line and determining where the function is positive if it's a greater than you know if the function is greater than zero where is it positive and what are the X values that are associated with that we've also done it by factoring let's say a polynomial or a line or whatever and using those factors as finding the x-intercepts and then splitting the number line up into sections where we test points to see if the inequality is true or not that's that's the second way we could do that this is the third way and this sign analysis will carry you into calculus this will be important in calculus you need to know how to do this in calculus so this is the time we need to show you about this teach you about this so it involves factoring our expression or function that we're given listing the factors and charting where the factors are positive and negative relative to the numbers on a number line that's the that's the um Crux of this whole uh concept here the most important highlighted part is where we see in a chart where the signs what the signs are for each individual factor and then as we analyze those signs and do the multiplication whatever we see where the function is positive or negative so I will obviously do an example here right away but that's what sign analysis involves so the previous two methods which I did just explain I'll show you as well on the board you've already seen this if we graph uh graph the function and so if we wanted to solve for this inequality we would graph the related function and then we would determine where the graph is in this case I want to know where the graph is less than or equal to zero so where is the graph below the x-axis and those X values that are associated with that negative part of the graph those are the solutions right there so that's graphing um roots and test points are where we Factor the related equation find out where the intercepts are negative one and three and then in visualizing at least on a number line the three sections that this creates so these are intercepts and then where is the graph positive where is it negative and you pick X values in one or more sections to determine whether the equality is true or not okay so again I've been through this in the lesson this is just a review for things we've already done look previous videos or ask me about this if you're not understanding those two so sign analysis again a little more detailed here what we want to do the process includes moving all terms if they're not already to one side of the inequality sign so that you have a zero on the other side so it's going to be important for us to have a full expression of the um the the related equation there is an expression of the polynomial let's say most of them are polynomials for you and then either we have our inequality less than or equal or equal to or greater than and then we have zero on the other side so that's going to be important you have to have this once you do that you want to factor the expression okay so on to step two let's go ahead and do that now so Step One is already done step two I want to factor this expression and it's a trinomial so I'll expect that I should be able to factor this as two binomials so X should be the first term and it looks like we have negative three and positive one and the reason why I would guess that is because negative 3 times positive 1 is negative 3 and negative 3 plus 1 is negative two so that should be that in factored form okay um now what we want to do is we want to chart or list the factors on a chart and then decide where these factors are positive and negative let me show you how to do that so the chart looks like this it's kind of an elongated l like this and oh gosh let me just okay now it's moving so this bottom part of the owl here is actually the number line so I'm going to put little arrows there I would encourage you to do that as well and we're going to list the factors in doesn't matter what order but we're going to list the factors on the side of this elongated L and like I say as we move through math classes you're going to have polynomials that might have three factors or four or five factors you might have rational functions where you have factors on the top and factors on the bottom where you have to divide and multiply all of this can happen with sign analysis sine analysis covers all of those complicated Expressions as well so in this case we have just two factors most of the what you're going to deal with is just two factors so what we do now is we okay now we have to see what we're doing is we're charting where the factors are positive and negative and actually where there's zero as well so watch this am I still frozen I thought I hit that done freeze button is that better all right sorry sorry okay I'll let you catch up there I don't know what how long it's been frozen I hit the wrong button I think so uh good now so elongated L bottom is the uh number line what I'm going to do is I'm going to identify First where these factors are zero so where does this Factor equal zero for what value of x does this Factor equal zero at x equals three okay when X is 3 this factor is zero so I'm going to put 3 on my number line over here and in the row that x minus 3 is I'm going to put a 0 in the row of this factor and above 3 just like that so this says that this factor is equal to zero when X is 3. okay let's do the same for the other one where is this Factor equal to zero at what x value very good negative one so I'm going to put negative 1 here and what's important is that you have the numbers properly in order like least to greatest on this number line doesn't matter where they are as long as they're in order relative to each other so I'm going to put negative 1 over here and I'm going to put a 0 above that in line with this Factor are we good so far okay so we want to label the zeros now we chart the signs so what this means let's go back to the first one if x is less than 3 any number that's less than three and I insert that number into X here like let's say it's a 0 because 0 is less than three what is the sign on this Factor it's negative so all I'm doing is I'm saying that and here's uh here's zero right I'll just put zero here that's a good test point because it's less than 3 here so when zero minus three that's a negative 3 that this factor is equal to so that's a negative here and it just so happens that every number I pick on this side of x equals 3 is going to give us a 0 a negative value for this Factor now if the number was greater than 3 like 10 10 minus 3 is a positive number so guess what we put plus signs in the same row but on the other side of the zero okay so I'm kind of visualizing where this factor is going to be positive or negative and the reason why this is helpful is because we can do it to any number of factors that we have listed here and and then I'll show you what we do at the end we just simply kind of find out the regions where it's positive or negative so here if I choose a number that's less than negative 1 like negative 10 well negative 10 plus 1 gives me a negative value for this whole expression so that's negative on this side and similarly if I choose zero zero plus one is a positive one so positive number so it's positive for every value on this side any any questions are so far on this or are you understanding this so far okay now it seems like a lot of work like to draw pictures here but trust me this is really going to help you straighten things out okay now we've got all the factors going to we've got all of them in the chart now what we do is we say okay this is we're actually kind of testing the related function right so I I always put like the function here what is the function value so what we're going to do now is we're going to say well there is a split right here at this zero so at each zero there's kind of a split in the regions so I have a split here and I have a split here you see that so I'm gonna find out what is the sign when I multiply these two factors in this region a negative value for this one times a negative value gives me positive so we see that the function is overall positive here in this this first section now if we have one factor as a zero at negative one I have zero so you can put a zero there a negative number times 0 is 0. in the middle section we take okay this this top factor is a negative the bottom factor is a positive when I multiply a negative times a positive what I get I get a negative so the function overall is negative in this middle section and a zero times a positive number would be zero so we are zero here and that makes sense right the function should the Y value of the function should be zero at an intercept which these are intercepts so and then here positive times positive is positive all right so what we're focused on is this line down here that's the result that's what we're looking for and what this tells us is similar to the second method where you have test points intercepts and test points this is similar but you can see that where it's positive and negative without using a test Point okay so the function is positive when X is less than 1. the function is also positive when X is greater than 3 because you see that here and here the function is negative between negative 1 and 3. so let's go back to our original question where is this function less than or equal to zero well where is this function the related function here less than or equal to zero it's be between negative one and positive 3 inclusive so the solution set here we just go from this sign analysis chart and we say from negative one less than or equal to X less than or equal to 3. and that signifies that's everything between negative one and three and including negative one and three okay so it's slightly different but but trust me this is actually going to go real quick and when you have five factors in calculus and you're doing this you're going to want to do this you're not going to want to do test points and figure out all the intercepts and just you know test you're just going to want to do this chart and you're just going to go plus minus zero you know plus zero minus zero plus it's going to go real quick okay should we do one more example let's do one more example here all right so here's a question uh from your assignment it says use sign analysis to determine the solution to each inequality so let's do 5A together and we'll go through all the steps and I'll show you how um nice and easy this can be all right so hopefully you've got that written down uh first step is we got to get everything over to one side so that we have a zero on the other side that's going to be pretty easy to do I'm going to bring that 18 over to this side and it's going to be less than or equal to zero okay all right awesome it's it's nice to deal with a positive x squared coefficient here right that's always best if let's just say if you had negative x squared is less than or equal to zero if you wanted to get rid of this negative do you remember the small rule with inequalities when you multiply or divide by a negative they might remember what you have to do there's there's a you have to flip the sign right so just as an example if I wanted to do this if I want to divide both sides by negative 1 I would have to flip this sign okay so that goes for this too if you have a negative x squared and you don't like factoring a negative x squared which I would agree with then flip the sign and divide everything by negative one to get rid of it but you just have to remember to flip the sign okay all right so you might do that in some questions all right so let's Factor can we Factor this hopefully we have a nice easy method to factoring this if you if it's not easy what you might have to do is use the quadratic formula and you might have an irrational you know irrational Roots which is not the end of the world but let's see what we can do so I'm thinking 6 times 3 is 18 and then six and three have a difference of three so I think we can make this work right so let's go plus six and minus three that multiplies to negative 18 and adds up to positive three doesn't it all right so let's do sign analysis I'll put an X plus 6 over here and an x minus 3 on the side of my little chart put my number line down and I see that I have zeros at remember that's at negative six and positive three so let's put those on the number line in the right order negative 6 first and positive 3 towards the right I'll plop zeros down at the appropriate spot so zeros is where you want to kind of start it's a good idea to do zeros through first so zero is there and there because again at x equals negative six this factor is equal to zero everything before negative six so less than negative six like negative a hundred Okay negative a hundred plus six that's going to be negative 94. so we're definitely going to have a negative value for everything that's less than negative six and a number like a hundred while a hundred plus six is going to give us a positive value so most of the time and I'll tell I'll tell you this as a shortcut as if you can remember it but if this x value in this term if it's positive then it's always going to end up to be negatives on the left side of the zero and positives on the right when you get into the the case where we have 6 minus X if this is a negative value in front of it like a negative then it's flipped and that might be too much for you to remember right now so I would do test points for these to begin with just to think that through but anything less than negative positive 3 here let's say is going to give us a negative value for this Factor and that anything greater than 3 is going to give us a positive value okay all right now the function itself okay and again we don't see f of x anywhere in here but the related function right here that we're kind of examining that's the f of x that's what we're examining that's what we're interested in the sign there is going to be well multiply negative times negative we get positive oh multiply 0 times a negative we get zero positive times negative is negative positive times zero is zero my computer is starting to shut down I mean positive times positive is positive so use sign analysis to figure this out well again this is the original question we did one short small manipulation here to see that I'm looking for X values that make the related function negative you see less than or equal to zero so negative or zero so I would just kind of go from here go from this second line here and say yeah I'm looking for the negatives while it's in here so that is related to the values of negative six all the way up to positive 3 and including including them so inclusive so solution set equals funny brackets the variable is X so we'll use x in the middle there less than or equal to less than or equal to use those signs from negative 6 to positive 3. and that is the proper convention for stating that all the X values that are solutions in this question are greater than negative six or equal to as this at the same time as being less than or equal to positive three and that's how you signify all these in here all right good hopefully you can do that in your own now so that's that's the kind of the last little part for uh 9.2 so you've got all three methods now that you can use to solve inequalities in one variable good all right |
187966 | https://www.wyzant.com/resources/answers/944152/how-do-you-change-put-a-an-equation-into-standard-form | WYZANT TUTORING
Jada B.
How do you change/put a/an equation into standard form?
Solve the four equations and please show your work for how you solved them.
1.5x-6y=15
2.-8-7y=20
3.3x+5y=-6
4.3x+9y=13
3 Answers By Expert Tutors
James S. answered • 04/28/24
BEST HELP MATH TUTOR
The requirements for standard form of a linear equation in 2 variables (x and y, for instance) is that it be written in the form ax + by = c where:
a, b, and c are all integers;
a ≥ 0;
a, b, and c can have no common factor other than 1.
a, b, and c cannot all be 0. (Otherwise you have the trivial equation 0=0.)
a and b cannot both be zero if c is not equal to 0 also (Otherwise you have 0 equal to a non-zero number, which is not true.)
If either a, b, or c is irrational, requirement (1) above can be relaxed.
However all of these given equations meet those requirements, except for (2). The -8 should be subtracted from both sides resulting in:
-7y = 28
Or, if this was supposed to be -8x, then multiply both sides by -1, resulting in:
8x + 7y = -20
Rule (2) requires the coefficient of x to be non-negative.
These equations cannot be solved. There must be 2 equations in 2 unknowns to solve them.
You can convert them to slope-intercept form so that you have y = some expression involving x and a constant. You could also convert to x-intercept form so that you have x = some expression involving y and a constant.
Sherri M. answered • 05/01/24
MY STUDENTS PASS ALL THEIR SOLs
Hi, standard form in my understanding is your ability to graph each equation in the format y=mx+b which is slope intercept format
Equation 1: 5x-6y = 15
Step 1: Using the addition principle, subtract 5x from both sides to generate a new equation
5x-6y = 15
-5x -5x
new equation: -6y = -5x+15
Step 2: Using the division principle, divide both sides by -6 but using a fraction format in case you can simplify the equation
-6y/-6 = -5x/-6 +15/-6
Final answer: y = - (15-5x)/6 The negative sign is actually on the fraction bar since the denominator was negative.
Equation 2: -8x-7y = 20
Step 1: Using the addition principle, add 8x to both sides to generate a new equation
-8x-7y = 20
+8x +8x
new equation: -7y = 8x+20
Step 2: Using the division principle, divide both sides by -7 but using a fraction format in case you can simplify the equation
-7y/-7 = 8x/-7 +20/-7
Final answer: y = - (20+8x)/7 The negative sign is actually on the fraction bar since the denominator was negative.
Equation 3: 3x+5y = 6
Step 1: Using the addition principle, subtract 3x from both sides to generate a new equation
3x+5y = 6
-3x -3x
new equation: 5y = -3x+6
Step 2: Using the division principle, divide both sides by 5 but using a fraction format in case you can simplify the equation
5y/5 = -3x/5 +6/5
Final answer: y = -3/5 x + 6/5
Equation 4: 3x+9y = 13
Step 1: Using the addition principle, subtract 3x from both sides to generate a new equation
3x+9y = 13
-3x -3x
new equation: 9y = -3x+13
Step 2: Using the division principle, divide both sides by 9 but using a fraction format in case you can simplify the equation
9y/9 = -3x/9 +13/9
Additional step due to simplifying: y = -3/9 x + 13/9 The slope which is the number in front of the variable "x" can be simplified by dividing by 3. -3/9x simplifies by dividing by 3 turns into -1/3x
Final answer: y = -1/3x + 13/9
Dalton P.
answered • 04/28/24
Instructor Access To Webassign Assignments With 10+ Years Of Tutoring!
What is your definition of standard form? By the usual definition, all these equations are in standard form. In general, the equation ax+by=c for numbers a, b, and c is the standard form of a line. If you wish to solve for y, then, for example, equation 1. can be done as follows: 5x-6y=15, subtracting 5x from each side gives -5x+5x-6y=15-5x, since -5x+5x=0, this becomes 6y=15-5x. Dividing each side by 6, we get 6y/6=(15-5x)/6, since 6/6=1, this becomes y=(15-5x)/6.
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187967 | https://math.stackexchange.com/questions/1871717/clarification-on-what-this-author-means-by-logging-does-not-change-the-maximum | calculus - Clarification on what this author means by "logging does not change the maximum" of a function - Mathematics Stack Exchange
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Clarification on what this author means by "logging does not change the maximum" of a function
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I am self-studying for an actuarial exam and I encountered the following:
The author seems to suggest that if we want to find the maximum of a function f(x)f(x) with respect to x x:
We can drop any multiplicative constant.
We can take the natural log of f(x)f(x) and ln(f(x))ln(f(x)) has the same maximum as f(x)f(x).
Property 1 is intuitive. If f(x)=k⋅g(x)f(x)=k⋅g(x), then setting f′(x)=k⋅g′(x)=0 f′(x)=k⋅g′(x)=0 will result in g′(x)=0 g′(x)=0.
I'm not necessarily convinced of property 2.
Why is it that f(x)f(x) has the same maximum as ln(f(x))ln(f(x))? I took several semesters of calculus/analysis in university, and I don't recall this property.
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asked Jul 26, 2016 at 14:54
Joseph DiNataleJoseph DiNatale
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ln(x)ln(x) is strictly increasing. That is, x>y⟺ln x>ln y x>y⟺lnx>lny (trusting everything is defined, of course). Any strictly increasing function preserves max.lulu –lulu 2016-07-26 14:56:40 +00:00 Commented Jul 26, 2016 at 14:56
2 It doesn't have the same maximum, but it reaches a max at the same place, for ln ln is an increasing function.André Nicolas –André Nicolas 2016-07-26 14:57:20 +00:00 Commented Jul 26, 2016 at 14:57
Re: "drop multiplicative constant" - This holds only for positive constants. A negative constant would swap the roles of maxima and minima Hagen von Eitzen –Hagen von Eitzen 2016-07-26 15:12:11 +00:00 Commented Jul 26, 2016 at 15:12
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Good question, this follows because the function l o g l o g is strictly increasing. This means that whenever x>y x>y, l o g(x)l o g(x) is greater than l o g(y)l o g(y).
So in this case, if you can find the value x, such that f(x)f(x) is maximised. Then that same value x x, must be the maximum of l o g(f(x))l o g(f(x)).
I hope that makes sense (and apologies for the formatting, I'm figuring out how to do the mathematical format in the answer!),
Kevin
(note: the value of l o g(f(x))l o g(f(x)) is not equal to the value of f(x)f(x), but I the question means that both are maximised for the same value of x x)
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answered Jul 26, 2016 at 14:57
Kevin CKevin C
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Exponent Rules Graphic Organizer Cheat Sheet
Created by
Boldly Inspired Curriculum
This free reference sheet is an excellent resource for students who need a little help remembering the differences between all of the properties. Quick and easy resource to give to individual students, use during group work, or present to the whole class during independent work. You may also like: Exponent Properties Escape RoomDividing with Exponents Task Cards
9th - 11th
Algebra, Algebra 2
CCSS
HSN-RN.A.2
FREE
Rated 4.82 out of 5, based on 11 reviews
4.8 (11)
Properties of Exponents Reference Sheet for Interactive Notebooks
Created by
Thought Well Math
This is a reference sheet with all of the laws of exponents expressed algebraically in a table. The table also includes space for students to solve a numerical example for each property. It can easily be cut to fit into a composition notebook. I hope it's useful! Practice with JeopardyI just made this Jeopardy Game to review the laws of exponents with my students. Check it out!
8th - 9th
Math
CCSS
8.EE.A.1
FREE
Rated 4.69 out of 5, based on 13 reviews
4.7 (13)
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Laws of Exponents Cheat Sheet
CutiePiNomial
$1.50Original Price $1.50
Rated 4.88 out of 5, based on 8 reviews
4.9 (8)
Laws of Exponents and Scientific Notation Reference Sheet
The Math Standard
$1.75Original Price $1.75
Rated 5 out of 5, based on 4 reviews
5.0 (4)
Laws of Exponents Student Choice Project | Exponent Rules Reference Sheet
AbsoluteValueMath
$3.00Original Price $3.00
Rated 5 out of 5, based on 3 reviews
5.0 (3)
Rules of Exponents, Reference Sheet
Elizabeth Kissel
$1.50Original Price $1.50
Rated 4.85 out of 5, based on 38 reviews
4.9 (38)
Rules of Exponents Cheat Sheet DISTANCE LEARNING
Created by
MathHop by Jackie B
This is a handout for students with all of the rules of exponents that they need to know in 8th grade. I have my students keep this in their binder because properties and rules of exponents is something that they often forget. Connect with Me!Use Hundreds of my Math Videos in Your Class!Follow me on Facebook!Follow me on Instagram!Looking for resources for your specific class? Check out my custom course materials:Algebra 1 ResourcesAlgebra 2 ResourcesGeometry ResourcesPrecalculus ResourcesCalcu
8th - 10th
Math
FREE
Rated 4.83 out of 5, based on 6 reviews
4.8 (6)
Exponents Cheat Sheet
Created by
The Toolbox for Teachers
This is a PDF File. This Chart shows the values of 1 - 30 raised from the 2nd power to the 15th power. It is on a legal size sheet of paper. This is perfect for those that do not own or struggle to operate scientific calculators
6th - 12th, Adult Education, Higher Education
Algebra, Math
FREE
Rated 4.93 out of 5, based on 16 reviews
4.9 (16)
Exponent Properties Reference Sheet
Created by
Math with Mrs D a Thinking Classroom
This is a great resource for students. You can find this resource linked on the Title Slide and Objectives Slides of all our Algebra 2, Precalculus, and AP Precalculus virtual classrooms. These virtual classrooms can be found in any of our "FULL LESSON" listings. Check out our free download to see more: DOMAIN + RANGE + CONTINUITY FREE LESSON
9th - 12th
Algebra 2, Math, PreCalculus
FREE
Rated 3 out of 5, based on 1 reviews
3.0 (1)
Exponents Cheat Sheet
Created by
Seven and Eight Designs
This is a one page guide to exponent rules. This can be placed into interactive math journals, one your classroom walls, or used for individual students who may need an extra guide to help them along.
6th - 12th
Algebra, Math, Other (Math)
FREE
Rated 5 out of 5, based on 6 reviews
5.0 (6)
Exponent Rules Cheat Sheet
Created by
Emily Schalkham
Freebie exponent rules cheat sheet!
8th - 12th
Algebra, Algebra 2
CCSS
8.EE.A.1
, 8.EE.A.2
FREE
Exponent Rules Reference Sheet
Created by
Washington Science Hub
Exponent rules with examples for multiplication, powers, division, negative powers, zero power, power of one, and introduction to scientific notation.
8th - 12th
Algebra, Math
FREE
Rated 5 out of 5, based on 1 reviews
5.0 (1)
Exponents Cheat Sheet with negative exponents
Created by
DeBer's Bulletin
Grab this free resource today!
7th - 10th
Math
FREE
Exponent Rules Reference Sheet
Created by
MG Math
A reference of each exponent rule with description and example. I also offer the same sheet with more examples and practice for each rule!
7th - 10th
Algebra, Math, Math Test Prep
CCSS
8.EE.A.1
FREE
Math Reference Sheet - 5th Grade
Created by
Sarah Rafferty
Equip your students for math class with a resource they can reference throughout the year! This reference sheet covers various 5th grade math topics such as: -Place Value -Improper Fractions & Mixed Numbers -Adding, Subtracting, Multiplying, and Dividing Fractions -Adding, Subtracting, Multiplying, and Dividing Decimals -Exponents -Order of Operations -Perimeter, Area, and Volume Formulas -Geometry Vocabulary List -Customary & Metric Conversion Charts -Keyword Lists for Addition, Subtraction
4th - 6th
Math
FREE
Rated 4.77 out of 5, based on 13 reviews
4.8 (13)
TI-84+ Graphing Calculator Reference Sheet for Students
Created by
My Heart is Full of Math
Texas Instruments TI-84+ Calculator Tips makes a great reference sheet for students to use when learning or needing a refresher on how to perform tasks on their calculator. The focus of this reference sheet is what would be needed for Pre-Algebra or Algebra 1. You can also print these poster size to hang in your classroom. You might also like: TI-84 Graphing Calculator ExplorationTopics included:Turning the Calculator On/OffChanging the WindowClearing the MemoryVariable X2nd KeySquares and Squ
8th - 9th
Algebra, Math
FREE
Rated 4.69 out of 5, based on 49 reviews
4.7 (49)
Desmos Graphing Calculator Reference Sheet for Students
Created by
My Heart is Full of Math
Desmos Graphing Calculator Tips makes a great reference sheet for students to use when learning or needing a refresher on how to perform tasks on their calculator. The focus of this reference sheet is what would be needed for Pre-Algebra or Algebra 1. You can also print these poster size to hang in your classroom. You might also like: Desmos Graphing Calculator ExplorationTopics included:Starting Desmos Test Mode Using NegativesPutting in Mixed NumbersExponentsDecimal to a FractionVariablesAbso
8th - 9th
Algebra, Math
FREE
Rated 4.82 out of 5, based on 33 reviews
4.8 (33)
Properties and Laws of Exponents: Poster/Anchor Chart
Created by
Brooks Jones
Working with exponents can be confusing! Help your students master the properties and laws of exponents with this jam-packed one-page resource, which can be used as a cheat sheet, anchor chart, or classroom poster. Concepts included are the properties of multiplication, division, zero and negative exponents, as well as rational (fractional) exponents. Properties are color-coded, shown with variable symbols and explained in clear language. Aligns with Common Core State Math Standards CCSS 8.EE.1
8th - 9th
Algebra, Arithmetic, Math
CCSS
8.EE.A.1
, HSN-RN.A.1
FREE
Rated 4.88 out of 5, based on 8 reviews
4.9 (8)
Order of Operations Poster and Cheat Sheet Handout for Middle School Math Decor
Created by
Creative Access
Enhance your classroom environment and provide your students with a clear and colorful guide to mastering the order of operations. This cleanly designed poster is not only visually appealing but also a fantastic math reference for middle school students. This resource breaks down the complex world of order of operations into four easy-to-follow steps, empowering your students to tackle challenging mathematical expressions and preparing them for work with algebraic expressions and equations. Ste
5th - 8th
Basic Operations, Math, Order of Operations
Also included in: Order of Operations Bundle for Evaluating Expressions and Solving Equations
FREE
Rated 4.78 out of 5, based on 9 reviews
4.8 (9)
Doodle Exponential Functions Reference Sheet
Created by
Alexis O'Malley
This is a handwritten reference sheet that I created for my students. It includes... Equation of an exponential functionExponential Growth vs. DecayExponent RulesWord Problem ExampleConverting Percents to DecimalsConverting Decimals to PercentsWriting the equation from a table (x increasing by 1)Writing the equation from a table (x = 0, x = 2)
9th - 11th
Algebra
FREE
Rated 4.82 out of 5, based on 11 reviews
4.8 (11)
Exponent Rules Digital Poster
Created by
Lyliana Chavez
Have a list of exponent rules in your digital classroom! Use this poster as a reference sheet for all students to look back to. This product comes with a filled out and blank version of the rules and examples. Great to use as a review tool.
7th - 9th
Math
CCSS
8.EE.A.1
FREE
Rated 4.43 out of 5, based on 7 reviews
4.4 (7)
Unit 1 Honors Precalculus Reference Sheet (Functions & Graphs)
Created by
Hazel Marie
Great tool for studying for unit 1 test! 1.1 Introduction to Precalculus - factoring, exponent rules, how to use graphing calculator 1.2 Functions & Their Properties - domain, range, increasing/decreasing/constant, bounds, even/odd, continuity, extrema, intercepts 1.3 Parent Functions & Transformations 1.4 Composition of Functions - combining, composition, decomposition 1.5 Inverse Functions - solving & graphing
10th - 12th
Math, PreCalculus
CCSS
HSF-IF.A.1
, HSF-IF.A.2
, HSF-IF.A.3
+7
FREE
Rated 4.33 out of 5, based on 3 reviews
4.3 (3)
Expressions: Order of Operations & Evaluating | 8th Grade Math Reference Sheet
Created by
This 8th Grade Math Quick Reference Sheet explains what the order of operations is as well as how to evaluate algebraic expressions. This info sheet can be used for students to review after they've already learned the concept or as an already complete notes page to help teach the concept initially. This is also a great resource to give to parents or substitute teachers so they're better able to help their student(s). More 8th Grade Math Reference Sheets available in my store as a BUNDLE (for a
8th - 9th
Algebra, Math Test Prep
Also included in: 8th Grade Math Reference Sheets | Full Year Resource | Quick Reference Booklet
FREE
Rated 5 out of 5, based on 1 reviews
5.0 (1)
Algebra Cheat Sheet Bible: Every Formula, Strategy and Concept Simplified!
Created by
Calculus Catalyst
Unlock Algebra Success with This All-in-One Cheat Sheet!Stop the stress of scattered notes! This Algebra Mastery Cheat Sheet combines every critical formula, strategy, and concept students need into a single, easy-to-use resource. Perfect for exams, homework, or daily review, it’s designed by an expert educator to simplify complex topics and boost confidence. What’s Inside?✅ Core Concepts Made Easy:Linear Equations & InequalitiesQuadratic Functions (Standard + Vertex Forms)Exponential Growth/De
5th - 10th
Algebra, Calculus
FREE
FREE Order of Operations PEMDAS Cheat Sheets and Video Links
Created by
Instill Skills and Drills
Do your students get confused with the Order of Operations? Let me share my secret! Download this FREE Anchor Chart on Order of Operations, showcasing every possible rule in PEMDAS. It's a fantastic tool for helping students understand PEMDAS rules and provides a step-by-step guide to solving order of operations. Plus, don't forget to explore the FREE teaching videos for supplementary support and the best teaching aids. Display these anchor charts on your bulletin board for quick reference and
4th - 6th
Arithmetic, Basic Operations, Math
Also included in: Order of Operations Practice and Word Problem Worksheets & Video Links BUNDLE
FREE
Rated 5 out of 5, based on 1 reviews
5.0 (1)
Combining Like Terms Key Vocabulary Reference Sheet
Created by
Educated by Kay
Give your students the essential key terms they need to master combining like terms with this FREE vocabulary reference sheet! This resource includes clear definitions for important algebra terms such as variable, coefficient, constant, exponent, term, and like term—plus a labeled example to help students see each concept in action. ✨ Perfect for: ✔ Interactive notebooks ✔ Math word walls ✔ Quick reference during lessons ✔ Independent practice support This free download is also includ
6th - 8th
Algebra, Vocabulary
FREE
Prime Factorization Reference Sheet
Created by
Rachael Whittemore Reference Sheets
A perfect resource for students solving problems related to factor trees, prime factorization, and exponents.
6th
Math, Math Test Prep, Numbers
CCSS
FREE
Rated 5 out of 5, based on 1 reviews
5.0 (1)
Laws Of Exponents GRAPHIC ORGANIZER
Created by
Progressive Pedagogy
A wonderful, easy to read graphic organizer when introducing laws of exponents. This version is free to use. A great print out for your students to put in their binder, or for you to laminate as a reference sheet for small groups! Looks great in color or black and white. I hope you enjoy this product!
7th - 10th
Algebra, Math
CCSS
8.EE.A.1
FREE
Rated 5 out of 5, based on 1 reviews
5.0 (1)
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187969 | https://medium.com/@jasparquinton/the-strip-mall-and-suburban-apocalypse-5294f3bdd94a | The Strip Mall and Suburban Apocalypse | by Jaspar Quinton | Sep, 2025 | Medium
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The Strip Mall and Suburban Apocalypse
Jaspar Quinton
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9 min read
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Sep 18, 2025
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How American Civic Design Is Killing Our Collective Soul
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The reality of today’s civic landscape: strip malls, parking lots, chain stores. Our most common public architecture teaches commodification and disposability, not pride.
I just got back from San Francisco, where I spent days wandering among the Palace of Fine Arts, the Legion of Honor, and block after block of ornate Victorian homes. Even the ordinary apartment buildings there seem touched by grandeur — crowned with cornices, carved in detail, layered with history. The city overwhelms you with the sense that you are walking through a place that believes in itself, a civilization confident enough to build beauty into the everyday.
Then I returned home to Washington, to the South Puget towns of Olympia and Tacoma — and the contrast was crushing. Here, civic life happens in strip malls, beige subdivisions, and prefab complexes built to be forgotten. Olympia and Tacoma both have a nucleus of true dignity — Olympia with its Capitol dome rising over the Eastside, a scattering of classical civic buildings with weight and presence; Tacoma with its ornate early 20th-century city center — but both are surrounded on all sides by sprawl that radiates nihilism. The beauty is there, like a memory of who we could be, but it has been buried under a century of drywall and asphalt.
This is the American condition. We live in a country that once built cathedrals of democracy and infrastructure that made the future feel inevitable and beautiful — and now settles for parking lots and chain stores as its civic heart. Our built environment isn’t neutral. It seeps into us. It shapes our sense of destiny, or our lack of it. And right now, it’s teaching us despair.
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The Legion of Honor museum in San Francisco. Even cultural buildings here were designed as monuments, instead of utilitarian tools — declaring permanence and belief in civilization.
When We Believed in Ourselves
During the Great Depression, when unemployment hit 25% and the future looked hopeless, FDR launched the Works Progress Administration (WPA). On paper it was a jobs program. In reality, it was a civilizational statement.
The WPA didn’t just build roads and bridges — it built post offices, schools, libraries, fire stations, dams, theaters, airports, and courthouses. It commissioned tens of thousands of murals and sculptures. It sent artists, writers, and musicians across the country to record and glorify American life.
While it addressed economic issues like unemployment, it was so much more — it was a national ethos in the form of infrastructure. A child walking into a WPA school saw more than bricks — they saw permanence, ornament, a belief that their town was worth beauty and that their government agreed. A farmer mailing a letter in a WPA post office saw their labor painted as heroic in the murals above their head. Even the dams and highways were designed as monuments to progress, curving with grace, inscribed with carvings and plaques.
The message was unmistakable: you are part of something enduring, and this nation believes in its future and ideals enough to etch it in stone.
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A WPA Mural at a municipal building in Alameda, CA. The program treated ordinary Americans as subjects worthy of grandeur, weaving art into the fabric of everyday civic life.
What We Haven’t Built in Generations
Now ask yourself: when was the last time America built something like that? A monument that people will travel to a century later, not for utility but for awe?
The Lincoln and Jefferson Memorials? Mid-20th century. The Hoover Dam? The 1930s. The interstate system of the 1950s? Functionally impressive, but hardly sublime. Since then, we’ve built glass boxes, lifeless bureaucracies, and disposable suburbs.
Even the exceptions prove the rule. The Vietnam Veterans Memorial is deeply moving, but it is a monument of mourning, not destiny. Half a century later, what has followed? Where is the temple to the civil rights movement? Where is the plaza of American democracy in the digital age? Where are the cathedrals of knowledge and science for the 21st century?
We haven’t built them — because somewhere along the way, we stopped believing we deserved them.
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Oskar Hansen’s often forgotten “Winged Figures of the Republic "at Hoover Dam — WPA-era architecture as myth, where infrastructure and art fused into civilizational confidence.
The Architecture of Shame
San Francisco is awe-inspiring not just because of its landmarks, but because it was built to be something larger than itself. After the 1906 earthquake, the city’s leaders pitched the 1915 Panama-Pacific International Exposition as proof that San Francisco was not just reborn, but destined. In the run-up to the fair, boosters called it “the first city of the West,” positioning it as a civic and cultural capital for the Pacific, not just another port town.
The Palace of Fine Arts, the Legion of Honor, the Victorian neighborhoods, the bridges — they were designed to awe. They told the world: this city is a civilization, not a campsite. Even its ordinary housing stock carries the pride of that vision.
Most American cities were not built with this ambition. They were built like frontier outposts — temporary, disposable, interchangeable. That ethos — capitalist churn layered on top of settler-colonial disposability — still governs our civic design. Build quick, flip it, abandon it. Extraction, not permanence.
And this has consequences. America’s lack of beauty is not just an aesthetic failure — it’s a psychic one. Strip malls, subdivisions, and disposable buildings do more than depress the eye. They train us to feel that we have no national identity, no continuity, nothing worth inheriting.
This is why so many Americans describe our culture as hollow, or why patriotism so often curdles into rage or irony. The landscape around us teaches us to see ourselves as cheap and temporary. It feeds not pride, but shame and self-disgust.
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In short: we don’t just live in sprawl. We become sprawl.
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Bernard Maybeck’s Palace of Fine Arts, built for the 1915 Panama-Pacific International Exposition — San Francisco presenting itself as “the first city of the West,” a Rome for the Pacific.
Architecture as Psychic Infrastructure
Carl Jung wrote about the collective unconscious — the deep reservoir of archetypes and symbols that shapes human thought beneath awareness. But nations, too, have their collective unconscious, and it is written in stone, brick, wood, and steel.
San Francisco makes this obvious. To walk through a city where even the ordinary homes are adorned with ornament and history is to feel an unconscious pride: this is a civilization that endures and believes in itself. Olympia, by contrast, sends mixed signals. The Capitol campus is genuinely august — classical columns, granite steps, a dome rising like a promise. Tacoma still carries soul in its downtown — towering early skyscrapers, the remains of the industrial age turned into museums and even a university. For a moment, you feel what America once tried to be. But step outside those nuclei, and you’re in a wasteland of parking lots, strip malls, and subdivisions with no identity beyond square footage.
That contrast is more than aesthetic whiplash — it’s psychological violence. A courthouse that looks like a warehouse teaches you that justice is banal. A school that looks like a temporary office teaches you that knowledge is disposable. A town that builds strip malls as its civic commons teaches you that community itself is an afterthought.
Architecture is not background — it’s psychic infrastructure. Just as a neglected home corrodes a person’s spirit, a neglected built environment corrodes a people’s sense of destiny.
The Washington State Capitol dome in Olympia — one of the last gasps of American civic grandeur before sprawl overtook permanence. A reminder that even smaller cities once aspired to the sublime.
A Charter for the New Sublime
If the built environment is psychic infrastructure, then rebuilding America’s confidence means rebuilding America’s beauty. We cannot keep living in drywall and asphalt and expect to produce anything but despair. A civilization that believes in itself builds like it believes in itself.
Here’s what that means:
Civic dignity is a right.
Every courthouse, library, fire station, and town hall should be designed with dignity. No citizen should walk into a building of justice or knowledge and feel like they’ve entered a strip mall.
2. Permanence over disposability.
Our buildings should be made to last a century or more. The drywall wasteland we accept today is an insult to the generations who will inherit it. If it won’t inspire in 2125, don’t build it.
3. Regional vernaculars revived.
A nation this vast should not flatten itself into beige sameness. Cascadia deserves its cedar, basalt, and rain-soaked plazas. The Southwest deserves its adobe and sandstone. The Midwest deserves its muscular brick and steel. Regional soul creates national pride.
4. Art integrated into every project.
The WPA understood this: a building without art is incomplete. Murals, mosaics, carvings, and statues are not extras — they are the symbolic lifeblood of civic space.
5. Infrastructure should inspire awe.
A bridge or a dam or a train station is not just utility. It is the theater of civilization. The Hoover Dam made America believe in the future; our infrastructure today should do the same for a new century.
6. Commons treated as sacred.
Plazas, amphitheaters, pedestrian boulevards — spaces for the public to gather should be designed with the same care as cathedrals. Community is sacred. Its spaces should say so.
7. Architecture as psychoanalysis.
Every building tells the unconscious a story. Right now the story is despair. We need to consciously design structures that tell us pride, destiny, and belonging.
This is not a call to copy Rome or Paris, nor a retreat into naive nostalgia — it is a demand that America rediscover the courage to build like it believes in itself. To stop entombing our future in glass and drywall, and instead inscribe confidence into stone, wood, and steel.
Cascadia as a Case Study
If America wants to remember how to build with pride, it can start in my home of the Pacific Northwest. Cascadia already has the raw ingredients of a distinct civic vernacular — it just hasn’t had the courage to embrace them.
Picture civic halls built from cedar and Douglas fir, with exposed beams that echo the forests and carry the scent of resin into the air. Imagine salmon motifs carved into entrances and fountains, honoring the lifeblood of the region’s ecology and economy. Picture courthouses that integrate Indigenous art respectfully — not as token decoration, as a reminder that this landscape has always had a human story far deeper than the suburban grid.
Our pride in the timber industry doesn’t need to vanish into nostalgia — it can be built into our regional identity. A new generation of buildings could celebrate wood not as raw extraction, but as a renewable, sacred material. Rain gardens, green roofs, and cascading water features could embody the cycle of the region’s endless rain, turning climate into symbol rather than inconvenience. Naturalistic design — structures that curve like rivers, plazas lined with basalt, skylights that filter gray light like a forest canopy — could remind people that civic life here belongs to the mountains, the Sound, and the evergreen hills as much as to politics or commerce.
You can already see glimpses of this in Seattle’s libraries, in Portland’s timber-glass civic projects, in Tacoma’s repurposed industrial warehouses turned into museums. But these are fragments, drowned in a sea of vinyl-sided subdivisions and warehouse offices.
If Cascadia embraced this vernacular deliberately — if every town hall, courthouse, and school was designed as a cedar-and-stone shrine to the region’s forests, salmon, and rain — it would not only lift the spirit of those who live here. It would also become one thread in a national tapestry.
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The new additions to PDX is a prime example of what a Cascadian Architectural vernacular could look like.
You can already glimpse this future in the new timber hall at Portland International Airport. Its vast Douglas fir roof curves like a forest canopy, filtering light through lattices of wood. The floor is lined with natural materials, trees rise indoors, and the entire space feels less like a transit hub than a cathedral to Cascadia’s forests and rain. It shows that when we choose permanence, regional material, and symbolic design, even an airport concourse can become a work of civic pride.
Imagine a United States where each region built like this: the Southwest with adobe and desert stone, the Midwest with muscular brick and steel, New England with red-brick spires, the Deep South with verandas and courtyards. Together, they would form a pluralistic sublimity instead of bland uniformity — a country that takes pride in its differences while finding unity in the shared act of building beauty.
Right now, America is teaching itself despair through its architecture. Strip malls and subdivisions are not just ugly — they are monuments to nihilism, etched into the landscape for all to see.
The choice is simple: we either build beauty again, or we keep entombing our future in glass and drywall.
America
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Consumerism
Architecture
Philosophy
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Written by Jaspar Quinton -------------------------
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I'm interested in culture, power, and the physical world: what our cities, myths, and politics reveal about who we are — and what we can do about it.
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Contents
1 Reactions
1.1 Structure and properties
2 Production
2.1 Historical routes
2.2 Direct synthesis
2.3 Organic synthesis
2.4 Laboratory methods
3 Applications
4 History
5 Safety
6 See also
7 References
8 External links
Hydrogen chloride
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From Wikipedia, the free encyclopedia
"HCl" redirects here. For the aqueous solution, see Hydrochloric acid. For other uses, see HCL (disambiguation).
Hydrogen chloride
| | | | --- | | | | |
| Names |
| IUPAC name Hydrogen chloride |
| Systematic IUPAC name Chlorane |
| Other names Hydrochloric acid gasHydrochloric gasHydrochloride |
| Identifiers |
| CAS Number | 7647-01-0Y |
| 3D model (JSmol) | Interactive image |
| Beilstein Reference | 1098214 |
| ChEBI | CHEBI:17883Y |
| ChEMBL | ChEMBL1231821N |
| ChemSpider | 307Y |
| ECHA InfoCard | 100.028.723 |
| EC Number | 231-595-7 |
| Gmelin Reference | 322 |
| KEGG | D02057Y |
| MeSH | Hydrochloric+acid |
| PubChem CID | 313 |
| RTECS number | MW4025000 |
| UNII | QTT17582CBY |
| UN number | 1050 |
| CompTox Dashboard (EPA) | DTXSID2020711 |
| InChI InChI=1S/HCl/h1HN Key: VEXZGXHMUGYJMC-UHFFFAOYSA-NY InChI=1/HCl/h1H Key: VEXZGXHMUGYJMC-UHFFFAOYAT |
| SMILES Cl |
| Properties |
| Chemical formula | HCl |
| Molar mass | 36.46 g/mol |
| Appearance | Colorless gas |
| Odor | pungent; sharp and burning |
| Density | 1.49 g/L |
| Melting point | −114.22 °C (−173.60 °F; 158.93 K) |
| Boiling point | −85.05 °C (−121.09 °F; 188.10 K) |
| Solubility in water | 823 g/L (0 °C) 720 g/L (20 °C) 561 g/L (60 °C) |
| Solubility | soluble in methanol, ethanol, ether and water |
| Vapor pressure | 4352 kPa (at 21.1 °C) |
| Acidity (pKa) | −3.0; −5.9 (±0.4) |
| Basicity (pKb) | 17.0 |
| Conjugate acid | Chloronium |
| Conjugate base | Chloride |
| Refractive index (nD) | 1.0004456 (gas) 1.254 (liquid) |
| Viscosity | 0.311 cP (−100 °C) |
| Structure |
| Molecular shape | linear |
| Dipole moment | 1.05 D |
| Thermochemistry |
| Heat capacity (C) | 0.7981 J/(K·g) |
| Std molarentropy (S⦵298) | 186.902 J/(K·mol) |
| Std enthalpy offormation (ΔfH⦵298) | −92.31 kJ/mol |
| Std enthalpy ofcombustion (ΔcH⦵298) | −95.31 kJ/mol |
| Pharmacology |
| ATC code | A09AB03 (WHO) B05XA13 (WHO) |
| Hazards |
| Occupational safety and health (OHS/OSH): |
| Main hazards | Toxic, corrosive |
| GHS labelling: |
| Pictograms | |
| Signal word | Danger |
| Hazard statements | H314, H331 |
| Precautionary statements | P261, P280, P303+P361+P353, P304+P340+P310, P305+P351+P338, P410+P403 |
| NFPA 704 (fire diamond) | 3 0 1 COR |
| Lethal dose or concentration (LD, LC): |
| LD50 (median dose) | 238 mg/kg (rat, oral) |
| LC50 (median concentration) | 3124 ppm (rat, 1 h)1108 ppm (mouse, 1 h) |
| LCLo (lowest published) | 1300 ppm (human, 30 min)4416 ppm (rabbit, 30 min)4416 ppm (guinea pig, 30 min)3000 ppm (human, 5 min) |
| NIOSH (US health exposure limits): |
| PEL (Permissible) | C 5 ppm (7 mg/m3) |
| REL (Recommended) | C 5 ppm (7 mg/m3) |
| IDLH (Immediate danger) | 50 ppm |
| Safety data sheet (SDS) | JT Baker MSDS |
| Related compounds |
| Related compounds | Hydrogen fluorideHydrogen bromideHydrogen iodideHydrogen astatide |
| Except where otherwise noted, data are given for materials in their standard state (at 25 °C [77 °F], 100 kPa). N verify (what is YN ?) Infobox references |
Chemical compound
The compound hydrogen chloride has the chemical formula HCl and as such is a hydrogen halide. At room temperature, it is a colorless gas, which forms white fumes of hydrochloric acid upon contact with atmospheric water vapor. Hydrogen chloride gas and hydrochloric acid are important in technology and industry. Hydrochloric acid, the aqueous solution of hydrogen chloride, is also commonly given the formula HCl.
Reactions
[edit]
Hydrogen chloride is a diatomic molecule, consisting of a hydrogen atom H and a chlorine atom Cl connected by a polar covalent bond. The chlorine atom is much more electronegative than the hydrogen atom, which makes this bond polar. Consequently, the molecule has a large dipole moment with a negative partial charge (δ−) at the chlorine atom and a positive partial charge (δ+) at the hydrogen atom. In part because of its high polarity, HCl is very soluble in water (and in other polar solvents).
Upon contact, H2O and HCl combine to form hydronium cations [H3O]+ and chloride anions Cl− through a reversible chemical reaction:
: HCl + H2O → [H3O]+ + Cl−
The resulting solution is called hydrochloric acid and is a strong acid. The acid dissociation or ionization constant, Ka, is large, which means HCl dissociates or ionizes practically completely in water. Even in the absence of water, hydrogen chloride can still act as an acid. For example, hydrogen chloride can dissolve in certain other solvents such as methanol:
: HCl + CH3OH → [CH3OH2]+ + Cl−
Hydrogen chloride can protonate molecules or ions and can also serve as an acid-catalyst for chemical reactions where anhydrous (water-free) conditions are desired.
Because of its acidic nature, hydrogen chloride is a corrosive substance, particularly in the presence of moisture.
Structure and properties
[edit]
The structure of solid DCl, as determined by neutron diffraction of DCl powder at 77 K. DCl was used instead of HCl because the deuterium nucleus is easier to detect than the hydrogen nucleus. The extended linear structure is indicated by the dashed lines.
Frozen HCl undergoes a phase transition at 98.4 K (−174.8 °C; −282.5 °F). X-ray powder diffraction of the frozen material shows that the material changes from an orthorhombic structure to a cubic one during this transition. In both structures the chlorine atoms are in a face-centered array. However, the hydrogen atoms could not be located. Analysis of spectroscopic and dielectric data, and determination of the structure of DCl (deuterium chloride) indicates that HCl forms zigzag chains in the solid, as does HF (see figure on right).
Solubility of HCl (g/L) in common solvents
| Temperature (°C) | 0 | 20 | 30 | 50 |
| Water | 823 | 720 | 673 | 596 |
| Methanol | 513 | 470 | 430 | |
| Ethanol | 454 | 410 | 381 | |
| Ether | 356 | 249 | 195 | |
The infrared spectrum of gaseous hydrogen chloride, shown on the left, consists of a number of sharp absorption lines grouped around 2886 cm−1 (wavelength ~3.47 μm). At room temperature, almost all molecules are in the ground vibrational state v = 0. Including anharmonicity the vibrational energy can be written as:
To promote an HCl molecule from the v = 0 to the v = 1 state, we would expect to see an infrared absorption about νo = νe + 2xeνe = 2880 cm−1. However, this absorption corresponding to the Q-branch is not observed due to it being forbidden by symmetry. Instead, two sets of signals (P- and R-branches) are seen owing to a simultaneous change in the rotational state of the molecules. Because of quantum mechanical selection rules, only certain rotational transitions are permitted. The states are characterized by the rotational quantum number J = 0, 1, 2, 3, ... selection rules state that ΔJ is only able to take values of ±1.
The value of the rotational constant B is much smaller than the vibrational one νo, such that a much smaller amount of energy is required to rotate the molecule; for a typical molecule, this lies within the microwave region. However, the vibrational energy of HCl molecule places its absorptions within the infrared region, allowing a spectrum showing the rovibrational transitions of this molecule to be easily collected using an infrared spectrometer with a gas cell. The latter can even be made of quartz as the HCl absorption lies in a window of transparency for this material.
Naturally abundant chlorine consists of two isotopes, 35Cl and 37Cl, in a ratio of approximately 3:1. While the spring constants are nearly identical, the disparate reduced masses of H35Cl and H37Cl cause measurable differences in the rotational energy, thus doublets are observed on close inspection of each absorption line, weighted in the same ratio of 3:1.
Production
[edit]
Most hydrogen chloride produced on an industrial scale is used for hydrochloric acid production.
Historical routes
[edit]
In the 17th century, Johann Rudolf Glauber from Karlstadt am Main, Germany used sodium chloride salt and sulfuric acid for the preparation of sodium sulfate in the Mannheim process, releasing hydrogen chloride. Joseph Priestley of Leeds, England prepared pure hydrogen chloride in 1772, and by 1808 Humphry Davy of Penzance, England had proved that the chemical composition included hydrogen and chlorine.
Direct synthesis
[edit]
Hydrogen chloride is produced by combining chlorine and hydrogen:
: Cl2 + H2 → 2 HCl
As the reaction is exothermic, the installation is called an HCl oven or HCl burner. The resulting hydrogen chloride gas is absorbed in deionized water, resulting in chemically pure hydrochloric acid. This reaction can give a very pure product, e.g. for use in the food industry.
The reaction can also be triggered by blue light.
Organic synthesis
[edit]
The industrial production of hydrogen chloride is often integrated with the formation of chlorinated and fluorinated organic compounds, e.g., Teflon, Freon, and other CFCs, as well as chloroacetic acid and PVC. Often this production of hydrochloric acid is integrated with captive use of it on-site. In the chemical reactions, hydrogen atoms on the hydrocarbon are replaced by chlorine atoms, whereupon the released hydrogen atom recombines with the spare atom from the chlorine molecule, forming hydrogen chloride. Fluorination is a subsequent chlorine-replacement reaction, producing again hydrogen chloride:
: RH + Cl2 → RCl + HCl
: RCl + HF → RF + HCl
The resulting hydrogen chloride is either reused directly or absorbed in water, resulting in hydrochloric acid of technical or industrial grade.
Laboratory methods
[edit]
Small amounts of hydrogen chloride for laboratory use can be generated in an HCl generator by dehydrating hydrochloric acid with either sulfuric acid or anhydrous calcium chloride. Alternatively, HCl can be generated by the reaction of sulfuric acid with sodium chloride:
: NaCl + H2SO4 → NaHSO4 + HCl↑
This reaction occurs at room temperature. Provided there is NaCl remaining in the generator and it is heated above 200 °C, the reaction proceeds further:
: NaCl + NaHSO4 → Na2SO4 + HCl↑
For such generators to function, the reagents should be dry.
Hydrogen chloride can also be prepared by the hydrolysis of certain reactive chloride compounds such as phosphorus chlorides, thionyl chloride (SOCl2), and acyl chlorides. For example, cold water can be gradually dripped onto phosphorus pentachloride (PCl5) to give HCl:
: PCl5 + H2O → POCl3 + 2 HCl↑
Applications
[edit]
Most hydrogen chloride is consumed in the production of hydrochloric acid. It is also used in the production of vinyl chloride and many alkyl chlorides. Trichlorosilane, a precursor to ultrapure silicon, is produced by the reaction of hydrogen chloride and silicon at around 300 °C.
: Si + 3 HCl → HSiCl3 + H2
History
[edit]
Around 900, the authors of the Arabic writings attributed to Jabir ibn Hayyan (Latin: Geber) and the Persian physician and alchemist Abu Bakr al-Razi (c. 865–925, Latin: Rhazes) were experimenting with sal ammoniac (ammonium chloride), which when it was distilled together with vitriol (hydrated sulfates of various metals) produced hydrogen chloride. It is possible that in one of his experiments, al-Razi stumbled upon a primitive method to produce hydrochloric acid. However, it appears that in most of these early experiments with chloride salts, the gaseous products were discarded, and hydrogen chloride may have been produced many times before it was discovered that it can be put to chemical use.
One of the first such uses was the synthesis of mercury(II) chloride (corrosive sublimate), whose production from the heating of mercury either with alum and ammonium chloride or with vitriol and sodium chloride was first described in the De aluminibus et salibus ("On Alums and Salts"), an eleventh- or twelfth century Arabic text falsely attributed to Abu Bakr al-Razi and translated into Latin by Gerard of Cremona (1144–1187).
Another important development was the discovery by pseudo-Geber (in the De inventione veritatis, "On the Discovery of Truth", after c. 1300) that by adding ammonium chloride to nitric acid, a strong solvent capable of dissolving gold (i.e., aqua regia) could be produced.
After the discovery in the late sixteenth century of the process by which unmixed hydrochloric acid can be prepared, it was recognized that this new acid (then known as spirit of salt or acidum salis) released vaporous hydrogen chloride, which was called marine acid air. In the 17th century, Johann Rudolf Glauber used salt (sodium chloride) and sulfuric acid for the preparation of sodium sulfate, releasing hydrogen chloride gas (see production, above). In 1772, Carl Wilhelm Scheele also reported this reaction and is sometimes credited with its discovery. Joseph Priestley prepared hydrogen chloride in 1772, and in 1810 Humphry Davy established that it is composed of hydrogen and chlorine.
During the Industrial Revolution, demand for alkaline substances such as soda ash increased, and Nicolas Leblanc developed a new industrial-scale process for producing the soda ash. In the Leblanc process, salt was converted to soda ash, using sulfuric acid, limestone, and coal, giving hydrogen chloride as by-product. Initially, this gas was vented to air, but the Alkali Act 1863 prohibited such release, so then soda ash producers absorbed the HCl waste gas in water, producing hydrochloric acid on an industrial scale. Later, the Hargreaves process was developed, which is similar to the Leblanc process except sulfur dioxide, water, and air are used instead of sulfuric acid in a reaction which is exothermic overall. In the early 20th century the Leblanc process was effectively replaced by the Solvay process, which did not produce HCl. However, hydrogen chloride production continued as a step in hydrochloric acid production.
Historical uses of hydrogen chloride in the 20th century include hydrochlorinations of alkynes in producing the chlorinated monomers chloroprene and vinyl chloride, which are subsequently polymerized to make polychloroprene (Neoprene) and polyvinyl chloride (PVC), respectively. In the production of vinyl chloride, acetylene (C2H2) is hydrochlorinated by adding the HCl across the triple bond of the C2H2 molecule, turning the triple into a double bond, yielding vinyl chloride.
The "acetylene process", used until the 1960s for making chloroprene, starts out by joining two acetylene molecules, and then adds HCl to the joined intermediate across the triple bond to convert it to chloroprene as shown here:
This "acetylene process" has been replaced by a process which adds Cl2 to the double bond of ethylene instead, and subsequent elimination produces HCl instead, as well as chloroprene.
Safety
[edit]
Hydrogen chloride forms corrosive hydrochloric acid on contact with water found in body tissue. Inhalation of the fumes can cause coughing, choking, inflammation of the nose, throat, and upper respiratory tract, and in severe cases, pulmonary edema, circulatory system failure, and death. Skin contact can cause redness, pain, and severe chemical burns. Hydrogen chloride may cause severe burns to the eye and permanent eye damage.
The U.S. Occupational Safety and Health Administration and the National Institute for Occupational Safety and Health have established occupational exposure limits for hydrogen chloride at a ceiling of 5 ppm (7 mg/m3), and compiled extensive information on hydrogen chloride workplace safety concerns.
See also
[edit]
Gastric acid, hydrochloric acid secreted into the stomach to aid digestion of proteins
Chloride, salts of hydrogen chloride
Hydrogen bromide
Hydrochloride, organic salts of hydrochloric acid
Hydrochlorination, addition reaction with alkenes
References
[edit]
^ "hydrogen chloride (CHEBI:17883)". Chemical Entities of Biological Interest (ChEBI). UK: European Bioinformatics Institute.
^ Favre, Henri A.; Powell, Warren H., eds. (2014). Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013. Cambridge: The Royal Society of Chemistry. p. 131. ISBN 9781849733069.
^ Haynes, William M. (2010). Handbook of Chemistry and Physics (91 ed.). Boca Raton, Florida, USA: CRC Press. p. 4–67. ISBN 978-1-43982077-3.
^ Hydrogen Chloride. Gas Encyclopaedia. Air Liquide
^ Tipping, E.(2002) . Cambridge University Press, 2004.
^ Trummal, A.; Lipping, L.; Kaljurand, I.; Koppel, I. A.; Leito, I. "Acidity of Strong Acids in Water and Dimethyl Sulfoxide" J. Phys. Chem. A. 2016, 120, 3663-3669. doi:10.1021/acs.jpca.6b02253
^ a b c NIOSH Pocket Guide to Chemical Hazards. "#0332". National Institute for Occupational Safety and Health (NIOSH).
^ a b "Hydrogen chloride". Immediately Dangerous to Life or Health Concentrations (IDLH). National Institute for Occupational Safety and Health (NIOSH).
^ Ouellette, Robert J.; Rawn, J. David (2015). Principles of Organic Chemistry. Elsevier Science. pp. 6–. ISBN 978-0-12-802634-2.
^ Natta, G. (1933). "Struttura e polimorfismo degli acidi alogenidrici". Gazzetta Chimica Italiana (in Italian). 63: 425–439.
^ Sándor, E.; Farrow, R. F. C. (1967). "Crystal Structure of Solid Hydrogen Chloride and Deuterium Chloride". Nature. 213 (5072): 171–172. Bibcode:1967Natur.213..171S. doi:10.1038/213171a0. S2CID 4161132.
^ Hydrochloric Acid – Compound Summary. Pubchem
^ a b Austin, Severin; Glowacki, Arndt (2000). Hydrochloric Acid. doi:10.1002/14356007.a13_283. ISBN 3527306730.
^ Priestley J (1772). "Observations on different kinds of air [i.e., gases]". Philosophical Transactions of the Royal Society of London. 62: 147–264 (234–244). doi:10.1098/rstl.1772.0021. S2CID 186210131.
^ Davy H (1808). "Electro-chemical researches, on the decomposition of the earths; with observations on the metals obtained from the alkaline earths, and on the amalgam procured from ammonia". Philosophical Transactions of the Royal Society of London. 98: 333–370. Bibcode:1808RSPT...98..333D. doi:10.1098/rstl.1808.0023. S2CID 96364168. p. 343: When potassium was heated in muriatic acid gas [i.e., gaseous hydrogen chloride], as dry as it could be obtained by common chemical means, there was a violent chemical action with ignition; and when the potassium was in sufficient quantity, the muriatic acid gas wholly disappeared, and from one-third to one-fourth of its volume of hydrogene was evolved, and muriate of potash [i.e., potassium chloride] was formed. (The reaction was: 2HCl + 2K → 2KCl + H2)
^ Cramer, Chris. Hydrogen Chloride Cannon.
^ Francisco J. Arnsliz (1995). "A Convenient Way To Generate Hydrogen Chloride in the Freshman Lab". J. Chem. Educ. 72 (12): 1139. Bibcode:1995JChEd..72.1139A. doi:10.1021/ed072p1139. Archived from the original on 24 September 2009. Retrieved 6 May 2009.
^ Simmler, Walter (2000). "Silicon Compounds, Inorganic". Ullmann's Encyclopedia of Industrial Chemistry. doi:10.1002/14356007.a24_001. ISBN 978-3-527-30385-4.
^ Kraus, Paul (1942–1943). Jbir ibn Hayyn: Contribution à l'histoire des idées scientifiques dans l'Islam. I. Le corpus des écrits jbiriens. II. Jbir et la science grecque. Cairo: Institut Français d'Archéologie Orientale. ISBN 9783487091150. OCLC 468740510. {{cite book}}: ISBN / Date incompatibility (help) vol. II, pp. 41–42; Multhauf, Robert P. (1966). The Origins of Chemistry. London: Oldbourne. pp. 141–142.
^ Stapleton, Henry E.; Azo, R.F.; Hidayat Husain, M. (1927). "Chemistry in Iraq and Persia in the Tenth Century A.D." Memoirs of the Asiatic Society of Bengal. VIII (6): 317–418. OCLC 706947607. p. 333. The relevant recipe reads as follows: "Take equal parts of sweet salt, Bitter salt, Ṭabarzad salt, Andarānī salt, Indian salt, salt of Al-Qilī, and salt of Urine. After adding an equal weight of good crystallised Sal-ammoniac, dissolve by moisture, and distil (the mixture). There will distil over a strong water, which will cleave stone (sakhr) instantly." (p. 333) For a glossary of the terms used in this recipe, see p. 322. German translation of the same passage in Ruska, Julius (1937). Al-Rāzī's Buch Geheimnis der Geheimnisse. Mit Einleitung und Erläuterungen in deutscher Übersetzung. Quellen und Studien zur Geschichte der Naturwissenschaften und der Medizin. Vol. VI. Berlin: Springer. p. 182, §5. An English translation of Ruska 1937's translation can be found in Taylor, Gail Marlow (2015). The Alchemy of Al-Razi: A Translation of the "Book of Secrets". CreateSpace Independent Publishing Platform. ISBN 9781507778791. pp. 139–140.
^ Multhauf 1966, p. 142, note 79.
^ Multhauf 1966, pp. 160–163.
^ Karpenko, Vladimír; Norris, John A. (2002). "Vitriol in the History of Chemistry". Chemické listy. 96 (12): 997–1005. p. 1002.
^ Multhauf 1966, p. 208, note 29; cf. p. 142, note 79.
^ Hartley, Harold (1960). "The Wilkins Lecture. Sir Humphry Davy, Bt., P.R.S. 1778–1829". Proceedings of the Royal Society A. 255 (1281): 153–180. Bibcode:1960RSPSA.255..153H. doi:10.1098/rspa.1960.0060. S2CID 176370921.
^ Contaminants, National Research Council (US) Committee on Emergency and Continuous Exposure Guidance Levels for Selected Submarine (2009), "Hydrogen Chloride", Emergency and Continuous Exposure Guidance Levels for Selected Submarine Contaminants: Volume 3, National Academies Press (US), retrieved 23 April 2024
^ CDC – NIOSH Pocket Guide to Chemical Hazards
^ "Hydrogen Chloride". CDC - NIOSH Workplace Safety and Health Topic. 5 March 2012. Retrieved 15 July 2016.
External links
[edit]
Wikimedia Commons has media related to Hydrogen chloride.
International Chemical Safety Card 0163
Thames & Kosmos Chem C2000 Experiment Manual
| v t e Hydrogen compounds |
| H3AsO3 H3AsO4 HArF HAt HSO3F H[BF4] HBr HBrO HBrO2 HBrO3 HBrO4 HCl HClO HClO2 HClO3 HClO4 HCN HCNO H2CrO4/H2Cr2O7 H2CO3 H2CS3 HF HFO HI HIO HIO2 HIO3 HIO4 HMnO4 H2MnO4 H2MoO4 HNC NaHCO3 HNCO HNO HNO2 HNO3 H2N2O2 HNO5S H3NSO3 H2O H2O2 H2O3 H2O4 H2O5 H3PO2 H3PO3 H3PO4 H4P2O7 H5P3O10 H[AuCl4] H2[PtCl6] H2OsCl6 H2S H2S2 H2Se H2SeO3 H2SeO4 H4SiO4 H2[SiF6] HSCN HNCS H2SO3 H2SO4 H2SO5 H2S2O3 H3O H2S2O6 H2S2O7 H2S2O8 CF3SO3H H2Te H2TeO3 H6TeO6 H4TiO4 H2Po H[Co(CO)4] |
| v t e Molecules detected in outer space |
| Molecules | | | | | --- | Diatomic | Aluminium monochloride Aluminium monofluoride Aluminium(II) oxide Argonium Carbon cation Carbon monophosphide Carbon monosulfide Carbon monoxide Cyano radical Diatomic carbon Fluoromethylidynium Helium hydride ion Hydrogen chloride Hydrogen fluoride Hydrogen (molecular) Hydroxyl radical Imidogen Iron(II) oxide Magnesium monohydride Methylidyne radical Nitric oxide Nitrogen (molecular) Oxygen (molecular) Phosphorus monoxide Phosphorus mononitride Potassium chloride Silicon carbide Silicon monoxide Silicon monosulfide Sodium chloride Sodium iodide Sulfanyl Sulfur mononitride Sulfur monoxide Titanium(II) oxide | | | Triatomic | Aluminium(I) hydroxide Aluminium isocyanide Amino radical Carbon dioxide Carbonyl sulfide CCP radical Chloronium Diazenylium Dicarbon monoxide Disilicon carbide Ethynyl radical Formyl radical Hydrogen cyanide (HCN) Hydrogen isocyanide (HNC) Hydrogen sulfide Hydroperoxyl Iron cyanide Isoformyl Magnesium cyanide Magnesium isocyanide Methylene Methylidynephosphane N2H+ Nitrous oxide Nitroxyl Ozone Potassium cyanide Sodium cyanide Sodium hydroxide Silicon carbonitride c-Silicon dicarbide SiNC Sulfur dioxide Thioformyl Thioxoethenylidene Titanium dioxide Tricarbon Trihydrogen cation Water | | Fouratoms | Acetylene Ammonia Cyanoethynyl Formaldehyde Fulminic acid HCCN Hydrogen peroxide Hydromagnesium isocyanide Isocyanic acid Isothiocyanic acid Ketenyl Methyl cation Methyl radical Methylene amidogen Propynylidyne Protonated carbon dioxide Protonated hydrogen cyanide Silicon tricarbide Thiocyanic acid Thioformaldehyde Tricarbon monosulfide Tricarbon monoxide | | Fiveatoms | Ammonium ion Butadiynyl Carbodiimide Cyanamide Cyanoacetylene Cyanoformaldehyde Cyanomethyl Cyclopropenylidene Formic acid Isocyanoacetylene Ketene Methane Methoxy radical Methylenimine Propadienylidene Protonated formaldehyde Silane Silicon-carbide cluster | | Sixatoms | Acetonitrile Cyanobutadiynyl radical Cyclopropenone Diacetylene E-Cyanomethanimine Ethylene Formamide HC4N Ketenimine Methanethiol Methanol Methyl isocyanide Pentynylidyne Propynal Protonated cyanoacetylene | | Sevenatoms | Acetaldehyde Acrylonitrile + Vinyl cyanide Cyanodiacetylene Ethylene oxide Glycolonitrile Hexatriynyl radical Methyl isocyanate Methylamine Propyne Vinyl alcohol | | Eightatoms | Acetic acid Acrolein Aminoacetonitrile Cyanoallene Ethanimine Glycolaldehyde Hexapentaenylidene Methyl formate Methylcyanoacetylene | | Nineatoms | Acetamide Cyanohexatriyne Dimethyl ether Ethanethiol Ethanol Methyldiacetylene N-Methylformamide Octatetraynyl radical Propene Propionitrile | | Tenatomsor more | Acetone Benzene Benzonitrile Buckminsterfullerene (C60, C60+, fullerene, buckyball) Butyronitrile C70 fullerene Cyanodecapentayne Ethyl formate Ethylene glycol Heptatrienyl radical Methyl acetate Methyl-cyano-diacetylene Methyltriacetylene Propionaldehyde Pyrimidine | |
| Deuteratedmolecules | Ammonia Ammonium ion Formaldehyde Formyl radical Heavy water Hydrogen cyanide Hydrogen deuteride Hydrogen isocyanide N2D+ Propyne Trihydrogen cation |
| Unconfirmed | Anthracene Dihydroxyacetone Glycine Graphene H2NCO+ Hemolithin Linear C5 Methoxyethane Naphthalene cation Phosphine Pyrene Silylidyne |
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| Category:Astrochemistry Outer space portal Astronomy portal Chemistry portal |
| v t e Chlorine compounds |
| Chlorides and acids | HCl HClO HClO2 HClO3 HClO4 HSO3Cl BaClF BCl3 CCl4 SiCl4 TiCl4 C3H5Cl |
| Chlorine fluorides | ClF ClF3 ClF5 |
| Chlorine oxides | ClO ClO2 Cl2O Cl2O2 Cl2O3 Cl2O4 Cl2O5 Cl2O6 Cl2O7 ClO4 |
| Chlorine oxyfluorides | ClOF ClOF3 ClO2F ClOF5 (predicted) ClO2F3 ClO3F |
| Chlorine(I) derivatives | ClNO3 ClSO3F ClN3 Cl3N |
| v t e Salts and covalent derivatives of the chloride ion |
| | | | | | | --- --- | HCl | | | | He | | LiCl | BeCl2 | B4Cl4B12Cl12BCl3B2Cl4+BO3 | C2Cl2C2Cl4C2Cl6CCl4+C+CO3 | NCl3ClN3+N+NO3 | ClxOyCl2OCl2O2ClOClO2Cl2O4Cl2O6Cl2O7ClO4+O | ClFClF3ClF5 | Ne | | NaCl | MgCl2 | AlClAlCl3 | Si5Cl12Si2Cl6SiCl4 | P2Cl4PCl3PCl5+P | S2Cl2SCl2SCl4+SO4 | Cl2 | Ar | | KCl | CaClCaCl2 | | ScCl3 | TiCl2TiCl3TiCl4 | VCl2VCl3VCl4VCl5 | CrCl2CrCl3CrCl4 | MnCl2MnCl3 | FeCl2FeCl3 | CoCl2CoCl3 | NiCl2 | CuClCuCl2 | ZnCl2 | GaClGaCl3 | GeCl2GeCl4 | AsCl3AsCl5+As | Se2Cl2SeCl2SeCl4 | BrCl | Kr | | RbCl | SrCl2 | | YCl3 | ZrCl2ZrCl3ZrCl4 | NbCl3NbCl4NbCl5 | MoCl2MoCl3MoCl4MoCl5MoCl6 | TcCl3TcCl4 | RuCl2RuCl3RuCl4 | RhCl3 | PdCl2 | AgCl | CdCl2 | InClInCl2InCl3 | SnCl2SnCl4 | SbCl3SbCl5 | Te3Cl2TeCl2TeCl4 | IClICl3 | XeClXeCl2XeCl4 | | CsCl | BaCl2 | | LuCl3 | HfCl4 | TaCl3TaCl4TaCl5 | WCl2WCl3WCl4WCl5WCl6 | ReCl3ReCl4ReCl5ReCl6 | OsCl2OsCl3OsCl4OsCl5 | IrCl2IrCl3IrCl4 | PtCl2PtCl4PtCl2−6 | AuCl(Au[AuCl4])2AuCl3 | Hg2Cl2HgCl2 | TlClTlCl3 | PbCl2PbCl4 | BiCl3 | PoCl2PoCl4 | AtCl | Rn | | FrCl | RaCl2 | | LrCl3 | RfCl4 | DbCl5 | SgO2Cl2 | BhO3Cl | Hs | Mt | Ds | Rg | Cn | Nh | Fl | Mc | Lv | Ts | Og | | | | | | LaCl3 | CeCl3 | PrCl3 | NdCl2NdCl3 | PmCl3 | SmCl2SmCl3 | EuCl2EuCl3 | GdCl3 | TbCl3 | DyCl2DyCl3 | HoCl3 | ErCl3 | TmCl2TmCl3 | YbCl2YbCl3 | | | AcCl3 | ThCl3ThCl4 | PaCl4PaCl5 | UCl3UCl4UCl5UCl6 | NpCl3NpCl4 | PuCl3PuCl4PuCl2−6 | AmCl2AmCl3 | CmCl3 | BkCl3 | CfCl3CfCl2 | EsCl2EsCl3 | FmCl2 | MdCl2 | NoCl2 | |
| v t e Binary compounds of hydrogen |
| Alkali metal (Group 1) hydrides | LiH NaH KH RbH CsH |
| Alkaline (Group 2) earth hydrides | | | | --- | | Monohydrides | BeH MgH CaH SrH BaH | | | | --- | | Dihydrides | BeH2 MgH2 CaH2 SrH2 BaH2 | |
| Group 13 hydrides | | | | --- | | Boranes | BH3 BH B2H6 B2H2 B2H4 B4H10 B5H9 B5H11 B6H10 B6H12 B10H14 B18H22 | | | | --- | | Alanes | AlH3 Al2H6 | | | | --- | | Gallanes | GaH3 Ga2H6 | | | | --- | | Indiganes | InH3 In2H6 | | | | --- | | Thallanes | TlH3 Tl2H6 | | | | --- | | Nihonanes (predicted) | NhH NhH3 Nh2H6 NhH5 | |
| Group 14 hydrides | | | | --- | | Hydrocarbons | alkanes alkenes alkynes Cycloalkanes Cycloalkenes Cycloalkynes Annulenes | | | | CH CH2 CH3 C2H | | | | --- | | Silanes | SiH4 Si2H6 Si3H8 Si4H10 Si5H12 Si6H14 Si7H16 Si8H18 Si9H20 Si10H22 more... | | | | --- | | Silenes | Si2H4 | | | | --- | | Silynes | Si2H2 SiH | | | | --- | | Germanes | GeH4 Ge2H6 Ge3H8 Ge4H10 Ge5H12 | | | | --- | | Stannanes | SnH4 Sn2H6 | | | | --- | | Plumbanes | PbH4 | | | | --- | | Flerovanes (predicted) | FlH FlH2 FlH4 | |
| Pnictogen (Group 15) hydrides | | | | --- | | Azanes | NH3 N2H4 N3H5 N4H6 N5H7 N6H8 N7H9 N8H10 N9H11 N10H12 more... | | | | --- | | Azenes | N2H2 N3H3 N4H4 | | | | --- | | Phosphanes | PH3 P2H4 P3H5 P4H6 P5H7 P6H8 P7H9 P8H10 P9H11 P10H12 more... | | | | --- | | Phosphenes | P2H2 P3H3 P4H4 | | | | --- | | Arsanes | AsH3 As2H4 | | | | --- | | Stibanes | SbH3 | | | | --- | | Bismuthanes | BiH3 | | | | --- | | Moscovanes | McH3 (predicted) | HN3 NH HN5 (hypothetical) NH5 (hypothetical) |
| Hydrogen chalcogenides (Group 16 hydrides) | | | | --- | | Polyoxidanes | - H2O - H2O2 - H2O3 - H2O4 - H2O5 - more... | | | | --- | | Polysulfanes | H2S H2S2 H2S3 H2S4 H2S5 H2S6 H2S7 H2S8 H2S9 H2S10 more... | | | | --- | | Selanes | H2Se H2Se2 | | | | --- | | Tellanes | H2Te H2Te2 | | | | --- | | Polanes | PoH2 | | | | --- | | Livermoranes | LvH2 (predicted) | HO HO2 HO3 H2O+–O– (hypothetical) H2S+-S- HS HS2 HDO D2O T2O |
| Hydrogen halides (Group 17 hydrides) | - HF - HCl - HBr - HI - HAt - HTs (predicted) |
| Transition metal hydrides | ScH2 YH2 YH3 YH6 YH9 LuH2 LuH3 LrH3 (predicted) TiH2 TiH4 ZrH2 ZrH4 HfH2 HfH4 VH VH2 NbH NbH2 TaH TaH2 CrH CrH2 CrHx FeH FeH2 FeH5 CoH2 RhH2 IrH3 NiH PdHx (x < 1) PtHx (x< 1) DsH2 (predicted) CuH AgH AuH RgH (predicted) ZnH2 CdH2 HgH Hg2H2 HgH2 CnH2 (predicted) |
| Lanthanide hydrides | LaH2 LaH3 LaH10 CeH2 CeH3 PrH2 PrH3 NdH2 NdH3 SmH2 SmH3 EuH2 GdH2 GdH3 TbH2 TbH3 DyH2 DyH3 HoH2 HoH3 ErH2 ErH3 TmH2 TmH3 YbH2 |
| Actinide hydrides | AcH2 ThH2 ThH4 Th4H15 PaH3 UH3 UH4 NpH2 NpH3 PuH2 PuH3 AmH2 AmH3 CmH2 BkH2 BkH3 CfH2 CfH3 |
| Exotic matter hydrides | |
| Authority control databases |
| International | |
| National | United States Japan Czech Republic Israel |
| Other | Yale LUX |
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187971 | https://brainly.com/question/26693342 | [FREE] What expression shows 4 less than a number? A. n + 4 B. n - 4 C. 4 - n D. n(4) - brainly.com
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What expression shows 4 less than a number?
A. n + 4
B. n - 4
C. 4 - n
D. n(4)
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B
Explanation
Because the expression in b is saying whatever number n is say 9 then your subtracting 4 by N or like I said take 9 for an example which=5
hope it helps
Answered by rajohnson5404 •159 answers•30.6K people helped
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The expression that shows 4 less than a number is n−4, which is the correct mathematical representation of the phrase. Therefore, the answer is option B. This correctly signifies that we are subtracting 4 from the number n.
Explanation
To express '4 less than a number,' we need to identify how to mathematically denote this situation.
Let's denote the number we are talking about as n.
The phrase '4 less than a number' indicates that we need to subtract 4 from that number.
Therefore, we write this as: n−4.
Now, let's examine each of the options provided:
A. n+4 - This means adding 4 to the number, which is not what we want.
B. n−4 - This correctly means 4 less than the number.
C. 4−n - This means 4 minus the number, which is also not correct.
D. n(4) - This represents the number multiplied by 4, which does not convey the idea of subtracting 4.
Thus, the correct answer is B: n−4. This shows that if you start with any number, say 10, and subtract 4, you will have 6, which directly aligns with the idea of being 4 less than the original number.
Examples & Evidence
For example, if the number is 8, then 4 less than 8 is 8−4=4. If the number is 15, then 4 less than 15 is 15−4=11.
In mathematics, expressions are formulated to represent situations clearly and accurately. The expression n−4 directly indicates that 4 is being subtracted from n, confirming its accuracy in expressing '4 less than a number.'
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187972 | https://math.stackexchange.com/questions/1919432/every-point-of-a-grid-is-colored-in-blue-red-or-green-how-to-prove-there-is-a | Stack Exchange Network
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Every point of a grid is colored in blue, red or green. How to prove there is a monochromatic rectangle?
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$\begingroup$
I have a $3$-coloring of $\mathbb{Z}\times\mathbb{Z}$, i.e. a function $f:\mathbb{Z}\times\mathbb{Z}\to{\color{red}{\text{red}},\color{green}{\text{green}},\color{blue}{\text{blue}}}$.
I have to prove that there is a monochromatic rectangle with its sides being parallel to the axis, i.e. to prove that for some choice of $a,b,c,d\in\mathbb{Z}$ with $a\neq b$ and $c\neq d$, all the points $(a,c),(a,d),(b,c),(b,d)\,$ have the same color.
I tried to work by contradiction, without achieving much.
Additionally, can we prove some upper bound on $|a-b|$ and $|c-d|$?
combinatorics
coloring
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edited Sep 8, 2016 at 17:14
Jack D'Aurizio
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asked Sep 8, 2016 at 16:20
user366454user366454
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$\begingroup$ Every point of what is colored? $\endgroup$
Arthur
– Arthur
2016-09-08 16:22:48 +00:00
Commented Sep 8, 2016 at 16:22
1
$\begingroup$ Of a graph... You know the one where all spaces are squares... How do you call them? $\endgroup$
user366454
– user366454
2016-09-08 16:24:12 +00:00
Commented Sep 8, 2016 at 16:24
4
$\begingroup$ @Jack's edit makes this question! Wish I could upvote edits... $\endgroup$
alexis
– alexis
2016-09-08 20:17:35 +00:00
Commented Sep 8, 2016 at 20:17
1
$\begingroup$ The idea in Joffan's answer can be used to construct a $4 \times 18$ grid that doesn't have a monochromatic rectangle. And clearly there's such a $3 \times n$ grid for any $n$ (use one colour per row). So what's left is to solve this for grids of size $m \times n$ where $4 $\endgroup$
JiK
– JiK
2016-09-09 09:11:29 +00:00
Commented Sep 9, 2016 at 9:11
1
$\begingroup$ There is no monochromatic rectangle in an infinite chessboard except for the trivial case of size=1. Considering the answers and the rest of the question I would change "monochromatic rectangle" to "rectangle where the four corner points are monochromatic". $\endgroup$
Anonymous Coward
– Anonymous Coward
2016-09-09 14:05:22 +00:00
Commented Sep 9, 2016 at 14:05
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4 Answers 4
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$\begingroup$
Such a rectangle will occur in a grid of $19$ columns by $4$ rows.
By the pigeonhole principle, each column must have a repeated colour point. Ignoring any later repeats, classify the columns according to the first two repeat colour positions; there are $6$ options: $(1,2), (1,3), (1,4), (2, 3), (2, 4)$ and $(3,4)$. So since there are also $3$ colour options for the repeated colours there are only $6\times3=18$ different options for repeated colour by position. Therefore, by the pigeonhole principle again, in a $19\times 4$ grid we must have a suitable rectangle with identically coloured corners.
From an observation by Pere in comments:
For $n$ colours, using the same approach, we would adopt a grid of $(n+1)$ rows and then require $n{n+1 \choose 2}+1 = n(n+1)n/2 +1 = (n^3+n^2)/2 + 1$ columns to find a rectangle with same-coloured corners.
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edited Apr 13, 2017 at 12:21
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answered Sep 8, 2016 at 17:30
JoffanJoffan
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$\begingroup$ Pigeonhole principle FTW :-). One of my favorite tools in 'restricted counting' -type problems. $\endgroup$
Carl Witthoft
– Carl Witthoft
2016-09-08 19:42:34 +00:00
Commented Sep 8, 2016 at 19:42
$\begingroup$ Is there a general formula for the case of n-colors? $\endgroup$
Shufflepants
– Shufflepants
2016-09-08 20:30:08 +00:00
Commented Sep 8, 2016 at 20:30
3
$\begingroup$ @Shufflepants Following Joffan's reasoning, for an arbitrary n such a rectangle will occur in a grid of $(n^3+n^2)/2+1$ columns by $n+1$ rows. $\endgroup$
Pere
– Pere
2016-09-08 20:35:04 +00:00
Commented Sep 8, 2016 at 20:35
1
$\begingroup$ 4 rows is very obviously minimal, and one can also clearly avoid a monochromatic-cornered rectangle by taking the 18 distinct options you mention but can someone show that every grid with fewer than $194=76$ points can be coloured without a m~-c~ r~? Elegantly? (P.S. I am pleased to report this answer currently has $38=194/2$ (net) votes up but I am about to change that!) $\endgroup$
PJTraill
– PJTraill
2016-09-12 14:38:43 +00:00
Commented Sep 12, 2016 at 14:38
$\begingroup$ @PJTraill Interesting question. I had a different challenge in mind, whether there is a square 3-coloured grid smaller than $19 \times 19$ that also must have a mono-coloured rectangle. $\endgroup$
Joffan
– Joffan
2016-09-12 20:40:46 +00:00
Commented Sep 12, 2016 at 20:40
Add a comment |
5
$\begingroup$
For a slightly more quantitative proof, we may show that there is a monochromatic rectangle with its largest dimension being $\color{red}{\leq 82}$ and its smallest dimension being $\color{red}{\leq 4}$.
Step 1. We may give to any integer point $(x,0)$ on the $x$-axis a maxi-color in the set ${1,2,\ldots,80,81}$ depending on the actual colors of $(x,0),(x,1),(x,2),(x,3)$.
Step 2. By the Dirichlet box principle, among $82$ consecutive lattice points on the $x$-axis at least two of them, say $(a,0)$ and $(b,0)$, have the same maxi-color.
Step 3. Since we have only three actual colors, at least two points among $(a,0),(a,1),(a,2),(a,3)$ have the same color. Such points, coupled with the corresponding points on $x=b$, give the wanted monochromatic rectangle.
The $82$ above can be replaced by a smaller number, namely $1+\alpha(G)$, i.e. one plus the size of the largest independent set in a quite complicated graph (of the Haggkvist-Hell type) on $81$ vertices. So we may also improve the given bound by using some rather deep results about topological graphs (see Kneser's conjecture, proved by Lovasz in 1978 through the Borsuk-Ulam theorem).
However, an elementary proof that $\alpha(G)=18$ is presented above by Joffan.
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edited Sep 8, 2016 at 17:36
answered Sep 8, 2016 at 17:07
Jack D'AurizioJack D'Aurizio
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$\begingroup$
Theorem: Let $X,Y$ be infinite sets. If the points of $X\times Y$ are colored with a finite number of colors in such a way that there exists a set $C$ of $c$ colors such that for every $x\in X$ only colors in $C$ appear an infinite number of times among pairs of the form $(x,y)$, then there exist $x_1,x_2\in X,y_1,y_2\in Y$ so that $(x_1,y_1),(x_1,y_2),(x_2,y_1),(x_2,y_2)$ all have the same color.
Proof:
We proceed by induction on $c$, when $c=1$ it is clear. Take any $x_1,x_2$ and let $\alpha$ be the only color that appears an infinite number of times. Notice that the set of $A_{x_1}={y\in Y | (x_1,y) \text{ has color } \alpha}$ has finite complement, and so does the set $A_{x_2}={y\in Y | (x_2,y) \text{ has color } \alpha}$. This implies $A_{x_1}\cap A_{x_2}$ is infinite. Taking $x1,x2$ and any $y_1,y_2\in A_{x_1}\cap A_{x_2}$ does the trick.
Inductive step: Consider any $x\in X$, there must be an $\alpha \in Y$ such that the set $A_{x}={y\in Y | (x,y) \text{ has color } \alpha}$ is infinite. Now consider the set $X'=X\setminus x$ and $Y'=A_x$. If there is an $x'\in X'$ such that there is an infinite number of $y'\in Y'$ such that $(x',y')$ has color $\alpha$ then we are done. just take $x,x'$ and $y_1,y_2$ such that $(x',y_1)$ and $(x',y_2)$ have color $\alpha$. Otherwise notice that the set $C'=C\setminus\alpha$ has $c-1$ elements, and for every $x\in X'$ the only colors that appear an infinite number of times among the pairs $(x,y)$ are in $C'$. By the inductive hypothesis there exist $x_1,x_2\in X'$ and $y_1,y_2\in Y'$ such that $(x_1,y_1),(x_1,y_2),(x_2,y_1),(x_2,y_2)$ all have the same color.
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edited Sep 8, 2016 at 20:08
answered Sep 8, 2016 at 16:46
AsinomásAsinomás
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$\endgroup$
9
$\begingroup$ I think that I'm not quite understanding your language. Is $C$ the set of finite- or infinite-multiplicity colours? You seem to say the former, but your base case seems to assume the latter. $\endgroup$
LSpice
– LSpice
2016-09-08 19:22:17 +00:00
Commented Sep 8, 2016 at 19:22
$\begingroup$ Infinito, let me update $\endgroup$
Asinomás
– Asinomás
2016-09-08 19:27:48 +00:00
Commented Sep 8, 2016 at 19:27
$\begingroup$ Definitely a Carl Gauss type of proof :-) i.e. very precise and formal, and equally difficult to connect to the real-world implementation. Not complaining, just comparing w/ the approach Joffan took. $\endgroup$
Carl Witthoft
– Carl Witthoft
2016-09-08 19:44:26 +00:00
Commented Sep 8, 2016 at 19:44
1
$\begingroup$ Yeah right... I wish this was a Carl Gauss type of proof, thanks @CarlWitthoft $\endgroup$
Asinomás
– Asinomás
2016-09-08 20:12:47 +00:00
Commented Sep 8, 2016 at 20:12
1
$\begingroup$ Minor point: I think $\alpha \in Y$ in the first sentence of the inductive step should say $\alpha in C$. Slightly more important, because it requires more effort to work out what it's supposed to say: "for every $x \in X'$ the only colors that appear an infinite number of times among the pairs $(x,y)$ are in $C'$" is missing the constraint $y \in Y'$. $\endgroup$
Peter Taylor
– Peter Taylor
2016-09-09 10:04:38 +00:00
Commented Sep 9, 2016 at 10:04
| Show 4 more comments
1
$\begingroup$
We will decide how to choose rows first. So basically there are three colors so applying the pigeonhole principle if we choose the minimum number of rows to be 4 then it is bound at the worst case one color will repeat.
Now for choosing the minimum number of columns out of 4 rows we will choose any 2 rows and assign its color to be the same, either red blue, or green. This will lead us to the combination of 18 possible options (4C23 (for each color)).
So again applying pigeonhole there should be at least 19 (18+1) columns for the possibility of two colors in 4 rows to be the same
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answered Mar 23, 2021 at 11:31
nishant bhatnishant bhat
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4th grade math (TX TEKS)12 units · 150 skillsUnit 1 Place value of whole numbers and decimalsUnit 2 Addition and subtraction of whole numbers and decimalsUnit 3 Multiplication of whole numbersUnit 4 Division of whole numbersUnit 5 Applying all operationsUnit 6 FractionsUnit 7 Data representationUnit 8 MeasurementUnit 9 Points, lines, and anglesUnit 10 Measuring anglesUnit 11 Personal financial literacyUnit 12 Teacher resources
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4th grade math (TX TEKS)
Unit 1: Place value of whole numbers and decimals
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Place value tables
Write whole numbers in expanded form
Write numbers in written form
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Place value of whole numbers and decimals: Quiz 1
Multiply whole numbers by 10
Divide whole numbers by 10
Multiply and divide by 10
Place value when multiplying and dividing by 10
Place value of whole numbers and decimals: Quiz 2
Compare place values
Compare multi-digit numbers
Compare multi-digit numbers written in different forms
Compare multi-digit numbers word problems
Order whole numbers
Place value of whole numbers and decimals: Quiz 3
Write decimals and fractions shown on grids
Write decimals and fractions shown on number lines
Relate decimals to money
Decimals in expanded notation (hundredths)
Decimals in words
Relate decimals and fractions in words
Place value of whole numbers and decimals: Quiz 4
Decimals on the number line: tenths 0-1
Decimals on the number line: tenths
Decimals on the number line: hundredths 0-0.1
Decimals on the number line: hundredths
Place value of whole numbers and decimals: Quiz 5
Compare decimals visually
Order decimals visually
Place value of whole numbers and decimals: Unit test
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187974 | https://artofproblemsolving.com/wiki/index.php/2016_AIME_I_Problems/Problem_2?srsltid=AfmBOooMpH_po6HCL1jtGKsT1jlkpEXvzYUd8iea1xtep1MQS-ImcvOB | Art of Problem Solving
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1 Problem 2
2 Solution
3 Solution 2
4 Solution 3
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Problem 2
Two dice appear to be normal dice with their faces numbered from to , but each die is weighted so that the probability of rolling the number is directly proportional to . The probability of rolling a with this pair of dice is , where and are relatively prime positive integers. Find .
Solution
It is easier to think of the dice as sided dice with sixes, fives, etc. Then there are possible rolls. There are rolls that will result in a seven. The odds are therefore . The answer is
See also 2006 AMC 12B Problems/Problem 17
Solution 2
Since the probability of rolling any number is 1, and the problem tells us the dice are unfair, we can assign probabilities to the individual faces. The probability of rolling is because Next, we notice that 7 can be rolled by getting individual results of 1 and 6, 2 and 5, or 3 and 4 on the separate dice. The probability that 7 is rolled is now which is equal to . Therefore the answer is ~PEKKA
Solution 3
Since the probability of rolling a is , the probability of rolling a is the probability of rolling a is and so on, we can make a chart of probabilities and add them together. Note that we only need the probabilities of and , and , and and , and the rest is symmetry and the others are irrelevant.
We have: = . Therefore, the answer is =
~Arcticturn
See also
2016 AIME I (Problems • Answer Key • Resources)
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187975 | https://www.site.uottawa.ca/~rhabash/elg3311ch1 | 1-1 09/17/01 CHAPTER 1 MAGNETIC CIRCUITS 1.1 Basic Magnetic Circuit Analysis 1.1.1 Magnetic Circuit Definitions The simplest method of analyzing magnetic systems such as inductors, transformers, solenoids etc., is to derive a magnetic circuit model and then apply basic magnetic circuit analysis on the model. Basic magnetic circuit analysis is analogous to basic electric circuit analysis once some simple definitions are established. Figure 1.1 shows a simple magnetic system consisting of a magnetic structure with a winding, around it. A current, I, flows through the coil. There are N turns in the winding. The magnetic circuit model is shown in Figure 1.2a whereas the analogous electrical circuit model is shown in Figure 1.2b. N I Φ Figure 1.1 A Simple Magnetic System F Φ ℜ a) Magnetic Equivalent Circuit V I R b) Analogous Electrical Circuit Figure 1.2 Magnetic and Electrical Equivalent Circuit Models for the Magnetic System of Figure 1.1 In the analysis of the electrical circuit model the governing equation is: 1-2 09/17/01 V = IR Where: V is the electric force, in volts I is the electric current, in amperes R is the electrical resistance, in ohms In the analysis of the magnetic circuit model the governing equation is: F = Φ ℜ Where: F is the magnetomotive force, mmf, in ampere-turns Φ is the magnetic flux, in webers ℜ is the magnetic reluctance, in amper-turns/weber Also, in most cases, the magnetomotive force is produced by the electric current in the coil, as given by the equation: F = NI Where: N is the number of turns in a winding I is the current in the winding In electrical circuits the resistance, R, is usually known and V or I are to be determined. However in magnetic circuits the reluctance, ℜ, has to be determined from fundamental physical properties of the materials involved. The analysis of magnetic circuits is further complicated by the fact that the most relevant physical properties of magnetic materials are not functions of the bulk external variables, F or Φ, per se but of related internal variables, H and B, where H is the magnetic field intensity, (or magneto-motive force per unit length), and B is the magnetic flux density per unit of cross sectional area. The physical properties of the material will determine how much flux density will be produced by a given level of field intensity. In other words; B = µH Where µ is the permeability of the material and is usually expressed as: µ = µoµr 1-3 09/17/01 And: µo is the permeability of free space, 4π x 10-7 µr is the relative permeability of the specific material involved. µr = 1 for air 1 << µr < 10,000 for magnetically useful materials The fundamental relationships between B, H and Φ, F are as follows. Φ = B z ∂A Where B is the flux density in Webers/m2 , (or Tesla), and A is the cross sectional area of the material. This equation can be reduced in simple cases where B does not vary with area: Φ = B z ∂A = BA if B is constant with A Also: F = H z ∂λ Where Η is the magnetic field intensity in ampere-turns/meter and λ is the length of the magnetic path. This equation can be also reduced in simple cases where F does not vary with length: F = H z ∂λ = Hλ if H is constant with λ Expressions for the reluctance, ℜ, can now be determined from: ℜ = F Φ In geometrically simple cases, we can substitute for: F = Hλ and Φ = BA to obtain: ℜ = H BA λ Another substitution can be made for B = µH = µoµrH To obtain: 1-4 09/17/01 ℜ = λ µ0µrA This equation is valid only for uniform materials with geometrically simple (usually rectangular) shapes. 1-5 09/17/01 Example 1.1 Determine the magnetic flux in a toroid of inner radius r1 and outer radius r2 . The toroid has a rectangular cross section of width W. A winding of N turns is uniformly distributed over the entire toroid and carries a current of I amps. Φ I 2 r W N r1 Figure 1.3 A Magnetic System Consisting of a Toroid Solution to example 1.1 The toroid is shown in Figure 1.3. The magnetic flux can be determined from: Φ = B z ∂A= BW r r 1 2 ∂ z r = W B r r 1 2 ∂ zr Substitute for B = µoµrH to obtain: Φ = W H o r r r 1 2 µ µ ∂ z r = µ µ o r r r W H 1 2 ∂ zr Now we can derive H from: F = NI = H∂ zλ Where λ is the path length, or in this case, the circumference for a given radius r. Since the windings are uniformly distributed over the entire toroid, then H will not vary over any given path length, even though the path length does vary with radius. Thus H will vary with radius, r, but for any given r, H will be constant over 1-6 09/17/01 the entire circumference, λ . Thus: F = NI = H∂ zλ= Hλ because H is constant with λ Substitute for: λ = 2πr To obtain: F = NI = H2πr Rearrange to obtain: H = NI 2 r π Substitute for H into the last equation for Φ to obtain: Φ = µ µ o r r r W H 1 2 ∂ zr = µ µ π o r r r W NI 2 r 1 2 ∂ z r = µ µ π o r r r WNI 2 r 1 2 ∂ z r Solve the integral to obtain: Φ = µ µ π o r 2 1 WNI 2 r r lnL N MO Q P In examples such as this there is often no need to determine ℜ, nevertheless: ℜ = F Φ = NI Φ = 2 WN r r o r 2 1 π µ µ lnL N MO Q P 1-7 09/17/01 1.1.2 Series/Parallel Circuits Most practical magnetic circuits do not consist of simple, uniform magnetic paths. However, most of them can be broken down into a number of simple paths each of which can be approximately uniform. For example the magnetic circuit shown in Figure 1.4 can be represented by the equivalent circuit shown in Figure 1.5 which consists of a number of magnetic elements, ℜa , ℜb , ℜc , ℜd , ℜx , ℜy , ℜg1 , ℜg1 . N I Φ1 Φ2 Soft iron air gap air gap Figure 1.4 Magnetic System Consisting of Several Elements F Φ 1 Φ 2 ℜ ℜ ℜ ℜ ℜ ℜ ℜ a c g2 g1 d b y x ℜ Figure 1.5 Magnetic Equivalent Circuit for the System of Figure 1.4 1-8 09/17/01 Each of these elements has a simple, rectangular geometry and will have an approximately uniform magnetic flux density and field intensity. Where ℜa, ℜb, ℜc, ℜd, ℜx, ℜy represent the reluctance of the various iron members, and ℜg1, ℜg2 represent the reluctance of the air gaps. Therefore if; µr >> 5,000 Then, ℜa, ℜb, ℜc, ℜd, ℜx, ℜy<< ℜg1, ℜg2 Thus the equivalent circuit can be adequately represented by the circuit shown in Figure 1.6a in which only the air gap reluctances, ℜg1, ℜg2, are included. This equivalent circuit can be further simplified, by combining reluctances to produce the circuit shown in Figure 1.6c, which consists of a single reluctance, ℜ, where: ℜ = ℜℜ ℜ ℜ g1 g2 g1 g2 + F Φ1 Φ2 ℜg2 ℜg1 a) Ignoring Reluctance of Magnetic Elements Φ F 1 Φ2 ℜg2 ℜg1 b) Rearranging Reluctances b) Combining Reluctances F ℜ Φ 2 Φ 1+ Figure 1.6 Simplifying the Magnetic Circuit of Figure 1.5 1-9 09/17/01 1.2 Electrical Characteristics of Magnetic Circuits 1.2.1 Inductance In most electrical circuits an important parameter of magnetic elements is their inductance, L, which is defined as: L = λ I Where λ is the flux linkage, or magnetic charge, in the magnetic system and is analogous to the electrical charge, Q in a capacitor. The magnetic charge, λ, is defined as: λ = NΦ Substitute for λ to obtain: L = N I Φ Substitute for Φ, to obtain: L = N2 ℜ = N A 2 o r µ µ λ for uniform and geometrically simple structures It is often convenient to derive expressions for voltage, current, power, and energy as functions of inductance. For example, V = ∂λ ∂t = ∂ ∂ (LI t ) = L ∂ ∂ I t + I ∂ ∂ L t = L I ∂ ∂t for L constant Conversely, I = 1 L V∂ zt Also: 1-10 09/17/01 W = P∂ zt = (VI)∂ z t = (L ) ∂ ∂ ∂ z I t t = LI∂ z I W = 1 2 LI2 for L constant 1-11 09/17/01 1.2.2 Electrical Performance of Ideal Inductors Given an ideal inductor, of inductance L, with a current i(t) and a voltage v(t) as shown in the circuit of Figure 1.7a. i(t) v(t) L a) Ideal Inductor v(t) i(t) t t b) Voltage & Current for v(t) constant v(t) i(t) t t c) Voltage & Current for v(t) “one shot” pulse t= T Figure 1.7 Voltage and Current in an Ideal Inductor From the preceding analysis we can derive expressions for the voltage and current: v(t) = L i( ) ∂ ∂ t t i(t) = 1 L v( ) t t ∂ z a) Assume the applied voltage is a constant DC level, Vdc. Therefore: v(t) = Vdc and thus, i(t) = 1 L v( ) t t ∂ z = 1 L v( ) t t ∂ z = V L dct This expression indicates that the inductor current, i(t), will increase linearly with time as shown in Figure 1.7b. The current will continue to increase without bound. In real life the current will increase until something breaks, i.e. a fuse blows, or some external element limits the current. b) Assume the applied voltage is a "single shot" pulse, as shown in Figure 1.7c. 1-12 09/17/01 In this case the expression for v(t) is: v(t) = Vp for 0 < t < Τ and v(t) = 0 for Τ< t < ∞ From the preceding analysis we can determine that: i(t) = V L pt for 0 < t < Τ when v(t) = Vp Also at t = Τ: i(Τ) = V T L p And at t = Τ+: v(t) = 0 = L i( ) ∂ ∂ t t Thus: ∂ ∂ i( ) t t = 0 Which means that the inductor current will not change from its preceding value even though the applied voltage has gone to zero. In other words: i(t) = V T L p for Τ< t < ∞ even though v(t) = 0 This indicates that the current through an ideal inductor will not change if the applied voltage is zero. A noteworthy application of this principle is in super conducting magnets, in which the coils are cooled below the critical temperature such that the wire resistance disappears and the inductor becomes "ideal". Once a current is established in such an inductor it will continue without diminishing, and maintain a high magnetic field, even with no additional voltage or energy input, (provided the windings are maintained below their critical temperature). An equally noteworthy, though adverse, effect of this principle is that when an inductor is switched in a switching circuit, eg. a power supply, the inductor 1-13 09/17/01 current may not be zero even if its applied voltage has been removed. c) Assume the applied voltage is a sinewave: v(t) = Vsin(ωt) Therefore, i(t) = 1 L v( ) t t ∂ z = 1 L Vsin( ) ωt t ∂ z = -V L ω ω cos( ) t t t t = = 0 i(t) = V L - cos( t) ω ω 1 Thus the current through an ideal inductor can have a DC component even though the applied voltage is a pure sinewave. In an ideal inductor this DC current will continue indefinitely, but in a real inductor there will always be a winding resistance, R, that will cause the DC component of the current to decay with a time constant of R/L. 1-14 09/17/01 1.2.3 Electrical Interaction Between Two Ideal Inductors Assume that two inductors are connected as shown in the circuit of Figure 1.8 and the initial conditions, at t = 0- are: i1(0-) = 0 i2(0-) = I Switch S is closed 2 1 i (t) v(t) L 1 L S 2 i (t) Figure 1.8 Two Ideal Inductors Connected by a Switch At t=0 switch S is opened, which has the effect of forcing the current through each inductor to be the same, i.e.: i1(0+) = i2(0+) = i(0+ ) The new value of inductor current is determined by the conservation of flux linkage, (analogous to the conservation of charge in capacitors): ∑λ(0-) = ∑λ(0+) or: i1(0-) L1 + i2(0-) L2 = i1(0+) L1 + i2(0+) L2 0 + IL2 = i(0+ )L1 + i(0+ )L2 Solve for : i(0+ ) = IL L + L 2 1 2 1-15 09/17/01 Thus i1 goes from 0 to IL2 /(L1 +L2 ) instantaneously and i2 goes from I to IL2 /(L1 +L2 ) instantaneously, which seems contrary to the previous conclusion that the current through an ideal inductor cannot change instantaneously. However, in this case the overriding law is that the total flux linkages in the inductor pair cannot change, just as the total charge in a capacitor pair cannot change. 1-16 09/17/01 1.2.4 Magnetic vs Capacitive Analogies Magnetic Capacitive Physical Parameter L C Voltage v = L i ∂ ∂t v = 1 C i∂ z t Current i = 1 L v∂ z t i = C v ∂ ∂t Energy W = 1 2 LI2 W = 1 2 CV2 Charge λ = LI Q = CV Charge Derivative v = ∂λ ∂t i = ∂ ∂ Q t Conservation Equation ∑λ(0-) = ∑λ(0+) ∑Q(0-) = ∑Q(0+) 1-17 09/17/01 1.3 Non-Ideal Magnetic Systems 1.3.1 Electrical Performance of Real Inductors As a first approximation, the main difference between a real inductor and an ideal inductor can be represented by incorporating a resistance, R, in series with an ideal inductance, L, as shown in Figure 1.9. The value of R is usually the resistance of the wires in the inductor. R L i(t) v(t) S v L Figure 1.9 Circuit Model of a "Real" Inductor (First Approximation) The governing equation now becomes: v(t) = Ri(t) + L i ∂ ∂t Assume that: v(t) = Vdc constant DC level the switch S has been closed for a sufficiently long time such that steady state conditions have been reached and thus: i(t) = V/R Assume now that the switch is opened at t=0, forcing the inductor current to go to zero in a short time interval of ∆t. Thus: vL(0+) = L i ∂ ∂t ≈ L I t ∆ ∆ Thus if the time interval, ∆t, is sufficiently small, then the voltage across the inductor, vL , could be very large. This voltage is usually sufficient to break down 1-18 09/17/01 the air gap across a mechanical switch and cause arcing. It is also sufficient to break down the semiconductor in solid state switches and destroy them unless precautions are taken. Effectively, the energy stored in the inductor has to be dissipated in the switch or in radiated energy, or both. In real applications this results in excessive stresses on the switching components, usually transistors, and/or excessive radiated noise. 1-19 09/17/01 1.3.2 Non-Linear Magnetics Most practical magnetic applications use ferrous or ferrite materials that have non-linear magnetic characteristics. The non-linearity mainly applies to the B vs H curve as shown in Figure 1.10 which demonstrates saturation and Figure 1.11 which demonstrates hysteresis as well as saturation. B H Linear Material Non-Linear Material µ0 µ0 r µ µ0 Figure 1.10 B vs H Curve Demonstrating Saturation 1-20 09/17/01 B H c H Bmax Br max H c H -Br -Bmax -max H -µ0 a) Generalized Hysteresis Curve b) Piecewise Linear Approximation c) Simplified Approximation B H B H Figure1.11 B vs H Curves Demonstrating Saturation and Hysteresis In general, a non-linear magnetic characteristic means that the relative 1-21 09/17/01 permeability of the material, µr , is not constant but varies with H. Saturation means that the effective value of µr decreases abruptly to unity, i.e. the magnetic material becomes no more effective than free air. Hysteresis means that the magnetic flux density, B, can assume more than one value for a given value of field intensity, H. The actual value of B will be dependent on whether H is increasing or decreasing and on the preceding magnetic history of the material, i.e. hysteresis implies a magnetic memory effect. Some key points on the B vs. H curve of Figure 1.11 are: Bmax is called the saturation flux density and is defined as the flux density when µr goes to 1. Hmax is called the maximum field intensity and is defined as the magnetic field intensity when µr goes to 1. Br is called the residual flux density and is defined as the flux density when the magnetic field intensity, H, is zero. Hc is called the coercive force and is defined as the magnetic field intensity required to bring the flux density, B, to zero. As will be shown in section 1.4.1 that the energy dissipated in a magnetic material is given by the expression: WΦ = m H B 3 B B 1 2 ∂ z and, if there is hysteresis, then : WΦ = m H B 3 B B 1 2 ∂ z > 0 even if B1 = B2 Thus a certain amount of energy is dissipated in a magnetic material each time it is "cycled through" the B H curve. In AC systems this occurs once during each cycle. This energy is referred to as hysteresis losses and becomes very significant at high frequencies. It is important to note that all magnetic materials demonstrate saturation and hysteresis to varying degrees. The more ideal applications operate in the region where B is much less than Bmax and the more ideal materials demonstrate less hysteresis, eg, Br and Hc are close to zero. In choosing the right magnetic material for a given application various tradeoffs are required in selecting the magnetic characteristics of the material to suit the application. Some desirable characteristics for specific applications are: 1-22 09/17/01 Application Desirable characteristic permanent magnets high Br measuring instruments very low Br motors/generators high Bmax , low Hmax magnetic memories high Br , repeatable Hc high frequency operation low H B B B 1 2 ∂ z DC operation high Hmax 1-23 09/17/01 Example 1.1 Given the magnetic system shown in Figure 1.12, assume the soft iron element has the B vs H characteristic shown in Figure 1.11 and that the cross-sectional area of the iron is Af , and the effective cross-sectional area of the air gap is Ag , the mean length in the soft iron is λ and the air gap width is g. Note that µr is comparable to µ0 and λ is comparable to g. N I Φ Soft iron air gap g Figure 1.12 A Magnetic System with an Air Gap and Soft Iron having the B H Characteristics shown in Figure 1.11 Solution: The equivalent magnetic circuit for this system is shown in Figure 1.13 where ℜf represents the reluctance of the iron element and ℜg represents the reluctance of the air gap. F Φ ℜ ℜ ℜ ℜg ℜ ℜ ℜ ℜf F F f g Figure 1.13 Equivalent Magnetic Circuit of System Shown in Figure 1.12 The actual value of ℜf cannot be analytically determined because it is dependent on the µr of the iron which is a non-linear function of the field intensity applied to the iron. This non-linearity is represented by the non-linear B 1-24 09/17/01 vs. H curve for the iron. Nevertheless, since the B vs. H curve for the iron is available it is possible to determine magnetic parameters for this system graphically. The actual values of B and H for the iron will have to satisfy two curves: one curve being the B vs. H curve for the iron as shown in Figure 1.11a, the second curve being the B vs. H relationship for the magnetic system as a whole. This second curve can be determined from the equations derived from the magnetic equivalent circuit shown in Figure 1.13 as follows: The basic equations for the air gap are: Fg = Φ ℜg ℜg = g A g g µ Φ = Bg Ag And for the iron: Ff = Hf λ Φ = Bf Af Furthermore, the same flux, Φ, is in both the iron and the air gap, or: Φ = Bg Ag = Bf Af The basic equation for the system is: F = N I = Ff + Fg Substitute for Ff ,Fg , and Φ. into the above equation to obtain: NI = Hf λ + Bf Af ℜg Solve for: Bf = NI - H A f g λ b g ℜ This is the second B vs. H curve for the iron. It's actually a straight line representing the relationships imposed by the air gap. The characteristics of this line are: 1-25 09/17/01 Vertical intersect: B = H A f g λ ℜ Horizontal intersect: H = NI λ Slope is given by: ∂ ∂ B H = ∂ ℜ L N M M O Q P P ∂ NI - H A H f f λ g = - A f g λ ℜ = -µ0 Figure 1.14 shows the superposition of the two B vs. H curves. B H f H Bf µ0 Operating point Magnetic Circuit Equation soft iron curve NI λ B=(NI - H λ ) Afℜg NI AfRg Figure 1.14 B vs H curves for the soft iron and the magnetic system shown in Figure 1.12 1-26 09/17/01 1.4 Energy and Force in Magnetic Systems 1.4.1 Energy Stored in a Magnetic Field To determine the electrical energy stored in a magnetic element we have to determine the voltage applied to the element. We can utilize Lenz's Law which states that: VΦ = ∂λ ∂t = ∂ ∂ NΦ t = N ∂Φ ∂t The energy, WΦ , that was required to generate the magnetic flux is given by the equation: WΦ = V I Φ ∂ z t Substitute for VΦ to obtain: WΦ = N I ∂Φ ∂ ∂ z t t = NI∂Φ z Substitute for NI to obtain: WΦ = ℜΦ∂Φ z For linear systems in which ℜ is independent of Φ the integral simplifies to: WΦ = ℜΦ2 2 For nonlinear systems the integral can still be simplified by first substituting F for ℜΦ to obtain: WΦ = ℜΦ∂Φ z = F∂Φ z Furthermore, for geometrically simple cases, we can substitute for: F = Hλ and Φ = BA to obtain: 1-27 09/17/01 WΦ = Hλ∂ z BA b g = H A λ ∂ z B bg = λAH∂ z B Note, however, that the expression, λA, is the product of the length and cross-sectional area of the magnetic path. This is the volume of the magnetic path, eg. WΦ = volume H × ∂ z B = m H 3 ∂ zB or: W m3 Φ = H∂ zB Similarly, to change the magnetic flux density in a material from B1 to B2 will require an energy input of: W m3 Φ = H B B 1 2 ∂ zB It was shown in section 1.3.2 that in most magnetic materials, due to hysteresis: W m3 Φ = H B B 1 2 ∂ zB > 0 even if B1 = B2 1-28 09/17/01 1.3.2 Energy Stored in an Air Gapped System A simple magnetic system with two air gaps is shown in Figure 1.5. It consist of two iron elements: a horseshoe and a bar. The bar is separated from the horseshoe by a gap of length g. The effective cross-sectional area of the air gap is A. There is a coil of N turns carrying a current of I amps. I g soft iron Figure 1.15 Simple Horeshoe Magnet The inductance, Lg , of each air gap is: Lg = N A g 2µ µ 0 r The energy stored in each air gap can be determined from: W = 1 2 LI2 Thus: W g = 1 2 LgI2 = N I A 2g 2 2µ µ 0 r The preceding expression for energy in each air gap can be used to determine the force, F g , acting on the bar to try to close the air gap. The expression for force as a function of energy is given by: 1-29 09/17/01 F = ∂ ∂ W g Substitute for W g to obtain: F g = ∂ ∂ W g g = - N I A 2g 2 2 2 µ µ 0 r in each gap Thus the total force acting on the bar is: F g = - N I A g 2 2 2 µ µ 0 r (sum of both air gaps) The above expression is negative indicating that the force is acting to make g smaller. 1-30 09/17/01 1.4.3 Energy Dissipation Between Two Ideal Inductors. In the example shown in Figure 1.8 and discussed in section 1.2.3 it was shown that the law of conservation of flux linkage, λ, determines the resultant current when two inductors are connected together. That is to say that the total flux linkage in the system does not change, however, the total energy, W, stored in the inductors does change: W(0- ) = 1 2 L1 i 2 1(0-) + 1 2 L2 i 2 2(0-) = 1 2 L2 I2 However, W(0+ ) = 1 2 L1 i 2 1(0+) + 1 2 L2 i 2 2(0+) = 1 2 L1 IL2 L1+L2 2 + 1 2L2 IL2 L1+L2 2 = 1 2 L2 I2 L2 L1+L2 = W(0- ) L2 L1+L2 Thus: W(0+ ) < W(0- ) The energy lost in the system was dissipated in the contact resistance of the switch and as radiated energy if there is a high ∂i/∂t. 1-31 09/17/01 1.5 Transforming Magnetic Circuits to Electrical Circuits Magnetic equivalent circuits can be transformed into electrical equivalent circuits by a form of topological transformation in which magnetic circuit loops are "mapped" into electrical circuit loops. Assume we have a generalized magnetic system as shown in Figure 1.16, consisting of several magnetic paths and two coils. N2 N1 Figure 1.16 Generalized Magnetic System to be mapped into an Electrical Circuit. The steps are as follows: Step1: Draw the equivalent magnetic circuit. This is done in Figure 1.17. F ℜ ℜ ℜ 3 1 ℜ5 ℜ4 1 2 2 F Figure 1.17 Magnetic Equivalent Circuit for the system shown in Figure 1.16 Step 2: Assign a number to each circuit loop in the magnetic equivalent circuit, as shown in Figure 1.18a, with the outside loop assigned the number 0. 1-32 09/17/01 Step 3: Assign the corresponding numbers to each node in the electrical equivalent circuit, as shown in Figure 1.18b. Step 4: Identify each magnetic circuit element between any two magnetic loops as shown in Figure 1.18c. Step 5: Insert a corresponding electrical circuit element between the corresponding electrical nodes, as shown in Figure 1.18d, but changing all reluctances into "internal" inductors, L ' , and all coils into ideal transformers. The value of each internal inductor is determined by: L ' = 1 ℜ 1-33 09/17/01 a) Magnetic Equivalent Circuit with Loops Numbered F ℜ3 ℜ1 ℜ5 ℜ4 1 ℜ2 2 F 1 2 0 1 2 0 b) Electrical Circuit with Corresponding Nodes Numbered F ℜ3 ℜ1 ℜ5 ℜ4 1 ℜ2 2 F 1 2 0 c) Reluctances and Coils between Loops Identified d) Corresponding Inductors and Tranformers between Nodes N2 N1 1 2 0 L' L' L' L' L' 1 2 3 4 5 1: :1 T T 1 2 Figure 1.18 Transforming a Magnetic Circuit into an Electrical Circuit The "internal" inductors of Figure 1.18c can be "externalized" by "pulling" them 1-34 09/17/01 through one or the other transformers and changing the value by: L = N L 2 ' = N2 ℜ The resultant electrical equivalent circuit is shown in Figure 1.19. N2 N1 L' L' L' L' L' 1 2 3 4 5 1: :1 T T 1 2 Internal Components N2 N1 L L L L L 1 2 3 4 5 : T2 a) Electrical Circuit with "Internal" Magnetic Elements b) "Internal" Magnetic elements brought out through transformer T1 Figure 1.19 Electrical Equivalent Circuit for the Magnetic System of Figure 1.16 |
187976 | https://www.math-linux.com/mathematics/limits/article/proof-of-limit-of-tan-x-x-1-as-x-approaches-0 | Proof of limit of tan x / x = 1 as x approaches 0 | Math-Linux.com
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How to prove that limit of tan x / x = 1 as x approaches 0 ?
Requirement
lim x→0 sinx x=1
Proof of limit of sin x / x = 1 as x approaches 0
Proof
By definition of tan x:
tanx x=sinx x⋅cosx=sinx x×1 cosx lim x→0 tanx x=lim x→0(sinx x×1 cosx)=1×1=1
we conclude that:
lim x→0 tanx x=1
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Methyl orange
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Catalog Number:
01239
CAS Number:
547-58-0
Methyl orange
Grade:
ACS reagent
Synonym(s):
4-[4-(Dimethylamino)-phenylazo]benzenesulfonic acid sodium salt, Acid orange 52, CI 13025
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Methyl orange is a widely recognized pH indicator known for its vibrant color change from red to yellow as the pH transitions from acidic to neutral. This compound is particularly valuable in titrations and various laboratory applications, allowing researchers to easily determine the acidity or alkalinity of a solution. Its sensitivity to pH changes makes it an essential tool in educational settings, chemical analysis, and quality control processes in industries such as food and beverage, pharmaceuticals, and environmental monitoring.
In addition to its role as a pH indicator, methyl orange is utilized in dyeing processes and as a colorimetric reagent in analytical chemistry. Its unique properties allow for precise measurements in complex mixtures, making it a preferred choice among professionals seeking reliable results. Furthermore, methyl orange's stability and ease of use enhance its appeal, providing a practical solution for both laboratory and industrial applications.
General Information Properties Safety and Regulations Applications
Synonyms
4-[4-(Dimethylamino)-phenylazo]benzenesulfonic acid sodium salt, Acid orange 52, CI 13025
CAS Number
547-58-0
Grade
ACS reagent
Molecular Formula
C 14 H 14 N 3 O 3 SNa
Molecular Weight
327.3
MDL Number
MFCD00007502
Appearance
Orange-yellow solid
Conditions
Store at RT
Additional property information coming soon!
Hazmat
Yes
Antibiotic
No
DEA-regulated
No
Warnings
Methyl orange is widely utilized in research focused on:
pH Indicator: Commonly used in titrations, it changes color from red to yellow as the pH shifts from acidic to neutral, making it a reliable choice for chemists in laboratories.
Textile Industry: Employed as a dye, it provides vibrant colors for fabrics, enhancing the aesthetic appeal of clothing and home textiles.
Food Industry: Used as a food colorant, it helps in achieving specific hues in products, although its use is regulated in many countries due to safety concerns.
Biochemical Research: Serves as a model compound in studies related to azo dyes, aiding researchers in understanding dye degradation and environmental impact.
Educational Purposes: Frequently used in chemistry education to demonstrate acid-base reactions, helping students visualize chemical changes in real-time.
General Information
Synonyms
4-[4-(Dimethylamino)-phenylazo]benzenesulfonic acid sodium salt, Acid orange 52, CI 13025
CAS Number
547-58-0
Grade
ACS reagent
Molecular Formula
C 14 H 14 N 3 O 3 SNa
Molecular Weight
327.3
MDL Number
MFCD00007502
Appearance
Orange-yellow solid
Conditions
Store at RT
Properties
Additional property information coming soon!
Safety and Regulations
Hazmat
Yes
Antibiotic
No
DEA-regulated
No
Warnings
Applications
Methyl orange is widely utilized in research focused on:
pH Indicator: Commonly used in titrations, it changes color from red to yellow as the pH shifts from acidic to neutral, making it a reliable choice for chemists in laboratories.
Textile Industry: Employed as a dye, it provides vibrant colors for fabrics, enhancing the aesthetic appeal of clothing and home textiles.
Food Industry: Used as a food colorant, it helps in achieving specific hues in products, although its use is regulated in many countries due to safety concerns.
Biochemical Research: Serves as a model compound in studies related to azo dyes, aiding researchers in understanding dye degradation and environmental impact.
Educational Purposes: Frequently used in chemistry education to demonstrate acid-base reactions, helping students visualize chemical changes in real-time.
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Oxidation state of group 2 elements
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1
$\begingroup$
I have came across the following statement : "the oxidation state $+II$ is stable for the elements of the alkalin earth metals" , the same is true for the elements of the first group , i've looked for a possible explanation and heard that it has a relation with the charge densities of the elements , if that's true what is that relation ? otherwise , what could explain the oxidation states for each of the groups in the periodic table ?
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edited Sep 28, 2013 at 18:03
user22323user22323
asked Sep 27, 2013 at 16:03
user22323user22323
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$\endgroup$
2
3
$\begingroup$ Can you clarify "the same is true for elements of the first group"? If you mean "the oxidation state +2 is stable for the elements of the alkali metals" then the statement is not correct. $\endgroup$
bobthechemist
– bobthechemist
2013-09-27 20:04:02 +00:00
Commented Sep 27, 2013 at 20:04
$\begingroup$ @bobthechemist no , i mean that the oxidation states +1 and + 2 are stable for the alkali and the alkali earth metals respectively . $\endgroup$
user22323
– user22323
2013-09-28 11:49:48 +00:00
Commented Sep 28, 2013 at 11:49
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The 'preferred' oxidation states of s-block metals can be explained by looking at their electron configurations:
Note that for the alkali metals, each element has one more electron than a noble-gas configuration. This electron is very weakly bound to the nucleus as evidenced by the ver low ionization energy:
It is much harder to remove the 2nd electron, since it is contained within a completely filled shell. (Please ignore the 2nd IE for Fr, as it doesn't exist in my data source.)
Therefore, the alkali metals have a tendency to lose just one electron, forming the +1 oxidation state.
The same thought process can be applied to the alkaline earth metals. Here are the first three ionization eneriges for each:
Hopefully, it is clear that in the case of the alkaline earth metals, it is the 3rd electron that is in the completely filled shell, and that electron is extremely difficult to remove. Thus, the metals in this group tend to stay in the +2 oxidation state.
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answered Sep 28, 2013 at 13:58
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Home> SI Prefixes
SI Prefixes
SI prefixes are used with SI base units to form a new SI unit that is greater or less than the base unit by some multiple of 10. For example "centi" is a prefix that means 1/100. When "centi" is placed in front of a base unit, such as meter, the new unit is called a centimeter and has the value of 1/100th the length of a meter.
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Fox and Mcdonald's Introduction to Fluid Mechanics
byPritchard, Philip J.; Mitchell, John W.; Leylegian, John C. (CON)
Edition: 9th
ISBN13: 9781118921876
ISBN10: 1118921879
Format: Loose-leaf
Pub. Date: 2015-03-02
Publisher(s): John Wiley & Sons Inc
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Summary
Fox & McDonald’s Introduction to Fluid Mechanics 9 th Editionhas been one of the most widely adopted textbooks in the field. This highly-regarded text continues to provide readers with a balanced and comprehensive approach to mastering critical concepts, incorporating a proven problem-solving methodology that helps readers develop an orderly plan to finding the right solution and relating results to expected physical behavior. The ninth edition features a wealth of example problems integrated throughout the text as well as a variety of new end of chapter problems.
Author Biography
Philip J. Pritchard is the author of Fox and McDonald's Introduction to Fluid Mechanics, 9th Edition, published by Wiley.
John W. Mitchell is the author of Fox and McDonald's Introduction to Fluid Mechanics, 9th Edition, published by Wiley.
Table of Contents
CHAPTER 1 INTRODUCTION 1
1.1 Introduction to Fluid Mechanics 2
Note to Students 2
Scope of Fluid Mechanics 3
Definition of a Fluid 3
1.2 Basic Equations 4
1.3 Methods of Analysis 5
System and Control Volume 6
Differential versus Integral Approach 7
Methods of Description 7
1.4 Dimensions and Units 9
Systems of Dimensions 9
Systems of Units 10
Preferred Systems of Units 11
Dimensional Consistency and “Engineering” Equations 11
1.5 Analysis of Experimental Error 13
1.6 Summary 14
Problems 14
CHAPTER 2 FUNDAMENTAL CONCEPTS 17
2.1 Fluid as a Continuum 18
2.2 Velocity Field 19
One-, Two-, and Three-Dimensional Flows 20
Timelines, Pathlines, Streaklines, and Streamlines 21
2.3 Stress Field 25
2.4 Viscosity 27
Newtonian Fluid 28
Non-Newtonian Fluids 30
2.5 Surface Tension 31
2.6 Description and Classification of Fluid Motions 34
Viscous and Inviscid Flows 34
Laminar and Turbulent Flows 36
Compressible and Incompressible Flows 37
Internal and External Flows 38
2.7 Summary and Useful Equations 39
References 40
Problems 40
CHAPTER 3 FLUID STATICS 47
3.1 The Basic Equation of Fluid Statics 48
3.2 The Standard Atmosphere 51
3.3 Pressure Variation in a Static Fluid 52
Incompressible Liquids: Manometers 52
Gases 57
3.4 Hydrostatic Force on Submerged Surfaces 59
Hydrostatic Force on a Plane Submerged Surface 59
Hydrostatic Force on a Curved Submerged Surface 66
3.5 Buoyancy and Stability 69
3.6 Fluids in Rigid-Body Motion (www.wiley.com/college/pritchard) 72
3.7 Summary and Useful Equations 72
References 73
Problems 73
CHAPTER 4 BASIC EQUATIONS IN INTEGRAL FORM FOR A CONTROL VOLUME 82
4.1 Basic Laws for a System 84
Conservation of Mass 84
Newton’s Second Law 84
The Angular-Momentum Principle 84
The First Law of Thermodynamics 85
The Second Law of Thermodynamics 85
4.2 Relation of System Derivatives to the Control Volume Formulation 85
Derivation 86
Physical Interpretation 88
4.3 Conservation of Mass 89
Special Cases 90
4.4 Momentum Equation for Inertial Control Volume 94
Differential Control Volume Analysis 105
Control Volume Moving with Constant Velocity 109
4.5 Momentum Equation for Control Volume with Rectilinear Acceleration 111
4.6 Momentum Equation for Control Volume with Arbitrary Acceleration (on the Web) 117
4.7 The Angular-Momentum Principle 117
Equation for Fixed Control Volume 117
4.8 The First and Second Laws of Thermodynamics 121
Rate of Work Done by a Control Volume 122
Control Volume Equation 123
4.9 Summary and Useful Equations 128
Problems 129
CHAPTER 5 INTRODUCTION TO DIFFERENTIAL ANALYSIS OF FLUID MOTION 144
5.1 Conservation of Mass 145
Rectangular Coordinate System 145
Cylindrical Coordinate System 149
5.2 Stream Function for Two-Dimensional Incompressible Flow 151
5.3 Motion of a Fluid Particle (Kinematics) 153
Fluid Translation: Acceleration of a Fluid Particle in a Velocity Field 154
Fluid Rotation 160
Fluid Deformation 163
5.4 Momentum Equation 167
Forces Acting on a Fluid Particle 167
Differential Momentum Equation 168
Newtonian Fluid: Navier–Stokes Equations 168
5.5 Introduction to Computational Fluid Dynamics 176
The Need for CFD 176
Applications of CFD 177
Some Basic CFD/Numerical Methods Using a Spreadsheet 178
The Strategy of CFD 182
Discretization Using the Finite-Difference Method 183
Assembly of Discrete System and Application of Boundary Conditions 184
Solution of Discrete System 185
Grid Convergence 185
Dealing with Nonlinearity 186
Direct and Iterative Solvers 187
Iterative Convergence 188
Concluding Remarks 189
5.6 Summary and Useful Equations 190
References 192
Problems 192
CHAPTER 6 INCOMPRESSIBLE INVISCID FLOW 198
6.1 Momentum Equation for Frictionless Flow: Euler’s Equation 199
6.2 Bernoulli Equation: Integration of Euler’s Equation Along a Streamline for Steady Flow 202
Derivation Using Streamline Coordinates 202
Derivation Using Rectangular Coordinates 203
Static, Stagnation, and Dynamic Pressures 205
Applications 207
Cautions on Use of the Bernoulli Equation 212
6.3 The Bernoulli Equation Interpreted as an Energy Equation 213
6.4 Energy Grade Line and Hydraulic Grade Line 217
6.5 Unsteady Bernoulli Equation: Integration of Euler’s Equation Along a Streamline (on the Web) 219
6.6 Irrotational Flow 219
Bernoulli Equation Applied to Irrotational Flow 219
Velocity Potential 220
Stream Function and Velocity Potential for Two-Dimensional, Irrotational, Incompressible Flow: Laplace’s Equation 221
Elementary Plane Flows 223
Superposition of Elementary Plane Flows 225
6.7 Summary and Useful Equations 234
References 235
Problems 236
CHAPTER 7 DIMENSIONAL ANALYSIS AND SIMILITUDE 244
7.1 Nondimensionalizing the Basic Differential Equations 245
7.2 Nature of Dimensional Analysis 246
7.3 Buckingham Pi Theorem 248
7.4 Significant Dimensionless Groups in Fluid Mechanics 254
7.5 Flow Similarity and Model Studies 256
Incomplete Similarity 258
Scaling with Multiple Dependent Parameters 263
Comments on Model Testing 266
7.6 Summary and Useful Equations 267
References 268
Problems 268
CHAPTER 8 INTERNAL INCOMPRESSIBLE VISCOUS FLOW 275
8.1 Internal Flow Characteristics 276
Laminar versus Turbulent Flow 276
The Entrance Region 277
PART A. FULLY DEVELOPED LAMINAR FLOW 277
8.2 Fully Developed Laminar Flow between Infinite Parallel Plates 277
Both Plates Stationary 278
Upper Plate Moving with Constant Speed, U 283
8.3 Fully Developed Laminar Flow in a Pipe 288
PART B. FLOW IN PIPES AND DUCTS 292
8.4 Shear Stress Distribution in Fully Developed Pipe Flow 293
8.5 Turbulent Velocity Profiles in Fully Developed Pipe Flow 294
8.6 Energy Considerations in Pipe Flow 297
Kinetic Energy Coefficient 298
Head Loss 298
8.7 Calculation of Head Loss 299
Major Losses: Friction Factor 299
Minor Losses 303
Pumps, Fans, and Blowers in Fluid Systems 308
Noncircular Ducts 309
8.8 Solution of Pipe Flow Problems 309
Single-Path Systems 310
Multiple-Path Systems 322
PART C. FLOW MEASUREMENT 326
8.9 Restriction Flow Meters for Internal Flows 326
The Orifice Plate 329
The Flow Nozzle 330
The Venturi 332
The Laminar Flow Element 332
Linear Flow Meters 335
Traversing Methods 336
8.10 Summary and Useful Equations 337
References 340
Problems 341
CHAPTER 9 EXTERNAL INCOMPRESSIBLE VISCOUS FLOW 353
PART A. BOUNDARY LAYERS 355
9.1 The Boundary-Layer Concept 355
9.2 Laminar Flat-Plate Boundary Layer: Exact Solution (www.wiley.com/college/pritchard) 359
9.3 Momentum Integral Equation 359
9.4 Use of the Momentum Integral Equation for Flow with Zero Pressure Gradient 363
Laminar Flow 364
Turbulent Flow 368
Summary of Results for Boundary-Layer Flow with Zero Pressure Gradient 371
9.5 Pressure Gradients in Boundary-Layer Flow 371
PART B. FLUID FLOW ABOUT IMMERSED BODIES 374
9.6 Drag 374
Pure Friction Drag: Flow over a Flat Plate Parallel to the Flow 375
Pure Pressure Drag: Flow over a Flat Plate Normal to the Flow 378
Friction and Pressure Drag: Flow over a Sphere and Cylinder 378
Streamlining 384
9.7 Lift 386
9.8 Summary and Useful Equations 400
References 402
Problems 403
CHAPTER 10 FLUID MACHINERY 412
10.1 Introduction and Classification of Fluid Machines 413
Machines for Doing Work on a Fluid 413
Machines for Extracting Work (Power) from a Fluid 415
Scope of Coverage 417
10.2 Turbomachinery Analysis 417
The Angular-Momentum Principle: The Euler Turbomachine Equation 417
Velocity Diagrams 419
Performance—Hydraulic Power 422
Dimensional Analysis and Specific Speed 423
10.3 Pumps, Fans, and Blowers 428
Application of Euler Turbomachine Equation to Centrifugal Pumps 428
Application of the Euler Equation to Axial Flow Pumps and Fans 429
Performance Characteristics 432
Similarity Rules 437
Cavitation and Net Positive Suction Head 441
Pump Selection: Applications to Fluid Systems 444
Blowers and Fans 455
10.4 Positive Displacement Pumps 461
10.5 Hydraulic Turbines 464
Hydraulic Turbine Theory 464
Performance Characteristics for Hydraulic Turbines 466
Sizing Hydraulic Turbines for Fluid Systems 470
10.6 Propellers and Wind-Power Machines 474
Propellers 474
Wind-Power Machines 482
10.7 Compressible Flow Turbomachines 490
Application of the Energy Equation to a Compressible Flow Machine 490
Compressors 491
Compressible-Flow Turbines 495
10.8 Summary and Useful Equations 495
References 497
Problems 499
CHAPTER 11 FLOW IN OPEN CHANNELS 507
11.1 Basic Concepts and Definitions 509
Simplifying Assumptions 509
Channel Geometry 511
Speed of Surface Waves and the Froude Number 512
11.2 Energy Equation for Open-Channel Flows 516
Specific Energy 518
Critical Depth: Minimum Specific Energy 521
11.3 Localized Effect of Area Change(Frictionless Flow) 524
Flow over a Bump 524
11.4 The Hydraulic Jump 528
Depth Increase across a Hydraulic Jump 531
Head Loss across a Hydraulic Jump 532
11.5 Steady Uniform Flow 534
The Manning Equation for Uniform Flow 536
Energy Equation for Uniform Flow 541
Optimum Channel Cross Section 543
11.6 Flow with Gradually Varying Depth 544
Calculation of Surface Profiles 545
11.7 Discharge Measurement Using Weirs 548
Suppressed Rectangular Weir 548
Contracted Rectangular Weirs 549
Triangular Weir 549
Broad-Crested Weir 550
11.8 Summary and Useful Equations 551
References 552
Problems 553
CHAPTER 12 INTRODUCTION TO COMPRESSIBLE FLOW 556
12.1 Review of Thermodynamics 557
12.2 Propagation of Sound Waves 563
Speed of Sound 563
Types of Flow—The Mach Cone 567
12.3 Reference State: Local Isentropic Stagnation Properties 570
Local Isentropic Stagnation Properties for the Flow of an Ideal Gas 571
12.4 Critical Conditions 577
12.5 Basic Equations for One-Dimensional Compressible Flow 577
Continuity Equation 577
Momentum Equation 578
First Law of Thermodynamics 578
Second Law of Thermodynamics 579
Equation of State 579
12.6 Isentropic Flow of an Ideal Gas: Area Variation 580
Subsonic Flow, M Supersonic Flow, M >1 583
Sonic Flow, M =1 583
Reference Stagnation and Critical Conditions for Isentropic Flow of an Ideal Gas 584
Isentropic Flow in a Converging Nozzle 589
Isentropic Flow in a Converging-Diverging Nozzle 593
12.7 Normal Shocks 598
Basic Equations for a Normal Shock 599
Normal-Shock Flow Functions for One-Dimensional Flow of an Ideal Gas 601
12.8 Supersonic Channel Flow with Shocks 605
12.8 Supersonic Channel Flow with Shocks
12.9 Flow in a Constant-Area Duct with Friction (www.wiley.com/college/pritchard) 607
12.10 Frictionless Flow in a Constant-Area Duct with Heat Exchange (www.wiley.com/college/pritchard) 607
12.11 Oblique Shocks and Expansion Waves(www.wiley.com/college/pritchard) 607
12.12 Summary and Useful Equations 607
References 610
Problems 610
APPENDIX A FLUID PROPERTY DATA 615
APPENDIX B VIDEOS FOR FLUID MECHANICS 627
APPENDIX C SELECTED PERFORMANCE CURVES FOR PUMPS AND FANS 629
APPENDIX D FLOW FUNCTIONS FOR COMPUTATION OF COMPRESSIBLE FLOW 640
APPENDIX E ANALYSIS OF EXPERIMENTAL UNCERTAINTY 643
APPENDIX F ADDITIONAL COMPRESSIBLE FLOW FUNCTIONS (WWW.WILEY.COM/
COLLEGE/PRITCHARD ONLINE) WF-1
APPENDIX G A BRIEF REVIEW OF MICROSOFT EXCEL (WWW.WILEY.COM/COLLEGE/ PRITCHARD) WG-1
Answers to Selected Problems 649
Index 656
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187981 | https://www.pasco.com/curriculum/essential-chemistry/BookInd-3417?srsltid=AfmBOootwt9M59osBpZYNcmT63CHmiJ8gSjtt4xJ29hWwGKMrbK2vcqp | Combustion Reactions - Essential Chemistry | PASCO
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Combustion reactions
Combustion reactionsCombustion reactions always involve the oxygen molecule, O 2 as a reactant. The general form of a combustion reaction is substance X reacts with oxygen to form a new XO compound. Most often, we think of fire and combustion as being the same thing. However, when iron rusts, solid iron reacts with O 2 in the atmosphere. Technically, rusting iron is a combustion reaction even though fire is not produced. Hydrocarbons A hydrocarbon is a molecule that only contains hydrogen and carbon. Methane (CH 4), propane (C 3 H 8), and butane (C 4 H 10) are examples of hydrocarbons. When hydrocarbons are burned, they always form carbon dioxide and water. Combustion reactions can be very challenging to balance! You often have to go through the balancing process many times. Solved problem Write the balanced reaction for the combustion of magnesium metal. Relationships In a combustion reaction, the given reactant combines with oxygen to form a product. Solve 1. Write the reactants. Magnesium does not form a diatomic molecule: Mg Oxygen forms a diatomic molecule: O 2 The reactants are: Mg + O 2 2. Find the charges of the reactants and criss-cross to form a compound. Magnesium = Mg 2+ and oxygen = O 2+ Criss-cross charges; in this case the charges cancel: MgO 3. Write the reaction: Mg + O 2 → MgO 4. Balance the reaction: 2Mg + O 2 → 2MgO Answer The balanced reaction is: 2Mg + O 2 → 2MgO Solved problem Write the balanced reaction for the combustion of pentane, C 5 H 12. Relationships This is the combustion of a hydrocarbon. The hydrocarbon will react with oxygen, and the products will be carbon dioxide and water. Solve 1. When a hydrocarbon undergoes combustion, it reacts with oxygen to produce carbon dioxide and water. No criss-crossing is necessary. C 5 H 12 + O 2 → CO 2 + H 2 O 2. Balance the reaction: Try balancing carbon first, then hydrogen, and balance oxygen last. Keep going back and forth until the reaction is balanced. C 5 H 12 + 8O 2 → 5CO 2 + 6H 2 O Answer The balanced reaction is: C 5 H 12 + 8O 2 → 5CO 2 + 6H 2 O
206
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Linear Formula:
H 2 O
CAS Number:
7732-18-5
Molecular Weight:
18.02
Beilstein:
2050024
EC Number:
231-791-2
MDL number:
MFCD00011332
UNSPSC Code:
41116107
PubChem Substance ID:
329798917
NACRES:
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| ###### This Item | EM3234 | 900682 | DENWAT3 |
--- --- |
| DENWAT Pure Water Density Standard Quick View | Supelco EM3234 Water, Deionized Distilled, ASTM Type II Quick View | Sigma-Aldrich 900682 Water Quick View | DENWAT3 Pure Water Density Standard Quick View |
| format mixture | format - | format - | format mixture |
| bp 100°C (lit.) | bp 100°C (lit.) | bp 100°C (lit.) | bp 100°C (lit.) |
| mp 0°C (lit.) | mp 0°C (lit.) | mp 0°C (lit.) | mp 0°C (lit.) |
| grade certified reference material | grade - | grade - | grade certified reference material |
| application(s) cleaning products cosmetics environmental flavors and fragrances food and beverages industrial qc personal care petroleum | application(s) - | application(s) food and beverages | application(s) cleaning products cosmetics environmental flavors and fragrances food and beverages industrial qc personal care petroleum |
| manufacturer/tradename Paragon Scientific Ltd | manufacturer/tradename - | manufacturer/tradename - | manufacturer/tradename Paragon Scientific Ltd |
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1: Introduction - The Ambit of Chemistry
ChemPRIME (Moore et al.)
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Fri, 16 Jun 2023 22:12:35 GMT
1.10: Conversion Factors and Functions
49281
49281
Valerie Flood
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Contents
1. Home
2. Bookshelves
3. General Chemistry
4. ChemPRIME (Moore et al.)
5. 1: Introduction - The Ambit of Chemistry
6. 1.10: Conversion Factors and Functions
Expand/collapse global location
ChemPRIME (Moore et al.)
Front Matter
1: Introduction - The Ambit of Chemistry
2: Atoms, Molecules, and Chemical Reactions
3: Using Chemical Equations in Calculations
4: The Structure of Atoms
5: The Electronic Structure of Atoms
6: Chemical Bonding - Electron Pairs and Octets
7: Further Aspects of Covalent Bonding
8: Properties of Organic Compounds
9: Gases
10: Solids, Liquids and Solutions
11: Reactions in Aqueous Solutions
12: Chemistry of the Representative Elements
13: Chemical Equilibrium
14: Ionic Equilibria in Aqueous Solutions
15: Thermodynamics- Atoms, Molecules and Energy
16: Entropy and Spontaneous Reactions
17: Electrochemical Cells
18: Chemical Kinetics
19: Nuclear Chemistry
20: Molecules in Living Systems
21: Spectra and Structure of Atoms and Molecules
22: Metals
Back Matter
1.10: Conversion Factors and Functions
Last updated Jun 16, 2023
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1.9.1: Density Lecture Demonstrations
2: Atoms, Molecules, and Chemical Reactions
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Page ID 49281
Ed Vitz, John W. Moore, Justin Shorb, Xavier Prat-Resina, Tim Wendorff, & Adam Hahn
Chemical Education Digital Library (ChemEd DL)
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. An Important Caveat
1. Example 1.10.1: Volume of Ethanol
1. Solution
2. Example 1.10.2: Volume of Benzene/01:_Introduction_-_The_Ambit_of_Chemistry/1.10:_Conversion_Factors_and_Functions#Example_.5C(.5CPageIndex.7B2.7D.5C):_Volume_of_Benzene)
1. Solution/01:_Introduction_-_The_Ambit_of_Chemistry/1.10:_Conversion_Factors_and_Functions#Solution_2)
3. Example 1.10.3: Volume of Mercury/01:_Introduction_-_The_Ambit_of_Chemistry/1.10:_Conversion_Factors_and_Functions#Example_.5C(.5CPageIndex.7B3.7D.5C):_Volume_of_Mercury)
1. Solution/01:_Introduction_-_The_Ambit_of_Chemistry/1.10:_Conversion_Factors_and_Functions#Solution_3)
4. Example 1.10.1: Volume to Mass Conversion/01:_Introduction_-_The_Ambit_of_Chemistry/1.10:_Conversion_Factors_and_Functions#Example_.5C(.5CPageIndex.7B1.7D.5C):_Volume_to_Mass_Conversion)
1. Solution/01:_Introduction_-_The_Ambit_of_Chemistry/1.10:_Conversion_Factors_and_Functions#Solution_4)
Earlier we showed how unity factors can be used to express quantities in different units of the same parameter. For example, a density can be expressed in g/cm 3 or lb/ft 3. Now we will see how conversion factors representing mathematical functions, like D = m/v, can be used to transform quantities into different parameters. For example, what is the
volume
of a given mass of gold? Unity factors and conversion factors are conceptually different, and we'll see that the "dimensional analysis" we develop for unit conversion problems must be used with care in the case of functions.
When we are referring to the same object or sample of material, it is often useful to be able to convert one parameter into another. For example, in our discussion of fossil-fuel reserves we find that 318 Pg (3.18 × 10 17 g) of
coal
, 28.6 km 3 (2.68 × 10 10 m 3) of petroleum, and 2.83 × 10 3 km 3 (2.83 × 10 13 m 3) of natural gas (measured at normal atmospheric pressure and 15°C) are available. But none of these quantities tells us what we really want to know ― how much heat energy could be released by burning each of these reserves? Only by converting the mass of
coal
and the volumes of petroleum and natural gas into their equivalent energies can we make a valid comparison. When this is done, we find that the
coal
could release 7.2 × 10 21 J, , the petroleum 1.1 × 10 21 J, and the gas 1.1 × 10 21 J of heat energy. Thus the reserves of
coal
are more than three times those of the other two fuels combined. It is for this reason that more attention is being paid to the development of new ways for using
coal
resources than to oil or gas. Conversion of one kind of quantity into another is usually done with what can be called a conversion factor, but the conversion factor is based on a mathematical function (D = m / V) or mathematical equation that relates parameters. Since we have not yet discussed energy or the units (joules) in which it is measured, an example involving the more familiar quantities mass and
volume
will be used to illustrate the way conversion factors are employed. The same principles apply to finding how much energy would be released by burning a fuel, and that problem will be encountered later.
For helpful context about the above discussion, check out the following Crash Course
Chemistry
video:
Unit Conversion & Significant Figures : Crash Course Chemistry #2(opens in new window) [youtu.be]
Suppose we have a rectangular solid sample of gold which measures 3.04 cm × 8.14 cm × 17.3 cm. We can easily calculate that its
volume
is 428 cm 3 but how much is it worth? The price of gold is about 5 dollars per gram, and so we need to know the mass rather than the
volume
. It is unlikely that we would have available a scale or balance which could weigh accurately such a large, heavy sample, and so we would have to determine the mass of gold equivalent to a
volume
of 428 cm 3. This can be done by manipulating the equation which defines density, ρ = m / V. If we multiply both sides by V, we obtain
(1.10.1)V×ρ=m V×V=m
m=V×ρ
or
mass=
volume
×density
Taking the density of gold from a reference table(opens in new window), we can now calculate
Mass=m=Vρ=428 cm 3×10 0.32 g 1 cm 3=8.27×10 3g=8.27 kg
This is more than 18 lb of gold. At the price quoted above, it would be worth over 40 000 dollars!
The formula which defines density can also be used to convert the mass of a sample to the corresponding
volume
. If both sides of Eq. 1.10.1 are multiplied by 1/ρ, we have
1 ρ×m=Vρ×1 ρ=V
(1.10.2)V=m×1 ρ
Notice that we used the mathematical function D = m/V to convert parameters from mass to
volume
or vice versa in these examples. How does this differ from the use of unity factors to change units of one parameter?
An Important Caveat
A mistake sometimes made by beginning students is to confuse density with concentration, which also may have units of g/cm 3. By dimensional analysis, this looks perfectly fine. To see the error, we must understand the meaning of the function
C=m V
In this case, V refers to the
volume
of a solution, which contains both a solute and solvent.
Given a
concentration
of an alloy is 10 g gold in 100 cm 3 of alloy, we see that it is wrong (although dimensionally correct as far as conversion factors go) to incorrectly calculate the
volume
of gold in 20 g of the alloy as follows:
20 \text{g} \times \dfrac{\text{100 cm^3}}{\text{10 g}} = 200 \text{ cm}^{3} \nonumber
It is only possible to calculate the
volume
of gold if the density of the alloy is known, so that the
volume
of alloy represented by the 20 g could be calculated. This
volume
multiplied by the
concentration
gives the mass of gold, which then can be converted to a
volume
with the density function.
The bottom line is that using a simple unit cancellation method does not always lead to the expected results, unless the mathematical function on which the conversion factor is based is fully understood.
Example 1.10.1:
Volume
of Ethanol
A solution of ethanol with a
concentration
of 0.1754 g / cm 3 has a density of 0.96923 g / cm 3 and a freezing point of -9 ° F . What is the
volume
of ethanol (D = 0.78522 g / cm 3 at 25 °C) in 100 g of the solution?
Solution
The
volume
of 100 g of solution is
V=m÷D=100 g÷0.96923 g cm 3=103.17cm 3
The mass of ethanol in this
volume
is
m=V×C=103.17cm 3×0.1754 g /cm 3=18.097 g
The
volume
of ethanol=m÷D=18.097 g÷0.78522 g /cm 3=23.05cm 3
Note that we cannot calculate the
volume
of ethanol by
0.96923g cm 3×100cm 3 0.78522g cm 3=123.4cm 3
even though this equation is dimensionally correct.
Note: Note that this result required when to use the function C = m/V, and when to use the function D=m/V as conversion factors. Pure dimensional analysis could not reliably give the answer, since both functions have the same dimensions.
Example 1.10.2:
Volume
of Benzene
Find the
volume
occupied by a 4.73-g sample of benzene.
Solution
The density of benzene is 0.880 g cm–3. Using Eq. (2),
Volume
=V=m×1 ρ= 4.73 g×1 cm 3 0.880 g= 5.38 cm 3
Note: Note that taking the reciprocal of 0.880 g 1 cm 3 simply inverts the fraction ― 1 cm 3 goes on top, and 0.880 g goes on the bottom.
The two calculations just done show that density is a conversion factor which changes
volume
to mass, and the reciprocal of density is a conversion factor changing mass into
volume
. This can be done because the mathematical formula defining density relates it to mass and
volume
. Algebraic manipulation of this formula gave us expressions for mass and for
volume
[Eq. 1.10.1 and 1.10.2], and we used them to solve our problems. If we understand the function D = m/V and heed the caveat above, we can devise appropriate conversion factors by unit cancellation, as the following example shows:
Example 1.10.3:
Volume
of Mercury
A student weighs 98.0 g of mercury. If the density of mercury is 13.6 g/cm 3, what
volume
does the sample occupy?
Solution
We know that
volume
is related to mass through density.
Therefore
V=m×conversion factor
Since the mass is in grams, we need to get rid of these units and replace them with
volume
units. This can be done if the reciprocal of the density is used as a conversion factor. This puts grams in the denominator so that these units cancel:
V=m×1 ρ=98.0 g×1 cm 3 13.6 g=7.21 cm 3
If we had multiplied by the density instead of its reciprocal, the units of the result would immediately show our error:
V=98.0 g×13.6 g 1 cm 3=1.333 g 2/cm 3 (no cancellation!)
It is clear that square grams per cubic centimeter are not the units we want.
Using a conversion factor is very similar to using a unity factor — we know the conversion factor is correct when units cancel appropriately. A conversion factor is not unity, however. Rather it is a physical quantity (or the reciprocal of a physical quantity) which is related to the two other quantities we are interconverting. The conversion factor works because of the relationship [ie. the definition of density as defined by Eqs. 1.10.1 and 1.10.2 includes the relationships between density, mass, and
volume
], not because it is has a value of one. Once we have established that a relationship exists, it is no longer necessary to memorize a mathematical formula. The units tell us whether to use the conversion factor or its reciprocal. Without such a relationship, however, mere cancellation of units does not guarantee that we are doing the right thing.
A simple way to remember relationships among quantities and conversion factors is a “road map“of the type shown below:
Mass⟷density
volume
or m⟷ρ V
This indicates that the mass of a particular sample of
matter
is related to its
volume
(and the
volume
to its mass) through the conversion factor, density. The double arrow indicates that a conversion may be made in either direction, provided the units of the conversion factor cancel those of the quantity which was known initially. In general the road map can be written
First quantity⟷conversion factor second quantity
As we come to more complicated problems, where several steps are required to obtain a final result, such road maps will become more useful in charting a path to the solution.
Example 1.10.1:
Volume
to Mass Conversion
Black ironwood has a density of 67.24 lb/ft 3. If you had a sample whose
volume
was 47.3 ml, how many grams would it weigh? (1 lb = 454 g; 1 ft = 30.5 cm).
Solution
The road map
V→ρ m
tells us that the mass of the sample may be obtained from its
volume
using the conversion factor, density. Since milliliters and cubic centimeters are the same, we use the SI units for our calculation:
Mass=m=47.3cm 3×67.24 lb 1 ft 3
Since the
volume
units are different, we need a unity factor to get them to cancel:
m= 47.3 cm 3×(1 ft 30.5 cm)3×67.24 lb 1 ft 3= 47.3 cm 3×1 ft 3 30.5 3cm 3×67.24 lb 1 ft 3
We now have the mass in pounds, but we want it in grams, so another unity factor is needed:
m= 47.3 cm 3×1 ft 3 30.5 3cm 3×67.24 lb 1 ft 3×454 g 1 lb= 50 0.9 g
In subsequent chapters we will establish a number of relationships among physical quantities. Formulas will be given which define these relationships, but we do not advocate slavish memorization and manipulation of those formulas. Instead we recommend that you remember that a relationship exists, perhaps in terms of a road map, and then adjust the quantities involved so that the units cancel appropriately. Such an approach has the advantage that you can solve a wide variety of problems by using the same technique.
This page titled 1.10: Conversion Factors and Functions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Ed Vitz, John W. Moore, Justin Shorb, Xavier Prat-Resina, Tim Wendorff, & Adam Hahn.
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187985 | https://www.epa.gov/sites/default/files/2016-09/documents/dde.pdf | DDE DDE (1,1-DICHLORO-2,2-BIS(p-CHLOROPHENYL) ETHYLENE) (A) 72-55-9 Hazard Summary 1,1-Dichloro-2,2-bis(p-chlorophenyl) ethylene (DDE) is a breakdown product of DDT, which was used in the past as an insecticide. No information is available on the acute (short-term) or chronic (long-term) effects of DDE. Acute, oral exposure to high doses of DDT in humans results in central nervous system (CNS) effects, such as headaches, nausea, and convulsions. The only effect noted in epidemiologic studies of workers exposed to DDT and other pesticides was an increase in activity of liver enzymes. Animal studies have reported effects on the liver, immune system, and CNS from chronic oral exposure to DDT. Human studies are inconclusive regarding DDE and cancer. Animal studies have reported an increased incidence of liver tumors in mice and hamsters, and thyroid tumors in female rats from oral exposure to DDE. EPA has classified DDE as a Group B2, probable human carcinogen.
Please Note: The main source of information for this fact sheet is the Agency for Toxic Substances and Disease Registry's (ATSDR's) Toxicological Profile for 4,4-DDT, 4,4-DDE, and 4,4-DDD (1) and EPA's Integrated Risk Information System (IRIS) (5), which contains information on the carcinogenic effects of DDE including the unit cancer risk for oral exposure. Uses DDT was extensively used in the past for the control of malaria, typhus, and other insect-transmitted diseases. It was banned for use in the United States in 1972, except in the case of a public health emergency. (1) DDE is a breakdown product of DDT and has no uses. (1) Sources and Potential Exposure DDE is found in the environment as a result of the breakdown of DDT, an insecticide. (1) Human exposure to DDE appears to be primarily through food; in the United States in 1981, consumption of DDE in foods was estimated to be 0.001 parts per million per day (ppm/d). However, the levels of DDE in foods have been decreasing and are expected to continue to decrease. (1) Levels of DDE in air and water samples are very low. (1) DDE has been listed as a pollutant of concern to EPA's Great Waters Program due to its persistence in the environment, potential to bioaccumulate, and toxicity to humans and the environment (2).
Assessing Personal Exposure DDE can be detected in fat, blood, urine, semen, and breast milk. (1) Health Hazard Information Acute Effects: Acute Effects: No studies are available on the acute effects of DDE in humans. (1) Acute oral exposure to high doses of DDT in humans results in CNS effects, such as headaches, nausea, and convulsions. (1) Case reports in humans have noted that doses as high as 285 milligrams DDT per kilogram body weight per day (mg/kg/d) have been ingested accidentally with no fatal results. (1) Tests involving acute exposure of rats, guinea pigs, and rabbits have shown DDT to have moderate acute toxicity from oral exposure. (3) Chronic Effects (Noncancer): The only effect noted in epidemiologic studies of workers exposed to DDT and other pesticides was an increase in activity of liver enzymes. No adverse effects on the blood, liver, heart, or CNS were noted. (1) Animal studies have reported effects on the liver, immune system, and CNS from chronic oral administration of DDT. (1,4,9) EPA has not established a Reference Concentration (RfC) or a Reference Dose (RfD) for DDE. (5) EPA has established an RfD of 0.0005 milligrams per kilogram body weight per day (mg/kg/d) for DDT based on liver effects in rats. The RfD is an estimate (with uncertainty spanning perhaps an order of magnitude) of a continuous ingestion exposure to the human population (including sensitive subgroups), that is likely to be without appreciable risk of deleterious noncancer effects during a lifetime. It is not a direct estimator of risk but rather a reference point to gauge the potential effects. At exposures increasingly greater than the RfD, the potential for adverse health effects increases. Lifetime exposure above the RfD does not imply that an adverse health effect would necessarily occur. (5) Reproductive/Developmental Effects: No information is available on the reproductive or developmental effects of DDT or DDE in humans via inhalation exposure. (1) No studies are available on the developmental effects in humans after oral exposure to DDT or DDE. However, DDT and DDE have been found in human blood, placental tissue, and umbilical cord blood. (1) Epidemiologic studies did not find an association between DDT maternal blood levels and miscarriages or premature rupture of fetal membranes in humans. (1) Oral animal studies have reported reproductive effects, such as reduced fertility, adverse effects on spermatogenesis, and decreased testicular and ovarian weights from DDT exposure. Developmental effects, such as embryotoxicity and fetotoxicity, but not teratogenicity (birth defects) have also been observed in oral animal studies. (1) DDT has been shown to elicit estrogenic activity in rats after oral exposure (1).
Cancer Risk: Studies of workers exposed to DDT have yielded conflicting results. Three studies reported that tissue levels of DDT and DDE were higher in cancer victims than in those dying of other diseases. In other studies, no such relationship was seen. (5,9) Animal studies have reported an increased incidence of liver tumors in mice and hamsters and thyroid tumors in female rats from oral exposure to DDE. (5) EPA has classified DDE as a Group B2, probable human carcinogen. (5) EPA uses mathematical models, based on animal studies to estimate the probability of a person developing cancer from ingesting water containing a specified concentration of a chemical. EPA has calculated an oral cancer slope factor of 0.34 (mg/kg/d) -1 and a unit risk estimate of 9.7 × 10 -6 (µg/L) -1 . EPA estimates that, if an individual were to continuously ingest water containing an average of DDE at 0.1 µg/L over his or her entire lifetime, that person would theoretically have no more than a one-in-a-million increased chance of developing cancer as a direct result of ingesting water containing this chemical. Similarly, EPA estimates that ingesting water containing 1.0 µg/L would result in not greater than a one-in-a-hundred-thousand increased chance of developing cancer, and water containing 10.0 µg/L would result in not greater than a one-in-ten thousand increased chance of developing cancer. For a detailed discussion of confidence in the one-in-ten thousand increased chance of developing cancer. For a detailed discussion of confidence in the potency estimates, please see IRIS. (5) Physical Properties DDE is also known as 1,1-dichloro-2,2-bis(p-chlorophenyl) ethylene and p,p-dichlorodiphenyldichloroethylene.
DDE is a white crystalline solid. (1) The odor threshold for DDE is not available. (1) The chemical formula for DDE is C 14H 8Cl 4, and the molecular weight is 318.03 g/mol. (1) The vapor pressure for DDE is 6.5 × 10 -6 torr at 20 °C, and it has a log octanol/water partition coefficient (log K ow) of 7.0. (1) Conversion Factors: To convert concentrations in air (at 25 °C) from ppm to mg/m 3 : mg/m 3 = (ppm) × (molecular weight of the compound)/(24.45). For DDE: 1 ppm = 13.0 mg/m 3 ; for DDT: 1 ppm = 14.5 mg/m 3 . Health Data from Inhalation Exposure ACGIH TLV --American Conference of Governmental and Industrial Hygienists' threshold limit value expressed as a ACGIH TLV --American Conference of Governmental and Industrial Hygienists' threshold limit value expressed as a time-weighted average; the concentration of a substance to which most workers can be exposed without adverse effects. NIOSH IDLH --National Institute of Occupational Safety and Health's immediately dangerous to life or health limit; NIOSH recommended exposure limit to ensure that a worker can escape from an exposure condition that is likely to cause death or immediate or delayed permanent adverse health effects or prevent escape from the environment. NIOSH REL --NIOSH's recommended exposure limit; NIOSH-recommended exposure limit for an 8- or 10-h time-weighted-average exposure and/or ceiling. OSHA PEL--Occupational Safety and Health Administration's permissible exposure limit expressed as a time-weighted average; the concentration of a substance to which most workers can be exposed without adverse effect averaged over a normal 8-h workday or a 40-h workweek.
All health and regulatory numbers are for DDT.
The health and regulatory values cited in this fact sheet were obtained in December 1999.
a Health numbers are toxicological numbers from animal testing or risk assessment values developed by EPA. b Regulatory numbers are values that have been incorporated in Government regulations, while advisory numbers are nonregulatory values provided by the Government or other groups as advice. OSHA numbers are regulatory, whereas NIOSH and ACGIH numbers are advisory.
Summary created in April 1992, updated January 2000 References 1. Agency for Toxic Substances and Disease Registry (ATSDR). Toxicological Profile for 4,4-DDT, 4,4-DDE, and 4,4-DDD. Public Health Service, U.S. Department of Health and Human Services, Atlanta, GA. 1994.
2. U.S. Environmental Protection Agency. Deposition of Air Pollutants to the Great Waters. EPA-453/R-93-055. First Report to Congress. Office of Air Quality Planning and Standards, Research Triangle Park, NC.
1994.
3. U.S. Department of Health and Human Services. Registry of Toxic Effects of Chemical Substances (RTECS, online database). National Toxicology Information Program, National Library of Medicine, Bethesda, MD.
1993.
4. U.S. Department of Health and Human Services. Hazardous Substances Data Bank (HSDB, online database).
National Toxicology Information Program, National Library of Medicine, Bethesda, MD. 1993.
5. U.S. Environmental Protection Agency. Integrated Risk Information System (IRIS) on p,p-Dichlorodiphenyldichloroethylene. National Center for Environmental Assessment, Office of Research and Development, Washington, DC. 1999.
6. Occupational Safety and Health Administration (OSHA). Occupational Safety and Health Standards, Toxic and Hazardous Substances. Code of Federal Regulations. 29 CFR 1910.1000. 1998.
7. American Conference of Governmental Industrial Hygienists (ACGIH). 1999 TLVs and BEIs. Threshold Limit Values for Chemical Substances and Physical Agents. Biological Exposure Indices. Cincinnati, OH. 1999.
8. National Institute for Occupational Safety and Health (NIOSH). Pocket Guide to Chemical Hazards. U.S.
Department of Health and Human Services, Public Health Service, Centers for Disease Control and Prevention. Cincinnati, OH. 1997.
9. U.S. Environmental Protection Agency. Integrated Risk Information System (IRIS) on p,p-Dichlorodiphenyltrichloroethane. National Center for Environmental Assessment, Office of Research and Development, Washington, DC. 1999.
A. This fact sheet focuses on the health effects of DDE. However, since DDE is a breakdown product of DDT, in those cases where no information is available on DDE and there is information on DDT, the information on DDT is presented. |
187986 | https://askfilo.com/user-question-answers-smart-solutions/11-given-f-x-sin-x-check-if-function-f-is-one-one-for-i-0-pi-3138373634323134 | Question asked by Filo student
Views: 5,213 students
Updated on: Jan 7, 2025
Text SolutionText solutionverified iconVerified
Concepts:
One-one function, Sine function, Intervals
Explanation:
To determine if the function f(x)=sinx is one-one (injective) on a given interval, we need to check if f(a)=f(b) implies a=b for all a,b in that interval.
Step by Step Solution:
Step 1
Consider the function f(x)=sinx.
Step 2
For the interval (0,π), assume f(a)=f(b) for some a,b∈(0,π).
Step 3
This implies sina=sinb. Since sinx is strictly increasing on (0,π), it follows that a=b.
Step 4
Therefore, f(x)=sinx is one-one on (0,π).
Step 5
For the interval (−π/2,π/2), assume f(a)=f(b) for some a,b∈(−π/2,π/2).
Step 6
This implies sina=sinb. Since sinx is strictly increasing on (−π/2,π/2), it follows that a=b. Therefore, f(x)=sinx is one-one on (−π/2,π/2).
Final Answer:
The function f(x)=sinx is one-one on both (0,π) and (−π/2,π/2).
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| Question Text | 11. Given f(x)=\sin x check if function f is one-one for (i) (0, \pi) (ii) (-\pi / 2, \pi / 2). |
| Updated On | Jan 7, 2025 |
| Topic | All topics |
| Subject | Smart Solutions |
| Class | Class 12 |
| Answer Type | Text solution:1 |
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187987 | https://www.anderson.k12.ky.us/Downloads/04_Lecture_Presentation2.pdf | CAMPBELL BIOLOGY IN FOCUS © 2016 Pearson Education, Inc.
URRY • CAIN • WASSERMAN • MINORSKY • REECE Lecture Presentations by Kathleen Fitzpatrick and Nicole Tunbridge, Simon Fraser University SECOND EDITION 4 A Tour of the Cell Overview: The Fundamental Units of Life All organisms are made of cells The cell is the simplest collection of matter that can be alive All cells are related by their descent from earlier cells Though cells can differ substantially from one another, they share common features © 2016 Pearson Education, Inc.
Figure 4.1-1 © 2016 Pearson Education, Inc.
Figure 4.1-2 © 2016 Pearson Education, Inc.
▲Paramecium caudatum 40 mm Concept 4.1: Biologists use microscopes and the tools of biochemistry to study cells Most cells are too small to be seen by the unaided eye © 2016 Pearson Education, Inc.
Microscopy Scientists use microscopes to observe cells too small to be seen with the naked eye In a light microscope (LM), visible light is passed through a specimen and then through glass lenses Lenses refract (bend) the light, so that the image is magnified © 2016 Pearson Education, Inc.
Three important parameters of microscopy Magnification, the ratio of an object’s image size to its real size Resolution, the measure of the clarity of the image, or the minimum distance between two distinguishable points Contrast, visible differences in parts of the sample © 2016 Pearson Education, Inc.
LMs can magnify effectively to about 1,000 times the size of the actual specimen Various techniques enhance contrast and enable cell components to be stained or labeled Most subcellular structures, including organelles (membrane-enclosed compartments), are too small to be resolved by light microscopy © 2016 Pearson Education, Inc.
Figure 4.2 © 2016 Pearson Education, Inc.
Human height Length of some nerve and muscle cells Chicken egg Frog egg Human egg Most plant and animal cells Nucleus Most bacteria Mitochondrion Smallest bacteria Viruses Ribosomes Super-resolution microscopy Proteins Lipids Small molecules Atoms 0.1 nm 1 nm 10 nm 100 nm 1 mm 10 mm 100 mm 1 mm 1 cm 0.1 m 1 m 10 m Unaided eye LM EM Figure 4.2-1 © 2016 Pearson Education, Inc.
Human height Length of some nerve and muscle cells Chicken egg Frog egg Human egg 100 mm 1 mm 1 cm 0.1 m 1 m 10 m Unaided eye LM Figure 4.2-2 © 2016 Pearson Education, Inc.
© 2016 Pearson Education, Inc.
Most plant and animal cells Nucleus Most bacteria Mitochondrion Smallest bacteria Viruses Ribosomes Super-resolution microscopy Proteins Lipids Small molecules Atoms 0.1 nm 1 nm 10 nm 100 nm 1 mm 10 mm LM EM 100 mm Figure 4.2 © 2016 Pearson Education, Inc.
Human height Length of some nerve and muscle cells Chicken egg Frog egg Human egg Most plant and animal cells Nucleus Most bacteria Mitochondrion Smallest bacteria Viruses Ribosomes Super-resolution microscopy Proteins Lipids Small molecules Atoms 0.1 nm 1 nm 10 nm 100 nm 1 mm 10 mm 100 mm 1 mm 1 cm 0.1 m 1 m 10 m Unaided eye LM EM Two basic types of electron microscopes (EMs) are used to study subcellular structures Scanning electron microscopes (SEMs) focus a beam of electrons onto the surface of a specimen, producing images that look three-dimensional Transmission electron microscopes (TEMs) focus a beam of electrons through a specimen TEM is used mainly to study the internal structure of cells © 2016 Pearson Education, Inc.
Figure 4.3 © 2016 Pearson Education, Inc.
Light Microscopy (LM) Electron Microscopy (EM) Brightfield (unstained specimen) Brightfield (stained specimen) Phase-contrast Differential-interference contrast (Nomarski) Confocal (without technique) Confocal (with technique) Longitudinal section of cilium Cross section of cilium Cilia Scanning electron microscopy (SEM) 2 mm 2 mm Transmission electron microscopy (TEM) Fluorescence 10 mm 50 mm 50 mm Figure 4.3-1 © 2016 Pearson Education, Inc.
Light Microscopy (LM) Brightfield (unstained specimen) Brightfield (stained specimen) Phase-contrast Differential-interference contrast (Nomarski) 50 mm Figure 4.3-1a © 2016 Pearson Education, Inc.
Brightfield (unstained specimen) 50 mm Figure 4.3-1b © 2016 Pearson Education, Inc.
Brightfield (stained specimen) 50 mm Figure 4.3-1c © 2016 Pearson Education, Inc.
Phase-contrast 50 mm Figure 4.3-1d © 2016 Pearson Education, Inc.
Differential-interference contrast (Nomarski) 50 mm Figure 4.3-2 © 2016 Pearson Education, Inc.
Light Microscopy (LM) Confocal (with technique) Fluorescence 10 mm 50 mm Confocal (without technique) Figure 4.3-2a © 2016 Pearson Education, Inc.
Fluorescence 10 mm Figure 4.3-2b © 2016 Pearson Education, Inc.
50 mm Confocal (without technique) Figure 4.3-2c © 2016 Pearson Education, Inc.
Confocal (with technique) 50 mm Figure 4.3-3 © 2016 Pearson Education, Inc.
Longitudinal section of cilium Cross section of cilium Cilia Scanning electron microscopy (SEM) 2 mm 2 mm Transmission electron microscopy (TEM) Electron Microscopy (EM) Figure 4.3-3a © 2016 Pearson Education, Inc.
Cilia Scanning electron microscopy (SEM) 2 mm Figure 4.3-3b © 2016 Pearson Education, Inc.
Longitudinal section of cilium Cross section of cilium 2 mm Transmission electron microscopy (TEM) Recent advances in light microscopy Labeling molecules or structures with fluorescent markers improves visualization of details Confocal and other types of microscopy have sharpened images of tissues and cells New techniques and labeling have improved resolution so that structures as small as 10–20 m can be distinguished © 2016 Pearson Education, Inc.
Cell Fractionation Cell fractionation breaks up cells and separates the components, using centrifugation Cell components separate based on their relative size Cell fractionation enables scientists to determine the functions of organelles Biochemistry and cytology help correlate cell function with structure © 2016 Pearson Education, Inc.
Concept 4.2: Eukaryotic cells have internal membranes that compartmentalize their functions The basic structural and functional unit of every organism is one of two types of cells: prokaryotic or eukaryotic Organisms of the domains Bacteria and Archaea consist of prokaryotic cells Protists, fungi, animals, and plants all consist of eukaryotic cells © 2016 Pearson Education, Inc.
Comparing Prokaryotic and Eukaryotic Cells Basic features of all cells Plasma membrane Semifluid substance called cytosol Chromosomes (carry genes) Ribosomes (make proteins) © 2016 Pearson Education, Inc.
In a eukaryotic cell most of the DNA is in the nucleus, an organelle that is bounded by a double membrane Prokaryotic cells are characterized by having No nucleus DNA in an unbound region called the nucleoid No membrane-bound organelles Both types of cells contain cytoplasm bound by the plasma membrane © 2016 Pearson Education, Inc.
Figure 4.4 © 2016 Pearson Education, Inc.
Fimbriae Nucleoid Ribosomes Plasma membrane Cell wall Capsule Flagella Bacterial chromosome A typical rod-shaped bacterium (a) A thin section through the bacterium Corynebacterium diphtheriae (colorized TEM) (b) 0.5 mm Figure 4.4-1 © 2016 Pearson Education, Inc.
Nucleoid Ribosomes Plasma membrane Cell wall A thin section through the bacterium Corynebacterium diphtheria (colorized TEM) (b) 0.5 mm Eukaryotic cells are generally much larger than prokaryotic cells Typical bacteria are 1–5 m in diameter Eukaryotic cells are typically 10–100 m in diameter © 2016 Pearson Education, Inc.
The plasma membrane is a selective barrier that allows sufficient passage of oxygen, nutrients, and waste to service the volume of every cell The general structure of a biological membrane is a double layer of phospholipids © 2016 Pearson Education, Inc.
Figure 4.5 © 2016 Pearson Education, Inc.
Outside of cell (a) TEM of a plasma membrane Inside of cell 0.1 mm Carbohydrate side chains Hydrophilic region Hydrophobic region Hydrophilic region Phospholipid Proteins (b) Structure of the plasma membrane Figure 4.5-1 © 2016 Pearson Education, Inc.
Outside of cell Inside of cell 0.1 mm (a) TEM of a plasma membrane Metabolic requirements set upper limits on the size of cells The ratio of surface area to volume of a cell is critical As the surface area increases by a factor of n2, the volume increases by a factor of n3 Small cells have a greater surface area relative to volume © 2016 Pearson Education, Inc.
Figure 4.6 © 2016 Pearson Education, Inc.
Surface area increases while total volume remains constant 1 1 5 6 150 750 Total surface area [sum of the surface areas (height ✕width) of all box sides ✕number of boxes] Total volume [height ✕width ✕length ✕number of boxes] Surface-to-volume (S-to-V) ratio [surface area ÷ volume] 1 125 125 6 1.2 6 A Panoramic View of the Eukaryotic Cell A eukaryotic cell has internal membranes that divide the cell into compartments—organelles The plasma membrane and organelle membranes participate directly in the cell’s metabolism © 2016 Pearson Education, Inc.
Animation: Tour of an Animal Cell © 2016 Pearson Education, Inc.
Animation: Tour of a Plant Cell © 2016 Pearson Education, Inc.
Figure 4.7-1 © 2016 Pearson Education, Inc.
ENDOPLASMIC RETICULUM (ER) Smooth ER Flagellum Centrosome CYTOSKELETON: Microfilaments Intermediate filaments Microtubules Microvilli Peroxisome Mitochondrion Lysosome Golgi apparatus Ribosomes (small brown dots) Plasma membrane NUCLEUS Nuclear envelope Nucleolus Chromatin Rough ER Figure 4.7-2 © 2016 Pearson Education, Inc.
Nuclear envelope Nucleolus Chromatin NUCLEUS Golgi apparatus Rough endoplasmic reticulum Smooth endoplasmic reticulum Central vacuole Microfilaments Microtubules CYTO-SKELETON Mitochondrion Peroxisome Plasma membrane Cell wall Wall of adjacent cell Plasmodesmata Chloroplast Ribosomes (small brown dots) Figure 4.7-3 © 2016 Pearson Education, Inc.
Cell Nucleus Nucleolus 10 mm Human cells from lining of uterus (colorized TEM) Figure 4.7-4 © 2016 Pearson Education, Inc.
5 mm Parent cell Buds Yeast cells budding (colorized SEM) Figure 4.7-5 © 2016 Pearson Education, Inc.
1 mm Cell wall Vacuole Nucleus Mitochondrion A single yeast cell (colorized TEM) Figure 4.7-6 © 2016 Pearson Education, Inc.
Cell Cell wall Chloroplast Mitochondrion Nucleus Nucleolus Cells from duckweed (colorized TEM) 5 mm Figure 4.7-7 © 2016 Pearson Education, Inc.
Chlamydomonas (colorized SEM) 8 mm Figure 4.7-8 © 2016 Pearson Education, Inc.
Chlamydomonas (colorized TEM) 1 mm Flagella Nucleus Nucleolus Vacuole Chloroplast Cell wall Concept 4.3: The eukaryotic cell’s genetic instructions are housed in the nucleus and carried out by the ribosomes The nucleus contains most of the DNA in a eukaryotic cell Ribosomes use the information from the DNA to make proteins © 2016 Pearson Education, Inc.
The Nucleus: Information Central The nucleus contains most of the cell’s genes and is usually the most conspicuous organelle The nuclear envelope encloses the nucleus, separating it from the cytoplasm The nuclear membrane is a double membrane; each membrane consists of a lipid bilayer © 2016 Pearson Education, Inc.
Nuclear pores regulate the entry and exit of molecules The shape of the nucleus is maintained by the nuclear lamina, which is composed of protein © 2016 Pearson Education, Inc.
Figure 4.8 © 2016 Pearson Education, Inc.
Nucleus Nucleolus Chromatin Nuclear envelope: Inner membrane Outer membrane Nuclear pore 1 mm Surface of nuclear envelope (TEM) Pore complex Ribosome Close-up of nuclear envelope 0.25 mm Pore complexes (TEM) 0.5 mm Nuclear lamina (TEM) Chromatin Rough ER Nucleus Figure 4.8-1 © 2016 Pearson Education, Inc.
Nucleolus Chromatin Nuclear envelope: Inner membrane Outer membrane Nuclear pore Pore complex Close-up of nuclear envelope Chromatin Rough ER Nucleus Ribosome Figure 4.8-2 © 2016 Pearson Education, Inc.
Nuclear envelope: Inner membrane Outer membrane Nuclear pore 1 mm Surface of nuclear envelope (TEM) Figure 4.8-3 © 2016 Pearson Education, Inc.
0.25 mm Pore complexes (TEM) Figure 4.8-4 © 2016 Pearson Education, Inc.
0.5 mm Nuclear lamina (TEM) In the nucleus, DNA is organized into discrete units called chromosomes Each chromosome is one long DNA molecule associated with proteins The DNA and proteins of chromosomes together are called chromatin Chromatin condenses to form discrete chromosomes as a cell prepares to divide The nucleolus is located within the nucleus and is the site of ribosomal RNA (rRNA) synthesis © 2016 Pearson Education, Inc.
Ribosomes: Protein Factories Ribosomes are complexes of ribosomal RNA and protein Ribosomes carry out protein synthesis in two locations In the cytosol (free ribosomes) On the outside of the endoplasmic reticulum or the nuclear envelope (bound ribosomes) © 2016 Pearson Education, Inc.
Figure 4.9 © 2016 Pearson Education, Inc.
Ribosomes 0.25 mm ER Free ribosomes in cytosol Endoplasmic reticulum (ER) Ribosomes bound to ER Large subunit Small subunit TEM showing ER and ribosomes Diagram of a ribosome Computer model of a ribosome Figure 4.9-1 © 2016 Pearson Education, Inc.
0.25 mm Free ribosomes in cytosol Endoplasmic reticulum (ER) Large subunit Small subunit TEM showing ER and ribosomes Diagram of a ribosome Ribosomes bound to ER Figure 4.9-1a © 2016 Pearson Education, Inc.
0.25 mm Free ribosomes in cytosol Endoplasmic reticulum (ER) TEM showing ER and ribosomes Ribosomes bound to ER Figure 4.9-2 © 2016 Pearson Education, Inc.
Large subunit Small subunit Computer model of a ribosome Concept 4.4: The endomembrane system regulates protein traffic and performs metabolic functions in the cell Components of the endomembrane system Nuclear envelope Endoplasmic reticulum Golgi apparatus Lysosomes Vacuoles Plasma membrane These components are either continuous or connected through transfer by vesicles © 2016 Pearson Education, Inc.
The Endoplasmic Reticulum: Biosynthetic Factory The endoplasmic reticulum (ER) accounts for more than half of the total membrane in many eukaryotic cells The ER membrane is continuous with the nuclear envelope There are two distinct regions of ER Smooth ER: lacks ribosomes Rough ER: surface is studded with ribosomes © 2016 Pearson Education, Inc.
Video: Endoplasmic Reticulum © 2016 Pearson Education, Inc.
Video: ER and Mitochondria © 2016 Pearson Education, Inc.
Figure 4.10 © 2016 Pearson Education, Inc.
Smooth ER Rough ER ER lumen Cisternae Ribosomes Transport vesicle Transitional ER Nuclear envelope Smooth ER Rough ER 0.2 mm Figure 4.10-1 © 2016 Pearson Education, Inc.
Smooth ER Rough ER 0.2 mm Functions of Smooth ER The smooth ER Synthesizes lipids Metabolizes carbohydrates Detoxifies drugs and poisons Stores calcium ions © 2016 Pearson Education, Inc.
Functions of Rough ER The rough ER Has bound ribosomes, which secrete glycoproteins (proteins covalently bonded to carbohydrates) Distributes transport vesicles, proteins surrounded by membranes Is a membrane factory for the cell © 2016 Pearson Education, Inc.
The Golgi Apparatus: Shipping and Receiving Center The Golgi apparatus consists of flattened membranous sacs called cisternae Functions of the Golgi apparatus Modifies products of the ER Manufactures certain macromolecules Sorts and packages materials into transport vesicles © 2016 Pearson Education, Inc.
Video: ER to Golgi Traffic © 2016 Pearson Education, Inc.
Video: Golgi 3-D © 2016 Pearson Education, Inc.
Video: Golgi Secretion © 2016 Pearson Education, Inc.
Figure 4.11 © 2016 Pearson Education, Inc.
Golgi apparatus cis face (“receiving” side of Golgi apparatus) Cisternae trans face (“shipping” side of Golgi apparatus) TEM of Golgi apparatus 0.1 mm Figure 4.11-1 © 2016 Pearson Education, Inc.
TEM of Golgi apparatus 0.1 mm Lysosomes: Digestive Compartments A lysosome is a membranous sac of hydrolytic enzymes that can digest macromolecules Lysosomal enzymes work best in the acidic environment inside the lysosome The three-dimensional shape of lysosomal proteins protects them from digestion by lysosomal enzymes © 2016 Pearson Education, Inc.
Some types of cell can engulf another cell by phagocytosis; this forms a food vacuole A lysosome fuses with the food vacuole and digests the molecules Lysosomes also use enzymes to recycle the cell’s own organelles and macromolecules, a process called autophagy © 2016 Pearson Education, Inc.
Animation: Lysosome Formation © 2016 Pearson Education, Inc.
Video: Phagocytosis © 2016 Pearson Education, Inc.
Video: Paramecium Vacuole © 2016 Pearson Education, Inc.
Figure 4.12 © 2016 Pearson Education, Inc.
Nucleus 1 mm Lysosome Digestive enzymes Lysosome Plasma membrane Digestion Food vacuole Figure 4.12-1 © 2016 Pearson Education, Inc.
Nucleus 1 mm Figure 4.13 © 2016 Pearson Education, Inc.
Vesicle containing two damaged organelles 1 mm Mitochondrion fragment Peroxisome fragment Lysosome Peroxisome Vesicle Mitochondrion Digestion Figure 4.13-1 © 2016 Pearson Education, Inc.
Vesicle containing two damaged organelles 1 mm Mitochondrion fragment Peroxisome fragment Vacuoles: Diverse Maintenance Compartments Vacuoles are large vesicles derived from the endoplasmic reticulum and Golgi apparatus The solution inside a vacuole differs in composition from the cytosol © 2016 Pearson Education, Inc.
Food vacuoles are formed by phagocytosis Contractile vacuoles, found in many freshwater protists, pump excess water out of cells Central vacuoles, found in many mature plant cells, hold organic compounds and water Certain vacuoles in plants and fungi carry out enzymatic hydrolysis like lysosomes © 2016 Pearson Education, Inc.
Figure 4.14 © 2016 Pearson Education, Inc.
Central vacuole Cytosol Nucleus Cell wall Chloroplast Central vacuole 5 mm Figure 4.14-1 © 2016 Pearson Education, Inc.
Cytosol Nucleus Cell wall Chloroplast Central vacuole 5 mm The Endomembrane System: A Review The endomembrane system is a complex and dynamic player in the cell’s compartmental organization © 2016 Pearson Education, Inc.
Figure 4.15 © 2016 Pearson Education, Inc.
Smooth ER cis Golgi trans Golgi Rough ER Nucleus Plasma membrane Concept 4.5: Mitochondria and chloroplasts change energy from one form to another Mitochondria are the sites of cellular respiration, a metabolic process that uses oxygen to generate ATP Chloroplasts, found in plants and algae, are the sites of photosynthesis Peroxisomes are oxidative organelles © 2016 Pearson Education, Inc.
The Evolutionary Origins of Mitochondria and Chloroplasts Mitochondria and chloroplasts display similarities with bacteria Enveloped by a double membrane Contain ribosomes and multiple circular DNA molecules Grow and reproduce somewhat independently in cells © 2016 Pearson Education, Inc.
The endosymbiont theory An early ancestor of eukaryotic cells engulfed a nonphotosynthetic prokaryotic cell, which formed an endosymbiont relationship with its host The host cell and endosymbiont merged into a single organism, a eukaryotic cell with a mitochondrion At least one of these cells may have taken up a photosynthetic prokaryote, becoming the ancestor of cells that contain chloroplasts © 2016 Pearson Education, Inc.
Video: ER and Mitochondria © 2016 Pearson Education, Inc.
Video: Mitochondria 3-D © 2016 Pearson Education, Inc.
Figure 4.16 © 2016 Pearson Education, Inc.
Endoplasmic reticulum Nuclear envelope Ancestor of eukaryotic cells (host cell) Engulfing of photosynthetic prokaryote Mitochondrion Chloroplast At least one cell Mitochondrion Nonphotosynthetic eukaryote Engulfing of oxygen-using nonphotosynthetic prokaryote, which, becomes a mitochondrion Nucleus Photosynthetic eukaryote Mitochondria: Chemical Energy Conversion Mitochondria are in nearly all eukaryotic cells They have a smooth outer membrane and an inner membrane folded into cristae The inner membrane creates two compartments: intermembrane space and mitochondrial matrix Some metabolic steps of cellular respiration are catalyzed in the mitochondrial matrix Cristae present a large surface area for enzymes that synthesize ATP © 2016 Pearson Education, Inc.
Figure 4.17 © 2016 Pearson Education, Inc.
Mitochondrion Intermembrane space Outer membrane DNA Inner membrane Cristae Matrix Free ribosomes in the mitochondrial matrix (a) Diagram and TEM of mitochondrion 0.1 mm (b) Network of mitochondria in Euglena (LM) Mitochondrial DNA 10 mm Mitochondria Nuclear DNA Figure 4.17-1 © 2016 Pearson Education, Inc.
Intermembrane space Outer membrane DNA Inner membrane Cristae Matrix Free ribosomes in the mitochondrial matrix (a) Diagram and TEM of mitochondrion 0.1 mm Figure 4.17-1a © 2016 Pearson Education, Inc.
Outer membrane Inner membrane Cristae Matrix 0.1 mm Figure 4.17-2 © 2016 Pearson Education, Inc.
(b) Network of mitochondria in Euglena (LM) Mitochondrial DNA 10 mm Mitochondria Nuclear DNA Chloroplasts: Capture of Light Energy Chloroplasts contain the green pigment chlorophyll, as well as enzymes and other molecules that function in photosynthesis Chloroplasts are found in leaves and other green organs of plants and in algae © 2016 Pearson Education, Inc.
Chloroplast structure includes Thylakoids, membranous sacs, stacked to form a granum Stroma, the internal fluid The chloroplast is one of a group of plant organelles called plastids © 2016 Pearson Education, Inc.
Figure 4.18 © 2016 Pearson Education, Inc.
Chloroplast Stroma Ribosomes Inner and outer membranes Granum DNA Thylakoid Intermembrane space (a) Diagram and TEM of chloroplast 1 mm (b) Chloroplasts in an algal cell Chloroplasts (red) 50 mm Figure 4.18-1 © 2016 Pearson Education, Inc.
Stroma Ribosomes Inner and outer membranes Granum DNA Thylakoid Intermembrane space (a) Diagram and TEM of chloroplast 1 mm Figure 4.18-1a © 2016 Pearson Education, Inc.
Stroma Inner and outer membranes Granum 1 mm Figure 4.18-2 © 2016 Pearson Education, Inc.
(b) Chloroplasts in an algal cell Chloroplasts (red) 50 mm Peroxisomes: Oxidation Peroxisomes are specialized metabolic compartments bounded by a single membrane Peroxisomes produce hydrogen peroxide and then convert it to water Peroxisomes perform reactions with many different functions © 2016 Pearson Education, Inc.
Figure 4.19 © 2016 Pearson Education, Inc.
Peroxisome Mitochon-drion Chloroplasts 1 mm Concept 4.6: The cytoskeleton is a network of fibers that organizes structures and activities in the cell The cytoskeleton is a network of fibers extending throughout the cytoplasm It organizes the cell’s structures and activities © 2016 Pearson Education, Inc.
Video: Cytoskeleton in Neuron © 2016 Pearson Education, Inc.
Video: Microtubule Transport © 2016 Pearson Education, Inc.
Video: Organelle Movement © 2016 Pearson Education, Inc.
Video: Organelle Transport © 2016 Pearson Education, Inc.
Figure 4.20 © 2016 Pearson Education, Inc.
10 mm Roles of the Cytoskeleton: Support and Motility The cytoskeleton helps to support the cell and maintain its shape It provides anchorage for many organelles and molecules It interacts with motor proteins to produce motility Inside the cell, vesicles and other organelles can “walk” along the tracks provided by the cytoskeleton © 2016 Pearson Education, Inc.
Figure 4.21 © 2016 Pearson Education, Inc.
(a) Motor proteins “walkˮ vesicles along cytoskeletal fibers.
(b) Two vesicles move toward the tip of a nerve cell extension called an axon (SEM) Microtubule Vesicles 0.25 mm Microtubule of cytoskeleton Motor protein (ATP powered) Receptor for motor protein Vesicle ATP Figure 4.21-1 © 2016 Pearson Education, Inc.
(b) Two vesicles move toward the tip of a nerve cell extension called an axon (SEM) Microtubule Vesicles 0.25 mm Components of the Cytoskeleton Three main types of fibers make up the cytoskeleton Microtubules are the thickest of the three components of the cytoskeleton Microfilaments, also called actin filaments, are the thinnest components Intermediate filaments are fibers with diameters in a middle range © 2016 Pearson Education, Inc.
Video: Actin Cytoskeleton © 2016 Pearson Education, Inc.
Video: Actin in Crawling Cell © 2016 Pearson Education, Inc.
Video: Actin in Neuron © 2016 Pearson Education, Inc.
Video: Chloroplast Movement © 2016 Pearson Education, Inc.
Video: Cytoplasmic Stream © 2016 Pearson Education, Inc.
Video: Microtubule Movement © 2016 Pearson Education, Inc.
Video: Microtubules © 2016 Pearson Education, Inc.
Figure 4.T01 © 2016 Pearson Education, Inc.
Figure 4.T01-1 © 2016 Pearson Education, Inc.
Microtubules (Tubulin Polymers) Hollow tubes 25 nm with 15-nm lumen Tubulin, a dimer consisting of a-tubulin and b-tubulin Maintenance of cell shape; cell mo-tility; chromosome movements in cell division; organelle movements 10 mm Column of tubulin dimers Tubulin dimer a b 25 nm Figure 4.T01-1a © 2016 Pearson Education, Inc.
10 mm 10 mm Figure 4.T01-2 © 2016 Pearson Education, Inc.
Microfilaments (Actin Filaments) Two intertwined strands of actin 7 nm Actin Maintenance of cell shape; changes in cell shape; muscle contraction; cy-toplasmic streaming (plant cells); cell motility; cell division (animal cells) Actin subunit 7 nm 10 mm Figure 4.T01-2a © 2016 Pearson Education, Inc.
10 mm 10 mm Figure 4.T01-3 © 2016 Pearson Education, Inc.
Intermediate Filaments Fibrous proteins coiled into cables 8–12 nm One of several different proteins (such as keratins) Maintenance of cell shape; anchor-age of nucleus and certain other organelles; formation of nuclear lamina Keratin proteins Fibrous subunit (keratins coiled together) 8–12 nm 5 mm Figure 4.T01-3a © 2016 Pearson Education, Inc.
5 mm 5 mm Microtubules Microtubules are hollow rods constructed from globular protein dimers called tubulin Functions of microtubules Shape and support the cell Guide movement of organelles Separate chromosomes during cell division © 2016 Pearson Education, Inc.
Centrosomes and Centrioles In animal cells, microtubules grow out from a centrosome near the nucleus The centrosome is a “microtubule-organizing center” The centrosome has a pair of centrioles, each with nine triplets of microtubules arranged in a ring © 2016 Pearson Education, Inc.
Video: Chlamydomonas © 2016 Pearson Education, Inc.
Animation: Cilia Flagella © 2016 Pearson Education, Inc.
Video: Ciliary Motion © 2016 Pearson Education, Inc.
Video: Flagella in Sperm © 2016 Pearson Education, Inc.
Video: Flagellum Microtubule © 2016 Pearson Education, Inc.
Video: Sperm Flagellum © 2016 Pearson Education, Inc.
Video: Paramecium Cilia © 2016 Pearson Education, Inc.
Figure 4.22 © 2016 Pearson Education, Inc.
Centrosome Centrioles Microtubule Cilia and Flagella Microtubules control the beating of cilia and flagella, microtubule-containing extensions projecting from some cells Flagella are limited to one or a few per cell, while cilia occur in large numbers on cell surfaces Cilia and flagella also differ in their beating patterns © 2016 Pearson Education, Inc.
Cilia and flagella share a common structure A group of microtubules sheathed by the plasma membrane A basal body that anchors the cilium or flagellum A motor protein called dynein, which drives the bending movements of a cilium or flagellum © 2016 Pearson Education, Inc.
Figure 4.23 © 2016 Pearson Education, Inc.
0.1 mm Outer microtubule doublet Motor proteins (dyneins) Central microtubule Radial spoke Cross-linking protein between outer doublets (b) Cross section of motile cilium (c) Cross section of basal body Microtubules Plasma membrane Basal body 0.5 mm (a) Longitudinal section of motile cilium 0.1 mm Triplet Plasma membrane Figure 4.23-1 © 2016 Pearson Education, Inc.
Microtubules Plasma membrane 0.5 mm (a) Longitudinal section of motile cilium Basal body Figure 4.23-2 © 2016 Pearson Education, Inc.
0.1 mm Outer microtubule doublet Motor proteins (dyneins) Central microtubule Radial spoke Cross-linking protein between outer doublets (b) Cross section of motile cilium Plasma membrane Figure 4.23-2a © 2016 Pearson Education, Inc.
0.1 mm Outer microtubule doublet Motor proteins (dyneins) Central microtubule Radial spoke Cross-linking protein between outer doublets (b) Cross section of motile cilium Figure 4.23-3 © 2016 Pearson Education, Inc.
(c) Cross section of basal body 0.1 mm Triplet Figure 4.23-3a © 2016 Pearson Education, Inc.
(c) Cross section of basal body 0.1 mm Triplet How dynein “walking” moves flagella and cilia Dynein arms alternately contact, move, and release the outer microtubules The outer doublets and central microtubules are held together by flexible cross-linking proteins Movements of the doublet arms cause the cilium or flagellum to bend © 2016 Pearson Education, Inc.
Microfilaments (Actin Filaments) Microfilaments are thin solid rods, built from molecules of globular actin subunits The structural role of microfilaments is to bear tension, resisting pulling forces within the cell Bundles of microfilaments make up the core of microvilli of intestinal cells © 2016 Pearson Education, Inc.
Figure 4.24 © 2016 Pearson Education, Inc.
Microfilaments (actin filaments) 0.25 mm Plasma membrane Microvillus Microfilaments that function in cellular motility interact with the motor protein myosin For example, actin and myosin interact to cause muscle contraction, amoeboid movement of white blood cells, and cytoplasmic streaming in plant cells © 2016 Pearson Education, Inc.
Intermediate Filaments Intermediate filaments are larger than microfilaments but smaller than microtubules Intermediate filaments are only found in the cells of some animals, including vertebrates They support cell shape and fix organelles in place Intermediate filaments are more permanent cytoskeleton elements than the other two classes © 2016 Pearson Education, Inc.
Concept 4.7: Extracellular components and connections between cells help coordinate cellular activities Most cells synthesize and secrete materials that are external to the plasma membrane These extracellular materials are involved in many cellular functions © 2016 Pearson Education, Inc.
Cell Walls of Plants The cell wall is an extracellular structure that distinguishes plant cells from animal cells The cell wall protects the plant cell, maintains its shape, and prevents excessive uptake of water Plant cell walls are made of cellulose fibers embedded in other polysaccharides and protein © 2016 Pearson Education, Inc.
Plant cell walls may have multiple layers Primary cell wall: relatively thin and flexible Middle lamella: thin layer between primary walls of adjacent cells Secondary cell wall (in some cells): added between the plasma membrane and the primary cell wall Plasmodesmata are channels between adjacent plant cells © 2016 Pearson Education, Inc.
Video: Collagen Model © 2016 Pearson Education, Inc.
Video: Extracellular Matrix © 2016 Pearson Education, Inc.
Video: Fibronectin © 2016 Pearson Education, Inc.
Figure 4.25 © 2016 Pearson Education, Inc.
Central vacuole Cytosol Plasma membrane Plant cell walls Plasmodesmata Secondary cell wall Primary cell wall Middle lamella 1 mm Figure 4.25-1 © 2016 Pearson Education, Inc.
Secondary cell wall Primary cell wall Middle lamella 1 mm The Extracellular Matrix (ECM) of Animal Cells Animal cells lack cell walls but are covered by an elaborate extracellular matrix (ECM) The ECM is made up of glycoproteins such as collagen, proteoglycans, and fibronectin ECM proteins bind to receptor proteins in the plasma membrane called integrins © 2016 Pearson Education, Inc.
Figure 4.26 © 2016 Pearson Education, Inc.
Collagen Fibronectin Plasma membrane Micro-filaments CYTOPLASM Integrins Proteoglycan molecule Core protein Carbo-hydrates Polysaccharide molecule A proteoglycan complex: EXTRACELLULAR FLUID Figure 4.26-1 © 2016 Pearson Education, Inc.
Collagen Fibronectin Plasma membrane Micro-filaments CYTOPLASM Integrins A proteoglycan complex: EXTRACELLULAR FLUID Figure 4.26-2 © 2016 Pearson Education, Inc.
Proteoglycan molecule Core protein Carbo-hydrates Polysaccharide molecule Cell Junctions Neighboring cells in an animal or plant often adhere, interact, and communicate through direct physical contact There are several types of intercellular junctions that facilitate this Plasmodesmata Tight junctions Desmosomes Gap junctions © 2016 Pearson Education, Inc.
Plasmodesmata in Plant Cells Plasmodesmata are channels that perforate plant cell walls Through plasmodesmata, water and small solutes (and sometimes proteins and RNA) can pass from cell to cell © 2016 Pearson Education, Inc.
Tight Junctions, Desmosomes, and Gap Junctions in Animal Cells Animal cells have three main types of cell junctions Tight junctions Desmosomes Gap junctions All are especially common in epithelial tissue © 2016 Pearson Education, Inc.
Animation: Desmosomes © 2016 Pearson Education, Inc.
Animation: Gap Junctions © 2016 Pearson Education, Inc.
Animation: Tight Junctions © 2016 Pearson Education, Inc.
Figure 4.27 © 2016 Pearson Education, Inc.
Tight junctions prevent fluid from moving across a layer of cells.
Tight junction TEM 0.5 mm TEM 1 mm Tight junction Intermediate filaments Desmosome Gap junction Ions or small molecules Plasma membranes of adjacent cells Space between cells Extracellular matrix 0.1 mm TEM Figure 4.27-1 © 2016 Pearson Education, Inc.
Tight junctions prevent fluid from moving across a layer of cells.
Tight junction Intermediate filaments Desmosome Gap junction Ions or small molecules Plasma membranes of adjacent cells Space between cells Extracellular matrix Figure 4.27-2 © 2016 Pearson Education, Inc.
Tight junction TEM 0.5 mm Figure 4.27-3 © 2016 Pearson Education, Inc.
TEM 1 mm Desmosome Figure 4.27-4 © 2016 Pearson Education, Inc.
Gap junction 0.1 mm TEM The Cell: A Living Unit Greater Than the Sum of Its Parts None of the components of a cell work alone For example, a macrophage’s ability to destroy bacteria involves the whole cell, coordinating components such as the cytoskeleton, lysosomes, and plasma membrane Cellular functions arise from cellular order © 2016 Pearson Education, Inc.
Figure 4.28 © 2016 Pearson Education, Inc.
10 mm Figure 4.UN01-1 © 2016 Pearson Education, Inc.
Budding cell Mature parent cell 1 mm Figure 4.UN01-2 © 2016 Pearson Education, Inc.
V = p r3 4 3 r d Figure 4.UN02 © 2016 Pearson Education, Inc.
5 mm Nucleus Figure 4.UN03 © 2016 Pearson Education, Inc.
Cell Component Structure Function Nucleus Ribosome (ER) Two subunits made of ribosomal RNA and proteins; can be free in cytosol or bound to ER Surrounded by nuclear envelope (double membrane) perforated by nuclear pores; nuclear envelope continuous with endoplasmic reticulum (ER) Houses chromosomes, which are made of chromatin (DNA and proteins); contains nucleoli, where ribosomal subunits are made; pores regulate entry and exit of materials Protein synthesis Figure 4.UN04 © 2016 Pearson Education, Inc.
Cell Component Structure Function Endoplasmic reticulum (Nuclear envelope) Extensive network of membrane-bounded tubules and sacs; membrane separates lumen from cytosol; continuous with nuclear envelope Rough ER: aids in synthesis of secretory and other proteins from bound ribosomes; adds carbohydrates to proteins to make glycoproteins; produces new membrane Golgi apparatus Stacks of flattened membranous sacs; has polarity (cis and trans faces) Modification of proteins, carbohydrates on proteins, and phospholipids; synthesis of many polysaccharides; sorting of Golgi products, which are then released in vesicles Lysosome Vacuole Membranous sac of hydrolytic enzymes (in animal cells) Large membrane-bounded vesicle Breakdown of ingested substances, cell macromolecules, and damaged organelles for recycling Digestion, storage, waste disposal, water balance, plant cell growth and protection Smooth ER: synthesis of lipids, metabolism of carbohydrates, Ca2+ storage, detoxification of drugs and poisons Figure 4.UN05 © 2016 Pearson Education, Inc.
Cell Component Structure Function Mitochondrion Chloroplast Peroxisome Bounded by double membrane; inner membrane has infoldings (cristae) Typically two membranes around fluid stroma, which contains thylakoids stacked into grana (in cells of photosynthetic eukaryotes, including plants) Specialized metabolic compartment bounded by a single membrane Cellular respiration Photosynthesis Contains enzymes that transfer hydrogen atoms from certain molecules to oxygen, producing hydrogen peroxide (H2O2) as a by-product; H2O2 is converted to water by another enzyme Figure 4.UN06 © 2016 Pearson Education, Inc.
Epithelial cell |
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Rounding calculator to round numbers up or down to any decimal place. Choose ones to round a number to the nearest dollar. Choose hundredths to round an amount to the nearest cent.
Rounding Numbers
Say you wanted to round the number 838.274. The final value depends on which place value you want to round to. Rounding 838.274:
Rounding to the nearest hundred is 800
Rounding to the nearest ten is 840
Rounding to the nearest one is 838
Rounding to the nearest tenth is 838.3
Rounding to the nearest hundredth is 838.27
Basic Rules of Rounding
When you "round to the nearest _____" regardless of what goes in the blank the steps are nearly always the same:
Identify which place value you are rounding to. The smaller the place value, the more accurate the final result.
Look to the next smallest place value, the digit to the right of the place value you're rounding to. For example, if you want to round to the nearest ten look at the ones place.
If the digit in the next smallest place value is less than five (0, 1, 2, 3, or 4), leave the digit you want to round to as-is. Any digits after that number become zeros, or drop-off if they're located after the decimal point. This is called rounding down.
If the next smallest place value is greater than or equal to five (5, 6, 7, 8, or 9), increase the value of the digit you're rounding to by one (+1). Just like before, any remaining digits before the decimal point become zeros, and any after the decimal point drop off. This is called rounding up.
How this Calculator Rounds Numbers
This Rounding Numbers Calculator uses the rounding method called Round Half Away From Zero. Visit the CalculatorSoup
Rounding Methods Calculator for an explanation of this and other popular rounding methods.
Rounding Half Away From Zero
Round halfway values away from zero. When the digit to the right is 5 it's at the halfway point and is important in deciding whether to round up or down.
For the following examples let's round to the ones place for values between 2 and 3 where the halfway point is 2.5. In the Rounding Half Away from Zero method, whether you round a positive or a negative number, the absolute value of the result is the same because you're rounding
away from 0.
Rounding Positive Numbers
Numbers less than the halfway point between 2 and 3, which is 2.5, round down toward 0. Numbers greater than or equal to the halfway point between 2 and 3 round up away from 0.
2.4 rounds down to 2
2.48 rounds down to 2
2.5 rounds up to 3 halfway, up and away from 0
2.52 rounds up to 3
2.6 rounds up to 3
Rounding Negative Numbers:
Numbers greater than the halfway point between -3 and -2, which is -2.5, round up toward 0. Numbers less than or equal to the halfway point of -2.5 round down away from 0.
-2.4 rounds up to -2
-2.48 rounds up to -2
-2.5 rounds down to -3 halfway, down and away from 0
-2.52 rounds down to -3
-2.6 rounds down to -3
Round to the Nearest Hundred: 3250
Identify the hundreds digit: the 2 in 3250
Identify the next smallest place value: the 5 in 3250
Is that digit greater than or equal to five? Yes, so round up.
Increase the hundreds digit by one, so 2 becomes 3. Every digit after becomes a zero.
3250 rounded to the nearest hundred is 3300
Round to the Nearest Ten: 323.5
Identify the tens digit: the 2 in 323.5
Identify the next smallest place value: the 3 in 323.5
Is that digit greater than or equal to five? No, so round down.
The tens digit stays the same at 2. Every digit after becomes a zero. Digits after the decimal point are dropped.
323.5 rounded to the nearest ten is 320
Round to the Nearest Ten: 499
Identify the tens digit: the first 9 in 499
Identify the next smallest place value: the second 9 in 499
Is that digit greater than or equal to five? Yes, so round up.
The tens digit increases by one. Since 9+1=10, you need to carry the 1 and add it to the digit in the hundreds place. Each digit after becomes a zero.
499 rounded to the nearest ten is 500
Round to the Nearest Tenth: 0.74
Identify the tenths digit: 7 in 0.74
Identify the next smallest place value: the 4 in 0.74
Is that digit greater than or equal to five? No, so round down.
The tenths digit stays the same at 7. Since the remaining digits are after the decimal point you just drop them.
0.74 rounded to the nearest tenth is 0.7
Round to the Nearest Hundredth: 3.141
Identify the hundredths digit: the 4 in 3.141
Identify the next smallest place value: the last 1 in 3.141
Is that digit greater than or equal to five? No, so round down.
The hundredths digit stays the same at 4. Drop the digits to the right of 4.
3.141 rounded to the nearest hundredth is 3.14
Related Calculators
To learn how to round a number to the nearest multiple see our
Round to Nearest Multiple Calculator. Round to multiples of whole numbers or decimals such as tenths, hundredths, thousandths, etc.
Further Reading
About.com on
How to Round Numbers
Math.com has
lessons on estimating and rounding.
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Inguinal canal
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From Wikipedia, the free encyclopedia
Human abdominal anatomy
| Inguinal canal | |
--- |
| Front of abdomen, showing surface markings for arteries and inguinal canal. (Inguinal canal is tube at lower left.) | |
| The scrotum. On the left side (image right side), the cavity of the tunica vaginalis has been opened; on the right side (image left side), only the layers superficial to the cremaster have been removed. (Right inguinal canal visible at upper left.) | |
| Details | |
| Identifiers | |
| Latin | canalis inguinalis |
| MeSH | D007264 |
| TA98 | A04.5.01.026 |
| TA2 | 2381 |
| FMA | 19928 |
| Anatomical terminology [edit on Wikidata] | |
The inguinal canal is a passage in the anterior abdominal wall on each side of the body (one on each side of the midline), which in males, convey the spermatic cords and in females, the round ligament of the uterus. The inguinal canals are larger and more prominent in males.
Structure
[edit]
The inguinal canals are situated just above the medial half of the inguinal ligament. The canals are approximately 4 to 6 cm long, angled anteroinferiorly and medially. In males, its diameter is normally 2 cm (±1 cm in standard deviation) at the deep inguinal ring.[notes 1]
A first-order approximation is to visualize each canal as a cylinder.
Walls
[edit]
To help define the boundaries, these canals are often further approximated as boxes with six sides. Not including the two rings, the remaining four sides are usually called the "anterior wall", "inferior wall ("floor")", "superior wall ("roof")", and "posterior wall". These consist of the following:
| | |
--- |
| | superior wall (roof): Medial crus of aponeurosis of external oblique Musculoaponeurotic arches of internal oblique and transverse abdominal Transversalis fascia conjoint tendon |
| anterior wall: aponeurosis of external oblique fleshy part of internal oblique (lateral third of canal only) superficial inguinal ring (medial third of canal only) | (inguinal canal) | posterior wall: transversalis fascia conjoint tendon (Inguinal falx, reflected part of inguinal ligament, medial third of canal only) deep inguinal ring (lateral third of canal only) |
| | inferior wall (floor): inguinal ligament lacunar ligament (medial third of canal only) iliopubic tract (lateral third of canal only) | |
Deep inguinal ring
[edit]
The deep inguinal ring (internal or deep abdominal ring, abdominal inguinal ring, internal inguinal ring, annulus abdominalis) is the entrance to the inguinal canal.
Location
[edit]
The surface marking of the deep inguinal ring is classically described as half an inch above the midpoint of the inguinal ligament.
However, the surface anatomy of the point is disputed. In a recent study, it was found to be in a region between the mid-inguinal point (situated midway between the anterior superior iliac spine and the pubic symphysis) and the midpoint of the inguinal ligament (i.e. midway between the anterior superior iliac spine and the pubic tubercle). Traditionally, either one of these two sites was claimed as its location. However, this claim is based upon the study's dissection of 52 cadavers, and may not reflect the live in vivo anatomy.
Some sources state that it is at the layer of the transversalis fascia.
Description
[edit]
The deep inguinal ring is an opening in the transversalis fascia. It is of an oval form, the long axis of the oval being vertical; it varies in size in different subjects, and is much larger in the male than in the female. It is bounded, above and laterally, by the arched lower margin of the transversalis fascia; below and medially, by the inferior epigastric vessels. It transmits the spermatic cord in the male and the round ligament of the uterus in the female.
From its circumference, a thin funnel-shaped membrane, the infundibuliform fascia, is continued around the cord and testis, enclosing them in a distinct covering.
Superficial inguinal ring
[edit]
The superficial inguinal ring (subcutaneous inguinal ring or external inguinal ring) is an anatomical structure in the anterior wall of the mammalian abdomen. It is a triangular opening that forms the exit of the inguinal canal, which houses the ilioinguinal nerve, the genital branch of the genitofemoral nerve, and the spermatic cord (in men) or the round ligament (in women). At the other end of the canal, the deep inguinal ring forms the entrance.
It is found within the aponeurosis of the external oblique, immediately above the pubic crest, 1 centimeter above and superolateral to the pubic tubercle.
It has the following boundaries—medial crura by pubic crest, lateral crura by pubic tubercle and inferiorly by inguinal ligament.
Contents
[edit]
The structures which pass through the canals differ between males and females:
in males: the spermatic cord and its coverings, and the ilioinguinal nerve.
in females: the round ligament of the uterus, and the ilioinguinal nerve.
The classic description of the contents of the spermatic cords in the male are:
3 arteries: artery to vas deferens (or ductus deferens), testicular artery, cremasteric artery;
3 fascial layers: external spermatic, cremasteric, and internal spermatic fascia;
3 other structures: pampiniform plexus, vas deferens (ductus deferens), testicular lymphatics;
3 nerves: genital branch of the genitofemoral nerve (L1/2), sympathetic and visceral afferent fibres, ilioinguinal nerve (N.B. outside spermatic cord but travels next to it)
Note that the ilioinguinal nerve passes through the superficial ring to descend into the scrotum, but does not formally run through the canal.
Development
[edit]
In males
[edit]
During development, each testicle descends from the starting point on the posterior abdominal wall (para-aortically) from the labioscrotal swellings near the kidneys, down the abdomen, and through the inguinal canals to reach the scrotum. This way, each testicle descends through the abdominal wall into the scrotum behind[clarification needed] the processus vaginalis (which later obliterates).
Clinical significance
[edit]
See also: Inguinal hernia
Abdominal contents (potentially including intestine) can be abnormally displaced from the abdominal cavity. Where these contents exit through the inguinal canal, having passed through the deep inguinal ring, the condition is known as an indirect or oblique inguinal hernia. This can also cause infertility. This condition is far more common in males than in females, owing to the inguinal canal's small size in females.
A hernia that exits the abdominal cavity directly through the deep layers of the abdominal wall, thereby bypassing the inguinal canal, is known as a direct inguinal hernia.
In males with strong presentation of the cremasteric reflex, the testes can—during supine sexual activity or manual manipulation—partially or fully retract into the inguinal canal for a short period of time. In juveniles and adults with inguinal injury, retraction can be prolonged and potentially lead to overheating-related infertility.
The superficial ring is palpable under normal conditions. It becomes dilated in a condition called athletic pubalgia. Abdominal contents may protrude through the ring in inguinal hernia.
Thus lymphatic spread from a testicular tumour is to the para-aortic nodes first, and not the inguinal nodes.
Additional images
[edit]
Wikimedia Commons has media related to Inguinal canal.
The spermatic cord in the inguinal canal
Inguinal fossae
The abdominal inguinal ring
The relations of the femoral and abdominal inguinal rings, seen from within the abdomen. Right side.
Diagram of an indirect, scrotal inguinal hernia (median view from the left)
Superficial inguinal ring
Anterior abdominal wall. Intermediate dissection. Anterior view.
See also
[edit]
Superficial inguinal ring
Inguinal hernia
Tucking
Notes
[edit]
^ The diameter has been estimated to be ±2.2cm ±1.08cm in Africans, and 2.1 cm ±0.41cm in Europeans.
References
[edit]
^ Tuma, Faiz; Lopez, Richard A.; Varacallo, Matthew (2023). "Anatomy, Abdomen and Pelvis: Inguinal Region (Inguinal Canal)". StatPearls. StatPearls Publishing. PMID 29261933. Retrieved 18 June 2023.
^ Mitura, Kryspin; Kozieł, Sławomir; Pasierbek, Michał (2018). "Ethnicity-related differences in inguinal canal dimensions between African and Caucasian populations and their potential impact on the mesh size for open and laparoscopic groin hernia repair in low-resource countries in Africa". Videosurgery and Other Miniinvasive Techniques. 13 (1): 74–81. doi:10.5114/wiitm.2018.72579. ISSN 1895-4588. PMC 5890843. PMID 29643962.
^ "Gross Anatomy Image". Archived from the original on 2007-11-11. Retrieved 2007-11-20.
^ Adam Mitchell; Drake, Richard; Gray, Henry David; Wayne Vogl (2005). Gray's anatomy for students. Elsevier/Churchill Livingstone. p. 260. ISBN 0-443-06612-4.
^ Jump up to: a b Dalley, Arthur F.; Moore, Keith L. (2006). Clinically oriented anatomy. Hagerstown, MD: Lippincott Williams & Wilkins. pp. 217. ISBN 0-7817-3639-0.
^ Jump up to: a b c d Arthur F., II Dalley; Anne M. R. Agur (2005). Grant's Atlas of Anatomy. Hagerstown, MD: Lippincott Williams & Wilkins. p. 102. ISBN 0-7817-4255-2.
^ Susan Standring (2004). Gray's Anatomy: The Anatomical Basis of Medicine and Surgery. Churchill-Livingstone. p. 1098. ISBN 0-443-07168-3.
^ Koliyadan S, Narayan G, Balasekran P (2004). "Surface marking of the deep inguinal ring". Clin Anat. 17 (7): 554–7. doi:10.1002/ca.10257. PMID 15376291. S2CID 30726776.
^ Jump up to: a b Kyung Won, PhD. Chung (2005). Gross Anatomy (Board Review). Hagerstown, MD: Lippincott Williams & Wilkins. p. 198. ISBN 0-7817-5309-0.
^ Sinnatamby, Chummy S. (2011). Last's Anatomy (12th ed.). Elsevier Australia. pp. 226–227. ISBN 978-0-7295-3752-0.
^ James Harmon, M.D., Ph.D., Lecture 13. Human Gross Anatomy. University of Minnesota. September 4, 2008.
^ "Anatomy Tables - Inguinal Region". Archived from the original on 2007-11-21. Retrieved 2007-11-20.
^ Mayo Clinic Staff. "Retractile testicle". Mayo Clinic. Mayo Foundation for Medical Education and Research. Retrieved 10 February 2018.
^ Moore & Agur, Essential Clinical Anatomy (2007)
Adam Mitchell; Drake, Richard; Gray, Henry David; Wayne Vogl (2010). Gray's anatomy for students. Elsevier/Churchill Livingstone. pp. 286. ISBN 0-443-06612-4.
External links
[edit]
Anatomy photo:36:01-0102 at the SUNY Downstate Medical Center
Anatomy figure: 36:01-03 at Human Anatomy Online, SUNY Downstate Medical Center - "The inguinal canal and derivation of the layers of the spermatic cord."
Anatomy image:7362 at the SUNY Downstate Medical Center
Anatomy photo:35:09-0101 at the SUNY Downstate Medical Center - "Anterior Abdominal Wall: Borders of the Superficial Inguinal Ring"
Anatomy figure: 36:01-13 at Human Anatomy Online, SUNY Downstate Medical Center - "The inguinal canal and derivation of the layers of the spermatic cord."
Atlas image: abdo_wall63 at the University of Michigan Health System - "The Male & Female Inguinal Canal"
Diagram at nurseminerva.co.uk
inguinalregion at The Anatomy Lesson by Wesley Norman (Georgetown University)
Atlas image: abdo_wall65 at the University of Michigan Health System - "The Coverings of the Inguinal Canal, External & Internal Oblique & Transversus Abdominis Removed"
| Muscles and ligaments of abdomen and pelvis | |
--- |
| Abdominal wall | | | | | | | | | | | | --- --- --- --- --- | | Anterior/ lateral | | | | --- | | Muscle | Abdominal external oblique Transverse abdominal + Conjoint tendon Rectus sheath + Rectus abdominis + Pyramidalis + Arcuate line Tendinous intersection Cremaster Abdominal internal oblique | | Fascia | | | | --- | | Abdominal fascia + Panniculus adiposus + Fascia of Camper Membranous layer + Fascia of Scarpa Transverse fascia + Interfoveolar ligament Linea alba Linea semilunaris | | | Inguinal | Inguinal triangle Inguinal canal + Deep inguinal ring + Superficial inguinal ring + Intercrural fibers + Crura of superficial inguinal ring Inguinal ligament + Pectineal ligament + Lacunar ligament + Reflected ligament | | | | Posterior | | | | --- | | Muscle | Quadratus lumborum Iliopsoas + Psoas major + Psoas minor + Iliacus | | Fascia | Iliac fascia Iliopectineal arch | | |
| Pelvis | | | | --- | | Muscle | Levator ani + Iliococcygeus + Pubococcygeus + Puborectalis Coccygeus Rectococcygeus | | Fascia | Pelvic fascia Visceral + Rectovaginal fascia + Rectoprostatic fascia Parietal + Obturator fascia + Tendinous arch + Piriformis fascia Floor + Superior fascia + Pubovesical ligament + Puboprostatic ligament + Inferior fascia Anococcygeal body | |
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Mastering the Burette: A Complete Guide to Using Burets in Chemistry
Accuracy is necessary in any well-equipped chemical lab. The burette, sometimes spelt buret, is one of the most essential instruments for accurate liquid measurement. The burette, which is well-known for its high degree of accuracy, is necessary for titration and other analytical processes. Everything you need to know about burettes, including how to use them, how to effectively read them, and how to pick the ideal one for your lab requirements, will be covered in this blog.
What is a Burette?
A graded glass or plastic tube with a tap (stopcock) at one end is called a burette. Usually used in titration research where accurate measurements are essential, it is made to dispense accurate amounts of liquids. Depending on the type, burettes can measure quantities up to 50 millilitres with graduations as small as 0.1 or 0.01 millilitres.
Types of Burettes There are multiple types of burettes that are each intended for a certain use:
Glass Burettes: These are the most popular, and they provide good visibility and chemical resistance. For most titration work, they are perfect.
Plastic Burettes: These are more robust and lighter since they are made of materials like polypropylene, however they might not work with all chemicals.
Digital Burettes: These offer a digital screen and enable very accurate application. They are more costly, but they are simpler to use for repetitive operations and decrease human error.
Burette vs. Buret The same piece of equipment is referred to by the words burette and buret. In American English, "buret" is preferred, although "burette" is more frequently used in British English. The structure and function are the same regardless of spelling.
How to Read a Burette Correctly In chemistry, accurate readings are essential, and the burette is no different. The procedures to guarantee accurate readings are as follows:
Position Your Eye Level: Make sure your eye is level with the meniscus at all times to prevent a parallax mistake.
Read the Bottom of the Meniscus: It is best to read the liquid's curved surface (meniscus) at its lowest point.
Note the Initial Volume: As you start your titration, note the initial volume.
Dispense Carefully: To manage the flow and release the liquid, slowly open the stopcock.
Record Final Volume: Determine the amount of liquid used by recording the final volume after dispensing.
Common Mistakes When Using a Burette
Once the liquid has been dispensed, note the final volume to determine how much was used.
Not washing the burette with the specified solution.
Air bubbles are left in the burette's tip.
Using the wrong eye level, I misread the meniscus.
Using a dirty burette, which produces unreliable results
Step by Step: How to Use a Burette in Titration
Setup: Make sure the burette is securely clamped vertically on a stand.
Rinse: Before using the fluid that the burette will dispense, rinse it with purified water.
Fill the Burette: Over the zero mark, pour the titrant solution into the burette.
Remove Bubbles: To release some solution and get care of any air bubbles, open the stopcock.
Initial Reading: Take note of the liquid's initial position.
Perform Titration: Swirl the flask as you gradually add the titrant to the analyte solution.
Observe Endpoint: When the colour shift signifies that the reaction is finished, stop.
Final Reading: To determine the volume consumed, note the final volume and compute the difference.
Applications of Burettes in Chemistry The most typical application for burettes is titration, which is the process of gradually adding a solution with a known concentration (the titrant) to a solution with an unknown concentration until the chemical reaction is finished. Moreover, they are utilised in:
Acid-base titrations
Redox reactions
Complexometric titrations
Precipitation titrations
Tips for Choosing the Right Burette The accuracy and convenience of your tests may be affected by the burette you use. Think about the following:
Material: Consider chemical compatibility while choosing between glass and plastic.
Capacity: 25 mL and 50 mL are standard alternatives; select based on your usual volume needs.
Graduation Precision: More accurate readings are possible with finer graduations.
Valve Type: Because they operate smoothly and are resistant to chemical damage, PTFE stopcocks are recommended.
Digital Features:Digital burettes are perfect for labs that need to reduce mistakes or have a high throughput.
Why Choose ChemScience for Your Burette Needs? We have a large selection of burettes at ChemScience, ranging from basic glass versions to cutting edge digital burettes. Every product undergoes quality and accuracy testing to guarantee dependable performance in any laboratory environment. Our customer service representatives are available at all times to assist you in selecting the ideal burette for your requirements.
Conclusion
An essential piece of equipment in every chemistry lab, the burette offers the level of accuracy and control required for sensitive analytical work. Your experiments will be more dependable and successful if you know how to choose, read, and use a burette correctly. Discover our whole selection of burettes and burets at ChemScience INC. Right now, to outfit your lab with instruments that satisfy the strictest requirements for accuracy and robustness.
Frequently Asked Questions
1. What is a burette used for in chemistry?
A. In titration investigations, a burette is mostly used to precisely measure and dispense liquids. Chemists can use it to figure out exactly how much titrant is needed to finish a chemical reaction.
Why is a burette more accurate than a measuring cylinder?
A. Burettes are far more accurate for delivering precise quantities of liquid because they contain finer graduations and a stopcock for regulated dispensing.
3. What is the typical capacity of a burette?
A. Although there are specific sizes available, most burettes have a capacity of 25 or 50 millilitres.
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Before you can determine the state of your skin barrier and how to go about fixing it, you first need to know what your skin barrier is. And while the term itself sounds a little foreboding, it’s actually pretty straightforward.
“Your skin barrier is basically your outer layer of the skin,” Dr. Piliang explains. “It’s dead, not alive, and it’s made up of dead cells, lipids, proteins and fats that help protect your skin from the environment.”
The rest of your skin is a living organism. But your skin barrier’s role is to keep that living organism well-protected from things that could harm it. Think of it like armor, protecting all the delicate skin just beneath the surface. “It works to keep water in and to keep chemicals and infectious bacteria out,” she adds. “So it’s very important for our skin health.”
Your skin barrier is part of your stratum corneum, your top layer of skin. And in terms of structure, it’s often likened to a brick wall:
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Corneocytes are the tough skin cells that make up the “bricks” of your skin barrier.
Lipids are natural fats in your skin, and in this analogy, they’re the “mortar” that fills in the gaps between corneocytes. These same lipids — like ceramides, cholesterol and fatty acids — are important ingredients in skin care products.
What can damage your skin barrier?
Let’s go back to that armor analogy. Imagine your once-sturdy armor has taken some hits, and it’s not in the same shape it used to be. Maybe it’s thinner than before or the material has warped. Maybe it even has some kind of visible damage, like a hole or a tear. It once kept you fully protected, but now it leaves you vulnerable to harm.
If your skin barrier is damaged due to an underlying skin condition, you should work with a dermatologist to make sure you’re doing what’s right for your specific needs. Medical conditions associated with having a poor skin barrier include:
Adult acne.
Atopic dermatitis (eczema).
Ichthyosis.
Rosacea.
Psoriasis.
But sometimes, the damage comes from things you do (or don’t do) to your skin. “There are many things that can break down your skin barrier and make it not work,” Dr. Piliang notes, like:
Using harsh chemicals or soaps.
Over-exfoliating or scrubbing your skin.
Not using a moisturizer.
Signs your skin barrier is damaged
In short, if you’re experiencing some sort of issue with your skin, it’s likely that your skin barrier has sustained some damage. That damage may be evident just based on the way your skin looks and feels, including:
Acne.
Dry, scaly and/or flaky skin.
Infection.
Inflammation and irritation.
Itchiness.
Rough patches.
Stinging, especially when you apply skin care products.
Tenderness or sensitivity.
How to prevent and heal damage to your skin barrier
To get (and keep) your skin barrier intact, tread lightly. “You want to be very gentle when you care for your skin,” Dr. Piliang cautions.
She shares tips for how to give your skin barrier the TLC it needs.
1. Wash with warm (but not scalding) water
While your skin barrier acts like armor for the delicate skin underneath, it’s also like a layer of fat on your skin. To better understand what that means and how it reacts to water temperature, imagine for a moment that it’s a pat of butter.
“The way I like to think about it is this: If you have butter on a knife and you put it under hot water, the butter melts away instantly,” Dr. Piliang illustrates. “The same is true when you’re cleaning your skin. Hot water washes away all those healthy natural oils.”
2. Use a soap-free cleanser
Soap sounds like a good thing, right? In reality, though, soap can strip your skin of its natural oils and wash away good bacteria, which damages your skin barrier. Soap can’t tell the difference between the good stuff and the bad stuff, so it just gets rid of it all.
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Instead, Dr. Piliang suggests using a mild, soap-free cleanser (with warm water, of course). “Look for soap-free cleansers that are formulated for sensitive skin and fragrance-free,” she advises.
3. Exfoliate gently
There was a time when exfoliants with “microbeads” were all the rage — little balls of plastic that rubbed against your skin and left you feeling freshly scrubbed. But they’re actually pretty problematic, as they create tiny tears in your skin barrier (and they’re bad for the environment, too).
A chemical exfoliant with a mild, alpha hydroxy acid is your best bet, Dr. Piliang says. “And use it regularly over time. That slowly exfoliates the skin and releases your glow underneath.”
4. Don’t over-cleanse
Some TikTokkers swear by the so-called #60secondrule, saying you should cleanse your face for at least a full minute in order to reap its benefits. But Dr. Piliang says this could actually damage your skin barrier.
“The ideal way to cleanse is to massage on your cleanser and then rinse it off,” she says. “There’s no benefit to massaging it in longer, and in fact, you may do damage because the fatty layer of your skin barrier can begin to break down.”
5. Keep skin moisturized
Moisturizers nourish your skin and help replace your skin barrier and keep it healthy. “Look for products that include ceramides, fatty acids and lipids, which naturally make up a large percentage of the skin barrier,” Dr. Piliang says.
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6. Wear sunscreen!
Wearing sunscreen is one of the absolute best and most important ways to protect your skin. There are lots of kinds of sunscreen, so spend a little time figuring out which one is right for you.
“The best sunscreen you can get is the one you’ll actually put on your skin,” Dr. Piliang notes. “Find a product that you like and will use and can afford, then use it every day.” (And don’t forget some of these often-overlooked sunscreen spots!)
7. Treat pimples with care
By now, you know the rules: Don’t pick! Picking at and popping pimples causes trauma to your skin and can lead to inflammation and scars.
But if you already have broken skin around a zit, don’t load up the area with a bunch of products to try to clean things up. “Back down on your acne treatments for a bit so you can focus on gentle skin care and letting it heal,” Dr. Piliang recommends.
8. Keep your skin’s pH balanced
“pH” stands for “potential hydrogen,” and in a nutshell, your pH number indicates how acidic your skin is. Skin’s normal pH is about 5.5, but if yours is higher or lower, it can throw your skin barrier out of whack.
There’s a lot of science behind your skin’s pH, which makes it tough to figure out what yours is and what products will help balance it. The best way to determine your needs is to talk to a dermatologist to see what they recommend.
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Can your skin barrier be permanently damaged?
It’s no surprise that as you age, your skin ages with you. After all, anti-aging skin care products are a hot commodity.
“As we age, our skin barrier does not replace itself as well or as quickly,” Dr. Piliang confirms. That means that it just doesn’t bounce back like it used to after a day in the sun or using a harsh product.
“A young person can take a shower three times a day with hot water and soap, and in six or seven hours, their skin barrier will replace itself enough that their skin doesn’t dry out,” she explains. “Once you’re in your 50s, 60s or 70s, though, your skin barrier doesn’t replace itself as quickly.”
Your skin care routine should change with you as you age. You may need to start using different products and focusing more than ever on hydration and moisturization.
How long does it take to repair your skin barrier?
We know, we know… You’re feeling pretty impatient about this whole skin barrier thing. Why isn’t your skin gorgeous and glowing yet?! It can be hard, but wait it out, as it typically takes three or four months to start seeing the benefits of your efforts.
“It can be frustrating because you want to see the effects of your treatments immediately,” Dr. Piliang says, “but remember that it took a long time for your skin to get to this point. You have to be patient and give your products and habits time to work.”
To learn more from Dr. Piliang on related topics, listen to the Health Essentials Podcast episode, “Skin Care Tips, Tricks and Trends.” New episodes of the Health Essentials Podcast publish every Wednesday.
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187992 | https://math.stackexchange.com/questions/444887/solving-the-trigonometric-equation-a-cos-x-b-sin-x-c | trigonometry - Solving the trigonometric equation $A\cos x + B\sin x = C$ - Mathematics Stack Exchange
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Solving the trigonometric equation A\cos x + B\sin x = C [duplicate]
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\begingroup
This question already has answers here:
Solving trigonometric equations of the form a\sin x + b\cos x = c (7 answers)
Closed 10 years ago.
I have a simple equation which i cannot solve for x:
A\cos x + B\sin x = C
Could anyone show me how to solve this. Is this a quadratic equation?
trigonometry
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edited Jul 17, 2013 at 12:51
dajoker
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asked Jul 16, 2013 at 10:01
71GA71GA
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\begingroup See this answer for the general approach and an example.\endgroup Ayman Hourieh –Ayman Hourieh 2013-07-16 10:12:41 +00:00 Commented Jul 16, 2013 at 10:12
\begingroupMy picture-answer to a related question may be helpful.\endgroup Blue –Blue 2013-07-16 11:19:50 +00:00 Commented Jul 16, 2013 at 11:19
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\begingroup
A\cos x+B\sin x=C so if A\neq 0, B\neq 0 then \frac{A}{\sqrt{A^2+B^2}}\cos x+\frac{B}{\sqrt{A^2+B^2}}\sin x=\frac{C}{\sqrt{A^2+B^2}} in which \frac{A}{\sqrt{A^2+B^2}}\le1,~~\frac{B}{\sqrt{A^2+B^2}}\le1,~~\frac{C}{\sqrt{A^2+B^2}}\le1 This means you can suppose there is a \xi such that \cos(\xi)=\frac{A}{\sqrt{A^2+B^2}},\sin(\xi)=\frac{B}{\sqrt{A^2+B^2}} and so...
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edited Jul 16, 2013 at 10:21
answered Jul 16, 2013 at 10:09
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\begingroup+1 . I think the condition needed here is merely \,A\neq 0\;\;or\;\;B\neq 0\; , though if one of them is zero then the equation is much easier.\endgroup DonAntonio –DonAntonio 2013-07-16 10:18:36 +00:00 Commented Jul 16, 2013 at 10:18
\begingroup@DonAntonio: Yes dear Don. It is my wrong using of notation. I'll fix. Sorry for being Lazy in typing.\endgroup Mikasa –Mikasa 2013-07-16 10:20:45 +00:00 Commented Jul 16, 2013 at 10:20
\begingroup Are there no easier solutions?\endgroup 71GA –71GA 2013-07-16 10:51:35 +00:00 Commented Jul 16, 2013 at 10:51
1 \begingroup@71GA: I don't know :) But there is a great comment above under question. Follow that.\endgroup Mikasa –Mikasa 2013-07-16 10:53:20 +00:00 Commented Jul 16, 2013 at 10:53
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\begingroup
HINT:
We can also utilize Weierstrass substitution (1, 2), which will convert the given equation to a Quadratic equation in \tan \frac x2
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answered Jul 16, 2013 at 10:15
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187993 | https://facultystaff.richmond.edu/~ggilfoyl/research/gilfoyle/node2.html | The Transfer Matrix Method
The Transfer Matrix Method
To investigate quantum mechanical tunneling one must extract the transmission coefficient from the solution to the one-dimensional, time-independent Schrödinger equation. The transmission coefficient is the ratio of the flux of particles that penetrate a potential barrier to the flux of particles incident on the barrier. It is related to the probability that tunneling will occur. Consider a rectangular potential barrier of height V 0 as shown in Figure 1. The general solution of the Schrödinger equation in each region is
=(2)
=(3)
=(4)
where and . This solution can be rewritten as a vector dot product so, for example, in Region 2 of Figure 1
(5)
where the are the coefficient vectors representing the wave function in each region. To generate relationships among the coefficients in Equations 2-4 one requires the wave function and its first derivative to be continuous at the boundaries of each region in Figure 1. At x=0 this leads to two expressions.
A+B=(6)
i k 1 A - i k 1 B=i k 2 C - i k 2 D(7)
In matrix notation Equations 6-7 can be expressed in the following way.
=(8)
=(9)
Using the definition of the matrix inverse yields the following result
(10)
which can be expressed as
(11)
The matrix, , is known as the discontinuity matrix and `connects' the coefficient vectors and in Region 1 and Region 2.
The wave function and its derivative must also be continuous at x=a. At this point, consider a new coordinate system such that the transition from Region 2 to Region 3 takes place at . By analogy with Equation 11 one can show that,
(12)
where the are the coefficient vectors in the new coordinate system in the equivalent regions of the potential energy curve and has the same form as in Equation 3 except for the interchange of the indices. To exploit this result, one must relate the coefficient vectors in the primed coordinate system to the ones in the original coordinates. The original coordinate system is transformed such that . The new wave function in Region 2 must satisfy
(13)
which can be written in matrix form and rearranged to yield the following result (recall Equation 5).
(14)
The row vector on the right hand side of Equation 14 is the same as the row vector in Equation 5 so the coefficient vectors representing and are related by
(15)
where is the propagation matrix in Region 2. Similarly, one can show that
Image 33: \begin{displaymath}\phi^\prime_3
\pmatrix{e^{ik_1a} & 0 \cr
0 &e^{-ik_1a} \cr}
\pmatrix{F\cr G\cr}
=
{\bf p_{-1}} \phi_3
\end{displaymath}(16)
where shifts the wave function back to the original coordinate system. Combining Equations 11,12, 15, and 16 and setting G=0 since there are no incoming waves in Region 3 one obtains
(17)
where is the transfer matrix relating the coefficient vectors in Regions 1 and 3. The transmission coefficient is then
(18)
This treatment of the rectangular potential barrier problem can be extended to potential barriers of arbitrary shape. Consider the radioactive -decay of . The potential barrier is shown as the solid curve in Figure 2.
Figure: 50pt 50pt Potential energy curve for an particle in the force field of , the daughter nucleus of the decay.
The central portion of the curve is the Coulomb potential V(x) = Z 1 Z 2 e 2/x where x is the distance between the nuclear centers and the product of the charges is Z 1 Z 2 e 2. It is divided into a sequence of adjacent barriers (the dot-dashed lines)lying between the nuclear radius, x 0, and the classical turning point, x max, for an particle of total energy, E. The potential energy is taken to be zero outside these limits in the manner of Condon and Gurney.12 One can now use the propagation and discontinuity matrices to relate the wave function inside the barrier to the wave function outside (x>x max). For the configuration shown in Figure 2 one chooses the origin at the nuclear radius and the two wave functions are related by
(19)
where is the discontinuity matrix between the region where V(x) = 0 and V(x) = V 1, is the propagation matrix where V(x) = V 1, and so on. The propagation matrix returns the wave function to the appropriate coordinate system. The last matrix is unnecessary for calculating the transmission coefficient since it changes the coordinates, but does not change the ratio of the coefficients. The adequacy of treating the potential energy curve in Figure 2 as a sequence of adjacent rectangular barriers will improve as the number of barriers increases and should converge to some limiting value. The transmission coefficient will be extracted from the transfer matrix using Equation 18.
1998-09-14 |
187994 | https://clopendebate.wordpress.com/2024/06/23/venn-diagram-puzzles-algebra/ | Venn Diagram Puzzles –Algebra – CLopen Mathdebater
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Venn Diagram Puzzles –Algebra
June 23, 2024cluzniak
I recently added some Venn Diagram puzzles I’ve made with some teachers this year for Geometry in this post. I wanted to add some Algebra ones here! Though most were created with certain answers in mind, we welcome creative ideas from students. So there may be more than one answer!
Some are good for vocab.
Some involve solving for x first.
Or looking at graphs.
Some are focused on what process you might prefer to use…
We made some “challenging” ones that may not always have answers for every question mark.
I’d love to hear what you think or how it goes if you try one with your students. Feel free to share any you make!
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187995 | http://msuperl.org/wikis/pcubed/doku.php?id=184_notes:b_flux | 184_notes:b_flux [Projects & Practices in Physics]
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Table of Contents
Changing Magnetic Flux
Magnetic Flux
Changing Magnetic Flux
Examples
Section 22.2 in Matter and Interactions (4th edition)
Changing Magnetic Flux
In these notes, we will start thinking the right hand side of Faraday's Law (the d Φ B d t d Φ B d t part) and what it means to have a changing magnetic flux. Let's start by defining what flux is.
In general, any sort of flux is how much of something goes through an area. For example, we could think of a child's bubble wand in terms of the air flux (from you blowing) through the circle (with the bubble solution in it). If you wanted to make bigger bubbles or make many more bubbles, you could do two things: increase the air flow or get a bubble wand with a bigger circle. Both of these actions (increasing the area and increasing the amount of air) will result in a larger “air flux” through the bubble wand. It's probably worth mentioning that we have assumed that you are holding the bubble wand so the circle is perpendicular to the air flow. If instead you rotate the wand 90 degrees, you will not get any bubbles since there is no air that is actually going through the circle part of the bubble wand. So the air flux not only depends on the amount of air and the area of circle, but also on how those two are oriented relative to each other. The idea of flux can be useful in many different contexts (i.e. fluids, electricity, air, etc.), but for any kind of flux, these are still the three conditions that matter: (1) the strength/amount, (2) the area, and (3) the orientation.
Therefore magnetic flux is the strength of the magnetic field on a surface area or rather the amount of the magnetic field that goes through an area. For magnetic flux, we need to consider: the strength of the magnetic field, the area that the field goes through, and the orientation of the magnetic field relative to the area. These notes will introduce the mathematics behind magnetic flux, which we will use in Faraday's Law.
Magnetic Flux
Area Vector is perpendicular to the surface.
Just like we did before with electric flux, we will say that magnetic flux is the strength of the magnetic field over an area or rather the amount of magnetic field that goes through an area. How the magnetic field is oriented relative to the area does matter so we will again want to treat the area as vector. This is very similar to how we defined the electric flux before. Mathematically, we represent the magnetic flux as:
Φ B=B⃗∙A⃗Φ B=B→∙A→
Flux through a horizontal plane (where dA is parallel to the magnetic field)
where Φ B Φ B is the magnetic flux (with units of T⋅m 2 T⋅m 2), B⃗B→ is the magnetic field and A⃗A→ is the area vector. The area vector in this case is the vector that has the same magnitude as the area (i.e. length times width for a rectangular area or π r 2 π r 2 for a circular area) and has a direction that is perpendicular to the area, which is represented by the green arrow in the figure to the right. (This is exactly the same as with electric flux). You may notice that there are actually two different vectors that are perpendicular to the gray surface. We've drawn the green arrow in the +y+y direction, but we could have also picked the vector that points in the −y−y direction. For an open surface, it doesn't matter which of the perpendicular vectors you pick, as long as you are consistent after you pick one.
The dot product between the B⃗B→ and the A⃗A→ then tells you about the direction the magnetic field points relative to the area. Mathematically, we can simplify the dot product by saying:
Φ B=|B⃗||A⃗|c o s(θ)Φ B=|B→||A→|c o s(θ)
where θ θ is the angle between the magnetic field and the area vector. This should also make some physical sense. As shown in the top figure to the left, if the magnetic field (shown in blue arrows) and area vector (red arrow) point in the same direction, then the dot product turns into a simple multiplication and c o s(θ)=1 c o s(θ)=1. If we look at the figure, there are lots of magnetic field arrows poking through the surface, so we'd expect a large magnetic flux. As shown in the bottom figure to the left, if the magnetic field (blue arrows) points perpendicular to the area vector (green arrow), then the dot product gives a zero since c o s(90)=0 c o s(90)=0. If we look at the figure, we see this result as well because none of the magnetic field arrows actually go through the surface.
Flux through a vertical plane (where dA is perpendicular to the magnetic field)
(As a side note, this technically gives you the magnitude of the flux. Flux can be positive or negative and that depends on how the area vector points relative to the magnetic field vectors. This is discussed further in the page of notes on directions.)
In writing the magnetic flux above, we have made an _assumption that the area vector points in the same direction everywhere on the surface_ (or rather that the area we are considering is flat rather than curved). If instead the surface is curved we would need to write the magnetic flux in terms of a very small area and then add the flux through each of those small areas together. Mathematically, we would do this with an integral:
Φ B=∫B⃗∙d A⃗Φ B=∫B→∙d A→
Changing Magnetic Flux
As you saw in the demo video, just having a magnetic flux is not enough though - to drive a current, the magnetic flux must be changing. Mathematically, we write this change as a change in the magnetic flux over a change in time. Namely:
d Φ B d t d Φ B d t
_If we assume that the change is happening at a constant rate_, we can write this in terms of large changes (using deltas) - Δ Φ B Δ t Δ Φ B Δ t. In this case, we only care about the initial and final flux relative to the initial and final time:
Δ Φ B Δ t=Φ B f−Φ B i t f−t i Δ Φ B Δ t=Φ B f−Φ B i t f−t i
If the change in flux is not happening at a constant rate, we then have to consider the instantaneous change in flux using the differential form: d Φ B d t d Φ B d t.
There are many ways that the magnetic flux could be changing - the size of the magnetic field could be increasing/decreasing, the size of the area could be getting bigger/smaller, the angle between the magnetic field and the area could be changing, or even some combination of these factors. The examples below go through a couple of these cases to show how you could calculate the change in the magnetic flux.
Examples
Review of Flux through a Loop
Flux Through a Changing, Rotating Shape
184_notes/b_flux.txt
Last modified: 2021/06/17 16:08
by bartonmo
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9: Solutions
CHE 124: General Chemistry for the Health Professions (Morsch and Andrews)
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9.4: Properties of Solutions
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Contents
1. Home
2. Campus Bookshelves
3. University of Illinois, Springfield
4. CHE 124: General Chemistry for the Health Professions (Morsch and Andrews)
5. 9: Solutions
6. 9.4: Properties of Solutions
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CHE 124: General Chemistry for the Health Professions (Morsch and Andrews)
Front Matter
1: Chemistry, Matter, and Measurement
2: Elements, Atoms, and the Periodic Table
3: Ionic Bonding and Simple Ionic Compounds
4: Covalent Bonding and Simple Molecular Compounds
5: Introduction to Chemical Reactions
6: Quantities in Chemical Reactions
7: Energy and Chemical Processes
8: Solids, Liquids, and Gases
9: Solutions
10: Acids and Bases
11: Nuclear Chemistry
12: Organic Chemistry - Alkanes & Halogenated Hydrocarbons
13: Unsaturated and Aromatic Hydrocarbons
14: Organic Compounds of Oxygen
15: Organic Acids and Bases and Some of Their Derivatives
16: Carbohydrates
17: Lipids
18: Amino Acids, Proteins, and Enzymes
19: Nucleic Acids
20: Energy Metabolism
Back Matter
9.4: Properties of Solutions
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9.3: The Dissolution Process
9.E: Solutions (Exercises)
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Table of contents
1. Colligative Properties
2. Vapor Pressure Depression
3. Boiling Point and Freezing Point Effects
4. Osmotic Pressure
5. To Your Health: Dialysis
6. Concept Review Exercises
7. Answers
8. Key Takeaway
9. Answers
Skills to Develop
To describe how the properties of solutions differ from those of pure solvents.
Solutions are likely to have properties similar to those of their major component—usually the solvent. However, some
solution
properties differ significantly from those of the solvent. Here, we will focus on liquid solutions that have a solid
solute
, but many of the effects we will discuss in this section are applicable to all solutions.
Colligative Properties
Solutes affect some properties of solutions that depend only on the
concentration
of the dissolved particles. These properties are called
colligative properties
. Four important
colligative properties
that we will examine here are
vapor pressure depression
,
boiling point elevation
,
freezing point depression
, and
osmotic pressure
.
Molecular compounds separate into individual molecules when they are dissolved, so for every 1 mol of molecules dissolved, we get 1 mol of particles. In contrast,
ionic compounds
separate into their constituent ions when they dissolve, so 1 mol of an ionic
compound
will produce more than 1 mol of dissolved particles. For example, every mole of NaCl that dissolves yields 1 mol of Na+ ions and 1 mol of Cl− ions, for a total of 2 mol of particles in
solution
. Thus, the effect on a
solution
’s properties by dissolving NaCl may be twice as large as the effect of dissolving the same amount of moles of glucose (C 6 H 12 O 6).
Vapor Pressure Depression
All liquids evaporate. In fact, given enough
volume
, a liquid will turn completely into a vapor. If enough
volume
is not present, a liquid will evaporate only to the point where the rate of evaporation equals the rate of vapor condensing back into a liquid. The
pressure
of the vapor at this point is called the
vapor pressure
of the liquid.
The presence of a dissolved solid lowers the characteristic
vapor pressure
of a liquid so that it evaporates more slowly. (The exceptions to this statement are if the
solute
itself is a liquid or a gas, in which case the
solute
will also contribute something to the evaporation process. We will not discuss such solutions here.) This property is called
vapor pressure depression
and is depicted in Figure 9.4.1.
_Figure 9.4.1:
Vapor Pressure Depression
. The presence of
solute
particles blocks some of the ability for liquid particles to evaporate. Thus, solutions of solid solutes typically have a lower
vapor pressure
than the pure solvent._
Boiling Point
and Freezing Point Effects
A related property of solutions is that their boiling points are higher than the
boiling point
of the pure solvent. Because the presence of
solute
particles decreases the
vapor pressure
of the liquid solvent, a higher temperature is needed to reach the
boiling point
. This phenomenon is called
boiling point elevation
. For every mole of particles dissolved in a liter of water, the
boiling point
of water increases by about 0.5°C.
Some people argue that putting a pinch or two of salt in water used to cook spaghetti or other pasta makes a
solution
that has a higher
boiling point
, so the pasta cooks faster. In actuality, the amount of
solute
is so small that the
boiling point
of the water is practically unchanged.
The presence of
solute
particles has the opposite effect on the freezing point of a
solution
. When a
solution
freezes, only the solvent particles come together to form a solid
phase
, and the presence of
solute
particles interferes with that process. Therefore, for the liquid solvent to freeze, more
energy
must be removed from the
solution
, which lowers the temperature. Thus, solutions have lower freezing points than pure solvents do. This phenomenon is called
freezing point depression
. For every mole of particles in a liter of water, the freezing point decreases by about 1.9°C.
Both
boiling point elevation
and
freezing point depression
have practical uses. For example, solutions of water and ethylene glycol (C 2 H 6 O 2) are used as coolants in automobile engines because the
boiling point
of such a
solution
is greater than 100°C, the normal
boiling point
of water. In winter, salts like NaCl and CaCl 2 are sprinkled on the ground to melt ice or keep ice from forming on roads and sidewalks (Figure 9.4.2). This is because the
solution
made by dissolving sodium chloride or calcium chloride in water has a lower freezing point than pure water, so the formation of ice is inhibited.
_Figure 9.4.2: Effect of
Freezing Point Depression
. The salt sprinkled on this sidewalk makes the water on the sidewalk have a lower freezing point than pure water, so it does not freeze as easily. This makes walking on the sidewalk less hazardous in winter. © Thinkstock_
Example 9.4.1
Which
solution
’s freezing point deviates more from that of pure water—a 1 M
solution
of NaCl or a 1 M
solution
of CaCl 2?
SOLUTION
Colligative properties
depend on the number of dissolved particles, so the
solution
with the greater number of particles in
solution
will show the greatest deviation. When NaCl dissolves, it separates into two ions, Na+ and Cl−. But when CaCl 2 dissolves, it separates into three ions—one Ca 2+ ion and two Cl− ions. Thus, mole for mole, CaCl 2 will have 50% more impact on
freezing point depression
than NaCl.
Exercise 9.4.1
Which
solution
’s
boiling point
deviates more from that of pure water—a 1 M
solution
of CaCl 2 or a 1 M
solution
of MgSO 4?
Osmotic Pressure
The last colligative property of solutions we will consider is a very important one for biological systems. It involves osmosis, the process by which solvent molecules can pass through certain membranes but
solute
particles cannot. When two solutions of different
concentration
are present on either side of these membranes (called semipermeable membranes), there is a tendency for solvent molecules to move from the more dilute
solution
to the more concentrated
solution
until the concentrations of the two solutions are equal. This tendency is called
osmotic pressure
. External
pressure
can be exerted on a
solution
to counter the flow of solvent; the
pressure
required to halt the osmosis of a solvent is equal to the
osmotic pressure
of the
solution
.
Osmolarity
(osmol) is a way of reporting the total number of particles in a
solution
to determine
osmotic pressure
. It is defined as the
molarity
of a
solute
times the number of particles a
formula unit
of the
solute
makes when it dissolves (represented by i):
(9.4.1)osmol=M×i
If more than one
solute
is present in a
solution
, the individual osmolarities are additive to get the total
osmolarity
of the
solution
. Solutions that have the same
osmolarity
have the same
osmotic pressure
. If solutions of differing osmolarities are present on opposite sides of a semipermeable membrane, solvent will transfer from the lower-
osmolarity
solution
to the higher-
osmolarity
solution
. Counterpressure exerted on the high-
osmolarity
solution
will reduce or halt the solvent transfer. An even higher
pressure
can be exerted to force solvent from the high-
osmolarity
solution
to the low-
osmolarity
solution
, a process called reverse osmosis. Reverse osmosis is used to make potable water from saltwater where sources of fresh water are scarce.
Example 9.4.2
A 0.50 M NaCl aqueous
solution
and a 0.30 M Ca(NO 3)2 aqueous
solution
are placed on opposite sides of a semipermeable membrane. Determine the
osmolarity
of each
solution
and predict the direction of solvent flow.
SOLUTION
The solvent will flow into the
solution
of higher
osmolarity
. The NaCl
solute
separates into two ions—Na+ and Cl−—when it dissolves, so its
osmolarity
is as follows:
osmol (NaCl) = 0.50 M × 2 = 1.0 osmol
The Ca(NO 3)2
solute
separates into three ions—one Ca 2+ and two NO 3−—when it dissolves, so its
osmolarity
is as follows:
osmol [Ca(NO 3)2] = 0.30 M × 3 = 0.90 osmol
The
osmolarity
of the Ca(NO 3)2
solution
is lower than that of the NaCl
solution
, so water will transfer through the membrane from the Ca(NO 3)2
solution
to the NaCl
solution
.
Exercise 9.4.2
A 1.5 M C 6 H 12 O 6 aqueous
solution
and a 0.40 M Al(NO 3)3 aqueous
solution
are placed on opposite sides of a semipermeable membrane. Determine the
osmolarity
of each
solution
and predict the direction of solvent flow.
To Your Health: Dialysis
The main function of the kidneys is to filter the blood to remove wastes and extra water, which are then expelled from the body as urine. Some diseases rob the kidneys of their ability to perform this function, causing a buildup of waste materials in the bloodstream. If a kidney transplant is not available or desirable, a procedure called dialysis can be used to remove waste materials and excess water from the blood.
In one form of dialysis, called hemodialysis, a patient’s blood is passed though a length of tubing that travels through an artificial kidney machine (also called a dialysis machine). A section of tubing composed of a semipermeable membrane is immersed in a
solution
of sterile water, glucose, amino acids, and certain electrolytes. The
osmotic pressure
of the blood forces waste molecules and excess water through the membrane into the sterile
solution
. Red and white blood cells are too large to pass through the membrane, so they remain in the blood. After being cleansed in this way, the blood is returned to the body.
_A patient undergoing hemodialysis depends on osmosis to cleanse the blood of waste
products
that the kidneys are incapable of removing due to disease. Image used with permission from Wikipedia._
Dialysis is a continuous process, as the osmosis of waste materials and excess water takes time. Typically, 5–10 lb of waste-containing fluid is removed in each dialysis session, which can last 2–8 hours and must be performed several times a week. Although some patients have been on dialysis for 30 or more years, dialysis is always a temporary
solution
because waste materials are constantly building up in the bloodstream. A more permanent
solution
is a kidney transplant.
Cell walls are semipermeable membranes, so the osmotic pressures of the body’s fluids have important biological consequences. If solutions of different
osmolarity
exist on either side of the cells, solvent (water) may pass into or out of the cells, sometimes with disastrous results. Consider what happens if red blood cells are placed in a hypotonic
solution
, meaning a
solution
of lower
osmolarity
than the liquid inside the cells. The cells swell up as water enters them, disrupting cellular activity and eventually causing the cells to burst. This process is called hemolysis. If red blood cells are placed in a hypertonic
solution
, meaning one having a higher
osmolarity
than exists inside the cells, water leaves the cells to dilute the external
solution
, and the red blood cells shrivel and die. This process is called crenation. Only if red blood cells are placed in isotonic solutions that have the same
osmolarity
as exists inside the cells are they unaffected by negative effects of
osmotic pressure
. Glucose solutions of about 0.31 M, or sodium chloride solutions of about 0.16 M, are isotonic with blood plasma.
The
concentration
of an isotonic sodium chloride (NaCl)
solution
is only half that of an isotonic glucose (C 6 H 12 O 6)
solution
because NaCl produces two ions when a
formula unit
dissolves, while molecular C 6 H 12 O 6 produces only one particle when a
formula unit
dissolves. The osmolarities are therefore the same even though the concentrations of the two solutions are different.
Osmotic pressure
explains why you should not drink seawater if you are abandoned in a life raft in the middle of the ocean. Its
osmolarity
is about three times higher than most bodily fluids. You would actually become thirstier as water from your cells was drawn out to dilute the salty ocean water you ingested. Our bodies do a better job coping with hypotonic solutions than with hypertonic ones. The excess water is collected by our kidneys and excreted.
Osmotic pressure
effects are used in the food industry to make pickles from cucumbers and other vegetables and in brining meat to make corned beef. It is also a factor in the mechanism of getting water from the roots to the tops of trees!
Career Focus: Perfusionist
A perfusionist is a medical technician trained to assist during any medical procedure in which a patient’s circulatory or breathing functions require support. The use of perfusionists has grown rapidly since the advent of open-heart surgery in 1953.
Most perfusionists work in operating rooms, where their main responsibility is to operate heart-lung machines. During many heart surgeries, the heart itself must be stopped. In these situations, a heart-lung machine keeps the patient alive by aerating the blood with oxygen and removing carbon dioxide. The perfusionist monitors both the machine and the status of the blood, notifying the surgeon and the anesthetist of any concerns and taking corrective action if the status of the blood becomes abnormal.
Despite the narrow parameters of their specialty, perfusionists must be highly trained. Certified perfusion education programs require a student to learn anatomy, physiology, pathology, chemistry, pharmacology, math, and physics. A college degree is usually required. Some perfusionists work with other external artificial organs, such as hemodialysis machines and artificial livers.
Concept Review Exercises
What are the
colligative properties of solutions?
2. Explain how the following properties of solutions differ from those of the pure solvent:
vapor pressure , boiling point , freezing point, and osmotic pressure .
Answers
Colligative properties are characteristics that a solution has that depend on the number, not the identity, of solute particles.
In solutions, the
vapor pressure is lower, the boiling point is higher, the freezing point is lower, and the osmotic pressure is higher.
Key Takeaway
`Exercises
In each pair of aqueous systems, which will have the lower
vapor pressure ?
1. pure water or 1.0 M NaCl
2. 1.0 M NaCl or 1.0 M C 6 H 12 O 6
3. 1.0 M CaCl 2 or 1.0 M (NH 4)3 PO 4
In each pair of aqueous systems, which will have the lower
vapor pressure ?
1. 0.50 M Ca(NO 3)2 or 1.0 M KBr
2. 1.5 M C 12 H 22 O 11 or 0.75 M Ca(OH)2
3. 0.10 M Cu(NO 3)2 or pure water
In each pair of aqueous systems, which will have the higher
boiling point ?
1. pure water or a 1.0 M NaCl
2. 1.0 M NaCl or 1.0 M C 6 H 12 O 6
3. 1.0 M CaCl 2 or 1.0 M (NH 4)3 PO 4
In each pair of aqueous systems, which will have the higher
boiling point ?
1. 1.0 M KBr
2. 1.5 M C 12 H 22 O 11 or 0.75 M Ca(OH)2
3. 0.10 M Cu(NO 3)2 or pure water
Estimate the
boiling point of each aqueous solution . The boiling point of pure water is 100.0°C.
1. 0.50 M NaCl
2. 1.5 M Na 2 SO 4
3. 2.0 M C 6 H 12 O 6
Estimate the freezing point of each aqueous
solution . The freezing point of pure water is 0.0°C.
1. 0.50 M NaCl
2. 1.5 M Na 2 SO 4
3. 2.0 M C 6 H 12 O 6
Explain why salt (NaCl) is spread on roads and sidewalks to inhibit ice formation in cold weather.
Salt (NaCl) and calcium chloride (CaCl 2) are used widely in some areas to minimize the formation of ice on sidewalks and roads. One of these
ionic compounds is better, mole for mole, at inhibiting ice formation. Which is that likely to be? Why?
9.
What is the
osmolarity of each aqueous solution ?
1. 0.500 M NH 2 CONH 2
2. 0.500 M NaBr
3. 0.500 M Ca(NO 3)2
What is the
osmolarity of each aqueous solution ?
1. 0.150 M KCl
2. 0.450 M (CH 3)2 CHOH
3. 0.500 M Ca 3(PO 4)2
A 1.0 M
solution of an unknown soluble salt has an osmolarity of 3.0 osmol. What can you conclude about the salt?
12. A 1.5 M NaCl
solution and a 0.75 M Al(NO 3)3 solution exist on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and the direction of solvent flow, if any, across the membrane.
Answers
1.0 M NaCl
1.0 M NaCl
1.0 M (NH 4)3 PO 4
1.0 M NaCl
1.0 M NaCl
1.0 M (NH 4)3 PO 4
100.5°C
102.3°C
101°C
NaCl lowers the freezing point of water, so it needs to be colder for the water to freeze.
0.500 osmol
1.000 osmol
1.500 osmol
It must separate into three ions when it dissolves.
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9.1: SolutionsSolutions form because a solute and a solvent experience similar intermolecular interactions.
9.2: ConcentrationVarious concentration units are used to express the amounts of solute in a solution. Concentration units can be used as conversion factors in stoichio...
9.3: The Dissolution ProcessWhen a solute dissolves, its individual particles are surrounded by solvent molecules and are separated from each other.
9.E: Solutions (Exercises)Problems and select solutions to this chapter.
9.S: Solutions (Summary)To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself...
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Spermatogenesis is the process whereby spermatozoa are formed. Spermatogenesis has two stages: spermatocytogenesis and spermiogenesis.
Spermatocytogenesis is the first stage of spermatogenesis, whereby spermatogonia develop into spermatocytes and eventually spermatids. Spermatogonia are the germ cells of spermatozoa. Type A dark (Ad) spermatogonia undergo mitotic division at irregular intervals giving rise to either a pair of cells of the same types, or a pair of type A pale (Ap) spermatogonia. Type Ap spermatogonia continue through many mitotic divisions and establish a population of progenitor cells, the type B spermatogonia. Type B spermatogonia undergo mitotic division to produce primary spermatocytes. Primary spermatocytes undergo two meiotic cell divisions so that the resultant haploid cells, called spermatids, contain half the original chromosome number and amount of DNA. Small cells, called secondary spermatocytes, arise from the first meiotic division. These cells are very sparse in the seminiferous epithelium because they rapidly enter the second meiotic division and give rise to spermatids.
The second stage of spermatogenesis is spermiogenesis. This involves the transformation of spermatids into spermatozoa. Unlike spermatocytogenesis, no cell division occurs during spermiogenesis. There are three phases including the Golgi, acrosomal, and maturation phase. At the end of the maturation phase, spermatozoa are released into the seminiferous tubule lumen.
It takes about 64 days from the commencement of the division of type B spermatogonia to the release of spermatozoa into the lumen of the seminiferous tubule. Spermatogenesis does not occur as a synchronous event along the length of the seminiferous tubule. Therefore, any given view of a section of the tubule will reveal different phases of spermatogenesis.
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187999 | https://www.biointeractive.org/professional-learning/educator-voices/modifying-biointeractive-resources-anatomy-and-physiology | Modifying BioInteractive Resources for an Anatomy and Physiology Course
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Modifying BioInteractive Resources for an Anatomy and Physiology Course
By: Melissa Haswell
09.21.18
Teaching Topic
AssessmentLesson Planning
Science Topic
Biochemistry & Molecular BiologyAnatomy & Physiology
Level
College — 2-yearCollege — 4-year
HHMI BioInteractive offers a rich compilation of resources in the fields of molecular biology, genetics, and evolution. But if you teach anatomy and physiology courses as I do, you may think at first glance that these materials do not apply to your courses. The students in my courses are primarily nursing students who often do not see the direct relevance of learning basic biological concepts such as the central dogma, osmosis, diffusion, cell division, genetics, and natural selection.
As instructors, we know that these basic biological principles are essential to understanding physiological processes. However, students may not have learned basic biological principles within the appropriate context required to apply them to physiological concepts.
I have used a vetted concept inventory from the Genetics Society of America that provides insight into my students’ genetics literacy (www.genetics.org/content/178/1/15). Nursing students who have previously completed their entire sequence of required introductory biology courses struggle to appropriately use and apply genetics concepts in an upper level pathophysiology course. Students seem to have the most problems with the central dogma, meiosis, genetic inheritance and mutations, evolutionary principles of genetics, and genetic engineering.
Why emphasize these concepts if your students, like mine, are not going to major in biology? Interestingly, the standard nursing education curriculum does not have a required component in genetics. Based on my teaching experience and a review of the literature, most nursing students only receive a cursory introduction to these disciplines either in an introductory biology course or in some cases in their anatomy and physiology courses.
A basic understanding of genomic medicine is essential for 21 st century nursing practice because nurses will most likely work with patients who have had some type of genetic test for a disease, prenatal genetic testing, or pharmacogenomic therapy. In addition, nurses often have the most contact with patients when doing patient intakes and in providing health education materials for patients. Understanding genetics would also better equip nurses during patient intake to recognize a possible genetic disorder or to have the ability to answer patient questions related to their diagnosis.
This is where HHMI BioInteractive resources that cover basic biological principles can be used, with minor adaptations, to provide nursing students with an exciting opportunity to immerse themselves in genomics research. The following is one of my favorite activities which I use in the second semester of a two-semester anatomy and physiology course.
Lactase persistence module: I use this in the second semester of my anatomy and physiology course to introduce students to the digestive system and enzymatic digestion. Students apply previously learned knowledge of acid-base balance via blood and respiratory system to digestive system processes. These resources also allow students to apply osmotic principles to physiological processes and reinforce genetic inheritance and natural selection concepts. For this activity, I use the following BioInteractive resources:
Pre-class assignment completed individually:
In-class group activity using clickers or color-coded ABCD student response cards:
Students complete data analysis portion in class and then are assessed individually using several additional critical thinking questions that I have created in which students apply principles of osmosis, acid-base balance, and enzyme activity to the signs and symptoms of lactose intolerance as related to the digestive and respiratory systems: www.hhmi.org/biointeractive/lactase-film-quiz
For homework, I assign the following questions which are based on the illustration below, as part of the summative assessment for the unit. I also include versions of a few of these on their unit exam.
Why does fermentation occur in the large intestine and not the small intestine?
Explain the role of the brush border (microvilli) of the small intestine in digesting lactose in a person who is not lactose intolerant.
Explain in your own words how the creation of hydrogen gas leads to bloating.
Explain in a short paragraph, or diagram, how the hydrogen gas moves from the large intestine to the lungs so that it can be exhaled. Be sure to include the role of the blood.
If lactose is consumed in large quantities, undigested lactose also builds up in the large intestine, leading to other recognizable side effects of lactase-deficient patients, such as diarrhea. In the space below, explain how the unabsorbed lactose pulls fluid into the large intestine. Be sure to include the following terms in your answer: osmosis and osmolality. (I instruct students to review these definitions in their textbooks.)
After completing this activity, students should have working knowledge of the following:
Identify and explain the genetic basis of lactase persistence.
Relate the role of the respiratory system in pH balance using the hydrogen breath test for lactose intolerance as an example.
Explain the process of osmosis as it applies to fluid movement between body compartments using lactose-induced intestinal problems as a model.
Analyze data and use it to create a graph.
Understand that mutations are changes in an organism’s DNA that occur randomly.
Understand the effect of a mutation on an organism’s traits based on the type of mutation and its location.
Explain how a mutation can lead to a change in physiological processes.
I recommend that you peruse all of the HHMI BioInteractive resources to see how you could create a meaningful activity for your anatomy and physiology students!
Come discuss course structure and more in our Facebook group!
Melissa Haswell is an associate professor in the science department at Davenport University in Michigan. She has been teaching anatomy and physiology, as well as pathophysiology, for pre-nursing students for the last 14 years. Melissa also conducts science education research, which focuses on social justice aspects of science education and the implementation of various learning modalities such as case-based learning, dialogue education as applied to biology education, and genetic literacy of nursing students. Her favorite hobbies include hiking with her Dalmatian, Chloe, as well as backpacking and traveling with her husband, Jim.
Use This With
Video Resource
Lactose Digestion in Infants
Video Resource
The Making of the Fittest: Got Lactase? The Co-evolution of Genes and Culture
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