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16200	https://cs.uwaterloo.ca/journals/JIS/VOL24/Trulen/trulen5.pdf	23 11 Article 21.6.1 Journal of Integer Sequences, Vol. 24 (2021), 2 3 6 1 47 2-Adic Valuations of Quadratic Sequences Will Boultinghouse Kentucky Wesleyan College Division of Natural Sciences and Mathematics 3000 Frederica Street Owensboro, KY 42301 USA Jane Long Stephen F. Austin State University Department of Mathematics and Statistics Nacogdoches, TX 75962-3040 USA Olena Kozhushkina Ursinus College Department of Mathematics and Computer Science Collegeville, PA 19426 USA Justin Trulen1 Kentucky Wesleyan College Division of Natural Sciences and Mathematics 3000 Frederica Street Owensboro, KY 42301 USA jtrulen@kwc.edu Abstract We determine properties of the 2-adic valuation sequences for general quadratic polynomials with integer coeﬃcients directly from the coeﬃcients. These properties in-clude boundedness or unboundedness, periodicity, and valuations at terminating nodes. 1Corresponding author. 1 We completely describe the periodic sequences in the bounded case. Throughout, we frame results in terms of trees and sequences. 1 Introduction For p prime and n ∈N = {0, 1, 2, 3, . . .}, the exponent of the highest power of p that divides n is called the p-adic valuation of n, which we denote νp(n). The valuation of 0 is deﬁned to be +∞. Formally, the valuation of a positive integer n of the form n = pkd, where k ∈N and d is an integer not divisible by p, is νp(n) = k. We can ﬁnd p-adic valuations of sequences by ﬁnding the valuation of each successive term. The present work considers 2-adic valuations of sequences generated from the natural numbers by evaluating quadratic functions of the form f(n) = an2 + bn + c where a, b, c ∈Z and a ̸= 0. Information about sequences of valuations can be viewed in two diﬀerent ways: in terms of sequences and in terms of trees. We let (ν2(f(n)))n≥0 denote the sequence of 2-adic valuations for the quadratic function f(n). Since every positive natural number n can be written in the form n = 2kd, where d is not divisible by 2, in many cases, we can determine the valuations of outputs of the quadratic function f(n) using characteristics of the coeﬃcients a, b, and c. The main results are given in Theorems 1 and 2; one would anticipate these results can be extended to odd primes with some modiﬁcations, which will be addressed in future work. Theorem 1. Let f(n) = an2 + bn + c where a, b, c ∈Z with a ̸= 0 and, without loss of generality, a, b, c are not all even. Then 1. If a and b are even and c is odd, then ν2(f(n)) = 0 for all n ∈N. 2. If a is even and b is odd, then (ν2(f(n)))n≥0 is an unbounded sequence. 3. If a is odd and b is even, then (a) if b2 −4ac = 0, then (ν2(f(n)))n≥0 is an unbounded sequence; (b) if b2 −4ac = 4ℓ∆for ℓ∈Z+ as large as possible and ∆≡1 (mod 8), then (ν2(f(n)))n≥0 is an unbounded sequence; (c) if b2 −4ac = 4ℓ∆for ℓ∈Z+ as large as possible, ∆≡m (mod 8), and m ∈ {2, 3, 5, 6, 7}, then the sequence (ν2(f(n)))n≥0 is bounded and its minimal period length equals 2ℓ. 4. If a and b are odd and c is even, then (ν2(f(n)))n≥0 is an unbounded sequence. 5. If a, b, and c are odd, then ν2(f(n)) = 0 for all n ∈N. Theorem 1 is proved in Section 3. Henceforward, we will refer to the minimal period length simply as the period. In Case 3, we use the discriminant to determine whether roots to f(n) = 0 lie in the 2-adic numbers Q2 or the ring of 2-adic integers Z2. Corollary 10 2 takes care of Case 3(a). Even though the statement of this theorem only classiﬁes these sequences as unbounded, the proofs of Cases 2 and 4 reveal more information about the 2-adic valuations. Theorem 1 represents a complete answer to when ν2(f(n))n≥0 is bounded or unbounded using only the coeﬃcients of the quadratic polynomial. Furthermore, Theorem 1 gives an explicit period length for the bounded sequences which can be determined by the coeﬃcients of the quadratic polynomial. In the unbounded cases we are able to describe what possible valuations will be for certain subsequences. Such statements are easier to frame in the sense of trees, which are discussed in Section 2. Theorem 2, proved in Sections 4 and 5, completely determines all valuations in the non-trivial bounded case (3(c) of Theorem 1). Theorem 2. Let f(n) = an2+bn+c where a, b, c ∈Z. If a is odd and b is even and b2−4ac = 4ℓ∆for ℓ∈Z+ as large as possible with ℓ≥2, ∆≡m (mod 8), and m ∈{2, 3, 5, 6, 7}, then the sequence (ν2(f(n)))n≥0 is bounded with period equal to 2ℓ. Furthermore, we have the following 2-adic valuations: ν2(f(n)) =                          0, if n ≡a−1 1 −b 2 (mod 2); 2(i −1), if n ≡a−1 2i−1 −b 2 (mod 2i) with 2 ≤i < ℓ; 2(ℓ−1), if n ≡a−1 2ℓ−1 −b 2 (mod 2ℓ) and m = 6, 2; 2ℓ−1, if n ≡a−1 2ℓ−1 −b 2 (mod 2ℓ) and m = 7, 3; 2ℓ, if n ≡a−1 2ℓ−1 −b 2 (mod 2ℓ) and m = 5; 2ℓ−1, if n ≡a−1 2ℓ−b 2 (mod 2ℓ) and m = 6, 2; 2(ℓ−1), if n ≡a−1 2ℓ−b 2 (mod 2ℓ) and m = 7, 5, 3; where a−1 is the inverse of a (mod 2ℓ). The case ℓ= 1 is covered by Lemma 13. In this case, the sequence is periodic with period equal to 2. Theorem 2 is proved in Proposition 12 and Corollary 24. Both of these results are an extension of the work by Byrnes et al. , which only considered quadratics of the form f(n) = an2 + c. The work of Medina et al. details conditions under which these sequences are bounded or unbounded for general primes but we extend these results for p = 2 by providing the exact conditions on the coeﬃcients of quadratic equations. Furthermore, we provide a closed form giving the exact valuation for the bounded sequences relying only on the coeﬃcients of the quadratic function. Boundedness of p-adic valuations of polynomial sequences is also discussed in Bell’s work . 2 Parity and trees Consider a quadratic function of the form f(n) = an2 + bn + c, where a, b and c are integers and a is nonzero. To prove the results stated in Theorems 1 and 2, we consider the eight possible cases based on the parity of the coeﬃcients a, b, and c. In the case where a, b, and c are all even, there exists an i ∈N such that 2i divides a, b, and c but 2i+1 does not. Then 3 ν2(f(n)) ν2(f(2q + 1)) ν2(f(4q + 3)) ν2(f(4q + 1)) ν2(f(2q)) ν2(f(4q + 2)) ν2(f(4q)) 2q + 1 2q 4q + 3 4q + 1 4q + 2 4q Figure 1: Levels 0, 1, and 2 of a tree. f(n) = 2i(a0n2 + b0n + c0) and it follows that ν2(f(n)) = i + ν2(a0n2 + b0n + c0). Hence, this case can be reduced to one of the other seven cases. So we assume, unless stated otherwise, that a, b, and c are not all even. Two more cases of Theorem 1 are trivial (Case 1 where a, b are even, and Case 5, where a, b, and c are odd), since ν2(f(n)) = 0 for all n ∈N. For the remaining ﬁve cases, we classify the behavior using trees. In the case that a is odd and b is even we show, with the help of the discriminant, that f(n) = 0 has a root in Q2. We must take some care since some quadratics may not have a zero in Q2. As discussed in Section 1, we can present information about the sequence of valuations using a tree. We begin the construction of the tree with the top node representing the valuation of the quadratic f(n) evaluated at any natural number n. If the 2-adic valuation is constant for every n in this node, then we stop the construction, as ν2(n) is completely determined for the sequence. If ν2(n) is not constant, this node splits into two branches, where one branch represents all numbers of the form n = 2q and the other branch represents all numbers of the form n = 2q + 1, where in both cases q ∈N. We then repeat this step as necessary to create the tree. The nodes correspond to the sets {2iq + ri−1\|q ∈N} where ri−1 = i−1 X k=0 αk2k, (1) for ﬁxed coeﬃcients αk ∈{0, 1}. This process does not always terminate. If it terminates, we say that the tree is ﬁnite; otherwise, the tree is inﬁnite. We say a node is non-terminating if (ν2(f(n)))n≥0 is non-constant for every n in that equivalence class. We say a node is terminating if (ν2(f(n)))n≥0 is constant for every n in that equivalence class. In practice, we label the node with this constant valuation. 4 For each of the remaining ﬁve nontrivial cases on the parity of the coeﬃcients a, b and c, either (ν2(f(n)))n≥0 produces a ﬁnite tree or an inﬁnite tree. We say a ﬁnite tree has ℓlevels if there exists ℓ∈Z+ such that for all rℓ−1 ∈{0, 1, 2, . . . , 2ℓ−1} we have (ν2(f(2ℓq+rℓ−1)))q≥0 constant for all q ∈N, and ℓis the smallest possible value. Every node at level ℓin a ﬁnite tree has a constant valuation, which depends on rℓ−1. Each node of a tree represents a subsequence of the sequence of 2-adic valuations. A ﬁnite tree of ℓlevels represents a sequence with period equal to 2ℓ. In the literature, these ﬁnite trees are also called ﬁnite automata. The sequences gen-erated via the 2-adic valuation are called 2-automatic sequences and, in particular, the se-quences f(2iq + r) are known as the 2-kernel sequences. See Allouche and Shallit’s book and Bell’s paper for more details. 2.1 2-adic numbers and selected lemmas First, we state several well-known lemmas. The ﬁrst is a well-established fact about the p-adic valuation, which can also be deﬁned on the set Q and extends to Q2; see Lemma 3.3.2 in . Lemma 3. Let x, y ∈Q, then νp(xy) = νp(x) + νp(y). An element n in Q2 can be represented in the form n = ∞ X i=k αi2i (2) where k = −ν2(n) and αi ∈{0, 1} for all i; it is well-known that this representation is unique. Lemma 3 and the construction of Q2 are well-known . Medina et al. provide a useful characterization of the sequence of 2-adic valuations of a polynomial. Before we state the result, we recall the following characterization of the ring of 2-adic integers Z2 = ( n ∈Q2 : n = ∞ X i=0 αi2i where αi ∈{0, 1} ) . Lemma 4. (, Theorem 2.1) Let f(n) ∈Z[n] be a polynomial that is irreducible over Z. Then (ν2(f(n)))n≥0 is either periodic or unbounded. Moreover, (ν2(f(n)))n≥0 is periodic if and only if f(n) has no zeros in Z2. In the periodic case, the minimal period length is a power of 2. We assume that the quadratic f(n) is irreducible because, if not, by Lemma 3, νp(f(n)) = νp(g(n) · h(n)) = νp(g(n)) + νp(h(n)), where g(n) and h(n) are irreducible. 5 Therefore, to determine whether (ν2(f(n)))n≥0 is periodic or unbounded, it suﬃces to determine if f(n) has zeros in Q2 and then determine whether the zeros are also in Z2. The following lemmas will be used in Section 3 to identify when the square root of a number is in Z2. The version of Hensel’s lemma stated below determines when a polynomial in Z2[x] has zeros in Z2. Lemma 6, which follows from Lemma 5, speciﬁcally determines whether the polynomial f(x) = x2 −a has zeros in Z2. Lemma 5. (Hensel’s lemma, [7, Sec. 6.4]) Assume that P ∈Z2[x] and x0 ∈Z2 satisﬁes P(x0) ≡0 (mod 2n) If φ = ν2(P ′(x0)) < n/2, then there exists a unique zero ξ of P in Z2 such that ξ ≡x0 (mod pn−φ) and ν2(P ′(ξ)) = ν2(P ′(x0)) Lemma 6. ([7, Sec. 6.6]) The function f(x) = x2 −a has a zero in Z× 2 , the set of invertible elements of Z2, if and only if a ≡1 (mod 8). 3 Proof of Theorem 1: unbounded cases and inﬁnite trees We now prove Theorem 1. The main idea is to describe the roots to f(n) = 0 in Q2 simply using the quadratic formula, the parity of the coeﬃcients, and the lemmas presented in Section 2.1. Moreover, according to Lemma 4, if a zero exists in Z2, it manifests as an inﬁnite branch in the tree. We begin with Cases 2 and 4. To this end, note that in Case 2, we can write a = 2r and b = 2k + 1 for some r, k ∈Z. Then an2 + bn + c = 0 has roots of the form x = −2k −1 ± p 1 −8(rc −β) 4r , (3) where β = (k2 + k)/2. Set j = rc −β. Also, in Case 4, we can write a = 2r + 1, b = 2k + 1, and c = 2p. Then an2 + bn + c = 0 has roots of the form x = −2k −1 ± p 1 −8((2r + 1)p −β) 2(2r + 1) , (4) where β = (k2 + k)/2. Set j = (2r + 1)p −β. Observe that in either case the roots contain √1 −8j where j ∈Z. Since √1 −8j is a zero of the function g(x) = x2 −(1 −8j), by Lemma 6 the zero is in Z2. Notice that both roots (3) and (4) have an even denominator. We still need to check if these roots are in Q2 or Z2. Therefore, in light of Lemma 4, Case 2 (Proposition 7) and Case 4 (Proposition 8) are proved by an inductive argument on the behavior of the tree. It turns out that, in Case 2, f(n) has exactly one zero in Z2 and in Case 4, f(n) has two zeros in Z2. See Figure 3 in the Appendix for an example of a tree with one inﬁnite branch and Figure 4 for an example of a tree with two inﬁnite branches. 6 Proposition 7. If a is even and b is odd, then the 2-adic valuation tree of f(n) = an2+bn+c has exactly one inﬁnite branch. Furthermore, the valuation of the terminating node at the ith level is i −1. Proof. Note that this Proposition corresponds to Case 2 of Theorem 1. Substituting a = 2r and b = 2k + 1 for some r, k ∈Z, we get an2 + bn + c = 2(rn2 + kn) + n + c. Now suppose that c is even. If n is even, then 2(rn2 + kn) + n + c is divisible by 2 and so ν2(f(2n)) ≥1. If n is odd, then 2(rn2 + kn) + n + c is not divisible by 2 and so ν2(f(2n + 1)) = 0. An analogous argument shows that, for c odd, ν2(f(2n)) = 0 and ν2(f(2n + 1)) ≥1. Therefore, the conclusion of the proposition is valid at the initial step. Now, arguing inductively, suppose that n = 2iq + ri−1 is the non-terminating node, that is ν2(f(n)) ≥i. So f(n) ≡0 (mod 2i) or a(2iq +ri−1)2 +b(2iq +ri−1)+c = 2iβ where β ∈Z. Consider f(n) evaluated at the next level: a(2i+1q + ri−1)2 + b(2i+1q + ri−1) + c ≡ar2 i−1 + bri−1 + c ≡2iβ (mod 2i+1), and a(2i+1q + 2i + ri−1)2 + b(2i+1q + 2i + ri−1) + c ≡ar2 i−1 + 2ib + bri−1 + c ≡2iβ + 2ib ≡2i(β + b) (mod 2i+1). Since b is odd it follows that the valuation of one node is i and the other is greater than i depending on if β is odd or even. Therefore one node terminates and the other is non-terminating. Proposition 8. If a and b are odd, and c is even, then the 2-adic valuation tree of f(n) = an2 + bn + c has two inﬁnite branches. Furthermore, the valuation of the terminating node at the ith level is i. Proof. This proposition addresses Case 4 of Theorem 1. Write a = 2r + 1, b = 2k + 1, and c = 2p for some integers r, k, and p. First note that both a(2q)2 + b(2q) + c and a(2q + 1)2 + b(2q + 1) + c are congruent to 0 (mod 2). We now verify that the proposition holds at the initial step. In the 2q case, we check 4q and 4q + 2. Note that a(4q)2 + b(4q) + c ≡c (mod 4) and a(4q + 2)2 + b(4q + 2) + c ≡2b + c (mod 4). If c ≡0 (mod 4), then 2b + c ̸≡0 (mod 4). If c ̸≡0 (mod 4) then c = 2p with p odd and 2b + c = 2(b + p) ≡0 (mod 4). That is, either ν2(f(4q)) ≥2 and ν2(f(4q + 2)) = 1, or 7 ν2(f(4q)) = 1 and ν2(f(4q + 2)) ≥2. For the 2q + 1 case, we check 4q + 1 and 4q + 3. Note that a(4q + 1)2 + b(4q + 1) + c ≡a + b + c (mod 4) and a(4q + 3)2 + b(4q + 3) + c ≡a + 3b + c (mod 4). By hypothesis, a + b + c = 2(r + k + p) and a + 3b + c = 2(r + 3k + p + 2). But note that r + 3k + p + 2 = (r + k + p + 1) + (2k + 1). Now it is clear that r + 3k + p + 2 is even (odd) if and only if r + k + p + 1 is odd (even). Again, either ν2(f(4q + 1)) ≥2 and ν2(f(4q + 3)) = 1, or ν2(f(4q + 1)) = 1 and ν2(f(4q + 3)) ≥2. For the inductive step, now suppose that n = 2iq + ri−1 and n = 2iq + r∗ i−1 are the non-terminating nodes where ri−1 = Pi−1 k=1 αk2k + 1 (the odd side branch) and r∗ i−1 = Pi−1 k=1 αk2k (the even side branch) where αk ∈{0, 1}. The fact that these branches are non-terminating follows from the same argument as in the proof of Proposition 7. We now consider Case 3(b) of Theorem 1. Proposition 9. Let a be odd, b be even and b2 −4ac = 4ℓ∆for some ℓ∈Z+ as large as possible and ∆≡1 (mod 8), then the 2-adic valuation tree of f(n) = an2 + bn + c has two inﬁnite branches. Proof. Let a be odd and b = 2k for some k ∈Z. Fix ℓ∈Z+. Then an2 + bn + c = 0 has roots of the form x = −k± √ k2−ac a . By the hypothesis 4k2 −ac = 22ℓ∆where ∆≡1 (mod 8). If ∆< 0 then we can naturally write ∆= 1 −8j where j ∈{1, 2, 3, . . .}. If ∆> 0, then we can write ∆= 1 + 8j = 1 −8(−j) where j ∈N. Thus in either case ∆= 1−8j where j ∈Z. Then it follows that √ 4k2 −4ac = 2ℓ√1 −8j. By Lemma 6, √1 −8j is in Z2. Furthermore, since the denominator of x is odd this also guarantees that x ∈Z2. Therefore, there are two inﬁnite branches, one corresponding to each root. Corollary 10. Under the conditions of Proposition 9, if b2 −4ac = 0, the tree has one inﬁnite branch. Proof. In this case (3(a) of Theorem 1) roots take the form x = −b 2a. Since b = 2k, then x = −k a which has 2-adic form x = P∞ i=0 αi2i where αi is either 0 or 1. This guarantees that the one branch is inﬁnite. 8 Remark 11. Note the connection between subsequences of (ν2(f(n)))n≥0 and the inﬁnite branches of a tree. Proposition 7 asserts that for all i ∈Z+ there exists exactly one subse-quence of the form n = 2iq + ri−1 such that ν2(f(n)) ≥i and exactly one subsequence of the form n = 2iq +r∗ i−1 with ν2(f(n)) = i−1. Similarly, Proposition 8 asserts that for all i ∈Z+ there are exactly two subsequences corresponding to n = 2iq+ri−1 such that ν2(f(n)) ≥i+1 and exactly two subsequences of the form n = 2iq + r∗ i−1 with ν2(f(n)) = i. For ri−1 and r∗ i−1, the representations presented in equation (1) of Section 2 equate the coeﬃcients αk and α∗ k for all 0 ≤k ≤i −2, and meanwhile α∗ i−1 ≡αi−1 + 1 (mod 2). As for the cases of Proposition 9 and Corollary 10, we can apply Lemma 4 to conclude that these sequences are unbounded. Much like Propositions 7 and 8, we can say that the results of Proposition 9 yield that for all i ∈N there are exactly two subsequences of the form n = 2iq + ri−1, where (ν2(f(n)))n≥0 is not constant, while Corollary 10 asserts there is exactly one such subsequence. 4 Bounded cases and ﬁnite trees In this section, we prove Case 3(c) of Theorem 1 and the ﬁrst part of Theorem 2. The coeﬃcients of these quadratics satisfy the following: a is odd and b is even, and b2−4ac = 4ℓ∆, where ℓ∈Z+ is as large as possible, ∆≡m (mod 8), and m ∈{2, 3, 5, 6, 7}. Their trees are ﬁnite with ℓlevels. We can again apply the reasoning of the proof of Proposition 9. If ∆< 0 we can naturally write ∆= m −8j where j ∈N and if ∆> 0 then we write ∆= m + 8j = m −8(−j) where j ∈N or j = 0. Henceforth, we will write ∆= m −8j where j ∈Z. Again, by Lemma 6 functions of the form g(x) = x2 −(m −8j) do not have a zero in Z2. By Lemma 4, the corresponding valuation sequences are periodic. Figures 5 and 6 in the Appendix illustrate examples of ﬁnite trees arising from functions f3(n) = 15n2 + 1142n + 25559 and f4(n) = 5n2 + 106n + 1125. We should take a moment to note why we only need to consider these ﬁve values of m. First note that in Cases 3(b) and 3(c) of Theorem 1, where a is odd and b is even, we have the condition that ℓis as large as possible. This corresponds to factoring out as many powers of 4 as possible, ruling out the possibilities m ∈{0, 4}. Now if m = 1 (Case 3(b), covered in Section 3), an inﬁnite tree is created. This leaves the cases m ∈{2, 3, 5, 6, 7}. As discussed above, the zeros of these quadratic functions are not elements of Q2; therefore, their trees must be ﬁnite. The proofs of the next two propositions follow the proofs of Propositions 7 and 8. Proposition 12. If a is odd and b is even, and b2 −4ac = 4ℓ∆where ℓ∈Z+ is as large as possible, ∆≡m (mod 8), and m ∈{2, 3, 5, 6, 7}, then the 2-adic valuation tree of f(n) is ﬁnite with ℓlevels. The proof of this proposition is broken down into Lemmas 13, 14, and 16. Unless stated otherwise, let b = 2k for some k ∈Z. Lemma 13 covers the case ℓ= 1, in which the 2-adic valuation tree has exactly one level. Lemmas 14 and 16 describe valuations for ﬁnite 9 trees with more than one level; Lemma 16 describes the valuation at the ﬁnal level and Lemma 14 describes the other levels. Under the assumptions of Proposition 12, with a odd and b even, we complete the square and use properties of the p-adic valuation to obtain ν2(an2 + bn + c) = ν2((an + k)2 −k2 + ac). Lemma 13. Let ℓ= 1, i.e., b2 −4ac = 4∆, ∆≡m (mod 8), and m ∈{2, 3, 5, 6, 7}. If m ∈{2, 7} and b ≡0 (mod 4) or if m ∈{3, 6} and b ≡2 (mod 4), then ν2(an2 + bn + c) = ( 0, if n even; 1, if n odd. If m ∈{3, 6} and b ≡0 (mod 4) or if m ∈{2, 7} and b ≡2 (mod 4), then ν2(an2 + bn + c) = ( 1, if n even; 0, if n odd. If m = 5 and b ≡0 (mod 4), then ν2(an2 + bn + c) = ( 0, if n even; 2, if n odd. If m = 5 and b ≡2 (mod 4), then ν2(an2 + bn + c) = ( 2, if n even; 0, if n odd. Proof. Using the convention that ∆= m −8j where j ∈Z and m ∈{2, 3, 5, 6, 7}, consider the case where m = 7 and b ≡2 (mod 4). Then, since b = 2k, we have k odd. If n is even, then (an + k)2 ≡k2 (mod 2) and so it follows that ((an + k)2 −k2 + ac) ≡k2 −7 ≡−6 ≡0 (mod 2), but ((an + k)2 −k2 + ac) ≡k2 −7 ≡−6 ≡2 (mod 4). Therefore ν2(an2 + bn + c) = 1 when n is even. Similarly, when m = 7 and b ≡2 (mod 4) if n is odd, then (an + k)2 is even. Therefore, (an + k)2 −k2 + ac ≡−7 ≡1 (mod 2). Thus ν2(an2 + bn + c) = 0 when n is odd. Now consider the case where m = 7 and b ≡0 (mod 4). We have b = 2k with k even. Thus, if n is odd we have (an + k)2 −k2 + ac ≡−6 ≡0 (mod 2) and (an + k)2 −k2 + ac ≡k2 −7 ≡−6 ≡2 (mod 4). 10 Thus ν2(an2 + bn + c) = 1 when n is odd. When n is even we have (an + k)2 −k2 + ac ≡ −7 ≡1 (mod 2). Thus ν2(an2 + bn + c) = 0 when n is even. The cases of m ∈{2, 3, 6} when b ≡0 (mod 4) or b ≡2 (mod 4) can be handled in the same fashion. For m = 5, the valuations are slightly diﬀerent. Consider the case where m = 5. Recall that b = 2k for some k ∈Z. Note that b2 −4ac = 4(5 −8j), and hence k2 −ac = 5 −8j. Thus, (an + k)2 −k2 + ac = (an + k)2 −5 + 8j. If (an + k) is even, which is the case when both n and k are even or both n and k are odd, then (an + k)2 −5 + 8j is odd. Thus, ν2(an2 + bn + c) = 0. Now suppose that (an+k) is odd, which is true when n and k have diﬀerent parity. Then (an + k)2 ≡1 (mod 4), and this implies (an + k)2 −5 + 8j ≡1 −5 + 8j ≡−4 + 8j ≡0 (mod 4). Thus, ν2(an2 + bn + c) ≥2. Since (an + k) is odd, let an + k = 2d + 1, for some d ∈Z. Then, (an + k)2 −5 + 8j = (2d + 1)2 −5 + 8j ≡4(d2 + d −1) (mod 8). Observe that d2+d−1 is odd, regardless of whether d is even or odd. Thus, ν2(an2+bn+c) < 3. Therefore, ν2(an2 + bn + c) = 2. Lemma 14. Under the assumptions of Proposition 12 (Case 3(c) of Theorem 1) let ℓ≥ 2 and suppose 0 < i < ℓ. At the ith level there is one terminal and one non-terminal node. Furthermore, the terminal node has valuation 2(i −1) and the non-terminal node has valuation at least 2i. First we need: Claim 15. Let a, k ∈Z with a odd. Let g(n) = an + k, then (ν2(g(n)))n≥0 creates an unbounded sequence. Proof. First note that the root of ax+k = 0 is x = −k a. Also note that ν2(x) = ν2(−k)−ν2(a). Since a is odd, ν2(a) = 0. Therefore ν2(x) = ν2(−k) ≥0. By equation (2), x ∈Z2, so Lemma 4 implies that (ν2(g(n)))n≥0 is an unbounded sequence. Proof. To prove Lemma 14, we proceed by an inductive argument on i. Again, using the convention that ∆= m−8j where j ∈Z, for the base case i = 1, note that b2−4ac ≡4ℓ(m− 8j) ≡0 (mod 4). Recall that b = 2k. First, assume that k is even. If n is even, then an+k is 11 even and so (an+k)2−k2+ac ≡0 (mod 4). Thus ν2(an2+bn+c) = ν2((an+k)2−k2+ac) ≥2 by Claim 15. If n is odd, then (an+k)2 −k2 +ac ≡1 (mod 2), and again using the technique of completing the square, ν2(an2 + bn + c) = 0. If k is odd, a similar argument shows that ν2(an2 + bn + c) ≥2 when n is odd. Observe also that Claim 15 can be used to show that (ν2((an+k)2))n≥0 forms an unbounded sequence therefore ν2((an+k)2 −k2 +ac) ≥2. Thus, the claim is true for i = 1. For the inductive step, notice that since i < ℓ, it follows that b2 −4ac ≡4ℓ(m −8j) ≡0 (mod 22i). Suppose there exists an i −1 ≥0 such that n = 2i−1q + ri−2 splits into two nodes: one node terminating with valuation 2(i−1) and the other node having valuation of at least 2i. We let n = 2iq + ri−1 denote the non-terminating node, where ri−1 = Pi−1 h=0 αh2h with αh ∈{0, 1}, for all 0 ≤h ≤i −2, and q ∈Z. Then we have (an + k)2 −k2 + ac ≡(a(2iq + ri−1) + k)2 ≡0 (mod 22i), so ν2(an2 + bn + c) ≥2i. This also implies that a(2iq + ri−1) + k ≡0 (mod 2i). Thus ari−1 + k = 2iβ where β ∈Z. Now suppose that k is even. (The proof for k odd can be handled in the same fashion, and thus is omitted.) Since k is even, then ri−1 must be even. Consider the (i + 1)st level where i + 1 < ℓ. Here again we have b2 −4ac = 4ℓ(m −8j) (mod 22(i+1)) ≡0. Moving to the next level, in the case n = 2i+1q + ri−1 we have ν2((an + k)2 −4ℓ−1(m −8j)) = ν2((a(2i+1q + ri−1) + k)2 −4ℓ−1(m −8j)) = ν2((2i+1aq + ari−1 + k)2 −4ℓ−1(m −8j)) = ν2((2i+1aq + 2iβ)2 −4ℓ−1(m −8j)) = ν2(22i(2aq + β)2 −22(ℓ−1)(m −8j)), and in the case n = 2i+1q + 2i + ri−1 we have ν2((an + k)2 −4ℓ−1(m −8j)) = ν2((a(2i+1q + 2i + ri−1) + k)2 −4ℓ−1(m −8j)) = ν2((2i+1aq + 2ia + ari−1 + k)2 −4ℓ−1(m −8j)) = ν2((2i+1aq + 2ia + 2iβ)2 −4ℓ−1(m −8j)) = ν2(22i(2aq + a + β)2 −22(ℓ−1)(m −8j)). Since β ∈Z either 2aq+β or 2aq+a+β is odd and the other is even. As long as i+1 < ℓthen in the odd case the valuation is 2i and in the even case the valuation is at least 2(i + 1). 12 Lemma 16. If a is odd and b is even with b = 2k for k ∈Z, and b2 −4ac = 4ℓ∆where ℓ∈Z+ is as large as possible, ∆≡m (mod 8), and m ∈{2, 3, 5, 6, 7}, then at the ℓth level the nodes of the 2-adic valuation tree terminate with valuations of 2(ℓ−1), 2ℓ−1 or 2ℓ. Suppose that n = 2ℓq + rℓ−2. If an + k ≡0 (mod 2ℓ), then ν2(f(n)) = ( 2(ℓ−1), if m = 7, 5, 3; 2ℓ−1, if m = 6, 2; and if an + k ̸≡0 (mod 2ℓ), then ν2(f(n)) =      2(ℓ−1), if m = 6, 2; 2ℓ−1, if m = 7, 3; 2ℓ, if m = 5. Suppose that n = 2ℓq + 2ℓ−1 + rℓ−2. If an + k ≡0 (mod 2ℓ), then ν2(f(n)) =      2(ℓ−1), if m = 6, 2; 2ℓ−1, if m = 7, 3; 2ℓ, if m = 5; and if an + k ̸≡0 (mod 2ℓ), then ν2(f(n)) = ( 2(ℓ−1), if m = 7, 5, 3; 2ℓ−1, if m = 6, 2. Proof. By Lemma 14 there exists a non-terminating node n = 2ℓ−1q + rℓ−2 with q ∈Z and ν2((an + k)2 −k2 + ac) ≥2(ℓ−1). Consider n = 2ℓq + rℓ−2 with q ∈Z. By the same argument as in Lemma 14 and using the convention that ∆= m −8j where j ∈Z, we have (an + k)2 −k2 + ac = (2ℓaq + 2ℓ−1β)2 −22(ℓ−1)(m −8j) = 22(ℓ−1)((2aq + β)2 + 8j −m), where β ∈Z. Recall that a is odd. Then depending on whether β is even or odd, simple calculations show the ﬁrst two results. In the case when n = 2ℓq + 2ℓ−1 + rℓ−2 with q ∈Z we have (an + k)2 −k2 + ac = 22(ℓ−1)((2aq + a + β)2 + 8j −m), where β ∈Z. Then again depending on whether β is odd or even, it is straightforward to show the last two results. 13 5 Structure of ﬁnite trees The section describes the overall structure of ﬁnite trees, continuing the discussion of Case 3(c) of Theorem 1, in which a is odd, b is even, b2 −4ac = 4ℓ∆where ∆≡m (mod 8), and m ∈{2, 3, 5, 6, 7}. Throughout this section, we make use of several operators. The operators allow us to track changes from very easily described trees, which we call type (ℓ, 1), to more complicated trees. Deﬁnition 17 (Translation operator, ). For quadratics of the form f(n) = an2 + bn + c we deﬁne τ s(f)(n) = f(n −s) for s ∈R, namely τ s(f)(n) = a(n −s)2 + b(n −s) + c = an2 + (b −2as)n + (c + as2 −bs). Proposition 18. Let the assumptions of Proposition 12 hold for the function f(n) = an2 + bn + c and suppose s ∈Z. Then we have the following relationship ν2(f(2iq + ri−1)) = ν2(τ sf(2iq + (ri−1 + s) mod 2i)). That is the valuations ν2(f(n)) at the node of the form n = 2iq + ri−1 are moved to the node of the form n = 2iq + (ri−1 + s) (mod 2i) under the operation τ s. Proof. Note that ﬁnite trees with ℓlevels correspond to periodic sequences with a period equal to 2ℓ. Since τ s is a translation operator, every element in the sequence (ν2(f(n)))n≥0 is moved over s spaces. Deﬁnition 19 (S-operator). Let a be a positive, odd integer. For quadratics of the form f(n) = n2 + bn + ac we deﬁne Sa(f)(n) = an2 + bn + c. Likewise, for quadratics of the form f(n) = an2 + bn + c deﬁne Sa−1(f)(n) = n2 + bn + ac. In general, the S-operator need not output a quadratic function with an integer constant term. However, the present work only applies Sa to functions whose output has integer coeﬃcients. Deﬁnition 20 (Dilation operator, ). For quadratics of the form f(n) = an2 + bn + c we deﬁne δs(f)(n) = f(sn) for s ∈R, namely δs(f)(n) = a(sn)2 + b(sn) + c. Lemma 21. Under the assumptions of Proposition 12 the trees created by f(n) = n2+bn+ac and Sa(f)(n) where a ∈Z have the same number of levels. Similarly, the trees created by g(n) = an2 + bn + c and τ s(g)(n) where s ∈Z have the same number of levels. Proof. The assumptions of Proposition 12 represent Cases 3(b) and 3(c) of Theorem 1. Simple calculations show that the discriminants of f(n) and Sa(f)(n) are the same, and that the discriminants of g(n) and τ s(g)(n) are the same. The conclusions then follow directly from Proposition 12. 14 Proposition 22. Let the assumptions of Proposition 12 hold and suppose f(n) = n2+bn+ac. Then we have the following relationship ν2(f(2iq + ri−1)) = ν2(Sa(f(2iq + a−1 · ri−1))). That is the valuation ν2(f(n)) at the node the form n = 2iq + ri−1 is moved to the node of the form of n = 2iq + (a−1 · ri−1) (mod 2i) under the operation Sa. In this context a−1 is the inverse of a (mod 2i). Proof. Since a is odd, note that ν2(Sa(f)(n)) = ν2((an2 + bn + c)) = ν2((an)2 + b(an) + ac) = ν2(δa(f)(n)), where δa(f)(n) = f(an) is the dilation operator. Thus, the valuation of f(n) for n = 2iq+ri−1 is the same as the valuation of n′ = 2i(a−1q) + a−1 · ri−1 after the Sa-operator is applied. Suppose that f(n) = an2 + bn + c creates a ﬁnite tree. We say that this tree is type (ℓ, 1), for ℓ≥2, if at every level the non-terminating node is of the form n = 2q or n = 2iq + 2i−2 + · · · + 21 + 20 for i < ℓand the tree has ℓlevels. We also say that a quadratic function is type (ℓ, 1) if it creates an (ℓ, 1) tree. That is, f(n) creates a ﬁnite tree of the following form: Figure 2: The form of trees of type (ℓ, 1). Here, we suppose that ℓ≥2 because ℓ= 1 creates a tree with one level, see Lemma 13, and the directional behavior we seek to classify is not deﬁned. The conditions 4a2−4ac = 4ℓ∆ for ℓ∈Z+ as large as possible, ∆≡m (mod 8), and m ∈{2, 3, 5, 6, 7} imply c must be odd. 15 Proposition 23. Under the assumptions of Proposition 12, if c is odd and ℓ≥2 is an integer, then a quadratic of the form f(n) = an2 +2an+c creates a tree that is of type (ℓ, 1). Furthermore, we have ν2(f(n)) =                          0, if n ≡0 (mod 2); 2(i −1), if n ≡Pi−2 k=0 2k (mod 2i) with 2 ≤i < ℓ; 2(ℓ−1), if n ≡Pℓ−2 k=0 2k (mod 2ℓ) and m = 6, 2; 2ℓ−1, if n ≡Pℓ−2 k=0 2k (mod 2ℓ) and m = 7, 3; 2ℓ, if n ≡Pℓ−2 k=0 2k (mod 2ℓ) and m = 5; 2ℓ−1, if n ≡Pℓ−1 k=0 2k (mod 2ℓ) and m = 6, 2; 2(ℓ−1), if n ≡Pℓ−1 k=0 2k (mod 2ℓ) and m = 7, 5, 3. Proof. In light of Lemma 14, we know that if a node is non-terminating, then it produces two nodes that either both terminate (i.e., these nodes are at the ℓth level) or one node is non-terminating and the other is terminating. So in order to show that the tree is of type (ℓ, 1), we only need to conﬁrm that nodes corresponding to n = 2iq + 2i−1 + · · · + 21 + 20, where 1 ≤i ≤ℓ, are always non-terminating. Since a is odd, completing the square and using the convention that 4a2 −4ac = 4ℓ∆where ∆= m −8j where j ∈Z gives ν2(f(n)) = ν2(an2 + 2an + c) = ν2(a(n + 1)2 −a + c) = ν2(a2(n + 1)2 −a2 + ac) = ν2(a2(n + 1)2 −4ℓ−1(m −8j)) = ν2(a2(2iq + 2i−1 + 2i−2 + · · · + 2 + 1 + 1)2 −4ℓ−1(m −8j)) = ν2(a2(2iq + 2i)2 −4ℓ−1(m −8j)) = ν2(a24i(q + 1)2 −4ℓ−1(m −8j)). If q is odd, then n = 2iq+2i−1+· · ·+21+20 is the non-terminating node, provided i < ℓ, and produces two nodes one of which does not terminate. If i = ℓ, then both nodes terminate by Proposition 12. The nodes that terminate are of the form n = 2iq + 2i−2 + · · · + 21 + 20 when 1 ≤i < ℓ. The case when n = 2q is handled by the proof of Lemma 21. For the case 1 < i < ℓ, by the same calculation as above we have ν2(f(n)) = ν2(a222(i−1)(2q + 1)2 −4ℓ−1(m −8j)) Since 2q + 1 is odd and i < ℓthe valuation must be 2(i −1). In the case when n = 2ℓq + 2ℓ−2 + · · · + 21 + 20 we have ν2(f(n)) = ν2(a222(ℓ−1)(2q + 1)2 −4ℓ−1(m −8j)) Thus the valuation must be 2(ℓ−1) if m = 6, 2, or 2ℓ−1 if m = 7, 3 or 2ℓif m = 5. Finally if n = 2ℓq+2ℓ−2+· · ·+21+20 we have ν2(f(n)) = ν2(a222ℓ(q+1)2−4ℓ−1(m−8j)). Thus the valuation must be 2(ℓ−1) if m = 7, 5, 3 or 2ℓ−1 if m = 6, 2. 16 If the function f(n) = an2 + bn + c meets the assumptions of Proposition 12 (Case 3(c) of Theorem 1) note if we deﬁne the function g(n) = n2 + 2n − 1 −b 2 2 + 2 1 −b 2 + ac, then it follows that Sa(τ 1−b 2(g))(n) = f(n). Therefore, by Propositions 18, 22, and 23 we immediately have the following corollary. Corollary 24. If f(n) = an2 + bn + c meets the assumptions of Proposition 12 (Case 3(c)) with ℓ≥2, then ν2(f(n)) =                          0, if n ≡a−1 1 −b 2 (mod 2); 2(i −1), if n ≡a−1 2i−1 −b 2 (mod 2i) with 2 ≤i < ℓ; 2(ℓ−1), if n ≡a−1 2ℓ−1 −b 2 (mod 2ℓ)and m = 6, 2; 2ℓ−1, if n ≡a−1 2ℓ−1 −b 2 (mod 2ℓ) and m = 7, 3; 2ℓ, if n ≡a−1 2ℓ−1 −b 2 (mod 2ℓ) and m = 5; 2ℓ−1, if n ≡a−1 2ℓ−b 2 (mod 2ℓ) and m = 6, 2; 2(ℓ−1), if n ≡a−1 2ℓ−b 2 (mod 2ℓ) and m = 7, 5, 3; where a−1 is the inverse of a (mod 2ℓ). Proof. Simply note that g is type (ℓ, 1) and recall the ways in which the operators aﬀect the function g. Each terminating node, under the operators, moves from n = 2iq + ri−2 to n = 2iq + a−1 ri−2 + 1 −b 2 (mod 2i). In the case of type (ℓ, 1) we have ri−2 = Pi−2 k=0 2k. Thus ri−2 + 1 = 2i−1 in each case. 6 Acknowledgments The authors would like to thank Dr. Victor Moll for suggesting this topic. We would also like to thank following institutions for providing support to collaborate: ICERM, AIM, Kentucky Wesleyan College, Ursinus College, and Stephen F. Austin State University. We are grateful to our other colleagues for their ongoing support: Dr. Maila Brucal-Hallare, Dr. Jean-Claude Pedjeu, and Dr. Bianca Thompson. And, ﬁnally, we are very grateful to the reviewer who provided many very helpful and insightful comments. Appendix: ﬁgures illustrating trees and tables of values for 2-adic valuation sequences of some quadratic func-tions In the following tree representations, a closed circle indicates a terminating node and an open circle indicates a non-terminating node. 17 n 0 1 2 3 4 5 6 7 8 9 10 11 f1(n) −25 −8 17 50 91 140 197 262 335 416 505 602 ν2(f12(n)) 0 3 0 1 0 2 0 1 0 5 0 1 n 12 13 14 15 16 17 18 19 20 21 f1(n) 707 820 941 1070 1207 1352 1505 1666 1835 2012 ν2(f1(n)) 0 2 0 1 0 3 0 1 0 2 Figure 3: The 2-adic valuation tree for f1(n) = 4n2 + 13n −25. Theorem 1 predicts that (ν2(f1(n)))n≥0 is an unbounded sequence, as it satisﬁes Case 2. 18 n 0 1 2 3 4 5 6 7 8 9 10 f2(n) −28 −3 48 125 228 357 512 693 900 1133 1392 ν2(f2(n)) 2 0 4 0 2 0 9 0 2 0 4 n 11 12 13 14 15 16 17 18 19 f2(n) 1677 1988 2325 2688 3077 3492 3933 4400 4893 ν2(f2(n)) 0 2 0 7 0 2 0 4 0 Figure 4: The 2-adic valuation tree and data for f2(n) = 13n2 + 12n −28. Notice that Theorem 1 predicts that (ν2(f2(n)))n≥0 is an unbounded sequence, as it satisﬁes Case 3(a) since 122 −4 · 13(−28) = 43(1 −8(−3)). 19 n 0 1 2 3 4 5 6 7 f3(n) 25559 26716 27903 29120 30367 31644 32951 34288 ν2(f3(n)) 0 2 0 6 0 2 0 4 n 8 9 10 11 12 13 14 15 f3(n) 35655 37052 38479 39936 41423 42940 44487 46064 ν2(f3(n)) 0 2 0 10 0 2 0 4 Figure 5: The 2-adic valuation tree and data for f3(n) = 15n2 + 1142n + 25559. Notice that Theorem 1 predicts that (ν2(f3(n))n≥0 is a bounded sequence, as it satisﬁes Case 3(c) since 11422 −4 · 15 · 25559 = 47(2 −8 · 2). 20 n 0 1 2 3 4 5 6 7 8 9 f4(n) 1125 1236 1357 1488 1629 1780 1941 2112 2293 2484 ν2(f4(n)) 0 2 0 4 0 2 0 6 0 2 n 10 11 12 13 14 15 16 17 18 19 f4(n) 2685 2896 3117 3348 3589 3840 4101 4372 4653 4944 ν2(f4(n)) 0 4 0 2 0 8 0 2 0 4 Figure 6: The 2-adic valuation tree and data for f4(n) = 5n2 + 106n + 1125. Notice that Theorem 1 predicts that (ν2(f4(n))n≥0 is a bounded sequence, as it satisﬁes Case 3(c) since 1062 −4 · 5 · 1125 = 45(5 −8 · 2). References J.-P. Allouche and J. Shallit, Automatic Sequences, Theory, Applications, Generaliza-tions, Cambridge University Press, 2003. J. P. Bell, p-adic valuations and k-regular sequences, Discrete Math. 307 (2007), 3070– 3075. A. N. Byrnes, J. Fink, G. Lavigne, I. Nogues, S. Rajasekaran, A. Yuan, L. Almodovar, X. Guan, A. Kesarwani, L. A. Medina, E. Rowland, and V. H. Moll, A closed-form solution might be given by a tree. Valuations of quadratic polynomials, Sci. Ser. A Math. Sci. 29 (2019), 11–28. 21 L. Grafakos, Classical Fourier Analysis, Second Edition, Springer, 2008. F. Gouvˆ ea, p-Adic Numbers: An Introduction, Springer-Verlag, 1997. L. A. Medina, V. H. Moll, and E. Rowland, Periodicity in the p-adic valuation of a polynomial, J. Number Theory 180 (2017), 139–153. A. M. Robert, A Course in p-adic Analysis, Springer, 2000. 2010 Mathematics Subject Classiﬁcation: Primary 11B50; Secondary 11S05. Keywords: p-adic valuation, polynomial sequences, p-adic tree. Received June 15 2020; revised versions received June 30 2020; March 12 2021; May 5 2021; May 14 2021. Published in Journal of Integer Sequences, May 17 2021. Return to Journal of Integer Sequences home page. 22
16201	https://math.stackexchange.com/a/1999458	functions - Why is $f:\mathbb{R} \rightarrow [-1,1]$ , $x$ mapped to $\sin(x)$, surjective? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Why is f:R→[−1,1]f:R→[−1,1] , x x mapped to sin(x)sin⁡(x), surjective? [closed] Ask Question Asked 8 years, 11 months ago Modified8 years, 11 months ago Viewed 128 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Closed. This question is off-topic. It is not currently accepting answers. This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. Closed 8 years ago. Improve this question I'm a little confused with the x x mapped to sin(x)sin⁡(x) and the [−1,1][−1,1]. functions trigonometry Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Nov 4, 2016 at 17:26 Lee Mosher 137k 7 7 gold badges 105 105 silver badges 218 218 bronze badges asked Nov 4, 2016 at 16:54 James MitchellJames Mitchell 491 1 1 gold badge 5 5 silver badges 17 17 bronze badges 2 Because ... for every −1≤y≤1−1≤y≤1 there is a θ θ so that sin θ=y sin⁡θ=y? I don't understand what you are asking.fleablood –fleablood 2016-11-04 17:03:35 +00:00 Commented Nov 4, 2016 at 17:03 I edited your tags; this has nothing to do with "discrete geometry".Lee Mosher –Lee Mosher 2016-11-04 17:27:06 +00:00 Commented Nov 4, 2016 at 17:27 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Note that f:R→[−1,1]f:R→[−1,1] defines by x↦sin(x)x↦sin⁡(x) means that f f is the function taking a real number x x to the number sin(x)sin⁡(x) whose value lies in the interval [−1,1][−1,1]. The intermediate value theorem states that if a continuous function, f f, with an interval, [a,b][a,b], as its domain, takes values f(a)f(a) and f(b)f(b) at each end of the interval, then it also takes any value between f(a)f(a) and f(b)f(b) at some point within the interval. If you can convince yourself that sin(x)sin⁡(x) is continuous on R R, than you just need to take a:=π 2 a:=π 2 and b:=3 π 2 b:=3 π 2. You can check that f(a)=1 f(a)=1 and f(b)=−1 f(b)=−1. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 4, 2016 at 17:02 M.GM.G 3,799 22 22 silver badges 43 43 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Perhaps a review of the geometric definition of the sine function would help. As is done in the picture linked, I'll use θ θ as the input angle (x x is used for a different purpose, as the first coordinate in the Cartesian plane). From the picture linked above, if you draw a line segment O P¯¯¯¯¯¯¯¯O P¯ with one endpoint at the origin O=(0,0)O=(0,0) and opposite endpoint at P=(x,y)P=(x,y) on the unit circle, and if θ θ is equal to the signed angle from the positive x x-axis to this line segment, then sin(θ)=y sin⁡(θ)=y. From the picture linked, you can see that any y y-coordinate between −1−1 and +1+1 can be obtained in this way from some appropriate choice of θ θ: draw the horizontal line L L with that y y-coordinate, take P P to be the point where L L intersects the right half of the unit circle, and take θ θ to be the signed angle from the positive x x-axis to the segment O P¯¯¯¯¯¯¯¯O P¯. In other words, for each y∈[−1,+1]y∈[−1,+1] there exists θ θ such that sin(θ)=y sin⁡(θ)=y, which proves surjectivity. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 4, 2016 at 17:26 Lee MosherLee Mosher 137k 7 7 gold badges 105 105 silver badges 218 218 bronze badges Add a comment\| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions functions trigonometry See similar questions with these tags. 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16202	https://mathinsight.org/local_minima_maxima_refresher	Local minima and maxima (First Derivative Test) - Math Insight Skip to navigation (Press Enter) Skip to main content (Press Enter) Home Threads Index About Math Insight Page Navigation Top In threads Calculus Refresher Links Similar pages Contact us To create your own interactive content like this, check out our new web site doenet.org! Local minima and maxima (First Derivative Test) A function f has a local maximum or relative maximum at a point x o if the values f(x) of f for x ‘near’ x o are all less than f(x o). Thus, the graph of f near x o has a peak at x o. A function f has a local minimum or relative minimum at a point x o if the values f(x) of f for x ‘near’ x o are all greater than f(x o). Thus, the graph of f near x o has a trough at x o. (To make the distinction clear, sometimes the ‘plain’ maximum and minimum are called absolute maximum and minimum.) Yes, in both these ‘definitions’ we are tolerating ambiguity about what ‘near’ would mean, although the peak/trough requirement on the graph could be translated into a less ambiguous definition. But in any case we'll be able to execute the procedure given below to find local maxima and minima without worrying over a formal definition. This procedure is just a variant of things we've already done to analyze the intervals of increase and decrease of a function, or to find absolute maxima and minima. This procedure starts out the same way as does the analysis of intervals of increase/decrease, and also the procedure for finding (‘absolute’) maxima and minima of functions. To find the local maxima and minima of a function f on an interval [a,b]: Solve f′(x)=0 to find critical points of f. Drop from the list any critical points that aren't in the interval [a,b]. Add to the list the endpoints (and any points of discontinuity or non-differentiability): we have an ordered list of special points in the interval: a=x o<x 1<…<x n=b Between each pair x i<x i+1 of points in the list, choose an auxiliary point t i+1. Evaluate the derivative f′ at all the auxiliary points. For each critical point x i, we have the auxiliary points to each side of it: t i<x i<t i+1. There are four cases best remembered by drawing a picture!: if f′(t i)>0 and f′(t i+1)<0 (so f is increasing to the left of x i and decreasing to the right of x i, then f has a local maximum at x o. if f′(t i)<0 and f′(t i+1)>0 (so f is decreasing to the left of x i and increasing to the right of x i, then f has a local minimum at x o. if f′(t i)<0 and f′(t i+1)<0 (so f is decreasing to the left of x i and also decreasing to the right of x i, then f has neither a local maximum nor a local minimum at x o. if f′(t i)>0 and f′(t i+1)>0 (so f is increasing to the left of x i and also increasing to the right of x i, then f has neither a local maximum nor a local minimum at x o. The endpoints require separate treatment: There is the auxiliary point t o just to the right of the left endpoint a, and the auxiliary point t n just to the left of the right endpoint b: At the left endpoint a, if f′(t o)<0 (so f′ is decreasing to the right of a) then a is a local maximum. At the left endpoint a, if f′(t o)>0 (so f′ is increasing to the right of a) then a is a local minimum. At the right endpoint b, if f′(t n)<0 (so f′ is decreasing as b is approached from the left) then b is a local minimum. At the right endpoint b, if f′(t n)>0 (so f′ is increasing as b is approached from the left) then b is a local maximum. The possibly bewildering list of possibilities really shouldn't be bewildering after you get used to them. We are already acquainted with evaluation of f′ at auxiliary points between critical points in order to see whether the function is increasing or decreasing, and now we're just applying that information to see whether the graph peaks, troughs, or does neither around each critical point and endpoints. That is, the geometric meaning of the derivative's being positive or negative is easily translated into conclusions about local maxima or minima. Find all the local (=relative) minima and maxima of the function f(x)=2 x 3−9 x 2+1 on the interval [−2,2]: To find critical points, solve f′(x)=0: this is 6 x 2−18 x=0 or x(x−3)=0, so there are two critical points, 0 and 3. Since 3 is not in the interval we care about, we drop it from our list. Adding the endpoints to the list, we have −2<0<2 as our ordered list of special points. Let's use auxiliary points −1,1. At −1 the derivative is f′(−1)=24>0, so the function is increasing there. At +1 the derivative is f′(1)=−12<0, so the function is decreasing. Thus, since it is increasing to the left and decreasing to the right of 0, it must be that 0 is a local maximum. Since f is increasing to the right of the left endpoint −2, that left endpoint must give a local minimum. Since it is decreasing to the left of the right endpoint +2, the right endpoint must be a local minimum. Notice that although the processes of finding absolute maxima and minima and local maxima and minima have a lot in common, they have essential differences. In particular, the only relations between them are that critical points and endpoints (and points of discontinuity, etc.) play a big role in both, and that the absolute maximum is certainly a local maximum, and likewise the absolute minimum is certainly a local minimum. For example, just plugging critical points into the function does not reliably indicate which points are local maxima and minima. And, on the other hand, knowing which of the critical points are local maxima and minima generally is only a small step toward figuring out which are absolute: values still have to be plugged into the function! So don't confuse the two procedures! (By the way: while it's fairly easy to make up story-problems where the issue is to find the maximum or minimum value of some function on some interval, it's harder to think of a simple application of local maxima or minima). Exercises Find all the local (=relative) minima and maxima of the function f(x)=(x+1)3−3(x+1) on the interval [−2,1]. Find the local (=relative) minima and maxima on the interval [−3,2] of the function f(x)=(x+1)3−3(x+1). Find the local (relative) minima and maxima of the function f(x)=1−12 x+x 3 on the interval [−3,3]. Find the local (relative) minima and maxima of the function f(x)=3 x 4−8 x 3+6 x 2+17 on the interval [−3,3]. Thread navigation Calculus Refresher Previous: Minimization and maximization refresher Next: An algebra trick for finding critical points Similar pages Minimization and maximization refresher Minimization and maximization problems Solutions to minimization and maximization problems Maximization and minimization Introduction to local extrema of functions of two variables Two variable local extrema examples An algebra trick for finding critical points The idea of the derivative of a function Derivatives of polynomials Derivatives of more general power functions More similar pages Cite this as Garrett P, “Local minima and maxima (First Derivative Test).” From Math Insight. Keywords: critical points, extrema, maxima, minima Send us a message about “Local minima and maxima (First Derivative Test)” Name: Email address: Comment: If you enter anything in this field your comment will be treated as spam: Local minima and maxima (First Derivative Test) by Paul Garrett is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. For permissions beyond the scope of this license, please contact us. Credits The page is based off the Calculus Refresher by Paul Garrett.
16203	https://www.ncbi.nlm.nih.gov/books/NBK441884/	Genital Warts - StatPearls - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Search term Genital Warts Stephen W. Leslie; Hussain Sajjad; Sandeep Kumar. Author Information and Affiliations Authors Stephen W. Leslie 1; Hussain Sajjad 2; Sandeep Kumar 3. Affiliations 1 Creighton University School of Medicine 2 RMU and Allied Hospitals 3 VA Medical Center Last Update: May 30, 2023. Go to: Continuing Education Activity Genital warts (condyloma acuminatum) are a sexually transmitted infection caused by the human papillomavirus (HPV) types 6 and 11. These are spread by skin-to-skin contact, usually during sex. These present in clusters or separately and can be found in the genital or anal area. This activity describes the evaluation and management of genital warts and explains the role of the interprofessional team in improving care for patients with this condition. Objectives: Outline the etiology of genital warts. Review the importance of biopsy in the evaluation of genital warts. Explain the use of physically destructive therapies and topical agents in the management of genital warts. Summarize the importance of collaboration and communication among the interprofessional team members to educate patients on the importance of the HPV vaccine, which will enhance the delivery of care for patients with genital warts. Access free multiple choice questions on this topic. Go to: Introduction Genital warts (condyloma acuminatum) are the clinical manifestations of a sexually transmitted infection caused by some types of human papillomavirus (HPV). Warts are a recognized symptom of genital HPV infections. About 90% of those exposed who contract HPV will not develop genital warts. Only about 10% who are infected will transmit the virus. HPV types 6 and 11 cause genital warts. There are over 100 different known types of HPV viruses.HPV is spread through direct skin-to-skin contact with an infected individual, usually during sex. While some types of HPV cause cervical and anal cancer, these are not the same viral types that cause genital warts. It is possible to be infected with different types of HPV at the same time. Go to: Etiology HPV is transmitted primarily through penetrative sex. While HPV also can be transmitted via non-penetrative sexual activity, it is less common. There is conflicting evidence about the effect of condoms on prevention. About 30% of genital warts will disappear within four months of their initial appearance. Most genital warts will recur within three months of completion of initial therapy, even if therapy was followed correctly. Recurrence rates depend on the patient's general health and immune status, previous HPV vaccinations, specific HPV strain, number of inoculations (sexual frequency with an infected partner), use of condoms, and the viral load. Smoking increases the risk of getting genital warts. Approximately three out of four unaffected partners of patients with warts develop them within eight months of contact. Although 90% of HPV infections are cleared within two years of infection, it is possible for a latency period to occur, with the first occurrence or a recurrence happening months or even years later. Latent HPV is transmissible, and if an individual has unprotected sex with an infected partner, there is a 70% chance they will become infected. In individuals with a prior HPV infection, the appearance of new warts may be either from a new exposure or a recurrence. Anal or genital warts may be transmitted during birth and may be an indicator of sexual abuse. Genital warts may sometimes result from autoinoculation by warts elsewhere on the body, such as from the hands. Go to: Epidemiology Genital HPV infections have an estimated prevalence of 10% to 20%, with clinical manifestations in 1%. The incidence of HPV infection has been increasing. About 80% of those infected are between the ages of 17 and 33 years, with the peak age group being 20 to 24. It has been estimated that 2.9% of the US male population will have genital HPV DNA. Although treatments can remove warts, they do not remove HPV. Warts may sometimes spontaneously regress. Traditional theories postulate that the virus remains in the body for a lifetime. However, it is now believed that the virus may be either cleared or suppressed to levels below what polymerase chain reaction (PCR) tests can measure. HPV infection appears to be the cause of most cases of anal cancer (about 90%) and virtually all cases of cervical cancer in women, with HPV type 16 accounting for about 50% of these. (Cervical cancer is the fourth most common cancer in women.) Some vulvar cancers have been linked to HPV infections (29% to 43%),while vaginal cancer is associated with HPV infections about 70% of the time (HPV Types 16 and 18). In men,Bowen disease of the penis and about 35% to 40% of all penile cancers are associated with HPV infections. Risk factors for HPV persistence include age, smoking, immunosuppression, and simultaneous infection with multiple HPV types. HPV Vaccinations An HPV vaccination is available. For previously unvaccinated adults, the CDC suggests vaccinations for those 27 to 45 years of age. For adolescents, the CDC suggests the vaccine be given at ages 11 or 12 years, but may be started as early as 9. Go to: Histopathology Genital warts are typically diagnosed visually, with confirmatory biopsy generally unnecessary.These exophytic lesions form due to enlargement of the dermal papillae and are lined by hyperplastic squamous epithelium that shows koilocytes, which are squamous epithelial cells characterized by an acentric, hyperchromatic nucleus displaced by a large perinuclear vacuole. Go to: History and Physical Genital warts may occur separately or in clusters. They may be found in the anal or genital area, including the penile shaft, scrotum, vagina, or labia majora. They can also be found on internal surfaces of the vagina and the anus. They can be small (5 mm or less in diameter) or spread into large masses in the genital or anal area. Their color is variable but tends to be skin-colored or darker, and they may occasionally bleed spontaneously. Sometimes warts may cause itching, redness, or discomfort. An outbreak of genital warts may also cause psychological distress. In most cases, the only identifiable symptoms of an HPV infection are warts. Go to: Evaluation The diagnosis of genital warts is usually made visually, although a biopsy may be necessary for confirmation. Small warts may sometimes be confused with molluscum contagiosum. Genital warts typically rise above the skin surface, have parakeratosis, and demonstrate nuclear changes typical of HPV infections (nuclear enlargement with perinuclear clearing). Because genital warts are caused by low-risk HPV types, DNA tests should not be used for diagnosis or in low-risk HPV infections. Some practitioners use an acetic acid solution to help identify small warts and affected skin areas, but this practice is controversial. A biopsy is recommended if there is uncertainty about the diagnosis or if the patient is immunocompromised. Pigmented and ulcerated lesions should also be considered for biopsy. Cystoscopy should be considered in patients where the glans is involved, the patient has lower urinary tract symptoms, or there are significant urethral symptoms. In patients who have no symptoms, some experts have suggested waiting until any glans lesions have healed to avoid possible transfer of the HPV virus into the urethra. Go to: Treatment / Management There is no cure for HPV. Removing visible warts does not necessarily reduce the transmission of the underlying HPV infection. About 80% of individuals with HPV will clear the infection spontaneously within 18 to 24 months. Treatment varies depending on the number, size, and location of warts. Treatment can cause permanent depigmentation, itching, pain, and scarring. Urethral meatus warts are best treated with surgery to minimize long-term complications. The American Urological Association does not recommend treating sub-clinical (invisible) lesions. Treatments are either ablative (vaporization, resection, coagulation, or excision) or involve the use of topical agents. Physically ablative treatments are more effective at wart removal; but in many cases, topical agents are preferred by patients as initial therapy, especially for smaller lesions. Topical Agents Topical agents may be very effective, are self-applied by patients, and are less traumatic than surgical intervention. However, patient compliance is spotty and recurrences are common. Podophyllotoxin solution 0.15% to 0.5% in a gel or cream can be applied to the affected area and is not washed off. Podofilox (an anti-mitotic drug) appears to be safer than podophyllin.It works by binding microtubular subunits.The gel form is easier for patients to apply than the liquid with equal efficacy.The recommended application schedule (apply BID for three days followed by four days off, then repeat) can be confusing for some patients. Side effects include localized burning, itching, pain, and inflammation. Should not be used in pregnancy. The original precursor topical therapy, podophyllin, is no longer recommended by the CDC due to its high level of mutagens. Imiquimod is a topical immune response cream applied to the affected area but may cause fungal infections and flu-like symptoms. Imiquimod is an immune enhancer and increases cytokines (such as tumor necrosis factor-alpha (TNF-a) and interferon alfa). This causes an enhanced cytotoxic immune reaction based on T cell-mediated response factors.Wart recurrence rates are better than podophyllin-based therapies and it tends to cause non-scarring healing.The recommended application schedule for imiquimod 5% cream is three times per week.Usage of imiquimod 3.75% cream has been recommended as being equally effective while minimizing local side effects such as pain, burning, inflammation, itching, and erythema.Usage may also be limited by cost as the medication is relatively expensive if not covered by insurance. Sinecatechins are an ointment of catechins extracted from green tea that appear to have a higher wart clearance rate than podophyllotoxin and imiquimod while causing less local irritation, but clearance takes longer than with imiquimod. The overall wart clearance rate is reported as high as 57.2%. Sinecatechins are available as a 15% ointment. The exact mechanism of action is unknown, but they have immuno-stimulatory, anti-proliferative and anti-tumor properties and appear to work by reducing HPV gene products E6 and E7.Side effects include local redness, inflammation, and pain. Isotretinoinis a medication typically used for acne. It acts to reduce sebum, shrink sebaceous glands, provides an anti-inflammatory effect, and has anti-bacterial benefits. Isotretinoin has also shown significant efficacy when used as an adjunct to standard treatment of genital warts in immunocompromised patients, where the condyloma are extensive or if the lesions have proven resistant to initial therapy.The dosage is 0.5 to 1 mg/kg/day. Topical therapy is continued during treatment. There are many potential side effects to isotretinoin therapy so it must be used cautiously. In particular, it can cause very severe birth defects and so should absolutely not be used in pregnancy. For women of childbearing age, this means two pregnancy tests initially and monthly while on medication. They must also use two separate forms of birth control. Side effects include dry skin, chapped lips, frequent nosebleeds, dry eyes, dry mouth, and severe sun sensitivity while on the medication. There may also be night blindness, hair thinning, muscle aches, arthralgias, rashes, stomach problems, and higher cholesterol levels. Liver damage, urethritis, and hydrocephalus have been reported but are quite rare.Side effects are typically dose-dependent. There are a few reported cases of severe depression and suicide associated with the use of isotretinoin as well as possibly exacerbation of inflammatory bowel disease. For these reasons, isotretinoin is only dosed in 30-day intervals and it should be used cautiously. Trichloroacetic acidis not as effective as cryosurgery and should be avoided on the vagina, cervix, or urinary meatus.It tends to have a high recurrence rate and should be applied only by a healthcare provider due to potential injury to surrounding tissues. Skin erosion and pain are most commonly reported with imiquimod and sinecatechins. Physical (Surgical) Removal or Destruction Direct surgical excision or physically destructive therapies are considered more effective on keratinized warts, especially if they are larger in size. Simple surgical excision under local anesthesia is simple and direct but will leave a scar and requires a small surgical procedure. Liquid nitrogen cryosurgery ablationis inexpensive, considered safe for use during pregnancy, and does not usually cause much scarring but requires cryosurgical equipment and training. It may require anesthesia due to pain and multiple treatments are often necessary. Electrocauterization is considered effective but causes scarring and requires some level of anesthesia. Laser vaporization has minimal bleeding but may be somewhat less effective than other ablative techniques. It is typically used for extensive areas of genital wart involvement, is relatively expensive, and may cause a plume of virus-containing smoke. Surgical removal under general anesthesia may be necessary for more extensive lesions, intra-anal warts, or in children. Photodynamic therapy with a photosensitizing agent (such as aminolevulinic acid) has demonstrated efficacy in eliminating external warts. The aminolevulinic acid is applied topically or directly intralesionally. The photosensitizing agent is absorbed quickly into the most rapidly growing cells. Light exposure activates the aminolevulinic acid releasing free oxygen singlet radicals resulting in the destruction of the wart by direct oxidative injury.This technique is currently considered off-label. Discontinued 5-fluorouracil (5-FU) 5% cream is no longer considered acceptable due to side effects. Interferon intralesional injections initially showed moderate efficacy with a complete response rate of 36% - 63% as monotherapy, but it has largely been supplanted by other treatments. Might still be considered for intractable cases as an adjunctive therapy. Podophyllin, podofilox, and especially isotretinoin should be avoided during pregnancy. Go to: Differential Diagnosis Condyloma lata or secondary syphilis Familial benign pemphigus Herpes simplex infection Benign nevi Vulvar neurofibromatosis Go to: Prognosis A large number of cases of genital warts fail to respond to treatment and often recur, especially with repeated infections from sexual contact or the long-incubation period of HPV. Verifying patient compliance with therapy, changing the therapeutic agent, and adding isotretinoin can help.Morbidity associated with the disease is due to pruritus, bleeding, and the psychosocial burden of genital lesions, while mortality is due to its malignant transformation to squamous cell carcinoma. Immunocompromised patients are likely to have more resistant lesions than the general population with more frequent recurrences. They are also more likely for their lesions to develop a malignant transformation into squamous cell carcinoma.In general, immunocompromised patients generally benefit from a combination of therapies, early addition of isotretinoin, a longer duration of treatment, and earlier implementation of surgery. Go to: Complications Local complications with disfigurement are the most common complications of this disease. With untreated and advanced-stage disease, there is a risk of malignant transformation, which is the most feared complication. The current standard of care emphasizes treatment and primary prevention strategies, including vaccination, to prevent this devastating outcome. Go to: Deterrence and Patient Education Gardasil is a vaccine used to protect against human papillomavirus types 6, 11, 16, and 18. Types 16 and 18 cause an estimated 70% of cervical cancers, and 6 and 11 cause an estimated 90% of genital warts. The vaccine prevents the disease but is not therapeutic. The vaccine must be given before exposure to the virus type to be effective. The vaccine was approved by the United States Food and Drug Administration (FDA) in 2006 for use in children as early as nine years of age, primarily for its prophylactic activity against cervical cancer. Gardasil 9 was FDA approved in 2014 to protect against the four HPV strains covered by the first generation of Gardasil as well as five other HPV strains responsible for 20% of cervical cancers (HPV-31, HPV-33, HPV-45, HPV-52, and HPV-58). Vaccines are preventative and should not be considered therapeutic. Quadrivalent or 9-valent vaccines are recommended and generally preferred over bivalent vaccines. According to the Advisory Committee for Immunization Practices (ACIP), routine HPV vaccination is recommended for women 9 to 26 years of age, but it has shown high efficacy up to age 45. The CDC advises that unvaccinated adults above 26 years to age 45 may be given the vaccine after discussion with their provider. The ACIP recommends routine male HPV quadrivalent vaccinations at age 11-12. If not previously given or incomplete (the vaccines are a three-dose series), the vaccine should be given up to age 21. From ages 22 to 26, the vaccine is considered optional. In other words, the optimal age for male HPV vaccination is 11 to 12 years, but it may be given up to age 45 years. It remains to be seen if the more extensive use of vaccines can reduce the prevalence and penetration of HPV exposure, infections, and complications. Go to: Pearls and Other Issues Experimentally, a nitric acid/zinc complex solution for topical application has been successfully used on difficult to treat warts. The solution includes dilute nitric acid, zinc, copper, and organic acids. The application of this solution causes destruction and desiccation of the wart through a combination of denaturation and active protein coagulation actions. So far, it appears to be well tolerated and effective. The use of antivirals such as topical cidofovir also appears promising, but more study is needed before it can be safely utilized clinically. Curcumin, a derivative of turmeric, has shown efficacy against genital warts anecdotally, but more study is needed to determine its true safety and effectiveness. Go to: Enhancing Healthcare Team Outcomes Genital warts are very common in clinical practice. Because of the risk of cancer, there is now a vaccine available to prevent these warts. Healthcare workers, including nurse practitioners, physician assistants, and primary care physicians, need to work in an interprofessional effort to educate patients about the importance of the HPV vaccine as it can prevent a variety of genital cancers. The ACIP recommends routine male HPV quadrivalent vaccinations at age 11-12. If not previously given or incomplete (the vaccines are a three-dose series), the vaccine should be given up to age 21. From ages 22 to 26, the vaccine is considered optional. In other words, the optimal age for male HPV vaccination is 11 to 12 years, but it may be given up to age 26 years. Go to: Review Questions Access free multiple choice questions on this topic. Click here for a simplified version. Comment on this article. Figure Genital Warts, Female DermNet New Zealand Figure Genital Warts, Male DermNet New Zealand Go to: References 1. 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Czelusta AJ, Evans T, Arany I, Tyring SK. A guide to immunotherapy of genital warts: focus on interferon and imiquimod. BioDrugs. 1999 May;11(5):319-32. [PubMed: 18031142] 19. Berman B, Wolf J. The role of imiquimod 3.75% cream in the treatment of external genital warts. Skin Therapy Lett. 2012 Apr;17(4):5-7. [PubMed: 22491804] 20. Rosen T, Nelson A, Ault K. Imiquimod cream 2.5% and 3.75% applied once daily to treat external genital warts in men. Cutis. 2015 Oct;96(4):277-82. [PubMed: 26682290] 21. Tatti S, Swinehart JM, Thielert C, Tawfik H, Mescheder A, Beutner KR. Sinecatechins, a defined green tea extract, in the treatment of external anogenital warts: a randomized controlled trial. Obstet Gynecol. 2008 Jun;111(6):1371-9. [PubMed: 18515521] 22. Schöfer H, Tatti S, Lynde CW, Skerlev M, Hercogová J, Rotaru M, Ballesteros J, Calzavara-Pinton P. Sinecatechins and imiquimod as proactive sequential therapy of external genital and perianal warts in adults. Int J STD AIDS. 2017 Dec;28(14):1433-1443. [PubMed: 28566057] 23. Jha AK, Sonthalia S, Ganguly S. Oral isotretinoin as an adjunctive treatment for recurrent genital warts. J Am Acad Dermatol. 2018 Feb;78(2):e35-e36. [PubMed: 29332721] 24. Nickle SB, Peterson N, Peterson M. Updated Physician's Guide to the Off-label Uses of Oral Isotretinoin. J Clin Aesthet Dermatol. 2014 Apr;7(4):22-34. [PMC free article: PMC3990537] [PubMed: 24765227] 25. Yew YW, Pan JY. Complete remission of recalcitrant genital warts with a combination approach of surgical debulking and oral isotretinoin in a patient with systemic lupus erythematosus. Dermatol Ther. 2014 Mar-Apr;27(2):79-82. [PubMed: 24703263] 26. Oren-Shabtai M, Snast I, Lapidoth M, Sherman S, Noyman Y, Mimouni D, Hodak E, Levi A. Topical and Systemic Retinoids for the Treatment of Genital Warts: A Systematic Review and Meta-Analysis. Dermatology. 2021;237(3):389-395. [PubMed: 33279886] 27. Paredes-Bhushan V, Rezaee ME, Chavez DR. Isotretinoin induced urethritis: A case report & review of the literature. Urol Case Rep. 2020 Mar;29:101109. [PMC free article: PMC6939063] [PubMed: 31908966] 28. Godley MJ, Bradbeer CS, Gellan M, Thin RN. Cryotherapy compared with trichloroacetic acid in treating genital warts. Genitourin Med. 1987 Dec;63(6):390-2. [PMC free article: PMC1194123] [PubMed: 3323028] 29. Fichman Y, Levi A, Hodak E, Halachmi S, Mazor S, Wolf D, Caplan O, Lapidoth M. Efficacy of pulsed dye laser treatment for common warts is not influenced by the causative HPV type: a prospective study. Lasers Med Sci. 2018 May;33(4):773-777. [PubMed: 29218494] 30. Iranmanesh B, Khalili M, Zartab H, Amiri R, Aflatoonian M. Laser therapy in cutaneous and genital warts: A review article. Dermatol Ther. 2021 Jan;34(1):e14671. [PubMed: 33314577] 31. Xie F, Yu HS, Wang R, Wang D, Li YM, Wen HY, Du JB, Ba W, Meng XF, Yang J, Lin BW, Li HJ, Li CX, Zhang LG, Fang XD, Zhao H. Photodynamic Therapy for Genital Warts Causes Activation of Local Immunity. J Cutan Med Surg. 2019 Jul/Aug;23(4):370-379. [PubMed: 31010295] 32. Tu P, Zhang H, Zheng H, Gu H, Xu J, Tao J, Wang H, Zhu X, Wang X. 5-Aminolevulinic photodynamic therapy versus carbon dioxide laser therapy for small genital warts: A multicenter, randomized, open-label trial. J Am Acad Dermatol. 2021 Mar;84(3):779-781. [PubMed: 31374308] 33. Kechichian E, Helou E, Sarkis J, Hayek C, Labaki C, Nemr E, Tomb R. The place of 5-aminolaevulinic acid-photodynamic therapy in the treatment landscape of urethral warts: A systematic review. Photodiagnosis Photodyn Ther. 2021 Mar;33:102204. [PubMed: 33529745] 34. Zhang H, Shi L, Zhang Y, Wang P, Zhang G, Cao Y, Zhou Z, Wang X. Modified photodynamic therapy to minimize pain in the treatment of condylomata acuminata: A prospective, randomized, self-controlled study. Photodiagnosis Photodyn Ther. 2020 Dec;32:101915. [PubMed: 32634656] 35. Czelusta A, Yen-Moore A, Van der Straten M, Carrasco D, Tyring SK. An overview of sexually transmitted diseases. Part III. Sexually transmitted diseases in HIV-infected patients. J Am Acad Dermatol. 2000 Sep;43(3):409-32; quiz 433-6. [PubMed: 10954653] 36. Nebesio CL, Mirowski GW, Chuang TY. Human papillomavirus: clinical significance and malignant potential. Int J Dermatol. 2001 Jun;40(6):373-9. [PubMed: 11589741] 37. Orlando G, Fasolo MM, Beretta R, Merli S, Cargnel A. Combined surgery and cidofovir is an effective treatment for genital warts in HIV-infected patients. AIDS. 2002 Feb 15;16(3):447-50. [PubMed: 11834957] 38. Kaliterna V, Barisic Z. Genital human papillomavirus infections. Front Biosci (Landmark Ed). 2018 Mar 01;23(9):1587-1611. [PubMed: 29293452] 39. Cusini M, Micali G, Lacarrubba F, Puviani M, Barcella A, Milani M. Efficacy and tolerability of nitric-zinc complex in the treatment of external genital warts and "difficult-to-treat" warts: a "proof of concept", prospective, multicentre, open study. G Ital Dermatol Venereol. 2015 Dec;150(6):643-8. [PubMed: 26513041] 40. Pontini P, Mastorino L, Gaspari V, Granger C, Ramoni S, Delmonte S, Evangelista V, Cusini M. A Multicentre, Randomised Clinical Trial to Compare a Topical Nitrizinc® Complex Solution Versus Cryotherapy for the Treatment of Anogenital Warts. Dermatol Ther (Heidelb). 2020 Oct;10(5):1063-1073. [PMC free article: PMC7477018] [PubMed: 32734366] 41. Muffarrej D, Khattab E, Najjar R. Successful treatment of genital warts with cidofovir cream in a pediatric patient with Fanconi anemia. J Oncol Pharm Pract. 2020 Jul;26(5):1234-1236. [PubMed: 31718429] 42. Hengge UR, Tietze G. Successful treatment of recalcitrant condyloma with topical cidofovir. Sex Transm Infect. 2000 Apr;76(2):143. [PMC free article: PMC1758281] [PubMed: 10858720] 43. Schürmann D, Bergmann F, Temmesfeld-Wollbrück B, Grobusch MP, Suttorp N. Topical cidofovir is effective in treating extensive penile condylomata acuminata. AIDS. 2000 May 26;14(8):1075-6. [PubMed: 10853999] 44. Blaizot R, Dutkiewicz AS, Guillet S, Pham-Ledard A, Beylot-Barry M. Intravenous cidofovir for diffuse genital warts in the setting of multifactorial immunosuppression. J Eur Acad Dermatol Venereol. 2017 Mar;31(3):e162-e163. [PubMed: 27527116] 45. Psomiadou V, Iavazzo C, Douligeris A, Fotiou A, Prodromidou A, Blontzos N, Karavioti E, Vorgias G. An Alternative Treatment for Vaginal Cuff Wart: a Case Report. Acta Medica (Hradec Kralove). 2020;63(1):49-51. [PubMed: 32422116] Disclosure:Stephen Leslie declares no relevant financial relationships with ineligible companies. Disclosure:Hussain Sajjad declares no relevant financial relationships with ineligible companies. Disclosure:Sandeep Kumar declares no relevant financial relationships with ineligible companies. Continuing Education Activity Introduction Etiology Epidemiology Histopathology History and Physical Evaluation Treatment / Management Differential Diagnosis Prognosis Complications Deterrence and Patient Education Pearls and Other Issues Enhancing Healthcare Team Outcomes Review Questions References 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. 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16204	https://nvlpubs.nist.gov/nistpubs/Legacy/IR/nistir6453.pdf	NISTIR 6453 Algebraic constraints implying monotonicity for cubics Charles Fenimore and John Libert, NIST Michael Brill, Sarnoff Corporation U.S. DEPARTMENT OF COMMERCE Technology Administration National Institute of Standards and Technology Electricity Division 100 Bureau Drive, Stop 8114 Gaithersburg, MD 20899-81 14 GC 100 .U56 NO. 6453 2000 NIST NISTIR 6453 Algebraic constraints implying monotonicity for cubics Charles Fenimore and John Libert, NIST Michael Brill, Sarnoff Corporation U.S. DEPARTMENT OF COMMERCE Technology Administration National Institute of Standards and Technology Electricity Division TOO Bureau Drive. Stop 8114 Gaithersburg, MD 20899-81 14 January 2000 U.S. DEPARTMENT OF COMMERCE William M. Daley, Secretary TECHNOLOGY ADMINISTRATION Dr. Cheryl L. Shavers. Under Secretary of Commerce for Technology NATIONAL INSTITUTE OF STANDARDS AND TECHNOLOGY Raymond G. Kammer, Director Abstract While it is straightforward to formulate constraints which ensure a cubic polynomial is monotonic on an interval, such constraints may not be in a form which is suitable for use with standard optimization techniques and software. The MATLAB package is typi-cal: the required constraints are a series of simultaneous inequalities. In what follows, two simultaneous algebraic inequalities on the coefficients of a cubic polynomial are shown to be necessary and sufficient to assure monotonicity on an interval. This transformation of interval constraints to simultaneous algebraic constraints requires the application of basic logic and analysis. The constraints are applied to a problem arising in analyzing the performance of several video quality measurement models , 1. Introduction The monotonicity of a differentiable function on an interval is equivalent to a condition on its derivative over the interval, namely the derivative must not change sign at any point. For polynomials, this condition over the interval is equivalent to conditions at the end points and inflection points in the interval, namely the derivatives at all such points must have the same sign. In the case of a cubic polynomial the sole inflection point can be expressed in a closed algebraic form and monotonicity is shown to be equivalent to a pair of algebraic inequalities on the coefficients of the polynomial. In this form, monotonicity is readily incorporated as a constraint in classical optimization techniques and available scientific software. A number of investigators have considered monotonic regression, particularly for splines. Andersson and Elfving have studied monotone interpolation and approximation by cubic splines. Ramsay has considered spline-based monotonic regression. In the case of cubic polynomials, there does not appear to be a basic package for monotonic regression. This paper describes a mathematical formulation of the constraints which enables the use of available software to solve the cubic problem. This study in low order fitting arises from a quality measurement problem. In recent years, as digital imaging and video technology has replaced analog systems, much attention has been focused on measures of video and image quality. The main new impetus for such attention is the use of lossy compression techniques to reduce data rates. Lossy means data is not recoverable. Such losses are hopefully imperceptible. In an international comparison of video quality measurement computational models, the computed results are to be compared with the results of subjective tests , The measurements for such subjective assessment may saturate at the extremes of the scale. The design of the test permits non-linear adjustments of the computed results. These "adjustments" include remapping the computed (or "objective") data with (among other functions) a monotonic cubic polynomial which best fits the objective and subjective 4 This article refers to commercial products by name in order to specify the means by which the computed results were obtained. Such mention is not meant to imply the product is the best available for the purpose nor is it an endorsement. 2 scores. A cubic polynomial is included because it is the lowest degree polynomial which can have an inflection point, and can thus compensate for either saturation or steepening at the extremes of the interval. On the other hand, rank ordering of the subjective scores is locally determined and the design of the test expresses confidence in this ranking by imposing monotonicity on the cubic. Doing so avoids reordering the objective scores. 2. Theory Without loss of generality, we consider the fitting problem over the unit interval [0,1 ] for a cubic polynomial, p, to data (tj,d,). Other intervals can be scaled to the unit interval. p(t) = x 0 + x, • t + x 2 • r + a" 3 • r' 0 < r < 1 To achieve monotonicity one must avoid a change in sign of the derivative, p'. A necessary, but insufficient, condition is the derivative of p must have the same sign at 0 and 1 . That is, the function a defined below must be nonnegative: A : p\0)p\) = a = jc, • (x, + 2x2 + 3x,)> 0 The location of the single inflection point, th of p(t) t\ = - 2 /33 is the second piece of information needed to establish monotonicity. When f, is not in the open unit interval, (0,1), Condition A implies that p is monotonic on [0,1]. This is a consequence of the following Lemma. Lemma 2.1. Assume the function/is twice continuously differentiable on the interval [ot,p. If/ has the same sign at a and f3 and /' does not change sign in (a,/), then/is monotonic on [a,p. Proof. Assume that /' > 0 on For any t in (a,/), f'(cc) < f\t) = /’(a)+ f f"(s)ds < /’(/). Ja As a consequence the derivative,/, never changes sign on [a,p] and /is monotonic. If instead,/' < 0, -/would satisfy the above conditions and/would be monotonic. \|\| Lemma 2.2 shows that in the case that r, is in (0, 1 ), monotonicity of a cubic is equivalent to p’ having the same sign at the inflection point as at the endpoints of the interval. 3 ' . Lemma 2.2. A cubic polynomial, p(t), is monotonic on the interval [0,1] if and only if A : p'( 0) pX 1 ) = a = x x (x , + 2x2 + ?>x3 ) > 0 and B : t ] e (0,l) implies C: (P X0) + pXl))-pXt ] )>0 Proof: A direct calculation shows that monotonicity implies the two conditions. Monotonicity of p on [0,1] can be established by considering three cases. I. For U & (0,1), the conditions of Lemma 2.1 are met and p is monotonic. II. For r, e (0,1) and p'(t{) = 0, p' has a zero of order 2 at th p'(t) = 3 • xj (t-r,) 2 . Because its derivative does not change sign on the interval, p is monotonic. III. For f, g (0,1) and p\t x ) ^ 0, the conditions of Lemma 2.2 require that p'(0), p'(l), and p'(ti) agree in sign. To see this, first observe that Condition A implies that p'(0) and p\ 1) have the same sign and as a consequence p'(0) + p\ 1) also agrees in sign. Condition C implies that p\t,) also has the same sign. Lemma 2.1 assures that p is monotonic on each of the subintervals [0,r,] and [rT , 1 ], p having a single inflection point. The non-zero derivative, p'(t t ), assures that p is monotonic in the same sense on both subintervals and therefore on the entire unit interval. II So, two constraints which are necessary and sufficient to assure monotonicity are: A and (B implies C). Now, recast the logical implication of the second constraint in Lemma 2.2 as a single inequality. Since Condition A is already an inequality, doing so exhibits simultaneous inequalities which are equivalent to the interval constraint for monotonicity. Theorem 2.3: The cubic polynomial, p(t) = xo + x\ t + x-i f + X?, r\ is monotonic on the interval [0,1] if and only if jc, (jc, +2jc 2 +3x 3 )>0 and 4 Proof. It is sufficient to replace the implication in Lemma 2.2 with the second inequality. First, recast B and C as polynomial expressions b and c, assuming X3 0. - x, 1 3x 3 2 ] 2 < — <=> b = x, + 3x ? x, < 0 2 " B : e (0,l) <=> and ? 2 2 C : (/7 , ( 0 ) + /7 '(l))/7 / (r,) = ( 2 x, + 2 x, + 3x,)-(x, + 3 x 3 3x 3 = ( 2 x, + 2 x, + 3x 3 ) ( 3x, -x\ - x 2 -x 3 ) 3x > 0 <=> c = (2 jc , + 2 jc 2 + 3 jc 3 )(3 • x, • x 3 - x\ x 3 ) > 0 Second, observe that (B implies C) is equivalent to (b < 0 implies c > 0). • In the case X3 ^ 0, the equivalence is a consequence of the preceding paragraph. • In the case x3 = 0, B is false so (B implies C) holds vacuously and b > 0 so (b < 0 implies c > 0) holds vacuously. The statements are equivalent. Third, convert the previous implication to an inequality. By logical calculus, the statement (8 implies C) is equivalent to (C or not 8) [5 - page7], ]. As a consequence, in the plane of vectors (b , c), (b < 0 implies c > 0) is satisfied by vectors with (b > 0 or c > 0). This is the entire plane except for the third quadrant, (b < 0 and c < 0). The vectors are at an angle of more than tt/4 to the vector (-1,-1). Noting that the dot product of two unit vectors is the cosine of the angle between them (b,c) t (-1 ,-1) ^ 1 \|\|(6,c)\|\| V2 V2 ’ equivalently, ~Jb 2 + c 2 +b + c> 0. Replacing b and c with the corresponding polynomials in the coefficients x } yields the second inequality. II Note that the region satisfying the constraints is not convex, but it is the union of convex regions. These regions are defined by algebraic constraints which can be found by factoring the constraints described in Lemma 2.2 and Theorem 2.3. 5 3. Application In an international comparison of several video quality measurement models , it was required to fit a monotonic cubic polynomial to the output of each model. The Figures are the results of two least squares fits to a typical data set. Figure 1 displays the results of unconstrained fitting. The best fitting cubic is not monotonic on [0,1]. The two critical points are a local maximum at t s 0.82 and a local minimum at t = 5.28. Polynomial Fit of Order 3 Figure 1: Unconstrained regression of cubic polynomial. The fitted curve is not monotonic Parameter Values: x0 = -0.07591 1 , x, = 2.824086, x2 = -1 .99055, x3 = 0.21 7546. A local maximum occurs at t = 0.82 and a local minimum at t = 5.28. 6 Figure 2 displays the best fitting cubic which is monotonic on [0,1]. Again, there are two critical points: a local maximum, t =1.00, and a local minimum, t =1.21. The constrained solution "sweeps" the interior critical point to the boundary of the interval. The results were obtained with the MATLAB Optimization Package using the CONSTR function with the constraints defined here. Monotonic Cubic Fit Figure 2: Monotonic regression of cubic polynomial. The fitted curve is monotonic on [0,1]. Parameter Values: x0 = -0.092407, Xt = 3.06557, x2 = -2.7999538, x3 = 0.844523. Although it is monotonic on [0,1], the cubic has a local maximum at t = 1 .00 and a local minimum at t = 1.21. 7 References 1 . L. Andersson and T. Elfving, Interpolation and Approximation by Monotone Cubic Splines, J. Approx. Theory 66 (1991) 302-333. 2. MATLAB Constrained Optimization Software, 3. J. O. Ramsay, Estimating smooth monotone functions, J. Royal Statistical Soc. Series B 60 (1998) 365-375. 4. The Video Quality Experts Group (VQEG), 5. A.N. Whitehead and B. Russell, "Principia Mathematica", Cambridge U. P., Cambridge, 1967. 8
16205	https://math.stackexchange.com/questions/3277220/find-number-of-distinct-integer-terms-in-the-sequence	Find number of distinct integer terms in the sequence - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Find number of distinct integer terms in the sequence Ask Question Asked 6 years, 3 months ago Modified3 years, 3 months ago Viewed 2k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. We have to find the number of distinct integer terms in the sequence ⌊i 2 2005⌋⌊i 2 2005⌋ from i=1 i=1 to 2005 2005 where ⌊.⌋⌊.⌋ represents the floor function. Initially I calculated the first few terms and saw all the integers from 0 0 coming up so I assumed this to be the norm and answered as 2006 2006. But when I put it on WolframAlpha, I saw that there didn't seem to be a pattern especially towards the end. Can someone explain any method to determine and predict this trend without using a calculator? sequences-and-series Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jun 28, 2019 at 17:46 lulu 77.4k 6 6 gold badges 89 89 silver badges 140 140 bronze badges asked Jun 28, 2019 at 17:43 infinite-blank-infinite-blank- 1,056 1 1 gold badge 8 8 silver badges 17 17 bronze badges 4 Note: I changed your notation, [n][n] into ⌊n⌋⌊n⌋ as being, I think the more standard notation for the floor function these days. If you prefer it the other way, just roll back my edit.lulu –lulu 2019-06-28 17:47:42 +00:00 Commented Jun 28, 2019 at 17:47 @lulu It's OK. Actually I wanted that notation too but didn't know how to do it.infinite-blank- –infinite-blank- 2019-06-28 17:49:00 +00:00 Commented Jun 28, 2019 at 17:49 No problem, if you click on the thing that says "edited X mins ago" you can see the syntax I used.lulu –lulu 2019-06-28 17:53:51 +00:00 Commented Jun 28, 2019 at 17:53 1 To your question, the values for the floor change every time i 2 i 2 crosses over a number of the form 2005 n−−−−−√2005 n for some integer n n. Thus the first 1 1 you get comes at i=45 i=45, where 2005−−−−√≈44.78 2005≈44.78 and the first 2 2 you get comes at i=64 i=64 where 2×2005−−−−−−−√≈63.32 2×2005≈63.32. The problems start around i=523 i=523, since 523×2005−−−−−−−−−√≈1024.02 523×2005≈1024.02 and 524×2005−−−−−−−−−√≈1024.99756 524×2005≈1024.99756 and of course it gets worse from there.lulu –lulu 2019-06-28 18:00:41 +00:00 Commented Jun 28, 2019 at 18:00 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. You're only guaranteed a distinct value going from i i to i+1 i+1 if (i+1)2−i 2=2 i+1≥2005.(i+1)2−i 2=2 i+1≥2005. You may get a distinct value for values of i i less that that, but not necessarily. However, this implies that you are guaranteed to hit every non-negative integer value up to that point. You may get the values more than once, but you only count them once. The crossover point happens at 2 i+1=2005 2 i+1=2005 or i=1002 i=1002. The value of the floor function is ⌊1002 2/2005⌋=500⌊1002 2/2005⌋=500. So you have all of the integers from 0 0 to 500 500, plus one distinct integer apiece for every value of i>1002 i>1002, which amounts to 1003 1003 integers. So, the list contains 501+1003=1504 501+1003=1504 distinct integers. (This agrees with the number of integers WA came up with.) Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jun 29, 2019 at 17:22 answered Jun 28, 2019 at 18:18 JohnJohn 26.6k 3 3 gold badges 41 41 silver badges 63 63 bronze badges 2 2 Since we get ⌊1002 2 2005⌋=500⌊1002 2 2005⌋=500 then that means that we have 0−500 0−500 till that point. Then doesn't that mean that we get a new distinct integer from 1003−2005 1003−2005. So that would be 501+1003=1504 501+1003=1504 distinct integers?infinite-blank- –infinite-blank- 2019-06-29 16:01:30 +00:00 Commented Jun 29, 2019 at 16:01 Thanks for re-checking. I counted i=1002 i=1002 twice. And when I pulled WA's results into Excel to remove the duplicates, I forgot to take out the blank that was counted. You're correct.John –John 2019-06-29 17:24:11 +00:00 Commented Jun 29, 2019 at 17:24 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. S⊂{0,1,2,...,2005}S⊂{0,1,2,...,2005} Let f:{1,2,3,⋯,2005}→S f:{1,2,3,⋯,2005}→S be such that f(x)=⌊x 2 2005⌋f(x)=⌊x 2 2005⌋. Claim : f f is onto function in {1,2,⋯,1002}{1,2,⋯,1002} and one to one function in {1003,1004,⋯,2005}{1003,1004,⋯,2005}. Proof : Let a=⌊x 2 2005⌋a=⌊x 2 2005⌋ where a∈S a∈S. From the property of greatest integer function, it can be concluded that a≤x 2 2005<a+1⟺2005 a≤x 2<2005(a+1)⟺2005 a−−−−−√≤x<2005(a+1)−−−−−−−−−−√a≤x 2 2005<a+1⟺2005 a≤x 2<2005(a+1)⟺2005 a≤x<2005(a+1) Now for f f to be onto, for every a a there exist x x, i.e., there must be at least one integer between 2005(a+1)−−−−−−−−−−√2005(a+1) and 2005 a−−−−−√2005 a. This is possible when 2005(a+1)−−−−−−−−−−√−2005 a−−−−−√≥1⟹2005 a≤1002 2⟹a≤⌊1002 2 2005⌋=500□2005(a+1)−2005 a≥1⟹2005 a≤1002 2⟹a≤⌊1002 2 2005⌋=500◻ For f f to be one to one function, no value in S S may be taken by f(x)f(x) more than once. Therefore, ⌊x 2 2005⌋≠⌊(x+1)2 2005⌋⌊x 2 2005⌋≠⌊(x+1)2 2005⌋ (x+1)2 2005=x 2 2005+2 x+1 2005(x+1)2 2005=x 2 2005+2 x+1 2005 Therefore when 2 x+1 2005>1⟺(2 x+1)>2005 2 x+1 2005>1⟺(2 x+1)>2005 , or x>1002 x>1002, f f is one to one □◻. Final Solution : \|S\|=501+(2005−1002)=1504\|S\|=501+(2005−1002)=1504 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jun 30, 2022 at 3:08 S DasS Das 521 4 4 silver badges 16 16 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions sequences-and-series See similar questions with these tags. 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16206	https://www.chegg.com/homework-help/questions-and-answers/draw-structure-consistent-following-1h-nmr-spectra-case-molecular-formula-provided-hi-help-q6609699	Solved Draw a structure that is consistent with each of the \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Science Chemistry Chemistry questions and answers Draw a structure that is consistent with each of the following 1H NMR spectra. In each case, the molecular formula is provided. Hi can you help me with these NMR questions in Organic chemistry? please help me! Thank you so much ! Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: Draw a structure that is consistent with each of the following 1H NMR spectra. In each case, the molecular formula is provided. Hi can you help me with these NMR questions in Organic chemistry? please help me! Thank you so much ! Draw a structure that is consistent with each of the following 1 H NMR spectra. In each case, the molecular formula is provided. Hi can you help me with these NMR questions in Organic chemistry? please help me! Thank you so much ! Here’s the best way to solve it.Solution 100%(4 ratings) Share Share Share done loading Copy link Here’s how to approach this question This AI-generated tip is based on Chegg's full solution. Sign up to see more! Calculate the degree of unsaturation using the formula C−(H 2)+1 to determine the presence of rings or double bonds in the molecular structure. (1) . The molecular formula = C8H10O Degree of unsaturation = C-(H/2)+1= 8-(10/2)+1 = 4 four degrees of unsaturation which is highly suggestive of an aromatic ring. To determine the relative integration values, we divide each of the integration valu… View the full answer Previous questionNext question Not the question you’re looking for? Post any question and get expert help quickly. Start learning Chegg Products & Services Chegg Study Help Citation Generator Grammar Checker Math Solver Mobile Apps Plagiarism Checker Chegg Perks Company Company About Chegg Chegg For Good Advertise with us Investor Relations Jobs Join Our Affiliate Program Media Center Chegg Network Chegg Network Busuu Citation Machine EasyBib Mathway Customer Service Customer Service Give Us Feedback Customer Service Manage Subscription Educators Educators Academic Integrity Honor Shield Institute of Digital Learning © 2003-2025 Chegg Inc. All rights reserved. Cookie NoticeYour Privacy ChoicesDo Not Sell My InfoGeneral PoliciesPrivacy PolicyHonor CodeIP Rights
16207	https://brainly.com/question/12451336	[FREE] Calculate the molar volume of an ideal gas at 25∘C and 1 atm pressure. - brainly.com 2 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +63,5k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +26,6k Ace exams faster, with practice that adapts to you Practice Worksheets +6,5k Guided help for every grade, topic or textbook Complete See more / Chemistry Textbook & Expert-Verified Textbook & Expert-Verified Calculate the molar volume of an ideal gas at 25∘C and 1 atm pressure. 2 See answers Explain with Learning Companion NEW Asked by miltontwo4027 • 04/09/2019 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 23801746 people 23M 5.0 2 Upload your school material for a more relevant answer Considering the definition of molar volume and ideal gas law, the molar volume of an ideal gas at 25∘C and 1 atm pressure is 24.436m o l e L. An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures. An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases: P×V = n× R× T On the other hand, molar volume is a property that indicates how much space a mole of a certain substance or compound occupies. The expression to calculate the molar volume is: M o l a r V o l u m e=n V where V is the volume it occupies, and n is the amount of the species in moles. Then, rearranging the expression for the ideal gas law, you obtain: n V=P R x T M o l a r V o l u m e=P R x T In this case, you know: R= 0.082 m o l K a t m L T= 25 C= 298 K P= 1 atm Replacing: M o l a r V o l u m e=1 a t m 0.082 m o l K a t m Lx 298 K Solving: MolarVolume= 24.436 m o l e L In summary, the molar volume of an ideal gas at 25∘C and 1 atm pressure is 24.436m o l e L. Learn more: brainly.com/question/13255185?referrer=searchResults brainly.com/question/4147359?referrer=searchResults Answered by CintiaSalazar •4.1K answers•23.8M people helped Thanks 2 5.0 (4 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 23801746 people 23M 5.0 2 Fundamentals of General Organic and Biological Chemistry - Mcmurry Et Al. Structure and Properties - Tro Introductory Chemistry Online! - Paul R. Young Upload your school material for a more relevant answer The molar volume of an ideal gas at 25°C and 1 atm pressure is approximately 24.475 L/mol. This is calculated using the ideal gas law, rearranging it to find the volume per mole. The calculation incorporates the ideal gas constant and the temperature converted to Kelvin. Explanation To calculate the molar volume of an ideal gas at 25°C and 1 atm pressure, we can use the ideal gas law, which is defined as: P V=n RT Where: P is the pressure in atm, V is the volume in liters, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Convert the temperature to Kelvin: The temperature at 25°C is converted as follows: T(K)=25°C+273.15=298.15 K Use the ideal gas constant: The ideal gas constant R is commonly used as 0.0821 m o l×K L×a t m. Rearrange the ideal gas law to find molar volume: We can express molar volume (V/n) as: M o l a r V o l u m e=P RT Substituting the known values: Substituting in R, T, and P: M o l a r V o l u m e=1 a t m 0.0821 m o l×K L×a t m×298.15 K Calculating the molar volume: M o l a r V o l u m e=1 24.475 m o l L≈24.475 L/m o l Hence, the molar volume of an ideal gas at 25°C and 1 atm pressure is approximately 24.475 L/m o l. This value shows that under these conditions, one mole of an ideal gas occupies about 24.475 liters, which is typical for gases at room temperature and pressure. Examples & Evidence For example, if we take one mole of oxygen gas (O2) at these conditions, it would occupy approximately 24.475 liters. This approximation allows us to use it in various gas-related calculations in chemistry. The calculations and results align with the principles outlined in the ideal gas law, which describes how gases behave under various conditions of pressure and temperature. Thanks 2 5.0 (3 votes) Advertisement Community Answer This answer helped 172096788 people 172M 5.0 1 24.5 liter / mol Explanation Data: a) T = 25°C = 25 + 273.15K = 298.15 K b) P = 1 atm c) assumption: ideal gas Formulae and principles a) Ideal gas equation: p V = n R T, where: p = pressure V = volume n = number of moles R = universal constant of gases = 0.0821 atm-liter / K-mol T = absolute temperature (Kelvin) b) Molar volume: ν = V / n, where: ν = molar volume V = volume n = number of moles Solution: p V = n R T⇒ V/n = ν = RT / p = = 0.0821 atm-liter / mol-K × 298.15 K / 1 atm = 24.5 liter / mol Answered by Edufirst •6.6K answers•172.1M people helped Thanks 1 5.0 (2 votes) Advertisement ### Free Chemistry solutions and answers Community Answer 4.2 19 A drink that contains 4 1/2 ounces of a proof liquor… approximately how many drinks does this beverage contain? Community Answer 5.0 7 Chemical contamination is more likely to occur under which of the following situations? When cleaning products are not stored properly When dishes are sanitized with a chlorine solution When raw poultry is stored above a ready-to-eat food When vegetables are prepared on a cutting board that has not been sanitized Community Answer 4.3 189 1. Holding 100mL of water (ebkare)__2. Measuring 27 mL of liquid(daudgtear ldnreiyc)____3. Measuring exactly 43mL of an acid (rtube)____4. Massing out120 g of sodium chloride (acbnela)____5. Suspending glassware over the Bunsen burner (rwei zeagu)____6. Used to pour liquids into containers with small openings or to hold filter paper (unfenl)____7. Mixing a small amount of chemicals together (lewl letpa)____8. Heating contents in a test tube (estt ubet smalcp)____9. Holding many test tubes filled with chemicals (estt ubet karc) ____10. Used to clean the inside of test tubes or graduated cylinders (iwer srbuh)____11. Keeping liquid contents in a beaker from splattering (tahcw sgasl)____12. A narrow-mouthed container used to transport, heat or store substances, often used when a stopper is required (ymerereel kslaf)____13. Heating contents in the lab (nuesnb bneurr)____14. Transport a hot beaker (gntos)____15. Protects the eyes from flying objects or chemical splashes(ggloges)____16. Used to grind chemicals to powder (tmraor nda stlepe) __ Community Answer Food waste, like a feather or a bone, fall into food, causing contamination. Physical Chemical Pest Cross-conta Community Answer 8 If the temperature of a reversible reaction in dynamic equilibrium increases, how will the equilibrium change? A. It will shift towards the products. B. It will shift towards the endothermic reaction. C. It will not change. D. It will shift towards the exothermic reaction. Community Answer 4.8 52 Which statements are TRUE about energy and matter in stars? Select the three correct answers. Al energy is converted into matter in stars Only matter is conserved within stars. Only energy is conserved within stars. Some matter is converted into energy within stars. Energy and matter are both conserved in stars Energy in stars causes the fusion of light elements Community Answer 4.5 153 The pH of a solution is 2.0. Which statement is correct? Useful formulas include StartBracket upper H subscript 3 upper O superscript plus EndBracket equals 10 superscript negative p H., StartBracket upper O upper H superscript minus EndBracket equals 10 superscript negative p O H., p H plus P O H equals 14., and StartBracket upper H subscript 3 upper O superscript plus EndBracket StartBracket upper O upper H superscript minus EndBracket equals 10 to the negative 14 power. Community Answer 5 Dimensional Analysis 1. I have 470 milligrams of table salt, which is the chemical compound NaCl. How many liters of NaCl solution can I make if I want the solution to be 0.90% NaCl? (9 grams of salt per 1000 grams of solution). The density of the NaCl solution is 1.0 g solution/mL solution. Community Answer 8 For which pair of functions is the exponential consistently growing at a faster rate than the quadratic over the interval mc015-1. Jpg? mc015-2. Jpg mc015-3. Jpg mc015-4. Jpg mc015-5. Jpg. Community Answer 4.8 268 How did Mendeleev come up with the first periodic table of the elements? (1 point) A He determined the mass of atoms of each element. B He estimated the number of electrons in atoms of each element. C He arranged the elements by different properties to find a pattern. D He organized the elements by their atomic number. New questions in Chemistry Which type of reaction is M g+S→M g S? A. polymerization B. synthesis C. replacement D. decomposition Reactants undergo chemical reaction to form products. This chemical equation represents one such reaction. The coefficient for one of the reactants or products is incorrect. Which part of the chemical equation is incorrect? 2 C 4H 1010 O 2⋯8 C O 2+10 H 2O Which of the samples most likely had the highest solubility? \| Sample \| Name \| Chemical formula \| Temperature of water (∘C) \| :---: :---: \| \| 1 \| Table sugar \| C 12H 22O 11 \| 80 \| \| 2 \| Table sugar \| C 12H 22O 11 \| 45 \| \| 3 \| Table salt \| NaCl \| 55 \| \| 4 \| Table salt \| NaCl \| 63 \| \bigcirc 1 \bigcirc 2 \bigcirc 3 \bigcirc 4 Which of the samples most likely had the highest solubility? \| Sample \| Name \| Chemical formula \| Temperature of water ( ∘C ) \| :--- :--- \| \| 1 \| Table sugar \| C 12H 22O 11 \| 80 \| \| 2 \| Table sugar \| C 12H 22O 11 \| 45 \| \| 3 \| Table salt \| NaCl \| 55 \| \| 4 \| Table salt \| NaCl \| 63 \| A. 1 B. 2 C. 3 D. 4 \ \hline 1 & Table sugar & $C {12} H {22} O {11}$ & 80 \ \hline 2 & Table sugar & $C {12} H {22} O {11}$ & 45 \ \hline 3 & Table salt & NaCl & 55 \ \hline 4 & Table salt & NaCl & 63 \ \hline \end{tabular} A. 1 B. 2 C. 3 D. 4") Consider the chemical equations shown here. 2 H 2(g)+O 2(g)→2 H 2O(g)Δ H 1=−483.6 k J+2=−241.8 k J/m o l 3 O 2(g)→2 O 3(g)Δ H 2=284.6 k J+2=142.3 k J/m o l What is the overall enthalpy of reaction for the equation shown below? 3 H 2(g)+O 3(g)→3 H 2O(g)□ kJ Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. 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16208	https://en.wikipedia.org/wiki/Quantum_chromodynamics_binding_energy	Quantum chromodynamics binding energy - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Source of mass 2 Gluon energy 3 See also 4 References Quantum chromodynamics binding energy [x] 3 languages العربية עברית Türkçe Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia Energy that binds quarks into hadrons Quantum chromodynamics binding energy (QCD binding energy), gluon binding energy or chromodynamic binding energy is the energy binding quarks together into hadrons. It is the energy of the field of the strong force, which is mediated by gluons. Motion-energy and interaction-energy contribute most of the hadron's mass. Source of mass [edit] Most of the mass of hadrons is actually QCD binding energy, through mass–energy equivalence. This phenomenon is related to chiral symmetry breaking. In the case of nucleons —protons and neutrons— QCD binding energy forms about 99% of the nucleon's mass. The kinetic energy of the hadron's constituents, moving at near the speed of light, contributes greatly to the hadron mass; otherwise most of the rest is actual QCD binding energy, which emerges in a complex way from the potential-like terms in the QCD Lagrangian. For protons, the sum of the rest masses of the three valence quarks (two up quarks and one down quark) is approximately 9.4MeV/c 2, while the proton's total mass is about 938.3 MeV/c 2. In the standard model, this "quark current mass" can nominally be attributed to the Higgs interaction. For neutrons, the sum of the rest masses of the three valence quarks (two down quarks and one up quark) is approximately 11.9 MeV/c 2, while the neutron's total mass is about 939.6 MeV/c 2. Considering that nearly all of the atom's mass is concentrated in the nucleons, this means that about 99% of the mass of everyday matter (baryonic matter) is, in fact, chromodynamic binding energy. Gluon energy [edit] While gluons are massless, they still possess energy — chromodynamic binding energy. In this way, they are similar to photons, which are also massless particles carrying energy — photon energy. The amount of energy per single gluon, or "gluon energy", cannot be directly measured, though a distribution can by inferred from deep inelastic scattering (DIS) experiments (see ref for an old but still valid introduction.) Unlike photon energy, which is quantifiable, described by the Planck–Einstein relation and depends on a single variable (the photon's frequency), no simple formula exists for the quantity of energy carried by each gluon. While the effects of a single photon can be observed, single gluons have not been observed outside of a hadron. A hadron is in totality composed of gluons, valence quarks, sea quarks and other virtual particles. The gluon content of a hadron can be inferred from DIS measurements. Again, not all of the QCD binding energy is gluon interaction energy, but rather, some of it comes from the kinetic energy of the hadron's constituents. Currently, the total QCD binding energy per hadron can be estimated through a combination of the factors mentioned. In the future, studies into quark–gluon plasma will better complement the DIS studies and improve our understanding of the situation. See also [edit] Gluon Quark Current quark and constituent quark Hadron Strong force Quantum chromodynamics Chiral symmetry breaking Photon energy Invariant mass and relativistic mass Binding energy References [edit] ^ Jump up to: abStrassler, Matt (15 April 2013). "Protons and Neutrons: The Massive Pandemonium in Matter". Of Particular Significance. Retrieved 30 May 2016. ^Cho, Adrian (2 April 2010). "Mass of the Common Quark Finally Nailed Down". Science Magazine. AAAS. Retrieved 30 May 2016. ^Decomposition of the proton mass (Lattice QCD) ° Halzen, Francis and Martin, John, "Quarks and Leptons:An Introductory Course in Modem Particle Physics", John Wiley & Sons (1984). Retrieved from " Categories: Physical quantities Hadrons Quantum chromodynamics Binding energy Hidden categories: Articles with short description Short description is different from Wikidata This page was last edited on 22 August 2024, at 21:10(UTC). 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16209	https://www.youtube.com/watch?v=xNYiB_2u8J4	Lewis Diagrams and VSEPR Models Bozeman Science 1420000 subscribers 4127 likes Description 523671 views Posted: 22 Aug 2013 022 - Lewis Diagrams and VSEPR Models In this video Paul Andersen explains how you can use Lewis Diagrams and VSEPR Models to make predictions about molecules. The Lewis diagrams are a two-dimensional representations of covalent bonds and the VSEPR models show how the molecule could exist in three dimensional space. Pi bonding and odd valence electrons require an extension of this model. Music Attribution Title: String Theory Artist: Herman Jolly All of the images are licensed under creative commons and public domain licensing: AJ. A Blue Balloon on a Ribbon, July 18, 2008. Open clip Art Library image's page. "File:Acetic Acid Atoms.svg." Wikipedia, the Free Encyclopedia. Accessed August 18, 2013. "File:AX4E0-3D-balls.png." Wikipedia, the Free Encyclopedia. Accessed August 16, 2013. "File:AX6E0-3D-balls.png." Wikipedia, the Free Encyclopedia. Accessed August 18, 2013. "File:Bent-3D-balls.png." Wikipedia, the Free Encyclopedia. Accessed August 17, 2013. "File:Linear-3D-balls.png." Wikipedia, the Free Encyclopedia. Accessed August 17, 2013. "File:Nitrate-ion-resonance-2D.png." Wikipedia, the Free Encyclopedia. Accessed August 17, 2013. "File:Nitrogen-dioxide-3D-vdW.png." Wikipedia, the Free Encyclopedia. Accessed August 17, 2013. "File:Pi-Bond.svg." Wikipedia, the Free Encyclopedia. Accessed August 18, 2013. "File:Sigma Bond.svg." Wikipedia, the Free Encyclopedia. Accessed August 18, 2013. "File:Tetrahedral-3D-balls.png." Wikipedia, the Free Encyclopedia. Accessed August 17, 2013. "File:Trigonal-3D-balls.png." Wikipedia, the Free Encyclopedia. Accessed August 17, 2013. "File:Trigonal-bipyramidal-3D-balls.png." Wikipedia, the Free Encyclopedia. Accessed August 18, 2013. 138 comments Transcript: Introduction [Music] Hi, it's Mr. Anderson and this is Chemistry Essentials video 22. It's on Lewis diagrams and vesper models. And it's a powerful model if we can use these two together to make predictions about the structure of something like methane over here. Works great if we're looking at atoms that are bonded covealently and also in ions. And so, um, we could take methane for example, look at its constituent parts. So carbon and hydrogen, we could figure out their Lewis dot diagrams. If you don't know how to do that, I'll put a little video up here so you can see that. Then you could create a Lewis structure. And finally, we can create a Vesper model. And once we have that, we can learn all of these things about a molecule such as methane. And so it works on molecules. We start with Lewis diagrams. And I'll give you a method for drawing Lewis diagrams. Um, what we can do is compare different diagrams that we come up with by looking at their formal charges. We can look alternatives. And lots of times we'll find multiple alternatives and then we have to show resonance. Another model is to use a vesper model. Vesper model is simply looking at where the pairs of electrons are and they're going to repel each other. This is simply going to be Kulom's law. And so based on that repulsion, we can create these three-dimensional models of what those Lewis diagrams look like. And those so these two together is a pretty powerful model that you can use in chemistry. What we can use is use that to predict geometry, bond angles, bond energy, bond length, and we can even predict uh polarity based on that as well. Lots of times we're going to have to extend it. It's just a model. It's not actually what the atoms look like. And so we can do extensions on this based on overlap of bonds and also sometimes it's confusing when we get odd veence electrons. And so um I'll present another theory. It's called molecular orbital theory. Won't go into detail, but I'll kind of explain where that would be useful. And so first thing we Drawing Lewis Structures have to do is to know how to draw Lewis structures. There's lots of different methods on how to do this out there. I couldn't find any good pneummonic devices. And so we're going to use Vesper twice. Vesper is going to be our model of repulsion, but it's also an easy way to remember how you draw Lewis structures. And so if we're drawing the Lewis structure, the first thing you want to do is add up all of the veence electrons. Next thing you want to do is just sketch out the skeleton of where the atoms might be connected. Next, you're going to add the electrons. In general, you should add it to the most electrogative atoms first and kind of work from the outside to the inside. After you've done that, we can look for pairs of bonds and then finally we're going to review the formal charges. Now, I know this seems confusing, but we'll work through a number of different molecules and I think it'll make sense. So, let's say we're starting with this. This is hydrogen hydrogen cyanide. If we're going to draw the le structure of it, the first thing we want to do is draw and add up the total number of veence electrons. So, to figure out veence electrons, we're ignoring the metals. You can see I've removed that d block right here. Um, but what we've got is if you're in this first column, then you're going to have one veence electron. So, hydrogen. If you're burillium, you're going to have two. If you're carbon, you're going to have four. And so, we go up to this molecule and we're going to add up all the veence electrons. And so, that's 1 + 4 + 5 is 10. So, we start by V, adding the veence electron. Next thing we do is simply sketch out the skeleton. An easy way to do that is to just sketch it out by drawing bonds between each of the atoms as they are in the molecule itself. And so we've done that. Now since these are pairs of electrons, we had to remove four of those electrons. So now we're down to six. Next thing we do is we're going to add the electrons. And so we should add that to the most electrogative or generally the one on the outside of the molecule. We'll get to carbon in a second. So I'm going to add that to satisfy the octet rule. Remember nitrogen wants eight electrons. So, it's going to have 2, four, six, and then it's sharing the one. So, it's going to have two electrons here. So, nitrogen at this point is totally happy. But you should see that carbon's not so happy. And so, now we go to P. We're going to look for pairs of bonds to try to make carbon a little bit happier. So, what we could do is we could share one of those electrons from nitrogen or one of those pairs. And now nitrogen still satisfies the octet rule, but carbon's closer. Now, it has six around it. And so, we could try that again. And now we've got a Lewis structure for hydrogen cyanide. What's the last thing we're going to do is review the formal charges. A lot of people don't know what a formal charge is. So we're going to start and go across from the left to the right with hydrogen. How many electrons does hydrogen have? How many veence electrons does it have? Well, it has one. How many have I assigned it in this model? Well, I've assigned it one electron. Remember, it's sharing one of its electrons with carbon, but one of the electrons is really it. And so we're going to have the one veence electron it has minus the one assigned. And so it's going to have a formal charge of zero. Let's go to carbon. How many veence electrons? It's going to have four. How many is it sharing? Well, it's sharing one here, one here, one here, one here. So it's sharing four. So it also has a formal charge of zero. Let's go to nitrogen. Nitrogen has five veence electrons. It's got these two just to itself, but it's also sharing these three over here. So it's got a formal charge of zero. If I got formal charges of zero all the way across, I'm happy. That means that I've got a pretty good model. So, let's go on to another one. Let's look at carbon dioxide. What do I do first? I'm going to add up the veence electrons. So, that's four for carbon plus six for each of the oxygen. So, I get 16. Now, I'm going to sketch out the skeletons. So, this is a quick tip that'll help you. If you have an atom written by itself and then a number of other atoms after that, generally this one's going to be in the middle. And so, when I sketch out the skeleton, it's going to look like that with the carbon in the middle. Remember I've used four of my electrons. So now I'm back to 12 veence electrons. So I'm going to add electrons. I'm going to add them to the electronegative oxygen on the outside. So how many have I added? I've added all of my electrons. I'm back down to zero. But you can see this carbon's not happy. So what I can do is I can do that pair switching right here. And now I've got carbon. It looks happier at this point. Now if we look at review the formal charges. Let's go from left to the right. If we look at oxygen, oxygen is going to have six. How many does it have? Well, you can see that it has 6 minus one. So, it's actually going to have a formal charge of -1. So, that's kind of a not a great sign on my model so far. If we look at carbon, it's zero. And if we look at oxygen, it's going to have a formal charge of + one because six veence electrons - 1 2 3 4 5. And so, it's going to have + one. And so could you put together this model in a different way so that I would have lower formal charges? For sure. What I could do is I could switch and get a pair on one side. I could switch and get a pair on one side. And now I got formal charges of 0 0 and zero. And so this is what it looks like. Now I should probably jimmy this a little bit. I should switch these pair of bonds over here and these over here. But this is Drawing Ozone Structures pretty close. Now let's go to ozone 03. So I'm going to add the veence electrons. Sketch out the skeleton. Let me add those electrons. You can see I've got an extra one compared to when I did carbon dioxide. And so now if we draw that, I've got one structure like this. Could draw my formal charges. So that's okay. But can you visualize this that I could kind of invert it the other way and I could also go get those formal charges. Remember formal charges, the closer they are to zero, the more stable kind of that structure is. But these ones have the exact same formal charges. And so what am I going to draw? uh I'm going to draw resonance. I'm going to draw this arrow in the middle, this two-way arrow. And that means that this structure or this structure are properly. So we have to show both of those structures. Now let's get to the VSEPR Model vesper model. So vesper model is the veence shell electron pair repulsion. What does that mean? It's very simple. It simply means that when you have pair of electrons, they have this negative charge and they're going to push on other pairs of electrons and they're going to push on themselves. And so think of it like a balloon. If you were to hold one balloon, it would look like this. But if I were to give you two balloons, they're going to move apart from each other if we hold it in the middle. And so that tells us what the electrons are going to do and therefore what the structure is going to be. So imagine if we have carbon dioxide like this. It's going to be the same where have these pair electrons, these pair of electrons. Your hand is kind of holding it in the middle. And so what shape three-dimensional shape do we think it's going to have? It's going to have this kind of a shape. We call that linear. Linear is in a line. And I know it doesn't look three-dimensional, but it will get there in just a second. What's going to be the bond angle between the two? We call that 180°. And this is a linear model. So, you can see now how Vesper allows us to start kind of visualizing these in three dimensionals um or in three dimensions. One quick note, if you're doing organic chemistry, lots of times they'll call this SP hybridized at this point. Not going to go into what that means in AP Chem. It's not important, but it is important that you understand the techn the terminology. Let's go to three electrons now. Excuse me, three pair of electrons. So, we got three balloons. It's going to organize itself like this. What's something that looks like that? Maybe nitrate. And so, if we're looking at nitrate like this, it's going to have the nitrogen in the middle, those electrons around the outside. And so, this is going to be that vesper model. It's going to be trional planer. So, it's going to have these three um electrons coming out or these three atoms coming out. It's going to be 120° angle between the two. Um but it's also going to be flat at this point. Again, quick organic chemistry note. We call this sp2 hybridized. So again, we haven't got to this third structure yet. We haven't got to this third dimension, but we're quickly going to get there. Um, let's say we're looking at something like ozone. Ozone, you might think, is going to be linear, but if you look at this pair of electrons out here, what kind of a structure are we going to get from ozone? We're going to get a bent kind of a structure. And now, let's get to four pair of electrons. with four pair of electrons. Methane is that example I gave you at the beginning. What's that going to look like? Well, if you hold four balloons in your hand, it's going to have this tetrahedral shape. The bond angle is going to be 109.5 and it's going to really rise out of the paper itself. We call this sp3 hybridized. And that's kind of where it ends. Uh in AP Chem, your knowledge of all these bond angles and geometry, know that we can also have more pairs of electrons. And as we do, we get to what's called if we let's say we're looking at uh PCL4, that's going to be trigonal by pyramidal. And so they're going to branch off like this. What you really have is a trigonal planer here in the middle. And then you're going to have a linear structure right down here through the middle. Why do we call it pyramidal? If I were to connect those with lines, you can see how we kind of have two pyramids on top of each other. Octa is when we're going to have six pair of electrons coming off. Again, bond angle. Don't have to know it, but you do have to know that we call this an octaedral at this point. And so what are some Extensions extensions? What are going to be some things that we have to build on this Vesper model? Well, one thing we're going to have to do is understand that not all the bonds are going to be the same. Single bonds like this are going to be different than double bonds like this. And that's because the orbitals are going to start to overlap. And so in a single bond, what we have is called a sigma bond. Sigma bond is where they're sharing those electrons between the nuclei. But as we start to add more electrons that are being shared, what we're using is an overlap of orbitals. And so we got get what are called pi bonds. And so pi bonds, so this would be a sigma bond and then a pi bond. Or I could say this is a sigma and this is a pi bond. Those pi bonds as we add them are going to be less powerful, but they're also going to kind of lock the molecule in itself. So you get formation of isomers. Also sometimes uh the math doesn't add up. In other words, we're going to have a odd number of electrons. So how do we add these pairs if we're adding one single electron? So nitrogen dioxide is actually going to have a le structure like this, but it's going to have one single electron out here. So we call that a free radical. And so there Molecular Orbital Model are extensions. And so we can build a better model. It's called the molecular orbital model or orbital theory. It's based on quantum theory and mathematics. It's just a different model. And we can answer a lot of these questions, but it's probably a little too intense and a little too sophisticated for AP camp. And so again, did you learn the following to use Lewis geometry and vesper models to predict geometry, hybridization, polarity? If you did, you did great. Try one on your own. Let's say I give you water. Could you draw the Lewis structure? Could you draw the uh vesper model? Watch out for those electrons. And I hope that was helpful. [Music]
16210	https://www.youtube.com/watch?v=6f72q3JTyU8	Calculating With Upper & Lower Bounds \| Number \| Maths \| FuseSchool FuseSchool - Global Education 894000 subscribers 360 likes Description 50446 views Posted: 6 May 2020 Calculating With Upper & Lower Bounds \| Number \| Maths \| FuseSchool In this video we are going to look at how to calculate with upper and lower bounds. To find the upper bound of an addition or of an area, you would want to multiply the upper bounds of both measurements, as this would give the largest possible sum / area. Whereas to get the lower bound of a addition or multiplication (such as area) you would multiply the lower bounds of both measurements. For subtraction we need to think a little more carefully. To get the upper bound of a subtraction, you would want to take the upper bound of the first measurement and subtract the lower bound of the second, to give the largest possible outcome. Whereas to get the lower bound of a subtraction you would want to take a big value (upper bound second value) away from the smallest possible value (lower bound first value). We also need to think more carefully about division. To get an upper bound answer, we would want to divide a big number (upper bound) by a small number (lower bound). Whereas a lower bound division would be the opposite; divide a small number (lower bound) by the biggest possible value (upper bound). CREDITS Animation & Design: Martin Mukhanu Narration: Lucy Billings Script: Lucy Billings SUBSCRIBE to the FuseSchool YouTube channel for many more educational videos. Our teachers and animators come together to make fun & easy-to-understand videos in Chemistry, Biology, Physics, Maths & ICT. VISIT us at www.fuseschool.org, where all of our videos are carefully organised into topics and specific orders, and to see what else we have on offer. Comment, like and share with other learners. You can both ask and answer questions, and teachers will get back to you. These videos can be used in a flipped classroom model or as a revision aid. Find all of our Chemistry videos here: Find all of our Biology videos here: Find all of our Physics videos here: Find all of our Maths videos here: Instagram: Facebook: Twitter: Access a deeper Learning Experience in the FuseSchool platform and app: www.fuseschool.org Follow us: Befriend us: This is an Open Educational Resource. If you would like to use the video, please contact us: info@fuseschool.org 36 comments Transcript: Intro [Music] we discovered how to find the upper and lower bounds in an earlier video in this video we're going to look at how to calculate with bounds when the limits of accuracy are combined we have to really carefully consider whether to use the upper or lower bounds you're probably wondering what on earth am i on about so let's just jump straight in with an example Example a football pitch is 110 by 70 meters both measured to the nearest 10 meters what are the upper and lower bounds of the area well this question isn't too tricky for the upper bound of the area we need to multiply the upper bounds of both dimensions and the lower bound of the area we multiply the lower bounds of both measurements but what about when we're dividing for the upper bound we want a divided by Upper Bound b to be as big as possible so we want a to be big which is then divided by the smallest possible b so a must be its upper bound and b must be its lower bound Lower Bound your turn now find the lower bound pause the video work out the answer and click play when you're ready did you get it right so here are some questions for you to do pause the video answer the questions and click play when you're ready how did you get on [Music] so there we have bounds just make sure you take your time to think about what the calculation is asking for and you'll have no problems make sure you show all of your working [Music]
16211	https://cdn.kutasoftware.com/Worksheets/Alg2/Arithmetic%20and%20Geometric%20Means.pdf	©P T2d0v1E1i MKDuAtHab 8SkoIfytrwjaqrdes DLDLnCY.6 h aAMlNlU ir8iXgbh8t9s0 7rSeJsge5rMvDekdd.o 5 jMcatdSe8 YwriptUhk UIbn2feiTnziYtNec 0ABlSgYepbnrrad K2h.P Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Name_____ Period_ Date___ Arithmetic and Geometric Means Find the missing term or terms in each arithmetic sequence. 1) ..., 29, , −31, ... 2) ..., −20, , −40, ... 3) ..., 13, , 7, ... 4) ..., −23, , −3, ... 5) ..., 9, , 3, ... 6) ..., 36, , 56, ... 7) ..., −16.3, , −20.5, ... 8) ..., −30, , 30, ... 9) ..., −4 7, , −68 21, ... 10) ..., 32, , , , 68, ... 11) ..., 3 5, , , , 29 15, ... 12) ..., 10, , , , 130, ... 13) ..., −7.7, , , −16.4, ... 14) ..., 19, , , −71, ... -1-©o Q2D0z1j1l gKEu7tHaK 2SGoGfEtBwyaqrFeX oLSLgCN.W F fArlFlJ 4r2iGgqhdths9 GrEe3sneXrnvPeodl.Z q hMnaGdoez EwRiVtohM ZILnFfYibn6iCtuem BAnlngTeObsrbak E2j.o Worksheet by Kuta Software LLC Find the missing term or terms in each geometric sequence. 15) ..., −3, , −108, ... 16) ..., −2, , −18, ... 17) ..., −3, , −75, ... 18) ..., 2, , 18, ... 19) ..., −2, , −8, ... 20) ..., −1, , −4, ... 21) ..., 1.5, , 6, ... 22) ..., −1 2, , −1 50, ... 23) ..., −1, , −16, ... 24) ..., 3125, , , , 5, ... 25) ..., −25 8 , , , , −32 25, ... 26) ..., 1, , , , 16, ... 27) ..., 3, , , 648, ... 28) ..., −1, , , −27, ... -2-©A P2h0t1r1n 7KUuktZa9 XSCozfhtvwCayrJeY FLhLNC4.j k JAklxlZ HrGiYgJhUtGsh FrQeksIeBrdvQeUdC.u 1 vMGaMdNe8 Awei1tEhM lIbnHfLiNnHiRtOej OAglxgvegb6rYag q21.C Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Name_____ Period Date____ Arithmetic and Geometric Means Find the missing term or terms in each arithmetic sequence. 1) ..., 29, , −31, ... −1 2) ..., −20, , −40, ... −30 3) ..., 13, , 7, ... 10 4) ..., −23, , −3, ... −13 5) ..., 9, , 3, ... 6 6) ..., 36, , 56, ... 46 7) ..., −16.3, , −20.5, ... −18.4 8) ..., −30, , 30, ... 0 9) ..., −4 7, , −68 21, ... −40 21 10) ..., 32, , , , 68, ... 41, 50, 59 11) ..., 3 5, , , , 29 15, ... 14 15, 19 15, 8 5 12) ..., 10, , , , 130, ... 40, 70, 100 13) ..., −7.7, , , −16.4, ... −10.6, −13.5 14) ..., 19, , , −71, ... −11, −41 -1-©I i220x1b1g ZKkuptEaR BSyoqfPtvweayrVew fL7LmCl.j G dAKlKlz TrOimghhWtts7 1rlePsueZrov1eUd8.Q G 9Mya0dxee HwkiQtXhI gIfnKf5ijntiLtBe6 3AVlEgaeob9rLaN X2D.k Worksheet by Kuta Software LLC Find the missing term or terms in each geometric sequence. 15) ..., −3, , −108, ... −18 16) ..., −2, , −18, ... −6 17) ..., −3, , −75, ... −15 18) ..., 2, , 18, ... 6 19) ..., −2, , −8, ... −4 20) ..., −1, , −4, ... −2 21) ..., 1.5, , 6, ... 3 22) ..., −1 2, , −1 50, ... −1 10 23) ..., −1, , −16, ... −4 24) ..., 3125, , , , 5, ... 625, 125, 25 25) ..., −25 8 , , , , −32 25, ... −5 2, −2, −8 5 26) ..., 1, , , , 16, ... 2, 4, 8 27) ..., 3, , , 648, ... 18, 108 28) ..., −1, , , −27, ... −3, −9 -2-Create your own worksheets like this one with Infinite Algebra 2. Free trial available at KutaSoftware.com
16212	https://pubmed.ncbi.nlm.nih.gov/8515788/	Clozapine-induced agranulocytosis. Incidence and risk factors in the United States - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Incidence and risk factors in the United States J M Alvir1,J A Lieberman,A Z Safferman,J L Schwimmer,J A Schaaf Affiliations Expand Affiliation 1 Department of Psychiatry, Hillside Hospital, Division of Long Island Jewish Medical Center, Glen Oaks, N.Y. 11004. PMID: 8515788 DOI: 10.1056/NEJM199307153290303 Item in Clipboard Clozapine-induced agranulocytosis. Incidence and risk factors in the United States J M Alvir et al. N Engl J Med.1993. Show details Display options Display options Format N Engl J Med Actions Search in PubMed Search in NLM Catalog Add to Search . 1993 Jul 15;329(3):162-7. doi: 10.1056/NEJM199307153290303. Authors J M Alvir1,J A Lieberman,A Z Safferman,J L Schwimmer,J A Schaaf Affiliation 1 Department of Psychiatry, Hillside Hospital, Division of Long Island Jewish Medical Center, Glen Oaks, N.Y. 11004. PMID: 8515788 DOI: 10.1056/NEJM199307153290303 Item in Clipboard Cite Display options Display options Format Abstract Background: Clozapine is an atypical antipsychotic agent that is more effective than standard neuroleptic drugs in the treatment of patients with refractory schizophrenia. Unlike classic neuroleptic agents, clozapine is not associated with the development of acute extrapyramidal symptoms or tardive dyskinesia. The main factor limiting its use is the risk of potentially fatal agranulocytosis, estimated to occur in 1 to 2 percent of treated patients. After clozapine was approved by the Food and Drug Administration, it became available for marketing in the United States in February 1990 only as part of a special surveillance system (the Clozaril Patient Management System, or CPMS), in which a weekly white-cell count was required for the patient to receive a supply of the drug. Methods: We evaluated the CPMS data for February 1990 through April 1991 by survival analysis to determine the incidence of agranulocytosis and the effects of potential risk factors such as age and sex. Data were available for 11,555 patients who received clozapine during the period after marketing began. Results: Agranulocytosis developed in 73 patients, resulting in death from infectious complications in 2 patients. Episodes of agranulocytosis occurred in 61 patients within three months after they began treatment. The cumulative incidence of this side effect was 0.80 percent (95 percent confidence interval, 0.61 to 0.99) at 1 year and 0.91 percent (95 percent confidence interval, 0.62 to 1.20) at 1 1/2 years. The risk of agranulocytosis increased with age and was higher among women. Conclusions: The occurrence of agranulocytosis is a substantial hazard of the administration of clozapine, but this hazard can be reduced by monitoring the white-cell count. The increasing risk of agranulocytosis with age and the reduced incidence after the first six months of treatment provide additional guidelines for the prescription and monitoring of clozapine treatment in the future. PubMed Disclaimer Comment in Clozapine--deciphering the risks.Gerson SL.Gerson SL.N Engl J Med. 1993 Jul 15;329(3):204-5. doi: 10.1056/NEJM199307153290312.N Engl J Med. 1993.PMID: 8515794 No abstract available. Similar articles Long-term patient monitoring for clozapine-induced agranulocytosis and neutropenia in Korea: when is it safe to discontinue CPMS?Kang BJ, Cho MJ, Oh JT, Lee Y, Chae BJ, Ko J.Kang BJ, et al.Hum Psychopharmacol. 2006 Aug;21(6):387-91. doi: 10.1002/hup.779.Hum Psychopharmacol. 2006.PMID: 16933201 Agranulocytosis: incidence and risk factors.Alvir JM, Lieberman JA.Alvir JM, et al.J Clin Psychiatry. 1994 Sep;55 Suppl B:137-8.J Clin Psychiatry. 1994.PMID: 7961558 Do white-cell count spikes predict agranulocytosis in clozapine recipients?Alvir JM, Lieberman JA, Safferman AZ.Alvir JM, et al.Psychopharmacol Bull. 1995;31(2):311-4.Psychopharmacol Bull. 1995.PMID: 7491384 Reducing clozapine-related morbidity and mortality: 5 years of experience with the Clozaril National Registry.Honigfeld G, Arellano F, Sethi J, Bianchini A, Schein J.Honigfeld G, et al.J Clin Psychiatry. 1998;59 Suppl 3:3-7.J Clin Psychiatry. 1998.PMID: 9541331 Review. Clozapine: the commitment to patient safety.Alphs LD, Anand R.Alphs LD, et al.J Clin Psychiatry. 1999;60 Suppl 12:39-42.J Clin Psychiatry. 1999.PMID: 10372610 Review. See all similar articles Cited by Risk factors for neutropenia in clozapine-treated children and adolescents with childhood-onset schizophrenia.Maher KN, Tan M, Tossell JW, Weisinger B, Gochman P, Miller R, Greenstein D, Overman GP, Rapoport JL, Gogtay N.Maher KN, et al.J Child Adolesc Psychopharmacol. 2013 Mar;23(2):110-6. doi: 10.1089/cap.2011.0136.J Child Adolesc Psychopharmacol. 2013.PMID: 23510445 Free PMC article. G-CSF mediated neutrophil augmentation in a unique case of comorbid idiopathic Parkinson's disease and treatment-resistant schizophrenia on clozapine.Morrow O, Gibson L, Bhamra M, David AS, Posporelis S.Morrow O, et al.Ther Adv Psychopharmacol. 2020 Sep 16;10:2045125320956414. doi: 10.1177/2045125320956414. eCollection 2020.Ther Adv Psychopharmacol. 2020.PMID: 32973999 Free PMC article. Genetic risk factors for clozapine-induced neutropenia and agranulocytosis in a Dutch psychiatric population.van der Weide K, Loovers H, Pondman K, Bogers J, van der Straaten T, Langemeijer E, Cohen D, Commandeur J, van der Weide J.van der Weide K, et al.Pharmacogenomics J. 2017 Oct;17(5):471-478. doi: 10.1038/tpj.2016.32. Epub 2016 May 10.Pharmacogenomics J. 2017.PMID: 27168101 Distinctive pattern of neutrophil count change in clozapine-associated, life-threatening agranulocytosis.Taylor D, Vallianatou K, Whiskey E, Dzahini O, MacCabe J.Taylor D, et al.Schizophrenia (Heidelb). 2022 Mar 14;8(1):21. doi: 10.1038/s41537-022-00232-0.Schizophrenia (Heidelb). 2022.PMID: 35288577 Free PMC article. Clozapine in drug induced psychosis in Parkinson's disease: a randomised, placebo controlled study with open follow up.Pollak P, Tison F, Rascol O, Destée A, Péré JJ, Senard JM, Durif F, Bourdeix I.Pollak P, et al.J Neurol Neurosurg Psychiatry. 2004 May;75(5):689-95. doi: 10.1136/jnnp.2003.029868.J Neurol Neurosurg Psychiatry. 2004.PMID: 15090561 Free PMC article.Clinical Trial. See all "Cited by" articles Publication types Research Support, U.S. Gov't, P.H.S. 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16213	https://mathworld.wolfram.com/CartesianEquation.html	Cartesian Equation -- from Wolfram MathWorld TOPICS AlgebraApplied MathematicsCalculus and AnalysisDiscrete MathematicsFoundations of MathematicsGeometryHistory and TerminologyNumber TheoryProbability and StatisticsRecreational MathematicsTopologyAlphabetical IndexNew in MathWorld Algebra Homological Algebra MathWorld Contributors Barile Cartesian Equation Download Wolfram Notebook An equation representing a locus in the -dimensional Euclidean space. It has the form (1) where the left-hand side is some expression of the Cartesian coordinates, ..., . The -tuples of numbers fulfilling the equation are the coordinates of the points of . For example, the locus of all points in the Euclidean plane lying at distance 1 from the origin is the circle that can be represented using the Cartesian equation (2) Similarly, the locus of all points of the three-dimensional Euclidean space lying at distance 1 from the origin is a sphere of radius 1 centered at the origin can be represented using the Cartesian equation (3) Often the letters , , are used instead of indexed coordinates , , . The intersection of two loci and is the set of points whose coordinates fulfil the system of equations (4) (5) For example, the system (6) (7) represents the intersection of the coordinate plane (the set of points for which ) with the coordinate plane (the set of points for which ). The result is the set of points , i.e., the above system represents the -axis. In general, in the three-dimensional Euclidean space, a single linear Cartesian equation represents a plane, whereas an algebraic surface of order is given by a polynomial equation of degree . Curves are represented as the intersection of two surfaces. For example, lines are represented as the intersection of two planes, circles as the intersection of a sphere and a plane (or of two spheres). Of course, a given curve can be realized by intersection in infinitely many ways, which correspond to infinitely many different equivalent systems of equations representing the same curve. In any case two equations are needed since a single Cartesian equation can represent a curve only in the plane. An alternative way to represent a locus is to use parametric equations. Cartesian equations of lines can be derived from parametric ones by algebraic elimination of the parametric variable(s). See also Affine Variety, Cartesian, Cartesian Coordinates, Cartesian Geometry, Coordinate System This entry contributed by Margherita Barile Explore with Wolfram\|Alpha More things to try: Cartesian ovals cycloid Cartesian equation Taubin's heart surface Cartesian equation Cite this as: Barile, Margherita. "Cartesian Equation." From MathWorld--A Wolfram Resource, created by Eric W. Weisstein. Subject classifications Algebra Homological Algebra MathWorld Contributors Barile About MathWorld MathWorld Classroom Contribute MathWorld Book wolfram.com 13,278 Entries Last Updated: Sun Sep 28 2025 ©1999–2025 Wolfram Research, Inc. Terms of Use wolfram.com Wolfram for Education Created, developed and nurtured by Eric Weisstein at Wolfram Research Created, developed and nurtured by Eric Weisstein at Wolfram Research
16214	https://www.khanacademy.org/science/up-class-11-chemistry/xa8e73391a9c6d31a:chemical-bonding-and-molecular-structure	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
16215	https://www.cs.utep.edu/vladik/slides/chiangmai17olga.pdf	CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 1 of 21 Go Back Full Screen Close Quit How to Explain Ubiquity of Constant Elasticity of Substitution (CES) Production and Utility Functions Without Explicitly Postulating CES Olga Kosheleva1, Vladik Kreinovich1, and Thongchai Dumrongpokaphan2 1University of Texas at El Paso, El Paso, TX 79968, USA olgak@utep.edu, vladik@utep.edu 2Department of Mathematics, Faculty of Science Chiang Mai University, Thailand, tcd43@hotmail.com CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 2 of 21 Go Back Full Screen Close Quit 1. Outline • The dependence of production on various factors is of-ten described by CES functions. • These functions are usually explained by postulating two requirements: – that the formulas should not change if we change a measuring unit, and – a less convincing CES requirement. • In this paper, we show that the CES requirement can be replaced by a more convincing requirement: – that the combined eﬀect of all the factors – should not depend on the order in which we com-bine these factors. CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 3 of 21 Go Back Full Screen Close Quit 2. CES Production Functions and CES Utility Functions Are Ubiquitous • Most observed data about production y is well de-scribed by the CES production function y = n X i=1 ai · xr i !1/r . • Here xi are the numerical measures of the factors that inﬂuence production, such as: – amount of capital, – amount of labor, etc. • A similar formula describes how the person’s utility y depends on diﬀerent factors xi such as: – amounts of diﬀerent types of consumer goods, – utilities of other people, etc. CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 4 of 21 Go Back Full Screen Close Quit 3. How This Ubiquity Is Explained Now • The current explanation for the empirical success of CES function is based on two requirements. • The ﬁrst requirement is that the corresponding func-tion y = f(x1, . . . , xn) is homogeneous: f(λ · x1, . . . , λ · xn) = λ · f(x1, . . . , xn). • Meaning: we can describe diﬀerent factors by using diﬀerent monetary units. • The results should not change if we replace the original unit by a one which is λ times smaller. • After this replacement, the numerical value of each fac-tor changes from xi to λ · xi and y is replace by λ · y. • So, we get exactly the above requirement. CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 5 of 21 Go Back Full Screen Close Quit 4. Second Requirement • The second requirement is that f(x1, . . . , xn) should provide constant elasticity of substitution (CES). • The requirement is easier to explain for the case of two factors n = 2. • In this case, this requirement deals with “substitution” situations in which: – we change x1 and then – change the original value x2 to the new value x2(x1) – so that the overall production or utility remain the same. • The corresponding substitution rate can then be cal-culated as s def = dx2 dx1 . CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 6 of 21 Go Back Full Screen Close Quit 5. Second Requirement (cont-d) • The substitution function x2(x1) is explicitly deﬁned by the equation f(x1, x2(x1)) = const, then s = −f,1(x1, x2) f,2(x1, x2), where f,i(x1, x2) def = ∂f ∂xi (x1, x2). • The requirement is that: – for each percent of the change in ratio x2 x1 , – we get the same constant number of percents change in s: ds d x2 x1 = const. • Problem: the CES condition is too mathematical to be convincing for economists. • We provide: more convincing arguments. CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 7 of 21 Go Back Full Screen Close Quit 6. Main Idea Behind a New Explanation • In our explanation, we will use the fact that in most practical situations, we combine several factors. • We can combine these factors in diﬀerent order. For example: – we can ﬁrst combine the eﬀects of capital and labor into a single characteristic, – and then combine it with other factors. • Alternatively: – we can ﬁrst combine capital with other factors, – and only then combine the resulting combined fac-tor with labor, etc. • The result should not depend on the order in which we perform these combinations. • We show that this idea implies the CES functions. CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 8 of 21 Go Back Full Screen Close Quit 7. Derivation of the CES Functions from the Above Idea • Let us denote a function that combines factors i and j into a single quantity xij by fi,j(xi, xj). • Similarly, let’s denote a function that combines xij and xkℓinto a single quantity xijkℓby fij,kℓ(xij, xkℓ). • In these terms, the requirement that the resulting val-ues do not depend on the order means that f12,34(f1,2(x1, x2), f3,4(x3, x4)) = f13,24(f1,3(x1, x3), f2,4(x2, x4)). • In both production and utility situations, for each i and j, fi,j(xi, xj) is increasing in xi and xj. • It is also reasonable to require that: – the function fi,j(xi, xj) is continuous, and – when one of the factors tends to inﬁnity, the result also tends to inﬁnity. CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 9 of 21 Go Back Full Screen Close Quit 8. Derivation (cont-d) • Under these assumptions, f(a, b) is invertible: – for every a ∈A and for every c ∈C, there exists a unique value b ∈B for which c = f(a, b); – for every b ∈B and for every c ∈C, there exists a unique value a ∈A for which f(a, b) = c. • It is known that: – for every set of invertible operations that satisfy the generalized associativity requirement, – there exists an Abelian group G and 1-1 mappings ri : Xi →G, rij : Xij →G and rX : X →G – for which, for all xi ∈Xi and xij ∈Xij, we have fij(xi, xj) = r−1 ij (g(ri(xi), rj(xj))) and fij,kl(xij, xkℓ) = r−1 X (g(rij(xij), rkℓ(xkℓ))). CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 10 of 21 Go Back Full Screen Close Quit 9. Groups and Abelian Groups: Reminder • A set G with an associative operation g(a, b) and a unit element e (g(a, e) = g(e, a) = a) is called a group – if every element is invertible, i.e., – if for every a, there exists an a′ for which g(a, a′) = e. • A group in which the operation g(a, b) is commutative is known as Abelian. CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 11 of 21 Go Back Full Screen Close Quit 10. Discussion • All continuous 1-D Abelian groups with order-preserving operations are isomorphic to (I R, +). • Here, (I R, +) is the additive group of real numbers, with g(a, b) = a + b. • Thus, we can conclude that all combining operations have the form fij(xi, xj) = r−1 ij (ri(xi) + rj(xj)). • Equivalently, fij(xi, xj) = y means that rij(y) = ri(xi) + rj(xj). CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 12 of 21 Go Back Full Screen Close Quit 11. Let Us Use Homogeneity • We will now prove that homogeneity leads exactly to the desired CES combinations. • Homogeneity means that if rij(y) = ri(xi)+rj(xj), then for every λ: rij(λ · y) = ri(λ · xi) + rj(λ · xj). • For each x′ i = xi + ∆xi, let us ﬁnd x′ j = xj + ∆xj for which ri(xi) + rj(xj) remains the same (thus, the combined value y remains the same): ri(x′ 1)+rj(x′ j) = ri(xi+∆xi)+rj(xj+∆xj) = ri(xi)+rj(xj). • For small ∆xi, ∆xj = k · ∆xi + o(∆xi) for some k. • Here, ri(xi+∆xi) = ri(xi)+r′ i(xi)·∆xi+o(∆xi), where, as usual, f ′ denotes the derivative. • Similarly, rj(xj + ∆xj) = rj(xj + k · ∆xi + o(∆xi)) = rj(xj) + k · r′ j(xj) · ∆xi + o(∆xi). CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 13 of 21 Go Back Full Screen Close Quit 12. Let Us Use Homogeneity (cont-d) • Thus, the above condition takes the form ri(xi)+rj(xj)+(r′ i(xi)+k·r′ j(xj))·∆xi+o(∆xi) = ri(xi)+rj(xj). • Thus, r′ i(xi) + k · r′ j(xj) = 0, and k = −r′ i(xi) r′ j(xj). • For re-scaled values, we similarly get k = −r′ i(λ · xi) r′ j(λ · xj), so r′ i(λ · xi) r′ i(xi) = r′ j(λ · xj) r′ j(xj) . • The right-hand side does not depend on xi, the left-hand side does not depend on xj, so: r′ i(λ · xi) r′ i(xi) = c(λ) for some c(λ). CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 14 of 21 Go Back Full Screen Close Quit 13. Let Us Use Homogeneity (cont-d) • Thus, the derivative Ri(xi) def = r′ i(xi) satisﬁes the func-tional equation: Ri(λ · xi) = Ri(xi) · c(λ) for all λ and xi. • It is known that every continuous solution to this equa-tion has the form r′ i(xi) = Ri(xi) = Ai · xαi i . • For diﬀerentiable functions, this can be proven if we diﬀerentiate both sides by λ and take λ = 1. • Then, we get xi · dRi dci = c · Ri. • Separating variables, we get dRi Ri = c · dxi xi . • Integration leads to ln(Ri) = c · ln(xi) + C1 and thus, to the desired formula r′ i(xi) = Ai · xαi i . CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 15 of 21 Go Back Full Screen Close Quit 14. Let Us Use Homogeneity (cont-d) • Integrating the above expression for r′ i(xi), we get ri(xi) = ai·xβi i +Ci and similarly, rj(xj) = aj ·xβj j +Cj. • One can easily check that homogeneity implies that βi = βj and Ci + Cj = 0, so ri(xi) + rj(xj) = ai · xr i + aj · xr j. • By considering a similar substitution between xi and y, we conclude that rij(y) = const · yr. • So, we indeed get the desired formula rij(xi, xj) = (ai · xr i + aj · xr j)1/r. • By using similar formulas to combine xij with xk, etc., we get the desired CES combination function. CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 16 of 21 Go Back Full Screen Close Quit 15. Possible Application to Copulas • A 1-D probability distribution of a random variable X can be described by its cdf FX(x) def = Prob(X ≤x). • A 2-D distribution of a random vector (X, Y ) can be similarly described by its 2-D cdf FXY (x, y) = Prob(X ≤x & Y ≤y). • It turns out that we can always describe F(x, y) as FXY (x, y) = CXY (FX(x), FY (y)) for some CXY (a, b). • This function CXY is called a copula. • For a joint distribution of several random variables X, Y , . . . , Z, we can similarly write FXY ...Z(x, y, . . . , z) def = Prob(X ≤x & Y ≤y & . . . & Z ≤z) = CXY ...Z(FX(x), FY (y), . . . , FZ(z)) for some CXY ...Z. CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 17 of 21 Go Back Full Screen Close Quit 16. Copulas (cont-d) • When we have many (n ≫1) random variables, then we need to describe a function of n variables. • Even if we use two values for each variable, we get 2n combinations. • For large n this is astronomically large. • Thus, a reasonable idea is to approximate the multi-D distribution. • A reasonable way to approximate is to use 2-D copulas. CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 18 of 21 Go Back Full Screen Close Quit 17. Copulas (cont-d) • For example, to describe a joint distribution of three variables X, Y , and Z: – we ﬁrst describe the joint distribution of X and Y as FXY (x, y) = CXY (FX(x), FY (y)), – and then use a copula CXY,Z to combine it with FZ(z): FXY Z(x, y, z) ≈CXY,Z(FXY (x, y), FZ(z)) = CXY,Z(CXY (FX(x), FY (y), FZ(z)). • Such an approximation, when copulas are applied to one another like a vine, are known as vine copulas. • It is reasonable to require that: – the result of the vine copula approximation – should not depend on the order in which we com-bine the variables. CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 19 of 21 Go Back Full Screen Close Quit 18. Copulas (cont-d) • In particular, for X, Y , Z, and T, we should get the same result in the following two situations: – if we ﬁrst combine X with Y , Z and T, and then combine the two results; or – if we ﬁrst combine X with Z, Y with T, and then combine the two results. • Thus, we require that for all possible real numbers x, y, z, and t, we get CXY,ZT(CXY (FX(x), FY (y)), CZT(FZ(z), FT(t))) = CXZ,Y T(CXZ(FX(x), FZ(z)), CY T(FY (y), FT(t))). • If we denote a = FX(x), b = FY (y), c = FZ(z), and d = FT(t), we conclude that for every a, b, c, and d: CXY,ZT(CXY (a, b), CZT(c, d)) = CXZ,Y T(CXZ(a, c), CY T(b, d)). CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 20 of 21 Go Back Full Screen Close Quit 19. Copulas: Conclusion • We have: CXY,ZT(CXY (a, b), CZT(c, d)) = CXZ,Y T(CXZ(a, c), CY T(b, d)). • This is exactly the above generalized associativity re-quirement; thus: – if we extend copulas to invertible operations, – then we can conclude that copulas can be re-scaled to associative operations ◦: C(a, b) = f(g(a) ◦h(b)) for some f, g, h. CES Production . . . How This Ubiquity Is . . . Second Requirement Main Idea Behind a . . . Derivation of the CES . . . Groups and Abelian . . . Discussion Let Us Use Homogeneity Possible Application to . . . Home Page Title Page ◀◀ ▶▶ ◀ ▶ Page 21 of 21 Go Back Full Screen Close Quit 20. Acknowledgments This work was supported in part: • by the National Science Foundation grants: – HRD-0734825 and HRD-1242122 (Cyber-ShARE Center of Excellence) and – DUE-0926721, • and by an award from the Prudential Foundation.
16216	https://nagyzoli.web.elte.hu/Thesis.pdf	Institute of Mathematics Ph.D. thesis Applications of the Combinatorial Nullstellensatz Zoltán Lóránt Nagy Doctoral School: Mathematics Director: Miklós Laczkovich, D.Sc. Professor, Member of the Hungarian Academy of Sciences Doctoral Program: Pure Mathematics Director: András Szűcs, D.Sc. Professor, Corresp. Member of the Hungarian Academy of Sciences Supervisors: András Gács, Ph.D. Gyula Károlyi, D.Sc. Tamás Szőnyi, D.Sc. Acknowledgments I am truly grateful to my supervisors, András Gács, Gyula Károlyi and Tamás Szőnyi for all their help and guidance. Their profound knowledge of combinatorics combined with generous and supporting attitude helped me in many ways: in moti-vation, careful writing, establishing a theory and most of all, learning how to study mathematical problems rather than simply solving interesting exercises and enjoying this kind of research work. I wish to thank all my co-authors for the inspiring work together. It was a really valuable experience for me to share ideas and study problems with Péter Csikvári, Tamás Héger, Balázs Patkós, Dömötör Pálvölgyi, Fedor Petrov, Ágnes Tóth and Máté Vizer. I am also thankful for the pleasant atmosphere provided under my PhD years espe-cially for Péter Csikvári, Tamás Héger, György Kiss, Erika Renáta Kovács, Marcella Takáts, Ágnes Tóth and Péter Sziklai. I received a diﬀerent kind of support too, from OTKA (number 81310). I am grateful to Kovács Csongorné and Sándor Dobos, my high-school maths teach-ers, for all their eﬀorts and patience. I was fortunate to participate in the splendid mathematical camps of Lajos Pósa, which were substantial experience in teamwork and in learning to ask good questions. The encouragement of László Lovász, Gyula Katona, Miklós Simonovits, András Gyárfás and János Barát meant me a lot. I am indebted to Dávid Kunszenti-Kovács and Tamás Héger for giving many valuable suggestions and thus improving the presentation of the thesis. I would like to thank my parents, grandparents, my siblings, all my friends and Anna for their constant love and support. 2 Contents 1 List of deﬁnitions and notations 5 2 Introduction 7 2.1 On the background and context of the Combinatorial Nullstellensatz of Alon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.2 Preliminary theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 3 Range of polynomials over ﬁnite ﬁelds and related problems 17 3.1 Introduction and combinatorial number theory point of view . . . . . 17 3.2 A result about polynomials of prescribed range . . . . . . . . . . . . . 19 3.3 A consequence about hyperplanes of a vector space over GF(q) . . . . 20 3.4 Proof of Theorem 3.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . 22 3.4.1 Easy combinatorial observations . . . . . . . . . . . . . . . . . 22 3.4.2 The algebraic tool . . . . . . . . . . . . . . . . . . . . . . . . . 25 3.4.3 The essential part of the proof . . . . . . . . . . . . . . . . . . 26 3.4.4 Proof for even q . . . . . . . . . . . . . . . . . . . . . . . . . . 34 3.5 Final remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 4 Extension to cyclic groups 38 4.1 Introduction and background . . . . . . . . . . . . . . . . . . . . . . 38 3 4.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 4.3 The case of odd order . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 4.4 Special cases of Theorem 4.1.3 . . . . . . . . . . . . . . . . . . . . . . 48 4.5 The case of even order . . . . . . . . . . . . . . . . . . . . . . . . . . 51 4.6 Abelian groups and sumsets - related topics . . . . . . . . . . . . . . 55 5 Quantitative Nullstellensatz and applications concerning Dyson-type polynomials and their q-analogues 60 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 5.2 The Dyson-identity and the q-analogue . . . . . . . . . . . . . . . . . 61 5.3 Generalizations and q-analogues . . . . . . . . . . . . . . . . . . . . . 66 5.4 The proof of the general identity . . . . . . . . . . . . . . . . . . . . . 71 5.4.1 The choice for the multisets Ai . . . . . . . . . . . . . . . . . 72 5.4.2 The combinatorics . . . . . . . . . . . . . . . . . . . . . . . . 73 5.4.3 The computation . . . . . . . . . . . . . . . . . . . . . . . . . 75 5.4.4 The rationality result . . . . . . . . . . . . . . . . . . . . . . . 77 5.5 Remarks and variations . . . . . . . . . . . . . . . . . . . . . . . . . . 80 6 Summary 83 7 Összefoglaló - in Hungarian 85 8 Bibliography 87 4 Chapter 1 List of deﬁnitions and notations N: the set of non-negative integers. Z+: the set of positive integers. Zm: the cyclic group of order m. Fp, GF(p) : the ﬁnite ﬁeld of order p, where p is a prime. Fq, GF(q) the ﬁnite ﬁeld of order q, where q is a prime power. char(F): the characteristics of the ﬁeld F G: an abelian group. (a, b): the greatest common divisor of the integers a and b. [a, b]: The set of integers between a and b inclusive, where a and b are both integers and a ≤b. (Sec 5.) I: an interval of type [a, b]. (Sec 5.) A: the complement of the set A with respect to the groundset. qA := q{a1, a2, . . . an} := {qa1, qa2, . . . , qan}, for a set A = {a1, a2, . . . an}. \|A\|: the cardinality of the (multi)set A. m(A): the greatest multiplicity in the multiset A. m(a), a ∈A: the multiplicity of the element a in the multiset A. k-set: a set of k elements 5 x + A: the translation of the multiset A of a ﬁeld K with an element x ∈K, that is, x + A = {x + ai : i ∈[1, n]} if A = {a1, . . . , an}. xA: the dilation of the multiset A of a ﬁeld K with an element x ∈K, that is, xA = {x · ai : i ∈[1, n]} if A = {a1, . . . , an}. A + B: the sumset of the (multi)sets A and B, A + B := {a + b : a ∈A, b ∈B}. ×Ai (i ∈[1, n]): the set of n-tuples (a1, . . . , an) : ai ∈Ai. Symn: the symmetric group on [1, n], i.e. the group whose elements are all the permutations of the n symbols, and whose group operation is the composition of such permutations. x: the vector x = (x1, . . . , xn). 1: the vector 1 = (1, 1, . . . , 1) (of suitable length). B: the matrix B = ((βij)). P(x, B): the Laurent polynomial P(x, B) := Q 0 di, there exists a k-tuple (s1, s2, . . . sk) ∈×Ai for which P(s1, s2, . . . sk) ̸= 0. For the sake of completeness we include a short proof. Proof. Suppose the result is false. We may assume that \| Ai \|= di + 1. Deﬁne the polynomials gi(xi) := Q s∈Ai(xi −s). Theorem 2.1.1 implies that P = Pk i=1 higi, where deg(hi) ≤Pk i=1 di −deg(gi) holds for all indices i. Consider the coeﬃcient of the monomial Qk i=1 xdi i , whose degree equals the total degree of P. On the one hand, it is nonzero. On the other hand, if the degree of a summand higi = hi Q s∈Ai(xi−s) is equal to the degree of P, it should be divisible by xdi+1 i , which leads to a contra-diction. Remark 2.1.3. [68, 73] The non-vanishing lemma can be stated in a slightly stronger form, as follows. We may also consider any monomial Qk i=1 xdi i of the polynomial P = P(x1, . . . , xk) with nonzero coeﬃcient, if there is no monomial Qk i=1 xδi i of P = P(x1, . . . , xk) such that δi ≥di for all i, aside from Q xdi i . That is, instead of taking a monomial of maximum degree, we may consider any maximal monomial with respect to the natural partial order associated to the exponent sequence of the ordered variables x1, x2, . . . , xk. In ordinary cases, this variant is used to show lower bounds on the size of some combinatorial objects, or at least an existence of a combinatorial object. Usually the application of the method does not provide a suitable structure, only shows the existence. To describe the phenomenon, we point out the key steps in order to 9 apply the theorem, and prove the Erdős-Ginzburg-Ziv theorem (prime case) using this technique, which will be a reference point later on. The Erdős-Ginzburg-Ziv theorem can be interpreted as a starting point of zero-sum theory, and many more general research. For more details, we refer to [17, 23, 40]. Theorem 2.1.4 (Erdős-Ginzburg-Ziv). Let (a1, a2, . . . , a2p−1) be a sequence of 2p−1 elements of Zp, where p is a prime. Then there exists a subsequence of length p, in which the sum of the elements equals zero. Proof. Step 1. To the contrary, assume that some set, associated to the problem, is small. In this case, the set in view is the set S of element of Fp admitted by the sum of a subsequence of p elements. That is, S is the set of possible sum values, and we suppose that 0 is not admitted. Step 2. Associate variables to the problem, and ﬁnd a polynomial which is vanishing on the set described in Step 1. Let variable yi take value 1 if ai appears among the summands of the sum, and take value 0 if ai does not appear. Hence, the polynomials f1(y1, . . . , y2p−1) := 2p−1 X i=1 yi and f2(y1, . . . , y2p−1) := 2p−1 X i=1 aiyi takes value zero, if and only if the number of summands is divisible by p, and if the sum is divisible by p, respectively. Since xp−1 = 1 for each x ∈Fp\0, the polynomial 1 −f p−1 1 (y1, . . . , y2p−1) 1 −f p−1 2 (y1, . . . , y2p−1) − 2p−1 Y i=1 (1 −yi) takes value zero, if and only if the number of summands is p and the sum is zero (mod p). Step 3. Determine the total degree, and a suitable monomial whose degree equals the total degree. Note that the degree of our polynomial is 2p −1. Clearly, the monomial Q2p−1 i=1 yi has coeﬃcient diﬀerent from zero. Step 4. Set a suitable set system Ai and apply Theorem 2.1.2 to the polynomial. The condition on any set Ai is to contain more than 1 element, hence the choice Ai = {0, 1} suits the requirements. Then Theorem 2.1.2 implies that the polynomial cannot vanish on the Cartesian product {0, 1}2p−1. This is a contradiction, which means that there must be a subsequence of p elements and of sum zero. 10 To emphasize the eﬃciency of this method, we note that it recently helped to solve the well known Erdős-Heilbronn conjecture (see ), one of Snevily’s conjectures (see ) and the ﬁnite ﬁeld Kakeya conjecture (see ). Another easy, yet a bit more delicate consequence of Theorem 2.1.1 is the following Corollary 2.1.5. If a polynomial P = P(Y1, . . . , Yk) over a ﬁnite ﬁeld Fq vanishes for all substitutions, then it can be written in the following form: P(Y1, . . . , Yk) = h1(Y q 1 −Y1) + · · · + hk(Y q k −Yk), where the hi are polynomials in Y1, . . . , Yk of total degree at most deg(P) −q. Proof. Applying Theorem 2.1.1 with Ai := Fq and gi(Yi) := Q s∈Fq(Y −s) = Y n −Y implies the statement. The general form of the Nullstellensatz is rather rarely used. It was ﬁrst applied in a paper of Károlyi . In Section 3., we contribute to the applications of Theorem 2.1.5. As it was pointed out in several papers and surveys [1, 72, 92], Theorem 2.1.1 may be considered as some generalization of the Lagrange interpolation. Indeed, the following facts are well known. Proposition 2.1.6. If a polynomial P ∈F[x] is a non-zero polynomial with deg(P) ≤d (d ≥1 is an integer), then P vanishes at at most d elements of F (has at most d roots in F). Conversely, for any given set A of cardinality at most d, there exists a polynomial of degree at most d that vanishes on the whole set A. That is, to obtain an upper bound on the size of a one-dimensional set A, it would suﬃce to exhibit a non-zero low-degree polynomial that vanishes on A. On the other hand, to bound the size of a set A from below, one would have to show that the only low-degree polynomial that vanishes on A is the zero polynomial. Proposition 2.1.7. [Lagrange interpolation] Suppose P is a polynomial of one vari-able over F, and deg(P) ≤d. If one knows its values P(s) at d + 1 distinct points 11 s1, s2, . . . , sd+1, then P(x) is determined by the formula P(x) = d+1 X i=1 P(si) d+1 Y j=1 j̸=i x −sj si −sj . Clearly, the former proposition’s ﬁrst assertion is a consequence of that of the lat-ter. The non-vanishing polynomial lemma (Theorem 2.1.2) may be considered as a generalization of Proposition 2.1.6 for multivariate polynomials. Our next aim is to introduce another lemma for multivariate polynomials, which can be consid-ered as an analogue of the Lagrange interpolation technique, and was ﬁrst observed independently by Lasoń ; and Karasev and Petrov . This form easily im-plies the non-vanishing form of the Nullstellensatz (Theorem 2.1.2) and will play a fundamental part in the proofs of constant term identities in Section 5. Lemma 2.1.8 (Quantitative Nullstellensatz). Let F be a ﬁeld, and let P ∈F[x1, x2, . . . , xm], be a multivariate polynomial for which deg(P) ≤d1+d2+. . .+dm. Take an arbitrary set system A1, A2, . . . , Am such that Ai ⊆F and \|Ai\| = di + 1. Then the coeﬃcient of Q xdi i is X z1∈A1 X z2∈A2 X zm∈Am P(z1, z2, . . . zm) φ′ 1(z1)φ′ 2(z2) · · · φ′ m(zm), where φi(x) = Q a∈Ai(x −a). Proof. Assume that P is a monomial. This implies the lemma for arbitrary P, as the coeﬃcient, and the formula as well is linear. Indeed, X z∈×Ai (P1 + P2)(z1, z2, . . . zm) φ′ 1(z1) · · · φ′ m(zm) = X z∈×Ai P1(z1, z2, . . . zm) φ′ 1(z1) · · · φ′ m(zm) + X z∈×Ai P2(z1, z2, . . . zm) φ′ 1(z1) · · · φ′ m(zm). Observe that if n = 1, and \|A\| := d + 1 = deg(P) + 1, then P(x) = X si∈A P(si) d+1 Y j=1,j̸=i x −sj si −sj = X si∈A P(si) φ′(si) · φ(x) (x −si) according to the Lagrange interpolation formula (Proposition 2.1.7), thus the coef-ﬁcient of xd is clearly X si∈A P(si) φ′(si). 12 If we choose P to be an m-variate monomial Q xdi i , and set system A1, A2, . . . , Am such that Ai ⊆F and \|Ai\| = di + 1, we obtain X z∈×Ai P(z1, z2, . . . zm) φ′ 1(z1)φ′ 2(z2) · · · φ′ m(zm) = X z∈×Ai zd1 1 zd2 2 · · · zdm m φ′ 1(z1)φ′ 2(z2) · · · φ′ m(zm) = = m Y i=1 X zi∈Ai zdi i φ′ i(zi). Apply the n = 1 case, and we get that this is exactly the coeﬃcient of the product xd1 1 · xd2 2 · · · · xdm m . Remark 2.1.9. This lemma implies the statement of Theorem 2.1.2. Indeed, if a polynomial P of degree d = d1 + d2 + . . . + dm vanishes on a Cartesian product ×Ai, \|Ai\| = di + 1, then each summand is zero in the expression of the coeﬃcient of Q xdi i , a contradiction. Lemma 2.1.8 seems somewhat weaker in higher dimension than the (multivariate version of the) Lagrange interpolation (see e.g. in ), as it provides only the leading coeﬃcient. However, it assures a straightforward way to determine the leading coeﬃcient of arbitrary n-variate polynomial, with no particular restriction on the interpolation subsets Ai - except their cardinality. Finally, we end this section by another generalization of Theorem 2.1.2 and Lemma 2.1.8. The main idea is the following. So far, we associated subsets Ai of F to each variable xi, to express that the zero locus of a polynomial can ( or can not ) contain the corresponding Cartesian product ×Ai. Instead of subsets, we may consider here multisets as well, that is, multiplicities for each common zero. Although the original non-vanishing lemma can not deal with it, recent papers of Kós, Rónyai, and Kós, Mészáros, Rónyai [70, 71] extend the result. Before we state it, we introduce some notions. For each s ∈Ai, mi(s) will denote its multiplicity in Ai. (Hence, the sum of the multiplicities equals the cardinality of the multiset.) It is well known that for an arbitrary s ∈Fn, we can express any polynomial P(x) ∈ F[x1, . . . , xn] as 13 P(x) = X u∈Nn Cu,s n Y i=1 (xi −si)ui, where the coeﬃcients Cu,s are uniquely determined by P, u and s. Note that if Pn i=1 ui > deg(P) holds then Cu,s is zero, while in case of equality, that is, Pn i=1 ui = deg(P), then Cu,s denotes the coeﬃcient of Qn i=1 xiui in P which is independent of s. Furthermore, observe that C0,s = P(s) for u = 0. Theorem 2.1.10. Let F be a ﬁeld, P = P(x1, . . . , xn) ∈F[x1, . . . , xn] be a polynomial of degree Pn i=1 di, where each di is a nonnegative integer. Assume that the coeﬃcient of the monomial Qn i=1 xdi i is nonzero in P. Suppose further that A1, A2, . . . An are multisets of F such that for the size \|Ai\| > di (i = 1, . . . , n). Then there exists a point s = (s1, . . . , sn) ∈×Ai and an exponent vector u = (u1, . . . , un) ∈ Nn with ui < mi(si) for each i, such that Cu,s ̸= 0. Remark 2.1.11. If each multiplicity is 1, the theorem gives back the non-vanishing theorem 2.1.2. Indeed, that would mean there exists a point s = (s1, . . . , sn) ∈×Ai such that Cu,s = P(s) ̸= 0. Theorem 2.1.10 can be generalized in the spirit of Lemma 2.1.8. Theorem 2.1.12 (Multiplicity version of the Quantitative Nullstellensatz). Let P ∈F[x1, . . . , xn] be a multivariate polynomial such that no monomial majorizes the term Q xdi i in P. Let A1, . . . , An be arbitrary multisets in F with corresponding multiplicity functions m1, . . . , mn such that \|Ai\| = di + 1 for every i. Assume that either char(F) = 0 or char(F) ≥max mi(c) (i ≤n, c ∈F). Then the coeﬃcient of Q xdi i in P can be evaluated as X s1∈A1 X u1<m1(s1) · · · X sn∈An X un<mn(sn) n Y i=1 κ(Ai, si, ui) ∂u1+···+unP ∂xu1 1 . . . ∂xun n (s1, . . . , sn), where κ(Ai, si, ui) = 1 ui! · (mi(si) −1 −ui)! · 1 Q c∈Ai{si}(x −c)ui(c) !(mi(si)−1−ui) x=si . 14 Consequently, if the coeﬃcient of Qn i=1 xdi i in P is not zero, then there exists a system of representatives si ∈Ai and multiplicities ui < ui(si) such that ∂m1+···+mnF ∂xm1 1 . . . ∂xmn n (s1, . . . , sn) ̸= 0. The latter two theorems can be considered as the generalization of the non-vanishing lemma built on Hermite interpolation. For more details and applications of the Combinatorial Nullstellensatz, we refer to [1, 72, 92]. We also mention further generalizations of the Nullstellensatz, see . 2.2 Preliminary theorems Here we introduce some basic facts and theorems, that will be needed later. Lemma 2.2.1 (Lucas). Let the p-adic expansion of n and k be n = Pr i=1 nipi−1 and k = Pr i=1 kipi−1, respectively. Then n k ≡ n1 k1 · · · nr kr (mod p). For a proof, see . We will use this often without explicitly referring to it. Lemma 2.2.2 (About the power sums of the elements of GF(q)). For arbitrary ﬁnite ﬁeld GF(q), P x∈GF(q) xk = 0 when 1 ≤k ≤q −2, and P x∈GF(q) xq−1 = −1. Polynomials over a ﬁnite ﬁeld may be considered as polynomials with a bounded degree. Indeed, if P(x) ∈Fq[x] has the form P(x) = cnxn +cn−1xn−1 +. . .+c1x+c0 and n ≥q, then P(x) and P(x) −cn(xq −x)xn−q are identical as functions, since xq −x = 0 for all x ∈Fq. Thus any polynomial over the ﬁeld GF(q) can be represented by a polynomial of degree at most q−1. In fact, any function over GF(q) can be represented by a polynomial of degree at most q −1 and this representation is unique. The number of functions over GF(q) is qq, which is equal to the number of polynomials of degree at most q −1. It is clear that these polynomials cannot represent the same function, which conﬁrms the statement. 15 For any function f ∈Fq[x], the corresponding polynomial is called the reduced polynomial, and its degree is called the reduced degree of f. Lemma 2.2.3. Suppose that f(x) = cq−1xq−1 + · · · + c0 is a polynomial over GF(q). Then P x∈GF(q) f(x) = −cq−1 and P x∈GF(q) xf(x) = −cq−2. Proof. Apply Lemma 2.2.2 to P x∈GF(q) f(x) = P i≤q−1 P x∈GF(q) cixi and to P x∈GF(q) xf(x) = P i≤q−1 P x∈GF(q) cixi+1. 16 Chapter 3 Range of polynomials over ﬁnite ﬁelds and related problems 3.1 Introduction and combinatorial number theory point of view This section is devoted to a result formulated in three diﬀerent terminologies. We start with a result in combinatorial number theory which might resemble Snevily’s conjecture . Then we derive two consequences (which are essentially equivalent to the original result), one about the range of polynomials over a ﬁnite ﬁeld, and one about hyperplanes in a vector space over a ﬁnite ﬁeld fully lying in the union of certain ﬁxed hyperplanes. Although perhaps the consequence about the range of polynomials solves a more natural question, our proof is most easily formulated in the additive combinatorial terminology, so we start with this result. It was motivated by a result of Stéphane Vinatier . Theorem 3.1.1. Let {a1, a2, . . . , ap} be a multiset in the ﬁnite ﬁeld GF(p), p prime. Then after a suitable permutation of the indices, either P i iai = 0, or the multiset consists of a (p −2 times), a + b and a −b (each once) for some ﬁeld elements a and b, b ̸= 0. 17 In the paper Vinatier proves a similar result (though with a slightly diﬀerent terminology) with the extra assumption that a1, . . . , ap, when considered as integers, satisfy a1 + · · · + ap = p. Before going further, let us recall that Snevily’s conjecture states that for any abelian group G of odd order (written multiplicatively), and positive integer n ≤\|G\|, for any sets {a1, . . . , an} and {b1, . . . , bn} of elements of G, there is a permutation π of the indices such that the elements a1bπ(1), a2bπ(2), . . . ,anbπ(n) are diﬀerent. Alon proved this for groups of prime degree and later Dasgupta, Károlyi, Serra and Szegedy for cyclic groups. Alon’s result is in fact more general: he only assumes that {a1, . . . , an} is a multiset. Let us remark that if this general version were true for cyclic groups (it is obviously not), then there would be no exception in Theorem 3.1.1, and the proof would easily follow from this general version. Theorem 3.1.1 will follow from the following more general result, where p is replaced by an arbitrary prime power q. Theorem 3.1.2. Let {a1, a2, . . . , aq} be a multiset in the ﬁnite ﬁeld GF(q). There are no distinct ﬁeld elements b1, b2, . . . , bq such that P i aibi = 0 if and only if after a suitable permutation of the indices, a1 = a2 = · · · = aq−2 = a, aq−1 = a + b, aq = a −b for some ﬁeld elements a and b, b ̸= 0. Note that if we let q = p, p prime in Theorem 3.1.2, then we get Theorem 3.1.1 (since q diﬀerent elements are in fact all the elements in some permutation). We may formulate it in a more general form which follows easily from the n = q case. Corollary 3.1.3. Let {a1, a2, . . . , an} be a multiset in the ﬁnite ﬁeld GF(q), with n ≤q. Then one can ﬁnd distinct ﬁeld elements b1, b2, . . . , bn such that P i aibi = 0, unless one of the following holds: (i) n = q and after a suitable permutation of the indices, a1 = a2 = · · · = aq−2 = a, aq−1 = a + b, aq = a −b for some ﬁeld elements a and b, b ̸= 0. (ii) n = q −1, and after a suitable permutation of the indices, a1 = a2 = · · · = aq−2 = a, aq−1 = 2a for a ﬁeld element a ̸= 0. 18 (iii) n < q −1 and after a suitable permutation of the indices, a1 = a2 = · · · = an−2 = 0, an−1 = b, an = −b for a ﬁeld element b ̸= 0. (iiii) n = q −2, q is even, and after a suitable permutation of the indices, a1 = a2 = · · · = an−2 = a for a ﬁeld element a ̸= 0. Proof. If n < q, then extend the set of ais to a set of size q with an+1 = · · · = aq = 0, then apply the theorem. In Subsections 2 and 3 we derive two consequences of Theorem 3.1.2. The proof of the theorem will be given in Section 4. Finally, Section 5 is devoted to remarks and open problems. Finally, let us recall the version of the Combinatorial Nullstellensatz that will be a key ingredient in this section, namely, the formerly introduced Theorem 2.1.5. Theorem 3.1.4. If a polynomial G(Y1, . . . , Yk) over the ﬁnite ﬁeld GF(q) vanishes for all substitutions, then it can be written in the following form: G(Y1, . . . , Yk) = (Y q 1 −Y1)f1 + · · · + (Y q k −Yk)fk, where the fis are polynomials in Y1, . . . , Yk of degree at most deg(G) −q. 3.2 A result about polynomials of prescribed range In this section we give another formulation of Theorem 3.1.2. Although it might seem to be a consequence, it is essentially equivalent to the original result. For a multiset M of size q of ﬁeld elements we say that M is the range of the polynomial f if M = {f(x) : x ∈GF(q)} as a multiset (that is, not only values, but also multiplicities need to be the same). Suppose we have a multiset M and wish to ﬁnd a low degree polynomial with range M. By Lemma 2.2.3, if the sum of elements of M is not zero, then every reduced polynomial of this range will have reduced degree q −1 and vice versa, if the sum is zero, then a reduced polynomial of range M will automatically have degree at most q −2. 19 Theorem 3.2.1. Let M = {a1, . . . , aq} be a multiset in GF(q), with a1+· · ·+aq = 0. There is no polynomial with range M of reduced degree at most q −3 if and only if M consists of q −2 a’s, one a + b and one a −b for some ﬁeld elements a and b, b ̸= 0. Proof. By Lemma 2.2.3, polynomials with range M have reduced degree q −1 if and only if P ai ̸= 0. Since P ai = 0, the second statement of Lemma 2.2.3 shows that a polynomial f with range M has reduced degree at most q −3 if and only if P x xf(x) = 0. On the other hand, there is a bijection between polynomials with range M and the ordered sets (b1, . . . , bq) (that is, permutations) of GF(q): a permutation corresponds to the function f(bi) = ai. Under this correspondence the condition P x xf(x) = 0 translates to P aibi = 0. Hence our claim follows from Theorem 3.1.2. Though the statement of the above theorem looks very innocent, it seems that one needs the whole machinery of Section 4 for the proof. After this result, the natural question is to look for polynomials of degree lower than q −3 with prescribed range. One might conjecture that the only reason for a multiset (with sum equal to zero) not to be the range a polynomial of degree less than q −k is that there is a value of multiplicity at least q −k. (Note that a value of multiplicity m ≤q −1 in the range guarantees that any polynomial of this range has degree at least m, since the corresponding reduced polynomial f is such that f −a has m roots in GF(q)). We will get back to this at the end of the Chapter. 3.3 A consequence about hyperplanes of a vector space over GF(q) In this section we prove a result about vector spaces over ﬁnite ﬁelds, which is again essentially equivalent to Theorem 3.1.2 Let q denote a prime power and denote by V the vector space of dimension n over the ﬁnite ﬁeld GF(q) consisting of all n-tuples (x1, x2, . . . , xn). Finally, denote by Hij the hyperplane with equation xi = xj (i ̸= j). We are interested in hyperplanes 20 fully contained in ∪i̸=jHij. Note that if n > q, then by the pigeon-hole principle the whole space is contained in this union, so the problem is non-trivial only for n ≤q. Our main result is the following. Theorem 3.3.1. Suppose that n ≤q and H ⊆∪i̸=jHij is a hyperplane in V , H ̸= Hij for any i ̸= j. Then one of the following holds: (i) n = q, H = {(x1, . . . , xn) : c(xj −xk) + P i xi = 0} for a ﬁeld element c ̸= 0 and indices j ̸= k; (ii) n = q −1, H = {(x1, . . . , xn) : xj + P i xi = 0} for an index j. Proof. Let H =< (a1, . . . , an) >⊥. The condition that H is contained in ∪i̸=jHij translates to the condition that whenever a1x1 + · · · + anxn = 0, necessarily xi = xj for an i ̸= j, or equivalently, there are no distinct elements x1, . . . , xn such that a1x1 + · · · + anxn = 0. Hence we are in (i) or (ii) or (iii) of Corollary 3.1.3. It is easy to see that Corollary 3.1.3 (i) implies (i) of the theorem being proved. If we have (ii) from Corollary 3.1.3, then (ii) holds here, ﬁnally, from 3.1.3 (iii) we get that H = Hij for an i and j, a contradiction. It is not diﬃcult to see that the hyperplanes given in (i) and (ii) are really contained in the union. Finally we show that aﬃne hyperplanes only give one more example. Theorem 3.3.2. All aﬃne hyperplanes contained in ∪i̸=jHij are linear (for n ≤q), except when n = q and the hyperplane is a translate of (1, . . . , 1)⊥. Proof. Suppose the aﬃne hyperplane {(x1, . . . , xn) : a1x1 + · · · + anxn = c} is con-tained in ∪i̸=jHij. First choose arbitrary distinct ﬁeld elements x1, . . . , xn. Let d = a1x1 + · · · + anxn. By the assumption, d ̸= c. If d ̸= 0, then ( c dx1, . . . , c dxn) is in our hyperplane, a contradiction, unless c = 0, which is what we wanted to prove. If d = 0, then interchange the values of two coordinates, xi and xj say, to obtain a1x1+· · ·+anxn = (ai−aj)(xj −xi). This is non-zero for well-chosen i and j (unless all the ais are the same), so we can use the above trick to prove c = 0. 21 Finally, if all the ais are the same, say a1 = · · · = an = 1, then one can easily ﬁnd distinct xis to obtain a1x1 + · · · + anxn ̸= 0 (and use the above trick), unless n = q, which was the exceptional case in the claim. 3.4 Proof of Theorem 3.1.2 The proof will be carried out in several steps. We will assume q ≥11. Small cases can be handled easily. We will also suppose q is odd in general, though our combinatorial observations in the ﬁrst section hold also for q even, except Lemma 3.4.2. For the proof of the even case (which is relatively easier) see the last part of the present section. In Subsection 1 we make some easy observations (with elementary combinatorial proofs). As we will see, the theorem easily follows from the n = q case (that is why results in Sections 2 and 3 are essentially equivalent to the result being proved). In Subsection 2, using algebraic methods, we will derive an identity about a poly-nomial that will reﬂect the combinatorial properties of a multiset {a1, . . . , ak} for which one cannot ﬁnd distinct ﬁeld elements b1, . . . , bk such that a1b1+· · ·+akbk = 0. The proof will be another application of the Combinatorial Nullstellensatz, in the spirit of Károlyi’s approach . The essential part of the proof of Theorem 3.1.2 will be carried out in Subsection 3, where (after supposing that one cannot ﬁnd distinct ﬁeld elements b1, . . . , bq such that a1b1 + · · · + aqbq = 0), we will use the information gained in Subsection 2 to deduce ﬁrst that most of the ais are equal, and later that exactly q −2 of them are equal. Subsection 4 will be devoted to the q even case. 3.4.1 Easy combinatorial observations Lemma 3.4.1. If for a multiset {a1, . . . , aq} there is no ordering b1, . . . , bq of the elements of GF(q) such that P aibi = 0, then the same holds for any translation {a1 + c, . . . , aq + c} and any non-zero multiple {ca1, . . . , caq}. 22 Proof. Straightforward. Note that if the ais are diﬀerent, then it is easy to ﬁnd a suitable ordering for which P i biai = 0 holds (for instance let bi = ai). Hence by the previous lemma, we can suppose that 0 is not among the ais. Lemma 3.4.2. Theorem 3.1.2 is true if n = q odd and the ais admit at most 3 diﬀerent values. Proof. If all the ais are the same, then any ordering results in P i aibi = 0, so suppose there are at least two values. After transformation suppose that 0 is the value with largest multiplicity and the remaining two values are 1 and a (here a = 1 is possible). First suppose a = 1 and that the 1-s are a1 = · · · = am = 1. We determine an appropriate ordering recursively. Let b1 ̸= 0 be arbitrary, b2 = −b1, b3 any non-zero value, which has not been used, b4 = −b3,... If m is even, then after we determined the ﬁrst m bis, the rest of the values are arbitrary. If m is odd, then bm = 0 and the rest is arbitrary. Next suppose a ̸= 1 and that a1 = · · · = am = 1, am+1 = · · · = am+l = a, and the rest is zero. If at most one of m and l is odd, then we can do the same as above. If m and l are both odd, then we can get rid of one 1 and one a by letting b1 = −a and bm+1 = 1 and do the same trick as above for the rest of the values (note that q is large enough and m + l < 2q/3). This does not work if a = −1. If m = l = 1, then we have that our set is q −2 zeros, a 1 and a −1, this is the exceptional case of the claim of the theorem. If one of them, m say, is at least 3, then b1 = A, b2 = B, b3 = C, bm+1 = A + B + C with well-chosen A, B and C, and the same trick can be applied again. In subsection 3, using algebraic tools we will be able to prove equations of the form (a1 −a2)(a2 −a3)... = 0 for any permutation of the indices. From this, we will try to deduce that most of the ais are the same. The following easy observations will be very useful tools for this. 23 Lemma 3.4.3. Suppose the multiset {a1, . . . , ak} contains at least 3 diﬀerent values and denote by l the maximal multiplicity in the set. Let m1, m2 and m3 be natural numbers with m1 + 2m2 + 3m3 = k. Then one can partition the ais into m3 classes of size 3, m2 classes of size 2 and m1 classes of size 1 in such a way, that elements in the same class are pairwise diﬀerent, provided we have one of the following cases. (i) m2 = 0, m1 = 1, l ≤m3; (ii) m2 = 1, m1 = 0, l ≤m3 + 1; (iii) m3 = 0, l ≤m1 + m2; (iv) m3 = 1, m2 = 0, l ≤m1; (v) m3 = 1, m2 = 1, l ≤m1 + 1. Proof. First permute the ais in such a way that equal elements have consecutive indices. This implies that if \|i −j\| ≥l, then ai and aj are diﬀerent. (i) We have k = 3m3 + 1 and l ≤m3. Let the i-th class consist of ai, ai+m3 and ai+2m3 for i = 1, . . . , m3; and let ak be the last class (of size 1). (ii) We have k = 3m3 + 2 and l ≤m3 + 1. Let the i-th class consist of ai, ai+m3+1 and ai+2m3+2 for i = 1, . . . , m3; and let am3+1 and a2m3+2 form the last class (of size 2). (iii) We have k = 2m2 + m1 and l ≤m1 + m2. Let the i-th class consist of ai and ai+m1+m2 for i = 1, . . . , m2; and the rest of the classes (of size 1) is arbitrary. (iv) We know that our multiset has at least three diﬀerent values, that is all we need for this case. (v) If we have at least 4 diﬀerent values, then it is easy to see that the arrangement is possible. If there are exactly 3 diﬀerent values, then at least two values occur at least twice because we have at least 5 elements, and it is again easy to ﬁnd the desired arrangement. 24 3.4.2 The algebraic tool After the above easy observations, we introduce the main tool of the proof. Theorem 3.4.4. Suppose a1, . . . , ak are non-zero ﬁeld elements with the property that there are no distinct ﬁeld elements b1, . . . , bk such that P i aibi = 0. Deﬁne the following polynomial: G(Y1, . . . , Yk) = (Y1 + · · · + Yk)q−1 −1 D(Y1, . . . , Yk), where D is the following determinant: ak−1 1 ak−2 1 Y1 ak−3 1 Y 2 1 . . . Y k−1 1 . . . . . . . . . . . . . . . ak−1 k ak−2 k Yk ak−3 k Y 2 k . . . Y k−1 k Then G(Y1, . . . , Yk) = k X i=1 (Y q i −Yi)fi, where the fis are polynomials in Y1, . . . , Yk of degree at most the degree of G minus q. Proof. First we will prove that the following polynomial vanishes for all substitu-tions: F(X1, . . . , Xk) = (a1X1 + · · · + akXk)q−1 −1 Y 1≤i<j≤k (Xi −Xj). Note that Q 1≤i 9 for an odd prime p and h > 1, and that there is no ordering b1, . . . , bq of the elements of GF(q) such that P i aibi = 0. Then at least q+3 2 of the ais are the same. Proof. The proof is similar to the previous one, but it will be much more diﬃcult to determine the coeﬃcient of the appropriate term in G. After transformation suppose 0 is not among the ais. Consider the polynomial G from Theorem 3.4.4 with k = q. By 3.4.4, terms of maximal degree of G have at least one Yi with degree at least q. The term to give information about the ais this time is the following: ( q Y i=1 Y i−1 i ) · (Y1Y3Y5 · · · Y2p−3)(Y2p−1Y2p · · · Y3p−3)p(Yp2+1Yp2+2 · · · Yp2+p−1)p2 · · · · · · (Yph−1+1Yph−1+2 · · · Yph−1+p−1)ph−1 The degree of this term is 1+2+· · ·+(q−1)+(p−1)(1+p+p2+· · ·+ph−1) = q 2 +q−1, this is the degree of G. A little calculation shows that all Yis have degree at most q −1 in this term. It is easy to see that one way to get this term in G is to take Qq i=1 Y i−1 i from the Vandermonde part and the rest from (Y1 + · · · + Yq)q−1. We will prove that besides this, the only way to get this term with a non-zero coeﬃcient is to interchange the role of some pairs of variables with the same degree. These pairs are: Y1 and Y2 (both of degree 1), Y3 and Y4 (both of degree 3),..., Y2p−3 and Y2p−2 (both of degree 2p −3); Y2p−1 and Y3p−1 (both of degree 3p −2), Y2p and Y3p (both of degree 3p −1),...,Y3p−3 29 and Y4p−3 (both of degree 4p −4); Yp2+1 and Y2p2+1 (both of degree 2p2), Yp2+2 and Y2p2+2 (both of degree 2p2 + 1),...,Yp2+p−1 and Y2p2+p−1 (both of degree 2p2 + p −2); ...; Yph−1+1 and Y2ph−1+1 (both of degree 2ph−1), Yph−1+2 and Y2ph−1+2 (both of degree 2ph−1 + 1),...,Yph−1+p−1 and Y2ph−1+p−1 (both of degree 2ph−1 + p −2). Let us look for the term in question. From the Vandermonde part, all terms are of the form Y 0 π(1) · · · Y q−1 π(q) for a permutation π of the indices. In the term in question, we have only two Yis of degree less than 2: Y1 and Y2, hence {π(1), π(2)} = {1, 2}. Similarly we get that {π(2k −1), π(2k)} = {2k −1, 2k} for k ≤p −1. This shows that the ﬁrst part of the term coming from (Y1 +· · ·+Yq)q−1 is Yπ(1)Yπ(3) · · · Yπ(2p−3). The coeﬃcient of such a term in (Y1 + · · · + Yq)q−1 is (q −1)(q −2) · · · (q −p + 1) times something depending on the degrees of the rest of the Yis. If the degree of any of the rest of the Yis is not divisible by p, then (by Lucas’ theorem) the coeﬃcient is zero, since it is divisible by (q −1)(q −2) · · · (q −p+1) q−p k with a k not divisible by p. Hence we only have to consider those possibilities, when the term coming from (Y1 + · · · + Yq)q−1 starts with Yπ(1)Yπ(3) · · · Yπ(2p−3) and continues with all the Yis having degree divisible by p. So far we have identiﬁed all Yis come from the Vandermonde part of degree at most 2p−3. After this in the term in question we have (Y2p−1Y3p−1)3p−2(Y2pY3p)3p−1 · · · (Y3p−3Y4p−3)4p−4. These should come from the Vandermonde part from the terms of degrees between 2p −2 and 4p −4. Since we know that the corresponding terms of the part coming from (Y1 + · · · + Yq)q−1 all need to have degree divisible by p, the only possibil-ity is that we have {π(2p −1), π(3p −1)} = {2p −1, 3p −1}, {π(2p), π(3p)} = {2p, 3p},...,{π(3p −3), π(4p −3)} = {3p −3, 4p −3}. After this there are terms with unique degrees, hence the Vandermonde part has to have this part: Y 4p−3 4p−2 Y 4p−2 4p−1 · · · Y p2−1 p2 . Hence we already know that the part coming from (Y1 + · · · + Yq)q−1 starts with p−1 terms of degree 1, then p−1 terms of degree p. This means that the rest of the Yis have to have degree divisible by p2, since otherwise we would get a coeﬃcient starting with 30 (q −1)(q −2) · · · (q −p + 1) q −p p q −2p p · · · q −(p −1)p p q −p2 k , where k is not divisible by p2, but this is zero. One can continue by induction on i to show that the part coming from the Vander-monde determinant has to have the following form: q Y i=1 Y i−1 π(i) , where (as we promised above) π is a permutation of the indices such that π(i) = i, except for a couple of values: {π(1), π(2)} = {1, 2}, {π(3), π(4)} = {3, 4},..., {π(2p −3), π(2p −2)} = {2p −3, 2p −2}; {π(2p −1), π(3p −1)} = {2p −1, 3p −1}, {π(2p), π(3p)} = {2p, 3p},...,{π(3p − 3), π(4p −3)} = {3p −3, 4p −3}; {π(p2 + 1), π(2p2 + 1)} = {p2 + 1, 2p2 + 1}, {π(p2 + 2), π(2p2 + 2)} = {p2 + 2, 2p2 + 2},...,{π(p2 + p −1), π(2p2 + p −1)} = {p2 + p −1, 2p2 + p −1}; ... {π(ph−1 + 1), π(2ph−1 + 1)} = {ph−1 + 1, 2ph−1 + 1}, {π(ph−1 + 2), π(2ph−1 + 2)} = {ph−1+2, 2ph−1+2},...,{π(ph−1+p−1), π(2ph−1+p−1)} = {ph−1+p−1, 2ph−1+p−1}. This means that apart form a non-zero constant (including powers of those ai for which we did not have a choice for π(i)), the term coming from the Vandermonde part is the product of 2 × 2 determinants of the form aq−1−k i Y k i aq−1−k−pm i Y k+pm i aq−1−k j Y k j aq−1−k−pm j Y k+pm j . Dividing such a term with the non-zero (aiaj)q−1−k−pm and using that x →xpm is an automorphism of the ﬁeld, we end up in a situation similar to the prime case: (a1 −a2)(a3 −a4) · · · (a2p−3 −a2p−2)· (a2p−1 −a3p−1)(a2p −a3p) · · · (a3p−3 −a4p−3)· 31 (ap2+1 −a2p2+1)(ap2+2 −a2p2+2) · · · (ap2+p−1 −a2p2+p−1)· ... (aph−1+1 −a2ph−1+1)(aph−1+2 −a2ph−1+2) · · · (aph−1+p−1 −a2pp−1+p−1) = 0 Similarly to the prime case, this is true after any permutation of the indices. The number of brackets here is h(p −1), so by Lemma 3.4.3 (iii), we only need q −p(h − 1) ≥q+1 2 , this is true for q > 9 odd. Let N denote the maximal multiplicity in the multiset {a1, ..., aq}. By the previous two claims N is large. After translation, suppose the value in question is zero. We need to show that if there is no ordering bi of the ﬁeld elements achieving P i aibi = 0, then N = q −2. The plan is to use the same machinery for the remaining non-zero ais. Lemma 3.4.8. Suppose a1, . . . , ak are non-zero elements of GF(q) with k < 2q/3 if q = p prime and k ≤q−3 2 if q = ph, h ≥2, admitting at least 3 diﬀerent values and with the property that no value occurs more than q −k times. Either there are diﬀerent elements b1, . . . , bk such that P aibi = 0 or k = 3. Proof. Consider the polynomial G from Theorem 3.4.4. By 3.4.4, terms of maximal degree of G have at least one Yi with degree at least q. Just like previously, we look for appropriate terms in G to gain information about the ais. If 4 ≤k ≤q+3 2 holds, then consider the following term (of maximal degree): Y (q−5)/2+k 1 Y (q−5)/2+k 2 Y k−3 3 Y k−3 4 Y k−5 5 Y k−6 6 · · · Y 0 k . It is easy to see that there are only four terms coming from (Y1 + · · · + Yq)q−1 that (multiplied by the appropriate term coming from the Vandermonde part) can contribute to this term. These four terms are YiY q−1 2 j Y q−3 2 k , where i = 3 or 4 and {j, k} = {1, 2}. Each of them comes with coeﬃcient (q −1) q−2 (q−1)/2 ̸= 0. Hence we have (a1 −a2)(a3 −a4) = 0. Just like previously, this is true for any permutation of the indices. By Lemma 3.4.3, this implies that there is a value among the ais with 32 multiplicity at least k −1 contradicting the assumption that the ais admit at least 3 values. Now consider the k > q+3 2 case, and note that this case can occur only if q = p prime. We have to distinguish between two cases according to whether p ≡1 or 2 (mod 3). If 3\|p −1, then consider the following term (of maximal degree): Y k+(p−7)/3 1 Y k+(p−7)/3 2 Y k+(p−7)/3 3 Y k−4 4 Y k−5 5 · · · Y 0 k . It is easy to see that the coeﬃcient is a non-zero term times (a1 −a2)(a2 −a3)(a3 −a1), implying (by Lemma 3.4.3) that there is a value among the ais with multiplicity at least k −2. This contradicts the assumption that no value has multiplicity more than q −k. If 3\|p + 1, then one should consider the following term (of maximal degree): Y k+(p−8)/3 1 Y k+(p−8)/3 2 Y k+(p−8)/3 3 Y k−4 4 Y k−4 5 Y k−6 6 Y k−7 7 · · · Y 1 k−1Y 0 k . Here the coeﬃcient is essentially (a1 −a2)(a2 −a3)(a3 −a1)(a4 −a5). It is not diﬃcult to see that similarly to the previous case, this leads to contradiction. Proof. (of Theorem 3.1.2) By Lemma 3.4.2 that there are at least 4 diﬀerent values among the ais. Suppose there is no ordering b1, . . . , bq of the elements of GF(q) giving P i aibi = 0. We have to ﬁnd a contradiction. After transformation (by Lemma 3.4.1 and the sentence after its proof) suppose 0 is not among the ais. Apply Lemma 3.4.5 or 3.4.7 to get that a signiﬁcant part of the elements must be identical. Apply a transformation to make this value zero and apply Lemma 3.4.8 for the rest of the ais. We cannot have diﬀerent bis for these indices such that P aibi = 0 (here the 33 sum is only for those i-s, for which ai ̸= 0), because otherwise the bis could be easily extended to an ordering of the ﬁeld such that P i aibi = 0. Hence we have k = 3, that is, the multiset {a1, . . . , aq} contains q −3 zeros and 3 distinct non-zero elements, a, b and c say. Suppose a + b ̸= 0. Then ba + (−a)b + 0c = 0, a contradiction. 3.4.4 Proof for even q The proof is similar for q even. We can use Lemma 3.4.1 (the proof presented works for q even). Lemma 3.4.2 should be replaced by the following. Lemma 3.4.9. If our multiset has only 1 or 2 diﬀerent values and n = q is even, then Theorem 3.1.2 is true. Proof. If our set has only one value (of multiplicity q) then any ordering of GF(q) is good, so suppose we have two values. After transformation we can achieve that 0 is the value with multiplicity larger than q/2 and 1 is the other value with multiplicity smaller than q/2. Hence all we need is that for any m ≤q/2, there are distinct ﬁeld elements b1, . . . , bm such that b1 + · · · + bm = 0. Denote by G an additive subgroup of GF(q) of index 2. Let b1, · · · , bm−1 be arbitrary distinct elements of G. If b1 + · · · + bm−1 is distinct from all the bis, then let bm = b1 +· · ·+bm−1 and we have the m elements we were looking for. If b1 + · · · + bm−1 equals one of the bis, bm−1 say, then we have b1 + · · · + bm−2 = 0. Let a ∈GF(q) \ G. Replace bm−2 with bm−2 + a, keep bm−1, and let bm = bm−1 + a. It is easy to see that the bis are distinct and their sum is zero. Lemma 3.4.3 and Theorem 3.4.4 are true for q even (the proofs presented did not assume q is odd). Lemma 3.4.7 should be replaced by the following. Lemma 3.4.10. Suppose q = 2h > 8, and that there is no ordering b1, . . . , bq of the elements of GF(q) such that P i aibi = 0. Then at least q+3 2 of the ais are the same. Proof. After transformation suppose 0 is not among the ais. Consider the polyno-mial G from Theorem 3.4.4 with k = q. By 3.4.4, terms of maximal degree of G have at least one Yi with degree at least q. 34 Consider the following term: Y1 h−1 Y i=1 Y 2h−i i+2 h Y i=1 Y i−1 i Similarly to the proof of Lemma 3.4.7, one can use Lucas’ theorem to ﬁnd the coeﬃcient of this term. This is the sum of some subterms from G(Y1, . . . , Yq). If a subterm from G(Y1, . . . , Yq) is non-zero, then from the ((Y1 + · · · + Yk)q−1 −1) part we need h variables on powers 1, 2, 4, . . . , 2h−1. Using similar observations as before, we can conclude that this must imply that the coeﬃcient of our term (apart from the usual non-zero constant) is (a1 −a2) h−1 Y i=1 (ai+2 −ai+2+2h−1). Thus this number must equal zero for any permutation of the indices which implies that one of the ais has multiplicity q −h + 1 because of Lemma 3.4.3 (iii). Instead of Lemma 3.4.8, one can immediately prove the following. Lemma 3.4.11. Suppose a1, . . . , ak are non-zero elements of GF(q), q even with 1 < k < q/2. Either there are diﬀerent elements b1, . . . , bk such that P aibi = 0 or all the ais are the same. Proof. Consider the polynomial G from Theorem 3.4.4. By 3.4.4, terms of maximal degree of G have at least one Yi with degree at least q. Considering the following term: (YkYk−1)q/2+k−2 · Y k−3 k−2 · · · Y 1 2 Y 0 1 . It is easy to see that there are only two possibilities to get this term and the coeﬃcient we have (apart from a non-zero constant) is ak −ak−1. This implies ak−1 = ak and, since we can permute the indices at the beginning, that all the ais are the same. After these lemmas, the proof is easy. 35 3.5 Final remarks At the end of this chapter, we discuss some possible generalizations of the presented results. Theorem 3.2.1, concerning range of polynomials over ﬁnite ﬁelds, raises natural questions. We deduced that if M = {a1, a2, . . . , aq} is a multiset over GF(q) with a1 + · · · + aq ̸= 0, then every reduced degree polynomial whose range is M has degree q −1. Notation 3.5.1. Let ∆q(M) denote the least integer d such that there exists a polynomial of degree d over GF(q) whose range is M. Is there a natural property that describes whether ∆q(M) is small or large? First, we have to assume that the elements of the multiset add up to zero, otherwise ∆q(M) is clearly q −1. On the one hand, it is easy to see that the greatest multiplicity m(M) in M - if it is less than q - gives a lower bound on ∆q(M). Indeed, if an element a has multiplicity m(a), then clearly f −a has m(a) roots in GF(q) for any polynomial f whose range is M, which shows that deg(f) ≥m(a). On the other hand, Theorem 3.2.1 implies that, given a1 + · · · + aq = 0, ∆q(M) is q −2 only if there is an element with multiplicity q −2. (In this case, the multiset must have the structure {a, . . . a, a+b, a−b}.) Furthermore, ∆q(M) = 1 if and only if every element in M has multiplicity 1, i.e, M is the set of all elements of the ﬁeld. These observations would suggest that if a1+· · ·+aq = 0, then ∆q(M) only depends on m(M), the largest multiplicity in M and probably ∆q(M) = m(M). However, this is not the case. Let us suppose that q = p is prime and deﬁne the multiset as 1 taken with multi-plicity m, p −m taken with multiplicity 1, and 0 taken with multiplicity p −m −1. By a result of Biró , all polynomials of this range have degree at least roughly 3p/4, unless m = p−1 2 or p−1 3 or 2p−1 3 . This shows that in the q = p prime case if the greatest multiplicity of M is smaller than c · p with c < 3/4, then it might hap-pen that ∆q(M) is bigger than the greatest multiplicity. Moreover, the diﬀerence between ∆q(M) and the greatest multiplicity m(M) of M can be linear in p. 36 The papers of Muratović-Ribić and Wang [78, 79] reveals that for every large enough q, the relation ∆q(M) > m(M) may hold for many possible values of m(M): Theorem 3.5.2. [78, 79] For every m with q 2 ≤m < q −3 there exists a multiset M with highest multiplicity m(M) = m whose elements add up to zero, such that every polynomial over GF(q) with the prescribed range M has degree greater than m, that is, ∆q(M) ≥m(M) + 1. Corollary 3.5.3. Among all multisets M with ∆q(M) = q −3, there must be some in which the greatest multiplicity is less than q −3. The proof is based on a delicate enumeration argument. However, it is still open to determine ∆q(M) in general or even provide good bounds on it. Another way to generalize the main result is presented in the Chapter 4. 37 Chapter 4 Extension to cyclic groups 4.1 Introduction and background In the previous chapter we considered multisets over ﬁnite ﬁelds. We exploited the ﬁeld structure when we applied the Combinatorial Nullstellensatz. However, the formulation of the problem itself, at least in the prime ﬁeld case, only requires the abelian group structure. This motivates the forthcoming investigation. The aim is to ﬁnd a diﬀerent kind of generalization of the prime case of Theorem 3.1.1, more combinatorial in nature, which refers only to the group structure. First we extend the result to cyclic groups of odd order. Theorem 4.1.1. Let {a1, a2, . . . , am} be a multiset in the Abelian group Zm = Z/mZ, where m is odd. Then after a suitable permutation of the indices, either P i iai = 0, or a1 = a2 = · · · = am−2 = a, am−1 = a + b, am = a −b for some elements a and b, (b, m) = 1. The situation is somewhat diﬀerent if the order of the group is even. In this case we have to deal with two types of exceptional structures. The following statements are easy to check. Proposition 4.1.2. Let m be an even number represented as m = 2kn, where n is odd. 38 (i) If a multiset M = {a1, a2, . . . , am} of Zm consists of elements having the same odd residue c mod 2k, then M has no permutation for which P i iai = 0 holds. (ii) If M = {a, a, . . . , a + b, a −b} mod m, where a is even and (b, m) = 1 holds, then M has no permutation for which P i iai = 0 holds. These two diﬀerent kinds of structures we call homogeneous and inhomogeneous exceptional multisets, respectively. Theorem 4.1.3. Let M = {a1, a2, . . . , am} be a multiset in the Abelian group Zm, m even. If M is not an exceptional multiset as deﬁned in Proposition 4.1.2, then after a suitable permutation of the indices P i iai = 0 holds. The presented results might be extended in diﬀerent directions. One may ask whether there exists a permutation of the elements of a given multiset M of Zm (consisting of m elements), for which the sum P i iai is equal to a prescribed ele-ment of Zm. This question is related to a conjecture of Britnell and Wildon, see [22, p. 20], which can be reformulated as follows. Given a multiset M = {a1, a2, . . . , am} of Zm, all elements of Zm are admitted as the value of the sum Pm i=1 iaπ(i) for an appropriate permutation π from the symmetric group Symm, unless one of the following holds: • M = {a, . . . , a, a + b, a −b}, • there exists a prime divisor p of m such that all elements of M are the same mod p. Our result may in fact be considered as a major step towards the proof of their conjecture, which would provide a classiﬁcation of values of determinants associated to special types of matrices. When m is a prime, the conjecture is an immediate consequence of Theorem 3.1.1 and Lemma 4.2.2 (ii). Indeed, if only one value was admitted, then the multiset would consist of a single element m times. On the other hand, if there was an admitted element w ̸= 0, all nonzero elements would be admitted via Lemma 4.2.2 (ii). Thus the value 0 is the crucial one, and it was investigated in Theorem 3.1.1. 39 The chapter is organized the following way. In Section 4.1, we collect several simple observations that are used frequently throughout the paper and sketch our proof strategy. Section 4.2 is devoted to the proof of Theorem 4.1.1. In Section 4.3 we will verify Theorem 4.1.3 for some particular cases, whose proofs do not exactly ﬁt into the general framework (and may be skipped at a ﬁrst reading). The complete proof, which is more or less parallel to that of Theorem 4.1.1, is carried out in Section 4.4. Finally, we discuss related problems and conjectures in Section 4.5. 4.2 Preliminaries Definition 4.2.1. Let M = {a1, . . . , am} be a multiset in Zm. A permutational sum of the elements of M is any sum of the form Pm i=1 iaπ(i), π ∈Symm. If, after some rearrangement, we ﬁx the order of the elements of M, then the permutational sum of M considered as a sequence (a1, . . . , am) is simply Pm i=1 iai. Accordingly, the aim is to determine which multisets admit a zero permutational sum. This property is invariant under certain transformations. Lemma 4.2.2. Let m be odd, and M be a multiset in Zm of cardinality m. (i) If no permutational sum of M admits the value 0, then the same holds for any translate M + c of M, and also for any dilate cM in case (c, m) = 1. (ii) If the permutational sums of M admit a value w, then they also admit the value kw for every integer k with (m, k) = 1. As a consequence, if (m, w) = 1, then the permutational sums take at least ϕ(m) diﬀerent values. (iii) Assume that M has the exceptional structure, i.e. M = {a, . . . , a, a + b, a −b} where (b, m) = 1. Then the permutational sums of M admit each element of Zm except zero. Proof. Parts (i) and (iii) are straightforward, for 1 + 2 + . . . + m ≡0 (mod m). Part (ii) follows from the fact that π ∈Symm holds for the function π deﬁned by π(i) = ki . 40 Remark 4.2.3. Part (ii) holds also if m is even, but this is not true in general for Parts (i) and (iii). The reason is that 1 + 2 + . . . + m ̸≡0 (mod m) if m is even, but 1 + 2 + . . . + m ≡m/2 (mod m). The sumset or Minkowski sum C + D of two subsets C and D of an Abelian group G written additively is C + D = {c + d \| c ∈C, d ∈D}. \|C\| denotes the cardinality of the set C. The following statement is folklore. Lemma 4.2.4. For C, D ⊆Zm, \|C\| + \|D\| > m implies C + D = Zm. Proof. Fix an arbitrary element g ∈Zm. If \|C\| + \|D\| > m, then the set −C + g (containing elements of form −c + g : c ∈C) and D can not be disjoint, conﬁrming the statement. In the remaining part of this section, we sketch the proof of Theorem 4.1.1, thus from now on, m is assumed to be odd in this section. Meanwhile, we will also use the deﬁnitions and notions introduced below in the even case. Recall that the arithmetic function Ω(n) represents the total number of prime fac-tors of n. Similarly to the classical result in zero-sum combinatorics due to Erdős, Ginzburg and Ziv , we proceed by induction on Ω(m). The initial case is covered by Theorem 3.1.1, so in the sequel we assume that m is a composite number and ﬁx a prime divisor p of m and write m = pkn, where (p, n) = 1. The proof is carried out in several steps (of which the ﬁrst two will be quite similar to the beginning of the proof of Theorem 4.1.3). 4.2.1. First step: choosing an initial order and partitioning into blocks We introduce the notion of initial order as follows. Definition 4.2.5. Let s = (b1, b2, . . . , bm) be any sequence in Zm. (i) A cyclic translate of s is any sequence of the form (bi, bi+1, . . . , bm, b1, . . . , bi−1). (ii) The sequence s is separable (relative to the prime divisor p of m) if equivalent elements mod pl are consecutive for every 1 ≤l ≤k. 41 (iii) A sequence (aπ(1), . . . , aπ(m)), where π ∈Symm, is an initial order for the multiset M = {a1, . . . , am} of Zm, if some cyclic translate of (aπ(1), . . . , aπ(m)) is separable. Thus separability means that for 1 ≤i < j ≤m and every l ≤k, ai ≡aj (mod pl) implies ai ≡ah (mod pl) for every i < h < j. Note that one can always order the elements of M into a separable sequence. A useful property of such an ordering is summarized in the following lemma whose proof is straightforward. Lemma 4.2.6. Consider a sequence of m elements in Zm, which admits a separable cyclic translate. Partition the elements into d ≥3 consecutive blocks T1, . . . , Td. If for an integer l, a certain residue r mod pl occurs in every block, then at most two of the blocks may contain an element having a residue diﬀerent from r. The same conclusion holds if the elements are rearranged inside the individual blocks. Let (a1, . . . am) be an initial order. Form p consecutive blocks of equal size, denoted by T1, T2, . . . , Tp, each containing m∗:= m/p consecutive elements. More precisely, Ti = {a(i−1)m∗+1, a(i−1)m∗+2, . . . , aim∗}. Si denotes the sum of the elements in Ti, while Ri denotes the permutational sum of the block Ti (as an ordered multiset), that is, Ri = Pm∗ j=1 jaj+(i−1)m∗. Writing R = Pp i=1 Ri, the permutational sum of M takes the form Φ = m X j=1 jaj = p X i=1 (Ri + m∗(i −1)Si) = R + m∗ p−1 X i=0 iSi+1. 4.2.2. Second step: divisibility by m∗:= m/p Our aim here is to ensure that m∗\| Φ holds after a well structured rearrangement of the elements. That is, we want to achieve that m∗\| R holds. To this end we allow reordering of the elements inside the individual blocks. Such a permutation will be referred to as a block preserving permutation. We distinguish three diﬀerent cases. 42 First, if there is no exceptionally structured block mod m∗, then by the inductional hypothesis the elements in each block Ti can be rearranged so that m∗divides Ri. Thus, after a block preserving permutation, m∗\| R. Next, if there is an exceptionally structured block Ti, then the permutational sums over Ti take m∗−1 diﬀerent values mod m∗, see Lemma 4.2.2 (iii). If there are at least two exceptionally structured blocks, then it follows from Lemma 4.2.4 that there is a block preserving permutation that ensures m∗\| R. Finally, if there is exactly one exceptionally structured block {a, . . . , a, a + b, a −b} (mod m∗), then a permutational sum of this block can take any value except 0 mod m∗. So after a block preserving permutation we are done, unless zero is the only value that the other blocks admit, that is, all elements must be the same in each block mod m∗. This latter case can be avoided by a suitable choice of the initial order in the ﬁrst step. Indeed, translating the initial order cyclically so that it starts with an appro-priate element from the exceptional block will break down this structure. 4.2.3. Third step To complete the proof, based on the relation m∗\| Φ we further reorganize the elements to achieve a zero permutational sum, or else to conclude that we are in (one of) the exceptional case(s). Here we only give an outline of the strategy of the proof, as the following section is devoted to the detailed discussion. Set R′ := R m∗. As a ﬁrst approximation, we try to change the order of the blocks to obtain p−1 X i=0 iSi+1 ≡−R′ (mod p), which would imply m \| Φ. One is tempted to argue that the case R′ ≡0 (mod p) would be easy to resolve applying Theorem 3.1.1 for the multiset {S1, . . . , Sp}. As it turns out, the main diﬃculty is to handle exactly this case, since the multiset {S1, . . . , Sp} may have the exceptional structure. A remedy for this is what we call the ‘braid trick’. The main idea of this tool will be to consider the transposition of a pair of elements whose indices diﬀer by a ﬁxed number x (typically a multiple of m∗). By this kind of transposition of a pair (ai, ai+x), the permutational sum 43 increases by x(ai −ai+x), providing a handy modiﬁcation. 4.3 The case of odd order In this section we complete the proof of Theorem 4.1.1, m = pkn where (p, n) = 1 will denote an odd integer throughout the section. We continue with the details of the third step outlined in the previous section. We distinguish two cases according to whether R′ is divisible by p or not. 4.3.1. R′ is not divisible by p. Note that Pp−1 i=0 iSi+1 can be viewed as a permutational sum of the multiset S = {S1, S2, . . . , Sp}. If there are two elements Si ̸≡Sj (mod p), then their transposi-tion changes the value of the permutational sum of S mod p. In particular, the permutational sums of S admit a nonzero value mod p. From Lemma 4.2.2 (ii) it follows that they admit each nonzero element of Zp and in particular −R′ too. Otherwise, we have S1 ≡S2 ≡. . . ≡Sp (mod p). We use the braid trick: we look at the pairs (ai, ai+m∗) for every i. The elements ai and ai+m∗occupy the same position in two consecutive blocks Tj, Tj+1, hence their transposition leaves R intact, and thus R′ does not change either. On the other hand, if they have diﬀerent residues mod p, Sj and Sj+1 change whereas each other Si remains the same, therefore the previous argument can be applied. Finally, we have to deal with the case when ai ≡ai+lm∗(mod p) holds for every possible i and l. This is the point where we exploit the separability property. The initial order has changed only inside the blocks during the second step. Since the number of blocks is at least three, it follows from Lemma 4.2.6 that ai ≡aj (mod p) for all 1 ≤i < j ≤m in M. In this case we prove directly that M has a zero permu-tational sum. In view of Lemma 4.2.2 (i), we may suppose that every ai is divisible by p. Consider M ∗:= { a1 p , a2 p , . . . , am p }. Apply the ﬁrst two steps for this multiset M ∗. It follows that M ∗has a zero permutational sum mod m∗, which implies that 44 M has a zero permutational sum mod m. 4.3.2. R′ is divisible by p Here our aim is to prove that p \| Pp−1 i=0 iSi+1 holds for a well chosen permutation of the multiset S := {S1, . . . , Sp}. This is exactly the problem that we solved in Theorem 3.1.1, which implies that we can reorder the blocks (and hence the multiset M itself) as required, except when the multiset S has the form {A, A, . . . , A, A + B, A −B}, with the condition (B, p) = 1. Once again, we apply the braid trick. If ai and ai+lm∗have diﬀerent residues mod p, then we try to transpose them in order to destroy this exceptional structure {A, A, . . . , A, A + B, A −B}. As in Subsection 4.3.1, R does not change. We call a pair of elements exchangeable if their indices diﬀer by a multiple of m∗. Thus, a zero permutational sum of M is obtained unless no transposition of two exchangeable elements destroys the exceptional structure of S. The following lemma gives a more detailed description of this situation. Lemma 4.3.1. Suppose that no transposition of two exchangeable elements destroys the exceptional structure of S. Then either this exceptional structure can be destroyed by two suitable transpositions, or M contains only three distinct elements mod p: t, t + B, t −B for some t with the following properties: • t + B occurs only in one block, and only once; • t −B occurs only in one block, and only once; • t + B and t −B occupy the same position in their respective blocks. Proof. Denote by T + and T −the blocks for which the sum of the elements is A + B and A −B, respectively. Apart from elements from T + and T −, two exchangeable elements must have the same residue mod p. Furthermore, if a transposition between aj ∈T −and aj+lm∗/ ∈T + does not change the structure of S, that means aj ≡aj+lm∗ (mod p) or aj ≡aj+lm∗−B (mod p). Similar proposition holds for T +. Consider now a set of pairwise exchangeable elements. One of the following describes their structure: either they all have the same residue mod p, or they have the same 45 residue t mod p except the elements from T + and T −, for which the residues are t + B and t −B, respectively. Observe that both cases must really occur since the sums Si of the blocks are not uniformly the same. In particular, there is a full set of p pairwise exchangeable elements having the same residue mod p. Since the number of blocks is at least 3, we can apply Lemma 4.2.6. We only used block-preserving permutations so far, hence it follows that all elements have the same residue mod p — let us denote it by t — except some (t + B)’s in T +, and the same number of (t −B)’s in T −, in the very same position relative to their blocks. We claim that this number of diﬀerent elements in T + and T −must be one, oth-erwise we can destroy the exceptional structure with two transpositions. Indeed, by contradiction, suppose that there exist two distinct set of exchangeable elements where the term corresponding to T + and T −is t + B and t −B, respectively. Pick a block diﬀerent from T + and T −and denote it by T. Then transpose t + B ∈T + and t ∈T in the ﬁrst set, and t −B ∈T −and t ∈T in the second set. The new structure of S′ obtained this way is not exceptional any more. Lemma 4.3.2. Suppose that M contains only three distinct elements mod p: t, t + B, t −B for some t with the following properties: • t + B occurs only in one block, and only once; • t −B occurs only in one block, and only once; • t + B and t −B occupy the same position in their respective blocks. Then either a suitable zero permutational sum exists or the conditions on M hold mod pl for every l ≤k, with a suitable B = Bl not divisible by p. Proof. We proceed by induction on l. Evidently, it holds for l = 1. According to Lemma 4.2.2 (i) we may assume that t ≡0 (mod p). Let a+ and a−denote the elements of T + and T −for which a+ ≡B (mod p) and a−≡−B (mod p). Note that their position is the same in their blocks. Suppose that l ≥2 and the conditions hold mod pl−1. Consider the residues of the elements mod pl now. We use again the braid trick. Suppose that there exist 46 ai, aj ̸∈{a−, a+} such that i −j is divisible by pk−l but not by pk−l+1, and ai ̸≡aj (mod pl). After we transpose them, (the residue of) R does not change mod pk−1, but it changes by (i−j)(aj −ai) ̸= 0 (mod pk). For the new permutational sum thus obtained, R′ ̸≡0 (mod p) holds, while the multiset S may change, but certainly it does not become homogeneous mod p. Thus M has a zero permutational sum, as in Subsection 3.1. Otherwise, in view of Lemma 4.2.6 it is clear that all the residues must be the same mod pl, and we may suppose they are zero, except the residues of a+ and a−. In addition, a+ + a−≡0 (mod pl) must hold too, since R′ ≡0 (mod p). This completes the inductive step. Lemma 4.3.2 applied for l = k completes the proof of Theorem 4.1.1 when m = pk is a prime power. In the sequel we assume that n ̸= 1. Let m = pk1 1 pk2 2 . . . pkr n be the canonical form of m. Note that the whole argument we had so far is valid for any prime divisor p of m. Therefore, to complete the proof , we may assume that M has the exceptional structure mod pki i as described in Lemma 4.3.2 for every p = pi. Lemma 4.3.3. The conclusion of Theorem 4.1.1 holds if M has exceptional structure modulo each pki i . Proof. We look at the permutational sums of M leaving the elements of M in a ﬁxed order a1, a2, . . . , am while permuting the coeﬃcients 1, 2, . . . , m. According to Lemma 4.2.2 (i) we may assume that all elements, except two, are divisible by pk1 1 ; all elements, except two, are divisible by pk2 2 , and so on. It follows that at least m−2r elements are zero mod m, so their coeﬃcients are irrelevant. So we only have to assign diﬀerent coeﬃcients to the nonzero elements xi of M. For any 0 ̸= x ∈M, we choose its coeﬃcient cx to be either m (m,x) or − m (m,x), ensuring that cxx = 0 in Zm. If such an assignment is possible, the permutational sum will be zero . First, observe that m (m,x) and − m (m,x) are the same if and only if (m, x) = 1. Note that for each i, pi divides (m, xi) for all xi, except two. Hence there is no triple x1, x2, x3 of the elements for which (m, xi) would be the same. Thus we can assign a diﬀerent coeﬃcient to each xi ̸= 0, except when there exist two of them, for which (m, xi) = 1. But this is exactly the exceptional case M = {0, 0, . . . , 0, c, −c}, where (c, m) = 1. 47 4.4 Special cases of Theorem 4.1.3 In this section we prove that Theorem 4.1.3 holds for some specially structured multisets. Lemma 4.4.1. Let m = 2n, n > 1 odd, and let M be a multiset in Zm consisting of two blocks of size n in the form T1 = {a, . . . , a, a + b, a −b} and T2 = {c, . . . , c} (mod n), where (b, n) = 1. If one of the blocks contains elements from only one parity class then Theorem 4.1.3 holds. Proof. First we obtain a permutation for which n divides the permutational sum of M. We choose an element c∗from T2. Assume that c∗̸= a −b (mod n) and exchange c∗with a −b ∈T1. (If the assumption does not hold then we pick a + b instead of a −b and continue the proof similarly.) This way we get two blocks T ′ 1 and T ′ 2, which do not have the exceptional structure mod n. Thus there exists a block preserving permutation ensuring that n divides the obtained permutational sums of T ′ 1 and T ′ 2, thus n also divides the permutational sum of M. We assume that both odd and even elements occur in M, otherwise either 2 is trivially a divisor of Φ or M has exceptional homogeneous structure. If the relation m \| Φ does not hold, then we apply the braid trick by looking at the pairs (xi, xi+n). If a pair consists of an odd and an even element, then we may transpose them and the proof is done. Otherwise the exchangement of c∗and a −b must have destroyed the property of having a uniform block mod 2 among T1 and T2, that is, c∗and a −b have diﬀerent parity. Since the choice of c∗∈T2 was arbitrary, we may assume that T2 = {c, . . . , c} (mod 2n). Moreover, since the braid trick did not help us, every element in T1 congruent to a mod n must have the same parity as the elements c, and the parity of element a + b must coincide with that of a −b. In this remaining case consider the blocks {a, . . . , a, c, c} and {c, . . . , c, a + b, a −b}. First, if c ̸≡a (mod n), then neither block is exceptional as a multiset in Zn, hence an appropriate block preserving permutation ensures that n divides the permuta-tional sum. If the permutational sum happens to be odd, then a suitable trans-position via the braid trick will increase its value by n, for the ﬁrst block contains 48 elements from the same parity class in contrast to the second. Finally, if c ≡a (mod n), then either c = a is even, providing that M has inhomogeneous excep-tional structure, or c = a is odd, in which case the permutational sum will be zero if we set an = a + b and a2n = a −b (mod n). Lemma 4.4.2. Let m = 2kn > 4, n odd and k > 1. Let M be a multiset of Zm, consisting of two even elements and m −2 odd elements having residue c mod 2k−1. Then the permutational sum of M admits the value zero. Proof. Denote the even elements by q1 and q2. We distinguish the elements having residue c mod 2k−1 according to their residues mod 2k, which are c and c∗≡c+2k−1 (mod 2k). We may suppose that the number of elements c is greater than or equal to the number of elements c∗. First we solve the case n = 1 meaning m = 2k, k > 2. Taking am/2 = q1, am = q2, the permutational sum will be divisible by m/2. If there is no element c∗, then the permutational sum is in fact divisible by m. If there exist some elements c∗among the odd elements and the permutational sum is not yet divisible by m, then a trans-position between two elements c and c∗whose indices diﬀer by an odd number will result in a zero permutational sum mod m. Turning to the general case n > 1, we initially order the elements as follows. Even elements precede the others, elements c mod 2k precede the elements c∗mod 2k, and equivalent elements mod m are consecutive. Form 2k blocks of equal size n. With an argument similar to the one in Section 4.2.2 we arrive at two cases. Either we obtain a permutational sum congruent to zero mod n after a block preserving permutation, or the structures of the blocks are as follows: there is exactly one exceptional block (as a multiset in Zn) and the other blocks only admit a zero per-mutational sum mod n meaning that each of them consists of equivalent elements mod n. Case 1) Consider the block preserving permutation, which results in a permuta-tional sum Φ0 divisible by n. We modify this permutation, if necessary, to get one 49 corresponding to a zero permutational sum mod 2k, while the divisibility by n is preserved. We denote by f and g the indices of q1 and q2 in the considered permutation. Thus Φ0 ≡c2k(2k −1) 2 + f(q1 −c) + g(q2 −c) (mod 2k−1). (∗) Note that {ln : l = 0, 1, . . . , 2k −1} is a complete system of residues mod 2k. Let l be the solution of the congruence (q1 −c)ln ≡−Φ0 (mod 2k). Thus transposing q1 = af with af+ln implies that Φ1 ≡ ( 0 (mod 2k) if af+ln ≡c (mod 2k) 2k−1 (mod 2k) if af+ln ≡c∗ (mod 2k). The relation n \| Φ1 still holds. So in the case when af+ln ≡c (mod 2k) we are done, and if af+ln ≡c∗(mod 2k) we have to increase the value of the permutational sum by 2k−1n mod m. Recall that each element in the second block is c mod 2k. Therefore transposing af ≡c∗(mod 2k) with af+n ≡c (mod 2k) in this latter case does the job. Case 2) One of the blocks (not necessarily the ﬁrst one) has the exceptional struc-ture, while every other is homogeneous mod n. We can still argue as in the previous case if, performing the following operation, we can destroy the exceptional struc-ture without changing the position of the even elements q1, q2 and the entire second block. Namely, we try to transpose two nonequivalent elements mod n, one from the exceptional block and one from another block. If this is not possible with the above mentioned constraints, then the exceptional block must be among the ﬁrst two. Furthermore, every element congruent to c mod 2k in the ﬁrst two blocks must be equivalent mod n. Thus we only have to deal with the following structure: the ﬁrst block is the exceptional one, q1 and q2 correspond to a + b and a −b in the exceptional structure, all the other elements contained in the ﬁrst two blocks are equivalent mod m (and congruent to c mod 2k), and the remaining blocks are all homogeneous mod n. 50 Exchanging q2 with any element from the second block destroys the exceptional structure of the ﬁrst block, which means that after a suitable block preserving per-mutation the permutational sum of each block becomes 0 mod n, ensuring n \| Φ for the multiset. At this point the indices of the even elements are n and 2n. Next, keeping the order inside the blocks we rearrange them so that the ﬁrst and second blocks become the 2k−1th and 2kth, that is, am/2 = q1 and am = q2. Hence, maintaining n \| Φ we also achieve 2k−1 \| Φ via equality (∗). Either we are done or Φ ≡2k−1n (mod m). The latter can only happen if there exists an element of type c∗. If a block contains both elements of type c and c∗, then a transposition of a consecutive pair of them within that block increases Φ by 2k−1n. Otherwise there must exist a block containing only elements of type c∗. This implies the existence of a pair of c and c∗whose position diﬀers by n. Their transposition increases Φ by 2k−1n2 ≡2k−1n (mod m), solving the case. 4.5 The case of even order One main diﬀerence between the odd and the even order case is due to the fact that Lemma 4.2.2 (i) does not hold if m is even, for 1 + 2 + . . . + m is not divisible by m. That explains the emergence of the exceptional structure, see Proposition 4.1.2. Remark 4.5.1. It is easy to check that after a suitable permutation of the indices, P i iai ≡m/2 (mod m) holds for the exceptionally structured multisets. In order to prove Theorem 4.1.3, we ﬁx the notation m = 2kn, where n is odd and k > 0. Since the cases m = 2 and m = 4 can be checked easily, we assume that m > 4 and prove the theorem by induction on k. Initial step We have m = 2n, where n > 1 according to our assumption. Take the multiset M = {a1, . . . , am} of Zm. Arrange the elements in such a way that both the odd 51 and the even elements are consecutive. Form two consecutive blocks of equal size, denoted by T1 and T2, each containing n elements. Using the notation of Section 2, the permutational sum of M is Φ = m X j=1 jaj = R1 + R2 + m 2 S2 = R + nS2. Our ﬁrst aim is to ensure that n \| Φ holds after a well structured rearrangement of the elements. To this end, we may take an appropriate block preserving permutation providing that n \| Ri holds for i = 1, 2. Such a permutation exists, except when at least one of the blocks are exceptional mod n. However it is enough to obtain a block preserving permutation for which n \| R, and such a permutation exists via Lemma 4.2.2 (iii), unless one of the blocks has exceptional structure (mod n) and the other consists of equivalent elements (mod n). This latter case was fully treated in Lemma 4.4.1. The next step is to modify the block preserving permutation such that 2 \| Φ also holds. If it does not hold, then we try to transpose a pair (ai, ai+n) for which ai and ai+n have diﬀerent parity, according to the braid trick. The permutational sum would change by n (mod m) and we are done. If all pairs have the same parity, then all elements have the same parity. Therefore either Φ is automatically even or M has homogeneous exceptional structure. This completes the initial step. Inductive step Assume that k > 1 and Theorem 4.1.3 holds for every even proper divisor of m. Recalling Deﬁnition 2.4, we choose a separable sequence relative to the prime divisor 2 of m as an initial order. Partition the multiset into two blocks of equal size, T1 and T2. Introduce m∗:= m/2 = 2k−1n, and assume ﬁrst that m∗\| R1 + R2 can be achieved by a suitable block preserving permutation. By induction, we can do it if both blocks as multisets have a structure diﬀerent from the ones mentioned in Proposition 4.1.2. If both blocks as multisets have exceptional structure mod m∗, then in view of Remark 4.5.1 there exists a block preserving permutation for each 52 block such that P i iai ≡m/4 (mod m∗), thus m∗\| R1 + R2 holds. Finally, we can also achieve this relation if exactly one of the blocks has exceptional structure, and the permutational sum of the other block admits the value m/4 mod m∗. Suppose that m \| R1 + R2 does not hold, otherwise we are done. Apply the braid trick and consider the pairs (ai, ai+2k−1n). They must have the same parity, otherwise transposing them would make Φ divisible by m, which would complete the proof. Due to the separability of the initial order, all elements must have the same parity. Consider now the pairs (ai, ai+2k−2n). Either we can transpose the elements of such a pair to achieve a zero permutational sum, or the elements must have the same residue mod 22. Apply this argument consecutively with exponent s = 1, 2, . . . k, for pairs (ai, ai+2k−sn) and modulo 2k−s, respectively. Either m \| Φ is obtained during this process by a suitable transposition of a pair (ai, ai+2k−sn) or all elements must have the same residue r mod 2k. If r is odd, then M has homogeneous exceptional structure described in Proposition 4.1.2. If r is even, then 2k would divide Φ, for Φ ≡r 2k(2k−1) 2 (mod 2k). Thus the conclusion of the theorem holds in this case. The remaining part of the proof is the case when only one of the blocks is excep-tional mod m∗, and the permutational sum of the other block does not admit the value m/4 (mod m∗). We refer to this latter condition by (), and we may suppose that the second block is the exceptional one (otherwise we reverse the sequence). According to Proposition 4.1.2, there are two cases to consider. 4.5.1. The inhomogeneous case T2 = {a, a, . . . , a, q1 = a + b, q2 = a −b} mod m∗, where a is even and (b, m) = 1. Note that T2 contains both even and odd elements. Due to the separability of the initial order, all elements in T1 have the same parity. If T1 consists of odd elements, then we exchange a pair of diﬀerent odd elements mod m∗, one from each block. This way T2 becomes non-exceptional. Moreover, an appropriate choice from {q1, q2} ensures that T1 does not become exceptional either. 53 Thus m∗will be a divisor of the permutational sum after a suitable block preserving permutation. If m \| Φ does not hold, we apply the braid trick for a pair (ai, ai+m∗) for which their parity diﬀers and we are done. If all elements of T1 are even, then we try to transpose a pair of diﬀerent even elements mod m∗, one from each block. Note that if it is possible, T1 will not become exceptional. Hence after a block preserving permutation m∗will be a divisor of the permutational sum. If m \| Φ did not hold, we apply the braid trick for a pair (ai, ai+m∗) for which their parity diﬀers and we are done. Assume that no appropriate transposition exists, that is, T1 must consist of even elements having the same residue a mod m∗. It may occur that M has the inhomo-geneous exceptional structure. Otherwise either q1 + q2 = 2a + m∗, or there exists a pair a(1) ̸≡a(2) (mod m) in M such that a(1) ≡a(2) ≡a (mod m∗). We set the permutation now for these cases. Let q1 and q2 be in the positions 1 and 1 + m∗. Fix arbitrary positions for the rest of elements supposing that if a pair of type {a(1), a(2)} exists, then the elements of such a pair are consecutive. Hence either we are done, or Φ ≡m∗(mod m). In the latter case, note that there must exist a pair of type {a(1), a(2)} that is arranged consecutively. Their transposition provides a zero permutational sum which completes the proof. 4.5.2. The homogeneous case T2 = {c, c, . . . , c} mod 2k−1 where c is odd and () holds for T1. Subcase 1) Every odd element c′ ∈T1 is congruent to c mod 2k−1. Since T1 is not exceptional mod m∗, it must contain some even elements. Thus T1 consists of even elements and possibly also some odd elements having residue c mod 2k−1. Choose an even element q1 from T1 and transpose it with c in T2. Since () holds for T1, neither T1 nor T2 become exceptional by this transposition. Take a permutation of each block for which the permutational sum is zero mod m∗. Either we are done or Φ ≡m∗(mod m) holds. Look at the pairs (ai, ai + m∗) according to the braid trick. If a pair takes diﬀerent residues mod 2, then their 54 transposition makes the permutational sum divisible by m and we are done. Oth-erwise we must have two even elements, and the others have residue c mod 2k−1. Hence Lemma 4.4.2 completes the proof. Subcase 2) There exists an odd c′ ∈T1 for which c′ ̸≡c (mod 2k−1). We transpose c and c′ to obtain T ′ 2 = {c′, c, . . . , c} (mod 2k−1). We claim that m∗\| Φ holds for the new blocks T ′ 1 and T ′ 2 after a suitable block preserving permutation. The permutational sum of T ′ 2 admits the value m/4 mod m∗. Indeed, it has a non-exceptional structure, hence it admits the value zero mod m∗, and then one transposition between c′ and another element is suﬃcient. Thus, neither () holds for T ′ 2 nor has it exceptional structure. Hence we may suppose that m∗\| Φ holds for the new blocks T ′ 1 and T ′ 2. Either we are done or Φ ≡m∗(mod m). In the latter case we need a transposition in T ′ 2 between c′ and another element congruent to c mod 2k−1, for which the permutation sum changes by m∗mod m. Such a transposition clearly exists. 4.6 Abelian groups and sumsets - related topics In combinatorial number theory, one of the classical subﬁelds is the so called zero-sum theory. Let G be a ﬁnite abelian group written additively. A typical zero-sum problem studies conditions which guarantee that a given multiset M of group elements have a non-empty sub-multiset (for which some extra conditions may hold) such that the sum of the elements of the sub-multiset is zero. Probably the most natural question in this area is related to the Davenport constant. For a ﬁnite abelian group G, D(G) denotes the least integer l for which any multiset {g1, . . . , gl} of G contains a sub-multiset where the sum of the elements equals zero. In spite of its relevance in algebraic number theory , the exact behavior of D(G) is still not known in general. Another variant of this problem allows to consider only sets instead of multisets, and hence studies the least integer l for which any set {g1, . . . , gl} of G contains a subset where the sum of the elements equals zero. This number is the so-called Olson constant. 55 For more details we refer to [4, 16, 24, 31, 41]. A classical result in this area is the Erdős-Ginzburg-Ziv theorem , mentioned already in the ﬁrst introductory chapter, in which the zero-sum sub-multisets has prescribed size \|G\|. Theorem 4.6.1 (Erdős-Ginzburg-Ziv). Let G be an arbitrary ﬁnite abelian group of order n. Then every multiset M of size 2n−1 contains a sub-multiset of size n such that the sum of the elements equals zero. The condition on the size of M is tight. We note that this theorem was originally proved for cyclic groups only, but it is not diﬃcult to deduce the generalization. The inverse problem is to characterize all multisets of size 2n −2 without a zero-sum sub-multiset of size n. Grynkiewicz in went much further: he proved that in an s element multiset in Zn, 2n −2 ≤s ≤61 3n there exist at least s 2 n + s 2 n sub-multisets zero-sum sub-multisets of size n. This was conjectured by Bialostocki for any s, proved by Kisin for prime powers , and known asymptotically for ﬁxed n and s →∞due to Füredi and Kleitman . Notice that this would be sharp in general, since the multiset containing 0s and 1s, ⌊s/2⌋and ⌈s/2⌉times, respectively, attains the bound. This inverse approach inspires the following generalization. Problem 4.6.2. Let R be a commutative ring and P(x1, . . . , xn) be an n-variate (homogeneous) polynomial over R. Determine the minimal cardinality m(R, P) for which the following holds: for every multiset M of R, \|M\| = m, probably under some extra conditions, there exists a sub-multiset {a1, . . . , an} for which P(a1, . . . , an) = 0. For R = Zn, P = x1 + . . . + xn, the Erdős-Ginzburg-Ziv theorem claims that the corresponding minimal cardinality is m = 2n −1. Another conjecture of Bialostocki is strongly related to Theorem 4.1.3 and the Erdős-Ginzburg-Ziv theorem as well. 56 Conjecture 4.6.3. Suppose that a1, . . . , an, and b1, . . . , bn are sequences of ele-ments from Zn, satisfying Pn i=1 ai = Pn i=1 bi = 0. If n is even, then there exists a permutation α such that Pn i=1 aibα(i) = 0. Clearly, the condition on the parity of n is necessary: Theorem 4.1.1 shows that if n is odd and the sequence (bi) consists of distinct elements then Pn i=1 bi = 0 holds, whereas there is no such permutation for the exceptional multisets introduced in Section 4.1. On the other hand, the sequence consisting of distinct elements does not fulﬁll the condition if n is even. If true, this conjecture would imply a recent theorem of Grynkiewicz (at least in the case when n is even), which can be considered as a remarkable generalization of the Erdős-Ginzburg-Ziv theorem and a special case of Problem 4.6.2. This was conjectured before by Caro . Theorem 4.6.4. [Grynkiewicz, weighted Erdős-Ginzburg-Ziv] Let b1, . . . , bn be ele-ments of Zn such that Pn i=1 bi = 0 holds. If M is a multiset of Zn of size 2n −1, then there exists a sub-multiset M ′ = {a1, . . . , an} ⊂M and a permutation α of the elements of M ′ such that Pn i=1 aibα(i) = 0. Note that in the formulation of Problem 4.6.2, this theorem gives an upper bound on the minimal cardinality for all linear n-variate polynomials P(x) over Zn for which P(1) = 0. This conjecture of Caro turned considerable attention to various weighted subse-quence sum questions, which provides a natural formulation for our main result, too. Let M = {a1, a2, . . . , an} be a multiset of Zn and consider the elements wi : i ∈[1, n] as integer (not necessarily distinct) weights. The sum X ai∈M aiwπ(i) is called a weighted sum of M via the permutation π ∈Sym(n). Since a weighted sum depends only on the weights’ congruence classes (mod n), we may suppose that actually wi ∈[0, n −1] for all i. Hence the most obvious weight set to con-sider (besides the constant weight set, which is essentially equivalent to the Erdős-Ginzburg-Ziv Theorem) is the set of elements of Zn. The result of this chapter 57 describes those multisets, where zero cannot be achieved as a value of the weighted sum. Very recently, Grynkiewicz, Philipp and Ponomarenko proved a theorem, which can be viewed as an extension of Theorem 4.1.1. They considered arbitrary ﬁnite abelian groups G, and asked for necessary and suﬃcient conditions on multisets for which any g ∈G is attained as a value of a weighted sum of W = {0, 1, . . . \|G\| −1}. Theorem 4.6.5 (Grynkiewicz, Philipp, Ponomarenko). If some g ∈G is not at-tained as a value of a weighted sum of W = {0, 1, . . . \|G\| −1}, then either • every element of M comes from a coset of a proper subgroup of G, or • G is the Klein group and M consists of all elements of the group, or • G is cyclic and M is of the form {a, a, . . . , a, a+b, a−b} where b is a generator of G. Our Theorem 4.1.3 reﬁnes and completes the above characterization when G is a cyclic group of even order. On the other hand, in the odd order case, Theorem 4.1.1 easily implies the following corollary concerning Problem 4.6.2. Corollary 4.6.6. Let R = Zn, P(x) = x1 + 2x2 + . . . + nxn. For every multiset M of R, \|M\| = m(Zn, P) = n + 1, there exists a sub-multiset {a1, . . . , an} for which P(a1, . . . , an) = 0. (That is, the minimal cardinality is n + 1.) We supplement this chapter with a list of some open problems, more precisely, with a generalization of Bialostocki’s Conjecture 4.6.3, which was closely related to the chapter’s main result. Consider a complete bipartite graph Kn,n. We associate an element of Zn to each vertex. The weight of an edge is simply the product of the two values associated to the endvertices, and a weight of a matching is the sum of the weights of the edges in the matching. We call x ∈Zn permitted if there exists a perfect matching (PM) of weight x. General Problem: describe the structure of the set H consisting of the elements of Zn that are permitted in the above sense. 58 More speciﬁc problems are the following ones. Give conditions which imply that a) H is the whole set Zn b) H contains 0. c) Kn,n can be partitioned into n PMs which have distinct weights. Note that if one vertex color class of Kn,n is the set Zn, we get back to Theorem 4.6.5 and Theorems 4.1.1, 4.1.3, in case a) and case b), respectively. On the other hand, Conjecture 4.6.3 asserts that if n is even and Pn i=1 ai = Pn i=1 bi = 0 (mod n) holds for the elements associated to the color classes, then b) holds. Case c) appears to be widely open. It is easy to give necessary conditions such as P ai P bi = P g∈Zn g must hold. That is, if n is odd, P ai P bi = 0 (mod n), if n is even, P ai P bi = n/2 (mod n). If n is a prime, it seems reasonable to think that polynomial techniques may help to deduce a necessary and suﬃcient condition on the multisets at least in the case when one color class of Kn,n is associated to the set of elements of Zn. 59 Chapter 5 Quantitative Nullstellensatz and applications concerning Dyson-type polynomials and their q-analogues 5.1 Introduction In this chapter, we summarize an approach which turned out to be eﬀective in solving conjectures concerning constant term identities, and seems to be useful in several other ﬁelds as well. Our main tool will be a variant of the Combinatorial Nullstellensatz, which enables us to determine the coeﬃcients of maximal monomials of a multivariate polynomial exactly by a simple sum formula, if the polynomial is evaluated on a Cartesian product of large enough sets. This formula appeared recently in the papers of Lasoń and of Karasev and Petrov , later generalized in . The statement, already formulated and proved in the second chapter (Theorem 2.1.8 ) is the following. Theorem 5.1.1. Let F be a ﬁeld, and let P ∈F[x1, x2, . . . , xm] be a multivariate polynomial for which deg(P) ≤d1 + d2 + . . . + dm. Take an arbitrary set system A1, A2, . . . , Am such that Ai ⊆F and \|Ai\| = di + 1. Then the coeﬃcient of Q xdi i is X z1∈A1 X z2∈A2 X zm∈Am P(z1, z2, . . . zm) φ′ 1(z1)φ′ 2(z2) · · · φ′ m(zm), 60 where φi(x) = Q a∈Ai(x −a). Although we will use this form, we mention that a similar theorem still holds if we want to express the coeﬃcient of any maximal monomial Q xdi i of a polynomial P. That is, for monomials where there is no monomial Qk i=1 xδi i of P such that δi ≥di for all i aside from Q xdi i itself, just as in Remark 2.1.3. The key to apply Theorem 5.1.1 eﬀectively is to reduce the seemingly diﬃcult eval-uation of the sum which is equal to the coeﬃcient. To this end, some combinatorial observations enable us to choose the arbitrary set system A1, A2, . . . , Am in such a way that the vast majority of the summands vanish. In fact, optimal choice of the sets provides that all but one of the summands vanish in several cases when some symmetries in the polynomial P can be exploited combinatorially. To present the phenomenon, we introduce the theory of q-analogue identities and provide a very short proof for an ex-conjecture of Andrews , ﬁrst proven by Bres-soud and Zeilberger in an essentially combinatorial however slightly complicated way. Then we demonstrate the strength of the method by reproving many known constant term identities, a long-standing conjecture and their common generaliza-tion. 5.2 The Dyson-identity and the q-analogue Let x1, . . . , xn denote independent variables, each xi associated with a nonnega-tive integer ai. In 1962, motivated by a problem in statistical physics Dyson formulated the hypothesis that the constant term of the Laurent polynomial Y 1≤i̸=j≤n 1 −xi xj ai is equal to a certain multinomial coeﬃcient. Theorem 5.2.1 (Dyson-identity). The constant term of PD(x, a) := Y 1≤i̸=j≤n 1 −xi xj ai 61 is equal to (a1 + a2 + · · · + an)! a1! · a2! · · · an! . Independently Gunson [unpublished] and Wilson conﬁrmed the statement in the same year, then Good gave an elegant proof using Lagrange interpolation. (Wil-son later received the Nobel Prize for his outstanding contributions to mathematical physics.) Let q denote yet another independent variable. In 1975 Andrews suggested the following q-analogue of Dyson’s conjecture: Theorem 5.2.2 (q-Dyson-identity). The constant term of the Laurent polynomial QD(x, a) := Y 1≤i<j≤n xi xj ai qxj xi aj ∈Q(q)[x, x−1] must be (q)a1+a2+···+an (q)a1 (q)a2 . . . (q)an . Here (t)k denotes the q-shifted factorial, also known as the q-Pochhammer-symbol, that is, (t)k = (1 −t)(1 −tq) . . . (1 −tqk−1) with (t)0 deﬁned to be 1. Recall that the Pochhammer symbol of parameter k is simply the falling factorial x(x −1) · · · (x −k + 1) in strong connection with the binomial coeﬃcient x k and with hypergeometric series. Specializing at q = 1, Andrews’ conjecture gives back that of Dyson. Despite several attempts [57, 87, 88] the problem remained unsolved until 1985, when Zeilberger and Bressoud found a combinatorial proof. Shorter proofs for the equal parameter case a1 = a2 = . . . = an are due to Habsieger , Kadell and Stembridge . A shorter proof of the Zeilberger–Bressoud theorem, manipulating formal Laurent series, was given by Gessel and Xin . Following up a recent idea of Karasev and Petrov we present a very short combinatorial proof. First note that if ai = 0 for some i, then we may omit all factors that include the variable xi without aﬀecting the constant term of QD. Accordingly, we may 62 assume that each ai is a positive integer. Let σ denotes the sum of all ais, that is, σ = Pn i=1 ai. Consider the homogeneous polynomial F(x1, x2, . . . , xn) = Y 1≤i<j≤n ai−1 Y t=0 (xj −xiqt) · aj Y t=1 (xi −xjqt) ! ∈Q(q)[x]. Clearly, the constant term of QD(x) is equal to the coeﬃcient of Q i xσ−ai i in the polynomial F(x). Since F is homogeneous, this term will be of maximal degree. Now we are to apply Theorem 5.1.1. The idea is to take F = Q(q) with a suitable choice of the sets Ai such that F vanishes for all but one element in A1×· · ·×An. To this end, we want to choose Ai so that the cardinality is \|Ai\| = σ −ai +1, and would like to guarantee that the product Q 1≤i i, either αj −αi ≥ai, or αi −αj ≥aj + 1. In other words, αj −αi ≥ai holds for every pair j ̸= i, with strict inequality if j < i. In particular, all of the αi are distinct. Consider the unique permutation π satisfying απ(1) < απ(2) < · · · < απ(n). Adding up the inequalities απ(i+1) −απ(i) ≥aπ(i) for i = 1, 2 . . . , n −1 we obtain απ(n) −απ(1) ≥ n−1 X i=1 aπ(i) = σ −aπ(n). 63 Given that απ(1) ≥0 and απ(n) ≤σ−aπ(n), strict inequality is excluded in all of these inequalities. It follows that π must be the identity permutation and αi = απ(i) = Pi−1 j=1 aπ(j) = σi must hold for every i = 1, 2, . . . , n. This proves the statement. This way ﬁnding the constant term of QD is reduced to the evaluation of F(qσ1, qσ2, . . . , qσn) φ′ 1(qσ1)φ′ 2(qσ2) . . . φ′ n(qσn), where φi(z) = (z −1)(z −q) . . . (z −qσ−ai). Thus one rather simple calculation will imply the result of Theorem 5.2.2. F(qσ1, qσ2, . . . , qσn) = Y 1≤i 0, ℜ(β) > 0, ℜ(γ) > −min{1/n, ℜ(α)/(n −1), ℜ(β)/(n −1)}. The continued interest in the Selberg integral, demonstrated for example by the most recent article , is due to its role in random matrix theory, statistical mechanics, special function theory among other ﬁelds; see the comprehensive exposition . Properly speaking, the motivation of Dyson came from statistical physics, as he proposed to replace Wigner’s classical Gaussian-based random matrix models by 67 what now is known as the circular ensembles. The study of their joint eigenvalue probability density functions led him to the Dyson-identity. One of the most important classical mechanical systems (of ﬁnite dimensional phase space) is the so called Calogero-Moser-Sutherland model, describing a quantum many-body system. It inﬂuenced the study of further Dyson type identities. We ﬁrst recall the constant term identity of Morris which has turned out to be equivalent to the Selberg integral. It can be interpreted as a generalization of a special case of the Dyson-identity, where an additional variable x0 is also considered in the Laurent polynomial, while the exponents ai = k for each i ∈[1, n]. It asserts that if we consider the Laurent polynomial PM(x0, x; a, b, k) : = n Y j=1 1 −xj x0 a1 −x0 xj b · PD(x; k · 1) = n Y j=1 1 −xj x0 a1 −x0 xj b Y 1≤i̸=j≤n 1 −xi xj k of nonnegative integer parameters a, b, k, the constant term can be determined as CT [PM(x0, x; a, b, k)] = n−1 Y j=0 (a + b + kj)!(kj + k)! (a + kj)!(b + kj)!k! . Using M(n; a, b, k) := n−1 Y j=0 (a + b + kj)!(kj + k)! (a + kj)!(b + kj)!k! , and the matrix BM BM =             0 b b b . . . b a 0 k k . . . k a k 0 k . . . k a k k 0 . . . k . . . . . . . . . . . . ... . . . a k k k . . . 0             associated to the Laurent polynomial PM, we get 68 Theorem 5.3.2 (Morris-identity, ). CT [P(x0, x; BM)] = M(n; a, b, k). Since then, several generalizations and extensions were revealed. In 1987, introducing an extra t1 · · · tm factor into the integrand Aomoto proved an extension of the Selberg integral. Based on the fundamental theorem of calculus, it yields besides Anderson’s one of the simplest known proofs of the Selberg integral itself. Turned into a constant term identity, Aomoto’s integral reads as CT n Y j=1 1 −xj x0 a+χ(j≤m)1 −x0 xj b · PD(x; k · 1) = n−1 Y j=0 (a + b + kj + χ(j ≥n −m))!(kj + k)! (a + kj + χ(j ≥n −m))!(b + kj)!k! , where χ(S) is equal to 1 if the statement S is true and 0 otherwise. Introducing the corresponding matrix BA =                0 b . . . b b . . . b a 0 . . . k k . . . k . . . . . . ... . . . . . . ... . . . a k . . . 0 k . . . k a + 1 k . . . k 0 . . . k . . . . . . ... . . . . . . ... . . . a + 1 k . . . k k . . . 0                , where the last m rows/columns are separated, we may formulate Theorem 5.3.3 (Aomoto-identity, ). CT [P(x0, x; BA)] = n−1 Y j=0 (a + b + kj + χ(j ≥n −m))!(kj + k)! (a + kj + χ(j ≥n −m))!(b + kj)!k! . The q-analogue of the above identity which also implies a q-version of Selberg’s integral formula conjectured by Askey was ﬁrst established by Kadell . Forrester, examining the wavefunction of a generalized Calogero-Moser-Sutherland model, initiated the study of a diﬀerent extension of the Morris-identity 5.3.2. Con-sider the Laurent polynomial PF(x0, x; n0; a, b, k) = PM(x0, x; a, b, k) · Y n0n0)(j−n0)+χ(j≥n−m)(q)kj+χ(j>n0)(j−n0)+k (q)a+kj+χ(j>n0)(j−n0)+χ(j≥n−m)(q)b+kj+χ(j>n0)(j−n0)(q)k × n−n0 Y j=1 1 −q(k+1)j 1 −qk+1 . When m = 0, this proves Baker and Forrester’s conjecture [13, Conjecture 2.1], and further specializing at q = 1, Forrester’s original conjecture as well. The n0 = n case gives the following q-analogue of Aomoto’s identity. Corollary 5.4.3. [q-Aomoto identity] Let n be a positive integer. For arbitrary nonnegative integers a, b, k and m ≤n, CT[Q(x0, x; BA)] = n−1 Y j=0 (q)a+b+kj+χ(j≥n−m)(q)kj+k (q)a+kj+χ(j≥n−m)(q)b+kj(q)k . We will follow the guidance of the proof of the q-Dyson identity, Section 5.1. Let us introduce the homogeneous polynomial Fq(x0, x; B) := Y 0≤in0) j is a monomial of maximum degree in the polynomial Fq(x0, x; BAF), hence we can apply Theorem 5.1.1 with a suitably chosen system of sets {Aj, j = 0 . . . n} for which \|Aj\| = Bj + 1 for every j. For some technical reasons, we assume ﬁrst that k > a. 5.4.1 The choice for the multisets Ai Notation 5.4.4. Let γi be deﬁned as γi = βin for 0 ≤i < n and let ∆t = Pt i=0 γi. We introduce the intervals It denoting It = [∆t −γt + 1, ∆t] = [∆t−1 + 1, ∆t]. 72 Observe that γ0 = b, γ1 = · · · = γn0 = k, γn0+1 = · · · = γn−1 = k + 1 and βij = γmin{i,j} for 1 ≤i ̸= j ≤n. Note also, that the intervals I0, I1, . . . , In−1 are mutually disjoint. The sets Ai are deﬁned in the form Aj = {q0} ∪ n−1 [ t=0 q[∆t −γmin{t,j} + 1, ∆t] ⊆{q0} ∪ n−1 [ t=0 qIt = q[0, ∆n−1] for 1 ≤j ≤n and A0 = {q0} ∪ n−1 [ t=0 q[∆t −b + 1, ∆t −b + βt+1,0]. Since we assume that k > a, A0 is an ordinary set (as well as the other Ais). Then \|Ai\| = Bi + 1 holds for every 0 ≤i ≤n. 5.4.2 The combinatorics We start with an easy observation on the structure of polynomial Fq(x0, x; B) := Y 0≤i i. Then αj ≥αi implies αj ≥αi+βij, and αi > αj implies αi ≥αj +βji+1. Both statements are valid even if the corresponding entry in B is zero. We are to show that F(c0, . . . , cn) = 0 holds for F = Fq(.; BAF) for all but one selection of elements ci ∈Ai, namely when c0 = 1, ci = q∆i−1 for 1 ≤i ≤n. This statement is veriﬁed by the juxtaposition of the following two lemmas. 73 Lemma 5.4.6. Let α0 = 0. If F(c0, c1, . . . , cn) ̸= 0, then αi = ∆i−1 for every 1 ≤i ≤n. Lemma 5.4.7. If α0 ̸= 0, then F(c0, . . . , cn) = 0. One key to each is the following consequence of Claim 5.4.5. Lemma 5.4.8. Suppose that F(c0, . . . , cn) ̸= 0. Then for every 1 ≤t ≤n −1 there is at most one index 1 ≤i ≤n such that αi ∈It. Proof. Assume that, on the contrary, there is a pair 1 ≤i ̸= j ≤n such that αi, αj ∈It. Let αj ≥αi, then it follows from Claim 5.4.5 that αj −αi ≥k. The length of It is γt ∈{k, k +1}. Thus, it must be γt = k +1, αi = ∆t −k and αj = ∆t. Consequently, t > n0, i < j and i ≤n0. Therefore ∆t −γmin{t,i} + 1 = ∆t −k + 1 and αi ̸∈Ai, a contradiction. Proof of Lemma 5.4.6. For every 1 ≤i ≤n we have αi ≥α0, therefore αi ≥β0i = b by Claim 5.4.5. Moreover, k > a ≥0 implies that α1, . . . , αn are all distinct, thus it follows from Lemma 5.4.8 that each of the intervals I0, I1, . . . , In−1 contains precisely one of them. Let π ∈Symn denote the unique permutation for which απ(1) < · · · < απ(n), then απ(i) ∈Ii−1. By Claim 5.4.5 we have απ(i+1) ≥απ(i) + βπ(i),π(i+1) + χ(π(i) > π(i + 1)). Consequently, απ(n0+1) ≥b + kn0 + n0 X i=1 χ(π(i) > π(i + 1)) ≥∆n0 + n0 X i=1 χ(π(i) > π(i + 1)). Since απ(n0+1) ≤∆n0, it follows that απ(1) = b, π(1) < · · · < π(n0 + 1), and βπ(i),π(i+1) = k for 1 ≤i ≤n0. This in turn implies that π(n0) ≤n0, thus π(i) = i and αi = ∆i−1 for 1 ≤i ≤n0. Now for n0 < i < n we have π(i), π(i + 1) > n0 and thus βπ(i),π(i+1) = k + 1. Restricting π to the set [n0 + 1, n] and starting with απ(n0+1) = ∆n0, a similar argument completes the proof. Proof of Lemma 5.4.7. Assume that, contrary to the statement, F(c0, . . . , cn) ̸= 0. 74 Since α0 ̸= 0, α0 ∈Sn−1 t=0 q[∆t −b + 1, ∆t −b + βt+1,0], that is, ∆u −b + 1 ≤α0 ≤∆u −b + βu+1,0. It is implied by Lemma 5.4.8 that at most n−1−u of the distinct numbers α1, . . . , αn can lie in the interval [∆u+m0+1, ∆n−1]. Thus, at least u+1 of the numbers α1, . . . , αn satisfy αj ≤∆u. Furthermore, if αj lies in the interval Tuj = [∆u −b + βu+1,0 −βj0, ∆u] for some 1 ≤j ≤n, then α0 −βj0 ≤αj ≤α0 + β0j −1 and there is a term of the form xj −qtx0 or x0 −qtxj in F which attains 0 when evaluated at the point (c0, c). It follows that at least u+1 of the numbers αj satisfy αj ≤∆u −b+βu+1,0 −βj0 −1. This is clearly impossible if u+1 ≤n−m, for then ∆u −b+βu+1,0 −βj0 −1 ≤uk−1 in view of n −m ≤n0, and on the other hand the diﬀerence between any two such αj is at least k in view of Claim 5.4.5. Thus, u ≥n−m and βu+1,0 = a+1. Consider αν(1) < · · · < αν(u+1) ≤∆u −b + βu+1,0 −βν(u+1),0 −1 ≤∆u −b. If u ≤n0, then it must be αν(i) = (i −1)k and ν(1) < · · · < ν(u + 1), but then ν(u + 1) ≥u + 1 > n −m, βν(u+1),0 = a + 1, implying αν(u+1) ∈Tu,ν(u+1), which is absurd. This means that u ≥n0 + 1. It is easy to see that αν(i+1) −αν(i) ≥γν(i) for i ≤u, thus αν(u+1) ≥Pu i=1 γν(i) ≥∆u −b. Therefore Pu i=1 γν(i) = ∆u −b, which implies that {ν(1), . . . , ν(u)} ⊇{1, . . . , n0}. Consequently, ν(u+1) ≥n0+1 > n−m, which leads to a contradiction as before. 5.4.3 The computation It only remains to evaluate Fq(q0, q∆0, . . . , q∆n−1; B) ψ0ψ1 . . . ψn , (4.1) where ψj = Y α∈Aj{∆j−1}(q∆j−1 −qα) 75 for j = 1, . . . n, and with the shorthand notation ∆v u = γu + · · · + γv = ∆v −∆u−1, ψ0 = n−1 Y t=0 ∆t 1+βt+1,0 Y α=∆t 1+1 (1 −qα) = n Y j=1 ∆j−1 1 + 1, ∆j−1 1 + βj0 q . (4.2) From now on, [u, v]q := (1−qu) . . . (1−qv) = (q)v/(q)u−1, with [u, u]q abbreviated as [u]q. Both the numerator and the denominator in (4.1) is the product of factors in the form ±qu(1−qv) with some non-negative integers u, v. More precisely, collecting factors of a similar nature together we ﬁnd that the numerator is the product of the factors (−1)γ0 × q1+···+(γ0−1) × ∆j−1 1 + 1, ∆j−1 0 + βj0 q for 1 ≤j ≤n, (4.3) (−1)γi × q∆i−1+···+(∆i−1+γi−1) × ∆j−1 i −γi + 1, ∆j−1 i q for 1 ≤i < j ≤n, (4.4) and qγi∆i−1 × ∆j−1 i + 1, ∆j−1 i + γi q for 1 ≤i < j ≤n. (4.5) In the denominator, besides (4.2) we have the factors (−1) × [∆j−1]q × ψj< × ψj= × ψj> for 1 ≤j ≤n, (4.6) where ψj< = j−2 Y t=0 (−1)γt × q(∆t−γt+1)+···+∆t × ∆j−1 t+1, ∆j−1 t+1 + γt −1 q , (4.7) ψj= = (−1)γj−1−1 × q(∆j−1−γj−1+1)+···+(∆j−1−1) × [1, γj−1 −1]q , (4.8) and ψj> = n−1 Y t=j qγj∆j−1 × ∆t j −γj + 1, ∆t j q . (4.9) Now the powers of −1 and q cancel out due to the simple identity nγ0 + X 1≤i<j≤n γi = n + X 0≤tn0)(j−n0)+χ(j≥n−m) (q)a+kj+χ(j>n0)(j−n0)+χ(j≥n−m) . (4.10) As for the rest, the contribution from (4.5) and (4.7) with the substitution t + 1 = i gives Y 1≤i\ 0. Combining (4.11) with the contribution of the factors [∆j−1]q = 1 −q∆j−1 from (4.6) and the factors [1, γj−1 −1]q = (q)γj−1−1 from (4.8), shifting indices we obtain n−1 Y j=1 (q)∆j 1+γ1 (q)∆j 1+γ0 × n−1 Y j=0 1 (q)γj × n−n0 Y j=2 (1 −q(k+1)j) !χ(n0>0) , in agreement with (q)kj+χ(j>n0)(j−n0)+k (q)b+kj+χ(j>n0)(j−n0)(q)k × n−n0 Y j=1 1 −q(k+1)j 1 −qk+1 . (4.12) Putting together (4.10) and (4.12) completes the proof of Theorem 5.4.2 in the case k > a. Avoiding this restriction, we extend it in the next subsection. 5.4.4 The rationality result The extension of the result that includes all non-negative integers for the parameter k depends on the following rationality lemma, inspired by [44, Proposition 2.4]. 77 Recall that QD(x; k · 1) denotes the q-analogue of the Laurent polynomial PD(x; k·a) corresponding to the special case when ai = k for all i,in the Dyson identity (and also the special case a = b = 1 of the Morris identity). Lemma 5.4.9. Fix nonnegative integers ri, si for 1 ≤i ≤n, satisfying P ri = P si. There exists a rational function R = R(z) ∈Q(q)(z) that depends only on n and the numbers ri, si such that CT xr1 1 . . . xrn n xs1 1 . . . xsn n QD(x; k · 1) = R(qk)(q)nk (q)n k . Proof. The constant term of QD(x; k · 1) equals the coeﬃcient of Q x(n−1)k i in the polynomial Fq(x) = Y 1≤i π(i + 1) ) holds for 1 ≤i ≤n −1 with the unique permutation π = πc ∈Symn satisfying απ(1) < · · · < απ(n). Consequently, αi = (π−1(i) −1)k + ǫi for some ǫi = ǫi(c) ∈[0, sπ(n)]. 78 Set C = {c ∈A∗ 1 × · · · × A∗ n \| F ∗ q (c) ̸= 0}, and write s = max si. It follows that \|C\| ≤n! s + n n . Moreover, the set S = {(πc, ǫ1(c), . . . ǫn(c)) \| c ∈C} is independent of k; it depends only on n and the numbers si. It follows from Lemma 2.1.8 that, using the notation τ = π−1, CT xr1 1 . . . xrn n xs1 1 . . . xsn n Dq(x; k) = X c∈C n Y i=1 q((τ(i)−1)k+ǫi)ri F(. . . , q(τ(i)−1)k+ǫi, . . .) ψ∗ 1ψ∗ 2 . . . ψ∗ n where ψ∗ π(i) = Y 0≤j≤(n−1)k+sπ(i),j̸=(i−1)k+ǫπ(i)(q(i−1)k+ǫπ(i) −qj). One readily checks that for each Σ = (π, ǫ1, . . . , ǫn) ∈S there exist rational functions Ri ∈Q(q)(z) that depend only on n, the numbers rj, sj and the sequence Σ such that n Y i=1 q((τ(i)−1)k+ǫi)ri = R0(qk), ψi ψ∗ π(i) = Ri(qk) and F(. . . , q(τ(i)−1)k+ǫi, . . .) F(q0, qk, . . . , q(n−1)k) = Rn+1(qk). The result follows. Proof of Theorem 5.4.2, generalizing from a < k to arbitrary k. Notice that the q-Laurent polynomial in view, Q(x0, x; BAF) = n Y j=1 (qxj)a+χ(j≤m)(1/xj)b Y n0n0)(j−n0)+χ(j≥n−m)(q)kj+χ(j>n0)(j−n0)+k (q)a+kj+χ(j>n0)(j−n0)+χ(j≥n−m)(q)b+kj+χ(j>n0)(j−n0)(q)k × n−n0 Y j=1 1 −q(k+1)j 1 −qk+1 can be written in the form R′(qk)(q)nk (q)n k with a rational function R′ ∈Q(q)(z) which also depends only on n, m, n0, a, b, for k ≥a + 1. Since R′(qk) = R(qk) for every k ≥a + 1, it follows that R ≡R′, completing the proof. 5.5 Remarks and variations We take an overview on related problems and results. First, let us look back on the framework we used to conﬁrm the identities through q-analogues. If we consider a Laurent polynomial P(x0, x1, . . . , xn, B) := Y 0≤i̸=j≤n 1 −xi xj βij, then obviously the order of the variables does not aﬀect the constant term. However, this is not the case with the q-analogue version, where an asymmetry appears along the order of xis. Indeed, taking xi xj βij xj xi βji instead of xi xj βij xj xi q βji would result in two essentially identical terms, (1 −xi xj ) and (1 −xj xi ): they both vanish if and only if xi = xj, providing the same combinatorial information. This explains the convenience of xi xj βij xj xi q βji, where no such overlapping occurs. Several natural questions and problems may arise. First, all the mentioned results correspond to constant terms of Laurent polynomials of form P(x, B). Essentially this is due to symmetry, since this term clearly has a special role compared to other coeﬃcients. It provides a wider range for applications and typically easier ways for the proofs at the same time. However, several papers studied the evaluation of diﬀerent coeﬃcients of Laurent-polynomials, mostly concerning the Laurent poly-nomial of Dyson [75, 76, 84]. We should point out here that following the approach of the rationality result (Subsection 5.3.4) is useful in general. Indeed, Lemma 5.4.9 80 presents a way to express other coeﬃcients using again the Quantitative Nullstel-lensatz. Generally, one cannot rely on getting only one non-vanishing term as we do get in the proofs, but the number of terms is bounded in terms of the considered de-gree sequence corresponding to the variables. Note that this idea was independently developed by Doron Zeilberger in . Another problem to consider is to describe the Laurent polynomials, or rather the corresponding matrices B, where the approach is applicable. So far we do not know the limits of our method; we do not have a general argument, which, given B as an input, would tell whether the corresponding constant term can be easily evaluated this way. However, we were able to handle basically all matrices, where the combinatorial approach of Zeilberger and Bressoud – built on an improvement of Good’s diﬀerence-equation proof idea for the Dyson-identity – or the method of Gessel, Lv, Xin and Zhou [43, 44, 96] – based on the idea of proving polynomial identities by pointing out enough values where the two polynomials are equal – were applied. Generally, our approach provides short proofs, which are easy to follow. In addition to the solution to Forrester’s problem, it can be also applied to prove various conjectures of Kadell, c.f. [58, 59, 68, 66, 100]. In the proof of Theorem 5.4.2, and in the method of Gessel, Lv, Xin and Zhou as well, one might have to come up with a rationality result, see Lemma 5.4.9 or Proposition 2.4 in , to complete the proof. This follows from the fact that we do not allow Ai to be general multisets, only sets in the Quantitative Nullstellensatz. Indeed, A0 would contain duplicated elements if k ≤a holds. However, the Quantitative Nullstellensatz can be extended to be applicable under these circumstances as well due to Lemma 2.1.12 introduced in the second chapter. For more details we refer to . Finally we present further applications of the Quantitative Nullstellensatz, related to additive combinatorics. Dias da Silva and Hamidoune conﬁrmed the long-standing conjecture of Erdős and Heilbronn . Later, Alon, Nathanson and Ruzsa obtained a proof via the polynomial method, see [1, 6]. For a collection of sets A1, . . . , An ⊆Zp, consider the following restricted sumset: ^ SAi = {a1 + · · · + an \| ai ∈Ai, aj −ai ̸∈Sij for i < j} . 81 The theorem can be formulated as follows. Theorem 5.5.1. ^ SAi ≥min p, n\|A\| −n2 + 1 , A far-reaching generalization was obtained by Hou and Sun . Theorem 5.5.2. Let A1, . . . , An be subsets of a ﬁeld F such that \|Ai\| = k for 1 ≤i ≤ n and assume that Sij ⊆F satisfy \|Sij\| ≤s for 1 ≤i < j ≤n. If either char(F) = 0 or char(F) > max {n⌈s/2⌉, n(k −1) −n(n −1)⌈s/2⌉} , then ^ SAi ≥n(k −1) −n(n −1)⌈s/2⌉+ 1. Clearly, s = 0 and s = 1 gives back the conditions of the Cauchy-Davenport and the Erdős- Heilbronn theorem, respectively. This extended result can also be proved by the Quantitative Nullstellensatz . In fact, Lilu Zhao pointed out that this can even be strengthened in the following way. Theorem 5.5.3. Let A1, . . . , An be subsets of a ﬁeld F such that \|Ai\| ∈{k, k + 1} for 1 ≤i ≤n and assume that Sij ⊆F satisfy \|Sij\| ≤s for 1 ≤i < j ≤n. If either char(F) = 0 or char(F) > max ( n⌈s/2⌉, n X i=1 (\|Ai\| −1) −n(n −1)⌈s/2⌉ ) , then ^ SAi ≥ n X i=1 (\|Ai\| −1) −n(n −1)⌈s/2⌉+ 1. His proof is based on the Combinatorial Nullstellensatz and the Aomoto identity, see Theorem 5.4.2. 82 Chapter 6 Summary Our work is based on applications of the original strong, and the quantitative ver-sions of Noga Alon’s Combinatorial Nullstellensatz. The Combinatorial Nullstellen-satz as a delicate, widely applicable polynomial method asserts that the zero locus of a (multivariate) polynomial cannot vanish on a large enough well structured point set. Alon pointed out that this approach is very fruitful and provides elegant proofs in many ﬁelds of combinatorics, including additive combinatorics, combinatorial or ﬁnite geometry, graph theory and extremal set theory, by ﬁnding a connection between the structure of the object in view and the zero locus of corresponding polynomials. In Section 3, we present a result based on a joint work with András Gács, Tamás Héger and Dömötör Pálvölgyi . The main theorem asserts a connection between the degree and the range of polynomials over a ﬁnite ﬁeld. More precisely, all multisets M ∈GF(q) of size q are described which cannot be a range of a polynomial of degree at most q−1 or q−2. This statement can be formulated also in the language of ﬁnite geometries and additive number theory. As for the additive number theory version, one may investigate extensions by study-ing the problem for multisets over abelian groups or cyclic groups Zn rather than the problem for multisets over cyclic groups Zp of prime order p. The result resembles to the one for the former problem: only some well characterized multisets provide exceptions in the corresponding structure theorem. Their description can be ﬁnd in Section 4, on the basis of . 83 The next problem is connected to the theory of q-analogues. Generally speaking, q-analogue versions are extensions of statements by the introduction of a new pa-rameter q, where the limit transition q →1 gives back the original assertion. q-analogue identities date back to Euler and turned out to be very eﬃcient in var-ious combinatorial problems and in applications in statistical physics as well. For example, the Dyson identity has an important role in a quantum many body system model, and its q-analogue is proven by Zeilberger and Bressoud in . The identity of Dyson asserts the following : The constant term of Y 1≤i̸=j≤n 1 −xi xj ai is (a1 + a2 + · · · + an)! a1!a2! · · · an! . Section 5 is based on a joint work with Gyula Károlyi. We studied the q-analogue of the Dyson-identity, and several similar constant term identities [67, 68]. The main beneﬁt of the proofs is the application of a quantitative version of the Nullstellensatz by determining the constant term of a given multivariate polynomial as an expo-nentially large sum of substitution values of a function; a sum whose summands can be chosen to vanish for all but one (or a few) substitution values. The method successfully solved some well-studied conjectures such as the Forrester conjecture . 84 Chapter 7 Összefoglaló - in Hungarian Munkánkban középpontjában a Noga Alon Kombinatorikus Nullhelytételének , il-letve ezen eredmény erősebb változatainak alkalmazása áll. A Kombinatorikus Null-helytétel, mint egy speciális, ugyanakkor általánosan alkalmazható módszer arra épül, hogy számos kombinatorikus struktúra szerkezetét többváltozós polinomok eltűnési helyeivel lehet leírni. Alon rámutatott, hogy a számelmélet, additív kom-binatorika, gráfelmélet, halmazrendszerek, véges geometria és más területek számos központi kérdése kezelhető elegáns egyszerű módszerének segítségével. Az első általunk vizsgált problémát a cikk írja le, ami Gács Andrással, Héger Tamással és Pálvölgyi Dömötörrel közös - ez képezi a 3. fejezet alapját. A főtétel az additív számelmélet, illetve a polinomok elméletének nyelvén is megfogalmazható, és a véges test feletti polinomok foka és értékkészlete közötti összefüggésre mutat rá, nevezetesen leírja azon GF(q) feletti q elemű M multihalmazokat, amelyekhez nem létezik legfeljebb q −1 illetve legfeljebb q −2 fokú polinom, melyek esetén M az értékkészlet multihalmaza. A kérdés egy másik irányú általánosítását kapjuk, ha a probléma additív számelméleti megfogalmazásában Zn (n > 1 egész) feletti multihalmazokat vizsgálunk a prím-rendű GF(p) testbeliek helyett. A problémára adható válasz az előzőhöz hasonló: néhány könnyen karakterizálható multihalmaz jelent kivételes struktúrát a struk-túratételben, ennek leírását a cikk alapján a 4. fejezetben találhatjuk. A másik probléma a q-analógiák elméletekhez kapcsolódik. Általánosan tekintve ez 85 állítások kiterjesztésére vonatkozik, ahol egy új q paraméter bevezetésével q →1 határátmenetben kapjuk az eredeti tételt. A q-analóg azonosságok a kombinatorikában is rendkívül hasznosak lehetnek, és Eu-lerre vezethetőek vissza, azonban számos alkalmazásuk van a statisztikus ﬁzikában is. A cikkben például a Dyson által vizsgált statisztikai ﬁzikus modellben kulcs-szerepet játszó azonosság q- változatát bizonyítja igen komplex módon Zeilberger és Bressoud. A Dyson-azonosság a következőt állítja : Y 1≤i̸=j≤n 1 −xi xj ai konstans tagja (a1 + a2 + · · · + an)! a1!a2! · · · an! . Károlyi Gyulával közös cikkeinken alapuló 5. fejezetben ezen állítás q-analógját, valamint hasonló konstans együtthatós azonosságokat vizsgáltunk [67, 68]. 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16217	https://study.com/skill/learn/solving-a-word-problem-involving-the-average-rate-of-change-explanation.html	Solving a Word Problem Involving the Average Rate of Change \| Algebra \| Study.com Log In Sign Up Menu Plans Courses By Subject College Courses High School Courses Middle School Courses Elementary School Courses By Subject Arts Business Computer Science Education & Teaching English (ELA) Foreign Language Health & Medicine History Humanities Math Psychology Science Social Science Subjects Art Business Computer Science Education & Teaching English Health & Medicine History Humanities Math Psychology Science Social Science Art Architecture Art History Design Performing Arts Visual Arts Business Accounting Business Administration Business Communication Business Ethics Business Intelligence Business Law Economics Finance Healthcare Administration Human Resources Information Technology International Business Operations Management Real Estate Sales & Marketing Computer Science Computer Engineering Computer Programming Cybersecurity Data Science Software Education & Teaching Education Law & Policy Pedagogy & Teaching Strategies Special & Specialized Education Student Support in Education Teaching English Language Learners English Grammar Literature Public Speaking Reading Vocabulary Writing & Composition Health & Medicine Counseling & Therapy Health Medicine Nursing Nutrition History US History World History Humanities Communication Ethics Foreign Languages Philosophy Religious Studies Math Algebra Basic Math Calculus Geometry Statistics Trigonometry Psychology Clinical & Abnormal Psychology Cognitive Science Developmental Psychology Educational Psychology Organizational Psychology Social Psychology Science Anatomy & Physiology Astronomy Biology Chemistry Earth Science Engineering Environmental Science Physics Scientific Research Social Science Anthropology Criminal Justice Geography Law Linguistics Political Science Sociology Teachers Teacher Certification Teaching Resources and Curriculum Skills Practice Lesson Plans Teacher Professional Development For schools & districts Certifications Teacher Certification Exams Nursing Exams Real Estate Exams Military Exams Finance Exams Human Resources Exams Counseling & Social Work Exams Allied Health & Medicine Exams All Test Prep Teacher Certification Exams Praxis Test Prep FTCE Test Prep TExES Test Prep CSET & CBEST Test Prep All Teacher Certification Test Prep Nursing Exams NCLEX Test Prep TEAS Test Prep HESI Test Prep All Nursing Test Prep Real Estate Exams Real Estate Sales Real Estate Brokers Real Estate Appraisals All Real Estate Test Prep Military Exams ASVAB Test Prep AFOQT Test Prep All Military Test Prep Finance Exams SIE Test Prep Series 6 Test Prep Series 65 Test Prep Series 66 Test Prep Series 7 Test Prep CPP Test Prep CMA Test Prep All Finance Test Prep Human Resources Exams SHRM Test Prep PHR Test Prep aPHR Test Prep PHRi Test Prep SPHR Test Prep All HR Test Prep Counseling & Social Work Exams NCE Test Prep NCMHCE Test Prep CPCE Test Prep ASWB Test Prep CRC Test Prep All Counseling & Social Work Test Prep Allied Health & Medicine Exams ASCP Test Prep CNA Test Prep CNS Test Prep All Medical Test Prep College Degrees College Credit Courses Partner Schools Success Stories Earn credit Sign Up Solving a Word Problem Involving the Average Rate of Change Algebra 1 Skills Practice Click for sound 6:29 You must c C reate an account to continue watching Register to access this and thousands of other videos Are you a student or a teacher? I am a student I am a teacher Try Study.com, risk-free As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it risk-free It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time. Already registered? Log in here for access Back What teachers are saying about Study.com Try it risk-free for 30 days Already registered? Log in here for access 00:04 Solving a word problem… 03:39 Solving a word problem… Jump to a specific example Speed Normal 0.5x Normal 1.25x 1.5x 1.75x 2x Speed Dana Hansen, Marisa Bjorland Instructors Dana Hansen Dana has taught high school mathematics for over 9 years. She has a masters in mathematics education from CU Denver. She is certified to teach both mathematics and physics in both Colorado as well as Iowa. She loves getting outdoors as much as possible hopefully with her two dogs. Dana has tutored with study.com for over a year, and loves the opportunity to work one on one with students to help them develop their content knowledge. View bio Marisa Bjorland Marisa Bjorland has taught math at the community college level since 2008, including remote and online courses. She has a master's degree in mathematics from University of California, Santa Cruz and a bachelor's degree in mathematics from University of California, San Diego. View bio Example SolutionsPractice Questions How to Solve a Word Problem Involving Average Rate of Change Step 1: Write the values from your word problems as points. Your independent variable should be your x value and your dependent variable is your y value. Step 2: Use the slope equation to calculate the average rate of change. Slope is just another way of saying average rate of change! Step 3: Simplify your answer. Make sure to follow the order of operations as you do this. Step 4: Write your answer as a sentence using units. How to Solve a Word Problem Involving Average Rate of Change Vocabulary Slope equation: This allows us to calculate the average rate of change of a function over a given interval. The equation for slope is: Y 2−Y 1 X 2−X 1 Independent Variable: This variable is graphed on the x-axis, it is considered the variable that we have control over. Dependent Variable: This variable is graphed on the y-axis, it is considered the variable that changes as a result of changing our independent variable. Order of Operations: This is the order with which we evaluate an expression. This order is: parentheses, exponents, multiplication, division, addition, and then subtraction. So, let's try using these steps to solve a word problem involving an average rate of change, in the following two examples! In the first example, we will use whole numbers and in the second example, we will get a decimal answer. How to Solve a Word Problem Involving Average Rate of Change: Example 1 You are taking a road trip with your family, and according to your map, you have 275 miles to travel today. If you drive for 5 hours what is the average rate of change of your travel during the trip? Step 1: Write the values from your word problems as points. Your independent variable should be your x value and your dependent variable is your y value. In this example, distance is is our dependent variable and time is our independent variable. That gives us the points: (0, 0) and (5, 45) Step 2: Use the slope equation to calculate the average rate of change. It doesn't matter what order you plug in your values as long as the order is the same on the top as the bottom. 0−275 0−5 Step 3: Simplify your answer. −275−5=55 Step 4: Write your answer as a sentence using units. The rate of change of travel that day was 55 miles per hour. How to Solve a Word Problem Involving Average Rate of Change: Example 2 You are going on a hike with your friend and you bring some Trail Mix as a snack. Between the two of you, 7 handfuls of Trail Mix have been eaten in the first 560 feet. What is the average rate at which you two are eating the Trail Mix? Step 1: Write the values from your word problems as points. Your independent variable should be your x value and your dependent variable is your y value. In this example, distance is is our dependent variable and a handful of food is our independent variable. That gives us the points: (0, 0) and (7, 560) Step 2: Use the slope equation to calculate the average rate of change. Slope is just another way of saying average rate of change! 0−7 0−560 Step 3: Simplify your answer. Make sure to follow the order of operations as you do this. −7−560=0.0125 Step 4: Write your answer as a sentence using units. In the first 560 feet, you are eating Trail Mix at a rate of 0.0125 handfuls per foot. Get access to thousands of practice questions and explanations! Create an account Table of Contents How to Solve a Word Problem Involving Average Rate of Change How to Solve a Word Problem Involving Average Rate of Change Vocabulary Example 1 Example 2 Test your current knowledge Practice Solving a Word Problem Involving the Average Rate of Change Recently updated on Study.com Videos Courses Lessons Articles Quizzes Concepts Teacher Resources The First & Second Balkan Wars \| Background & Consequences Ethnic Groups in Indonesia \| Demographics & People Gods of the Winter Solstice Holes by Louis Sachar \| Themes, Quotes & Analysis Aurangzeb \| Empire, Achievements & Failures Libya Ethnic Groups \| Demographics, Population & Cultures Cold War Lesson for Kids: Facts & Timeline The Chronicles of Narnia Series by C.S. Lewis \| Overview... 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16218	https://www.youtube.com/watch?v=meONGezEXlI	How to Find the Equation of a Tangent Line with Derivatives (NancyPi) NancyPi 713000 subscribers 14907 likes Description 858853 views Posted: 15 May 2018 MIT grad shows how to find the tangent line equation using a derivative (Calculus). To skip ahead: 1) For a BASIC example, skip to time 0:44. 2) For an example that uses the QUOTIENT RULE, skip to time 5:58. 3) For an example that uses TRIG FUNCTIONS and the CHAIN RULE for finding the equation of a tangent line, skip to time 11:35. Nancy formerly of MathBFF explains the steps. For HOW TO FIND DERIVATIVES, jump to: Follow me on Instagram! Follow me on Twitter: What is a tangent line? It is a line that is tangent to a curve at one point. You can use calculus for finding the tangent line equation, by taking the derivative of the given function. Finding the EQUATION OF A TANGENT LINE using a DERIVATIVE: There are four main steps to find the equation of a tangent line with derivatives: 1) TAKE DERIVATIVE: The first step is to take the derivative of the given equation, with respect to x. For instance, to find the equation of the line tangent to f(x) = x^2 + 3 at x = 4, first take the derivative of f(x), which is 2x, by the power rule. 2) PLUG IN X-VALUE INTO DERIVATIVE TO GET SLOPE: The second step is to plug the given x-value into the derivative of f(x). The value you get is the slope of the tangent line, m. 3) FIND Y-VALUE OF POINT WITH ORIGINAL EQUATION: The third step is to find the y-value of the point. You get this by plugging the given x-value into the original equation to find the corresponding y-value. Since you want the tangent line at x = 4, plug x = 4 into the original f(x), and you get a y-value of 7. Now that you have both the x and y coordinates, this means that the full point is (4,7). 4) PLUG THE SLOPE AND X,Y POINT VALUES INTO POINT-SLOPE EQUATION: The last step is to put the x and y point values, as well as the slope, m, into the point slope equation of a line. The point slope form is x - x1 = m (y - y1). Plug the x and y point values into this equation for x1 and y1. Plug the slope you found in for m. This is the equation of the tangent line. The equation will still have x and y variables in it. You can also rearrange the equation you get so that it is in "y equals" or slope intercept form, if you want. You may need to find the derivative with other derivative rules, such as the quotient rule or chain rule. Even if you need to use other rules when taking derivatives, or even if you need to differentiate something like a trigonometry function (ex. secant) or a fraction form with a numerator and denominator (rational function), the four main steps to find the equation of the tangent line will be the same. For more math help, check out: 634 comments Transcript: Intro Hi guys! I'm Nancy.. and I'm going to show you how to find the equation of a tangent line using derivatives. What is a tangent line? Well, it just means if you have a graph... with some curve on it the tangent line is the line that touches your graph at one point exactly. It's a linear line that is tangent to your curve and sort of grazes it and touches at one point. So tangent line problems can come in many forms but thankfully you can use the same four steps in all of them so let me show you some examples. Problem OK. Say you have a problem that says to find the tangent line to Y equals X cubed plus 2X at a point at the point: 1, 3. How do you start this problem? Well it's always gonna be the same four steps in these tangent line problems. The first step is to take the derivative of the equation that you're given. This Y equals is also F of X. It's also a function. You want to take the derivative of the equation you're given, so F prime of X is your first step. F prime of X and then you take the derivative of this expression. The derivative of X cubed... remember for the power rule, you bring the power down and then decrease that power by 1. so for the first term, we have 3X squared for the second term, the derivative of 2X is just 2. The X goes away. You have 3X squared plus 2. So that's your derivative, your F prime of X. The second step is always to take the X coordinate that you're given in the point... and plug it in to F prime of X. You can call that F prime of 1. Since you're X coordinate is 1. So you take that number, and you're going to plug it in to what you just found, the derivative. And this is just notation for that, for plugging in 1. Anyway, when you plug in 1 for X, you have 3 times 1 squared plus 2... which simplifies to 3 times 1 plus 2 which is just 5, so F prime of 1... equals 5. This is going to be your slope number that you'll use. The third step is... to find the Y coordinate of your point, if you're not already given it so sometimes you're not given the Y. In this problem, you're given the whole point, which could happen. So you have Y equals 3, so you don't need to find Y in this case. You can move on to the fourth step, the last step, is to plug in the numbers that you've found into the point-slope formula for a line. The point-slope formula... in general looks like Y minus Y1 equals M times X minus X1. M is just your slope and that's going to be the F prime value that you found when you plugged in X so your M in this problem is 5. Which you found right here. X1 and Y1 are the coordinates of the point that you found or that you were given. So this here you can also call X1, Y1. You have both of those numbers. Now you want an equation for the tangent line. That means that in your equation you need to leave an X variable and Y variable in it. So it's OK to leave those there. In fact you want it. You can go ahead and you can write your equation by plugging in values for M, for X1, and for Y1. So your X1 is the coordinate 1. Your Y1 is the Y coordinate of the point. The original point, which is 3. So if you rewrite this, your equation is Y - 3 equals the slope 5, times (X - 1). Notice you're leaving X and Y in the equation because you want it to be an equation for Y in terms of X. So this... this is correct. This is a correct answer. Sometimes it's considered simpler to move it around and change it to be in slope intercept form or Y = MX + B form. So you can do that with some simple algebra if you were to multiply, distribute out, the... 5X - 5, and rearrange the numbers you could get into Y equals form. if you were to multiply, distribute out, the... 5X - 5, and rearrange the numbers you could get into Y equals form. Y = 5X - 5 + 3... which is just minus 2. So this.. Y = 5X - 2 is your equation of the tangent line to this F of X at that point. Solution Alright, say you have F(X) = X / (X - 1) and you have to find the equation of the tangent line at X = 2. This looks a lot more complicated but the truth is it's the same four steps no matter how complicated your F(X), your equation gets. but the truth is it's the same four steps no matter how complicated your F(X), your equation gets. This could be ridiculous.. but as long as you know how to take F prime of X, the derivative, you can still do all the same steps and get the equation of the linear tangent line. So the first step is still, find F of X. Find the derivative of your given equation. You want F prime of X.. Now since your equation is X / (X - 1) this is a fraction. This is a quotient. so you'll have to use the quotient rule to take the derivative. Which is a little more complicated than you might like.. but the steps are pretty straightforward. Think back to doing the quotient rule.. It's the bottom of the function times the derivative of the top.. derivative of X is just 1.. ..minus the top expression, X, as is.. times the derivative of the bottom, which is 1 from the X and nothing from the -1, so it's just 1 there. And then it's all over the bottom squared. The lower function squared. (X - 1) squared. OK. So this is the derivative, but you should simplify it because things will cancel. Try simplifying within the top first. Within the numerator. This is X - 1 - X ... another X... all over (X - 1) ^ 2. Great news is that something cancels on top. This becomes simpler. X - X cancels and you're just left with -1 on top. So, the simplest way to write your derivative F prime of X is.. -1 / (X - 1)^2 Remember F prime just means the derivative. Alright. Second step. Again, plug in your X value, into the derivative F prime of X. What you can call that is F prime of 2, because you're plugging in 2. So the notation for that is.. F prime of 2 means you're putting 2 in place of every X in your derivative expression. So in this case, that means -1 on top... and (2 - 1)^2 on the bottom. This simplifies to... -1 over 1^2, which is just 1. So your slope value... your M is -1. I'll make that clear. OK. Third step. Find the original Y value if you don't already have it.. if you weren't given it. You weren't a full point, you were only half, just the X coordinate. So you have to find Y. How do you do that? All you need to is take the original X, plug it in to your original equation, your original F... and find Y, find the Y value. So you can call that F of 2 That will give you your Y. F of 2 means you're plugging in 2 for X, so you have 2 / (2 - 1)... which is 2 / 1, which is 2. Alright, so that is your Y. So if you had to write a full point for the original point... it would be 2 for X and 2 for Y, or (2,2). That's important, you're going to use both of those numbers next. Because the last step is still to plug in your M, your Y, and your X... all into the point-slope form for the equation of a line. And that point-slope form looks like... Y minus your Y coordinate, your Y1, which is 2... equals M, your slope, which is -1... times parentheses X, the X you want that in the equation. Otherwise it's not a full equation... minus the X coordinate, or X1, which is also 2 in this problem. Alright, so you know, this is technically correct, you can leave your answer in point-slope form. Unless you have to rearrange it to get it into Y = MX + B form, but technically this is correct so this is the equation for the tangent line to this graph at X = 2. OK. One final problem. Final Problem I'm giving you one that has a trig function in it. F of X equals (sec X)^2 There's also a power in here, so this might look kind of scary, maybe... but just rely on what you know about derivatives, differentiating. In this case you're going to have to use the Chain Rule.. ..yay chain rule.. because there's a function inside of a power. but even if this is a more complicated derivative you're still going to do the same steps. The first step is still... find F prime of X. Find the derivative. F prime of X Like I said, this is a Power Rule with another function inside, so... use the Chain Rule because you don't just have X inside. You have some other X expression inside a power, you have to use the Chain Rule. First, you take the derivative of the power using the Power Rule. So the 2 power comes down out front as a coefficient, 2. You can still put parentheses. Sec X is still there... this power 2 is reduced by 1, so the power is 1. We don't need to write that. If we don't write it, it's assumed that the power is 1. You're not done. You are not done, because it's not just X inside. Since there's a sec X, you also need to multiply by the derivative of sec X.. the inside function. You did the derivative of the outside function. Now you need to multiply by the derivative of the inside function. What is the derivative of sec X? Well you can get that from a derivative table with the trig derivatives, common trig derivatives, or maybe you just know it. The derivative of sec X is... Sec X... times Tan X Sec X Tan X This is your F prime of X. Alright, second step. Plug in the X value that you're given. You're given pi over 3 for your X value. Plug it in to F prime. So you have F prime of pi over 3. It's a weird pi number for your X value.. but that's totally valid, especially for a trig function to have a pi/3X value that you're plugging in. so this would be 2 times sec(pi/3)... times another sec(pi/3)... tangent... of pi/3. Alright, this can be simplified. This should come out to be just a number, a constant so you don't want to leave it like this. As for secant... It's always easier to find cosine first and then take the reciprocal, flip it. It's easier to evaluate secant if you first do cosine because secant is 1 / cos, so we have 1 / cos(pi/3)... times another 1 / cos(pi/3)... then you can just leave tan(pi/3). Unless you prefer to write sin over cos. But leave it as tan(pi/3). Now you need to use what you know about the unit circle... which you may remember cos(pi/3)... pi/3 is up here. cos(pi/3) is the x coordinate of pi/3 which is one half. So this becomes 2 times... 1 over (1/2)... times 1 over (1/2)... times tan(pi/3). The tangent of pi/3... is root 3. So it's times square root of 3. These values - one half, one half, root 3 - you can also get them from your calculator if you're allowed to use a calculator. Just make sure that you're in radian mode before you plug in cos(pi/3). This simplifies because you have 2... 1 over (1/2) Dividing by a fraction is the same as multiplying by its reciprocal so it's the same as multiplying by 2. So this is 2, and this is 2. So that's 4. 2... times 2. Four. And then you have a root 3 left over. So your final F prime of pi/3, slope, value is 8 root 3. You can combine the 2 times 4 And that right there is your M, slope, value. Alright, that was step two. Step three, remember, you probably already have figured this out.. step three is to find the Y value, original Y value if you're not given it. But here you were given the full point, so Y is 4. Your Y1 in the point-slope form is 4. You have your X1... and you have your M, so now you're ready to plug into the point-slope form as your final step. And that is Y minus Y1, which is 4 equals M, which is 8 root 3, you can just write that. You don't want to write the decimal version of that. 8 root 3 times X, you want X in your equation minus X1, which is pi/3. You can just write that, pi/3. And this right here is a totally valid way of writing your final answer. It's in point-slope form. Rearrange it if you have to, but this is a valid answer for the equation of the tangent line to this F of X, at that point. So I hope this video helped you figure out how to write the equation of a tangent line using derivatives. Look.. anything that makes calculus easier is a good thing. So if this video helped you.. please click 'like' or subscribe!
16219	https://pmc.ncbi.nlm.nih.gov/articles/PMC2788073/	Role of vesicle tethering factors in the ER-Golgi membrane traffic - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Published in final edited form as: FEBS Lett. 2009 Nov 1;583(23):3770–3783. doi: 10.1016/j.febslet.2009.10.083 Search in PMC Search in PubMed View in NLM Catalog Add to search Role of vesicle tethering factors in the ER-Golgi membrane traffic Elizabeth Sztul Elizabeth Sztul 1 Department of Cell Biology, University of Alabama at Birmingham, Birmingham, Alabama 35294 Find articles by Elizabeth Sztul 1, Vladimir Lupashin Vladimir Lupashin 2 Department of Physiology and Biophysics, University of Arkansas for Medical Sciences, Little Rock, AR 72205 Find articles by Vladimir Lupashin 2 Author information Article notes Copyright and License information 1 Department of Cell Biology, University of Alabama at Birmingham, Birmingham, Alabama 35294 2 Department of Physiology and Biophysics, University of Arkansas for Medical Sciences, Little Rock, AR 72205 ✉ corresponding author: Elizabeth Sztul, Department of Cell Biology, University of Alabama at Birmingham, 1918 University Boulevard, Birmingham, AL 35294, Phone: 205-934-1465, Fax: 205-975-9131, esztul@uab.edu Issue date 2009 Dec 3. © 2009 Federation of European Biochemical Societies. Published by Elsevier B.V. All rights reserved. PMC Copyright notice PMCID: PMC2788073 NIHMSID: NIHMS160277 PMID: 19887069 The publisher's version of this article is available at FEBS Lett Abstract Tethers are a diverse group of loosely related proteins and protein complexes grouped into 3 families based on structural and functional similarities. A well-accepted role for tethering factors is the initial attachment of transport carriers to acceptor membranes prior to fusion. However, accumulating evidence indicates that tethers are more than static bridges. Tethers have been shown to interact with components of the fusion machinery and with components involved in vesicle formation. Tethers belonging to the 3 families act at the same stage of traffic, suggesting that they mediate distinct events during vesicle tethering. Thus, multiple tether-facilitated events are required to provide selectivity to vesicle fusion. In this review, we highlight findings that support this model. Keywords: Tethering, Rabs, SNAREs, coats, ARFs, ARF guanine nucleotide exchange factors (GEFs), GBF1, membrane traffic, coil-coil tethers, tethering complexes 1. INTRODUCTION Tethering factors are a diverse group of peripherally associated membrane proteins and protein complexes that bridge newly formed transport vesicles (as well as other types of intracellular transport carriers) with acceptor membranes to ensure correct docking and fusion. Historically, tethers were assigned to two general classes: multi-subunit complexes and coiled-coil proteins. More recent phylogenetic and structural evidence suggest the existence of three distinct functional classes: 1) oligomeric complexes that bind to SNAREs and typically act as Rab effectors (the DCGE group that includes Dsl1 complex, Conserved Oligomeric Golgi (COG) complex, Golgi-associated retrograde protein (GARP) complex, and Exocyst), 2) oligomeric complexes that function as guanine nucleotide exchange factors (GEFs) for Rab proteins (Transport Protein Particle TRAPP I and TRAPP II complexes and HOPS (HOPS is both, a GEF and a Rab effector)); and 3) coiled-coil tethers. Tethers with confirmed roles in membrane trafficking are listed in Table 1. Table 1. Golgi tethering proteins and protein complexes \| Tether \| Components /references \| Other names \| Interacting proteins related to trafficking \| :--- :--- \| \| yeast \| mammals \| coats \| Small GTPases \| SNAREs \| others \| \| HOPS \| Pep3[152,153] Pep5 VAM6[155,156] Vps16 Vps33 Vps41[155,156] \| Vps18, Vpt18, VAM18 End1, Vam1, Vpl9, Vps11, Vpt11 CVT4, VPL18, VPL22, VPS39 SVL6, VAM9, VPT16 CLS14, MET27, PEP14, SLP1, VAM5, VPL25, VPT33 \| VPS18, hVps18 VPS11, hVps11 VPS39, hVam6, TLP hVps16 VPS33A, VPS33B hVps41 \| \| Ypt7 \| Vam7 , Vam3 , Nyv1, Vam2 \| \| \| TRAPP I and TRAPP II \| Bet3 Bet5 Trs85 Trs65 Trs20 Trs23 Trs31 Trs33 Trs120 Trs130 \| GSG1 Kre11 \| TRAPPC3, mBet3 TRAPPC1, mBet5 ? ? TRAPPC2, SEDL TRAPPC4, mTrs23 TRAPPC5, mTrs31 TRAPPC6A, Trs33A,Trs33B TRAPPC9, NIBP TRAPPC10, mTrs130 \| Sec23 (TRAPPI), γ-COPI (TRAPPII) \| Ypt1[61,62 ], Rab1, Ypt31/32 \| \| \| \| COG \| COG1 COG2 COG3 COG4 COG5 COG6 COG7 COG8 \| Cod3, Sec36, Tfi1 Sec35 Sec34, Grd20 Cod1, Sgf1, Sec38, Tfi3 Cod4 Cod2, Sec37, Tfi2 Cod5 Dor1 \| LdlB LdlC hSec34 hCod1 GTC-90 hCod2 hCod5 hDor1 \| γ-COPI, β-COPI \| Ypt1 Ypt6 Rab1, Rab6, Rab30, Rab41 \| Gos1/ Ykt6/Sed5[ 13], Sed5, GS28, Syntaxin 5[32,34,54] \| Sly1 P115, GM130[32,56] \| \| Dsl1/ZW10 \| Dsl1 Tip20 Dsl3 \| Tip1 \| ZW10 RINT-1 NAG \| δ-COP, α-COP \| \| Sec20/Use1 /Ufe1[174, 175], Syntaxin 18[25,176] \| \| \| Exocyst \| Sec3 Sec5 Sec6 Sec8 Sec10 Sec15 Exo70 Exo84 \| \| SEC3L1, EXOC1 SEC5L1, EXOC2 SEC6L1, EXOC3 SEC8L1, EXOC4 SEC10L1, EXOC5 SEC15L1, EXOC6, SEC15L2 EXOC7, SEC70 EXOC8, SEC84 \| \| Sec4, Rho1, Rho3, Cdc42p, RalA[186- 190] \| Sec9 \| Sec1 \| \| GARP \| Vps51[192,193] Vps52 Vps53 Vps54 \| API3, VPS67, WHI6 SAC2 CGP1, LUV1, TCS3 \| ---- ARE1, SAC2, SACM2L, hVPS52 FLJ10979, hVps53L HCC8, SLP-8p, VPS54L, hVps54L \| \| Ypt6 \| Tlg1[192,193], Syntaxin 10 Syntaxin6, Syntaxin16, VAMP4 \| \| \| CASP \| \| COY1 \| \| hSec23 \| \| \| Golgin- 84 \| \| GCC88 \| [199,200] \| \| \| \| Arl1 \| \| \| \| GCC185 \| \| \| GCC2, REN53, RANBP2L4 \| \| Rab1/ Rab2/ Rab6, Arl1[202,203], \| Syntaxin16 \| CLASP \| \| GCP60 \| \| \| ACBD3, GOCAP1, GOLPH1 \| \| \| \| Giantin, Golgin16 0, \| \| Giantin \| [99,208-210] \| \| GOLGB1, GCP372. Macrogolgin \| \| Rab1, Rab6, \| \| p115[96,99], Gcp60 \| \| GM130 \| [95,101,110- 112,125,130,208,211-213] \| Bug1 \| GOLGA2, Golgin-95 \| \| Rab1[129,130], Rab2, Rab33b \| Syntaxin5 \| p115[95,96], Giantin,G RASP65, COG[32,56], ZFPL1 \| \| GMAP210 \| \| Rud3 \| TRIP11, TRIP230, CEV14 \| \| Arf1[216,217] \| \| \| \| Golgin-45 \| [124,218] \| \| BLZF1, JEM-1 \| \| Rab2 \| \| \| \| Golgin-84 \| [127,128] \| \| GOLGA5 \| \| Rab1[127,128] \| \| CASP \| \| Golgin-97 \| \| \| GOLGA1 \| \| Arl1 Rab6, Rab19, Rab30 \| \| \| \| Golgin-160 \| \| GRP1 \| GOLGA3, MEA-2, GCP170 \| \| \| \| PIST \| \| Golgin-245 \| [208,223-230] \| Imh1 \| GOLGA4, p230, tGolgin-1 \| \| Arl1, Rab2, Rab30 \| \| \| \| GRASP55 \| [124,232-236] \| \| GORASP2, GOLPH6, GRS2 \| \| Rab2 \| \| Golgin- 45, p24 \| \| GRASP65 \| [101,111,112, 225] \| Grh1 \| GORASP1, P65; GOLPH5 \| Sec23/24 \| \| \| GM130, p24 \| \| P115 \| [80- 82,84,87,88,94,95,97,237- 240] \| Uso1 \| VDP, TAP \| β-COP \| Rab1/Y pt1, dRab30 \| GS28, Syntaxin5[ 88] \| Giantin, GM130, COG2[56 ], GBF1 \| Open in a new tab Distinct tethers localize to different compartments of the secretory and endocytic pathways and have either been shown to or are proposed to function therein. While some tethers appear selective for a single step of traffic, others seem more general and localize to multiple compartments. It is possible that the same tether may function at distinct stages of traffic by binding to distinct Rabs and/or SNAREs. A single step of traffic may utilize a tether from each of the three functional classes: ER-Golgi traffic appears to require the activity of Dsl1 and COG (both are members of the DCGE class), TRAPP I (a GEF tether) and p115 (a coiled-coil tether). This suggests that different tethers might perform distinct functions in the overall transport process. In this review, we discuss a model of vesicular tethers working together with SNAREs, Rabs, coats and GEFs to ensure correct docking of vesicular membranes prior to vesicle fusion. We focus exclusively on tethers involved in ER-Golgi traffic. 1.1. DCGE TETHERS INVOLVED IN ER-GOLGI TRAFFIC: DSL1 AND COG The DCGE sub-family of tethers (Dsl1, COG, GARP and exocyst complexes ) has been initially grouped together based on the finding that subunits of the Dsl1p, COG, GARP and exocyst complexes share limited sequence homology [1,2] that suggests convergent evolution . Emerging structural information shows significant similarities between folding of key subunits within Dsl1, COG, GARP and the exocyst [3,4]. Partial structures of two COG subunits and four exocyst subunits have been reported [3,5-10]; these structures contain one (Cog2p), two (Sec15, Exo84p), three (Sec6p), or four (Exo70p) helix bundle domains. In each of the exocyst structures, the tandem helical domains form an extended, rod-like array . Crystal structure of two subunits of Dsl1 complex, Tip20p and Dsl1p, resemble known subunits of the exocyst complex . Recently, the X-ray structure of Cog4 establishes an unambiguous link between the COG, exocyst and Dsl1p complexes, providing strong evidence that they are derived from a common evolutionary precursor . The structural similarities suggest a mechanistic similarity in the function of the DCGE tethers. Structural studies of the TRAPPI multisubunit tethering complex , demonstrate that it is structurally unrelated to the DCGE family. All members of the DCGE family have been shown to interact with SNAREs (Table 1). However, their interactions with other components of the trafficking machinery vary: interactions with known vesicular coats have been observed for Dsl1, COG and GARP, but not the exocyst. Interactions with Rabs have been observed and in some cases shown to be essential for COG, GARP and the exocyst, but not for Dsl1 [13-15]. Whether this reflects lack of knowledge or real differences in mechanisms of action of these tethers remains to be defined. Dsl1 complex The yeast Dsl1 trimeric complex is composed of essential subunits, Dsl1p, Tip20p and Dsl3/Sec39 (Table 1). According to the most recent model, the Dsl1p subunit lies at the center of a ternary Sec39p-Dsl1p-Tip20p complex . The Dsl1 complex localizes to the ER . It is stably associated with ER membranes, and this association is independent of Rab GTPases. The ER localization may be mediated, at least in part, by two independent interactions: between Tip20p and the ER SNARE protein Sec20p [16-18] and between Sec39p and the ER SNARE protein Use1 [4,16]. Dsl1 is essential for recycling traffic into the ER. The strongest evidence comes from studies of the dsl1-22 mutation that causes defects in Golgi-to-ER retrieval of ER-resident SNARE proteins, BiP/Kar2p and integral membrane proteins harboring a C-terminal KKXX retrieval motif. Cells with defects in Dsl1 complex function show massive accumulation of COPI-coated vesicles . Together, the data suggest that Dsl1 is required for fusion of recycling COPI vesicles with the ER. This function of Dsl1 is further supported by the findings that Dsl1p binds the δ-COP and α-COP subunits of the COPI vesicle coat in vitro. The δ-COP belongs to the F subcomplex of COPI, while α-COP is part of the B subcomplex . Both are involved in binding cargo during vesicle formation. Interestingly, Dsl1p binds to COPI subunits at sites identical to those involved in interactions between COPI subunits that stabilize the COPI coat [19,23]. Thus, it is possible that Dsl1 acts to tether COPI vesicles and to destabilize the COPI coat to expose proximal SNAREs and allow close apposition of membranes before fusion. Like all DCGE tethers, the yeast Dsl1 complex interacts with SNAREs, the three ER Q-SNAREs, Ufe1, Sec20 and Use1p . Dsl1 appears to promote SNARE complex stability , suggesting that it also may promote SNARE interactions during fusion. However, direct evidence of Dsl1 catalyzing SNARE complex formation is unavailable. There is also no evidence that Dsl1 complex directly interacts with Rab GTPases. In mammalian cells, a Dsl1-like complex composed of ZW10, RINT-1 and NAG has been identified [24,25]. Depletion of ZW10 or RINT-1 results in central, disconnected cluster of Golgi elements and inhibition of Golgi protein recycling to the ER. No vesicle accumulation was observed at the light microscopy level. Mammalian Dsl1 complex interacts with ER-localized t-SNAREs Syntaxin-18, BNIP1 and p31 [24,25]. It is not known whether mammalian Dsl1/ZW10 complex directly interacts with Rab GTPases, but both, knock-down of Golgi-localized Rab6, or the expression of the dominant-negative GDP-restricted Rab6 suppressed Golgi disruption induced by ZW10 knockdown . COG complex COG is an octameric complex composed of Cog1-8 subunits and is found in all organisms from yeast to mammalian cells (Table 1) . Based on genetic and biochemical studies [1,28-30] and images of quick-freeze/deep-etch/rotary-shadow purified COG complexes , the eight subunits of the COG complex are grouped into two sub-complexes: Cog 1–4 (Lobe A) and Cog 5–8 (Lobe B) that are assembled in a bi-lobed structure. Protein-protein interaction studies suggest that each lobe contains a heterotrimeric COG sub-complex (COG2–4 and COG5–7). The two lobes of the complex are bridged by the COG1/COG8 heterodimer[28-30]. Secondary structure algorithms for COG subunits suggest that COG subunits are predominantly α-helical. Light microscopy revealed that COG complex is preferentially associated with cis- and medial-Golgi membranes [1,31,32]. Electron microscopy in mammalian cells showed Cog1 localized to the tips and rims of the Golgi cisternae as well as on vesicles . Thus, COG is positioned to mediate tethering of vesicles trafficking to the Golgi and within the Golgi. The exact mechanism that restricts COG localization to Golgi cisternae is currently unknown. The COG complex interacts genetically and physically with subunits of the COPI coat. Yeast Cog2 interacts with γ-COPI in yeast two hybrid assay, while antibodies to human Cog3p co-precipitate β-COPI. These findings led to hypothesis that COG acts as a tether that connects COPI vesicles with cis-Golgi and perhaps medial-Golgi membranes during retrograde traffic . This model is consistent with findings that cog2 and cog3 temperature-sensitive yeast mutants accumulate multiple vesicles at the non-permissive temperature . Additional support for this function of COG comes from mammalian cells where siRNA-mediated depletion of Cog3p leads to the accumulation of vesicles carrying Golgi glycosylation enzymes, the Golgi SNAREs GS15 and GS28 and the cis-Golgi glycoprotein GPP130 . Importantly, COG-depleted cells do not support the retrograde trafficking of Shiga and SubAB toxins [34,36], even though they are competent for anterograde trafficking of VSVG protein to the PM. Further support for COG being involved in retrograde membrane trafficking of Golgi resident proteins is provided by data showing that mutations in COG subunits lead in humans to congenital disorders of glycosylation (CDG Type II)[3,37-45]. Mutations in COG subunits severely alter the localization and function of Golgi glycosylation machinery [1,13,37,39,46-48], resulting in defects in the processing of N-linked glycan chains (for review, see ). Such alterations in glycosylation cause type II CDGs that lead to a wide variety of neurological and developmental abnormalities (for review, see ). Genetic and physical interactions have been detected between the COG complex and Golgi SNARE proteins [13,32,34,50-53]. Yeast COG complex interacts in vitro with the SNARE domain of the Golgi t-SNARE Sed5p and preferentially binds to the quaternary Sed5p-containing SNARE complexes. The mammalian hCog4p subunit directly binds to Syntaxin-5a, the mammalian homologue of Sed5p. The same subunit is simultaneously interacting with Syntaxin-5 partner Sly1p that belongs to the Sec1/Munc18 (SM) protein family . Defects in the function of mammalian COG complex lead to a significant decrease in Golgi SNARE mobility, an accumulation of uncomplexed Syntaxin-5, and a decrease in the steady-state level of intra-Golgi SNARE complexes. Together, the data suggest that the COG complex enhances formation and/or stability of intra-Golgi SNARE complexes . Thus, COG complex appears to have dual functions of mediating vesicle recognition via interactions with COPI coat and directly facilitating SNARE-catalyzed fusion events. These two activities are likely to reside in different subunits of the COG complex: Cog2p binds to COPI coat, while Cog4p interacts with SNARE and SM proteins. COG complex is an effector of multiple Golgi-localized Rab proteins. Purified yeast COG complex preferentially binds to Ypt1-GTP, and to a lesser extent to Ypt6-GTP . Mammalian Cog6p binds to GTP-restricted Rab1, Rab6 and Rab41, while Cog4p preferentially interacts with Rab30-GTP . Recently, a novel direct interaction between Cog2p and the coiled-coil tether p115 was uncovered and is described below . This interaction was shown to be essential for Golgi ribbon structure. 1.2. GEF TETHERS INVOLVED IN ER-GOLGI TRAFFIC: TRAPPI AND TRAPPII TRAPPI complex in yeast contains seven subunits (Table 1) [53,57-59]. Additional three subunits (trs130, trs120 and trs65; Table 1) are present in the TRAPP II complex . The yeast 300 kDa TRAPPI complex has been shown to have a dumbbell shape by single-particle electron microscopy . One of the lobes contains the trs20-trs31-bet3 heterotrimer, while the other lobe contains the bet3-trs33-bet5 heterotrimer. The two lobes of TRAPPI are bridged by the trs23 subunit (. TRAPP complexes are different from other tethers, most of which are Rab effectors, because they uniquely act as Rab activators. In yeast, TRAPP complexes have been shown to act as guanine nucleotide exchange factors for the Rab GTPases Ypt1p and Ypt31/32p [61,62]. Initially, TRAPPI and TRAPPII have been shown to specifically act as GEFs for Ypt1p[60,61]. The GEF activity can be reconstituted in vitro with only five of the seven TRAPPI subunits [12,63]. Interestingly, all these subunits are also present in TRAPPII. Subsequent studies have suggested that the addition of the TRAPPII-specific subunits (Trs120p/Trs130p) changes the substrate specificity of the GEF and allows it to activate Ypt31p/Ypt32p. Further work will be needed to unequivocally define whether TRAPPII is a GEF for Ypt1p alone, Ypt31p/Ypt32p alone, or both. Mammalian TRAPPII specifically activates Rab1 . Biochemical characterization of yeast TRAPPs suggests that these complexes are stably anchored to Golgi membranes . TRAPPI co-fractionates with early Golgi compartments and appears to be required for fusion of ER-derived COPII vesicles with the Golgi . The GEF activity is essential for TRAPPI function in membrane traffic, and in cells expressing subunits compromised in GEF activity secretion is inhibited at the non-permissive temperature [57,59,66]. The TRAPPII subunit Trs120p localizes to the late Golgi, and TRAPPII appears to regulate intra-Golgi traffic and traffic from the early endosome to the late Golgi in vivo [60,67]. In mammalian cells, only one eight-subunit TRAPP complex could be identified according to its size . Human Bet3 is mostly a cytosolic protein that is found both as a monomer and a part of a TRAPP complex . The membrane bound fraction of Bet3 is localized to the transitional ER, and to some extent to endosomes . This is consistent with Bet3 activity at both the ER-Golgi interface and at the late Golgi-endosomal interface. The function of mammalian Bet3 in ER-Golgi traffic was explored in semi-intact cell transport assays . Staging experiments with cytosols depleted of specific components suggest that COPII>Bet3>Rab1>α-SNAP> GS28 SNARE act sequentially to facilitate ER to Golgi traffic of cargo proteins. Interfering with mBet3 function still allowed the budding of COPII vesicles, but blocked formation of pre-Golgi compartments. Thus, Bet3 is required for homotypic fusion of COPII vesicles, which is essential to maintain the structure and integrity of the VTCs and the Golgi network. The key function of TRAPPI in traffic appears to be activation of Ypt1p/Rab1. However, TRAPPI might also have bona-fide tethering function. Interestingly, the Bet3 subunit of TRAPPs interacts with the Sec23 subunit of COPII . COPII vesicles are formed by the initial recruitment of Sec23/Sec24 complex, followed by the recruitment of the Sec13/Sec31 complex. The binding of Bet3 to the Sec23/Sec24 “interior” layer of the coat might suggest that Bet3 (and thus the TRAPPI complex) interacts with COPII vesicles while they are forming. Alternatively, TRAPPI might bind after COPII vesicles partially uncoat. Importantly, a copy of Bet3 is present in each of the two lobes of the TRAPPI complex. Thus, a single TRAPPI complex could simultaneously bind two COPII vesicles and mediate their tethering. Mammalian TRAPPII is enriched on COPI coated vesicles and buds, but not Golgi cisternae. The depletion of mTrs130 by shRNA leads to increased number of vesicles in the vicinity of the Golgi and the accumulation of cargo in an early Golgi compartment . Mammalian TRAPPII specifically activates Rab1, and its subunits Trs120 and Trs130 co-immunoprecipitates with γ-COP, but not ε-COPI . Intriguingly, Bet3 does not bind to COP, indicating that only a sub-complex of TRAPPII interacts with the inner layer of COPI coat. The binding to COPI components might mediate the association of TRAPPII with COPI vesicles and would ensure activation of Rab required for the recruitment of other tethers (such as p115 and/or COG) to mediate fusion of COPI vesicles with Golgi cisternae. Mutation in the human homologue of the Trs20p subunit of both TRAPPs is the cause of the human X-linked disease spondyloepiphyseal dysplasia (SEDL) [71,72]. SEDL causes bone growth defects, resulting in short stature, barrel chest and degenerative joint disease . It is likely that the disease results from defects in the trafficking of type II collagen through the Golgi. 1.3. COILED-COIL TETHERS INVOLVED IN ER-GOLGI TRAFFIC: P115 AND GM130 A family of proteins characterized by extensive coiled-coil domains has been shown to have tethering function. The proteins are (or are predicted to be) highly extended, making them ideal for linking membranes over a relatively large distance . The current model is that coiled-coil proteins mediate long-range tethering (>300 nm) that is subsequently followed by close-range (<12 nm) recognition via the SNAREs. It is unknown how the tethered membranes are brought into close apposition, but tethers may actively participate in this process. Most coiled-coil tethers contain regions of high flexibility “hinges” between the rod-like coiled-coil domains, and it has been suggested that an accordion-like collapse of the tethers might bring the membranes close together. Such collapse or conformational changes in tether structure could be regulated through binding to Rab GTPases, as all coiled-coil tethers have been shown to bind Rabs. In addition to acting as membrane bridges, coiled-coil tethers have been shown to interact with SNAREs and might facilitate SNARE complex formation to promote fusion. Coiled-coil tethers also interact with coat components, perhaps to provide an initial level of recognition. By interacting with both, coats and SNAREs, coiled-coil tethers might provide increasingly stringent tethering. P115 P115 was shown to be a myosin-like homodimer with an N-terminal globular head (~70 kDa), containing two evolutionary-conserved domains (Homology Region, HR1 and HR2), and a smaller (~30 kDa) C-terminal, extended coil-coiled domain containing four heptad repeats (CC1-CC 4) terminated by a short acidic domain [75-77]. The recent crystal structure of the N-terminal domain of p115 shows a novel architecture constructed of 10–12 armadillo-like motifs that are decorated by elongated loops and carry a C-terminal non-canonical repeat [78,79]. The terminal repeat folds into the armadillo superhelical groove and allows homodimeric association . P115 is peripherally associated with multiple compartments of the ER-Golgi interface, including COPII vesicles, Vesicular Tubular Clusters (VTCs), cis-Golgi and COPI vesicles. The yeast homologue, Uso1p, is associated with yeast COPII vesicles. Both, p115 and Uso1p are essential for secretory trafficking [75,80-82]. Both proteins have been shown to tether COPII vesicles, either directly to the cis-Golgi in the case of Uso1p , or to other COPII vesicles and pre-Golgi intermediates in the case of p115 . In addition, p115 is required for cargo traffic from pre-Golgi intermediates to the cis-Golgi, as shown by inhibition in traffic at the VTC stage by anti-p115 antibodies . P115 also has been detected on COPI vesicles, suggesting that it participates in traffic between Golgi cisternae [80,86-88]. Precisely how p115 facilitates tethering and fusion of distinct types of vesicles to distinct membranes is unknown, but might involve associations with distinct proteins (such as GM130, giantin and COG) at distinct compartments. P115 interacts with Rab1-GTP and this interaction mediates the recruitment of p115 to membranes . It has been initially proposed that CC1 domain of p115 (residues 650–780) is primarily responsible for Rab1 binding , but the recent structural work indicates that the N-terminal HR1 domain of p115 (residues 21–54) is also involved . The identification of the H1 domain as a binding partner for Rab1 raises the possibility that two different Rab1 binding regions may have different functions in the sequential events that direct tethering and t-SNARE assembly. Because activated Rab1 is required for p115 recruitment to membranes, the GEF activity of TRAPPI and TRAPPII are required on COPII vesicles, VTCs, Golgi membranes and COPI vesicles to promote p115 association with membranes. Thus, a cascade of tethers is required to facilitate traffic. P115 association with membranes is extremely dynamic and p115 undergoes rapid cycles of association and dissociation that is regulated by activated Rab1 and free SNAREs . Membrane association of p115 is probably also influenced by binding to GM130: GM130 binding induces a conformational change in p115 that releases the autoinhibitory folding of p115 C-terminus and allows p115 to interact with Rab1 . Drosophila p115 has been shown to bind Rab30 (a possible homologue of mammalian Rab33b) , but the functional significance of this interaction is currently unknown. P115 interacts with a number of ER-Golgi SNAREs (Table 1). P115 appears to preferentially associate with free SNAREs [89,92] and CC1 domain of p115 has been shown to facilitate SNARE-SNARE (Syntaxin5-GS28) pairing in vitro . The interaction of p115 with unassembled SNAREs might ensure that p115 is only recruited to sites that can participate in trans-SNARE complex formation. The function of p115 in facilitating SNARE complex formation might be regulated by Rab1, because binding of Rab1 to p115 causes rearrangements in p115 coiled-coil domains, one of which is essential for SNARE interactions . In support for a fundamental role for p115 in SNARE events, “replacement” experiments in which wild-type p115 was replaced with p115 mutant lacking the SNARE-binding domain show Golgi disruption, suggesting that p115-SNARE interaction is essential . P115 has been shown to bind the coiled-coil proteins GM130 and giantin. The binding site for both proteins lies within the same C-terminal domain of p115, and GM130 and giantin compete for binding to p115 . The functional significance of the interactions between p115 and those two proteins is unclear. Perturbing the p115-GM130 interaction with competing peptides or antibodies inhibits ER to Golgi trafficking and Golgi assembly in vitro [95-99]. Similarly, expression of mutant GM130 lacking the p115 binding site, or proteolytic removal of the p115 acidic tail that mediates interaction with GM130 and giantin causes defects in protein trafficking in vivo [77,100-102]. However, “replacement” experiments in which wild-type p115 was replaced with p115 mutant lacking the GM130/giantin-binding domain show normal trafficking, suggesting that the p115-GM130/giantin interactions are not essential in that system . A possible explanation for the latter finding is that the p115-GM130/giantin interaction is regulatory and can be bypassed by removal of the C-terminus of p115 . Alternatively, the interaction between p115 and GM130/giantin may increase traffic efficiency as opposed to an absolute requirement in vesicle tethering. Studies described below suggest that GM130 is required for traffic under certain, but not all conditions. P115 also binds the C-terminal domain (residues 613–669) of Cog2 subunit of the octomeric COG and co-precipitates with other subunits, suggesting that p115 binds the entire COG complex in vivo[32,56]. Replacement of wild-type p115 with p115 unable to bind Cog2 in p115-depleted cells results in alterations in Golgi architecture and delayed traffic, indicating that the p115-COG interaction is essential for the maintenance of Golgi structure, presumably by facilitating normal trafficking to the Golgi . The Cog2-binding region on p115 was mapped to residues 200–247, encompassing the Homology Region 2 (HR2) domain of p115. The interaction of p115 and Cog2 is dispensable for the proper targeting of COG complex to the Golgi apparatus . In addition to facilitating tethering and fusion, events that consume vesicles, p115 also may participate in vesicle formation. Uso1p is required for cargo sorting during the formation of COPII vesicles in yeast . Similarly, in mammalian cells depletion of p115 results in the retention of GOS27 SNARE at ER exit sites . A function in cargo selection is not unique to p115, and golgin-160 also has been shown to mediate Golgi sorting of the GLUT4 glucose transporter in adipocytes , and in trafficking of the 1-adrenergic receptor and the ROMK channel to the plasma membrane [105,106]. Thus, coiled-coil tethers may participate in cargo sorting during carrier formation. Such function raises the intriguing possibility that tethers may be involved in both, vesicle fusion and vesicle formation. Additional support for a role in vesicle formation comes from the finding that p115 interacts with the guanine nucleotide exchange factor GBF1 . GBF1 is a guanine nucleotide exchange factor for ARF GTPases and has been shown to be essential for COPI recruitment to membranes [87,107,108]. The p115-GBF1 interaction appears functionally significant because expression of competing peptide in vivo leads to Golgi disruption, possibly due to inhibition in traffic . The interaction is not required for the membrane association of either p115 of GBF1. It is more likely that p115-GBF1 binding occurs after both proteins are recruited to membranes and acts to position GBF1 at sites of p115-mediated fusion. Fusion sites receive incoming membranes and need to form recycling vesicles to compensate for membrane volume. Thus, p115 could act as a central toggle to simultaneously mediate the fusion of incoming vesicles and the formation of recycling vesicles. Interestingly, p115 interacts with the β-COP component of COPI . The interaction occurs between the conserved WF motif in the appendage domain of β-COP and the E 19 E 21 residues in the Homology Region 1 (HR1) in the head domain of p115. The interaction appears important because studies in which wild-type p115 was replaced with p115 unable to bind β-COP show Golgi disruption. Thus, p115 may participate in COPI-mediated retrograde traffic within the Golgi and from the Golgi to the ER. The interaction may function as a tethering mechanism for linking COPI vesicles to acceptor membranes prior to fusion. However, it can also be proposed that p115 binds free coatomer and positions it adjacent to the site at which GBF1 has just activated ARF. Thus, p115 can be considered a platform that brings into close proximity a GEF to activate ARF (GBF1) and coatomer to promote vesicles formation. At the same time, the p115 platform is positioned at vesicle fusion sites through p115 interactions with Rab1, SNAREs and COG. GM130 One of the most studied coiled-coil proteins is GM130 . GM130 binds stably to GRASP65 that is anchored to the membrane via an N-terminal myristoyl group . The GRASP65-GM130 complex localization is restricted to the cis-Golgi [101,112]. Although GM130 interaction with membranes is stable and GM130 is recovered exclusively with membranes during fractionation, FRAP experiments in live cells indicate that GRASP65-GM130 is extremely dynamic and undergoes rapid cycles of association and dissociation from the membrane . This might be expected of a protein complex that regulates a highly dynamic trafficking process. The function of GM130 in secretory traffic has been predominantly inferred from studies exploring its interaction with p115 (described above) and is still controversial. Addition of anti-GM130 antibodies to permeabilized cells arrests VSV-G traffic at the Golgi . However, GM130 depletion studies report a delay in secretory trafficking, and this has only been observed in some [114,115] but not in other studies [116-118]. Similarly, depletion of GRASP (and thus preventing GM130 association with membranes) doesn’t cause a secretory defect, suggesting either a minor or redundant function in traffic, at least under the analyzed conditions [116,119]. Depletion of GM130 in mammalian cells causes the disruption of the Golgi ribbon, but maintains correct stacking of the resulting Golgi mini-stacks[115,116]. It is possible that GM130 functions to tether lateral Golgi elements in what would be considered a homotypic fusion . The same Golgi fragmentation into mini-stacks is observed in cells depleted of p115 [56,120], suggesting that lateral tethering might involve both GM130 and p115. Alternatively, GM130 might work with p115 to facilitate tethering and fusion of VTCs at the cis-Golgi. It can be envisioned that slowing down the rate of incorporation of new membranes at the cis-face of the Golgi could eventually result in the shortening of the cisterna and subsequent breaks between the fenestrated areas to generate mini-stacks [101,115]. This model would be consistent with the role of the yeast orthologue of GM130, Bug1p in trafficking at the cis-Golgi . Thus, GM130 could have a kinetic rather than an absolute function in trafficking. This suggestion is supported by studies on the ldlG cell line that lacks GM130. In these cells, Golgi architecture and traffic are normal at low temperatures (32–34°C), but Golgi is highly fragmented and traffic in arrested when the temperature is elevated to 39°C . Thus, GM130 function can be bypassed under normal conditions but is essential under conditions that would require increased membrane flow. GM130 might function to improve efficiency of membrane traffic and is not essential in cultured cells within minimal secretory load, but might be important when trafficking rate has to be optimized. The Saccharomyces cerevisiae GRASP orthologue Grh1p is also stably associated with the membrane via an acetylated amphipathic helix and binds the yeast version of GM130, Bug1p . Deletion of Grh1p or Bug1p does not affect yeast cell viability or growth, suggesting a minor or redundant role in trafficking. However, both proteins show genetic and biochemical interactions with ER to Golgi trafficking proteins, supporting a role in ER-Golgi traffic. Grh1 interacts with the Sec23/24 component of the COPII coat . GM130 binds to other proteins implicated in ER-Golgi traffic (Table 1). GM130 binds Rab1, Rab2 and Rab33b [124,125], but the functional significance of these interactions is unclear. GM130 binds syntaxin-5 and appears to directly regulate SNARE complex formation by preventing syntaxin-5 from forming complexes with GOS28 . Binding of p115 to GM130 causes a conformational change in GM130, such that it no longer binds syntaxin-5. Syntaxin-5 is then able to interact with GOS28 and perhaps other SNAREs . GM130 interacts with syntaxin-5 through coiled-coils 4–6, a region distinct from coiled-coil 3 involved in binding Rab1, the N-terminus involved in binding p115, and the C-terminus involved in binding GRASP65 . Thus, GM130 may bind many of its partner proteins simultaneously in a nexus of tether-SNARE, tether-Rab and tether-tether interactions to promote tethering and fusion of vesicles during trafficking. 2. INTERACTIONS OF TETHERS WITH OTHER COMPONENTS OF TRAFFICKING MACHINERY 2.1. Tether interactions with Rabs Almost all known tethers interact with Rabs (Table 1). So far, 11 Rabs have been identified in yeast, and there are more than 60 mammalian Rabs (Reviewed in ). Like all GTPases, Rabs cycle between an inactive GDP-bound state and an active GTP-bound form that interacts with effector proteins. Inactive cytosolic Rabs exist in a complex with a guanine nucleotide dissociation inhibitor (GDI). Rabs are prenylated and this lipid modification mediates their association with membranes. Membrane-bound Rabs are activated through a GDP/GTP exchange. Rabs have slow intrinsic GDP/GTP exchange ability and in cells this reaction is facilitated by Rab guanine nucleotide exchange factors (GEFs). Three complexes (TRAPPI, TRAPPII and HOPS) have been shown to function as Rab GEFs. All three are first recruited to newly formed Rab-less membranes of vesicles and traffic intermediates. Their GEF activity results in membranes of vesicles and acceptor membranes that are loaded with specific GTP-bound Rab molecules, thus allowing the subsequent recruitment of other oligomeric complexes and coiled-coil tethers. Thus, the activity of GEF tethers must precede the function of other oligomeric and coiled-coil tethers that are Rab effectors. This implies that the positioning of GEF tethers on vesicular membranes is the most upstream event in tethering. The only known tethering complex not yet found to interact with Rab proteins is the ER-localized Dsl1 complex. A possible explanation is that Dsl1 is the most ancient complex of the DCGE tethers subfamily, and is stably associated with ER membrane . Because ER is not generated through continuous membrane flow, the on-off membrane cycling of this complex is not essential for its function. The other three complexes, COG, GRASP and exocyst and the coiled-coil tethers are associated with highly mobile membranes, and their interaction with membranes is transient and highly regulated. The role of activated Rab in recruiting oligomeric and coiled-coil tethers to membranes is well documented and binding to Rab1-GTP is required for membrane association of p115. However, other coiled-coil tethers bind Rab-GTP even though their membrane association or localization doesn’t depend on the interaction. The transmembrane proteins golgin-84 and giantin bind to Rab1 [90,127,128], as does GM130, which is recruited to membranes via GRASP65, independently of Rabs [124,129,130]. Thus, it is likely that binding of tethers to activated Rabs has multiple functions, including specific capture and retention of Rab-containing membranes . While tether-Rab interaction may mediate membrane association, it may also trigger conformational changes that regulate tether function. Initial models suggested that Rabs provide specificity to compartment recognition during vesicular traffic. However, a single Rab can localize to multiple compartments of the pathway and function therein, and thus can’t be the sole determinant of recognition or specificity. Ypt1p has been shown to function at multiple stages of the secretory pathway including COPII vesicle fusion, intra-Golgi traffic and endosomal trafficking [131-136]. Similarly, mammalian Rab1 localizes to COPII vesicles and is required for their fusion in vitro [84,130], localizes to VTCs and is required for their trafficking [85,137,138] and localizes to the early Golgi where it is required for intra-Golgi traffic . Presumably, Ypt1p/Rab1 function at distinct stages of traffic by interacting with distinct effectors, including tethering proteins. Tethering may well be a combinatorial “coincidence detection” - thus binding to the Rab would work together with binding to SNAREs, other tethers and/or additional yet unknown membrane receptors to mediate the destination of a vesicle. 2.2. Tether interactions with SNAREs Many tethers have been shown to interact with SNAREs (Table 1). This interaction might impart targeting information and ensure that only the correct vesicle containing the appropriate SNARE zip code will tether to and fuse with a given acceptor membrane. However, it is likely that tethers can actively promote trans-SNARE pairing in vitro and facilitate SNARE complex formation in vivo. The coiled-coil tether p115 has been shown to interact with and promote SNARE pin formation through one of its coiled-coil domains that bears limited homology to a typical SNARE motif . Dsl1/ZW10, COG and GARP complexes have been shown to actively interact with step-specific SNARE complexes and are implicated in regulation of their assembly and/or stability [21,24,32,54,140]. GM130 regulates the availability of syntaxin-5 to participate in SNARE complex formation . Recently, an evolutionary conservation in sequence has been detected between the MUN domain found in Munc13 and the Sec6 subunit of the exocyst and the Vps53 subunit of GARP . In addition, MUN domains were also detected in Sec3 and Exo70 subunits of the exocyst, the RINT1 subunit of the Dsl1 complex and the COG4 subunit of the COG complex. The MUN domain has been shown to bind SNAREs and is postulated to have a role in promoting assembly of SNARE complexes [142,143]. Thus, it is likely that the MUN domain within the tethering factors also participate in SNARE complex assembly. It should be noted that the tethering complexes containing MUN domains (Dsl1, COG, GARP and exocyst) are evolutionarily related. In addition to directly binding to and promoting SNARE complex assembly, tethers might also influence fusion by binding to and alleviating inhibitory effects of Sec1/Munc18 family proteins [54,144]. 2.3. Tether interactions with coats Numerous tethers have been shown to interact with coat components (Table 1). The oligomeric Dsl1 and COG complexes bind COPI subunits. These interactions are likely to promote the tethering of COPI vesicles to the ER (for Dsl1) and to Golgi cisterna (for COG). The GEF tethers also bind coats: TRAPPI binds Sec23/Sec24 components of the COPII coat and promotes homotypic COPII vesicle tethering . TRAPPII binds the γ-COP subunit of COPI and may tether COPI vesicles to Golgi membranes. Coiled-coil tethers also bind coats. Grh1p (yeast homologue of GRASP65) binds the Sec23p/Sec24p subcomplex of COPII coat , and might tether COPII vesicles to Golgi membranes in yeast. P115 has been shown to bind β-COPI and may tether COPI vesicles at VTCs and Golgi. In vitro studies suggest the existence of at least two distinct classes of Golgi-derived COPI-coated vesicle. One class, enriched in Giantin and p24/p25 proteins was efficiently bound to p115 tether, while the second class, enriched in Mannosidase I and II was preferentially interacting with CASP and golgin-84 tethers . The observed interactions between tethers and coats suggest that the engagement of tethers occurs prior to vesicle uncoating. Thus, a model emerges in which tethers use multiple proof reading mechanisms to recognize incoming vesicles - tethers might first interact with coats to ensure that a vesicle of either the anterograde or retrograde type is coming, followed by tether interactions with the SNARE machinery to provide a more stringent level of recognition. Dsl1p binds COPI subunits at sites involved in interactions that stabilize the COPI complex. Thus, Dsl1p binding may act to destabilize the COPI coat . Similarly, TRAPP interacts with subunits (Sec23/Sec24) of the internal layer of COPII coat, consistent with the suggestion that it might destabilize the outer (Sec13/Sec31) layer of the coat. Thus, it is possible that tethers may perform two coat-related functions: bind coats as part of the recognition process and trigger or facilitate uncoating of vesicles to ensure unhampered SNARE pairing prior to fusion. 3. ESSENTIAL STEPS IN VESICULAR TRAFFIC Cargo transport between compartments is mediated by carriers, and their formation and fusion have been extensively studied (reviewed in [145-147]. Cargo is first separated from resident proteins into subdomains of the donor compartment that subsequently bud off the compartment. Coats participate in cargo sorting by binding the cytoplasmic tails of cargo proteins and of cargo receptors and by assembling a lattice that deforms the membrane into a bud. All coats are cytoplasmic components that are recruited onto membrane by small GTPases of the ARF/Sar1 family. Coat recruitment is mediated through the active GTP-bound ARF/Sar1 and the GDP/GTP exchange factors that activate ARFs/Sar1 are critical components of the coating process. The GEFs define the sites of ARF/Sar1 activation, and spatially and temporally initiate budding events. Newly formed vesicles and larger carriers are transported from the donor compartment to the acceptor compartment either by diffusion or via microtubules and motor-mediated process. Both, dynein and kinesin motors have been implicated in ER-Golgi traffic (reviewed in)). It has been assumed that vesicles uncoat during transport and that tethering at the acceptor membranes involves uncoated vesicles. Findings that tethers interact with coat components suggest that vesicles remain partially coated during transit and the coat may participate during tethering to confer specificity. Following tethering and uncoating, the membranes undergo fusion mediated by cognate SNAREs on the incoming vesicle and the acceptor membranes. The particulars of SNARE-mediated fusion have been extensively studied and are reviewed in [149,150]. Following fusion, membrane and components of the trafficking machinery are recycled back to the donor compartment by the same cycle of vesicle budding, transport and fusion. To maintain a constant size of compartments despite continuous flow of membranes, cells must coordinate fusion and fission of transport intermediates. A key question is the molecular mechanism that ensures integration of anterograde and retrograde traffic. A system must be in place that coordinately regulates the rate of budding and fusion. From an engineering standpoint, an efficient means to generate such a dual proofreader is to use a single component to regulate both, budding and fusion. A dual regulatory mechanism would initiate the formation of a recycling vesicle at the same time that an incoming vesicle was fusing into the compartment. We suggest that tethering factors may function to integrate anterograde and retrograde trafficking between compartments. 4. SPECULATION: TETHERS COUPLE ANTEROGRADE AND RETROGRADE TRAFFIC We propose that tethers from the three functional groups work together to mediate a coordinated round of anterograde and retrograde traffic. At the ER-Golgi interface, TRAPPI, Dsl1, COG, p115 and GM130 may act in a sequence of events diagramed in Figure 1. We suggest that the cycle is initiated by the interaction between TRAPPI complex and COPII vesicles. This interaction is mediated by direct binding of Bet3 subunit of TRAPPI to the Sec23 subunit of the COPII coat . COPII-bound TRAPPI functions as a GEF and activates (and likely participates in loading) Rab1 GTPase on COPII vesicles. The active form of Rab1-GTP recruits its effector p115 to COPII vesicles (step 1). P115 tethers COPII vesicles to each other and directly promotes formation of trans-SNARE complexes to facilitate homotypic fusion of COPII vesicles to generate VTC. TRAPPI, Rab-GTP and p115 are also present on VTCs, and p115 also facilitates fusion between the newly generated VTCs and additional COPII vesicles (step 2). Figure 1. Open in a new tab At the same time, p115 on VTCs interacts with GBF1 and positions it at sites that have just received an input of membranes and presumably need to form recycling carriers. GBF1 activates ARF, leading to subsequent recruitment of COPI coat and the formation of COPI vesicles for recycling to the ER (step 3). The p115-GBF1 interaction also may underlie the observed exchange of COPII coats for COPI coats at VTCs . COPI vesicles budded from VTCs would be recognized and tethered to the ER membrane through the Dsl1 tethering complex binding to subunits of COPI (step 4). Dsl1 also interacts with SNAREs and might directly facilitate the fusion of COPI vesicles with the ER. The newly formed VTCs traffic towards the microtubule-organizing center (MTOC) and undergo fusions at the cis-Golgi (step 5). Recognition of incoming membranes to pattern on the existing Golgi might be mediated by p115 on VTCs interacting with GM130 and COG on Golgi membranes (dotted lines). The VTCs are predominantly COPI-coated. P115 has been shown to bind β-COP and may mediate tethering of COPI-coated VTCs at the cis-Golgi . P115 also facilitates SNARE complex formation, leading to fusion. During VTC transport and fusion, p115 on VTCs membranes continues to interact with GBF1 to facilitate production of recycling COPI vesicles (step 3). Following VTC fusions at the cis-Golgi, the newly formed compartment may receive COPI vesicles from more distal Golgi cisterna (step 6). This event may be mediated through TRAPPI, p115 and COG tethers. TRAPPI is required to provide activated Rab1 to recruit p115 and COG to Golgi membranes. The COPI vesicles may be tethered to Golgi membranes through interactions with COG which binds γ and/or β subunits of COPI coat (dotted lines). Binding of COPI vesicles may also be stabilized through p115 that interacts with β-COP (dotted lines). P115-mediated tethering may involve p115 bound to Golgi membranes through SNARE interactions, or through p115 bound to GM130. Both, p115 and COG directly interact with Syntaxin5 Qa-SNARE, suggesting that both could directly facilitate fusion of COPI vesicles with Golgi membranes. Throughout the process of tethering and fusion of COPII vesicles and COPI-coated vesicles, p115 may continue to interact with GBF1 and position it at specific sites to promote formation of recycling COPI vesicles (step 7). It is very tempting to suggest that while p115 is facilitating fusion of membranes, it already programs the membranes with GBF1 to nucleate formation of recycling COPI vesicles. Interestingly, COG interacts with p115 and this may ensure that membrane fusion mediated by COG also is coupled to GBF1 and the recycling pathway. Overall, the available data are consistent with a model in which tethers interact with members of the fusion and the budding machinery to integrate delivery of new membranes into a compartment, while at the same time initiating the removal of membrane from the same compartment. This speculative model will need further experimental evidence to test it. 5. CONCLUDING REMARKS Models of tethering are constantly evolving. The initial model of tethers as static fibrous bridges between membranes has been expanded to include active participation in the fusion reaction. Tethers are emerging as molecular platforms that facilitate multiple events through interactions with various partner proteins. Tethers are now known to bind coats, other tethers and SNAREs to facilitate membrane recognition and membrane fusion. In addition, tethers have been shown to participate in cargo selection and at least one tether (p115) binds components of the budding machinery (GBF1). Thus, a new model can be suggested in which tethers mediate both, vesicle fusion and vesicle formation. In that model, tethers can be viewed as a mechanism to integrate forward and recycling flow of membranes by engaging the fusion machinery to ensure vesicular input, while simultaneously nucleating the budding machinery for membrane removal. Such coordinated process would ensure a balance in membrane flow through a compartment and maintain compartment size. Many questions regarding tether function remain. Some of the key issues are: what is the relationship between tethers and coats? do tethers facilitate uncoating? how are tethers regulated to initiate uncoating at a particular stage during transport? Some tethers are specific to a single compartment while others localize to and function at multiple compartments. How are “one love” tethers recruited to only a single membrane? How are “promiscuous” tethers regulated at different membranes? What additional components regulate their specificity to recognize the correct vesicle? Perhaps tethers that show wide localization perform a less selective role in vesicle recognition and instead function in a general process, such as facilitating SNARE complex formation. It appears that a single step of traffic may engage numerous mechanistically distinct tethers. What is the order of molecular events at each stage of traffic? Are the functions of distinct tethers in tandem or parallel? Do tethers have multiple functions, depending on what partner they bind at a particular time and site during traffic? Why are different tethers needed at the same step of traffic? Maybe the requirement reflects different populations of transport carriers? Oligomeric tethers exist in both assembled and partially disassembled forms. It is possible to imagine that the assembly process itself is an important step in vesicle tethering. Is the assembly/disassembly cycle important for the function of these tethers? How the assembly/disassembly cycle is controlled? What is the role of this cycle in the recycling of tethers themselves? How the interaction between different tethers is regulated? Biochemical, genetic, molecular and structural studies are rapidly expanding our knowledge of the principles that govern membrane traffic. It is likely that many of these questions will be answered soon. ACKNOWLEDGEMENTS We apologize to all colleagues whose work was not discussed due to space limitations. We thank Tomasz Szul for the art-work and Kathryn Howell and Martin Lowe for insightful comments. The authors are supported by grants from the NIH (ES and VL), the NSF (ES and VL) and the AHA (ES). Abbreviations siRNA short (or small) interfering RNA PM plasma membrane ER endoplasmic reticulum COPI vesicle coat protein complex I COPII vesicle coat protein complex II VTC vesicular-tubular cluster GEF guanine nucleotide exchange factor TRAPP transport protein particle COG conserved oligomeric Golgi complex GRASP Golgi re-assembly stacking protein TGN trans-Golgi network SM Sec1/Munc18-like proteins ARF ADP-ribosylation factor GARP Golgi-associated retrograde protein complex Footnotes Publisher's Disclaimer: This is a PDF file of an unedited manuscript that has been accepted for publication. 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16220	https://brainly.com/question/10843286	[FREE] put the numbers 1 through 8 in the circles. Use each number only once. All rows must have a sum of 12 - brainly.com Advertisement Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +77,7k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +28,3k Ace exams faster, with practice that adapts to you Practice Worksheets +8,8k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified put the numbers 1 through 8 in the circles. Use each number only once. All rows must have a sum of 12 1 See answer Explain with Learning Companion NEW Asked by immamuele • 08/26/2018 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 28323393 people 28M 3.0 0 Upload your school material for a more relevant answer The missing number can be placed so the sum of the number in each row is 12. What is an arithmetic operation? It is defined as the operation in which we do the addition of numbers, subtraction, multiplication, and division. It has a basic four operators that is +, -, ×, and ÷. It is given that: The blank circle is given in the picture. As we know, from the arithmetic operation, the operation in which we do the addition of numbers, subtraction, multiplication, and division. A number is a mathematical entity that can be used to count, measure, or name things. For example, 1, 2, 56, etc. are the numbers. The numbers are in the blank: 1 + 8 + 3 = 12 5 + 7 = 12 6 + 4 + 2 = 12 Thus, the missing number can be placed so the sum of the number in each row is 12. Learn more about the arithmetic operation here: brainly.com/question/20595275 SPJ2 Answered by maheshpatelvVT •7.9K answers•28.3M people helped Thanks 0 3.0 (1 vote) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 28323393 people 28M 3.0 0 Quantitative Methods for Plant Breeding - Walter Suza Open Education International Perspectives in Higher Education - Patrick Blessinger; TJ Bliss A Treatise of Probability - John Maynard Keynes Upload your school material for a more relevant answer To sum to 12 using the numbers 1 to 8, we can create combinations like (1, 5, 6), (2, 4, 6), and (3, 4, 5). It's important to ensure that each row in your arrangement meets the sum requirement while using each number exactly once. Arrange these combinations accordingly and verify each row's total to ensure accuracy. Explanation To solve the problem of placing the numbers 1 to 8 in circles with the requirement that each row sums to 12, we can follow these steps: Understand the conditions: We need to arrange the numbers in such a way that the sum of the numbers in each row equals 12. Since we have the numbers from 1 to 8, we cannot repeat any number, and each number must be used exactly once. Think about combinations: We need to find combinations of three numbers (if we are assuming three circles per row) that together sum to 12. The combinations must include different sets of numbers from 1 to 8 only. Finding combinations: One possible combination could be (1, 5, 6) because: 1+5+6=12 Another combination is (2, 4, 6): 2+4+6=12 A third combination can be (3, 4, 5): 3+4+5=12 Placement: Place these combinations into rows as required. If your circles form a grid or a specific shape, ensure that numbers in the adjacent rows also add up to 12. Testing the arrangement: After placing the numbers, check each row to ensure that they all equal 12. If one row does not add up to 12, you will need to rearrange the numbers accordingly. This exercise helps enhance your understanding of addition, combinations, and systematic problem-solving. Feel free to experiment with different placements and combinations to achieve the correct sums! Examples & Evidence Examples of valid combinations of numbers 1 to 8 that sum to 12 are (1, 5, 6), (2, 4, 6), and (3, 4, 5). You can arrange these sets in circles or rows as needed to meet the sum condition. These combinations were derived from simple arithmetic addition, ensuring that the sum of three distinct numbers is 12, as required by the problem. Thanks 0 3.0 (1 vote) Advertisement immamuele has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics If 24 out of 32 students prefer iPhones, how many out of 500 students in the school would be expected to prefer them? Choose an equivalent expression for 1 2 3⋅1 2 9⋅1 2 4⋅1 2 2. A. 1 2 4 B. 1 2 18 C. 1 2 35 D. 1 2 216 Choose an equivalent expression for 1 0 6÷1 0 4. A. 1 0 2 B. 1 0 3 C. 1 0 10 D. 1 0 24 How would you write 1 2−3 using a positive exponent? 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16221	https://link.springer.com/article/10.1007/s10231-025-01615-7	Monotone properties of the second even Neumann eigenfunction in symmetric domains \| Annali di Matematica Pura ed Applicata (1923 -) Loading [MathJax]/jax/output/HTML-CSS/config.js Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Log in Menu Find a journalPublish with usTrack your research Search Cart Search Search by keyword or author Search Navigation Find a journal Publish with us Track your research Home Annali di Matematica Pura ed Applicata (1923 -) Article Monotone properties of the second even Neumann eigenfunction in symmetric domains Published: 27 September 2025 (2025) Cite this article Annali di Matematica Pura ed Applicata (1923 -)Aims and scopeSubmit manuscript Hongbin Chen1, Ke Wu2,3& Ruofei YaoORCID: orcid.org/0000-0003-2010-84094 Abstract This paper investigates symmetric Neumann eigenfunctions on a class of planar domains that possess a single line of symmetry. We focus on the Neumann eigenfunctions corresponding to the second eigenvalue among functions that are symmetric with respect to the axis of symmetry of the domain. By employing the method of continuity, we establish the monotonicity of these eigenfunctions in the direction parallel to the axis of symmetry. Furthermore, we show that the corresponding eigenfunction is a Morse function on the boundary with exactly two critical points, and that the associated eigenvalue is simple within the space of symmetric functions. This is a preview of subscription content, log in via an institution to check access. Access this article Log in via an institution Subscribe and save Springer+ from $39.99 /Month Starting from 10 chapters or articles per month Access and download chapters and articles from more than 300k books and 2,500 journals Cancel anytime View plans Buy Now Buy article PDF USD 39.95 Price excludes VAT (USA) Tax calculation will be finalised during checkout. Instant access to the full article PDF. 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ArticleGoogle Scholar Li, R., Yao, R.: Monotonicity of positive solutions to semilinear elliptic equations with mixed boundary conditions in triangles. arXiv:2401.17912 (2024) Mai, J., Yao, R.: Symmetry and monotonicity of positive solutions to elliptic equations with mixed boundary conditions in a kite. Discrete Contin. Dyn. Syst. 46(1), 305-330 (2025). ArticleGoogle Scholar Milnor, J.: Morse theory, annals of mathematics studies, vol. 51. Princeton University Press, Princeton, NJ (1963). BookGoogle Scholar Miyamoto, Y.: The “hot spots’’ conjecture for a certain class of planar convex domains. J. Math. Phys. 50(10), 103530 (2009). ArticleMathSciNetGoogle Scholar Miyamoto, Y.: A planar convex domain with many isolated “hot spots’’ on the boundary. Jpn. J. Ind. Appl. Math. 30(1), 145–164 (2013). ArticleMathSciNetGoogle Scholar Pascu, M.N.: Scaling coupling of reflecting Brownian motions and the hot spots problem. Trans. Amer. Math. Soc. 354(11), 4681–4702 (2002). ArticleMathSciNetGoogle Scholar Polymath: Polymath project 7 research thread 5: the hot spots conjecture, June 3, 2012 through August 9, 2013. Pütter, R.: On the nodal lines of second eigenfunctions of the fixed membrane problem. Comment. Math. Helv. 65(1), 96–103 (1990). ArticleMathSciNetGoogle Scholar Pütter, R.: Bounds for Neumann eigenvalues of (n)-dimensional balls and second eigenfunctions on ellipsoids. In: Progress in partial differential equations: elliptic and parabolic problems (Pont-à-Mousson, 1991), Pitman Res. Notes Math. Ser., vol.266, pp. 220–231. Longman Sci. Tech., Harlow (1992) Rauch, J.: Five problems: an introduction to the qualitative theory of partial differential equations. In: Partial differential equations and related topics (Program, Tulane Univ., New Orleans, La., 1974), Lecture Notes in Math., vol. Vol. 446, pp. 355–369. Springer, Berlin-New York (1975). Rohleder J. A new approach to the hot spots conjecture. arXiv preprint arXiv:2106.05224 (2021) Serrin, J.: A symmetry problem in potential theory. Arch. Rational Mech. Anal. 43, 304–318 (1971). ArticleMathSciNetGoogle Scholar Siudeja, B.: Hot spots conjecture for a class of acute triangles. Math. Z. 280(3–4), 783–806 (2015). ArticleMathSciNetGoogle Scholar Yao, R., Chen, H., Gui, C.: Symmetry of positive solutions of elliptic equations with mixed boundary conditions in a super-spherical sector. Calc. Var. Partial. Differ. Equ. 60(4), 130 (2021). ArticleMathSciNetGoogle Scholar Yao, R., Chen, H., Li, Y.: Symmetry and monotonicity of positive solutions of elliptic equations with mixed boundary conditions in a super-spherical cone. Calc. Var. Partial. Differ. Equ. 57(6), 154 (2018). ArticleMathSciNetGoogle Scholar Download references Acknowledgements The authors would like to thank the anonymous referee for careful reading and helpful suggestions which led to improvements of our original manuscript. The research of Ke Wu is supported by the National Natural Science Foundation of China (No. 12401264). The research of Ruofei Yao is supported by Guangdong Basic and Applied Basic Research Foundation (No. 2025A1515011856) and the National Natural Science Foundation of China (No. 12001543). Author information Authors and Affiliations School of Mathematics and Statistics, Xi’an Jiaotong University, Xi’an, 710049, People’s Republic of China Hongbin Chen Department of Mathematics, Chinese University of Hong Kong, Shatin, NT, Hong Kong Ke Wu School of Mathematics, Yunnan Normal University, Kunmin, 650500, People’s Republic of China Ke Wu School of Mathematics, South China University of Technology, Guangzhou, 510641, People’s Republic of China Ruofei Yao Authors 1. Hongbin ChenView author publications Search author on:PubMedGoogle Scholar 2. Ke WuView author publications Search author on:PubMedGoogle Scholar 3. Ruofei YaoView author publications Search author on:PubMedGoogle Scholar Corresponding author Correspondence to Ruofei Yao. Ethics declarations Conflict of interest On behalf of all authors, the corresponding author states that there is no Conflict of interest. Additional information Publisher's Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Rights and permissions Springer Nature or its licensor (e.g. a society or other partner) holds exclusive rights to this article under a publishing agreement with the author(s) or other rightsholder(s); author self-archiving of the accepted manuscript version of this article is solely governed by the terms of such publishing agreement and applicable law. Reprints and permissions About this article Cite this article Chen, H., Wu, K. & Yao, R. Monotone properties of the second even Neumann eigenfunction in symmetric domains. Annali di Matematica (2025). Download citation Received: 05 July 2025 Accepted: 04 September 2025 Published: 27 September 2025 DOI: Share this article Anyone you share the following link with will be able to read this content: Get shareable link Sorry, a shareable link is not currently available for this article. Copy shareable link to clipboard Provided by the Springer Nature SharedIt content-sharing initiative Keywords Monotonicity Hot spots Neumann eigenfunctions Continuity method Mathematics Subject Classification Primary 35P05 Secondary 58J50 35B50 35J05 Access this article Log in via an institution Subscribe and save Springer+ from $39.99 /Month Starting from 10 chapters or articles per month Access and download chapters and articles from more than 300k books and 2,500 journals Cancel anytime View plans Buy Now Buy article PDF USD 39.95 Price excludes VAT (USA) Tax calculation will be finalised during checkout. Instant access to the full article PDF. Institutional subscriptions Sections Figures References Abstract References Acknowledgements Author information Ethics declarations Additional information Rights and permissions About this article Advertisement Fig. 1 View in article Fig. 2 View in article Atar, R., Burdzy, K.: On Neumann eigenfunctions in lip domains. J. Amer. Math. Soc. 17(2), 243–265 (2004). ArticleMathSciNetGoogle Scholar Bañuelos, R., Burdzy, K.: On the “hot spots’’ conjecture of. J. Rauch. J. Funct. Anal. 164(1), 1–33 (1999). ArticleMathSciNetGoogle Scholar Bañuelos, R., Pang, M.M.: An inequality for potentials and the “hot-spots’’ conjecture. Indiana Univ. Math. J. 53(1), 35–47 (2004). ArticleMathSciNetGoogle Scholar Bañuelos, R., Pang, M.M., Pascu, M.N.: Brownian motion with killing and reflection and the “hot-spots’’ problem. Probab. Theory Relat. Fields 130(1), 56–68 (2004). ArticleMathSciNetGoogle Scholar Bass, R.F., Burdzy, K.: Fiber Brownian motion and the “hot spots’’ problem. Duke Math. 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ArticleMathSciNetGoogle Scholar de Dios Pont, J.: Convex sets can have interior hot spots. arXiv:2412.06344 (2024) Evans, L.C.: Partial differential equations, graduate studies in mathematics, second edn. American Mathematical Society, Providence, RI (2010). BookGoogle Scholar Gidas, B., Ni, W.M., Nirenberg, L.: Symmetry and related properties via the maximum principle. Comm. Math. Phys. 68(3), 209–243 (1979). ArticleMathSciNetGoogle Scholar Gilbarg D, Trudinger N.S. Elliptic partial differential equations of second order, Grundlehren der mathematischen Wissenschaften [Fundamental principles of mathematical sciences], vol.224. Springer-Verlag, Berlin, second edn. (1983). Helffer, B., Persson-Sundqvist, M.: On nodal domains in euclidean balls. Proc. Amer. Math. Soc. 144(11), 4777–4791 (2016). ArticleMathSciNetGoogle Scholar Jerison, D., Nadirashvili, N.: The “hot spots’’ conjecture for domains with two axes of symmetry. J. Amer. Math. Soc. 13(4), 741–772 (2000). ArticleMathSciNetGoogle Scholar Judge, C., Mondal, S.: Euclidean triangles have no hot spots. Ann. of Math. (2) 191(1), 167–211 (2020). ArticleMathSciNetGoogle Scholar Judge, C., Mondal, S.: Erratum: euclidean triangles have no hot spots. Ann. of Math. (2) 195(1), 337–362 (2022). ArticleMathSciNetGoogle Scholar Kawohl, B. Rearrangements and convexity of level sets in PDE, lecture notes in mathematics, vol.1150. Springer-Verlag, Berlin (1985). Kleefeld, A.: The hot spots conjecture can be false: some numerical examples. Adv. Comput. Math. 47(6), 85 (2021). ArticleMathSciNetGoogle Scholar Laugesen, R.S., Siudeja, B.A.: Minimizing Neumann fundamental tones of triangles: an optimal poincaré inequality. J. Differ. Equ. 249(1), 118–135 (2010). ArticleGoogle Scholar Li, R., Yao, R.: Monotonicity of positive solutions to semilinear elliptic equations with mixed boundary conditions in triangles. arXiv:2401.17912 (2024) Mai, J., Yao, R.: Symmetry and monotonicity of positive solutions to elliptic equations with mixed boundary conditions in a kite. Discrete Contin. Dyn. Syst. 46(1), 305-330 (2025). ArticleGoogle Scholar Milnor, J.: Morse theory, annals of mathematics studies, vol. 51. Princeton University Press, Princeton, NJ (1963). BookGoogle Scholar Miyamoto, Y.: The “hot spots’’ conjecture for a certain class of planar convex domains. J. Math. Phys. 50(10), 103530 (2009). ArticleMathSciNetGoogle Scholar Miyamoto, Y.: A planar convex domain with many isolated “hot spots’’ on the boundary. Jpn. J. Ind. Appl. Math. 30(1), 145–164 (2013). ArticleMathSciNetGoogle Scholar Pascu, M.N.: Scaling coupling of reflecting Brownian motions and the hot spots problem. Trans. Amer. Math. Soc. 354(11), 4681–4702 (2002). ArticleMathSciNetGoogle Scholar Polymath: Polymath project 7 research thread 5: the hot spots conjecture, June 3, 2012 through August 9, 2013. Pütter, R.: On the nodal lines of second eigenfunctions of the fixed membrane problem. Comment. Math. Helv. 65(1), 96–103 (1990). ArticleMathSciNetGoogle Scholar Pütter, R.: Bounds for Neumann eigenvalues of (n)-dimensional balls and second eigenfunctions on ellipsoids. In: Progress in partial differential equations: elliptic and parabolic problems (Pont-à-Mousson, 1991), Pitman Res. Notes Math. Ser., vol.266, pp. 220–231. Longman Sci. Tech., Harlow (1992) Rauch, J.: Five problems: an introduction to the qualitative theory of partial differential equations. In: Partial differential equations and related topics (Program, Tulane Univ., New Orleans, La., 1974), Lecture Notes in Math., vol. Vol. 446, pp. 355–369. Springer, Berlin-New York (1975). Rohleder J. A new approach to the hot spots conjecture. arXiv preprint arXiv:2106.05224 (2021) Serrin, J.: A symmetry problem in potential theory. Arch. Rational Mech. Anal. 43, 304–318 (1971). ArticleMathSciNetGoogle Scholar Siudeja, B.: Hot spots conjecture for a class of acute triangles. Math. Z. 280(3–4), 783–806 (2015). ArticleMathSciNetGoogle Scholar Yao, R., Chen, H., Gui, C.: Symmetry of positive solutions of elliptic equations with mixed boundary conditions in a super-spherical sector. Calc. Var. Partial. Differ. Equ. 60(4), 130 (2021). ArticleMathSciNetGoogle Scholar Yao, R., Chen, H., Li, Y.: Symmetry and monotonicity of positive solutions of elliptic equations with mixed boundary conditions in a super-spherical cone. Calc. Var. Partial. Differ. Equ. 57(6), 154 (2018). 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16222	https://en.wikipedia.org/wiki/MRI_contrast_agent	Jump to content MRI contrast agent العربية Français 한국어 کٲشُر Magyar 日本語 Українська Edit links From Wikipedia, the free encyclopedia Types of contrast agents used for magnetic resonance imaging MRI contrast agents are contrast agents used to improve the visibility of internal body structures in magnetic resonance imaging (MRI). The most commonly used compounds for contrast enhancement are gadolinium-based contrast agents (GBCAs). Such MRI contrast agents shorten the relaxation times of nuclei within body tissues following oral or intravenous administration. Due to safety concerns, these products carry a Black Box Warning in the US. Theory of operation [edit] In MRI scanners, sections of the body are exposed to a strong magnetic field causing primarily the hydrogen nuclei ("spins") of water in tissues to be polarized in the direction of the magnetic field. An intense radiofrequency pulse is applied that tips the magnetization generated by the hydrogen nuclei in the direction of the receiver coil where the spin polarization can be detected. Random molecular rotational oscillations matching the resonance frequency of the nuclear spins provide the "relaxation" mechanisms that bring the net magnetization back to its equilibrium position in alignment with the applied magnetic field. The magnitude of the spin polarization detected by the receiver is used to form the MR image but decays with a characteristic time constant known as the T1 relaxation time. Water protons in different tissues have different T1 values, which is one of the main sources of contrast in MR images. A contrast agent usually shortens, but in some instances increases, the value of T1 of nearby water protons thereby altering the contrast in the image. Most clinically used MRI contrast agents work by shortening the T1 relaxation time of protons inside tissues via interactions with the nearby contrast agent. Thermally driven motion of the strongly paramagnetic metal ions in the contrast agent generate the oscillating magnetic fields that provide the relaxation mechanisms that enhance the rate of decay of the induced polarization. The systematic sampling of this polarization over the spatial region of the tissue being examined forms the basis for construction of the image. MRI contrast agents may be administered by injection into the blood stream or orally, depending on the subject of interest. Oral administration is well suited to gastrointestinal tract scans, while intravascular administration proves more useful for most other scans. MRI contrast agents can be classified by their: Chemical composition Administration route Magnetic properties Biodistribution and applications: Extracellular fluid agents (intravenous contrast agents) Blood pool agents (intravascular contrast agents) Organ specific agents (gastrointestinal contrast agents and hepatobiliary contrast agents) Active targeting/cell labeling agents (tumor-specific agents) Responsive (smart or bioactivated) agents pH-sensitive agents Gadolinium(III) [edit] Gadolinium(III) containing MRI contrast agents (often termed simply "gado" or "gad") are the most commonly used for enhancement of vessels in MR angiography or for brain tumor enhancement associated with the degradation of the blood–brain barrier (BBB). Over 450 million doses have been administered worldwide from 1988 to 2017. For large vessels such as the aorta and its branches, the dose can be as low as 0.1 mmol/kg of body mass. Higher concentrations are often used for finer vasculature. At much higher concentration, there is more T2 shortening effect of gadolinium, causing gadolinium brightness to be less than surrounding body tissues. However at such concentration, it will cause greater toxicity to bodily tissues. Gd3+ chelates are hydrophilic and do not readily cross the intact blood–brain barrier. Thus, they are useful in enhancing lesions and tumors where the blood–brain barrier is compromised and the Gd(III) leaks out.[a] In the rest of the body, the Gd3+ initially remains in the circulation but then distributes into the interstitial space or is eliminated by the kidneys. Available gadolinium-based contrast agents (GBCAs) (brand names, approved for human use by EMA[when?] and by the FDA in 1988; (standard dose)): Extracellular fluid agents [edit] Macrocyclic ionic gadoterate (Dotarem, Clariscan): EMA, FDA (SD: 0.1 mmol/kg) non-ionic gadobutrol (Gadovist [EU] / Gadavist [US]): EMA, FDA (SD: 0.1 mmol/kg) gadoteridol (ProHance): EMA, FDA (SD: 0.1 mmol/kg) gadopiclenol (Elucirem, Vueway): EMA, FDA (SD: 0.05 mmol/kg) Linear (suspended by EMA) ionic gadopentetate (Magnevist, EU: Magnegita, Gado-MRT ratiopharm): FDA (SD: 0.1 mmol/kg) gadobenate (MultiHance): FDA, EMA (liver) (SD: 0.1 mmol/kg) gadopentetic acid dimeglumine (Magnetol) gadoxetate (Eovist, EU: Primovist): FDA (SD: 0.025 mmol/kg) non-ionic gadoversetamide (OptiMARK): FDA (SD: 0.1 mmol/kg) gadodiamide (Omniscan): FDA (SD: 0.1 mmol/kg) Blood pool agents [edit] Albumin-binding gadolinium complexes gadofosveset (Ablavar, formerly Vasovist): FDA (SD: 0.1 mmol/kg) gadocoletic acid Polymeric gadolinium complexes gadomelitol gadomer 17 Hepatobiliary (liver) agents [edit] gadoxetic acid (Primovist [EU] / Eovist [US]) is used as a hepatobiliary agent as 50% is taken up and excreted by the liver and 50% by the kidneys. Safety [edit] \| \| \| --- \| \| \| This section needs to be updated. Please help update this article to reflect recent events or newly available information. (August 2021) \| Main article: Nephrogenic systemic fibrosis The use of Gd3+ chelates in persons with acute or chronic kidney disease can cause nephrogenic systemic fibrosis (NSF), a rare but severe systemic disease resembling scleromyxedema and to some extent scleroderma. It may occur months after contrast injection. Patients with severely deteriorated kidney function are more at risk for NSF, with dialysis patients being more at risk than patients with mild chronic kidney disease. NSF can be caused by linear and macrocyclic gadolinium-containing MRI contrast agents, although macrocyclic ionic compounds have been found the least likely to release the Gd3+. While NSF is a severe form of disease, gadolinium deposition disease (GDD) is a mild variant with pain (e.g. headache), fatigue, and / or gadolinium depositions. As a free solubilized aqueous ion, gadolinium(III) is highly toxic, but the chelated compounds are relatively safe for individuals without kidney disease. Free Gd3+ has a median lethal dose of 0.34 mmol/kg (IV, mouse) or 100–200 mg/kg, but the LD50 is increased by a factor of 31 times when Gd3+ is chelated. The spectrum of adverse drug reactions is greater with gadolinium-based contrast agents than with iodinated contrast agents (radiocontrast agents). Gadolinium has been found to remain in the brain, heart muscle, kidney, liver, and other organs after one or more injections of a linear or macrocyclic gadolinium-based contrast agents, even after a prolonged period of time. The amount differs with the presence of kidney injury at the moment of injection, the molecular geometry of the ligand, and the dose administered.[citation needed] In vitro studies have found gadolinium-based contrast agents to be neurotoxic, and a study found signal intensity in the dentate nucleus of MRI (indicative of gadolinium deposition) to be correlated with lower verbal fluency. Confusion is often reported as a possible clinical symptom. The FDA has asked doctors to limit the use of gadolinium contrast agents to examinations where necessary information is obtained only through its use. Intrathecal injections of doses higher than 1 mmol are associated with severe neurological complications and can lead to death. The glymphatic system could be the main access of GBCA to the brain in intravenous injection. Continuing evidence of the retention of gadolinium in brain and other tissues following exposure to gadolinium containing contrast media, led to a safety review by the Committee for Medicinal Products for Human Use (CHMP) which led the EMA to restrict or suspend authorization for the intravenous use of most brands of linear gadolinium-based media, in which Gd3+ has a lower binding affinity, in 2017. In the United States, the research has led the FDA to revise its class warnings for gadolinium-based contrast media. It is advised that the use of gadolinium-based media should be based on careful consideration of the retention characteristics of the contrast, with extra care being taken in patients requiring multiple lifetime doses, pregnant, and paediatric patients, and patients with inflammatory conditions. They also advise minimizing repeated GBCA imaging studies when possible, particularly closely spaced MRI studies, but not avoiding or deferring necessary GBCA MRI scans. In December 2017, the FDA announced that it was requiring these warnings to be included on all GBCAs. The FDA also called for increased patient education and requiring gadolinium contrast vendors to conduct additional animal and clinical studies to assess the safety of these agents. The French health authority recommends to use the lowest possible dose of a GBCA and only when essential diagnostic information cannot be obtained without it. The World Health Organization issued a restriction on use of several gadolinium contrast agents in November 2009 stating that "High-risk gadolinium-containing contrast agents (Optimark, Omniscan, Magnevist, Magnegita, and Gado-MRT ratiopharm) are contraindicated in patients with severe kidney problems, in patients who are scheduled for or have recently received a liver transplant, and in newborn babies up to four weeks of age." In magnetic resonance imaging in pregnancy, gadolinium contrast agents in the first trimester is associated with a slightly increased risk of a childhood diagnosis of several forms of rheumatism, inflammatory disorders, or infiltrative skin conditions, according to a retrospective study including 397 infants prenatally exposed to gadolinium contrast. In the second and third trimester, gadolinium contrast is associated with a slightly increased risk of stillbirth or neonatal death, by the same study. Guidelines from the Canadian Association of Radiologists are that dialysis patients should receive gadolinium agents only where essential and that they should receive dialysis after the exam. If a contrast-enhanced MRI must be performed on a dialysis patient, it is recommended that certain high-risk contrast agents be avoided but not that a lower dose be considered. The American College of Radiology recommends that contrast-enhanced MRI examinations be performed as closely before dialysis as possible as a precautionary measure, although this has not been proven to reduce the likelihood of developing NSF. The FDA recommends that potential for gadolinium retention be considered when choosing the type of GBCA used in patients requiring multiple lifetime doses, pregnant women, children, and patients with inflammatory conditions. Anaphylactoid reactions are rare, occurring in about 0.03–0.1%. Iron oxide: superparamagnetic [edit] Main article: Superparamagnetic relaxometry Two types of iron oxide contrast agents exist: superparamagnetic iron oxide (SPIO) and ultrasmall superparamagnetic iron oxide (USPIO). These contrast agents consist of suspended colloids of iron oxide nanoparticles and when injected during imaging reduce the T2 signals of absorbing tissues. SPIO and USPIO contrast agents have been used successfully in some instances for liver lesion evaluation. Feridex I.V. (also known as Endorem and ferumoxides). This product was discontinued by AMAG Pharma in November 2008. Resovist (also known as Cliavist). This was approved for the European market in 2001, but production was abandoned in 2009. Sinerem (also known as Combidex). Guerbet withdrew the marketing authorization application for this product in 2007. Lumirem (also known as Gastromark). Gastromark was approved by the FDA in 1996 and was discontinued by its manufacturer in 2012. Clariscan (also known as PEG-fero, Feruglose, and NC100150). This iron based contrast agent was never commercially launched and its development was discontinued in early 2000s due to safety concerns. In 2017 GE Healthcare launched a macrocyclic extracellular gadolinium based contrast agent containing gadoteric acid as gadoterate meglumine under the trade name Clariscan. Iron platinum: superparamagnetic [edit] Superparamagnetic iron–platinum particles (SIPPs) have been reported and had significantly better T2 relaxivities compared with the more common iron oxide nanoparticles. SIPPs were also encapsulated with phospholipids to create multifunctional SIPP stealth immunomicelles that specifically targeted human prostate cancer cells. These are, however, investigational agents which have not yet been tried in humans. In a recent study, multifunctional SIPP micelles were synthesized and conjugated to a monoclonal antibody against prostate-specific membrane antigen. The complex specifically targeted human prostate cancer cells in vitro, and these results suggest that SIPPs may have a role in the future as tumor-specific contrast agents.[citation needed] Manganese [edit] Manganese(II) chelates such as Mn-DPDP (mangafodipir) enhance the T1 signal. The chelate dissociates in vivo into manganese and DPDP; the manganese is excreted in bile, while DPDP is eliminated via kidney filtration. Mangafodipir has been used in human neuroimaging clinical trials, including for neurodegenerative diseases such as multiple sclerosis. Manganese(II) ions are often used as a contrast agent in animal studies, often called MEMRI (manganese-enhanced MRI). Because Mn2+ ions can enter cells through calcium transport channels, it has been used for functional brain imaging. Manganese(III) chelates with porphyrins and phthalocyanines have also been studied. Unlike the other well-studied iron oxide-based nanoparticles, research on Mn-based nanoparticles is at a relatively early stage. Oral administration [edit] A wide variety of oral contrast agents can enhance images of the gastrointestinal tract. They include gadolinium and manganese chelates, or iron salts for T1 signal enhancement. SPIO, barium sulfate, air and clay have been used to lower T2 signal. Natural products with high manganese concentration such as blueberry and green tea can also be used for T1 increasing contrast enhancement. Perflubron, a type of perfluorocarbon, has been used as a gastrointestinal MRI contrast agent for pediatric imaging. This contrast agent works by reducing the number of hydrogen ions in a body cavity, thus causing it to appear dark in the images. Protein-based MRI contrast agents [edit] See also: Enzyme-activated MR contrast agents Newer research suggests the possibility of protein based contrast agents, based on the abilities of some amino acids to bind with gadolinium. See also [edit] Lanthanide probes Footnotes [edit] ^ "Disruption of the BBB tight junctions is thought to be an early or initiating event in new MS lesion formation. T1-w MRI in combination with low molecular weight gadolinium-based contrast agents (GBCA) is most typically used to characterize BBB compromise in MS. MRI GBCAs do not readily cross cellular membranes, are avid extracellular space markers, and are thought to enter the brain from the blood by diffusive transport between endothelial cells (ie, via intercellular pathways). Although it is widely believed that the MRI GBCAs do not cross the BBB under homeostatic conditions, there is substantial evidence that they do, albeit with very small volume transfer rate constants." — Bagnato, Gauthier, Laule, et al. (2020) References [edit] ^ Rinck, Peter A. (2024). "Magnetic resonance contrast agents". Magnetic Resonance in Medicine. A critical introduction (14th ed.). TRTF – The Round Table Foundation / EMRF – European Magnetic Resonance Forum. "Magnetic resonance contrast agents". Magnetic Resonance in Medicine (www.magnetic-resonance.org) (e-Textbook). ^ Geraldes, Carlos F.G.C.; Laurent, Sophie (2009). "Classification and basic properties of contrast agents for magnetic resonance imaging". 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"MRI contrast agents". magnetic-resonance.org. \| Contrast media (V08) \| \| --- \| \| X-ray and CT \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| Iodinated, Water soluble \| \| \| \| --- \| \| Nephrotropic, high osmolar \| Diatrizoic acid# Metrizoic acid Iodamide Iotalamic acid Ioxitalamic acid Ioglicic acid Acetrizoic acid Iocarmic acid Methiodal Diodone \| \| Nephrotropic, low osmolar \| Metrizamide Iohexol# Ioxaglic acid Iopamidol Iopromide Iotrolan Ioversol Iopentol Iodixanol Iomeprol Iobitridol Ioxilan \| \| Hepatotropic \| Iodoxamic acid Iotroxic acid# Ioglycamic acid Adipiodone Iobenzamic acid Iopanoic acid Iocetamic acid Sodium iopodate Tyropanoic acid Calcium iopodate \| \| \| Iodinated, Water insoluble \| Ethiodized oil (= Ethyl esters of iodised fatty acids, lipiodol) Iopydol Propyliodone Iofendylate‡ \| \| Non-iodinated \| Barium sulfate# \| \| \| MRI \| \| \| \| --- \| \| Paramagnetic \| Gadolinium-based: Gadobenic acid Gadobutrol Gadodiamide Gadofosveset Gadolinium Gadopentetic acid Gadopiclenol Gadoteric acid Gadoteridol Gadoversetamide Gadoxetic acid Other: Ferric ammonium citrate Mangafodipir‡ \| \| Superparamagnetic \| Ferumoxsil Ferristene Iron oxide, nanoparticles \| \| Other \| \| \| \| Ultrasound \| Microspheres of human albumin Microparticles of galactose Perflenapent Microspheres of phospholipids Sulfur hexafluoride \| \| #WHO-EM ‡Withdrawn from market Clinical trials: + †Phase III + §Never to phase III \| \| Retrieved from " Category: MRI contrast agents Hidden categories: CS1 French-language sources (fr) Source attribution All articles with dead external links Articles with dead external links from May 2025 Articles with short description Short description is different from Wikidata Use dmy dates from July 2023 All articles with vague or ambiguous time Vague or ambiguous time from April 2022 Wikipedia articles in need of updating from August 2021 All Wikipedia articles in need of updating All articles with unsourced statements Articles with unsourced statements from May 2023 Commons category link is on Wikidata
16223	https://www.empathia.ai/drug/obstetrics-gynecology-obgyn/tetracycline-isotretinoin-drug-interaction	Tetracycline and Isotretinoin Interaction: Risks and Management\| empathia.ai Product AI Scribe & Clinical Assistant CPT Billing Codes Specialties Cardiology Neurology Primary Care Orthopedics Psychiatry Pediatrics Internal Medicine Allergy & Immunology Emergency Medicine Oncology ENT Dentists Plastic Surgery General Surgery OBGyn & Midwifery Pricing Blog About us Log InTry Free Log InTry Free Tetracycline and Isotretinoin Drug Interaction Summary The combination of tetracycline and isotretinoin is contraindicated due to an increased risk of pseudotumor cerebri (benign intracranial hypertension). Both medications can independently cause elevated intracranial pressure, and their concurrent use significantly amplifies this serious adverse effect. Introduction Tetracycline is a broad-spectrum antibiotic belonging to the tetracycline class, commonly used to treat bacterial infections including acne vulgaris, respiratory tract infections, and sexually transmitted diseases. Isotretinoin is a powerful oral retinoid medication primarily prescribed for severe, treatment-resistant acne. Both medications are frequently used in dermatology, making awareness of their interaction crucial for healthcare providers treating acne patients. Mechanism of Interaction The mechanism underlying this interaction involves both drugs' ability to increase intracranial pressure through different pathways. Tetracycline can cause pseudotumor cerebri by potentially interfering with cerebrospinal fluid absorption or by direct toxic effects on the central nervous system. Isotretinoin similarly increases intracranial pressure through mechanisms that may involve alterations in cerebrospinal fluid dynamics and vitamin A toxicity effects. When used together, these mechanisms appear to have an additive or synergistic effect, dramatically increasing the risk of developing benign intracranial hypertension. Risks and Symptoms The primary risk of combining tetracycline and isotretinoin is the development of pseudotumor cerebri (benign intracranial hypertension), a serious condition characterized by increased pressure within the skull. Symptoms include severe headaches, visual disturbances, papilledema (optic disc swelling), nausea, vomiting, and potential permanent vision loss if left untreated. This interaction is considered clinically significant and potentially life-threatening. The risk appears to be dose-independent and can occur even with short-term concurrent use. Patients may experience symptoms within days to weeks of starting combination therapy. Management and Precautions The combination of tetracycline and isotretinoin should be avoided entirely. If both medications are deemed necessary for a patient, they should not be used concurrently. Healthcare providers should maintain a minimum washout period between discontinuing one medication and starting the other, typically 2-4 weeks. Alternative antibiotics such as erythromycin, clindamycin, or trimethoprim-sulfamethoxazole should be considered for patients requiring antibiotic therapy while on isotretinoin. Patients should be counseled about the signs and symptoms of increased intracranial pressure and advised to seek immediate medical attention if they experience severe headaches, vision changes, or other neurological symptoms. Regular monitoring and patient education are essential components of safe prescribing practices. Tetracycline interactions with food and lifestyle Tetracycline should be taken on an empty stomach, at least 1 hour before or 2 hours after meals, as food significantly reduces absorption. Avoid dairy products (milk, cheese, yogurt), calcium supplements, iron supplements, magnesium-containing antacids, and aluminum-containing antacids within 2-3 hours of taking tetracycline, as these can form chelation complexes that dramatically reduce drug absorption. Avoid alcohol consumption during tetracycline treatment as it may increase the risk of liver toxicity and reduce the effectiveness of the antibiotic. Specialty: Obstetrics & Gynecology (ObGyn) \| Last Updated: August 2025 Ready to Streamline Your Chart Prep? Empathia AI highlights drug risks and flags interactions right inside your intake summaries—before or during the visit. Trusted by thousands of clinicians. Try for Free Email: support@empathia.ai Phone: +1 626-217-2048 (US) +1 437-781-9194 (Canada) Book a Demo @2025 Empathia AI, Inc. All rights reserved. Empathia is your clinical assistant to augment your daily work. Compliance Trust Center Data Security GDPR PIPEDA HIPAA DPA Product Pricing Features Download Request a Demo Company About Us Publications Success Stories Resources Tutorials FAQs Help Center Patient Consent Template Legal Privacy Policy Terms of Service Cookies Policy
16224	https://zhidao.baidu.com/question/244811475.html	下列各式中与lg(cosx-1)^2相等的式子是（） A.[lg(cosx-1)]^2 B.2lg(cosx-1) C.2lgcosx D.4lg\|sin(x/2)\|+_百度知道百度首页商城注册登录网页资讯视频图片知道文库贴吧采购地图更多搜索答案我要提问下列各式中与lg(cosx-1)^2相等的式子是（） A.[lg(cosx-1)]^2 B.2lg(cosx-1) C.2lgcosx D.4lg\|sin(x/2)\|+ 首页用户认证用户认证团队合伙人热推榜单企业媒体政府其他组织商城法律手机答题我的百度知道> 无分类下列各式中与lg(cosx-1)^2相等的式子是（） A.[lg(cosx-1)]^2 B.2lg(cosx-1) C.2lgcosx D.4lg\|sin(x/2)\|+ 下列各式中与lg(cosx-1)^2相等的式子是（）A.[lg(cosx-1)]^2B.2lg(cosx-1)C.2lgcosxD.4lg\|sin(x/2)\|+2lg2求... 下列各式中与lg(cosx-1)^2相等的式子是（） A.[lg(cosx-1)]^2 B.2lg(cosx-1) C.2lgcosx D.4lg\|sin(x/2)\|+2lg2 求详细解题过程。谢谢。展开分享复制链接新浪微博微信扫一扫举报 1个回答 #热议#生活中有哪些实用的心理学知识？俺安人1W 2011-03-28 · TA获得超过9.5万个赞知道大有可为答主回答量：1.1万采纳率：0% 帮助的人：1.7亿我也去答题访问个人页关注展开全部 B.2lg(cosx-1) lg a^b=blg a 这个是公式本回答由提问者推荐 1已赞过已踩过< 你对这个回答的评价是？评论分享复制链接新浪微博微信扫一扫举报收起推荐律师服务：若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询为你推荐：特别推荐 “网络厕所”会造成什么影响？癌症的治疗费用为何越来越高？新生报道需要注意什么？华强北的二手手机是否靠谱？百度律临—免费法律服务推荐超3w专业律师，24H在线服务，平均3分钟回复免费预约随时在线律师指导专业律师一对一沟通完美完成等你来答  换一换一年级数学考80多分正常吗，暑假需要补课吗？等 50299 人想问  我来答  公立制和私立制教学的结果有什么优劣？等 17158 人想问  我来答  婴儿一般带到几个月时，大人会觉得轻松些？等 23298 人想问  我来答  情感的突如其来怎猝不及防的再见？等 10959 人想问  我来答  糖尿病人早上喝粥好不好？等 8860 人想问  我来答  健身剧烈跑步会很伤膝盖吗？等 22772 人想问  我来答帮助更多人  下载百度知道APP，抢鲜体验使用百度知道APP，立即抢鲜体验。你的手机镜头里或许有别人想知道的答案。扫描二维码下载 × 个人、企业类侵权投诉违法有害信息,请在下方选择后提交类别色情低俗涉嫌违法犯罪时政信息不实垃圾广告低质灌水我们会通过消息、邮箱等方式尽快将举报结果通知您。说明 0/200 提交取消领取奖励我的财富值 0 兑换商品 -- 去登录我的现金 0 提现下载百度知道APP 在APP端-任务中心提现我知道了 -- 去登录做任务开宝箱累计完成 0 个任务 10任务略略略略… 50任务略略略略… 100任务略略略略… 200任务略略略略… 任务列表加载中... 新手帮助如何答题获取采纳使用财富值玩法介绍知道商城合伙人认证您的账号状态正常感谢您对我们的支持投诉建议意见反馈账号申诉非法信息举报京ICP证030173号-1 京网文【2023】1034-029号 ©2025Baidu使用百度前必读\|知道协议\|企业推广辅助模式返回顶部
16225	https://math.stackexchange.com/questions/2531034/composition-of-piecewise-functions-with-even-odd-conditions	composition of piecewise functions with even/odd conditions - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more composition of piecewise functions with even/odd conditions Ask Question Asked 7 years, 10 months ago Modified7 years, 10 months ago Viewed 901 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. I have the functions: f,g∈F(Z,Z):f,g∈F(Z,Z): f(n)={n 2,n+1,if n is even if n is odd f(n)={n 2,if n is even n+1,if n is odd g(n)={n−1,2 n,if n is even if n is odd g(n)={n−1,if n is even 2 n,if n is odd I am asked to find f g f g and g f g f, the notation that my textbook uses for the compositions f(g(n))f(g(n)) and g(f(n))g(f(n)) respectively. My professor gave me the hint that I will have three cases for at least one of the compositions. --update-- For the compositions I got these answers: f g(n)={n,n,if n is even if n is odd=n f g(n)={n,if n is even n,if n is odd=n g f(n)=⎧⎩⎨n,n 2−1,n,if n≠4 t,t∈Z if n=4 t,t∈Z if n is odd g f(n)={n,if n≠4 t,t∈Z n 2−1,if n=4 t,t∈Z n,if n is odd Do these appear to be the correct answers? function-and-relation-composition piecewise-continuity Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Nov 21, 2017 at 17:13 wzbillingswzbillings asked Nov 21, 2017 at 16:37 wzbillingswzbillings 557 5 5 silver badges 21 21 bronze badges 1 Almost. Notice that if n n is odd, then f(n)f(n) is even, and then g(f(n))=n g(f(n))=n. Also, if n n is an even multiple of 4 4, then f(n)f(n) is even and thus g(f(n))=n 2−1 g(f(n))=n 2−1. Then the last case if that n n is even but not a multiple of 4 4, then f(n)f(n) is odd and then g(f(n))=n g(f(n))=n.Bergson –Bergson 2017-11-21 17:05:43 +00:00 Commented Nov 21, 2017 at 17:05 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. I will do it for f(g(n))f(g(n)). The other composition is left to you ;) If n n is even, g(n)=n−1 g(n)=n−1 is odd. Then, f(g(n))=f(n−1)=n−1+1=n.f(g(n))=f(n−1)=n−1+1=n. On the other hand if n n is odd, g(n)=2 n g(n)=2 n is even. Then, f(g(n))=f(2 n)=2 n 2=n.f(g(n))=f(2 n)=2 n 2=n. It follows that f(g)f(g) is the identity. You should apply an analogous reasoning to compute g(f(n))g(f(n)), paying now some extra care when n n is even (hint: there is a difference on whether 4 4 divides it or not). Can you take it from here? Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 21, 2017 at 16:46 Paolo IntuitoPaolo Intuito 1,194 7 7 silver badges 21 21 bronze badges 3 I believe so! I got the identity for f g f g so that makes me a little more confident. Would you mind to check the answer for g f g f that I just added to my original question? Thank you!wzbillings –wzbillings 2017-11-21 16:55:25 +00:00 Commented Nov 21, 2017 at 16:55 If n n is odd, f(n)=n+1 f(n)=n+1 is even. Then, g(f(n))=n+1−1=n g(f(n))=n+1−1=n. I think you write the n=4 t n=4 t and n n even but n≠4 t n≠4 t the other way round.Paolo Intuito –Paolo Intuito 2017-11-21 17:02:21 +00:00 Commented Nov 21, 2017 at 17:02 You're right; I went back over my work and fixed my errors. Thank you!wzbillings –wzbillings 2017-11-21 17:13:49 +00:00 Commented Nov 21, 2017 at 17:13 Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Further hint when you compute g(f(n))g(f(n)): The three cases are: When n n is odd, then f(n)f(n) is even. Use the appropriate rule for g g on f(n)f(n). When n n is a multiple of 4 4, then f(n)f(n) is also even, use the appropriate rule for g g on f(n)f(n). When n n is even but not a multiple of 4 4, then f(n)f(n) is odd, use the appropriate rule for g g on f(n)f(n). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 21, 2017 at 16:43 Siong Thye GohSiong Thye Goh 155k 20 20 gold badges 94 94 silver badges 158 158 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions function-and-relation-composition piecewise-continuity See similar questions with these tags. 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16226	https://www.ncbi.nlm.nih.gov/sites/books/NBK545255/	Anatomy, Thorax, Superior Vena Cava - StatPearls - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Search term Anatomy, Thorax, Superior Vena Cava Hunter J. White; Michael P. Soos. Author Information and Affiliations Authors Hunter J. White 1; Michael P. Soos 2. Affiliations 1 Alabama College of Osteopathic Medicine 2 McLaren Greater Lansing Last Update: September 4, 2023. Go to: Introduction The superior vena cava (SVC) is a large, significant vein responsible for returning deoxygenated blood collected from the body to the right atrium. It is present within the superior and middle mediastinum. The superior vena cava handles the venous return of blood from structures located superior to the diaphragm. In contrast, its counterpart, the inferior vena cava, handles venous return from the portion of the body inferior to the diaphragm. The impedance of flow in the superior vena cava by thrombosis or external compression can cause hemodynamic compromise depending on the severity and acuity of the obstruction. Go to: Structure and Function The right and left brachiocephalic veins, also known as the innominate veins, join to form the superior vena cava posterior to the inferior border of the first right costal cartilage. These veins carry blood from the chest wall, esophagus, lungs, neck, and pericardium. The superior vena cava descends downwards vertically, running posterior to the second and third intercostal spaces. At the level of the right second costal cartilage, the SVC receives the azygos vein via the azygos arch, having looped over the right lung hilum from posterior to anterior to reach the SVC.At this point, the superior vena cava also penetrates the fibrous pericardium and continues within it until its termination. Within the superior mediastinum, the superior vena cava is related to the parietal pleura of the right lung. The right phrenic nerve runs directly laterally along its surface, and the right vagus nerve descends with the trachea posteromedially to the SVC.The right lung root is directly posterior. The SVC drains into the right atrium at the level of the third costal cartilage via a valveless orifice at the superior part of the sinus venarum, a smooth part of the posterior wall of the right atrium. The superior vena cava terminates in the superior portion of the sinus venarum, a quadrangular smooth-walled space posterior to the crista terminalis that receives both the superior and inferior vena cava. The orifice of the SVC lies directly posterolateral to the auricle of the right atrium. In adults, the superior vena cava is 7 cm long and approximately 2 cm wide. In developing children, studies have shown that the superior vena cava's size most closely corresponds with the child's height, reaching adult dimensions at approximately 10 to 12 years of age. Go to: Embryology The superior vena cava derives from the proximal portion of the right anterior cardinal vein and the right common cardinal vein at a point that is caudal to the transverse anastomosis in the embryo. During week eight of development, a large anastomosis formed via the thymic and thyroid veins shunts blood from the left anterior cardinal vein towards the right.Superior to this anastomosis, the anterior cardinal veins become the internal jugular veins. The cardinal veins are responsible for venous return; they represent the primary drainage system during embryonic development. The anterior cardinal veins drain the cephalic portion of the embryo, while the posterior cardinal veins drain the remaining embryonic blood. The anterior and posterior cardinal veins merge before entering the sinus venosus. At this point, they become the short common cardinal veins. The right brachiocephalic vein forms from the right anterior cardinal vein at a point that is cranial to the transverse anastomosis. The left brachiocephalic vein forms from a portion of the left anterior cardinal vein and transverse intercardinal anastomosis. During embryonic development, several malformations may occur, including left superior vena cava, double superior vena cava, as well as improper pulmonary venous drainage into the super vena cava. Go to: Blood Supply and Lymphatics The right and left brachiocephalic, or innominate, veins converge to form the superior vena cava at the level of the right first costal cartilage. These veins form at the venous angle from the subclavian and jugular veins merging posteriorly to the sternoclavicular joints. The brachiocephalic veins carry deoxygenated blood returning to the heart from the pericardium, lungs, esophagus, chest wall, and neck. The left brachiocephalic vein is more horizontal in its orientation and is longer than the more vertical and shorter right brachiocephalic vein. The right brachiocephalic vein is approximately 2 to 3 cm long and sits anterior to the innominate artery. The left brachiocephalic vein is approximately 6 cm long and lies anterior to the left subclavian vein and common carotid arteries, posterior to the manubrium, and superior to the aortic arch. The left superior intercostal vein joins with the left brachiocephalic vein; the azygos vein joins with the right brachiocephalic vein. The vertebral, internal thoracic, and inferior thyroid veins are all tributaries of the brachiocephalic veins. The azygos vein is another vein that drains into the superior vena cava. This vein is unpaired. It runs along the right aspect of the thoracic vertebral column and enters into the thorax at the aortic hiatus of the diaphragm. The azygos vein forms from the joining of the right subcostal and ascending lumbar veins. The hemiazygos veins, bronchial veins, pericardial veins, esophageal veins, and posterior right intercostal veins all drain into the azygos vein. Go to: Nerves Sympathetic axons originating from the right stellate ganglion descend along the superior vena cava. Most of these axons travel to the posterior surface of the heart. However, some extend onto the anterior side of the heart. These axons provide sympathetic innervation for the sinoatrial node. Cardiac parasympathetic nerves also run along the length of the superior vena cava. Efferent vagal nerve fibers travel through a pad of fat located in between the proximal aorta and medial superior vena cava. Go to: Muscles As with other arteries and veins, the superior vena cava is made up of three total layers of muscle. The innermost endothelial lining of the superior vena cava is known as the tunica intima. The middle layer, comprised of smooth muscle tissue, is known as the tunica media. This is the layer that receives input from the nervous system. The outermost layer is the tunica adventitia. It consists of collagen and elastic connective tissue. This outmost layer is primarily responsible for the superior vena cava’s flexibility. Go to: Physiologic Variants Persistent Left Superior Vena Cava (PLSVC) PLSVC is the anomalous presence of a persistent embryological left anterior cardinal vein that remains as a left superior vena cava and is present in approximately 0.3% of the population.In patients with congenital heart disease, its prevalence is increased, up to 10%.However, this anomaly may be undetected throughout life or only be diagnosed during cardiac investigation or intervention. When the PLSVC drains aberrantly into the left atrium, this can also be asymptomatic but, if causing significant right to left cardiac shunting, may present with cyanosis, respiratory distress, and progressive fatigue. Double Superior Vena Cava Double superior vena cava, or SVC duplication, is the commonest subset of PLSVC in which there remains a morphologically normal right superior vena cava to accompany the left superior vena cava.For this reason, given that the right superior vena cava will have normal drainage into the right atrium, it is even more likely to not be symptomatic during life and be identified only incidentally. Go to: Surgical Considerations Obstruction of the superior vena cava from a mediastinal mass or thrombus may require surgical intervention. Trauma to the thoracic region, damage from long-term central line catheterization, or damage from chronic use of pacing electrodes can also result in a need for surgical repair of the superior vena cava. Great care is necessary when clamping off the superior vena cava for surgical repair, as venous return from the head and neck will be impaired under these circumstances, and such impairment could potentially lead to poor patient outcomes.Access to the superior vena cava for open surgical bypass of obstruction can be achieved through a combination of a right neck incision and median sternotomy. Go to: Clinical Significance Complete or partial obstruction of the superior vena cava can occur, resulting in a condition known as superior vena cava syndrome. Causes of the obstruction may include intrinsic stenosis, extrinsic compression, or thrombosis of the vein. Malignancy is the single most common cause, with cancer of the lung, lymphoma, or metastatic cancers resulting in the majority of superior vena cava obstructions. The clinical presentation of superior vena cava syndrome may vary widely and depends on a variety of factors, including the severity, speed, and location of the obstruction. Patients with a slowly developing obstruction may show few or no symptoms, as collateral veins from the azygos system compensate to some degree.The location of these collateral channels depends upon the location of the blockage. If the obstruction occurs in the area of the superior vena cava located before the azygos vein, the right superior intercostal veins will serve as the primary collateral pathway for azygos vein drainage. This pathway is often the most asymptomatic. If the obstruction occurs at the azygos vein, collateral channels will form between the superior vena cava and inferior vena cava via the internal mammary veins, superior and inferior epigastric veins, and the iliac vein. If the obstruction occurs past the point at which the azygos vein drains into the superior vena cava, the blood will move via channels from the azygos and hemiazygos veins into the ascending lumbar and lumbar veins, which will drain into the inferior vena cava and allow return into the right atrium. When the superior vena cava suffers more acute obstruction, it can result in upstream effects causing tributary veins in the head and neck to appear enlarged. It can also result in clinical symptomology, including shortness of breath, cough, angina, facial flushing, headache, bilateral upper extremity swelling, and dysphagia. Patients with an obstructed superior vena cava may have a positive Pemberton sign - described as a facial plethora and venous engorgement upon raising both arms above the head.Neurological symptoms such as syncope, dizziness, and confusion are possible but less common presenting signs. Treatment of SVC syndrome depends on the cause. In the context of cancer, chemotherapy or radiation are options, as well as surgical resection of the mass if possible. In these cases, endovascular stenting may also be trialed however, although providing good short-term outcomes, it has been shown to likely require multiple repeat future interventions.In the event of a thrombus, anticoagulation may be indicated to prevent further clot formation. Stent placement is considered a first-line treatment option here. CT is the imaging modality of choice for diagnosing a superior vena cava obstruction. Plain radiographs may also be used to detect indirect signs of a mediastinal mass but do not provide sufficient 3D anatomical accuracy for treatment planning. Go to: Review Questions Access free multiple choice questions on this topic. Comment on this article. Figure Anatomy of the Heart. This illustration shows the anatomic relationships between the pulmonary valve, the anterior cusp of the tricuspid valve, chordae tendineae, papillary muscles, the valve of the coronary sinus, the valve of inferior vena cava, the (more...) Figure Veins and Arteries of the Neck. Superior vena cava, left and right innominate vein, left and right subclavian vein, left and right internal jugular vein, innominate arteries, hyoid bone, thyroid gland, and trachea. Henry Vandyke Carter, Public Domain, (more...) Figure Superior vena cava Image courtesy O.Chaigasame Figure Superficial chest wall collateral venous engorgement secondary to superior vena cava obstruction Contributed By EMAHkempny - Own work, CC BY-SA 4.0, Go to: References 1. Nascimbene A, Angelini P. Superior vena cava thrombosis and paradoxical embolic stroke due to collateral drainage from the brachiocephalic vein to the left atrium. Tex Heart Inst J. 2011;38(2):170-3. [PMC free article: PMC3066820] [PubMed: 21494530] 2. Piciucchi S, Barone D, Sanna S, Dubini A, Goodman LR, Oboldi D, Bertocco M, Ciccotosto C, Gavelli G, Carloni A, Poletti V. The azygos vein pathway: an overview from anatomical variations to pathological changes. Insights Imaging. 2014 Oct;5(5):619-28. [PMC free article: PMC4195836] [PubMed: 25171956] 3. Fell SC. Surgical anatomy of the diaphragm and the phrenic nerve. Chest Surg Clin N Am. 1998 May;8(2):281-94. [PubMed: 9619305] 4. Furlow PW, Mathisen DJ. Surgical anatomy of the trachea. Ann Cardiothorac Surg. 2018 Mar;7(2):255-260. [PMC free article: PMC5900092] [PubMed: 29707503] 5. McKay T, Thomas L. Prominent crista terminalis and Eustachian ridge in the right atrium: Two dimensional (2D) and three dimensional (3D) imaging. Eur J Echocardiogr. 2007 Aug;8(4):288-91. [PubMed: 16621718] 6. Sanjeev S, Karpawich PP. Superior vena cava and innominate vein dimensions in growing children : an aid for interventional devices and transvenous leads. Pediatr Cardiol. 2006 Jul-Aug;27(4):414-9. [PubMed: 16830087] 7. Nadesan T, Keough N, Suleman FE, Lockhat Z, van Schoor AN. Apprasial of the surface anatomy of the Thorax in an adolescent population. Clin Anat. 2019 Sep;32(6):762-769. [PubMed: 30758865] 8. Vuillemin M, Pexieder T. Normal stages of cardiac organogenesis in the mouse: II. Development of the internal relief of the heart. Am J Anat. 1989 Feb;184(2):114-28. [PubMed: 2712003] 9. Sondermeijer BM, Macgillavry MR, Tan HL. Left superior vena cava, a remnant of embryological development. Neth Heart J. 2008 May;16(5):173-4. [PMC free article: PMC2431163] [PubMed: 18566698] 10. Schauerte P, Mischke K, Plisiene J, Waldmann M, Zarse M, Stellbrink C, Schimpf T, Knackstedt C, Sinha A, Hanrath P. Catheter stimulation of cardiac parasympathetic nerves in humans: a novel approach to the cardiac autonomic nervous system. Circulation. 2001 Nov 13;104(20):2430-5. [PubMed: 11705820] 11. Qin M, Zhang Y, Liu X, Jiang WF, Wu SH, Po S. Atrial Ganglionated Plexus Modification: A Novel Approach to Treat Symptomatic Sinus Bradycardia. JACC Clin Electrophysiol. 2017 Sep;3(9):950-959. [PubMed: 29759719] 12. Manousiouthakis E, Mendez M, Garner MC, Exertier P, Makita T. Venous endothelin guides sympathetic innervation of the developing mouse heart. Nat Commun. 2014 May 29;5:3918. [PMC free article: PMC4080092] [PubMed: 24875861] 13. Phillippi JA. On vasa vasorum: A history of advances in understanding the vessels of vessels. Sci Adv. 2022 Apr 22;8(16):eabl6364. [PMC free article: PMC9020663] [PubMed: 35442731] 14. Gustapane S, Leombroni M, Khalil A, Giacci F, Marrone L, Bascietto F, Rizzo G, Acharya G, Liberati M, D'Antonio F. Systematic review and meta-analysis of persistent left superior vena cava on prenatal ultrasound: associated anomalies, diagnostic accuracy and postnatal outcome. Ultrasound Obstet Gynecol. 2016 Dec;48(6):701-708. [PubMed: 26970258] 15. Batouty NM, Sobh DM, Gadelhak B, Sobh HM, Mahmoud W, Tawfik AM. Left superior vena cava: cross-sectional imaging overview. Radiol Med. 2020 Mar;125(3):237-246. [PubMed: 31823296] 16. Azizova A, Onder O, Arslan S, Ardali S, Hazirolan T. Persistent left superior vena cava: clinical importance and differential diagnoses. Insights Imaging. 2020 Oct 15;11(1):110. [PMC free article: PMC7561662] [PubMed: 33057803] 17. Demos TC, Posniak HV, Pierce KL, Olson MC, Muscato M. Venous anomalies of the thorax. AJR Am J Roentgenol. 2004 May;182(5):1139-50. [PubMed: 15100109] 18. Albay S, Cankal F, Kocabiyik N, Yalcin B, Ozan H. Double superior vena cava. Morphologie. 2006 Mar;90(288):39-42. [PubMed: 16929820] 19. Farazi-Chongouki C, Dalianoudis I, Ninos A, Diamantopoulos P, Filippou D, Pierrakakis S, Skandalakis P. Double superior vena cava: presentation of two cases and review of the literature. Acta Chir Belg. 2019 Oct;119(5):316-321. [PubMed: 29458311] 20. Raut MS, Das S, Sharma R, Daniel E, Motihar A, Verma A, Kar S, Maheshwari A, Shivnani G, Kumar A. Superior vena cava clamping during thoracic surgery: Implications for the anesthesiologist. Ann Card Anaesth. 2018 Jan-Mar;21(1):85-87. [PMC free article: PMC5791501] [PubMed: 29336403] 21. De Raet JM, Vos JA, Morshuis WJ, van Boven WJ. Surgical management of superior vena cava syndrome after failed endovascular stenting. Interact Cardiovasc Thorac Surg. 2012 Nov;15(5):915-7. [PMC free article: PMC3480597] [PubMed: 22843656] 22. Wilson LD, Detterbeck FC, Yahalom J. Clinical practice. Superior vena cava syndrome with malignant causes. N Engl J Med. 2007 May 03;356(18):1862-9. [PubMed: 17476012] 23. Yu JB, Wilson LD, Detterbeck FC. Superior vena cava syndrome--a proposed classification system and algorithm for management. J Thorac Oncol. 2008 Aug;3(8):811-4. [PubMed: 18670297] 24. Katabathina VS, Restrepo CS, Betancourt Cuellar SL, Riascos RF, Menias CO. Imaging of oncologic emergencies: what every radiologist should know. Radiographics. 2013 Oct;33(6):1533-53. [PubMed: 24108550] 25. Sheth S, Ebert MD, Fishman EK. Superior vena cava obstruction evaluation with MDCT. AJR Am J Roentgenol. 2010 Apr;194(4):W336-46. [PubMed: 20308479] 26. Keshvani N, Yek C, Johnson DH. Pemberton's sign in SVC syndrome from metastatic renal cell carcinoma. BMJ Case Rep. 2018 Mar 28;2018 [PMC free article: PMC5878269] [PubMed: 29593001] 27. Kalra M, Gloviczki P, Andrews JC, Cherry KJ, Bower TC, Panneton JM, Bjarnason H, Noel AA, Schleck C, Harmsen WS, Canton LG, Pairolero PC. Open surgical and endovascular treatment of superior vena cava syndrome caused by nonmalignant disease. J Vasc Surg. 2003 Aug;38(2):215-23. [PubMed: 12891100] 28. Niazi AK, Reese AS, Minko P, O'Donoghue D, Ayad S. Superior Vena Cava Syndrome and Otorrhagia During Cardiac Surgery. Cureus. 2019 May 05;11(5):e4602. [PMC free article: PMC6609288] [PubMed: 31309025] 29. Nunnelee JD. Superior vena cava syndrome. J Vasc Nurs. 2007 Mar;25(1):2-5; quiz 6. [PubMed: 17324762] 30. Fagedet D, Thony F, Timsit JF, Rodiere M, Monnin-Bares V, Ferretti GR, Vesin A, Moro-Sibilot D. Endovascular treatment of malignant superior vena cava syndrome: results and predictive factors of clinical efficacy. Cardiovasc Intervent Radiol. 2013 Feb;36(1):140-9. [PubMed: 22146975] Disclosure:Hunter White declares no relevant financial relationships with ineligible companies. Disclosure:Michael Soos declares no relevant financial relationships with ineligible companies. Introduction Structure and Function Embryology Blood Supply and Lymphatics Nerves Muscles Physiologic Variants Surgical Considerations Clinical Significance Review Questions References 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: NBK545255 PMID: 31424839 Share on Facebook Share on Twitter Views PubReader Print View Cite this Page In this Page Introduction Structure and Function Embryology Blood Supply and Lymphatics Nerves Muscles Physiologic Variants Surgical Considerations Clinical Significance Review Questions References Related information PMCPubMed Central citations PubMedLinks to PubMed Similar articles in PubMed Scimitar syndrome with an accessory diaphragm and an absent right superior vena cava.[Jpn J Surg. 1986]Scimitar syndrome with an accessory diaphragm and an absent right superior vena cava.Kimura M, Asao M, Kawano Y, Inakazu T, Hamamoto K, Oda T. Jpn J Surg. 1986 Jul; 16(4):284-7. Review Unique Medley of Cardinal Veins: Duplicated Superior and Inferior Venae Cavae With Left Renal Agenesis and Hemiazygos Continuation of Left Inferior Vena Cava With Drainage Into Left Atrium.[Vasc Endovascular Surg. 2022]Review Unique Medley of Cardinal Veins: Duplicated Superior and Inferior Venae Cavae With Left Renal Agenesis and Hemiazygos Continuation of Left Inferior Vena Cava With Drainage Into Left Atrium.Vignesh S, Bhat TA. Vasc Endovascular Surg. 2022 Apr; 56(3):330-334. Epub 2022 Feb 7. The superior vena cava syndrome caused by malignant disease. Imaging with multi-detector row CT.[Eur J Radiol. 2006]The superior vena cava syndrome caused by malignant disease. Imaging with multi-detector row CT.Eren S, Karaman A, Okur A. Eur J Radiol. 2006 Jul; 59(1):93-103. Epub 2006 Feb 14. Superior vena cava obstruction: a venographic classification.[AJR Am J Roentgenol. 1987]Superior vena cava obstruction: a venographic classification.Stanford W, Jolles H, Ell S, Chiu LC. AJR Am J Roentgenol. 1987 Feb; 148(2):259-62. Review Epidural venous plexus enlargements presenting with radiculopathy and back pain in patients with inferior vena cava obstruction or occlusion.[Spine (Phila Pa 1976). 2004]Review Epidural venous plexus enlargements presenting with radiculopathy and back pain in patients with inferior vena cava obstruction or occlusion.Paksoy Y, Gormus N. Spine (Phila Pa 1976). 2004 Nov 1; 29(21):2419-24. See reviews...See all... Recent Activity Clear)Turn Off)Turn On) Anatomy, Thorax, Superior Vena Cava - StatPearlsAnatomy, Thorax, Superior Vena Cava - StatPearls Your browsing activity is empty. Activity recording is turned off. Turn recording back on) See more... Follow NCBI Connect with NLM National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov PreferencesTurn off External link. Please review our privacy policy. Cite this Page Close White HJ, Soos MP. Anatomy, Thorax, Superior Vena Cava. [Updated 2023 Sep 4]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Available from: Making content easier to read in Bookshelf Close We are experimenting with display styles that make it easier to read books and documents in Bookshelf. Our first effort uses ebook readers, which have several "ease of reading" features already built in. The content is best viewed in the iBooks reader. You may notice problems with the display of some features of books or documents in other eReaders. Cancel Download Share Share on Facebook Share on Twitter URL
16227	https://english.stackexchange.com/questions/373758/a-perfect-use-of-a-past-participle-without-the-auxiliary-verb-have	A perfect use of a past participle without the auxiliary verb "have" - English Language & Usage Stack Exchange Join English Language & Usage By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more A perfect use of a past participle without the auxiliary verb "have" Ask Question Asked 8 years, 7 months ago Modified8 years, 7 months ago Viewed 5k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. An article titled "Hulk Hogan lawyer tells SN: Gawker sex tape a 'massive' invasion of wrestler's privacy" has this: Hogan's $100 million lawsuit against Gawker for posting a private video of him having sex with a friend's wife heads to trial at a Florida state court March 7. If Hogan (aka Terry Bollea) wins, Gawker could be ruined financially — or forced to sell. Now fallen on hard times after his WWE glory days, Hogan is looking forward to his day in court with Florida jurors who might take a jaundiced eye toward a New York media outlet that traffics in gossip and rumors. In the last sentence, the past participle "fallen" is a perfect use, and I think you can start the sentence with "having" as follows: Having now fallen on hard times after his WWE glory days, Hogan is looking forward to... Am I right? Is this dropping of "having" commonplace? Can such a dropping be done to other verbs than "fall"? If it can, I'd like to have some examples of a perfect use of a past participle (other than "fall") without the auxiliary verb "have". past-participles perfect-aspect Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited Jun 15, 2020 at 7:40 CommunityBot 1 asked Feb 15, 2017 at 4:55 JK2JK2 7,751 5 5 gold badges 55 55 silver badges 120 120 bronze badges 3 That is not necessarily the "dropping" of having. It can be fallen used simply as an adjective. Compare the gone girl is still missing. There's no "dropped" having from that.Arm the good guys in America –Arm the good guys in America 2017-02-15 05:52:48 +00:00 Commented Feb 15, 2017 at 5:52 @Clare "Fall on hard times" is a verb phrase, as shown here: dictionary.cambridge.org/us/dictionary/english/… So, I wonder how you could claim that "fallen" is not a verb (past participle) but an adjective.JK2 –JK2 2017-02-15 06:07:17 +00:00 Commented Feb 15, 2017 at 6:07 2 Participles (present or past) as a word, phrase and clause can serve as adjectives; nothing wrong in it. The gerunds (-ing form of verbs) serve as nouns, don't they?mahmud k pukayoor –mahmud k pukayoor 2017-02-15 07:10:37 +00:00 Commented Feb 15, 2017 at 7:10 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Let's look at a similar pair: Now gorged on my blood, the mosquito stopped buzzing around my ear and settled on the wall. This means now stuffed, the mosquito etc. Having just gorged on my blood, the mosquito buzzed more excitedly around my ear. This means that the mosquito just finished biting me (perhaps more than once), and now is buzzing more excitedly. Do you see? The meaning is slightly different. Or maybe I should say the point of view. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Feb 15, 2017 at 6:39 aparente001aparente001 21.7k 13 13 gold badges 59 59 silver badges 94 94 bronze badges 2 1 Is your first example natural English? If so, I need you to answer my first question as to whether the dropping of "having" is commonplace and can be done to any verb. If not, how do I know if the dropping works for a certain verb or not?JK2 –JK2 2017-02-15 16:42:38 +00:00 Commented Feb 15, 2017 at 16:42 1 @JK2 - Well, I don't see it as dropping the having. Let's look at another example: "Having eaten my fill, I gave Mary my full attention." It would not work to say, "Eaten, I gave Mary my full attention." I see that according to the way you've been looking at this, it feels like "having" got dropped, but that's really not what's going on.aparente001 –aparente001 2017-02-16 19:14:16 +00:00 Commented Feb 16, 2017 at 19:14 Add a comment\| This answer is useful -2 Save this answer. Show activity on this post. From GrammarBank.com ( Participle Clauses are used to shorten sentences. Past Participle Contrary to popular misuse, a past participle doesn't have a past meaning; but instead, it has a similar usage to present participle but in passive form. The little girl was taken to the nearest hospital after she was attacked by a dog. Attacked by a dog, the little girl was taken to the nearest hospital. The museum, which was built in 1953, needs renovation. Built in 1953, the museum needs renovation. The new night club, which is located on the beach side, attracts the attention of all ages. Located on the beach side, the new night club attracts the attention of all ages. Perfect Participle Indicates an action that happens long before the action in the main clause. After he had spent ten years in Italy, he could speak Italian fluently. Having spent ten years in Italy, he could speak Italian fluently. Because Tom had attended this course before, he knew what to expect. Having attended this course before, Tom knew what to expect. Note: to get passive form in perfect participle, we add "been" after "having". Because he had been fired, he didn't attend the meeting. Having been fired, he didn't attend the meeting. Because he hadn't been invited to the wedding, he didn't come. Not having been invited to the wedding, he didn't come. The following example sentences are from Michael Swan's Practical English Usage - (Participles): Served with milk and sugar, it makes a delicious breakfast. Used economically, one tin will last for six weeks. Hands held high, the dancers circle to the right. Rejected by all his friends, he decided to become a monk. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Feb 15, 2017 at 6:56 mahmud k pukayoormahmud k pukayoor 8,559 5 5 gold badges 22 22 silver badges 39 39 bronze badges 1 2 What's the point of listing all these examples that seem irrelevant to the question?JK2 –JK2 2017-02-15 16:44:55 +00:00 Commented Feb 15, 2017 at 16:44 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions past-participles perfect-aspect See similar questions with these tags. 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16228	https://pmc.ncbi.nlm.nih.gov/articles/PMC6686151/	Revisiting Trade-offs between Rubisco Kinetic Parameters - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer \| PMC Copyright Notice Biochemistry . 2019 Jul 1;58(31):3365–3376. doi: 10.1021/acs.biochem.9b00237 Search in PMC Search in PubMed View in NLM Catalog Add to search Revisiting Trade-offs between Rubisco Kinetic Parameters Avi I Flamholz Avi I Flamholz †Department of Molecular and Cell Biology, University of California, Berkeley, California 94720, United States Find articles by Avi I Flamholz †, Noam Prywes Noam Prywes ‡Innovative Genomics Institute, University of California, Berkeley, California 94704, United States Find articles by Noam Prywes ‡, Uri Moran Uri Moran §Department of Plant and Environmental Sciences, Weizmann Institute of Science, Rehovot 76100, Israel Find articles by Uri Moran §, Dan Davidi Dan Davidi §Department of Plant and Environmental Sciences, Weizmann Institute of Science, Rehovot 76100, Israel Find articles by Dan Davidi §, Yinon M Bar-On Yinon M Bar-On §Department of Plant and Environmental Sciences, Weizmann Institute of Science, Rehovot 76100, Israel Find articles by Yinon M Bar-On §, Luke M Oltrogge Luke M Oltrogge †Department of Molecular and Cell Biology, University of California, Berkeley, California 94720, United States Find articles by Luke M Oltrogge †, Rui Alves Rui Alves ∥Institute of Biomedical Research of Lleida, IRBLleida, 25198 Lleida, Catalunya, Spain ⊥Departament de Ciències Mèdiques Bàsiques, University of Lleida, 25198 Lleida, Catalunya, Spain Find articles by Rui Alves ∥,⊥, David Savage David Savage †Department of Molecular and Cell Biology, University of California, Berkeley, California 94720, United States Find articles by David Savage †, Ron Milo Ron Milo §Department of Plant and Environmental Sciences, Weizmann Institute of Science, Rehovot 76100, Israel Find articles by Ron Milo §, Author information Article notes Copyright and License information †Department of Molecular and Cell Biology, University of California, Berkeley, California 94720, United States ‡Innovative Genomics Institute, University of California, Berkeley, California 94704, United States §Department of Plant and Environmental Sciences, Weizmann Institute of Science, Rehovot 76100, Israel ∥Institute of Biomedical Research of Lleida, IRBLleida, 25198 Lleida, Catalunya, Spain ⊥Departament de Ciències Mèdiques Bàsiques, University of Lleida, 25198 Lleida, Catalunya, Spain E-mail: ron.milo@weizmann.ac.il. Received 2019 Mar 18; Revised 2019 Jun 28; Issue date 2019 Aug 6. Copyright © 2019 American Chemical Society This is an open access article published under an ACS AuthorChoice License, which permits copying and redistribution of the article or any adaptations for non-commercial purposes. PMC Copyright notice PMCID: PMC6686151 PMID: 31259528 Abstract Rubisco is the primary carboxylase of the Calvin cycle, the most abundant enzyme in the biosphere, and one of the best-characterized enzymes. On the basis of correlations between Rubisco kinetic parameters, it is widely posited that constraints embedded in the catalytic mechanism enforce trade-offs between CO 2 specificity, S C/O, and maximum carboxylation rate, k cat,C. However, the reasoning that established this view was based on data from ≈20 organisms. Here, we re-examine models of trade-offs in Rubisco catalysis using a data set from ≈300 organisms. Correlations between kinetic parameters are substantially attenuated in this larger data set, with the inverse relationship between k cat,C and S C/O being a key example. Nonetheless, measured kinetic parameters display extremely limited variation, consistent with a view of Rubisco as a highly constrained enzyme. More than 95% of k cat,C values are between 1 and 10 s–1, and no measured k cat,C exceeds 15 s–1. Similarly, S C/O varies by only 30% among Form I Rubiscos and <10% among C 3 plant enzymes. Limited variation in S C/O forces a strong positive correlation between the catalytic efficiencies (k cat/K M) for carboxylation and oxygenation, consistent with a model of Rubisco catalysis in which increasing the rate of addition of CO 2 to the enzyme–substrate complex requires an equal increase in the O 2 addition rate. Altogether, these data suggest that Rubisco evolution is tightly constrained by the physicochemical limits of CO 2/O 2 discrimination. Ribulose-1,5-bisphosphate carboxylase/oxygenase (Rubisco) is the primary carboxylase of the Calvin–Benson–Bassham (CBB) cycle, the carbon fixation cycle that is responsible for growth throughout the green lineage and many other autotrophic taxa, and the ultimate source of nearly all carbon atoms entering the biosphere.1 Typically, 20–30% of total soluble protein in C 3 plant leaves is Rubisco.2 As Rubisco is so strongly expressed and plants are the dominant constituents of planetary biomass,3 it is often said that Rubisco is the most abundant enzyme on Earth.1,4 Because Rubisco is ancient (>2.5 billion years old) and abundant and remains central to biology, one might expect it to be exceptionally fast, but Rubisco is not fast.5−8 Typical central metabolic enzymes have a turnover number (k cat) of ≈80 s–1.7 However, >95% of Rubisco carboxylation k cat,C values are between 1 and 10 s–1, and no measured k cat,C values exceed 15 s–1. In addition to relatively low k cat,C values, Rubisco reacts with O 2 in a process called oxygenation (Figure1A). Although both carboxylation and oxygenation of the five-carbon substrate ribulose 1,5-bisphosphate (RuBP) are energetically favorable,9 carboxylation is the productive reaction for incorporating carbon from CO 2 into precursors that generate biomass (Figure1B). While it may play a role in sulfur, nitrogen, and energy metabolism,10,11 oxygenation is often considered counterproductive as it occupies Rubisco active sites and yields a product, 2-phosphoglycolate (2PG), that is not part of the CBB cycle and must be recycled through metabolically expensive photorespiration at a partial loss of carbon.12 As such, oxygenation can substantially reduce the net rate of carboxylation by Rubisco, depending on CO 2 and O 2 concentrations and the kinetic parameters of the particular enzyme. There are at least four distinct Rubisco isoforms in nature,13 but all isoforms catalyze carboxylation and oxygenation of RuBP through the multistep mechanism described in panels A and C of Figure1.14,15 Even though many autotrophs depend on Rubisco carboxylation for growth, all known Rubiscos are relatively slow carboxylases and fail to exclude oxygenation. Figure 1. Open in a new tab Description of the catalytic mechanism of Rubisco. The “middle-out” diagram in panel A shows the ordered mechanisms of carboxylation and oxygenation. Circles represent carbon atoms. RuBP is isomerized to an enediolate before carboxylation or oxygenation. Addition of CO 2 or O 2 to the enediolate of RuBP is considered irreversible as are the subsequent hydration and cleavage steps of the carboxylation and oxygenation arms. (B) Carboxylation displays effective Michaelis–Menten kinetics (maximum catalytic rate k cat,C, half-maximum CO 2 concentration K M = K C) with competitive inhibition by O 2 (assuming half-maximum inhibitory O 2 concentration K i = K O). Carboxylation results in net addition of one carbon to the five-carbon RuBP, producing two 3PG molecules. 3PG is part of the CBB cycle and can therefore be used to continue the cycle and produce biomass. Oxygenation also displays effective Michaelis–Menten kinetics (k cat,O, K M = K O, half-maximum inhibitory CO 2 concentration K I = K C). Oxygenation of RuBP produces one 3PG and one 2PG. Rates of carboxylation (R C) and oxygenation (R O) are calculated from kinetic parameters and the CO 2 and O 2 concentrations. The reaction coordinate diagram in panel C describes carboxylation and oxygenation as a function of two “effective” barriers.6 The first effective barrier includes enolization and gas addition, while the second includes hydration and cleavage. (D) Given standard assumptions (Supporting Information), catalytic efficiencies (k cat/K M) are related to the height of the first effective barrier while k cat s are related to the second. The first barrier to oxygenation is drawn higher than for carboxylation because oxygenation is typically slower than carboxylation. Net reactions of RuBP carboxylation and oxygenation are both quite thermodynamically favorable.9 The fastest-carboxylating Rubisco observed (at 25 °C) is from the cyanobacterium Synechococcus elongatus PCC 7942.16 This enzyme has a maximum per-active site carboxylation rate (k cat,C) of 14 s–1. However, because the present-day atmosphere contains abundant O 2 and relatively little CO 2 (≈21% O 2 and ≈0.04% CO 2), PCC 7942 Rubisco carboxylates at a rate 20-fold below maximum under ambient conditions [R C ≈ 0.7 s–1 per active site (rate law in Figure1A)]. Due to its relatively low CO 2 specificity, PCC 7942 Rubisco will also oxygenate RuBP appreciably under ambient conditions (R O ≈ 0.3 s–1), necessitating substantial photorespiratory flux to recycle 2PG. As downstream processing of 2PG by the C 2 photorespiratory pathway leads to the loss of one carbon for every two 2PGs,11,12 every two oxygenations “undo” a carboxylation. In ambient air, therefore, the net rate of carboxylation by PCC 7942 Rubisco would be f = R C – R O/2 ≈ 0.6 s–1, or ≈4% of k cat,C. Given the kinetics of PCC 7942 Rubisco, it is not surprising that all known cyanobacteria use a CO 2-concentrating mechanism to ensure Rubisco functions in a CO 2-rich environment. An elevated level of CO 2 ensures that oxygenation is competitively inhibited and that carboxylation proceeds at a near-maximum rate.17 Thirtyfold enrichment of CO 2 above ambient increases the carboxylation rate of PCC 7942 Rubisco to 8.9 s–1 and suppresses the oxygenation rate to 0.14 s–1, giving a net carboxylation rate f = 8.8 s–1 per active site (≈60% of k cat,C). For comparison, the Rubisco from spinach leaves (Spinacia oleracea) is characteristic of plant Rubiscos in having a lower k cat,C of ≈3 s–1 and a CO 2 affinity much greater than that of the S. elongatus enzyme (spinach half-maximum CO 2 concentration K C of ≈12 μM, PCC 7942 K C of ≈170 μM). As a result, the spinach enzyme outperforms the cyanobacterial one in ambient air, with an R C of ≈1.2 s–1, an R O of ≈0.4 s–1, and a net carboxylation rate f ≈1 s–1. Spinach is a C 3 plant, meaning it does not have a CO 2-concentrating mechanism, which may explain why it employs a slow-but-specific Rubisco. Still, most central metabolic enzymes catalyze far more than one reaction per second,7 leading many to wonder if Rubisco catalysis could be improved. Improved Rubisco carboxylation might increase C 3 crop yields,18,19 but a substantially improved enzyme has evaded bioengineers for decades.20 The repeated evolution of diverse CO 2-concentrating mechanisms, which modulate the catalytic environment rather than Rubisco itself, raises further doubts about whether Rubisco catalysis can be strictly improved.21 Various nomenclatures have been used to describe the kinetics of Rubisco carboxylation and oxygenation since its discovery in the 1950s.5,6,22,23 Here we use k cat,C and k cat,O to denote turnover numbers (maximum rates per active site, units of inverse seconds) for carboxylation and oxygenation, respectively. K C and K O denote the Michaelis constants (half-saturation concentrations in micromolar) for carboxylation and oxygenation. Specificity factor S C/O = (k cat,C/K C)/(k cat,O/K O) is a unitless measure of the relative preference for CO 2 over O 2 (Figure1D). Because S C/O relates only to the ratio of kinetic parameters, a higher S C/O does not necessarily imply higher carboxylation rates. Rather, absolute carboxylation and oxygenation rates depend on CO 2 and O 2 concentrations (Figure1B), which can vary between organisms and environments (Supporting Information). As data on bacterial, archaeal, and plant carboxylases have accumulated over the decades, many researchers have noted that fast-carboxylating Rubiscos are typically less CO 2-specific.24−26 In other words, Rubiscos with high k cat,C values were observed to have lower S C/O values due to either a lower CO 2 affinity (high K C) or more efficient oxygenation (higher k cat,O/K O). A negative correlation between k cat,C and S C/O is often cited to motivate the idea that a trade-off between carboxylation rate and specificity constrains Rubisco evolution.5,6,26,27 It is worth pausing to clarify the concepts of “trade-off”, “constraint”, and “correlation” (Figure2). Correlation indicates an apparent linear (or log–log, etc.) relationship between two kinetic parameters. Correlations between enzyme kinetic parameters can result from a “trade-off” due to two distinct kinds of underlying constraints (Figure2).6 In the “mechanistic coupling” scenario, the enzymatic mechanism forces a strict quantitative relationship between two kinetic parameters such that varying one forces the other to vary in a defined manner. This results in a situation in which the value of one parameter strictly determines the other and vice versa [i.e., an “equality constraint” (Figure2A)]. This could arise for Rubisco, for example, if a single catalytic step [e.g., enolization of RuBP (Figure1A)] determines the rates of both CO 2 and O 2 entry. Figure 2. Open in a new tab Scenarios that produce strong correlations between enzyme kinetic parameters. As the logs of the kinetic parameters are linearly related to energy barriers, linear energetic trade-offs should manifest as log–log correlations between kinetic parameters (power laws). Panel A describes a situation in which two kinetic parameters are inextricably linked by the enzyme mechanism, diagrammed here as negative coupling between k cat,C and S C/O as an example. These couplings take the form of “equality constraints” in which one parameter determines the other within measurement error. Correlation is expected as long as diverse enzymes are measured. In panel A, selection moves enzymes along the blue curve but cannot produce enzymes off the curve (gray) because they are not feasible. Panel B diagrams an alternative scenario in which the enzyme mechanism imposes an upper limit on two parameters (an inequality constraint). In the “selection within limits” scenario, effective selection is required for correlation to emerge because suboptimal enzymes (e.g., ancestral sequences) are feasible. In the examples plotted, different environmental CO 2 and O 2 concentrations should select for different combinations of rate (k cat,C) and affinity (S C/O), resulting in present-day enzymes occupying distinct regions of the plots in panels A and B. In the “selection within limits” model (Figure2B), in contrast, the catalytic mechanism imposes an upper limit on kinetic parameters, i.e., an inequality constraint. A clear correlation between parameters will emerge only if there is sufficient selection to reach the boundary. To highlight the difference between these models, consider the kinetics of ancestral enzymes. In the “mechanistic coupling” model, kinetic parameters of ancestors should lie along the same curve as present-day enzymes because the gray regions off the curve are disallowed. Selection could act by moving enzymes along the line of mechanistic coupling, e.g., from a region of high selectivity and low rate toward higher rate and lower selectivity (Figure2A). According to the “selection within limits” model, in contrast, ancestral enzymes can lie beneath the upper limit determined by the catalytic mechanism (Figure2B). This second model requires selection to produce a situation in which the kinetics of present-day Rubiscos extracted from various organisms trace out a curve determined by the upper limit enforced by the mechanism.28 Previous research advanced two distinct families of mechanistic models to explain correlations between Rubisco kinetic parameters.5,6 The first model, which we term “k cat,C–K C coupling”, hypothesizes a trade-off between the rate and affinity of carboxylation that leads to a negative correlation between k cat,C and S C/O (Figure S2).5 A second model, which was advanced in a study in which the last author of this work participated,6 hypothesizes that multiple trade-offs constrain Rubisco such that kinetic parameters can vary only along a one-dimensional curve. In addition to k cat,C–K C coupling, this work hypothesized a trade-off between catalytic efficiencies for carboxylation and oxygenation (coupling k cat,C/K C and k cat,O/K O) wherein improving carboxylation efficiency also improves oxygenation efficiency. These mechanistic models are substantively different. Though both models imply limitations on the concurrent improvement of k cat,C and S C/O, “k cat,C–K C coupling” relates only to carboxylation kinetics, leaving the possibility that oxygenation kinetics are unconstrained. Coupling between k cat,C/K C and k cat,O/K O, in contrast, relates to both reaction pathways. While these models appeal to physical and chemical intuition, they are based on data from only ≈20 organisms. Moreover, “mechanistic coupling” and “selection within limits” could plausibly underlie either model (Figure2).6 Here we take advantage of the accumulation of new data to revisit correlations and trade-offs between Rubisco kinetic parameters. We collected and curated literature measurements of ≈300 Rubiscos. Though diverse organisms are represented, the Form I Rubiscos of C 3 plants make up the bulk of the data [>80% (Figure3A)]. Most previously reported correlations between Rubisco kinetic parameters are substantially attenuated in this data set, with the negative correlation between k cat,C and specificity S C/O being a key example. Weakened k cat,C–S C/O and k cat,C–K C correlations imply that these parameters are not straightforwardly mechanistically coupled, suggesting that models of k cat,C–K C coupling should be revisited in future experiments. Overall, weakened correlations call into question previous claims that (i) Rubisco kinetics are constrained to evolve on a one-dimensional line and (ii) natural Rubiscos are optimized to suit environmental CO 2 and O 2 concentrations.5,6 Figure 3. Open in a new tab Summary of the full extended data set. We collected measurements of Rubisco kinetic parameters from a variety of organisms (A) representing four classes of Rubisco isoforms (B). Form I enzymes from plants, cyanobacteria, and algae make up the bulk of the data (A and B). (C) Rubisco kinetic parameters display a narrow dynamic range. The box plot and gray points describe the distribution of Form I Rubiscos, while data for Form II Rubiscos are colored yellow. Colored boxes give the range of the central 50% of FI values, and the notch indicates the median. N is the number values, and σ gives the geometric standard deviation of Form I data. σ < 3 for all parameters, meaning a single standard deviation varies by <3-fold. All data are from wild-type Rubiscos measured at 25 °C and near pH 8. More detailed histograms are given in Figure S4. Despite weakened correlations, Rubisco kinetic parameters display extremely limited variation. k cat,C varies by only 50% among Form I Rubiscos, and S C/O varies even less than that [≈30% (Figure3C)]. Limited variation in S C/O forces a strong positive power-law correlation between the catalytic efficiencies for carboxylation (k cat,C/K C) and oxygenation (k cat,O/K O).6 We propose a simple model of mechanistic coupling that explains how constraints on the Rubisco mechanism could restrict variation in S C/O. In this model, variation in catalytic efficiency (k cat,C/K C and k cat,O/K O) derives solely from gating access of the substrate to the active site complex, which could help explain why Rubisco has been so recalcitrant to improvement by mutagenesis and rational engineering. Materials and Methods Data Collection and Curation We reviewed the literature to find Rubisco kinetic data measured at 25 °C and near pH 8. Ultimately, 61 primary literature studies were included, yielding 335 S C/O, 284 k cat,C, 316 K C, and 254 K o values for Rubiscos from 304 distinct organisms (Data Sets S1 and S2). We also recorded 58 measurements of the Michaelis constant for RuBP (K RuBP). The experimental error was recorded for all of these values (when reported) along with the pH, temperature, and other metadata. Data were filtered as described in the Supporting Information. k cat,O is usually not measured directly29 but is rather inferred as k cat,O = (k cat,C/K C)/(S C/O/K O). We used 10 4-fold bootstrapping to estimate 199 k cat,O values and 95% confidence intervals thereof. We used an identical procedure to estimate k cat,C/K C and k cat,O/K O and confidence intervals thereof (Supporting Information). Altogether, we were able to calculate 274 k cat,C/K C and 199 k cat,O/K O values. Data Sets S1 and S2 provide all source and inferred data, respectively. Fitting Power Laws Certain model Rubiscos are measured frequently. For example, we found 12 independent measurements of the spinach Rubisco. In such cases, the median measured value was used to avoid bias in correlation and regression analyses. In contrast to textbook examples with one independent and one dependent variable, there is experimental error associated with both variables in all scatter plots shown here (e.g., plotting k cat,C vs K C in Figure5B). As such, we used total least-squares linear regression on a log scale to fit relationships between Rubisco parameters. Because R 2 values of total least-squares fits do not convey the explained fraction of Y axis variance, they are challenging to interpret. We instead report the degree of correlation as Pearson R values of log-transformed values. Bootstrapping was used to determine 95% confidence intervals for the Pearson correlation coefficient, power-law exponents, and prefactors (i.e., the slopes and intercepts of linear fits on a log–log scale). In each iteration of the bootstrap, data were subsampled to 90% with replacement. Total least-squares regression was applied to each subsample, and the procedure was repeated 10 4 times to determine 95% confidence intervals. Python source code is available at github.com/flamholz/rubisco. Figure 5. Open in a new tab Focal correlations of previous analyses are not robust to new data. Points with black outlines are from ref (6), and dashed gray lines represent the best fit to FI Rubisco data. Histograms for k cat,C, S C/O, and K C are plotted on parallel axes. Panel A plots k cat,C against S C/O. k cat,C and S C/O correlate with an R of approximately −0.6 among FI Rubiscos as compared to ≈0.9 previously.5,6 The 95% confidence intervals are (−4.0, −2.0) for the fit exponent and (3 × 10 4, 2 × 10 8) for the prefactor (slope and intercept on a log–log scale, respectively), indicating that the form of k cat,C–S C/O correlation is very uncertain. Notably, S C/O displays very limited variation overall and especially within physiological groupings with sufficient data. Median S C/O values are 177 for red algae (σ = 1.2; N = 6), 98 for C 3 plants (σ = 1.1; N = 162), 80 for C 4 plants (σ = 1.1; N = 35), and 48 for cyanobacteria (σ = 1.1; N = 16). Panel B plots k cat,C against K C. Here, the R is ≈0.5 as compared to ≈0.9 previously.6 This fit is more robust, with 95% confidence intervals of (0.3, 0.5) and (0.8, 1.5) for the fit exponent and prefactor, respectively. Results An Extended Data Set of Rubisco Kinetic Parameters To augment existing data, we collected literature data on ≈300 Rubiscos, including representatives of clades and physiologies that had been poorly represented in earlier data sets, e.g., diatoms, ferns, CAM plants, and anaerobic bacteria (Figure3A). We collected kinetic parameters associated with carboxylation and oxygenation (S, K C, k cat,C, K O, and k cat,O) as well as measurements of the RuBP Michaelis constant (half-maximum RuBP concentration, K RuBP) and experimental uncertainty for all values where available. All data considered below were measured at 25 °C and near pH 8 to ensure that measured values are comparable (Supporting Information). Notably, Rubisco assays are challenging to perform, and variation in measurements across laboratories is expected. Some of the spread in the data may come from systematic differences between laboratories and assay methods. The Rubisco activation state, for example, may differ between methods and preparations.15 Though we cannot resolve this issue here, we were careful to review each study’s methods, document a small number of problematic measurements, and record experimental error when reported (Data Set S1). The resulting data set contains Rubisco measurements from a total of 304 distinct species, including 335 S C/O values, 284 k cat,C values, 316 K C values, 254 K O values, and 199 k cat,O values (Figure3B). k cat,O values are rarely measured directly (Supporting Information) and are typically inferred as k cat,O = (k cat,C/K C)/(S C/O/K O).29 The Michaelis constant for RuBP (K RuBP) is measured infrequently, and only 58 values were extracted. We were able to estimate catalytic efficiencies for carboxylation (k cat,C/K C) in 274 cases and for oxygenation (k cat,O/K O) in 199 cases (Materials and Methods). Though the data include measurements of some Form II, III, and II/III Rubiscos, they remain highly focused on the Form I Rubiscos found in cyanobacteria, diatoms, algae, and higher plants, which make up >95% of the data set (Figure3B). As such, we focus here on the kinetic parameters of Form I Rubiscos (abbreviated FI Rubisco). Rubisco kinetic parameters display a very narrow dynamic range (Figure3C). The geometric standard deviation (σ_) expresses multiplicative variability in the data set and is well below one order of magnitude (_σ ≪ 10) for all parameters. Rubisco displays a particularly low variation in k cat,C (σ = 1.5) as compared to other enzymes for which 20 or more _k cat measurements are available [median _σ ≈ 7 (Figure S5)]. Specificity S C/O displays the least variation of all parameters (σ = 1.3). This is due in part to overrepresentation of C 3 plants in the data set, which occupy a narrow S C/O range of ≈80–120. Nonetheless, measurements of S C/O for FI and FII enzymes are clearly distinct, with values ranging from 7 to 15 for FII measurements and from ≈50 to 200 for FI (Figure3C). Energetic Trade-offs Tend To Produce Power-Law Correlations All kinetic parameters (S C/O, k cat,C, K C, k cat,O, and K O) are mathematically related to the microscopic rate constants of the Rubisco mechanism. Given common assumptions about irreversible and rate-limiting steps, this multistep mechanism can be simplified so that logarithms of measured kinetic parameters are proportional to effective transition state barriers (Figure1C,D and Supporting Information). As such, correlations between kinetic parameters will emerge if effective transition state barriers vary together (Figure2). If, for example, lowering the effective transition state barrier to CO 2 addition (Δ G 1,C) requires an increase in the effective barrier to the subsequent hydration and cleavage steps of carboxylation (Δ G 2,C), then we should observe a negative linear correlation ΔG 1,C ∝ −ΔG 2,C. Because k cat,C/K C is related to the first effective carboxylation barrier [k cat,C/K C ∝ exp(−ΔG 1,C/RT)] and k cat,C to the second [k cat,C ∝ exp(−ΔG 2,C/RT)], a linear correlation between transition state barrier heights translates to a log scale correlation between kinetic parameters such that ln(k cat,C/K C) ∝ −ln(k cat,C). These relationships are known as power laws and motivate us and others to investigate the kinetic parameters on a log–log scale. We expect to observe strong power-law correlations between pairs of kinetic parameters in two cases. (i) The associated energy barriers co-vary because they are linked by the enzymatic mechanism [“mechanistic coupling” (Figure2A)], or (ii) the mechanism imposes an upper bound on the sum (or difference) of two barrier heights. In case ii, strong selection favors the emergence of enzymes at or near the imposed limit [“selection within limits” (Figure2B)]. As Rubisco is the central enzyme of photoautotrophic growth, it likely evolved under selection pressure toward maximizing the net rate of carboxylation in each host, so either of these scenarios is plausible a priori. Notably, host physiology and growth environments can affect the catalytic environment. Rubiscos from different organisms will experience different temperatures, pHs, and prevailing CO 2 and O 2 concentrations due, for example, to an anaerobic host or a CO 2 concentrating mechanism increasing the level of CO 2.6 Different conditions should favor different combinations of kinetic parameters (Figure2). Correlations between Kinetic Parameters of Form I Rubiscos We performed a correlation analysis to investigate relationships between kinetic parameters of FI Rubiscos. Figure4 gives log scale Pearson correlations between parameters that are measured directly: k cat,C, K C, K O, S C/O, and K RuBP. Linear scale correlations are reported in Figure S7. Figure 4. Open in a new tab Correlations between measured kinetic parameters are attenuated by the addition of new data. This figure gives Pearson correlations (R) between pairs of log-transformed kinetic parameters of Form I Rubiscos. When multiple measurements of the same enzyme were available, the median value was used (Materials and Methods). S C/O–K C, S C/O–k cat,C, and K C–k cat,C correlations are of particular interest because they were highlighted in previous works, which found R values of 0.8–0.95. None of these pairs have R values exceeding 0.7 in the extended data set. Overall, correlations are weaker in the extended data set than documented in previous studies of smaller data sets.5,6 Nonetheless, we observed modestly strong, statistically significant correlations between k cat,C and S C/O (R = −0.56; p< 10–10), k cat,C and K C (R = 0.48; p< 10–10), K C and S C/O (R = −0.66; p< 10–10), and K C and K O (R = 0.56; p< 10–10). Because Rubisco kinetic parameters are mathematically interrelated through the microscopic mechanism as it is commonly understood, some level of correlation is expected. For example, when we derive expressions for k cat,C and K C from the Rubisco mechanism, they share common factors that should produce some correlation even in the absence of underlying coupling (Supporting Information). Similarly, S C/O is defined as (k cat,C/K C)/(k cat,O/K O) and could correlate negatively with K C for this reason. Because a modest correlation is expected irrespective of underlying trade-offs, the correlations in Figure4 do not necessarily support any particular trade-off model. Correlations between k cat,C and S C/O as well as k cat,C and K C were previously highlighted to support particular mechanistic trade-off models.5,6 However, these correlations are substantially attenuated by the addition of new data (Figures4 and 5). Plotting k cat,C against S C/O (Figure5A) shows that these parameters are modestly correlated, with an R of ≈0.6 compared to an R of ≈0.9 in previous analyses.5,6Figure5A also highlights the extremely limited and stereotyped variation in S C/O where Rubiscos from organisms sharing a particular physiology (e.g., C 3 plants or cyanobacteria) occupy a very narrow range of S C/O values. Multiplicative standard deviations (σ) of S C/O are <1.25 in all groups. Plotting k cat,C against K C (Figure5B) shows that this correlation is also weakened, with an R of ≈0.5 compared to ≈0.9 previously.6 We interpret weakened correlations as evidence that previously proposed trade-off models should be revisited. We therefore proceed to evaluate the correlations predicted by specific trade-off models, with an eye toward understanding the restricted variation in S C/O shown in Figure5A. Re-Evaluation of Proposed Trade-off Models Two distinct mechanistic trade-off models have been advanced.5,6 The first model, which we term k cat,C–K C coupling, posits that increased specificity toward CO 2 necessitates a slower maximum carboxylation rate, k cat,C.5,6 It was proposed that this trade-off is due to stabilization of the first carboxylation transition state (TS).5 Under this model, a stable Rubisco–TS complex produces high CO 2 specificity but slows the subsequent carboxylation steps and limits k cat,C (Figure S2). This proposal can be cast in energetic terms by relating the measured catalytic parameters to effective transition state barrier heights (Figure1D and Supporting Information). This model can be construed in energetic terms as follows. Lowering the effective barrier to CO 2 addition (Δ G 1,C in Figure6A) will make Rubisco more CO 2-specific even if oxygenation kinetics remain unchanged.6k cat,C–K C coupling posits a negative coupling between CO 2 addition and the subsequent carboxylation steps of hydration and bond cleavage (effective barrier height Δ G 2,C diagrammed in Figure6A). Therefore, the energetic interpretation of this model predicts a negative correlation between Δ G 1,C and Δ G 2,C and, as a result, a negative power-law correlation between k cat,C and k cat,C/K C.6 Figure 6. Open in a new tab Negative power-law correlation between k cat,C and k cat,C/K C is not supported by the extended data set. In the model diagrammed in panel A, CO 2-specific Rubiscos have low barriers to enolization and CO 2 addition (first effective carboxylation barrier Δ G 1,C), but lowering the first effective barrier necessarily increases the second effective barrier (Δ G 2,C), reducing k cat,C. In this view, stabilizing the first carboxylation TS also enhances selectivity but also slows carboxylation (Figure S2). Δ G 1,C and Δ G 2,C should be negatively correlated, which would manifest as negative power-law correlation between k cat,C and k cat,C/K C under certain assumptions (Supporting Information). (B) The extended data set does not evidence the expected correlation (for Form I enzymes, R = 0.02 and p = 0.8). While previous analyses gave an R of approximately −0.9,6 the 95% confidence interval for R now includes 0.0. Restricting our focus to particular physiologies like C 3 plants does not result in the expected correlation. In previous work, k cat,C and k cat,C/K C were found to correlate strongly on a log–log scale.6 The reported correlation, however, is not strongly supported by our data set (Figure6B). The true barrier height to CO 2 addition depends on the CO 2 concentration, which could partially explain the apparent lack of correlation. However, correlation is not improved by restricting focus to C 3 plants for which data are abundant and for which measured leaf CO 2 concentrations vary by only 20–30% due to variation in CO 2 conductance and Rubisco activity.30,31 The absence of correlation does not necessarily imply the absence of an underlying mechanistic limitation. Rather, if the Rubisco mechanism limits the joint improvement of k cat,C and k cat,C/K C, a much decreased correlation over the extended data set (R< 0.4) could result from several factors, including measurement error, undersampling of Rubiscos with high k cat,C (e.g., from cyanobacteria), or, alternatively, insufficient selection pressure. Diminished correlation, with many points observed below the previous correlation line, suggests that the “mechanistic coupling” model is less likely than “selection within limits” in this case (Figure2). The second mechanistic model, wherein faster CO 2 addition entails faster O 2 addition,6 is well-supported by the extended data set (Figure7). This model was previously supported by a power-law correlation between catalytic efficiencies for carboxylation and oxygenation [k cat,C/K C ∝ (k cat,O/K O)2]. As k cat,C/K C is exponentially related to the first effective carboxylation barrier [k cat,C/K C ∝ exp(−Δ G 1,C)] and k cat,O/K O to the first effective oxygenation barrier [k cat,O/K O ∝ exp(−Δ G 1,O)], correlation was taken to imply that lowering the barrier to CO 2 addition also lowers the barrier to O 2 addition (Figure7A). Our data set supports a similar power law, albeit with an exponent of ≈1.0 instead of ≈2.0. Figure 7. Open in a new tab Second mechanistic proposal that is remarkably well-supported by the extended data set. (A) In this proposal, mutations increasing the rate of addition of CO 2 to the Rubisco–RuBP complex also increase the rate of O 2 addition. In energetic terms, lowering the effective barrier to enolization and CO 2 addition (Δ G 1,C) lowers the first effective barrier to O 2 addition (Δ G 1,O), as well. Given this model, barrier heights should be positively correlated, which would manifest as a positive linear correlation on a log–log plot of k cat,C/K C against k cat,O/K O. (B) S C/O displays limited variation within physiological groups such as C 3 and C 4 plants for which we have substantial data. Dashed lines give the geometric mean of S C/O values. The multiplicative standard deviation, σ, sets the width of the shaded region. (C) S C/O = (k cat,C/K C)/(k cat,O/K O), so restricted S C/O variation implies a power-law relationship (k cat,C/K C) = S C/O(k cat,O/K O). k cat,C/K C is strongly correlated with k cat,O/K O on a log–log scale (R = 0.94; p< 10–10). Fitting FI measurements gives k cat,C/K C = 119(k cat,O/K O)1.04. A 95% confidence interval for the exponent is (0.94, 1.13), which includes 1.0. The geometric mean of measured S C/O values predicts k cat,O/K O = (k cat,C/K C)/S C/O and vice versa. This simple approach accurately predicts the k cat,O/K O for FI Rubiscos (prediction R 2 = 0.80), C 3 plants (R 2 = 0.84), C 4 plants (R 2 = 0.96), and cyanobacteria (R 2 = 0.79). Other groups, e.g., red algae, are omitted because of insufficient data. Again, we found that S C/O varies little among FI Rubiscos (Figure3C) and even less within C 3 plants, cyanobacteria, and other physiological groupings (Figures5A and 7B). S C/O = (k cat,C/K C)/(k cat,O/K O) by definition, so the fact that S C/O is approximately constant forces a positive power-law relationship of log(k cat,C/K C) = log(k cat,O/K O) + log(S C/O). Indeed, Form I enzymes display a remarkably high-confidence power-law relationship between k cat,C/K C and k cat,O/K O (R = 0.94; p< 10–10). Because S C/O is the only free parameter in this equation and is nearly constant, the geometric mean of S C/O measurements (≈90 for Form I Rubiscos) can be used to predict k cat,O/K O as S C/O–1(k cat,C/K C). This simple approach, which uses a power-law exponent of 1.0 and a prefactor of S C/O–1, predicts Form I k cat,O/K O values with an R 2 of 0.80, nearly as accurate as fitting both the prefactor and exponent as free parameters (R 2 = 0.81). As shown in Figure7C, predictions of k cat,O/K O = S C/O–1(k cat,C/K C) generally improve when considering specific physiological groupings like C 3 and C 4 plants because S C/O varies so little within these groups. Assuming a roughly constant S C/O forces a 1:1 relationship Δ G 1,C = Δ G 1,O + C, meaning that decreasing CO 2 addition barrier Δ G 1,C is associated with an equal decrease in O 2 addition barrier Δ G 1,O. Implications for the Mechanism of CO 2/O 2 Discrimination by Rubisco A 1:1 relationship between effective barriers to CO 2 and O 2 addition suggests that a single factor controls both barriers. We offer a simple model based on the mechanism of Rubisco that can produce a 1:1 correlation between barrier heights and constant S C/O. In this model, the RuBP-bound active site fluctuates between reactive and unreactive states (Figures8A). The fraction of enzyme in the reactive state is denoted ϕ. In the unreactive state, neither oxygenation nor carboxylation can proceed. In the reactive state, either gas reacts at its intrinsic rate, which does not vary across Rubiscos of the same class (Figure8B). Because RuBP must undergo enolization for carboxylation or oxygenation to occur, ϕ may be determined by the degree of enolization of RuBP (Supporting Information). Figure 8. Open in a new tab A power-law relationship between k cat,C/K C and k cat,O/K O can be explained by an active site that fluctuates between “reactive” and “unreactive” states. (A) In this model, CO 2 and O 2 react with bound RuBP only when the enzyme is in the reactive state, which has an occupancy φ. (B) φ can vary between related enzymes. In the reactive state, CO 2 and O 2 react with the bound RuBP with intrinsic reactivities ΔG1,C and ΔG1,O that do not vary between related Rubiscos. If the difference in intrinsic reactivities (ΔG1,O – ΔG1,C) is constant, we derive a power-law relationship between k cat,C/K C and k cat,O/K O with an exponent of 1.0. This relationship requires a constant S C/O (Supporting Information). This model can be phrased quantitatively as ∝ ϕ exp(−Δ G1,C/RT) and ∝ ϕ exp(−Δ G1,O/RT) where ΔG1,C and ΔG1,O are the intrinsic reactivities of the RuBP enediolate to CO 2 and O 2, respectively. Under this model, S C/O should be roughly constant, which forces a power-law relationship between k cat,C/K C and k cat,O/K O with an exponent of 1.0 (Figure7C). Variation in k cat,C/K C and k cat,O/K O implies that ϕ can vary between related Rubiscos, perhaps by evolutionary tuning of the equilibrium constant for RuBP enolization. S C/O is independent of the equilibrium fraction of on-enzyme RuBP enolization, so variation in enolization should affect k cat,C/K C and k cat,O/K O without altering S C/O. Rather, S C/O is determined by the difference ΔG1,O – ΔG1,C, so changes to the conformation of the RuBP enediolate might explain characteristic differences between the S C/O of C 3 plant and cyanobacterial Rubiscos.5,8 See the Supporting Information for a derivation of this model and further discussion of its implications. Discussion We collected and analyzed literature measurements of ≈300 Rubiscos (Figure3A). The literature is very phylogenetically biased, with the readily purified plant Rubiscos making up >80% of the data (Figure3B). Despite incomplete coverage, some trends are clear. Rubisco kinetic parameters display an extremely limited dynamic range, with multiplicative standard deviations being <3-fold in all cases (Figure3C). k cat,C and S C/O appear to be particularly constrained. Rubisco displays much less k cat variability than any other enzyme for which sufficient data are available (Figure S5); 97% of k cat,C values are between 1 and 10 s–1, and the highest k cat,C measured at 25 °C (14 s–1, S. elongatus PCC 794216) is only ≈20 times greater than the lowest reported Form I value (0.8 s–1 from the diatom Cylindrotheca N132). Altogether, these data suggest that there is some limitation of the maximum rate of carboxylation by Rubisco in the presence of O 2. Focusing on O 2, we find that measured Rubiscos oxygenate slowly. More than half of k cat,O measurements are <1 s–1, and k cat,C is 4 times greater than k cat,O on average (Figure3C and Figure S4A). Similarly, the O 2 affinity is quite low in general. The median K O is ≈470 μM, nearly double the Henry’s law equilibrium of water with a 21% O 2 atmosphere (≈270 μM at 25 °C). With a multiplicative standard deviation of 1.3, S C/O displays the least variation all Rubisco kinetic parameters (Figure3C and Figure S4A). Figures5A and 7B highlight the stereotyped variation in S C/O, where C 3 plant, C 4 plant, cyanobacterial, and red algal enzymes display very limited variation around characteristic S C/O values. All groups have multiplicative standard deviations (σ) of <1.25. Nonetheless, FI Rubiscos are approximately 1 order of magnitude more CO 2-specific than the few characterized Form II, III, and II/III enzymes (Figure7B and Supporting Information). This might be explained by the prevalence of FII, FIII, and FII/III enzymes in bacteria and archaea that fix CO 2 under anaerobic conditions, where it is doubtful that oxygenation affects organismal fitness. We note, however, that there is substantial variation among measurements of the model FII Rubisco from Rhodospirillum rubrum (Figure S4B). This and the paucity of data on non-Form I Rubiscos (Figure3B) indicate that more measurements are required to evaluate FII, FII/III, and FIII enzymes. As such, we focused here on FI Rubiscos, for which data are abundant. Rubisco kinetics were previously argued to vary in a one-dimensional landscape6 and hypothesized to be “nearly perfectly optimized”.5 Overall, FI Rubiscos appear to be less constrained than previously supposed. Figure4 documents an overall reduction in correlation between FI Rubisco kinetic parameters, and the data set is no longer well-approximated as one-dimensional (Figure S8). Many natural Rubiscos appear to be suboptimal in plots of k cat,C against S C/O because other enzymes have roughly equal S C/O values but higher k cat,C values (Figure5A and Figure S6). Weakened correlations could be due to measurement error and systematic biases, though we find this explanation unlikely because (i) measurements of Form I Rubiscos from similar organisms are broadly consistent (Figure S6), (ii) some correlations remain strong and statistically significant across the entire data set, (iii) systematic bias toward C 3 plants would tend to increase correlations, and (iv) standardization of Rubisco assays using stoichiometric inhibitors to quantify active sites should improve data quality over time (Supporting Information). Reduced correlations therefore lead us to reject the notion that Rubisco kinetics vary in a strictly one-dimensional landscape and to revisit previous models of mechanistic trade-off. The mechanistic models described in Figures6 and 7 are based on a simple chemical intuition: that the intrinsic difficulty of discriminating CO 2 and O 2 requires the enzyme to differentiate between carboxylation and oxygenation transition states. The requirement of transition state discrimination is a direct consequence of two assumptions supported by experimental evidence.22 Briefly, it is assumed that addition of either gas is irreversible and that there is no binding site for CO 2 or O 2 and, thus, no “Michaelis complex” for either gas.5,6,22,33,34 If CO 2 bound a specific site on Rubisco before reacting, it might be possible to modulate K C by mutation without substantially affecting the kinetics of subsequent reaction steps. In the unlikely case that gas addition is substantially reversible,34,35 we might expect to find Rubiscos that evolved enhanced selectivity by energy-coupled kinetic proofreading. Energy coupling can enable amplification of selectivity due to differential CO 2 and O 2 off rates.36 The fact that no such Rubiscos have been found suggests that gas addition is irreversible or that CO 2 and O 2 off rates are incompatible with kinetic proofreading in some other way.6,37 As Rubisco likely does not bind CO 2 directly, it was hypothesized that high CO 2 specificity (large S C/O) is realized by discriminating between the first carboxylation and oxygenation transition states, i.e., between the developing carboxyketone and peroxyketone (Figures S1 and S2).5 A late carboxylation transition state would be maximally discriminable because the developing carboxylic acid is distinguishable from the peroxyl group of the oxygenation intermediate. The extraordinarily tight binding of the carboxyketone analogue CABP to plant Rubisco provides strong support for a late carboxylation transition state.5 Because a late transition state resembles the carboxyketone intermediate, it was argued that CO 2-specific Rubiscos must tightly bind the intermediate, slowing the subsequent reaction steps and restricting k cat,C (Figure S2).5 As k cat,C/K C is related to the effective barrier to enolization and CO 2 addition (Δ G 1,C) and k cat,C is related to the effective barrier to hydration and cleavage [Δ G 2,C (Figure1D)], an energetic framing of this model argues that decreasing Δ G 1,C (increasing k cat,C/K C) entails increasing Δ G 2,C [decreasing k cat,C (Figure6A)].6 Despite nuanced differences, we collectively term these models k cat,C–K C coupling due to the hypothesized coupling of carboxylation kinetics. Though these models are motivated by the need to discriminate between CO 2 and O 2, they invoke a trade-off between carboxylation steps only. That is, specificity requires tighter binding of the carboxylation intermediate, which slows downstream processing of that same intermediate, irrespective of O 2. Three correlations previously supported k cat,C–K C coupling, correlations between k cat,C and S C/O, between k cat,C and K C, and between k cat,C and k cat,C/K C. k cat,C and S C/O remain negatively correlated in our larger data set but more weakly than previously observed (Figure5A). The same is true for k cat,C and K C (Figure5B) and for k cat,C–k cat,C/K C (Figure6B). Rather than arguing for strong coupling of carboxylation kinetics, Figure5 highlights the stereotyped variation in S C/O described above. We interpret weakened correlations as implying that carboxylation kinetics are not strictly coupled. Considering residuals of the k cat,C–K C fit (Figure S9) shows that outliers include recent measurements of cyanobacterial38 and diatom39 Rubiscos, which fall well below the fit line. This is consistent with a “selection within limits” view of k cat,C–K C coupling (Figure2B). The second mechanistic trade-off model posits that faster addition of CO 2 to the Rubisco–RuBP complex necessarily allows faster O 2 addition. This model was previously supported by a positive power-law correlation between the catalytic efficiencies for carboxylation and oxygenation (k cat,C/K C and k cat,O/K O, respectively),6 which can be understood as a positive coupling of the effective barriers to enolization and gas addition for CO 2 and O 2 [Δ G 1,C and Δ G 1,O (Figure7A)]. We showed that extremely limited and stereotyped variation in S C/O = (k cat,C/K C)/(k cat,O/K O) necessitates a power-law correlation with of exponent of 1.0 between k cat,C/K C and k cat,O/K O (Figure7B,C). An exponent of 1.0 implies that decreasing Δ G 1,C (enabling faster carboxylation) requires a roughly equal decrease in Δ G 1,O (enabling faster oxygenation, as well). Although several research groups have attempted to isolate improved Rubisco mutants, none of the mutants examined so far exceed wild-type enzymes on these axes (Figure S11). A power-law relation with an exponent of 1.0 can be seen as resulting from an active site that fluctuates between a reactive and unreactive state (Figure8A). This coarse-grained model is motivated by the Rubisco mechanism in two ways. Because Rubisco likely does not bind CO 2 or O 2 directly, active site concentrations are determined by solution concentrations (e.g., in the chloroplast stroma). Rubisco could close the active site to diffusion to limit O 2 entry,40 but this would also slow carboxylation. Similarly, RuBP must enolize for oxygenation or carboxylation to proceed (Figure1A), so modulating the degree of enolization would affect both reaction pathways equally.14,15 In either case, the average occupancy of the reactive state mechanistically couples the rates of CO 2 and O 2 addition (Figure2A) and throttles the subsequent steps of carboxylation and oxygenation equally (Figure8). In previous work, where Rubisco kinetics were thought to vary in a one-dimensional landscape, setting k cat,C determined all other kinetic parameters.6 In this setting, it was argued that Rubisco kinetic parameters are determined by the prevailing CO 2 and O 2 concentrations because a unique choice of parameters on the one-dimensional curve maximizes the net rate of carboxylation.6 Because the data are no longer clearly one-dimensional, we cannot argue that Rubisco is “perfectly optimized” to match prevailing concentrations. Moreover, the model presented in Figure8 sets no upper limit on k cat,C, suggesting that selection for an increased level of carboxylation in the absence of O 2 could produce Rubisco mutants with superlative k cat,C values (i.e., k cat,C ≫ 15 s–1). Such enzymes might be found in anaerobic bacteria and would be of interest in probing the limits of Rubisco catalysis. The prospect of engineering an improved Rubisco is tantalizing, not only because it could plausibly improve crop yields18 but also because the task tests our understanding of enzymes on a very basic level. It is clear from the data presented here that there is some evolutionary constraint on Rubisco catalysis. Surely, a superlative Rubisco would have arisen if it were mutationally accessible from existing enzymes. More detailed biochemical investigation of naturally occurring Rubiscos will help delineate the evolutionary constraints imposed on Rubisco kinetics. Still, the Rubisco large subunit displays extremely limited sequence variation.41 Perhaps exploring a wider swath of sequence space via protein engineering techniques42−44 would enable strict improvements to Rubisco kinetics? We argue that biochemical and bioengineering techniques should be used in concert to probe the limits of Rubisco catalysis and propose several avenues of future research to evaluate the prospects of Rubisco engineering. First, the kinetics of non-plant Rubiscos should be characterized more thoroughly. These should include the Form II, III, and II/III enzymes of bacteria and archaea as well as FI enzymes of bacteria and diverse eukaryotic autotrophs.13,39 Ideally, these enzymes would be chosen in a manner that maximizes sequence and phylogenetic diversity45 and characterized for their binding (e.g., of RuBP and CABP) and catalytic activity (measuring k cat,C, K C, k cat,O, K O, and S C/O) as a function of temperature and pH.29,46,47 A facile assay for direct measurement of oxygenation would also reduce the number of assumptions made in measuring and analyzing Rubisco kinetics.29 These data would help resolve whether Rubisco isoforms display characteristic differences in catalytic potential. It is possible, for example, that non-Form I enzymes are subject to different constraints than FI Rubiscos and might serve as useful chassis for engineering. It is also important to revisit the classic experiments undergirding our understanding of the Rubisco catalytic mechanism, especially those supporting the central assumptions that (i) there is no Michaelis complex for CO 2 or O 2 and (ii) gas addition is irreversible.22,34,35 These assumptions substantially constrain CO 2 specificity. If we were to find Rubiscos for which these assumptions are relaxed, they might serve as a basis for engineering a fast-and-selective carboxylase. On the other hand, all Rubiscos may share the same limitations. Because these limitations are likely described as couplings between transition state barriers (as in Figures7 and 8), measurements of barrier heights for a wide variety of Rubiscos would enable more direct testing of trade-off models. One avenue for drawing inferences about barrier heights is by measuring the binding energies of intermediate and transition state analogues.5,48 Kinetic isotope effects for CO 2 and O 2 report indirectly on the relevant barriers49 and can be measured by mass spectrometry.50 Investigating the relationship between transition state barriers and kinetic parameters will help delineate which reaction steps limit carboxylation and oxygenation in different Rubisco lineages.5 Some disagreement about the precise ordering of carboxylation steps remains,5,14,15 and the mechanism of oxygenation is not well understood.48 Chemical reasoning about the mechanisms of Rubisco carboxylation and oxygenation would benefit from progress in structural biology. Intermediate and transition state analogues should be used to capture the active site at various points along the reaction trajectory.14,40,48,51 If experiments and structural analyses confirm that the assumptions described above hold for all Rubiscos, it would greatly limit our capacity to engineer Rubisco and strongly suggest that alternative strategies for improving carbon fixation should be pursued.19,52−54 If, however, these assumptions are invalidated, many enzyme engineering strategies would become viable. Such data and analyses will be instrumental in guiding the engineering of carbon fixation for the next decade. Acknowledgments The authors thank Uri Alon, Kapil Amarnath, Doug Banda, Arren Bar-Even, Cecilia Blikstad, Jack Desmarais, Woodward Fischer, Vahe Galstyan, Laura Helen Gunn, Itai Halevy, Oliver Mueller-Cajar, Robert Nichols, Elad Noor, Jeremy Roop, Yonatan Savir, Patrick Shih, Daniel Stolper, Dan Tawfik, Guillaume Tcherkez, Tsvi Tlusty, and Renee Wang for helpful conversations and comments on the manuscript. Supporting Information Available The Supporting Information is available free of charge on the ACS Publications website at DOI: 10.1021/acs.biochem.9b00237. Supplementary text, tables, figures, and references (PDF) Data set S1 (XLSX) Data set S2 (XLSX) Data set S3 (XLSX) Data set S4 (XLSX) This research was supported by the U.S. National Science Foundation (NSF) (Grant MCB-1818377), the European Research Council (Project NOVCARBFIX 646827), the Israel Science Foundation (ISF) (Grant 740/16), the ISF-NRF Singapore joint research program (Grant 7662712), the Beck-Canadian Center for Alternative Energy Research, Dana and Yossie Hollander, the Ullmann Family Foundation, the Helmsley Charitable Foundation, the Larson Charitable Foundation, the Wolfson Family Charitable Trust, Charles Rothschild, and Selmo Nussenbaum. R.M. is the Charles and Louise Gartner professional chair. A.I.F. was supported by a National Science Foundation Graduate Research Fellowship. Y.M.B.-O. is an Azrieli fellow. The authors declare no competing financial interest. 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16229	https://chengzhaoxi.xyz/8ba327db.html	【连载】概率面试题 \| 潮汐朝夕潮汐朝夕的生活实验室 \Doge 陪伴一个算法工程师的职业生涯 Home AI业务企业服务多媒体处理广告系统数据分析数据可视化数据挖掘数据采集游戏娱乐特征工程量化投资风控系统 AI工程 AI系统分布式系统工具链数据库系统系统运维计算机系统软件系统 AI模型 NLP 人工智能可解释性图模型多模态序列模型强化学习无监督机器学习概率模型深度学习计算机视觉任意门反作弊大风控泛产品泛内容泛商业泛金融生活实验室社会科学自然科学读书笔记软技能连载合集吐槽回忆墙围棋挖坑区改进日志碎碎念算法栏解决方案转载钢笔图鉴面试门数代数优化信息论几何分析博弈图论应用拓扑拾遗数论方程概率物理组合数学统计计算数学随机过程集合论算法 leetcode 动态规划图算法基础算法大数据算法字符串搜索数据结构模拟疑难杂症算术算法分析算法设计编译原理计算理论编程 C++ C语言 Go Java Python Shell About 任意门连载合集【连载】概率面试题字数统计: 1.3k字 \| 阅读时长: 4分 2612 2021-06-26 概率 67, 连载 14, 面试 25 摘要: 关于概率面试题的文章【对算法，数学，计算机感兴趣的同学，欢迎关注我哈，阅读更多原创文章】我的网站：潮汐朝夕的生活实验室我的公众号：算法题刷刷我的知乎：潮汐朝夕我的github：FennelDumplings 我的leetcode：FennelDumplings 面试为什么考概率最近几年在程序员面试中，尤其是校招面试中，经常会问到一些与概率、期望相关的问题，一方面是考查候选人的数学素养，另一方面是概率在计算机中的应用确实很多。算法工程师、算法研究员、策略产品经理、数据分析师、以数据驱动为核心的公司的风控/营销相关的岗位等等，都是最近几年新出现并且比较热门的岗位，而概率在这些岗位中是通用基础，面试中基本上都会问到。此外，很多程序员会去金融科技工作，而金融业中的技术岗位，概率是必考的，并且会重点考。准备这些岗位的同学，有时间的话都可以练习本专栏，将来一定会有用。这些题零零散散出现在各个平台的面经，周围同学朋友的口述，以及自己的面试实践中，但是没有一个系统的梳理。算法工程师的概率我工作至今的职位一直都是算法工程师，这里我从算法工程师的角度谈一谈概率。最近几年人工智能、数据科学已成为一项推动业务发展的重要技术。而进入这个领域的人，基本上一定会翻阅领域内的文章，以及参与相关任务。那么你会发现与概率有关的问题基本上绕不开：过滤垃圾邮件，不具备贝叶斯思维的话恐怕不行要从文本中提取出名称实体，需要研究概率图模型要做语音识别，那么理解随机过程中的隐马尔科夫模型要通过样本推断出某类对象的总体特征，估计理论和大数定理的思想是必须建立的在统计推断过程中，有一类应用广泛的采样方法，蒙特卡洛方法以及马尔科夫过程的稳定性需要研究可以看到，概率对于程序员来说除了是一项经常在面试中出现的技术，还能够解决很多有意思的实际问题。关于这个连载专栏由于概率是很多岗位的通用基础，经常考到。同时我本人对概率还比较感兴趣。因此开启了本专栏的写作，并且长期连载。这是我陆陆续续更新的一个专栏，本专栏收录的都是面试中常见的概率题，目的就是详细梳理程序员面试中常见的知识点与题目。当然还是以题目为主。主要来源是一些书中见到的好题(例如陈希孺的《概率论与数理统计》，Ross的《概率论基础教程》中的一些精彩例题等等)、我和周围朋友面试遇到的题以及网上的零散面经中觉得还不错的题。对于每道题，首先给出思路参考，从概率论的角度推公式求解，然后用蒙特卡洛模拟的方式验证结果。注意有些题目的思路有时不止一种，而我给出的是基本上都是我自己做的时候的方法，大家如果有不同思路也可以讨论。以下为持续更新的每道题的链接，另外在 github 的 probability_puzzle 仓库中也有同步更新。 \| 编号 \| 网站链接 \| 备注 \| --- \| 1 \| 抽屉中的袜子 \| 2 \| 系列赛中连续获胜 \| 3 \| 轻率的陪审团 \| 4 \| 三人循环赛 \| 5 \| 试验直到第一次成功 \| 6 \| 公司从用户一次游戏取得的收入 \| 全期望公式、期望DP \| \| 7 \| 向正方形区域扔硬币 \| 8 \| 祝你好运 \| 9 \| 资助赌徒 \| 10 \| 完美手牌 \| 11 \| 双骰子赌博 \| 12 \| 收集优惠券 \| 设计随机变量、全期望公式 \| \| 13 \| 一排座位 \| 全期望公式、期望DP \| \| 14 \| 第二强的选手是否拿亚军 \| 15 \| 孪生骑士 \| 条件概率 \| \| 16 \| 萨缪尔·佩皮斯的问题 \| 全概率公式、概率DP \| \| 17 \| 三人枪战 \| 18 \| 不公平的硬币 \| 贝叶斯公式 \| \| 19 \| 不公平的硬币2 \| 贝叶斯公式 \| \| 20 \| 不公平的硬币3 \| 贝叶斯公式 \| \| 21 \| 不公平的硬币4 \| 贝叶斯公式 \| \| 22 \| 有放回抽样还是无放回抽样 \| 贝叶斯公式 \| \| 23 \| 圆周上随机取3个点形成锐角三角形 \| 几何概型 \| \| 24 \| 选票盒 \| 概率DP \| \| 25 \| 系列赛中不出现平局 \| 概率DP \| \| 26 \| 仓促的决斗 \| 几何概型 \| \| 27 \| 较短的一节木棍 \| 连续型随机变量 \| Share follow: newer 【Puzzle】资助赌徒older Python性能优化-极速数据处理 -- Numba和Pandas Catalog 面试为什么考概率算法工程师的概率关于这个连载专栏 recents 算法leetcode Ad-Hoc问题：在字符串中删改字符 2025-03-05 算法leetcode leetcode第一题，两数之和 2025-03-04 算法动态规划频繁查询子串是否回文：区间DP 2025-03-01 算法字符串频繁查询子串是否回文：Manacher预处理后 O(1)O(1)响应 2025-03-01 算法动态规划状态转移方程描述最优子结构 \| 优化状态表示 \| 单调队列优化DP 2025-01-24 Random 算法疑难杂症按位单独处理的技巧 2025-01-12 任意门泛内容原公司注销，将公众号迁移至新公司 2024-05-29 编程Python 让lru_cache忽略某些参数 2023-12-08 任意门泛内容小红书的内容与栏目 2022-02-11 吐槽挖坑区最小生成树2 2020-10-12 categories AI业务43 企业服务1 多媒体处理4 广告系统3 数据分析7 数据可视化12 数据挖掘2 数据采集2 游戏娱乐1 特征工程3 量化投资2 风控系统6 AI工程70 AI系统13 分布式系统1 工具链21 数据库系统14 系统运维15 计算机系统1 软件系统5 AI模型55 NLP4 人工智能1 可解释性3 图模型1 多模态1 序列模型3 强化学习18 无监督5 机器学习6 概率模型4 深度学习4 计算机视觉5 任意门189 反作弊5 大风控7 泛产品5 泛内容20 泛商业4 泛金融11 生活实验室106 社会科学1 自然科学2 读书笔记6 软技能4 连载合集18 吐槽357 回忆墙70 围棋19 挖坑区14 改进日志37 碎碎念40 算法栏40 解决方案25 转载17 钢笔图鉴76 面试门19 数212 代数14 优化11 信息论12 几何16 分析20 博弈4 图论12 应用3 拓扑1 拾遗7 数论8 方程5 概率45 物理5 组合数学24 统计12 计算数学7 随机过程2 集合论4 算法535 leetcode128 动态规划80 图算法36 基础算法69 大数据算法14 字符串31 搜索25 数据结构76 模拟6 疑难杂症29 算术9 算法分析22 算法设计2 编译原理7 计算理论1 编程114 C++22 C语言2 Go1 Java37 Python47 Shell5 archives March 20254 January 202513 December 202412 November 20244 October 20247 September 20246 August 20249 July 202412 June 20249 May 202426 April 202421 March 202411 February 20248 January 20241 December 202314 November 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16230	https://www.facebook.com/groups/2239445476353952/posts/3130919267206564/	MATH \| Prove that the sum of the first n odd numbers equals n² (Provide a short algebraic or visual proof.) \| Facebook Log In Log In Forgot Account? Is sum of first n odd numbers equal to n2? Summarized by AI from the post below MATH · Join Maths Solution Platform · 5d · Prove that the sum of the first n odd numbers equals n² (Provide a short algebraic or visual proof.) #Proofs#MathBeauty#NumberTheory#MathLesson#Induction Like Comment Share See more on Facebook See more on Facebook Email or phone number Password Log In Forgot password? or Create new account
16231	https://www.sciencedirect.com/topics/biochemistry-genetics-and-molecular-biology/convergent-evolution	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Development of T Cell Immunity 2010, Progress in Molecular Biology and Translational ScienceMasanori Kasahara D Convergent Evolution or Common Ancestry? In evolutionary biology, convergent evolution is defined as the process whereby distantly related organisms independently evolve similar traits to adapt to similar necessities. VLRs and TCRs/BCRs both serve as antigen receptors, but are evolutionarily unrelated. Thus, the use of distinct receptors in jawed and jawless vertebrates can be regarded as a prime example of convergent evolution.13 However, the overall design of the AIS in jawed and jawless vertebrates seems too similar to be accounted for solely by convergent evolution. Particularly striking is the observation that both jawed and jawless vertebrates have two major populations of lymphoid cells presumed to have similar specialized immune functions145 (Fig. 6). To account for this, it seems more reasonable to assume that VLRA+ cells and T cells evolved from a common ancestor and that, likewise, VLRB+ cells and B cells shared common ancestry; most likely, a common ancestor of all vertebrates was equipped with two lineages of lymphoid cells.7 Recent evidence indicates that, contrary to a commonly held belief, T cells and B cells do not share an immediate common ancestor, but differentiate from myeloid-T and myeloid-B progenitors, respectively.149,150 If T cells and B cells are distantly related as suggested by these studies, it is not surprising if the two lineages of lymphoid cells diverged at an earlier stage in evolution than previously thought.151 In summary, authentic T cells and B cells, as defined by surface expression of TCRs and BCRs, are unique to jawed vertebrates (Fig. 1). However, jawless vertebrates have at least two populations of lymphoid cells that likely share common ancestry with T cells and B cells of jawed vertebrates. View chapterExplore book Read full chapter URL: Book series2010, Progress in Molecular Biology and Translational ScienceMasanori Kasahara Review article Protein functional epitopes: hot spots, dynamics and combinatorial libraries 2001, Current Opinion in Structural BiologyBuyong Ma, ... Ruth Nussinov Convergent evolution has also been achieved in simulations [34•], consistently illustrating higher conservation of polar residues on the protein surface [10••]. View article Read full article URL: Journal2001, Current Opinion in Structural BiologyBuyong Ma, ... Ruth Nussinov Chapter Convergent Evolution 2013, Brenner's Encyclopedia of Genetics (Second Edition)L. Gabora Abstract Convergent evolution occurs when organisms that are not closely related evolve traits that are similar, or analogous, as an adaptive response to similar environmental pressures. The genes responsible for the two versions of the trait may be quite different. Examples include the evolution of wings in the bat, the bird, and the pterodactyl, and the evolution of glycoproteins with antifreeze properties in the Arctic and Antarctic fishes. View chapterExplore book Read full chapter URL: Reference work2013, Brenner's Encyclopedia of Genetics (Second Edition)L. Gabora Review article Advances of Genetics Research in China: A Special Issue to Celebrate the 40th Anniversary of GSC 2018, Journal of Genetics and GenomicsHuizhong Fan, ... Fuwen Wei 3.5.1 Convergent evolution Phenotype convergence generally refers to describe the parallel evolution of identical or similar traits in distantly related species (Christin et al., 2010; Stern, 2013; Storz, 2016). Convergent evolution has always been a topic of intense interest in adaptive evolution and an important source of information on the relationship between phenotype and genotype. A representative study of comparative genomics is the genome-wide convergent evolution between bamboo-eating red pandas and giant pandas (Hu et al., 2017b). Based on the improved giant panda genome and de novo sequencing of red panda genome, Hu et al. (2017b) used the amino acid convergence and positive selective strategies to detect adaptive convergence genes from 14,254 orthologous genes among eight mammal genomes. Two limb development genes DYNC2H1 and PCNT, which may be important candidate genes for pseudothumb development, were found to undergo adaptive convergence. In addition, some genes involved in the digestion and utilization of essential nutrients from bamboo were also identified to undergo adaptive convergence. View article Read full article URL: Journal2018, Journal of Genetics and GenomicsHuizhong Fan, ... Fuwen Wei Review article Evolution in a Community Context: On Integrating Ecological Interactions and Macroevolution 2017, Trends in Ecology & EvolutionMarjorie G. Weber, ... Blake Matthews Likewise, ignoring local community context gives us at best an incomplete picture of the factors influencing trait evolution in clades, and at worst can also be positively misleading. For example, convergent evolution is a pattern of longstanding interest to evolutionary biologists, and new analytical tools have recently heightened interest in testing for this pattern in clade-level phylogenies . However, the mechanisms underlying convergence may differ widely depending on the distribution and coexistence of the clade members. If convergent forms are always found allopatrically, this suggests that similar selective pressures and/or resource distributions have led to the repeated evolution of similar ecotypes (e.g., ). In contrast, if convergent forms are found sympatrically, this could suggest that local interactions may be promoting the evolution of similar phenotypes. This is especially powerful if patterns of sympatry per se can be disentangled from specific shared abiotic conditions. By integrating clade-and community-based perspectives, we can move beyond pattern detection approaches, and towards evaluating the likelihood of different ecological mechanisms underlying clade-based patterns in trait evolution (see also Box 1). View article Read full article URL: Journal2017, Trends in Ecology & EvolutionMarjorie G. Weber, ... Blake Matthews Review article Hypoxia Inducible Factor pathway proteins in high-altitude mammals 2024, Trends in Biochemical SciencesFrank S. Lee Convergent evolution The possibility of convergent evolution is raised by the observation of natural selection acting on the HIF2A locus in multiple species, including humans (Tibetans and Andeans), North American deer mice, Andean horses, and Tibetan dogs, goats, horses, and pigs (Table 1). Likewise, there is evidence for natural selection acting on the PHD2 locus in humans (Tibetans and Andeans) and possibly Tibetan cattle. Importantly, there is evidence that convergent evolution can act by different molecular mechanisms (Figure 3). For example, high-altitude North American deer mouse T755M Hif-2α shows a defective interaction with Cbp in vitro (Figure 3B) , and deer mice with the Hif2a allele shows impaired VAH , consistent with a partial loss of function. Andean H194R HIF-2α displays weakened heterodimerization with ARNT in vitro (Figure 3C), and mice with a knockin Andean Hif2a mutation (H194R) display resistance to hypoxia-induced pulmonary hypertension; both observations are consistent with a partial loss of function . Tibetan HIF2A likewise displays phenotypic properties consistent with a partial loss of function (reviewed in ). However, in contrast to the situation with high-altitude deer mouse or Andean HIF-2α, there are no coding sequence mutations in Tibetan HIF-2α, and current evidence is consistent with it being due to decreased transcription (or possibly splicing) of the HIF2A gene (Figure 3D). It should be noted that core HIF pathway genes are not invariably targets of natural selection in high-altitude mammals. For example, high-altitude Ethiopian gelada monkeys, like Tibetans, do not exhibit elevated hemoglobin concentrations . However, in contrast to Tibetans, they do not display evidence of natural selection acting on the Hif2a gene. View article Read full article URL: Journal2024, Trends in Biochemical SciencesFrank S. Lee Mini review Convergent evolution for antibiotic biosynthesis in bacteria and animals 2023, Trends in GeneticsNicolas Papon, ... Marnix H. Medema Abstract Convergent evolution has been described for several metabolic pathways across the kingdoms of life. However, there is hitherto no evidence for such an interkingdom process for antimicrobials. A new report suggests that marine animals have evolved the ability to biosynthesize antimicrobial polyketides, in parallel with bacteria. View article Read full article URL: Journal2023, Trends in GeneticsNicolas Papon, ... Marnix H. Medema Review article Structural Bioinformatics 2012, Journal of Structural BiologyChristof Winter, ... Michael Schroeder 2.1.3 Convergent evolution The analyses by Kim et al. (2006a,b) show that sometimes binding sites may be re-used by different binding partners. The faces of these different partners can hence be said to have converged in evolution to bind the common partner. In general, convergent evolution of interfaces can be observed when two unrelated proteins that do not share a common ancestor exhibit a remarkably similar interface structure. As noted above, a classic example of convergent evolution is that of the serine proteases chymotrypsin, a pancreatic digestive enzyme found in vertebrates, and subtilisin, found in the prokaryote Bacillus subtilis. Although the tertiary structures of the two enzymes bear no resemblance, the active site residues are in almost exactly the same position (Wright et al., 1969). Both proteins have independently evolved to carry out the same enzymatic function. Convergent evolution can be found in particular in viruses mimicking native interfaces (Davey et al., 2011). With comprehensive structural protein interaction databases, it became possible to systematically search for examples of convergent evolution of protein–protein interfaces. Henschel et al. (2006) screened the Protein Data Bank for pairs of protein–protein interactions A–B and A–C where B and C are unrelated (different SCOP superfamilies) and bind to equivalent sites of A. This was named the ABAC2 method. One ABAC example from Henschel et al. (2006) is shown in Fig. 4. The crystal structures of the complexes of caspase with the p35 protein from baculoviruses (Xu et al., 2001) and of caspase with XIAP, an endogenous inhibitor of the IAP family (Riedl et al., 2001) were solved by two independent groups. Both p35 and XIAP act as apoptotic suppressors. Aligning the two complexes using the common partner caspase reveals that both p35 and XIAP bind to the same surface patch of the caspase, which involves the active site of the caspase. Fig. 4 shows the remarkable structural similarity of the interacting motifs of p35 (red) and XIAP (blue), while the rest of the structures adopts a completely different fold. Interestingly, the interacting motifs are also different on the sequence level, with only one aspartate residue being identical. This aspartate residue (marked with an arrowhead in Fig. 4) is contacting a well-conserved potential hot spot on the caspase surface consisting of two tryptophane and one arginine residues. In total, 914 such examples were identified with the ABAC method (Henschel et al., 2006). Most of them concern enzymes. View article Read full article URL: Journal2012, Journal of Structural BiologyChristof Winter, ... Michael Schroeder Chapter Mammals, Biodiversity of 2001, Encyclopedia of BiodiversityJoshua R. Ginsberg IV.E. Convergent Evolution Mammalian evolution has been rich with novelty, whether in the evolution of flight in bats, the suite of adaptations evolved by whales that have brought mammals into the sea to live, or more specifically in one lineage the well–developed venom gland found in the hind legs of the platypus. Throughout mammalian evolutionary history, however, some of the more remarkable evolutionary patterns have involved the repeated evolution of derived characters in widely divergent mammal lineages. Convergent evolution spans a wide array of adaptations, from the suite of carnivorous Miocene South American marsupials that resemble extinct and contemporaneous placental carnivores to the striking similarity of burrowing forms in the marsupial mole (family Notoryctidae), the golden moles, (family Chrysochloridae), and true moles (family Talpidae). Throughout their evolutionary history, mammals have shown trends to increasing specialization in many morphological and correlated ecological functions. For instance, as grasslands became widespread in the Miocene, ungulate tooth morphology shifted from a dominance of low-crowned, or brachyodont, teeth, which are easily ground down through the animal's lifetime, to a dominance of hypsodonty, or high-crown teeth, which last longer when an animal eats siliceous grasses covered in dust and dirt. This evolution has been complemented by an increased complexity of molar teeth to enable herbivores to more thoroughly grind their food. This pattern is observed in both artiodactyls and perrisodactyls. Rodents have solved the problem differently, evolving ever-growing teeth, while proboscideans roll their teeth out of their jaw, each tooth having a limited useful life. Convergent evolution may address the same problems but find different solutions. The expansion of grasslands and a high-cellulose diet have led to two different solutions to digesting an essentially indigestible substance. Ruminant digestion, widespread in the artiodactyls, uses the fore-gut, or rumen, as a fermentation chamber in which bacteria break down cellulose, making the grass cell contents available for absorption and converting the cellulose into digestible material. Because the passage of materials through the rumen limits intake, artiodactyls tend to be relatively selective, choosing grasses of high quality. In perissodactyls and some kangaroos, a similar solution has evolved—bacterial breakdown of cellulose—but the site of bacterial digestion is in the hindgut and the process is less efficient. While this means that hindgut fermenters cannot draw out as many nutrients from a given pulse of food, by rapid processing of food, hindgut fermenters can gain sustenance from larger volumes of lower quality forage. Table III lists a few of the more widespread patterns of convergent evolution in mammals. Table III. Examples of Convergent Evolution in Mammalsa \| Geographic location \| \| \| \| \| \| --- --- --- \| \| Adaptation \| Holarctic \| Africa \| Madagascar \| South America \| Australia \| \| Carnivory \| Carnivora \| Carnivora \| Carnivora \| Borhyaenidae \| Dasyuroid \| \| \| Many families \| Many families \| Viverrids \| \| Phalangeroid \| \| Pursuit predators \| Carnivora \| Carnivora \| \| Didelphoid \| Dasyuroid \| \| \| Dogs \| Hyenas \| \| Borhyaenid \| Thylacinus \| \| \| Dog bears \| Cheetah \| \| \| \| \| \| Bear dogs \| \| \| \| \| \| Saber-tooth carnivores \| Smilodon \| Felids \| Viverrids \| Borhyaenid \| Phalangeroid \| \| \| Felids \| \| Fossa \| Thylacosmilus \| Thlacoleo \| \| \| Nimravids \| \| \| \| \| \| Semiaquatic \| Insectivora \| Insectivora \| Insectivora \| Didelphoid \| Monotreme \| \| \| Desmans \| Otter shrew \| Otter tenrec \| Water opossum \| Platypus \| \| \| Water voles \| \| \| \| \| \| \| Pantolestids \| \| \| \| \| \| \| Rodentia \| \| \| \| Rodentia \| \| \| Musk rat \| \| \| \| Water rat \| \| \| Carnivora \| \| \| \| \| \| \| Otters \| \| \| \| \| \| Aquatic \| Carnivora \| \| \| \| \| \| \| Sea lions/seals \| \| \| \| \| \| \| Cetacea \| \| \| \| \| \| \| Whales \| \| \| \| \| \| Herbivores with cellulose fermentation \| Ungulates \| Ungulates \| \| Ungulates \| Phalangeroids \| \| \| Artiodactyla \| Proboscidians \| \| Notoungulates \| \| \| \| Perissodactyla \| Hyracoids \| \| Lipoterns \| \| \| \| Some rodents \| Arsinotheres \| \| Pyrotheres \| \| \| \| \| Some rodents \| \| \| \| \| Hornlike structures \| Perissodactyls \| Arsinotheres \| \| Notoungulates \| \| \| \| Rhinos \| \| \| Toxodontids \| \| \| \| Brontotheres \| \| \| \| \| \| \| Artiodactyls \| \| \| \| \| \| \| Protoceratids \| \| \| \| \| \| \| Oreodonts \| \| \| \| \| \| \| Pigs \| \| \| \| \| \| \| Ruminants \| \| \| \| \| \| \| Archaic ungulates \| \| \| \| \| \| \| Uinthatheres \| \| \| \| \| \| \| Rodents \| \| \| \| \| \| \| Mylagaulids \| \| \| \| \| \| Anteaters \| Pholidota \| Pholidota \| \| Edentates \| Dayroids \| \| \| Pangolins \| Pangolins \| \| Anteaters \| Numbat \| \| \| \| \| \| Armadillos \| Monotremes \| \| \| \| \| \| \| Echidna \| \| Gliding \| Rodents \| Rodents \| \| \| Phalangers \| \| \| Flying squirrels \| Scaly-tailed squirrels \| \| \| Flying possums \| \| \| Dermopterans \| \| \| \| \| \| \| Flying lemur \| \| \| \| \| \| \| Paromomyidae \| \| \| \| \| a : From Janis and Damuth (1990). Asterisks indicate extinct taxa. View chapterExplore book Read full chapter URL: Reference work2001, Encyclopedia of BiodiversityJoshua R. Ginsberg Review article Convergence in Multispecies Interactions 2016, Trends in Ecology & EvolutionLeonora S. Bittleston, ... Anne Pringle Convergence in Evolution and Ecology The word convergence typically describes convergent evolution, the independent evolution of similar traits in different lineages resulting from strong selective pressures: ‘[a]nimals, belonging to two most distinct lines of descent, may readily become adapted to similar conditions, and thus assume a close external resemblance’ . Although convergent evolution is primarily a descriptor of morphological features of animals and plants, it can be used to describe microbes and physiological processes as well (e.g., convergent evolution of transcriptional regulation of gene circuits in bacteria and fungi; see ). Convergence is also recognized in ecological assemblages; for example, in high-altitude plant communities of the Andes, Alps, and Himalayas . The homogeneity of vegetation in geographically distant biomes was discussed early in the history of ecology [4,5]. The resemblance of high-altitude plant communities, or whole communities of plants, birds, and lizards in the Mediterranean climates of California, Chile, South Africa, and the Mediterranean Basin, are examples of community convergence, defined as the physiognomic similarity of assemblages of co-occurring plants or animals resulting from comparable physical and biotic selective pressures [6–8]. Community convergence focuses on community structure and functional traits but does not explicitly investigate interactions among species. View article Read full article URL: Journal2016, Trends in Ecology & EvolutionLeonora S. Bittleston, ... Anne Pringle Related terms: Genetics Anabolism Amino Acids Enzyme Phylogeny Shigella Cladistics Taxon Infectious Agent Mouse View all Topics We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. You can manage your cookie preferences using the “Cookie Settings” link. For more information, see ourCookie Policy Cookie Preference Center We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors. You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. See the different category headings below to find out more or change your settings. 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16232	https://emedicine.medscape.com/article/798397-overview	Tools & Reference>Ophthalmology Periorbital Infections Updated: Dec 26, 2024 Author: Bobak Zonnoor , MD, MMM; Chief Editor: Gil Z Shlamovitz, MD, FACEP, FAMIA more...;) Share Print Feedback Close Facebook Twitter LinkedIn WhatsApp Email Sections Periorbital Infections Sections Periorbital Infections Overview Background Anatomy Etiology Prognosis Pathophysiology Epidemiology Show All Presentation History Physical Examination Show All DDx Workup Approach Considerations Imaging Studies Jones Dye Test Other Tests Show All Treatment Approach Considerations Complications Show All Medication Medication Summary Antibiotics, Other Show All Media Gallery;) References;) Overview Background Periorbital infections consist of a group of infections that can be broadly classified into 2 distinct groups. [1, 2] One group consists of infections of the dermis and associated tissues around the eyes. The other group consists of infections of the lacrimal system. Periorbital cellulitis Infections of the superficial skin around the eyes are called periorbital, or preseptal cellulitis. These infections are limited to the area anterior to the orbital septum. Periorbital cellulitis is predominantly, although not exclusively, a pediatric disease. [1, 2, 3, 4] Periorbital cellulitis. This image shows an 8-year-old patient who presented with unilateral eyelid swelling and erythema. View Media Gallery) Periorbital cellulitis represents the first of the 5 nonprogressive types of orbital infections. These are classified as follows [1, 2, 5] : Group 1 - Preseptal cellulitis Group 2 - Orbital cellulitis Group 3 - Subperiosteal abscess Group 4 - Orbital abscess Group 5 - Cavernous sinus thrombosis Lacrimal system infections Infections of the lacrimal system are classified based on the location of the infection; they include the following : Blepharitis - Inflammation of the lid margins; anterior blepharitis affects the area of the lid where the eyelashes attach; posterior blepharitis affects the inner portion of the eyelid margin that is in contact with the eye [6, 7] ; blepharitis primarily affects older persons (mean age 40-50 years) Canaliculitis – Inflammation of the canaliculi Dacryoadenitis - Inflammation of the lacrimal gland Dacryocystitis - Inflammation of the lacrimal duct or sac; obstructed lacrimal ducts causing dacryocystitis are common in infants and usually resolve by age 9-12 months [11, 12] Acute dacryocystitis. View Media Gallery) Next: Anatomy Anatomy Orbital septum The orbital septum is a fibrous membrane that extends from the periosteum of the orbit as the arcus marginalis and lies just deep to the orbicularis oculi muscle. In the upper lids, the septum fuses with the levator aponeurosis. In the lower lids, the septum fuses with the capsulopalpebral fascia. The orbital septum acts as a physical barrier to the spread of infection. Upper eyelid anatomy. View Media Gallery) Lower eyelid anatomy. View Media Gallery) Lacrimal system The lacrimal system includes the structures involved in the production and drainage of tears. The lacrimal gland is located in the lateral upper lid margin. It produces about 10 mL of secretions per day. In the process of blinking, the eyes close from the lateral edge to the medial edge, pushing the tear film across the surface of the eye. Most of the tear volume is lost through evaporation. A small portion is drained from the lacrimal lake, located at the inner canthus, through the puncta and into the superior and inferior canaliculi. Tears then flow into the common canaliculus and lacrimal sac. The lacrimal duct, which lies within the bone, connects the lacrimal sac with the eventual site of egress, the inferior meatus of the nose. Eye and lacrimal duct, anterior view. View Media Gallery) Previous Next: Anatomy Etiology Periorbital cellulitis Data regarding the causes of preorbital cellulitsi is limited. Periorbital cellulitis can occur by several mechanisms, including the following [1, 4, 14] : Infection as a result of local trauma, including insect bites Infection as a result of spread from contiguous structures, as in conjunctivitis, hordeolum, lacrimal system infections, and impetigo Infections secondary to hematogenous spread during bacteremia due to nasopharyngeal pathogens Infection secondary to sinusitis, causing venous and lymphatic congestion - Sinusitis may be of odontogenic origin ; thorough examination of dentition may be warranted Infectious organisms When associated with trauma, periorbital cellulitis can be caused by the following bacterial species : Staphylococcus aureus - This may include methicillin-resistant S aureus ; treatment should be tailored to local incidence of infection Streptococcus pyogenes (group A streptococci) In the absence of trauma, periorbital cellulitis can be caused by the following microbes: Streptococcus pneumoniae Haemophilus influenzae type b - Was the predominant cause of periorbital cellulitis prior to the advent of the Hib vaccine but has now been shown to cause only rare cases [3, 17, 18, 19] Other unusual causes of periorbital cellulitis include the following : Neisseria gonorrhoeae Neisseria meningitidis [21, 22] Vaccinia virus - Has been reported in a laboratory worker; autoinoculation resulting in periorbital infection in patients receiving the vaccine has been reported Herpes simplex virus Mycobacterium tuberculosis Bacillus anthracis Taenia solium – Periorbital cellulitis can develop secondary to orbital cysticercosis caused by this organism Blepharitis Anterior blepharitis is usually secondary to infection or seborrheic in nature, or else it is a combination of both. If the pilosebaceous glands of Zeiss and Moll become infected, an abscess may occur. This abscess is known as an external hordeolum, or stye. Cell-mediated immunologic mechanisms have been implicated in the development of chronic blepharitis. [29, 30] Posterior blepharitis is caused by Meibomian gland dysfunction. The Meibomian gland secretes the oily layer of the tear film. This oily layer is responsible for preventing excessive evaporation of the aqueous layer of the tear film. If the secretions become inspissated, causing plugging of the gland, a chalazion may develop. A chalazion is a noninfectious, granulomatous reaction. If there is infection secondary to plugging, an internal hordeolum develops. The use of eye makeup, especially eyeliner, can cause acute exacerbations of blepharitis by plugging the glands. Anterior blepharitis Causative organisms in anterior blepharitis include the following: Infectious etiology usually due to Staphylococcus species Other bacteria include Propionibacterium acnes,Moraxella species, and Corynebacterium species [12, 32] Helicobacter pylori - Associated with blepharitis, but cause and effect have not been established Viruses - Herpes simplex virus, herpes zoster virus, and human papillomavirus Mites -Demodex folliculorum and Demodex brevis Lice -Phthirus pubis, which causes a condition known as phthiriasis palpebrarum Noninfectious entities such as ocular rosacea and seborrheic dermatitis also cause anterior blepharitis Posterior blepharitis Etiologic characteristics of posterior blepharitis include the following: Meibomian gland dysfunction leading to increased evaporation of the protective tear layer Most frequently associated with rosacea Pityrosporum fungal infection is associated with Meibomian gland dysfunction Dacryoadenitis Dacryoadenitis is caused by local infection of the lacrimal gland by bacteria or viruses. Chronic dacryoadenitis associated with inflammation and swelling of the salivary glands of unknown origin is called Mikulicz disease. When associated with other entities such as tuberculosis, sarcoidosis, or lymphoma, it is termed Mikulicz syndrome. This was previously considered a subtype of Sjögren syndrome, although now differences between the 2 entities have been determined. Bacteria Dacryoadenitis is most often caused by gram-positive cocci, usually staphylococci. It may also be caused by S pneumoniae. Viruses Prior to increased immunization rates, the mumps virus was most often implicated in the development of dacryoadenitis. Now, the Epstein-Barr virus is most often associated with chronic dacryoadenitis. Dacryocystitis Dacryocystitis is caused by inflammation of the lacrimal sac; this usually occurs in the setting of obstruction of the lacrimal apparatus. The obstruction may be congenital or secondary to infection, tumor, or trauma. Infectious causes include gram-positive isolates (Staphylococcus and Streptococcus species), in 71-78% of cases, and gram-negative isolates, in 22-29% of cases. [40, 41] Rarely, dacryocystitis may result from mucormycosis. Canaliculitis Canaliculitis is caused by infection of the canaliculi; often, it is chronic. The disorder may also be iatrogenic, occurring after the use of instrumentation or the placement of silicone plugs in the treatment of dry eyes. It is classically taught that the most common pathogens of canaliculitis are Actinomyces israelii and Nocardia (formerly known as Streptothrix) species. Case reviews, however, have shown mixed flora associated with infection. Species isolated include Staphylococcus species, Escherichia coli, Haemophilus species, Pseudomonas aeruginosa, Klebsiella oxytocia, Arcanobacterium (previously Corynebacterium) haemolyticum, and M chelonae. Previous Next: Anatomy Prognosis Prognoses in periorbital infections are as follows: Periorbital cellulitis - With appropriate antibiotics, the prognosis is good Blepharitis - Typically, this is a chronic disease with waxing and waning of symptoms Dacryoadenitis - Prognosis of the acute form is excellent, as it is a self-limiting process; prognosis of the chronic form depends on the underlying disease process Dacryocystitis - Success rates for dacryocystorhinostomy are good; external procedures fare better due to the ability to create a larger ostium. Canaliculitis - Prognosis is excellent with definitive treatment by canaliculotomy with curettage. Complications Periorbital cellulitis Complications of periorbital cellulitis include the following: Recurrent periorbital cellulitis (RPOC) - Three periorbital infections occurring within 1 year, spaced by at least 1 month of convalescence Toxic shock syndrome Eschar formation leading to scarring Meningitis Orbital cellulitis Intracranial extension of infection Periorbital necrotizing fasciitis - A potential complication with devastating results; the mortality rate is reported to be 14.4% Blepharitis Complications of blepharitis include the following: Hordeolum Chalazion Scarring of the cornea, which could lead to blindness Trichiasis (misdirection of eyelashes) Corneal infection Endophthalmitis - Patients with blepharitis undergoing intraocular surgery may have an increased risk of developing this condition Dacryocystitis Dacryocystitis may be associated with periorbital cellulitis. Rarely, it may be associated with orbital cellulitis or abscess formation. This is usually prevented by the orbital septum, which surrounds the lacrimal sac. Complications may occur during a dacryocystorhinostomy, including hemorrhage, infection, and cerebrospinal fluid (CSF) leakage. Canaliculitis Epiphora may result from failure of tear film to appropriately drain through the lacrimal system. Previous Next: Anatomy Pathophysiology S aureus, Staphylococcus epidermidis, and S pyogenes account for approximately 75% of pediatrics periorbital infections. Staphylococcus and Streptococcus are the most two common pathogens responsible for pediatric orbital cellulitis. Previous Next: Anatomy Epidemiology Periorbital cellulitis is primarily a pediatric disease, occurring mostly in children younger than 5 years. Periorbital cellulitis is almost three times more common than orbital cellulitis. While orbital cellulitis is more common in the pediatric population, it can affect all age groups. In one retrospective analysis, the average age of affected patients was 6.8 years (ranging from age 1 week to 16 years). Worldwide, orbital cellulitis occurs more often in winter. It is associated with upper respiratory tract infections, and most cases have a unilateral presentation. [52, 53] Periorbital cellulitis has no sexual predilection. Orbital infections have a male-to-female ratio of 2:1. Previous Clinical Presentation References Baiu I, Melendez E. Periorbital and Orbital Cellulitis. JAMA. 2020 Jan 14. 323 (2):196. [QxMD MEDLINE Link]. [Full Text]. Allen RC. Preseptal and Orbital Cellulitis. Porter RE. The Merck Manual of Diagnosis and Therapy. Rahway, NJ: Merck & Co Inc; Reviewed/Revised July 2024. [Full Text]. Givner LB, Mason EO Jr, Barson WJ, et al. Pneumococcal facial cellulitis in children. Pediatrics. 2000 Nov. 106(5):E61. [QxMD MEDLINE Link]. Moubayed SP, Vu TT, Quach C, Daniel SJ. 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Arch Ophthalmol. 2003 Apr. 121(4):491-9. [QxMD MEDLINE Link]. Lamoreau KP, Fanciullo LM. Pott’s Puffy Tumor Mimicking Preseptal Cellulitis. Clin Exp Opt 2007; Published early online. Accessed 15 March, 2008. Morley AM. Pott's puffy tumour: a rare but sinister cause of periorbital oedema in a child. Eye (Lond). 2009 Apr. 23(4):990-1. [QxMD MEDLINE Link]. Swann PG, Weir J. Is it blepharitis?. Clin Exp Optom. 2005 Mar. 88(2):113-4. [QxMD MEDLINE Link]. Boruchoff SA, Boruchoff SE. Infections of the lacrimal system. Infect Dis Clin North Am. 1992 Dec. 6(4):925-32. [QxMD MEDLINE Link]. Charles NC, Lisman RD, Mittal KR. Carcinoma of the lacrimal canaliculus masquerading as canaliculitis. Arch Ophthalmol. 2006 Mar. 124(3):414-6. [QxMD MEDLINE Link]. Howe L, Jones NS. Guidelines for the management of periorbital cellulitis/abscess. Clin Otolaryngol Allied Sci. 2004 Dec. 29(6):725-8. [QxMD MEDLINE Link]. Rudloe TF, Harper MB, Prabhu SP, Rahbar R, Vanderveen D, Kimia AA. 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Fusidic Acid Resistance Rates and Prevalence of Resistance Mechanisms Among Staphylococcus spp. isolated in North America and Australia (2007-2008). Antimicrob Agents Chemother. 2010 Jun 21. [QxMD MEDLINE Link]. Knezevic T, Ivekovic R, Astalos JP, Novak Laus K, Mandic Z, Matejcic A. Botulinum toxin A injection for primary and recurrent chalazia. Graefes Arch Clin Exp Ophthalmol. 2009 Jun. 247(6):789-94. [QxMD MEDLINE Link]. Gao YY, Di Pascuale MA, Elizondo A, Tseng SC. Clinical treatment of ocular demodecosis by lid scrub with tea tree oil. Cornea. 2007 Feb. 26(2):136-43. [QxMD MEDLINE Link]. Pinckney J 2nd, Cole P, Vadapalli SP, Rosen T. Phthiriasis palpebrarum: a common culprit with uncommon presentation. Dermatol Online J. 2008. 14(4):7. [QxMD MEDLINE Link]. Aburn NS, Sullivan TJ. Infectious mononucleosis presenting with dacryoadenitis. Ophthalmology. 1996 May. 103(5):776-8. [QxMD MEDLINE Link]. Greenberg MF, Pollard ZF. The red eye in childhood. 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Eye and lacrimal duct, anterior view. of 5 Tables Back to List Contributor Information and Disclosures Author Bobak Zonnoor , MD, MMM Assistant Professor of Emergency Medicine, SUNY Downstate School of Medicine; Director, ED Observation Unit, Department of Emergency Medicine, Kings County Hospital; Volunteer Assistant Clinical Professor, University of California, Los Angeles, David Geffen School of Medicine; Clinical Faculty, University of California, Riverside, School of Medicine Bobak Zonnoor , MD, MMM is a member of the following medical societies: American College of Emergency Physicians, Emergency Medicine Residents' Association Disclosure: Nothing to disclose. Coauthor(s) Elizabeth Fiedler, MD Clinical Instructor, Department of Emergency Medicine, Montefiore Medical Center-Weiler Division Elizabeth Fiedler, MD is a member of the following medical societies: American Medical Association Disclosure: Nothing to disclose. Richard H Sinert, DO Professor of Emergency Medicine, Clinical Assistant Professor of Medicine, Research Director, State University of New York College of Medicine; Consulting Staff, Vice-Chair in Charge of Research, Department of Emergency Medicine, Kings County Hospital Center Richard H Sinert, DO is a member of the following medical societies: American College of Physicians, Society for Academic Emergency Medicine Disclosure: Nothing to disclose. Chief Editor Gil Z Shlamovitz, MD, FACEP, FAMIA Professor of Clinical Emergency Medicine, Keck School of Medicine of the University of Southern California; Chief Medical Information Officer, Keck Medicine of USC Gil Z Shlamovitz, MD, FACEP, FAMIA is a member of the following medical societies: American College of Emergency Physicians Disclosure: Nothing to disclose. Additional Contributors Robert E O'Connor, MD, MPH Professor and Chair, Department of Emergency Medicine, University of Virginia Health System Robert E O'Connor, MD, MPH is a member of the following medical societies: American Academy of Emergency Medicine, American College of Emergency Physicians, American Heart Association, American Medical Association, National Association of EMS Physicians, Society for Academic Emergency Medicine Disclosure: Nothing to disclose. Zach Kassutto, MD, FAAP Director, Pediatric Emergency Medicine, Capital Health System; Associate Professor of Pediatrics and Emergency Medicine, Drexel University College of Medicine; Attending Physician, St Christopher's Hospital for Children Zach Kassutto, MD, FAAP is a member of the following medical societies: American Academy of Pediatrics Disclosure: Nothing to disclose. R Gentry Wilkerson, MD, FACEP, FAAEM Assistant Professor, Coordinator for Research, Department of Emergency Medicine, University of Maryland School of Medicine R Gentry Wilkerson, MD, FACEP, FAAEM is a member of the following medical societies: American Academy of Emergency Medicine, American College of Emergency Physicians Disclosure: Nothing to disclose. Edmond A Hooker II, MD, DrPH, FAAEM Assistant Professor, Department of Emergency Medicine, University of Cincinnati College of Medicine Edmond A Hooker II, MD, DrPH, FAAEM is a member of the following medical societies: American Academy of Emergency Medicine, American Public Health Association, Society for Academic Emergency Medicine, and Southern Medical Association Disclosure: Nothing to disclose. Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference Disclosure: Medscape Salary Employment encoded search term (Periorbital Infections) and Periorbital Infections What to Read Next on Medscape Log in or register for free to unlock more Medscape content Unlimited access to our entire network of sites and services Log in or Register
16233	https://www.cuemath.com/calculus/differentiable/	Differentiable A differentiable function is a function in one variable in calculus such that its derivative exists at each point in its entire domain. The tangent line to the graph of a differentiable function is always non-vertical at each interior point in its domain. A differentiable function does not have any break, cusp, or angle. A differentiable function is always continuous but every continuous function is not differentiable. In this article, we will explore the meaning of differentiable, how to use differentiability rules to find if the function is differentiable, understand the importance of limits in differentiability, and discover other interesting aspects of it. \| \| \| --- \| \| 1. \| What is Differentiable? \| \| 2. \| Rules for Differentiable Functions \| \| 3. \| Some Common Differentiability Formulas \| \| 4. \| Limit Formula for Differentiable Functions \| \| 5. \| Difference Between Differentiable and Continuous Function \| \| 6. \| FAQs on Differentiable \| What is Differentiable? A function is said to be differentiable if the derivative of the function exists at all points in its domain. Particularly, if a function f(x) is differentiable at x = a, then f′(a) exists in the domain. Let us look at some examples of polynomial and transcendental functions that are differentiable: Rules for Differentiable Functions If f, g are differentiable functions, then we can use some rules to determine the derivatives of their sum, difference, product and quotient. Here are some differentiability formulas used to find the derivatives of a differentiable function: Example Let's use the differentiability rules to find the derivative of the function f(x) = (2x+1)3 df/dx = d(2x+1)3/dx = d(8x3 + 12x2 + 6x + 1)/dx = 24x2 + 24x + 6 = 6(2x+1)2 Some Common Differentiability Formulas In calculus, differentiation of differentiable functions is a mathematical process of determining the rate of change of the functions with respect to the variable. Some common differentiability formulas that we use to solve various mathematical problems are: Limit Formula for Differentiable Functions There is an alternative way to determine if a function f(x) is differentiable using the limits. A function f(x) is differentiable at the point x = a if the following limit exists: [\lim_{h\rightarrow 0}\dfrac{f(c+h)-f(c)}{h}] Example: Consider the absolute value function given by f(x) = \|x\| We will determine if this function is differentiable at c = 0 or not. Let's find the limit (\begin{align}\lim_{h\rightarrow 0}\dfrac{f(c+h)-f(c)}{h}\end{align}). [\begin{align}\lim_{h\rightarrow 0}\dfrac{f(c+h)-f(c)}{h}&=\lim_{h\rightarrow 0}\dfrac{f(0+h)-f(0)}{h}\&=\lim_{h\rightarrow 0}\dfrac{\|h0\|}{h}\&=\lim_{h\rightarrow 0}\dfrac{\|h\|}{h}\end{align}] What happens when (h) approaches 0 from left? Let's see the behavior of the function as h becomes closer to 0 from the negative x - axis. [\begin{align}\lim_{h\rightarrow 0^{-}}\dfrac{\|h\|}{h}&=\lim_{h\rightarrow 0}\dfrac{-h}{h}\&=-1\end{align}] What happens when h approaches 0 from right? Now, let's see the behavior of the function as h becomes closer to 0 from the positive x - axis. [\begin{align}\lim_{h\rightarrow 0^{+}}\dfrac{\|h\|}{h}&=\lim_{h\rightarrow 0}\dfrac{h}{h}\&=1\end{align}] Did you observe that limits are different? This means that the limit (\begin{align}\lim_{h\rightarrow 0}\dfrac{f(c+h)-f(c)}{h}\end{align}) does not exists at c = 0 for f(x) = \|x\|. This implies that the absolute value function f(x) = \|x\| is not differentiable at x = 0. Difference Between Differentiable and Continuous Function We say that a function is continuous at a point if its graph is unbroken at that point. A differentiable function is always a continuous function but a continuous function is not necessarily differentiable. Example We already discussed the differentiability of the absolute value function. Clearly, there are no breaks in the graph of the absolute value function. The function is continuous everywhere. Particularly, the function is continuous at x=0 but not differentiable at x=0. Hence the main difference between a differentiable and continuous function is that a differentiable function is always a continuous function but a continuous function may not be differentiable. Tips and Tricks for Differentiable Functions Important Notes on Differentiable Differentiable functions are those functions whose derivatives exist. If a function is differentiable, then it is continuous. If a function is continuous, then it is not necessarily differentiable. The graph of a differentiable function does not have breaks, corners, or cusps. Related Topics on Differentiable Differentiation of Trigonometric Functions Derivative of ln x Derivative Formula Differentiation Differentiable Examples Example 1: Use the differentiability rules to determine the derivative of f(x) = (2x + 1)/x3 Solution: We will use the quotient rule for differentiable functions to determine the derivative of f(x). df/dx = d((2x + 1)/x3)/dx = [2x3 - 3x2.(2x + 1)]/x6 = [2x3 - 6x3 - 3x2]/x6 = -(4x3 + 3x2)/x6 Answer: The derivative of f(x) = (2x + 1)/x3 is -(4x3 + 3x2)/x6 Example 2: Find out where the given function f(x) = \|x + 2\| is not differentiable using graph and limit definition. Solution: Clearly, there is a sharp corner at point x = -2. The function is not differentiable at x = -2. Now, let's use the limit definition of differentiable functions. We already observed that the limits are different for absolute value of function. [\begin{align}\lim_{h\rightarrow 0}\dfrac{f(c+h)-f(c)}{h}&=\lim_{h\rightarrow 0}\dfrac{f(-2+h)-f(-2)}{h}\&=\lim_{h\rightarrow 0}\dfrac{\|h\|-0}{h}\&=\lim_{h\rightarrow 0}\dfrac{\|h\|}{h}\end{align}] This means that the limit (\begin{align}\lim_{h\rightarrow 0}\dfrac{f(c+h)-f(c)}{h}\end{align}) does not exists at c = -2 for f(x) = \|x+2\|. go to slidego to slide Book a Free Trial Class Differentiable Practice Questions go to slidego to slide FAQs on Differentiable What is Differentiable in Calculus? A function is said to be differentiable if the derivative of the function exists at all points in its domain. Is the Cubic Function Differentiable? Yes, the cubic function is differentiable. For example, the function f(x) = x3 is differentiable and its derivative is f′(x) = 3x2 What does Twice Differentiable Mean? If a function is twice differentiable, then it means that the second derivative of the function exists. Why is the Absolute Value Function not Differentiable at 0? The absolute value function is not differentiable at 0 because the graph of the function has a sharp corner at this point. What are Some Common Differentiable Formulas? When is a Function Differentiable? A differentiable function is a function in one variable in calculus such that its derivative exists at each point in its entire domain. How to Prove a Function is Differentiable? A function can be proved differentiable if its left-hand limit is equal to the right-hand limit and the derivative exists at each interior point of the domain.
16234	https://www.studocu.com/en-us/messages/question/2743013/if-the-demand-curve-is-horizontal-then-consumer-surplus-is-always-equal-to-zero-in-market	[Solved] If the demand curve is horizontal then consumer surplus is always - Introduction To Economics I: Macroeconomics (ECON 201-CN) - Studocu Skip to main content Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes Home My Library AI Notes Ask AI AI Quiz New Recent If the demand curve is horizontal then consumer surplus is always equal to zero in market equilibrium. T or F?Introduction To Economics I: Macroeconomics (ECON 201-CN) My Library Courses You don't have any courses yet. Add Courses Studylists You don't have any Studylists yet. Create a Studylist Northwestern University Introduction To Economics I: Macroeconomics Question If the demand curve is horizontal then consumer surplus is always Northwestern University Introduction To Economics I: Macroeconomics Question Anonymous Student 3 years ago If the demand curve is horizontal then consumer surplus is always equal to zero in market equilibrium. T or F? Like 0 Related documents Intro to Macro Ch 4 - Government Intervention, Price Ceilings and Price Floors, Consumer and Producer Introduction To Economics I: Macroeconomics Lecture notes None Econ 201 Equilibrium of Supply & Demand - Google Docs Introduction To Economics I: Macroeconomics Summaries 100%(1) Intro To Macro Ch 3 - Supply and Demand, Competitive Markets, Shifts in Demand and Supply, Complements Introduction To Economics I: Macroeconomics Lecture notes None PSET 5 - Solutions - PSET 5 Introduction To Economics Ii: Microeconomics Assignments 100%(3) PSET 3 - Solutions - PSET 3 Introduction To Economics Ii: Microeconomics Assignments 100%(1) Week 1 Homework Questions: Market Dynamics & Demand Analysis Introduction To Economics Ii: Microeconomics Coursework None Microeconomics 202 - Exam #2 Instructions and Questions - Winter 2020 Introduction To Microeconomics Assignments None Answer Created with AI 3 years ago Solution Correct answer- True The horizontal D (or, demand ) curve is a perfectly elastic D curve. The perfectly elastic D curve implies that consumers are willing to purchase the commodity at a prevailing market P (or, Price) and makes the quantity demanded fall to zero even on a slight increase in market P. Consumers are very sensitive to changes in the P of the commodity when D curve is perfectly elastic or horizontal. CS (or, Consumer Surplus) is the amount the consumer is WTP (or, Willing To Pay) minus the amount he actually paid for the commodity. CS is estimated as the area of the triangle that the D curve forms when combined with the P line and the y-axis. As the perfectly elastic D curve is horizontal, the D curve coincides with the P line. The area between the D curve and the P line will be zero as the P line and the D curve are over-lapping each other. The consumer will be earning no surplus and the entire surplus will be acquired by the produces known as the PS (or, Producer Surplus). Another reason for the zero consumer surplus is that when the D is perfectly elastic, consumers are always trying to align their willingness to pay with the market P and hence, the WTP is not higher than the P for any consumer in the market. The WTP being equal to the P therefore would result in a zero CS. Like 0 AI answers may contain errors. Please double check important information and use responsibly. Ask a new question Discover more from: Introduction To Economics I: Macroeconomics ECON 201-CN Northwestern University 63 Documents Go to course 3 Ch. 21 - Taught by Professor Guido Lorenzoni Introduction To Economics I: Macroeconomics 100%(6) 3 Lastset 201 - Problem Sets Introduction To Economics I: Macroeconomics 100%(2) 3 Secondset - Problem Sets Introduction To Economics I: Macroeconomics 80%(5) 10 Intro to Macro Ch 13-16 - Fiscal Policy, Expansionary and Contractionary Fiscal Policy, Automatic Stabilizers,Introduction To Economics I: Macroeconomics 100%(1) Discover more from: Introduction To Economics I: Macroeconomics ECON 201-CNNorthwestern University63 Documents Go to course 3 Ch. 21 - Taught by Professor Guido Lorenzoni Introduction To Economics I: Macroeconomics 100%(6) 3 Lastset 201 - Problem Sets Introduction To Economics I: Macroeconomics 100%(2) 3 Secondset - Problem Sets Introduction To Economics I: Macroeconomics 80%(5) 10 Intro to Macro Ch 13-16 - Fiscal Policy, Expansionary and Contractionary Fiscal Policy, Automatic Stabilizers,Introduction To Economics I: Macroeconomics 100%(1) 5 Production Possibilities Frontiers - Introduction To Economics I: Macroeconomics Introduction To Economics I: Macroeconomics 100%(1) 7 Lecture 13-16 - Taught by Professor Guido Lorenzoni Introduction To Economics I: Macroeconomics 100%(1) Related Answered Questions 9 months ago are there any books on this syllabus Introduction To Economics I: Macroeconomics (ECON 201-CN) 11 months ago Answer the question based on the frame below/ Beantwoord die volgende vraag op grond van die inligting in die raam hieronder Consider a Keynesian model where: full employment output = R80 million exports = R5 million the marginal propensity to import out of income = 0,25 autonomous imports = R5 million the tax rate = 0,25 investment = R20 million autonomous consumption = R15 million the marginal propensity to consume = 0,6. At the full employment level of output, the government budget will be in _, and the balance of payments on the current account will be in _./ By die volle-indiensnemingsopbrengspeil toon die owerheidsbegroting 'n _ en die betalingsbalans op die lopende rekening 'n _. A. surplus; surplus/ surplus; surplus B. deficit; surplus/ tekort; surplus C. surplus; deficit/ surplus; tekort D. deficit; deficit/ tekort; tekort E. This is not possible to determine from the information given Introduction To Economics I: Macroeconomics (ECON 201-CN) 3 years ago If savers become more ‘patient’ and place more weight on the future, the supply curve in the market for loanable funds will shift left. T or F? Introduction To Economics I: Macroeconomics (ECON 201-CN) 3 years ago The nominal interest rate can be thought of as the intertemporal price of money. T or F? Introduction To Economics I: Macroeconomics (ECON 201-CN) 3 years ago n a closed economy, if the government is running a budget deficit then investment I exceeds private savings S^p = Y − C − T Introduction To Economics I: Macroeconomics (ECON 201-CN) 3 years ago If China’s GDP grows faster than the USA’s GDP, China’s output will eventually exceed that of the USA. T or F? Introduction To Economics I: Macroeconomics (ECON 201-CN) Ask AI Home My Library Discovery Discovery Universities High Schools High School Levels Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI English (US) United States Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. 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16235	https://www.sciencedirect.com/topics/computer-science/contingency-table	Skip to Main content My account Sign in Contingency Table In subject area:Computer Science A contingency table is a summary of information about multiple discrete random variables. It represents the relationship between two variables and is often used in probability and statistics. The table includes row and column marginal totals, which correspond to marginal probability mass functions, and each row and column represents conditional probability mass functions. AI generated definition based on: Introduction to Statistical Machine Learning, 2016 How useful is this definition? Add to Mendeley Also in subject areas: Mathematics Nursing and Health Professions Discover other topics Chapters and Articles You might find these chapters and articles relevant to this topic. Multidimensional Probability Distributions 2016, Introduction to Statistical Machine LearningMasashi SugiyamaMasashiSugiyama 5.3Contingency Table A contingency table summarizes information of multiple discrete random variables. An example of the contingency table is given in Table 5.1: is the random variable representing students’ likes and dislikes of probability and statistics, while is the random variable representing their drowsiness during the lecture. Table 5.1. Example of Contingency Table \| \| Sleepy during the Lecture \| Not Sleepy during the Lecture \| Total \| --- --- \| \| Like statistics and probability \| \| \| \| \| Dislike statistics and probability \| \| \| \| \| Total \| \| \| \| A contingency table corresponds to a probability mass function. The right-most column is called the row marginal total, the bottom row is called the column marginal total, and the right-bottom cell is called the grand total. The row marginal total and the column marginal total correspond to marginal probability mass functions, while each row and each column correspond to conditional probability mass functions. Hypothesis testing using the contingency table will be explained in Section 10.4. View chapterExplore book Read full chapter URL: Book2016, Introduction to Statistical Machine LearningMasashi SugiyamaMasashiSugiyama Chapter Handbook of Chemometrics and Qualimetrics: Part B 1998, Data Handling in Science and TechnologyB.G.M. Vandeginste, ... J. Smeyers-Verbeke 32.1Contingency table Table 32.1 describes 30 persons who have been observed to use one of four available therapeutic compounds for the treatment of one of three possible disorders. The four compounds in this measurement table are the benzodiazepine tranquillizers Clonazepam (C), Diazepam (D), Lorazepam (L) and Triazolam (T). The three disorders are anxiety (A), epilepsy (E) and sleep disturbance (S). In this example, both measurements (compounds and disorders) are defined on nominal scales. Measurements can also be defined on ordinal scales, or on interval and ratio scales in which case they need to be subdivided in discrete and non-overlapping categories. Table 32.1. Measurement table describing the use of four compounds (C, D, L, T) for the treatment of three disorders (A, E, S) by 30 persons. For explanation of symbols, see text. \| # \| Compound \| Disorder \| --- \| 1 \| L \| A \| \| 2 \| D \| E \| \| 3 \| C \| S \| \| 4 \| T \| S \| \| 5 \| L \| A \| \| 6 \| D \| A \| \| 7 \| C \| E \| \| 8 \| C \| S \| \| 9 \| D \| A \| \| 10 \| C \| E \| \| 11 \| D \| S \| \| 12 \| L \| A \| \| 13 \| D \| S \| \| 14 \| C \| E \| \| 15 \| C \| S \| \| 16 \| D \| A \| \| 17 \| L \| S \| \| 18 \| D \| E \| \| 19 \| L \| S \| \| 20 \| L \| A \| \| 21 \| C \| E \| \| 22 \| T \| S \| \| 23 \| D \| A \| \| 24 \| T \| S \| \| 25 \| D \| E \| \| 26 \| T \| A \| \| 27 \| T \| S \| \| 28 \| C \| E \| \| 29 \| D \| E \| \| 30 \| D \| A \| Each column of a measurement table can be expanded into an indicator table. The rows of an indicator table refer to the same objects and in the same order as in the measurement table. The columns of the indicator table represent non-overlapping categories of the selected measurement. Table 32.1 has been expanded into the indicator Table 32.2 for compounds and into the indicator Table 32.3 for disorders. In the indicator table for compounds, a value of one in a particular row is recorded if a person has used the corresponding compound. In the indicator table for disorders, one in a particular row indicates that the person has been treated for the corresponding disorder. All other elements of the row are set to zero. Note that the order of the columns in the indicator tables is not relevant. Table 32.2. Indicator table for compounds, from Table 32.1 \| \| \| \| \| \| --- --- \| # \| Compound \| Empty Cell \| Empty Cell \| Empty Cell \| \| C \| D \| L \| T \| \| 1 \| 0 \| 0 \| 1 \| 0 \| \| 2 \| 0 \| 1 \| 0 \| 0 \| \| 3 \| 1 \| 0 \| 0 \| 0 \| \| 4 \| 0 \| 0 \| 0 \| 1 \| \| 5 \| 0 \| 0 \| 1 \| 0 \| \| 6 \| 0 \| 1 \| 0 \| 0 \| \| 7 \| 1 \| 0 \| 0 \| 0 \| \| 8 \| 1 \| 0 \| 0 \| 0 \| \| 9 \| 0 \| 1 \| 0 \| 0 \| \| 10 \| 1 \| 0 \| 0 \| 0 \| \| 11 \| 0 \| 1 \| 0 \| 0 \| \| 12 \| 0 \| 0 \| 1 \| 0 \| \| 13 \| 0 \| 1 \| 0 \| 0 \| \| 14 \| 1 \| 0 \| 0 \| 0 \| \| 15 \| 1 \| 0 \| 0 \| 0 \| \| 16 \| 0 \| 1 \| 0 \| 0 \| \| 17 \| 0 \| 0 \| 1 \| 0 \| \| 18 \| 0 \| 1 \| 0 \| 0 \| \| 19 \| 0 \| 0 \| 1 \| 0 \| \| 20 \| 0 \| 0 \| 1 \| 0 \| \| 21 \| 1 \| 0 \| 0 \| 0 \| \| 22 \| 0 \| 0 \| 0 \| 1 \| \| 23 \| 0 \| 1 \| 0 \| 0 \| \| 24 \| 0 \| 0 \| 0 \| 1 \| \| 25 \| 0 \| 1 \| 0 \| 0 \| \| 26 \| 0 \| 0 \| 0 \| 1 \| \| 27 \| 0 \| 0 \| 0 \| 1 \| \| 28 \| 1 \| 0 \| 0 \| 0 \| \| 29 \| 0 \| 1 \| 0 \| 0 \| \| 30 \| 0 \| 1 \| 0 \| 0 \| \| All \| 8 \| 11 \| 6 \| 5 \| Table 32.3. Indicator table for disorders, from Table 32.1 \| \| \| \| \| --- --- \| \| # \| Disorder \| Empty Cell \| Empty Cell \| \| A \| E \| S \| \| 1 \| 1 \| 0 \| 0 \| \| 2 \| 0 \| 1 \| 0 \| \| 3 \| 0 \| 0 \| 1 \| \| 4 \| 0 \| 0 \| 1 \| \| 5 \| 1 \| 0 \| 0 \| \| 6 \| 1 \| 0 \| 0 \| \| 7 \| 0 \| 1 \| 0 \| \| 8 \| 0 \| 0 \| 1 \| \| 9 \| 1 \| 0 \| 0 \| \| 10 \| 0 \| 1 \| 0 \| \| 11 \| 0 \| 0 \| 1 \| \| 12 \| 1 \| 0 \| 0 \| \| 13 \| 0 \| 0 \| 1 \| \| 14 \| 0 \| 1 \| 0 \| \| 15 \| 0 \| 0 \| 1 \| \| 16 \| 1 \| 0 \| 0 \| \| 17 \| 0 \| 0 \| 1 \| \| 18 \| 0 \| 1 \| 0 \| \| 19 \| 0 \| 0 \| 1 \| \| 20 \| 1 \| 0 \| 0 \| \| 21 \| 0 \| 1 \| 0 \| \| 22 \| 0 \| 0 \| 1 \| \| 23 \| 1 \| 0 \| 0 \| \| 24 \| 0 \| 0 \| 1 \| \| 25 \| 0 \| 1 \| 0 \| \| 26 \| 1 \| 0 \| 0 \| \| 27 \| 0 \| 0 \| 1 \| \| 28 \| 0 \| 1 \| 0 \| \| 29 \| 0 \| 1 \| 0 \| \| 30 \| 1 \| 0 \| 0 \| \| All \| 10 \| 9 \| 11 \| A contingency table X can be constructed by means of the matrix product of two indicator tables: (32.1) where I is the indicator table for the first measurement and J is the indicator matrix for the second measurement. The rows of I and J must refer to the same objects and in the same order. The dimensions of a contingency table are defined by the number of categories of the two measurements. Each element xij in an n × p contingency table X represents the number of objects that can be associated simultaneously with category i of the first measurement and with category j of the second measurement. The element xij can be interpreted as the number of occurrences of i contingent (or conditional) upon j, or equivalently as the number of instances of j contingent upon i. The 4 × 3 contingency Table 32.4 is the result of multiplying the 30 × 4 indicator Table 32.2 for compounds with the 30 × 3 indicator Table 32.3 for disorders. Table 32.4. Contingency table X derived from the indicator Tables 32.2 and 32.3 \| \| \| \| \| \| --- --- \| Empty Cell \| Anxiety (A) \| Epilepsy (E) \| Sleep (S) \| Sum \| \| Clonazepam (C) \| 0 \| 5 \| 3 \| 8 \| \| Diazepam (D) \| 5 \| 4 \| 2 \| 11 \| \| Lorazepam(L) \| 4 \| 0 \| 2 \| 6 \| \| Triazolam (T) \| 1 \| 0 \| 4 \| 5 \| \| Sum \| 10 \| 9 \| 11 \| 30 \| In this chapter we deal explicitly with contingency tables that result from combining two measurements and which are generally referred to as two-way contingency tables. From this point of view, it can be regarded as a generalization of Chapter 16 to the case of more than two categories for each of the two measurements. It is possible to cross multiple measurements, which then results in a multi-way contingency table. For example, when three measurements with respectively n, p and q categories are crossed with one another, this results in an n × p × q three-way contingency table. Recently there has been a growing interest in the analysis of this type of data structure. In this chapter, however, we only deal with two-way contingency tables. Summation row-wise (horizontally) of the elements of a contingency table produces the vector of row-sums with elements xi+. Summation column-wise (vertically) yields the vector of column-sums with elements x+ j. The global sum is denoted by x++. These marginal sums are defined as follows: (32.2) View chapterExplore book Read full chapter URL: Book series1998, Data Handling in Science and TechnologyB.G.M. Vandeginste, ... J. Smeyers-Verbeke Chapter Handbook of Chemometrics and Qualimetrics: Part B 1998, Data Handling in Science and TechnologyB.G.M. Vandeginste, ... J. Smeyers-Verbeke 32.6.1Historical background The origin of the analysis of contingency tables can be traced back to the definition by Pearson in 1900 of the chi-square statistic as a measure for goodness of fit . The partition of the chi-square into main effects and interaction has been described around 1950 by Lancaster . One of the first practical applications of multivariate analysis to contingency tables is due to Hill who developed the method of reciprocal averaging. This method is used for the ordination of plant species from tabulated surveys carried out in various plots of land yielding a species × plots matrix. The result allows one to order species according to a latent vector which can be ascribed to environmental characteristics, such as density of the soil, abundance of sunlight, thickness of cover, etc. Reciprocal averaging is considered as the forerunner of correspondence factor analysis (CFA) which has been developed by Benzécri [6,7] and a group of French statisticians . It is often also referred to as correspondence analysis (CA) for short. Originally, the term correspondence means association between rows and columns of a table, i.e. the number of joint occurrences of the categories represented by the rows and columns of a contingency table. The French term ‘analyse des correspondances’ points to the analysis of the interaction between rows and columns. The French school also used a geometrical approach for the interpretation of its results, especially by means of the biplot which has been designed by Gabriel . In the discussion of measurement tables in Chapter 31, we have defined biplots as the joint representation of rows and columns of a table in a low-dimensional space spanned by latent vectors. The concept of latent vectors will be generalized to the case of weighted metrics in the following section. Initially, CFA has been applied in linguistic studies. Later on, applications have been described in many diverse fields of the empirical sciences. The method of CFA can be extended to the analysis of tables with non-negative elements, not necessarily counts, provided that these have been recorded with a common unit of measurement. Such a table is called homogeneous, as opposed to a heterogeneous table in which the rows or columns are defined with different units. Heterogeneous tables can be analyzed by means of CFA, after subdivision of the measurements into discrete categories (e.g., in the case of lipophilicity: strongly lipophilic, weakly lipophilic or hydrophilic, strongly hydrophilic). CFA has played a decisive role in the development of multivariate data analysis, especially by means of its graphic aspect. It also has stimulated interest in related areas such as principal components analysis and cluster analysis. The position of CFA, as well as that of its related methods, is that of a preliminary step to statistical inference. In this respect, it has been regarded as an inductive approach . This means that, first, hypotheses have to be formulated from an analysis of empirical evidence. After this exploratory phase, the newly formulated hypotheses have to be tested more rigorously using appropriate statistical methods of inference. Correspondence factor analysis can be described in three steps. First, one applies a transformation to the data which involves one of the three types of closure that have been described in the previous section. This step also defines two vectors of weight coefficients, one for each of the two dual spaces. The second step comprises a generalization of the usual singular value decomposition (SVD) or eigenvalue decomposition (EVD) to the case of weighted metrics. In the third and last step, one constructs a biplot for the geometrical representation of the rows and columns in a low-dimensional space of latent vectors. View chapterExplore book Read full chapter URL: Book series1998, Data Handling in Science and TechnologyB.G.M. Vandeginste, ... J. Smeyers-Verbeke Review article A four-phase model of the evolution of clinical decision support architectures 2008, International Journal of Medical InformaticsAdam Wright, Dean F. Sittig Two years after Ledley and Lusted published their paper, Warner of the LDS Hospital and the University of Utah published a mathematical model for diagnosing congenital heart defects . This model used contingency tables to map symptoms and signs to diagnoses, based on the frequency of manifestation for each symptom or sign given an underlying diagnosis. The system was evaluated by comparing its diagnoses with gold-standard surgical diagnoses, and was found to compare favorably with experienced cardiologists. View article Read full article URL: Journal2008, International Journal of Medical InformaticsAdam Wright, Dean F. Sittig Chapter Correlations 2005, Encyclopedia of Social MeasurementAndrew B. Whitford Contingency Coefficient (C) and Cramer's V The contingency coefficient (C) and Cramer's V are appropriate when relating two variables measured on the unordered polychotomous scale. The formula for C for a population is based on the χ2 test of independence across the cells of an r × c contingency table: where Here, Aij is the actual count of cases in the ijth cell and Eij is the expected count when the null of independence is true and is calculated based on the marginals; C is not bounded +1.0, but instead depends on the number of categories for the row and column variables. This limitation makes all C values calculated across data sets incomparable unless the coding scheme for the two data sets is exactly the same. Cramer's V is intended to limit this deficiency by ranging only from 0 to 1. The formula is again based on the calculated value of χ2 but now adjusted by the smallest number of rows or columns in the table: View chapterExplore book Read full chapter URL: Reference work2005, Encyclopedia of Social MeasurementAndrew B. Whitford Chapter Graphical Models: Overview 2001, International Encyclopedia of the Social & Behavioral SciencesN. Wermuth, D.R. Cox 2Models for Contingency Tables and the Need for Extensions For a q-dimensional contingency table, i.e., for counts on q categorical variables, the most complex undirected graph model has no constraints. It is called the saturated model and is represented by a graph in q vertices or nodes, with each node pair having exactly one edge, i.e., being connected by a full line. Each full line represents the conditional association of the connected variable pair given all remaining q − 2 variables. Removing a line from the graph introduces a particular independence constraint for the corresponding variable pair: the pair are to be conditionally independent given all remaining variables. In a loglinear model formulation this is achieved by setting the two-factor and all higher-order interaction terms involving this variable pair to zero. If all full lines are removed, the graph of q disconnected nodes results. It represents the simplest model in the class, the one of mutual independence of all q variables. This development built on previous work (quoted in monographs on graphical models) by M. S. Bartlett, M. W. Birch, L. A. Goodman, Y. Y. Bishop, S. E. Fienberg and P. Holland and much earlier work by A. A. Markov, who used the notion of conditional independence around 1900 to formulate simple multivariate models, now called Markov chains. A Markov chain for q categorical variables can be represented by an undirected graph which is a single path of full lines, i.e., a sequence of nodes i1, i2,…, iq where just each consecutive pair is connected by a full line. Compared to the larger class of loglinear models the graphical models are attractive for three main reasons: (a) the interpretation of a model can be much simpler if independencies are taken into account compared to only loglinear interactions; (b) parameter estimation can be strengthened by basing it on a sequence of possible small contingency tables which depends on the unique decomposition of a given graph into its prime graphs, i.e., into graphs which in a graphical sense cannot be split any further; and (c) a powerful separation criterion permits one to read all independence statements implied by a model directly off the graph. This separation criterion for undirected graphs takes any three nonoverlapping subsets a, b, c of nodes in the graph of which the set c may be empty. Then the corresponding variables Xa are conditionally independent of Xb given Xc if every path from a to b has a node in c. This holds provided two conditions are satisfied: all pairwise independence statements specified with the missing edges of the undirected graph hold, and a technical positivity condition is fulfilled. One sufficient condition for checking the latter from an observed contingency table is that there are no zero frequencies. In graphical models represented by undirected graphs all variables are treated on an equal footing, i.e., response variables and possible explanatory variables are not distinguished. This is appropriate for various different types of data, for instance for symptoms of a given disease, for items which are to measure slightly varying aspects of a particular attitude or of a behavior of persons, or for several aspects which all might contribute to a specific risk factor. However, for much multivariate research some response variables are of primary interest. There is in addition a set of background or context variables, and there are possibly sequences of intermediate variables which play the role of potential response variables to some and of potential explanatory variables to other variables under investigation. Typically, not all variables are only categorical; some are nominal, some are ordinally scaled, and still others are quantitative measurements. Subsets of variables may have to be considered as joint responses instead of univariate responses. View chapterExplore book Read full chapter URL: Reference work2001, International Encyclopedia of the Social & Behavioral SciencesN. Wermuth, D.R. Cox Chapter Handbook of Chemometrics and Qualimetrics: Part B 1998, Data Handling in Science and TechnologyB.G.M. Vandeginste, ... J. Smeyers-Verbeke 32.2Chi-square statistic Each element xij of a contingency table X can be thought of as a random variate. Under the assumption that all marginal sums are fixed, we can derive the expected values E(xij) for each of the random variates xij : (32.3) Table 32.5 presents the expected values of the elements in the contingency Table 32.4. Note that the marginal sums in the two tables are the same. There are, however, large discrepancies between the observed and the expected values. Small discrepancies between the tabulated values of our illustrations and their exact values may arise due to rounding of intermediate results. Table 32.5. Table of expected values E(X) computed from Table 32.4 \| \| \| \| \| \| --- --- \| Empty Cell \| Anxiety \| Epilepsy \| Sleep \| Sum \| \| Clonazepam \| 2.67 \| 2.40 \| 2.93 \| 8 \| \| Diazepam \| 3.67 \| 3.30 \| 4.03 \| 11 \| \| Lorazepam \| 2.00 \| 1.80 \| 2.20 \| 6 \| \| Triazolam \| 1.67 \| 1.50 \| 1.83 \| 5 \| \| Sum \| 10 \| 9 \| 11 \| 30 \| For example, the observed value for Clonazepam in anxiety has been recorded as 0 in Table 32.4. The corresponding expected value in Table 32.5 has been computed from eq. (32.3) as follows: where x11 is the element in the first row and first column of the table, where x1 +, x+ 1 are the corresponding marginal sums, and where x++ is the global sum. The generalization of Pearson’s chi-square statistic χ2 for 2 × 2 contingency tables, which has been discussed in Section 16.2.3, can be written as: (32.4) with (n − 1) (p − 1) degrees of freedom. The chi-square statistic is a measure for the global deviation of the observed values in a contingency table from their expected values. The number of degrees of freedom of the chi-square statistic defines the number of elements of the table that can be varied independently of each other. Since n row-sums and p column-sums are held fixed and since both add up to the same global sum, the number of degrees of freedom of an n × p contingency table is np minus n + p − 1 or (n − 1)(p − 1). The chi-square statistic can be tested for significance using tabulated critical values of the statistic for a fixed level of significance (e.g. 0.05, 0.01 or 0.001) and for the specified degrees of freedom. In the case of the 4 × 3 contingency Table 32.4 we obtain a chi-square value of 15.3 with 6 degrees of freedom, which is significant at the 0.05 level of probability as it exceeds the critical value of 12.6. The probability corresponding with a chi-square value exceeding 15.3 equals 0.02. From this result we may be led to conclude that there are significant differences between the prescription patterns of the four compounds, as well as between the treatment patterns of the three disorders. One should be cautious, however, in drawing such a conclusion, because of three considerations. First, the statistical evidence of this positive outcome is not overwhelming, as it leaves a two percent chance of being false. Secondly, the chi-square test is not quite appropriate in this case, since all but two cell frequencies are smaller than five and three of them are even zero. Finally, a statistically significant result does not necessarily mean that the differences between compounds and disorders are of practical relevance. These considerations should not preclude, however, an exploratory analysis of the data in order to search for meaningful correlations and trends that might be the object of more detailed studies in the future. In the case of a negative outcome of the chi-square test, one might have assumed that there are no differences between the drugs and between the disorders. This conclusion also might be erroneous in view of the lack of power that is inherent to a small sample (30 persons in this example). In this case, an exploratory analysis may still yield relevant associations and tendencies which can guide the design of future studies. View chapterExplore book Read full chapter URL: Book series1998, Data Handling in Science and TechnologyB.G.M. Vandeginste, ... J. Smeyers-Verbeke Chapter Yule, George Udny 2005, Encyclopedia of Social MeasurementLeslie Hepple Contingency Tables The second arena in which Yule made a contribution to quantitative social research was the analysis of categorical variables and contingency tables. The statistical advances of the 1890s, including Yule's own work, had focused on interval or continuous-scale variables, but by the turn of the century, attention was also being given to attribute or categorical variables. Yule gave the example of mortality from some disease with and without the administration of a new antitoxin: an individual died or lived, had the antitoxin or did not, but there were no continuous scales of variation involved. In other cases, as Yule noted, there might be gradation possible, but the actual data were categorical (blind or not blind, deaf or not deaf). For the two-category, two-variable case, a simple contingency table (as in Table I) can represent the data: thus a is the number of people who had the antitoxin and survived, c is those who had antitoxin but still died, and a + c is the total number of people who had the antitoxin; b and d are defined similarly for the “no antitoxin” groups, and N (= a + b + c + d) gives the total number of people involved. How were correlations or other measures of association to be constructed for such cases? Table I. A Simple Two-by-Two Contingency Table \| Outcome \| Antitoxin \| No antitoxin \| Total \| --- --- \| \| Survived \| a \| b \| a + b \| \| Died \| c \| d \| c + d \| \| Total \| a + c \| b + d \| N \| Yule and Pearson came up with different perspectives on this situation, leading to serious, and sometimes intense, controversy, described in the literature as “the politics of the contingency table.” Pearson wanted to treat all such categorical data as simplifications from an underlying normal frequency distribution, so accommodating them within his biometric correlation frame. This is plausible for some cases, but not for others, such as Yule's mortality example. By contrast, Yule took the categories as given, and attempted to construct measures of association relevant to the actual case. A coefficient, calculated directly from the table, should be zero if the two attributes were independent of each other, +1 or −1 if, and only if, they were completely associated in a positive or negative sense. Yule's Q-coefficient, defined as Q = (ad − bc)/(ad + bc), did this, but it had no special justification, and other measures, equally valid, could be constructed but did not always give identical inferences. The controversy between Yule and Pearson, and that between their allies, was seen as exemplary of a deeper division in terms of attitudes to the biometric program and eugenics. This perhaps overinterprets the differences, but certainly it reflected Yule's more pragmatic, flexible approach to inference and Pearson's allegiance to the model of normal frequency distribution. Both perspectives had their supporters. Pearson's measures were adopted within the emerging field of psychometrics whereas Yule's measures became popular in sociology. The field of contingency table and categorical data analysis has moved on a long way since this early work, and overall has built on Yule's insights rather than on Pearson's more rigid assumptions. View chapterExplore book Read full chapter URL: Reference work2005, Encyclopedia of Social MeasurementLeslie Hepple Chapter Bivariate descriptive statistics 2023, Data Science, Analytics and Machine Learning with RLuiz Paulo Fávero, ... Rafael de Freitas Souza Introduction In this chapter we study the associations between two qualitative variables using contingency tables and measures of association, such as chi-square statistics (used for nominal and ordinal qualitative variables. We also study the correlation between two quantitative variables using joint frequency distribution tables, as well as graphical representations such as a scatter plots. R-based language packages used for this chapter : library(sjPlot) : library(DescTools) Don't forget to define the R working directory (the location where your datasets are installed): : setwd("C:/book/chapter5") View chapterExplore book Read full chapter URL: Book2023, Data Science, Analytics and Machine Learning with RLuiz Paulo Fávero, ... Rafael de Freitas Souza Chapter Multivariate Analysis: Discrete Variables (Overview) 2001, International Encyclopedia of the Social & Behavioral SciencesA. Agresti 1Traditional Analyses of Discrete Multivariate Data Most commonly, discrete variables are categorical in nature. That is, the measurement scale is a set of categories, such as (liberal, moderate, conservative) for political philosophy. In multivariate analyses with two or more categorical variables, a contingency table displays the counts for the cross-classification of categories. For instance, a contingency table might display counts indicating the number of subjects classified in each of the 9=3×3 combinations of responses on political philosophy and political party orientation (Democrat, Republican, Independent). Those combinations form the cells of the table. Two categorical variables are independent if the probability of response in any particular category of one variable is identical for each category of the other variable. This is an idealized situation that is rarely true in practice, although like any statistical model it is sometimes adequate for describing reality. For instance, some surveys of American college students have shown that opinion about whether abortion should be legal is essentially independent of gender; that is, the percentage favoring legalization is about the same for male and female students. Treating independence as a simple model for a contingency table, we can test its fit using statistics that compare the observed cell counts in the table to those expected (predicted) by that model. The most commonly used test statistics are (1) These both have large-sample chi-squared distributions. For r rows and c columns, the degrees of freedom (df) for the test equal df=(r−1)(c−1), and the P-value is the right-tail probability above the observed test statistic value. These chi-squared statistics also can compare counts to expected values for more complex models than independence. In conducting goodness-of-fit tests, one does not seriously entertain the possibility that the model truly holds; any model is false and provides a simple representation that only approximates reality. However, when the chi-squared statistics do not show significant lack of fit, benefits of model parsimony accrue from using the model fit, which smooths the sample cell proportions somewhat. Interpretations are simpler, and estimates of population characteristics such as the true cell proportions are likely to be better using the model-based estimates. Multidimensional contingency tables result from cross-classifying more than two categorical variables, such as for studying the association separately at levels of control variables race, gender, or educational level. The model of conditional independence states that two categorical variables are independent at each level of a third; that is, ordinary independence applies in each partial table. Large-sample chi-squared tests of fit of this model, such as the Cochran–Mantel–Haenszel test, combine the information from the partial tables. Most are designed to be relatively powerful when the strength of the association is similar in each partial table. The model of homogeneous association states that the strength of association between two variables is identical at each level of a third variable. Goodness-of-fit tests of these models were originally formulated for stratified two-by-two tables, but generalizations exist for stratified r-by-c tables with ordered or unordered categories. Goodness-of-fit tests have limited use, and in many circumstances it is more informative to estimate the strength of the associations. Goodman and Kruskal (1979) discussed a variety of summary measures. Those for ordinal variables, such as their gamma measure, are correlation-like indices that fall between −1 and +1. More attention has been focused on measures that arise as parameters in models. The odds ratio, discussed in the next section, is a parameter in logistic regression and loglinear models and is applicable for two-way or multi-way tables. For estimating parameters in models for discrete data, the method of maximum likelihood estimation is the standard. The probability function for the data (e.g., Poisson, binomial), expressed as a function of the parameters after observing the data, is called the likelihood function. The maximum likelihood estimates are values for the parameters for which this achieves its maximum; that is, these estimates are the parameter values for which the probability of the observed data takes its maximum value. Except in a few simple cases, numerical methods are needed to find the maximum likelihood estimates and their standard errors. Software is reasonably well developed in the major statistical packages. It is often most natural to analyze discrete data using procedures for generalized linear models that permit non-normal responses, such as the Poisson and binomial. Examples include PROC GENMOD in SAS and the ‘glm’ function in S-Plus. For instance, for SAS see Allison (1999) and Stokes et al. (1995), for S-Plus see Venables and Ripley (1999), and for Stata see Rabe-Hesketh and Everitt (2000). View chapterExplore book Read full chapter URL: Reference work2001, International Encyclopedia of the Social & Behavioral SciencesA. Agresti Related terms: Ontology Analysis of Variance Categorical Data Categorical Variable Conditional Probability Singular Value Loglinear Model Response Variable Square Statistic Expected Frequency View all Topics
16236	http://web.mit.edu/smcs/8.02/	\| \| \| \| \| \| \| --- \| \| Electricity and MagnetismSpring 2002Lecturer: Prof. Walter Lewin Return to 8.02 Homepage \| \| \| 8.02 Spring 2002 Lecture 1 What holds our world together? Electric charges (historical) Polarization Electric Force Coulomb's Law Recorded on 02/06/02 56K 80K 220K 8.02 Spring 2002 Lecture 2 Electric Field Field Lines Superposition Inductive Charging Dipoles Induced Dipoles Recorded on 02/08/02 56K 80K 220K 8.02 Spring 2002 Lecture 3 Electric Flux Gauss's Law Examples Recorded on 02/11/02 56K 80K 220K 8.02 Spring 2002 Lecture 4 Electrostatic Potential Electric Energy eV Conservative Field Equipotential Surfaces Recorded on 02/13/02 56K 80K 220K 8.02 Spring 2002 Lecture 5 E = -grad V More on Equipotential Surfaces Conductors Electrostatic Shielding (Faraday cage) Recorded on 02/15/02 56K 80K 220K 8.02 Spring 2002 Lecture 6 High-voltage breakdown Lightning Sparks - St. Elmo's Fire Recorded on 02/19/02 56K 80K 220K 8.02 Spring 2002 Lecture 7 Capacitance Field energy Recorded on 02/20/02 56K 80K 220K 8.02 Spring 2002 Lecture 8 Polarization Dielectrics The Van de Graaff More on Capacitors Recorded on 02/22/02 56K 80K 220K 8.02 Spring 2002 Lecture 9 Currents Resistivity Ohm's Law Recorded on 02/25/02 56K 80K 220K 8.02 Spring 2002 Lecture 10 Batteries EMF Energy Conservation Power Kirchhoff's Rules Circuits Kelvin Water Dropper Recorded on 02/27/02 56K 80K 220K 8.02 Spring 2002 Lecture 11 Magnetic field Lorentz force Torques Electric Motors (DC) Cathode Ray Tube Oscilloscope Recorded on 03/01/02 56K 80K 220K 8.02 Spring 2002 Lecture 12 Review Exam 1 (Secret Top!) 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Recorded on 03/20/02 56K 80K 220K \| 8.02 Spring 2002 Lecture 19 Vacation Special How do Magicians levitate women? 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16237	https://www.quora.com/Is-an-empty-set-a-proper-subset-of-all-sets	Something went wrong. Wait a moment and try again. Difference Between Proper... Zermelo-Fraenkel Set Theo... Sets (mathematics) Subset Relationship Improper Subset Void Set Number of Subsets 5 Is an empty set a proper subset of all sets? Henk Brozius Studied Mathematics · Upvoted by Keith Ramsay , PhD Mathematics, Harvard University (1991) and Terry Moore , M.Sc. Mathematics, University of Southampton (1968) · Author has 3.6K answers and 2.2M answer views · 4y There is only one empty set. The empty set is a subset of every set. The empty set is not a proper subset of the empty set. The empty set is a proper subset of every non-empty set. Related questions What is the proper subset of an empty set? Is the following statement true: "the empty set is a subset of every set"? Is a set containing an empty set a proper subset of any set? If the empty set is a proper subset of every set besides itself, why doesn't it count as an element of every set besides itself? Is an empty set a proper subset or an improper subset of a non-empty set A? David Joyce Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979) · Upvoted by Patrick T , PhD Mathematics, University of California, Davis (2018) · Author has 9.9K answers and 68.4M answer views · Updated 3y Originally Answered: Is a set containing an empty set a proper subset of any set? · No set is a proper subset of every set. A set S is a subset of a set T when every element of S is also an element of T. A subset S is a proper subset of T when there’s some element of T that is not an element of S, that is, when S≠T. Every set is a subset of itself, but it is not a proper subset of itself. On the other hand, every set is a proper subset of some set. Just add some element that’s not in the first set. So, for example, the set containing an empty set, which is the set {∅}, is a proper subset of the set {∅,{∅}}. Assistant Bot · 1y Yes, the empty set (denoted as ∅) is considered a proper subset of all sets. Explanation: Definition of Proper Subset: A set A is a proper subset of a set B if all elements of A are in B, and A is not equal to B (i.e., A≠B). Empty Set as a Subset: The empty set is a subset of any set B because there are no elements in the empty set that could violate the condition of being a subset. Proper Subset Condition: Since the empty set contains no elements, it cannot be equal to any non-empty set B. Therefore, it satisfies the condition of being a proper subset of any non-empty set. Conclusion: The em Yes, the empty set (denoted as ∅) is considered a proper subset of all sets. Explanation: Definition of Proper Subset: A set A is a proper subset of a set B if all elements of A are in B, and A is not equal to B (i.e., A≠B). Empty Set as a Subset: The empty set is a subset of any set B because there are no elements in the empty set that could violate the condition of being a subset. Proper Subset Condition: Since the empty set contains no elements, it cannot be equal to any non-empty set B. Therefore, it satisfies the condition of being a proper subset of any non-empty set. Conclusion: The empty set is a proper subset of every non-empty set. It is also a subset of itself, but in that case, it is not a proper subset because it does equal the set. In summary, the empty set is a proper subset of all non-empty sets. Jim Hollerman Studied Mathematics · Upvoted by Dwight House , Ph. D. Mathematics, Duke University (1972) · Author has 1.4K answers and 1.2M answer views · Updated 4y Originally Answered: Is the following statement true: "the empty set is a subset of every set"? · “Is the following statement true: "the empty set is a subset of every set"?” Yes, it is true. Definition: Set A is a subset of set B if and only if for every x, if x∈A, then x∈B An if ____, then ____ statement is false only when the antecedent (the first part) is true and the consequent is false. If A is empty, then x∈A is false for all x, so the conditional is true. If that’s not clear, note that A fails to be a subset of B if and only if there is an x in A that is not also in B. If A is empty then there is no possibility of such an x. Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Related questions Is the empty set ∅ a subset of itself? Is an empty set the subset of a set containing the empty set? Why is the empty set a subset of every set? If an empty set is a proper subset of any non-empty set A, how can we explain that every element of the empty set is an element of A? What is the set in which a subset is empty called? Eugene Bucamp Knows French · Author has 4.1K answers and 1.1M answer views · 5y Originally Answered: Is the following statement true: "the empty set is a subset of every set"? · It is true or it is false, which at least is always true. So, if for I don’t know what reason you want to prove formally that it is true, then you have to make assumptions. Otherwise, no, it isn’t necessarily true. It is just… true or false. Your assumptions, no mine… But we don’t have to assume that there is such a thing as an empty set. Essentially, this is a convention adopted by some mathematicians and subsequently accepted by most mathematicians. An assumed definition. An assumption. Or, rather the consequence of an earlier assumption. Today’s mathematicians, well, mostly, they learned the blo It is true or it is false, which at least is always true. So, if for I don’t know what reason you want to prove formally that it is true, then you have to make assumptions. Otherwise, no, it isn’t necessarily true. It is just… true or false. Your assumptions, no mine… But we don’t have to assume that there is such a thing as an empty set. Essentially, this is a convention adopted by some mathematicians and subsequently accepted by most mathematicians. An assumed definition. An assumption. Or, rather the consequence of an earlier assumption. Today’s mathematicians, well, mostly, they learned the bloody thing at school, so there won’t be any debate here. Thus, rather than say that there is no element satisfying this or that relation, people say the set is empty, or it is the null set. And then, one thing brings another, and suddenly you need to accept that the null set is subset of each and every set in the Creation. Somewhat like saying that when you have something in your pocket, you also have something else which is nothing. Which, then, is just an inevitable and unfortunate consequence of your assumptions. And then the empty set has to be a subset of the empty set… Yeah! So, you can never be done counting empty subsets. Mathematical emptiness is a Matryoshka doll. Me, when I look into my wallet and find nothing, I don’t go thinking, well, that nothing is still something, a something which is nothing, which is something, which is nothing, and so on, for ever and ever and ever. Well, seems the ontological implications of your careless assumption are just a little staggering. Well, think of zero. Zero is zero plus zero. Which is therefore zero plus zero plus zero plus zero. And, four is four plus zero. And four minus zero, too. This suddenly gives another meaning to infinity. Not only is there an infinity of points in a line, but there is an infinity of points in one point, and an infinity of lines in one line, and an infinity of planes in one plane. Who ever said infinity couldn’t possibly exist?! We defined it into existence. Because of some assumption someone once made. Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Alan Bustany Trinity Wrangler, 1977 IMO · Upvoted by Dwight House , Ph. D. Mathematics, Duke University (1972) and Erik Bergland , PhD Mathematics, Brown University (2024) · Author has 9.8K answers and 58.5M answer views · 6y Related The empty set is a subset of every set. Therefore the empty set is a subset of the empty set. So, empty set is not empty. Am I right or wrong? You are wrong. Your reasoning depends on “{} is a subset of S” therefore “S is not empty”. That is a non sequitur. If you think otherwise you need to explain your rationale. For S to be non-empty it must have a member. Showing that A is a subset of S and that A has some members would do the trick. Your problem is that you have A={} and, by definition, the empty set has no members (even though, as you point out, it does have a subset). In common parlance "subset" is often taken to mean what mathematicians call a proper subset — a subset that is not the set itself. The empty set does not have a You are wrong. Your reasoning depends on “{} is a subset of S” therefore “S is not empty”. That is a non sequitur. If you think otherwise you need to explain your rationale. For S to be non-empty it must have a member. Showing that A is a subset of S and that A has some members would do the trick. Your problem is that you have A={} and, by definition, the empty set has no members (even though, as you point out, it does have a subset). In common parlance "subset" is often taken to mean what mathematicians call a proper subset — a subset that is not the set itself. The empty set does not have any proper subsets. English is also imprecise with the preposition "in" as we might say a proper subset is (contained) in a set as well as saying an element is in the set (meaning is a member of the set) although the two meanings are quite distinct. It is this confusion that is probably at the source of the question. Danya Rose I'm an Aussie feller, I've been one all my life. It may make me seem funny... · Author has 2K answers and 7M answer views · Updated 8y Related Is an empty set the subset of a set containing the empty set? First, in a given set theory, there is only one empty set. Thus the indefinite article “an” is incorrect. Second, there is only one set which has exactly one subset; the empty set itself. Using the definite article “the” to refer to “the subset of a set” implies that the set you’re asking about is the empty set. This is because , if we use \|X\| to denote (in some sense) the “size” or “number of elements” of X, the number of subsets of X is 2\|X\|. Thus 2\|∅\|=20=1. Any set X containing the empty set (in the sense that ∅∈X) has at least two subsets, because it has at leas First, in a given set theory, there is only one empty set. Thus the indefinite article “an” is incorrect. Second, there is only one set which has exactly one subset; the empty set itself. Using the definite article “the” to refer to “the subset of a set” implies that the set you’re asking about is the empty set. This is because , if we use \|X\| to denote (in some sense) the “size” or “number of elements” of X, the number of subsets of X is 2\|X\|. Thus 2\|∅\|=20=1. Any set X containing the empty set (in the sense that ∅∈X) has at least two subsets, because it has at least one member. Regardless of even that fact, the empty set is a subset of every set. This is because of the definition of a subset. We say that X⊆Y if and only if for each x∈X it is true that x∈Y. Because there is no x such that x∈∅, this statement is vacuously true of the empty set: every element of ∅ is indeed a member of every definable set, so the empty set is a subset of every set. Sponsored by CDW Corporation How well do your distributed teams work together? Enable seamless and secure communication for everyone with Cisco collaboration solutions from CDW. Rishabh Narula 3y Originally Answered: I have some reasons for why the empty set is a subset of every set. Are they correct? I think they are. · I think the empty set being a subset of every set is like a notion of the idea that if everything in the universe is seen just as number of things, then the number representing collection of smallest number of things would be there in the number representing any collection of all things. or another way of looking at it is if any collection of things is considered as a universe itself,then of course you could keep taking things away from it to make the collection smaller,and conclude this collection of things was inside what you originally had and conclude that nothing must be in it by the end o I think the empty set being a subset of every set is like a notion of the idea that if everything in the universe is seen just as number of things, then the number representing collection of smallest number of things would be there in the number representing any collection of all things. or another way of looking at it is if any collection of things is considered as a universe itself,then of course you could keep taking things away from it to make the collection smaller,and conclude this collection of things was inside what you originally had and conclude that nothing must be in it by the end of the process. hence nothing is in everything. One more idea is if it is not a subset of some set, then there must be something in it that is not there in the other set. there is nothing in it. Joseph Rosenstein Ph.D. in Mathematics, Cornell University (Graduated 1966) · Author has 215 answers and 768.9K answer views · 7y Originally Answered: Is a set containing an empty set a proper subset of any set? · The empty set is a subset of every set, since each element of the empty set (there are none!) is an element of every set. The empty set is a proper subset of every set … except that it is not a proper subset of itself. A set S containing an empty set is not a subset of any set that does not contain the empty set. Since the empty set is an element of S, it would have to be an element of any set T of which S is a subset. Most sets don’t contain the empty set, so a set S that contains an empty set would not be a subset, let alone a proper subset of such sets. For example, the empty set is not an el The empty set is a subset of every set, since each element of the empty set (there are none!) is an element of every set. The empty set is a proper subset of every set … except that it is not a proper subset of itself. A set S containing an empty set is not a subset of any set that does not contain the empty set. Since the empty set is an element of S, it would have to be an element of any set T of which S is a subset. Most sets don’t contain the empty set, so a set S that contains an empty set would not be a subset, let alone a proper subset of such sets. For example, the empty set is not an element of {1}, or of {1,2,3,4,5,6,7}, or of the set N of natural numbers. By the way, it is easy to prove that there is only one empty set, so your question should begin, “Is a set containing the empty set …” Sponsored by CDW Corporation Ready for the latest in AI-powered capability? AMD Ryzen™ and Windows 11-powered x86 PCs from CDW drive business productivity with AI intuition. Keith Ramsay Ph.D. in mathematics · Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) and Anton Fahlgren , M.S. Mathematics, Stockholm University (2018) · Author has 5.4K answers and 8.2M answer views · Updated 7y Related Is there an intuitionist (constructive) proof that the empty set is a subset of every set? The are other answers that correctly point out that the usual classical proof is also a valid proof using intuitionistic logic. The rule that we’re using is “ex falso quodlibet”, which is that F→A when F is the false statement and A is any statement. That x is a member of the empty set is false, and therefore that x is a member of the empty set implies that x is a member of S (regardless of the x and the S). Some sources write misleadingly about the avoidance of “proof by contradiction” in intuitionistic logic. Using “ex falso quodlibet” may seem like a form of proof by contradiction. The are other answers that correctly point out that the usual classical proof is also a valid proof using intuitionistic logic. The rule that we’re using is “ex falso quodlibet”, which is that F→A when F is the false statement and A is any statement. That x is a member of the empty set is false, and therefore that x is a member of the empty set implies that x is a member of S (regardless of the x and the S). Some sources write misleadingly about the avoidance of “proof by contradiction” in intuitionistic logic. Using “ex falso quodlibet” may seem like a form of proof by contradiction. So does a proof that shows a statement A implies both B and B (not B), and concludes that A is false (i.e. A). But both of those are valid in intuitionistic logic. The form of “proof by contradiction” that is not allowed in general is to derive a contradiction by assuming that A is false, and use that to conclude that A is true. Intuitionistic logic allows us to conclude only A, that A is not false. If A has some constructive content (like asserting that a number with some property exists) then we may be able to prove A without being able to construct the number, and thus without being able to prove A. There is a type of logic called minimal logic in which “ex falso quodlibet” is not a rule (and which otherwise is similar to intuitionistic logic). Without this rule, “the false” becomes much less different from other propositions, and so it’s a lot like intuitionistic logic without negation. Errett Bishop commented in his Foundations of Constructive Analysis that he had found avoiding the use of negation had improved some of his proofs. What he actually said was stronger than just that, but I don’t remember exactly how he put it. I don’t find that mathematicians working constructively pursue it as an end in itself, but like it for the valuable qualities it brings to mathematics. So although this result may be fine, if one is using this kind of reasoning as part of a longer proof, constructive mathematics might naturally lead you to ask whether you are using negation unnecessarily. Stating your theorem in a more “positive” way is often a good idea, for not merely following the rules but for strengthening your result. Alexander Farrugia Ph.D. in Mathematics, University of Malta (Graduated 2016) · Upvoted by Nathan Hannon , Ph. D. Mathematics, University of California, Davis (2021) and Tom McFarlane , M.S. Mathematics, University of Washington (1994) · Author has 3.2K answers and 27.5M answer views · 6y Related The empty set is a subset of every set. Therefore the empty set is a subset of the empty set. So, empty set is not empty. Am I right or wrong? The sentence “I ate every vegetable in my plate.” does not imply that there were any vegetables in my plate. Likewise, the sentence “The empty set is a subset of the empty set.” does not imply that the empty set is non-empty. Mathivanan Palraj Thinker · Author has 1.7K answers and 3.9M answer views · 3y Originally Answered: I have some reasons for why the empty set is a subset of every set. Are they correct? I think they are. · In mathematics nothing is also a thing. Why the word ‘nothing’ is included in the dictionary. It has a meaning and therefore it is listed. Moreover, math works well when it is included. Zero has no value but it facilitates counting. And counting rule obeys when ‘nothing’ is included as a subset. For the question ‘how many have you? ‘, ‘nothing’ is also an answer. Originally Answered: Is a set containing an empty set a proper subset of any set? · The wording of your question is ambiguous. For all sets A, is {{}} a proper subset of A? No. For example, {{}} is not a subset (proper or improper) of {2}. Does there exist a set A for which {{}} is a proper subset of A? Yes. For example, {{}} is a proper subset of {{},2} Related questions What is the proper subset of an empty set? Is the following statement true: "the empty set is a subset of every set"? Is a set containing an empty set a proper subset of any set? If the empty set is a proper subset of every set besides itself, why doesn't it count as an element of every set besides itself? Is an empty set a proper subset or an improper subset of a non-empty set A? Is the empty set ∅ a subset of itself? Is an empty set the subset of a set containing the empty set? Why is the empty set a subset of every set? If an empty set is a proper subset of any non-empty set A, how can we explain that every element of the empty set is an element of A? What is the set in which a subset is empty called? How can you show that all subsets of a set are either proper or equal to the original set or empty? What are the additional subsets of an empty set? What are all subsets of {1, empty set}? Is an empty set an element of every set? How can the empty set be a subset of every set, including itself? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
16238	https://en.wikipedia.org/wiki/Indefinite_sum	Jump to content Search Contents (Top) 1 Fundamental theorem of discrete calculus 2 Definitions 2.1 Laplace summation formula 2.2 Newton's formula 2.3 Faulhaber's formula 2.4 Mueller's formula 2.5 Euler–Maclaurin formula 3 Choice of the constant term 4 Summation by parts 5 Period rules 6 Alternative usage 7 List of indefinite sums 7.1 Antidifferences of rational functions 7.2 Antidifferences of exponential functions 7.3 Antidifferences of logarithmic functions 7.4 Antidifferences of hyperbolic functions 7.5 Antidifferences of trigonometric functions 7.6 Antidifferences of inverse hyperbolic functions 7.7 Antidifferences of inverse trigonometric functions 7.8 Antidifferences of special functions 8 See also 9 References 10 Further reading Indefinite sum Bahasa Indonesia 日本語 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia In discrete calculus the indefinite sum operator (also known as the antidifference operator), denoted by or , is the linear operator, inverse of the forward difference operator . It relates to the forward difference operator as the indefinite integral relates to the derivative. Thus More explicitly, if , then If F(x) is a solution of this functional equation for a given f(x), then so is F(x)+C(x) for any periodic function C(x) with period 1. Therefore, each indefinite sum actually represents a family of functions. However, due to the Carlson's theorem, the solution equal to its Newton series expansion is unique up to an additive constant C. This unique solution can be represented by formal power series form of the antidifference operator: . Fundamental theorem of discrete calculus [edit] Indefinite sums can be used to calculate definite sums with the formula: Definitions [edit] Laplace summation formula [edit] The Laplace summation formula allows the indefinite sum to be written as the indefinite integral plus correction terms obtained from iterating the difference operator, although it was originally developed for the reverse process of writing an integral as an indefinite sum plus correction terms. As usual with indefinite sums and indefinite integrals, it is valid up to an arbitrary choice of the constant of integration. Using operator algebra avoids cluttering the formula with repeated copies of the function to be operated on: In this formula, for instance, the term represents an operator that divides the given function by two. The coefficients etc., appearing in this formula are the Gregory coefficients, also called Laplace numbers. The coefficient in the term is where the numerator of the left hand side is called a Cauchy number of the first kind, although this name sometimes applies to the Gregory coefficients themselves. Newton's formula [edit] : where is the falling factorial. Faulhaber's formula [edit] Main article: Faulhaber's formula Faulhaber's formula provides that the right-hand side of the equation converges. Mueller's formula [edit] If then Euler–Maclaurin formula [edit] Main article: Euler–Maclaurin formula Choice of the constant term [edit] Often the constant C in indefinite sum is fixed from the following condition. Let Then the constant C is fixed from the condition or Alternatively, Ramanujan's sum can be used: or at 1 respectively Summation by parts [edit] Main article: Summation by parts Indefinite summation by parts: Definite summation by parts: Period rules [edit] If is a period of function then If is an antiperiod of function , that is then Alternative usage [edit] Some authors use the phrase "indefinite sum" to describe a sum in which the numerical value of the upper limit is not given: In this case a closed form expression F(k) for the sum is a solution of which is called the telescoping equation. It is the inverse of the backward difference operator. It is related to the forward antidifference operator using the fundamental theorem of discrete calculus described earlier. List of indefinite sums [edit] This is a list of indefinite sums of various functions. Not every function has an indefinite sum that can be expressed in terms of elementary functions. Antidifferences of rational functions [edit] : For positive integer exponents the misnamed Faulhaber's formula can be used. Note that in the result of Faulhaber's formula must be replaced with due to the offset caused by the indefinite sum being defined the inverse operation of the forward difference operator rather than the inverse operation of the backwards difference operator. For negative integer exponents , which is the sum over reciprocal integer powers, we have, : where is the polygamma function can be used. : More generally, : where is the Hurwitz zeta function and is the Digamma function. This is related to the Generalized harmonic number. As the generalized harmonic number uses reciprocal powers, must be substituted for , and the most common form uses the inverse of the backward difference offset: : For further information, refer to Balanced polygamma function and Hurwitz zeta function#Special cases and generalizations. Further generalization comes from use of the Lerch transcendent: : Which generalizes the Generalized harmonic number as when taking the inverse operator of the backwards difference rather than the forward difference. Additionally, the partial derivative is given by : For the more specialized partial derivative with respect to of the indefinite sum of see the Bernoulli polynomials: : Setting gives , the Bernoulli numbers. The relation to the Riemann zeta function is located here. Antidifferences of exponential functions [edit] Antidifferences of logarithmic functions [edit] Antidifferences of hyperbolic functions [edit] : where is the q-digamma function. Antidifferences of trigonometric functions [edit] : where is the q-digamma function. : where is the normalized sinc function. Antidifferences of inverse hyperbolic functions [edit] Antidifferences of inverse trigonometric functions [edit] Antidifferences of special functions [edit] : where is the incomplete gamma function. : where is the falling factorial. : (see super-exponential function) See also [edit] Indefinite product Time scale calculus List of derivatives and integrals in alternative calculi References [edit] ^ Man, Yiu-Kwong (1993), "On computing closed forms for indefinite summations", Journal of Symbolic Computation, 16 (4): 355–376, doi:10.1006/jsco.1993.1053, MR 1263873 ^ Goldberg, Samuel (1958), Introduction to difference equations, with illustrative examples from economics, psychology, and sociology, Wiley, New York, and Chapman & Hall, London, p. 41, ISBN 978-0-486-65084-5, MR 0094249, If is a function whose first difference is the function , then is called an indefinite sum of and denoted by {{citation}}: ISBN / Date incompatibility (help); reprinted by Dover Books, 1986 ^ "Handbook of discrete and combinatorial mathematics", Kenneth H. Rosen, John G. Michaels, CRC Press, 1999, ISBN 0-8493-0149-1 ^ a b c Merlini, Donatella; Sprugnoli, Renzo; Verri, M. Cecilia (2006), "The Cauchy numbers", Discrete Mathematics, 306 (16): 1906–1920, doi:10.1016/j.disc.2006.03.065, MR 2251571 ^ Markus Müller. How to Add a Non-Integer Number of Terms, and How to Produce Unusual Infinite Summations Archived 2011-06-17 at the Wayback Machine (note that he uses a slightly alternative definition of fractional sum in his work, i.e. inverse to backwards difference, hence 1 as the lower limit in his formula) ^ Bruce C. Berndt, Ramanujan's Notebooks Archived 2006-10-12 at the Wayback Machine, Ramanujan's Theory of Divergent Series, Chapter 6, Springer-Verlag (ed.), (1939), pp. 133–149. ^ Éric Delabaere, Ramanujan's Summation, Algorithms Seminar 2001–2002, F. Chyzak (ed.), INRIA, (2003), pp. 83–88. ^ Algorithms for Nonlinear Higher Order Difference Equations, Manuel Kauers Further reading [edit] "Difference Equations: An Introduction with Applications", Walter G. Kelley, Allan C. Peterson, Academic Press, 2001, ISBN 0-12-403330-X Markus Müller. How to Add a Non-Integer Number of Terms, and How to Produce Unusual Infinite Summations Markus Mueller, Dierk Schleicher. Fractional Sums and Euler-like Identities S. P. Polyakov. Indefinite summation of rational functions with additional minimization of the summable part. Programmirovanie, 2008, Vol. 34, No. 2. "Finite-Difference Equations And Simulations", Francis B. Hildebrand, Prenctice-Hall, 1968 Retrieved from " Categories: Mathematical analysis Mathematical tables Finite differences Linear operators in calculus Non-Newtonian calculus Hidden categories: CS1 errors: ISBN date Webarchive template wayback links Indefinite sum Add topic
16239	https://www.cut-the-knot.org/arithmetic/SevenTermSequence.shtml	Site... What's new Content page Front page Index page About Privacy policy Help with math Subjects... Arithmetic Algebra Geometry Probability Trigonometry Visual illusions Articles... Cut the knot! What is what? Inventor's paradox Math as language Problem solving Collections... Outline mathematics Book reviews Interactive activities Did you know? Eye opener Analogue gadgets Proofs in mathematics Things impossible Index/Glossary Simple math... Fast Arithmetic Tips Stories for young Word problems Games and puzzles Our logo Make an identity Elementary geometry Seven Terms Periodic Sequence A simple sequence of arithmetic operations leads to a curious result. First, to demonstrate the idea, I'll show an example, and afterwards prove a general result. Take two numbers, say, 3 and 5. Perform the following sequence of operations: \| \| \| --- \| \| Add 1 to the second number and divide the sum by the first \| (5 + 1)/3 = 2 \| \| Add 1 to the third number and divide the sum by the second \| (2 + 1)/5 = 3/5 \| \| Add 1 to the fourth number and divide the sum by the third \| (3/5 + 1)/2 = 4/5 \| \| Add 1 to the fifth number and divide the sum by the fourth \| (4/5 + 1)/(3/5) = 3 \| \| Add 1 to the sixth number and divide the sum by the fifth \| (3 + 1)/(4/5) = 5 \| Note that the sixth and the seventh numbers coincide respectively with the first and the second, such that, should we continue the process, we would repeat the previous steps. This is a general result: for any starting numbers a and b, the sequence repeats itself from the sixth term on: \| \| \| \| \| --- --- \| \| (1) \| a \| \| \| \| (2) \| b \| \| \| \| (3) \| (b + 1)/a \| \| \| \| (4) \| ((b + 1)/a + 1)/b \| = (a + b + 1)/ab \| \| \| (5) \| ((a + b + 1)/ab + 1)/((b + 1)/a) \| = (a + 1)(b + 1)/ab/((b + 1)/a) \| = (a + 1)/b \| \| (6) \| (a + 1)/b + 1)/((a + b + 1)/ab) \| = (a + b + 1)/b/((a + b + 1)/ab) \| = a \| \| (7) \| (a + 1)/((a + 1)/b) \| = b \| \| The derivation was greatly simplified on step (5) by an observation that (a + b + 1 + ab) = (a + 1)(b + 1). If one does not use algebra, but repeats the computations with different pairs of two numbers, how long would it take to notice that the same period always turns up? Meanwhile, one gets a good exercise practicing addition and multiplication. It may be instructive to present the problem in a slightly different form. Define sequence Un, n = 0, 1, 2, ... recursively U0 = a, U1 = b, Un+1 = (1 + Un)/Un-1, n > 0. The gist of the discussion then is reduced to the claim that For any selection of positive a and b, Un+5 = Un. References W. W. Rouse Ball and H. S. M. Coxeter, Mathematical Recreations and Essays, Dover, 1987 \|Contact\| \|Front page\| \|Contents\| \|Algebra\| Copyright © 1996-2018 Alexander Bogomolny
16240	https://fiveable.me/ap-chem/unit-5/intro-rate-law/study-guide/oq5mJS35IadrTLHLjdqZ	printables 🧪AP Chemistry Unit 5 Review 5.2 Introduction to Rate Law 🧪AP Chemistry Unit 5 Review 5.2 Introduction to Rate Law Written by the Fiveable Content Team • Last updated September 2025 Verified for the 2026 exam Verified for the 2026 exam•Written by the Fiveable Content Team • Last updated September 2025 🧪AP Chemistry Unit & Topic Study Guides Unit 5 Overview: Kinetics 5.1 Reaction Rates 5.2 Introduction to Rate Law 5.3 Concentration Changes Over Time 5.4 Elementary Reactions 5.5 Collision Model 5.6 Reaction Energy Profile 5.7 Introduction to Reaction Mechanisms 5.8 Reaction Mechanism and Rate Law 5.9 Pre-Equilibrium Approximation 5.10 Multistep Reaction Energy Profile 5.11 Catalysis In chemistry, when discussing kinetics, we know that when the concentration of a reactant rises, the rate of the reaction similarly increases. This makes sense since it logically follows that if we have more reactants in the same volume compared to a lesser amount of reactant, there will be a quicker reaction. However, how do we quantitatively determine how much faster the rate will be? Well, this is where a rate law comes into play. 👉 Be sure to review reaction rates and factors that influence it, such as concentration as we spoke about here. What is a Rate Law? In chemistry, a rate law is an equation that describes the relationship between the rate of a chemical reaction and the concentrations of the reactants. A rate law is defined by saying: R = k[A]^n[B]^m... where: R is the rate of the reaction (sometimes also notated as Δ[]/Δt, which we will delve more in-depth into in the next section), k is the rate constant, [A] and [B] represent the concentrations of reactants, and n and m are reaction orders for each reactant (A, B, etc). This rate law is generalized, which is why there is a ... following. A reaction could hypothetically have 3, 4, or 5 reactants, though for the AP exam you often won't see more than 2. It's actually quite rare for a reaction with 3+ reactants since it would require three atoms/molecules to bump into each other just right for a reaction to take place. It occurs, but not often and not quickly. more resources to help you study practice questionscheatsheetscore calculator Animation Courtesy of GIPHY; We'll delve deeper into collisions later in this unit. What Does Reaction Order Mean? n and m define what is called each reactant's reaction order. Reaction order brings us back to the initial question we asked: how do we determine quantitatively how concentration changes the rate of reaction? The reaction order is the answer! It describes how the rate of the reaction changes as the concentration of each reactant changes. For example, let's say the imaginary rate law for the reaction A + B → C is R = k[A]²[B]¹. This can tell us that as we increase the concentration of A (assuming a constant [B]), the rate will increase quadratically. For example, if we double the concentration of A, the rate will quadruple. Similarly, if we double the concentration of B, the rate will double, since the order of B is 1. The same applies for orders of 3, 4, etc. (if we double [], R goes up by 8 times and 16 times respectively). The overall reaction order for the full reaction is the sum of the orders for each reactant. In our imaginary example, the overall reaction order would be 3, since the reaction order of reactant A is 2 and the reaction order of reactant B is 1. Quadratic vs. Linear Relationships (x^2 vs x) Using Experiments to Determine a Rate Law There's one important thing to note about rate laws: they can only be determined experimentally. What a chemist will do is run a ton of tests at different concentrations and then find the corresponding rates for each test. With this data, they can determine the rate of the reaction. Let's take a look at an example to find out how we mathematically figure this out: Example Courtesy of Khan Academy Here we have a reaction: 2NO + 2H₂ → N₂ + 2H₂O. We can see three experiments done with different concentrations for each reactant. Let's take a look at the first two experiments where [NO] changes. We see the concentration of [NO] double from the first experiment to the second, and then we see the rate increase from 1.25 10⁻⁵ to 5.00 10⁻⁵. This is a change by a factor of 4 (5 10⁻⁵/1.25 10⁻⁵ = 4). Therefore, since doubling concentration makes the rate quadruple, the reaction is 2nd order with regard to NO. Note that in order to find the reaction order with respect to NO, we had to choose two experiments where the [NO] changed, but [H₂] was constant. Let's take a look at H₂ now. For experiments 2 and 3, the concentration of H₂ doubles like it did for NO, but the rate increases from 5 10⁻⁵ to 1 10⁻⁴, by a factor of 2 (1 10⁻⁴/⁵ 10⁻⁵ = 2). Therefore, the reaction is first order in H₂. Now, we can put together the rate law by putting all of this together: R = k[NO]²[H₂]. As an exercise, pick one of the experiments and plug in the correct numbers to figure out the value of k, and then read the next section and figure out the right units for k. (You should get k = 250 M⁻²s⁻¹). If you were to assume the reaction orders just by looking at the chemical equation, you'd get a rate law of R = k[NO]²[H₂]². This is why it is important to use experimental data to write the correct rate law: R = k[NO]²[H₂]. Understanding k, the Rate Constant The rate constant, k, is a tricky thing to understand. Essentially, it serves as a proportionality constant for the reaction to take place. It makes a bit more sense if you understand the calculus behind kinetics (which we will describe in the next section, though it is by no means required for the AP exam), but essentially all you need to know is that k is a constant that quantifies the rate of each reaction and that it is temperature specific. This means that for the same reaction at different temperatures, the rate constant is different! Another important aspect of the rate constant is that its units change depending on the overall reaction order. Let's see if we can figure some of them out. Rate is always in M/s, and concentration is always in M (M = mol/L). Thus it follows that for certain reactions: Zeroth Order If the overall reaction order is 0: The rate law is R = k[A]⁰, which simplfies to R = k k is in units M/s. First Order If the overall reaction order is 1: R = k[A]¹, which you can think of as M/s = k M k is in s⁻¹ (per seconds) Second Order If the overall reaction order is 2: R = k[A]², which you can think of as M/s = k M² k is in M⁻¹s⁻¹ (1/Ms) 🎥 Watch AP Chemistry teacher Mónica Gracida review reaction rates and rate laws in unit five of AP Chemistry: Kinetics. Frequently Asked Questions What is a rate law and how do I write one? A rate law is an equation that links reaction rate to reactant concentrations: rate = k[A]^m[B]^n… Here k is the rate constant (temperature dependent), m and n are the reaction orders for A and B (they’re exponents you find from data). To write one from experiments, use the method of initial rates: run experiments with different starting concentrations, compare how the initial rate changes, and solve for the exponents. Example: if doubling [A] doubles rate (with [B] constant), m = 1; if doubling [B] quadruples rate, n = 2. The overall order is m + n. Units of k depend on overall order (e.g., M·s⁻1 for first order, M⁻1·s⁻1 for second order). The AP exam expects you to represent experimental data with a consistent rate-law expression and use initial-rate comparisons (CED 5.2.A). For a focused review and examples, see the Topic 5.2 study guide ( the Unit 5 overview ( and extra practice problems ( Why do we need to determine reaction order experimentally instead of just using the balanced equation? Because the balanced equation just shows overall stoichiometry, not how molecules actually collide and react. The rate law depends on the reaction mechanism—which elementary steps and collisions make the slow (rate-determining) step—so the exponents (orders) come from molecular-level steps, not coefficients in the balanced equation except for elementary reactions. That’s why you must determine orders experimentally (e.g., method of initial rates, spectrophotometric or conductometric monitoring) and find the rate law r = k[A]^m[B]^n. The rate constant k is temperature dependent, and the units of k tell you the overall order (CED 5.2.A.2–5.2.A.4). On the AP exam you’ll be asked to use initial-rate data or propose mechanisms that match observed orders, so practice those skills (see the Topic 5.2 study guide: For more practice, try the unit problems ( or the large practice set ( How do you find the rate constant k from experimental data? First find the reaction’s rate law (the orders) from experiments, then solve for k. 1) Use the method of initial rates (CED 5.2.A.5): compare experiments where one reactant’s concentration changes while others are constant to determine each reactant’s order (m, n). 2) Once you have the form rate = k[A]^m[B]^n, pick one experimental trial and plug in the measured initial rate and concentrations. Solve k = rate / ([A]^m[B]^n). Repeat with other trials and average k—k should be the same for all trials at the same temperature. 3) Check units: units of k depend on overall order (e.g., M s⁻¹ for zero-order, s⁻¹ for first-order, M⁻1 s⁻¹ for second-order). Remember k depends on temperature (CED 5.2.A.4). 4) Alternate approach: if you know the reaction is zero-, first-, or second-order in one reactant, use integrated rate laws (plot [A] vs t, ln[A] vs t, or 1/[A] vs t)—the slope gives k directly. For more practice and the AP-aligned summary, see the Topic 5.2 study guide ( or the unit overview ( For lots of problems, try Fiveable practice ( What's the difference between reaction order and overall order? Reaction order (or “order with respect to” a reactant) is the exponent on that reactant’s concentration in the rate law—it tells you how the rate changes when that reactant’s concentration changes. Overall order is the sum of those exponents for all reactants in the rate law. Example: rate = k[A]^2[B]^1 → order with respect to A = 2, with respect to B = 1, overall order = 3. The orders come from experiment (e.g., method of initial rates); they’re not just the stoichiometric coefficients unless the step is elementary (CED 5.2.A.2–5.2.A.5). The units of k depend on the overall order (for a third-order rate constant, units are M^-2 s^-1). On the AP exam you’ll be asked to write consistent rate laws from data and use initial rates to find orders (Topic 5.2). For a quick refresher see the Topic 5.2 study guide ( and try practice problems ( I'm confused about how to use initial rates to find the order of reaction - can someone explain this step by step? Start with the rate law form: rate = k [A]^m [B]^n. Using initial-rate experiments you keep things simple: compare two experiments where only one reactant concentration changes. Then the ratio of initial rates equals the concentration ratio raised to the unknown exponent. Step-by-step: 1. Pick two trials where [B] is the same but [A] changes. Write rate2/rate1 = ([A]2/[A]1)^m. 2. Plug numbers and solve for m. If rate doubles when [A] doubles, m = 1; if rate quadruples, m = 2; if rate doesn’t change, m = 0. 3. Repeat for B using trials where [A] is constant to find n. 4. Once m and n are known, find k using k = rate / ([A]^m [B]^n) with any initial-rate data. Calculate units of k from overall order (CED: 5.2.A.2–5.2.A.4). 5. If concentration changes for multiple reactants at once, use rate ratios and take logs: log(rate2/rate1) = m·log([A]2/[A]1)+ n·log([B]2/[B]1). This is the AP “method of initial rates” (CED 5.2.A.5). For guided examples and practice, check the Topic 5.2 study guide ( and try problems at Fiveable’s practice page ( How do you know what units the rate constant should have? You get the units of k from the rate law and the fact that rate is in concentration/time (usually M s⁻¹). Write rate = k [A]^m [B]^n, where overall order = m + n. Solve for k: k = (rate) / ([A]^m[B]^n). So the units are M·s⁻¹ divided by M^(m+n) = M^(1−(m+n))·s⁻¹. Quick examples: - Zero order (overall 0): units = M·s⁻¹ (e.g., M/s) - First order (overall 1): units = s⁻¹ - Second order (overall 2): units = M⁻¹·s⁻¹ Remember k is temperature dependent and its units always reflect the overall reaction order (CED 5.2.A.4). If you want practice identifying units from given rate laws, check the Topic 5.2 study guide ( or more unit review at ( For extra problems try Fiveable’s practice set ( What does it mean when a reaction is zero order vs first order vs second order? Zero-, first-, and second-order describe how the rate depends on concentration in the rate law: rate = k[A]^n (CED 5.2.A.2–3). - Zero order (n = 0): rate = k. The rate is constant and independent of [A]. Units of k are M·s⁻¹. A plot of [A] vs. time is linear. - First order (n = 1): rate = k[A]. Rate is proportional to [A]; doubling [A] doubles the rate. Units of k are s⁻¹. The integrated form is ln[A] = −kt + ln[A]0, so ln[A] vs. time is linear; the half-life t½ = 0.693/k is constant. - Second order (n = 2): common forms are rate = k[A]^2 or k[A][B]. Doubling [A] quadruples the rate (for A^2). Units of k are M⁻¹·s⁻¹. For A^2, 1/[A] vs. time is linear and t½ depends on [A]0. To find orders experimentally use the method of initial rates (CED 5.2.A.5) or test linearity of the integrated plots. For more practice and the Topic 5.2 study guide, see Fiveable ( and the Unit 5 overview ( For lots of problems, try Fiveable practice ( Why does temperature affect the rate constant but not the reaction order? Reaction order (the exponents in a rate law) is an experimental description of how rate depends on concentrations—you determine it by comparing initial rates (CED 5.2.A, method of initial rates). It reflects the reaction mechanism (which elementary steps involve which species) and so is set by how collisions/steps control rate, not by how fast those steps happen. Temperature changes how often and how energetically molecules collide and how many collisions exceed the activation energy. That shows up in the rate constant k (CED 5.2.A.4) via the Arrhenius relation k = A e^(−Ea/RT): raising T increases k exponentially. k’s value (and units—which depend on overall order) is temperature dependent; the numeric reaction orders (exponents) stay the same unless the mechanism itself changes at a different T. For AP review: practice applying initial-rates data to find orders and remember k varies with T (see the Topic 5.2 study guide) ( For more unit review and tons of practice problems go to ( and ( How do I compare initial rates to determine the order with respect to each reactant? Compare initial rates by holding one reactant’s concentration effectively the same between two experiments and seeing how the rate changes—that gives the order for the reactant that changed. Steps (method of initial rates): 1. Pick two experiments where [B], [C], etc. are constant and only [A] changes. 2. Write the rate law form: rate = k[A]^m[B]^n... 3. Divide the two rate expressions to cancel k and constant concentrations: rate2/rate1 = ([A]2/[A]1)^m. 4. Solve for m (the order in A). Repeat for each reactant. Quick rules-of-thumb: - If doubling [A] doubles rate → first order (m = 1). - If doubling [A] leaves rate unchanged → zero order (m = 0). - If doubling [A] quadruples rate → second order (m = 2). After finding each exponent, sum them for overall order and use one experiment to solve for k. This is exactly what the AP wants you to do on initial-rate problems (Topic 5.2). For walkthroughs and practice, see the Topic 5.2 study guide ( and more practice questions ( What's the relationship between the exponents in the rate law and the coefficients in the balanced equation? Short answer: there’s no guaranteed connection—the exponents in a rate law (the orders) come from experiments, not from the balanced stoichiometric coefficients, unless the reaction step shown is an elementary (single-step) reaction. Why: the CED (Topic 5.2) says the rate law is rate = k[A]^m[B]^n and m, n are determined by data (method of initial rates), not by the overall balanced equation. Only for elementary reactions do the molecularities equal the exponents (e.g., a bimolecular elementary step gives 2nd order). The sum of the exponents gives the overall order and affects k’s units and how you analyze initial-rate problems on the AP exam. For practice: review the Topic 5.2 study guide ( and try method-of-initial-rates problems from the unit page ( or the practice bank ( Can you explain how to set up a rate law experiment in the lab? Pick a clean, easy reaction (e.g., A + B → products) and a measurable signal (absorbance, conductivity, gas volume). Run at least three trials that change only one reactant’s initial concentration while keeping temperature constant and other concentrations large/constant. Monitor concentration vs. time and determine the initial rate (slope at t ≈ 0 or during first ~10% conversion). Use the method of initial rates: compare how rates change when [A] or [B] changes to get orders m and n (rate ∝ [A]^m[B]^n). Once orders are known, solve for k from rate = k[A]^m[B]^n and report units matching overall order. Repeat trials for precision, calibrate your instrument, control temperature, and follow safety rules. AP tip: the exam expects using initial-rate comparisons to find orders and k (CED 5.2.A, method of initial rates). For a refresher and sample setups, see the Topic 5.2 study guide ( and more practice problems ( I don't understand how to calculate reaction order when the concentration doubles but the rate increases by 4 - what am I missing? You're close—use the rate law idea: rate ∝ [A]^n. If [A] doubles and rate increases by 4, plug into the ratio: rate2 / rate1 = (2)^n = 4 → 2^n = 4 → n = 2. So the reaction is second order with respect to that reactant. This is exactly the method of initial rates (compare initial rates while changing one concentration)—orders are determined experimentally, not from the balanced equation. Remember: if doubling gave a 2× change, n = 1 (first order); if no change, n = 0 (zero order). Also note the overall order (sum of exponents) sets the units of k and is tested on the AP (Topic 5.2: method of initial rates). For more worked examples and AP-aligned practice, see the Topic 5.2 study guide ( and the AP practice problems ( What happens to the rate law if one of the reactants is in large excess? If one reactant is present in large excess, its concentration stays essentially constant during the time you measure the rate. That lets you treat the rate law as if that reactant were part of the rate constant—a pseudo-order simplification. For example, if rate = k[A][B] and [B] ≫ [A], then rate ≈ k′[A] where k′ = k[B] (constant). The true reaction orders (the exponents in the real rate law) don’t change—you’re just measuring an apparent order that’s easier to analyze. This is why experiments often use an excess reactant to get pseudo-first-order kinetics and determine the order with respect to the limiting species using the method of initial rates (an AP CED technique). For more on this and practice problems, see the Topic 5.2 study guide ( and Unit 5 resources ( How do you determine if a reaction is first order or second order just by looking at the data? Look at how rate changes when you change concentration (method of initial rates). If you double [A] and: - rate doubles → rate ∝ [A]^1 → first order in A. - rate quadruples → rate ∝ [A]^2 → second order in A. - rate doesn’t change → zero order. For more rigorous checks use linear plots of time-course data: - First order: ln[A] vs t is straight line (slope = −k). - Second order (one reactant): 1/[A] vs t is straight line (slope = +k). On the AP exam they expect you to compare initial rates or use these integrated-rate plots to identify orders (CED 5.2.A, method of initial rates). If you want step-by-step examples and quick practice, see the Topic 5.2 study guide ( and try problems at Fiveable’s practice page ( Why can't you determine the rate law just from looking at the balanced chemical equation? You can’t get the rate law just from the balanced equation because the balanced equation only shows overall stoichiometry, not the microscopic steps that actually make the reaction happen. Rate laws depend on the mechanism: the elementary steps (and the slow, rate-determining step) determine how rate depends on concentrations. For many reactions the overall equation comes from multiple elementary steps, so the exponents (orders) in rate = k[A]^m[B]^n must be measured experimentally (e.g., by the method of initial rates). The rate constant k is also temperature-dependent and not shown by the equation. On the AP exam you’re expected to represent experimental data with a consistent rate law and use initial-rate comparisons to find orders (CED 5.2.A, keywords: rate law, reaction order, rate constant, method of initial rates). For a clear review, see the Topic 5.2 study guide ( and try practice problems (
16241	https://www.youtube.com/watch?v=Dom591uT0j8	Physics 4A - OpenStax University Physics Vol 1, Chapter 10, Problem 70 (integration practice) CoA Physics 1360 subscribers 1 likes Description 192 views Posted: 17 Aug 2021 Recording from Spring 2020 PHYS 4A class. Transcript: it's not cheating but you know if you just cared about the answer and um and you you just want to get the answer quickly here you don't um you don't care to follow the actual directions that are given here you know calculate the moment of inertia by direct integration then then you can quickly obtain the answer this way which is um so let me show you the shortcut way first just so that we can get to the answer and then i'll do it properly which is um so it's good to have memorized some of the rotational inertia formulas i have this formula memorized rotational inertia of a rod or about its endpoint is equal to one-third mass of the rod times its length squared so i can actually apply this i can use a superposition principle think of this like a two rod separated one here and you know another there and i just plug in the correct values of mass and length so lengths are already there i can use that and mass is where i have to be careful because it's a 14 m for the entire thing so it's gonna be um 14 over 3 m for this and 28 over 3 m for this so all right let's write down the total rotational inertia that's going to be left to the end of rod which will be one third times the mass there 14 over 3 m times the length squared which is the third over l so l over three squared plus the rotational shot from the right hand side so that's one third uh mass there 28 over three and and then the length there 2l over 3 squared you can see that when you carry out the calculation here 1 and l squared is going to factor out so the rest is doing the numerical calculation in between to get this reduce the fraction a over b so uh let me let's just do that so it's gonna be the first term is gonna be 14 divided by three times 3 times 9 so up 81 plus um it's gonna be 28 times 4 so sorry i'm kind of confusing myself four times eight is i think 32 right why am i not 36 right no 32 32 what's right okay uh so 32 plus um so one one two let me just do this in calculator to be sure 28 times four sorry i'm getting a little bit tired and cranky probably oh yeah one one two good all right uh divided by oh i guess 81 again because it's 3 times 3 times 9 divided by 81. okay that makes it easier to combine this fraction it's gonna be um one two six one two six over eighty one looks like both are divisible by nine so let's do that [Music] to 9 and then 1 4. all right so it should be 14 over 9 and when you do you'll get that to be the correct answer uh but you know i won't end this there because um really the point of this question wasn't i mean you know this is actually a good technique to know which is why i did it but if you're doing this question properly it is asking you to do this by direct integration not to use a superposition principle although you could and my opponent has no way of knowing so let me do this one more time one last time by direct integration calculate the moment of inertia by direct integration of a thin rod of some mass and length l about an axis through the rod at l over 3 as shown below okay so where do i start so this is uh there's a reason we put some time some emphasis into this in our physics foray and it has to do with something that we are hoping to build you up to in anticipation of what i think we have you have to do in physics 4b so and and there is an ability to kind of look at a complex problem recognize that it can be broken down into a smaller manageable chunk express the smaller manageable chunk and then use that to work out a solution for the whole big thing you've seen an example of this when you show me a lecture on the potent spring potential energy because you have a variable force work done by the variable force can be done just as simple as force times distance so what you have to do is you break up the distance interval into an infinite decimal pieces you express the infinite decimal worker for done for the infinite decimal piece and then you add it up using integral and that's what you have to do here so the very first step that you have to do here is to express that infinite decimal piece um i like to do it as a kind of uh representative piece first so i have this entire rod and the idea here is that i can be looking at basically any portion of this rod so let me just take a piece here and um i can parameterize this piece and then i can refer to that piece by its coordinate position location x and this is what i can say about rotational inertia of this tiny little piece if i knew the mass of that tiny little piece i can say that for this tiny little piece its rotational inertia about this axis here is gonna be so the tiny little rotational inertia of the tiny little piece it's small enough that i can treat it like a point mass so i write down the formula for rotational inertia of a point mass mass times the distance squared so this is the starting place and this is a kind of a representative piece in that the x is a variable you can let x vary all the way from x equals minus l over 3 to all the way to x equals 2 l over 3. that identifies position of each one of those pieces and using that using this variable you can address the entire rod you can add this is how you we are going to do the integration then there's a bit of a hurdle you have to pass which is that this dm here is not expressed in terms of your coordinate variable it's kind of a dm it's a schematical symbol it it's uh um like when you integrate what do you do with that i don't know um so what you need to do is you need to express this dm in terms of this coordinate variable and this is where you kind of think through it and then you realize oh it doesn't say it but it says a thin rod so i think uniform is implied so you can associate the amount of mass that's in that portion with the kind of the size of that little piece dx so if this rod has a linear mass density lambda then this mass dm here can be represented this way dm will be the density times the small interval dx so when you have that now it's something you can you recognize as something you can integrate oops um t i is equal to lambda dx x squared so it's an integral you can form an integral in terms of x and lambda is going to be a constant let's work out the constant and just write it into our expression so linear mass density that's the amount of mass 14 m per length l and if it's uniform then it's this ratio is going to be the density throughout and just so you know um the way we assess if you know how to go through this process is a i will give you a non-uniform rod so i'm doing this integral is the only way you can figure out the rotational inertia once again not for example but uh for a future example i'll do that so um here's the linear mass density so let me just plug that in so this is kind of a beginning setup of your integral d i the the infinite decimal rotation inertia due to that infinitesimal segment dx is a 14 m over l dx x squared and let me swap this order here so that it's kind of clean sorry i need to do this with the mouse um let me swap this order here so that it's kind of clearer what's in the integrand what's the variable of integration and you can recall the formulas that are relevant here okay so i'm integrating with respect to x and when i do integrate what i'm doing basically is i'm integrating over the entire rod when i do that i should get the rotational inertia of the rod that's what i should get so what does that mean to integrate over the rod well if this is kind of why we want the coordinate variable you can do it in terms of the coordinate variable as x varies from minus l over three to twelve over three you have to address the entire rod so i let x go from minus l over three to two l over three and when i work out this right hand side that should give me the answer and now if you're looking at um i mean is this minus okay like am i doing something um unphysical um this is something you will begin to see more and more as you go to higher levels which is where you write down your expressions so that negative signs are meaningful and this kind of a weird looking integration limit is perfectly valid there's nothing wrong with it when we work out the answer you will see that i got the correct answer um it's uh it's kind of a more formal way of setting up problem solving we don't do a lot of that in physics for a and you will begin to see kind of shifting into that as you're going to higher levels of physics engineering and math so um let me factor out all the constants that don't matter so it's a 14 m over l it doesn't matter in the integration so um i have this relatively simple uh integrand that i use power rule so the the anti-derivative of that out to be one-third x cubed you can always double check by imagine taking derivative of this see if you get x squared back if you do great move on so having this anti-derivative you evaluated at the limit of 2l over 3 to minus l over 3. so evaluating at the limits it means you plug it in and you take the difference of the upper limit to lower limit so let's write it out 14 m over l times plugging in the upper limit it's a one-third sorry too many cubes two cubed is eight times l cubed over 3 cubed is 27 minus 1 3 and i plug in minus l over 3 cubed so i'm going to get minus still l cubed over 3 cubed is 27. all right uh minus the signs cancel so i get plus this seems reasonable i should be adding stuff some factors of l cancel out so i get end up with l squared which is good um i don't need the n l squared and looks like i have that and l squared double factor out so the rest is a matter of working out oh if 14 is what i'm gonna get i hope all this adds up to one over nine so it's eight over eighty-one plus one over eighty-one so all of this will add up to [Music] over 81 yeah that adds up to 1 over 9. so 14 times 1 over 9 gives me that so yeah if i get the same answer i get the same right answer so yeah this is uh once you kind of get the hang of it this is actually not very difficult but we've seen many students struggle with it because it's a kind of it's conceptually new it's the kind of thing i don't think you really learn it that thoroughly even in your calculus 2 class um and like when you get good at setting up integrals like this is when you you know you're getting good at physics problem solving because many people do struggle with it
16242	https://ecampusontario.pressbooks.pub/introductorychemistry/back-matter/appendix-standard-reduction-potentials-by-value-2/	Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. Appendix: Standard Reduction Potentials by Value \| Standard Cathode (Reduction) Half-Reaction \| Standard Reduction Potential E° (volts) \| --- \| \| Li+(aq) + e– ⇌ Li(s) \| –3.040 \| \| Ba2+ + 2e−⇌ Ba(s) \| –2.92 \| \| Rb+ + e– ⇌ Rb (s) \| -2.98 \| \| K+(aq) + e– ⇌ K(s) \| -2.93 \| \| Cs+(aq) + e– ⇌ Cs(s) \| -2.92 \| \| Ba2+(aq) + 2e- ⇌ Ba(s) \| -2.91 \| \| Sr2+(aq) + 2e- ⇌ Sr(s) \| -2.89 \| \| Ca2+(aq) + 2e- ⇌ Ca(s) \| –2.84 \| \| Na+(aq) + e– ⇌ Na(s) \| –2.713 \| \| Mg(OH)2(s) + 2e−⇌ Mg(s) + 2OH− \| –2.687 \| \| La3+ + 3e−⇌ La(s) \| –2.38 \| \| Mg2+(aq) + 2e– ⇌ Mn(s) \| -2.356 \| \| Ce3+ + 3e−⇌ Ce(s) \| –2.336 \| \| Al(OH)4− + 3e−⇌ Al(s) + 4OH− \| –2.310 \| \| AlF63− + 3e−⇌ Al(s) + 6F− \| –2.07 \| \| Be2+ + 2e−⇌ Be(s) \| –1.99 \| \| B(OH)4− + 3e−⇌ B(s) + 4OH− \| –1.811 \| \| U3+ + 3e−⇌ U(s) \| –1.66 \| \| Al3+(aq) + 3e– ⇌ Al(s) \| –1.676 \| \| SiF62− + 4e−⇌ Si(s) + 6F− \| –1.37 \| \| Zn(CN)42− + 2e−⇌ Zn(s) + 4CN \| –1.34 \| \| Zn(OH)42− + 2e−⇌ Zn(s) + 4OH− \| –1.285 \| \| Mn2+ + 2e−⇌ Mn(s) \| –1.17 \| \| V2+ + 2e−⇌ V(s) \| –1.13 \| \| 2SO32− + 2H2O(l) + 2e−⇌ S2O42− + 4OH− \| –1.13 \| \| Zn(NH3)42+ + 2e−⇌ Zn(s) + 4NH3 \| –1.04 \| \| O2 (aq) + e– ⇌ O2–(aq) \| -1.0 \| \| Cd(CN)42− + 2e−⇌ Cd(s) + 4CN− \| –0.943 \| \| MoO42− + 4H2O(l) + 6e−⇌Mo(s) + 8OH− \| –0.913 \| \| SiO2(s) + 4H+ + 4e−⇌ Si(s) + 2H2O(l) \| –0.909 \| \| SO42− + H2O(l) + 2e− ⇌ SO32−+ 2OH− \| –0.936 \| \| Cr2+ + 2e−⇌ Cr(s) \| –0.90 \| \| B(OH)3 + 3H+ + 3e−⇌ B(s) + 3H2O(l) \| –0.890 \| \| 2H2O(l) + 2e– ⇌ H2(g) + 2OH–(aq) \| -0.828 \| \| Zn2+(aq) + 2e– ⇌ Zn(s) \| –0.7618 \| \| Co(OH)2(s) + 2e−⇌ Co(s) + 2OH− \| –0.746 \| \| Cr3+(aq) + 3e– ⇌ Cr(s) \| –0.424 \| \| Ni(OH)2 + 2e−⇌ Ni(s) + 2OH− \| –0.72 \| \| Ag2S(s) + 2e−⇌ 2Ag(s) + S2− \| –0.71 \| \| Se(s) + 2e−⇌ Se2− \| –0.67 in 1 M NaOH \| \| Cd(NH3)42+ + 2e−⇌ Cd(s) + 4NH3 \| –0.622 \| \| 2SO32− + 3H2O(l) + 4e−⇌S2O32− + 6OH− \| –0.576 in 1 M NaOH \| \| U4+ + e−⇌ U3+ \| –0.52 \| \| SiO2(s) + 8H+ + 8e−⇌ SiH4(g) + 2H2O(l) \| –0.516 \| \| Sb + 3H+ + 3e−⇌ SbH3(g) \| –0.510 \| \| H3PO3+ 2H+ + 2e−⇌ H3PO2 + H2O(l) \| -0.50 \| \| Ni(NH3)62+ + 2e−⇌ Ni(s) + 6NH3 \| –0.49 \| \| 2CO2(g) + 2H+ +2e−⇌ H2C2O4 \| –0.481 \| \| Cr3+ + e−⇌ Cr2+ \| –0.424 \| \| Fe2+(aq) + 2e– ⇌ Fe(s) \| -0.44 \| \| S(s) + 2e−⇌ S2− \| –0.407 \| \| Cd2+(aq) + 2e– ⇌ Cd(s) \| –0.4030 \| \| Ag(NH3)2+ + e−⇌ Ag(s) + 2NH3 \| –0.373 \| \| Ti3+ + e−⇌ Ti2+ \| –0.37 \| \| PbSO4(s) + 2e−⇌ Pb(s) + SO42− \| –0.356 \| \| Co2+(aq) + 2e– ⇌ Co(s) \| –0.277 \| \| 2SO42− + 4H+ + 2e−⇌ S2O62− + 2H2O(l) \| –0.25 \| \| N2(g) + 5H+ + 4e−⇌ N2H5+ \| –0.23 \| \| H3PO4 + 2H+ + 2e−⇌ H3PO3 + H2O(l) \| -0.28 \| \| Ni2+(aq) + 2e– ⇌ Ni(s) \| –0.257 \| \| V3+ + e−⇌ V2+ \| –0.255 \| \| As + 3H+ + 3e−⇌ AsH3(g) \| –0.225 \| \| CO2(g) + 2H+ + 2e−⇌ HCO2H \| –0.20 \| \| Mo3+ + 3e−⇌ Mo(s) \| –0.2 \| \| Sn2+ + 2e−⇌ Sn(s) \| –0.19 in 1 M HCl \| \| Ti2+ + 2e−⇌ Ti(s) \| –0.163 \| \| MoO2(s) + 4H+ + 4e−⇌ Mo(s) + 2H2O(l) \| –0.152 \| \| AgI(s) + e−⇌ Ag(s) + I− \| –0.152 \| \| Sn2+(aq) + 2e– ⇌ Sn(s) \| -0.14 \| \| Pb2+(aq) + 2e– ⇌ Pb(s) \| -0.126 \| \| CrO42− + 4H2O(l) + 3e−⇌2Cr(OH)4− + 4OH− \| –0.13 in 1 M NaOH \| \| WO2(s) + 4H+ + 4e−⇌ W(s) + 2H2O(l) \| –0.119 \| \| Se(s) + 2H+ + 2e−⇌ H2Se(g) \| –0.115 \| \| CO2(g) + 2H+ + 2e−⇌ CO(g) + H2O(l) \| –0.106 \| \| WO3(s) + 6H+ + 6e−⇌ W(s) + 3H2O(l) \| –0.090 \| \| Hg2I2(s) + 2e−⇌ 2Hg(l) + 2I− \| –0.0405 \| \| Fe3+(aq) + 3e– ⇌ Fe(s) \| -0.037 \| \| 2H+(aq) + 2e– ⇌ H2(g) \| 0.00 \| \| P(s,white) + 3H+ + 3e−⇌ PH3(g) \| 0.06 \| \| AgBr(s) + e−⇌ Ag(s) + Br− \| 0.071 \| \| S4O62− + 2e−⇌2S2O32− \| 0.080 \| \| Co(NH3)63+ +e−⇌Co(NH3)62+ \| 0.1 \| \| Ru(NH3)63+ + e−⇌ Ru(s) + Ru(NH3)62+ \| 0.10 \| \| S(s) + 2H+ + 2e−⇌ H2S \| 0.144 \| \| Sn4+(aq) + 2e– ⇌ Sn2+(aq) \| 0.154 \| \| Cu2+(aq) + e– ⇌ Cu+(aq) \| 0.159 \| \| UO22+ + e−⇌UO2+ \| 0.16 \| \| Co(OH)3(s) + e−⇌ Co(OH)2(s) + OH− \| 0.17 \| \| ClO4–(aq) + H2O(l) + 2e- ⇌ ClO3–(aq) + 2OH–(aq) \| 0.17 \| \| SO42− + 4H+ + 2e−⇌ H2SO32−+ H2O(l) \| 0.172 \| \| BiCl4− + 3e−⇌ Bi(s) + 4Cl− \| 0.199 \| \| SbO++ 2H+ + 3e−⇌ Sb(s) + H2O(l) \| 0.212 \| \| AgCl(s) + e– ⇌ Ag(s) + Cl–(aq) \| 0.2223 \| \| HCHO + 2H+ + 2e−⇌ CH3OH \| 0.2323 \| \| HAsO2 + 3H+ + 3e−⇌ As(s) + 2H2O(l) \| 0.240 \| \| Ru3+ + e−⇌ Ru2+ \| 0.249 \| \| IO3−+ 3H2O(l) + 6e−⇌ I− + 6OH− \| 0.257 \| \| Hg2Cl2(s) + 2e−⇌ 2Hg(l) + 2Cl− \| 0.2682 \| \| UO2+ + 4H+ + e−⇌U4+ + 2H2O(l) \| 0.27 \| \| Bi3+ + 3e−⇌ Bi(s) \| 0.317 \| \| UO22+ + 4H+ + 2e−⇌ U4+ + 2H2O(l) \| 0.327 \| \| VO2+ + 2H+ +e−⇌V3+ + H2O(l) \| 0.337 \| \| Cu2+(aq) + 2e– ⇌ Cu(s) \| 0.3419 \| \| ClO3–(aq) + H2O(l) + 2e– ⇌ ClO2–(aq) + 2OH–(aq) \| 0.35 \| \| Fe(CN)63− + e−⇌ Fe(CN)64− \| 0.356 \| \| O2(g) + 2H2O(l) + 4e−⇌ 4OH− \| 0.401 \| \| ClO− + H2O(l) + e−⇌ ½Cl2(g) + 2OH− \| 0.421 in 1 M NaOH \| \| Ag2C2O4(s) + 2e−⇌ 2Ag(s) + C2O42− \| 0.47 \| \| Cu+(aq) + e– ⇌ Cu(s) \| 0.52 \| \| I2(s) + 2e– ⇌ 2I–(aq) \| 0.5355 \| \| I3− + 2e−⇌ 3I− \| 0.536 \| \| Ga3+ + 3e−⇌ Ga(s) \| -0.56 \| \| Cu2+ + Cl− + e−⇌ CuCl(s) \| 0.559 \| \| S2O62− + 4H+ + 2e−⇌ 2H2SO3 \| 0.569 \| \| H3AsO4 + 2H+ + 2e−⇌ HAsO2 + 2H2O(l) \| 0.560 \| \| ClO2–(aq) + H2O(l) + 2e– ⇌ ClO–(aq) + 2OH–(aq) \| 0.59 \| \| MnO4− + 2H2O(l) + 3e−⇌ MnO2(s) + 4OH− \| 0.60 \| \| Sb2O5(s) + 6H+ + 4e−⇌ 2SbO+ + 3H2O(l) \| 0.605 \| \| PtCl62− + 2e−⇌ PtCl42- + 2Cl− \| 0.68 \| \| RuO2(s) + 4H+ + 4e−⇌ Ru(s) + 2H2O(l) \| 0.68 \| \| O2(g) + 2H+ + 2e−⇌ H2O2 \| 0.695 \| \| PtCl42− + 2e−⇌ Pt(s) + 4Cl− \| 0.73 \| \| H2SeO3 + 4H+ + 4e−⇌ Se(s) + 3H2O(l) \| 0.74 \| \| Tl3+ + 3e−⇌ Tl(s) \| 0.742 \| \| Fe3+(aq) + e– ⇌ Fe2+(aq) \| 0.771 \| \| Hg22+(aq) + 2e– ⇌ 2Hg(l) \| 0.7960 \| \| Ag+(aq) + e- ⇌ Ag(s) \| 0.7996 \| \| Hg2+(aq) + 2e– ⇌ Hg(l) \| 0.8535 \| \| Cu2+ + I− +e−⇌ CuI(s) \| 0.86 \| \| Ru(CN)63− +e−⇌ Ru(s) + Ru(CN)64− \| 0.86 \| \| ClO− + H2O(l) + 2e−⇌ Cl− + 2OH− \| 0.890 in 1 M NaOH \| \| 2Hg2+(aq) + 2e– ⇌ Hg22+(aq) \| 0.911 \| \| HgO(s) + 2H+ + 2e−⇌ Hg(l) + H2O(l) \| 0.926 \| \| NO3− + 3H+ + 2e−⇌ HNO2 + H2O(l) \| 0.94 \| \| MnO2(s) + 4H+ + e– ⇌ Mn3+ (aq) + H2O(I) \| 0.95 \| \| NO3–(aq) + 4H+(aq) + 3e– ⇌ NO(g) + 2H2O(l) \| 0.96 \| \| HIO + H+ + 2e−⇌ I− + H2O(l) \| 0.985 \| \| HNO2 + H+ +e−⇌ NO(g) + H2O(l) \| 0.996 \| \| VO22+ + 2H+ +e−⇌ VO2+ + H2O(l) \| 1.000 \| \| AuCl4− + 3e−⇌ Au(s) + 4Cl− \| 1.002 \| \| NO2 (g) + H+ (aq) + e– ⇌ HNO2 (aq) \| 1.07 \| \| Br2(l) + 2e– ⇌ 2Br–(aq) \| 1.087 \| \| Fe(phen)63+ +e−⇌ Fe(phen)62+ \| \| \| SeO43− + 4H+ + e−⇌ H2SeO3 + H2O(l) \| \| \| ClO3− + 2H+ + e−⇌ ClO2(g) + H2O \| \| \| ClO3− + 3H+ + 2e−⇌ HClO2 + H2O \| \| \| IO3− + 6H+ + 5e−⇌ ½I2(s) + 3H2O(l) \| \| \| Pt2+ + 2e−⇌ Pt(s) \| \| \| ClO4− + 2H+ + 2e−⇌ ClO3− + H2O \| \| \| O2(g) + 4H+(aq) + 4e– ⇌ 2H2O(l) \| \| \| MnO2(s) + 4H+ + 2e−⇌ Mn2+ + 2H2O(l) \| \| \| Tl3+ + 2e−⇌ Tl+ \| 0.77 in 1 M HCl \| \| 2HNO2 + 4H+ + 4e−⇌ N2O(g) + 3H2O(l) \| \| \| HOBr + H+ + 2e−⇌ Br− + H2O(l) \| \| \| Cr2O72-(aq) + 14H+(aq) + 6e– ⇌ 2Cr3+(aq) + 7H2O(l) \| \| \| Cr2O72− + 14H+ + 6e−⇌ 2Cr3+ + 7H2O(l) \| 1.36 \| \| Cl2(g) + 2e– ⇌ 2Cl–(aq) \| \| \| Au3+ + 2e−⇌ Au+ \| 1.36 \| \| Hg2Br2(s) + 2e−⇌ 2Hg(l) + 2Br− \| \| \| Ce4+(aq) + e– ⇌ Ce3+(aq) \| \| \| PbO2(s) + 4H+ +2e– ⇌ Pb2+(aq) + 2H2O(l) \| \| \| BrO3− + 6H+ + 6e−⇌ Br− + 3H2O \| \| \| Mn3+ + e−⇌ Mn2+ \| \| \| MnO4–(aq) + 8H+(aq) + 5e– ⇌ Mn2+(aq) + 4H2O(l) \| \| \| BrO3− + 6H+ + 5e−⇌ ½Br2(l) + 3H2O \| 1.5 \| \| Au3+ + 3e−⇌ Au(s) \| \| \| 2NO(g) + 2H+ + 2e−⇌ N2O(g) + H2O(l) \| \| \| HOBr + H+ + e−⇌ ½Br− + H2O(l) \| \| \| HClO2 + 2H+ + 2e−⇌ HOCl + H2O \| \| \| PbO2(s) + 4SO42− + 4H+ + 2e−⇌ PbSO4(s) + 2H2O(l) \| \| \| MnO4− + 4H+ +3e−⇌ MnO2(s) + 2H2O(l) \| \| \| Ce4+ + e−⇌ Ce3+ \| \| \| N2O(g) + 2H+ + 2e−⇌ N2(g) + H2O(l) \| \| \| H2O2(aq) + 2H+(aq) + 2e– ⇌ 2H2O(l) \| \| \| Au+ + e−⇌ Au(s) \| \| \| Co3+(aq) + e– ⇌ Co2+(aq) \| \| \| S2O82− + 2e−⇌2SO42− \| \| \| O3(g) + 2H+(aq) + 2e– ⇌ O2(g) + H2O(l) \| \| \| BaO(s) + 2H+ + 2e−⇌ Ba(s) + H2O(l) \| \| \| F2(g) + 2e– ⇌ 2F–(aq) \| \| \| F2(g) + 2H+ + 2e−⇌ 2HF \| \| From UC Davis Chem Wiki (creative commons licence): UC Davis GeoWiki by University of California, Davis. CC-BY-NC-SA-3.0 Original source materials: Bard, A. J.; Parsons, B.; Jordon, J., eds. Standard Potentials in Aqueous Solutions, Dekker: New York, 1985; Milazzo, G.; Caroli, S.; Sharma, V. K. Tables of Standard Electrode Potentials, Wiley: London, 1978; Swift, E. H.; Butler, E. A. Quantitative Measurements and Chemical Equilibria, Freeman: New York, 1972. License Introductory Chemistry- 1st Canadian Edition Copyright © 2014 by Jessie A. Key is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted. Share This Book
16243	https://www.mathway.com/popular-problems/Trigonometry/320412	Find the Other Trig Values in Quadrant II sin(2x)=-1/2 \| Mathway Enter a problem... [x] Trigonometry Examples Popular Problems Trigonometry Find the Other Trig Values in Quadrant II sin(2x)=-1/2 sin(2 x)=−1 2 sin(2 x)=-1 2 Step 1 Use the definition of sine to find the known sides of the unit circle right triangle. The quadrant determines the sign on each of the values. sin(2 x)=opposite hypotenuse sin(2 x)=opposite hypotenuse Step 2 Find the adjacent side of the unit circle triangle. Since the hypotenuse and opposite sides are known, use the Pythagorean theorem to find the remaining side. Adjacent=−√hypotenuse 2−opposite 2 Adjacent=-hypotenuse 2-opposite 2 Step 3 Replace the known values in the equation. Adjacent=−√(2)2−(−1)2 Adjacent=-(2)2-(-1)2 Step 4 Simplify inside the radical. Tap for more steps... Step 4.1 Negate √(2)2−(−1)2(2)2-(-1)2. Adjacent =−√(2)2−(−1)2=-(2)2-(-1)2 Step 4.2 Raise 2 2 to the power of 2 2. Adjacent =−√4−(−1)2=-4-(-1)2 Step 4.3 Multiply−1-1 by (−1)2(-1)2 by adding the exponents. Tap for more steps... Step 4.3.1 Multiply−1-1 by (−1)2(-1)2. Tap for more steps... Step 4.3.1.1 Raise −1-1 to the power of 1 1. Adjacent =−√4+(−1)(−1)2=-4+(-1)(-1)2 Step 4.3.1.2 Use the power rule a m a n=a m+n a m a n=a m+n to combine exponents. Adjacent =−√4+(−1)1+2=-4+(-1)1+2 Adjacent =−√4+(−1)1+2=-4+(-1)1+2 Step 4.3.2 Add 1 1 and 2 2. Adjacent =−√4+(−1)3=-4+(-1)3 Adjacent =−√4+(−1)3=-4+(-1)3 Step 4.4 Raise −1-1 to the power of 3 3. Adjacent =−√4−1=-4-1 Step 4.5 Subtract 1 1 from 4 4. Adjacent =−√3=-3 Adjacent =−√3=-3 Step 5 Find the value of cosine. Tap for more steps... Step 5.1 Use the definition of cosine to find the value of cos(2 x)cos(2 x). cos(2 x)=adj hyp cos(2 x)=adj hyp Step 5.2 Substitute in the known values. cos(2 x)=−√3 2 cos(2 x)=-3 2 Step 5.3 Move the negative in front of the fraction. cos(2 x)=−√3 2 cos(2 x)=-3 2 cos(2 x)=−√3 2 cos(2 x)=-3 2 Step 6 Find the value of tangent. Tap for more steps... Step 6.1 Use the definition of tangent to find the value of tan(2 x)tan(2 x). tan(2 x)=opp adj tan(2 x)=opp adj Step 6.2 Substitute in the known values. tan(2 x)=−1−√3 tan(2 x)=-1-3 Step 6.3 Simplify the value of tan(2 x)tan(2 x). Tap for more steps... Step 6.3.1 Dividing two negative values results in a positive value. tan(2 x)=1√3 tan(2 x)=1 3 Step 6.3.2 Multiply 1√3 1 3 by √3√3 3 3. tan(2 x)=1√3⋅√3√3 tan(2 x)=1 3⋅3 3 Step 6.3.3 Combine and simplify the denominator. Tap for more steps... Step 6.3.3.1 Multiply 1√3 1 3 by √3√3 3 3. tan(2 x)=√3√3√3 tan(2 x)=3 3 3 Step 6.3.3.2 Raise √3 3 to the power of 1 1. tan(2 x)=√3√3√3 tan(2 x)=3 3 3 Step 6.3.3.3 Raise √3 3 to the power of 1 1. tan(2 x)=√3√3√3 tan(2 x)=3 3 3 Step 6.3.3.4 Use the power rule a m a n=a m+n a m a n=a m+n to combine exponents. tan(2 x)=√3√3 1+1 tan(2 x)=3 3 1+1 Step 6.3.3.5 Add 1 1 and 1 1. tan(2 x)=√3√3 2 tan(2 x)=3 3 2 Step 6.3.3.6 Rewrite √3 2 3 2 as 3 3. Tap for more steps... Step 6.3.3.6.1 Use n√a x=a x n a x n=a x n to rewrite √3 3 as 3 1 2 3 1 2. tan(2 x)=√3(3 1 2)2 tan(2 x)=3(3 1 2)2 Step 6.3.3.6.2 Apply the power rule and multiply exponents, (a m)n=a m n(a m)n=a m n. tan(2 x)=√3 3 1 2⋅2 tan(2 x)=3 3 1 2⋅2 Step 6.3.3.6.3 Combine 1 2 1 2 and 2 2. tan(2 x)=√3 3 2 2 tan(2 x)=3 3 2 2 Step 6.3.3.6.4 Cancel the common factor of 2 2. Tap for more steps... Step 6.3.3.6.4.1 Cancel the common factor. tan(2 x)=√3 3 2 2 tan(2 x)=3 3 2 2 Step 6.3.3.6.4.2 Rewrite the expression. tan(2 x)=√3 3 tan(2 x)=3 3 tan(2 x)=√3 3 tan(2 x)=3 3 Step 6.3.3.6.5 Evaluate the exponent. tan(2 x)=√3 3 tan(2 x)=3 3 tan(2 x)=√3 3 tan(2 x)=3 3 tan(2 x)=√3 3 tan(2 x)=3 3 tan(2 x)=√3 3 tan(2 x)=3 3 tan(2 x)=√3 3 tan(2 x)=3 3 Step 7 Find the value of cotangent. Tap for more steps... Step 7.1 Use the definition of cotangent to find the value of cot(2 x)cot(2 x). cot(2 x)=adj opp cot(2 x)=adj opp Step 7.2 Substitute in the known values. cot(2 x)=−√3−1 cot(2 x)=-3-1 Step 7.3 Simplify the value of cot(2 x)cot(2 x). Tap for more steps... Step 7.3.1 Dividing two negative values results in a positive value. cot(2 x)=√3 1 cot(2 x)=3 1 Step 7.3.2 Divide√3 3 by 1 1. cot(2 x)=√3 cot(2 x)=3 cot(2 x)=√3 cot(2 x)=3 cot(2 x)=√3 cot(2 x)=3 Step 8 Find the value of secant. Tap for more steps... Step 8.1 Use the definition of secant to find the value of sec(2 x)sec(2 x). sec(2 x)=hyp adj sec(2 x)=hyp adj Step 8.2 Substitute in the known values. sec(2 x)=2−√3 sec(2 x)=2-3 Step 8.3 Simplify the value of sec(2 x)sec(2 x). Tap for more steps... Step 8.3.1 Move the negative in front of the fraction. sec(2 x)=−2√3 sec(2 x)=-2 3 Step 8.3.2 Multiply 2√3 2 3 by √3√3 3 3. sec(2 x)=−(2√3⋅√3√3)sec(2 x)=-(2 3⋅3 3) Step 8.3.3 Combine and simplify the denominator. Tap for more steps... Step 8.3.3.1 Multiply 2√3 2 3 by √3√3 3 3. sec(2 x)=−2√3√3√3 sec(2 x)=-2 3 3 3 Step 8.3.3.2 Raise √3 3 to the power of 1 1. sec(2 x)=−2√3√3√3 sec(2 x)=-2 3 3 3 Step 8.3.3.3 Raise √3 3 to the power of 1 1. sec(2 x)=−2√3√3√3 sec(2 x)=-2 3 3 3 Step 8.3.3.4 Use the power rule a m a n=a m+n a m a n=a m+n to combine exponents. sec(2 x)=−2√3√3 1+1 sec(2 x)=-2 3 3 1+1 Step 8.3.3.5 Add 1 1 and 1 1. sec(2 x)=−2√3√3 2 sec(2 x)=-2 3 3 2 Step 8.3.3.6 Rewrite √3 2 3 2 as 3 3. Tap for more steps... Step 8.3.3.6.1 Use n√a x=a x n a x n=a x n to rewrite √3 3 as 3 1 2 3 1 2. sec(2 x)=−2√3(3 1 2)2 sec(2 x)=-2 3(3 1 2)2 Step 8.3.3.6.2 Apply the power rule and multiply exponents, (a m)n=a m n(a m)n=a m n. sec(2 x)=−2√3 3 1 2⋅2 sec(2 x)=-2 3 3 1 2⋅2 Step 8.3.3.6.3 Combine 1 2 1 2 and 2 2. sec(2 x)=−2√3 3 2 2 sec(2 x)=-2 3 3 2 2 Step 8.3.3.6.4 Cancel the common factor of 2 2. Tap for more steps... Step 8.3.3.6.4.1 Cancel the common factor. sec(2 x)=−2√3 3 2 2 sec(2 x)=-2 3 3 2 2 Step 8.3.3.6.4.2 Rewrite the expression. sec(2 x)=−2√3 3 sec(2 x)=-2 3 3 sec(2 x)=−2√3 3 sec(2 x)=-2 3 3 Step 8.3.3.6.5 Evaluate the exponent. sec(2 x)=−2√3 3 sec(2 x)=-2 3 3 sec(2 x)=−2√3 3 sec(2 x)=-2 3 3 sec(2 x)=−2√3 3 sec(2 x)=-2 3 3 sec(2 x)=−2√3 3 sec(2 x)=-2 3 3 sec(2 x)=−2√3 3 sec(2 x)=-2 3 3 Step 9 Find the value of cosecant. Tap for more steps... Step 9.1 Use the definition of cosecant to find the value of csc(2 x)csc(2 x). csc(2 x)=hyp opp csc(2 x)=hyp opp Step 9.2 Substitute in the known values. csc(2 x)=2−1 csc(2 x)=2-1 Step 9.3 Divide 2 2 by −1-1. csc(2 x)=−2 csc(2 x)=-2 csc(2 x)=−2 csc(2 x)=-2 Step 10 This is the solution to each trig value. sin(2 x)=−1 2 sin(2 x)=-1 2 cos(2 x)=−√3 2 cos(2 x)=-3 2 tan(2 x)=√3 3 tan(2 x)=3 3 cot(2 x)=√3 cot(2 x)=3 sec(2 x)=−2√3 3 sec(2 x)=-2 3 3 csc(2 x)=−2 csc(2 x)=-2 s i n(2 x)=−1 2 s i n⁡(2 x)=-1 2 1 2 x 2 1 2 x 2 1 2 x 1 2 x,, ( ( ) ) \| \| [ [ ] ] √ √   ≥ ≥   ° °       7 7 8 8 9 9       ≤ ≤   θ θ       4 4 5 5 6 6 / / ^ ^ × ×   π π       1 1 2 2 3 3 - - + + ÷ ÷ < <          , , 0 0 . . % %  = =     ⎡⎢⎣x 2 1 2√π∫x d x⎤⎥⎦[x 2 1 2 π∫⁡x d x] Please ensure that your password is at least 8 characters and contains each of the following: a number a letter a special character: @$#!%?& Do Not Sell My Personal Information When you visit our website, we store cookies on your browser to collect information. 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16244	https://sburer.github.io/papers/039-smooth.pdf	SIAM J. OPTIM. c ⃝2014 Society for Industrial and Applied Mathematics Vol. 24, No. 2, pp. 598–620 A FIRST-ORDER SMOOTHING TECHNIQUE FOR A CLASS OF LARGE-SCALE LINEAR PROGRAMS∗ JIEQIU CHEN† AND SAMUEL BURER‡ Abstract. We study a class of linear programming (LP) problems motivated by large-scale machine learning applications. After reformulating the LP as a convex nonsmooth problem, we apply Nesterov’s primal-dual excessive-gap technique. The iteration complexity of the excessive-gap technique depends on a parameter θ that arises because we must bound the primal feasible set, which is originally unbounded. We also dynamically update θ to speed up the convergence. The application of our algorithm to two machine learning problems demonstrates several advantages of the excessive-gap technique over existing methods. Key words. excessive-gap technique, large-scale linear programming, nonsmooth optimization, machine learning AMS subject classiﬁcations. 65K05, 65K10 DOI. 10.1137/110854400 1. Introduction. We investigate a ﬁrst-order smoothing technique to solve large-scale instances of the following linear programming (LP) problem: min α,ξ cT α + wT ξ (P) s. t. Aα −b ≤ξ, α, ξ ≥0, where α ∈ℜn and ξ ∈ℜm are the decision vectors and A ∈ℜm×n, b ∈ℜm, c ∈ℜn +, and w ∈ℜm + are the data. The optimal value of (P) is bounded below by zero, and thus an optimal solution exists. The variable ξ may be interpreted as an error, allowing some of the constraints “Aα ≤b” to be violated, and the objective term wT ξ then serves to penalize such violations. Expressed diﬀerently, if ξ were ﬁxed to 0, then the above formulation would be similar to an LP with c ≥0. Problem (P) is motivated by several machine learning problems. One such exam-ple is the LP-based ranking algorithm (Ataman and Ataman, Street, and Zhang . Another example is the 1-norm support vector machine (Zhu et al. , Mangasarian ). In section 5, we will demonstrate how to cast such machine learning problems as (P). We are also particularly interested in large-scale instances of (P) for two reasons: (i) while small instances can be solved eﬃciently by current LP solvers, for problems ∗Received by the editors November 8, 2011; accepted for publication (in revised form) January 7, 2014; published electronically April 3, 2014. The submitted manuscript has been created by the UChicago Argonne, LLC, Operator of Argonne National Laboratory (“Argonne”) under Contract DE-AC02-06CH11357 with the U.S. Department of Energy. The U.S. Government retains for itself, and others acting on its behalf, a paid-up, nonexclusive, irrevocable worldwide license in said article to reproduce, prepare derivative works, distribute copies to the public, and perform publicly and display publicly, by or on behalf of the Government. This research was supported in part by NSF grant CCF-0545514. †This manuscript was prepared while the author was a postdoctoral fellow at Mathematics and Computer Science Division, Argonne National Laboratory, Argonne, IL 60439 (jieqiu0808@ gmail.com). This author’s research was supported in part by the Oﬃce of Advanced Scientiﬁc Com-puting Research, Oﬃce of Science, U.S. Department of Energy, under contract DE-AC02-06CH11357. ‡Department of Management Sciences, University of Iowa, Iowa City, IA 52242-1994 (samuel-burer@uiowa.edu). 598 A SMOOTHING METHOD FOR A CLASS OF LARGE-SCALE LPs 599 where A is large and dense, using simplex or interior-point methods might not be feasible because of memory limits; and (ii) the machine learning problems above are often applied to large datasets, making A large. In applications that involve kernel matrices, such as the popular radial basis function (RBF) kernels (Hsu, Chang, and Lin ), A is also usually completely dense. Several approaches can be used to solve (P). Standard approaches are the simplex method or interior-point methods. As mentioned in the previous paragraph, however, (P) is often too large in machine learning applications. For example, Ataman reported that a moderately sized ranking problem formulated as (P) caused CPLEX to run out of memory on a standard PC. Mangasarian formulated a class of LPs, which includes (P) as a special case, and proposed an alternative method. He posed the problem as the unconstrained minimization of a convex diﬀerentiable piecewise-quadratic objective function and solved it using a generalized Newton method. As we will see in section 5, however, this method may not be suﬃciently robust in some cases. Another approach is to treat (P) as the equivalent nonsmooth problem min cT α + wT (Aα −b)+ (NS0) s. t. α ≥0, where (Aα−b)+ denotes the nonnegative part of the vector Aα−b, and then to solve (NS0) using nonsmooth techniques such as the standard subgradient method. Com-pared with the simplex method and interior-point methods, the subgradient method requires much less memory (basically the memory to store A). It also has low com-putational costs at each iteration (basically A times a vector). The main drawback of the subgradient method is slow convergence: its worst-case iteration complexity is O(1/ϵ2) , where ϵ is the error tolerance of a calculated solution. In this paper, we use Nesterov’s ﬁrst-order excessive-gap method [18, 17] to solve (NS0). For bounded feasible sets, the excessive-gap method has worst-case itera-tion complexity O(1/ϵ), which is an order of magnitude faster than the subgradient method. At the same time, its computational cost per iteration and memory require-ments are comparable to the subgradient method. Researchers have successfully applied the excessive-gap and related methods to many large-scale problems. Banerjee et al. considered the problem of ﬁtting a large-scale covariance matrix to multivariate Gaussian data. Hoda et al. and Gilpin et al. approximate Nash equilibria of large sequential two-player zero-sum games. Becker, Bobin, and Candes and Aybat and Iyengar demonstrate that the excessive-gap method is ideally suited for solving large-scale compressed sensing reconstruction problems. Smoothing techniques also have applications in semideﬁnite programming (Nesterov , d’Aspremont ) and general convex optimization (Lan, Lu, and Monteiro ). Nesterov’s ﬁrst-order excessive-gap method applies to the following generic prob-lem: (1) min{f(x) : x ∈Q}, where f is a continuous convex function with a certain structure and Q is a compact convex set (see section 2). To apply the excessive-gap method to (NS0), we will need the following assumption. Assumption 1.1. Deﬁne B := {i ∈{1, . . ., n} : ci = 0}. We assume αB is bounded in (P), where αB is deﬁned as the vector composed of entries of α that are indexed 600 JIEQIU CHEN AND SAMUEL BURER by B. Speciﬁcally, we assume knowledge of h ∈R\|B\| ++ such that αB ≤h is valid for at least one optimal solution of (P). This assumption allows us to bound the primal feasible set as follows. First, the assumption just mentioned bounds the subvector αB. Next, the remaining components of α are naturally bounded by any ﬁxed upper bound on the optimal value of (NS0). In particular, let θ∗be the optimal value of (NS0), α∗be an optimal solution, and θ > 0 be any upper bound on θ∗. We have cT α∗≤cT α∗+ wT (Aα∗−b)+ = θ∗≤θ, and so cT α ≤θ is valid for the optimal set of (NS0). Therefore, (NS0) is equivalent to the following problem over a compact convex set: min α≥0 cT α + wT (Aα −b)+ (NS) s. t. cT α ≤θ, αB ≤h. Based on (NS0) and (NS), we prove that the iteration complexity of Nesterov’s excessive-gap method is a function of θ, and we show that the algorithm can be improved by dynamically updating θ as it generates better bounds on θ∗. Independently, considered applying Nesterov’s technique to several support vector machine (SVM) problems, including the 1-norm SVM that we will study in section 5. However, it is not clear to us how they circumvented the unboundedness issue of the primal feasible set since they provide no discussion of this issue. Moreover, they use Nesterov’s primal-only smoothing method whereas our algorithm is based on the primal-dual excessive-gap technique . Our treatment of speciﬁc ingredients of the method is also diﬀerent, as will be shown in section 3. This paper is organized as follows. We summarize the major ingredients of the excessive-gap technique in section 2 to facilitate later discussion. In section 3, we show how to convert (P) into a problem of the form (1), and we specify the major components of the excessive-gap technique, such as the choice of the so-called prox-functions and derivations of various parameters. In section 4, we design a strategy to dynamically update the upper bound θ and show how it speeds up the excessive-gap technique. In section 5, we present two machine learning applications of the excessive-gap technique. In particular, we compare the excessive-gap technique with two existing methods, respectively, and demonstrate that our algorithm has several advantages. 2. Nesterov’s excessive-gap technique. In this section, we review the major ingredients of Nesterov’s excessive-gap technique: the excessive-gap condition and convergence rate. We focus on the concepts and results that will be used in our study and leave out technical details. First, we introduce some notation and deﬁnitions that will be used throughout the paper. 2.1. Notation and terminology. Let E denote a ﬁnite-dimensional real vector space, possibly with an index. This space is equipped with a norm ∥·∥, which has the same index as the corresponding space. Let ˆ A be a linear operator from E1 to E∗ 2, that is, ˆ A : E1 →E∗ 2, where E∗ 2 is the dual space of E2. Deﬁne the operator norm of ˆ A, induced by the norms ∥· ∥1 and ∥· ∥2, as ∥ˆ A∥1,2 = max ∥x∥1=1 max ∥u∥2=1⟨ˆ Ax, u⟩, A SMOOTHING METHOD FOR A CLASS OF LARGE-SCALE LPs 601 where ⟨·, ·⟩refers to the regular inner product. Note ∥·∥1 and ∥·∥2 do not necessarily represent the standard l1-norm and l2-norm; the subscripts are indices only. The operator norm has the following property: ∥ˆ A∥1,2 = max ∥x∥1=1 ∥ˆ Ax∥∗ 2 = max ∥u∥2=1 ∥ˆ AT u∥∗ 1, where ∥· ∥∗denotes the dual norm associated with ∥· ∥and is deﬁned as ∥z∥∗:= max{⟨z, x⟩: ∥x∥≤1}. We use ˆ Ai to denote the ith row of ˆ A and ˆ Aj the jth column of ˆ A. Similarly, we use ˆ AI ( ˆ AJ ) to denote the rows (columns) of ˆ A indexed by the set I (J ). For a vector v ∈ℜn, v−1 ∈ℜn denotes the vector whose components are the inverses of the components of v. We let Diag(v) represent the diagonal matrix with diagonal v. We use e to represent a vector of all ones. The dimension of e may diﬀer but should be clear from the context. 2.2. Excessive-gap technique. Consider the following functions f(x) and φ(u): f(x) = ˆ f(x) + max u∈Q2{⟨ˆ Ax, u⟩−ˆ φ(u)}, (2) φ(u) = −ˆ φ(u) + min x∈Q1{⟨ˆ Ax, u⟩+ ˆ f(x)}, (3) where Q1 and Q2 are (simple) compact convex sets in ﬁnite-dimensional Euclidean spaces E1 and E2, respectively; ˆ A is a linear operator mapping E1 to E∗ 2; and ˆ f(x) and ˆ φ(u) are continuous convex functions on Q1 and Q2, respectively. Thus, f(x) is convex and φ(u) is concave, but they are not necessarily diﬀerentiable. For any ¯ x ∈Q1 and ¯ u ∈Q2 we have (4) φ(¯ u) ≤f(¯ x) because φ(¯ u) = −ˆ φ(¯ u) + min x∈Q1{⟨ˆ Ax, ¯ u⟩+ ˆ f(x)} ≤−ˆ φ(¯ u) + ⟨ˆ A¯ x, ¯ u⟩+ ˆ f(¯ x) ≤ˆ f(¯ x) + max u∈Q2{⟨ˆ A¯ x, u⟩−ˆ φ(u)} = f(¯ x). Nesterov’s excessive-gap technique uses a primal-dual approach to solve the fol-lowing optimization problems simultaneously: min{f(x) : x ∈Q1}, (5) max{φ(u) : u ∈Q2}. (6) Note that by Fenchel duality (Borwein and Lewis ), (6) is the dual problem of (5) and there is no duality gap. Because of the nondiﬀerentiability, the primal-dual approach does not directly deal with f(x) and φ(u). Instead it works with the following “smoothed” versions: fμ2(x) = ˆ f(x) + max u∈Q2{⟨ˆ Ax, u⟩−ˆ φ(u) −μ2d2(u)}, (7) φμ1(u) = −ˆ φ(u) + min x∈Q1{⟨ˆ Ax, u⟩+ ˆ f(x) + μ1d1(x)}, (8) 602 JIEQIU CHEN AND SAMUEL BURER where μi is a positive smoothness parameter and di(·) is a prox-function on the set Qi, which means di(·) is continuous and strongly convex on Qi. A strongly convex function d(·) has the following property for some σ > 0: d(x) ≥d(x∗) + 1 2σ∥x −x∗∥2 for all x ∈Q, where x∗= arg minx∈Q d(x). The purpose of introducing these prox-functions is to smooth f(x) and φ(u). The resultant fμ2(x) and φμ1(u) are diﬀerentiable, and their gradients are Lipschitz continuous. When μ1 and μ2 are small, fμ2 ≈f and φμ1 ≈φ. By deﬁnition we have fμ2(x) ≤f(x) and φ(u) ≤φμ1(u). For suﬃciently large μ1 and μ2, one can show that there exist some ¯ x ∈Q1 and ¯ u ∈Q2 satisfying the following excessive-gap condition (EGC): fμ2(¯ x) ≤φμ1(¯ u). (EGC) (EGC) is like a switched version of (4), which ensures that the primal-dual gap is bounded above, as stated in the following lemma. Lemma 2.1 (Nesterov ). Let ¯ x ∈Q1 and ¯ u ∈Q2 satisfy (EGC). Then 0 ≤ f(¯ x)−φ(¯ u) ≤μ1D1 + μ2D2, where D1 := maxx∈Q1 d1(x) and D2 := maxu∈Q2 d2(u). In addition to Lemma 2.1, the excessive-gap technique features three other im-portant ingredients: (i) a procedure that calculates an initial (x0, u0, μ0 1, μ0 2) satisfying (EGC), namely, fμ0 2(x0) ≤φμ0 1(u0); (ii) given (xk, uk, μk 1, μk 2) satisfying (EGC), a procedure that generates (xk+1, uk+1, μk+1 1 , μk+1 2 ) satisfying (EGC) as well; (iii) μk+1 i ≤μk i , i = 1, 2, where one of the two inequalities is strict and μk i →0, i = 1, 2. Ingredients (i) and (ii) generate a sequence {(xk, uk, μk 1, μk 2)} that satisﬁes (EGC) for each k. Because of Lemma 2.1 and (iii), the primal-dual gap goes to zero as k increases; that is, (9) 0 ≤f(xk) −φ(uk) ≤μk 1D1 + μk 2D2 →0. As long as (EGC) is maintained, (9) will hold for all k. Theorem 2.2 (Nesterov ). Given ϵ > 0, there is an algorithm based on the excessive-gap technique that produces a pair (xN, uN) ∈Q1 × Q2 such that 0 ≤f(xN) −φ(uN) ≤ϵ in N = 4∥ˆ A∥1,2 ϵ D1D2 σ1σ2 iterations. In each iteration of the algorithm, one needs to update (xk, uk, μk 1, μk 2), a process that requires solving several subproblems in the form of the inner max problem in (7) or the inner min problem in (8). Therefore, the solutions of these max and min problems should be easily computable. We omit the generic scheme here; we will describe the speciﬁc algorithm with respect to our problem in section 3.3. 3. Applying the excessive-gap technique. In this section, we show how to convert (NS) into the standard form required by the excessive-gap technique. After the conversion, we specify all ingredients, including our choice of the prox-functions, the calculation of parameters for the excessive-gap technique, and the iteration complexity for solving (NS). In the last subsection, we detail the algorithm. A SMOOTHING METHOD FOR A CLASS OF LARGE-SCALE LPs 603 3.1. Reformulation. We deﬁne two notations that will be used throughout this section. Let □m denote the standard box of dimension m: □m := {u ∈ℜm : 0 ≤u ≤e}. Let △n denote the standard simplex of dimension n: △n := {x ∈ℜn : x ≥0, ⟨e, x⟩= 1} . For notational convenience, we will drop the dimension subscript when the dimension is clear from context. We ﬁrst reformulate the nonsmooth part of the objective function, ⟨w, (Aα−b)+⟩, as a maximization problem. Since w is nonnegative, we have ⟨w, (Aα −b)+⟩= e, Diag(w) (Aα −b)+ = e, Diag(w)(Aα −b) + = max u∈□m {⟨Diag(w)(Aα −b), u⟩} = max u∈□m {⟨Diag(w)Aα, u⟩−⟨Diag(w)b, u⟩} . Next, we transform the primal feasible set {α ∈ℜn + : αB ≤h, ⟨c, α⟩≤θ} with θ > 0 into a simpler set by changing variables. The purpose of the transformation is to simplify the problem presentation, especially to facilitate the application of Nesterov’s method. Recall that cB = 0 and thus ⟨c, α⟩= ⟨c ¯ B, α ¯ B⟩, where ¯ B := {1, . . ., n}\B. Deﬁne a new variable x ∈ℜn+1 as follows: xB := Diag h−1 αB, xS := 1 θ Diag (c ¯ B) α ¯ B θ −⟨c ¯ B, α ¯ B⟩ , (10) where S := ¯ B ∪{n + 1}. Because 0 ≤αB ≤h, we see that xB is inside a box with dimension \|B\|. Because α ≥0 and ⟨c, α⟩≤θ, xS resides in a simplex with dimension \|S\| = n + 1 −\|B\|. So the primal feasible set becomes {(xS, xB) ∈ℜn+1 : xS ∈△, xB ∈□} =: FP . The last step of the reformulation involves deﬁning a new set of data ˆ A ∈ ℜm×(n+1), ˆ b ∈ℜm, and ˆ e ∈ℜn+1 as follows: ˆ AS := Diag(w)A ¯ B Diag(c−1 ¯ B ) 0 , (11) ˆ AB := 1 θ Diag(w)AB Diag(h), (12) ˆ b := Diag(w)b, ˆ eB := 0, ˆ e ¯ B := e, ˆ en+1 := 0. With the above deﬁnition, one can easily verify that the objective function of (NS) can be expressed in terms of x: (13) ⟨c, α⟩+ ⟨w, (Aα −b)+⟩= θ ⟨ˆ e, x⟩+ max u∈□ ⟨ˆ Ax, u⟩−1 θ⟨ˆ b, u⟩ =: θf(x; θ), and thus (NS) is equivalent to the following problem: min {f(x; θ) : x ∈FP } . (SP) 604 JIEQIU CHEN AND SAMUEL BURER Based on the primal-dual structure of (2) and (3), we immediately have the dual problem max {φ(u; θ) : u ∈□} , (SD) where φ(u; θ) := −1 θ⟨ˆ b, u⟩+ min xS∈Δ,xB∈□ ⟨ˆ Ax, u⟩+ ⟨ˆ e, x⟩ . The excessive-gap technique we described in section 2.2 can be used to solve (SP) and (SD). For any primal feasible solution ¯ x obtained from the excessive-gap technique, it is easy to recover a feasible solution ¯ α to (NS) by just reversing the transformation. Note that by (13), the objective f(¯ x; θ) needs to be scaled by θ in order to recover the objective value of (NS). In fact, this property will inﬂuence the iteration complexity of the excessive-gap technique. In particular, the primal error (the absolute diﬀerence between the primal objective value and the optimal value) of (NS) is the primal error of (SP) scaled by θ. Lemma 3.1. Deﬁne p(α) := ⟨c, α⟩+ ⟨w, (Aα −b)+⟩, and let x∗be an optimal solution of (SP). Recall θ∗denotes the optimal value of (NS). Suppose ¯ x is feasible to (SP) and ¯ α is the corresponding feasible solution to (NS). Then (14) p(¯ α) −θ∗= θ f(¯ x; θ) −f(x∗; θ) . Proof. ¯ α (α∗) can be obtained by ¯ x (x∗) through the relationship (10). By (13), p(¯ α) = θf(¯ x; θ) and p(α∗) = θ∗= θf(x∗; θ), and the result follows. 3.2. Speciﬁcations. In this subsection, we discuss in detail each ingredient of the excessive-gap technique. The choice of norms and prox-functions is critical. We select the l1-norm and the entropy distance function as the prox-function for the primal space, similar to what did for the simplex: (15) ∥x∥1 := n+1 i=1 \|xi\|, d1(x) := ln(\|S\|) + \|B\| · exp(−1) + n+1 i=1 xi ln xi, and so d1(·) is minimized at the prox-center x◦ i = 1/\|S\| if i ∈S, exp(−1) if i ∈B. The constant term ln(\|S\|) + \|B\| · exp(−1) in d1(·) is present such that d1(·) evaluated at x◦is 0. Based on the selected prox-function, we calculate the parameter D1 = max x {d1(x) : x ∈FP } = ln(\|S\|) + \|B\| · exp(−1), where d1(x) is maximized when xS lies in one corner point of the simplex and xB = 0 since n+1 i=1 xi ln xi is nonnegative when x ∈FP . The convexity parameter of d1(x) is σ1 = 1 1+\|B\| because ⟨d′′(x)h, h⟩= n+1 i=1 (hi)2 xi and, from the Cauchy–Schwarz inequal-ity, ∥h∥2 1 ≤ i∈S xi + i∈B xi n+1 i=1 (hi)2 xi ≤(1 + \|B\|)⟨d′′(x)h, h⟩. A SMOOTHING METHOD FOR A CLASS OF LARGE-SCALE LPs 605 For the dual space, we select the l2-norm and a “distance squared” quadratic function (16) ∥u∥2 := m j=1 (uj)2, d2(u) := 1 2 m j=1 uj −1 2 2 . It is easy to verify that D2 = maxu{d2(u) : u ∈□m} = m 8 , σ2 = 1, and d2(·) achieves its minimum at u◦= 1 2e. As can be seen from Theorem 2.2, larger D1 and D2 correspond to a larger iteration bound N, and thus we select the above prox-function and centers in order to achieve small D1 and D2. For example, u◦= 1 2e because D1 is smallest by selecting u◦as the center of the box. The operator norm of ˆ A is calculated as follows: ∥ˆ A∥1,2 = max u ∥ˆ AT u∥∗ 1 : ∥u∥2 = 1 = max u max i∈{1,...,n+1} ⟨ˆ Ai, u⟩ : ∥u∥2 = 1 = max i∈{1,...,n} max u ⟨ˆ Ai, u⟩: ∥u∥2 = 1 = max i∈{1,...,n} ∥ˆ Ai∥2, where the second equality follows because the dual norm of the l1-norm is the l∞-norm, and the third equality follows from the fact that ˆ A’s last column is zero, as shown by (11). With all the parameters computed, we are ready to state the iteration complexity of solving (SP) and (SD). Proposition 3.2. Using Nesterov’s excessive-gap technique, for any ϵ > 0, we obtain a pair of solutions (xN, uN) to (SP) and (SD) such that 0 ≤f(xN; θ) −φ(uN; θ) ≤ϵ in (17) N := N(θ) := 1 ϵ max i∈{1,...,n} ∥ˆ Ai∥2 2 (1 + \|B\|) [ln(\|S\|) + \|B\| · exp(−1)] · m iterations. Proof. The result is obtained by applying Theorem 2.2. By (12), ˆ AB is dependent on 1/θ, and so the number of iterations of the excessive-gap technique depends on θ. We thus write N(θ) as a function of θ to reﬂect this dependence. We comment that diﬀerent combinations of norms and prox-functions other than (15) and (16) may lead to diﬀerent parameter values and thus diﬀerent iteration complexities. We considered several choices for the dual space, and the choice (16) gives us the lowest iteration complexity among those considered. For example, one could choose the l1-norm and the same prox-function as d1(x) for the dual space; the resultant iteration complexity is O(√m) times larger than (17), which is much worse than the current one if m is large. In Proposition 3.2, the iteration complexity is stated with respect to (SP). Now we state the iteration complexity with respect to (NS). 606 JIEQIU CHEN AND SAMUEL BURER Proposition 3.3. Using Nesterov’s excessive-gap technique, for any ϵ > 0, we obtain a solution ¯ α to (NS) such that 0 ≤p(¯ α) −θ∗≤ϵ in N(θ) := θN(θ) iterations, where N(θ) is given by (17). Proof. By Proposition 3.2, in N(θ) iterations, we obtain a solution (¯ x, ¯ u) such that the primal-dual gap is small enough: 0 ≤f(¯ x; θ) −φ(¯ u; θ) ≤ϵ θ . We then can construct ¯ α feasible to (NS). Thus, by (14), we have 0 ≤p(¯ α) −θ∗= θ f(¯ x; θ) −f(x∗; θ) ≤θ f(¯ x; θ) −φ(¯ u; θ) ≤ϵ. From this proposition, the iteration complexity of solving (NS) is a function of the upper bound θ. This is not so surprising because the primal feasible set is originally unbounded, and the hidden upper bound θ is discovered by exploiting the structure of the objective function. Using (17), we see that N(θ) = θN(θ) is a nondecreasing function in θ. More precisely, let i∗:= argmaxi∈{1,...,n} ∥ˆ Ai∥2, and it follows from the deﬁnition of ˆ A, (11) and (12), that N(θ) does not depend on θ when i∗∈S and is proportional to 1/θ when i∗∈B. Thus we see N(θ) is an increasing function of θ when i∗∈S and does not depend on θ when i∗∈B. In the former case, the smaller θ is, the better. One could always apply some heuristics to get a good θ. In section 4, we instead introduce a strategy that dynamically updates θ and consequently reduces the iteration complexity for solving (NS) as the algorithm progresses. Next we discuss the subproblems associated with our choice of prox-functions. These subproblems will be solved repeatedly in the algorithm presented in section 3.3, and thus it is important to have closed-form solutions for them. The subproblems are the max and min problems presented within the following smoothed versions of our primal and dual objective functions (recall (7) and (8)): fμ2(x; θ) = ⟨ˆ e, x⟩+ max u∈□m ⟨ˆ Ax, u⟩−1 θ⟨ˆ b, u⟩−μ2d2(u) , φμ1(u; θ) = −1 θ⟨ˆ b, u⟩+ min xS∈Δ, xB∈□ ⟨ˆ Ax, u⟩+ ⟨ˆ e, x⟩+ μ1d1(x) , where d1(x) and d2(u) are given in (15) and (16). Consider the min subproblem ﬁrst. Its solution is sargmin(d1, −1 μ1 ( ˆ AT u + ˆ e)), where sargmin(d1, s) := arg min x∈Q1 {−⟨s, x⟩+ d1(x)} = arg min xS∈Δ,xB∈□ −⟨s, x⟩+ n+1 i=1 xi ln xi . The following lemma establishes a closed form for sargmin(d1, s). Lemma 3.4. Given s, the solution sargmin(d1, s) is given by [sargmin(d1, s)]i = exp(si) \|S\| j=1 exp(sj), i ∈S, proj[0,1] (exp (si)) , i ∈B, where proj0,1 projects y to the nearest point between 0 and 1. A SMOOTHING METHOD FOR A CLASS OF LARGE-SCALE LPs 607 Proof. The objective function is separable in S and B. For optimizing this function over the simplex, see Lemma 4 in . For optimizing over the box, we compute the point at which the ﬁrst-order derivative vanishes and then project that point back to the feasible region. Now consider the max subproblem. In a similar manner, its solution is sargmin(d2, 1 μ2 ( ˆ Ax −ˆ b/θ)), where sargmin(d2, s) := arg min u∈Q2 {−⟨s, u⟩+ d2(u)} = arg min u∈□ ⎧ ⎨ ⎩−⟨s, u⟩+ m j=1 uj −1 2 2 ⎫ ⎬ ⎭. This problem also has a closed-form solution: Lemma 3.5. Given s, the solution sargmin(d2, s) is given by [sargmin(d2, s)]j = proj[0,1] 1 2(1 + sj) , j = 1, . . . , m, where proj0,1 projects y to the nearest point between 0 and 1. Proof. Observe that −⟨s, u⟩+ m j=1 uj −1 2 2 = m j=1 u2 j −(sj + 1) uj + 1 4 . So the problem reduces to solving m one-dimensional quadratic problems whose so-lutions are as stated. 3.3. Algorithm. The algorithmic scheme presented by is generic, and the problem-speciﬁc parameters for our problem have been calculated in sections 3.1 and 3.2. In this subsection, we explicitly state the scheme with respect to (SP) and (SD). In this algorithm, there are three functions: (i) Initial initializes all the parameters and the primal-dual solution (x0, u0) satisfying (EGC); (ii) Update1 is the primal update; (iii) Update2 is the dual update. Update1 and Update2 are symmetric but entail diﬀerent subproblems. Here is the algorithm Initial. Algorithm 1. Initial. Input: Data (m, n, ˆ A,ˆ b, ˆ e, θ) Output: Initialized parameters (μ0 1, μ0 2) and solutions (x0, u0) that satisfy (EGC) 1: D1 = ln(\|S\|) + \|B\| · exp(−1), D2 = m/8, σ1 = 1 1+\|B\|, σ2 = 1, ∥ˆ A∥1,2 = maxi∈{1,...,n} ∥ˆ Ai∥2 2: μ0 1 = 2 ∥ˆ A∥1,2 # D2 σ1σ2D1 , μ0 2 = ∥ˆ A∥1,2 # D1 σ1σ2D2 3: ¯ x = sargmin(d1, 0) 4: u0 = sargmin d2, 1 μ0 2 ( ˆ A¯ x −ˆ b/θ) 5: x0 = sargmin d1, ln(¯ x) + e − μ0 2 ∥ˆ A∥2 1,2 (AT u0 + ˆ e) 608 JIEQIU CHEN AND SAMUEL BURER Here is the primal update (when k is even). Algorithm 2. Update1: primal update. Input: Current solution (x, u) and parameters (μ1, μ2, τ, θ) Output: (x+, u+, μ+ 1 , μ+ 2 ) that satisfy (EGC) 1: ¯ x = sargmin d1, −1 μ1 ( ˆ AT u + ˆ e) 2: ˆ x = (1 −τ)x + τ ¯ x 3: ¯ u = sargmin d2, 1 μ2 ( ˆ A ˆ x −ˆ b/θ) 4: ˜ x = sargmin d1, ln(¯ x) + e − τ (1−τ)μ1 ( ˆ AT ¯ u + ˆ e) 5: x+ = (1 −τ)x + τ ˜ x 6: u+ = (1 −τ)u + τ ¯ u 7: μ+ 1 = (1 −τ)μ1, μ+ 2 = μ2 Here is the dual update (when k is odd). Algorithm 3. Update2: dual update. Input: Current solution (x, u) and parameters (μ1, μ2, τ, θ) Output: (x+, u+, μ+ 1 , μ+ 2 ) that satisfy (EGC) 1: ¯ u = sargmin d2, 1 μ2 ( ˆ A x −ˆ b/θ) 2: ˆ u = (1 −τ)u + τ ¯ u 3: ¯ x = sargmin d1, −1 μ1 ( ˆ AT ˆ u + ˆ e) 4: ˜ u = sargmin d2, τ (1−τ)μ2 ( ˆ A ¯ x −ˆ b/θ) + ¯ u −1 2e 5: x+ = (1 −τ)x + τ ¯ x 6: u+ = (1 −τ)u + τ ˜ u 7: μ+ 2 = (1 −τ)μ2, μ+ 1 = μ1 The entire Smooth algorithm is as follows. Algorithm 4. Smooth. Input: (i) Data (m, n, ˆ A,ˆ b, ˆ e, θ, N(θ)); (ii) Subroutines sargmin(d1, ·) and sargmin(d2, ·) Output: (xN(θ), uN(θ)) 1: (x0, u0, μ0 1, μ0 2) = Initial (m, n, ˆ A,ˆ b, ˆ e, θ) 2: for k = 0, 1, . . . , N(θ) −1 do 3: τ = 2 k+3 4: if k is even then 5: (xk+1, uk+1, μk+1 1 , μk+1 2 ) = Update1(xk, uk, μk 1, μk 2, τ, θ) 6: else 7: (xk+1, uk+1, μk+1 1 , μk+1 2 ) = Update2(xk, uk, μk 1, μk 2, τ, θ) A SMOOTHING METHOD FOR A CLASS OF LARGE-SCALE LPs 609 We have two comments regarding the Smooth algorithm. First, according to Lemmas 3.4 and 3.5, the subproblems have closed-form solutions and can be solved quickly. The most time-consuming operations are thus the matrix-vector multipli-cations ˆ A¯ x and ˆ AT ¯ u. Second, θ is treated as a ﬁxed parameter for Update1 and Update2. In the next section, we will discuss a simple procedure that dynamically updates θ, and the algorithm will be valid even with changing values of θ. 4. Speeding up the convergence. Recall θ > 0 is an upper bound on the optimal value θ∗of (NS). As shown in Proposition 3.3 and its subsequent discus-sion, the iteration complexity for obtaining an ϵ-solution of (NS) is directly related to θ/ϵ, when the parameter ∥ˆ A∥1,2 does not depend on θ. This happens when argmaxi∈{1,...,n} ∥ˆ Ai∥∈S and is the case we consider in this section. As an in-put parameter, the smaller θ is, the better the iteration complexity, and the best possible θ is the optimal value θ∗. We will of course obtain new information on θ∗as Algorithm 4 progresses; in particular, an improved primal objective value will give a better upper bound on θ∗than θ. To take advantage of this information, we consider updating θ dynamically within Algorithm 4. Suppose we have a better bound θ+ ∈[θ∗, θ) available after running Algorithm 4 for K iterations, where K < N(θ) = θN(θ). Recall N(θ), as deﬁned in (17), depends on the error tolerance ϵ. With θ+ on hand, it may be worthwhile to restart the algorithm and input θ+ for the new run, if the number of iterations required by the new run plus the number of iterations already run is smaller than the iteration estimate under θ, that is, if K + θ+N(θ+) < θN(θ) holds. Instead of restarting the algorithm, however, we update θ dynamically. In fact, we will show that it is possible to improve the iteration complexity to θ+N(θ)—without restarting the algorithm—provided that the condition (EGC) is carefully maintained when updating the parameter θ to θ+. To achieve this result, let θ > θ∗be the current upper bound on the optimal value, and let k be the iteration counter. The following lemma shows, under mild conditions, the existence of θ+ ∈(θ∗, θ) and k such that (xk, uk, μk 1, μk 2) satisﬁes (EGC) with respect to θ+. Lemma 4.1. Within Algorithm 4, suppose (xk, uk, μk 1, μk 2) satisﬁes (EGC) strictly with respect to θ; that is, fμk 2 (xk; θ) < φμk 1 (uk; θ). Suppose also that f(xk; θ) < 1. Then there exists θ+ ∈(θ∗, θ) such that (xk, uk, μk 1, μk 2) also satisﬁes (EGC) for θ+: (18) fμk 2(xk; θ+) ≤φμk 1 (uk; θ+). Proof. Deﬁne θk := θf(xk; θ). Based on the relationship (14), we know θk ≥θ∗, i.e., θk is a valid upper bound on θ∗. Then f(xk; θ) < 1 implies θk ∈[θ∗, θ). If fμk 2(xk; θk) ≤φμk 1 (uk; θk) holds, simply set θ+ := θk. Otherwise, g(τ) := fμk 2(xk; τ) − φμk 1 (uk; τ) is a continuous function of τ, and we have g(θk) > 0 and g(θ) < 0. So there exists θ+ ∈(θk, θ) such that g(θ+) ≤0. In other words, (18) holds. We remark that, intuitively, if the error tolerance ϵ is small enough, the condition f(xk; θ) < 1 in the lemma eventually holds for large enough k since f(x∗; θ) = θ∗/θ < 1. From now on, we use pk to represent the primal value p(αk) for notational con-venience. By (9), we have f(xk; θ) −φ(uk; θ) ≤μk 1D1 + μk 2D2 =: U k, k = 1, 2, . . . , where the bounding sequence $ U k%∞ k=1 depends on θ as {μk 1} and {μk 2} depend on θ. According to (14), the primal error pk −θ∗at iteration k is bounded above by θU k 610 JIEQIU CHEN AND SAMUEL BURER because (19) pk −θ∗= θ f(xk; θ) −f(x∗; θ) ≤θ f(xk; θ) −φ(uk; θ) ≤θU k, k = 1, 2, . . .. On the other hand, the assumption of Lemma 2.1 is satisﬁed because (EGC) holds for θ+, as shown by Lemma 4.1. Thus applying Lemma 2.1 we have (20) f(xk; θ+) −φ(uk; θ+) ≤U k. Therefore, we obtain (21) pk −θ∗≤θ+ f(xk; θ+) −φ(uk; θ+) ≤θ+U k, where the ﬁrst inequality follows from (14) and the second inequality follows from (20). So, instead of θU k, the improved upper bound θ+U k is shown to hold at iteration k. Since the excessive-gap technique guarantees that (EGC) is maintained in the next iteration as long as it holds at the current iteration, immediately switching θ to θ+ allows us to bound the primal error from the current iteration onwards by the sequence {θ+U l}∞ l=k. The above analysis leads to the following proposition. Proposition 4.2. If, during the course of Algorithm 4 at iteration k, a new upper bound θ+ ∈[θ∗, θ) satisfying (18) is employed for Algorithm 4, then the bound for primal error can be improved from ϵ to θ+ θ ϵ < ϵ, where ϵ denotes the original bound for primal error at iteration k. Proof. It follows immediately from (19) and (21). We have a few comments. (i) It is worth emphasizing that Uk depends on θ and does not change while updating θ to θ+ and thus the improvement of the bounds is completely due to the improved θ+. (ii) The primal-dual solutions generated in subsequent iterations are diﬀerent than those that would have been generated with the parameter θ because the parameter θ+ aﬀects the subroutines Update1 and Update2. (iii) We have demonstrated the existence of θ+ in Lemma 4.1, but there appears to be no closed-form formula for it. One simple procedure to obtain θ+ is shown in Algorithm 5. Lines 2 and 3 of Algorithm 5 do not aﬀect other parts of the Smooth algorithm and can be put into the procedure conveniently. Thus Update3 can be performed periodically, say every 50 iterations, to improve the bound on the number of iterations. Algorithm 5. Update3: bound update. Input: θ Output: θ+ satisﬁes (EGC) 1: if fμ2(xk; θ) ≤φμ1(uk; θ) then 2: θ+ = θf(xk; θ) 3: while fμ2(xk; θ+) > φμ1(uk; θ+) do 4: θ+ = 1/2(θ+ + θ) (iv) To maintain the excessive-gap condition, one can easily see that θ+ should be chosen in a neighborhood of θ, and so the reduction of the number of iterations by updating θ may not be big. However, the procedure of updating θ can be performed repeatedly. As the algorithm converges, the updated A SMOOTHING METHOD FOR A CLASS OF LARGE-SCALE LPs 611 Fig. 1. Comparison of running the Smooth algorithm under cases (i)–(iii). The left subplot shows the change of the primal error p(α)−θ∗over time; the right subplot shows how the primal-dual gap changes over time and compares it with the theoretical upper bounds. parameter value becomes a better and better approximation of θ∗, and thus the cumulative improvements might be signiﬁcant. (v) After several updates to the upper bound, say, from the intial θ to θ+ to θ++, one can see that the complexity bound updates from θN(θ) to θ+N(θ) to θ++N(θ). Hence, a lower bound for the best iteration complexity of our method is θ∗N(θ), where θ is the initial upper bound and θ∗is the optimal value. We close this section with an example that illustrates the eﬀects of dynamically updating θ. The example is an instance from the ﬁrst application that will be discussed in section 5.1 and is created with the dataset dermatology from Asuncion and Newman . The data matrix A in (P) has dimension 6760 × 358. We did three runs of the Smooth algorithm with the following variations: (i) Initial input of trivial θ = ⟨w, (−b)+⟩, and the Smooth algorithm is run without updating θ; (ii) initial input of trivial θ = ⟨w, (−b)+⟩, and the Smooth algorithm is run with the dynamic update of θ; (iii) initial input of θ = θ∗, where the optimal value θ∗has been calculated using a standard LP solver, and there is no update on θ because the bound is already optimal. While case (iii) is not realistic because θ∗is the idealized best upper bound, which is gotten by presolving the LP, it serves as a best-case comparison for cases (i) and (ii). Figure 1 shows the results in log-log scale. The ﬁrst subplot shows how the primal error pk −θ∗changes over time. One can see that dynamically updating θ reduces the error signiﬁcantly and its performance is almost as good as the ideal case (iii). We also comment that, in this particular example, θ∗≪⟨w, (−b)+⟩and thus the starting value of the error for case (iii) is far smaller than for (i) and (ii). Note also that (i) and (ii) are identical until the iteration when θ is ﬁrst updated. The second subplot demonstrates how the primal-dual gap θ(f(xk; θ)−φ(uk; θ)) changes over time for the three cases. It also plots the upper bound sequence θU k to show how θ aﬀects the 612 JIEQIU CHEN AND SAMUEL BURER primal-dual gap. Note that for this example, the upper bound associated with case (ii) changes over time because θ is being updated repeatedly. 5. Applications and computational experiments. In this section, we study the application of Algorithm 4 for solving two machine learning problems: (i) an LP-based ranking method [3, 2]; (ii) 1-norm SVMs (see, for example, for the for-mulation of the 1-norm SVM and more references). In the following two subsections, we ﬁrst brieﬂy describe the two applications and reformulate them as (P). We then conduct computational experiments to compare Algorithm 4 with existing algorithms for the respective applications. 5.1. LP-based ranking method. The LP-based ranking method proposed in is designed to train a scoring function that ranks all positive points higher than all negative points (from data that are assumed to have binary output). This ranking method is reported to perform better than SVM-based ranking algorithms. Let (xl, yl) be an instance in the training set X ∋l. The class label is yl ∈{1, −1}. Let X+, X−represent the set of points with positive and negative labels, respectively. Let K(·, ·) denote a chosen kernel function, for example, the RBF kernel function. At its core, the ranking method is the following optimization problem: min l∈X αl + C i∈X+, j∈X− wi,jξi,j (LPR) s. t. l∈X yl [K(xi, xl) −K(xj, xl)] αl ≥1 −ξi,j ∀i ∈X+, j ∈X−, α ≥0, ξ ≥0, where α ∈ℜ\|X\| and ξ ∈ℜ\|X+\|×\|X−\| are the decision vectors and all else are data. It is assumed that C > 0 and wi,j ≥0 for all i ∈X+, j ∈X−. To put (LPR) in the form of (P), we deﬁne Ai×j,l := −yl [K(xi, xl) −K(xj, xl)] ∀i × j ∈X+ × X−, l ∈X, m := \|X+\| × \|X−\|, n := \|X\| = \|X+\| + \|X−\|, such that A ∈ℜm×n. Now (LPR) is readily modeled as (P) with data c = e ∈ℜn, w = Ce ∈ℜm, b = −e ∈ℜm. Notice that m grows quadratically as the number of data points grows, and thus A becomes large scale even for medium-sized datasets. In addition, since every entry of A is the diﬀerence of two kernel function evaluations, A is usually fully dense. In , the subgradient method is proposed for (LPR) because the method is memory eﬃcient and thus is able to solve large-scale instances. In our computational study, we compare Algorithm 4 with the subgradient method implemented in , which is essentially the incremental subgradient method (Nedic and Bertsekas ). We collected 14 datasets from the UCI Machine Learning repository (Asuncion and Newman ) and prepare them in the same way as in . Table 1 describes the processed data A. We point out that among the problems in Table 1, cancer and diabetes cannot be solved by a commercial LP solver, like CPLEX (via either primal A SMOOTHING METHOD FOR A CLASS OF LARGE-SCALE LPs 613 Table 1 Statistics of A: dimension, percentage of nonzeros, and storage size in MATLAB format (in-stances are ordered increasingly by the number of nonzeros). Instance m n Nonzeros (%) Size (MB) wine 6240 178 11% 0.2 iris 5000 150 100% 5.4 glass 5365 214 100% 8.3 ntyroid 5550 215 98% 7.3 sonar 10767 208 100% 16.4 derma 6760 358 93% 6.6 heart 18000 270 100% 32.9 ecoli 14768 336 100% 36.3 spectf 24638 351 93% 20.0 ion 28350 351 100% 72.7 liver 29000 345 100% 60.6 boston 21984 506 100% 50.2 cancer 75684 569 100% 314.6 diabetes 134000 768 96% 409.1 or dual LP formulation), on a machine with 4-GB RAM without perhaps some special handling of the memory. One beneﬁt of the excessive-gap technique is the availability of the primal-dual gap, which can serve as a good stopping criterion. On the contrary, it is not easy to obtain a primal-dual gap for the subgradient method.1 So, in order to compare the two methods without a common stopping criterion, we perform the computational experiments in the following way. First, we solve each instance via Algorithm 4 with error tolerance ϵ = 1 and obtain the best objective value found by the excessive-gap technique. Second, we run the subgradient method until it ﬁnds at least as good an objective value as the excessive-gap technique, or until it reaches the time limit. We set a time limit of 18,000 seconds for both methods. All computations were performed on a Pentium D running at 3.2 GHz under the Linux operating system with 4-GB RAM. Table 2 presents the CPU times and the best objective values found when the algorithms terminate. Except for the two largest instances, we have the optimal values of the ranking problems available for gauging the quality of the solutions, as listed under the column θ∗. The subgradient method was able to ﬁnd as good a solution as the excessive-gap technique in 5 out of the 14 instances within the time limit and is faster in 3 out of those 5 instances. For the remaining 9 instances, the excessive-gap technique either is faster than the subgradient method in achieving the same quality solution or obtains better solutions in the same amount of time. We also plot the primal objective value errors of both methods versus the CPU times for one instance, glass, in Figure 2. We comment that Figure 2 reﬂects the typical behavior of the two methods. The subgradient method ﬁnds good solutions rapidly, but its convergence is slow; the excessive-gap technique’s initial objective value is usually not as good but converges faster, as predicted by theory. This obser-vation suggests that the subgradient method might be a better choice if obtaining a good solution in a short time is important. On the other hand, if the accuracy of the optimization problem is critical, the excessive-gap technique is the better choice for large-scale problems. 1Nesterov proposes a primal-dual subgradient scheme, and we have implemented it for com-parison. However, empirically we found the estimate (upper bound) of the primal-dual gap of the subgradient method (see (3.3) in ) to be pessimistic and usually much larger than the gap of the excessive-gap technique when both methods achieve similar primal values. 614 JIEQIU CHEN AND SAMUEL BURER Table 2 Comparison of Algorithm 4 ( Smooth) and the subgradient method ( Subg.) when applied to 14 linear ranking problems. θ∗is the optimal value of the ranking problem, but it is not available for the two largest instances. Times are in seconds and rounded to the nearest integers. Here t represents that the time limit (18000 sec.) was exceeded. Time (in sec.) Best Obj. Dataset Smooth Subg. θ∗ Smooth Subg. wine 111 39 77.82 78.82 78.76 iris 1104 7001 3317.72 3318.72 3318.72 glass 1296 t 2850.45 2851.45 2852.00 ntyroid 1232 4153 1694.65 1695.65 1695.65 sonar 2136 1055 191.85 192.85 192.85 derma 119 7 28.71 29.67 29.36 heart 5687 t 1240.92 1241.92 1249.52 ecoli t t 8421.05 8422.60 8453.03 spectf 11203 t 1580.99 1587.17 1634.22 ion t t 2578.49 2583.15 2634.61 liver t t 11339.9 11370.30 11641.30 boston t t 889.88 890.94 899.03 cancer t t — 17989.09 27520.79 diabetes t t — 14327.73 28070.29 Fig. 2. Error versus time (log-log scale): comparison of the excessive-gap technique and the subgradient method. 5.2. 1-norm support vector machines. SVMs are popular techniques for classiﬁcation, and 1-norm SVMs are known to be eﬀective in reducing input space features. Existing solution approaches for the 1-norm SVM include solving them as LPs and using a generalized Newton method or its variants (Fung and Mangasarian , Mangasarian ). We show in this subsection that Algorithm 4 can also be applied to solve 1-norm SVMs. First, we introduce the standard form of the 1-norm SVM and reformulate it as (P). Second, we compare Algorithm 4 with the generalized Newton method proposed in for solving linear 1-norm SVM and nonlinear kernel 1-norm SVM problems. From now on, we will refer to the method of as Newton for short. A SMOOTHING METHOD FOR A CLASS OF LARGE-SCALE LPs 615 A standard 1-norm SVM is the following optimization problem: min (x,γ,ξ) ∥x∥1 + C∥ξ∥1 (1-norm SVM) s. t. D( ˜ Ax −eγ) + ξ ≥e, (22) ξ ≥0, where ˜ A ∈ℜm×n represents m points in ℜn to be separated by a hyperplane (23) aT x = γ. D ∈ℜm×m is a diagonal matrix with element Dii ∈{−1, 1} representing the label for the ith data point, and C > 0 is a trade-oﬀparameter. We claim that the variables (x, γ, ξ) are implicitly bounded. In fact, since the problem is a minimization problem, there exists M > 0 such that ∥x∗∥1+C∥ξ∗∥1 < M, where (ξ∗, x∗) is an optimal solution to (1-norm SVM). Thus both x∗and ξ∗are bounded. At optimality, (22) can be rewritten as De γ∗≤D ˜ A x∗+ ξ∗−e, and it follows from the boundedness of (x∗, ξ∗) that the right-hand side of the above inequality is bounded. Therefore, there exists a scalar h > 0 such that De γ∗≤h e, which implies that γ∗∈[−h, h] because De’s components are either 1 or −1. To actually compute h, note that the absolute value of each component of x∗and ξ∗is also bounded by M, and thus one can compute an upper (if Dii > 0) or lower (if Dii < 0) estimate of the ith component of D ˜ Ax∗+ ξ∗−e as follows: β(i) = ( ˜ Ai)+(M e) + (M −1) if Dii = 1, −(−˜ Ai)+(M e) + 1 if Dii = −1 ∀i = 1, . . . , m, where ˜ Ai is ith row of ˜ A and M can be easily estimated. Now we can calculate h = maxi=1,...,m{\|β(i)\|}. To reformulate (1-norm SVM) as (P), we ﬁrst express the variables x and γ as the diﬀerence of two nonnegative variables, that is, x = x+ −x−, γ = γ+ −γ−. Because of the boundedness of γ, we may enforce γ+, γ−∈[0, h]. Deﬁne the matrix A := (De, −De, −D ˜ A, D ˜ A) and α := (γ+; γ−; x+; x−) ≥0, such that (22) can be equivalently expressed as Aα + e ≤ξ. Then (1-norm SVM) is equivalent to min (α,ξ) cT α + C eT ξ (1-norm SVM′) s. t. Aα + e ≤ξ, α, ξ ≥0, where c is a vector of ones except that the ﬁrst two entries are zero. Note that Assumption 1.1 is satisﬁed since αB = ( γ+ γ−) ≤h e with B = {1, 2}. Therefore, Algorithm 4 is applicable to (1-norm SVM′). In the following, we compare Algorithm 4 with Newton for solving two types of 1-norm SVMs: the linear 1-norm SVM and the nonlinear kernel 1-norm SVM. The major diﬀerence is that the former seeks a separating hyperplane such as (23) while the latter solves for a nonlinear separating surface. According to , the nonlinear 616 JIEQIU CHEN AND SAMUEL BURER Table 3 Dimensions of ˜ A and K( ˜ A, ˜ AT ). ˜ A K( ˜ A, ˜ AT ) Dataset m n 1st Set, m 2nd Set, m a6a 10000 122 5000 8000 covtype 10000 54 5000 8000 ijcnn1 10000 22 5000 8000 mushrooms 8124 112 5000 8000 usps 7291 256 5000 7291 w5a 9888 300 5000 8000 kernel SVM can be modeled readily by (1-norm SVM′) with a simple replacement of the data ˜ A as follows: ˜ A ← −K( ˜ A, ˜ AT )D, where K(·, ·) is a kernel function. We discuss the details of the computational exper-iment in the following paragraphs. We selected six classiﬁcation datasets from LIBSVM data2 and used them to create both ˜ A and K( ˜ A, ˜ AT ). For the linear 1-norm SVM problem, we randomly sampled from each dataset 10000 or the maximum number of data points, whichever was smaller, to form ˜ A. For the nonlinear kernel 1-norm SVM problem, we created two sets of K( ˜ A, ˜ AT ) with diﬀerent sizes. In particular, we randomly sampled 5000 data points to create the ﬁrst set of kernel matrices. We then randomly sampled 8000 or the maximum number of data points, whichever was smaller, to form the second set of kernel matrices. We choose the RBF kernel and followed the procedure as prescribed in to prepare K( ˜ A, ˜ AT ). The dataset and the sizes of the data matrices are presented in Table 3. Note that the kernel matrices are square matrices and thus we only show the sizes of their ﬁrst dimensions. In SVMs, C is a user-speciﬁed parameter that often takes a set of diﬀerent values in tuning for the best parameter setting. In our experiment, we tested C = 1 and C = 0.01 for both the linear SVM and nonlinear kernel SVM. The stopping criterion for our excessive-gap technique is that the relative primal-dual gap r = p−d max{1,1/2(\|p\|+\|d\|)} should be smaller than 0.01, where p and d represent the primal and dual objective values, respectively. For Newton, an unconstrained optimization is solved, and its gradient is zero at optimality. Thus we set the stopping criterion for Newton to be when the gradient has norm smaller than ϵ = 1e −5.3 We also set a time limit of 18000 seconds for all runs. Both Algorithm 4 and Newton are implemented in MATLAB. For the Newton method, we adopted the code from the author’s web site.4 All computations were performed on the same machine mentioned in the previous subsection. We summarize the computational results for the linear 1-norm SVM in Tables 4 and 5, corresponding to C = 1 and C = 0.01, respectively. We compare the excessive-gap technique and the Newton method in terms of best objective values, CPU times, and optimality. Note that the measures of optimality are diﬀerent for the two methods and so cannot be compared directly. The excessive-gap technique was able to solve all 2Available at 3We observe in our computational experiments that if the gradient is not suﬃciently small, the resulting solution of the Newton method is infeasible. Therefore, we set a small ϵ. 4 A SMOOTHING METHOD FOR A CLASS OF LARGE-SCALE LPs 617 Table 4 Linear 1-norm SVM with C = 1: comparison of Algorithm 4 ( Smooth) and the Newton method ( Newton) in terms of best objective values, CPU times, and measures of optimality. Objectives Times Optimality Data Name Smooth Newton Smooth Newton Gap Gradient a6a 3566.9 87283.5 1627.4 t 1.0e−2 7.5e+1 covtype 5778.1 5916.7 869.0 t 1.0e−2 6.7e−2 ijcnn1 1741.2 1746.7 51.3 1.6 1.0e−2 1.0e−5 mushrooms 16.2 18.7 1117.0 605.4 1.0e−2 2.0e−6 usps 127.5 37177.5 3595.6 t 1.0e−2 2.1e+2 w5a 324.7 333.2 3509.6 2 1.0e−2 1.0e−5 Table 5 Linear 1-norm SVM with C = 0.01: comparison of Algorithm 4 ( Smooth) and the Newton method ( Newton) in terms of best objective values, CPU times, and measures of optimality. Objectives Times Optimality Data Name Smooth Newton Smooth Newton Gap Gradient a6a 43.1 28.5 11.5 t 9.0e−03 2.0e+1 covtype 79.5 15.0 8.1 3757 1.0e−02 1.0e−6 ijcnn1 19.7 0.8 3.3 0.3 9.0e−03 2.0e−6 mushrooms 8.5 17.7 9.7 0.5 9.0e−03 2.0e−6 usps 9.6 88.5 150.7 t 1.0e−02 1.1e+2 w5a 5.7 100.4 6.3 0.1 1.0e−02 1.0e−5 Table 6 Nonlinear kernel 1-norm SVM with C = 1, ﬁrst set of data: comparison of Algorithm 4 and the Newton method in terms of best objective values, CPU times, and measures of optimality. Objectives Times Optimality Data Name Smooth Newton Smooth Newton Gap Gradient a6a 1791.2 1791.3 15521.6 3863.6 1.0e−2 0.0e+0 covtype 2844.7 62739 10925.6 t 1.0e−2 6.7e−1 ijcnn1 963.1 962 t 170.8 1.6e−2 0.0e+0 mushrooms 378.4 376.6 t 4698.5 1.8e−2 0.0e+0 usps 117.5 17053.3 11974.7 t 1.0e−2 1.3e+0 w5a 319.2 314 10454.4 1422.4 1.0e−2 0.0e−0 instances except for one (usps) to within the error tolerance in the given time. The Newton method, however, fails to reduce the gradient to within the error tolerance in three out of six instances when C = 1, and in two out of six instances when C = 0.01. On the other hand, the Newton method is usually fast when it does converge. Overall, we see that the excessive-gap technique is eﬃcient and robust compared with the Newton method in solving the linear 1-norm SVM problem. In each table, t signiﬁes that the time limit (18000 sec.) was exceeded. The results for the ﬁrst set of nonlinear kernel 1-norm SVM problems are sum-marized in Tables 6 and 7, corresponding to C = 1 and C = 0.01, respectively. The kernel matrices for this set of problems all have dimension 5000 × 5000. When C = 1, both methods ran out of time on two instances each, but Newton returned much worse objective values when it ran out of time. The Newton method is faster than the excessive-gap technique on problems solved by both methods to within the error tolerances. For C = 0.01, the excessive-gap technique solved all problems within the given time while the Newton method ran out of time on four out of the six instances. The computational results for the second set of nonlinear kernel 1-norm SVM are presented in Tables 8 and 9, corresponding to C = 1 and C = 0.01, respectively. From 618 JIEQIU CHEN AND SAMUEL BURER Table 7 Nonlinear kernel 1-norm SVM with C = 0.01, ﬁrst set of data: comparison of Algorithm 4 and the Newton method in terms of best objective values, CPU times, and measures of optimality. Objectives Times Optimality Data Name Smooth Newton Smooth Newton Gap Gradient a6a 23.8 2171.6 6110.5 t 1.0e−2 3.1e−1 covtype 45.7 3499.2 2090.7 t 1.0e−2 6.1e−1 ijcnn1 9.8 9.6 10343.6 118.4 1.0e−2 0.0e+0 mushrooms 35.5 1471.2 5571.6 t 1.0e−2 1.1e−1 usps 13.1 2841.6 2180.6 t 1.0e−2 6.9e−1 w5a 3.2 3.1 t 2915.6 2.0e−2 0.0e+0 Table 8 Nonlinear kernel 1-norm SVM with C = 1, second set of data: comparison of Algorithm 4 and the Newton method in terms of best objective values, CPU times, and measures of optimality. Objectives Times Optimality Data Name Smooth Newton Smooth Newton Gap Gradient a6a 5943.2 2980.2 t 12341.2 4.7e−1 9.0e−6 covtype 1991.3 22851.5 t t 5.7e−2 8.4e+1 ijcnn1 3005.8 1542 t 172.2 3.3e−1 1.0e−5 mushrooms 470.1 433.6 t t 1.5e−1 5.9e−5 usps 166.9 7081.2 t t 4.3e−2 2.9e+2 w5a 786.8 542.5 t t 2.7e−1 1.1e−5 Table 9 Nonlinear kernel 1-norm SVM with C = 0.01, second set of data: comparison of Algorithm 4 and the Newton method in terms of best objective values, CPU times, and measures of optimality. Objectives Times Optimality Data Name Smooth Newton Smooth Newton Gap Gradient a6a 38.3 1008.5 t t 1.5e−2 2.1e+1 covtype 23.9 2292.5 14100.2 t 1.0e−2 5.3e+1 ijcnn1 16.0 15.4 t 758.4 2.8e−2 1.0e−5 mushrooms 44.0 1291 5018.8 t 1.0e−2 3.5e+1 usps 16.3 1544 6773.7 t 1.0e−2 2.4e+2 w5a 5.8 5.6 t 5462.1 2.7e−2 6.0e−6 Table 3, we know that the sizes of these problems are larger than the corresponding linear problems and the ﬁrst set of nonlinear kernel problems, and thus they are more diﬃcult to solve. When C = 1, the excessive-gap technique was able to solve only one out of six instances to within the error tolerance in the given amount of time and the Newton method was able to solve two out of six. When C = 0.01, the excessive-gap technique was able to solve four out of six, while the Newton method was able to solve two out of six. In terms of optimality, we ﬁnd that the optimality gap delivered by the excessive-gap technique is more consistent (particularly when C = 0.01), whereas the Newton method often fails to converge, as indicated by large gradient values. For both the linear and nonlinear kernel problems, we observe that the excessive-gap technique is faster with smaller C values, a result that is consistent with the theoretical result in Proposition 3.3 because a larger C value corresponds to a larger θ value. Our computational experiments also suggest that the excessive-gap technique is a favorable choice to solve the 1-norm SVM problem when the parameter C is relatively small. The Newton method is fast when it does converge but is not as robust as the excessive-gap technique. A SMOOTHING METHOD FOR A CLASS OF LARGE-SCALE LPs 619 6. Conclusion. In this paper, we have developed a ﬁrst-order excessive-gap technique for solving (P) and the equivalent problem (NS). We show that the iteration complexity of this technique depends on the parameter θ, which arises when bounding the feasible set. We estimate θ as an upper bound on θ∗, the optimal value of (NS). Since a smaller θ means a better iteration complexity, we have designed a strategy that dynamically updates the value of θ as the algorithm obtains more information about θ∗, resulting in faster convergence. This idea could be extended to other convex nonsmooth problems with unbounded feasible sets. Our method is designed for large-scale instances of (P), and we have applied it to two problems in machine learning: the LP problem and the 1-norm SVMs. We demon-strate the eﬀectiveness of our method by comparing it with two existing methods: the subgradient method for solving the ranking problem and the Newton method for solv-ing the 1-norm SVM, respectively. Our computational experience indicates that under many circumstances the excessive-gap technique is a more attractive method because of its balanced eﬃciency and reliability. Acknowledgments. The authors are in debt to two anonymous referees for many helpful suggestions that have improved the paper signiﬁcantly. REFERENCES A. Asuncion and D.J. Newman, UCI Machine Learning Repository edu/ml/ (2007). K. Ataman, Learning to Rank by Maximizing the AUC with Linear Programming for Problems with Binary Output, Ph.D. thesis, University of Iowa, Iowa City, IA, 2007. K. Ataman, W.N. Street, and Y. Zhang, Learning to rank by maximizing auc with linear programming, in International Joint Conference on Neural Networks, IEEE, Piscataway, NJ, 2006, pp. 123–129. N. S. Aybat and G. Iyengar, A ﬁrst-order smoothed penalty method for compressed sensing, SIAM J. Optim., 21 (2011), pp. 287–313. O. Banerjee, L. El Ghaoui, A. d’Aspremont, and G. Natsoulis, Convex optimization techniques for ﬁtting sparse Gaussian graphical models, in Proceedings of the 23rd Inter-national Conference on Machine Learning, ACM, New York, 2006, pp. 89–96. S. Becker, J. Bobin, and E. J. Cand es, Nesta: A fast and accurate ﬁrst-order method for sparse recovery, SIAM J. Imag. Sci., 4 (2011), pp. 1–39. J.M. Borwein and A.S. Lewis, Convex analysis and nonlinear optimization, 2nd ed., CMS Books Math/Ouvrages de Math´ ematiques de la SMC 3, Springer, New York, 2006. A. d’Aspremont, Smooth optimization with approximate gradient, SIAM J. Optim., 19 (2008), pp. 1171–1183. G. Fung and O.L. Mangasarian, A feature selection Newton method for support vector ma-chine classiﬁcation, Comput. Optim. Appl., 28 (2004), pp. 185–202. A. Gilpin, S. Hoda, J. Pe˜ na, and T. Sandholm, Gradient-based algorithms for ﬁnding Nash equilibria in extensive form games, in Internet and Network Economics (WINE), Lecture Notes in Comput. Sci. 4848, Springer, Berlin, 2007, pp. 57–69. S. Hoda, A. Gilpin, J. Pe` oa, and T. Sandholm, Smoothing techniques for computing Nash equilibria of sequential games, Math. Oper. Res., 35 (2010), pp. 494–512. C.W. Hsu, C.C. Chang, and C.J. Lin, A Practical Guide to Support Vector Classiﬁcation, Technical report, Department of Computer Science and Information Engineering, National Taiwan University, Taipei, 2003. G. Lan, Z. Lu, and R.D.C. Monteiro, Primal-dual ﬁrst-order methods with o(1/ϵ) iteration-complexity for cone programming, Math. Program., (2009), pp. 1436–4646. O.L. Mangasarian, Exact 1-norm support vector machines via unconstrained convex diﬀer-entiable minimization, J. Mach. Learn. Res., 7 (2006), pp. 1517–1530. A. Nedic and D.P. Bertsekas, Incremental subgradient methods for nondiﬀerentiable opti-mization, SIAM J. Optim., 12 (2001), pp. 109–138. Y. Nesterov, Introductory Lectures on Convex Optimization: A Basic Course, Kluwer Aca-demic, Boston, MA, 2004. 620 JIEQIU CHEN AND SAMUEL BURER Yu. Nesterov, Excessive gap technique in nonsmooth convex minimization, SIAM J. Optim., 16 (2005), pp. 235–249. Y. Nesterov, Smooth minimization of non-smooth functions, Math. Program., 103 (2005), pp. 127–152. Y. Nesterov, Smoothing technique and its applications in semideﬁnite optimization, Math. Program., 110 (2007), pp. 245–259. Y. Nesterov, Primal-dual subgradient methods for convex problems, Math. Program., 120 (2009), pp. 221–259. T. Zhou, D. Tao, and X. Wu, Nesvm: A fast gradient method for support vector machines, in IEEE International Conference on Data Mining, IEEE, Piscataway, NJ, 2010, pp. 679–688. J. Zhu, S. Rosset, T. Hastie, and R. Tibshirani, 1-norm support vector machines, in Neural Information Processing Systems, Vol. 16, MIT Press, Cambridge, MA, 2003.
16245	https://physics.stackexchange.com/questions/497692/why-are-there-fixed-target-experiments	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Why are there fixed target experiments? Ask Question Asked Modified 6 years, 1 month ago Viewed 1k times 6 $\begingroup$ Collider experiments put all their energy into $ \sqrt{s} = 2E $ while target experiments only provide $ \sqrt{s} = \sqrt{2Em+m^²} $. Yet, there are fixed target experiments. Why? particle-physics particle-detectors particle-accelerators Share Improve this question edited Aug 19, 2019 at 23:07 Qmechanic♦ 222k5252 gold badges636636 silver badges2.6k2.6k bronze badges asked Aug 19, 2019 at 21:21 BenBen 1,55711 gold badge1717 silver badges4646 bronze badges $\endgroup$ 9 1 $\begingroup$ Well, they are a lot easier to arrange for random mixes of incident particle and target. Not quite sure how Geiger and Marsden would have arranged a collider experiment. $\endgroup$ Jon Custer – Jon Custer 2019-08-19 21:28:26 +00:00 Commented Aug 19, 2019 at 21:28 $\begingroup$ :D I doubt that the Rutherford experiment (didn't hear it as Geiger and Marsden before) is a "particle accelerator", seen in a current technical manner ;) But yet, it is true $\endgroup$ Ben – Ben 2019-08-19 21:35:33 +00:00 Commented Aug 19, 2019 at 21:35 2 $\begingroup$ Luminosity, cost, the option of continuous operation (no fill-n-spill), ... If your exposure to particle physics had been mostly through the popular press then you probably have a very warped impression of what particle physics consists of. $\endgroup$ dmckee --- ex-moderator kitten – dmckee --- ex-moderator kitten 2019-08-19 21:35:49 +00:00 Commented Aug 19, 2019 at 21:35 2 $\begingroup$ The experimental paper was authored by Geiger and Marsden, who worked in Rutherford's lab. en.wikipedia.org/wiki/Geiger%E2%80%93Marsden_experiment $\endgroup$ Jon Custer – Jon Custer 2019-08-19 21:38:34 +00:00 Commented Aug 19, 2019 at 21:38 1 $\begingroup$ As an aside, the limitations of using natural alpha sources were the motivation for Cockroft and Walton, as well as Van de Graaff, to develop high voltage electrostatic accelerators. Ditto for Wideroe (linac) and Lawrence (cyclotron). Note that none of those were originally colliders yet lots of useful nuclear and particle physics were done with them. $\endgroup$ Jon Custer – Jon Custer 2019-08-19 21:56:39 +00:00 Commented Aug 19, 2019 at 21:56 \| Show 4 more comments 1 Answer 1 Reset to default 9 $\begingroup$ The target density is much greater, i.e you don't have a bullet trying to hit another bullet but rather a single bullet trying to hits lots of (more) densely packed targets. As a result, the luminosity (i.e. the reaction rate) is greater. Share Improve this answer edited Aug 19, 2019 at 21:38 answered Aug 19, 2019 at 21:36 ZeroTheHeroZeroTheHero 49.4k2121 gold badges7272 silver badges149149 bronze badges $\endgroup$ 5 $\begingroup$ Is there really such a difference? I thought the emittance(?) would be good enough in colliding experiments respectively quite similar? $\endgroup$ Ben – Ben 2019-08-19 21:38:14 +00:00 Commented Aug 19, 2019 at 21:38 3 $\begingroup$ there is a huge difference.... several orders of magnitude depending on the targets. $\endgroup$ ZeroTheHero – ZeroTheHero 2019-08-19 21:38:50 +00:00 Commented Aug 19, 2019 at 21:38 $\begingroup$ Plus, colliders may be hard to arrange. It is easy to accelerate protons at a gold target. Much harder to arrange a collider with proton and gold beams. And then do protons on to silver next hour, then alphas on to gold, then silver... $\endgroup$ Jon Custer – Jon Custer 2019-08-19 21:58:35 +00:00 Commented Aug 19, 2019 at 21:58 $\begingroup$ @Jon One of my grad school buddies ended up at RHIC for about a decade. That's what their life was like, but with added gold-on-gold, lead-on-lead, ... $\endgroup$ dmckee --- ex-moderator kitten – dmckee --- ex-moderator kitten 2019-08-19 22:01:44 +00:00 Commented Aug 19, 2019 at 22:01 1 $\begingroup$ Maybe you can develop an intuition for the difference in density by considering that an accelerator is not filled with drops of, say, liquid hydrogen. If I have the numbers correct the LHC contains 10^14 protons at any given time. 1g (~ 15mL) of liquid hydrogen contains 10^23 protons -- 10^9 times more. One reason for the difference is of course the same reason why I differentiated between hydrogen and the protons which are what actually circulates around the LHC: in a fixed target you have electrically neutral substances which can be packed much more tightly -- but which can't be accelerated. $\endgroup$ tobi_s – tobi_s 2019-08-20 09:19:16 +00:00 Commented Aug 20, 2019 at 9:19 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions particle-physics particle-detectors particle-accelerators See similar questions with these tags. 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16246	https://app.achievable.me/study/gre-v2/learn/quantitative-reasoning-special-triangles-45-45-90	2.4.5 Right triangles (45-45-90) Achievable GRE Quantitative reasoning 2.4. Geometry Right triangles (45-45-90) 4 min read FontDiscussShareFeedback A triangle with angle measurements of 45, 45, and 90 degrees is called a 45-45-90 triangle. The relative measurements of the sides and angles will always be in the same ratio as the figure below. Regardless of the size of the triangle, these ratios will always be consistent. The two sides adjacent to the right angle are always equal, and the side opposite the right angle (the hypotenuse) is always equal to the smaller side length x multiplied by 2, i.e., x2. Let’s consider a concrete example. Imagine if the two smaller side lengths were 2, and we used the Pythagorean Theorem to find the length of the hypotenuse. a2+b222+224+4822=c2=c2=c2=c2=c It’s the same! So while you could derive these equations using the Pythagorean Theorem, it’s better to simply memorize this x:x:x2 ratio of the 3 sides of a 45-45-90 triangle. Example 45-45-90 triangle question Let’s try a question that involves 45-45-90 triangles. Square A has half the area of Square B. Quantity A: Twice the length of the diagonal of Square AQuantity B: The length of the diagonal of Square B Give it a try! (spoiler) Answer: Quantity A is greater At a glance, it may seem likely that the two quantities are equal because Quantity A doubles the diagonal of the square with half the area. However, this logic isn’t entirely accurate, and you can check that by plugging in numbers for the area of the two squares. Let’s imagine that Square A has an area of 1, which means that Square B has an area of 2. If Square A has an area of 1, its side lengths would also be 1, since 1(1)=1. The diagonal of a square is the hypotenuse of a 45-45-90 triangle, which has the side ratios of x:x:x2. Given that the short side lengths x are 1, the length of the diagonal must be 12, which is simply 2. Now let’s take a look at Square B. If Square B has an area of 2, and the area is A=x2, we can plug in 2=x2 to get the side length of x=2. And again, given that the side lengths of the square are also the short side lengths of a 45-45-90 triangle, the length of the diagonal must be x∗2=2∗2=2. We know that was a lot of steps in a row, especially written out in text, but it’s essential that you know how to break down these questions. If you’re unclear, please take a moment to slowly re-read the last two paragraphs to understand how we reached these values for the diagonals: Diagonal of Square A: 2 Diagonal of Square B: 2 Now that we’ve defined the diagonals, we just need to come back to the quantities. Quantity A: Twice the length of the diagonal of Square A =22 Quantity B: The length of the diagonal of Square B =2 Clearly, 22 is greater than 2, so Quantity A is greater. The final step is to ask if this is true in every case, and the answer is yes. The length of the diagonals is always based on the ratio of the side lengths, and the ratio of the side lengths is always consistent with the constraints of the question. This means double the diagonal of Square A will always be greater than the diagonal of Square B. Read more now by creating a free account Plus, take quizzes and track your progress All rights reserved ©2016 - 2025Achievable, Inc.
16247	https://www.cuemath.com/ncert-solutions/in-an-ap-i-given-a-5-d-3-an-50-find-n-and-sn-ii-given-a-7-a13-35-find-d-and-s13-iii-given-a12-37-d-3-find-a-and-s12-iv-given-a3-15-s10-125-find-d-and-a10/	In an AP: (i) Given a = 5, d = 3, aₙ = 50 , find n and Sₙ (ii) Given a = 7, a₁₃ = 35 , find d and S₁₃ (iii) Given a₁₂ = 37, d = 3, find a and S₁₂ (iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀ (v) Given d = 5, S₉ = 75 , find a and a₉ (vi) Given a = 2, d = 8, Sₙ = 90 , find n and aₙ (vii) Given a = 8, aₙ = 62, Sₙ = 210 , find n and d (viii) Given aₙ = 4, d = 2, Sₙ = - 14 , find n and a (ix) Given a = 3, n = 8, S = 192 , find d (x) Given l = 28, S = 144 and there are total 9 terms. Find a Solution: The sum of the first n terms of an AP is given by Sₙ = n/2 [2a + (n - 1) d] or Sₙ = n/2 [a + l], and the nth term of an AP is aₙ = a + (n - 1)d Here, a is the first term, d is a common difference and n is the number of terms and l is the last term. (i) Given a = 5, d = 3, aₙ = 50 , find n and Sₙ. Given, aₙ = a + (n - 1)d 50 = 5 + (n - 1)3 45 = (n - 1)3 15 = n - 1 n = 16 Sustitute the value of n, to find sum of the n terms. Sₙ = n/2 [a + l] S₁₆= 16/2 [ 5 + 50] = 8 × 55 = 440 (ii) Given a = 7, a₁₃ = 35 , find d and S₁₃. Given, aₙ = a + (n - 1)d a₁₃ = a + (13 - 1) d 35 = 7 + 12d 35 - 7 = 12d d = 28/12 d = 7/3 Sₙ = n/2 [a + l] S₁₃= 13/2 [7 + 35] = 13/2 × 42 = 13 × 21 = 273 (iii) Given a₁₂ = 37, d = 3, find a and S₁₂. Given, aₙ = a + (n - 1)d a₁₂ = a + (12 - 1) 3 37 = a + 33 a = 4 Substitute the value of a to find the sum of n terms. Sₙ = n / 2 [a + l] S₁₂= 12 / 2 [4 + 37] = 6 × 41 S₁₂= 246 (iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀. Given, aₙ = a + (n - 1)d a₃ = a + (3 - 1) d 15 = a + 2d - - - - - Equation (i) Sₙ = n/2 [2a + (n - 1) d] S₁₀ = 10/2 [2a + (10 - 1) d] 125 = 5 [2a + 9d] 25 = 2a + 9d - - - - - Equation (ii) On multiplying equation (i) by 2, we obtain 30 = 2a + 4d - - - - - Equation (iii) On subtracting equation (iii) from Equation (ii), we obtain 5 = 5d d = - 1 From equation (i), 15 = a + 2 (- 1) 15 = a - 2 a = 17 a₁₀ = a + (10 - 1) d a₁₀ = 17 + 9 (- 1) a₁₀ = 17 - 9 a₁₀ = 8 (v) Given d = 5, S₉ = 75 , find a and a₉. Given, Sₙ = n/2 [2a + (n - 1) d] 75 = 9/2 [2a + (9 - 1) 5] 75 = 9/2 (2a + 40) 25 = 3(a + 20) 25 = 3a + 60 a = - 35 / 3 We know that nth term of the AP is given by formula aₙ = a + (n - 1)d a₉ = a + (9 - 1) × 5 = - 35 / 3 + 8 × 5 = - 35 / 3 + 40 = (- 35 + 120) / 3 = 85 / 3 (vi) Given a = 2, d = 8, Sₙ = 90 , find n and an. Given, Sₙ = n/2 [2a + (n - 1) d] 90 = n/2 [4 + (n - 1) 8] 90 = n [2 + (n - 1)4] 90 = n [2 + 4n - 4] 90 = n [4n - 2] 90 = 4n² - 2n 4n² - 2n - 90 = 0 4n² - 20n + 18n - 90 = 0 4n (n - 5) + 18(n - 5) = 0 (n - 5)(4n + 18) = 0 Either (n - 5) = 0 or (4n + 18) = 0 n = 5 or n = - 9/2 However, n can neither be negative nor fractional. Therefore, n = 5 aₙ = a + (n - 1) d a₅ = 2 + (5 - 1)8 a₅ = 2 + 4 × 8 a₅ = 2 + 32 a₅ = 34 (vii) Given a = 8, aₙ = 62, Sₙ = 210 , find n and d. Given, Sₙ = n / 2 [a + l] 210 = n / 2 [8 + 62] 210 = n × 35 n = 6 We know that nth term of the AP is given by aₙ = a + (n - 1)d 62 = 8 + (6 - 1) d 62 - 8 = 5d 54 = 5d d = 54/5 (viii) Given aₙ = 4, d = 2, Sₙ = - 14 , find n and a Given, We know that nth term of AP is aₙ = a + (n - 1)d 4 = a + (n - 1)2 4 = a + 2n - 2 a = 6 - 2n .... (1) Sₙ = n/2 [a + l] 14 = n/2 [6 - 2n + 4] [from(1)] 14 = n (5 - n) 14 = 5n - n² n² - 5n - 14 = 0 n² - 7n + 2n -14 = 0 n (n - 7) + 2 (n - 7) = 0 (n - 7)(n + 2) = 0 Either n - 7 = 0 or n + 2 = 0 n = 7 or n = - 2 However, n can neither be negative nor fractional. Therefore, n = 7 From equation (1), we obtain a = 6 - 2n a = 6 - 2 × 7 a = 6 - 14 a = - 8 (ix) Given a = 3, n = 8, S = 192 , find d. Given, Sₙ = n/2 [2a + (n - 1) d] 192 = 8/2 [2 × 3 + (8 - 1) d] 192 = 4[6 + 7d] 48 = 6 + 7d 42 = 7d d = 6 (x) Given l = 28, S = 144 and there are total 9 terms. Find a. Given, Sₙ = n/2 [a + l] 144 = 9/2 (a + 28) 32 = a + 28 a = 4 ☛ Check: NCERT Solutions Class 10 Maths Chapter 5 Video Solution: In an AP (i) Given a = 5, d = 3, aₙ = 50 , find n and Sₙ. (ii) Given a = 7, a₁₃ = 35 , find d and S₁₃. (iii) Given a₁₂ = 37, d = 3, find a and S₁₂. (iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀. (v) Given d = 5, S₉ = 75 , find a and a₉. (vi) Given a = 2, d = 8, Sₙ = 90 , find n and aₙ. (vii) Given a = 8, aₙ = 62, Sₙ = 210 , find n and d. (viii) Given aₙ = 4, d = 2, Sₙ = - 14 , find n and a. (ix) Given a = 3, n = 8, S = 192 , find d. (x) Given l = 28, S = 144 and there are total 9 terms. Find a Class 10 Maths NCERT Solutions Chapter 5 Exercise 5.3 Question 3 : Summary: In an AP (i) Given a = 5, d = 3, aₙ = 50, n = 16, S₁₆= 400 (ii) Given a = 7, a₁₃ = 35 , d = 7/3, S₁₃= 273. (iii) Given a₁₂ = 37, d = 3, a = 4, S₁₂= 246 (iv) Given a₃ = 15, S₁₀ = 125, d = - 1, a₁₀ = 8, (v) Given d = 5, S₉ = 75 , a = - 35 / 3, a₉ = 85/3, (vi) Given a = 2, d = 8, Sₙ = 90 , n = 5, a₅ = 34, (vii) Given a = 8, aₙ = 62, Sₙ = 210 , n = 6, d = 54/5, (viii) Given aₙ = 4, d = 2, Sₙ = - 14 , n = 7, a = - 8, (ix) Given a = 3, n = 8, S = 192 , d = 6, d = 6, (x) Given l = 28, S = 144 and there are total 9 terms, a = 4 ☛ Related Questions:
16248	https://dwello.in/converter/square-meter-to-hectare	Published Time: 01-01-2024T05:03+0000 Convert Square Meter to Hectare - Dwello Select City Search Home Home Converters ConvertersKnowledge BasePin code Convert Square Meter to Hectare Convert Square Meter to HectareConvert Square Meter to Square YardConvert Square Meter to DecimalConvert Square Meter to Square MileConvert Square Meter to BighaConvert Square Meter to ChatakConvert Square Meter to CentConvert Square Meter to KathaConvert Square Meter to GroundConvert Square Meter to Square Feet Home Converters ConvertersKnowledge BasePin code Convert Square Meter to Hectare Convert Square Meter to HectareConvert Square Meter to Square YardConvert Square Meter to DecimalConvert Square Meter to Square MileConvert Square Meter to BighaConvert Square Meter to ChatakConvert Square Meter to CentConvert Square Meter to KathaConvert Square Meter to GroundConvert Square Meter to Square Feet Square Meter to Hectare 1 Sq. M. = 0.000100 Hectare AreaUnit Square Meter Convert Square Meter to Hectare Converting from Square Meters to Hectares is a practical and crucial calculation in various fields, particularly in real estate, agriculture, and land planning. A hectare is a unit of area commonly used in the metric system, and it's equivalent to 10,000 square meters. In real estate, where large plots of land are often involved, expressing the area in hectares provides a more manageable and comprehensible figure. This is especially true when dealing with extensive farmlands, commercial developments, or nature reserves. The Square Meter to Hectare converter serves as a handy tool for individuals working in these fields. It simplifies the conversion process, ensuring accuracy and efficiency in dealing with land areas. Users can input the area in square meters and promptly receive the equivalent measurement in hectares, facilitating quick decision-making and analysis. About Square Meter Square Meter is a unit of area measurement in the metric system, widely used across the globe for expressing the size of plots, living spaces, and various land areas. One square meter is equal to the area of a square with sides that are each one meter long. It is abbreviated as "m²" and is a fundamental unit in the International System of Units (SI). In real estate, the square meter is a key measurement used to quantify the size of properties. Residential apartments, houses, commercial spaces, and land parcels are often advertised and transacted based on their square meter measurements. Buyers and sellers rely on this unit to quickly understand the spatial dimensions and make informed decisions. What are the general uses of Square Meter? The general uses of the square meter are diverse, and this unit of measurement is widely employed in various fields. One of the primary uses of the square meter is in real estate. Property listings, whether for residential or commercial spaces, often include the total area measured in square meters. Buyers and sellers use this information to assess the size of the property and compare it with others. Architects and builders use square meters in designing and planning structures. Floor plans, building layouts, and construction blueprints specify dimensions in square meters, facilitating accurate planning and utilization of space. Property managers use square meters to assess and manage the space within commercial or residential buildings. This includes determining rental values, maintenance needs, and occupancy planning. How to measure Square Meter to Hectare? hectares = square meters ÷ 10,000 Example Square Meter to HectareThe process to convert Square Meter to Hectare is quite simple. Here's how to convert 5,000 square meters to hectares using the formula above. Hectares = (5,000 sq m ÷ 10,000) = 0.5 ha About Hectare A hectare is a unit of measurement for area, and it plays a significant role in various fields, particularly in agriculture, land management, and environmental conservation. A hectare is a metric unit of area equal to 10,000 square meters or 2.471 acres. It is derived from the metric system, which is widely adopted worldwide for its ease of use and standardization. Hectares are frequently employed in urban planning, real estate, and land development. When large parcels of land are involved, such as in the creation of residential developments or industrial zones, hectares provide a practical and standardized measurement. This aids in zoning decisions, infrastructure planning, and the allocation of space for various purposes. How to measure Hectare to Square Meter? square meters = hectares × 10,000 Square Meter to Hectare Conversion Table \| Square Meters (m 2) \| Hectares (ha) \| --- \| \| 1 m 2 \| 0.0001 hectares \| \| 2 m 2 \| 0.0002 hectares \| \| 3 m 2 \| 0.0003 hectares \| \| 4 m 2 \| 0.0004 hectares \| \| 5 m 2 \| 0.0005 hectares \| \| 6 m 2 \| 0.0006 hectares \| \| 7 m 2 \| 0.0007 hectares \| \| 8 m 2 \| 0.0008 hectares \| \| 9 m 2 \| 0.0009 hectares \| \| 10 m 2 \| 0.001 hectares \| \| 20 m 2 \| 0.002 hectares \| \| 30 m 2 \| 0.003 hectares \| \| 40 m 2 \| 0.004 hectares \| \| 50 m 2 \| 0.005 hectares \| \| 75 m 2 \| 0.0075 hectares \| \| 100 m 2 \| 0.01 hectares \| \| 250 m 2 \| 0.025 hectares \| \| 500 m 2 \| 0.05 hectares \| \| 750 m 2 \| 0.075 hectares \| \| 1,000 m 2 \| 0.1 hectares \| \| 2,500 m 2 \| 0.25 hectares \| \| 5,000 m 2 \| 0.5 hectares \| \| 7,500 m 2 \| 0.75 hectares \| \| 10,000 m 2 \| 1 hectares \| \| 25,000 m 2 \| 2.5 hectares \| \| 50,000 m 2 \| 5 hectares \| \| 75,000 m 2 \| 7.5 hectares \| \| 100,000 m 2 \| 10 hectares \| SqMeter to Other Unit Convert Square Meter to Hectare Convert Square Meter to Square Yard Convert Square Meter to Decimal Convert Square Meter to Square Mile Convert Square Meter to Bigha Convert Square Meter to Chatak Convert Square Meter to Cent Convert Square Meter to Katha Convert Square Meter to Ground Convert Square Meter to Square Feet Frequently asked questions (FAQs) What is the conversion factor from square meters to hectares? The conversion factor is 1 hectare = 10,000 square meters. To convert square meters to hectares, divide the area in square meters by 10,000. When is it preferable to use hectares instead of square meters? How do I convert square meters to hectares? Quick Convert Convert Lakh to Crore Convert Billion to Crore Convert Trillion to Crore Convert Yard to Meter Convert Billion to Million Convert Crore to Billion Convert Crore to Million Convert Crore To Lakh Convert Fluid Ounce to Milliliter Convert Million to Crore Explore further Search projects By City GurugramBengaluruPuneDelhiMumbaiDubai Collections Affordable HomesBy Top Rated DevelopersLaunched RecentlyLuxury HomesReady Possession Homes Resources ConvertersKnowledge BasePin code NEED HELP? 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16249	https://www.youtube.com/watch?v=ozWzqDBL4JQ	Specific Heat Capacity - Finding the Thermal Equilibrium Tempreture Harold Walden 10300 subscribers 32 likes Description 4540 views Posted: 3 Mar 2017 Hot Lead Bar + Cold Water = Final Temperature? Use the formula mcΔT = -mcΔT to show that heat gained = heat lost and solve for whatever "x" is. In this case, it's the final temperature. 5 comments Transcript: Intro g'day guys a student of mine is having trouble with this physics question so I thought I'd make a video and work through it so we have a 200 gram block of lead at 80 degrees which is placed in an insulated container of 500 grams of water at 20 degrees what is the final temperature the combination when they reach thermal equilibrium assume no heat is lost to the surroundings okay so Formula basically what we're going to have here is we're going to have a object with a greater temperature put in and into water which is of a lower temperature and the temperature of the water is going to rise and the temperature of the lead obviously is going to have to fall so because of this assumption that we make in the question we can come up with the following like equality we can say that the heat that is going to be lost by the lid is going to be equal to the heat that is gained by the water now this will come in handy for when we're solving this problem a little bit later so let's just go over the formula that we're going to need to use specifically this heat formula so hopefully you guys are aware that we can work out this heat in joules which is equal to M in kilos the mass times the specific heat of the object times the change in temperature cool so this change in temperature we're going to convert is supposed to be in Kelvin so we're going as you can see here so we're going to convert both of these temperatures to Kelvin where's the other temperature there so let's just quickly do that underneath so we have 80 degrees Celsius is going to be equal to 353 K and we have 20 degrees Celsius which is equal to 293 okay now because there's a one-to-one relationship between degrees Celsius and Kelvin we could we're not really interested in the actual degrees here we're just interested in the change in the degrees Celsius or the change in the degrees Kelvin so like it's almost irrelevant at what we put these values as so let's start this question now getting back to this equality that we came up with just before what we're going to do is we're going to set the Q of the water equal to the Q of the lead so what we're going to see is the mass of the lead times the specific heat capacity of lead times the change in temperature of the lead so this is for the lead side it's going to be equal to the mass of the water times the specific heat capacity of water times the change in temperature of the water now so that's just this is going to be of the lead and this is going to be of the water so what we can do is let's enter in the information that we know and see what we have left to work out so we know that the mass of the lead in kilos is 0.2 the specific heat capacity of lead is equal to 129 and obviously we don't know the change in the temperature however we do know what the initial temperature is so the change in temperature is going to be well let's just quickly just write a side note here sorry for going off on a couple of tangents here we can write that the change in temperature is equal to temperature final subtract temperature initial and that's going to be our delta T so if we call temperature final X we can set up an equation for this so we're going to have temperature final which is X subtract the initial temperature which is 353 so that's of the lead and this has to be equal to the masses of water which is 0.5 kilos times the much larger specific heat capacity of water four one eight zero times by delta T which again temperature final and because it's going to be thermal equilibrium the temperature of the water and the temperature of the lead are going to be the same so we can say that the final temperature is going to be equal to X minus the initial temperature of the water which is 293 now voila hopefully Solving you realize that because we only have one variable in this equation here we can solve this so let's just get straight to solving it so what I'm going to do is I'm going to just multiply the two coefficients or the mass and the specific heat by each other just to make the equation a little bit shorter on this side we have twenty five point eight X minus 353 is equal to half of four one eight zero or 2090 in brackets X minus 293 cool so like we would normally do if we were going to try and solve this we are going to simply multiply their brackets so we're gonna have twenty five point eight X minus nine thousand one hundred and seven point four is equal to two thousand and ninety X minus 10 this is gonna be a big number six hundred and twelve thousand three hundred and seventy now all I'm doing here this is not physics anymore this is just simple algebra that you would learn when you're 13 I guess years old so we combine our like terms on either side of the equality so I'm gonna move my just my digits over to the left and I'm going to move my x values to the right and I get six one two three seven zero minus nine one zero seven point four is equal to two oh nine oh X minus twenty five point eight X so what we end up with let's just continue this up here what we end up with is two thousand and sixty four point two X is equal to six hundred and three thousand two hundred and sixty two point six and then we can simply divide both sides by two thousand and sixty four point two and we get an x value of two hundred and ninety two point five and that is going to be a temperature file so what we can see is because of the larger mass and the way larger specific heat capacity if we drop an 80 degrees Celsius block of lead into half a kilo of water it's only going to rise the water temperature by two point two and a quarter degrees but it's going to lower the lead temperature by what would that be 353 to 292 so 53 plus seven so we've got about sixty degrees there so you can see that because of the larger mass and the larger specific heat of the water the final temperature is much closer to the temperature the initial temperature of the water than of the lead so basically what we did here guys is we using the assumption that no heat is lost to the surroundings we were able to come up with this equality here once we had that equality what we could do is we could enter in the data that we have been given into this Q equals MC delta T formula yeah once we've done that what we could then do is we'd find out once we break down delta T into temperature final take temperature initial what we could do is we could find out that this entire equality only has one unknown X so what we then did is we just use quite basic algebra to make sure so doctor make sure to solve for X and then that pops out our final temperature value of 292 and a quarter degrees Kelvin so guys I hope the video helped I tried to go through quick leads so not to waste any of your time so if it did help just give it a thumbs up you know subscribe to my channel I put out new videos on maths and science all the time but until next time guys enjoy your physics
16250	https://stackoverflow.com/questions/19085937/finding-intervals-of-a-set-that-are-overlapping	algorithm - Finding intervals of a set that are overlapping - Stack Overflow Join Stack Overflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google Sign up with GitHub OR Email Password Sign up Already have an account? Log in Skip to main content Stack Overflow 1. About 2. Products 3. 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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Finding intervals of a set that are overlapping Ask Question Asked 12 years ago Modified7 years, 11 months ago Viewed 6k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. So I have a set containing the endpoints of intervals. For example, Set s = {(1,4),(3,7),(5,8),(14,17),(0,2),(11,14)} I need a way to find out how many overlapping intervals are there. In the above case, the answer would be 5 since (1,4) --> (3,7), (0,2) (3,7) --> (5,8),(0,2) (5,8) --> (14,17) --> (11,14) I need an algorithm that takes O(N log N) time to find out the sum. Now if I sort all the starting points and apply the answer suggested here on every point Find number range intersection I get an O(N^2) solution. Any clue on what kind of data structure I might need in addition to a set? Thanks! algorithm data-structures Share Share a link to this question Copy linkCC BY-SA 3.0 Improve this question Follow Follow this question to receive notifications edited May 23, 2017 at 12:32 CommunityBot 1 1 1 silver badge asked Sep 30, 2013 at 3:14 as3rdaccountas3rdaccount 3,971 12 12 gold badges 45 45 silver badges 66 66 bronze badges 2 1 This is related to interval treeqxixp –qxixp 2013-09-30 03:34:41 +00:00 Commented Sep 30, 2013 at 3:34 11 Wait -- why does (3,7) overlap with (0,2)?jacobm –jacobm 2013-09-30 04:05:27 +00:00 Commented Sep 30, 2013 at 4:05 Add a comment\| 5 Answers 5 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. Build a list of values (a, -1), (b, 1) for every interval [a, b]. Now sorting these lets you run through the array, counting how many intervals are currently open at each endpoint, just by aggregating the +1 and -1s. It's important that (b, 1) comes after (b, -1) in the sorted list; because we want to count the intersection even if the intersection is at a single point. Complete code here. def overlaps(intervals): es = [] for a, b in intervals: es.append((a, -1)) es.append((b, 1)) es.sort() result = 0 n = 0 for a, s in es: if s == -1: result += n n -= s return result Note, this is trivially O(n log n), and doesn't need any fancy data-structures. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Sep 30, 2013 at 7:18 Paul HankinPaul Hankin 58.7k 11 11 gold badges 99 99 silver badges 125 125 bronze badges 4 Comments Add a comment justhalf justhalfOver a year ago This is great!! Sorry, my previous comment was confused because you use -1 to begin and 1 to end an interval. You should have used the other way around. 2013-09-30T07:26:25.567Z+00:00 0 Reply Copy link kasitan kasitanOver a year ago Great, the simplest and the fastest one. But it should be result += n - 1 because no need to count the interval with itself. EDIT: oh, sorry too. I was confused by -1/+1 too. result += n is correct 2013-09-30T07:31:55.263Z+00:00 0 Reply Copy link Paul Hankin Paul HankinOver a year ago I think the +1 and -1 are the right way round, and result += n is right. n is updated after result, so an interval never counts against itself. If you run it on the intervals given in the question it produces the correct answer 4. 2013-09-30T07:33:20.24Z+00:00 1 Reply Copy link stackoverflowuser2010 stackoverflowuser2010Over a year ago @PaulHankin: Consider the intervals: intervals = []; intervals.append((1, 10)); intervals.append((5, 20)); intervals.append((15, 25)). Your code returns 2, but it should be 3 since all three intervals have some overlap, just not all at the same time. If you change the last interval to be (12, 25), then your code returns 3 instead. 2018-08-03T01:05:02.163Z+00:00 0 Reply Copy link Add a comment This answer is useful 2 Save this answer. Show activity on this post. First I assume from your example that (0,1) and (1,2) overlap. Btw, your example is incorrect, (3,7) does not overlap with (0,2) One way to solve this is: Sort the intervals based on the starting point Iterate from lowest starting point to highest point 2a. Count the number of previous endpoints larger or equal than current starting point. 2b. Add one to the count of current end point. Step 1 can be done in O(n log n) Step 2 is iterating over all the intervals while doing the count. So it's O(n X) where X is the complexity to do the counting. With Fenwick Tree, we can do that in O(log n)1, so step 2 can be completed in O(n log n) also. So the overall complexity is O(n log n). Pseudocode: ``` def fenwick_tree.get(num): return the sum from counter to counter[num] def fenwick_tree.add(num, val): add one to counter[num] intervals = [...] sort(intervals) # using the starting point as the key init_fenwick_tree() sum = 0 count = 0 for (starting_point, end_point) in intervals: sum = sum + (count - fenwick_tree.get(starting_point-1)) fenwick_tree.add(end_point,1) return sum ``` Python code (only works when the interval points are non-negative integers): ``` MAX_VALUE = 220-1 f_arr = MAX_VALUE def reset(): global f_arr, MAX_VALUE f_arr[:] = MAX_VALUE def update(idx,val): global f_arr while idx<MAX_VALUE: f_arr[idx]+=val idx += (idx & -idx) def read(idx): global f_arr if idx <= 0: return 0 result = 0 while idx > 0: result += f_arr[idx] idx -= (idx & -idx) return result intervals = [(1,4),(3,7),(5,8),(14,17),(0,2),(11,14)] intervals = sorted(intervals, key=lambda x: x) reset() total = 0 for processed, interval in enumerate(intervals): (start, end) = interval total += processed - read(start-1) update(end, 1) print total ``` Will print 4, which comes from these overlaps: (0,2) - (1,4) (1,4) - (3,7) (3,7) - (5,8) (11,14) - (14,17) Note that we can't get the overlapping intervals, since in the worst case there will be O(n^2) overlaps, which cannot be printed in O(n log n) time. 1 Actually Fenwick tree does the counting in O(log M) time, where M is the largest possible number of distinct values in the interval. Note that M <= 2n since there can only be that many distinct values. So it's also correct to say that Fenwick Tree does the counting in O(log n) time. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Sep 30, 2013 at 4:22 answered Sep 30, 2013 at 3:58 justhalfjusthalf 9,135 4 4 gold badges 51 51 silver badges 79 79 bronze badges Comments Add a comment This answer is useful 1 Save this answer. Show activity on this post. Quick idea: First sort your intervals. Now go through your intervals, adding each to a min-heap ordered by endpoint. For each interval, remove everything from the heap that is smaller than that interval's start point. Every endpoint remaining in the heap represents an interval that starts before this interval and that overlaps it, so increment an overlaps by the size of the heap. Now add the current interval to the heap and continue. In pseudocode: ``` Sort(intervals) (firstStart, firstEnd) = intervals currentIntervals = MinHeap() overlaps = 0 for each (start, end) in intervals[1:]: remove from currentIntervals all numbers < start overlaps += Size(currentIntervals) HeapPush(end, currentIntervals) return overlaps ``` I haven't tested this, but it seems at least plausible. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications edited Sep 30, 2013 at 4:21 answered Sep 30, 2013 at 4:03 jacobmjacobm 14.1k 1 1 gold badge 27 27 silver badges 29 29 bronze badges 6 Comments Add a comment justhalf justhalfOver a year ago It will only count at most one overlap per interval, where the OP wants to have all possible interval counts. Your algorithm will produce output 2 to the input {(0,3),(0,2),(1,3)}, where it should be 3. 2013-09-30T04:13:18.517Z+00:00 0 Reply Copy link jacobm jacobmOver a year ago Ah, okay. Updated to use a heap rather than a simple counter. 2013-09-30T04:21:29.047Z+00:00 1 Reply Copy link justhalf justhalfOver a year ago Looks good, although it is O(n^2) in worst case due to imbalanced heap. Say for this input: (0,10), (1,11), (2,12), ..., (10, 20) the heap will be linear, so the HeapPush operation will be O(n), resulting in O(n^2) total. You can use self-balancing binary tree such as AVL tree. 2013-09-30T04:25:32.507Z+00:00 0 Reply Copy link jacobm jacobmOver a year ago I don't think that's right. Binomial heaps can't be imbalanced, they're always as bushy as possible. 2013-09-30T04:27:17.857Z+00:00 1 Reply Copy link justhalf justhalfOver a year ago Oops, yes, sorry, I mixed that up with binary search tree. 2013-09-30T05:45:52.49Z+00:00 0 Reply Copy link Add a comment\|Show 1 more comment This answer is useful 0 Save this answer. Show activity on this post. This can simply be done using Greedy technique. just follow the following steps: Sort the intervals based on the starting point Iterate from lowest starting point to highest point Count the number of previous endpoints larger or equal than current starting point. Increment the count of current end point. Share Share a link to this answer Copy linkCC BY-SA 3.0 Improve this answer Follow Follow this answer to receive notifications answered Sep 30, 2013 at 7:00 Shubham ChawlaShubham Chawla 131 1 1 silver badge 10 10 bronze badges Comments Add a comment This answer is useful 0 Save this answer. Show activity on this post. Paul's answer is really smart. I don't think I could come up with that idea if it is an interview. Here I have another version which is also O(nlog(n)). ``` import heapq def countIntervals(s): s.sort() end = [s] res, cur = 0, 0 for l in s[1:]: if l>heapq.nsmallest(1, end): heapq.heappop(end) cur = len(end) res += cur end.append(l) return res ``` We maintain a min heap that stores upcoming endpoints. Every time when a new interval comes in, we should compare its start point with the smallest end point so far. If the start point is bigger, it means the smallest end point (represents that interval) will never cause more overlaps. Hence we pop it out. If the start point is smaller, it means all the end points (corresponding intervals) are overlapping with the new coming interval. Then the number of end points ("cur") is the number of overlaps that brings by this new coming interval. After we updated the result, we add the new coming interval's endpoint to the heap. 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16251	http://clubztutoring.com/ed-resources/science/what-is-additive-adjacent/	Home / Educational Resources / Science Resources / Additive Adjacent Additive Adjacent: Definitions, Examples, and Quiz Additive adjacent refers to a mathematical concept that involves combining numbers that are next to each other in a sequence or list. This concept is often taught to students in elementary and middle school as a way to help them develop their understanding of basic arithmetic and problem-solving skills. Definitions There are a few key terms that are important to understand when it comes to additive adjacent: Adjacent: Adjacent refers to two numbers that are next to each other in a sequence or list. For example, in the sequence 2, 4, 6, 8, 10, the numbers 4 and 6 are adjacent. Additive: Additive refers to the process of combining two or more numbers using addition. For example, if you have the numbers 2 and 3, you can add them together to get the sum of 5. Additive adjacent: Additive adjacent refers to the process of combining two adjacent numbers using addition. For example, if you have the sequence 2, 4, 6, 8, 10, you could add the adjacent numbers 4 and 6 together to get a sum of 10. Examples Here are five examples of how you can use additive adjacent to solve problems: Find the sum of a sequence of numbers: Let’s say you have the following sequence of numbers: 2, 4, 6, 8, 10. Using additive adjacent, you can find the sum of this sequence by adding each pair of adjacent numbers together. The sum would be: 2 + 4 = 6, 6 + 6 = 12, and 12 + 8 = 20. The final sum of the sequence would be 20. Find the missing number in a sequence: Suppose you have the following sequence of numbers: 3, 7, ?, 15, 19. Using additive adjacent, you can find the missing number by adding the adjacent numbers 7 and 15 together and dividing the result by 2. The missing number would be 11. Determine if a sequence of numbers is increasing or decreasing: By comparing the sum of adjacent numbers in a sequence, you can determine if the sequence is increasing or decreasing. For example, if you have the sequence 2, 4, 6, 8, 10 and you add the adjacent numbers together, you will get the following: 2 + 4 = 6, 4 + 6 = 10, and 6 + 8 = 14. Since the sum of the adjacent numbers is increasing, you can conclude that the sequence is also increasing. Find the average of a sequence of numbers: You can use additive adjacent to find the average of a sequence of numbers by adding all of the adjacent numbers together and then dividing the result by the total number of numbers in the sequence. For example, if you have the sequence 2, 4, 6, 8, 10, you can find the average by adding the adjacent numbers together and dividing the result by 5: (2 + 4)/5 = 6/5 = 1.2, (4 + 6)/5 = 10/5 = 2, and (6 + 8)/5 = 14/5 = 2.8. The average of the sequence would be 2. Find the right fit or it’s free. 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16252	https://www.youtube.com/watch?v=tJFaZz0tN_U	Long Division with 2-Digit Divisors \| Dividing 3-Digit Numbers by 2-Digit Numbers Lampofilm 138000 subscribers 2946 likes Description 202040 views Posted: 28 Jan 2023 Making Math Fun! Learn to divide a 3-digit number by a 2-digit number. Long Division Step-by-step lessons, great for distance learning and school lessons. Please check out my other math videos and we would love it if you could subscribe to our channel. education #maths #teacher #math Please help support our channel by subscribing and looking at our Amazon links. We thank you for your continued support. We hope you have a GREAT day. Amazon Link Please help support our channel by subscribing and looking at our Amazon links. We thank you for your continued support. We hope you have a GREAT day. Amazon Link If you have a Product you would like reviewed, please contact us by email. We are very interested in reviewing and testing Robotics, STEM TOYS, and New Technology. 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Please Donate 25 cents to us on PayPal Blog WebSite Facebook Transcript: hey welcome to the show today in the show we are going to be dividing a three-digit number but I had two digit number okay let's get started here hey when you get a big number like that don't worry we're going to take it step by step easy cheesy lemon squeezy all right let's get started 25 has to go into 600 all right we have to kind of figure out how we're going to do that first of all can we put 25 into 6 can't do it can we so then we move it over to 60. we're just going to be working just with 60 for right now so sometimes I like to take a piece of paper and cover that up so it doesn't look too scary at all so how many times can I put 25 into 60. well 25 is like a quarter you got 25 cents and then two quarters is 50 Cents three quarters to 75 cents and four quarters is one dollar I think I can put 25 into 60 what do you think two times all right let's try it so I'm going to put 2 right here 2 times 25 is like two quarters would be 50 cents or 50. all right so then what I do is I divide multiply and now I have to subtract all right so I'm going to put a line right underneath here and put my subtraction sign here and we're going to go ahead and subtract what is 6 minus 5 1 or 60 minus 50 is 10. so put 10 right here now what I do is bring it down drop it down drop that next number right down right over here so now we have 100 remember when I was talking about quarters when you have four quarters that equals a dollar that's the same thing right here 4 times 25 if you work it out right here 4 times 25 5 times 4 is 20 carry the two four times two is eight plus the 2 is 100. so I put the 4 right here 4 times 25 equals 100. and I have no remainder don't put a remainder of zero just just leave it blink wasn't that fun let's do another one yeah all right welcome back hey anytime you need to pause the video go ahead and pause the video try to work out this problem and then come right back [Music] all right let's take a look at it now remember before we were working with 25 and 2 times 25 is 50. so sometimes what you have to do is you kind of have to estimate kind of have to do a little bit of guessing to try to figure out how many times I can get this number into this number remember we're going to worry about 60 right now so I know if I took 25 plus 25 would be 50 let's see what 27 times 2 would be so I'm going to go ahead and put 27 times 2 right here so 7 times 2 would be 14 carry the 1 2 times 1 is 4 4 plus the 154. I definitely could put 54 into 60. let's let's raise it up one more I just want to see what happens 27 times 3. so 7 times 3 21 put the 2 up here 3 times 2 is 6 plus the 2. all right so that is 81. three times two is six seven eight kind of have to always check myself when you're doing problems like this big problems dividing you're going to have a lot of little uh multiplication problems you're going to be writing on the side sometimes you might want to use a scratch piece of paper for that and it might come around later on where you're going to need one of these so right now I'm going to stick with this one right here 72 or 72 27 times 2. so I'm going to put the 2 right here 2 times 27 was yeah 54. all right so what I did is I did a little multiplication right now and now I'm going to go and subtract I'm going to put my subtraction sign here put a line underneath there and go ahead and subtract well we've got I gotta start remember the subtraction with the borrowing and all that we got borrowing regrouping that's what we have to do here so I'm going to put a line through the six that'll be a five and then this turns into 10. so 10 minus four what is it four five six seven eight nine ten six five minus five zip zero all right so what do we do next bring it down drop it down drop it down bring this number all the way down here drop it down now we have to decide how many times can we put 27 into 60. let's take a look at our figures here well we know that 2 times 27 is 54 and 3 times 27 is 81. too much bring it down we're going to be using this number here all right so let's put another 2 right here so 2 times 27 what was it same thing 54. track that now remember I I did the same thing right there all right so I put 5 10 10 minus four six Pickup Sticks zero all right now I don't have any more numbers to drop down we're at the end this is our remainder this is what is left over so we put an R right here RMA T and we put a 6 right there so our answer is 22 with a remainder of 6. ready to try one last one yeah let's do it yeah all right welcome back whoa there's there's a lot more numbers in there there's no more zeros this might be a little bit of a challenge but I know you can do it my math ninjas all right let's get started first of all we need to figure out how many times we can put 23 3 into 58. now remember if you want to take that little piece of paper there and cover it up so I'm going to do is I put a piece of paper right here so it doesn't look too scary just like I would if there was a spider crawling around so you say spider oh okay all right all right calm down calm down okay all right let's see here 23 goes into 58. let's pretend that let's round that to 20. all right so 20 and then 20 plus 2 would be 40 and 20 times 3 would be 60. it's getting really close there isn't it so I I I'm putting it putting my bet that it's going to be probably two but let's try to do a little figurine right here 23 times 2 all right 3 times 2 is 6. 2 times 2 is 4. all right still a little low let's see what 23 times three is gotta put my uh multiplication symbol there three times three is nine three times two is six Pickup Sticks so we got 69. oh that's too much we gotta drop it down drop it down to 46. all right so 23 times 2 put the 2 right there 2 times 23 46 . all right now we're going to be doing some subtraction I think I can move this paper over here a little bit all right not too scary all right let's do some subtraction eight minus six uno dos one two two five minus four uno one all right I can't put 23 into 12. can I no so I need to bring that next number down drop it down drop it down all right you can add those sound effects those are pretty cool aren't they all right now 23 goes into 123. do I have anything like that no so then I need to start making a couple more uh problems here don't I got to figure out how many times I can put 23 in there so I know that if I had if that was 20 let me see 20 40 60 80. that might be four or five let's try the number four all right so I'm going to put 23 it's not 3 because 3 is is 69. so let's try 4 so 3 times 4 is 12. put the 2 here carry the one and when four times two is eight and plus the one would be nine let's see if we can push it one more 23 times 5. see what I mean talking about a scratch piece of paper you're going to have a lot of little problems on the side here that's the only way to do it five times three is fifteen carry the one five times two is ten plus the one is eleven I think I'm pretty close if I went to six I'm I'm really sure it's going to be over so let's go ahead and use five so I'm going to put five right here all right 5 times 23 what do we have 115 so I put 115 right here subtract I'm going to have to do a little regrouping don't worry we can do it borrow one from the two that'll be a one this turns into a 13 what is 13 minus five we see 8 9 10 11 12 13. put the 8 right here one minus one zero you don't even have to put zero all right it might get a little confusing that's it this is it this is your remainder there's no more numbers to drop down that is it we are finished with a remainder of eight it is 25 remainder eight yes you did it okay I wonder if there's a way we can high five on the camera just put your hand on the let's Boop yeah we did it yes I'm so happy hey check out my other videos I'm gonna have a lot more videos on problems like this you can do some examples and we got other a lot of other fun videos on this channel I would really appreciate if you hit subscribe hit that like button send us cookies and have a fun day adios foreign [Music]
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Log out Cancel Tools & Reference>Infectious Diseases Cellulitis Updated: Dec 30, 2024 Author: Thomas E Herchline, MD; Chief Editor: Michael Stuart Bronze, MD more...;) Share Print Feedback Facebook Twitter LinkedIn WhatsApp Email Sections Cellulitis Overview Practice Essentials Background Pathophysiology Etiology Epidemiology Prognosis Patient Education Show All Presentation History Physical Examination Show All DDx Workup Approach Considerations Moderate to Severe Cases and Systemic Symptoms Ultrasonography, CT Scanning, and MRI Aspiration, Dissection, and Biopsy Histologic Findings Show All Treatment Approach Considerations Outpatient Care IV Antibiotic Therapy Surgical Examination and Drainage Impetigo MSSA and MSRA SSTIs Erysipelas and Cellulitis Necrotizing Infections Animal- or Human-Bite Infections Surgical Site Infections Infections in the Immunocompromised Host Show All Guidelines Guidelines Summary Nonpurulent Cellulitis Purulent Cellulitis Staphylococcal and Streptococcal Skin and Soft Tissue Infections Show All Medication Medication Summary Penicillins, Natural Penicillins, Amino Penicillins, Penicillinase Resistant Penicillins, Extended-Spectrum Cephalosporins, Other Macrolides Carbapenems Fluoroquinolones Antibiotics, Other Tetracyclines Antifungal Agents Show All Media Gallery;) Tables) References;) Practice Essentials The term cellulitis commonly is used to indicate a nonnecrotizing inflammation of the skin and subcutaneous tissues, usually from acute infection. Cellulitis usually follows a breach in the skin, although a portal of entry may not be obvious; the breach may involve microscopic skin changes or invasive qualities of certain bacteria. Patient with cellulitis of the left ankle. This cellulitis was caused by community-acquired methicillin-resistant Staphylococcus aureus (CA-MRSA). (Photo courtesy of Texas Dept. of Public Health.) View Media Gallery) Signs and symptoms Nonpurulent cellulitis is associated with the four cardinal signs of infection, as follows: Physical examination findings that suggest the most likely pathogen include the following: Skin infection without underlying drainage, penetrating trauma, eschar, or abscess is most likely caused by streptococci; Staphylococcus aureus, often community-acquired MRSA, is the most likely pathogen when these factors are present Violaceous color and bullae suggest more serious or systemic infection with organisms such as Vibrio vulnificus or Streptococcus pneumoniae The following findings suggest severe infection: Malaise, chills, fever, and toxicity Lymphangitic spread (red lines streaking away from the area of infection) Circumferential cellulitis Pain disproportionate to examination findings Indications for emergent surgical evaluation are as follows [3, 4] : Violaceous bullae Cutaneous hemorrhage Skin sloughing Skin anesthesia Rapid progression Gas in the tissue Hypotension See Clinical Presentation for more detail. Diagnosis Generally, no workup is required in uncomplicated cases of cellulitis that meet the following criteria: Limited area of involvement Minimal pain No systemic signs of illness (eg, fever, altered mental status, tachypnea, tachycardia, hypotension) No risk factors for serious illness (eg, extremes of age, general debility, immunocompromise) The Infectious Disease Society of America (IDSA) recommends the following blood tests for patients with skin or soft tissue infection (SSTI) who have signs and symptoms of systemic toxicity : Blood cultures CBC with differential Levels of creatinine, bicarbonate, creatine phosphokinase, and C-reactive protein (CRP) Blood cultures should also be done in the following circumstances : Moderate to severe disease (eg, cellulitis complicating lymphedema ) Cellulitis of specific anatomic sites (eg, facial and especially ocular areas) Patients with a history of contact with potentially contaminated water Patients with malignancy who are receiving chemotherapy Neutropenia or severe cell-mediated immunodeficiency Animal bites Other tests to consider are as follows: Mycologic investigations are advisable if recurrent episodes of cellulitis are suspected to be secondary to tinea pedis or onychomycosis Creatinine levels help assess baseline renal function and guide antimicrobial dosing Imaging studies Ultrasonography may play a role in the detection of occult abscess and direction of care Ultrasonographic-guided aspiration of pus can shorten hospital stay and fever duration in children with cellulitis If necrotizing fasciitis is a concern, CT imaging typically is used in stable patients; MRI can be performed, but MRI typically takes much longer than CT scanning Strong clinical suspicion of necrotizing fasciitis should prompt surgical consultation without delay for imaging Aspiration, Dissection, and Biopsy Needle aspiration should be performed only in selected patients or in unusual cases, such as in cases of cellulitis with bullae or in patients who have diabetes, are immunocompromised, are neutropenic, are not responding to empiric therapy, or have a history of animal bites or immersion injury [10, 11, 12] Aspiration or punch biopsy of the inflamed area may have a culture yield of 2-40% and is of limited clinical value in most cases Gram stain of aspiration or biopsy specimens has a low yield and is unnecessary in most cases, unless purulent material is draining or bullae or abscess is present; however, Gram stain and culture after incision and drainage of an abscess yields positive results in more than 90% of cases Dissection of the underlying fascia to assess for necrotizing fasciitis may be determined by surgical consultation or indicated following initial evaluation and imaging studies Skin biopsy is not routine but may be performed in an attempt to rule out a noninfectious entity Hospital admission The IDSA recommends considering inpatient admission in patients with hypotension and/or the following laboratory findings : Elevated creatinine level Elevated creatine phosphokinase level (2-3 times the upper limit of normal) CRP level >13 mg/L (123.8 mmol/L) Low serum bicarbonate level Marked left shift on the CBC with differential See Workup for more detail. Management Treatment of cellulitis is as follows: Antibiotic regimens are effective in more than 90% of patients All but the smallest of abscesses require drainage for resolution, regardless of the pathogen Drainage only, without antibiotics, may suffice if the abscess is relatively isolated, with little surrounding tissue involvement In cases of cellulitis without draining wounds or abscess, streptococci continue to be the likely etiology, and beta-lactam antibiotics are appropriate therapy, as noted in the following: In mild cases of cellulitis treated on an outpatient basis: dicloxacillin, amoxicillin, or cephalexin In patients who are allergic to penicillin: clindamycin or a macrolide (clarithromycin or azithromycin) An initial dose of parenteral antibiotic with a long half-life (eg, ceftriaxone) followed by an oral agent Treatment of recurrent disease (usually related to venous or lymphatic obstruction) is as follows: The cellulitis is most often due to Streptococcus species; daily amoxicillin or a macrolide may be effective for prevention of recurrences. If tinea pedis is suspected to be the predisposing cause, treat with topical or systemic antifungals. Compressive therapy has been shown to decrease risk for recurrence in patients with chronic edema and recurrent cellulitis. Patients with severe cellulitis require parenteral therapy, such as the following: Cefazolin, cefuroxime, ceftriaxone, nafcillin, or oxacillin for presumed staphylococcal or streptococcal infection Clindamycin or vancomycin for penicillin-allergic patients Broad gram-positive, gram-negative, and anaerobic coverage for cases associated with diabetic ulcers Coverage for MRSA, until culture and sensitivity information become available, for severe cellulitis apparently related to a furuncle or an abscess For cellulitis involving wounds sustained in an aquatic environment, recommended antibiotic regimens vary with the type of water involved, as follows : Saltwater or brackish water: doxycycline and ceftazidime, or a fluoroquinolone Freshwater: a third- or fourth-generation cephalosporin (eg, ceftazidime or cefepime) or a fluoroquinolone (eg, ciprofloxacin or levofloxacin) Lack of response to an appropriate antibiotic regimen should raise suspicion for Mycobacterium marinum infection and suggest wound biopsy for mycobacterial stains and culture See Treatment and Medication for more detail. Next: Background The term cellulitis is commonly used to indicate a nonnecrotizing inflammation of the skin and subcutaneous tissues, a process usually related to acute infection that does not involve the fascia or muscles. Cellulitis is characterized by localized pain, swelling, tenderness, erythema, and warmth. Cellulitis has been classically considered to be an infection without formation of abscess (nonpurulent), purulent drainage, or ulceration. At times, cellulitis may overlap with other conditions, so that the macular erythema coexists with nodules, areas of ulceration, and frank abscess formation (purulent cellulitis) (see Presentation). Mild cellulitis with a fine lacelike pattern of erythema. This lesion was only slightly warm and caused minimal pain, which is typical for the initial presentation of mild cellulitis. View Media Gallery) Swelling seen in cellulitis involving the hand. In a situation with hand cellulitis, always rule out deep infection by imaging studies or by obtaining surgical consultation. View Media Gallery) Severe cellulitis of the leg in a woman aged 80 years. The cellulitis developed beneath a cast and was painful and warm to the touch. Significant erythema is evident. The margins are irregular but not raised. An ulcerated area is visible in the center of the photograph. View Media Gallery) Burns complicated by cellulitis. The larger lesion is a second-degree burn (left), and the smaller lesion is a first-degree burn (right), each with an expanding zone of erythema consistent with cellulitis. View Media Gallery) Streptococcal species are the most common causes of erysipelas and diffuse cellulitis or nonpurulent cellulitis that is not associated with a defined portal. S aureus is the usual causative organism in purulent cellulitis associated with furuncles, carbuncles, or abscesses. Previous Next: Pathophysiology Cellulitis usually follows a breach in the skin, such as a fissure, cut, laceration, insect bite, or puncture wound. In some cases, there is no obvious portal of entry and the breach may be due to microscopic changes in the skin or invasive qualities of certain bacteria. Organisms on the skin and its appendages gain entrance to the dermis and multiply to cause cellulitis. Facial cellulitis of odontogenic origin may also occur. Patients with toe-web intertrigo and/or tinea pedis as well as those with lymphatic obstruction, venous insufficiency, pressure ulcers, and obesityare particularly vulnerable to recurrent episodes of cellulitis. [10, 20, 21, 22] The vast majority of cases of cellulitis are likely caused by Streptococcus pyogenes and, to a lesser degree, by Staphylococcus aureus. In rare cases, cellulitis results from the metastatic seeding of an organism from a distant focus of infection, especially in immunocompromised individuals. Distant seeding is particularly common in cellulitis due to S pneumoniae (pneumococcus) and marine Vibriospecies. Neisseria meningitidis, Pseudomonas aeruginosa, Brucella species, and Legionella species have also been reported as rare causes of cellulitis resulting from hematogenous spread. [19, 23] Previous Next: Etiology Host factors Certain host factors predispose to severe infection. The elderly and individuals with diabetes mellitus are at risk for more severe disease. In addition, patients with diabetes, immunodeficiency, cancer, venous stasis, chronic liver disease, peripheral arterial disease, and chronic kidney disease appear to be at higher risk for recurrent infection because of an altered host immune response. Local control of immune function through interleukin-driven neutrophil recruitment, protective action of antimicrobial peptides, and the integrity of the cutaneous barrier have significant effects on the hosts defense against infection. [19, 25] Cellulitis due to lymphatic obstruction or venectomy may be caused by nongroup A streptococci (ie, groups B, C, and G). [26, 27] Postvenectomy status following saphenous vein stripping can also result in cellulitis. Lymphadenectomy following tumor excision, such as mastectomy, is also a predisposing factor for cellulitis. Immunogenetic factors may play a role in some families who have an underlying susceptibility to an infection progressing to cellulitis. Other factors that affect host immunity and predispose to cellulitis include concurrent intravenous or subcutaneous skin popping drug use; infections in this setting may be polymicrobial, but community-acquired methicillin-resistant S aureus (CA-MRSA) is the most common pathogen in these patients. Patient with cellulitis of the left ankle. This cellulitis was caused by community-acquired methicillin-resistant Staphylococcus aureus (CA-MRSA). (Photo courtesy of Texas Dept. of Public Health.) View Media Gallery) Abscess and associated cellulitis caused by community-acquired methicillin-resistant Staphylococcus aureus (CA-MRSA). (Photo courtesy of Texas Dept. of Public Health.) View Media Gallery) In individuals with normal host defenses, the most common causative organisms are group A streptococci (GAS) and S aureus. Group B Streptococcus cellulitis occurs in infants younger than 6 months because their immune responses are not fully developed, and it may also be seen in adults with comorbidities such as diabetes or liver disease. For infantile cellulitis, presentations may include sepsis. [19, 28] Historically, facial cellulitis in children was frequently associated with H influenzae type B and S pneumoniae, but this is now generally considered a rarity because of routine H influenza e type B and pneumococcal vaccines. However, a study of 500,000 pediatric hospitalizations demonstrated that, although bacterial meningitis and epiglottitis diminished as a result of immunization for H influenzae type B and S pneumoniae, the incidence of facial cellulitis was unaffected. Nonetheless, another study noted that 96% of the serotypes that cause facial cellulitis were included in the heptavalent-conjugated pneumococcal vaccine that was routinely used at the time of the study. Impetigo is commonly caused by strains of S aureus and/or S pyogenes, and erysipelas (acute infection of the upper dermis, characterized by a sharply demarcated, raised border) is more commonly caused by streptococcal species such as S pyogenes. Immunocompromised hosts may become infected from nontraditional cellulitis organisms, including gram-negative rods (eg, Pseudomonas, Proteus, Serratia, Enterobacter, Citrobacter), anaerobes, and others (eg, Helicobacter cinaedi, Fusarium species). Although fungi (eg, Cryptococcus) and herpes simplex virus may also cause cellulitis, these causes are rare. Pneumococci may cause a particularly malignant form of cellulitis that is frequently associated with tissue necrosis, suppuration, and bloodstream invasion. Two distinct syndromes are recognized: the first is marked by involvement of the extremities in patients with diabetes or substance abuse, and the second is marked by involvement of the head, neck, and upper torso in patients with systemic lupus erythematosus, nephrotic syndrome, or hematologic disorders. Mycobacterial infections may present as cellulitis. In contradistinction to the usual bacterial cellulitis, these presentations often range from subacute to chronic and are typically unresponsive to short courses of antibioticswhich should then prompt further investigation. The diagnosis is made on the basis of the presence of granulomas, multinucleated giant cells, and acid-fast bacilli (AFB) from biopsy specimens or mycobacterial culture. [31, 32, 33] S aureus is the leading cause of SSTIs in injection drug users, followed by Streptococcus species. Gram-negative bacteria may cause bullous cellulitis in patients with cirrhosis. Early recognition is vital, because the course of the disease is rapid, typically progressing to septic shock and death. Gram stain and culture of fluid aspirated from the bullae may aid in management. Recurrent staphylococcal cellulitis may occur in otherwise immunologically normal patients with nasal carriage of staphylococci and those with Job syndrome. Hospital-acquired infections Various hospital-acquired infections following soft tissue trauma may lead to cellulitis. It is unusual to have infection occur in areas around surgical wounds fewer than 24 hours postoperatively, but if there is such a clinical problem, group A beta-hemolytic Streptococcus [GABHS] or Clostridium perfringens (which produces gas that may be appreciated as crepitus on examination) usually is the cause. Acinetobacter baumannii is an emerging multidrug-resistant pathogen in these scenarios. Cellulitis due to lymphatic obstruction or venectomy may be caused by nongroup A streptococci (ie, groups B, C, and G). [26, 27] Postvenectomy status following saphenous vein stripping can also result in cellulitis. Cellulitis may also be associated with tinea pedis, and in such cases, culture of toe-web spaces may help identify a bacterial pathogen. Lymphadenectomy following tumor excision, such as mastectomy, is also a predisposing factor for cellulitis. Varicella Cellulitis can complicate varicella and may be identified by larger margins of erythema surrounding the vesicles. One study identified patients with invasive GAS cellulitis complicating varicella. The median onset of GAS infection was Day 4 of varicella, with fever, vomiting, and localized swelling reported. This condition mandates antibiotic treatment and careful clinical follow-up. Untreated cellulitis in association with varicella may progress to severe necrotizing SSTIs requiring surgical intervention. [40, 4] MRSA Although cellulitis can be complicated by abscess formation, it typically develops from an abscessogenic focus. One maxim in microbiology is the following: "The hallmark of staph infection is abscess formation." This has become a significant concern because of changing patterns of antibiotic resistance of S aureus, particularly MRSA. MRSA was first reported in 1968 ; for years, MRSA infections were identified only in patients with recent hospitalization, surgery, renal dialysis, residence in long-term-care facilities, or IV drug use. However, in the 1990s, isolates of S aureus were found in patients without risk factors for nosocomial disease. These isolates, which mostly maintain susceptibility to antibiotics such as trimethoprim-sulfamethoxazole or tetracycline, have been termed CA-MRSA to distinguish them from the previously identified hospital or healthcare-associated MRSA (HA-MRSA). Patient with cellulitis of the left ankle. This cellulitis was caused by community-acquired methicillin-resistant Staphylococcus aureus (CA-MRSA). (Photo courtesy of Texas Dept. of Public Health.) View Media Gallery) Abscess and associated cellulitis caused by community-acquired methicillin-resistant Staphylococcus aureus (CA-MRSA). (Photo courtesy of Texas Dept. of Public Health.) View Media Gallery) Although reports have indicated that MRSA causes the majority of SSTIs, these studies are plagued by variability in case-finding methodologies. Furthermore, in the context of cellulitis, the finding is misleading in that these reports come from analysis of wound cultures in cases in which abscess formation occurred. Cultures in cellulitis are difficult to perform and frequently do not yield positive results; therefore, these tests are rarely done clinically. Consequently, the results of these studies cannot be generalized to cellulitis without abscess formation. Studies are under way to determine the incidence of S aureus in particular, CA-MRSA in SSTI in which there is no identifiable abscess. However, until results of those studies are available, treatment decisions must be made on clinical grounds. Because treatment failures after empiric treatment may often occur, because of the emergence of resistant strains, microbiologic investigations are strongly recommended. Bite wounds, lacerations, and puncture wounds Mammalian bite wounds represent a specific subset of cellulitis with unique pathogens, and the infections are usually polymicrobial. Human, dog, cat, and wild animal bites all predispose to cellulitis with unique pathogens, but dog bites are the most commonly encountered bite wound in both the primary care and the emergency setting. Several organisms are of particular interest in animal bites, including the following : Capnocytophaga canimorsus (dog) Eikenella corrodens (human) Pasteurella multocida (dog or cat) Streptobacillus moniliformis (rat) Puncture wounds, especially through the bottom of athletic shoes, may cause Pseudomonas osteomyelitis and/or cellulitis. However, lacerations and puncture wounds sustained in an aquatic environment (eg, oceans, lakes, streams) may be contaminated with bacteria not typically found in land-based injuries, including Aeromonas hydrophila, Pseudomonas and Plesiomonas species, Vibrio species, Erysipelothrix rhusiopathiae, and Mycobacterium marinum. Individuals with chronic liver disease are particularly susceptible to V vulnificus infections. Cellulitis due to documented Vibrio vulnificus infection. (Image courtesy of Kepler Davis.) View Media Gallery) Previous Next: Epidemiology Because cellulitis is not a reportable disease, the exact prevalence is uncertain; however, it is a relatively common infection, affecting all racial and ethnic groups. There is no statistically significant difference in the incidence of cellulitis in men and women, and usually no age predilection is described. Nonetheless, studies have found a higher incidence of cellulitis in individuals older than 45 years. [21, 50, 51] Cellulitis was more common in geriatric patients in a retrospective study of international travelers by the GeoSentinel Surveillance Network. Certain age groups are at higher risk in some unique scenarios, such as the following: Historically, buccal cellulitis caused by H influenzae type B was more common in children younger than 3 years; vaccination against this organism may have decreased the incidence of buccal cellulitis, but recent data suggest that this source remains a consideration, even in vaccinated cohorts Facial cellulitis is more common in adults older than 50 years; however, pneumococcal facial cellulitis occurs primarily in young children who are at risk for pneumococcal bacteremia [30, 53] Perianal cellulitis, usually with group A beta-hemolytic Streptococcus (GABHS), occurs in children younger than 3 years Elderly patients with cellulitis are predisposed to thrombophlebitis A study of an insurance database in Utah found an incidence rate of 24.6 cases per 1000 person-years. The incidence was higher in males and in those individuals aged 45-64 years. In a large epidemiologic hospital-based study on skin, soft tissue, bone, and joint infections, 37.3% patients were identified as having cellulitis. Overall rates of visits increased for SSTIs from 32.1 to 48.1 visits per 1000 population and reached 14.2 million by 2005, and visits for abscess and cellulitis increased from 17.3 to 32.5 visits per 1000 population and accounted for more than 95% of the increase, according to the National Ambulatory Medical Care Survey and National Hospital Ambulatory Medical Care Survey. The study provided data regarding visits by patients with SSTIs to physician offices, hospital outpatient departments, and emergency departments in the United States. Cellulitis was found to account for approximately 3% of emergency medical consultations at one United Kingdom district general hospital. Previous Next: Prognosis Many cases of cellulitis and SSTI can be treated on an outpatient basis with oral antibiotics and do not result in lasting sequelae. Most patients conditions respond well to oral antibiotics. When outpatient therapy is unsuccessful, or for patients who require admission initially, IV antibiotics are usually effective. Cellulitis may progress to serious illness by uncontrolled contiguous spread, including via the lymphatic or circulatory systems. Associated conditions or complications include lymphangitis, abscess formation, and, rarely, gangrenous cellulitis or necrotizing fasciitis. Certain species, most notably group A beta-hemolytic Streptococcus (GABHS) and S aureus, produce toxins that may mediate a more severe systemic infection, leading to septic shock and death. [58, 59] Previous Next: Patient Education Depending on the location of the affected area, the patient should decrease physical activity and elevate the extremity, if possible. They may take over-the-counter (OTC) pain medication such as acetaminophen (Tylenol) or ibuprofen (Advil, Motrin) for pain, if approved by their physician. Patients should call their doctor's office or seek urgent evaluation if they have any of the following features: Fever (>100.5°F), especially when associated with chills Cellulitis with surrounding soft, fluctuant areas that are suggestive of abscess formation Red streaking from an area of cellulitis or a fast-spreading area of redness, which indicates that the infection may need closer observation, change in antibiotic treatment, or inpatient supportive care Significant pain not relieved by acetaminophen or ibuprofen Inability to move an extremity or joint because of pain Although any cellulitis infection may be severe, patients with diabetes, cancer, chronic lymphedema, or immunosuppression should be made aware that they are more predisposed to serious infection. Patients with an underlying genetic condition, such as an immunodeficiency disease, are also at especially high risk for minor skin infections to progress to cellulitis. 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Surgical implications of necrotizing fasciitis in children with chickenpox. J Pediatr Surg. Aug 1996. 31(8):1138-41. Lowy FD. Staphylococcus aureus infections. N Engl J Med. Aug 20 1998. 339(8):520-32. Barrett FF, McGehee RF Jr, Finland M. Methicillin-resistant Staphylococcus aureus at Boston City Hospital. Bacteriologic and epidemiologic observations. N Engl J Med. Aug 29 1968. 279(9):441-8. Four Pediatric Deaths from Community-Acquired Methicillin-Resistant Staphylococcus aureus -- Minnesota and North Dakota, 1997-1999. CDC. August 20, 1999. 707-10. [Full Text]. Furuya EY, Cook HA, Lee MH, Miller M, Larson E, Hyman S. Community-associated methicillin-resistant Staphylococcus aureus prevalence: how common is it? A methodological comparison of prevalence ascertainment. Am J Infect Control. Aug 2007. 35(6):359-66. [QxMD MEDLINE Link]. Brook I. Microbiology and management of human and animal bite wound infections. Prim Care. Mar 2003. 30(1):25-39. Dendle C, Looke D. Review article: Animal bites: an update for management with a focus on infections. Emerg Med Australas. Dec 2008. 20(6):458-67. Noonburg GE. Management of extremity trauma and related infections occurring in the aquatic environment. J Am Acad Orthop Surg. Jul-Aug 2005. 13(4):243-53. Dechet AM, Yu PA, Koram N, Painter J. Nonfoodborne Vibrio infections: an important cause of morbidity and mortality in the United States, 1997-2006. Clin Infect Dis. Apr 1 2008. 46(7):970-6. McNamara DR, Tleyjeh IM, Berbari EF, et al. Incidence of lower-extremity cellulitis: a population-based study in Olmsted county, Minnesota. Mayo Clin Proc. Jul 2007. 82(7):817-21. Ellis Simonsen SM, van Orman ER, Hatch BE, et al. Cellulitis incidence in a defined population. Epidemiol Infect. Apr 2006. 134(2):293-9. [QxMD MEDLINE Link]. Lamagni TL, Darenberg J, Luca-Harari B, et al. Epidemiology of severe Streptococcus pyogenes disease in Europe. J Clin Microbiol. Jul 2008. 46(7):2359-67. [QxMD MEDLINE Link]. Lederman ER, Weld LH, Elyazar IR, et al. Dermatologic conditions of the ill returned traveler: an analysis from the GeoSentinel Surveillance Network. Int J Infect Dis. Nov 2008. 12(6):593-602. Givner LB, Mason EO Jr, Barson WJ, et al. Pneumococcal facial cellulitis in children. Pediatrics. Nov 2000. 106(5):E61. Kokx NP, Comstock JA, Facklam RR. Streptococcal perianal disease in children. Pediatrics. Nov 1987. 80(5):659-63. Lipsky BA, Weigelt JA, Gupta V, Killian A, Peng MM. Skin, soft tissue, bone, and joint infections in hospitalized patients: epidemiology and microbiological, clinical, and economic outcomes. Infect Control Hosp Epidemiol. Nov 2007. 28(11):1290-8. [QxMD MEDLINE Link]. Hersh AL, Chambers HF, Maselli JH, Gonzales R. National trends in ambulatory visits and antibiotic prescribing for skin and soft-tissue infections. Arch Intern Med. Jul 28 2008. 168(14):1585-91. [QxMD MEDLINE Link]. Davies HD, McGeer A, Schwartz B, et al. Invasive group A streptococcal infections in Ontario, Canada. Ontario Group A Streptococcal Study Group. N Engl J Med. Aug 22 1996. 335(8):547-54. [QxMD MEDLINE Link]. Bisno AL, Cockerill FR 3rd, Bermudez CT. The initial outpatient-physician encounter in group A streptococcal necrotizing fasciitis. Clin Infect Dis. Aug 2000. 31(2):607-8. [QxMD MEDLINE Link]. Francis JS, Doherty MC, Lopatin U, Johnston CP, Sinha G, Ross T. Severe community-onset pneumonia in healthy adults caused by methicillin-resistant Staphylococcus aureus carrying the Panton-Valentine leukocidin genes. Clin Infect Dis. Jan 1 2005. 40(1):100-7. [QxMD MEDLINE Link]. Durupt F, Dalle S, Ronger S, Thomas L. Does erysipelas-like rash after hip replacement exist?. Dermatology. 2006. 212(3):216-20. Simon MS, Cody RL. Cellulitis after axillary lymph node dissection for carcinoma of the breast. Am J Med. Nov 1992. 93(5):543-8. Masmoudi A, Maaloul I, Turki H, et al. Erysipelas after breast cancer treatment (26 cases). Dermatol Online J. Dec 1 2005. 11(3):12. Zippel D, Siegelmann-Danieli N, Ayalon S, Kaufman B, Pfeffer R, Zvi Papa M. Delayed breast cellulitis following breast conserving operation. Eur J Surg Oncol. May 2003. 29(4):327-30. El Saghir NS, Otrock ZK, Bizri AR, Uwaydah MM, Oghlakian GO. Erysipelas of the upper extremity following locoregional therapy for breast cancer. Breast. Oct 2005. 14(5):347-51. Lazzarini L, Conti E, Tositti G, de Lalla F. Erysipelas and cellulitis: clinical and microbiological spectrum in an Italian tertiary care hospital. J Infect. Dec 2005. 51(5):383-9. Spear RM, Rothbaum RJ, Keating JP, Blaufuss MC, Rosenblum JL. Perianal streptococcal cellulitis. J Pediatr. Oct 1985. 107(4):557-9. Markham RB, Polk BF. Seal finger. Rev Infect Dis. May-Jun 1979. 1(3):567-9. Crum NF, Higginbottom PA, Fehl FC, Graham BS. Sweet's syndrome masquerading as facial cellulitis. Cutis. Jun 2003. 71(6):469-72. Jenkins TC, Knepper BC, Sabel AL, Sarcone EE, Long JA, Haukoos JS, et al. Decreased Antibiotic Utilization After Implementation of a Guideline for Inpatient Cellulitis and Cutaneous Abscess. Arch Intern Med. 2011 Jun 27. 171(12):1072-9. [QxMD MEDLINE Link]. Perl B, Gottehrer NP, Raveh D, Schlesinger Y, Rudensky B, Yinnon AM. Cost-effectiveness of blood cultures for adult patients with cellulitis. Clin Infect Dis. Dec 1999. 29(6):1483-8. Liu C, Bayer A, Cosgrove SE, et al. Clinical practice guidelines by the Infectious Diseases Society of America for the treatment of methicillin-resistant Staphylococcus aureus infections in adults and children. Clin Infect Dis. 2011 Feb. 52:1-38. [QxMD MEDLINE Link]. Lipsky BA, Berendt AR, Cornia PB, et al. 2012 Infectious Diseases Society of America Clinical Practice Guideline for the Diagnosis and Treatment of Diabetic Foot Infections. Clin Infect Dis. 2012. 54(12):3132-3172. [Full Text]. Brown T. Recurrent Cellulitis: Penicillin Effective for Prevention. Medscape Medical News, May 1, 2013. Available at Accessed: May 8, 2013. Thomas KS, Crook AM, Nunn AJ, Foster KA, Mason JM, Chalmers JR, et al. Penicillin to prevent recurrent leg cellulitis. N Engl J Med. 2013 May 2. 368(18):1695-703. [QxMD MEDLINE Link]. Moran GJ, Krishnadasan A, Gorwitz RJ, Fosheim GE, McDougal LK, Carey RB, et al. Methicillin-resistant S. aureus infections among patients in the emergency department. N Engl J Med. Aug 17 2006. 355(7):666-74. Cenizal MJ, Skiest D, Luber S, Bedimo R, Davis P, Fox P, et al. Prospective randomized trial of empiric therapy with trimethoprim-sulfamethoxazole or doxycycline for outpatient skin and soft tissue infections in an area of high prevalence of methicillin-resistant Staphylococcus aureus. Antimicrob Agents Chemother. Jul 2007. 51(7):2628-30. Daum RS. Clinical practice. Skin and soft-tissue infections caused by methicillin-resistant Staphylococcus aureus. N Engl J Med. Jul 26 2007. 357(4):380-90. Stryjewski ME, Chambers HF. Skin and soft-tissue infections caused by community-acquired methicillin-resistant Staphylococcus aureus. Clin Infect Dis. Jun 1 2008. 46 Suppl 5:S368-77. Davis SL, McKinnon PS, Hall LM, Delgado G Jr, Rose W, Wilson RF. Daptomycin versus vancomycin for complicated skin and skin structure infections: clinical and economic outcomes. Pharmacotherapy. Dec 2007. 27(12):1611-8. Krige JE, Lindfield K, Friedrich L, Otradovec C, Martone WJ, Katz DE. Effectiveness and duration of daptomycin therapy in resolving clinical symptoms in the treatment of complicated skin and skin structure infections. Curr Med Res Opin. Sep 2007. 23(9):2147-56. Hepburn MJ, Dooley DP, Skidmore PJ, Ellis MW, Starnes WF, Hasewinkle WC. Comparison of short-course (5 days) and standard (10 days) treatment for uncomplicated cellulitis. Arch Intern Med. Aug 9-23 2004. 164(15):1669-74. Wong CH, Khin LW, Heng KS, Tan KC, Low CO. The LRINEC (Laboratory Risk Indicator for Necrotizing Fasciitis) score: a tool for distinguishing necrotizing fasciitis from other soft tissue infections. Crit Care Med. 2004 Jul. 32(7):1535-41. [QxMD MEDLINE Link]. Michael Y, Shaukat NM. Erysipelas. 2022 Jan. [QxMD MEDLINE Link]. [Full Text]. King MD, Humphrey BJ, Wang YF, Kourbatova EV, Ray SM, Blumberg HM. Emergence of community-acquired methicillin-resistant Staphylococcus aureus USA 300 clone as the predominant cause of skin and soft-tissue infections. Ann Intern Med. Mar 7 2006. 144(5):309-17. Halilovic J, Heintz BH, Brown J. Risk factors for clinical failure in patients hospitalized with cellulitis and cutaneous abscess. J Infect. 2012 Mar 21. [Epub ahead of print]. [QxMD MEDLINE Link]. Byl B, Clevenbergh P, Jacobs F, Struelens MJ, Zech F, Kentos A. Impact of infectious diseases specialists and microbiological data on the appropriateness of antimicrobial therapy for bacteremia. Clin Infect Dis. Jul 1999. 29(1):60-6; discussion 67-8. Chuang YC, Yuan CY, Liu CY, Lan CK, Huang AH. Vibrio vulnificus infection in Taiwan: report of 28 cases and review of clinical manifestations and treatment. Clin Infect Dis. Aug 1992. 15(2):271-6. Fernandez JM, Serrano M, De Arriba JJ, Sanchez MV, Escribano E, Ferreras P. Bacteremic cellulitis caused by Non-01, Non-0139 Vibrio cholerae: report of a case in a patient with hemochromatosis. Diagn Microbiol Infect Dis. May 2000. 37(1):77-80. Bush LM, Vazquezquez-Pertejo MT. Gram-Negative Bacilli. Porter RE. The Merck Manual of Diagnosis and Therapy. Rahway, NJ: Merck & Co Inc; Reviewed/Revised June 2024. [Full Text]. Media Gallery Mild cellulitis with a fine lacelike pattern of erythema. This lesion was only slightly warm and caused minimal pain, which is typical for the initial presentation of mild cellulitis. Swelling seen in cellulitis involving the hand. In a situation with hand cellulitis, always rule out deep infection by imaging studies or by obtaining surgical consultation. Severe cellulitis of the leg in a woman aged 80 years. The cellulitis developed beneath a cast and was painful and warm to the touch. Significant erythema is evident. The margins are irregular but not raised. An ulcerated area is visible in the center of the photograph. Burns complicated by cellulitis. The larger lesion is a second-degree burn (left), and the smaller lesion is a first-degree burn (right), each with an expanding zone of erythema consistent with cellulitis. Cellulitis due to documented Vibrio vulnificus infection. (Image courtesy of Kepler Davis.) A case of cellulitis without associated purulence in an infant. Note the presence of lymphedema, a risk factor for cellulitis.(Photo courtesy of Amy Williams.) Patient with cellulitis of the left ankle. This cellulitis was caused by community-acquired methicillin-resistant Staphylococcus aureus (CA-MRSA). (Photo courtesy of Texas Dept. of Public Health.) Abscess and associated cellulitis caused by community-acquired methicillin-resistant Staphylococcus aureus (CA-MRSA). (Photo courtesy of Texas Dept. of Public Health.) Guidelines for the management of patients who require hospitalization for cellulitis or cutaneous abscess. AFB = acid-fast bacilli; BID = twice daily; CRP = C reactive protein; CT = computed tomography scanning; DS = double strength; DM = diabetes mellitus; ESR = erythrocyte sedimentation rate; ESRD = end-stage renal disease; HIV = human immunodeficiency virus; ICU = intensive care unit; I&D = incision and drainage; ID = infectious disease; IDU = injection drug user; IV = intravenous; LRINEC = Laboratory Risk Indicator for Necrotizing Fasciitis; MRI = magnetic resonance imaging; MSRA = methicillin-resistant Staphylococcus aureus; NSAIDS = nonsteroidal anti-inflammatory drugs; PO = by mouth; SSTI = skin and soft-tissue infections; TID = 3 times daily. Adapted from Jenkins TC, Knepper BC, Sabel AL, et al. Decreased antibiotic utilization after implementation of a guideline for inpatient cellulitis and cutaneous abscess. Arch Intern Med. 2011;171(12):1072-9. A male patient with orbital cellulitis with proptosis, ophthalmoplegia, and edema and erythema of the eyelids. The patient also exhibited pain on eye movement, fever, headache, and malaise. A male patient with orbital cellulitis with proptosis, ophthalmoplegia, and edema and erythema of the eyelids. The patient also exhibited chemosis and resistance to retropulsion of the globe. Gross photograph of complicated cellulitis. Instead of the presence of yellow fat, the tissue is hemorrhagic and necrotic. Hematoxylin and eosin (H&E) stain, high power. This image shows deeper subcutaneous tissue involved in a case of cellulitis, with acute inflammatory cells and fat necrosis. Hematoxylin and eosin (H&E) stain, high power. This image shows cellulitis caused by herpes simplex virus, with the multinucleated organism in the center of the picture. of 14 Tables Table 1. Empiric Antibiotic Therapy for Cellulitis by Etiology and Anatomic Location;) Table 1. Empiric Antibiotic Therapy for Cellulitis by Etiology and Anatomic Location \| \| \| \| \| \| \| \| --- --- --- \| Location \| Likely Organisms \| Other Organisms \| Complication/ Discussion \| Antibiotic Regimen -- Oral/ Outpatient \| Indication for Hospitalization \| Antibiotic Regimen -- Parenteral/ Hospitalized \| \| Uncomplicated cellulitis \| Group A streptococci much more likely than Staphylococcus aureus \| \| \| Cephalexin or dicloxacillin or clindamycin \| \| Cefazolin or oxacillin or nafcillin \| \| Cellulitis, concern for methicillin-resistant S aureus is a concern \| Group A streptococci and S aureus \| \| \| [(Cephalexin or dicloxacillin or clindamycin) plus trimethoprim/ sulfamethoxazole] or Clindamycin \| \| Vancomycin Daptomycin Ceftaroline \| \| Dog bite \| Pasteurella species (50% of wounds) S aureus Streptococcus pyogenes \| Staphylococci, streptococci Aerobes --Moraxella and Neisseria Anaerobes --Fusobacterium, Bacteroides, Porphyromonas, and Prevotella \| Capnocytophaga canimorsus may cause sepsis in patients with asplenia/hepatic disease. Avoid first-generation cephalosporins/ erythromycin/ dicloxacillin. High likelihood of infection Prophylactic antibiotics indicated for the following wounds: deep puncture, hands, requiring surgical repair, immunocompromised host, venous or lymphatic compromise, crush injury. Requires close follow-up care within 24-48 h. \| Amoxicillin/ clavulanate Penicillin allergic: Moxifloxacin \| Deep wounds or severe wounds; infections not responding to oral antibiotics \| Third-generation cephalosporin (ceftriaxone [Rocephin]) plus metronidazole or beta-lactam/beta-lactamase inhibitor (eg, ampicillin/sulbactam) or fluoroquinolone plus metronidazole or carbapenem (ertapenem) \| \| Human bite \| Eikenella corrodens (gram-negative facultative anaerobe, 29% of wounds) Aerobic gram-positive cocci, anaerobes \| \| Clenched fist lacerations over metacarpophalangeal joints should be considered human bites; anesthetize wounds and irrigate; reevaluate within 24-48 h. Intercanine distance >3 cm is likely bite from adult; if wound to child, consider abuse. \| Amoxicillin/ clavulanate Penicillin allergic: Moxifloxacin or (Clindamycin or metronidazole) plus (doxycycline or cefuroxime or trimethoprim/ sulfamethoxazole) \| \| Third-generation cephalosporin (Rocephin) plus metronidazole or beta-lactam/beta-lactamase inhibitor (eg, ampicillin/sulbactam) or fluoroquinolone plus metronidazole or carbapenem (ertapenem) \| \| Cat bite \| Pasteurella multocida and P septica (75% of wounds) \| Staphylococci, streptococci, Bacteroides, Peptostreptococcus, Actinomyces, Fusobacterium, Porphyromonas, and Veillonella parvula \| Avoid first-generation cephalosporins/ erythromycin/ dicloxacillin High likelihood of infection -- Prophylactic antibiotics indicated for the following wounds: deep puncture, hands, requiring surgical repair, immunocompromised host, venous or lymphatic compromise. Requires close follow-up care within 24-48 h. \| Amoxicillin/ clavulanate Penicillin allergic -- Moxifloxacin or (Clindamycin or metronidazole) plus (doxycycline or cefuroxime or trimethoprim/ sulfamethoxazole) \| Deep wounds or severe wounds; infections not responding to oral antibiotics \| Third-generation cephalosporin (Rocephin) plus metronidazole or beta-lactam/beta-lactamase inhibitor (eg, ampicillin/sulbactam) or fluoroquinolone plus metronidazole or carbapenem (ertapenem) \| \| Preseptal (periorbital) cellulitis \| Haemophilus influenzae type b, Streptococcus pneumoniae, S aureus, other streptococcal species, and anaerobes \| Nocardia brasiliensis, Bacillus anthracis, Pseudomonas aeruginosa, Neisseria gonorrhoeae, Proteus species, Pasteurella multocida, Mycobacterium tuberculosis \| Largest study indicates that H influenzae type b and S pneumoniae not diminished in facial cellulitis as a result of immunizations \| Amoxicillin-clavulanate, cefpodoxime, cefdinir \| Age < 1 y/ more severe disease require intravenous antibiotic \| Third-generation cephalosporin (Rocephin) \| \| Lower extremity -- Complicating saphenous venectomy site after coronary bypass grafting \| No pathogen identifiable in most infections, but it is likely to be streptococcal (> staphylococcal) Non-group A beta-hemolytic streptococci most likely organism; S aureus less common \| \| Recurrent episodes common; may be associated with rigors, extreme fatigue, myalgias, and hypotension; some associated with tinea pedis (toe web cultures may be useful in establishing probable pathogen) \| Dicloxacillin or cephalexin. Add trimethoprim/ sulfamethoxazole or tetracycline or clindamycin if concern for methicillin-resistant S aureus \| \| First-generation cephalosporin (cefazolin); clindamycin; vancomycin \| \| Breast/arm - - (not mastitis) Complicating breast cancer surgery/lymph node dissection \| No pathogen identifiable in most infections Group A or Non-group A beta-hemolytic streptococci most likely organisms \| \| \| Dicloxacillin, cephalexin. Add trimethoprim/ sulfamethoxazole or tetracycline or clindamycin if concern for methicillin-resistant S aureus \| Fever, recent chemotherapy, neutropenia \| Multiple regimens, none clearly superior Piperacillin/tazobactam or ceftazidime plus aminoglycoside; or ciprofloxacin plus beta-lactam or monotherapy with piperacillin/tazobactam or cefepime \| \| Aquatic environment -- Fresh water/ salt water/ brackish water/ swimming pools/ aquarium Puncture/ laceration \| Aeromonas hydrophila, Pseudomonas and Plesiomonas species, Vibrio species, Erysipelothrix rhusiopathiae, Mycobacterium marinum, and others \| \| A hydrophila and Vibrio vulnificus may produce rapidly progressive soft tissue infection and sepsis \| Fluoroquinolone (eg, ciprofloxacin or levofloxacin) Note: For M marinum infection, use clarithromycin plus either ethambutol or rifampin \| \| Third- or fourth-generation cephalosporin (eg, ceftazidime or cefepime) or fluoroquinolone (eg, ciprofloxacin or levofloxacin) \| \| Clenched-fist injury \| E corrodens (gram-negative anaerobe, 29 % of wounds); aerobic gram-positive cocci, anaerobes \| \| Lacerations over metacarpophalangeal joints should be considered human bites; anesthetize wounds and irrigate; reevaluate within 24-48 h Lacerations of extensor tendon \| Amoxicillin/ clavulanate; penicillin allergic: Moxifloxacin or (clindamycin or metronidazole) plus (doxycycline or cefuroxime or trimethoprim/ sulfamethoxazole) \| Failure to respond to oral therapy marked by increasing pain and swelling or purulent drainage \| Beta-lactam/beta-lactamase inhibitor (eg, ampicillin/sulbactam) \| \| Odontogenic facial cellulitis \| Aerobic and facultative organisms: group A beta-hemolytic streptococci, Neisseria and Eikenella species Anaerobes: Prevotella and Peptostreptococcus species \| \| Require extraction or root canal \| Amoxicillin-clavulanate or clindamycin \| \| Beta-lactam/beta-lactamase inhibitor (eg, ampicillin/sulbactam) or clindamycin \| Back to List Contributor Information and Disclosures Author Thomas E Herchline, MD Professor of Medicine, Wright State University, Boonshoft School of MedicineThomas E Herchline, MD is a member of the following medical societies: Alpha Omega Alpha, Infectious Diseases Society of America, Infectious Diseases Society of OhioDisclosure: Nothing to disclose. Coauthor(s) Subramanian Swaminathan, DNB, MD Fellow, Department of Infectious Diseases, Detroit Medical Center, Wayne State University School of MedicineSubramanian Swaminathan, DNB, MD is a member of the following medical societies: American College of Physicians, Infectious Diseases Society of America, International AIDS Society, HIV Medicine AssociationDisclosure: Nothing to disclose. Pranatharthi Haran Chandrasekar, MBBS, MD Professor, Chief of Infectious Disease, Department of Internal Medicine, Wayne State University School of MedicinePranatharthi Haran Chandrasekar, MBBS, MD is a member of the following medical societies: American College of Physicians, American Society for Microbiology, International Immunocompromised Host Society, Infectious Diseases Society of AmericaDisclosure: Nothing to disclose. Chief Editor Michael Stuart Bronze, MD David Ross Boyd Professor and Chairman, Department of Medicine, Stewart G Wolf Endowed Chair in Internal Medicine, Department of Medicine, University of Oklahoma Health Science Center; Master of the American College of Physicians; Fellow, Infectious Diseases Society of America; Fellow of the Royal College of Physicians, LondonMichael Stuart Bronze, MD is a member of the following medical societies: Alpha Omega Alpha, American College of Physicians, American Medical Association, Association of Professors of Medicine, Infectious Diseases Society of America, Oklahoma State Medical Association, Southern Society for Clinical InvestigationDisclosure: Nothing to disclose. Barry E Brenner, MD, PhD, FACEP Professor of Emergency Medicine, Professor of Internal Medicine, Program Director, Emergency Medicine, Case Medical Center, University Hospitals, Case Western Reserve University School of Medicine Barry E Brenner, MD, PhD, FACEP is a member of the following medical societies: Alpha Omega Alpha, American Academy of Emergency Medicine, American College of Chest Physicians, American College of Emergency Physicians, American College of Physicians, American Heart Association, American Thoracic Society, Arkansas Medical Society, New York Academy of Medicine, New York Academy ofSciences,andSociety for Academic Emergency Medicine Disclosure: Nothing to disclose. David F Butler, MD Professor of Dermatology, Texas A&M University College of Medicine; Chair, Department of Dermatology, Director, Dermatology Residency Training Program, Scott and White Clinic, Northside Clinic David F Butler, MD is a member of the following medical societies: Alpha Omega Alpha, American Academy of Dermatology, American Medical Association, American Society for Dermatologic Surgery, American Society for MOHS Surgery, Association of Military Dermatologists, and Phi Beta Kappa Disclosure: Nothing to disclose. Dennis Cunningham, MD Assistant Professor of Pediatrics, Section of Infectious Diseases, Children's Hospital, Ohio State College of Medicine Disclosure: Nothing to disclose. Danny Lee Curtis, MD Clinical Assistant Professor of Medicine, University of South Florida College of Medicine; Consulting Staff, James A Haley Veterans Hospital Danny Lee Curtis, MD is a member of the following medical societies: American Academy of Emergency Medicine Disclosure: Nothing to disclose. Vinod K Dhawan, MD, FACP, FRCP(C), FIDSA Professor, Department of Clinical Medicine, University of California, Los Angeles, David Geffen School of Medicine; Chief, Division of Infectious Diseases, Rancho Los Amigos National Rehabilitation Center Vinod K Dhawan, MD, FACP, FRCP(C), FIDSA is a member of the following medical societies: American College of Physicians, American Society for Microbiology, American Society of Tropical Medicine and Hygiene, Infectious Diseases Society of America, and Royal College of Physicians and Surgeons of Canada Disclosure: Pfizer Inc Honoraria Speaking and teaching Robert Edelman, MD Professor, Associate Director for Clinical Research, Department of Medicine, Division of Geographic Medicine, Center for Vaccine Development, University of Maryland School of Medicine Disclosure: Nothing to disclose. Dirk M Elston, MD Director, Ackerman Academy of Dermatopathology, New York Dirk M Elston, MD is a member of the following medical societies: American Academy of Dermatology Disclosure: Nothing to disclose. Eric S Halsey, MD Head, Virology Department, Naval Medical Research Center Detachment-Peru (NMRCD-Peru); Assistant Professor of Medicine, Uniformed Services University of the Health Sciences Eric S Halsey, MD is a member of the following medical societies: Armed Forces Infectious Diseases Society, HIV Medicine Association of America, and Infectious Diseases Society of America Disclosure: Nothing to disclose. Isaac P Humphrey, MD Assistant Professor of Internal Medicine, Uniformed Services University of the Health Sciences; Clinical Assistant Professor of Internal Medicine, Wright State University Boonshoft School of Medicine Isaac P Humphrey, MD is a member of the following medical societies: American College of Physicians Disclosure: Nothing to disclose. William D James, MD Paul R Gross Professor of Dermatology, Vice-Chairman, Residency Program Director, Department of Dermatology, University of Pennsylvania School of Medicine William D James, MD is a member of the following medical societies: American Academy of Dermatology and Society for Investigative Dermatology Disclosure: Elsevier Royalty Other Sungnack Lee, MD Vice President of Medical Affairs, Professor, Department of Dermatology, Ajou University School of Medicine, Korea Sungnack Lee, MD is a member of the following medical societies: American Dermatological Association Disclosure: Nothing to disclose. Fred A Lopez, MD Associate Professor and Vice Chair, Department of Medicine, Assistant Dean for Student Affairs, Louisiana State University School of Medicine Fred A Lopez, MD is a member of the following medical societies: Alpha Omega Alpha, American College of Physicians-American Society of Internal Medicine, Infectious Diseases Society of America, and Louisiana State Medical Society Disclosure: Nothing to disclose. Mark Louden, MD, FACEP Assistant Medical Director, Emergency Department, Duke Raleigh Hospital Mark Louden, MD, FACEP is a member of the following medical societies: American Academy of Emergency Medicine and American College of Emergency Physicians Disclosure: Nothing to disclose. Giuseppe Micali, MD Head, Professor, Department of Dermatology, University of Catania School of Medicine, Italy Giuseppe Micali, MD is a member of the following medical societies: American Academy of Dermatology Disclosure: Nothing to disclose. Christen M Mowad, MD Associate Professor, Department of Dermatology, Geisinger Medical Center Christen M Mowad, MD is a member of the following medical societies: Alpha Omega Alpha, American Academy of Dermatology, and Phi Beta Kappa Disclosure: Nothing to disclose. Maria R Nasca, MD, PhD Assistant Professor, Department of Dermatology, University of Catania School of Medicine, Italy Disclosure: Nothing to disclose. Charles V Sanders, MD Edgar Hull Professor and Chairman, Department of Internal Medicine, Professor of Microbiology, Immunology and Parasitology, Louisiana State University School of Medicine at New Orleans; Medical Director, Medicine Hospital Center, Charity Hospital and Medical Center of Louisiana at New Orleans; Consulting Staff, Ochsner Medical Center Charles V Sanders, MD is a member of the following medical societies: Alliance for the Prudent Use of Antibiotics, Alpha Omega Alpha, American Association for the Advancement of Science, American Association of University Professors, American Clinical and Climatological Association, American College of Physician Executives, American College of Physicians, American Federation for Medical Research, American Foundation for AIDS Research, AmericanGeriatricsSociety, American Lung Association, American Medical Association, American Society for Microbiology, American Thoracic Society, American Venereal Disease Association, Association for Professionals in Infection Control and Epidemiology, Association of American Medical Colleges, Association of American Physicians, Association of Professors of Medicine, Infectious Disease Society for Obstetrics and Gynecology, InfectiousDiseases Societyof America, Louisiana State Medical Society, Orleans Parish Medical Society, Royal Society of Medicine, Sigma Xi, Society of General Internal Medicine, Southeastern Clinical Club, Southern Medical Association, Southern Society for Clinical Investigation, and Southwestern Association of Clinical Microbiology Disclosure: Baxter International and Johnson & Johnson Royalty Other Barry J Sheridan, DO Chief Warrior in Transition Services, Brooke Army Medical Center Barry J Sheridan, DO is a member of the following medical societies: American Academy of Emergency Medicine Disclosure: Nothing to disclose. Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference Disclosure: Medscape Salary Employment Close;) What would you like to print? What would you like to print? Print this section Print the entire contents of Print the entire contents of article Sections Cellulitis Overview Practice Essentials Background Pathophysiology Etiology Epidemiology Prognosis Patient Education Show All Presentation History Physical Examination Show All DDx Workup Approach Considerations Moderate to Severe Cases and Systemic Symptoms Ultrasonography, CT Scanning, and MRI Aspiration, Dissection, and Biopsy Histologic Findings Show All Treatment Approach Considerations Outpatient Care IV Antibiotic Therapy Surgical Examination and Drainage Impetigo MSSA and MSRA SSTIs Erysipelas and Cellulitis Necrotizing Infections Animal- or Human-Bite Infections Surgical Site Infections Infections in the Immunocompromised Host Show All Guidelines Guidelines Summary Nonpurulent Cellulitis Purulent Cellulitis Staphylococcal and Streptococcal Skin and Soft Tissue Infections Show All Medication Medication Summary Penicillins, Natural Penicillins, Amino Penicillins, Penicillinase Resistant Penicillins, Extended-Spectrum Cephalosporins, Other Macrolides Carbapenems Fluoroquinolones Antibiotics, Other Tetracyclines Antifungal Agents Show All Media Gallery;) Tables) References;) encoded search term (Cellulitis) and Cellulitis What to Read Next on Medscape Related Conditions and Diseases Diabetic Foot Infections Stevens-Johnson Syndrome Upper Respiratory Tract Infection Otitis Externa Fast Five Quiz: Tattoos Impetigo Group A Streptococcal (GAS) Infections News & Perspective EMA Backs New Antibiotic for Complex Bacterial Infections Infections Down in Acute Care Hospitals Acute Bacterial Arthritis in Children: New Guidelines What's New in Psoriasis? 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16254	https://rheumatology.org/clinical-practice-guidelines	Global Community Join Log In Go Home Clinical Tools & Guidelines Clinical Practice Guidelines Call for Letters of Interest Axial Spondyloarthritis Glucocorticoid-Induced Osteoporosis Gout Integrative RA Treatment Interstitial Lung Disease Juvenile Idiopathic Arthritis Lupus Lyme Disease Optimal Timing of Total Hip and Knee Arthroplasty Osteoarthritis Perioperative Management Polymyalgia Rheumatica Psoriatic Arthritis Reproductive Health in Rheumatic Diseases Rheumatoid Arthritis Vaccinations Vasculitis Clinical Guidance Guidance Documents COVID-19 Guidance Musculoskeletal Ultrasound Clinical Tools Medication Guides Quality Measurement Pediatric to Adult Rheumatology Care Transition Criteria Call for Criteria Proposals Clinical Practice Guidelines The American College of Rheumatology (ACR) places a high priority on developing methodologically rigorous, evidence-based clinical practice guidelines that take into consideration the expertise and viewpoints of multiple stakeholders in a transparent fashion. Individuals and organizations interested in quality care for patients with rheumatic disease are encouraged to review the information on this page and provide input into ACR guideline activities. About ACR Clinical Practice Guidelines Clinical practice guidelines and recommendations are developed to: Reduce inappropriate care Minimize geographic variations in practice patterns Enable effective use of health care resources Guidelines and recommendations developed and/or endorsed by the ACR are intended to provide guidance for patterns of practice and not to dictate the care of a particular patient. The ACR considers adherence to these guidelines and recommendations to be voluntary, with the ultimate determination regarding their application to be made by the clinician in light of each patient's individual circumstances. Guidelines and recommendations are intended to promote beneficial or desirable outcomes but cannot guarantee any specific outcome. Guidelines and recommendations developed and/or endorsed by the ACR are subject to periodic revision as warranted by the evolution of medical knowledge, technology, and practice. Available ACR Guidelines Axial SpondyloarthritisGlucocorticoid-Induced OsteoporosisGoutIntegrative RA TreatmentInterstitial Lung DiseaseJuvenile Idiopathic ArthritisLupusLyme DiseaseOptimal Timing of Total Hip/Knee ArthroplastyOsteoarthritisPerioperative ManagementPolymyalgia RheumaticaPsoriatic ArthritisReproductive HealthRheumatoid ArthritisVaccinationsVasculitis Clinical Practice Guideline Project Steps Step 1: Define project scope and identify team Step 2: Identify important clinical questions and outcomes Step 3: Obtain feedback on project plan via public comment Step 4: Conduct literature review Step 5: Consider evidence in light of clinical expertise and experience and draft recommendations Step 6: Draft guideline manuscript Step 7: Peer review by ACR Step 8: Post guideline summary Step 9: Peer review by journal Step 10: Final guideline published and disseminated Step 11: Update literature searches and reevaluate the need for updating and revising Guideline Policies and Procedures To promote transparency and assist ACR guideline development groups in their work, the ACR has outlined its policies and procedures for guideline development and maintenance in the ACR Guideline Manual. Patient Involvement Patients are active participants in ACR guideline development. For each guideline, a Patient Panel is convened and provides input just before the Voting Panel meets to decide the guideline recommendations. At least two patients from the Patient Panel sit on the Voting Panel, tasked with sharing patient perspectives during the discussions and decision-making. Learn more about the ACR’s initial pilot project to test its current process for patient involvement in guideline work. When Patients Write the Guidelines: Patient Panel Recommendations for the Treatment of Rheumatoid Arthritis When Patients Write the Guidelines Appendix: Evidence Reports Patient Care Roles of the Interdisciplinary Rheumatology Team To work together effectively, clinicians and rheumatology professionals need to understand the roles of every member of the interdisciplinary team. Explore our new series of fact sheets that describe the modern roles of members of the interdisciplinary care team. Explore Fact Sheets We use cookies on our website to improve our service to you and for security purposes. By continuing to use our site without changing your browser cookie settings, you agree to our cookie policy and the use of cookies. See ACR Policies
16255	https://www.nature.com/articles/s41467-023-44165-3	Anomalous enhancement of thermoelectric power factor in multiple two-dimensional electron gas system \| Nature Communications Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Thank you for visiting nature.com. You are using a browser version with limited support for CSS. 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Advertisement View all journals Search Search Search articles by subject, keyword or author Show results from Search Advanced search Quick links Explore articles by subject Find a job Guide to authors Editorial policies Log in Explore content Explore content Research articles Reviews & Analysis News & Comment Videos Collections Subjects Follow us on Facebook Follow us on Twitter Sign up for alerts RSS feed About the journal About the journal Aims & Scope Editors Journal Information Open Access Fees and Funding Calls for Papers Editorial Values Statement Journal Metrics Editors' Highlights Contact Editorial policies Top Articles Publish with us Publish with us For authors For Reviewers Language editing services Open access funding Submit manuscript Sign up for alerts RSS feed nature nature communications articles article Anomalous enhancement of thermoelectric power factor in multiple two-dimensional electron gas system Download PDF Download PDF Article Open access Published: 16 January 2024 Anomalous enhancement of thermoelectric power factor in multiple two-dimensional electron gas system Yuto UematsuORCID: orcid.org/0009-0001-8161-62581, Takafumi IshibeORCID: orcid.org/0000-0002-8662-875X1, Takaaki ManoORCID: orcid.org/0000-0002-6955-260X2, Akihiro OhtakeORCID: orcid.org/0000-0002-3519-46132, Hideki T. MiyazakiORCID: orcid.org/0000-0003-4152-11712, Takeshi KasayaORCID: orcid.org/0000-0002-1976-87602& … Yoshiaki NakamuraORCID: orcid.org/0000-0002-5387-16301 Show authors Nature Communicationsvolume 15, Article number:322 (2024) Cite this article 5018 Accesses 11 Citations 76 Altmetric Metrics details Abstract Toward drastic enhancement of thermoelectric power factor, quantum confinement effect proposed by Hicks and Dresselhaus has intrigued a lot of researchers. There has been much effort to increase power factor using step-like density-of-states in two-dimensional electron gas (2DEG) system. Here, we pay attention to another effect caused by confining electrons spatially along one-dimensional direction: multiplied 2DEG effect, where multiple discrete subbands contribute to electrical conduction, resulting in high Seebeck coefficient. The power factor of multiple 2DEG in GaAs reaches the ultrahigh value of ~100 μWcm−1 K−2 at 300 K. We evaluate the enhancement rate defined as power factor of 2DEG divided by that of three-dimensional bulk. The experimental enhancement rate relative to the theoretical one of conventional 2DEG reaches anomalously high (~4) in multiple 2DEG compared with those in various conventional 2DEG systems (~1). This proposed methodology for power factor enhancement opens the next era of thermoelectric research. Similar content being viewed by others Competing electronic states emerging on polar surfaces Article Open access 25 July 2022 High-precision digital terahertz phase manipulation within a multichannel field perturbation coding chip Article 09 August 2021 Transport behaviors of topological band conduction in KTaO 3’s two-dimensional electron gases Article Open access 29 December 2022 Introduction Human beings have been seeking a powerful solution to the energy crisis. Thermoelectric (TE) material, which enables the direct conversion between waste heat and electricity, is attracting worldwide interests as one of the sustainable power sources1,2. The TE performance is quantified by a dimensionless figure-of-merit ZT; ZT = S 2 σT/κ, where S is Seebeck coefficient, σ is electrical conductivity, κ is thermal conductivity, T is absolute temperature, and S 2 σ is power factor (PF). The ZT increase has been done by two approaches: κ reduction or PF enhancement3,4,5,6,7,8,9,10,11,12,13,14,15. In 2000s, nanostructuring approach intensified interface phonon scattering, decreasing κ drastically. Some studies achieved 100–200 times smaller κ by introducing nanostructures, making a big impact on TE research3,4,5,6,7,8. On the other hand, in the ever-reported methodologies of PF enhancement, the enhancement rate of PF is achieved to be several times (2–3 times for energy filtering effect10,11,12, 1.5–2 times for resonant scattering effect13,14). Epoch-making methodologies for PF enhancement have been expected for further increase in thermoelectric performance. In 1993, Hicks and Dresselhaus proposed the concept of PF enhancement by quantum confinement effect16; e.g. step-like density of states (DOS) in two-dimensional electron gas (2DEG) system increases S (step-like DOS effect) (Supplementary Note1). Since then, much effort has been made to experimentally demonstrate PF enhancement by quantum confinement effect17,18,19,20,21,22. In 2018, Zhang et al. experimentally observed an evident feature of 2DEG in SrTiO 323: the phenomenon of S enhancement brought by decreasing t 2DEG/λ, where t 2DEG is 2DEG channel thickness and λ is the thermal de Broglie wavelength23,24,25. Furthermore, PF enhancement has been tried by a combination of step-like DOS effect for high S and modulation doping effect for high carrier mobility μ (Fig.1a)17,21. Toward further high enhancement rate R 2D/3D defined as R 2D/3D = PF 2DEG/PF 3D, where PF 2DEG is PF of 2DEG and PF 3D is PF of three-dimensional (3D) bulk, it is strongly demanded to obtain more drastic increase of R 2D/3D as a function of t 2DEG/λ than theoretical function R 2D/3D ((R 2D/3D)th) reported in the previous study24 (Fig.1d). Although step-like DOS effect by quantum confinement has been spotlighted so far, we pay attention to another effect caused by quantum confinement effect: multiple discrete subbands with step-like DOS. Provided that multiple subbands with step-like DOS at higher energy, which are formed by quantum confinement in two-dimensional electron gas (2DEG) systems26,27, contributed to electrical conduction, S would be substantially enhanced because the participation rate of higher-energy carriers in the carrier conduction becomes larger (Fig.1b, c); we call multiplied two-dimensional electron gas effect (M2DE). In this study, we choose GaAs as a material to demonstrate M2DE. Therein, quantum confinement effect easily appears because t 2DEG/λ of GaAs becomes sufficiently small for 2DEG even in relatively large t 2DEG due to its relatively long λ. In addition, GaAs, which is applied to photonic devices such as vertical cavity surface emitting laser for smart phone, is an ideal material in terms of social application. Fig. 1: Power factor S 2 σ enhancement by multiplied 2DEG effect (M2DE). a Schematic illustration of single 2DEG (S-2DEG) in rectangular quantum well (RQW), where modulation doping effect increases carrier mobility μ and step-like density-of-states (DOS) effect originated in quantum confinement effect increases Seebeck coefficient S. b Schematic illustration of multiple 2DEG (M-2DEG) in triangular quantum well (TQW), where M2DE bringing high S appears in addition to modulation doping effect for high μ and step-like DOS effect for high S. c Schematic illustration of S 2 σ enhancement by three effects: modulation doping effect, step-like DOS effect, and M2DE. d The enhancement rate of S 2 σ (R 2D/3D) as a function of the 2DEG channel thickness/de Broglie wavelength. The solid triangles, the solid squares, and the open marks are R 2D/3D values of M-2DEG with M2DE (This study), S-2DEG without M2DE or with almost no M2DE (This study), and 2DEG without M2DE or with almost no M2DE (Preceding studies by other groups: PbTe RQW17 (the open circles), PbTe RQW18 (the open triangles), Si RQW19 (the open diamonds), SiGe RQW21 (the open squares)), respectively. The solid line represents the theoretical R 2D/3D without M2DE (R 2D/3D)th24 which is consistent with the data of S-2DEG and preceding data by the other groups. The broken line denotes R 2D/3D = 1 corresponding to the performance of 3D materials. e Experimental R 2D/3D (R 2D/3D)ex divided by theoretical R 2D/3D without M2DE (R 2D/3D)th. In this work, GaAs TQW (the red star) and GaAs RQW (the red square). In the preceding results, GaN TQW30,31,32 (the blue triangles), PbTe RQW17 (the purple square), SiGe RQW21 (the brown square), ZnO TQW33 (the light blue triangle), SrTiO 3 TQW20 (the pink triangle), and SrTiO 3 RQW20,22 (the pink squares). Full size image Here, we demonstrate that M2DE brings drastic PF enhancement as follows. We form GaAs triangular quantum well (TQW) with M2DE in addition to modulation doping effect and step-like DOS effect (Fig.1b, c)28. TQW samples exhibit higher S than rectangular quantum well (RQW) samples without M2DE or with almost no M2DE when comparing S values under the situation that the channel width t ch of TQW is equal to the well width t well of RQW. This indicates that multiple 2DEG (M-2DEG) in TQW with M2DE is more promising than conventional single 2DEG (S-2DEG) in RQW without M2DE. The PF of M-2DEG reaches the maximum value of ~100 μW cm−1 K−2 at n of ~1×10 18 cm−3 at 300 K, which is in a class of ultrahigh PF. Thanks to M2DE, M-2DEG shows more drastic increase of R 2D/3D with decreasing t 2DEG/λ than S-2DEG (Fig.1d). The experimental R 2D/3D ((R 2D/3D)ex) relative to the theoretical R 2D/3D without M2DE ((R 2D/3D)th) is anomalously high in M-2DEG compared with those in various conventional 2DEG systems (~1) (Fig.1e)17,20,21,22,23,24,29,30,31,32,33. Therein, the layered materials are excluded owing to the difficulty in discussing the contribution of M2DE in the layered materials because the electronic band structure related to the layer number34,35 influences on the TE properties. This proposed methodology for PF enhancement opens the next era of TE research. Results Sample structures and calculated energy band diagrams The TQW and RQW samples were formed for M-2DEG and S-2DEG, respectively, using molecular beam epitaxy (MBE). Illustrations of sample structures and simple band diagrams are shown in Fig.2a, b, where conduction band bottom of 3D GaAs (E c), carrier energy (E) and the bottom energy of i- th subband (E i). The index i (i = 1, 2,…) is the subband number, where the subband bottom with the smaller number of i positions at the lower energy level. In general RQW, the energy difference between discrete subband bottoms (E i+1-E i) is monotonically increasing with increase in the i value. Therefore, unlike TQW, it is expected that one subband (or two subbands) can only exist in the present AlGaAs/GaAs/AlGaAs RQW with ~0.2 eV barrier height when the step-like DOS appears due to the sufficiently small t well, indicating S-2DEG system (Supplementary Note2). Modulation doping was performed for both samples by inserting Si-doped Al 0.3 Ga 0.7 As layers as carrier suppliers. In TQW and RQW, 2DEG channels were formed at the interfaces of undoped GaAs/Al 0.3 Ga 0.7 As spacer and in the quantum well of GaAs layers sandwiched by two Al 0.3 Ga 0.7 As layers, respectively. In TQW, electron Hall concentration n values of channels were tuned by controlling the thicknesses of spacer layers t sp (0, 2, 30, 60, and 90 nm). The control of t sp also changed the energy band structure36, bringing the t ch variation from 8 to 18 nm. In RQW, t well was controlled from 3 to 12 nm. Fig. 2: Sample structure illustrations, calculated energy band diagrams, and theoretical demonstration of multiplied 2DEG effect (M2DE). a, b Illustrations of sample structures and energy band diagrams of rectangular quantum well (RQW) (a) and triangular quantum well (TQW) samples (b)48. c–f Calculated energy band diagrams of RQW with the well width t well = 4 nm (c), 12 nm (d) and TQW samples with the channel width t ch = 8 nm (e), 15 nm (f). The solid black lines: conduction band bottom of 3D GaAs (E c), the broken black lines: Fermi energy E F, the solid red lines: the bottom energy of i-th subband (E i) (for simplicity, E i with i< 7 are displayed), and the solid blue line: the calculated carrier distribution n cal as a function of z. z is the distance from the interface of undoped GaAs/AlGaAs spacer along the direction perpendicular to the sample surface. g The carrier occupation ratio R O as a function of E i-E F in the TQW sample with t ch = 15 nm. h Calculated Seebeck coefficient S as a function of i m in the TQW sample with t ch = 15 nm, when the contribution of the i- th subband is considered until i m. Full size image We reveal that multiple subbands can contribute to electrical conduction in TQW, not in RQW. As examples of calculation model, we consider the samples of RQW with t well = 4, 12 nm (Fig.2c, d), and TQW with t ch = 8, 15 nm (Fig.2e, f). The energy band diagrams and the calculated carrier distribution, n cal were obtained by self-consistent computation using one-dimensional Poisson-Schrödinger equation37. It was found that the TQW is formed at the interface of undoped GaAs/AlGaAs spacer. Therein, multiple subbands locate near Fermi energy E F. For example, some subbands locate in the range of E-E F< ~ 0.1 eV (Fig.2e, f) in the TQW unlike only one or two subbands in the RQW (Fig.2c, d, Supplementary Note2). Theoretical demonstration of M2DE To clarify the contribution of carrier existing at each subband to electrical conduction, we calculated the occupation ratio R O defined as R O=n i/n t, where n i is sheet carrier concentration at the i- th subband and n t is the sum of n i. The expressions for n i and n t are described as follows: $${n}{{{{{{\rm{i}}}}}}}={\int }{{E}{{{{{{\rm{i}}}}}}}}^{\infty }{f}{0}\left(E\right){D}_{{{{{{\rm{i}}}}}}}\left(E\right){{{{{{\rm{d}}}}}}E}$$ (1) $${n}{{{{{{\rm{t}}}}}}}=\mathop{\sum}\limits{{{{{{\rm{i}}}}}}}{n}_{{{{{{\rm{i}}}}}}}$$ (2) where f 0(E) is the Fermi–Dirac distribution function. D i(E) is DOS at the i- th subband, which is described as m/π ℏ 2. Therein, m is effective mass of carrier and ℏ is Dirac constant. It is found that R O is the function of E i from Eqs.(1) and (2). Figure2g shows R O at the i- th subband. As E i-E F increased, R O decreased nearly exponentially, which is coming from the energy dependence of the Fermi–Dirac distribution function. However, some R O values at the i- th subband (i> 1) seem to be relatively high. This implies that multiple subbands can contribute to electrical conduction38. Thus, it is expected that M2DE can appear in TQW. We theoretically demonstrate S enhancement by M2DE in TQW. As an example of calculation model, we consider the sample with t ch = 15 nm. Theoretical S was calculated under parabolic band for 2DEG and bulk, and relaxation time approximations on the basis of Boltzmann transport theory (details available in Methods). In the summation of i-th subband in the calculation, it is enough to consider up to the maximum i- th subband contributing to electrical conduction although it is ideal to consider up to infinity. Therefore, when the contribution of the i- th subband is summated until i m, we investigated the relationship between S and i m (Fig.2h), which was calculated using physical parameters39,40 displayed in Table1. S was saturated in the range of i m> 20 because of less contribution of subbands with i> 20 to electrical conduction. This saturation indicates that i m of 20 is critical value (i C) to calculate S accurately. Namely, it is enough to calculate S using the i of less than i C (in this case, 20). In this study, calculations of TE properties for TQW samples with various t ch were also performed with i m ~ i C for sufficient calculation accuracy (Supplementary Note3). In the sample with t ch = 15 nm (Fig.2h), the saturated S value in the calculation with multiple subbands (i m> 20) was ~1.7 times higher than that in the calculation with single subband (i = 1), namely the calculation without M2DE. This theoretically proves that M2DE substantially enhances S. Table 1 Parameters used in the calculation Full size table Thermoelectric properties Experimental and calculated TE properties of M-2DEG and S-2DEG are shown in Fig.3. Therein, theoretical calculation of S and μ was performed under parabolic band and relaxation time approximations on the basis of Boltzmann transport theory41. The details of carrier scattering models and used parameters are written in the section of Numerical calculation and Table1 respectively. Figure3a, b show S and μ as a function of n at 300 K, respectively. When estimating n of M-2DEG in TQW, we defined the t ch as FWHM of the carrier concentration distribution along the perpendicular direction to substrate surface (Supplementary Note4)20. In M-2DEG (TQW), n was tuned by controlling t sp. As shown in Fig.3a, when decreasing t sp (t ch), n was increased because of increase of carrier supply from Si-doped Al 0.3 Ga 0.7 As layers. The S values of M-2DEG (the solid red triangles) were compared with that of 3D GaAs film (the solid black circle) that does not have modulation doping effect, step-like DOS effect, and M2DE. We plotted the calculation curve of 3D GaAs42 which reproduces the experimental value of 3D GaAs film. When comparing them at the same n, M-2DEG exhibited higher S than the calculation curve of 3D GaAs. To demonstrate S enhancement by M2DE experimentally, we measured S values of conventional S-2DEG in RQW samples (the solid blue squares) without M2DE or with almost no M2DE for comparison. When varying t well from 12 to 3 nm, S values of S-2DEG were gradually increased because of step-like DOS effect. This tendency was well reproduced by the S calculation for S-2DEG (the open blue squares). Thus, not only M2DE but also step-like DOS effect causes S enhancement, making it difficult to understand the physical mechanism of S enhancement. To discuss the difference between the two effects, let us compare S values of M-2DEG with those of S-2DEG. At almost the same n, the M-2DEG with t ch of ~8 nm exhibited higher S than S-2DEG with t well of ~8 nm, while in the stronger confinement case of small width (t well ~ 3 nm) in RQW, high S was obtained to be comparable to that in the case of 8 nm width in TQW. This is because S enhancement appears in M-2DEG (TQW) over a wide range of confinement width, unlike S-2DEG (RQW) with strong t well dependence, which is also confirmed by the calculation (Supplementary Note5). Furthermore, the S calculation (the open yellow triangles) including M2DE in M-2DEG agreed with the experimental n-S data (Fig.3a) and T–S data (Supplementary Note6), which is the theoretical evidence that M2DE appears. Thus, S enhancement by M2DE was demonstrated both experimentally and theoretically. Fig. 3: Thermoelectric properties. a, b Carrier concentration n dependences of Seebeck coefficient S (a) and carrier mobility μ (b) measured at 300 K in multiple 2DEG (M-2DEG) in triangular quantum well (TQW) with multiplied 2DEG effect (M2DE) (the solid red triangles), single 2DEG (S-2DEG) in rectangular quantum well (RQW) without M2DE or with almost no M2DE (the solid blue squares), respectively. The calculation data for M-2DEG (the open yellow triangles) and S-2DEG (the open blue squares) are also plotted simultaneously. For comparison with the data in 3D GaAs without 2DEG, the experimental value (the solid black circle) and calculation curves (the broken lines) of 3D GaAs are simultaneously plotted. The channel width t ch of M-2DEG and the well width t well of S-2DEG are displayed around the experimental data points. The inset in (a) is an enlarged n-S plot: experimental and calculated n dependences of S in S-2DEG. c Temperature T dependences of μ in M-2DEG with t ch = 8 (the solid diamonds) and 15 nm (the solid triangles), 3D GaAs film without 2DEG (the solid circles). We also simultaneously plotted the calculated T-μ curves of M-2DEG with t ch = 8 (the open diamonds) and 15 nm (the open triangles). d, en dependences of electrical conductivity σ (d) and power factor S 2 σ (e) at 300 K in M-2DEG with M2DE (the solid triangles), S-2DEG without M2DE or with almost no M2DE (the solid squares), respectively. The experimental data (the solid circles) and the calculation curves (the broken lines) of 3D GaAs are simultaneously plotted. The dotted curves in (e) are eye-guides for M-2DEG (red) and S-2DEG (blue). In (e), the t ch of M-2DEG and t well of S-2DEG are displayed around the experimental data points. The insets show the density of states (DOS) of M-2DEG (with M2DE) and S-2DEG (without M2DE). Full size image As well as S, μ values of M-2DEG were compared with the calculation curve of 3D GaAs (Fig.3b). When comparing them at the same n, M-2DEG with modulation doping effect exhibited higher μ than the calculation curve of 3D GaAs without modulation doping effect which reproduces the experimental value of 3D GaAs film. Furthermore, the experimental μ data agreed with the μ calculation including M2DE in addition to modulation doping effect and step-like DOS effect for M-2DEG. We also obtained T–μ data in the T range of 80-300 K (Fig.3c). The experimental T-μ data of M-2DEG with t ch of 8 nm were compared with those of 3D GaAs, where compared samples had almost the same n of ~1 × 10 18 cm−3 at 300 K. Then, μ values of the M-2DEG drastically increased as T decreased, while μ of 3D GaAs film did not depend on the T. The tendency of experimental data in the M-2DEG was explained by the theoretical T–μ curve of M-2DEG (the open marks in Fig.3c), where the dominant scattering is polar optical phonon scattering due to the almost no ionized impurity scattering unlike 3D GaAs with ionized impurities. The orders of magnitude higher mobility at low temperature is reported as the result from modulation doping effect43,44. These results strongly support that the modulation doping effect, step-like DOS effect, and M2DE appear in M-2DEG. On the other hand, as shown in Fig.3b, S-2DEG with modulation doping effect also exhibited higher μ than the calculation curve of 3D GaAs without modulation doping effect at the same n. When varying t well from 12 to 3 nm, μ values of S-2DEG were monotonically decreased because of increase of interface carrier scattering rate. This tendency was well reproduced by the μ calculation for S-2DEG. Thus, S-2DEG has a trade-off relationship between S and μ with respect to t well, making it difficult to realize ultrahigh PF. In contrast, high μ values of M-2DEG did not depend on the t ch within the range of 8–18 nm. Therefore, M2DE is expected along with the high μ of ~6000 cm 2 V−1 s−1. Namely, M-2DEG has a high potentiality of exhibiting ultrahigh PF by simultaneous enhancement of S and μ. Figure3d shows σ as a function of n at 300 K. M- and S-2DEG exhibited higher σ values than 3D GaAs at almost the same n because of higher μ. As for the σ tendency against n, there was a significant difference between M- and S-2DEG; σ of M-2DEG increased as n increased, while σ of S-2DEG did not depend on the n. The increasing σ tendency of M-2DEG is explained by constant μ tendency against n (Fig.3b). On the other hand, constant σ tendency of S-2DEG is attributed to the drastically-decreasing μ tendency against n. When σ values of M-2DEG with t ch of 8 nm were compared with those of S-2DEG with t well of 3 nm, where both samples exhibited the equivalent S values at almost the same n, M-2DEG had approximately 3 times higher σ than S-2DEG. This indicates that M-2DEG is more promising than S-2DEG in terms of simultaneous realization of high S and high σ. Experimentally observed anomalous power factor enhancement in M-2DEG Figure3e shows PF as a function of n at 300 K. Both M- and S-2DEG exhibited higher PF than 3D GaAs at almost the same n. When decreasing t well from 12 to 3 nm in RQW, PF values of S-2DEG increased monotonically because S was substantially increased by step-like DOS effect. A remarkable fact is that M-2DEG always exhibited much higher PF than S-2DEG because of M2DE. In Fig.3a, b and e, at the ~(1-2) × 10 18 cm−3, higher PF in M-2DEG comes from higher μ in TQW, where S values of TQW and RQW are comparable, while at ~4 × 10 17 cm−3, higher PF in M-2DEG is due to higher S in TQW, where μ values of TQW and RQW are comparable. This is because there is a trade-off relationship between S and μ in RQW. On the other hand, S and μ are simultaneously enhanced in TQW with M2DE. As a result, the maximum PF of M-2DEG reached ~100 μW cm−1 K−2 at n of ~1 × 10 18 cm−3 at 300 K, which is in a class of ultrahigh PF. Thanks to the S enhancement by M2DE along with the high μ, M-2DEG showed more drastic increase of R 2D/3D with t 2DEG/λ decrease than S-2DEG (Fig.1d). As a result, M-2DEG exhibited the highest (R 2D/3D)ex/(R 2D/3D)th among various 2DEG systems (Fig.1e), which was anomalously high like singularity compared with those in various 2DEG systems (~1). This highlights that M2DE can bring ultrahigh PF beyond conventional 2DEG. Discussion In summary, we demonstrated that M2DE caused by the quantum confinement effect brings drastic PF enhancement. M-2DEG with M2DE in addition to modulation doping effect and step-like DOS effect exhibited higher S than conventional S-2DEG without M2DE or with almost no M2DE over a wide range of confinement width. The PF of M-2DEG reached the maximum value of ~100 μW cm−1 K−2 at n of ~1 × 10 18 cm−3 at 300 K, which is in a class of ultrahigh S 2 σ. Thanks to M2DE, M-2DEG exhibited the highest (R 2D/3D)ex/(R 2D/3D)th among various conventional 2DEG systems except for the layered materials with the electronic band structure depending on the layer number (Fig.1e). This value was anomalously high like singularity compared with those in various conventional 2DEG systems. This study presented the methodology enabling the drastic PF enhancement based on quantum confinement effect, which opens the next era of TE research. Methods Sample preparation TQW samples were formed using MBE in the following process. To obtain clean surfaces of undoped GaAs(001) substrates, undoped GaAs (300 nm) initial layers were grown on the GaAs substrates. Subsequently, as the buffer layers, GaAs/Al 0.3 Ga 0.7 As superlattice layers were grown on the undoped GaAs (300 nm)/GaAs substrates by alternately depositing GaAs (10 nm) and Al 0.3 Ga 0.7 As (10 nm) 20 times. On the buffer layers, undoped GaAs (1000 nm) layers with high crystallinity were grown. These layers were grown at 893 K. After the growth of Al 0.3 Ga 0.7 As spacer layers on the undoped GaAs (1000 nm) layers at 893 K, Si-doped Al 0.3 Ga 0.7 As (dopant concentration: 5 × 10 17 cm−3, thickness: 80 nm) layers were grown at 823 K to supply carrier to the interface of undoped GaAs (1000 nm)/Al 0.3 Ga 0.7 As spacer. The n was tuned by controlling t sp (0, 2, 30, 60, and 90 nm). Finally, to prevent the oxidation of samples, the sample surfaces were capped by depositing GaAs layers (10 nm) at 823 K. For reference, RQW samples without M2DE or with almost no M2DE were formed. Undoped GaAs (500 nm) layers were grown on the GaAs(001) substrates. Subsequently, as the buffer layers, GaAs/AlAs superlattice layers were grown on the undoped GaAs (500 nm)/GaAs substrates by alternately depositing GaAs (2 nm) and AlAs (2 nm) 100 times. On the buffer layers, undoped Al 0.3 Ga 0.7 As (20 nm) barrier layers, GaAs (3, 4, 5, 6, 8, 10, and 12 nm) layers, and Al 0.3 Ga 0.7 As (2, 10 nm) spacer layers were grown in a sequential order. The growths of these layers were carried out at 873 K. After that, as the carrier suppliers to GaAs wells, Si-doped Al 0.3 Ga 0.7 As (dopant concentration: 7 × 10 17 cm−3, thickness: 80 nm) layers were grown at 813 K on the Al 0.3 Ga 0.7 As (2 or 10 nm) spacer layers. Finally, to prevent the oxidation of samples, capping GaAs layers (10 nm) were formed at 813 K. Thermoelectric property measurements The stacked structures of AuGe/Ni/Au were formed on the samples as electrodes. To make ohmic contact, the samples were annealed at 723 K for 90 s. Sheet electrical conductivity and sheet carrier concentration were measured using the van der Pauw method and Hall effect measurement, respectively. σ and n are obtained by dividing measured sheet electrical conductivity and sheet carrier concentration by t well or t ch20. In our Hall effect measurement, we used 2401 sourcemeter (Keithley) as source measure unit, and the range of magnetic field is from −0.5 T to 0.5 T. Therein, the errors of n and μ are about 13%. S was measured using ZEM-3 (ADVANCE RIKO Inc.)45,46, where the temperature difference is applied along the in-plane direction, and the differences of temperatures and the electric voltages between two points on the films were obtained by thermocouple probes. The contribution of Si-doped Al 0.3 Ga 0.7 As layer to electrical conduction was removed using the parallel conduction model (Supplementary Note7). Numerical calculation Theoretical T–μ, n–μ, and n–S curves were calculated under effective mass and relaxation time approximations on the basis of Boltzmann transport theory as follows: $$S=-\frac{1}{{eT}}\frac{{\sum }{{{{{{\rm{i}}}}}}}{\int }{-\infty }^{\infty }\left(E-{E}{{{{{{\rm{i}}}}}}}\right)\left(E-{E}{F}\right)\frac{\partial {f}{0}}{\partial E}{\tau }{{{{{{\rm{i}}}}}}}\left(E-{E}{{{{{{\rm{i}}}}}}}\right){D}{{{{{{\rm{i}}}}}}}\left(E-{E}{{{{{{\rm{i}}}}}}}\right){{{{{{\rm{d}}}}}}E}}{{\sum }{{{{{{\rm{i}}}}}}}{\int }{-\infty }^{\infty }\left(E-{E}{{{{{{\rm{i}}}}}}}\right)\frac{\partial {f}{0}}{\partial E}{\tau }{{{{{{\rm{i}}}}}}}\left(E-{E}{{{{{{\rm{i}}}}}}}\right){D}{{{{{{\rm{i}}}}}}}\left(E-{E}_{{{{{{\rm{i}}}}}}}\right){{{{{{\rm{d}}}}}}E}}$$ (3) $$\mu=\frac{e}{m}\frac{\mathop{\sum}\limits_{{{{{{\rm{i}}}}}}}{\int }{-\infty }^{\infty }\left(E-{E}{{{{{{\rm{i}}}}}}}\right)\frac{\partial {f}{0}}{\partial E}{\tau }{{{{{{\rm{i}}}}}}}\left(E-{E}{{{{{{\rm{i}}}}}}}\right){D}{{{{{{\rm{i}}}}}}}\left(E-{E}{{{{{{\rm{i}}}}}}}\right){{{{{{\rm{d}}}}}}E}}{\mathop{\sum}\limits{{{{{{\rm{i}}}}}}}{\int }{-\infty }^{\infty }\left(E-{E}{{{{{{\rm{i}}}}}}}\right)\frac{\partial {f}{0}}{\partial E}{D}{{{{{{\rm{i}}}}}}}\left(E-{E}_{{{{{{\rm{i}}}}}}}\right){{{{{{\rm{d}}}}}}E}}$$ (4) where e is the elementary charge and τ i is the total carrier relaxation time at the i- th subband. D i(E) was simply assumed as a step function. 1/τ i is described as the sum of each scattering rate at the i- th subband through Matthissen’s rule as follows: 1/τ i = 1/τ POP + 1/τ ADP + 1/τ RII + 1/τ IFR, where 1/τ POP is polar optical phonon (POP) scattering rate, 1/τ ADP is acoustic deformation potential (ADP) scattering rate, 1/τ RII is remote ionized impurity (RII) scattering rate, 1/τ IFR_rec is interfacial roughness (IFR) scattering rate in RQW, and 1/τ IFR_tri is IFR scattering rate in TQW. Inter-subband and intra-subband scatterings are both considered by choosing proper wave functions in POP and ADP scattering calculations. Each scattering rate is described as follows39,47,48,49,50: $$\frac{1}{{\tau }{{{{{{\rm{POP}}}}}}}}=\frac{{e}^{2}m\hslash {\omega }{{{{{{\rm{LO}}}}}}}}{8{\pi }^{2}{\hslash }^{3}{\varepsilon }{0}}\left(\frac{1}{{\kappa }{\infty }}-\frac{1}{{\kappa }{0}}\right)\frac{1}{1-{f}{0}\left(E\right)}\mathop{\sum}\limits_{{{{{{\rm{j}}}}}}}\left(\left[1-{f}{0}\left(E+\hslash {\omega }{{{{{{\rm{LO}}}}}}}\right)\right]{N}{q}\int \frac{{{\|I}\left({q}{z}\right)\|}^{2}}{{q}{+}^{2}+{q}{z}^{2}}d{q}{z} \right. \ \left.+\left[1-{f}{0}\left(E-\hslash {\omega }{{{{{{\rm{LO}}}}}}}\right)\right]u\left(E-\hslash {\omega }{{{{{{\rm{LO}}}}}}}\right)({N}{q}+1)\int \frac{{{\|I}\left({q}{z}\right)\|}^{2}}{{q}{-}^{2}+{q}{z}^{2}}d{q}_{z}\right)$$ (5) $$\frac{1}{{\tau }{{{{{{\rm{ADP}}}}}}}}=\frac{{D}{{{{{{\rm{A}}}}}}}^{2}{k}{{{{{{\rm{B}}}}}}}T}{2\hslash {c}{{{{{{\rm{L}}}}}}}}\mathop{\sum}\limits_{{{{{{\rm{j}}}}}}}\left{{\int }{-\infty }^{\infty }{\varphi }{{{{{{\rm{i}}}}}}}\left(z\right){\varphi }{{{{{{\rm{j}}}}}}}\left(z\right){{{{{{\rm{d}}}}}}z}\right}{D}{{{{{{\rm{j}}}}}}}\left(E\right)$$ (6) $$\frac{1}{{\tau }{{{{{{\rm{RII}}}}}}}} !=! {\int }{{t}{{{{{{\rm{sp}}}}}}}}^{{t}{{{{{{\rm{sp}}}}}}}+{t}{{{{{{\rm{dope}}}}}}}}\left[\frac{1}{2}{n}{{{{{{\rm{imp}}}}}}}\left(\frac{m}{\pi {\hslash }^{3}}\right){\left(\frac{{e}^{2}}{2{\kappa }{0}{\varepsilon }{0}}\right)}^{2}{\int }{0}^{2\pi }\frac{\exp \left(-2q\cdot z\right)}{{\left(q+{q}{{{{{{\rm{TF}}}}}}}\right)}^{2}}\left(1-\cos \theta \right){{{{{\rm{d}}}}}}\theta \right]{{{{{{\rm{d}}}}}}z}$$ (7) $$\frac{1}{{\tau }{{{{{{\rm{IFR}}}}}}_{{{{{\rm{rec}}}}}}}}=\frac{4\pi m{E}{{{{{{\rm{i}}}}}}}^{2}{\Delta }^{2}{\Lambda }^{2}}{{\left(t+\sqrt{\frac{2{\hslash }^{2}}{m\left({V}{0}-{E}{{{{{{\rm{i}}}}}}}\right)}}\right)}^{2}{\hslash }^{3}}\cdot \frac{1}{2\pi }{{{{{\rm{exp}}}}}} \left(-\frac{{\Lambda }^{2}\left(\frac{2m\left(E-{E}{{{{{{\rm{i}}}}}}}\right)}{{\hslash }^{2}}\right)\left(1-\cos \theta \right)}{2}\right)\left(1-{{{{{\rm{cos}}}}}} \theta \right)\frac{q}{q+\frac{2}{{a}{{{{{{\rm{B}}}}}}}}}$$ (8) $$\frac{1}{{\tau }{{{{{{\rm{IFR}}}}}}_{{{{{\rm{tri}}}}}}}}=\frac{m{\Delta }^{2}{\Lambda }^{2}{e}^{2}{\left(e\left(\frac{n}{2}+{n}{{{{{{\rm{depl}}}}}}}\right)\right)}^{2}}{2{\hslash }^{3}{\left({\kappa }{0}{\varepsilon }{0}+\frac{1}{q}\cdot \frac{{e}^{2}m}{2\pi {\hslash }^{2}}\cdot F\left(q\right)\right)}^{2}}{{{{{\rm{exp}}}}}} \left(-\frac{{\Lambda }^{2}{q}^{2}}{4}\right)\left(1-{{{{{\rm{cos}}}}}} \theta \right)$$ (9) where ℏ ω LO is the longitudinal optical phonon energy, ε 0 is the vacuum dielectric constant, κ∞ is the relative high-frequency dielectric constant, κ 0 is the relative static dielectric constant, N q is the distribution function of optical phonon. \|I(q z)\|2 is the form factor due to the quantized wave function; (I\left({q}{z}\right)={\int }{-\infty }^{\infty }{\varphi }{{{{{{\rm{i}}}}}}}\left(z\right){\varphi }{{{{{{\rm{j}}}}}}}\left(z\right)\exp (i{q}{z}z){{{{{{\rm{d}}}}}}z}), where _φ i(z) is the wave function at i- th subband, z is the distance along the direction perpendicular to the sample surface (z = 0 is defined as the interface position of undoped GaAs/AlGaAs spacer.), qz is the scattering wave vector in the z direction. q described as q = k2-k1 is a two-dimensional scattering wave vector from initial state k1 to the final state k2 in the elastic collisions. q+ and q- are two-dimensional scattering wave vectors in the phonon absorption and the phonon emission, respectively, as follows: $${q}{+}=\sqrt{2\cdot \left(\frac{2m\left(E-{E}{{{{{{\rm{i}}}}}}}\right)}{{\hslash }^{2}}\right)+\frac{2m\hslash {\omega }{{{{{{\rm{LO}}}}}}}}{{\hslash }^{2}}-2\sqrt{\frac{2m\left(E-{E}{{{{{{\rm{i}}}}}}}\right)}{{\hslash }^{2}}}\cdot \sqrt{\frac{2m\left(E-{E}{{{{{{\rm{i}}}}}}}\right)}{{\hslash }^{2}}+\frac{2m\hslash {\omega }{{{{{{\rm{LO}}}}}}}}{{\hslash }^{2}}}\cos \theta }$$ (10) $${q}{-}=\sqrt{2\cdot \left(\frac{2m\left(E-{E}{{{{{{\rm{i}}}}}}}\right)}{{\hslash }^{2}}\right)-\frac{2m\hslash {\omega }{{{{{{\rm{LO}}}}}}}}{{\hslash }^{2}}-2\sqrt{\frac{2m\left(E-{E}{{{{{{\rm{i}}}}}}}\right)}{{\hslash }^{2}}}\cdot \sqrt{\frac{2m\left(E-{E}{{{{{{\rm{i}}}}}}}\right)}{{\hslash }^{2}}-\frac{2m\hslash {\omega }{{{{{{\rm{LO}}}}}}}}{{\hslash }^{2}}}\cos \theta }$$ (11) $$q=2\sqrt{\frac{2m\left(E-{E}_{{{{{{\rm{i}}}}}}}\right)}{{\hslash }^{2}}}\sin \frac{\theta }{2}$$ (12) where θ is the scattering angle between k1 and k2. D A is the deformation potential constant, k B is Boltzmann constant, c L is the longitudinal elastic constant, t dope is the thickness of Si-doped Al 0.3 Ga 0.7 As layer, n imp is the concentration of impurity atoms, q TF is Thomas-Fermi wave number, V 0 is the energy barrier height, Δ and Λ are the mean interface roughness values at the z direction and at the perpendicular direction to z, respectively (in this calculation, these parameters are fixed at Δ=0.5 nm and Λ=5 nm), a B is the effective Bohr radius, n depl is the charge density of the depletion layer, and F(q) = ∫d z∫d z′\|φ(z)\|2\|φ(z′)\|2 exp(−q\|z − z′\|). In the calculation, m value shown in Table1 was used for each subband under the assumption that the non-parabolicity effect on m is negligible51. The energy band diagram was computed using 1D Poisson solver developed by G. Snider; wavefunction and carrier concentration distribution were self-consistently computed using the Poisson–Schrodinger equation. 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Miyazaki&Takeshi Kasaya Authors 1. Yuto UematsuView author publications Search author on:PubMedGoogle Scholar 2. Takafumi IshibeView author publications Search author on:PubMedGoogle Scholar 3. Takaaki ManoView author publications Search author on:PubMedGoogle Scholar 4. Akihiro OhtakeView author publications Search author on:PubMedGoogle Scholar 5. Hideki T. MiyazakiView author publications Search author on:PubMedGoogle Scholar 6. Takeshi KasayaView author publications Search author on:PubMedGoogle Scholar 7. Yoshiaki NakamuraView author publications Search author on:PubMedGoogle Scholar Contributions Y.N. conceived the idea. Y.U. measured thermoelectric properties of the samples. Y.U. and T.I. performed numerical calculation. Y.U., T.I., and Y.N. discussed the physics and wrote the paper. Y.N. is the principal investigator of this work. T.M. and A.O. formed the samples. H.T.M. and T.K. formed the electrode. All authors discussed the results and contributed to the revision of the final manuscript. Corresponding author Correspondence to Yoshiaki Nakamura. Ethics declarations Competing interests The authors declare no competing interests. Peer review Peer review information Nature Communications thanks Je-Hyeong Bahk, Yue Lin and the other, anonymous, reviewer(s) for their contribution to the peer review of this work. A peer review file is available. Additional information Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. 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To view a copy of this licence, visit Reprints and permissions About this article Cite this article Uematsu, Y., Ishibe, T., Mano, T. et al. Anomalous enhancement of thermoelectric power factor in multiple two-dimensional electron gas system. Nat Commun15, 322 (2024). Download citation Received: 09 June 2023 Accepted: 03 December 2023 Published: 16 January 2024 DOI: Share this article Anyone you share the following link with will be able to read this content: Get shareable link Sorry, a shareable link is not currently available for this article. Copy shareable link to clipboard Provided by the Springer Nature SharedIt content-sharing initiative Subjects Thermoelectric devices and materials Two-dimensional materials Download PDF Sections Figures References Abstract Introduction Results Discussion Methods Data availability References Acknowledgements Author information Ethics declarations Peer review Additional information Supplementary information Rights and permissions About this article Advertisement Fig. 1: Power factor S 2 σ enhancement by multiplied 2DEG effect (M2DE). View in articleFull size image Fig. 2: Sample structure illustrations, calculated energy band diagrams, and theoretical demonstration of multiplied 2DEG effect (M2DE). View in articleFull size image Fig. 3: Thermoelectric properties. View in articleFull size image Shi, X.-L., Zou, J. & Chen, Z.-G. Advanced thermoelectric design: from materials and structures to devices. Chem. Rev.120, 7399–7515 (2020). ArticleCASPubMedGoogle Scholar Tan, G., Zhao, L.-D. & Kanatzidis, M. G. Rationally designing high-performance bulk thermoelectric materials. Chem. Rev.116, 12123–12149 (2016). ArticleCASPubMedGoogle Scholar Venkatasubramanian, R., Siivola, E., Colpitts, T. & O’Quinn, B. Thin-film thermoelectric devices with high room-temperature figures of merit. Nature413, 597–602 (2001). ArticleCASPubMedADSGoogle Scholar Hochbaum, A. I. et al. Enhanced thermoelectric performance of rough silicon nanowires. Nature451, 163–167 (2008). ArticleCASPubMedADSGoogle Scholar Bux, A. K. et al. Nanostructured bulk silicon as an effective thermoelectric material. Adv. Funct. Mater.19, 2445–2452 (2009). ArticleCASGoogle Scholar Nakamura, Y. et al. Anomalous reduction of thermal conductivity in coherent nanocrystal architecture for silicon thermoelectric material. Nano Energy12, 845–851 (2015). ArticleCASGoogle Scholar Nakamura, Y. 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16256	https://www.nationalgeographic.com/animals/birds/facts/african-gray-parrot	Animals Photo Ark African gray parrot Common Name: : African gray parrots Scientific Name: : Psittacus erithacus Diet: : Herbivore Group Name: : Flock Average Life Span: : Not known Size: : 12.9 inches Weight: : 2.5 pounds IUCN Red List Status: : Endangered LC NT VU EN CR EW EX Least Concern Extinct Current Population Trend: : Decreasing What is a gray parrot? Gray parrots, commonly called African grays, are native to rainforests of central Africa, ranging in a band across the continent from Côte d’Ivoire to western Kenya. The largest parrot in Africa, this species has silver feathers, a white mask, and a bright, reddish tail. Males and females are very similar in appearance. Their colors may be less stunning than other parrots, but African grays are bright in other ways: They’re among the smartest birds in the world and the greatest mimic of human speech among the 350 or so known parrot species. Research has shown that the birds possess cognitive skills equal to that of a five-year-old child. They will also help members of their species, even complete strangers, without expecting their altruism to be reciprocated. Behavior and mating African grays are highly social species, flying through the sky in noisy flocks and roosting in big groups amid the treetops each evening.They feed in smaller groups of about 30, eating foods like oil palm nuts and the berries of the cola plant, grasping them in their claws and tearing them open with their strong beak. The birds will also sometimes raid human crops, such as maize. The monogamous parrots, which mate for life, begin searching for mates between three and five years of age. A pair will seek out pre-existing tree cavities in which to make a nest, lay a clutch of about three to four eggs, which are incubated by the female. Parents are attentive, building well-made nests and feeding their chicks together. The pet threat Because of their intelligence and ability to mimic human speech, African grays are the most popular pet bird in the world. The birds breed well in captivity, and at least 1.3 million gray parrots that have been exported legally from Africa over the past four decades, particularly to countries in the Middle East. (Learn more: Have parrots become too popular for their own good?) However, hundreds of thousands of others—maybe more—have died in transit or been snatched illegally from the forests of West and Central Africa as part of the illegal wildlife trade. Because grays are gregarious and social, it makes them relatively easy to catch. Trappers, particularly in the Democratic Republic of the Congo and the Republic of the Congo, tear down the trees to pluck babies from nests or put out wooden sticks covered with glue to ensnare roosting adults en masse. The majority of wild-caught grays probably die in transit. (Read about the illegal trade in gray parrots.) In 2016, the Convention on International Trade in Endangered Species, or CITES, which monitors international trade in rare species, made the controversial decision to ban all international trade in wild African grays, except in “exceptional circumstances.” In 2018, the bird was listed as endangered by the International Union for Conservation of Nature. African grays also suffer from habitat loss, in particular because they prefer to nest in tall trees that are targeted by loggers. Legal Terms of Use Privacy Policy Your US State Privacy Rights Children's Online Privacy Policy Interest-Based Ads About Nielsen Measurement Do Not Sell or Share My Personal Information Our Sites Nat Geo Home Attend a Live Event Book a Trip Buy Maps Inspire Your Kids Shop Nat Geo Visit the D.C. Museum Watch TV Learn About Our Impact Support Our Mission Masthead Press Room Advertise With Us Join Us Subscribe Customer Service Renew Subscription Manage Your Subscription Work at Nat Geo Sign Up for Our Newsletters Contribute to Protect the Planet Follow us Copyright © 1996-2015 National Geographic SocietyCopyright © 2015-2025 National Geographic Partners, LLC. All rights reserved
16257	https://www.youtube.com/watch?v=4LN1CJ211SM	How do Ammeters and Voltmeters in Circuits work? Dan the Tutor 8780 subscribers 48 likes Description 5022 views Posted: 21 Mar 2024 Explanation of ammeters and voltmeters in circuits. Ammeters have a very low resistance and must be placed in series with a resistor. Voltmeters have a very high resistance and must be placed in parallel with a resistor. Patreon Link (recurring donations): patreon.com/user?u=84517584 Paypal Link (one-time donations): 9 comments Transcript: hello everyone how's it going my name is Dan the tutor in today's video we're going to be talking about ammeters and voltmeters so first let's get started with a little T chart to explain what the difference is and what they both do so the biggest difference is that ammeters will measure current and voltmeters measure voltage the second big difference is how they're placed ammers are going to be placed in series with the resistor you want to look at while voltmeters are going to be placed in parallel around the resistor you want to look at and the reason why is because of this third Point ammeters have a very low resistance and voltmeters have a very high resistance and if you're wondering why that makes it good that ammers are in series if it's very low resistance and volt are good for parallel cuz it's very high resistance if you want to know why that is just think about it think about ohms law and if you're still confused comment below and I'll answer your questions so now let's look at some questions the first one I have here I have a circuit connected to two resistors like this both of these resistors are going to be 6 ohms the battery is 12 volts and I'm going to place a voltmeter here around this 6 Ohm resistor and the question is what's the reading on the voltmeter but another way I could ask this is what is the voltage on this resistor and so there's many ways that you can find this voltage on the six Ohm resistor in this problem I would recommend using voltage division which only works when the resistors are all in series and what's going to look like is V1 where V1 is the voltage on the 6 ohm that I want is equal to V total time resistor 1 over resistor total so in this case V1 equal the total which is 12 resistor 1 which is 6 / R total which is 6 + 6 and so that becomes 12 that's just 12 and so we'll get a voltage of 6 volts that's the voltage on the resistor we wanted and since the question was asking about the voltmeter reading the answer is just 6 volts so that's it for that first one for the second circuit I have I have an ammeter here that is in series with two resistors who are in parallel like this I'll say the voltage source is 8 volts the left resistor is 12 ohms and the right resistor is 24 ohms and I want the reading on that ameter right there so in other words we can find this by finding the current at the battery or you could also add the two currents across these two resistors but I think that would take a little bit longer so what I'm going to do is if I want to find the current as fast as possible the first thing I would do is I would combine these two resistors who R in parallel and the way we add resistors in parallel is R total equals quantity 1 / R1 + 1 / R2 to the 1st power and so that means it's 1 12 + 1 over4 to the 1st power you plug this in the calculator and you will get 8 ohms so that's the total resistance in other words if I were to redraw this circuit still the 8vt battery connected to the ammeter also connected to the 8 Ohm resistor right there and now if I want the current I can pretend that the ammeter is not even in the circuit it's like not even there it's invisible and I just want to find the current to do that I'm using ohms law v = i R so voltage 8 equals current unknown times resistor 8 divide both sides by 8 current is just 1 amp very easy for that one and another one is done and then the last example we're going to look at today is the hardest one so here's my circuit and I want to find the ammeter and voltmeter readings the ammeter is connected to the 2 ohm in series and the voltmeter is connected to the 4 ohm in parallel so it doesn't matter which one I find first I think it's going to be easier to find the ammeter reading and to find that current I need to remember that this blue path is in parallel with this red path and the important thing about that is that if they're in parallel that means both of these paths have the same voltage in this case both paths have a total of 6 volts and since the blue path is really just one resistor it means that all 6 volts belong to that 2 ohms that's great because now I can do ohms law v = i voltage is 6 equal I which I'm solving for time resistor this is the 2 ohm resistor here so divide both sides by two and we get a current of 3 amps there we already found the ammeter reading now the volt meter reading here a little more challenging the first thing I'll say is that all three resistors together together are 6 volts cuz again they're in parallel with the blue path and the battery so they're all going to have 6 volts total now since just these three resistors are in series in other words let me group them together like this the three of these resistors are in series it means I can use voltage division the same equation I used for the last one except this time the numbers will be a little bit different so V1 the vol voltage I'm looking for it's on the 4 ohm resistor is the total voltage which we said was six which by the way it will not always be the battery for instance if I had another resistor right here this resistor would be eating up some of the voltage so it would be less than six for the other two branches and if that was the case then I'd have to do a whole lot more combining resistors and finding different voltages to solve the problem luckily I don't have to do that cuz that resistor does not actually go there so anyway V1 equals the total 6 resistor 1 which is 4 / the total resistance this is not the total resistance of the whole circuit this is just the total resistance of the three resistors in series so it's 4 + 3 + 1 or 8 ohms and that's going to be voltage 1 = 6 12 3 volts and there there's my readings on the ammeter which was 3 amps and the reading on the voltmeter is 3 volts and that is going to do it for this video thank you all for watching I hope you have a great rest of your day if you have any questions please post them in the comments below and I'll see you in the next video take care and bye-bye
16258	https://www.youtube.com/watch?v=BvNKHUICHYw	Worked example: separable differential equation (with taking exp of both sides) \| Khan Academy Khan Academy 9090000 subscribers 69 likes Description 5769 views Posted: 8 Nov 2017 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: Solving the equation dy/dx=x_y. This involves an extra step of raising _ by both sides of the resulting x-y equation, in order to isolate y. Practice this lesson yourself on KhanAcademy.org right now: Watch the next lesson: Missed the previous lesson? AP Calculus BC on Khan Academy: Learn AP Calculus BC - everything from AP Calculus AB plus a few extra goodies, such as Taylor series, to prepare you for the AP Test About Khan Academy: Khan Academy is a nonprofit with a mission to provide a free, world-class education for anyone, anywhere. We believe learners of all ages should have unlimited access to free educational content they can master at their own pace. We use intelligent software, deep data analytics and intuitive user interfaces to help students and teachers around the world. Our resources cover preschool through early college education, including math, biology, chemistry, physics, economics, finance, history, grammar and more. We offer free personalized SAT test prep in partnership with the test developer, the College Board. Khan Academy has been translated into dozens of languages, and 100 million people use our platform worldwide every year. For more information, visit www.khanacademy.org, join us on Facebook or follow us on Twitter at @khanacademy. And remember, you can learn anything. For free. For everyone. Forever. #YouCanLearnAnything Subscribe to Khan Academy's AP Calculus BC channel: Subscribe to Khan Academy: 5 comments Transcript: what we're going to do in this video is see if we can solve the differential equation the derivative of y with respect to x is equal to x times y pause this video and see if you can find a general solution here so the first thing that my brain likes to do when i see a differential equation is to say hey is this separable and when i say separable can i get all the y's and d y's on one side and all the x's and d x's on the other side and you can indeed do that if we treat our differentials like if we could treat them like algebraic variables which is fair game when you're dealing with differential equations like this we could multiply both sides by dx so multiply both sides by dx and we could divide both sides by y let me move this over a little bit so we have some space so we could also multiply both sides by 1 over y 1 over y and so what that does is these dx's cancel out and this y and 1 over y cancels out and we are left with let me write all the things in terms of y on the left hand side in blue so we have 1 over y d y is equal to and then i'll do all this stuff in orange we have is equal to we're just left with an x and a dx x dx and then we'll want to take the indefinite integral of both sides now what's the antiderivative of 1 over y well if we want it in the most general form this would be the natural log of the absolute value of y and then this is going to be equal to the anti-derivative of x is x squared over 2 and then we want to put a constant on either side i'll just put it on the right hand side plus c this is ensures that we're dealing with the general solution now if we want to solve explicitly for y we could raise e to both sides power another way to think about it is if this is equal to that then e to this power is going to be the same thing as e to that power now what does that do for us well what is e to the natural log of the absolute value of y well i'm raising e to the power that i would have to raise e2 to get to the absolute value of y so the left hand side here it simplifies to the absolute value of y and we get that as being equal to now we could use our exponent properties this over here is the same thing as e to the x squared over 2 times e to the c times e to the c i am just using our exponent properties here well e to the c we could just view that as some other type of constant this is just some constant c so we could rewrite this whole thing as c e e to the x squared x squared over 2. hopefully you see what i'm doing there i just use my exponent properties e to the sum of two things is equal to e to the first thing times e to the second thing and i just said well e to the the power of some constant c we could just relabel that as let's call that our blue c and so this simplifies to blue c times e to the x squared over 2. now we still have this absolute value sign here so this essentially means that y y could be equal to we could write it this way y could be equal to plus or minus c c e e to the x squared over 2 x squared over 2 but once again we don't know what this constant is i didn't say that it was positive or negative so when you say a plus or minus of a constant here you could really just subsume all of this i'll just call this with red c so we could say that y is equal to i'll just rewrite it over again for for fun y is equal to red c not the red c but a red z times e to the x squared over 2. this right over here is the general solution to the original separable differential equation
16259	https://www.learningfundamentals.com/exercises/math/timezone.php	Learning Fundamentals Exercises AppsSoftwareOrderExercisesWordsLogin Overview Addition Subtraction Multiplication Division Time Zones Converting UTC to local time The current time is: 19:18:40 UTC. The current time is: 12:18:40 or 12:18:40 pm in California. The current time is: 13:18:40 or 01:18:40 pm in Denver. The current time is: 14:18:40 or 02:18:40 pm in Chicago. The current time is: 15:18:40 or 03:18:40 pm in New York. Pilots and astronomers are two professions that use a standard time based on the time in Greenwich England, the home of the Royal Observatory and the prime meridian. Since 1884 the time in Greenwich was used as a standard and referred to as Greenwich Mean Time (GMT). The time zone at Greenwich was assigned the letter Z and so GMT is often referred to as Zulu time and shown with the letter Z. In 1928 Coordinated Universal Time (abbreviated UTC) was introduced by the International Astronomical Union to more precisely measure time. Details are at the US Naval Observatory Pilots often fly across multiple time zones in a single day and to avoid confusion use UTC time for filing flight plans, accessing weather, and calculating time en route. They need to be able to convert UTC time to local time and vice versa. The conversion from UTC to local is a two step process. First, subtract the appropriate offset from UTC and then covert to AM or PM. For example, if the time is 2300Z and daylight saving time is not in effect, the time in New York (Eastern Time Zone) is 2300 - 500 = 1800 or 6 PM. It gets a little more complicated if the time is 0200Z. Subtracting 5 from 2 gives negative 3. Then you need to subtract 3 from 12 to get 9 PM. I usually find that it is easier to add 12 first, then subtract. So in this case add 12 plus 2 to get 14 and then subtract 5 to get 9 PM. Also, for me, subtracting 7, 8, 9 is a bit difficult to do. It is easier for me to subtract 10 and then add 3, 2, or 1. So in the winter in California I’d convert 2300Z by subtracting 10 and adding 2: 23-10 = 13 + 2 = 15 or 3 PM. I’d do the second example the same way, 0200Z is 26 -10 = 16 + 2 = 18 or 6 PM.As I write this it is 0120Z. To convert to PDT I’d add 12 to get 1320, then subtract 10 to get 0320 and add 3 to get 06:20 PM. Practice with different time zones. Atlantic—UTC-4 Eastern—UTC-5 Central—UTC-6 Mountain—UTC-7 Western—UTC-8 Alaska—UTC-9 Hawaii-Aleutian—UTC-10 Practice with daylight savings time. Atlantic—UTC-3 Eastern—UTC-4 Central—UTC-5 Mountain—UTC-6 Western—UTC-7 Alaska—UTC-8 Hawaii-Aleutian—UTC-9 Download Learning Fundamentals apps for iPhone and iPad. Home \| Apps \| Software \| Support \| Order \| Purchase Apps Catalog \| Resources \| Exercises \| Downloads \| Words \| Contact © Copyright $2025 Learning Fundamentals 1130 Grove Street, San Luis Obispo, CA 93401 805.544.0775;
16260	https://www.ck12.org/c/geometry/surface-area-and-volume-of-cylinders/	Surface Area and Volume of Cylinders \| CK-12 Foundation AI Teacher Tools – Save Hours on Planning & Prep. Try it out! What are you looking for? Search Math Grade 6 Grade 7 Grade 8 Algebra 1 Geometry Algebra 2 PreCalculus Science Earth Science Life Science Physical Science Biology Chemistry Physics Social Studies Economics Geography Government Philosophy Sociology Subject Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? Science Grade K to 5 Earth Science Life Science Physical Science Biology Chemistry Physics Advanced Biology FlexLets Math FlexLets Science FlexLets English Writing Spelling Social Studies Economics Geography Government History World History Philosophy Sociology More Astronomy Engineering Health Photography Technology College College Algebra College Precalculus Linear Algebra College Human Biology The Universe Adult Education Basic Education High School Diploma High School Equivalency Career Technical Ed English as 2nd Language Country Bhutan Brasil Chile Georgia India Translations Spanish Korean Deutsch Chinese Greek Polski Explore EXPLORE Flexi A FREE Digital Tutor for Every Student FlexBooks 2.0 Customizable, digital textbooks in a new, interactive platform FlexBooks Customizable, digital textbooks Schools FlexBooks from schools and districts near you Study Guides Quick review with key information for each concept Adaptive Practice Building knowledge at each student’s skill level Simulations Interactive Physics & Chemistry Simulations PLIX Play. Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign InSign Up Support CK-12 Foundation is a non-profit organization that provides free educational materials and resources. FLEXIAPPS ABOUT Our missionMeet the teamPartnersPressCareersSecurityBlogCK-12 usage mapTestimonials SUPPORT Certified Educator ProgramCK-12 trainersWebinarsCK-12 resourcesHelpContact us BYCK-12 Common Core MathK-12 FlexBooksCollege FlexBooksTools and apps CONNECT TikTokInstagramYouTubeTwitterMediumFacebookLinkedIn v2.11.10.20250923073248-b88c97d744 © CK-12 Foundation 2025 \| FlexBook Platform®, FlexBook®, FlexLet® and FlexCard™ are registered trademarks of CK-12 Foundation. Terms of usePrivacyAttribution guide Curriculum Materials License Please wait... Please wait... Oops, looks like cookies are disabled on your browser. Click here to see how to enable them. X Student Sign Up Are you a teacher? Sign up here Sign in with Google Having issues? Click here Sign in with Microsoft Sign in with Apple or Sign up using email By signing up, I confirm that I have read and agree to the Terms of use and Privacy Policy Already have an account? Sign In No Results Found Your search did not match anything in . Got It
16261	https://pubmed.ncbi.nlm.nih.gov/15505110/	Practice standards for electrocardiographic monitoring in hospital settings: an American Heart Association scientific statement from the Councils on Cardiovascular Nursing, Clinical Cardiology, and Cardiovascular Disease in the Young: endorsed by the International Society of Computerized Electrocardiology and the American Association of Critical-Care Nurses - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Practice standards for electrocardiographic monitoring in hospital settings: an American Heart Association scientific statement from the Councils on Cardiovascular Nursing, Clinical Cardiology, and Cardiovascular Disease in the Young: endorsed by the International Society of Computerized Electrocardiology and the American Association of Critical-Care Nurses Barbara J Drew,Robert M Califf,Marjorie Funk,Elizabeth S Kaufman,Mitchell W Krucoff,Michael M Laks,Peter W Macfarlane,Claire Sommargren,Steven Swiryn,George F Van Hare;American Heart Association;Councils on Cardiovascular Nursing, Clinical Cardiology, and Cardiovascular Disease in the Young PMID: 15505110 DOI: 10.1161/01.CIR.0000145144.56673.59 Item in Clipboard Guideline Practice standards for electrocardiographic monitoring in hospital settings: an American Heart Association scientific statement from the Councils on Cardiovascular Nursing, Clinical Cardiology, and Cardiovascular Disease in the Young: endorsed by the International Society of Computerized Electrocardiology and the American Association of Critical-Care Nurses Barbara J Drew et al. Circulation.2004. Show details Display options Display options Format Circulation Actions Search in PubMed Search in NLM Catalog Add to Search . 2004 Oct 26;110(17):2721-46. doi: 10.1161/01.CIR.0000145144.56673.59. Authors Barbara J Drew,Robert M Califf,Marjorie Funk,Elizabeth S Kaufman,Mitchell W Krucoff,Michael M Laks,Peter W Macfarlane,Claire Sommargren,Steven Swiryn,George F Van Hare;American Heart Association;Councils on Cardiovascular Nursing, Clinical Cardiology, and Cardiovascular Disease in the Young PMID: 15505110 DOI: 10.1161/01.CIR.0000145144.56673.59 Item in Clipboard Full text links Cite Display options Display options Format Erratum in Circulation. 2005 Jan 25;111(3):378 Abstract The goals of electrocardiographic (ECG) monitoring in hospital settings have expanded from simple heart rate and basic rhythm determination to the diagnosis of complex arrhythmias, myocardial ischemia, and prolonged QT interval. Whereas computerized arrhythmia analysis is automatic in cardiac monitoring systems, computerized ST-segment ischemia analysis is available only in newer-generation monitors, and computerized QT-interval monitoring is currently unavailable. Even in hospitals with ST-monitoring capability, ischemia monitoring is vastly underutilized by healthcare professionals. Moreover, because no computerized analysis is available for QT monitoring, healthcare professionals must determine when it is appropriate to manually measure QT intervals (eg, when a patient is started on a potentially proarrhythmic drug). The purpose of the present review is to provide 'best practices' for hospital ECG monitoring. Randomized clinical trials in this area are almost nonexistent; therefore, expert opinions are based upon clinical experience and related research in the field of electrocardiography. This consensus document encompasses all areas of hospital cardiac monitoring in both children and adults. The emphasis is on information clinicians need to know to monitor patients safely and effectively. Recommendations are made with regard to indications, timeframes, and strategies to improve the diagnostic accuracy of cardiac arrhythmia, ischemia, and QT-interval monitoring. Currently available ECG lead systems are described, and recommendations related to staffing, training, and methods to improve quality are provided. PubMed Disclaimer Similar articles AHA scientific statement: practice standards for electrocardiographic monitoring in hospital settings: an American Heart Association Scientific Statement from the Councils on Cardiovascular Nursing, Clinical Cardiology, and Cardiovascular Disease in the Young: endorsed by the International Society of Computerized electrocardiology and the American Association of Critical-Care Nurses.Drew BJ, Califf RM, Funk M, Kaufman ES, Krucoff MW, Laks MM, Macfarlane PW, Sommargren C, Swiryn S, Van Hare GF; American Heart Association.Drew BJ, et al.J Cardiovasc Nurs. 2005 Mar-Apr;20(2):76-106. doi: 10.1097/00005082-200503000-00003.J Cardiovasc Nurs. 2005.PMID: 15855856 Review. Practice standards for ECG monitoring in hospital settings: executive summary and guide for implementation.Drew BJ, Funk M.Drew BJ, et al.Crit Care Nurs Clin North Am. 2006 Jun;18(2):157-68, ix. doi: 10.1016/j.ccell.2006.01.007.Crit Care Nurs Clin North Am. 2006.PMID: 16728301 Review. AHA/ACCF/HRS recommendations for the standardization and interpretation of the electrocardiogram: part IV: the ST segment, T and U waves, and the QT interval: a scientific statement from the American Heart Association Electrocardiography and Arrhythmias Committee, Council on Clinical Cardiology; the American College of Cardiology Foundation; and the Heart Rhythm Society: endorsed by the International Society for Computerized Electrocardiology.Rautaharju PM, Surawicz B, Gettes LS, Bailey JJ, Childers R, Deal BJ, Gorgels A, Hancock EW, Josephson M, Kligfield P, Kors JA, Macfarlane P, Mason JW, Mirvis DM, Okin P, Pahlm O, van Herpen G, Wagner GS, Wellens H; American Heart Association Electrocardiography and Arrhythmias Committee, Council on Clinical Cardiology; American College of Cardiology Foundation; Heart Rhythm Society.Rautaharju PM, et al.Circulation. 2009 Mar 17;119(10):e241-50. doi: 10.1161/CIRCULATIONAHA.108.191096. Epub 2009 Feb 19.Circulation. 2009.PMID: 19228821 Review.No abstract available. An algorithm for continuous real-time QT interval monitoring.Helfenbein ED, Zhou SH, Lindauer JM, Field DQ, Gregg RE, Wang JJ, Kresge SS, Michaud FP.Helfenbein ED, et al.J Electrocardiol. 2006 Oct;39(4 Suppl):S123-7. doi: 10.1016/j.jelectrocard.2006.05.018. Epub 2006 Aug 21.J Electrocardiol. 2006.PMID: 16920145 Recommendations for the standardization and interpretation of the electrocardiogram: part I: the electrocardiogram and its technology a scientific statement from the American Heart Association Electrocardiography and Arrhythmias Committee, Council on Clinical Cardiology; the American College of Cardiology Foundation; and the Heart Rhythm Society endorsed by the International Society for Computerized Electrocardiology.Kligfield P, Gettes LS, Bailey JJ, Childers R, Deal BJ, Hancock EW, van Herpen G, Kors JA, Macfarlane P, Mirvis DM, Pahlm O, Rautaharju P, Wagner GS, Josephson M, Mason JW, Okin P, Surawicz B, Wellens H; American Heart Association Electrocardiography and Arrhythmias Committee, Council on Clinical Cardiology; American College of Cardiology Foundation; Heart Rhythm Society.Kligfield P, et al.J Am Coll Cardiol. 2007 Mar 13;49(10):1109-27. doi: 10.1016/j.jacc.2007.01.024.J Am Coll Cardiol. 2007.PMID: 17349896 Review. See all similar articles Cited by ECG monitoring leads and special leads.Francis J.Francis J.Indian Pacing Electrophysiol J. 2016 May-Jun;16(3):92-95. doi: 10.1016/j.ipej.2016.07.003. Epub 2016 Jul 17.Indian Pacing Electrophysiol J. 2016.PMID: 27788999 Free PMC article.Review. Monitoring the Burden of Seizures and Highly Epileptiform Patterns in Critical Care with a Novel Machine Learning Method.Kamousi B, Karunakaran S, Gururangan K, Markert M, Decker B, Khankhanian P, Mainardi L, Quinn J, Woo R, Parvizi J.Kamousi B, et al.Neurocrit Care. 2021 Jun;34(3):908-917. doi: 10.1007/s12028-020-01120-0. Epub 2020 Oct 6.Neurocrit Care. 2021.PMID: 33025543 Free PMC article. Utilization of Continuous Cardiac Monitoring on Hospitalist-led Teaching Teams.Chen DW, Park R, Young S, Chalikonda D, Laothamatas K, Diemer G.Chen DW, et al.Cureus. 2018 Sep 13;10(9):e3300. doi: 10.7759/cureus.3300.Cureus. 2018.PMID: 30443470 Free PMC article. Over-monitoring and alarm fatigue: for whom do the bells toll?Feder S, Funk M.Feder S, et al.Heart Lung. 2013 Nov-Dec;42(6):395-6. doi: 10.1016/j.hrtlng.2013.09.001.Heart Lung. 2013.PMID: 24183197 Free PMC article.No abstract available. Association between QTc prolongation and mortality in patients with suspected poisoning in the emergency department: a transnational propensity score matched cohort study.Schade Hansen C, Pottegård A, Ekelund U, Kildegaard Jensen H, Lundager Forberg J, Brabrand M, Lassen AT.Schade Hansen C, et al.BMJ Open. 2018 Jul 7;8(7):e020036. doi: 10.1136/bmjopen-2017-020036.BMJ Open. 2018.PMID: 29982199 Free PMC article. See all "Cited by" articles Publication types Guideline Actions Search in PubMed Search in MeSH Add to Search Practice Guideline Actions Search in PubMed Search in MeSH Add to Search MeSH terms Arrhythmias, Cardiac / diagnosis Actions Search in PubMed Search in MeSH Add to Search Cardiology / education Actions Search in PubMed Search in MeSH Add to Search Electrocardiography / instrumentation Actions Search in PubMed Search in MeSH Add to Search Electrocardiography / methods Actions Search in PubMed Search in MeSH Add to Search Electrocardiography / standards Actions Search in PubMed Search in MeSH Add to Search Electrodes Actions Search in PubMed Search in MeSH Add to Search Hospitalization Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Long QT Syndrome / diagnosis Actions Search in PubMed Search in MeSH Add to Search Monitoring, Physiologic / standards Actions Search in PubMed Search in MeSH Add to Search Myocardial Ischemia / diagnosis Actions Search in PubMed Search in MeSH Add to Search Workforce Actions Search in PubMed Search in MeSH Add to Search Related information Cited in Books LinkOut - more resources Full Text Sources Atypon Ovid Technologies, Inc. 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16262	https://www.reddit.com/r/askmath/comments/1c5ihzm/finding_anotherdifferent_formulas_for_a_given/	Finding a(nother/different) formula(s) for a given series using induction : r/askmath Skip to main contentFinding a(nother/different) formula(s) for a given series using induction : r/askmath Open menu Open navigationGo to Reddit Home r/askmath A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to askmath r/askmath r/askmath This subreddit is for questions of a mathematical nature. Please read the subreddit rules below before posting. 207K Members Online •1 yr. ago Puzzleheaded_Bet14 Finding a(nother/different) formula(s) for a given series using induction Pre Calculus In class, we were taught how to prove that a series is this formula for all positive integers n using induction. Our professor is very vague and didn't even teach us how to prove divisibility using induction 💀. So, let's assume that I have basic knowledge of induction and I can see the pattern of the formula, which is (nn!). Some students, including me, initially thought that (nn!) was the answer; however, our professor pointed out that (nn!) won't work because it doesn't cover the terms before it, like (11!), (22!), etc... Is he wrong? Is there any other way? I think there is. Do we use S_(k+1)? Read more Archived post. New comments cannot be posted and votes cannot be cast. Share New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community Top Posts Reddit reReddit: Top posts of April 16, 2024 Reddit reReddit: Top posts of April 2024 Reddit reReddit: Top posts of 2024 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation TOPICS Internet Culture (Viral) Amazing Animals & Pets Cringe & Facepalm Funny Interesting Memes Oddly Satisfying Reddit Meta Wholesome & Heartwarming Games Action Games Adventure Games Esports Gaming Consoles & Gear Gaming News & Discussion Mobile Games Other Games Role-Playing Games Simulation Games Sports & Racing Games Strategy Games Tabletop Games Q&As Q&As Stories & Confessions Technology 3D Printing Artificial Intelligence & Machine Learning Computers & Hardware Consumer Electronics DIY Electronics Programming Software & Apps Streaming Services Tech News & Discussion Virtual & Augmented Reality Pop Culture Celebrities Creators & Influencers Generations & Nostalgia Podcasts Streamers Tarot & Astrology Movies & TV Action Movies & Series Animated Movies & Series Comedy Movies & Series Crime, Mystery, & Thriller Movies & Series Documentary Movies & Series Drama Movies & Series Fantasy Movies & Series Horror Movies & Series Movie News & Discussion Reality TV Romance Movies & Series Sci-Fi Movies & Series Superhero Movies & Series TV News & Discussion RESOURCES About Reddit Advertise Reddit Pro BETA Help Blog Careers Press Communities Best of Reddit Top Translated Posts Topics
16263	https://stats.stackexchange.com/questions/218156/what-are-some-of-the-most-common-misconceptions-about-linear-regression	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams What are some of the most common misconceptions about linear regression? Ask Question Asked Modified 5 years, 3 months ago Viewed 7k times $\begingroup$ I'm curious, for those of you who have extensive experience collaborating with other researchers, what are some of the most common misconceptions about linear regression that you encounter? I think can be a useful exercise to think about common misconceptions ahead of time in order to Anticipate people's mistakes and be able to successful articulate why some misconception is incorrect Realize if I am harboring some misconceptions myself! A couple of basic ones I can think of: Independent/Dependent variables must be normally distributed Variables must be standardized for accurate interpretation Any others? All responses are welcome. regression multiple-regression Improve this question edited Jun 10, 2016 at 8:48 community wiki 3 revs, 2 users 85%ST21 $\endgroup$ 3 1 $\begingroup$ A lot of people I know still insist on performing linearizations on their data and leaving it at that, even when the computing environment they use has good support for nonlinear regression. (The linearizations are of course useful as starting points for the nonlinear fits, but these people don't even realize that.) $\endgroup$ J. M. is not a statistician – J. M. is not a statistician 2016-06-09 21:35:53 +00:00 Commented Jun 9, 2016 at 21:35 $\begingroup$ Great answers, but most assume "other researchers" means people with statistics training. Many of the researchers I've worked with are from other disciplines and maybe had one basic stat course. Their misconceptions are much more fundamental. Like: correlation implies cause & effect, and extrapolating from the result will be accurate at values far from the source data. $\endgroup$ fixer1234 – fixer1234 2016-06-11 23:54:12 +00:00 Commented Jun 11, 2016 at 23:54 2 $\begingroup$ If God had made the world linear, there wouldn't be nonlinear regression. $\endgroup$ Mark L. Stone – Mark L. Stone 2016-06-14 20:26:05 +00:00 Commented Jun 14, 2016 at 20:26 Add a comment \| 12 Answers 12 Reset to default 43 $\begingroup$ False premise: A $\boldsymbol{\hat{\beta} \approx 0}$ means that there is no strong relationship between DV and IV.Non-linear functional relationships abound, and yet data produced by many such relationships would often produce nearly zero slopes if one assumes the relationship must be linear, or even approximately linear. Relatedly, in another false premise researchers often assumepossibly because many introductory regression textbooks teachthat one "tests for non-linearity" by building a series of regressions of the DV onto polynomial expansions of the IV (e.g., $Y \sim \beta_{0} + \beta_{X}X + \varepsilon$, followed by $Y \sim \beta_{0} + \beta_{X}X + \beta_{X^{2}}X^{2} + \varepsilon$, followed by $Y \sim \beta_{0} + \beta_{X}X + \beta_{X^{2}}X^{2} + \beta_{X^{3}}X^{3} + \varepsilon$, etc.). Just as straight line cannot well represent a non-linear functional relationship between DV and IV, a parabola cannot well represent literally an infinite number of nonlinear relationships (e.g., sinusoids, cycloids, step functions, saturation effects, s-curves, etc. ad infinitum). One may instead take a regression approach that does not assume any particular functional form (e.g., running line smoothers, GAMs, etc.). A third false premise is that increasing the number of estimated parameters necessarily results in a loss of statistical power. This may be false when the true relationship is non-linear and requires multiple parameters to estimate (e.g., a "broken stick" function requires not only the intercept and slope terms of a straight line, but requires point at which slope changes and a how much slope changes by estimates also): the residuals of a misspecified model (e.g., a straight line) may grow quite large (relative to a properly specified functional relation) resulting in a lower rejection probability and wider confidence intervals and prediction intervals (in addition to estimates being biased). Share Improve this answer edited Jun 30, 2020 at 1:35 community wiki 7 revs, 2 users 92%Alexis $\endgroup$ 8 6 $\begingroup$ (+1) Quibbles: (1) I don't think even introductory texts imply that all curves are polynomial functions, rather that they can be approximated well enough over a given range by polynomial functions. So they fall into the class of "regression approaches that do not assume any particular functional form", governed by a "hyperparameter" specifying wiggliness: the span for loess, the no. knots for regression on a spline basis, the degree for regression on a polynomial basis. (I'm not waving a flag for polynomials - it's well known they tend to flail around at the ends more than we'd like -, ... $\endgroup$ Scortchi – Scortchi ♦ 2016-06-10 09:49:50 +00:00 Commented Jun 10, 2016 at 9:49 2 $\begingroup$ ... just giving them their due.) (2) A sinusoid might well be fit as such, within the linear model framework; a saturation effect using a non-linear model (a rectangular hyperbola, say); &c. Of course you didn't say otherwise, but it's perhaps worth pointing out that if you know there's a cycle, or an asymptote, applying those constraints in your model will be helpful. $\endgroup$ Scortchi – Scortchi ♦ 2016-06-10 09:49:58 +00:00 Commented Jun 10, 2016 at 9:49 2 $\begingroup$ @Scortchi I could not agree more! (Indeed, given an infinite number of polynomials, any function can be perfectly represented.) Was aiming at concise. :) $\endgroup$ Alexis – Alexis 2016-06-10 17:13:00 +00:00 Commented Jun 10, 2016 at 17:13 3 $\begingroup$ @Alexis Try approximating Conway's base 13 function by polynomials. :) $\endgroup$ Reinstate Monica – Reinstate Monica 2016-06-12 02:05:06 +00:00 Commented Jun 12, 2016 at 2:05 2 $\begingroup$ Or $\chi_{\mathbb{Q}}$... $\endgroup$ Stephan Kolassa – Stephan Kolassa 2016-06-12 13:20:13 +00:00 Commented Jun 12, 2016 at 13:20 \| Show 3 more comments 24 +100 $\begingroup$ It's very common to assume that only $y$ data are subject to measurement error (or at least, that this is the only error that we shall consider). But this ignores the possibility - and consequences - of error in the $x$ measurements. This might be particularly acute in observational studies where the $x$ variables are not under experimental control. Regression dilution or regression attenuation is the phenomenon recognised by Spearman (1904) whereby the estimated regression slope in simple linear regression is biased towards zero by the presence of measurement error in the independent variable. Suppose the true slope is positive — the effect of jittering the points' $x$ co-ordinates (perhaps most easily visualised as "smudging" the points horizontally) is to render the regression line less steep. Intuitively, points with a large $x$ are now more likely to be so because of positive measurement error, while the $y$ value is more likely to reflect the true (error-free) value of $x$, and hence be lower than the true line would be for the observed $x$. In more complex models, measurement error in $x$ variables can produce more complicated effects on the parameter estimates. There are errors in variables models that take such error into account. Spearman suggested a correction factor for disattenuating bivariate correlation coefficients and other correction factors have been developed for more sophisticated situations. However, such corrections can be difficult — particularly in the multivariate case and in the presence of confounders — and it may be controversial whether the correction is a genuine improvement, see e.g. Smith and Phillips (1996). So I suppose this is two misconceptions for the price of one — on the one hand it is a mistake to think that the way we write $y = X\beta + \varepsilon$ means "all the error is in the $y$" and ignore the very physically real possibility of measurement errors in the independent variables. On the other hand, it may be inadvisable to blindly apply "corrections" for measurement error in all such situations as a knee-jerk response (though it may well be a good idea to take steps to reduce the measurement error in the first place). (I should probably also link to some other common error-in-variables models, in increasingly general order: orthogonal regression, Deming regression, and total least squares.) References Smith, G. D., & Phillips, A. N. (1996). "Inflation in epidemiology: 'the proof and measurement of association between two things' revisited". British Medical Journal, 312(7047), 16591661. Spearman, C. (1904). "The proof and measurement of association between two things." American Journal of Psychology 15: 72101. Share Improve this answer edited Jun 10, 2016 at 12:29 community wiki 3 revsSilverfish $\endgroup$ 4 1 $\begingroup$ On that note: this is one reason for the use of the technique that is called either "total least squares" or "orthogonal regression" (depending on the reference you are reading); it is significantly more complicated than plain least squares, but is worth doing if all your points are contaminated with error. $\endgroup$ J. M. is not a statistician – J. M. is not a statistician 2016-06-10 03:45:33 +00:00 Commented Jun 10, 2016 at 3:45 $\begingroup$ @J.M. Thanks - yes, in fact I'd originally meant to put in a link to TLS, but got distracted by the Smith and Phillips article! $\endgroup$ Silverfish – Silverfish 2016-06-10 12:30:22 +00:00 Commented Jun 10, 2016 at 12:30 3 $\begingroup$ +1 Great addition to this topic. I've often considered EIV models in my work. However, apart from their complexity or reliance on knowledge of "error ratios", there is a more conceptual issue to consider: Many regressions, especially in supervised learning or prediction, want to relate observed predictors to observed outcomes. EIV models, on the other hand, attempt to identify the underlying relationship between the mean predictor and mean response...a slightly different question. $\endgroup$ user75138 – user75138 2016-06-12 13:07:50 +00:00 Commented Jun 12, 2016 at 13:07 2 $\begingroup$ So, what one would call "dilution" of the "true" regression (in a scientific context) would be called "absence of predictive utility" or something like that in a prediction context. $\endgroup$ user75138 – user75138 2016-06-12 13:11:08 +00:00 Commented Jun 12, 2016 at 13:11 Add a comment \| 22 $\begingroup$ There are some standard misunderstandings that apply in this context as well as other statistical contexts: e.g., the meaning of $p$-values, incorrectly inferring causality, etc. A couple of misunderstandings that I think are specific to multiple regression are: Thinking that the variable with the larger estimated coefficient and/or lower $p$-value is 'more important'. Thinking that adding more variables to the model gets you 'closer to the truth'. For example, the slope from a simple regression of $Y$ on $X$ may not be the true direct relationship between $X$ and $Y$, but if I add variables $Z_1, \ldots, Z_5$, that coefficient will be a better representation of the true relationship, and if I add $Z_6, \ldots, Z_{20}$, it will be even better than that. Share Improve this answer answered Jun 9, 2016 at 20:01 community wiki gung - Reinstate Monica $\endgroup$ 1 15 $\begingroup$ Good stuff. This answer might be even more useful if it explained why the two are wrong and what one should do instead? $\endgroup$ D.W. – D.W. 2016-06-10 01:11:56 +00:00 Commented Jun 10, 2016 at 1:11 Add a comment \| 14 $\begingroup$ I'd say the first one you list is probably the most common -- and perhaps the most widely taught that way -- of the things that are plainly seen to be wrong, but here are some others that are less clear in some situations (whether they really apply) but may impact even more analyses, and perhaps more seriously. These are often simply never mentioned when the subject of regression is introduced. Treating as random samples from the population of interest sets of observations that cannot possibly be close to representative (let alone randomly sampled). [Some studies could instead be seen as something nearer to convenience samples] With observational data, simply ignoring the consequences of leaving out important drivers of the process that would certainly bias the estimates of the coefficients of the included variables (in many cases, even to likely changing their sign), with no attempt to consider ways of dealing with them (whether out of ignorance of the problem or simply being unaware that anything can be done). [Some research areas have this problem more than others, whether because of the kinds of data that are collected or because people in some application areas are more likely to have been taught about the issue.] Spurious regression (mostly with data collected over time). [Even when people are aware it happens, there's another common misconception that simply differencing to supposed stationary is sufficient to completely avoid the problem.] There are many others one could mention of course (treating as independent data that will almost certainly be serially correlated or even integrated may be about as common, for example). You may notice that observational studies of data collected over time may be hit by all of these at once... yet that kind of study is very common in many areas of research where regression is a standard tool. How they can get to publication without a single reviewer or editor knowing about at least one of them and at least requiring some level of disclaimer in the conclusions continues to worry me. Statistics is fraught with problems of irreproducible results when dealing with fairly carefully controlled experiments (when combined with perhaps not so carefully controlled analyses), so as soon as one steps outside those bounds, how much worse must the reproducibility situation be? Share Improve this answer edited Jun 9, 2016 at 23:42 community wiki 2 revsGlen_b $\endgroup$ 4 6 $\begingroup$ Closely related to some of your points might be the idea that "only $y$ data are subject to measurement error" (or at least, "this is the only error that we shall consider"). Not sure if that deserves shoe-horning in here, but it is certainly very common to ignore the possibility - and consequences - of random error in the $x$ variables. $\endgroup$ Silverfish – Silverfish 2016-06-10 00:55:28 +00:00 Commented Jun 10, 2016 at 0:55 2 $\begingroup$ @Silverfish I totally agree with you. $\endgroup$ Mark L. Stone – Mark L. Stone 2016-06-10 01:08:35 +00:00 Commented Jun 10, 2016 at 1:08 $\begingroup$ @Silverfish it's CW so you should feel extra-free to edit in a suitable addition like that. $\endgroup$ Glen_b – Glen_b 2016-06-10 01:28:26 +00:00 Commented Jun 10, 2016 at 1:28 $\begingroup$ @Silverfish there's a reason I didn't already add it myself when you mentioned it... I think it probably is worth a new answer $\endgroup$ Glen_b – Glen_b 2016-06-10 01:49:41 +00:00 Commented Jun 10, 2016 at 1:49 Add a comment \| 12 $\begingroup$ I probably wouldn't call these misconceptions, but maybe common points of confusion/hang-ups and, in some cases, issues that researchers may not be aware of. Multicollinearity (including the case of more variables than data points) Heteroskedasticity Whether values of the independent variables are subject to noise How scaling (or not scaling) affects interpretation of the coefficients How to treat data from multiple subjects How to deal with serial correlations (e.g. time series) On the misconception side of things: What linearity means (e.g. $y = ax^2 + bx + c$ is nonlinear w.r.t. $x$, but linear w.r.t. the weights). That 'regression' means ordinary least squares or linear regression That low/high weights necessarily imply weak/strong relationships with the dependent variable That dependence between the dependent and independent variables can necessarily be reduced to pairwise dependencies. That high goodness-of fit on the training set implies a good model (i.e. neglecting overfitting) Share Improve this answer answered Jun 9, 2016 at 20:13 community wiki user20160 $\endgroup$ 1 $\begingroup$ If the weights are zero, then this implies there's no LINEAR relationship between the IV and DV? If the weights are very small, I don't think this says anything about the linear relationship. $\endgroup$ user5965026 – user5965026 2020-06-29 21:51:48 +00:00 Commented Jun 29, 2020 at 21:51 Add a comment \| 7 $\begingroup$ In my experience, students frequently adopt the view the that squared errors (or OLS regression) are an inherently appropriate, accurate, and overall good thing to use, or are even without alternative. I have frequently seen OLS advertised along with remarks that it "gives greater weight to more extreme/deviant observations", and most of the time it is at least implied that this is a desirable property. This notion may be modified later, when the treatment of outliers and robust approaches are introduced, but at that point the damage is done. Arguably, the widespread use of squared errors has historically more to do with their mathematical convenience than with some natural law of real-world error costs. Overall, greater emphasis could be placed on the understanding that the choice of error function is somewhat arbitrary. Ideally, any choice of penalty within an algorithm should be guided by the corresponding real-world cost function associated with potential error (i.e., using a decision-making framework). Why not establish this principle first, and then see how well we can do? Share Improve this answer answered Jun 10, 2016 at 13:20 community wiki Ben $\endgroup$ 1 2 $\begingroup$ The choice is also application-dependent. OLS is useful for algebraic, y-axis fits but less so for geometric applications, where total least squares (or some other cost function based on orthogonal distance) makes more sense. $\endgroup$ user11284 – user11284 2016-06-10 23:25:10 +00:00 Commented Jun 10, 2016 at 23:25 Add a comment \| 5 $\begingroup$ Another common misconception is that the error term (or disturbance in econometrics parlance) and the residuals are the same thing. The error term is a random variable in the true model or data generating process, and is often assumed to follow a certain distribution, whereas the residuals are the deviations of the observed data from the fitted model. As such, the residuals can be considered to be estimates of the errors. Share Improve this answer edited Jun 15, 2016 at 11:12 community wiki 2 revsRobert Long $\endgroup$ 1 1 $\begingroup$ I bet people would be interested in explanation as to why this matters, or in what sorts of cases. $\endgroup$ rolando2 – rolando2 2018-03-27 21:41:31 +00:00 Commented Mar 27, 2018 at 21:41 Add a comment \| 4 $\begingroup$ The most common misconception I encounter is that linear regression assumes normality of errors. It doesn't. Normality is useful in connection with some aspects of linear regression e.g. small sample properties such as confidence limits of coefficients. Even for these things there are asymptotic values available for non-normal distributions. The second most common is a cluster of confusion with regards to endogeneity, e.g. not being careful with feedback loops. If there's a feedback loop from Y back to X it's an issue. Share Improve this answer edited Jun 20, 2016 at 16:00 community wiki 2 revsAksakal $\endgroup$ 0 Add a comment \| 4 $\begingroup$ The one I've often seen is a misconception on applicability of linear regression in certain use cases, in practice. For example, let us say that the variable that we are interested in is count of something (example: visitors on website) or ratio of something (example: conversion rates). In such cases, the variable can be better modeled by using link functions like Poisson (counts), Beta (ratios) etc. So using generalized model with more appropriate link function is more suitable. But just because the variable is not categorical, I've seen people starting with simple linear regression (link function = identity). Even if we disregard the accuracy implications, the modeling assumptions are a problem here. Share Improve this answer answered Jun 22, 2016 at 9:07 community wiki hssay $\endgroup$ 1 $\begingroup$ I think it would helpful to clarify what "more suitable" means more precisely. For example, let's say I have an experiment on visitors to a set of pages and I estimate a Poisson model with a binary X, and run OLS. The average marginal effect will be identical in both models (which will generally be true for fully saturated models). The Poisson also makes some strong assumptions about the mean-variance relationship, that are often restrictive. It also makes dealing with interactions and panel data models more complicated. $\endgroup$ dimitriy – dimitriy 2020-07-21 01:23:26 +00:00 Commented Jul 21, 2020 at 1:23 Add a comment \| 4 $\begingroup$ An error that I made is to assume a symmetry of X and Y in the OLS. For instance, if I assume a linear relation $$ Y = a \, X + b$$ with a and b given by my software using OLS, then I believe that assuming X as a function of Y will give using OLS the coefficients: $$ X = \frac{1}{a} \, Y - \frac{b}{a}$$ that is wrong. Maybe this is also related to the difference between OLS and total least square or first principal component. Share Improve this answer answered Mar 27, 2018 at 20:03 community wiki Jf Parmentier $\endgroup$ Add a comment \| 3 $\begingroup$ Here is one I think is frequently overlooked by researchers: Variable interaction: researchers often look at isolated betas of individual predictors, and often don't even specify interaction terms. But in real world things interact. Without proper specification of all possible interaction terms, you don't know how your "predictors" engage together into forming an outcome. And if you want to be diligent and specify all interactions, the number of predictors will explode. From my calculations you can investigate only 4 variables and their interactions with 100 subjects. If you add one more variable you can overfit very easily. Share Improve this answer answered Jun 14, 2016 at 19:15 community wiki user4534898 $\endgroup$ Add a comment \| 1 $\begingroup$ Another common misconception is that the estimates (fitted values) are not invariant to transformations, e.g. $$f(\hat{y}_i) \neq \widehat{f(y_i)}$$ in general, where $\hat{y}_i = \vec{x}_i ^T \hat{\beta}$, the fitted regression value based on your estimated regression coefficients. If this is what you want for monotonic functions $f(\cdot)$ not necessarily linear, then what you want is a quantile regression. The equality above holds in linear regression for linear functions but non-linear functions (e.g. $log(\cdot)$) this will not hold. However, this will hold for any monotonic function in quantile regression. This comes up all the time when you do a log transform of your data, fit a linear regression, then exponentiate the fitted value and people read that as the regression. This isn't the mean, this is the median (if things are truly log-normally distributed). Share Improve this answer answered Mar 28, 2018 at 16:55 community wiki Lucas Roberts $\endgroup$ 4 $\begingroup$ The exponentiated prediction is technically a geometric mean, which coincides with the median in lognormal data. But this is rarely the mean that people have in mind when they do this. $\endgroup$ dimitriy – dimitriy 2020-07-20 23:53:53 +00:00 Commented Jul 20, 2020 at 23:53 $\begingroup$ @DimitriyV.Masterov yes, what you are stating is (one of) the points of my answer-if you want invariance under monotonic transformations you'd be better served with a a quantile regression. After all, the question is about common misconceptions. $\endgroup$ Lucas Roberts – Lucas Roberts 2020-07-21 00:28:31 +00:00 Commented Jul 21, 2020 at 0:28 $\begingroup$ I like your answer very much (and just dealt with this at work today). I just wanted to add that it is "a mean" (regardless of distribution), but not "the [arithmetic] mean". $\endgroup$ dimitriy – dimitriy 2020-07-21 00:38:35 +00:00 Commented Jul 21, 2020 at 0:38 $\begingroup$ @DimitriyV.Masterov sure fair point, my use of "the" is a bit of a misnomer. I was thinking in terms of arithmetic mean when I wrote, "the mean". I could add that to the post if you think it would clarify things but your comments here serve the same purpose. $\endgroup$ Lucas Roberts – Lucas Roberts 2020-07-21 20:31:42 +00:00 Commented Jul 21, 2020 at 20:31 Add a comment \| Start asking to get answers Find the answer to your question by asking. 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16264	https://www.academia.edu/39170962/Entrepreneurship_2e	Academia.edu no longer supports Internet Explorer. To browse Academia.edu and the wider internet faster and more securely, please take a few seconds to upgrade your browser. Log In Sign Up About Press Papers Terms Privacy Copyright We're Hiring! Help Center Principal Findings from Gem Case Alison Barnard Finding Your Passion Case Jim Poss Finding the Waste Importance-Performance Analysis Opening Discussion (Introduction and Warm-Up) Product Demo/Introduction/Description Case Executive Summary Financial Statement Overview Build-Up Method Comparable Method Case P'Kolino Financials Case Dayone Question of Distribution Present Value of Future Cash Flows Future Earnings (Ni) 2 Case Bladelogic Presentation Questions Case Feed Resource Recovery Comparison of Equity Sharing Methods Design Patents Limitations on Licenses Case Lazybones An Overview Evaluating Results Conclusion Case Earthwatch Preparation Questions Entrepreneurship-2e sasi bhooshan Sign up for access to the world's latest research checkGet notified about relevant papers checkSave papers to use in your research checkJoin the discussion with peers checkTrack your impact Abstract AI This paper discusses the second edition of a textbook on entrepreneurship, emphasizing key topics such as team building, financing, legal issues, and social entrepreneurship. The updated content reflects contemporary challenges and incorporates case studies that illustrate practical applications of entrepreneurial concepts. It aims to equip students with essential skills for launching and growing their businesses while considering social impacts. Related papers Risk-Based New Venture Valuation Technique: Win-Win for Entrepreneur and Investor Whittington Vara Journal of Business Valuation and Economic Loss Analysis, 2013 downloadDownload free PDFView PDFchevron_right VENTURE CAPITAL IN INDIAN ECONOMY IJAR Indexing The Venture Capital Industry in India has grown significantly since last few years. The Venture capital sector is the most vibrant industry in the financial market today. Venture capitalists are professional investors who specialize in funding to companies and building young, innovative enterprises to support business expansion with strategically. Venture capitalists are long-term investors who take a hands-on approach with all of their investments and actively work with entrepreneurial management teams in order to build great companies which will have the potential to develop into significant economic contributors. The duration for entry and exit such venture capitalist for 5-7 years to maximize their profit and return back. Venture capital is an important source of equity for start-up companies. Venture capitalists are people or companies who pool financial resources from high net worth individuals, pension funds, corporate insurance companies, etc. to invest in high risk - high return ventures that are unable to source funds from regular channels like banks and capital markets. The venture capital industry in India has really taken off in and it unbelievable role in Indian Economy. Venture capitalists not only provide monetary resources but also help the entrepreneur with guidance in formalizing his ideas into a viable business venture with existing resources or import new technology. downloadDownload free PDFView PDFchevron_right Exploring the informal capital market in the Netherlands: characteristics, mismatches and causes Zulaicha Parastuty 2006 downloadDownload free PDFView PDFchevron_right What Are High-Flow and Low-Flow Oxygen Delivery Systems? Response Ritesh Singhal Stroke, 2005 downloadDownload free PDFView PDFchevron_right Psychology of Entrepreneurship Gorgievski Marjan uned.es downloadDownload free PDFView PDFchevron_right ANALYSIS OF FINANCING SOURCES FOR START-UP COMPANIES Chameera Liyanage This paper presents the development of start-up companies, their types and potential sources of financing with special emphasis on financing ventures in Croatia. The expected scientific contribution supports the defining stages of development for start-ups, as well as their financing sources at each stage. The goal of the research was to investigate whether Croatia has made a shift from traditional to newer methods of financing. Scientific and research contributions of the paper are reflected in the fact that there is a relatively small number of papers, especially in the domestic literature, that address these issues. Therefore, this research can contribute to a better understanding of the financing strategy of entrepreneurial ventures. Presented and interpreted results could be a useful basis and encouragement for a further research in this and similar topics related to the start-up scene at the local as well as the global level. downloadDownload free PDFView PDFchevron_right Loading Preview Sorry, preview is currently unavailable. You can download the paper by clicking the button above. Related papers The entrepreneurial spirit - towards an education model for entrepreneurial success in South Africa Riaan Steenberg This paper presents the results in improvement of business plans through an education model designed from working with 826 South African entrepreneurs over a period of 1 year and extracts the curriculum that contributed to the successful creation of bankable business plans. Initially, 13% of entrepreneurs had a business concept, often based on micro-enterprise while eventually 75% of entrepreneurs had a bankable business plan with a planned minimum turnover annually of R 2 million per annum. The unit of study is the entrepreneurial business concept and the objective is to link the critical requirements of South African entrepreneurs to a valid education programme. The study found that a program based on concepts of context, business models, ownership concepts including the entrepreneurial spirit, customer and product design and sustainability created the 470% increase in fundable business concepts. This link was confirmed through a broad national research study that targeted South African entrepreneurs and that evaluated their persona’s and both submission rates and quality of submissions was used to show the efficacy of a targeted education programme. downloadDownload free PDFView PDFchevron_right London School of Commerce MBA for Executives Management Skills & Entrepreneurship Yousef Hamad This report aims to discuss the supportive elements and gives advices that help people and potential entrepreneurs in understanding how to start-up small businesses, and what are the main stations on the way from the idea to the Break-Even. downloadDownload free PDFView PDFchevron_right An Overview of Research on Early Stage Venture Capital: Current Status and Future Directions Annaleena Parhankangas Handbook of Research on Venture Capital, 2007 downloadDownload free PDFView PDFchevron_right CROWDFUNDING: MORE THAN MONEY JUMPSTARTING UNIVERSITY ENTREPRENEURSHIP CJ Cornell Crowdfunding is a relatively recent phenomenon and is changing the nature of entrepreneurship, particularly for younger and firsttime entrepreneurs. 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16265	https://www.sciencedirect.com/science/article/abs/pii/S0190962285701399	Xanthomas and hyperlipidemias - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Patient Access Other access options Search ScienceDirect Article preview Abstract References (155) Cited by (153) Journal of the American Academy of Dermatology Volume 13, Issue 1, July 1985, Pages 1-30 Xanthomas and hyperlipidemias Author links open overlay panel Frank Parker M.D. Show more Add to Mendeley Share Cite rights and content Referred to by Correction Journal of the American Academy of Dermatology, Volume 13, Issue 4, October 1985, Pages 560 View PDF The ability to recognize diverse clinical forms of xanthomas, such as tuberous, planar, eruptive and tendinous, is important in the detection of underlying systemic disease. A variety of primary genetic disorders, as well as numerous secondary conditions such as diabetes, obstructive liver disease, thyroid disease, renal disease, and pancreatitis, can lead to hyperlipoproteinemia that results in the formation not only of xanthomas but also of life-threatening vascular atherosclerosis. An understanding of the pathogenesis of the underlying lipoprotein alterations provides a rational approach to therapy utilizing dietary manipulations and drugs. Such treatment is capable of correcting most disorders of lipid metabolism, and, if appropriate therapy is initiated at the first sign of xanthoma evolution, it may prevent progression of atherosclerosis, provide resolution of xanthomas, and in some instances prevent serious pancreatitis. Recommended articles References (155) I Matsuo et al. Phytosterolemia and Type IIa hyperlipoproteinemia with tuberous xanthomas J Am Acad Dermatol (1981) RJ Havel Approach to the patient with hyperlipidemia. Symposium on lipid disorders Med Clin North Am (1982) JC LaRosa et al. A specific apoprotein activator for lipoprotein lipase Biochem Biophys Res Commun (1970) WM Bortz### On the control of cholesterol synthesis: Metabolism (1973) PJ Scott et al. Low density lipoprotein accumulation in actively growing xanthomas J Atheroscler Res (1967) AD Sniderman et al. Association of hyperapobetalipoproteinemia with endogenous hypertriglyceridemia and atherosclerosis Ann Intern Med (1982) JD Brunzell et al. Evidence for diabetes mellitus and genetic forms of hypertriglyceridemia as independent entities Metabolism (1975) WR Hazzard et al. Abnormal lipid composition of chylomicrons in broad-beta disease (Type III hyperlipoproteinemia) J Clin Invest (1970) R Fleischmajer Familial hyperlipoproteinemia Type III Arch Dermatol (1969) JL Goldstein et al. Familial hypercholesterolemia, a genetic regulatory defect in cholesterol metabolism Am J Med (1975) O Stein et al. Binding, internalization and degradation of low density lipoprotein by normal human fibroblasts and by fibroblasts from a case of homozygous familial hypercholesterolemia Proc Natl Acad Sci USA (1976) MI New et al. The significance of blood lipid alteration in diabetes mellitus Diabetes (1963) AL Warshaw et al. Inhibition of serum and urine amylase activity in pancreatitis and hyperlipemia Ann Surg (1975) JB Marsh et al. Experimental reconstruction of metabolic pattern of lipid nephrosis: Key role of hepatic protein synthesis in hyperlipemia Metabolism (1960) F Parker et al. Ultrastructural and lipid biochemical comparisons of human eruptive, tuberous and planar xanthomas Isr J Med Sci (1973) F Parker Hyperlipoproteinemia and xanthomatosis R Fleischmajer Cutaneous and tendon xanthomas Dermatologica (1964) WR Harlan et al. Familial hypercholesterolemia: A genetic and metabolic study Medicine (1966) A Gattereau et al. An improved radiological method for the evaluation of achilles tendon xanthomatosis Can Med Assoc J (1973) AK Khachadurian Migratory polyarthritis in familial hypercholesterolemia (Type II hyperlipoproteinemia) Arthritis Rheum (1968) CJ Glueck et al. Acute tendinitis and arthritis. A presenting symptom of familial Type II hyperlipoproteinemia JAMA (1968) M Philippart et al. Cholestanolosis (cerebrotendinous xanthomatosis) Arch Neurol (1969) V Shore et al. Abnormal high density lipoproteins in cerebrotendinous xanthomatosis J Clin Invest (1981) AK Bbattacharyya et al. B-Sitosterolemia and xanthomatosis J Clin Invest (1974) MD Polano Xanthomatosis and hyperlipoproteinemia Dermatologica (1974) P Douste-Blazy et al. Increased frequency of Apo E-ND phenotype and hyperapobetalipoproteinemia in normolipidemic subjects and xan thelasmas of the eyelids Ann Intern Med (1982) RH Rosenman et al. Relation of corneal arcus to cardiovascular risk factors and the incidence of coronary disease N Engl J Med (1924) M Feiwel Xanthomatosis in cryoglobulinaemia and other paraproteinaemias with report of a case Br J Dermatol (1968) WC Roberts Classification and etiology of hyperlipoproteinemia Fed Proc (1971) JD Brunzel et al. Chylomicronemia syndrome. Symposium on lipid disorders Med Clin North Am (1983) MA Mishkel et al. Xanthoma disseminatum Arch Dermatol (1977) RJ Havel et al. Lipoproteins and lipid transport JP Kane Plasma lipoproteins: Structure and metabolism HB Brewer et al. Type III hyperlipoproteinemia: Diagnosis, molecular defects, pathology and treatment Ann Intern Med (1983) EJ Blanchette-Mackie et al. Sites of lipoprotein lipase activity in adipose tissue perfused with chylomicrons: Electron microscope cytochemical study J Cell Biol (1971) CH Hollenberg Effect of nutrition on activity and release of lipase from rat adipose tissue Am J Physiol (1959) JC Kessler Effect of diabetes and insulin on the activity of myocardial and adipose tissue lipoprotein lipase of rats J Clin Invest (1963) CJ Glueck et al. Estrogen-induced pancreatitis in patients with previously covert familial Type V hyperlipoproteinemia Metabolism (1972) MS Brown et al. Receptor-mediated control of cholesterol metabolism. Study of human mutants has disclosed how cells regulate a substance that is both vital and lethal Science (1976) JL Goldstein et al. The LDL receptor locus and the genetics of familial hypercholesterolemia Annu Rev Genet (1979) MS Brown et al. Lipoprotein receptors in the liver. Control signals for plasma cholesterol traffic J Clin Invest (1983) AK Bhattacharyya et al. Excretion of sterols from the skin of normal and hypercholesterolemic humans J Clin Invest (1972) SM Grundy Treatment of hypercholesterolemia by interference with bile acid metabolism Arch Intern Med (1972) JM Dietschy et al. Regulation of cholesterol metabolism N Engl J Med (1970) ### N Engl J Med (1970) ### N Engl J Med (1970) HB Brewer et al. JA Glomset The plasma lecithins: Cholesterol acyltransferase reaction J Lipid Res (1968) MK Polano et al. Xanthomata in primary hyperlipoproteinemia Arch Dermatol (1969) F Parker et al. Evidence for the chylomicron origin of lipids accumulating in diabetic eruptive xanthomas: A correlative lipid, biochem ical, histochemical and electron microscopic study J Clin Invest (1970) TF Whyne et al. Plasma apolipoprotein B and VLDL-, and LDL-, and HDL-cholesterol as risk factors in the development of coronary artery disease in male patients examined by angiography Atherosclerosis (1981) JJ Maciejko et al. Apolipoprotein A-1 as a marker of angiographically assessed coronary artery disease N Engl J Med (1983) View more references Cited by (153) Dermatologic Manifestations of Diabetes Mellitus. A Review. 2013, Endocrinology and Metabolism Clinics of North America Citation Excerpt : The histopathology shows an accumulation of lipid-laden histiocytic foam cells with a mixed infiltrate of lymphocytes and neutrophils in the dermis. Approximately one-third of all diabetic patients have serum lipoprotein abnormalities caused by insulinopenia, but the prevalence of xanthomatosis among this population is unclear.90 The reason for the increased frequency of eruptive xanthomatosis among individuals with diabetes has been well characterized. ### Dermal, subcutaneous, and tendon xanthomas: Diagnostic markers for specific lipoprotein disorders 1988, Journal of the American Academy of Dermatology Show abstract Many patients with lipoprotein disorders are at increased risk for the development of premature atherosclerosis and, less commonly, other disorders that cause systemic morbidity. In some of these patients, xanthomas also develop and provide cutaneous markers for the lipoprotein disorder. As advances in molecular biology refine our understanding of lipoprotein metabolism, it has become increasingly clear that several types of xanthomas are associated with specific disease states. This article presents a differential diagnosis of xanthomas that incorporates contemporary thinking about lipoprotein disorders and focuses on the relationship between abnormalities in lipoprotein metabolism, content, or structure and the development of specific xanthomas. ### Cerebrotendinous xanthomatosis: The spectrum of imaging findings and the correlation with neuropathologic findings 2000, Radiology ### Type III hyperlipoproteinemia and spontaneous atherosclerosis in mice resulting from gene replacement of mouse Apoe with human APOE2 1998, Journal of Clinical Investigation ### Probucol: A Reappraisal of its Pharmacological Properties and Therapeutic Use in Hypercholesterolaemia 1989, Drugs ### Normocholesterolemic Xanthomatosis 1986, Archives of Dermatology View all citing articles on Scopus The CME articles are made possible through an educational grant from Syntex Laboratories, Inc. View full text Copyright © 1985 Published by Mosby, Inc. 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16266	https://www.scribd.com/doc/314864489/NPTEL-Ocean-engineering	Naval Architecture: Hydrostatics Basics \| PDF \| Naval Architecture \| Ships Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 245 views 2 pages Naval Architecture: Hydrostatics Basics This NPTEL course covers hydrostatics and stability principles for naval architecture students. It begins with Archimedes' principle and the conditions for stability, studying metacentric he… Full description Uploaded by utkarshsabberwal AI-enhanced title and description Go to previous items Go to next items Download Save Save NPTEL Ocean engineering For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save NPTEL Ocean engineering For Later You are on page 1/ 2 Search Fullscreen NPTEL Syllabus Hydrostatics and Stability - Video course COURSE OUTLINE This is the first course for Naval Architecture students in 'Hydrostatics and Stability'. This course covers the basic principles of stability, starting from the Archimedes principle and going deeper into the conditions for stability of a vessel.Study of metacentric height and radius gives an insight into Naval Architectural stability problem. A thorough study of the various weight shiftings w hich give rise to various forms of heeling are investigated. Conditi ons of free-surface eff ect and inclining experiment are studied.This is followed by stability of the ship at larger angles of heel under various turning moments. The course then discusses about the various conditions of damaged stability, dry docking and launching calculations. A demonstration of the software 'SPAN' which performs these hydrostatic calculations will be demonstrated. Contents : Hull form definition of ships and ocean structures; Deadweight,capacity and tonnage measurement; Numerical Integration, Hydrostatic calculations; Hydrostatic curves, Initial stability, free surface effects, stability at large angles; Intact and damaged stability computations; IMO stability criteria;Damaged stability and its calculation by lost buoyancy and added weight methods; Subdivision and floodable length calculations; Launching and dry docking calculations; Stability of fully submerged body; Stability of multibody systems; Pressure integrat ion technique of computing hydrostatic and stabil ity.COURSE DETAIL S l N o. T o p i c/s H o u r s 1 I n t r o d u c t i o n 1 2 A r c h i m e d e s p r i n c i p l e 2 3 H y d r o s t a t i c s 3 4 N u m e r i c a l I n t e g r a t i o n 2 5 F r e e s u r f a c e e f f e c t 1 6 H y d r o s t a t i c c u r v e s 3 7 S t a b i l i t y a t l a r g e a n g l e s o f h e e l 3 8 D y n a m i c s t a b i l i t y 2 9 W e i g h t a n d t r i m c a l c u l a t i o n s 2 NPTEL Oc ean Engineering Pre-requisites: Mathem atics I Additional Reading: Resistance & propulsion Sea keeping and maneuvering Coordinators:Dr. Hari V. Warrior Department of Ocean Engineering &Naval A rchitectureII T Kharagpur adDownload to read ad-free 1 0 I n t a c t S t a b i l i t y R e g u l a t i o n s 3 1 1 P a r a m e t r i c r e s o n a n c e 2 1 2 F l o o d i n g a n d d a m a g e d s t a b i l i t y 4 1 3 D r y d o c k i n g 2 1 4 L o n g i t u d i n a l s t a b i l i t y 3 1 5 B e n d i n g m o m e n t, s h e a r f o r c e 2 1 6 R e s p o n s e i n w a v e s 2 1 7 T u r n i n g s t a b i l i t y 1 1 8 C o m p u t e r m e t h o d s 2 Total 40 References:Reference Books: Brian, A. Ship H ydrostatics and stability, Butterworth Heinemann.Lester A R, Merchant ship stability, Butterworths A j o i n t v e n t u r e b y I I S c a n d I I T s, f u n d e d b y M H R D, G o v t o f I n d i a h t t p://n p t e l.i i t m.a c.i n Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Seam 2 Trim, Stability and Stress 93% (29) Seam 2 Trim, Stability and Stress 141 pages MEP Handbook for Engineers 85% (27) MEP Handbook for Engineers 108 pages Introduction To Naval Architecture Gillmer 100% (9) Introduction To Naval Architecture Gillmer 338 pages SHIP HYDROSTATICS and Stability Prof Omar PDF 100% (4) SHIP HYDROSTATICS and Stability Prof Omar PDF 69 pages ABKOWITZ - Stability and Motion Control of Ocean Vehicles PDF No ratings yet ABKOWITZ - Stability and Motion Control of Ocean Vehicles PDF 362 pages Naval Architecture 1 Class Notes: Omar Bin Yaakob 100% (3) Naval Architecture 1 Class Notes: Omar Bin Yaakob 126 pages Jack Up and Semi Submersible Stability PDF No ratings yet Jack Up and Semi Submersible Stability PDF 33 pages UPVC Pipe No ratings yet UPVC Pipe 132 pages Upvc Pipe For Strom Water Above Ground Elevated Highway No ratings yet Upvc Pipe For Strom Water Above Ground Elevated Highway 75 pages Naval Ship Design 100% (1) Naval Ship Design 51 pages Lecture 01 No ratings yet Lecture 01 17 pages Naval Architecture - I For IMU Syllabus 2017 PDF 75% (4) Naval Architecture - I For IMU Syllabus 2017 PDF 571 pages Hydrostatics and Stability No ratings yet Hydrostatics and Stability 1 page Offshore Hydromechanics: Dr. Ir. 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16267	https://pubmed.ncbi.nlm.nih.gov/26590005/	Intracranial venous sinus thrombosis as a complication of otitis media in children: Critical review of diagnosis and management - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Intracranial venous sinus thrombosis as a complication of otitis media in children: Critical review of diagnosis and management Elisabetta Zanoletti1,Diego Cazzador2,Chiara Faccioli2,Marianna Sari2,Roberto Bovo2,Alessandro Martini2 Affiliations Expand Affiliations 1 Department of Neuroscience, Operative Unit of Otolaryngology, Padova University, via Giustiniani 2, Padova, Italy. Electronic address: elisabetta.zanoletti@tiscali.it. 2 Department of Neuroscience, Operative Unit of Otolaryngology, Padova University, via Giustiniani 2, Padova, Italy. PMID: 26590005 DOI: 10.1016/j.ijporl.2015.10.059 Item in Clipboard Intracranial venous sinus thrombosis as a complication of otitis media in children: Critical review of diagnosis and management Elisabetta Zanoletti et al. Int J Pediatr Otorhinolaryngol.2015 Dec. Show details Display options Display options Format Int J Pediatr Otorhinolaryngol Actions Search in PubMed Search in NLM Catalog Add to Search . 2015 Dec;79(12):2398-403. doi: 10.1016/j.ijporl.2015.10.059. Epub 2015 Nov 5. Authors Elisabetta Zanoletti1,Diego Cazzador2,Chiara Faccioli2,Marianna Sari2,Roberto Bovo2,Alessandro Martini2 Affiliations 1 Department of Neuroscience, Operative Unit of Otolaryngology, Padova University, via Giustiniani 2, Padova, Italy. Electronic address: elisabetta.zanoletti@tiscali.it. 2 Department of Neuroscience, Operative Unit of Otolaryngology, Padova University, via Giustiniani 2, Padova, Italy. PMID: 26590005 DOI: 10.1016/j.ijporl.2015.10.059 Item in Clipboard Full text links Cite Display options Display options Format Abstract Objectives: Otogenic lateral sinus thrombosis (LST) is a rare intracranial complication of acute otitis media (AOM), which can lead to severe neurological sequelae and death. The aim of this study was to analyze the clinical presentation, management and outcome of LST in children, investigating a possible correlation between clinical aspects, radiological findings and anatomical variations. Methods: At a tertiary Italian hospital, a retrospective review was conducted on the medical records of eight patients diagnosed with otogenic LST over a 3-year period. Four children were males and mean age was 4.7 years. Results: All patients had a history of otitis media at diagnosis and 4/8 presented also with more than one neurological sign or symptom. Mastoiditis signs were detected in 5/8 patients. Thrombosis was diagnosed by computed tomography, enhanced magnetic resonance and magnetic resonance venography. Treatment was medical, alone or combined with surgery. Medical treatment consisted in anticoagulants eventually combined with anti-edema medication on clinical basis. Mastoidectomy and/or myringotomy±trans-tympanic drainage placement were performed in 7/8 patients. Complete vessel recanalization was obtained in 6/8 children after a median follow-up time of 4.8 months. No complications, neither clinical sequelae occurred. In our series, neurological signs and symptoms were significantly associated with the presence of hypoplasia of the contralateral venous sinus (p=0.029). Conclusion: LST is a severe condition occurring even in absence of otological signs, and despite adequate antibiotic therapy for AOM, which should be ruled out and promptly treated. A dominant neurological presentation is associated in our series with anatomical variations of cerebral sinus venous drainage patterns. This should be carefully evaluated and considered in diagnosis, treatment planning and prognosis. Keywords: Acute mastoiditis; Children; Lateral sinus thrombosis; Otitis media; Venous sinus thrombosis. Copyright © 2015 Elsevier Ireland Ltd. All rights reserved. PubMed Disclaimer Similar articles Otitic hydrocephalus associated with lateral sinus thrombosis and acute mastoiditis in children.Kuczkowski J, Dubaniewicz-Wybieralska M, Przewoźny T, Narozny W, Mikaszewski B.Kuczkowski J, et al.Int J Pediatr Otorhinolaryngol. 2006 Oct;70(10):1817-23. doi: 10.1016/j.ijporl.2006.06.012. Epub 2006 Aug 8.Int J Pediatr Otorhinolaryngol. 2006.PMID: 16899305 Otogenic lateral sinus thrombosis: case series and controversies.Funamura JL, Nguyen AT, Diaz RC.Funamura JL, et al.Int J Pediatr Otorhinolaryngol. 2014 May;78(5):866-70. doi: 10.1016/j.ijporl.2014.03.002. Epub 2014 Mar 12.Int J Pediatr Otorhinolaryngol. 2014.PMID: 24680135 Lateral sinus thrombosis as a complication of otitis media: 10-year experience at the children's hospital of Philadelphia.Bales CB, Sobol S, Wetmore R, Elden LM.Bales CB, et al.Pediatrics. 2009 Feb;123(2):709-13. doi: 10.1542/peds.2008-0280.Pediatrics. 2009.PMID: 19171642 Paediatric otogenic lateral sinus thrombosis: therapeutic management, outcome and thrombophilic evaluation.Novoa E, Podvinec M, Angst R, Gürtler N.Novoa E, et al.Int J Pediatr Otorhinolaryngol. 2013 Jun;77(6):996-1001. doi: 10.1016/j.ijporl.2013.03.030. Epub 2013 Apr 29.Int J Pediatr Otorhinolaryngol. 2013.PMID: 23639339 Review. Pediatric otogenic lateral sinus thrombosis: role of anticoagulation and surgery.Sitton MS, Chun R.Sitton MS, et al.Int J Pediatr Otorhinolaryngol. 2012 Mar;76(3):428-32. doi: 10.1016/j.ijporl.2011.12.025. Epub 2012 Jan 25.Int J Pediatr Otorhinolaryngol. 2012.PMID: 22277267 Review. See all similar articles Cited by Cerebral sinovenous thrombosis as a complication of otitis media.Salloum S, Belzer K.Salloum S, et al.Clin Case Rep. 2018 Dec 4;7(1):186-188. doi: 10.1002/ccr3.1948. eCollection 2019 Jan.Clin Case Rep. 2018.PMID: 30656038 Free PMC article. Response to the Letter to the Editor: "Pediatric otogenic lateral sinus thrombosis: focus on the prognostic role of contralateral venous drainage".Scorpecci A, Marsella P, Giannantonio S, Zangari P, Longo D, Luciani M.Scorpecci A, et al.Eur Arch Otorhinolaryngol. 2019 Jun;276(6):1853-1854. doi: 10.1007/s00405-019-05372-1. Epub 2019 Mar 14.Eur Arch Otorhinolaryngol. 2019.PMID: 30874882 No abstract available. Otogenic Cerebral Sinus Thrombosis in Children: A Narrative Review.Kotowski M, Szydlowski J.Kotowski M, et al.Neurol Ther. 2023 Aug;12(4):1069-1079. doi: 10.1007/s40120-023-00499-0. Epub 2023 Jun 2.Neurol Ther. 2023.PMID: 37266793 Free PMC article.Review. Management of Otogenic Sigmoid Sinus Thrombosis: A Systematic Review on the Role of Anticoagulation and its Outcome.George M, Kolethekkat AA, Torrigiani E, Marston SS, Thomas CM, Raphael M.George M, et al.Indian J Otolaryngol Head Neck Surg. 2023 Jun;75(2):450-456. doi: 10.1007/s12070-022-03289-6. Epub 2022 Nov 28.Indian J Otolaryngol Head Neck Surg. 2023.PMID: 37275038 Free PMC article. Cavernous Sinus Thrombosis and Blindness Complicating Dental Infection.Ng EMC, Othman O, Chan LY, Bahari NA.Ng EMC, et al.Cureus. 2022 Jan 17;14(1):e21318. doi: 10.7759/cureus.21318. eCollection 2022 Jan.Cureus. 2022.PMID: 35186577 Free PMC article. See all "Cited by" articles MeSH terms Acute Disease Actions Search in PubMed Search in MeSH Add to Search Adolescent Actions Search in PubMed Search in MeSH Add to Search Anticoagulants / therapeutic use Actions Search in PubMed Search in MeSH Add to Search Child Actions Search in PubMed Search in MeSH Add to Search Child, Preschool Actions Search in PubMed Search in MeSH Add to Search Female Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Infant Actions Search in PubMed Search in MeSH Add to Search Lateral Sinus Thrombosis / diagnosis Actions Search in PubMed Search in MeSH Add to Search Lateral Sinus Thrombosis / etiology Actions Search in PubMed Search in MeSH Add to Search Lateral Sinus Thrombosis / therapy Actions Search in PubMed Search in MeSH Add to Search Male Actions Search in PubMed Search in MeSH Add to Search Mastoid / surgery Actions Search in PubMed Search in MeSH Add to Search Mastoiditis / diagnosis Actions Search in PubMed Search in MeSH Add to Search Mastoiditis / etiology Actions Search in PubMed Search in MeSH Add to Search Middle Ear Ventilation / adverse effects Actions Search in PubMed Search in MeSH Add to Search Otitis Media / complications Actions Search in PubMed Search in MeSH Add to Search Otitis Media / diagnosis Actions Search in PubMed Search in MeSH Add to Search Otitis Media / therapy Actions Search in PubMed Search in MeSH Add to Search Phlebography Actions Search in PubMed Search in MeSH Add to Search Prognosis Actions Search in PubMed Search in MeSH Add to Search Retrospective Studies Actions Search in PubMed Search in MeSH Add to Search Tomography, X-Ray Computed Actions Search in PubMed Search in MeSH Add to Search Substances Anticoagulants Actions Search in PubMed Search in MeSH Add to Search Related information MedGen LinkOut - more resources Full Text Sources Elsevier Science Full text links[x] Elsevier Science [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). 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16268	https://gunn-gatm.github.io/textbook/chapters/06%20Geometry%20of%20Complex%20Numbers.pdf	6 Geometry of Complex Numbers Thanks to Tristan Needham’s Visual Complex Analysis for many of the problems/examples and to Josh Zucker for most of the text. a b a b z iz θ ϕ 4 3 A z z z z iz iz iz 4z + 3iz = Az ϕ θ 4z 3iz Figure 1: iz is perpendicu-lar to z. Figure 2: The complex number A = 4 + 3i. Figure 3: Breaking up Az into its com-ponents, we can observe the geometry of complex multiplication. Last year, you (were supposed to have) mastered the art of manipulating complex numbers. We will build on that background in this section. Throughout the rest of the book, you can reinforce your skills with a healthy dose of "Vitamin i." There are at least two ways to think about the equation x3 = 1. One way is to factor the equation into (x −1)(x2 + x + 1) and find the solutions using the quadratic formula. The other way is to use DeMoivre’s theorem: (r1(cos θ + i sin θ))(r2(cos ϕ + i sin ϕ)) = (r1r2)(cos(θ + ϕ) + i sin(θ + ϕ)). Recall that cis θ = cos θ + i sin θ. z = r cis θ is a complex number r units away from the origin and making an angle θ with the x axis, taken counterclockwise. Let’s rewrite DeMoivre’s theorem using cis: (r1 cis θ)(r2 cis ϕ) = (r1r2)(cis(θ + ϕ)). Notice that the magnitudes are multiplied and the angles are added.6 By repeatedly applying DeMoivre’s theorem, we know that (r cis θ)n = rn cis nθ. If x = r cis θ, then x3 = r3 cis 3θ. Going back to our original x3 = 1, since 1 = cis(2πk) for any integer k, we find that r = 1 and 3θ = 2πk. This yields three solutions: 1 cis 0, 1 cis 2π 3 , 1 cis 4π 3 . Any other value of θ = 2kπ 3 reduces to one of these values because of the periodicity of cis. These correspond to k = 0, 1, 2; other values of k produce coterminal angles and are therefore duplicates. You can confirm that these three solutions are the same solutions that you obtain from factoring. You can prove DeMoivre’s theorem using the angle addition formulae for cos and sin. You can also under-stand it through pure geometry. Consider a complex number z = a + bi, being multiplied by A = 4 + 3i. z forms an angle of θ with the real axis, and A forms an angle of ϕ. In Figure 1, observe that iz is perpendicular to z for any z. Figure 2 depicts the complex number A. Finally, in Figure 3, you see the multiplication carried out: Az = (4 + 3i)z = 4z + 3iz. These two components, 4z and 3iz, are indicated. Combining the observation in Figure 1 and our knowledge from geometry, we know the triangles in Figure 2 and 3 are similar. Since the scaling is by a factor of \|z\|, multiplying A by z has the effect rotating z by the angle of A, and multiplying it by the length of A. This method of proving DeMoivre’s theorem for A = 4 + 3i works for all complex numbers A = a + bi. Some notation: The angle θ of a complex number z = a+bi is often called the argument, written as Arg z. The real part of z is written Re(z) = a, and the imaginary part of z is written Im(z) = b. Note that Im(z) is a real number b, not an imaginary number bi. Finally, the complex conjugate of z, in which the imaginary part is negated, is written with a bar on top: z = a −bi. 1. Explain why iz is perpendicular to z without using DeMoivre’s theorem. 2. How does Arg z relate to Arg z? (Hint: symmetry!) 3. Compute zz and relate it to the cis form of z. 4. Explain, using a picture, why tan(Arg z) = Im(z) Re(z). 6What else is added when you multiply? Exponents! In fact, cis θ = eiθ, but that’s another story. 14 5. Divide a+bi c+di by rationalizing the denominator. 6. Divide r1 cis θ r2 cis ϕ using DeMoivre’s theorem. 7. Compare and contrast the methods of division in Problems 5 and 6. Which is more convenient? Or does it depend on the circumstance? 8. (a) If z = r cis θ, what is 1 z ? (b) Explain how this result shows 1 a+bi = a−bi a2+b2 , without having to rationalize the denominator. (Hint: use Problems 3, 4, and 7.) 9. Compute (1 + i)13; pencil, paper, and brains only. No calculators! 10. Compute (1+i √ 3)3 (1−i)2 without a calculator. 11. Draw cis π 4 + cis π 2 . Use your picture to prove an expression for tan 3π 8 . (Hint: Add them as vectors.) 12. Solve z3 = 1, and show that its solutions under the operation of multiplication form a group, isomorphic to the rotation group of the equilateral triangle. Write a group table! 13. (a) Find multiplication groups of complex numbers which are isomorphic to the rotation groups for i. a non-square rectangle, and ii. a regular hexagon. (b) Make a table for each group. (c) Compare the regular hexagon’s group to the dihedral group of the equilateral triangle, D3. Con-sider: how are they the same? How are they different? Is the difference fundamental? 14. Which of the following sets is a group under (i) addition and (ii) multiplication? (a) {0} (b) {1} (c) {0, 1} (d) {−1, 1} (e) {1, −1, i, −i} (f) {naturals} (g) {integers} (h) {rationals}, Q (i) {Q without zero} (j) {complex numbers}, C (k) {C without zero} DeMoivre’s theorem is the “universal” trig identity, in the sense that it can be used to calculate every other trig identity. For example, suppose you want an identity for cos 3θ. For convenience, let c = cos θ and s = sin θ. Then we have: cis 3θ = (cis θ)3 [DeMoivre’s Theorem] = (c + is)3 [Definition of cis ] = c3 + 3c2si −3cs2 −s3i [Binomial expansion] cos 3θ + i sin 3θ = (c3 −3cs2) + i(3c2s −s3). [Combining like terms] Equating real parts on both sides, cos 3θ = cos3 θ −3 cos θ sin2 θ. 15. Prove that (r1 cis θ)(r2 cis ϕ) = r1r2 cis(θ +ϕ) using brute force and the angle-sum trig identities for cos and sin. Do you prefer this method or the one on the previous page? Which method gives you a better understanding of why the formula works? 16. Find an analogous identity for sin 3θ. Most of the work is already done for you! 17. Your friend’s textbook says cos 3θ = 4 cos3 θ −3 cos θ, different from our identity. Who’s right? 15 0 = A B C D P R Q S m n A B C D a a b b c c d d Figure 4: The quadrilateral with four squares. P a ia A Figure 5: 2(a + b + c + d) = 0. Figure 6: P = a + ia. Let’s apply complex numbers to a geometry problem. We want to prove that if we construct squares with centers P, Q, R, S on the sides of any quadrilateral ABCD, as shown in Figure 5, then (i) PR ⊥QS and (ii) PR ∼ = QS. In other words, segments joining centers of opposite squares are perpendicular and the same length. We represent all points in the figure as complex numbers. For convenience, let A = 0 be the origin. The edges of the quadrilateral can be thought of as vectors in the form of complex numbers, and are found using subtraction; for example, the edge from A to B is B −A. Similarly, the edge from B to C is C −B. Now, define complex numbers a = B −A 2 , b = C −B 2 , c = D −C 2 , d = A −D 2 . a is the vector halfway along − − → AB, b is halfway along − − → BC, etc.; see Figure 4. We also have a + b + c + d = B −A + C −B + D −C + A −D 2 = 0 2 = 0. More geometrically, this cancellation is because 2(a + b + c + d) = 2a + 2b + 2c + 2d is the sum of the vectors − − → AB, − − → BC, − − → CD, − − → DA, which is just − → AA = − → 0 (see Figure 5). P, Q, R, S are also complex numbers. Let m = R −P and n = Q −S be our two segments PR and QS. To prove that they are perpendicular, recall that z is perpendicular to iz for any complex z ̸= 0, so we just need to prove that n = ±im. We now need to relate P, Q, R, S back to a, b, c, d. Remembering that a is the vector halfway along − − → AB, we can see that P = a + ia. a takes you from the origin A to the midpoint of AB, then ia takes you to P. This shown in Figure 6. We can extend this logic to the other points, of course. 18. Now you can finish the rest of the proof. (a) Draw a, b, c, d, m, n for the quadrilateral on the previous page. (b) Why does showing n = ±im prove the segments are (i) perpendicular and (ii) the same length? (c) Explain why Q = 2a + b + ib. (d) Find formulae for R and S in terms of c and d. (e) Find m and n in terms of a, b, c, and d. (f) Check that n −im = 0, using the fact that a + b + c + d = 0. 16 19. In the previous problem, we drew squares outside a quadrilateral and connected their centers. Conjec-ture what happens if we draw equilateral triangles outside a triangle and connect their centers. Prove your conjecture using complex numbers. 20. The hard way to find an identity for tan 3θ is to divide the identity for sin and cos that we already found. Try it. Make sure your answer is in terms of tan only! 21. An easier way to get an identity for tan 3θ starts with setting z = 1 + i tan θ. (a) Why is Arg z = θ? (b) Why is tan 3θ = Im(z3) Re(z3)? (c) Use (b) to find an identity for tan 3θ. 22. Find multiplication groups of complex numbers isomorphic to rotation groups for (a) the regular octagon; (b) the regular pentagon. 23. Make tables for (a) the rotation group of the regular octagon; (b) the dihedral group of the square. (c) Is the difference between them fundamental? 24. Which of the following tables defines a group? Why or why not? (a) $ I A B C D I I A B C D A A C D B I B B I C D A C C D A I B D D B I A C (b) # I A B C D I I A B C D A A B C D I B B C D I A C C D I A B D D I A B C 25. Name some subsets of the complex numbers that are groups under multiplication. I can name an infinite number of both finite and infinite groups with this property, so after you list a few of each type, try to generalize. 26. Prove with a diagram that if \|z\| = 1, then Im z (z+1)2 = 0. 27. Prove geometrically that if \|z\| = 1, then \|1 −z\| = 2 sin Arg z 2 . 28. (a) Prove that if (z −1)10 = z10, then Re(z) = 1 2. (Hint: If two numbers are equal, they have the same magnitude.) (b) How many solutions does this equation have? 29. I claim that eiθ = cos θ + i sin θ = cis θ, for θ in radians. (a) Find e−iθ. (b) Find eiθ+e−iθ 2 . (c) Find eiθ−e−iθ 2i . 30. Use your new, complex definitions for cos and sin to find: (a) cos2 θ + sin2 θ (b) tan θ (c) cos 2θ (d) sin 2θ (e) What kind of group is generated by eiθ, e−iθ under the operation of multiplication if θ is an integer? A rational multiple of π? 31. You’ve used the quadratic equation throughout high school, but there’s also a cubic equation that finds the roots of any cubic. Let’s derive it, starting with the cubic x3 + bx2 + cx + d = 0. 17 (a) Make the substitution x = y −b 3. Combine like terms to create an equation of the form y3 −3py − 2q = 0, with p, q in terms of b, c, and d. (b) Rearrange this equation as y3 = 3py + 2q. (c) Make the substitution y = s + t into (b), and prove that y is a solution of the cubic in part (a) if st = p and s3 + t3 = 2q. (d) Eliminate t between these two equations to get a quadratic in s3. (e) Solve this quadratic to find s3. By symmetry, what is t3? (f) Find a formula for y in terms of p and q. What about a formula for x? (g) What if we started with ax3 + bx2 + cx + d = 0, with a coefficient in front of the x3 term as well? Can you come up with a formula for x? 32. Starting with the same cubic as in Problem 31b. (a) Let c = cos θ. Remember that cos 3θ = 4c3 −3c, as we proved. Substitute y = 2c√p into y3 = 3py + 2q to obtain 4c3 −3c = q p3/2 . (b) Provided that q2 ≤p3, show that y = 2√p cos 1 3(θ + 2πn) , where n is an integer. Why does this yield all three solutions? (c) Explain how you would find θ from p and q, and how we would use what we have found to solve an arbitrary cubic ax3 + bx2 + cx + d = 0. 18
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SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: 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SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: 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SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: SessionType: HTTP Cookie Storage Duration: PersistentType: HTML Local Storage optimizely_data$#$#$contextual_mabPendingMaximum Storage Duration: PersistentType: HTML Local Storage optimizely_data$pending_eventsThis cookie is set to make split-tests on the website, which optimizes the website's relevance towards the visitor – the cookie can also be set to improve the visitor's experience on a website.Maximum Storage Duration: PersistentType: HTML Local Storage v2/zaius.gifPendingMaximum Storage Duration: SessionType: Pixel Tracker optimizelyEndUserIdUsed to measure how selected users react to targeted changes to the website's content and functionality, in order to determine what variation is most efficacious in terms of converting users to customers.Maximum Storage Duration: 180 daysType: HTTP Cookie googe 1_ga_0EVGQ1TF2G part of the new GA4 measurement protocol, specifically a Google Analytics 4 Measurement ID-specific cookie.Maximum Storage Duration: 1 dayType: HTTP Cookie www.royalmint.com 1trm_sess_eventsTracks user interactions within sesionMaximum Storage Duration: SessionType: HTML Local Storage Marketing 91- [x] Marketing cookies are used to track visitors across websites. The intention is to display ads that are relevant and engaging for the individual user and thereby more valuable for publishers and third party advertisers. Meta Platforms, Inc.4Learn more about this providerlastExternalReferrerDetects how the user reached the website by registering their last URL-address.Maximum Storage Duration: PersistentType: HTML Local Storage lastExternalReferrerTimeDetects how the user reached the website by registering their last URL-address.Maximum Storage Duration: PersistentType: HTML Local Storage topicsLastReferenceTimeCollects data on the user across websites - This data is used to make advertisement more relevant.Maximum Storage Duration: PersistentType: HTML Local Storage _fbpUsed by Facebook to deliver a series of advertisement products such as real time bidding from third party advertisers.Maximum Storage Duration: 3 monthsType: HTTP Cookie Amazon 3Learn more about this providerjumbe_versionPendingMaximum Storage Duration: SessionType: HTTP Cookie z_idsyncsUSED BY THE OPTIMZELY DATA PLATFORM FOR MULTI VARIANT TESTINGMaximum Storage Duration: 2 yearsType: HTTP Cookie zaius_js_versionUSED BY THE OPTIMZELY DATA PLATFORM FOR MULTI VARIANT TESTINGMaximum Storage Duration: 2 yearsType: HTTP Cookie Appnexus 1Learn more about this providerpixiePresents the user with relevant content and advertisement. The service is provided by third-party advertisement hubs, which facilitate real-time bidding for advertisers.Maximum Storage Duration: SessionType: Pixel Tracker Google 5Learn more about this providerSome of the data collected by this provider is for the purposes of personalization and measuring advertising effectiveness. NIDRegisters a unique ID that identifies a returning user's device. The ID is used for targeted ads.Maximum Storage Duration: 6 monthsType: HTTP Cookie _gaRegisters a unique ID that is used to generate statistical data on how the visitor uses the website.Maximum Storage Duration: 2 yearsType: HTTP Cookie ga#Used by Google Analytics to collect data on the number of times a user has visited the website as well as dates for the first and most recent visit. Maximum Storage Duration: 2 yearsType: HTTP Cookie _gcl_auUsed by Google AdSense for experimenting with advertisement efficiency across websites using their services. Maximum Storage Duration: 3 monthsType: HTTP Cookie _gcl_lsTracks the conversion rate between the user and the advertisement banners on the website - This serves to optimise the relevance of the advertisements on the website. Maximum Storage Duration: PersistentType: HTML Local Storage HubSpot 1Learn more about this provider__ptq.gifSends data to the marketing platform Hubspot about the visitor's device and behaviour. Tracks the visitor across devices and marketing channels.Maximum Storage Duration: SessionType: Pixel Tracker JotForm 1Learn more about this provideruserRefererDetermines how the user accessed the website. This information is used by the website operator in order to measure the efficiency of their marketing. Maximum Storage Duration: 30 daysType: HTTP Cookie LinkedIn 2Learn more about this providerbcookieUsed by the social networking service, LinkedIn, for tracking the use of embedded services.Maximum Storage Duration: 1 yearType: HTTP Cookie lidcUsed by the social networking service, LinkedIn, for tracking the use of embedded services.Maximum Storage Duration: 1 dayType: HTTP Cookie Microsoft 5Learn more about this provider_clckPersists a user ID to link visits across sessions for Microsoft Clarity.Maximum Storage Duration: 1 yearType: HTTP Cookie _clskGroups multiple page views into a single session for Clarity analytics.Maximum Storage Duration: 1 yearType: HTTP Cookie MUIDUsed by Microsoft to identify users across websites for advertising, tracking, and performance reportingMaximum Storage Duration: 1 yearType: HTTP Cookie ANONCHKMicrosoft tracking cookie used for ad click tracking, fraud detection, and improving ad relevance in Microsoft Advertising and Clarity.Maximum Storage Duration: 1 yearType: HTTP Cookie CLIDStores a unique identifier to track user behavior across visits for analytics via Microsoft Clarity.Maximum Storage Duration: 1 yearType: HTTP Cookie Optimizely 11Learn more about this provideroptimizely_data$#$event_queuePendingMaximum Storage Duration: PersistentType: HTML Local Storage optimizely_data$#$layer_mapPendingMaximum Storage Duration: PersistentType: HTML Local Storage optimizely_data$#$layer_statesPendingMaximum Storage Duration: PersistentType: HTML Local Storage optimizely_data$#$session_statePendingMaximum Storage Duration: PersistentType: HTML Local Storage optimizely_data$#$tracker_optimizelyPendingMaximum Storage Duration: PersistentType: HTML Local Storage optimizely_data$#$variation_mapPendingMaximum Storage Duration: PersistentType: HTML Local Storage optimizely_data$#$visitor_profilePendingMaximum Storage Duration: PersistentType: HTML Local Storage optimizelyDomainTestCookie[x2]Tracks the conversion rate between the user and the advertisement banners on the website - This serves to optimise the relevance of the advertisements on the website. Maximum Storage Duration: 180 daysType: HTTP Cookie optimizelyOptOut[x2]Collects visitor data related to the user's visits to the website, such as the number of visits, average time spent on the website and what pages have been loaded, with the purpose of displaying targeted ads.Maximum Storage Duration: SessionType: HTTP Cookie Pinterest 3Learn more about this provider_pinterest_ct_uaUsed by Pinterest to track the usage of services.Maximum Storage Duration: 1 yearType: HTTP Cookie v3/Used by Pinterest to track the usage of services.Maximum Storage Duration: SessionType: Pixel Tracker _pin_unauthUsed by Pinterest to track the usage of services.Maximum Storage Duration: 1 yearType: HTTP Cookie Rakuten Marketing 4Learn more about this providerconsent/v3/pDetermines whether the user has accepted the cookie consent box. Maximum Storage Duration: SessionType: Pixel Tracker rmuidRakuten Marketing user ID used to track affiliate interactions and conversions.Maximum Storage Duration: 1 yearType: HTTP Cookie __rmcoNecessary for the implementation of Instagram picture-gallery onto the website. Maximum Storage Duration: PersistentType: HTML Local Storage __rmidPendingMaximum Storage Duration: PersistentType: HTML Local Storage Reddit 3Learn more about this providerrp.gifNecessary for the implementation of the Reddit.com's share-button function.Maximum Storage Duration: SessionType: Pixel Tracker _rdt_uuid[x2]Reddit - Used to track visitors on multiple websites, in order to present relevant advertisement based on the visitor's preferences.Maximum Storage Duration: 3 monthsType: HTTP Cookie StackAdapt 9Learn more about this providersa-user-id[x2]Used to track visitors on multiple websites, in order to present relevant advertisement based on the visitor's preferences. Maximum Storage Duration: 1 yearType: HTTP Cookie sa-user-idUsed to track visitors on multiple websites, in order to present relevant advertisement based on the visitor's preferences. Maximum Storage Duration: PersistentType: HTML Local Storage sa-user-id-v2Used to track visitors on multiple websites, in order to present relevant advertisement based on the visitor's preferences. Maximum Storage Duration: PersistentType: HTML Local Storage sa-user-id-v2[x2]Used to track visitors on multiple websites, in order to present relevant advertisement based on the visitor's preferences. Maximum Storage Duration: 1 yearType: HTTP Cookie sa-user-id-v3[x2]Collects data on the user across websites - This data is used to make advertisement more relevant.Maximum Storage Duration: 1 yearType: HTTP Cookie sa-user-id-v3Collects data on the user across websites - This data is used to make advertisement more relevant.Maximum Storage Duration: PersistentType: HTML Local Storage TikTok 8Learn more about this providertt_appInfoUsed by the social networking service, TikTok, for tracking the use of embedded services.Maximum Storage Duration: SessionType: HTML Local Storage tt_pixel_session_indexUsed by the social networking service, TikTok, for tracking the use of embedded services.Maximum Storage Duration: SessionType: HTML Local Storage tt_sessionIdUsed by the social networking service, TikTok, for tracking the use of embedded services.Maximum Storage Duration: SessionType: HTML Local Storage _tt_enable_cookieUsed by the social networking service, TikTok, for tracking the use of embedded services.Maximum Storage Duration: 1 yearType: HTTP Cookie _ttp[x2]Used by the social networking service, TikTok, for tracking the use of embedded services.Maximum Storage Duration: 1 yearType: HTTP Cookie ttcsidStores a unique TikTok Pixel session ID to link user interactions and attribute conversions within a single browsing session.Maximum Storage Duration: 1 yearType: HTTP Cookie ttcsid_CJ4HABBC77UC1837QFO0Stores a unique TikTok Pixel session ID to link user interactions and attribute conversions within a single browsing session.Maximum Storage Duration: 1 yearType: HTTP Cookie Twitter Inc.4Learn more about this provideri/adsct[x2]The cookie is used by Twitter.com in order to determine the number of visitors accessing the website through twitter advertisement content. Maximum Storage Duration: SessionType: Pixel Tracker muc_adsCollects data on user behaviour and interaction in order to optimize the website and make advertisement on the website more relevant. Maximum Storage Duration: 400 daysType: HTTP Cookie personalization_idThis cookie is set by Twitter - The cookie allows the visitor to share content from the website onto their Twitter profile. Maximum Storage Duration: 400 daysType: HTTP Cookie YouTube 17Learn more about this provider#-#[x2]Used to track user’s interaction with embedded content.Maximum Storage Duration: SessionType: HTML Local Storage iU5q-!O9@[#COOKIETABLE_ADVERTISING#]nbsp;[x2]Registers a unique ID to keep statistics of what videos from YouTube the user has seen.Maximum Storage Duration: SessionType: HTML Local Storage LAST_RESULT_ENTRY_KEY[x2]Used to track user’s interaction with embedded content.Maximum Storage Duration: SessionType: HTTP Cookie nextIdUsed to track user’s interaction with embedded content.Maximum Storage Duration: SessionType: HTTP Cookie requestsUsed to track user’s interaction with embedded content.Maximum Storage Duration: SessionType: HTTP Cookie TESTCOOKIESENABLED[x2]Used to track user’s interaction with embedded content.Maximum Storage Duration: 1 dayType: HTTP Cookie yt.innertube::nextIdRegisters a unique ID to keep statistics of what videos from YouTube the user has seen.Maximum Storage Duration: PersistentType: HTML Local Storage __Secure-YECYouTube (owned by Google) and is related to user authentication and session management.Maximum Storage Duration: SessionType: HTTP Cookie __Secure-YNIDPendingMaximum Storage Duration: 180 daysType: HTTP Cookie LogsDatabaseV2:V#\|\|LogsRequestsStoreUsed to track user’s interaction with embedded content.Maximum Storage Duration: PersistentType: IndexedDB remote_sidNecessary for the implementation and functionality of YouTube video-content on the website. Maximum Storage Duration: SessionType: HTTP Cookie VISITOR_INFO1_LIVETries to estimate the users' bandwidth on pages with integrated YouTube videos.Maximum Storage Duration: 180 daysType: HTTP Cookie YSCRegisters a unique ID to keep statistics of what videos from YouTube the user has seen.Maximum Storage Duration: SessionType: HTTP Cookie google 6__Secure-1PAPISID Google cookie used for ad personalization and user tracking across siteMaximum Storage Duration: 1 yearType: HTTP Cookie __Secure-1PSID Google cookie used for ad personalization and user tracking across siteMaximum Storage Duration: 1 yearType: HTTP Cookie __Secure-3PAPISID Google cookie used for ad personalization and user tracking across siteMaximum Storage Duration: 1 yearType: HTTP Cookie APISIDGoogle cookie used to create a user profile based on interests and display personalized ads across Google servicesMaximum Storage Duration: 1 yearType: HTTP Cookie IDE third-party marketing cookie set by Google DoubleClick to track and serve personalized ads across websites.Maximum Storage Duration: 1 yearType: HTTP Cookie SAPISIDSAPISID is a Google cookie used to build a profile of the user’s interests for personalized ads across Google services; itMaximum Storage Duration: 1 yearType: HTTP Cookie royalmint.com 1rmStore37343Part of Rakuten Advertising's tracking and affiliate system. 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16270	https://elementarymath.edc.org/resources/the-number-line-subtraction-and-measurement/	The Number Line – subtraction, and measurement – Elementary Math Contact but that is never its purpose. We don’t rely on the number line for getting answers — for that, we want the children to know basic facts and methods and use their heads — but we do use the number line to understand things about the operations (addition and subtraction) and to understand what the answers mean. For example, the answer to the subtraction problem 92 – 49 is the distance between those two numbers on the number line. That image can greatly help mental computation: 49 to 50 is one step, 50 to 90 is another forty steps, and 90 to 92 is another two steps, so altogether 43 steps. (See “Number line in addition and subtraction” below.) It also makes arithmetic with negative numbers a snap when students later learn about those numbers. And the number line is essential for full understanding of fractions and decimals. In fact, a ruler (a number line!) is one of the important places students encounter and need to use fractions. In counting Number lines are first used just to show sequence—numbers standing on line in order! At this stage, neither the straightness of the line nor the distance between numbers is mathematically important, though our images are always standard anyway. Some of the strategy we used was drawn from the work of Frédérique Papy, whose pre-K and K students would even draw these in spirals and zig-zags. Children will look at chunks of the line, not always starting at 1, and will work forwards or backwards from some number that is placed on the line. They are learning about sequence and order, and that develops somewhat independently from counting. As you know (and I mentioned over the phone to assure you that I knew, too) kids are often able to recite the numbers—“counting,” as their parents might say—before they can actually count objects correctly, one-to-one. In measurement K students also learn about intervals on the number line, but just begin that process. (I have only grades 1-5 with me, so I can’t say where in K.) That takes time, and is harder to develop than it was forty (or so) years ago. Kids used to gain the interval idea (in a slightly different form) from their experience playing board games. They knew that when they rolled a 5, they had to count their five spaces beginning with the next space. That is, they were counting moves, not positions. Today’s children—at all SE levels, to only slightly different degrees—have had far less board-game experience, so they need to get that idea in school. Measurement depends on it. Addition and subtraction with counters does not depend on it, but those operations on the number line do depend on it. In addition and subtraction If we can add and subtract with counters, why use the number line? To connect these operations with measurement, and also because the counters no longer suffice when we get to fractions, decimals, and negative numbers. Over time, kids will connect number line images with thermometers, clocks, rulers (with fractional inches)… Coordinate graphs are based on perpendicular number lines; even bar graphs require the measurement idea more than the count idea, although they can begin with count. Addition and subtraction, or comparison, of distance is also why we use Cuisenaire rods instead of Unifix cubes. (Unifix cubes emphasize counting—one can’t know relationships among Unifix rods without counting because there are no fixed lengths.) Part-part-whole in red rod + green rod = yellow rod is easy to see without caring what numbers they represent. And if R + G = Y then Y – R = G. (See, for example, Gr 2, Ch 8, Lesson 1, but kids see this in Gr 1 as well.) Adding distance is further developed on open number lines. Students develop many ways to subtract (in second grade, they learn to subtract 8 from anything by subtracting 10 and then compensating, and then they extend that idea to other additions and subtractions). In Grade 2, Ch 2, Lesson 8, they develop the idea on the number line and in third grade, they develop it further and use it to subtract much larger numbers. Its value as a model is that it continues to work for negative numbers as well. And in decimals and fractions Locating fractions on the number line has been noted by several mathematicians and researchers as essential for understanding them as numbers and not just parts of pizza. Students need to learn to locate a fraction on a number line. Many misconceptions about the meaning of “one half” become apparent when students are asked to locate it on a number line. As a number it is less than 1 and greater than 0, in fact “half” way between them, but it is quite common for students, the first time they are asked to place it on the number line, to locate it between 1 and 2, possibly because those are the two digits used in writing the symbol ‘‘ or possibly because of the commonly-taught strategy of thinking of as meaning “one out of two.” Students who know about negative numbers before first encountering the question “where would 1/2 fit on the number line” sometimes place it exactly on top of 0, because they see the zero as the mark that divides the number line “in half.” These are, respectively, strategies based on how the number is written, and how a fraction represents a “part of” a geometric object, both of which tend to be taught earlier and more intensively than messages about fractions being numbers, in their own right, and having magnitude and order (like all other real numbers). Addition and subtraction with fractions can also benefit from a number line image. The calculation 10 – 2 can be represented, as can any other subtraction problem, as distance on a number line. Having a clear sense of what the computation means makes it easier to interpret, and solve mentally, problems like the following, without the baggage of converting mixed numbers to improper fractions and converting back. The object is not merely to make the computation easier, nor is it to render obsolete (which it does not) the converting-mixed-numbers-to-improper-fractions method, but to make clear what the computation means so that one can better understand the algorithm in those cases where it is necessary because the numbers are, without conversion, too hard to manage. Developmental issues In Grade 1 (Chapter 2, lesson 1), students “record jumps” on a number line. The image presents considerable cognitive challenge to kids. What stands out visually is the dots, not the spaces between them. Some children see the arrows as skipping “4 dots” because they encompass four, or skipping “2 dots” because the skip over two. Seeing the intervals is harder, and takes time, but is essential for measurement. Many children will not be secure experts at the end of this lesson, but observation shows that even 6th graders who have not had this experience before will take a while to get it. It is not the only reason children typically do poorly on measurement tasks on tests—a major reason is that they don’t get time doing real things that require measuring—but this appears to be one important reason. In Grade 3 of Think Math!, students “zoom out” on the number line (Chap 4 Lesson 7), looking at large numbers, and setting the stage for “zooming in” on the number line (Gr4, Ch8 L3). Share This material is based upon work supported by the National Science Foundation under NSF Grant No. DRL-1934161 (Think Math+C), NSF Grant No. DRL-1741792 (Math+C), and NSF Grant No. ESI-0099093 (Think Math). Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation. © 2023 Education Development Center. All Rights Reserved. This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.
16271	https://www.quora.com/How-do-you-find-the-normal-vector-of-a-plane-and-its-direction	Something went wrong. Wait a moment and try again. Norm of a Vector Planes (geometry) 3D Linear Algebra Normal Equations Spatial Geometry Product of Vectors Geometry Class Unit Normal Vector 5 How do you find the normal vector of a plane and its direction? · To find the normal vector of a plane, you need to understand the equation of the plane and how the normal vector relates to it. Here’s a step-by-step guide: Identify the Equation of the Plane The general equation of a plane in three-dimensional space can be expressed in the form: Ax+By+Cz+D=0 where A, B, C, and D are constants. Extract the Normal Vector From the equation of the plane, the coefficients A, B, and C represent the components of the normal vector n of the plane. Thus, the normal vector can be written as: n=⎛⎜⎝ABC⎞⎟⎠ Direc To find the normal vector of a plane, you need to understand the equation of the plane and how the normal vector relates to it. Here’s a step-by-step guide: Identify the Equation of the Plane The general equation of a plane in three-dimensional space can be expressed in the form: Ax+By+Cz+D=0 where A, B, C, and D are constants. Extract the Normal Vector From the equation of the plane, the coefficients A, B, and C represent the components of the normal vector n of the plane. Thus, the normal vector can be written as: n=⎛⎜⎝ABC⎞⎟⎠ Direction of the Normal Vector The direction of the normal vector is determined by its components. The normal vector points perpendicular to the surface of the plane. If you want to find a unit normal vector (a normal vector of length 1), you can normalize the normal vector n: nunit=n∥n∥=⎛⎜⎝ABC⎞⎟⎠√A2+B2+C2 Example For the plane given by the equation 2x−3y+4z−5=0: Normal Vector: n=⎛⎜⎝2−34⎞⎟⎠ Direction: The vector points in the direction of (2,−3,4). Unit Normal Vector: ∥n∥=√22+(−3)2+42=√4+9+16=√29 Thus, nunit=1√29⎛⎜⎝2−34⎞⎟⎠ Summary The normal vector of a plane can be directly obtained from its equation. The direction of the normal vector is determined by its components. Normal vectors can be normalized to find unit normal vectors. Related questions How do I find the normal vector of a plane with 3 points? How do I find normal of any vector? How do you calculate the direction vector with up to three points on a plane? How do you identify and name planes using two or three direction vectors? A direction vector indicates the orientation of a line. Why must a normal vector be used? Instead of a single direction vector, to indicate the orientation of the plane? Maximilian Bernard RF Systems Engineer · Author has 85 answers and 141.6K answer views · Updated 4y Originally Answered: What is the normal vector of a plane? · If two vectors, A and B, lie in a plane; the cross product of these two will result in a vector which is perpendicular to that plane. Incidentally, A×B=ABsinθ^n where ^n is a unit vector that is perpendicular to the plane of A and B and θ is the angle between them. You may be curious as to how ^n has only one direction, since there are two directions perpendicular to the plane. This is resolved by using the “right-hand rule.” Point your fingers in the direction of the If two vectors, A and B, lie in a plane; the cross product of these two will result in a vector which is perpendicular to that plane. Incidentally, A×B=ABsinθ^n where ^n is a unit vector that is perpendicular to the plane of A and B and θ is the angle between them. You may be curious as to how ^n has only one direction, since there are two directions perpendicular to the plane. This is resolved by using the “right-hand rule.” Point your fingers in the direction of the first vector, and curl them in the direction of the second vector; your thumb will be in the direction of ^n. Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Shaizy FA in Children & Wanting and Making Money, Punjab, Pakistan (Graduated 1992) · 1y To find the normal vector of a plane, you can use the coefficients of the plane's equation. The normal vector is formed by the coefficients of x, y, and z in the equation. The direction of the normal vector indicates the direction perpendicular to the plane. LVPokerpro BSEE/MSEE in Electrial Engineering & Physics, California State University, Northridge (Graduated 1990) · Author has 3.1K answers and 1.6M answer views · Feb 8 To find the normal vector of a plane, you can use the cross product of two vectors that lie within the plane; this resulting vector will be perpendicular to the plane, which is the definition of a normal vector; the "direction" of the normal vector can be considered as pointing directly "outwards" from the plane, and can be reversed by negating the vector if necessary (pointing inwards). Related questions What is necessary in order to show the correct direction of a vector? What is the direction of a normal vector on a plane and why does it point in that direction? What will be the vector lying in the plane of a and b? How do I find a vector from a point along a line lying on a direction vector? Why are the direction ratios of a normal to a plane proportional to the direction ratios of that plane? Anonymous 5y Originally Answered: How do you find the unit normal vector of a plane? · Unit Normal Vector Any nonzero vector can be divided by its length to form a unit vector. Thus for a plane (or a line), a normal vector can be divided by its length to get a unit normal vector. Example: For the equation, x + 2y + 2z = 9, the vector A = (1, 2, 2) is a normal vector. Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Chaitanya Chavali Mathematical thinker · Author has 98 answers and 138.2K answer views · 5y Originally Answered: How do I compute the normal vector of a plane? · I am assuming that you know the equation of the plane. Let it be ax+by+cz+d=0 Then, λ(a^i+b^j+c^k) is normal to the plane. 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Igor PhD in Mechanical Engineering, Technion - Israel Institute of Technology (Graduated 1997) · Author has 3.1K answers and 1.3M answer views · 4y Originally Answered: How do I find a normal vector to a plane? · The components of the normal to a plane are the corresponding coefficients by x, y, and z in the linear equation of this plane, A x + B y + C z = 0. Vector (A, B, C) is perpendicular to any line in the plane Andrew Droffner Studied Mathematics at Rutgers University (Graduated 1995) · Author has 8.8K answers and 5.7M answer views · 5y Related How do I find a normal vector? A normal vector is orthogonal, or perpendicular, to another vector (or plane). There are a few ways to find a normal vector, depending on the available information. Dot Products A normal vector means the dot product is the number zero. A single known vector can be used to find another vector that is normal to it. There are algebraic and geometric equations to find the normal vector. →a∘→b=0 \|\|→a\|\|⋅\|\|→b\|\|cos(90∘)=0 →a∘→b=∑kak⋅bk=0 Cross Products The cross product has a normal vector as its answer. Two known vectors can A normal vector is orthogonal, or perpendicular, to another vector (or plane). There are a few ways to find a normal vector, depending on the available information. Dot Products A normal vector means the dot product is the number zero. A single known vector can be used to find another vector that is normal to it. There are algebraic and geometric equations to find the normal vector. →a∘→b=0 \|\|→a\|\|⋅\|\|→b\|\|cos(90∘)=0 →a∘→b=∑kak⋅bk=0 Cross Products The cross product has a normal vector as its answer. Two known vectors can be used to find the normal vector answer, or its anti-normal negative vector. Both product vectors →c,(−1)→c are perpendicular to the plane defined by →a×→b. →a×→b=→c⟺→a⊥→c,→b⊥→c and the anti-normal negative vector … (−1)→c⟺→a⊥→c,→b⊥→c Functions and Normal Vectors Find a unit vector →n, normal to a function’s curve. Read the link provided. Constructing a unit normal vector to curve (article) \| Khan Academy Sponsored by CDW Corporation How do updated videoconference tools support business goals? Upgrades with CDW ensure compatibility with platforms, unlock AI features, and enhance collaboration. Matt Warrell 9mo Related What is the direction of a normal vector on a plane and why does it point in that direction? The normal vector is, by definition, “normal” (perpendicular) to the plane. The reason for this is that, if it wasn’t perpendicular to the plane, there would be no way to know the exact orientation of the plane. John Fryer Author has 890 answers and 885.5K answer views · 4y Related How do I find the normal vector of a plane with 3 points? How do I find the normal vector of a plane with three points? Finding the normal to a plane Call the three points Q, R and S. They are defined by a unique set of coordinates ( x, y, z ) Imagine that the line QR represents a vector a And that the line QS represents a vector b The fact that the cross-product a x b is perpendicular to both a and b makes it very useful to find the normal vector. We will call this cross product the normal vector n of our plane. Example: Find the normal vector of a plane passing through the points: Q(−1,1,2), R(−4,2,2), S(−2,1,5). Let the line QR represent a vector a and the How do I find the normal vector of a plane with three points? Finding the normal to a plane Call the three points Q, R and S. They are defined by a unique set of coordinates ( x, y, z ) Imagine that the line QR represents a vector a And that the line QS represents a vector b The fact that the cross-product a x b is perpendicular to both a and b makes it very useful to find the normal vector. We will call this cross product the normal vector n of our plane. Example: Find the normal vector of a plane passing through the points: Q(−1,1,2), R(−4,2,2), S(−2,1,5). Let the line QR represent a vector a and the line QS represent a vector b a = ( -4–1, 2- 1, 2-2 ) = ( -3, 1, 0 ) b = ( -2–1, 1-1, 5-2 ) = ( -1, 0, 3 ) To calculate the cross-producta×b perhaps this is one method. Construct a table for i, j and k values in table form as below. Then repeat the first two values. This makes the cross product easier to do without error Thus start with the vector component i at top left and descend diagonally via the second or j component for vector a then continue to the k term of the second vectorb. We then subtract the other two components of vector a and band the best way to do this is to repeat the i and j components of our original vectors, so we can multiply diagonally starting with the second i with the third term of vector a and continue diagonally in the reverse direction arriving at the second component of vector b Repeat for the other vector components ie j and k in like fashion. We get: i x 1 x 3 - i x 0 x 0 = 3i j x 0 x -1 - j x -3 x 3 = 9j k x -3 x 0 - k x 1 x -1 = k Thus the cross-product = 3i+9j+k This is the normal vector to the plane ie n = [ 3, 9, 1 ] It is better to do the calculations with real examples than with unknown values or the results would be even more difficult to follow perhaps Vasant Barve SSC 11 in Mathematics & Science, Niphad (Graduated 1958) · Author has 360 answers and 243.2K answer views · 2y Related How do you find the magnitude and angle of a vector? A vector has magnitude and direction. Normally you are given these in a question or a problem. Else you measure those, viz to get Position Vector of Nagpur with respect to Mumbai, You measure distance between these cities on map. This is magnitude then measure the angle of line from Mumbai to Nagpur with respect to respect east or north . Thus you can get the vector. Here vector of Mumbai - Nagpur distance is 6.04 to some scale at 19.76 degrees North of East. You you are going in a vehicle, the speedometer tells you the magnitude of velocity vector and compass will tell you the direction thus y A vector has magnitude and direction. Normally you are given these in a question or a problem. Else you measure those, viz to get Position Vector of Nagpur with respect to Mumbai, You measure distance between these cities on map. This is magnitude then measure the angle of line from Mumbai to Nagpur with respect to respect east or north . Thus you can get the vector. Here vector of Mumbai - Nagpur distance is 6.04 to some scale at 19.76 degrees North of East. You you are going in a vehicle, the speedometer tells you the magnitude of velocity vector and compass will tell you the direction thus you get vector of velocity. Anirban Ghoshal Studied physics up till high school. · Upvoted by Justin Rising , PhD in statistics and David Vanderschel , PhD Mathematics & Physics, Rice (1970) · Author has 2.1K answers and 5.8M answer views · 9y Related If two planes are perpendicular, is the direction vector of the line where the two planes intersect simply the normal vector of either plane? No, the line along which they intersect is contained in both the planes. It isn't normal to the plane, but rather along either plane. Below is a diagram for reference. No, the line along which they intersect is contained in both the planes. It isn't normal to the plane, but rather along either plane. Below is a diagram for reference. Related questions How do I find the normal vector of a plane with 3 points? How do I find normal of any vector? How do you calculate the direction vector with up to three points on a plane? How do you identify and name planes using two or three direction vectors? A direction vector indicates the orientation of a line. Why must a normal vector be used? Instead of a single direction vector, to indicate the orientation of the plane? What is necessary in order to show the correct direction of a vector? What is the direction of a normal vector on a plane and why does it point in that direction? What will be the vector lying in the plane of a and b? How do I find a vector from a point along a line lying on a direction vector? Why are the direction ratios of a normal to a plane proportional to the direction ratios of that plane? What is the formula for a plane containing a point and a normal vector? How can one find the normal vector of a line using its direction vector? How do I find a vector which is in the same direction with the other vector? How do you verify which plane contains a line (vectors, vector analysis, and math)? How do I find the angle between a vector and a plane (let's say an xy plane)? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
16272	https://intech-gmbh.ru/en/heat_exchangers_calc_and_select/	Heat exchangers calculation and selection Together to success since 1997 "INTECH GmbH" EN About Company Corporate Policy LCC "INTECH GmbH" Merits and Advantages Our Team Our Company’s Ethics and Values Our Customers Our customers in chemical industry Our customers in metal and ore-dressing industries Supplied equipment for chemical industry for metallurgy Complex maintenance services Supply of spare parts and wear components for GE (General Electric) gas turbines Supply of wearing spare parts and components Asset Performance Management by IT (Information Technology) solutions Supply of spare parts and wear components for Siemens gas turbines Production of components and spare parts for turbines and aircraft engines using precision casting method Production and recovery of gas turbine blade Technical Information Contacts Your distributor in CIS Our partners Intech GmbH - About the Company Heat exchangers calculation and selection Heat exchangers calculation and selection Heat exchangers calculation Thermal Calculations Design Calculation Hydraulic Calculations Examples of Problems for Calculating and Selecting Heat Exchangers General Information about Heat Exchangers Heat exchangers calculation Thermal Calculations While setting the task of conducting a technical calculation of heat-exchange equipment, input data of the heat transfer medium (flow rate, initial and final temperature, physico-chemical properties) should be known. The missing values are determined during the thermal calculation. The aim of the thermal calculation is to determine the main technical data of heat exchange equipment, i.e. thermal load, heat transfer medium flows, average temperature difference of the heat transfer media, heat transfer coefficient. These parameters are calculated based on the heat balance equation. Below is an example of a general calculation of heat exchange equipment. In heat exchangers, heat energy is transferred from one process stream (heating medium) to another, resulting in the heating or cooling. Q = Q г = Q х Q – quantity of heat transferred or received by the heat transfer medium [W], Wherefrom: Q г = G г c г·(t гн – t гк) и Q х = G х c х·(t хк – t хн) Where G г,х – hot and cold heat transfer medium flow rates [kg/h]; с г,х – heat capacity of hot and cold heat transfer media [J/kg·deg]; t г,х н – initial temperature of hot and cold heat transfer media [°C]; t г,х к – final temperature of hot and cold heat transfer media [°C]; It should be taken into account that the amount of transmitted/ received heat depends on the aggregate state of the heat transfer media. If, in the course of heat exchange process, it does not change, then the above mentioned formula is used for calculation. If one (or both) heat transfer medium changes its aggregate state (steam heating), the following formula is used to calculate the amount of transmitted or received heat: Q = Gc п·(t п – t нас)+ Gr + Gc к·(t нас – t к) Where r – heat of condensation [J/kg]; с п,к – specific heat capacities of steam and condensate [J/kg·deg]; t к – condensate temperature at the unit outlet [°C]. If condensate does not cool, the first and third terms are excluded from the right side of the equation, and then it takes the following form: Q гор = Q конд = Gr The flow of the heat transfer media can be determined as follows: G гор = Q/c гор(t гн – t гк) или G хол = Q/c хол(t хк – t хн) In case of steam heating, its consumption is defined by the formula: G пара = Q/ Gr Where G – flow rate of corresponding heat transfer medium [kg/h]; Q – amount of heat [W]; с – specific heat of heat transfer media [J/kg·deg]; r – heat of condensation [J/kg]; t г,х н – initial temperature of hot and cold heat transfer media [°C]; t г,х к – final temperature of hot and cold heat transfer media [°C]. The difference between heat transfer media is a driving force of the heat exchange process. Since the temperature of the streams changes as they pass the path, the temperature difference also changes, so it is customary to use an averaged value for the calculation. The average temperature difference for the direct and counter flows is calculated as the average logarithmic value of: ∆t ср = (∆t б - ∆t м) / ln (∆t б/∆t м) Where ∆t б, ∆t м – greater and smaller average temperature difference of the heat transfer media at the unit inlet and outlet. The same formula, with the addition of the correction factor ∆t ср = ∆t ср ·f попр, is used during the cross and mixed flows of heat transfer media. Heat transfer coefficient can be determined as follows: 1/k = 1/α 1 + δ ст/λ ст + 1/α 2 + R заг In the equation: δ ст – wall thickness [mm]; λ ст – coefficient of thermal conductivity of the wall material [W/m·deg]; α 1,2 – heat transfer coefficients of the inner and outer sides of the wall [W/m 2·deg]; R заг – coefficient of wall contamination. Design Calculation Design calculation of heat exchange equipment is divided into the preliminary and detailed calculation. The preliminary calculation aims to choose approximate values of the heat transfer coefficient from reference materials, and to determine heat exchange surface and the flow path cross-section area for heat transfer media. The approximate heat exchange surface is calculated as follows: F = Q/ k·∆t ср [m 2] The flow path cross-section area of the heat transfer media is determined from the formula: S = G/(w·ρ) [m 2] Where G – heat transfer medium flow [kg/h]; (w·ρ) – mass flow rate of the heat transfer medium [kg/m 2·s]. For calculations, the flow velocity is assumed based on the type of heat transfer medium: Type of heat transfer mediumFlow velocity, m/s Viscous liquids<1 Low-viscous liquids 1-3 Dusty gases 5-10 Pure gases 10-15 Saturated steam 30-50 Based on the preliminary design calculation, one or more heat exchangers, which satisfy the required heat exchange surface, are selected. Then, detailed structural and thermal calculations are carried out under specified conditions for the selected units. During design calculations, additional indicators are determined for different heat exchangers. E.g., for shell-and-tube heat exchangers, a length or a number of tubes is to be found. l = F/ πdn Where: l – tube length [m]; n – number of tubes [pcs]; F – required heat exchange surface [m 2]; d – tube diameter [m]; Typically, during shell-and-tube heat exchanger calculations, the number and diameter of tubes is determined from reference materials. The inner diameter is determined as follows: D вн = s (b-1) + 4d п Where: D вн – inner diameter of the heat exchanger [m]; s – tube pitch [m] (to be assumed from 1.2 to 1.5 d н); d п – tube outer diameter [m]; b – number of tubes [m] (b = 2a-1, where a is the number of tubes on the side of the largest hexagon); Then, the tube and shell area is determined: S тр = (πd 2 в /4) n х Where: S тр – tube space area [m 2]; d 2 в – tube inner diameter [m]; n х – number of tubes in one pass; S мтр = (π/4) (D 2 - nd 2 п) Where: S мтр – shell space area [m 2]; D – shell inner diameter [m]; d в – tube outer diameter [m]; n – number of tubes in one pass; In a specific case, when longitudinal partitions are placed in the shell side to increase the heat exchange intensity, the area will be determined as follows: S мтр = (π/4) (D 2 - nd 2 п/ N) Where: N – number of passes when divided by partitions; During the design calculation of coil heat exchangers, the total length of the coil as well as the number of turns and sections are determined. L = F/ πd р Where: L – total length of the coil [m]; d р – design diameter of the coil tube [m]; n = L/ πd р Where: n – number of turns; Knowing the heat transfer medium flow rate and velocity in the coil tube, the number of coil sections can be determined: m = V сек/(π/4)d 2 w Where: V сек – flow rate [kg/h]; d – coil tube diameter [m]; w – heat transfer medium velocity in the coil tube [m/s]; In the spiral heat exchanger calculations, such data as the cross-section of channels as well as width, length, pitch, number of turns and outer diameter of the spiral, are determined. S = G/W Where S – channel cross-section [m 2]; G – heat transfer medium flow [kg/h]; W – heat transfer medium mass velocity [kg/m 2·sec]. Hydraulic Calculations When process streams pass through heat exchange equipment, a loss of head or pressure, caused by the hydraulic resistance of the units, occurs. The general formula for calculating the hydraulic resistance, created by heat exchangers, is as follows: ∆Р п = (λ·(l/d) + ∑ζ) · (ρw 2/2) Where ∆p п – pressure loss [Pa]; λ – coefficient of friction; l – tube length [m]; d – tube diameter [m]; ∑ζ – sum of local resistance coefficients; ρ – density [kg/m 3]; w – flow velocity [m/s]. Examples of Problems for Calculating and Selecting Heat Exchangers (Heat Exchange Equipment) with Solutions Problem 1 A hot product flow, leaving the reactor, should be cooled from the initial temperature t 1 н = 95°C to the final temperature t 1 к = 50°C; for this purpose, it is directed to a refrigerator, where water with the initial temperature of t 2 н = 20°C is supplied. Please calculate ∆t ср for the direct flow and counter flow conditions in the refrigerator. Solution: 1) Since the final temperature of the cooling water t 2 к for the direct flow of the heat transfer media can not exceed the value of the final temperature of the hot heat transfer medium (t 1 к = 50°C), so let’s assume that t 2 к = 40°C. Let’s calculate average temperatures at the refrigerator inlet and outlet: ∆t н ср = 95 - 20 = 75; ∆t к ср = 50 - 40 = 10 ∆t ср = 75 - 10 / ln(75/10) = 32.3 °C 2) For the counter flow conditions, let’s assume the final water temperature the same as for the direct flow of the heat transfer media, i.e. t 2 к = 40°C. ∆t н ср = 95 - 40 = 55; ∆t к ср = 50 - 20 = 30 ∆t ср = 55 - 30 / ln(55/30) = 41.3°C Problem 2 Using the conditions of Problem 1, please determine the required heat exchange surface (F) and the cooling water flow (G). The hot product flow G = 15000 kg/h and its heat capacity C = 3430 J/kg·grad (0.8 kcal·kg·deg). Cooling water parameters are as follows: heat capacity c = 4080 J/kg·grad (1 kcal·kg·grad), heat transfer coefficient k = 290 W/m 2·grad (250 kcal/m 2deg). Solution: Using the heat balance equation, we will obtain an expression for determining a heat flux when cold heat transfer medium is heated: Q = Q гт = Q хт Wherefrom: Q = Q гт = GC (t 1 н - t 1 к) = (15000/3600)·3430·(95 - 50) = 643125 W Assuming t 2 к = 40°C, we will find the flow rate of the cold heat transfer medium: G = Q/ c(t 2 к - t 2 н) = 643125/ 4080(40 - 20) = 7.9 kg/s = 28 500 kg/h Required heat exchange surface In case of a direct flow: F = Q/k·∆t ср = 643125/ 290·32.3 = 69 m 2 In case of a counter flow: F = Q/k·∆t ср = 643125/ 290·41.3 = 54 m 2 Problem 3 At a plant, gas is transported through a steel pipeline with an outer diameter d 2 = 1500 mm, wall thickness δ 2 = 15 mm, thermal conductivity λ 2 = 55 W/m·deg. From the inside, the pipeline is lined with fireclay bricks, whose thickness is δ 1 = 85 mm, thermal conductivity λ 1 = 0.91 W/m·deg. The coefficient of heat transfer from gas to the wall is α 1 = 12.7 W/m 2·deg; from the outer surface of the wall to air is α 2 = 17.3 W/m 2·deg. Please find the coefficient of heat transfer from gas to air. Solution: 1) Let’s determine the inner diameter of the pipeline: d 1 = d 2 - 2·(δ 2 + δ 1) = 1500 - 2(15 + 85) = 1300 mm = 1.3 m Average lining diameter: d 1 ср = 1300 + 85 = 1385 mm = 1.385 m Average pipeline wall diameter: d 2 ср = 1500 - 15 = 1485 mm = 1.485 m Let’s calculate the coefficient of heat transfer by the formula: k = [(1/α 1)·(1/d 1) + (δ 1/λ 1)·(1/d 1 ср)+(δ 2/λ 2)·(1/d 2 ср)+(1/α 2)]-1 = [(1/12.7)·(1/1.3) + (0.085/0.91)·(1/1.385)+(0.015/55)·(1/1.485)+(1/17.3)]-1 = 5.4 W/m 2·grad Problem 4 A single-pass shell-and-tube heat exchanger heats methanol with water from the initial temperature of 20 to 45°C. The water flow is cooled from 100 to 45°C. A tube bundle of the heat exchanger contains 111 tubes, the diameter of one tube is 25x2.5 mm. The velocity of the methanol flow through the tubes is 0.8 m/s (w). The heat transfer coefficient is 400 W/m 2·deg. Please, determine the total length of the tube bundle. Solution: Let’s determine the average temperature difference of the heat transfer media as the average logarithmic value. ∆t н ср = 95 - 45 = 50; ∆t к ср = 45 - 20 = 25 ∆t ср = 50 + 25 / 2 = 37.5°C Then, let’s determine the average temperature of the heat transfer medium flowing through the tube side space. ∆t ср = 45 + 20 / 2 = 32.5°C Let’s determine a mass flow of methanol. G сп = n·0.785·d вн 2·w сп·ρ сп = 111·0.785·0.02 2·0.8· = 21.8 ρ сп = 785 kg/m 3 – methanol density at 32.5°C, the value is taken the reference literature. Then, let’s determine the heat flux. Q = G сп с сп (t к сп – t н сп) = 21.8·2520 (45 – 20) = 1.373·10 6 W c сп = 2520 kg/m 3– heat capacity of methanol at 32.5°C, the value is taken the reference literature. Let’s determine the required heat exchange surface. F = Q/ K∆t ср = 1.373·10 6/ (400·37.5) = 91.7 m 3 Let’s calculate the total length of the tube bundle by the average diameter of the tubes. L = F/ nπd ср = 91.7/ 111·3.14·0.0225 = 11.7 m. In accordance with the recommendations, the total length of the tube bundle should be divided into several sections of the proposed standard size, with a required margin of the heat exchange surface to be provided. Problem 5 A plate heat exchanger is used to heat a 10% NaOH solution flow from 40°C to 75°C. The sodium hydroxide flow is 19000 kg/h. Water vapor condensate with a flow rate of 16000 kg/h and initial temperature of 95°C is used as a heating agent. Assume that heat transfer coefficient is 1400 W/m 2·deg. Please calculate basic parameters of the plate heat exchanger. Solution: Let's find the amount of heat transferred. Q = G р с р (t к р – t н р) = 19000/3600 · 3860 (75 – 40) = 713 028 W From the heat balance equation, let’s determine the final temperature of the condensate. t к х = (Q·3600/G к с к) – 95 = (713028·3600)/(16000·4190) – 95 = 56.7°C с р,к – heat capacity of the solution and condensate, the values are found in the reference materials. Let’s determine average temperatures of the heat transfer media. ∆t н ср = 95 - 75 = 20; ∆t к ср = 56.7 - 40 = 16.7 ∆t ср = 20 + 16.7 / 2 = 18.4°C Let’s determine the cross-section of the channels; for calculation, let’s assume the mass velocity of the condensate as Wk = 1500 kg/m 2·sec. S = G/W = 16000/3600·1500 = 0.003 m 2 Assuming the channel width b = 6 mm, we will find the width of the spiral. B = S/b = 0.003/ 0.006 = 0.5 m Based on the recommendations, let’s assume the width of the spiral according to the nearest larger tabulated value B = 0.58 m. Let’s refine the cross-section of the channel S = B·b = 0.58·0.006 = 0.0035 m 2 and mass velocity of the streams W р = G р/S = 19000/ 3600·0.0035 = 1508 kg/m 3·sec W к = G к/S = 16000/ 3600·0.0035 = 1270 kg/m 3·sec The heat transfer surface of the spiral heat exchanger is determined as follows. F = Q/K∆t ср = 713028/ (1400·18.4) = 27.7 m 2 Let’s determine the working length of the spiral L = F/2B = 27.7/(2·0.58) = 23.8 m Next, let’s determine a pitch of the spiral, while setting the sheet thickness as δ=5 mm. t = b + δ = 6 + 5 = 11 mm o calculate the number of turns of each spiral, the initial diameter of the spiral shall be assumed, based on the recommendations, as d = 200 mm. N = (√(2L/πt)+x 2) – x = (√(2·23.8/3.14·0,011)+8.6 2) – 8.6 = 29.5 where х = 0.5 (d/t - 1) = 0.5 (200/11 – 1) = 8.6 The required diameter of the spiral is determined as follows. D = d + 2Nt + δ = 200 + 2·29.5·11 + 5 = 860 mm. Problem 6 Please determine the hydraulic resistance of a heat transfer media, created in a four-way plate heat exchanger with a channel length of 0.9 m and an equivalent diameter of 7.5×10-3, when butyl alcohol is cooled with water. Butyl alcohol properties are as follows: flow rate G = 2.5 kg/s, velocity W = 0.240 m/s and density ρ = 776 kg/m 3 (Reynolds criterion Re = 1573 > 50). Cooling water properties are as follows: flow rate G = 5 kg/s, velocity W = 0.175 m/s and density ρ = 995 kg/m 3 (Reynolds criterion Re = 3101 > 50). Solution: Let’s determine the coefficient of local hydraulic resistance. ζ бс = 15/Re 0.25 = 15/1573 0.25 = 2.38 ζ в = 15/Re 0.25 = 15/3101 0.25 = 2.01 Let's clarify the velocity of alcohol and water in the fittings (assuming that d шт = 0.3m) W шт = G бс/ρ бс 0.785d шт 2 = 2.5/776 ·0.785·0.3 2 = 0.05 m/s is less than 2 m/s, therefore it can be ignored. W шт = G в/ρ в 0.785d шт 2 = 5/995 ·0.785·0.3 2 = 0.07 m/s is less than 2 m/s, therefore it can be ignored. Let's determine the hydraulic resistance for butyl alcohol and cooling water. ∆Р бс = хζ·(l/d) · (ρ бс w 2/2) = (4·2.38·0.9/ 0.0075)·(776·0.240 2/2) = 25532 Pa ∆Р в = хζ·(l/d) · (ρ в w 2/2) = (4·2.01·0.9/ 0.0075)·(995·0.175 2/2) = 14699 Pa. General Information about Heat Exchangers Heat exchange equipment is designed to transfer heat energy from one environment to another, i.e., for transferring heat from hot to cold heat transfer medium. The variety of units, differed in design, purpose and heat transfer methods, allows the implementation of customer-tailored processes. Heat exchange equipment can be used either as major equipment, or auxiliary (stand-alone) equipment. The fields of application of heat exchange equipment are as follows: Heat supply or removal during certain reactions; Process stream heating or cooling; Distillation; Adsorption and absorption; Melting of solids and crystallization of substances; Evaporation; Condensation. 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16273	https://math.stackexchange.com/questions/1561880/does-induction-find-all-solutions	Skip to main content Does induction find all solutions? Ask Question Asked Modified 9 years, 8 months ago Viewed 277 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Induction shows that an equality holds for all values of n. It doesn't show that this is the only equality or formula for the expression that may hold true, correct? For example, say a question asks to find an explicit formula for a functional equation given by f(n)=f(f(n−1))+f(n−f(n−1)) for n>1. You can prove by induction that f(n)=n is indeed an explicit expression for f(n), but it may not be the only one? If so, what steps must be taken to prove that it is the only one? Edit: Sorry, I meant to point out that f(1)=1 induction functional-equations Share CC BY-SA 3.0 Follow this question to receive notifications edited Dec 6, 2015 at 2:47 user19405892 asked Dec 6, 2015 at 2:36 user19405892user19405892 15.9k77 gold badges3838 silver badges121121 bronze badges 8 Could you post a the proof by induction that you're thinking of? I would imagine that it is a proof that there is only one solution - because you don't need induction to prove that f(n)=n is a solution; you just substitute it in. – Milo Brandt Commented Dec 6, 2015 at 2:40 1 f(x) = 0 is also a solution. – marty cohen Commented Dec 6, 2015 at 2:42 To prove by induction is quite simple. You just assume it is true for some k and then use the RHS expression for k+1 to show that it eventually simplifies to k+1. – user19405892 Commented Dec 6, 2015 at 2:43 @martycohen are you sure? – user19405892 Commented Dec 6, 2015 at 2:44 In this case n>1. – user19405892 Commented Dec 6, 2015 at 2:48 \| Show 3 more comments 2 Answers 2 Reset to default This answer is useful 2 Save this answer. Show activity on this post. Induction proves . . . whatever it proves. For instance: We might use an induction argument to show that some f is a solution to a functional equation E. In the absence of any other arguments, this won't say that f is the unique solution. On the other hand, we might use induction to show that f is the unique solution to a functional equation E. For a silly example, if E is f(n+1)=f(n)+1, we can prove by induction that every solution f to E is of the form f(x)=x+c for some fixed c. (Specifically, we would argue: let f satisfy E, and set c=f(0). Then by induction on n, we can show f(n)=n+c. So every solution to E has this form.) EDIT: in the specific case you're interested in, we can indeed use induction to prove that f(n)=n is the only solution. There's a few ways to phrase it, but here's one which is general enough to be useful in other contexts: We have an initial value, f(1)=1. (I'm assuming your natural numbers start with 1, not 0.) Informally: In the functional equation, each value of f(n) (for n>1) is determined by the values of f on numbers <n. Formally: By induction, we can show that if f and g are two solutions to the functional equation and f(1)=g(1), then f(n)=g(n) for every n; that is, f=g. (This is a good exercise.) So there is a unique solution. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Dec 6, 2015 at 2:51 answered Dec 6, 2015 at 2:45 Noah SchweberNoah Schweber 261k2222 gold badges385385 silver badges668668 bronze badges 2 Is it necessary to prove that this is the only solution for the purpose of this question or just proving by induction that it is a solution is okay? – user19405892 Commented Dec 6, 2015 at 2:52 1 I think the point of this answer is that there is nothing inherent in inductiom vs. any other type of proof that makes it better or worse in whether it proves a solution is sufficient or whether it is unique. – fleablood Commented Dec 6, 2015 at 3:10 Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. f(n)=f(f(n−1))+f(n−f(n−1)) If f(n)=an+b, an+b=f(a(n−1+b))+f(n−(a(n−1)+b))=a((a(n−1)+b))+b+a(n−(a(n−1)+b))+b=a(an−a+b)+2b+a(n−(an−a)−b))=a2n−a2+ab+2b+an−a2n+a2−ab=an+ab+2b−ab=an+2b so b=0, and f(n)=an for any a is a solution. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Dec 6, 2015 at 3:02 marty cohenmarty cohen 111k1010 gold badges8888 silver badges185185 bronze badges 1 Note that while this works for the OP's original question, their edit - adding the initial condition "f(1)=1" - changes things. – Noah Schweber Commented Dec 6, 2015 at 3:15 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions induction functional-equations See similar questions with these tags. Featured on Meta Community help needed to clean up goo.gl links (by August 25) Related 27 How does backwards induction work to prove a property for all naturals? 2 Proof by induction on {1,…,m} instead of N 2 Induction with negative step 5 Question on induction technique 48 Is it a new type of induction? 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16274	https://www.themathdoctors.org/how-many-different-meals-are-possible/	How Many Different Meals Are Possible? – The Math Doctors Loading web-font TeX/Size3/Regular Skip to content Main Menu Home Ask a Question Blog FAQ About The Math Doctors Contact Search Search for: How Many Different Meals Are Possible? December 5, 2018 February 24, 2024 / AQOTW, Probability / Alternatives, Counting / By Dave Peterson (An archive question of the week) While gathering combinatorics questions, there were several that stood out. This one will serve well to summarize the topic, showing multiple methods for counting, and contrasting other kinds of problems. The problem The question, from 2007, relates to an Arby’s promotion: How Many Different Dinners Can Be Ordered? I was trying to calculate the number of different meals you can get for the 5 for $5.95 at Arby's and wasn't confident in the way I was calculating it. Given 8 different choices, how many combinations of 5 are there? You do not need to choose different entrées. I am confused about the fact that you are calculating a combination with replacement. I thought that I would start this way... 8 C 1 for the possibility of choosing 5 entrées alike, 2(8 C 2) for choosing 4 alike and 1 different entrée. Multiplied by 2 because there are 2 ways to assign each pair of choices. 2(8 C 2) for choosing 3 alike and 2 alike and multiply by 2 because there are 2 ways to assign each pair. (8 C 3)(3 C 1) for choosing 2 alike, 2 alike and 1 different entrée and once the three are chosen, choose the one that is single. (8 C 3)(3 C 1) for choosing 3 alike, 1 different, 1 different. (8 C 4)(4 C 1) for choosing 2 alike, and 3 different entrées plus choosing which one is doubled up. (8 C 5) to choose all 5 entrées different. Then add each of the above together. Am I on the right track? Dave is representing the common notation \displaystyle nC_r = \frac{n!}{r!(n-r)!}n C r=n!r!(n−r)! (read as “_n choose r” ) for combinations. He has done very well. Doctor Wilko chose to expand on what he did: Thanks for writing to Ask Dr. Math! Yes, your approach to the problem looks excellent. I like this problem a lot. I think it's a good real-world problem. We often see many restaurants and pizza parlors trying to figure out how many different combinations of meals or pizzas they offer. The difficulty with these problems is that it is easy to calculate the number of combinations incorrectly. I'll present two solutions below to answer, "How many combinations of 5 items for $5.95 can one order from Arby's 8-item menu?" The first uses the same logic you applied in your work, and the second approaches it slightly differently. Then I'll also show some errors that are commonly made on this sort of problem. The first step, always, is to clarify the problem, restating it if necessary to expose all assumptions that are required: Recall, we are assuming: There are 8 choices on this particular menu. 5 items can be ordered (for $5.95). Duplicate items can be ordered (the 5 items don't have to be distinct). The last point is important. There are not 8 finite items to choose from; there are 8 choices on the menu and you can order anywhere from 0-5 of each of those items. So this is not a straight permutation or combination problem, both of which involve subsets of a given set, with no repetition allowed. In this case, the items are distinguishable but repeatable; order doesn’t matter. A good model for the choice of a meal would be a menu list, with a blank for each item in which you put how many of that item ordered; these numbers have to add up to 5: ____ Arby’s Melt ____ Small Drink ____ Ham Melt ____ Small Fries ____ Potato Cakes ____ Turnover ____ Chicken Sandwich ____ Shake First method: multiple cases The first is just a little different from Dave’s method, and has a slightly different order: For this solution, I'll try to explicitly enumerate the different ways I can order items from the menu and then apply the combinations formulas respectively. Last I'll add up all the answers to get the final answer. I'm basically partitioning the sample space into smaller subproblems, solving each subproblem, and then combining the results to answer the big problem. I can order 5 of the same item. There is only 1 choice. Which of the 8 should I choose? Mathematically, this can be written as, 8C1 = 8. There are 8 ways I can choose 1 item--once I pick it, then I can order a quantity of 5 of that 1 item. I can order 4 of one item and 1 other item. Here, there are only 2 choices. Out of the 8 menu items, which 2 should I choose? Mathematically, this can be written as, 8C2 = 28 But out of those two choices, I need to order 4 of one and 1 of the other. To figure out which of the items I order 4 of, I can multiply by 2C1 or 2. The final answer for this portion of the problem is, 8C2 2C1 = 28 2 = 56. Note, at this point, you could explicitly enumerate all the ways to order 4 of one item and 1 other item from the menu of 8 items. You'll see that there are 56 combinations. This case (like the others) could also be counted in other ways. For example, we could just use a permutation to choose two items in order, the first to be the 4 and the second to be the 1. This gives\displaystyle _8P_2 = \frac{8!}{6!} = 8\cdot7 = 56 8 P 2=8!6!=8⋅7=56 as before. If I continue with the same reasoning as above, the problem continues as follows: 3 same items, 2 other same items 8C2 2C1 = 56. 3 same items, 1 other item, 1 other item 8C3 3C1 = 168. 2 same items, 2 other same items, 1 other item 8C3 3C1 = 168. 2 same items, 1 other item, 1 other item, 1 other item 8C4 4C1 = 280. 5 different items 8C5 = 56. Total combinations of 5 items I can order from a menu of 8 items at Arby's is, 8 + 56 + 56 + 168 + 168 + 280 + 56 = 792. Apparently, Arby's boasts "Over 790 combinations!". Arby's claim gives me confidence in this answer. Doctor Wilko’s result can be written as _8C_1\ +\ _8C_2\ \cdot\ _2C_1\ +\ _8C_2\ \cdot\ _2C_1\ +\ _8C_3\ \cdot\ _3C_1\ +\ _8C_3\ \cdot\ _3C_1\ +\ _8C_4\ \cdot\ _4C_1\ +\ _8C_5. 8 C 1+8 C 2⋅2 C 1+8 C 2⋅2 C 1+8 C 3⋅3 C 1+8 C 3⋅3 C 1+8 C 4⋅4 C 1+8 C 5. Dave’s version was _8C_1\ +\ _8C_2\ \cdot\ 2\ +\ _8C_2\ \cdot\ 2\ +\ _8C_3\ \cdot\ _3C_1\ +\ _8C_3\ \cdot\ _3C_1\ +\ _8C_4\ \cdot\ _4C_1\ +\ _8C_5, 8 C 1+8 C 2⋅2+8 C 2⋅2+8 C 3⋅3 C 1+8 C 3⋅3 C 1+8 C 4⋅4 C 1+8 C 5, which is identical. My version, using permutations, is _8C_1\ +\ _8P_2\ +\ _8P_2\ +\ _8P_3 \div 2!\ +\ _8P_3 \div 2!\ +\ _8P_4 \div 3!\ +\ _8C_5. 8 C 1+8 P 2+8 P 2+8 P 3÷2!+8 P 3÷2!+8 P 4÷3!+8 C 5. These are all essentially the same idea. Finding the cases How do we know we haven’t missed a case? Dave made an orderly list, though it isn’t entirely clear what principle he was using to do so. Here is a list of the 7 cases used, in Doctor Wilko’s order, expressed by representing each kind of item with a letter: AAAAA AAAAB AAABB AAABC AABBC AABCD ABCDE We can see that he used the ordering principle of largest-first, where items are listed in decreasing order of frequency in each case, and the number of the first item mentioned decreases from case to case. In this order, it is easy to see that nothing was missed. This is the main benefit of Doctor Wilko’s work over Dave’s, and it is essential when you use cases. Keep in mind that here the letters don’t stand for specific items; the main work was to count the number of distinct ways in which a meaning could be assigned to each letter. Note that we could also list these cases as sums that total 5: 5 5 4+1 4+1 3+2 3+2 3+1+1 3+1+1 2+2+1 2+2+1 2+1+1+1 2+1+1+1 1+1+1+1+1 1+1+1+1+1 These are the “unordered partitions” of 5 (since we are not distinguishing different orders of the same number), which are harder to count (without listing as we have done here) than other things we’ve discussed. Fortunately, what we needed to do was to list them, not just to count! If you are curious, see here: Number of Unordered Partitions Second method: Stars and bars This will look familiar from last time. I am rewriting his to use stars and bars in the classic form, whereas he swapped the symbols, using the bars as tallies. Here we will directly use my menu list order form from above as a model. This solution is a little more elegant, but perhaps not as initially intuitive as trying explicit enumeration. I credit another one of our volunteers, Dr. Anthony, for reminding me of this technique. Let's label the menu items as A-H. You are ordering from the menu of 8 items. You may tally up your order like this: Item A \| B \| C \| D \| E \| F \| G \| H \| \| \| \| \| \| \| Meaning 2 of item A, 1 of item C, 1 of item E, and 1 of item H. There are other possibilities, too. For instance, Item A \| B \| C \| D \| E \| F \| G \| H \| \| \| \| \| \| \| Meaning 3 of item A and 2 of item C. You could do this for many different combinations, but should soon realize that every possible order can be made by placing 5 stars under the 8 items in any manner you choose. We could also think of each choice as a partition — this time as an ordered partition in which there must be 8 terms, any of which can be zero. We might have started a list like this: 5+0+0+0+0+0+0+0 5+0+0+0+0+0+0+0 4+1+0+0+0+0+0+0 4+1+0+0+0+0+0+0 4+0+1+0+0+0+0+0 4+0+1+0+0+0+0+0 … 4+0+0+0+0+0+0+1 4+0+0+0+0+0+0+1 3+2+0+0+0+0+0+0 3+2+0+0+0+0+0+0 … This would not be easy to list, would it? Nor is it obvious in this form how you could count them. But this time, there is a nice formula (as we saw last time)! In the world of partitions, you can be surprised by what is easy and what is hard. We just make the mental twist to see the first example above merely as \| \| \| \| \| \| \| : The "leap" with this method is to now see this as a permutations problem where you are arranging 5 stars and 7 bars, where 5 stars are identical and the 7 bars are identical. From probability, the permutations formula for this is, 12! ------- = 792. 5! 7! You can also think of it as you have 12 spaces and you can choose a place for the 5 \|'s in 12C5 ways. Likewise, you have 12 spaces and you can choose a place for the 7 's in 12C7 ways. Note, 12C5 = 12C7 = 792. This confirms our answer from above! He closes this with a comment that many of us have made: My favorite technique with combinations/permutations problems is to find more than one way to solve the problem. If I get confirmation from two different approaches, I usually feel more confident in my answer! In algebra, I always want to check my work by plugging the solution into the original problem; in combinatorics I never fully trust my work until I can get the same answer two ways. Incorrect methods Doctor Wilko then provided a catalog of basic formulas for different kinds of problems, showing why they don’t apply. The most important step in a combinatorics problem is determining what kind of problem it is, and what model or method applies. A mistake there ruins everything you do. Below are a couple of incorrect solutions and reasons why they are incorrect. 8^5 = 32,768 It seems logical at first, you have 8 choices for your first choice, 8 choices for your second choice, etc... I was actually guilty of thinking this was the answer at first! But then it hit me that this is a PERMUTATIONS solution. For instance, you would have ABBCC and BBACC. These are just two different permutations of the same combination of 5 items. In our problem, we want to buy a sack of 5 items from the 8-item menu. Order is irrelevant in purchasing the 5 items. In general, when order matters, and objects can be chosen more than once, the number of permutations is n^r (Permutations with Repetition) where n is the number of total objects and r is the number chosen. The model for this is _ _ _ _ _, where each of the 5 blanks (the items ordered) can be filled in any of 8 ways (the items on the menu), independently, so the total is 8\cdot 8\cdot 8\cdot 8\cdot 8 8⋅8⋅8⋅8⋅8. The trouble is that the blanks are distinguishable, so that order counts. For our problem, this would overcount, as we would be counting each meal multiple times. This is how you might count the ways for 5 distinct people to each choose one item, or to arrange 5 items in a 5-part tray. But we don’t want to distinguish the order of a meal. Note that we can’t correct for this by dividing by the number of arrangements of each meal, as we do in creating the formula for combinations, because that varies according to the number of duplicates. AAAAA has only one arrangement, while ABCDE has 60 arrangements. 8P5 = 6,720 This assumes 8 finite, distinct items, and you want all the different arrangements of 5 of them. ABCDE would be a different permutation than EDCBA. Remember, for the problem, we want the number of possible combinations a person could physically order from the menu. Order doesn't have anything to do with this. In general, when order matters, and each object can only be chosen once, the number of permutations is nPr = n!/(n-r)! (Permutations without Repetition) where n is the number of total objects and r is the number chosen. The model for this is _ _ _ _ _, where each blank must be filled with a different item, so the total is 8\cdot 7\cdot 6\cdot 5\cdot 4 8⋅7⋅6⋅5⋅4. This is wrong both because order counts (as it shouldn’t for our problem) and because repetition is not allowed (as it should be). Nothing is as we need it to be. 8C5 = 56 This assumes 8 finite, distinct items, and you are choosing 5 of them. This is not representative of the problem. This solution wouldn't let you order 5 roast beefs, for example. In general, when order doesn't matter, and each object can only be chosen once, the number of combinations is nCr = n!/[r!(n-r)!] (Combinations without Repetition) where n is the number of total objects and r is the number chosen. One way to derive this formula is to start with the 8\cdot 7\cdot 6\cdot 5\cdot 4 8⋅7⋅6⋅5⋅4 permutations, and divide by the 5\cdot 4\cdot 3\cdot 2\cdot 1 5⋅4⋅3⋅2⋅1 ways to arrange each combination. We do need combinations for our problem, because the order in which we choose items doesn’t matter; but we need to allow for repetitions. So, finally, that leads us back to our solution, which really is a "combinations with repetition" problem. In general, when order doesn't matter, and each object can be chosen more than once, the number of combinations is nPr = (n+r-1)!/[r!(n-1)!] (Combinations with Repetition) where n is the number of total objects and r is the number chosen. This is the formula we found last time, using n-1 n−1 bars (separating n bins) and r stars. The name nPr is probably a typo; there is a symbol for this (which I never learned), namely \displaystyle \left({8 \choose 5}\right). Post navigation ← Previous Post Next Post → Leave a Comment Cancel Reply Your email address will not be published.Required fields are marked Type here.. Name Email Website Δ This site uses Akismet to reduce spam. Learn how your comment data is processed. Have a question? Ask it here! We are a group of experienced volunteers whose main goal is to help you by answering your questions about math. To ask anything, just click here. Recent Blog Posts Is a Circle One-Dimensional or Two-Dimensional? Pigeonhole Principle II: Sets, Subsets, and Sums Pigeonhole Principle I: Paths, Penguins, and Points Sample Standard Deviation as an Unbiased Estimator Formulas for Standard Deviation: More Than Just One! 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16275	https://books.google.com/books/about/104_Number_Theory_Problems.html?id=BnKCjW8VroQC	104 Number Theory Problems: From the Training of the USA IMO Team - Titu Andreescu, Dorin Andrica, Zuming Feng - Google Books Sign in Hidden fields Try the new Google Books Books View sample Add to my library Try the new Google Books Check out the new look and enjoy easier access to your favorite features Try it now No thanks Try the new Google Books My library Help Advanced Book Search Good for: Web Tablet / iPad eReader Smartphone#### Features: Flowing text Scanned pages Help with devices & formats Learn more about books on Google Play EBOOK FROM $19.79 Get this book in print▼ Springer Shop Amazon.com Barnes&Noble.com Books-A-Million IndieBound Find in a library All sellers» My library My History 104 Number Theory Problems: From the Training of the USA IMO Team ================================================================= Titu Andreescu, Dorin Andrica, Zuming Feng Springer Science & Business Media, Apr 5, 2007 - Mathematics - 204 pages This book contains 104 of the best problems used in the training and testing of the U. S. International Mathematical Olympiad (IMO) team. It is not a collection of very dif?cult, and impenetrable questions. Rather, the book gradually builds students’ number-theoretic skills and techniques. The ?rst chapter provides a comprehensive introduction to number theory and its mathematical structures. This chapter can serve as a textbook for a short course in number theory. This work aims to broaden students’ view of mathematics and better prepare them for possible participation in various mathematical competitions. It provides in-depth enrichment in important areas of number theory by reorganizing and enhancing students’ problem-solving tactics and strategies. The book further stimulates s- dents’ interest for the future study of mathematics. In the United States of America, the selection process leading to participation in the International Mathematical Olympiad (IMO) consists of a series ofnational contests called the American Mathematics Contest 10 (AMC 10), the American Mathematics Contest 12 (AMC 12), the American Invitational Mathematics - amination (AIME), and the United States of America Mathematical Olympiad (USAMO). Participation in the AIME and the USAMO is by invitation only, based on performance in the preceding exams of the sequence. The Mathematical Olympiad Summer Program (MOSP) is a four-week intensive training program for approximately ?fty very promising students who have risen to the top in the American Mathematics Competitions. More » Preview this book » Selected pages Title Page Table of Contents Index References Contents Introductory Problems 75 Solutions to Introductory Problems 91 Solutions to Advanced Problems 131 Glossary 189 Index203 Copyright Other editions - View all 104 Number Theory Problems: From the Training of the USA IMO Team Titu Andreescu,Dorin Andrica,Zuming Feng No preview available - 2006 Common terms and phrases a₁ak+1Andreescuanswerarithmetic progressionassumeBézout's identityclaimcomplete setconclusion followscongruentconsecutive integersconsidercontradictionCorollarycubedenoteDiophantine Equationsdistinct primesdividesdivisibleelementsequalEuclidean algorithmEuler's theoremexactlyExampleexistsFengFermat numberFermat's little theoremFindfinitegcd(a About the author(2007) Titu Andreescu received his Ph.D. from the West University of Timisoara, Romania. The topic of his dissertation was "Research on Diophantine Analysis and Applications." Professor Andreescu currently teaches at The University of Texas at Dallas. He is past chairman of the USA Mathematical Olympiad, served as director of the MAA American Mathematics Competitions (1998–2003), coach of the USA International Mathematical Olympiad Team (IMO) for 10 years (1993–2002), director of the Mathematical Olympiad Summer Program (1995–2002), and leader of the USA IMO Team (1995–2002). In 2002 Titu was elected member of the IMO Advisory Board, the governing body of the world's most prestigious mathematics competition. Titu co-founded in 2006 and continues as director of the AwesomeMath Summer Program (AMSP). He received the Edyth May Sliffe Award for Distinguished High School Mathematics Teaching from the MAA in 1994 and a "Certificate of Appreciation" from the president of the MAA in 1995 for his outstanding service as coach of the Mathematical Olympiad Summer Program in preparing the US team for its perfect performance in Hong Kong at the 1994 IMO. Titu’s contributions to numerous textbooks and problem books are recognized worldwide. Dorin Andrica received his Ph.D. in 1992 from "Babes-Bolyai” University in Cluj-Napoca, Romania; his thesis treated critical points and applications to the geometry of differentiable submanifolds. Professor Andrica has been chairman of the Department of Geometry at "Babes-Bolyai" since 1995. He has written and contributed to numerous mathematics textbooks, problem books, articles and scientific papers at various levels. He is an invited lecturer at university conferences around the world: Austria, Bulgaria, Czech Republic, Egypt, France, Germany, Greece, Italy, the Netherlands, Portugal, Serbia, Turkey, and the USA. Dorin is a member of the Romanian Committee for theMathematics Olympiad and is a member on the editorial boards of several international journals. Also, he is well known for his conjecture about consecutive primes called "Andrica's Conjecture." He has been a regular faculty member at the Canada–USA Mathcamps between 2001–2005 and at the AwesomeMath Summer Program (AMSP) since 2006. Zuming Feng received his Ph.D. from Johns Hopkins University with emphasis on Algebraic Number Theory and Elliptic Curves. He teaches at Phillips Exeter Academy. Zuming also served as a coach of the USA IMO team (1997-2006), was the deputy leader of the USA IMO Team (2000-2002), and an assistant director of the USA Mathematical Olympiad Summer Program (1999-2002). He has been a member of the USA Mathematical Olympiad Committee since 1999, and has been the leader of the USA IMO team and the academic director of the USA Mathematical Olympiad Summer Program since 2003. Zuming is also co-founder and academic director of the AwesomeMath Summer Program (AMSP) since 2006. He received the Edyth May Sliffe Award for Distinguished High School Mathematics Teaching from the MAA in 1996 and 2002. Bibliographic information Title 104 Number Theory Problems: From the Training of the USA IMO Team AuthorsTitu Andreescu, Dorin Andrica, Zuming Feng Edition illustrated Publisher Springer Science & Business Media, 2007 ISBN 0817645616, 9780817645618 Length 204 pages SubjectsMathematics › Algebra › General Mathematics / Algebra / General Mathematics / Calculus Mathematics / History & Philosophy Mathematics / Logic Mathematics / Mathematical Analysis Mathematics / Number Theory Export CitationBiBTeXEndNoteRefMan About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home
16276	https://math.stackexchange.com/questions/4701260/solving-x-textsgnx-sqrtx-uniqueness-of-solutions-of-finite-duratio	real analysis - Solving $x'=-\text{sgn}(x)\sqrt{\|x\|}$: Uniqueness of solutions of finite duration - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Solving x′=−sgn(x)√\|x\|: Uniqueness of solutions of finite duration Ask Question Asked 2 years, 4 months ago Modified2 months ago Viewed 394 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. Show that x(t)=sgn(x(0))4(2√\|x(0)\|−t)2⋅θ(2√\|x(0)\|−t) is a solution to x′=−sgn(x)√\|x\|. Is the solution unique? This question was modified since I was able to find the answer to the first question. Here, θ(t) is the Heaviside step function, and sgn(x) is the Sign function. This is a continuation of this question. By using the answer given by @KBS, I could find solutions with the same "structure" as the mentioned answer if x(0)>0, but for initial conditions x(0)<0, I struggle because the term (2√x(0)−t)2 is, first, insensible to sign changes, and secondly, becomes complex, so I added an absolute value function to the initial condition, and later just pasted a term sgn(x(0)) in order to "make" the same solutions I find in the "slope-field" of wolfram-Alpha for negative initial values, leading to: x(t)=sgn(x(0))4(2√\|x(0)\|−t)2⋅θ(2√\|x(0)\|−t) which fulfill as I require: lim t→0 x(t)≡x(0) for any x(0)∈R (there is a discontinuity when arriving to x(0+) and x(0−) but since x(0+)≡x(0−)≡0 I think there is no issues at all: checked here and then here. It have a finite extinction time T=2√\|x(0)\|≥0 where the system stop moving at zero x(t)=0,∀t≥T. But when I insert the function x(t) into the equation: x′+sgn(x)√\|x\|=0 I cannot make the terms to become zero: from the differentiation side I got some Dirac delta functions as shown here, and from the nonlinear term I cannot match the exponent in the sign functions outside and within the square root as shown here, so I don't know if the solution that we have found is indeed a solution to the differential equation "formally" speaking (at every point and real-valued initial condition). Is the x(t) that we obtained indeed a formal solution to x′+sgn(x)√\|x\|=0? (SOLVED) Is this solution unique? at least in the domain t∈[0,T) Added Later Later I remembered this answer by @md2perpe where show that since x δ(x)=0 in this situation where solution x(t)=f(t)θ(t) is such as: (f θ)′=f′θ+f θ′=f′θ+f δ=f′θ since f(t) is a polynomial with the same argument as the Dirac's delta functionδ(t), it happens to be f δ≡(c−t)n δ(c−t)=0. its equivalent sgn(f θ)√\|f θ\|=sgn(f)√\|f\|θ Then I do could match the left-hand-sidef′θ with the right-hand-sidesgn(f)√\|f\|θ at least while t<T, meaning that x(t) is indeed a solution of finite duration (this, since the trivial solution x(t)=0 satisfies the differential equation after t≥T). You could see it's numerical verification on Desmos. Update 2 @CalvinKhor have sugest me this comment by @ThomasRichard in MathOverflow suggesting a real life physical model that fits the differential equation: It might be useful to mention that the ODE y′=−√y+, for which non uniqueness holds, actually models a real physical system. y(t) represents the water height in a pierced cylindrical bucket. And the non uniqueness of the Cauchy problem with initial condition y(0)=0 is natural in this model: if you know the bucket is empty at time 0, it is hard to tell if there was water in it before and when it got empty. And later, @CalvinKhor also shared the explanation in this other comment, which I like to share since I found it really interesting and suits here: Motivation(added due requirement by @RodrigodeAzevedo in the comments) About two years ago I accidentally found these papers about V. T. Haimo: Finite Time Differential Equations and Finite Time Controllers, and realized from them that no analytic function, so no non-piecewised defined power series, neither no linear ODE, could accurately represent a dynamic system that stop moving from its own, and as engineer I felt cheated since everything I was taught could become a power series, and daily life phenomena do stops at finite times. So I started to make myself the easier examples I can and try to solve them in close form, since very few info there is about these solutions of finite duration, at least grouped as one specific topic, and the mentioned papers don't show closed form solutions neither (this question example is a modification of an example on the first paper). Last update Recently in another question user @KBS shared in the comments the following text were a similar, but more general form of this equation is studied and present general conditions for a system y′=F(y,t) to show solutions of finite duration: real-analysis ordinary-differential-equations dynamical-systems nonlinear-dynamics finite-duration Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jul 17 at 10:25 Rodrigo de Azevedo 23.4k 7 7 gold badges 49 49 silver badges 116 116 bronze badges asked May 17, 2023 at 17:47 JoakoJoako 2,145 1 1 gold badge 10 10 silver badges 38 38 bronze badges 5 Where does this question come from?Rodrigo de Azevedo –Rodrigo de Azevedo 2025-04-15 09:03:12 +00:00 Commented Apr 15 at 9:03 @RodrigodeAzevedo About two years ago I accidentally found these papers about V. T. Haimo: Finite Time Differential Equations and Finite Time Controllers, and realized from them that no analytic function, so no non-piecewised defined power series, neither no linear ODE, could accurately represent a dynamic system that stop moving from its own, and as engineer I felt cheated since everything I was taught could become a power series, and daily life phenomena do stops at finite times. (...)Joako –Joako 2025-04-15 14:53:35 +00:00 Commented Apr 15 at 14:53 @RodrigodeAzevedo So I started to make myself the easier examples I can and try to solve them in closed form, since very few info there it is about these systems, at least grouped as one specific topic, and the mentioned papers don't show closed form solutions neither (this specific questions came as a variation of the equation studied in the first paper)Joako –Joako 2025-04-15 14:55:28 +00:00 Commented Apr 15 at 14:55 @RodrigodeAzevedo Sometimes I do it, but sometimes I do not, I have had questions closed or badly downvoted because of it extension (MSE sometimes is squizo)... in this case, at the beginning I cited another question where it is explained.Joako –Joako 2025-04-15 15:27:24 +00:00 Commented Apr 15 at 15:27 @RodrigodeAzevedo here you could see the perfect example of I was talking about... a well redacted and explained question get closed inmediatly because of their extension Joako –Joako 2025-04-18 03:13:53 +00:00 Commented Apr 18 at 3:13 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 6 Save this answer. Show activity on this post. IVPs for x′(t)=−s g n(x(t))√\|x(t)\| are uniquely solvable to the right, since x↦−s g n(x)√\|x\| is monotone decreasing on R. More general, let f:R→R be monotone decreasing, and let x 1,x 2:[0,T)→R be solutions of x′(t)=f(x(t)), x(0)=x 0. Set d(t):=(x 1(t)−x 2(t))2(t∈[0,T)). Then d′(t)=2(x 1(t)−x 2(t))(f(x 1(t))−f(x 2(t)))≤0(t∈[0,T)). Hence d is decreasing and ≥0, and d(0)=0. Thus d(t)=0, that is x 1(t)=x 2(t)(t∈[0,T)). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered May 19, 2023 at 11:46 GerdGerd 10.2k 1 1 gold badge 7 7 silver badges 27 27 bronze badges 2 thanks a lot! it's brilliant since it an easy to follow proof (in contrast to classic existence and uniqueness proofs like Arzela-Ascoli or the fixed point theorem). The only part I don't fully get it's why d′≤0, Could you elaborate on this?Joako –Joako 2023-05-19 19:43:25 +00:00 Commented May 19, 2023 at 19:43 2 Since f is decreasing we have: If x 1(t)≤x 2(t) then f(x 1(t))≥f(x 2(t)), and if x 1(t)≥x 2(t) then f(x 1(t))≤f(x 2(t)). Thus x 1(t)−x 2(t) and f(x 1(t))−f(x 2(t)) always have opposite sign (or are 0). Thus the product is ≤0.Gerd –Gerd 2023-05-19 20:03:17 +00:00 Commented May 19, 2023 at 20:03 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions real-analysis ordinary-differential-equations dynamical-systems nonlinear-dynamics finite-duration See similar questions with these tags. 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16277	https://www.studocu.com/en-us/document/purdue-university/general-chemistry-for-engineers/hesss-law-lecture-notes-chm11500-2023/53626727	Hess's Law Fundamentals - Lecture Notes CHM11500 2023 - Studocu Skip to document Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes New Home My Library AI Notes Ask AI AI Quiz Chats Recent You don't have any recent items yet. My Library Courses You don't have any courses yet. Add Courses Books You don't have any books yet. Studylists You don't have any Studylists yet. Create a Studylist Home My Library Discovery Discovery Universities High Schools High School Levels Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI Hess's Law Fundamentals - Lecture Notes CHM11500 2023 Lecture notes - Hess Law Chemistry 115 Original title: Hess's Law Lecture Notes CHM11500 2023 Course General Chemistry for Engineers (CHM 115) 184 documents University Purdue University Academic year:2023/2024 Uploaded by: Anonymous Student Purdue University Recommended for you 6 CHM 115 Lecture Notes E2 General Chemistry for Engineers Lecture notes 100% (1) 5 CHEM 115 Chapter 11 Notes General Chemistry for Engineers Summaries 100% (7) 2 Bal. Eq. #3 - Balancing Reactions Practice Problems Worksheet General Chemistry for Engineers Practice materials 100% (5) 7 Lab Safety Certification Quiz General Chemistry for Engineers Tutorial work 100% (4) 2 Honors Chemistry: Electromagnetic Spectrum Worksheet Insights General Chemistry for Engineers Practice materials 100% (2) Comments Please sign in or register to post comments. Report Document Students also viewed CHM11500 LAB 9 Notes on DNA extraction 2023 CHM11500 Polymerization Notes 2023 CHM11500 Lab 10 Terminology to Know 2023 CHM11500 Crosslinking Polymers Notes 2023 CHM11500 Nylon Lab 9 Notes 2023 CHM11500 Thermochemistry Simple Introduction Notes 2023 Related documents CHM115 Organic Chemistry Basics Notes for the Final Exam 2023 CHM115 Chemical Bonding - Course Notes Spring 2023 CHM115 Organic Chemistry Polymer Reaction Notes 2023 CHM11500 Purdue Nuclear Chemistry Notes 2023 CHM11500 Purdue Radiation and Decay Notes 2023 CHM11500 Dische Test LAB 9 Notes 2023 Preview text Purdue University 2023 CHM115 Lecture Notes Title: The Fundamentals of Hess’s Law Introduction: Hess’s law is a fundamental concept in thermochemistry that enables us to calculate the enthalpy change associated with a chemical reaction. It is named after Germain Hess, who established the law in 1840. This law states that the change in enthalpy during a chemical reaction is independent of the route taken from the initial to the final state, provided that the reaction occurs at constant pressure. Hess's Law of Constant Heat Summation: Hess's law establishes that if a chemical reaction occurs in stages or through different pathways, the overall enthalpy change of the reaction is equal to the sum of enthalpy changes of each of the individual steps. This principle is established as the law of constant heat summation. It is useful in predicting the enthalpy change of a chemical reaction that is difficult or impossible to measure directly. The enthalpy change in a chemical reaction is usually measured by conducting the reaction under a controlled environment and measuring the heat evolved or absorbed by the reaction. However, some reactions might not be possible to measure directly, either because they are too fast or too slow. In such cases, Hess's law can be used to calculate the enthalpy change of the reaction indirectly. Hess's law applies to both exothermic and endothermic reactions. It also applies to reactions that are in a solid, liquid, or gaseous state, as long as the reaction occurs at constant volume or pressure. Additionally, it is essential to consider quantum mechanics in determining the state of the reaction. The law of Hess provides a means to calculate enthalpy changes using standard enthalpies of formation. The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its constituent elements under standard conditions. In conclusion, Hess’s law provides a foundation for calculating enthalpy changes of chemical reactions that are difficult or impossible to measure directly. By utilising the principle of the law of constant heat summation, scientists can determine the enthalpy change of a reaction through several pathways. Moreover, standard enthalpies of formation can be used as a means of calculating enthalpy changes via Hess's law. It is a fundamental concept in thermochemistry that has vast applications in many scientific fields. Hess's Law Fundamentals - Lecture Notes CHM11500 2023 Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 0 0 Save Hess's Law Fundamentals - Lecture Notes CHM11500 2023 Course: General Chemistry for Engineers (CHM 115) 184 documents University: Purdue University Info More info Download Download AI Tools Ask AI Multiple Choice Flashcards Quiz Video Audio Lesson 0 0 Save Purdue University 2023 CHM1 15 Lecture Notes T itle:The Fundamentals of Hess’s Law Introduction:Hess’s law is a fundamental concept in thermochemistry that enables us to calculate the enthalpy change associated with a chemical reaction.It is named after Germain Hess,who established the law in 1840.This law states that the change in enthalpy during a chemical reaction is independent of the route taken from the initial to the final state,provided that the reaction occurs at constant pressure. Hess's Law of Constant Heat Summation: Hess's law establishes that if a chemical reaction occurs in stages or through different pathways,the overall enthalpy change of the reaction is equal to the sum of enthalpy changes of each of the individual steps.This principle is established as the law of constant heat summation.It is useful in predicting the enthalpy change of a chemical reaction that is difficult or impossible to measure directly. The enthalpy change in a chemical reaction is usually measured by conducting the reaction under a controlled environment and measuring the heat evolved or absorbed by the reaction.However,some reactions might not be possible to measure directly,either because they are too fast or too slow.In such cases,Hess's law can be used to calculate the enthalpy change of the reaction indirectly. Hess's law applies to both exothermic and endothermic reactions.It also applies to reactions that are in a solid,liquid,or gaseous state,as long as the reaction occurs at constant volume or pressure.Additionally,it is essential to consider quantum mechanics in determining the state of the reaction. The law of Hess provides a means to calculate enthalpy changes using standard enthalpies of formation.The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its constituent elements under standard conditions. In conclusion,Hess’s law provides a foundation for calculating enthalpy changes of chemical reactions that are difficult or impossible to measure directly.By utilising the principle of the law of constant heat summation,scientists can determine the enthalpy change of a reaction through several pathways.Moreover,standard enthalpies of formation can be used as a means of calculating enthalpy changes via Hess's law.It is a fundamental concept in thermochemistry that has vast applications in many scientific fields. 1 out of 1 Share Download Download More from:General Chemistry for Engineers(CHM 115) More from: General Chemistry for EngineersCHM 115Purdue University 184 documents Go to course 5 CHEM 115 Chapter 11 Notes General Chemistry for Engineers Summaries 100% (7) 2 Bal. Eq. #3 - Balancing Reactions Practice Problems Worksheet General Chemistry for Engineers Practice materials 100% (5) 7 Lab Safety Certification Quiz General Chemistry for Engineers Tutorial work 100% (4) 2 Honors Chemistry: Electromagnetic Spectrum Worksheet Insights General Chemistry for Engineers Practice materials 100% (2) More from: General Chemistry for EngineersCHM 115Purdue University184 documents Go to course 5 CHEM 115 Chapter 11 Notes General Chemistry for Engineers 100% (7) 2 Bal. Eq. #3 - Balancing Reactions Practice Problems Worksheet General Chemistry for Engineers 100% (5) 7 Lab Safety Certification Quiz General Chemistry for Engineers 100% (4) 2 Honors Chemistry: Electromagnetic Spectrum Worksheet Insights General Chemistry for Engineers 100% (2) 9 Chemistry 11500 - Fall 2023 Exam 1 Practice Questions General Chemistry for Engineers 100% (2) 2 Prelab 7 General Chemistry for Engineers 80% (5) Recommended for you 6 CHM 115 Lecture Notes E2 General Chemistry for Engineers Lecture notes 100% (1) 5 CHEM 115 Chapter 11 Notes General Chemistry for Engineers Summaries 100% (7) 2 Bal. Eq. #3 - Balancing Reactions Practice Problems Worksheet General Chemistry for Engineers Practice materials 100% (5) 7 Lab Safety Certification Quiz General Chemistry for Engineers Tutorial work 100% (4) 6 CHM 115 Lecture Notes E2 General Chemistry for Engineers 100% (1) 5 CHEM 115 Chapter 11 Notes General Chemistry for Engineers 100% (7) 2 Bal. Eq. #3 - Balancing Reactions Practice Problems Worksheet General Chemistry for Engineers 100% (5) 7 Lab Safety Certification Quiz General Chemistry for Engineers 100% (4) 2 Honors Chemistry: Electromagnetic Spectrum Worksheet Insights General Chemistry for Engineers 100% (2) 9 Chemistry 11500 - Fall 2023 Exam 1 Practice Questions General Chemistry for Engineers 100% (2) Students also viewed CHM11500 LAB 9 Notes on DNA extraction 2023 CHM11500 Polymerization Notes 2023 CHM11500 Lab 10 Terminology to Know 2023 CHM11500 Crosslinking Polymers Notes 2023 CHM11500 Nylon Lab 9 Notes 2023 CHM11500 Thermochemistry Simple Introduction Notes 2023 Related documents CHM115 Organic Chemistry Basics Notes for the Final Exam 2023 CHM115 Chemical Bonding - Course Notes Spring 2023 CHM115 Organic Chemistry Polymer Reaction Notes 2023 CHM11500 Purdue Nuclear Chemistry Notes 2023 CHM11500 Purdue Radiation and Decay Notes 2023 CHM11500 Dische Test LAB 9 Notes 2023 Get homework AI help with the Studocu App Open the App English United States Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. 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16278	https://www.scilag.net/problem/P-240322.2	Combinatorics - The Erdos Distinct Subset Sum Problem \| SciLag Sci L agPost a problem Log inRegister The Erdos Distinct Subset Sum Problem Paul Erdos Open Posted online: 2024-03-22 16:13:14Z by Stefan Steinerberger171 Cite as: P-240322.2 Combinatorics Follow Share Description Editorial history Remarks Problem's Description Let a 1≤a 2≤⋯≤a n a 1≤a 2≤⋯≤a n be a set of n n positive integers such that all 2 n 2 n subsets of the set are uniquely identified by the sum of their elements. An example of such a sequence are the powers of 2 (this is the uniqueness of the binary expansion in disguise). The question, due to Erdos in 1931 or 1932, is whether sets with such a property have to always have a large number. Problem. Does this property force a n≥c⋅2 n a n≥c⋅2 n? We note that the largest sum is a 1+⋯+a n a 1+⋯+a n and thus, if all the subsets are distinct, we have to have 2 n≤a 1+⋯+a n≤n⋅a n 2 n≤a 1+⋯+a n≤n⋅a n and thus a n≥2 n n. a n≥2 n n. The currently best result is due to Erdos-Moser a n≥c 2 n√n. a n≥c 2 n n−−√. Many people have worked on improving the constant c c. Conway-Guy showed that the powers of 2 are not extremal: there are examples with smaller a n a n. Elkies notes the equivalence to a completely analytic question: If a 1,…,a n a 1,…,a n are positive integers, then ∫1 0 n∏i=1 cos(2 π a i x)2 d x≥1 2 n with equality if and only if all subset sums are distinct. The question is thus: if the integral is small, does this force a n to be large?References 1. Conference Paper Problems and results in additive number theory Paul Erdos Colloque sur la Theorie des Nombres, Bruxelles, 1955 2. Article An improved lower bound on the greatest element of a sum-distinct set of fixed order. Noam Elkies Journal of Combinatorial Theory, Series A, 89-94, 1986 3. Article Sets of natural numbers with distinct sums J. Conway, R. K. Guy Notices of the American Mathematical Society, 345, 1968 No solutions added yet No remarks yet Created at: 2024-03-22 16:13:14Z Follow the problem Email Follow Social share Sci L ag Research problems Problem mining Announcements Our members About us User guide Contact us Contact: info@scilag.net Facebook Subscription: Customize Subscribe © 2025. All Rights Reserved — SciLag Terms of Service Privacy policy Get email updates when [x] New problem is posted [x] New solution is posted [x] All Subjects Subject area - [x] Algebraic Geometry - [x] Algebraic Topology - [x] Analysis of PDEs - [x] Category Theory - [x] Classical Analysis and ODEs - [x] Combinatorics - [x] Commutative Algebra - [x] Complex Variables - [x] Differential Geometry - [x] Dynamical Systems - [x] Functional Analysis - [x] General Mathematics - [x] General Topology - [x] Geometric Topology - [x] Group Theory - [x] History and Overview - [x] Information Theory - [x] K-Theory and Homology - [x] Logic - [x] Mathematical Physics - [x] Metric Geometry - [x] Number Theory - [x] Numerical Analysis - [x] Operator Algebras - [x] Optimization and Control - [x] Probability - [x] Quantum Algebra - [x] Representation Theory - [x] Rings and Algebras - [x] Spectral Theory - [x] Statistics Theory - [x] Symplectic Geometry Email Subscribe Advanced search Subject area Choose subject Problem status Choose status With all of the words With the exact phrase Authored by Posted by References contain Year of origin Year of origin Search Single or several? If you plan to formulate more than one problem all sharing the same background (e.g. they are all from the same paper) then please choose "Group", otherwise select "Single" option. SingleGroup Log in Affiliation Email Password [x] Remember me Forgot password? Log inRegister Email Get password reset token
16279	https://artofproblemsolving.com/wiki/index.php/Inequality?srsltid=AfmBOoqXP0mAQO_lky040Ry-MhJjOyVbDaw3xYKa0dPgYAb6lmmTmQxV	Art of Problem Solving Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Inequality The subject of mathematical inequalities is tied closely with optimization methods. While most of the subject of inequalities is often left out of the ordinary educational track, they are common in mathematics Olympiads. Contents [hide] 1 Overview 2 Solving Inequalities 2.1 Linear Inequalities 2.2 Polynomial Inequalities 2.3 Rational Inequalities 3 Complete Inequalities 4 List of Theorems 4.1 Introductory 4.2 Advanced 5 Problems 5.1 Introductory 5.2 Intermediate 5.3 Olympiad 6 Resources 6.1 Books 6.1.1 Intermediate 6.1.2 Olympiad 6.2 Articles 6.2.1 Olympiad 6.3 Classes 6.3.1 Olympiad 7 See also Overview Inequalities are arguably a branch of elementary algebra, and relate slightly to number theory. They deal with relations of variables denoted by four signs: . For two numbers and : if is greater than , that is, is positive. if is smaller than , that is, is negative. if is greater than or equal to , that is, is nonnegative. if is less than or equal to , that is, is nonpositive. Note that if and only if , , and vice versa. The same applies to the latter two signs: if and only if , , and vice versa. Some properties of inequalities are: If , then , where . If , then , where . If , then , where . Solving Inequalities In general, when solving inequalities, same quantities can be added or subtracted without changing the inequality sign, much like equations. However, when multiplying, dividing, or square rooting, we have to watch the sign. In particular, notice that although , we must have . In particular, when multiplying or dividing by negative quantities, we have to flip the sign. Complications can arise when the value multiplied can have varying signs depending on the variable. We also have to be careful about the boundaries of the solutions. In the example , the value does not satisfy the inequality because the inequality is strict. However, in the example , the value satisfies the inequality because the inequality is nonstrict. Solutions can be written in interval notation. Closed bounds use square brackets, while open bounds (and bounds at infinity) use parentheses. For instance, ![Image 49: $x \in 3,6)$ means . Linear Inequalities Linear inequalities can be solved much like linear equations to get implicit restrictions upon a variable. However, when multiplying/dividing both sides by negative numbers, we have to flip the sign. Polynomial Inequalities The first part of solving polynomial inequalities is much like solving polynomial equations -- bringing all the terms to one side and finding the roots. Afterward, we have to consider bounds. We're comparing the sign of the polynomial with different inputs, so we could imagine a rough graph of the polynomial and how it passes through zeroes (since passing through zeroes could change the sign). Then we can find the appropriate bounds of the inequality. Rational Inequalities A more complex example is . Here is a common mistake: The problem here is that we multiplied by as one of the last steps. We also kept the inequality sign in the same direction. However, we don't know if the quantity is negative or not; we can't assume that it is positive for all real . Thus, we may have to reverse the direction of the inequality sign if we are multiplying by a negative number. But, we don't know if the quantity is negative either. A correct solution would be to move everything to the left side of the inequality, and form a common denominator. Then, it will be simple to find the solutions to the inequality by considering the sign (negativeness or positiveness) of the fraction as varies. We will start with an intuitive solution, and then a rule can be built for solving general fractional inequalities. To make things easier, we test positive integers. makes a good starting point, but does not solve the inequality. Nor does . Therefore, these two aren't solutions. Then we begin to test numbers such as , , and so on. All of these work. In fact, it's not difficult to see that the fraction will remain positive as gets larger and larger. But just where does , which causes a negative fraction at and , begin to cause a positive fraction? We can't just assume that is the switching point; this solution is not simply limited to integers. The numerator and denominator are big hints. Specifically, we examine that when (the numerator), then the fraction is , and begins to be positive for all higher values of . Solving the equation reveals that is the turning point. After more of this type of work, we realize that brings about division by , so it certainly isn't a solution. However, it also tells us that any value of that is less than brings about a fraction that has a negative numerator and denominator, resulting in a positive fraction and thus satisfying the inequality. No value between and (except itself) seems to be a solution. Therefore, we conclude that the solutions are the intervals ![Image 78: $(-\infty,-5)\cup\frac{3}{2},+\infty)$. For the sake of better notation, define the "x-intercept" of a fractional inequality to be those values of that cause the numerator and/or the denominator to be .To develop a method for quicker solutions of fractional inequalities, we can simply consider the "x-intercepts" of the numerator and denominator. We graph them on the number line. Then, in every region of the number line, we test one point to see if the whole region is part of the solution. For example, in the example problem above, we see that we only had to test one value such as in the region , as well as one value in the region ![Image 83: $(-\infty,-5]$]( and ![Image 84: $\frac{3}{2},+\infty)$; then we see which regions are part of the solution set. This does indeed give the complete solution set. One must be careful about the boundaries of the solutions. In the example problem, the value was a solution only because the inequality was nonstrict. Also, the value was not a solution because it would bring about division by . Similarly, any "x-intercept" of the numerator is a solution if and only if the inequality is nonstrict, and every "x-intercept" of the denominator is never a solution because we cannot divide by . Complete Inequalities A inequality that is true for all real numbers or for all positive numbers (or even for all complex numbers) is sometimes called a complete inequality. An example for real numbers is the so-called Trivial Inequality, which states that for any real , . Most inequalities of this type are only for positive numbers, and this type of inequality often has extremely clever problems and applications. List of Theorems Here are some of the more useful inequality theorems, as well as general inequality topics. Introductory Arithmetic Mean-Geometric Mean Inequality Cauchy-Schwarz Inequality Titu's Lemma Chebyshev's Inequality Geometric inequalities Jensen's Inequality Nesbitt's Inequality Rearrangement Inequality Power mean inequality Triangle Inequality Trivial inequality Schur's Inequality Advanced Aczel's Inequality Callebaut's Inequality Carleman's Inequality Hölder's inequality Radon's Inequality Homogenization Isoperimetric inequalities Maclaurin's Inequality Muirhead's Inequality Minkowski Inequality Newton's Inequality Ptolemy's Inequality Can someone fix that Ptolemy's is in Advanced? Problems Introductory Practice Problems on Alcumus Inequalities (Prealgebra) Solving Linear Inequalities (Algebra) Quadratic Inequalities (Algebra) Basic Rational Function Equations and Inequalities (Intermediate Algebra) A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly . During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than . What's the largest number of matches she could've won before the weekend began? (1992 AIME Problems/Problem 3) Intermediate Practice Problems on Alcumus Quadratic Inequalities (Algebra) Advanced Rational Function Equations and Inequalities (Intermediate Algebra) General Inequality Skills (Intermediate Algebra) Advanced Inequalities (Intermediate Algebra) Given that , and show that . (weblog_entry.php?t=172070 Source) Olympiad See also Category:Olympiad Inequality Problems Let be positive real numbers. Prove that (2001 IMO Problems/Problem 2) Resources Books Intermediate Introduction to Inequalities Geometric Inequalities Olympiad Advanced Olympiad Inequalities: Algebraic & Geometric Olympiad Inequalities by Alijadallah Belabess. The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele. Problem Solving Strategies by Arthur Engel contains significant material on inequalities. Inequalities by G. H. Hardy, J. E. Littlewood, G. Pólya. Articles Olympiad Inequalities by MIT Professor Kiran Kedlaya. Inequalities by IMO gold medalist Thomas Mildorf. Classes Olympiad The Worldwide Online Olympiad Training Program is designed to help students learn to tackle mathematical Olympiad problems in topics such as inequalities. 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16280	https://www.yourschoolmatch.com/blog/criminal-justice/how-to-become-a-prison-warden-career-salary-information/	How To Become A Prison Warden: Career & Salary Information - YourSchoolMatch ___ YourSchoolMatch Home Schools Blog About Contact For StudentsFor Schools find criminal justice degrees near you FIND DEGREE BlogCriminal JusticeHow To Become A Prison Warden: Career & Salary Information How To Become A Prison Warden: Career & Salary Information find criminal justice degrees near you From fictional portrayals in movies, to the setting in real life, prison wardens hold a sense of power, and a sense of braveness and control that many aspire to possess. But being a prison warden goes far beyond just overseeing prisoners at a correctional facility, and the requirements to hold such a role are quite extensive. To pursue a career as a prison warden, there are basic requirements that you need to meet. You would need, at the very least, a bachelor's degree incriminal justice, law enforcement, or criminology. However, considering the duties and the requirements of a prison warden, obtaining a master's degree in criminal justice may make you a star candidate for the position and equip you with additional skills that will allow you to thrive and grow in this career. Let us take a deeper look at this sought-after career and how you could grow and thrive in the field. What Is A Prison Warden? If you ever saw yourself in a leadership role, in a top position, or as a governing force of an institute, then being a prison warden may be for you. In any correctional facility, the prison warden holds the highest-ranking role. One may wonder if there is a difference between a prison warden and a jail warden. Although done quite often, the two terms are used interchangeably, however, they technically do not hold the same meaning. A jail warden is someone who oversees the cells of a holding facility where an individual would be placed if they are awaiting trial and conviction. A jail is the facility where a person can be held temporarily while awaiting bail. It can also refer to the holding rooms within a police station. On the other hand, a prison warden oversees operations of a fully established correctional facility that houses criminals who have been convicted of a crime and are now in various stages of their incarceration. What Does A Prison Warden Do? Everything! But in all seriousness, the role of a prison warden is a weight that needs to be carried by extraordinarily strong shoulders. With the prison warden holding they highest ranking role in a prison, they are responsible for all operations pertaining to the prison and keeping it functional. The extent of the prison warden's responsibility will depend greatly on the size of the prison they are in charge of, but large or small, the warden oversees all the operations and responsibilities. From operations management, security management, budget management, and human resource management, the warden implements structures and processes to ensure that the prison staff and inmates are always correctly taken care of. Budgetary Management There are two types of prisons that exist – private and public prisons. While both are used for correctional purposes, one is a government owned entity, and the other is a privately owned entity that receives its funding from government contracts. The funding for private prisons depends on how many inmates are housed in the prison, and how long is each inmate's sentence. It therefore falls to the prison warden to accurately and effectively utilize the funding that the prison receives and to draft effective budgets while adhering to it as best as possible. Operations Management All operations that occur within and involving the prison is entirely overseen by the prison warden. The warden is responsible for the inmates that are housed in the prison, and they are responsible for ensuring that the basic humane care that the prisoners are entitled to, is effectively and accurately provided. The prison warden is also responsible for all the human resources that are working onsite of the prison. From prison guards, to doctors, cooks, and cleaners, all will report to the warden. The warden may find assistance from the head of security within the prison and from department heads, but for clearance purposes, the warden holds all authority within the prison. All security measures that are needed to be in place are signed off and approved by the warden. Additionally, the warden needs to authorize all attorneys, family, and outside visitors that come onto prison property, and they are liable for the safety of all persons on the prison site. The warden is also responsible for all administrative and recordkeeping structures that are set in place within the prison and that are used for legal purposes and legal proceedings regarding each inmate. It is here where you can see the intense requirements and responsibilities that lies in the hands of the prison warden. They are not only required to be skilled in law enforcement, criminal justice, and general legal proceedings, but they are also required to be skilled in budgetary and financial operations, security processes, and general administrative operations. Steps To Become A Prison Warden If you are seeking to pursue a career as a prison warden, there are some concrete steps that you can take to solidify your success. Step One: Obtain A High School Diploma The first step to becoming a prison warden would be deciding on your career path quite early in life. This will allow you to progress further, a lot quicker. The first step would be completing high school and obtaining your diploma or a GED. Step Two: Obtain A Degree Once you decide that the career of a prison warden is for you, you can pursue your bachelors or associates degree in the relevant field of criminal justice, criminology, or even social work. Step Three: Check Your Eligibility In order to be considered for the role of a prison warden, there are some minimum requirements that you would need to meet in the U.S. You would first need to be a U.S. citizen, over the age of twenty-one, have the appropriate qualifications, have a pristine criminal record, and your record would need to reflect that you are mentally, legally, and physically capable of possessing a firearm. You would also need soft skills such as superior people skills, no history of drug use or alcohol abuse, and high fitness capabilities. Any previous convictions and negative employment history records can disqualify you from consideration. Step Four: Accumulate Experiences Considering the high role that a prison warden holds, it can be easy to assume that it isn't a role that one easily walks into. You would need to work your way up the ranking to the position of prison warden after obtaining relevant experience in a related field. Experience that could greatly contribute to the role of prison warden would be police officers, government clerks, and correction officers. Step Five: Expand Your Skills Furthering your studies and pursuing relevant training courses is a sure way to assist you in climbing your career ladder a lot faster. Furthering your studies and pursuing your master's degree is also another way to grow into the role of warden. Step Six: Apply Your Skills A lot of the steps you need to become a warden, and even a successful one, begins well before you even start applying for such a role. Once you have gained enough experience, you have mastered the skills, and you are equipped with the tools you need to conduct the job successfully, you may begin applying for available roles as a prison warden. What Degree Do You Need To Be Prison Warden? You would need a minimum of a bachelor's or an associate degree in criminal justice, criminology, or justice administration. Obtaining a master's degree will be of greater benefit to your growth in the career and applying for a job with already such a qualification will lead to you entering on the higher end of the pay scale. What Is The Average Salary For A Prison Warden? Being in a job within correctional services is a strenuous task. Moving up in the career chain is often short lived as stress is a great motivating factor for law enforcement practitioners to leave their positions. It is for that reason that the pay range of a prison warden is expected to be quite high, given the nature, role, and responsibilities of the job. However, while theU.S. Bureau of Labor and Statisticsdoes not provide a pay range for the role of prison warden, it does state that the average pay for bailiffs and correctional officers stands at $47,920 per year or $23.04 per hour. What Makes A Good Prison Warden? Aside from having the relevant skills and expertise to manage a hostile and often dangerous work environment, for themselves and for others, prison wardens are expected to have great people skills, be well-mannered, resourceful, and flexible in terms of scheduling. Additionally, having the correct qualifications for the job is a minimum requirement, and nothing less would be accepted. Overseeing such large numbers of extremely different and even dangerous individuals, as well as being the go-between between the prison and the media, as a prison warden, you may need to develop a metaphorical thick skin, which will enable you to take the criticism that you may often be faced with. FAQs Is A Prison Warden A Good Job? While it is extremely stressful and takes place in quite a dangerous environment, the job of a prison warden is one that is greatly sought after because of its high salary opportunities and room for internal growth. What Is Another Name For A Prison Warden? While there are many correct and incorrect terms for a prison warden, some that stand out above the rest in accuracy would be custodian, guard, and keeper. What Is A Female Warden Called? A female prison warden may be called a prison Matron. What Is The Salary Of A Prison Warden? While not explicitly determined, roles below the ranking of prison warden will receive an annual compensation of about $47,920. This is according to data recorded by the BLS before May 2021. What Qualifications Do You Need To Be A Prison Warden? You will need a bachelor's degree or an associate degree in criminal justice, or other related fields, as well as a master's degree depending on the growth within the position that you are hoping to achieve. Conclusion Pursuing a career as a prison warden needs to come from both the head and the heart. It needs to be driven by a passion for the industry, as well as the skill and expertise that will allow you to successfully perform the duties of a prison warden. If this is the road you seek to pursue, let us get to work to make this dream a reality. find criminal justice degrees near you FIND DEGREE find criminal justice degrees near you FIND DEGREE Articles How To Become A Correctional Officer: Career & Salary Information How To Become A Crime Scene Photographer: Career & Salary Information How To Become A Crime Scene Technician: Career & Salary Information How To Become A Narcotics Detective: Career & Salary Information How To Become A Police Detective: Career & Salary Information How To Become A Police Officer: Career & Salary Information How To Become A Prison Warden: Career & Salary Information How To Become A Probation Officer: Career & Salary Information How To Become An Air Marshal in The U.S.: Career & Salary Information How To Become An Arson Investigator: Career & Salary Information How To Become An Insurance Fraud Investigator: Career & Salary Information How To Become An Intelligence Analyst: Career & Salary Information How to Become a Campus Police Officer: Career & Salary Information How to Become a Crime Scene Cleaner: Career & Salary Information How to Become a Criminal Profiler: Career & Salary Information How to Become a DEA Agent: Career & Salary Information How to Become a Fingerprint Technician: Career & Salary Information How to Become a Forensic Psychologist: Career & Salary Information How to Become a Paralegal: Career & Salary Information How to Become a Private Investigator: Career & Salary Information How to Become a State Trooper: Career & Salary Information The Three Branches Of Criminal Justice Types Of Criminal Justice Degrees Types Of Criminal Justice Jobs Types of Police Detectives: Criminal Justice Career Options © 2025 - yourschoolmatch.com - all rights reserved Do Not Sell My Personal Information YourSchoolMatch Facebook AccountYourSchoolMatch Twitter AccountYourSchoolMatch Instagram AccountYourSchoolMatch LinkedIn Account Explore Areas of Study About Us Contact Us Useful Links Privacy Policy Terms of Service FAQ Get Services For StudentsFor Schools
16281	https://www.quora.com/How-do-I-know-that-x-+-1-x-2-is-the-smallest-value-for-the-above-expression-I-mean-how-do-I-know-that-the-smallest-value-is-acquired-by-AM-GM	Something went wrong. Wait a moment and try again. Optimization Problems Inequality of Arithmetic ... Algebraic Proofs Mean (mathematics) Geometric Means Mathematical Inequalities Arithmetic Means 5 How do I know that x + 1/x = 2 is the smallest value for the above expression? I mean, how do I know that the smallest value is acquired by AM-GM? Terry Moore M.Sc. in Mathematics, University of Southampton (Graduated 1968) · Author has 16.6K answers and 29.3M answer views · 9mo How do I know that x + 1/x = 2 is the smallest value for the above expression? I mean, how do I know that the smallest value is acquired by AM-GM? Try x=−1. I think you’ll find −1+1/(−1)<2 and if x=−10 we get a smaller value still. For positive x, you can differentiate: the derivative is 1–1/x2 and this is zero at a local minimum (or maximum), that means x=1. To rule out a maximum note that the derivative is negative for any positive x less than 1 and positive for any x greater than 1. The arithmetic-geometric mean inequality says (for positive data) that GM≤AM. GM of x and 1/x is 1 and the A How do I know that x + 1/x = 2 is the smallest value for the above expression? I mean, how do I know that the smallest value is acquired by AM-GM? Try x=−1. I think you’ll find −1+1/(−1)<2 and if x=−10 we get a smaller value still. For positive x, you can differentiate: the derivative is 1–1/x2 and this is zero at a local minimum (or maximum), that means x=1. To rule out a maximum note that the derivative is negative for any positive x less than 1 and positive for any x greater than 1. The arithmetic-geometric mean inequality says (for positive data) that GM≤AM. GM of x and 1/x is 1 and the AM is (x+1/x)/2, so (x+1/x)/2≥1 and x+1/x≥2. That avoids all that calculus stuff (so long as you have proved the AM-GM inequality without calculus). And we can achieve 2 if we put x=1. Steven Smith Earned 98% or higher in all my math classes at UCMO. · Author has 3.4K answers and 9.1M answer views · 8y I don’t know what AM-GM stands for, but I will make a guess as to what you are asking. I will guess that AM stands for Absolute Minimum and GM stands for Global Minimum. This problem can be solved with calculus. y=x+1x y′=1−1x2 y′′=2x3 To find a max/min, you set the derivative equal to 0. 0=1−1x2 1=1x2 x2=1 x=±1 Notice that the function is always positive when x>0 and the function is always negative when x<0 I will assume you are o I don’t know what AM-GM stands for, but I will make a guess as to what you are asking. I will guess that AM stands for Absolute Minimum and GM stands for Global Minimum. This problem can be solved with calculus. y=x+1x y′=1−1x2 y′′=2x3 To find a max/min, you set the derivative equal to 0. 0=1−1x2 1=1x2 x2=1 x=±1 Notice that the function is always positive when x>0 and the function is always negative when x<0 I will assume you are only looking at positive values of x, because there are many smaller values if x<0. Based on that assumption, x=1 is a max or min. To know which, we look at the second derivative at that point. y′′=213=2 Since that second derivative is positive, the point is a local minimum. limx→0x+1x=∞ limx→∞x+1x=∞ Therefore, we have proven that the local minimum is the absolute minimum in the interval x=(0,∞) The value of the function at that point is: y=1+11=2 Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. 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If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Related questions How do I know that the smallest value of the expression x + 1/x = 2 is acquired by AM-GM and there is no smaller value of it? How does the GM/AM method work for solving inequalities? What role does the AM-GM inequality or other algebraic inequalities play in proving \ (\frac {a^2+b} {a+b} + \frac {b^2+1} {b+1} \geq 2)? What is your favorite proof using the AM-GM inequality? Can AM-GM inequalities be used to prove geometric inequalities? Anthony Hawken Author has 10.8K answers and 3.8M answer views · Jan 16 2 is not the smallest value for the expression, x + 1/x, but it is the local minimum. 2 is not the smallest value for the expression, x + 1/x, but it is the local minimum. Enrico Gregorio Associate professor in Algebra · Upvoted by Justin Rising , PhD in statistics · Author has 18.4K answers and 16M answer views · Jan 2 We know that, for x>0, the smallest value for the expression x+1x is 2 because we prove it. Assume x>0 and set y=x+1x so we have x2−xy+1=0 Thus the discriminant y2−4 is certainly nonnegative and, as y>0, we conclude that y≥2. For x=1 we get x+x−1=2, so the minimum is indeed 2. Or you can use AM-GM (after having proved it) and conclude that x+x−12≥√x⋅x−1 hence x+1x≥2 and x=1 provides the minimum. Assistant Bot · 1y To show that the expression x+1x achieves its minimum value of 2 using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we can follow these steps: Step 1: Apply AM-GM Inequality The AM-GM inequality states that for any non-negative real numbers a and b: a+b2≥√ab with equality when a=b. Step 2: Set a and b In our case, let a=x and b=1x. Both x and 1x are positive for x>0. Step 3: Apply AM-GM Applying AM-GM to x and 1x: x+1x2≥√x⋅1x=√1=1 Multiplying both sides by 2 gives: x+\fr To show that the expression x+1x achieves its minimum value of 2 using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we can follow these steps: Step 1: Apply AM-GM Inequality The AM-GM inequality states that for any non-negative real numbers a and b: a+b2≥√ab with equality when a=b. Step 2: Set a and b In our case, let a=x and b=1x. Both x and 1x are positive for x>0. Step 3: Apply AM-GM Applying AM-GM to x and 1x: x+1x2≥√x⋅1x=√1=1 Multiplying both sides by 2 gives: x+1x≥2 Step 4: Equality Condition The equality holds when x=1x. Solving this gives: x2=1⟹x=1(since x>0) Conclusion Thus, the minimum value of x+1x is 2, which occurs at x=1. Therefore, we can conclude that the smallest value of the expression is indeed acquired by the AM-GM inequality, and it is verified that the minimum value is 2 when x=1. 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Harshit Bajpai Former Research Intern at Washington State University (2018–2018) · Author has 53 answers and 119.2K answer views · 8y Originally Answered: How do I know that the smallest value of the expression x + 1/x = 2 is acquired by AM-GM and there is no smaller value of it? · Well smallest value is 2 only if x is positive. In actual its range is (−infinity,−2]U[2,infinity). However from your question’s point of view , we take x as positive. So applying AM ≥ GM we have x+1x2≥√x∗1x x+1x≥2 Equality holds for x=1x x2=1 x=+−1 We consider only the positive value of x here as stated earlier Ted Alper Math instructor, OHS; Director (coordinator) of the Stanford Math Circle · Author has 836 answers and 3.2M answer views · 8y Originally Answered: How do I know that the smallest value of the expression x + 1/x = 2 is acquired by AM-GM and there is no smaller value of it? · Assuming Aand B are both greater than or equal to 0, the AM-GM inequality says that A+B2≥√AB with equality only when A=B [I am assuming you are not asking for the proof of the AM-GM inequality, though there are many fine proofs, perhaps the simplest is to observe that (√A−√B)2≥0 and expand the binomial to see (√A)2−2√AB+(√B)2≥0, etc] Anyway, if you let A=x and B=1/x you’ll get your result, since √AB=1 Sergei Michailov Retired System Analyst (2009–present) · Author has 5.5K answers and 3.5M answer views · 8y Assuming x>0, we get arithmetic mean of x and 1/x. It is 1/2(x+1/x). It is more or equal than geometric mean which is sqrt(11/x) = 1 1/2(x+1/x)>= 1 x+1/x>=2. At x=1 1/x+x=2, but it can not be any smaller. Max Gretinski Studied Mathematics · Author has 6.5K answers and 2.5M answer views · 9mo Let f(x) = x + 1x for all x > 0. Then f is twice differentiable, with f’(x) = 1 - 1x2, and f”(x) = 2x3 f’(x) = 0 only when x = 1. For this value of x, we see f”(1) = 2 > 0, meaning that the value attained by f(x) at x = 1 is a minimum. f(1) = 2, as you indicate. Related questions How do I know that the smallest value of the expression x + 1/x = 2 is acquired by AM-GM and there is no smaller value of it? How does the GM/AM method work for solving inequalities? What role does the AM-GM inequality or other algebraic inequalities play in proving \ (\frac {a^2+b} {a+b} + \frac {b^2+1} {b+1} \geq 2)? What is your favorite proof using the AM-GM inequality? Can AM-GM inequalities be used to prove geometric inequalities? Let a, b, c be positive real numbers such that abc = 1. Prove that a/b + b/c + c/a ≥ a + b + c. How? What is the greatest and smallest integer value of (x-2) (x^2-9x+14)? What is the value value of x [x (2^x-3^x)] / (x^2+x+1) (x-1) >0? If 121 = 1 + x 1 + x 2 + x 3 + x 4 what is x ? How do i integrate ∫ ( x − 1 ) ( x + 1 ) √ x 3 + x 2 + x d x ? What is the value of the following expression, (A-X) (B-X) … (Y-X) (Z-X)? If x + 1 x = a , what is x 2 + 1 x 2 ? Can you provide some proofs for a generalized form of the AM-GM Inequality that do not involve calculus? How is 2 x + 2 x + 1 = 3 ( 2 x ) ? What is the value of x = ? Given expression 27^x = 1/x About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
16282	https://pressbooks.bccampus.ca/practicalphysicsphys1104/chapter/7-3-gravitational-potential-energy/	Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. Chapter 6 Work, Energy, and Energy Resources 6.3 Gravitational Potential Energy Summary Explain gravitational potential energy in terms of work done against gravity. Show that the gravitational potential energy of an object of mass m at height h on Earth is given by PEg = mgh. Show how knowledge of the potential energy as a function of position can be used to simplify calculations and explain physical phenomena. Work Done Against Gravity Climbing stairs and lifting objects is work in both the scientific and everyday sense—it is work done against the gravitational force. When there is work, there is a transformation of energy. The work done against the gravitational force goes into an important form of stored energy that we will explore in this section. Let us calculate the work done in lifting an object of mass m through a height h, such as in Figure 1. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh. We define this to be the gravitational potential energy (PEg)put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force. For convenience, we refer to this as the PEg gained by the object, recognizing that this is energy stored in the gravitational field of Earth. Why do we use the word “system”? Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs. Converting Between Potential Energy and Kinetic Energy Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to mgh on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem). We will find it more useful to consider just the conversion of PEg to KE without explicitly considering the intermediate step of work. (See Example 2.) This shortcut makes it is easier to solve problems using energy (if possible) rather than explicitly using forces. More precisely, we define the change in gravitational potential energy ΔPEg to be where, for simplicity, we denote the change in height by h rather than the usual Δh. Note that h is positive when the final height is greater than the initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00 m, then its change in gravitational potential energy is Note that the units of gravitational potential energy turn out to be joules, the same as for work and other forms of energy. As the clock runs, the mass is lowered. We can think of the mass as gradually giving up its 4.90 J of gravitational potential energy, without directly considering the force of gravity that does the work. Using Potential Energy to Simplify Calculations The equation ΔPEg = mgh applies for any path that has a change in height of h, not just when the mass is lifted straight up. (See Figure 2.) It is much easier to calculate mgh (a simple multiplication) than it is to calculate the work done along a complicated path. The idea of gravitational potential energy has the double advantage that it is very broadly applicable and it makes calculations easier. From now on, we will consider that any change in vertical position h of a mass m is accompanied by a change in gravitational potential energy mgh, and we will avoid the equivalent but more difficult task of calculating work done by or against the gravitational force. Figure 2. The change in gravitational potential energy (ΔPEg) between points A and B is independent of the path. ΔPEg=mgh for any path between the two points. Gravity is one of a small class of forces where the work done by or against the force depends only on the starting and ending points, not on the path between them. Example 1: The Force to Stop Falling A 60.0-kg person jumps onto the floor from a height of 3.00 m. If he lands stiffly (with his knee joints compressing by 0.500 cm), calculate the force on the knee joints. Strategy This person’s energy is brought to zero in this situation by the work done on him by the floor as he stops. The initial PEg is transformed into KE as he falls. The work done by the floor reduces this kinetic energy to zero. Solution The work done on the person by the floor as he stops is given by with a minus sign because the displacement while stopping and the force from floor are in opposite directions (cos θ = cos 180° = 1). The floor removes energy from the system, so it does negative work. The kinetic energy the person has upon reaching the floor is the amount of potential energy lost by falling through height h: The distance d that the person’s knees bend is much smaller than the height h of the fall, so the additional change in gravitational potential energy during the knee bend is ignored. The work W done by the floor on the person stops the person and brings the person’s kinetic energy to zero: Combining this equation with the expression for W gives Recalling that h is negative because the person fell down, the force on the knee joints is given by Discussion Such a large force (500 times more than the person’s weight) over the short impact time is enough to break bones. A much better way to cushion the shock is by bending the legs or rolling on the ground, increasing the time over which the force acts. A bending motion of 0.5 m this way yields a force 100 times smaller than in the example. A kangaroo’s hopping shows this method in action. The kangaroo is the only large animal to use hopping for locomotion, but the shock in hopping is cushioned by the bending of its hind legs in each jump.(See Figure 3.) Figure 3. The work done by the ground upon the kangaroo reduces its kinetic energy to zero as it lands. However, by applying the force of the ground on the hind legs over a longer distance, the impact on the bones is reduced. (credit: Chris Samuel, Flickr) Example 2: Finding the Speed of a Roller Coaster from its Height (a) What is the final speed of the roller coaster shown in Figure 4 if it starts from rest at the top of the 20.0 m hill and work done by frictional forces is negligible? (b) What is its final speed (again assuming negligible friction) if its initial speed is 5.00 m/s? Figure 4. The speed of a roller coaster increases as gravity pulls it downhill and is greatest at its lowest point. Viewed in terms of energy, the roller-coaster-Earth system’s gravitational potential energy is converted to kinetic energy. If work done by friction is negligible, all ΔPEg is converted to KE. Strategy The roller coaster loses potential energy as it goes downhill. We neglect friction, so that the remaining force exerted by the track is the normal force, which is perpendicular to the direction of motion and does no work. The net work on the roller coaster is then done by gravity alone. The loss of gravitational potential energy from moving downward through a distance h equals the gain in kinetic energy. This can be written in equation form as -ΔPEg = ΔKE. Using the equations for PEg and KE, we can solve for the final speed v, which is the desired quantity. Solution for (a) Here the initial kinetic energy is zero, so that The equation for change in potential energy states that ΔPEg = mgh. Since h is negative in this case, we will rewrite this as ΔPEg = –mg\|h\| to show the minus sign clearly. Thus, becomes Solving for v, we find that mass cancels and that Substituting known values, Solution for (b) Again -ΔPEg = ΔKE. In this case there is initial kinetic energy, so Thus, Rearranging gives This means that the final kinetic energy is the sum of the initial kinetic energy and the gravitational potential energy. Mass again cancels, and This equation is very similar to the kinematics equationbut it is more general—the kinematics equation is valid only for constant acceleration, whereas our equation above is valid for any path regardless of whether the object moves with a constant acceleration. Now, substituting known values gives Discussion and Implications First, note that mass cancels. This is quite consistent with observations made in Chapter 2.7 Falling Objects that all objects fall at the same rate if friction is negligible. Second, only the speed of the roller coaster is considered; there is no information about its direction at any point. This reveals another general truth. When friction is negligible, the speed of a falling body depends only on its initial speed and height, and not on its mass or the path taken. For example, the roller coaster will have the same final speed whether it falls 20.0 m straight down or takes a more complicated path like the one in the figure. Third, and perhaps unexpectedly, the final speed in part (b) is greater than in part (a), but by far less than 5.00 m/s. Finally, note that speed can be found at any height along the way by simply using the appropriate value of h at the point of interest. We have seen that work done by or against the gravitational force depends only on the starting and ending points, and not on the path between, allowing us to define the simplifying concept of gravitational potential energy. We can do the same thing for a few other forces, and we will see that this leads to a formal definition of the law of conservation of energy. MAKING CONNECTIONS: TAKE-HOME INVESTIGATION— CONVERTING POTENTIAL TO KINETIC ENERGY One can study the conversion of gravitational potential energy into kinetic energy in this experiment. On a smooth, level surface, use a ruler of the kind that has a groove running along its length and a book to make an incline (see Figure 5). Place a marble at the 10-cm position on the ruler and let it roll down the ruler. When it hits the level surface, measure the time it takes to roll one meter. Now place the marble at the 20-cm and the 30-cm positions and again measure the times it takes to roll 1 m on the level surface. Find the velocity of the marble on the level surface for all three positions. Plot velocity squared versus the distance traveled by the marble. What is the shape of each plot? If the shape is a straight line, the plot shows that the marble’s kinetic energy at the bottom is proportional to its potential energy at the release point. Figure 5. A marble rolls down a ruler, and its speed on the level surface is measured. Section Summary Work done against gravity in lifting an object becomes potential energy of the object-Earth system. The change in gravitational potential energy, ΔPEg, is ΔPEg = mgh, with h being the increase in height and g the acceleration due to gravity. The gravitational potential energy of an object near Earth’s surface is due to its position in the mass-Earth system. Only differences in gravitational potential energy, ΔPEg, have physical significance. As an object descends without friction, its gravitational potential energy changes into kinetic energy corresponding to increasing speed, so that ΔKE = -ΔPEg. Conceptual Questions 1: In Example 2, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m/s downhill. Suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m below the start. We would find in that case that it had the same final speed. Explain in terms of conservation of energy. 2: Does the work you do on a book when you lift it onto a shelf depend on the path taken? On the time taken? On the height of the shelf? On the mass of the book? Problems & Exercises 1: A hydroelectric power facility (see Figure 6) converts the gravitational potential energy of water behind a dam to electric energy. (a) What is the gravitational potential energy relative to the generators of a lake of volume 50.0 km3 mass = 5.00 × 1013 kg, given that the lake has an average height of 40.0 m above the generators? (b) Compare this with the energy stored in a 9-megaton fusion bomb. Figure 6. Hydroelectric facility (credit: Denis Belevich, Wikimedia Commons) 2: (a) How much gravitational potential energy (relative to the ground on which it is built) is stored in the Great Pyramid of Cheops, given that its mass is about 7 × 109 kg and its center of mass is 36.5 m above the surrounding ground? (b) How does this energy compare with the daily food intake of a person? 3: Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.5 m from the ground to a branch. (a) How much work did the bird do on the snake? (b) How much work did it do to raise its own center of mass to the branch? 4: In Example 2, we found that the speed of a roller coaster that had descended 20.0 m was only slightly greater when it had an initial speed of 5.00 m/s than when it started from rest. This implies that ΔPE>>KEi. Confirm this statement by taking the ratio of ΔPE to KEi. (Note that mass cancels.) 5: A 100-g toy car is propelled by a compressed spring that starts it moving. The car follows the curved track in Figure 7. Show that the final speed of the toy car is 0.687 m/s if its initial speed is 2.00 m/s and it coasts up the frictionless slope, gaining 0.180 m in altitude. Figure 7. A toy car moves up a sloped track. (credit: Leszek Leszczynski, Flickr) 6: In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic energy is small compared with the gain in gravitational potential energy on even small hills.) To demonstrate this, find the final speed and the time taken for a skier who skies 70.0 m along a 30° slope neglecting friction. Hint: you have to use trigonometry to find the height of the hill first. (a) Starting from rest. (b) Starting with an initial speed of 2.50 m/s. (c) Does the answer surprise you? Discuss why it is still advantageous to get a running start in very competitive events. Glossary gravitational potential energy : the energy an object has due to its position in a gravitational field Solutions Problems & Exercises 1: (a) 1.96 x 1016 J (b) The ratio of gravitational potential energy in the lake to the energy stored in the bomb is 0.52. That is, the energy stored in the lake is approximately half that in a 9-megaton fusion bomb. 3: (a) 1.8 J (b) 8.6 J 5: 6: h = 35.0 m = (70 m) sin 30.0o m g h = 1/ 2 m v2 so v = 26.2 m/s
16283	https://mathquestions.quora.com/What-is-the-equation-of-an-ellipse-whose-foci-are-3-0-and-3-0-such-that-for-any-point-on-it-the-sum-of-the-distan	What is the equation of an ellipse whose foci are (-3,0) and (3, 0) such that for any point on it, the sum of the distances from the distances of the foci is 10? - Math Questions - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Math Questions PROBABLY THE MOST ACTIVE MATH COMMUNITY ON THE WEB!! UP/DOWN VOTE GOOD/BAD Qs/As Follow · 379.1K 379.1K What is the equation of an ellipse whose foci are (-3,0) and (3, 0) such that for any point on it, the sum of the distances from the distances of the foci is 10? Submission accepted by Abu Khan · See parent question Answer Request Follow · 9 1 1 Answer Sort Recommended Enrico Gregorio Associate professor in Algebra · 3y · If the point has coordinates (x,y)(x,y), then we need √(x+3)2+y 2+√(x−3)2+y 2=10(x+3)2+y 2+(x−3)2+y 2=10 Move the first summand to the right-hand side and square to get x 2−6 x+9+y 2=100+x 2+6 x+9+y 2−20√x 2+6 x+9+y 2 x 2−6 x+9+y 2=100+x 2+6 x+9+y 2−20 x 2+6 x+9+y 2 Cancel equal terms, simplify the common factor 4 4 and isolate the radical in the left-hand side 5√x 2+6 x+9+y 2=3 x+25 5 x 2+6 x+9+y 2=3 x+25 Square again: 25 x 2+150 x+225+25 y 2=9 x 2+150 x+625 25 x 2+150 x+225+25 y 2=9 x 2+150 x+625 Rearrange: 16 x 2+25 y 2=400 16 x 2+25 y 2=400 Canonical form x 2 25+y 2 16=1 x 2 25+y 2 16=1 More generally, if the focal distance is 2 c 2 c and the fixed sum of distances is 2 a 2 a, then the equation is x 2 a 2+y 2 b 2=1 x 2 a 2+y 2 b 2=1 where b 2=a 2−c 2 b 2=a 2−c 2. In this case, a=5,c=3 a=5,c=3 and so b 2=25−9=16 b 2=25−9=16. 758 views · · View 4 other answers on parent question Related questions Can you help me solve (2-x) ^ (x²-1) =2? What is 5000x 454 x67 x90? If 5^2(x-1) × 5^x+1=0.04, what is the value of x? Can 41 books be placed on more than one shelf, provided that each shelf has the same number of books? Do you write 0.85 as a fraction? When we look at the equation 2=2, there is a 2 on the left side of the equal sign and a 2 on the right side of the equal sign, so how does the equal sign represent complete equality? Can we guess the correct prime number divisor of 90861 in our head? Does Euclidean Space only work because Gödel's first order completeness? Is Euclidean space an objective feature of the universe, or is it a mental abstraction produced by human perception? If Euclidean space is an abstraction derived from reality, can it ever capture the “completeness” of reality itself? And is this something set theory should have resolved? If Euclidean space is a cognitive encoding of perceptual deltas rather than an ontological substrate, does this imply that orthodox mathematics and physics are structurally flawed in their foundational assumptions? Did Gödel’s incompleteness theorems prove that Descartes’ Euclidean sandbox cannot prove or disprove reality, and that Plato’s assertions about eternal mathematical forms are therefore false? What is the explicit form of u n−1=3 a n−2 a n−1 u 1+(2 a+u 1)u n−1 u n−1=3 a n−2 a n−1 u 1+(2 a+u 1)u n−1? How do I figure it out the value of a a that the function f(x)=⎧⎨⎩a x+2;x≤2 x 3−2 x 2+2 x−4 x−2,x>2 f(x)={a x+2;x≤2 x 3−2 x 2+2 x−4 x−2,x>2 will be continous? For a=1 a=1 find the inverse function of f f and show that is one-to-one. The real sequence is defined recursively by b n+1=1 2(b n+3 b n)b n+1=1 2(b n+3 b n) with b 1=2 b 1=2. How do I show that this sequence is convergent and how to find its limit? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
16284	https://jdapm.org/DOIx.php?id=10.17245/jdapm.2021.21.4.357	:: JDAPM :: Journal of Dental Anesthesia and Pain Medicine Open Access, Peer-reviewed pISSN 2383-9309 eISSN 2383-9317 Home About About the JDAPM Editorial Board Journal Management Team Best Practice Journal Information Open Access Contact us Browse Current Issue Archive JDAPM on JDAPM on Most Read Most Cited For Contributors Instructions for Authors Research & Publication Ethics Article Processing Charges Author’s checklist Copyright Transfer Agreement Advertising Policies Journal of Dental Anesthesia and Pain Medicine About About the JDAPM Editorial Board Journal Management Team Best Practice Journal Information Open Access Contact us Browse Current Issue Archive JDAPM on JDAPM on Most Read Most Cited For Contributors Instructions for Authors Research & Publication Ethics Article Processing Charges Author’s checklist Copyright Transfer Agreement Advertising Policies Advanced Search Abstract INTRODUCTION CASE REPORT DISCUSSION Notes References Home Archive v.21(4);Aug 2021 10.17245/jdapm.2021.21.4.357 J Dent Anesth Pain Med. 2021 Aug;21(4):357-361. English. Published online Jul 30, 2021. Copyright © 2021 Journal of Dental Anesthesia and Pain Medicine Case Report Methemoglobinemia caused by a low dose of prilocaine during general anesthesia Makiko Shibuya,Takayuki Hojo,Yuri Hase,Yukifumi Kimura and Toshiaki Fujisawa Author information Author notes Copyright and License Department of Dental Anesthesiology, Faculty of Dental Medicine and Graduate School of Dental Medicine, Hokkaido University, Sapporo, Japan. Corresponding Author: Makiko Shibuya, Department of Dental Anesthesiology, Faculty of Dental Medicine and Graduate School of Dental Medicine, Hokkaido University, Kita 13, Nishi 7, Kita-ku, Sapporo 060-8586, Japan. Tel: +81-11-706-4336, Email: shibu@den.hokudai.ac.jp Received May 27, 2021; Revised July 09, 2021; Accepted July 11, 2021. This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited. Go to: AbstractINTRODUCTIONCASE REPORTDISCUSSIONNotesReferences Abstract Methemoglobinemia is a blood disorder in which an abnormal amount of methemoglobin is produced, and prilocaine is one of the drugs that can cause this disorder. The maximum recommended dose of prilocaine is 8 mg/kg. We report a case of methemoglobinemia caused by the administration of 4.2 mg/kg of prilocaine without other methemoglobinemia-inducing drugs during general anesthesia. A 17-year-old girl with hyperthyroidism and anemia was scheduled to undergo maxillary sinus floor elevation and tooth extraction. The patient's peripheral oxygen saturation (SpO 2) decreased from 100% at arrival to 95% after receiving prilocaine with felypressin following induction of general anesthesia. However, the fraction of inspired oxygen was 0.6. Blood gas analysis showed that the methemoglobin level was 3.8% (normal level, 1%–2%), fractional oxygen saturation was 93.9%, partial pressure of oxygen was 327 mmHg, and arterial oxygen saturation was 97.6%. After administration of 1 mg/kg of methylene blue, her SpO 2 improved gradually to 99%, and the methemoglobin value decreased to 1.2%. When using prilocaine as a local anesthetic, it is important to be aware that methemoglobinemia may occur even at doses much lower than the maximum recommended dose. Keywords Anesthesia, General; Methemoglobinemia; Oximetry; Oxyhemoglobins; Prilocaine Go to: AbstractINTRODUCTIONCASE REPORTDISCUSSIONNotesReferences INTRODUCTION Methemoglobinemia is a condition in which the proportion of methemoglobin (MetHb), which cannot bind and carry oxygen, increases to > 1%–2% of the total hemoglobin . Prilocaine is one of the drugs that cause methemoglobinemia, and a dose ≤ 8 mg/kg is recommended for healthy adults . However, there are no definitive criteria regarding the dosage specific to young people. Herein, we report a case of methemoglobinemia during general anesthesia in a teenager, caused by 4.2 mg/kg of prilocaine, which is much lower than the maximum recommended dose, without the administration of other methemoglobinemia-inducing drugs. Go to: AbstractINTRODUCTIONCASE REPORTDISCUSSIONNotesReferences CASE REPORT The patient and her mother provided written informed consent to publish this report. A 17-year-old girl (height, 170.8 cm; weight, 51.2 kg; body mass index, 17.55 kg/m 2) was scheduled to undergo maxillary sinus floor elevation and tooth removal under general anesthesia. The patient was previously diagnosed with jaw deformity, excessive resorption of the maxillary alveolar ridge, and bilateral impacted mandibular wisdom teeth. At the age of 15 years, the patient was diagnosed with hyperthyroidism which was being treated with 10 mg thiamazole orally. She had no reported drug or food allergies, but reported a history of administration of local anesthetic several times with no complications. Initially, the surgery was postponed because of complications related to hyperthyroidism. During the preoperative examination, she was in a clinically euthyroid state. Laboratory tests showed a hemoglobin level of 10.2 g/dL (Table 1). No abnormalities were noted on other laboratory tests or chest radiographs. A 12-lead preoperative electrocardiogram showed normal sinus rhythm. Table 1 Results of laboratory tests Click for larger image Click for full table Download as Excel file Given her hyperthyroidism, we planned to administer 3% prilocaine containing 0.03 IU/mL felypressin as a local anesthetic to avoid complications that may arise from adrenaline-containing drugs. Her medications were continued orally until 14 h before surgery. Upon arrival in the operating room, her peripheral oxygen saturation (SpO 2) on room air was 100%. General anesthesia was induced with 100 µg of fentanyl, target-controlled infusion of propofol with an initial target plasma concentration of 5.0 µg/mL, and 30 mg of rocuronium. The patient was intubated using a nasal endotracheal tube. Anesthesia was maintained using an intermittent bolus of fentanyl (total dose was 400 µg), propofol infusion (target plasma concentration of 2.8–3.2 µg/mL), and remifentanil (0.3–0.6 mg/h). Approximately 61 min after inducing general anesthesia, the patient was administered 3.6 mL prilocaine with felypressin for infiltration anesthesia; the same dose was added 43 min later, which led to a gradual decrease in SpO 2, even though the fraction of inspired oxygen was increased to 0.6 (Fig. 1). Fig. 1 Changes in SpO 2 after administration of prilocaine. Fifty-three minutes after the second administration of prilocaine, the SpO 2 level decreased to 95%. After 50 mg of methylene blue was gradually administered (in 10-mg doses), the SpO 2 value gradually improved to 99%. The SpO 2 value transiently decreased immediately after the administration of methylene blue and then improved because SpO 2 was affected by the color of methylene blue itself. SpO 2, peripheral oxygen saturation. Click for larger image Download as PowerPoint slide Ninety-six minutes after the first administration of prilocaine, the SpO 2 level decreased to 95%. Blood gas analysis showed that the MetHb value was 3.8% and fractional oxyhemoglobin was 93.9%, despite a partial pressure of oxygen of 327 mmHg (BGA#1 in Table 2). Next, 50 mg of methylene blue was administered slowly, 127 min after the first administration of prilocaine. The SpO 2 value improved gradually to 99%, and the MetHb value decreased to 1.2% (BGA#2 in Table 2). SpO 2 remained above 99%, and the MetHb value 2.5 h after the administration of methylene blue was 1.6% (BGA#3 in Table 2). No noteworthy adverse respiratory findings were observed during the perioperative period. Table 2 Blood gas analysis data Click for larger image Click for full table Download as Excel file Upon returning to her room, the patient did not present with any symptoms of methemoglobinemia, and SpO 2 remained above 99% for 2 h under oxygenation with a simple face mask at 3 L/min. After 2 h, she remained asymptomatic, and her SpO 2 remained at 98% after the oxygen mask was removed. From that point until the next day, her SpO 2 was measured intermittently and remained at 97%–98% on room air. The MetHb value 7 days after surgery was 1.2%, which is within normal limits, and congenital methemoglobinemia was ruled out. Go to: AbstractINTRODUCTIONCASE REPORTDISCUSSIONNotesReferences DISCUSSION The agents that produce methemoglobinemia include local anesthetics such as prilocaine, benzocaine, and lidocaine, and nitrites such as nitroglycerin and metoclopramide. Reports of similar cases suggest that a combination of two drugs might cause methemoglobinemia . Guay et al. reported that the highest number of incidents of methemoglobinemia due to local anesthetics was caused by benzocaine, followed by prilocaine . It has been reported that the incidence of methemoglobinemia due to prilocaine requiring treatment was 0.008% (2/2431) in pediatric patients . However, the incidence in teenagers or adults is unclear. In this case, we concluded that methemoglobinemia was caused by prilocaine because 1) prilocaine was the only drug that might have caused methemoglobinemia during general anesthesia, 2) SpO 2 decreased 10 min after the second administration of prilocaine, and 3) congenital methemoglobinemia was ruled out. When using prilocaine, methemoglobinemia may occur even at a dose much lower than the recommended maximum dose. The recommended maximum dose for healthy adults is 8 mg/kg (600 mg) or 6 mg/kg (400 mg) . Some reports suggest that lower doses should be administered in children and patients with renal failure [1, 2, 6]. It has also been reported that the dose of prilocaine should be limited to 5 mg/kg for safe use because, in some cases, a dose of 5.0–7.4 mg/kg of prilocaine can cause methemoglobinemia. In the present case, methemoglobinemia occurred at a dose of 216 mg (4.2 mg/kg), although the patient was not a child and did not have renal failure. However, the patient was thin; her body mass index was 17.55 kg/m 2. Consequently, a dose of 216 mg corresponded to only 3.4 mg/kg based on a standard body weight of 64.2 kg. The risk factors for the development of methemoglobinemia while using prilocaine in adults are as follows [1, 3, 7, 8]: 1) Renal failure 2) Younger age: A previous study reported that the predicted level of MetHb reduced by approximately 0.3% per decade in patients older than 18 years. 3) Female sex 4) Using a higher concentration of prilocaine 5) An environment where prilocaine is easily absorbed: A study reported that direct injection of prilocaine into a bleeding surgical site was a contributing factor. 6) Concomitant administration of other oxidizing drugs The patient presented three of the four factors for increased risk of MetHb formation reported in a prospective observational study : young age, female sex, and high concentration of prilocaine. The cause may also be related to the infiltration of the anesthetic near the maxillary sinus, where blood flow is abundant and drugs are easily absorbed. However, a large interindividual variability between the dose of prilocaine and the amount of MetHb generated has been reported, and reliable prediction of MetHb formation is difficult . Therefore, it was not possible to determine the cause in this patient. It has been reported that the MetHb level is approximately 20% in symptomatic patients and approximately 30% in asymptomatic patients before the administration of therapeutic agents . Further, it has been reported that the symptoms of methemoglobinemia can appear earlier and be more severe in patients with anemia, acidosis, respiratory failure, and heart failure . Therefore, these patients must be treated as early as possible. In this case, we administered methylene blue before emergence from anesthesia because 1) the patient was under general anesthesia and the presence or absence of symptoms could not be determined; 2) it was not possible to predict whether the elevation of blood concentration of MetHb would continue, and how long it would take for MetHb to be metabolized; and 3) a decrease in the reserve respiratory capacity should be avoided during emergence from anesthesia. In addition, we believe that early intervention was appropriate because the patient was anemic with a hemoglobin level of 10.2 g/dL, and had low oxygen-carrying capacity. A study reported that anemic patients might be more sensitive to symptoms of methemoglobinemia because of their lower functional hemoglobin reserve . Ascorbic acid administration can be considered as an alternative treatment. However, the slow onset of action may be problematic [11, 12]. If there are no factors that exacerbate methemoglobinemia, such as anemia, oxygen therapy with close monitoring may be preferred. There are no reports describing the criteria for the treatment of methemoglobinemia during general anesthesia. Therefore, additional case reports and studies are required. When MetHb levels increase, the SpO 2 value, measured by a conventional two-wavelength pulse oximeter, is higher than the actual oxygen saturation, resulting in an underestimation of hypoxemia. This type of pulse oximeter does not consider abnormal hemoglobin, including MetHb, while determining the percentage of oxyhemoglobin. However, conventional pulse oximeters can provide clues for detecting methemoglobinemia . Therefore, when SpO 2 gradually decreases after administration of prilocaine, methemoglobin levels should always be checked even if the dose is small. For the correct diagnosis of methemoglobinemia and accurate evaluation of hypoxemia, a blood gas analysis apparatus including a carbon monoxide oximeter or a multiwavelength pulse oximeter is needed. These devices can determine MetHb and fractional oxyhemoglobin levels, including the percentage of oxyhemoglobin in all types of hemoglobin, including abnormal hemoglobin. In conclusion, when using prilocaine as a local anesthetic, clinicians should be aware that methemoglobinemia may occur even at doses lower than the recommended maximum dose. Go to: AbstractINTRODUCTIONCASE REPORTDISCUSSIONNotesReferences Notes AUTHOR CONTRIBUTIONS: Makiko Shibuya: Conceptualization, Data curation, Visualization, Writing — original draft, Writing — review & editing. Takayuki Hojo: Conceptualization, Writing — review & editing. Yuri Hase: Writing — review & editing. Yukifumi Kimura: Conceptualization, Writing — review & editing. Toshiaki Fujisawa: Supervision, Writing — review & editing. DECLARATION OF INTERESTS:There are no conflicts of interest to declare. Go to: AbstractINTRODUCTIONCASE REPORTDISCUSSIONNotesReferences References Guay J. Methemoglobinemia related to local anesthetics: a summary of 242 episodes. Anesth Analg 2009;108:837–845. PubMed CrossRef Wilburn-Goo D, Lloyd LM. When patients become cyanotic: acquired methemoglobinemia. J Am Dent Assoc 1999;130:826–831. PubMed CrossRef Hojo T, Kimura Y, Ohiwa D, Fujisawa T. A case of methemoglobinemia thought to have been caused by the combined use of propitocaine and nitroglycerin during general anesthesia. Anesth Prog 2020;67:170–171. PubMed CrossRef Arslan D, Yıldız G, Şahin MO. The incidence of methemoglobinemia due to prilocaine use in circumcision. J Urol Surg 2019;6:38–41. CrossRef Becker DE, Reed KL. Local anesthetics: review of pharmacological considerations. Anesth Prog 2012;59:90–102. PubMed CrossRef Alanazi MQ. Drugs may be induced methemoglobinemia. J Hematol Thrombo Dis 2017;5:1–5. Kreutz RW, Kinni ME. Life-threatening toxic methemoglobinemia induced by prilocaine. Oral Surg Oral Med Oral Pathol 1983;56:480–482. PubMed CrossRef Vasters F, Eberhart L, Koch T, Kranke P, Wulf H, Morin A. Risk factors for prilocaine-induced methaemoglobinaemia following peripheral regional anaesthesia. Eur J Anaesthesiol 2006;23:760–765. PubMed CrossRef Wright RO, Lewander WJ, Woolf AD. Methemoglobinemia: etiology, pharmacology, and clinical management. Ann Emerg Med 1999;34:646–656. PubMed CrossRef Ash-Bernal R, Wise R, Wright SM. Acquired methemoglobinemia: a retrospective series of 138 cases at 2 teaching hospitals. Medicine 2004;83:265–273. PubMed CrossRef Lee KW, Park SY. High-dose vitamin C as treatment of methemoglobinemia. Am J Emerg Med 2014;32:936. PubMed CrossRef Boran P, Tokuc G, Yegin Z. Methemoglobinemia due to application of prilocaine during circumcision and the effect of ascorbic acid. J Pediatr Urol 2008;4:475–476. PubMed CrossRef Hurford WE, Kratz A. Case 23-2004: a 50-year-old woman with low oxygen saturation. N Engl J Med 2004;351:380–387. PubMed CrossRef Cite Article PDF Cited by Crossref 3 G o o g l e Scholar PubMed 3 Publication Types Case Report MeSH Terms Adolescent Anemia Anesthesia, General Anesthetics, Local Blood Gas Analysis Felypressin Female Humans Hyperthyroidism Maxillary Sinus Methemoglobin Methemoglobinemia Methylene Blue Oximetry Oxygen Oxyhemoglobins Partial Pressure Prilocaine Prilocaine Tooth Extraction × Since 2021/08/01 Metrics Page Views 345 PDF Downloads 65 Share Links to PubMed PubMed Central Related citations in PubMed Figures Show all... Previous Next 1 / 1 Tables Show all... Previous Next 1 / 2 ORCID IDs Makiko Shibuya Takayuki Hojo Yuri Hase Yukifumi Kimura Toshiaki Fujisawa Abstract INTRODUCTION CASE REPORT DISCUSSION Notes References Cited by Crossref Is Cited by the Following Articles in Close Cite Citation successfully copied. Copy and paste a formatted citation from below or use one of the hyperlinks at the bottom to download a file for import into a bibliography manager. Copy Download EndNote / ProCite / Reference Manager RIS Format Plain Text Download with Abstract EndNote / ProCite / Reference Manager RIS Format Plain Text PERMALINK Copy Permalink information copied to clipboard PublisherThe Korean Dental Society of AnesthesiologyContact usOpen AccessKOFST JDAPM onPubMedPubMed CentralGoogle ScholarCrossref InformationAims and ScopeEditorial BoardInstructions for AuthorsArticle Processing ChargesJournal Data BrowseCurrent IssuePast Issues © 2025
16285	https://www.reddit.com/r/leetcode/comments/130az0j/bulb_switcher_o1_solution_detailed_explanation/	Bulb Switcher O(1) Solution Detailed Explanation (Daily Question) : r/leetcode Skip to main contentBulb Switcher O(1) Solution Detailed Explanation (Daily Question) : r/leetcode Open menu Open navigationGo to Reddit Home r/leetcode A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to leetcode r/leetcode•2 yr. ago Alec-Reddit Bulb Switcher O(1) Solution Detailed Explanation (Daily Question) There are n bulbs that are initially off. You first turn on all the bulbs, then you turn off every second bulb. On the third round, you toggle every third bulb (turning on if it's off or turning off if it's on). For the ith round, you toggle every ith bulb. For the nth round, you only toggle the last bulb. Return the number of bulbs that are on after n rounds. All bulbs are initially off, so only bulbs toggled an odd number of times will be on At each round i, we toggle all bulbs with a factor of i → So the question boils down to which numbers have an odd number of factors Key intuition #2: all factors come in pairs. 10’s factors are (5,2) and(10,1) for example. So how can a number have an odd number of factors? → A number must be a perfect square to have an odd number of factors A perfect square has whatever other factors plus a pair of the same number, its square root. Ex. 9 has 3 factors: the pair (1,9), and the square root 3 O(n) - times out for very large numbers def bulbSwitch0(self, n: int) -> int: return sum([isqrt(i) == sqrt(i) for i in range(1, n+1)]) This is not fast enough for leetcode though. Can we get in constant time the number of perfect squares less than a number? Consider a number N. The square root function is strictly increasing so we know that all square roots of perfect squares less than N are less than sqrt(N). So the answer is floor(sqrt(N)) O(1) def bulbSwitch(self, n: int) -> int: return isqrt(n) Read more Share Related Answers Section Related Answers Detailed explanation of bulb switcher problem Best solutions for bulb switcher on LeetCode Top coding interview tips for beginners Best resources for mastering data structures How to approach system design interviews New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Is this how a hashtable looks like? Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of April 27, 2023 Reddit reReddit: Top posts of April 2023 Reddit reReddit: Top posts of 2023 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
16286	https://gamedev.stackexchange.com/questions/44720/line-intersection-from-parametric-equation	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Line Intersection from parametric equation Ask Question Asked Modified 12 years, 10 months ago Viewed 11k times 7 \$\begingroup\$ I'm sure this question has been asked before. However, I'm trying to connect the dots by translating an equation on paper into an actual function. I thought It would be interesting to ask here instead on the Math sites (since it's going to be used for games anyway ). Let's say we have our vector equation : x = s + Lr; where x is the resulting vector, s our starting point/vector. L our parameter and r our direction vector. The ( not sure it's called like this, please correct me ) normal equation is : x.n = c; If we substitute our vector equation we get: (s+Lr).n = c. We now need to isolate L which results in L = (c - s.n) / (r.n); L needs to be 0 < L < 1. Meaning it needs to be between 0 and 1. My question: I want to know what L is so if I were to substitute L for both vector equation (or two lines) they should give me the same intersection coordinates. That is if they intersect. But I can't wrap my head around on how to use this for two lines and find the parameter that fits the intersection point. Could someone with a simple example show how I could translate this to a function/method? 2d mathematics vector intersection Share asked Nov 30, 2012 at 10:56 SidarSidar 3,58722 gold badges1717 silver badges2828 bronze badges \$\endgroup\$ 1 \$\begingroup\$ x.n=c looks a bit like the point normal form of a plane (I can't think what else it might be), why are you using that? Are you trying to find the intersection of a line and a plane? \$\endgroup\$ Ken – Ken 2012-11-30 14:06:24 +00:00 Commented Nov 30, 2012 at 14:06 Add a comment \| 2 Answers 2 Reset to default 14 \$\begingroup\$ Given two parametric equations of lines; L=a+t.b // t is the paramerter, a & b are vectors M=c+u.d //u is parameter, c & d are vectors The the point of intersection is the one place in space where both these equations are equal(produce the same point). When the lines intersect at some value of t and some value of u, the equations are equal, because they resolve to the same point; L=M a+tb=c+u.d Now we have an equation with two unknowns (u & t). In order to find the point of intersection we need at least one of the unknowns. First step is to isolate one of the unknowns, in this case t; t=(c+u.d-a)/b but this is a 2D Vector equation, so it is really two equations, one in x and the other in y. Two equations is (usually) enough to solve a system with two unknowns. t=(cx+u.dx-ax)/bx t=(cy+u.dy-ay)/by So now we have two expressions, which equal the same value (t), so they must equal each other. That is why we isolated t, in order to have two equations which equal each other AND only one unknown on either side. (cx+u.dx-ax)/bx = t = (cy+u.dy-ay)/by //from above // equate and isolate u (cx+u.dx-ax)/bx=(cy+udy-ay)/by //note: `t` is gone Now, we have an equation with only one unknown, which we will solve by isolating the remaining unknown, u cx.by+u.dx.by-ax.by=cy.bx+u.dy.bx-ay.bx u.dx.by-u.dy.bx=cy.bx-ay.bx-cx.by+ax.by //result u=(cy.bx-ay.bx-cx.by+ax.by)/(dx.by-dy.bx) u=(bx(cy-ay) +by(ax-cx))/(dx.by-dy.bx) //tidied up slightly Calculating u and putting it back into M=c+u.d will give you the point of intersection of the two lines. of course you will need to check if 0<=u<=1 to see if the line segment intersects Note that you cannot use this value of u in the equation L=a+t.b, you have to use a different but similar equation to calculate t and check if 0<=t<=1. //derived in a similar fashion to u t=(dx(ay-cy) +dy(cx-ax))/(bx.dy-by.dx) //tidied up slightly u & t will have different values for the same intersection point, but if both are between 0 & 1 the the two line segments intersect. Note that if (dx.by-dy.bx) is zero, then the lines are parallel and never intersect. By the way, if you only have the start and end points of the line segments, you can still use these equations, using the following conversions. s1x,s1y // start point of line 1 e1x,e1y // end point of line 1 s2x,s2y // start point of line 2 e2x,e2y // end point of line 2 ax=s1x ay=s1y bx=e1x-s1x by=e1y-s1y cx=s2x cy=s2y dx=e2x-s2x dy=e2y-s2y Share edited Nov 30, 2012 at 16:51 answered Nov 30, 2012 at 14:28 KenKen 6,13633 gold badges3636 silver badges5151 bronze badges \$\endgroup\$ 2 \$\begingroup\$ So basically we isolate one parameter first (t). Then isolate the other(u)? which eventually can be substituted for the equation of M. \$\endgroup\$ Sidar – Sidar 2012-11-30 16:21:46 +00:00 Commented Nov 30, 2012 at 16:21 1 \$\begingroup\$ @Sidar, Kinda, I'll edit the answer to make that step clearer. \$\endgroup\$ Ken – Ken 2012-11-30 16:36:37 +00:00 Commented Nov 30, 2012 at 16:36 Add a comment \| 4 \$\begingroup\$ Let's see if I can clarify the maths a bit. You have two lines: Line r1 such that some point p1 = a + Lb where a is a position vector and b is the direction and L is a parameter. Line r2 such that some point p2 = c + Md where c is a position vector and d is the direction and M is a parameter. Each line has its own parameter. You want to find where these intersect. At the intersection point, p1 = p2. Unless r1 and r2 are parallel, there must exist some values for L and M for which p1 = p2 is true. You only need one of these values to find the intersection point. This is easiest to solve by splitting r1 and r2 from cartesian form into parametric form, where the x and y coordinates of the vectors are described separately: r1 is such that p1x = ax + Lbx and p1y = ay + Lby. r2 is such that p2x = cx + Mdx and p2y = cy + Mdy. You can then rearrange the equations to isolate and eliminate either L or M and solve for the other. The resultant equation (derivation here) is, M = ((cy - ay) bx - (cx - ax) by) / (dx by - dy bx). You could write your code around that. This tutorial might also help to understand the concepts. Share edited May 23, 2017 at 12:37 CommunityBot 1 answered Nov 30, 2012 at 14:28 AnkoAnko 13.5k1010 gold badges5656 silver badges8282 bronze badges \$\endgroup\$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions 2d mathematics vector intersection See similar questions with these tags. 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16287	https://www-mdp.eng.cam.ac.uk/web/library/enginfo/aerothermal_dvd_only/aero/fprops/cvanalysis/node8.html	\| \| \| Next: Eularian and Lagrangian approaches Up: Integral equations for the Previous: One, Two and Three Flow Description, Streamline, Pathline, Streakline and Timeline Streamline, pathline, streakline and timeline form convenient tools to describe a flow and visualise it. They are defined below. Figure 3.5 : Streamlines Figure 3.6: Streamline definition A streamline is one that drawn is tangential to the velocity vector at every point in the flow at a given instant and forms a powerful tool in understanding flows. This definition leads to the equation for streamlines. (3.7) where u,v, and w are the velocity components in x, y and z directions respectively as sketched. Figure 3.7 : Streamtube Hidden in the definition of streamline is the fact that there cannot be a flow across it; i.e. there is no flow normal to it. Sometimes, as shown in Fig.3.7 we pull out a bundle of streamlines from inside of a general flow for analysis. Such a bundle is called stream tube and is very useful in analysing flows. If one aligns a coordinate along the stream tube then the flow through it is one-dimensional. Figure 3.8: Pathlines Figure 3.9: Streaklines Figure 3.10: Timeline Pathline is the line traced by a given particle. This is generated by injecting a dye into the fluid and following its path by photography or other means (Fig.3.8). Streakline concentrates on fluid particles that have gone through a fixed station or point. At some instant of time the position of all these particles are marked and a line is drawn through them. Such a line is called a streakline (Fig.3.9). Timeline is generated by drawing a line through adjacent particles in flow at any instant of time. Fig.3.10 shows a typical timeline. In a steady flow the streamline, pathline and streakline all coincide. In an unsteady flow they can be different. Streamlines are easily generated mathematically while pathline and streaklines are obtained through experiments. The following animation illustrates the differences between a streakline and a pathline. Figure 3.11: Animation to illustrate Streaklines and Pathlines. Subsections Eularian and Lagrangian approaches System and Control Volume Differential and Integral Approach Next: Eularian and Lagrangian approaches Up: Integral equations for the Previous: One, Two and Three (c) Aerospace, Mechanical & Mechatronic Engg. 2005 University of Sydney \|
16288	https://bcrc.bio.umass.edu/courses/spring2019/biol/biol312section2/content/sonic-hedgehog-gene-snakes	Sonic Hedgehog gene in snakes \| Writing in Biology Skip to main content Writing in Biology BIOL312Section2;Spring2019 Brewer User login Username Password Request new password Main menu Home Image Gallery Perfect Paragraphs You are here Home » Blogs » jhussaini's blog Sonic Hedgehog gene in snakes Submitted by jhussaini on Tue, 04/02/2019 - 17:34 The Hedgehog signaling pathway plays a role in the loss of limbs in snakes. According to the fossil record, ancestral snakes had limbs, which became reduced over the course of evolution. Some species of snakes such as boas and pythons still have a rudimentary femur and claw, while in other species such as cobras, limbs have disappeared completely. This phenotypic difference suggests that limbs were lost over time.The Sonic Hedgehog gene is regulated by a limb-specific enhancer called the zone of polarizing activity regulatory sequence (ZRS). This enhancer is conserved in all invertebrates and accounts for differences in limb development in vertebrates. A team of researchers led by a genomicist named Visel found that by substituting the ZRS of mice with the ZRS of snakes caused truncated limbs. In contrast, substituting mice with human and fish enhancers caused legs to grow normally. Sequence analysis of the ZRS in snakes showed deletions, which resulted in weak expression of Sonic Hedgehog and truncated limbs. The varying degrees of expression of Sonic Hedgehog is a possible cause of limb reduction in snakes. Post: Perfect Paragraph jhussaini's blog Comments cslavin Tue, 04/02/2019 - 18:37 Permalink Comments To strengthen this paragraph, you could define what the hedgehog signalling pathway is. aspark Wed, 04/03/2019 - 17:40 Permalink Comment 2 I would move the first sentence to after the sentences about limb loss. It should be joined with the rest of the information about that specific gene. nalexandroum Thu, 04/04/2019 - 01:01 Permalink I agree with both the above I agree with both the above comments, and would also add that you could add more information about the sonic hedgehog gene itself Perfect Paragraphs Each week, post your own Perfect Paragraph and comment on three Perfect Paragraphs. Suggest improvements. Don't just say "Looks good." Recent comments Comments6 years 5 months ago Feedback6 years 5 months ago Descriptive words6 years 5 months ago Sentences6 years 5 months ago Suggestion6 years 5 months ago Suggestion6 years 5 months ago Suggestion6 years 5 months ago Feedback6 years 5 months ago Feedback6 years 5 months ago Feedback6 years 5 months ago Blogs Updated This Week afeltrin alanhu angelinamart aprisby aspark cbbailey cnwokemodoih cslavin cynthiaguzma danielnguyen ddoyleperkin ewinter jhussaini klaflamme kwarny lgarneau lpotter mqpham mscheller nalexandroum ncarbone rdigregorio rharrison sbrewer scasimir sditelberg This page is maintained by The University of Massachusetts Biology Department.University of Massachusetts Amherst•Site Policies•
16289	https://artofproblemsolving.com/wiki/index.php/Change_of_base_formula?srsltid=AfmBOoo5ppINlsEPTdSht-Q91MqvuUjL_V0RrT5dCBN42YrQO3Tk1gpN	Art of Problem Solving Change of base formula - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Change of base formula Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Change of base formula The change of base formula is a formula for expressing a logarithm in one base in terms of logarithms in other bases. For any positivereal numbers such that neither nor are , we have This allows us to rewrite a logarithm in base in terms of logarithms in any base . This formula can also be written Proof Let . Then . And, taking the of both sides, we get By the properties of logarithms, Substituting for y, Use for computations The change of base formula is useful for simplifying certain computations involving logarithms. For example, we have by the change of base formula that The formula can also be useful when calculating logarithms on a calculator. Many calculators have only functions for calculating base-10 and base-e logarithms. But you can still calculate logs in other bases, you just need to use the change of base formula to put in in base 10. For example, if you wanted to calculate , you would first convert it to the form . Then you would evaluate it using the base-10 log function on the calculator. Special cases and consequences Many other logarithm rules can be written in terms of the change of base formula. For example, we have that . Using the second form of the change of base formula gives . One consequence of the change of base formula is that for positive constants , the functions and differ by a constant factor, for all . This article is a stub. Help us out by expanding it. Retrieved from " Category: Stubs Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
16290	https://pubmed.ncbi.nlm.nih.gov/12849365/	Subcortical ischaemic vascular dementia - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Subcortical ischaemic vascular dementia Gustavo C Román1,Timo Erkinjuntti,Anders Wallin,Leonardo Pantoni,Helena C Chui Affiliations Expand Affiliation 1 University of Texas at San Antonio and the Audie L Murphy Memorial Veterans Hospital, San Antonio, Texas 78284-7883, USA. romang@uthscsa.edu PMID: 12849365 DOI: 10.1016/s1474-4422(02)00190-4 Item in Clipboard Review Subcortical ischaemic vascular dementia Gustavo C Román et al. Lancet Neurol.2002 Nov. Show details Display options Display options Format Lancet Neurol Actions Search in PubMed Search in NLM Catalog Add to Search . 2002 Nov;1(7):426-36. doi: 10.1016/s1474-4422(02)00190-4. Authors Gustavo C Román1,Timo Erkinjuntti,Anders Wallin,Leonardo Pantoni,Helena C Chui Affiliation 1 University of Texas at San Antonio and the Audie L Murphy Memorial Veterans Hospital, San Antonio, Texas 78284-7883, USA. romang@uthscsa.edu PMID: 12849365 DOI: 10.1016/s1474-4422(02)00190-4 Item in Clipboard Cite Display options Display options Format Abstract Vascular dementia is the second most common type of dementia. The subcortical ischaemic form (SIVD) frequently causes cognitive impairment and dementia in elderly people. SIVD results from small-vessel disease, which produces either arteriolar occlusion and lacunes or widespread incomplete infarction of white matter due to critical stenosis of medullary arterioles and hypoperfusion (Binswanger's disease). Symptoms include motor and cognitive dysexecutive slowing, forgetfulness, dysarthria, mood changes, urinary symptoms, and short-stepped gait. These manifestations probably result from ischaemic interruption of parallel circuits from the prefrontal cortex to the basal ganglia and corresponding thalamocortical connections. Brain imaging (computed tomography and magnetic resonance imaging) is essential for correct diagnosis. The main risk factors are advanced age, hypertension, diabetes, smoking, hyperhomocysteinaemia, hyperfibrinogenaemia, and other conditions that can cause brain hypoperfusion such as obstructive sleep apnoea, congestive heart failure, cardiac arrhythmias, and orthostatic hypotension. Cerebral autosomal dominant arteriopathy with subcortical infarcts and leucoencephalopathy (CADASIL)and some forms of cerebral amyloid angiopathy have a genetic basis. Treatment is symptomatic and prevention requires control of treatable risk factors. PubMed Disclaimer Similar articles Brain hypoperfusion: a critical factor in vascular dementia.Román GC.Román GC.Neurol Res. 2004 Jul;26(5):454-8. doi: 10.1179/016164104225017686.Neurol Res. 2004.PMID: 15265263 Review. Cerebral blood flow in ischemic vascular dementia and Alzheimer's disease, measured by arterial spin-labeling magnetic resonance imaging.Schuff N, Matsumoto S, Kmiecik J, Studholme C, Du A, Ezekiel F, Miller BL, Kramer JH, Jagust WJ, Chui HC, Weiner MW.Schuff N, et al.Alzheimers Dement. 2009 Nov;5(6):454-62. doi: 10.1016/j.jalz.2009.04.1233.Alzheimers Dement. 2009.PMID: 19896584 Free PMC article. Heterogeneity of vascular dementia: mechanisms and subgroups.Wallin A, Blennow K.Wallin A, et al.J Geriatr Psychiatry Neurol. 1993 Jul-Sep;6(3):177-88. doi: 10.1177/089198879300600307.J Geriatr Psychiatry Neurol. 1993.PMID: 8397762 Review. Sorting out the clinical consequences of ischemic lesions in brain aging: a clinicopathological approach.Gold G, Kovari E, Hof PR, Bouras C, Giannakopoulos P.Gold G, et al.J Neurol Sci. 2007 Jun 15;257(1-2):17-22. doi: 10.1016/j.jns.2007.01.020. Epub 2007 Feb 23.J Neurol Sci. 2007.PMID: 17321551 Review. Comparison of regional cerebral blood flow in two subsets of subcortical ischemic vascular dementia: statistical parametric mapping analysis of SPECT.Shim YS, Yang DW, Kim BS, Shon YM, Chung YA.Shim YS, et al.J Neurol Sci. 2006 Dec 1;250(1-2):85-91. doi: 10.1016/j.jns.2006.07.008. Epub 2006 Sep 22.J Neurol Sci. 2006.PMID: 16996088 See all similar articles Cited by Consensus statement for diagnosis of subcortical small vessel disease.Rosenberg GA, Wallin A, Wardlaw JM, Markus HS, Montaner J, Wolfson L, Iadecola C, Zlokovic BV, Joutel A, Dichgans M, Duering M, Schmidt R, Korczyn AD, Grinberg LT, Chui HC, Hachinski V.Rosenberg GA, et al.J Cereb Blood Flow Metab. 2016 Jan;36(1):6-25. doi: 10.1038/jcbfm.2015.172.J Cereb Blood Flow Metab. 2016.PMID: 26198175 Free PMC article.Review. Association of chronic obstructive pulmonary disease with cognitive decline in very elderly men.Zhou G, Liu J, Sun F, Xin X, Duan L, Zhu X, Shi Z.Zhou G, et al.Dement Geriatr Cogn Dis Extra. 2012 Jan;2(1):219-28. doi: 10.1159/000338378. Epub 2012 May 22.Dement Geriatr Cogn Dis Extra. 2012.PMID: 22719748 Free PMC article. Cerebrovascular disease, multiple sclerosis, or both? Case report and review of the challenging distinction between two potentially synergistic syndromes.Suarez P, Restrepo L.Suarez P, et al.Cereb Circ Cogn Behav. 2021 Feb 16;2:100006. doi: 10.1016/j.cccb.2021.100006. eCollection 2021.Cereb Circ Cogn Behav. 2021.PMID: 36324716 Free PMC article. Characteristics of MR infarcts associated with dementia and cognitive function in the elderly.Aggarwal NT, Schneider JA, Wilson RS, Beck TL, Evans DA, Carli CD.Aggarwal NT, et al.Neuroepidemiology. 2012;38(1):41-7. doi: 10.1159/000334438. Epub 2011 Dec 17.Neuroepidemiology. 2012.PMID: 22179433 Free PMC article. Pathology and pathogenesis of vascular cognitive impairment-a critical update.Jellinger KA.Jellinger KA.Front Aging Neurosci. 2013 Apr 10;5:17. doi: 10.3389/fnagi.2013.00017. eCollection 2013.Front Aging Neurosci. 2013.PMID: 23596414 Free PMC article. 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16291	https://www.youtube.com/watch?v=xqE8bu_dcDA	Stratified Random Sampling - Proportional Allocation Project BOLEN 792 subscribers 352 likes Description 44193 views Posted: 27 Aug 2020 This video shows how to allocate proportionally for stratified random sampling. Refer to the example we have presented in class. Note: This video is intended for my Statistics class as an added reference. You are also welcome to use this as your reference. I hope you find this useful. 23 comments Transcript: okay so to do the stratified random sampling uh what we have to do first is to determine the proportion of each stratum relative to the population in order to do that we have to allocate proportionally okay so the proportional allocation formula is given by the following as what you will see on screen and here based on the example that we have presented what we want is to determine uh the number of employees that we are going to take from each race so uh we take note that um there are 182 filipinos 51 chinese and 17 american and we're going to come up with a committee which is composed of 15 uh individuals coming from all three races indicated or included in the insurance company so how do we allocate proportionally all right so using the formula we are going to have let me make use of small n sub f to indicate that we are computing for the um proportional allocation uh coming from the philippine race so from our example the population or the total number of filipinos in the company is 182. so we will have 182 we divided by our population size or the total number of employees in the company and that is equal to 250. and we multiply it by the sample size that we want to uh have in this case we want to have a sample or a committee which is composed of 15 people okay so in that case we will have 100 divided by 250 and we multiply it by 15. okay so here what we will have is equal to a 10.92 okay or uh we are going to uh round this up to uh 11. or actually we are going to round it off okay so in that case of the 15 uh employees to be included in the committee there will be a total of um 11 filipinos okay so let's continue uh we will compute for the rest of um the number of individuals coming from the different races so um or for those who will be belonging in the chinese race so what we will have there's a total of 51 of them divided by 250 okay times uh 15 and in this case we will have that will be equal to so 51 divided by 250 or times a 15 that will give us 3.06 and we're going to round it off to three okay so that means to say three uh people uh will be part of the committee of 15 who are actually chinese all right let's do the same for the number of americans to be included in the committee but if you're going to look at this since our uh n uh the the sample size of people that we want to have is 15 if you're going to inspect this we already have 11 we already have 11 and 3 okay and that will give us a total of 14. so that means to say we only need one more coming from the american race to complete our sample or to complete the committee that we want to form which is composed of 15 individuals but to verify that indeed we will need one more or that what is lacking is only one we are going to compute using our formula so here we will have n sub a to indicate that we are computing for the number of employees who are americans to be included in the committee that we are forming so in this case the total number of americans is 17. we divide it by 250 and then we multiply it by a 15. and um computing how much will we have that is 17 divided by 250 times 15 indeed that will be equal to 1.02 and that is uh rounded off to one okay so therefore uh here we have verified that we only need one more individual coming from the american race to complete the committee which is uh 15. all right so um in this case the next step is once we have allocated proportionally okay how do we determine 11 coming from the 182 filipinos how do we choose the three chinese coming from the chinese race how do we choose only one american to be part of the committee coming from the american race okay so um you have the option either to make use of the simple random sampling or you can make use of the stratified random sampling so basically it's up to you uh which um which technique you are going to use right after you have allocated proportionally okay now if we're going to analyze this further if you notice since majority of the 250 individuals are filipinos we take note that there are 182 filipinos out of the 250 employees in the company it is only logical that of the 15 um employees to be chosen to be part of the committee majority of them okay because look at this one we have 11 filipinos to be part of the committee of 15. so majority will be coming from the filipino race it's because most of the employees are actually filipinos and that is followed by chinese and americans respectively okay so this is how we have allocated proportionally um in terms of the use of the stratified random sampling so i hope this is clear
16292	https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/97124709de1bc662ca39fd2acfee25c0_unit2_sept08.pdf	MIT OpenCourseWare 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: Lecture 9: Linear and Quadratic Approximations Unit 2: Applications of Diﬀerentiation Today, we’ll be using diﬀerentiation to make approximations. Linear Approximation y=f(x) y = b+a(x-x0) y x b = f(x0) ; x0 ,f(x0) ( ) a = f’(x0) Figure 1: Tangent as a linear approximation to a curve The tangent line approximates f(x). It gives a good approximation near the tangent point x0. As you move away from x0, however, the approximation grows less accurate. f(x) ≈ f(x0) + f 0(x0)(x − x0) Example 1. f(x) = ln x, x0 = 1 (basepoint) 1 f(1) = ln 1 = 0; f 0(1) = = 1 x x=1 ln x Change the basepoint: Basepoint u0 = x0 − 1 = 0. ≈ f(1) + f 0(1)(x − 1) = 0 + 1 · (x − 1) = x − 1 x = 1 + u = ⇒ u = x − 1 ln(1 + u) ≈ u 1 Lecture 9 18.01 Fall 2006 Basic list of linear approximations In this list, we always use base point x0 = 0 and assume that \|x\| << 1. 1. sin x ≈ x (if x ≈ 0) (see part a of Fig. 2) 2. cos x ≈ 1 (if x ≈ 0) (see part b of Fig. 2) x 3. e ≈ 1 + x (if x ≈ 0) 4. ln(1 + x) ≈ x (if x ≈ 0) 5. (1 + x)r ≈ 1 + rx (if x ≈ 0) Proofs Proof of 1: Take f(x) = sin x, then f 0(x) = cos x and f(0) = 0 f 0(0) = 1, f(x) ≈ f(0) + f 0(0)(x − 0) = 0 + 1.x So using basepoint x0 = 0, f(x) = x. (The proofs of 2, 3 are similar. We already proved 4 above.) Proof of 5: f(x) = (1 + x)r; f(0) = 1 f 0(0) = d (1 + x)r x=0 = r(1 + x)r−1 x=0 = r dx \| \| f(x) = f(0) + f 0(0)x = 1 + rx y = x sin(x) y=1 cos(x) (a) (b) Figure 2: Linear approximation to (a) sin x (on left) and (b) cos x (on right). To ﬁnd them, apply f (x) ≈ f (x0) + f0(x0)(x − x0) (x0 = 0) e−2x Example 2. Find the linear approximation of f(x) = near x = 0. √1 + x We could calculate f 0(x) and ﬁnd f 0(0). But instead, we will do this by combining basic approxi mations algebraically. u e−2x ≈ 1 + (−2x) (e ≈ 1 + u, where u = −2x) 2 Lecture 9 18.01 Fall 2006 √ 1 + x = (1 + x)1/2 ≈ 1 + 1 x 2 Put these two approximations together to get e−2x 1 − 2x x ≈ (1 − 2x)(1 + 1 x)−1 √1 + x ≈ 1 + 1 2 2 Moreover (1 + 1 x)−1 ≈ 1 − 1 x (using (1 + u)−1 ≈ 1 − u with u = x/2). Thus 1 2 2 e−2x 1 1 1 2 √1 + x ≈ (1 − 2x)(1 − 2 x) = 1 − 2x − 2 x + 2(2)x Now, we discard that last x2 term, because we’ve already thrown out a number of other x2 (and higher order) terms in making these approximations. Remember, we’re assuming that x << 1. 2 3 \| \| This means that x is very small, x is even smaller, etc. We can ignore these higher-order terms, because they are very, very small. This yields e−2x 1 5 √1 + x ≈ 1 − 2x − 2 x = 1 − 2 x Because f(x) ≈ 1 − 5 x, we can deduce f(0) = 1 and f 0(0) = −5 directly from our linear approxi 2 2 mation, which is quicker in this case than calculating f 0(x). Example 3. f(x) = (1 + 2x)10 . On the ﬁrst exam, you were asked to calculate lim (1 + 2x)10 − 1. The quickest way to do this with x→0 x the tools of Unit 1 is as follows. lim (1 + 2x)10 − 1 = lim f(x) − f(0) = f 0(0) = 20 x 0 x x 0 x → → (since f 0(x) = 10(1 + 2x)9 2 = 20 at x = 0) · Now we can do the same problem a diﬀerent way, namely, using linear approximation. (1 + 2x)10 ≈ 1 + 10(2x) (Use (1 + u)r ≈ 1 + ru where u = 2x and r = 10.) Hence, (1 + 2x)10 − 1 1 + 20x − 1 = 20 x ≈ x Example 4: Planet Quirk Let’s say I am on Planet Quirk, and that a satellite is whizzing overhead with a velocity v. We want to ﬁnd the time dilation (a concept from special relativity) that the clock onboard the satellite experiences relative to my wristwatch. We borrow the following equation from special relativity: T T 0 = q 1 − v c 2 2 1 1 1 1A shortcut to the two-step process √1 + x ≈ 1 + x ≈ 1 − 2 x is to write 2 1 1 √1 + x = (1 + x)−1/2 ≈ 1 − 2 x 3 Lecture 9 18.01 Fall 2006 me satellite (with velocity v) Figure 3: Illustration of Example 4: a satellite with velocity v speeding past “me” on planet Quirk. Here, T 0 is the time I measure on my wristwatch, and T is the time measured onboard the satellite. 2 −1/2 2 2 v 1 v v 1 T 0 = T 1 − c2 ≈ 1 + 2 c2 (1 + u)4 ≈ 1 + ru, where u = − c2 , r = − 2 2 If v = 4 km/s, and the speed of light (c) is 3 × 105 km/s, v ≈ 10−10 . There’s hardly any diﬀerence c2 between the times measured on the ground and in the satellite. Nevertheless, engineers used this very approximation (along with several other such approximations) to calibrate the radio transmitters on GPS satellites. (The satellites transmit at a slightly oﬀset frequency.) Quadratic Approximations These are more complicated. They are only used when higher accuracy is needed. f(x) ≈ f(x0) + f 0(x0)(x − x0) + f 00(x0)(x − x0)2 (x ≈ x0) 2 Geometric picture: A quadratic approximation gives a best-ﬁt parabola to a function. For example, let’s consider f(x) = cos(x) (see Figure 4). If x0 = 0, then f(0) = cos(0) = 1, and f 0(x) = − sin(x) = ⇒ f 0(0) = − sin(0) = 0 f 00(x) = − cos(x) = ⇒ f 00(0) = − cos(0) = −1 1 1 cos(x) ≈ 1 + 0 · x − 2 x 2 = 1 − 2 x 2 1 You are probably wondering where that in front of the x2 term comes from. The reason it’s 2 there is so that this approximation is exact for quadratic functions. For instance, consider f(x) = a + bx + cx 2; f 0(x) = b + 2cx; f 00(x) = 2c. Set the base point x0 = 0. Then, f(0) = a + b 0 + c 02 = a = f(0) · · ⇒ f 0(0) = b + 2c 0 = b = b = f 0(0) · ⇒ f 00(0) f 00(0) = 2c = c = ⇒ 2 4 Lecture 9 18.01 Fall 2006 cos(x) y x 1- x2/2 Figure 4: Quadratic approximation to cos(x). 0.0.1 Basic Quadratic Approximations : f(x) ≈ f(0) + f 0(0)x + f 00 2 (0) x 2 (x ≈ 0) 1. sin x ≈ x (if x ≈ 0) 2 x 2. cos x ≈ 1 − 2 (if x ≈ 0) 3. e x ≈ 1 1 + x + x 2 (if x ≈ 0) 2 4. ln(1 + x) ≈ x − 1 x 2 (if x ≈ 0) 2 5. (1 + x)r ≈ 1 + rx + r(r − 1) x 2 (if x ≈ 0) 2 Proofs: The proof of these is to evaluate f(0), f 0(0), f 00(0) in each case. We carry out Case 4 ⇒ f(x) = ln(1 + x) = f(0) = ln 1 = 0 1 f 0(x) = [ln(1 + x)]0 = = f 0(0) = 1 1 + x ⇒ 1 f 00(x) = 1 + 0 −1 x = (1 + x)2 = ⇒ f 00(0) = −1 Let us apply a quadratic approximation to our Planet Quirk example and see where it gives. 1 − v c2 2 −1/2 ≈ 1 + 2 1 v c2 2 + " ( − 2 1 )( − 2 2 1 − 1) − v c2 2 2 # Case 5 with x = − c v 2 2 , r = − 2 1 5 Lecture 9 18.01 Fall 2006 2 2 2 Since v ≈ 10−10, that last term will be of the order v ≈ 10−20 . Not even the best atomic c2 c2 clocks can measure time with this level of precision. Since the quadratic term is so small, we might as well ignore it and stick to the linear approximation in this case. e−2x Example 5. f(x) = √1 + x Let us ﬁnd the quadratic approximation of this expression. We can rewrite it as f(x) = e−2x(1 + x)−1/2 . Using the approximation of each factor gives 1 1 (− 1 2 )(− 1 2 − 1) 2 f(x) ≈ 1 − 2x + 2(−2x)2 1 − 2 x + 2 x 1 1 2 + 3 5 27 2 f(x) ≈ 1 − 2x − 2x + (−2)(− 2)x 2 + 2x 8 x 2 = 1 − 2 x + 8 x (Note: we drop the x3 and higher order terms. This is a quadratic approximation, so we don’t care about anything higher than x2.) 6 Lecture 9 18.01 Fall 2006 Lecture 10: Curve Sketching Goal: To draw the graph of f using the behavior of f 0 and f 00. We want the graph to be qualitatively correct, but not necessarily to scale. Typical Picture: Here, y0 is the minimum value, and x0 is the point where that minimum occurs. x0= critical point y0 Figure 1: The critical point of a function Notice that for x < x0, f 0(x) < 0. In other words, f is decreasing to the left of the critical point. For x > x0, f 0(x) > 0: f is increasing to the right of the critical point. Another typical picture: Here, y0 is the critical (maximum) value, and x0 is the critical point. f is decreasing on the right side of the critical point, and increasing to the left of x0. x0= critical point y0 f’(x) < 0 x > x0 Figure 2: A concave-down graph 1 Lecture 9 18.01 Fall 2006 Rubric for curve-sketching 1. (Precalc skill) Plot the discontinuities of f — especially the inﬁnite ones! 2. Find the critical points. These are the points at which f 0(x) = 0 (usually where the slope changes from positive to negative, or vice versa.) 3. (a) Plot the critical points (and critical values), but only if it’s relatively easy to do so. (b) Decide the sign of f 0(x) in between the critical points (if it’s not already obvious). 4. (Precalc skill) Find and plot the zeros of f. These are the values of x for which f(x) = 0. Only do this if it’s relatively easy. 5. (Precalc skill) Determine the behavior at the endpoints (or at ±∞). Example 1. y = 3x − x3 1. No discontinuities. 2. y0 = 3 − 3x2 = 3(1 − x2) so, y0 = 0 at x = ±1. 3. (a) At x = 1, y = 3 − 1 = 2. (b) At x = −1, y = −3 + 1 = −2. Mark these two points on the graph. 3 4. Find the zeros: y = 3x − x = x(3 − x2) = 0 so the zeros lie at x = 0, ± √ 3. 5. Behavior of the function as x → ±∞. As x →∞, the x3 term of y dominates, so y → −∞. Likewise, as x → −∞, y →∞. Putting all of this information together gives us the graph as illustrated in Fig. 3) (-√3,0) (√3,0) (-1,-2) (1,2) 2 1 -2 -1 3 Figure 3: Sketch of the function y = 3x − x . Note the labeled zeros and critical points Let us do step 3b (the sign of f 0) to double-check for consistency. y0 = 3 − 3x 2 = 3(1 − x 2) y0 > 0 when \|x\| < 1; y0 < 0 when \|x\| > 1. Sure enough, y is increasing between x = −1 and x = 1, and is decreasing everywhere else. 2 Lecture 9 18.01 Fall 2006 1 Example 2. y = . x This example illustrates why it’s important to ﬁnd a function’s discontinuities before looking at the properties of its derivative. We calculate y0 = − x2 1 < 0 Warning: The derivative is never positive, so you might think that y is always decreasing, and its graph looks something like that in Fig. 4. Figure 4: A monotonically decreasing function 1 But as you probably know, the graph of looks nothing like this! It actually looks like Fig. 5. In x 1 fact, y = is decreasing except at x = 0, where it jumps from −∞ to +∞. This is why we must x watch out for discontinuities. Figure 5: Graph of y = 1 . x 3 Lecture 10 18.01 Fall 2006 Example 3. y = x3 − 3x2 + 3x. y0 = 3x 2 − 6x + 3 = 3(x 2 − 2x + 1) = 3(x − 1)2 There is a critical point at x = 1. y0 > 0 on both sides of x = 1, so y is increasing everywhere. In this case, the sign of y0 doesn’t change at the critical point, but the graph does level out (see Fig. 6. 1 1 horizontal slope (1,1) 3 Figure 6: Graph of y = y = x − 3x2 + 3x ln x Example 4. y = (Note: this function is only deﬁned for x > 0) x What happens as x decreases towards zero? Let x = 2−n . Then, ln 2−n y = 2−n = (−n ln 2)2n → −∞ as n →∞ In other words, y decreases to −∞ as x approaches zero. Next, we want to ﬁnd the critical points. y0 = ln x 0 = x( x 1 ) − 1(ln x) = 1 − ln x x x2 x2 y0 = 0 = ⇒ 1 − ln x = 0 = ⇒ ln x = 1 = ⇒ x = e In other words, the critical point is x = e (from previous page). The critical value is ln e 1 y(x) \|x=e = e = e 4 Lecture 10 18.01 Fall 2006 Next, ﬁnd the zeros of this function: y = 0 ln x = 0 ⇔ So y = 0 when x = 1. What happens as x →∞? This time, consider x = 2+n . ln 2n n ln 2 n(0.7) y = = 2n 2n ≈ 2n So, y → 0 as n →∞. Putting all of this together gets us the graph in Fig. 7. e 1 1/e (e,1/e) Figure 7: Graph of y = ln x x Finally, let’s double-check this picture against the information we get from step 3b: y0 = 1 − ln x > 0 for 0 < x < e x2 Sure enough, the function is increasing between 0 and the critical point. 5 Lecture 10 18.01 Fall 2006 2nd Derivative Information When f 00 > 0, f 0 is increasing. When f 00 < 0, f 0 is decreasing. (See Fig. 8 and Fig. 9) slope < 0 slope = 0 slope > 0 Figure 8: f is convex (concave-up). The slope increases from negative to positive as x increases. Figure 9: f is concave-down. The slope decreases from positive to negative as x increases. Therefore, the sign of the second derivative tells us about concavity/convexity of the graph. Thus the second derivative is good for two purposes. 1. Deciding whether a critical point is a maximum or a minimum. This is known as the second derivative test. f 0(x0) f 00(x0) Critical point is a: 0 negative maximum 0 positive minimum 2. Concave/convex “decoration.” 6 Lecture 10 18.01 Fall 2006 The points where f 00 = 0 are called inﬂection points. Usually, at these points the graph changes from concave up to down, or vice versa. Refer to Fig. 10 to see how this looks on Example 1. Inflection point (where f” = 0) 3 Figure 10: Inﬂection point: y = 3x − x , y00 = −6x = 0, at x = 0. 7 Lecture 10 18.01 Fall 2006 Lecture 11: Max/Min Problems Example 1. y = ln x (same function as in last lecture) x x0=e 1/e Figure 1: Graph of y = ln x . x 1 What is the maximum value? Answer: y = . • e • Where (or at what point) is the maximum achieved? Answer: x = e. (See Fig. 1).) Beware: Some people will ask “What is the maximum?”. The answer is not e. You will get so used to ﬁnding the critical point x = e, the main calculus step, that you will forget to ﬁnd the maximum 1 1 value y = . Both the critical point x = e and critical value y = are important. Together, they e e 1 form the point of the graph (e, ) where it turns around. e Example 2. Find the max and the min of the function in Fig. 2 Answer: If you’ve already graphed the function, it’s obvious where the maximum and minimum values are. The point is to ﬁnd the maximum and minimum without sketching the whole graph. Idea: Look for the max and min among the critical points and endpoints.You can see from Fig. 2 that we only need to compare the heights or y-values corresponding to endpoints and critical points. (Watch out for discontinuities!) 1 Lecture 11 18.01 Fall 2006 max min Figure 2: Search for max and min among critical points and endpoints Example 3. Find the open-topped can with the least surface area enclosing a ﬁxed volume, V. r h Figure 3: Open-topped can. 1. Draw the picture. 2. Figure out what variables to use. (In this case, r, h, V and surface area, S.) 3. Figure out what the constraints are in the problem, and express them using a formula. In this example, the constraint is V = πr2h = constant We’re also looking for the surface area. So we need the formula for that, too: S = πr2 + (2πr)h Now, in symbols, the problem is to minimize S with V constant. 2 Lecture 11 18.01 Fall 2006 4. Use the constraint equation to express everything in terms of r (and the constant V ). h = V ; S = πr2 + (2πr) V 2πr πr2 5. Find the critical points (solve dS/dr = 0), as well as the endpoints. S will achieve its max and min at one of these places. dS 2V 3 V V 1/3 dr = 2πr − r2 = 0 = ⇒ πr3 − V = 0 = ⇒ r = π = ⇒ r = π We’re not done yet. We’ve still got to evaluate S at the endpoints: r = 0 and “r = ∞”. 2V S = πr2 + , 0 ≤ r < ∞ r 2 As r → 0, the second term, r , goes to inﬁnity, so S →∞. As r →∞, the ﬁrst term πr2 goes to inﬁnity, so S →∞. Since S = +∞ at each end, the minimum is achieved at the critical point r = (V/π)1/3, not at either endpoint. s r to ∞ to ∞ Figure 4: Graph of S We’re still not done. We want to ﬁnd the minimum value of the surface area, S, and the values of h. 1/3 −2/3 1/3 V V V V V V r = π ; h = πr2 = π V 2/3 = π π = π π 2/3 1/3 S = πr2 + 2 V = π V + 2V V = 3π−1/3V 2/3 r π π Finally, another, often better, way of answering that question is to ﬁnd the proportions of the can. In other words, what is h r ? Answer: h r = (V/π)1/3 (V/π)1/3 = 1. 3 Lecture 11 18.01 Fall 2006 Example 4. Consider a wire of length 1, cut into two pieces. Bend each piece into a square. We want to ﬁgure out where to cut the wire in order to enclose as much area in the two squares as possible. (1/4)x 0 x 1 (1/4)(1-x) Figure 5: Illustration for Example 5. 2 x x The ﬁrst square will have sides of length . Its area will be . The second square will have 2 4 16 sides of length 1− 4 x . Its area will be 1− 4 x . The total area is then x 2 1 − x 2 A = + 4 4 A0 = 2 16 x + 2(1 16 − x)(−1) = x 8 − 1 8 + x 8 = 0 = ⇒ 2x − 1 = 0 = ⇒ x = 1 2 So, one extreme value of the area is 1 2 1 2 1 A = 2 + 2 = 4 4 32 We’re not done yet, though. We still need to check the endpoints! At x = 0, A = 02 + 1 − 0 2 = 1 4 16 At x = 1, 2 1 1 A = + 02 = 4 16 4 Lecture 11 18.01 Fall 2006 By checking the endpoints in Fig. 6, we see that the minimum area was achieved at x = 1 2 . The maximum area is not achieved in 0 < x < 1, but it is achieved at x = 0 or 1. The maximum corresponds to using the whole length of wire for one square. 1/2 1 1/16 1/32 x Area Figure 6: Graph of the area function. Moral: Don’t forget endpoints. If you only look at critical points you may ﬁnd the worst answer, rather than the best one. 5 Lecture 11 18.01 Fall 2006 Lecture 12: Related Rates Example 1. Police are 30 feet from the side of the road. Their radar sees your car approaching at 80 feet per second when your car is 50 feet away from the radar gun. The speed limit is 65 miles per hour (which translates to 95 feet per second). Are you speeding? First, draw a diagram of the setup (as in Fig. 1): Road Car Police 30 D=50 x Figure 1: Illustration of example 1: triangle with the police, the car, the road, D and x labelled. Next, give the variables names. The important thing to ﬁgure out is which variables are changing. dD At D = 50, x = 40. (We know this because it’s a 3-4-5 right triangle.) In addition, = D0 = dt −80. D0 is negative because the car is moving in the −x direction. Don’t plug in the value for D yet! D is changing, and it depends on x. The Pythagorean theorem says 302 + x 2 = D2 Diﬀerentiate this equation with respect to time (implicit diﬀerentiation: d 2 2DD0 302 + x = D2 = 2xx0 = 2DD0 = x0 = dt ⇒ ⇒ 2x Now, plug in the instantaneous numerical values: 50 feet x0 = 40(−80) = −100 s This exceeds the speed limit of 95 feet per second; you are, in fact, speeding. 1 Lecture 12 18.01 Fall 2006 p There is another, longer, way of solving this problem. Start with D = 302 + x2 = (302 + x 2)1/2 d 1 dx D = (302 + x 2)−1/2(2x ) dt 2 dt Plug in the values: 1 dx −80 = (302 + 402)−1/2(2)(40) 2 dt and solve to ﬁnd dx feet = −100 dt s (A third strategy is to diﬀerentiate x = √ D2 − 302). It is easiest to diﬀerentiate the equation in its simplest algebraic form 302 + x2 = D2, our ﬁrst approach. The general strategy for these types of problems is: 1. Draw a picture. Set up variables and equations. 2. Take derivatives. 3. Plug in the given values. Don’t plug the values in until after taking the derivatives. Example 2. Consider a conical tank. Its radius at the top is 4 feet, and it’s 10 feet high. It’s being ﬁlled with water at the rate of 2 cubic feet per minute. How fast is the water level rising when it is 5 feet high? h r Figure 2: Illustration of example 2: inverted cone water tank. From Fig. 2), the volume of the tank is given by 1 V = πr2h 3 2 Lecture 12 18.01 Fall 2006 The key here is to draw the two-dimensional cross-section. We use the letters r and h to represent the variable radius and height of the water at any level. We can ﬁnd the relationship between r and h from Fig. 3) using similar triangles. 10 4 r h Figure 3: Relating r and h. From Fig. 3), we see that r 4 = h 10 or, in other words, 2 r = 5 h Plug this expression for r back into V to get 2 1 2 4 V = π h h = πh3 3 5 3(25) dV 4 = V 0 = πh2h0 dt 25 dV Now, plug in the numbers ( = 2, h = 5): dt 4 2 = π(5)2h0 25 1 h0 = 2π Related rates also arise on Problem Set 3 (Fig. 4). There’s a part II margin of error problem ΔL involving a satellite, where you’re asked to ﬁnd . Δh 3 Lecture 12 18.01 Fall 2006 h L satellite c Figure 4: Illustration of the satellite problem. L2 + c 2 = h2 2LL0 = 2hh0 ΔL L0 h Hence, Δh ≈ h0 = L There is also a parabolic mirror problem based on similar ideas (Fig. 5). Δa Δθ Figure 5: Illustration of the parabolic mirror problem. Δa Δθ Here, you want to ﬁnd either or . This type of sensitivity of measurement problem Δθ Δa matters in every measurement problem, for instance predicting whether asteroids will hit Earth. 4 Lecture 12 18.01 Fall 2006 Lecture 13: Newton’s Method and Other Applications Newton’s Method Newton’s method is a powerful tool for solving equations of the form f(x) = 0. Example 1. f(x) = x2 − 3. In other words, solve x2 − 3 = 0. We already know that the solution to this is x = √ 3. Newton’s method, gives a good numerical approximation to the answer. The method uses tangent lines (see Fig. 1). x0=1 x1 (1,-2) y = x2 -3 Figure 1: Illustration of Newton’s Method, Example 1. The goal is to ﬁnd where the graph crosses the x-axis. We start with a guess of x0 = 1. Plugging that back into the equation for y, we get y0 = 12 − 3 = −2, which isn’t very close to 0. Our next guess is x1, where the tangent line to the function at x0 crosses the x-axis. The equation for the tangent line is: y − y0 = m(x − x0) When the tangent line intercepts the x-axis, y = 0, so −y0 = m(x1 − x0) y0 − m = x1 − x0 y0 x1 = x0 − m Remember: m is the slope of the tangent line to y = f(x) at the point (x0, y0). 1 Lecture 13 18.01 Fall 2006 In terms of f: y0 = f(x0) m = f 0(x0) Therefore, f(x0) x1 = x0 − f 0(x0) x1 x0 x2 Figure 2: Illustration of Newton’s Method, Example 1. In our example, f(x) = x2 − 3, f 0(x) = 2x. Thus, (x0 2 − 3) 1 3 x1 = x0 − 2x = x0 − 2 x0 + 2x0 1 3 x1 = x0 + 2 2x0 The main idea is to repeat (iterate) this process: 1 3 x2 = x1 + 2 2x1 1 3 x3 = x2 + 2 2x2 and so on. The procedure approximates √ 3 extremely well. 2 Lecture 13 18.01 Fall 2006 Lecture 13, Version 3.0 18.01 Fall 2006 x y accuracy: \|y − √3\| x0 1 x1 2 3 × 10−1 x2 7 4 2 × 10−2 x3 7 8 + 6 7 10−4 x4 18,817 10,864 3 × 10−9 Notice that the number of digits of accuracy doubles with each iteration. Summary Newton’s Method is illustrated in Fig. 3 and can be summarized as follows: f(xk) xk+1 = xk − f 0(xk) xk = kth iterate (xk, yk) xk+1 y=f(x) Figure 3: Illustration of Newton’s Method. Example 1 considered the particular case of f(x) = x 2 − 3 f(xk) 1 3 xk+1 = xk − f 0(xk) = ... = 2xk + 2xk Now, we deﬁne x = lim xk (xk → x as k →∞) k→∞ To evaluate x in Example 1, take the limit as k →∞ in the equation 1 3 xk+1 = xk + 2 2xk 3 Lecture 13, Version 3.0 18.01 Fall 2006 This yields 1 3 1 3 1 3 x ¯ = 2 x ¯ + 2¯ x = ⇒ x − 2 x = 2x = ⇒ 2 x = 2x = ⇒ x 2 = 3 which is just what we hoped: x = √ 3. Warning 1. Newton’s Method can ﬁnd an unexpected root. Example: if you take x0 = −1, then xk → − √ 3 instead of + √ 3. This convergence to an unexpected root is illustrated in Fig. 4 y = x2-3 x0 x1 tangent to curve at x = x0 Figure 4: Newton’s method converging to an unexpected root. Warning 2. Newton’s Method can fail completely. This failure is illustrated in Fig. 5. In this case, x2 = x0, x3 = x1, and so forth. It repeats in a cycle, and never converges to a single value. x0 x1 (x1, y1) (x0, y0) Figure 5: Newton’s method converging to an unexpected root. 4 p p Ring on a String Consider a ring on a string 1 held ﬁxed at two ends at (0, 0) and (a, b) (see Fig. 6). The ring is free to slide to any point. Find the position (x, y) of the string. (a, b) (0, 0) (x, y) a-x x α β √(x2 +y2) √[(a-x)2 +(b-y2)] α = β Figure 6: Illustration of the Ring on a String problem. Physical Principle The ring settles at the lowest height (lowest potential energy), so the prob lem is to minimize y subject to the constraint that (x, y) is on the string. Constraint The length L of the string is ﬁxed: x2 + y2 + (x − a)2 + (y − b)2 = L The function y = y(x) is determined implicitly by the constraint equation above. We traced the constraint curve (possible positions of the ring) on the blackboard. This curve is an ellipse with foci at (0, 0) and (a, b), but knowing that the curve is an ellipse does not help us ﬁnd the lowest point. Experiments with the hanging ring show that the lowest point is somewhere in the middle. Since the ends of the constraint curve are higher than the middle, the lowest point is a critical point (a point where y0(x) = 0). In class we also gave a physical demonstration of this by drawing the horizontal tangent at the lowest point. To ﬁnd the critical point, diﬀerentiate the constraint equation implicitly with respect to x, p x + yy0 + p x − a + (y − b)y0 = 0 x2 + y2 (x − a)2 + (y − b)2 Since y0 = 0 a the critical point, the equation can be rewritten as p x = p a − x x2 + y2 (x − a)2 + (y − b)2 1 c 1999 and c 2007 David Jerison 5 Lecture 13 18.01 Fall 2006 p p From Fig. 6, we see that the last equation can be interpreted geometrically as saying that sin α = sin β where α and β are the angles the left and right portions of the string make with the vertical. Physical and geometric conclusions The angles α and β are equal. Using vectors to compute the force exerted by gravity on the two halves of the string, one ﬁnds that there is equal tension in the two halves of the string - a physical equilibrium. (From another point of view, the equal angle property expresses a geometric property of ellipses: Suppose that the ellipse is a mirror. A ray of light from the focus (0, 0) reﬂects oﬀ the mirror according to the rule angle of incidence equals angle of reﬂection, and therefore the ray goes directly to the other focus at (a, b).) Formulae for x and y We did not yet ﬁnd the location of (x, y). We will now show that p a b 1 x = 2 1 −√ L2 − a2 , y = 2 b − L2 − a2 Because α = β, p p x = x2 + y2 sin α; a − x = (x − a)2 + (y − b)2 sin α Adding these two equations, p p a a = x2 + y2 + (x − a)2 + (y − b)2 sin α = L sin α = ⇒ sin α = L The equations for the vertical legs of the right triangles are (note that y < 0): −y = x2 + y2 cos α; b − y = (x − a)2 + (y − b)2 cos β Adding these two equations, and using α = β, p p 1 b − 2y = x2 + y2 + (x − a)2 + (y − b)2 cos α = L cos α = ⇒ y = 2(b − L cos α) a Use the relation sin α = to write L cos α = L p 1 − sin2 α = √ L2 − a2. Then the formula for y is L 1 p y = 2 b − L2 − a2 Finally, to ﬁnd the formula for x, use the similar right triangles tan α = x = a − x = x(b − y) = (−y)(a − x) = (b − 2y)x = −ay −y b − y ⇒ ⇒ Therefore, x = = b − − ay 2y a 2 1 −√ L2 b − a2 Thus we have formulae for x and y in terms of a, b and L. I omitted the derivation of the formulae for x and y in lecture because it is long and because we got all of our physical intuition and understanding out of the problem from the balance condition that was the immediate consequence of the critical point computation. Final Remark. In 18.02, you will learn to treat constrained max/min problems in any number of variables using a method called Lagrange multipliers. 6 Lecture 13 18.01 Fall 2006 Lecture 14: Mean Value Theorem and Inequalities Mean-Value Theorem The Mean-Value Theorem (MVT) is the underpinning of calculus. It says: If f is diﬀerentiable on a < x < b, and continuous on a ≤ x ≤ b, then f(b) − f(a) = f 0(c) (for some c, a < c < b) b − a f(b) − f(a) Here, is the slope of a secant line, while f 0(c) is the slope of a tangent line. b − a secant line slope f’(c) a b c Figure 1: Illustration of the Mean Value Theorem. Geometric Proof: Take (dotted) lines parallel to the secant line, as in Fig. 1 and shift them up from below the graph until one of them ﬁrst touches the graph. Alternatively, one may have to start with a dotted line above the graph and move it down until it touches. If the function isn’t diﬀerentiable, this approach goes wrong. For instance, it breaks down for the function f(x) = \|x\|. The dotted line always touches the graph ﬁrst at x = 0, no matter what its slope is, and f 0(0) is undeﬁned (see Fig. 2). 1 Lecture 14 18.01 Fall 2006 Figure 2: Graph of y = \|x\|, with secant line. (MVT goes wrong.) Interpretation of the Mean Value Theorem You travel from Boston to Chicago (which we’ll assume is a 1,000 mile trip) in exactly 3 hours. At 1000 some time in between the two cities, you must have been going at exactly mph. 3 f(t) = position, measured as the distance from Boston. f(3) = 1000, f(0) = 0, a = 0, and b = 3. 1000 = f(b) − f(a) = f 0(c) 3 3 where f 0(c) is your speed at some time, c. Versions of the Mean Value Theorem There is a second way of writing the MVT: f(b) − f(a) = f 0(c)(b − a) f(b) = f(a) + f 0(c)(b − a) (for some c, a < c < b) There is also a third way of writing the MVT: change the name of b to x. f(x) = f(a) + f 0(c)(x − a) for some c, a < c < x The theorem does not say what c is. It depends on f, a, and x. This version of the MVT should be compared with linear approximation (see Fig. 3). f(x) ≈ f(a) + f 0(a)(x − a) x near a 2 Lecture 14 18.01 Fall 2006 The tangent line in the linear approximation has a deﬁnite slope f 0(a). by contrast formula is an exact formula. It conceals its lack of speciﬁcity in the slope f 0(c), which could be the slope of f at any point between a and x. (a,f(a)) (x,f(x)) y=f(a) + f’(a)(x-a) error Figure 3: MVT vs. Linear Approximation. Uses of the Mean Value Theorem. Key conclusions: (The conclusions from the MVT are theoretical) 1. If f 0(x) > 0, then f is increasing. 2. If f 0(x) < 0, then f is decreasing. 3. If f 0(x) = 0 all x, then f is constant. Deﬁnition of increasing/decreasing: Increasing means a < b f(a) < f(b). Decreasing means a < b = f(a) < f(b). ⇒ ⇒ Proofs: Proof of 1: a < b f(b) = f(a) + f 0(c)(b − a) Because f 0(c) and (b − a) are both positive, f(b) = f(a) + f 0(c)(b − a) > f(a) (The proof of 2 is omitted because it is similar to the proof of 1) Proof of 3: f(b) = f(a) + f 0(c)(b − a) = f(a) + 0(b − a) = f(a) Conclusions 1,2, and 3 seem obvious, but let me persuade you that they are not. Think back to the deﬁnition of the derivative. It involves inﬁnitesimals. It’s not a sure thing that these inﬁnitesimals have anything to do with the non-inﬁnitesimal behavior of the function. 3 Lecture 14 18.01 Fall 2006 Inequalities The fundamental property f 0 > 0 = f is increasing can be used to deduce many other inequali ⇒ ties. x Example. e x 1. e > 0 x 2. e > 1 for x > 0 x 3. e > 1 + x x Proofs. We will take property 1 (e > 0) for granted. Proofs of the other two properties follow: Proof of 2: Deﬁne f1(x) = ex −1. Then, f1(0) = e0 −1 = 0, and f 0(x) = ex > 0. (This last assertion 1 is from step 1). Hence, f1(x) is increasing, so f(x) > f(0) for x > 0. That is: e x > 1 for x > 0 . x Proof of 3: Let f2(x) = e − (1 + x). f 0(x) = e x − 1 = f1(x) > 0 (if x > 0). 2 Hence, f2(x) > 0 for x > 0. In other words, e x > 1 + x 2 2 x x Similarly, e x > 1 + x + 2 (proved using f3(x) = e x − (1 + x + 2 )). One can keep on going: 2 3 x x e x > 1 + x + + for x > 0. Eventually, it turns out that 2 3! 2 3 x x e x = 1 + x + + + (an inﬁnite sum) 2 3! · · · We will be discussing this when we get to Taylor series near the end of the course. 4 Lecture 14 18.01 Fall 2006 Lecture 15: Diﬀerentials and Antiderivatives Diﬀerentials New notation: dy = f 0(x)dx (y = f(x)) Both dy and f 0(x)dx are called diﬀerentials. You can think of dy = f 0(x) dx as a quotient of diﬀerentials. One way this is used is for linear approximations. Δy dy Δx ≈ dx Example 1. Approximate 651/3 Method 1 (review of linear approximation method) f(x) = x 1/3 1 f 0(x) = x−2/3 3 f(x) ≈ f(a) + f 0(a)(x − a) 1 x 1/3 ≈ a 1/3 + 3 a−2/3(x − a) A good base point is a = 64, because 641/3 = 4. Let x = 65. 1 1 1 1 651/3 = 641/3 + 64−2/3(65 − 64) = 4 + (1) = 4 + 48 ≈ 4.02 3 3 16 Similarly, 1 (64.1)1/3 ≈ 4 + 480 Method 2 (review) 1/3 1 1 1 651/3 = (64 + 1)1/3 = [64(1 + )]1/3 = 641/3[1 + ]1/3 = 4 1 + 64 64 64 1 1 Next, use the approximation (1 + x)r ≈ 1 + rx with r = 3 and x = 64. 1 1 1 651/3 ≈ 4(1 + ( )) = 4 + 3 64 48 This is the same result that we got from Method 1. 1 Lecture 15 18.01 Fall 2006 Z Z Z Z Z Z Z Method 3 (with diﬀerential notation) y = x 1/3\|x=64 = 4 1 1 1 1 dy = 3 x−2/3dx\|x=64 = 3 16 dx = 48 dx 1 We want dx = 1, since (x + dx) = 65. dy = when dx = 1. 48 1 (65)1/3 = 4 + 48 What underlies all three of these methods is y = x 1/3 dy 1 x−2/3 dx = 3 \|x=64 Anti-derivatives F (x) = f(x)dx means that F is the antiderivative of f. Other ways of saying this are: F 0(x) = f(x) or, dF = f(x)dx Examples: 1. sin xdx = − cos x + c where c is any constant. n+1 x 2. x ndx = n + 1 + c for n 6= −1. dx 3. x = ln \|x\| + c (This takes care of the exceptional case n = −1 in 2.) 4. sec2 xdx = tan x + c dx 1 5. √ 1 − x2 = sin−1 x + c (where sin−1 x denotes “inverse sin” or arcsin, and not sin x ) 6. dx = tan−1(x) + c 1 + x2 Proof of Property 2: The absolute value \|x\| gives the correct answer for both positive and negative x. We will double check this now for the case x < 0: ln \|x\| d dx ln(−x) = = ln(−x) d du ln(u) du dx where u = −x. d dx ln(−x) = 1 u (−1) = 1 −x (−1) = 1 x 2 Lecture 15 18.01 Fall 2006 Z Z Z Z Z Uniqueness of the antiderivative up to an additive constant. If F 0(x) = f(x), and G0(x) = f(x), then G(x) = F (x) + c for some constant factor c. Proof: (G − F )0 = f − f = 0 Recall that we proved as a corollary of the Mean Value Theorem that if a function has a derivative zero then it is constant. Hence G(x) − F (x) = c (for some constant c). That is, G(x) = F (x) + c. Method of substitution. Example 1. x 3(x 4 + 2)5dx Substitution: 1 u = x 4 + 2, du = 4x 3dx, (x 4 + 2)5 = u 5 , x 3dx = du 4 Hence, Z Z 1 u6 u6 1 x 3(x 4 + 2)5dx = u 5du = = + c = (x 4 + 2)6 + c 4 4(6) 24 24 x Example 2. dx √ 1 + x2 Another way to ﬁnd an anti-derivative is “advanced guessing.” First write x √ 1 + x2 dx = x(1 + x 2)−1/2dx Guess: (1 + x 2)1/2 . Check this. d 1 (1 + x 2)1/2 = (1 + x 2)−1/2(2x) = x(1 + x 2)−1/2 dx 2 Therefore, Z x(1 + x 2)−1/2dx = (1 + x 2)1/2 + c Example 3. e 6xdx Guess: e 6x . Check this: d e 6x = 6e 6x dx Therefore, Z 1 e 6xdx = e 6x + c 6 3 Lecture 15 18.01 Fall 2006 Z Z Z Z 2 Example 4. xe−x dx 2 Guess: e−x Again, take the derivative to check: d e−x 2 2 = (−2x)(e−x ) dx Therefore, Z 2 xe−x dx = − 2 1 2 e−x + c 1 Example 5. sin x cos xdx = sin2 x + c 2 Another, equally acceptable answer is sin x cos xdx = − 1 cos2 x + c 2 This seems like a contradiction, so let’s check our answers: d sin2 x = (2 sin x)(cos x) dx and d 2 cos x = (2 cos x)(− sin x) dx So both of these are correct. Here’s how we resolve this apparent paradox: the diﬀerence between the two answers is a constant. 1 sin2 x − (− 1 cos2 x) = 1(sin2 x + cos2 x) = 1 2 2 2 2 So, 1 sin2 x − 1 = 1(sin2 x − 1) = 1(− cos2 x) = − 1 cos2 x 2 2 2 2 2 The two answers are, in fact, equivalent. The constant c is shifted by 1 2 from one answer to the other. dx Example 6. (We will assume x > 0.) x ln x 1 Let u = ln x. This means du = dx. Substitute these into the integral to get x Z Z dx 1 = du = ln u + c = ln(ln(x)) + c x ln x u 4 Lecture 15 18.01 Fall 2006 Z Z Z Lecture 16: Diﬀerential Equations and Separation of Variables Ordinary Diﬀerential Equations (ODEs) Example 1. dy = f(x) dx Solution: y = f(x)dx. We consider these types of equations as solved. Example 2. d + x y = 0 or dy + xy = 0 dx dx d ( + x is known in quantum mechanics as the annihilation operator.) dx Besides integration, we have only one method of solving this so far, namely, substitution. Solving for dy gives: dx dy = −xy dx The key step is to separate variables. dy = −xdx y Note that all y-dependence is on the left and all x-dependence is on the right. Next, take the antiderivative of both sides: dy y = − xdx 2 x ln \|y\| = − 2 + c (only need one constant c) \|y\| = e c e−x 2/2 (exponentiate) 2 y = ae−x /2 (a = ±e c) c Despite the fact that e = 6 0, a = 0 is possible along with all a = 6 0, depending on the initial conditions. For instance, if y(0) = 1, then y = e−x 2/2 . If y(0) = a, then y = ae−x 2/2 (See Fig. 1). 1 Lecture 16 18.01 Fall 2006 Z Z −6 −4 −2 0 2 4 6 0 0.2 0.4 0.6 0.8 1 X Y 2 Figure 1: Graph of y = e− x 2 . In general: dy = f(x)g(y) dx dy = f(x)dx which we can write as g(y) 1 h(y)dy = f(x)dx where h(y) = . g(y) Now, we get an implicit formula for y: H(y) = F (x) + c (H(y) = h(y)dy; F (x) = f(x)dx) where H0 = h, F 0 = f, and y = H−1(F (x) + c) (H−1 is the inverse function.) In the previous example: 2 f(x) = x; F (x) = − 2 x ; 1 1 g(y) = y; h(y) = g(y) = y , H(y) = ln \|y\| 2 Lecture 16 18.01 Fall 2006 dy y Example 3 (Geometric Example). = 2 . dx x Find a graph such that the slope of the tangent line is twice the slope of the ray from (0, 0) to (x, y) seen in Fig. 2. (x,y) Figure 2: The slope of the tangent line (red) is twice the slope of the ray from the origin to the point (x, y). dy = 2dx (separate variables) y x ln \|y\| \|y\| = = 2 ln \|x\| + c (antiderivative) e c x 2 (exponentiate; remember, e 2 ln \|x\| = x 2 ) Thus, y = ax 2 Again, a < 0, a > 0 and a = 0 are all acceptable. Possible solutions include, for example, y = x 2 (a = 1) y = 2x 2 (a = 2) y = −x 2 (a = −1) y = 0x 2 = 0 (a = 0) y = −2y 2 (a = −2) y = 100x 2 (a = 100) 3 Lecture 16 18.01 Fall 2006 s Example 4. Find the curves that are perpendicular to the parabolas in Example 3. We know that their slopes, dy −1 −x = = dx slope of parabola 2y Separate variables: ydy = −xdx 2 Take the antiderivative: 2 2 2 2 y x x y 2 = − 4 + c = ⇒ 4 + 2 = c which is an equation for a family of ellipses. For these ellipses, the ratio of the x-semi-major axis to the y-semi-minor axis is √ 2 (see Fig. 3). Figure 3: The ellipses are perpendicular to the parabolas. Separation of variables leads to implicit formulas for y, but in this case you can solve for y. x2 y = ± 2 c − 4 Exam Review Exam 2 will be harder than exam 1 — be warned! Here’s a list of topics that exam 2 will cover: 1. Linear and/or quadratic approximations 2. Sketches of y = f(x) 3. Maximum/minimum problems. 4. Related rates. 5. Antiderivatives. Separation of variables. 6. Mean value theorem. More detailed notes on all of these topics are provided in the Exam 2 review sheet. 4 Lecture 16 18.01 Fall 2006 x 18.01 UNIT 2 REVIEW; Fall 2007 The central theme of Unit 2 is that knowledge of f 0 (and sometimes f 00) tells us something about f itself. This is even true of our ﬁrst topic, approximation. For instance, knowing that f(x) = e satisﬁes f(0) = 1 and f 0(0) = 1, we can say e x 1 + x provided x 0 x The linear function 1 + x is much simpler than e , so f(0) and f 0(0) give us a (very) simpliﬁed picture of our function, useful only near near 0. For more detail, use the quadratic approximation, x e 1 + x + x 2/2 provided x 0 (still only works well near 0) The second and third practice exams are actual tests from previous years. The exam this year is similar to the one from 2006 posted at our site. It has 6 questions covering the following topics. (No Newton’s method, but there is a seventh, extra credit problem.) 1. Linear and/or quadratic approximations 2. Sketch a graph y = f(x) 3. Max/min 4. Related rates 5. Find antiderivatives and solve a diﬀerential equation by separating variables 6. Mean value theorem. Remarks. 1. Recall that linear [and quadratic] approximation is f(x) f(a) + f 0(a)(x − a) [+(f 00(a)/2)(x − a)2] 2. You should expect to graph a function y = f(x), where f(x) is a rational function (ratio of polynomials). Warnings: a) When asked to label the critical point on the graph, ﬁnd and mark the point (a, b). In lecture we called x = a the critical point and y = b the critical value, and this is what is used in 18.02, and elsewhere. But for this exam (and this is just an inconsistency in language that you will have to tolerate) the words “critical point” refer to the point on the graph (a, b), not the number a and the point on the x-axis. The same applies to inﬂection points. b) y = 1/(x − 1) is decreasing on the intervals −≈ < x < 1 and 1 < x < ≈, but it is not decreasing on the interval −≈ < x < ≈. Draw the graph to see. You cannot just use the fact that y0 = −1/(x − 1)2 < 0 because there is a point in the middle at which y is not diﬀerentiable — and not even continuous. So the mean value theorem does not apply. c) Similarly, y = 1/(x − 1)2 is concave up on −≈ < x < 1 and 1 < x < ≈, but it is not concave up on the interval −≈ < x < ≈. Here y00 = 6/(x − 1)4 > 0, but there is a singularity in the middle. Plot the graph yourself to see. 1 3. The mean value theorem says that if f is diﬀerentiable, then for some c, a < c < x, f(x) = f(a) + f 0(c)(x − a) It is used as follows. Suppose that m < f 0(c) < M on the interval a < c < x, then f(x) = f(a) + f 0(c)(x − a) < f(a) + M(x − a) Similarly, f(x) = f(a) + f 0(c)(x − a) > f(a) + m(x − a) Put another way, if f = f(x) − f(a) and x = x − a, and m < f 0(c) < M for a < c < x, then mx < f < Mx More consequences of the mean value theorem. A function f is called increasing (also called strictly increasing) if x > a implies f(x) > f(a). The reasoning above with m = 0 shows that if f 0 > 0, then f is increasing. Similarly if f 0 < 0, then f is decreasing. We use these facts every time we sketch a graph of a function or ﬁnd a maximum or minimum. A similar discussion works when the inequality is not strict. If m f 0(c) M for a < c < x, then f(a) + m(x − a) f(x) f(a) + M(x − a) A function is called nondecreasing if x > a implies f(x) f(a). If f 0 0, then the inequality above shows that f is nondecreasing. Conversely, if the function is nondecreasing and diﬀerentiable, then f 0 0. Similarly, diﬀerentiable functions are nonincreasing if and only if they satisfy f 0 0. Key corollary to the mean value theorem: f 0 = g0 implies f − g is constant. In Unit 2, we have found that information about f 0 gives information about f. In particular, knowing a starting value for a function and its rate of change determines the function. A seemingly obvious example is that if f 0 = 0 for all x, then f is constant. If this were not true, then the mathematical notion of derivative would fail to coincide with our intuitive notion of what rate of change and cause and eﬀect mean. But this fundamental fact needs a proof. Derivatives are instantaneous quantities, obtained as limits. It is the mean value theorem that allows us to pass in rigorous mathematical fashion from the inﬁnitesimal to the practical, human scale. Here is the proof. If f 0 = 0, then one can take m = M = 0 in the inequalities above, and conclude that f(x) = f(a). In other words, f is constant. As an immediate consequence, if f 0 = g0, then f and g diﬀer by a constant. (Apply the previous argument to the function f − g, whose derivative is 0.) This basic fact will lead us shortly to what is known as the fundamental theorem of calculus. 2
16293	https://en.wikipedia.org/wiki/Couette_flow	Jump to content Search Contents (Top) 1 Planar Couette flow 1.1 Startup 1.2 Planar flow with pressure gradient 1.3 Compressible flow 1.4 Rectangular channel 2 Coaxial cylinders 2.1 Coaxial cylinders of finite length 3 See also 4 References 5 Sources 6 External links Couette flow Català Deutsch Español فارسی Français Italiano עברית 日本語 Русский Suomi Українська 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item Appearance From Wikipedia, the free encyclopedia Model of viscous fluid flow between two surfaces moving relative to each other In fluid dynamics, Couette flow is the flow of a viscous fluid in the space between two surfaces, one of which is moving tangentially relative to the other. The relative motion of the surfaces imposes a shear stress on the fluid and induces flow. Depending on the definition of the term, there may also be an applied pressure gradient in the flow direction. The Couette configuration models certain practical problems, like the Earth's mantle and atmosphere, and flow in lightly loaded journal bearings. It is also employed in viscometry and to demonstrate approximations of reversibility. It is named after Maurice Couette, a Professor of Physics at the French University of Angers in the late 19th century. Isaac Newton first defined the problem of Couette flow in Proposition 51 of his Philosophiæ Naturalis Principia Mathematica, and expanded upon the ideas in Corollary 2. Planar Couette flow [edit] Couette flow is frequently used in undergraduate physics and engineering courses to illustrate shear-driven fluid motion. A simple configuration corresponds to two infinite, parallel plates separated by a distance ; one plate translates with a constant relative velocity in its own plane. Neglecting pressure gradients, the Navier–Stokes equations simplify to where is the spatial coordinate normal to the plates and is the velocity field. This equation reflects the assumption that the flow is unidirectional — that is, only one of the three velocity components is non-trivial. If the lower plate corresponds to , the boundary conditions are and . The exact solution can be found by integrating twice and solving for the constants using the boundary conditions. A notable aspect of the flow is that shear stress is constant throughout the domain. In particular, the first derivative of the velocity, , is constant. According to Newton's Law of Viscosity (Newtonian fluid), the shear stress is the product of this expression and the (constant) fluid viscosity. Startup [edit] In reality, the Couette solution is not reached instantaneously. The "startup problem" describing the approach to steady state is given by subject to the initial condition and with the same boundary conditions as the steady flow: The problem can be made homogeneous by subtracting the steady solution. Then, applying separation of variables leads to the solution: : . The timescale describing relaxation to steady state is , as illustrated in the figure. The time required to reach the steady state depends only on the spacing between the plates and the kinematic viscosity of the fluid, but not on . Planar flow with pressure gradient [edit] A more general Couette flow includes a constant pressure gradient in a direction parallel to the plates. The Navier–Stokes equations are where is the dynamic viscosity. Integrating the above equation twice and applying the boundary conditions (same as in the case of Couette flow without pressure gradient) gives The pressure gradient can be positive (adverse pressure gradient) or negative (favorable pressure gradient). In the limiting case of stationary plates (), the flow is referred to as Plane Poiseuille flow, and has a symmetric (with reference to the horizontal mid-plane) parabolic velocity profile. Compressible flow [edit] In incompressible flow, the velocity profile is linear because the fluid temperature is constant. When the upper and lower walls are maintained at different temperatures, the velocity profile is more complicated. However, it has an exact implicit solution as shown by C. R. Illingworth in 1950. Consider the plane Couette flow with lower wall at rest and the upper wall in motion with constant velocity . Denote fluid properties at the lower wall with subscript and properties at the upper wall with subscript . The properties and the pressure at the upper wall are prescribed and taken as reference quantities. Let be the distance between the two walls. The boundary conditions are where is the specific enthalpy and is the specific heat. Conservation of mass and -momentum requires everywhere in the flow domain. Conservation of energy and -momentum reduce to where is the wall shear stress. The flow does not depend on the Reynolds number , but rather on the Prandtl number and the Mach number , where is the thermal conductivity, is the speed of sound and is the specific heat ratio. Introduce the non-dimensional variables In terms of these quantities, the solutions are where is the heat transferred per unit time per unit area from the lower wall. Thus are implicit functions of . One can also write the solution in terms of the recovery temperature and recovery enthalpy evaluated at the temperature of an insulated wall i.e., the values of and for which .[clarification needed] Then the solution is If the specific heat is constant, then . When and , then and are constant everywhere, thus recovering the incompressible Couette flow solution. Otherwise, one must know the full temperature dependence of . While there is no simple expression for that is both accurate and general, there are several approximations for certain materials — see, e.g., temperature dependence of viscosity. When and , the recovery quantities become unity . For air, the values are commonly used, and the results for this case are shown in the figure. The effects of dissociation and ionization (i.e., is not constant) have also been studied; in that case the recovery temperature is reduced by the dissociation of molecules. Rectangular channel [edit] One-dimensional flow is valid when both plates are infinitely long in the streamwise () and spanwise () directions. When the spanwise length is finite, the flow becomes two-dimensional and is a function of both and . However, the infinite length in the streamwise direction must be retained in order to ensure the unidirectional nature of the flow. As an example, consider an infinitely long rectangular channel with transverse height and spanwise width , subject to the condition that the top wall moves with a constant velocity . Without an imposed pressure gradient, the Navier–Stokes equations reduce to with boundary conditions Using separation of variables, the solution is given by When , the planar Couette flow is recovered, as shown in the figure. Coaxial cylinders [edit] Taylor–Couette flow is a flow between two rotating, infinitely long, coaxial cylinders. The original problem was solved by Stokes in 1845, but Geoffrey Ingram Taylor's name was attached to the flow because he studied its stability in a famous 1923 paper. The problem can be solved in cylindrical coordinates . Denote the radii of the inner and outer cylinders as and . Assuming the cylinders rotate at constant angular velocities and , then the velocity in the -direction is This equation shows that the effects of curvature no longer allow for constant shear in the flow domain. Coaxial cylinders of finite length [edit] The classical Taylor–Couette flow problem assumes infinitely long cylinders; if the cylinders have non-negligible finite length , then the analysis must be modified (though the flow is still unidirectional). For , the finite-length problem can be solved using separation of variables or integral transforms, giving: where are the Modified Bessel functions of the first and second kind. See also [edit] Laminar flow Stokes-Couette flow Hagen–Poiseuille equation Taylor–Couette flow Hagen–Poiseuille flow from the Navier–Stokes equations Ostroumov flow References [edit] ^ Zhilenko et al. (2018) ^ Guyon et al. (2001), p. 136 ^ Heller (1960) ^ Donnelly, Russell J. (1991-11-01). "Taylor-Couette Flow: The Early Days". Physics Today. 44 (11): 32–39. Bibcode:1991PhT....44k..32D. doi:10.1063/1.881296. ISSN 0031-9228. ^ Rowlands, Peter (2017). Newton – Innovation And Controversy. World Scientific Publishing. p. 162. ISBN 9781786344045. ^ Pozrikidis (2011), pp. 338–339 ^ Kundu et al. (2016), p. 415 ^ Lagerstrom (1996) ^ Liepmann et al. (1956, 1957) ^ Landau and Lifshitz (1987) ^ Stokes (1845) ^ Taylor (1923) ^ Guyon et al. (2001), pp. 163–166 ^ Wendl (1999) Sources [edit] Acheson, D.J. (1990). Elementary Fluid Dynamics. Oxford University Press. ISBN 0-19-859679-0. Batchelor, G.K. (2000) . An Introduction to Fluid Dynamics. Cambridge University Press. ISBN 0-521-66396-2. Guyon, Etienne; Hulin, Jean-Pierre; Petit, Luc; Mitescu, Catalin D. (2001). Physical Hydrodynamics. Oxford University Press. ISBN 0-19-851746-7. Heller, John P. (1960). "An Unmixing Demonstration". American Journal of Physics. 28 (4): 348–353. Bibcode:1960AmJPh..28..348H. doi:10.1119/1.1935802. ISSN 0002-9505. Illingworth, C. R. (1950). "Some solutions of the equations of flow of a viscous compressible fluid". Mathematical Proceedings of the Cambridge Philosophical Society. 46 (3): 469–478. Bibcode:1950PCPS...46..469I. doi:10.1017/S0305004100025986. ISSN 0305-0041. S2CID 122559614. Kundu, Pijush K.; Cohen, Ira M.; Dowling, David R. (2016). Fluid Mechanics (6th ed.). Elsevier. ISBN 978-0-12-405935-1. Lagerstrom, Paco (1996). Laminar flow theory. Princeton University Press. ISBN 978-0691025988. Landau, L. D.; Lifshitz, E.M. (1987). Fluid Mechanics (2nd ed.). Elsevier. ISBN 978-0-08-057073-0. Liepmann, H. W., and Z. O. Bleviss. "The effects of dissociation and ionization on compressible couette flow." Douglas Aircraft Co. Rept. SM-19831 130 (1956). Liepmann, Hans Wolfgang, and Anatol Roshko. Elements of gasdynamics. Courier Corporation, 1957. Pozrikidis, C. (2011). Introduction to Theoretical and Computational Fluid Dynamics. Oxford University Press. ISBN 978-0-19-975207-2. Richard Feynman (1964) The Feynman Lectures on Physics: Mainly Electromagnetism and Matter, § 41–6 Couette flow, Addison–Wesley ISBN 0-201-02117-X Stokes, George Gabriel (1880). "On the Theories of the Internal Friction of Fluids in Motion, and of the Equilibrium and Motion of Elastic Solids". Mathematical and Physical Papers. Cambridge University Press: 75–129. doi:10.1017/CBO9780511702242.005. ISBN 9780511702242. {{cite journal}}: ISBN / Date incompatibility (help) Taylor, Geoffrey I. (1923). "Stability of a viscous liquid contained between two rotating cylinders". Philosophical Transactions of the Royal Society of London. Series A, Containing Papers of a Mathematical or Physical Character. 223 (605–615): 289–343. Bibcode:1923RSPTA.223..289T. doi:10.1098/rsta.1923.0008. JSTOR 91148. Wendl, Michael C. (1999). "General solution for the Couette flow profile". Physical Review E. 60 (5): 6192–6194. Bibcode:1999PhRvE..60.6192W. doi:10.1103/PhysRevE.60.6192. ISSN 1063-651X. PMID 11970531. Zhilenko, Dmitry; Krivonosova, Olga; Gritsevich, Maria; Read, Peter (2018). "Wave number selection in the presence of noise: Experimental results". Chaos: An Interdisciplinary Journal of Nonlinear Science. 28 (5): 053110. Bibcode:2018Chaos..28e3110Z. doi:10.1063/1.5011349. hdl:10138/240787. ISSN 1054-1500. PMID 29857673. S2CID 46925417. External links [edit] AMS Glossary: Couette Flow A rheologists perspective: the science behind the couette cell accessory Retrieved from " Categories: Flow regimes Fluid dynamics Hidden categories: Articles with short description Short description is different from Wikidata Wikipedia articles needing clarification from December 2020 CS1 errors: ISBN date Couette flow Add topic
16294	https://www.utmb.edu/pedi_ed/Obesity/page_18.htm	Secondary Causes of Hypercholesterolemia Obesity, Overweight, and Dyslipidemia in Children and Adolescents Table of Contents OBESITY, OVERWEIGHT, AND DYSLIPIDEMIA IN CHILDREN AND ADOLESCENTS DEFINITIONS BODY MASS INDEX (BMI) EPIDEMIOLOGY ETIOLOGY ASSESSMENT CO-MORBIDITIES HYPERLIPIDEMIA Primary Disorders of Cholesterol Metabolism Secondary Causes of Hypercholesterolemia Cholesterol Metabolism Cholesterol Screening DIET AND EXERCISE RECOMMENDATIONS PHARMACOLOGIC INTERVENTIONS REFERRING TO SUBSPECIALISTS TAKE HOME POINTS RESOURCES & REFERENCES ABOUT THE MODULE Secondary Causes of Hypercholesterolemia (Starc, 1996) Endocrine disorders Diabetes mellitus Hypothyroidism Renal Disorders Nephrotic syndrome Renal failure Drugs Adrenal steroids Isotretinoin Thiazides Anticonvulsants Oral contraceptives Alcohol Hepatic Disease Obstructive liver disease Hepatitis Storage Diseases Gaucher disease Von Gierke disease Other Acute intermittent porphyria Anorexia nervosa Systemic lupus erythematosus Obesity Cigarette smoking Loni McCuistion Shepherd and Marney Gundlach. University of Texas Medical Branch. 2009
16295	https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut22_linineq.htm	Title \| \| \| \| --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- \| \| \| \| \| VML COLLEGE ALGEBRA INTERMEDIATE ALGEBRA BEGINNING ALGEBRA GRE MATH THEA/ACCUPLACER \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- \| College Algebra Tutorial 22: Linear Inequalities --- WTAMU > Virtual Math Lab > College Algebra Learning Objectives --- \| \| \| After completing this tutorial, you should be able to: 1. Use the addition, subtraction, multiplication, and division properties of inequalities to solve linear inequalities. 2. Solve linear inequalities involving absolute values. 3. Write the answer to an inequality using interval notation. 4. Draw a graph to give a visual answer to an inequality problem. \| \| In this tutorial we will be looking at solving linear inequalities. When solving linear inequalities, we use a lot of the same concepts that we use when solving linear equations (shown in Tutorial 14: Linear Equations in One Variable). Basically, we still want to get the variable on one side and everything else on the other side by using inverse operations. The difference is, when a variable is set equal to one number, that number is the only solution. But, when a variable is less than or greater than a number, there are an infinite number of values that would be a part of the answer. We will also revisit the definition of absolute value and how it applies to inequalities. If you need a review on absolute values go to Tutorial 21: Absolute Value Equations. You never know when you will need to know about inequalities, so you better get started. \| \| Inequality Signs \| \| \| \| Read left to right: a < b a is less than b a < b a is less than or equal to b a > b a is greater than b a > b a is greater than or equal to b \| \| \| \| Interval Notation \| \| \| \| --- \| \| Interval notation is a way to notate the range of values that would make an inequality true. There are two types of intervals, open and closed (described below), each with a specific way to notate it so we can tell the difference between the two. Note that in the interval notations (found below), you will see the symbol , which means infinity. \| \| \| Positive infinity () means it goes on and on indefinitely to the right of the number - there is no endpoint on the right. Negative infinity (-) means it goes on and on indefinitely to the left of the number - there is no endpoint to the left. Since we don’t know what the largest or smallest numbers are, we need to use infinity or negative infinity to indicate there is no endpoint in one direction or the other. \| In general, when using interval notation, you always put the smaller value of the interval first (on the left side), put a comma between the two ends, then put the larger value of the interval (on the right side). You will either use a curved end ( or ) or a boxed end [ or ], depending on the type of interval (described below). If you have either infinity or negative infinity on either end, you always use a curve for that end. This will indicate that there is no definite endpoint in that direction, it keeps going and going. \| \| \| \| Open Interval \| \| \| \| An open interval does not include where your variable is equal to the endpoint. To indicate this, we use a curved end as shown below. \| \| \| \| \| --- \| \| Interval Notation forOpen Intervals \| \| \| x > a \| (a, ) \| \| \| x < a \| (-, a) \| \| \| \| \| When you graph an open ended end point, you use the same curved end ( or ) on the graph as you do in the interval notation. Also, darken in the part of the graph that is the solution. For example, \| \| \| \| \| --- \| Inequality \| Interval Notation for Open Intervals \| Graph \| \| x > 4 \| (4, ) \| \| \| x < 4 \| (-, 4) \| \| \| \| \| Closed Interval \| \| \| \| A closed interval includes where your variable is equal to the endpoint. To indicate this, we use a boxed end as shown below. As mentioned above, even though a is included and has a boxed end, if it goes to either infinity or negative infinity on the other end, we will notate it with a curved end for that end only! \| \| \| \| \| --- \| Inequality \| Interval Notation for Closed Intervals \| \| \| x > a \| [a, ) \| \| \| x < a \| (-, a] \| \| \| \| \| When you graph a closed ended end point, you use the same boxed end [ or ] on the graph as you do in the interval notation. Also, darken in the part of the graph that is the solution. For example, \| \| \| \| \| --- \| Inequality \| Interval Notation for Closed Intervals \| Graph \| \| x > 4 \| [4, ) \| \| \| x < 4 \| (-, 4] \| \| \| \| \| Combining Open and Closed Intervals \| \| \| \| Sometimes one end of your interval is open and the other end is closed. You still follow the basic ideas described above. The closed end will have a [ or ] on it’s end and the open end will have a ( or ) on its end. \| \| \| \| \| --- \| Inequality \| Interval Notation for Combining Open and Closed Intervals \| \| \| a < x < b \| (a, b] \| \| \| a < x < b \| [a, b) \| \| \| \| \| \| --- \| Inequality \| Interval Notation for Combining Open and Closed Intervals \| Graph \| \| 3 < x < 6 \| [3, 6) \| \| \| \| \| Addition/Subtraction Property for Inequalities If a < b, then a + c < b + c If a < b, then a - c < b - c \| \| \| \| In other words, adding or subtracting the same expression to both sides of an inequality does not change the inequality. � Example 1: Solve, write your answer in interval notation and graph the solution set:. � View a video of this example \| \| \| \| --- \| \| Interval notation: Graph: \| Inv. of sub. 10 is add. 10 Open interval indicating all values less than 5 Visual showing all numbers less than 5 on the number line \| \| \| \| Note that the inequality stayed the same throughout the problem. Adding or subtracting the same value to both sides does not change the inequality. The answer 'x is less than 5' means that if we put any number less than 5 back in the original problem, it would be a solution (the left side would be less than the right side). As mentioned above, this means that we have more than just one number for our solution, there are an infinite number of values that would satisfy this inequality. Interval notation: We have an open interval since we are not including where it is equal to 5. x is less than 5, so 5 is the largest value of the interval, so it goes on the right. Since there is no lower endpoint (it is ALL values less than 5), we put the negative infinity symbol on the left side. The curved end on 5 indicates an open interval. Negative infinity always has a curved end because there is not an endpoint on that side. Graph: We use the same type of notation on the endpoint as we did in the interval notation, a curved end. Since we needed to indicate all values less than 5, the part of the number line that was to the left of 5 was darkened. \| � \| \| \| Example 2: Solve, write your answer in interval notation and graph the solution set: . View a video of this example \| \| \| \| --- \| \| Interval notation: [-3, ) Graph: \| Inv. of add 4 is sub. 4 Closed interval indicating all values greater than or = -3 Visual showing all numbers greater than or = to -3 on the number line. \| \| \| \| Note that the inequality stayed the same throughout the problem. Adding or subtracting the same value to both sides does not change the inequality. The answer 'x is greater than or equal to -3' means that if we put any number greater than or equal to -3 back in the original problem, it would be a solution (the left side would be greater than or equal to the right side). As mentioned above, this means that we have more than just one number for our solution, there are an infinite number of values that would satisfy this inequality. Interval notation: We have a closed interval since we are including where it is equal to -3. x is greater than or equal to -3, so -3 is our smallest value of the interval, so it goes on the left. Since there is no upper endpoint (it is ALL values greater than or equal to -3), we put the infinity symbol on the right side. The boxed end on -3 indicates a closed interval. Infinity always has a curved end because there is not an endpoint on that side. Graph: We use the same type of notation on the endpoint as we did in the interval notation, a boxed end. Since we needed to indicate all values greater than or equal to -3, the part of the number line that was to the right of -3 was darkened. \| \| \| \| Multiplication/Division Properties for Inequalities when multiplying/dividing by a positive value If a < b AND c is positive, then ac < bc If a < b AND c is positive, thena/c < b/c \| \| \| \| In other words, multiplying or dividing the same POSITIVE number to both sides of an inequality does not change the inequality. Example 3: Solve, write your answer in interval notation and graph the solution set: . View a video of this example \| \| \| \| --- \| \| Interval notation: (-, -3) Graph: \| Inv. of mult. by 3 is div. by 3 Open interval indicating all values less than -3 Visual showing all numbers less than -3 on the number line \| \| \| \| Note that the inequality sign stayed the same direction. Even though the right side was a -9, the number we were dividing both sides by, was a positive 3. Multiplying or dividing both sides by the same positive value does not change the inequality. Interval notation: We have an open interval since there we are not including where it is equal to -3. x is less than -3, so -3 is our largest value of the interval, so it goes on the right. Since there is no lower endpoint (it is ALL values less than -3), we put the negative infinity symbol on the left side. The curved end on -3 indicates an open interval. Negative infinity always has a curved end because there is not an endpoint on that side. Graph: We use the same type of notation on the endpoint as we did in the interval notation, a curved end. Since we needed to indicate all values less than -3, the part of the number line that was to the left of -3 was darkened. \| � \| \| \| Multiplication/Division Properties for Inequalities when multiplying/dividing by a negative value If a < b AND c is negative, thenac > bc If a < b AND c is negative, then a/c > b/c \| \| \| \| In other words, multiplying or dividing the same NEGATIVE number to both sides of an inequality reverses the sign of the inequality. The reason for this is, when you multiply or divide an expression by a negative number, it changes the sign of that expression. On the number line, the positive values go in a reverse or opposite direction than the negative numbers go, so when we take the opposite of an expression, we need to reverse our inequality to indicate this. \| \| \| \| Example 4: Solve, write your answer in interval notation and graph the solution set: . View a video of this example \| \| \| \| --- \| \| Interval notation: Graph: \| Inv. of div. by -4 is mult. by -4, so reverse inequality sign Open interval indicating all values less than -20 Visual showing all numbers less than -20 on the number line \| \| \| \| I multiplied by a -4 to take care of both the negative and the division by 4 in one step. In line 2, note that when I did show the step of multiplying both sides by a -4, I reversed my inequality sign. Interval notation: We have an open interval since we are not including where it is equal to -20. x is less than -20, so -20 is our largest value of the interval, so it goes on the right. Since there is no lower endpoint (it is ALL values less than -20), we put the negative infinity symbol on the left side. The curved end on -20 indicates an open interval. Negative infinity always has a curved end because there is not an endpoint on that side. Graph: We use the same type of notation on the endpoint as we did in the interval notation, a curved end. Since we needed to indicate all values less than -20, the part of the number line that was to the left of -20 was darkened. \| \| \| \| Example 5: Solve, write your answer in interval notation and graph the solution set: . View a video of this example \| \| \| \| --- \| \| Interval notation: Graph: \| Inv. of mult. by -2 is div. by -2, so reverse inequality sign Closed interval indicating all values greater than or = -5/2 Visual showing all numbers greater than or = -5/2 on the number line \| \| \| \| In line 2, note that when I did show the step of dividing both sides by a -2, that I reversed my inequality sign. Interval notation: We have a closed interval since we are including where it is equal to -5/2. x is greater than or equal to -5/2, so -5/2 is our smallest value of the interval so it goes on the left. Since there is no upper endpoint (it is ALL values greater than or equal to -5/2), we put the infinity symbol on the right side. The boxed end on -5/2 indicates a closed interval. Infinity always has a curved end because there is not an endpoint on that side. Graph: We use the same type of notation on the endpoint as we did in the interval notation, a boxed end. Since we needed to indicate all values greater than or equal to -5/2, the part of the number line that was to the right of -5/2 was darkened. \| \| \| \| Strategy for Solving a Linear Inequality \| \| \| \| --- \| \| Step 1: Simplify each side if needed. \| \| \| This would involve things like removing ( ), removing fractions, adding like terms, etc. \| Step 2: Use Add./Sub. Properties to move the variable term on one side and all other terms to the other side. Step 3: Use Mult./Div. Properties to remove any values that are in front of the variable. Note that it is the same basic concept we used when solving linear equations as shown in Tutorial 14: Linear Equations in One Variable. \| \| \| \| Example 6: Solve, write your answer in interval notation and graph the solution set: . View a video of this example \| \| \| \| --- \| \| Interval notation: Graph: \| Inv. of add. 5 is sub. 5 Inv. of mult. by -2 is div. both sides by -2, so reverse inequality sign Open interval indicating all values greater than -3 Visual showing all numbers greater than -3 on the number line \| \| \| \| Interval notation: We have an open interval since we are not including where it is equal to -3. x is greater than -3, so -3 is our smallest value of the interval so it goes on the left. Since there is no upper endpoint (it is ALL values less than -3), we put the infinity symbol on the right side. The curved end on -3 indicates an open interval. Infinity always has a curved end because there is not an endpoint on that side. Graph: We use the same type of notation on the endpoint as we did in the interval notation, a curved end. Since we needed to indicate all values greater than -3, the part of the number line that was to the right of -3 was darkened. \| \| \| \| Example 7: Solve, write your answer in interval notation and graph the solution set: . View a video of this example \| \| \| \| --- \| \| Interval notation: Graph: \| Distributive property Get x terms on one side, constants on the other side Inv. of sub. 3 is add. by 3 Open interval indicating all values less than -1/2 Visual showing all numbers less than -1/2 on the number line. \| \| \| \| Interval notation: Again, we have an open interval since we are not including where it is equal to 8. This time x is less than 8, so 8 is our largest value of the interval so it goes on the right. Since there is no lower endpoint (it is ALL values less than 8), we put the negative infinity symbol on the left side. The curved end on 8 indicates an open interval. Negative infinity always has a curved end because there is not an endpoint on that side. Graph: Again, we use the same type of notation on the endpoint as we did in the interval notation, a curved end. Since we needed to indicate all values less than 8, the part of the number line that was to the left of 8 was darkened. \| \| \| \| Example 8: Solve, write your answer in interval notation and graph the solution set: . View a video of this example \| \| \| \| --- \| \| Interval notation: Graph: \| Mult. both sides by LCD of 6 Get x terms on one side, constants on the other side Inv. of add. 3 is sub. by 3 Inv. of mult. by 10 is div. by 10 Closed interval indicating all values greater than or equal to -3/2 Visual showing all numbers greater than or equal to -3/2 on the number line. \| \| \| \| Interval notation: This time we have a closed interval since we are including where it is equal to -3/2. x is greater than or equal to -3/2, so -3/2 is our smallest value of the interval so it goes on the left. Since there is no upper endpoint (it is ALL values greater than or equal to -3/2), we put the infinity symbol on the right side. The boxed end on -3/2 indicates a closed interval. Infinity always has a curved end because there is not an endpoint on that side. Graph: Again, we use the same type of notation on the endpoint as we did in the interval notation, a boxed end this time. Since we needed to indicate all values greater than or equal to -3/2, the part of the number line that was to the right of -3/2 was darkened. \| \| \| \| Solving a Compound Inequality \| \| \| \| A compound linear inequality is one that has two inequalities in one problem. For example, 5 < x + 3 < 10 or -1 < 3x < 5. You solve them exactly the same way you solve the linear inequalities shown above, except you do the steps to three "sides" (or parts) instead of only two. \| \| \| \| Example 9: Solve, write your answer in interval notation and graph the solution set: . View a video of this example \| \| \| \| This is an example of an compound inequality \| \| \| \| --- \| \| Interval notation: Graph: \| Inv. of add. 2 is sub. by 2 Apply steps to all three parts All values between -6 and 8, with a closed interval at -6 (including -6) Visual showing all numbers between -6 and 8, including -6 on the number line. \| \| \| \| Interval notation: This time we have a mixed interval since we are including where it is equal to -6, but not equal to 8. x is between -6 and 8, including -6, so -6 is our smallest value of the interval so it goes on the left and 8 goes on the right. The boxed end on -6 indicates a closed interval on that side. A curved end on 8 indicates an open interval on that side. Graph: Again, we use the same type of notation on the endpoints as we did in the interval notation, a boxed end on the left and a curved end on the right. Since we needed to indicate all values between -6 and 8, including -6, the part of the number line that is in between -6 and 8 was darkened. \| \| \| \| Solving an Absolute Value Inequality \| \| \| \| Step 1: Isolate the absolute value expression. \| \| \| \| If there is a constant that is on the same side of the inequality that the absolute value expression is but is not inside the absolute value, use inverse operations to isolate the absolute value. \| \| \| \| Step 2: Use the definition of absolute value to set up the inequality without absolute values. \| \| \| \| A quick reminder, the absolute value measures the DISTANCE a number is away from the origin (zero) on the number line. No matter if the number is to the left (negative) or right (positive) of zero on the number line, the DISTANCE it is away from zero is going to be positive. Hence, the absolute value is always positive, (or zero if you are taking the absolute value of 0). If you need a review on absolute values, feel free to go to Tutorial 21: Absolute Value Equations. \| \| \| \| --- \| \| If d is POSITIVE and \|x\| < d, then -d < x < d The graph below illustrates all the values on the number line whose distance would be less than d units away from 0. It shows us why we set up the inequality, shown above, the way we do. \| If d is NEGATIVE and \|x\| < d, then there is no solution The absolute value is always positive, and any positive number is greater than any negative number, therefore it would be no solution. \| \| \| \| --- \| \| \| \| \| \| \| --- \| \| If d is POSITIVE and \|x\| > d, then x < -d OR x > d The graph, shown below, illustrates all the values on the number line whose distance would be greater than d units away from 0. It shows us why we set up the inequality, shown above, the way we do. \| If d is NEGATIVE and \|x\| > d, then x is all real numbers The absolute value is always positive, and any positive number is greater than any negative number, therefore all real numbers would work. \| \| \| \| \| \| \| \| Step 3: Solve the linear inequalities set up in step 2. \| \| \| \| You will solve these linear inequalities just like the ones shown above. \| \| \| \| Example 10: Solve, write your answer in interval notation and graph the solution set: . View a video of this example \| \| \| \| Step 1: Isolate the absolute value expression. \| \| \| \| The absolute value expression is already isolated. \| \| \| \| Step 2: Use the definition of absolute value to set up the inequality without absolute values. AND Step 3: Solve the linear inequalities set up in step 2. \| \| \| \| The distance that the expression x - 4 is away from the origin needs to be less than 7. All numbers between -7 and 7 are less than 7 units away from the origin. So, the expression x - 4 needs to be between -7 and 7. \| \| \| \| --- \| \| Interval notation: Graph: \| Inv. of sub. 4 is add. by 4 Apply steps to all three parts All values between -3 and 11 Visual showing all numbers between -3 and 11 \| \| \| \| Interval notation: This time we have an open interval since we are not including either endpoint. x is between -3 and 11, so -3 is our smallest value of the interval so it goes on the left and 11 goes on the right. The curved end on both numbers indicate an open interval on both sides. Graph: Again, we use the same type of notation on the endpoints as we did in the interval notation, a curved end on both ends. Since we needed to indicate all values between -3 and 11, the part of the number line that is in between -3 and 11 was darkened. \| \| \| \| Example 11: Solve, write your answer in interval notation and graph the solution set: . View a video of this example \| \| \| \| Step 1: Isolate the absolute value expression. \| \| \| \| The absolute value expression is already isolated. \| \| \| \| Step 2: Use the definition of absolute value to set up the inequality without absolute values. AND Step 3: Solve the linear inequalities set up in step 2. \| \| \| \| Be careful, since the absolute value (the left side) is always positive, and positive values are always greater than negative values, the answer is no solution. There is no value that we can put in for x that would make this inequality true. \| \| \| \| Example 12: Solve, write your answer in interval notation and graph the solution set: . View a video of this example \| \| \| \| Step 1: Isolate the absolute value expression. \| \| \| \| --- \| \| \| Inv. of add. 1 is sub. 1 Abs. value exp. isolated \| \| \| \| Step 2: Use the definition of absolute value to set up the inequality without absolute values. AND Step 3: Solve the linear inequalities set up in step 2. \| \| \| \| The distance that the expression(7 - 2y)/2 is away from the origin needs to be greater than or equal to 4. All numbers that are less than or equal to - 4 OR greater than or equal to 4 are greater than or equal to 4 units away from the origin. So the expression (7 - 2y)/2 needs to be less than or equal to - 4 OR greater than or equal to 4. \| \| \| \| --- \| \| OR Interval notation: Graph: \| First inequality, where it is less than or = to -4 Inv. of div. by 2 is mult. by 2 Inv. of mult. by -2 is div. by -2, so reverse inequality sign Second inequality, where it is greater than or = to 4 Inv. of div. by 2 is mult. by 2 Inv. of mult. by -2 is div. by -2, so reverse inequality signs All values less than or = to -1/2 or greater than or = to 15/2 Visual showing all numbers less than or = to -1/2 or greater than or = to 15/2 \| \| \| \| Interval notation: This time we have two closed intervals since we are including the endpoints -1/2 and 15/2. In the first interval, y is less than or equal to -1/2, so -1/2 is our largest value of the interval so it goes on the right. Since there is no lower endpoint of that first interval, we put negative infinity on the left side. The boxed end on -1/2 indicates a closed interval. Infinity always has a curved end because there is not an endpoint on that side. In the second, interval, y is greater than or equal to 15/2, so 15/2 is our smallest value of the interval so it goes on the left. Since there is no upper endpoint of that second interval, we put the infinity symbol on the right side. The boxed end on 15/2 indicates a closed interval. Infinity always has a curved end because there is not an endpoint on that side. Graph: Again, we use the same type of notation on the endpoints as we did in the interval notation, a boxed end on both y = -1/2 and y = 15/2. Since we needed to indicate all values less than or equal to -1/2 OR greater than or equal to 15/2, the parts of the number line that are to the left of -1/2 and to the right of 15/2 were darkened. \| \| \| \| Example 13: Solve, write your answer in interval notation and graph the solution set: . View a video of this example \| \| \| \| Step 1: Isolate the absolute value expression. \| \| \| \| The absolute value expression is already isolated. \| \| \| \| Step 2: Use the definition of absolute value to set up the inequality without absolute values. AND Step 3: Solve the linear inequalities set up in step 2. \| \| \| \| Again, be careful - since the absolute value (the left side) is always positive, and positive values are always greater than negative values, the answer is all real numbers. No matter what value you plug in for x, when you take the absolute value the left side will be positive. All positive numbers are greater than -2. \| Practice Problems --- \| \| \| These are practice problems to help bring you to the next level. It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. In fact there is no such thing as too much practice. To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer. \| Practice Problems 1a - 1c: Solve, write your answer in interval notation and graph the solution set. \| \| \| --- \| \| 1a. (answer/discussion to 1a) \| 1b. (answer/discussion to 1b) \| \| \| \| 1c. (answer/discussion to 1c) \| Practice Problems 2a - 2d: Solve, write your answer in interval notation and graph the solution set. \| \| \| --- \| \| 2a. (answer/discussion to 2a) \| 2b. (answer/discussion to 2b) \| \| \| \| --- \| \| 2c. (answer/discussion to 2c) \| 2d. (answer/discussion to 2d) \| Need Extra Help on these Topics? --- \| \| \| --- \| \| The following are webpages that can assist you in the topics that were covered on this page: \| \| \| This website helps you with linear inequalities. This website helps you with linear inequalities. This website helps you with linear inequalities. \| Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions. \| \| \| \| \| --- WTAMU > Virtual Math Lab > College Algebra Videos at this site were created and produced by Kim Seward and Virginia Williams Trice. Last revised on Dec. 17, 2009 by Kim Seward. All contents copyright (C) 2002 - 2010, WTAMU and Kim Seward. 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16296	https://www.sciencedirect.com/topics/agricultural-and-biological-sciences/apical-dominance	Apical Dominance - an overview \| ScienceDirect Topics Skip to Main content Journals & Books Apical Dominance In subject area:Agricultural and Biological Sciences Apical dominance is defined as the inhibitory control that the apical portions of the shoot exert over the growth of lateral buds below. This phenomenon can be demonstrated by the release of inhibition when the main vegetative shoot apex is removed, leading to the growth of the lateral buds. AI generated definition based on:Plant Growth and Development: Hormones and Environment, 2002 How useful is this definition? Press Enter to select rating, 1 out of 3 stars Press Enter to select rating, 2 out of 3 stars Press Enter to select rating, 3 out of 3 stars About this page Add to MendeleySet alert Also in subject area: Neuroscience Discover other topics 1. On this page On this page Definition Chapters and Articles Related Terms Recommended Publications Featured Authors On this page Definition Chapters and Articles Related Terms Recommended Publications Featured Authors Chapters and Articles You might find these chapters and articles relevant to this topic. Chapter Apical Dominance and Some Other Phenomena Illustrating Correlative Effects of Hormones 2002, Plant Growth and Development: Hormones and EnvironmentLalit M. Srivastava APICAL DOMINANCE IS A COMPLICATED PHENOMENON 319 1.1. Apical Dominance Varies with Species and Cultivars 319 1.2. Environmental Factors Affect Apical Dominance 319 1.3. Apical Dominance Changes with Developmental Status 320 View chapterExplore book Read full chapter URL: Book 2002, Plant Growth and Development: Hormones and EnvironmentLalit M. Srivastava Chapter Apical Dominance and Some Other Phenomena Illustrating Correlative Effects of Hormones 2002, Plant Growth and Development: Hormones and EnvironmentLalit M. Srivastava 1.APICAL DOMINANCE IS A COMPLICATED PHENOMENON Even though apical dominance can be defined in simple terms, it is a complicated process in which the genetic makeup of the plant, environmental factors, and developmental and hormonal signals all play a part. Most interestingly, although apical dominance is a phenomenon typically seen in shoots, roots have an important role in its regulation. Apical dominance determines the branching pattern and the shape of plants, but it does so not by regulating the initiation of lateral buds, but whether they subsequently grow out. Initiation of lateral buds, as explained in Chapter 1, is related to the activity of the shoot apical meristem and has little to do with their subsequent growth. Apical dominance is best seen in herbaceous plants and in trees in their first year of growth. In older trees, the phenomenon becomes more complex, being influenced by shading and resource allocation, but can still be seen in branches in their current years’ growth. Apical dominance has been better studied in herbaceous plants, and, unless otherwise stated, the following account pertains mostly to them. The state of arrest or quiescence of lateral buds is not to be confused with bud “dormancy,” which has an entirely different physiological basis. Bud dormancy applies to terminal as well as lateral buds and, in plants native to temperate and colder regions of earth, is regulated by temperature and possibly photoperiod. 1.1.Apical Dominance Varies with Species and Cultivars Apical dominance is genetically based and varies with species and cultivars. Some species show strong apical dominance [e.g., Helianthus annuus (sunflower), Tradescantia sp.]. Field-grown sunflower plants, 4 feet tall, may show little or no lateral branching in intact plants. In a nursery or greenhouse, by suitable gibberellin treatment, they can be made to grow 7-8 feet tall and still show no branching. Other species show an intermediate or partial apical dominance, i.e., there is some branching [e.g., Phaseolus vulgaris (common bean), Vicia faba (broad bean), Pisum sativum (pea), Ipomoea nil (Japanese morning glory)], whereas still others show very weak or no apical dominance— they show substantial and continuing branching in intact plants (e.g., Coleus , Arabidopsis). 1.2.Environmental Factors Affect Apical Dominance Various environmental factors, such as availability of nutrients, proximity of neighbors, light quantity, and length of photoperiod, regulate bud outgrowth and affect apical dominance. Experiments with Coleus, a plant with weak apical dominance, show that plants grown under low light conditions are much less branched, whereas those grown under high irradiance are branched and bushier (Fig. 14-20). Sign in to download full-size image FIGURE 14-20. Coleus plants grown under low light indoors (left) and under high-light intensity in a greenhouse (right). With permission from Cline (1996). Experiments with tall varieties of pea plants show that short photoperiods cause them to be bushier and more branched than long photoperiods. In pea, long photoperiods also promote a change from vegetative growth to flowering. 1.3.Apical Dominance Changes with Developmental Status Several developmental cues, including distance of the lateral bud from the terminal apex and a change to reproductive phase, affect bud outgrowth. In herbaceous forms with strong to moderate apical dominance, inhibition is greater for the lateral buds closer to the apex. As the distance from the apex increases, inhibition lessens and lateral buds are released from dominance. Thus, branching becomes more common in lower internodes. In contrast, in some other plants, the lower, older nodes may remain free of branching, whereas the younger nodes may show lateral outgrowths. Young trees of many species, e.g., Fraxinus (ash), Acer (sycamore), Alnus (alder), and Ginkgo (a gymnosperm), show strong apical dominance, but as the tree ages, lateral buds begin to grow and the tree becomes much more branched. The change to the reproductive phase brings about major changes in branching patterns in many plants. For example, lateral buds that were inhibited from growing in the vegetative phase may be released from inhibition with the onset of flowering (e.g., oat, Perilla). Fruiting and seed set may reimpose the inhibition (e.g., Phaseolus). Show more View chapterExplore book Read full chapter URL: Book 2002, Plant Growth and Development: Hormones and EnvironmentLalit M. Srivastava Chapter Apical Dominance and Some Other Phenomena Illustrating Correlative Effects of Hormones 2002, Plant Growth and Development: Hormones and EnvironmentLalit M. Srivastava 8.SECTION SUMMARY Apical dominance is a complex phenomenon, which is determined genetically. Some plants show strong apical dominance, others show moderate, and still others show weak dominance or none. In herbaceous plants that have been studied, apical dominance has been thought to be regulated by the relative concentrations of IAA and cytokinins at the bud site. The polar basipetal transport of IAA in the shoot and cytokinins transported upward in the xylem stream from roots play important roles in this phenomenon, but how the two hormones regulate bud quiescence or release from arrest is not clear. Once release has been affected, application of auxins as well as gibberellins promotes bud growth. Studies utilizing branching mutants and transgenic plants expressing altered levels of IAA or cytokinins generally indicate that the branching patterns, in the first analysis, are controlled by the genetic makeup and the developmental program of the plant. They also suggest that some factor other than cytokinin may be interacting with auxin in regulation of branching. A plant as an integrated whole is also subject to various environmental stimuli, such as photoperiod and proximity to neighbors, and resource allocation. There is a correlation between apical dominance of a shoot and a strong polar basipetal transport of IAA. Because apical dominance can be lost (by a dominant shoot on decapitation) or acquired (by a subordinate shoot), it suggests that the auxin efflux carriers, which are responsible for polar transport, can be lost from or randomized in the plasma membrane, and also selectively relocated at the basal ends of the plasma membrane of transport-competent cells. Moreover, the polar flow of IAA may itself provide the inducing signal for such relocation of efflux carriers. View chapterExplore book Read full chapter URL: Book 2002, Plant Growth and Development: Hormones and EnvironmentLalit M. Srivastava Chapter Apical Dominance and Some Other Phenomena Illustrating Correlative Effects of Hormones 2002, Plant Growth and Development: Hormones and EnvironmentLalit M. Srivastava SECTION V.APICAL DOMINANCE Apical dominance is defined as an inhibitory control exercised by the apical portions of the shoot over the growth of the lateral buds below. The phenomenon is easily demonstrated by decapitation. If the main vegetative shoot apex of a plant such as Helianthus (sunflower) is cut off, inhibition is released and the buds below respond by growing out (Fig. 14-19). Indeed, any mechanical wounding or chemical treatment that injuries the main apex causes lateral buds to grow. Sign in to download full-size image FIGURE 14-19. Apical dominance in Helianthus annuus (mammoth gray-striped sunflower). Plant showing lateral bud outgrowth approximately 1 week after decapitation. Courtesy of Morris Cline, Ohio State University, Columbus, OH. View chapterExplore book Read full chapter URL: Book 2002, Plant Growth and Development: Hormones and EnvironmentLalit M. Srivastava Review article Regulation of Shoot and Root Development through Mutual Signaling 2012, Molecular PlantJérôme Puig, ... Pascal Gantet APICAL DOMINANCE: A CASE STUDY OF SYSTEMIC NETWORKS Apical dominance refers to the inhibition of axillary bud growth by the active shoot tip. Apical dominance can be abolished by removing the shoot tip, which triggers the growth of one to several axillary buds. Recent studies have revealed that control of apical dominance involves various signaling molecules and occurs in different parts of the plant, including the root system. Early evidence regarding a role for auxin in apical dominance was provided by decapitation experiments followed by replacement of the shoot tip with an auxin-containing agar block. The artificial auxin source was able to restore dominance on the remaining axillary buds (Thimann and Skoog, 1933, 1934). Primarily synthesized in the shoot apex and young leaves, auxin is transported to other parts of the shoot and roots by a set of transporters, which includes the influx facilitator AUXIN INFLUX CARRIER PROTEIN1 (AUX1)/LIKE-AUX (LAX), p-GLYCOPROTEIN AUXIN EFFLUX CARRIERS (PGP), and PIN-FORMED AUXIN EFFLUX CARRIERS (PIN) (Blakeslee et al., 2005; Paponov et al., 2005; Geisler and Murphy, 2006). Though auxin is an essential signal for apical dominance, several experiments have shown that it does not directly inhibit axillary bud outgrowth on its own (Gocal et al., 1991; Booker et al., 2003). Analyses of mutants with aberrant regulation of bud outgrowth in Arabidopsis, pea, petunia, and rice have only recently deciphered a link between auxin and shoot branching (Beveridge et al., 1996; Napoli, 1996; Sorefan et al., 2003; Booker et al., 2004; Booker et al., 2005; Ishikawa et al., 2005). These mutants exhibit reduced height and a bunchy branching pattern. In A. thaliana, mutants exhibiting this phenotype have been identified as MORE AXILLARY GROWTH (MAX). MAX3 and MAX4 genes encode the CAROTENOID CLEAVAGE DIOXYGENASEenzymes CDD7 and CDD8, respectively (Sorefan et al., 2003; Booker et al., 2004). MAX1 encodes a cytochrome P450 enzyme (Booker et al., 2005). All of these enzymes are required for the biosynthesis of strigolactones from β-carotene (Booker et al., 2004; Schwartz et al., 2004; Booker et al., 2005; Auldridge et al., 2006). Interestingly, grafting of max1, max3, and max4 shoots onto WT (wild-type) rootstocks restores the WT shoot development phenotype (Turnbull et al., 2002; Sorefan et al., 2003; Booker et al., 2004). This suggests that strigolactones act as an inhibitory signal that is produced in the roots and can move acropetally to inhibit the growth of axillary buds. Expression of MAX4 in the root tip is down-regulated in response to shoot tip removal and is restored when the shoot tip is replaced by an artificial auxin source (Sorefan et al., 2003; Bainbridge et al., 2005). This indicates that strigolactone biosynthesis is stimulated in roots in response to auxin synthesized in the shoot tip. A fourth gene involved in axillary bud outgrowth, MAX2, encodes a conserved F-box protein that may contribute to the strigolactone signal transduction pathway (Stirnberg et al., 2002). MAX2 is primarily expressed in shoots and is a potential receptor for the strigolactone signal at the level of axillary buds (Stirnberg et al., 2007). CKs are also involved in regulating axillary bud outgrowth. CKs are known to promote axillary bud outgrowth (Sachs and Thimann, 1967; Li and Bangerth, 2003; Nordstrom et al., 2004). CKs are synthetized in the shoots and roots and are transported from root-to-shoot via the xylem (Chen et al., 1985). Auxin inhibits CK biosynthesis, but this inhibition is abolished by mutation of the AUXIN RESPONSE1 (AXR1) gene, which encodes an F-box protein that acts upstream of the auxin signal transduction pathway (Nordstrom et al., 2004). Axr1 mutants exhibit a bushy phenotype, which indicates a reduction in apical dominance (Stirnberg et al., 1999). It has been shown that removal of shoot tips induces translocation of CKs from roots and their accumulation in axillary buds; however, placing a source of auxin at the decapitated shoot tip prevents this CK accumulation (Li and Bangerth, 2003). This suggests that auxin produced in shoot tips inhibits root CK biosynthesis and translocation to axillary buds. Interestingly, max mutants, which exhibit enhanced axillary bud outgrowth, possess reduced levels of CKs in xylem sap relative to WT plants; this suggests that a feedback signal may also be produced by active axillary buds and inhibit CK biosynthesis in roots (Foo et al., 2007). It is likely that such interplay between auxin, strigolactone, and CK ensures developmental homeostasis of shoot branching via axillary bud growth control. The fact that MAX orthologous genes have also been functionally characterized in pea, petunia, and rice indicates that this pathway is conserved across plant species. Strigolactone-mediated control of axillary bud outgrowth is highlighted in orange in Figure 1. Sign in to download full-size image Figure 1. A Schematic Representation of Key Elements of the Local and Systemic Signaling Pathways between Shoot and Root and Their Hypothetical Interactions. Arrows represent positive regulatory actions of one element of the network on another. A T-shaped line represents a negative regulatory action of one element of the network on another. Dashed arrows represent the translocation of one molecule from one part of the plant to another one. Common systemic pathways involving auxin, cytokinins, and sucrose that are involved in different constitutive and adaptive developmental pathways are depicted in black. Elements specific to the MAX/strigolactone signaling pathway are depicted in orange. Elements specific to the bypass pathway are depicted in red. Elements specific to the nitrate responsive pathway are depicted in blue. Elements specific to the phosphate-responsive pathway are depicted in green. ANR1, Arabidopsis Nitrate Regulated 1; ARF8, Auxin Response Factor 8; ARF19, Auxin Response Factor 19; AUX, auxin; BSP, Bypass1-related signal; BSP1, bypass1; CK, cytokinins; CRE1, Cytokinin Response Element 1; GLU, glutamate; MAX1, More AXillary growth 1; MAX2, More AXillary growth 2; MAX3, More AXillary growth 3; MAX4, More AXillary growth 4; MYB62, Myeloblastosis 62; N, nitrate; NRT1.1, Nitrate Transporter 1.1; P, phosphate; PHO2, phosphate deficient 2; PHO3: phosphate deficient 3; PHR1, Phosphate Starvation Response 1; PHR2, Phosphate Starvation Response 2; RAM, Root Apical Meristem; SAM, Shoot Apical Meristem; SCR, Scarecrow; STRG, strigolactone; SUC, sucrose; SUC2, Sucrose Transporter 2; ZAT6, Zinc finger of Arabidopsis thaliana 6. Systemic signals are: AUX, BSP, CK, GLU, miR399, STRG, SUC; other components of the diagram are involved in local production or response to these systemic signals or involved in local N and P signal transduction pathways involved in adaptive root development. Show more View article Read full article URL: Journal2012, Molecular PlantJérôme Puig, ... Pascal Gantet Chapter Apical Dominance and Some Other Phenomena Illustrating Correlative Effects of Hormones 2002, Plant Growth and Development: Hormones and EnvironmentLalit M. Srivastava 1.1.Apical Dominance Varies with Species and Cultivars Apical dominance is genetically based and varies with species and cultivars. Some species show strong apical dominance [e.g., Helianthus annuus (sunflower), Tradescantia sp.]. Field-grown sunflower plants, 4 feet tall, may show little or no lateral branching in intact plants. In a nursery or greenhouse, by suitable gibberellin treatment, they can be made to grow 7-8 feet tall and still show no branching. Other species show an intermediate or partial apical dominance, i.e., there is some branching [e.g., Phaseolus vulgaris (common bean), Vicia faba (broad bean), Pisum sativum (pea), Ipomoea nil (Japanese morning glory)], whereas still others show very weak or no apical dominance— they show substantial and continuing branching in intact plants (e.g., Coleus , Arabidopsis). View chapterExplore book Read full chapter URL: Book 2002, Plant Growth and Development: Hormones and EnvironmentLalit M. Srivastava Chapter Physiology of the tea plant 2022, Scientific Perspectives of Tea Plant Horticulture and ProductivityL. Manivel PHD (UC DAVIS, CA., USA) 6.4.1 Overcoming apical dominance The apical dominance is broken by removing the tip, breaking the stem, pegging, debudding the main stem, or by using plant growth promoters like gibberellins. In weak plants and seedlings, induction of laterals is a bit difficult and erratic due to the genetic complex, where all these measures are used to induce lateral bud break, breaking the apical dominance. Tipping the shoots periodically induces branching and spread of the frame, building up the top hamper/canopy. By raising the tipping height (35, 40, 45, and 55 cm) yearly, the plucking table is formed at 90 cm a convenient height for the women workers to harvest the crop. Formative pruning is done in the third or fourth year depending on the growth, when final cleaning, desnagging, and opening the center, fine-tuning the frame. View chapterExplore book Read full chapter URL: Book 2022, Scientific Perspectives of Tea Plant Horticulture and ProductivityL. Manivel PHD (UC DAVIS, CA., USA) Chapter Vegetative Growth 2008, Physiology of Woody Plants (Third Edition) Dr.Stephen G. Pallardy Apical Dominance As mentioned in Chapter 2, the terminal leader of most gymnosperms elongates more than the lateral branches below it. This produces a more or less conical tree form, often described as having excurrent branching. A few gymnosperms lack strong apical dominance. For example, second-order branches of Norfolk Island pine apparently lack the inherent capacity to assume apical dominance, and removal of the apical shoot is not followed by formation of a new leader by one of the lateral branches as occurs in pine and spruce. Rooted cuttings of this species may grow horizontally for many years. Although many pines exhibit strong apical dominance for a long time, some species, such as Italian stone pine, often lose apical dominance rather early in life (Fig. 3.13). Sign in to download full-size image FIGURE 3.13. Ten-year-old Italian stone pine tree showing lack of apical dominance. Photo courtesy of A. De Phillips.Copyright © 1964 The physiological mechanism of apical dominance, an example of paradormancy, is not well understood. It is clear that apical dominance has both genetic and developmental influences, as species vary in manifestation of the phenomenon and respond to release from dominance by apical buds differently depending on age (Srivistava, 2001). Polar basipetal transport of auxin in the stem and cytokinins transported in xylem sap likely play significant roles maintaining or overcoming bud quiescence, respectively, but exactly how is not known. The occurrence of apical dominance is very important to foresters. When apical dominance is destroyed by invasion of the terminal leader of eastern white pine by the white pine weevil (Pissodes strobi), one of the lateral shoots of the first false-whorl eventually assumes dominance while other shoots in the same whorl are suppressed. However, because of competition among branches, considerable time elapses before one of the false-whorl shoots establishes dominance and others are suppressed. Meanwhile the tree is degraded as a potential source of logs because of the fork that develops in the stem as a result of the injury to the terminal leader. Loss of apical dominance in branches often is fostered by Christmas tree growers (Chapter 7, Kozlowski and Pallardy, 1997). Many conifers have long internodes in the main stem and branches and these give the tree a spindly appearance. By “shearing” or cutting back current-year shoots or by debudding shoots, Christmas tree growers inhibit shoot elongation and stimulate expansion of dormant buds as well as formation and expansion of new buds. Thus new lateral shoots form along branches and produce densely-leaved, high-quality Christmas trees. Show more View chapterExplore book Read full chapter URL: Book 2008, Physiology of Woody Plants (Third Edition) Dr.Stephen G. Pallardy Chapter ROOTSTOCKS \| Scion–Rootstock Relationships 2003, Encyclopedia of Rose ScienceD.P. de Vries Apical Dominance Apical dominance means that the growing apex of a plant controls the quiescence of buds sitting in lower positions along a shoot. This kind of bud inhibition is termed ‘correlative inhibition’. Correlative inhibition usually has a gradient: the lower on the shoot, the stronger the inhibition. In plants with strong apical dominance, the growth is very upright, with one dominant central axis, as for example in firs. Such trees are called ‘excurrent’. Low apical dominance is seen in the bushy growth of many shrubs, which are indicated as ‘decurrent’. Correlative inhibition is explained by the action of auxin, mainly indole-3-acetic acid, which is synthesized by the growing shoot apex. Auxin, which migrates slowly via the phloem in a basipetal direction, inhibits the sprouting of axillary buds sitting lower on the shoot. The correlative inhibition of trees is usually lifted after taking away a shoot's apex, or after bending the shoot into a horizontal position. In roses too, the habitus of plants is controlled by correlative inhibition. An important difference from trees is that correlative inhibition is only active as long as the apices of growing shoots produce auxin. When left unpruned (unharvested), auxin synthesis in a rose shoot terminates whenever the shoot apex transcends from the vegetative to the generative phase, which occurs when the apically placed flower bud is initiated. Correlative inhibition of a rose shoot is also lifted after decapitation. It is likely that, in contrast to the more bushy garden-rose varieties, apical dominance in the upright-growing cut roses is strongest. It has been shown in various ways that shoot yield in cut roses is determined by degree of bud-break. When grafted on the same rootstock, scion varieties show large genotypic variation for the number of bottom-breaks as well as for the number of axillary buds that break after each harvest. This means that genotypic variation for bud-break may be explained by a genotypic variation for apical dominance. However, there are differences in induced shoot yield of various rootstocks. Because rootstocks clearly affect bud-break of the scion variety, the question is: how do they manage? View chapterExplore book Read full chapter URL: Reference work 2003, Encyclopedia of Rose ScienceD.P. de Vries Review article Shoot branching 2004, Current Opinion in Plant BiologySally P Ward, Ottoline Leyser Apical dominance, the suppression of axillary meristem growth by a growing apical meristem, is a central mechanism in the regulation of plant shoot architecture . Dormant axillary buds are available to replace growing shoots if the apex is damaged or removed through disease, herbivore grazing or pruning. Classical decapitation studies in many plant species (first described in ) demonstrate that the removal of the growing shoot apex, a major site for auxin biosynthesis, allows the growth of dormant axillary buds. Addition of the plant hormone auxin to the cut stump reimposed the inhibition of bud growth, identifying auxin as the likely mediator of apical dominance. In Arabidopsis, excised nodal assays demonstrate a similar phenomenon, with apically applied auxin inhibiting axillary meristem outgrowth . The importance of auxin in apical dominance is supported by transgenic studies in which the endogenous levels of auxin are altered. Increasing auxin levels reduces branching whereas decreasing auxin levels increases branching . Although the role of auxin in apical dominance and thus the regulation of shoot branching is undeniable, several studies have shown that this effect is indirect; it involves controlling the concentrations or location of cytokinin and an as yet unidentified branch-inhibiting substance. New data regarding the identity of this substance will be discussed later. View article Read full article URL: Journal2004, Current Opinion in Plant BiologySally P Ward, Ottoline Leyser Related terms: Spadix Auxin Cytokinin Abscisic Acid Indole-3-Acetic Acid Strigolactone Plant Growth Meristem Axillary Bud Arabidopsis View all Topics Recommended publications Forest Ecology and ManagementJournal Scientia HorticulturaeJournal Current Opinion in Plant BiologyJournal Industrial Crops and ProductsJournal Browse books and journals Featured Authors Beveridge, Christine AnneThe University of Queensland, Brisbane, Australia Citations12,320 h-index55 Publications14 Wang, ZhanbiaoChinese Academy of Agricultural Sciences, Beijing, China Citations1,774 h-index25 Publications42 Nanba, KenjiFukushima University, Fukushima, Japan Citations1,718 h-index23 Publications44 Yoschenko, Vasyl I.Fukushima University, Fukushima, Japan Citations1,398 h-index22 Publications44 About ScienceDirect Remote access Advertise Contact and support Terms and conditions Privacy policy Cookies are used by this site. 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16297	https://www.ae.msstate.edu/tupas/SA2/chA14.7_text.html	Section III.4 S ection III.4 S hear F low A nalysis for S ymmetric B eams Shear Flow Calculation: To calculate the shear flow over a section of interest we must have the value of transverse shear force V that is acting along a principal axis. This force is either given or should be obtained from the shear diagram. Then we need to have the moment of inertia about an axis that is perpendicular to the direction of the transverse shear force. For example, if V is along y, we need to have I z, or if V is along z, we need to calculate I y. With V z/I y or V y/I z known, we calculate the first moment of area, Q. If V is along the y direction, we need to calculate Q about the z axis. To do this a segment of the cross section is isolated from the rest, and its moment about the z axis is calculated. The way we isolate a segment is by cutting it perpendicular to its thickness. We will see how this is done in the example problems at the end of this section. An example of a shear flow diagram is shown below. Notice that in this example, the transverse shear load is in the vertical direction. Thus, the moment of inertia about the horizontal centroidal axis is used for the calculation of shear flow. The shear flows along the top flange and the web are calculated as The direction of shear flow has to be consistent with that of the resultant shear force, in this case V y. We can also calculate the shear force acting in each member. To do this, we simply integrate the shear flow along each member. Key Observation: If we examine the internal shear force distribution we observe the following facts. Since the resultant shear force is in vertical direction, the two flange forces (in horizontal direction) will add up to zero as they should. However, if we compare the force in the vertical web, F 2 to V y we find that they are not the same. Although this may indicate that force equilibrium is violated, the fact is otherwise. We know from previous discussion that shear stress at a given point is represented in terms of its two perpendicular components. We also said that usually one component is much larger than the other. In this case the vertical component of shear stress in top and bottom flanges is much smaller than its horizontal component but not necessarily zero. If we were to consider the vertical component of shear stress in the two flanges and calculate the corresponding shear force, we will find that they added with F 2 will be equal to V y. The reason we don't go through the trouble of calculating these forces is because they are much smaller than F 2. In fact we can see that if b >> t, then F 2 is > 90% of V y. It is important to know this fact when doing bending analysis under transverse loads. Shear Center Calculation: The shear center is found using moment equilibrium. We show the resultant transverse shear force acting at the shear center which is at some distance e from the point of reference, usually the centroid of the cross section. We then write the moment produced by the resultant shear force V, set it equal to the sum of moments produced by individual internal force components, F 1, F 2 and F 3, and solve for the unknown distance, e z in this case. Paying attention to the direction of each moment, we can write Beam Sections Not Loaded Through the Shear Center: If the applied shear force does not pass through the shear center, it will force the beam to twist as it bends. This eccentricity produces a torque, that will cause an additional shear flow and shear stress. The analysis used for torsion of beams with open cross sections (I.4) can be used here to find the constant shear flow and corresponding shear stress at a desired point on the cross section. In analysis of such sections, the shear force is replaced by an equivalent force-couple at the shear center. The final shear stress diagram will be the superposition of the shear stress with the shear force passing through the shear center,and the shear stress induced by the associated torque about the shear center. An example of this kind of loading is shown below. Note that in this case, the maximum shear stress occurs at a point on the neutral axis which is on the left edge of the vertical flange. Shear Force in Fasteners: In many applications, beam sections consist of several pieces of material that are attached together in a number ways: bolts, rivets, nails, glue, weld, etc. In such so called built-up sections we are interested in knowing the amount of shear stress and the resulting shear force at the cross section of fasteners or over the glued surface. The figure shown below gives an example of two rectangular members that are attached by means of mechanical fasteners. In this case, we want to know the amount of shear stress as well as shear force carried by each fastener. The fasteners are spaced evenly at a distance of s. Each fastener has a cross-sectional area denoted by A f. Note that the surface of contact or joint is treated as a frictionless surface. Therefore, the shear flow is carried entirely by the fasteners. If at a given section, there are more than one fastener, the shear flow will maintain the same value, but the shear force and shear stress will change depending on the number and size of fasteners used. For example, if at a given section there are two identical fasteners as shown below, then the force in each is found as shown below. E XAMPLE P ROBLEMS Example 1 Shear flow distribution, shear center location, and max shear stress calculation for a simply-supported beam with doubly symmetric cross section Example 2 Shear flow distribution, shear center location, and max shear stress calculation for a simply-supported beam with a symmetric cross section Example 3 Shear flow distribution, shear center location, and max shear stress calculation for a simply-supported beam with a symmetric cross section. The transverse shear force does not pass through the shear center To Section III.5 To Section III.3 To Index Page of Transverse Shear Loading of Open Sections
16298	https://www.albert.io/blog/what-is-meiosis-ap-biology-review/	Skip to content ➜ AP® Biology What is Meiosis: AP® Biology Review The Albert Team Last Updated On: What We Review Introduction Meiosis is a fundamental concept in AP® Biology that students often find challenging but essential to master. Understanding this process will help you grasp how genetic information is passed from one generation to the next. In this review, we’ll explore what meiosis is, its purpose, and why it’s such a critical factor in sexual reproduction. By the end of this article, you’ll have a clearer picture of its role, especially in comparison to mitosis. Start Practicing AP® Biology on Albert now! What is Meiosis? Meiosis is a specialized type of cell division that reduces the chromosome number by half, creating four haploid cells from one diploid cell. This process is necessary for sexual reproduction and ensures that offspring have the correct number of chromosomes. Unlike mitosis, which primarily deals with growth and tissue repair, meiosis introduces genetic variation and produces gametes (egg and sperm cells in animals; pollen and ovules in plants) that unite during fertilization. The Purpose of Meiosis Formation of Haploid Gametes Meiosis creates sex cells (gametes) with half the number of chromosomes found in somatic (body) cells. When these gametes fuse, they form a zygote with a complete set of chromosomes. Ensuring Genetic Variation During meiosis, crossing over and independent assortment mix genetic material in ways that create genetically unique gametes. This diversity is vital for populations to adapt and evolve. Meiosis vs. Mitosis To better understand meiosis, it helps to compare it to mitosis, another form of cell division. Similarities Both involve the segregation of chromosomes. Both go through phases of prophase, metaphase, anaphase, and telophase. Differences Number of Daughter Cells Mitosis produces two diploid daughter cells; meiosis produces four haploid daughter cells. Genetic Content Mitosis forms genetically identical cells for growth and repair, while meiosis forms genetically diverse cells for reproduction. Meiosis Stages This process is divided into two rounds of cell division: A. Meiosis I Sometimes called the “reduction division” because it reduces the chromosome number by half. Homologous chromosomes pair up, allowing crossing-over to occur. B. Meiosis II Similar to mitosis but starts with haploid cells. Sister chromatids separate, creating four distinct haploid cells by the end. Ready to boost your AP® Biology scores? Explore our plans and pricing here! Detailed Breakdown of Each Stage It’s useful to break down the key steps within each sub-phase of meiosis. Meiosis I Prophase I Homologous chromosomes pair up (synapsis). Crossing-over occurs, exchanging genetic material between non-sister chromatids. Metaphase I Homologous chromosome pairs align along the metaphase plate. Independent assortment: chromosomes line up in random orientations. Anaphase I Homologous pairs separate and move to opposite poles. Telophase I Cytokinesis divides the cytoplasm into two haploid cells. Meiosis II Prophase II– Chromosomes condense again (in each haploid cell). Spindle fibers form and attach to chromosomes. Metaphase II Chromosomes align at the metaphase plate in a single file (like mitosis). Anaphase II Sister chromatids split and migrate to opposite poles. Telophase II Nuclear membranes reform, and cytokinesis produces four haploid daughter cells. Outcomes of Meiosis By the end of this process, a single diploid cell has morphed into four genetically unique haploid cells. This outcome gives rise to great genetic variation in populations, forming the basis of most evolutionary processes. With random fertilization and natural selection acting on these variations, species can adapt and survive changing environments. Conclusion Meiosis is the cornerstone of sexual reproduction, driving genetic diversity and ensuring each generation receives the correct amount of genetic material. From the formation of haploid gametes to the introduction of genetic variability, this process plays a critical role in the continuity of life. As you prepare for the AP® Biology exam, make sure you’re comfortable with each stage, how it differs from mitosis, and its purpose within the larger scope of genetics. With a thorough understanding of meiosis, you’ll be well on your way to success on the AP® Biology exam. Keep reviewing, practice often, and watch your confidence grow as you master this essential topic. Sharpen Your Skills for AP® Biology Are you preparing for the AP® Biology test? We’ve got you covered! Try our review articles designed to help you confidently tackle real-world math problems. You’ll find everything you need to succeed, from quick tips to detailed strategies. Start exploring now! Ready to boost your AP® Biology scores? Explore our plans and pricing here! Need help preparing for your AP® Biology exam? Albert has hundreds of AP® Biology practice questions, free response, and full-length practice tests to try out. Start practicing AP® Biology on Albert now! Interested in a school license? Bring Albert to your school and empower all teachers with the world's best question bank for: ➜ SAT® & ACT® ➜ AP® ➜ ELA, Math, Science, & Social Studies ➜ State assessments Options for teachers, schools, and districts. EXPLORE OPTIONS Popular Posts AP® Score Calculators Simulate how different MCQ and FRQ scores translate into AP® scores AP® Review Guides The ultimate review guides for AP® subjects to help you plan and structure your prep. 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16299	https://physics.umd.edu/courses/Phys260/ji/HW4.pdf	Course PHYSICS260 Assignment 4 Due at 11:00pm on Wednesday, February 27, 2008 A Simple Introduction to Interference Description: Interference is discussed for pulses on strings and then for sinusoidal waves. Learning Goal: To understand the basic principles underlying interference. One of the most important properties of waves is the principle of superposition. The principle of superposition for waves states that when two waves occupy the same point, their effect on the medium adds algebraically. So, if two waves would individually have the effect "+1" on a specific point in the medium, then when they are both at that point the effect on the medium is "+2." If a third wave with effect "-2" happens also to be at that point, then the total effect on the medium is zero. This idea of waves adding their effects, or canceling each other's effects, is the source of interference. First, consider two wave pulses on a string, approaching each other. Assume that each moves with speed meter per second. The figure shows the string at time . The effect of each wave pulse on the string (which is the medium for these wave pulses) is to displace it up or down. The pulses have the same shape, except for their orientation. Assume that each pulse displaces the string a maximum of meters, and that the scale on the x axis is in meters. Part A At time , what will be the displacement at point ? Express your answer in meters, to two significant figures. ANSWER: = Part B Choose the picture that most closely represents what the rope will actually look like at time . ANSWER: A B C D The same process of superposition is at work when we talk about continuous waves instead of wave pulses. Consider a sinusoidal wave as in the figure. Part C How far to the left would the original sinusoidal wave have to be shifted to give a wave that would completely cancel the original? The variable in the picture denotes the wavelength of the wave. Express your answer in terms of . ANSWER: = Part D In talking about interference, particularly with light, you will most likely speak in terms of phase differences, as well as wavelength differences. In the mathematical description of a sine wave, the phase corresponds to the argument of the sine function. For example, in the function , the value of at a particular point is the phase of the wave at that point. Recall that in radians a full cycle (or a full circle) corresponds to radians. How many radians would the shift of half a wavelength from the previous part correspond to? Express your answer in terms of . ANSWER: phase difference = radians Part E The phase difference of radians that you found in the previous part provides a criterion for destructive interference. What phase difference corresponds to completely constructive interference (i.e., the original wave and the shifted wave coincide at all points)? Express your answer as a number in the interval . ANSWER: phase difference = radians Part F Since sinusoidal waves are cyclical, a particular phase difference between two waves is identical to that phase difference plus a cycle. For example, if two waves have a phase difference of , the interference effects would be the same as if the two waves had a phase difference of . The complete criterion for constructive interference between two waves is therefore written as follows: Write the full criterion for destructive interference between two waves. Express your answer in terms of and . ANSWER: phase difference = The phase for a plane wave is a somewhat complicated expression that depends on both position and time. For most interference problems, you will work at a specific time and with coherent light sources, so that only geometric considerations are relevant. Consider two light rays propagating from point A to point B in the figure, which are apart. One ray follows a straight path, and the other travels at a angle to that path and then reflects off a plane surface to point B. Both rays have wavelength . Part G Find the phase difference between these two rays at point B. Part G.1 Find the difference in distance Find the difference in length between the direct path and the reflected path. You can use the fact that triangle ABC is an equilateral triangle. Express your answer in terms of . ANSWER: path length difference = Now that you have the difference in path length, convert that to radians. Recall that every cycle of radians is equivalent to one wavelength. Express your answer in terms of . ANSWER: phase difference = radians Part H Suppose that the reflected ray receives an extra half-cycle phase shift when it reflects. What is the new phase shift at point B? Hint H.1 How many radians in a half cycle? Since radians corresponds to a full cycle, a half cycle must correspond to radians. Express your answer in terms of . ANSWER: phase difference = radians Whenever light reflects from a transparent interface, moving from lower index of refraction to higher index of refraction, it gets an extra half cycle phase difference. Being able to accurately find the phase differences between waves at various points will be useful in both interference and diffraction problems. Normal Modes and Resonance Frequencies Description: Multiple choice questions about the definition and origin of normal modes. Then compute the frequency and wavelength of the first three normal modes in a string. Learning Goal: To understand the concept of normal modes of oscillation and to derive some properties of normal modes of waves on a string. A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general there are an infinite number of such modes, each one with a distinctive frequency and associated pattern of oscillation. Consider an example of a system with normal modes: a string of length held fixed at both ends, located at and . Assume that waves on this string propagate with speed . The string extends in the x direction, and the waves are transverse with displacement along the y direction. In this problem, you will investigate the shape of the normal modes and then their frequency. The normal modes of this system are products of trigonometric functions. (For linear systems, the time dependance of a normal mode is always sinusoidal, but the spatial dependence need not be.) Specifically, for this system a normal mode is described by Part A The string described in the problem introduction is oscillating in one of its normal modes. Which of the following statements about the wave in the string is correct? Hint A.1 Normal mode constraints The key constraint with normal modes is that there are two spatial boundary conditions, and , which correspond to the string being fixed at its two ends. ANSWER: The wave is traveling in the +x direction. The wave is traveling in the -x direction. The wave will satisfy the given boundary conditions for any arbitrary wavelength . The wavelength can have only certain specific values if the boundary conditions are to be satisfied. The wave does not satisfy the boundary condition . Part B Which of the following statements are true? ANSWER: The system can resonate at only certain resonance frequencies and the wavelength must be such that . must be chosen so that the wave fits exactly on the string. Any one of or or can be chosen to make the solution a normal mode. The key factor producing the normal modes is that there are two spatial boundary conditions, and , that are satisfied only for particular values of . Part C Find the three longest wavelengths (call them , , and ) that "fit" on the string, that is, those that satisfy the boundary conditions at and . These longest wavelengths have the lowest frequencies. Hint C.1 How to approach the problem The nodes of the wave occur where . This equation is trivially satisfied at one end of the string (with ), since . The three largest wavelengths that satisfy this equation at the other end of the string (with ) are given by , where the are the three smallest, nonzero values of that satisfy the equation . Part C.2 Values of that satisfy The spatial part of the normal mode solution is a sine wave. Find the three smallest (nonzero) values of (call them , , and ) that satisfy . Express the three nonzero values of as multiples of . List them in increasing order, separated by commas. ANSWER: , , = Hint C.3 Picture of the normal modes Consider the lowest four modes of the string as shown in the figure. The letter N is written over each of the nodes defined as places where the string does not move. (Note that there are nodes in addition to those at the end of the string.) The letter A is written over the antinodes, which are where the oscillation amplitude is maximum. Express the three wavelengths in terms of . List them in decreasing order of length, separated by commas. ANSWER: , , = The procedure described here contains the same mathematics that leads to quantization in quantum mechanics. Part D The frequency of each normal mode depends on the spatial part of the wave function, which is characterized by its wavelength . Find the frequency of the ith normal mode. Hint D.1 Propagation speed for standing waves Your expression will involve , the speed of propagation of a wave on the string. Of course, the normal modes are standing waves and do not travel along the string the way that traveling waves do. Nevertheless, the speed of wave propagation is a physical property that has a well-defined value that happens to appear in the relationship between frequency and wavelength of normal modes. Hint D.2 Use what you know about traveling waves The relationship between the wavelength and the frequency for standing waves is the same as that for traveling waves and involves the speed of propagation . Express in terms of its particular wavelength and the speed of propagation of the wave . ANSWER: = The frequencies are the only frequencies at which the system can oscillate. If the string is excited at one of these resonance frequencies it will respond by oscillating in the pattern given by , that is, with wavelength associated with the at which it is excited. In quantum mechanics these frequencies are called the eigenfrequencies, which are equal to the energy of that mode divided by Planck's constant . In SI units, Planck's constant has the value Part E Find the three lowest normal mode frequencies , , and . Express the frequencies in terms of , , and any constants. List them in increasing order, separated by commas. ANSWER: , , = Note that, for the string, these frequencies are multiples of the lowest frequency. For this reason the lowest frequency is called the fundamental and the higher frequencies are called harmonics of the fundamental. When other physical approximations (for example, the stiffness of the string) are not valid, the normal mode frequencies are not exactly harmonic, and they are called partials. In an acoustic piano, the highest audible normal frequencies for a given string can be a significant fraction of a semitone sharper than a simple integer multiple of the fundamental. Consequently, the fundamental frequencies of the lower notes are deliberately tuned a bit flat so that their higher partials are closer in frequency to the higher notes. Thin Film (Oil Slick) Description: This problem explores thin film interference for both transmission and reflection. A scientist notices that an oil slick floating on water when viewed from above has many different rainbow colors reflecting off the surface. She aims a spectrometer at a particular spot and measures the wavelength to be 750 (in air). The index of refraction of water is 1.33. Part A The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot? Hint A.1 Thin-film interference In thin films, there are interference effects because light reflects off the two different surfaces of the film. In this problem, the scientist observes the light that reflects off the air-oil interface and off the oil-water interface. Think about the phase difference created between these two rays. The phase difference will arise from differences in path length, as well as differences that are introduced by certain types of reflection. Recall that if the phase difference between two waves is (a full wavelength) then the waves interfere constructively, whereas if the phase difference is (half of a wavelength) the waves interfere destructively. Hint A.2 Path-length phase difference The light that reflects off the oil-water interface has to pass through the oil slick, where it will have a different wavelength. The total "extra" distance it travels is twice the thickness of the slick (since the light first moves toward the oil-water interface, and then reflects back out into the air). Hint A.3 Phase shift due to reflections Recall that when light reflects off a surface with a higher index of refraction, it gains an extra phase shift of radians, which corresponds to a shift of half of a wavelength. What used to be a maximum is now a minimum! Be careful, though; if two beams each reflect off a surface with a higher index of refraction, they will both get a half-wavelength shift, canceling out that effect. Express your answer in nanometers to three significant figures. 750 625 1.20 oil n n nm nm air air oil oil air oil air n n λ λ = = 2 oil t n λ λ = = λ = , where n is an integer. When n=1, t has minimum value 313 2 oil t nm λ = = ANSWER: = Part B Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now? Hint B.1 Phase shift due to reflections Keep in mind that when light reflects off a surface with a higher index of refraction, it gains an extra shift of half of a wavelength. What used to be a maximum is now a minimum! Be careful, though; if two beams reflect, they will both get a half-wavelength shift, canceling out that effect. Also, reflection off a surface with a lower index of refraction yields no phase shift. Express your answer in nanometers to three significant figures. 75 air oil air n n λ λ = = 0 500 1.50 oil nm nm = 2 ( 1/ 2) oil t n λ = + where n is an integer. When n=0, t has minimum value 125 4 oil t nm λ = = ANSWER: = Part C Now assume that the oil had a thickness of 200 and an index of refraction of 1.5. A diver swimming underneath the oil slick is looking at the same spot as the scientist with the spectromenter. What is the longest wavelength of the light in water that is transmitted most easily to the diver? Hint C.1 How to approach the problem For transmission of light, the same rules hold as before, only now one beam travels straight through the oil slick and into the water, while the other beam reflects twice (once off the oil-water interface and once again off the oil-air interface) before being finally transmitted to the water. Part C.2 Determine the wavelength of light in air Find the wavelength of the required light in air. Express your answer numerically in nanometers. The wave most easily transmitted into the water is the one most hard to reflect out of the water, i.e., find out the wave that has destructive effect with its reflected wave. Also note that index of refraction for oil is1.5, water is 1.33, so it gains an extra shift of half of a wavelength while reflecting on the air-water interface. 2 / /1.50 oil air air oil air t k kn n k λ λ λ = = = =1.502 600 air t nm , where n is an integer. The longest wavelength in air is for k=1. λ = ANSWER: = Hint C.3 Relationship between wavelength and index of refraction There is a simple relationship between the wavelength of light in one medium (with one index of refraction ) and the wavelength in another medium (with a different index of refraction ): . Express your answer in nanometers to three significant figures. ANSWER: = This problem can also be approached by finding the wavelength with the minimum reflection. Conservation of energy ensures that maximum transmission and minimum reflection occur at the same time (i.e., if the energy did not reflect, then it must have been transmitted to conserve energy), so finding the wavelength of minimum reflection must give the same answer as finding the wavelength of maximum transmission. In some cases, working the problem one way may be substantially easier, so you should keep both approaches in mind. Two Loudspeakers in an Open Field Description: Determine if constructive interference occurs at a certain point and then find the shortest distance you need to walk in order to experience destructive interference. Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound waves in phase at 688 . Take the speed of sound in air to be 344 . Part A If you are 3.00 from speaker A directly to your right and 3.50 from speaker B directly to your left, will the sound that you hear be louder than the sound you would hear if only one speaker were in use? Hint A.1 How to approach the problem The perceived loudness depends on the amplitude of the sound wave detected by your ear. When two sound waves arrive at the same region of space they overlap, and interference occurs. The resulting wave has an amplitude that can vary depending on how the two waves interfere. If destructive interference occurs, the total wave amplitude is zero and no sound is perceived; if constructive interference occurs, the total wave amplitude is twice the amplitude of a single wave, and sound is perceived as louder than what it would be if only one wave reached your ear. Hint A.2 Constructive and destructive interference Constructive interference occurs when the distances traveled by two sound waves differ by a integer number of wavelengths. If the difference in paths is equal to any half-integer number of wavelengths, destructive interference occurs. Part A.3 Find the wavelength of the sound What is the wavelength of the sound emitted by the loudspeakers? Hint A.3.a Relationship between wavelength and frequency In a periodic wave, the product of the wavelength and the frequency is the speed at which the wave pattern travels; that is, . Express your answer in meters. ANSWER: = ANSWER: yes no Part B What is the shortest distance you need to walk forward to be at a point where you cannot hear the speakers? Hint B.1 How to approach the problem You will not be able to hear the speakers if you are at a point of destructive interference. At a point of destructive interference, the lengths of the paths traveled by the sound waves differ by a half-integer number of wavelengths. Therefore, you can find the shortest distance you need to walk in the forward direction by determining the difference in distance from the two speakers that corresponds to the smallest possible half-integer multiple of the wavelength. Then, figure out how far forward you need to walk to obtain this path-length difference. Part B.2 Find the path-length difference at a point of destructive interference If is the distance between you and speaker A and is the distance between you and speaker B, by how much does differ from if you are now at the closest possible point of destructive interference? Hint B.2.a Condition for destructive interference Destructive interference occurs if the difference in paths traveled by sound waves is equal to any half-integer number of wavelengths. Therefore, the closest possible point of destructive interference corresponds to a path-length difference of half a wavelength. Express your answer in meters. ANSWER: = Now combine this result with the Pythagorean Theorem and solve for . Part B.3 Find your distance from speaker A If initially you were 3.00 from speaker A and then you walked forward the shortest possible distance needed to experience destructive interference, what is your new distance from that same speaker? Hint B.3.a Geometrical considerations Geometrically, your initial distance from one speaker and the distance you walked north represent the legs of a right triangle, whose hypothenuse is your new distance from that same speaker. If you apply the Pythagorean Theorem twice, you can write an expression that links to . This equation, combined with the relation previously found by imposing the condition of destructive interference, allows you to find . 2 2 2 2 3.5 3 0.25 d d + − + = d=5.625m Express your answer in meters to four significant figures. ANSWER: = Express your answers in meters to three significant figures. ANSWER: = Problem 21.75 Description: A flutist assembles her flute in a room where the speed of sound is 340 m/s. When she plays the note A, it is in perfect tune with a 440 Hz tuning fork. After a few minutes, the air inside her flute has warmed to where the speed of sound is v_2. Part A How many beats per second will she hear if she now plays the note A as the tuning fork is sounded? 1. Flute frequency in air 1 340 440 f Hz Hz λ = = 340 440 m λ = 2. Flute frequency in air 2 2 v f Hz λ = Beats frequency ( ) 2 2 2 1 340 440 340 340 beat v v f f f λ − ⋅ − = − = = ANSWER: beats Part B How far does she need to extend the "tuning joint" of her flute to be in tune with the tuning fork? Assuming after extending the tuning joint the wavelength of note A traveling in room2 is 2 2 1 ' 440 v v m f λ = = The original wavelength of note A is 340 440 m λ = 2 340 2 ' 440 440 v L m λ λ λ ⎛ ⎞ Δ = Δ = − = − ⎜ ⎟ ⎝ ⎠ 2 340 440 440 1000 2 v L mm ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ Δ = ANSWER: mm Problem 21.40 Description: A violinist places her finger so that the vibrating section of a mu string has a length of L, then she draws her bow across it. A listener nearby in a 20 degree(s) C room hears a note with a wavelength of lambda. (a) What is the tension in the string? A violinist places her finger so that the vibrating section of a 1.30 string has a length of 40.0 , then she draws her bow across it. A listener nearby in a room hears a note with a wavelength of 50.0 . Part A What is the tension in the string? The fundamental frequency of a vibrating string is 1 2 T L f μ = Wave traveling in the room where the listener stays v f λ = ( ) Therefore, 2 2 2 343 2 2 2 v T Lf L L μ μ μ λ λ ⎛ ⎞ ⎛ ⎞ = = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ANSWER: N Problem 21.53 Description: A L_1-long wire with a linear density of mu passes across the open end of an L_2-long open-closed tube of air. If the wire, which is fixed at both ends, vibrates at its fundamental frequency, the sound wave it generates excites the second vibrational mode of the tube of air. What is the tension in the wire? A 21.0 -long wire with a linear density of 10.0 passes across the open end of an 84.0 -long open-closed tube of air. If the wire, which is fixed at both ends, vibrates at its fundamental frequency, the sound wave it generates excites the second vibrational mode of the tube of air. What is the tension in the wire? Part A Assume . The fundamental frequency of the wire is 1 1 2 T f L = μ The second vibration of the open-closed tube of air is 2 2 3 4 v f f L = = 2 2 3 4 v f f L = = ( ) 2 2 1 1 2 3 340 2 2 4 T L f L L μ μ ⎛ ⎞ ⋅ = = ⎜ ⎟ ⎝ ⎠ ANSWER: N Problem 21.61 DescriptionTwo loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when speaker 2 is at the origin and speaker 1 is at 0.480 . If speaker 1 is slowly moved forward, the sound intensity decreases and then increases, reaching another maximum when speaker 1 is at 0.890 . Part A What is the frequency of the sound? Assume . Assuming the listener is at position s on the x-axis. 1 s x mλ − = ( ) 2 1 s x m λ − = − 2 1 x x λ − = = 340 / v m s f f = 2 1 340 f Hz x x = − ANSWER: Hz Part B What is the phase difference between the speakers? Consider while speaker1 is at x1. 1 2 2 x π ϕ + Δ = π λ Phase difference 1 1 2 1 2 2 -2 (1-) x π x rad x x ϕ π π λ = = − ANSWER: rad

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