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3000 | https://www.geeksforgeeks.org/chemistry/geometric-and-optical-isomerism/ | Geometric and Optical Isomerism
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Transition metals can generate a wide range of complex compounds by sharing electrons with a variety of anions or neutral molecules, such compounds are referred to as coordination compounds in present terminology.
New ideas on chemical bonds and molecular structure have shed light on how these substances act as essential parts of biological systems. The coordination compounds of magnesium, iron, and cobalt are chlorophyll, haemoglobin, and vitamin B12, respectively. Coordination compounds are used as analytical reagents, industrial catalysts, and various metallurgical processes. Additionally, there are numerous uses for coordination compounds in the electroplating, textile dyeing, and pharmaceutical sector.
What is Isomerism?
When two or more chemical compounds share the same molecular formula, this phenomenon is known as isomorphism. Greek terms isos, which means "equal," and meros, which means "part," are the roots of the word "isomerism." When two or more compounds have the same molecular formula but different chemical and physical properties, this is referred to as isomorphism. One of the causes of the existence of many organic molecules is isomerism, which is a common phenomenon in organic chemistry. It describes substances with the same molecular formula but differing physical or chemical properties
Isomerism in Coordination compounds
Two or more compounds with the same chemical formula but a distinct atom arrangement are said to be isomers. They have one or more different physical or chemical properties as a result of the varied atom arrangements. Due to the many bond types and challenging formulae of many coordination compounds, we find a variety of isomers.
If more than one isomer is available, it is essential to know which one we are working with because isomers typically have different physical and chemical properties. We shall observe that coordination compounds display many of the same isomers as organic compounds and some unique isomers.
Among coordination compounds, there are two main kinds of isomers that are identified.
Stereoisomerism
Geometrical Isomerism
Optical Isomerism
Structural isomerism
Linkage Isomerism
Coordination Isomerism
Ionization Isomerism
Solvate Isomerism
Stereoisomerism
Stereoisomers share the same chemical structure and chemical interactions, but they are arranged differently in space. Isomer refers to two or more compounds with the same formula but different atom arrangements. These isomers' properties differ from compound to compound. Isomerism is the phenomenon that results in a variety of isomers. One kind of stereoisomerism is optical isomerism, and the second is geometric isomers.
Geometric Isomerism
The molecules' spatial arrangement changes around the carbon atoms as a result of compounds having impeded rotation in space around a particular carbon-carbon double bond or a single bond. This type of isomerism is known as geometric isomerism, and the isomers are known as geometric isomers.
Geometrical isomerism is also a type of stereoisomerism in which the identical substituent is bonded differently to the carbon-carbon double bond. They have different physical characteristics.
Geometric Isomerism (in complex with 4 coordinates bond)
Tetrahedral or square planar geometry are two possible configurations for complexes with a core atom and a 4-coordination number. Tetrahedral complexes cannot show geometric isomerism since all four ligands are at neighboring (cis) positions to one another and all four bond angles are the same (109.5o).
Geometrical Isomerism in Square Planar Complexes
Trans-isomer is a square planar complex with two similar ligands at opposite locations (180o apart), whereas cis-isomer is a square planar complex with similar ligands at adjacent positions (90o apart).
Geometrical Isomerism (in complex with 6-Coordinate Bond) or Octahedral Complexes
We know that the shape of a complex compound with a core atom and a coordination number of 6 is octahedral. Complexes of type [MA4B2], where A and B are monodentate ligands, can exist in both cis and trans isomeric forms.
Ions are in opposite positions in trans isomers, while in cis isomers two ions occupy the adjacent sites of the octahedron.
Despite having the same chemical formula, these cis and trans isomers have different physical characteristics. Because of their symmetry, trans isomers have higher melting points than cis isomers. Likewise, cis isomers have higher boiling points than trans isomers because they contain polarity brought on by a dipole moment.
Optical Isomerism or (d-form, l-form)
Stereoisomerism known as optical isomerism refers to compounds with the same chemical structure and connectivity but a distinct spatial arrangement. They have the same physical characteristics. They do not display cis and trans isomers in optical isomerism. In nature, optical isomers are optically active. A plane polarized light's plane rotates through an angle, which can be to the left or to the right, when a solution of certain complicated substances is in its path. A compound is considered to be optically active if it has the ability to rotate the plane of polarized light, which is referred to as its optical activity.
When the plane of polarized light is rotated towards the right (in a clockwise direction) by the isomer, it is said to be dextro-rotatory or d-form.
Likewise, when the plane of polarized light is rotated towards the left (in an anti-clockwise direction) by the isomer, it is said to be levorotatory or l-form.
Optically inactive isomers are those that cannot rotate the plane of polarized light. Such an isomer is known as meso, racemic, or dl-form
The d- and l- forms are physically and chemically similar, yet they behave differently when exposed to polarized light.
Similar to how the left and right hands are mirror images of one another, the d and l shapes are the same.
The molecule must be asymmetric (have no plane of symmetry) and not be superimposable on its mirror image in order to exhibit optical isomerism. These two are the most important and essential requirements for this phenomenon.
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Isomerism
Isomerism in Coordination Compounds
Applications of Coordination Compounds
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3001 | https://www.youtube.com/watch?v=CXgeSLbx7RA | Function Point Analysis | Function Based Metrics | Function Point Analysis Example 1
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in the function based metrics this is a example we want to use the function point metric and this is a simple flow diagram for a user interaction function in say a safety security safe home system so there is a hardware when you whether you are going to key in and there will be a software inside it so the function manages user interaction what is the user interaction for example accepting a user password in order to activate or deactivate the system this allows inquiries on the status of the security zones and the various security sensors whether they are working not my parking is safe or not whether my kitchen sensor is working and then displaying of also display of some prompting messages also sending some appropriate control signal to various components of the security system so these are the function managing the user interaction but our idea is to compute the function point so what are the three external inputs we need to find out eis what are eis external inputs so what are the inputs password is an input the panic button is an input external input and the activate and deactivate is the external input so how many are there three so in front of external inputs we are going to put under count three [Music] three okay how these three has come now we have two external inquiries what are these the zone inquiry which area we are talking about and which sensor we are talking about so sensor inquiry these are two external inquiries so in place of e i e we are going to give the two number then ilf how many ilifs are there system configuration file so we have one ilf so let us give ilf as one two external outputs are there so these are messages which are going outside and sensor status so these are two external outputs being given so we'll put external outputs as two now what are eifs the external interface files for example your test sensor zone setting activate deactivate whether it is not done or not what is the test status and the alarm alert these are external interface files so we'll put 4 here now we have a count for each of these input parameters now we need count written so let us assume that it is simple we are putting the weighing factor for simple and these numbers are empirically already found and these are specific to certain application so average complex these are already already empirically predetermined value so we are not taking average and complex right now we are just saying that okay let us use use it for a simple functionality okay so we will not use the last two columns now we have to multiply the first two columns one by one three into three is nine two into four is eight two into three is 6 1 into 7 is 7 4 into 5 is 20. now we'll add them summation 50 50 is our count total this is our count total okay we need to go for sigma fi so let us say that it is 46 for a moderately complex product but i need to tell you how this six sigma fi has come so these are all the general system characteristic now you have to give a value from 0 to 5 depending upon the importance of these system characteristics in your software or a product in terms of usability for example data communication you give 2 online data entry give 4 5 usability 0 so every number you have to give between 0 and 5 and just add them and just assume that that number is 46 you can try with any other number 40 50 it will be below 70. okay so now this addition of these numbers each number between 0 to 5 assigned to these 14 system characteristics you add them and i'm just saying that it is 46 to make our computation easy so this is 50 this is how much this is 46 by adding all these and then the final computation 50 is the total count 46 is the sigma fi and the actual functional function point is 56 56 |
3002 | https://www.youtube.com/watch?v=2L7__8_6YUg | 05-05: Congruences with Non-Coprime Bases
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While the Chinese Remainder Theorem covers the situation of a system of linear congruences in one variable where the modular bases are pairwise relatively prime, what can we say about systems where the bases are not coprime?
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Transcript:
Intro so far in solving linear congruences and using the chinese remainder theorem we've had bases of the equivalences that were all co-primed to one another each uh pairwise relatively prime what happens if you have equivalences where the bases are not relatively prime THREE EXAMPLES so we want to solve a system of linear congruences in one variable but the bases might not be pairwise relatively prime anymore it's tempting to just say well then no solutions but that's not necessarily true can you find an x equivalent to 1 mod 10 and also equivalent to 1 mod 12. well the bases 10 and 12 are definitely not relatively prime but 1 is definitely an integer that solves both so there might be solutions even if the bases aren't relatively prime and then an objection that i would get frequently in class at this point is well sure i just had a 1 on both sides so of course there's solutions but what about this x equivalent to 5 mod 10 and x equivalent to 7 mod 12 is also got solutions for example 55 is a solution so just having the same number on the right hand side isn't really the issue okay i was able to find integers equivalent to 5 mod 10 and 7 mod 12 even though 10 and 12 are not relatively prime and even though 5 and 7 weren't just the same number and then some might say well maybe solutions always exist however this system x equivalent to one mod two and equivalent to zero mod four can't possibly have any solutions the first is x is odd equivalent to one mod two means a multiple of two plus one an odd number the second equivalent server says divisible by four and if you're divisible by four you're definitely divisible by two so you're even and a number cannot be both even and odd so this system can't possibly have any solutions THE THEOREM so how can we determine which systems do versus which systems do not have solutions at all the pair of congruences x equivalent to b1 mod n1 and equivalent to b2 mod m2 solutions if and only if the target values b1 and b2 are equivalent modulo the greatest common divisor of the two bases in this case there is exactly one solution modulo the least common multiple of the two bases if the two bases are relatively prime this is a result we already know if the two bases are relatively prime this is asking for something modulo one in other words is one a factor of b1 minus b2 well yes it is one is a factor of everything so b1 is always equivalent to b2 modulo one so if these two numbers are relatively prime there are always solutions and then in fact there is exactly one solution modulo the least common multiple but if the two numbers were relatively prime the least common multiple is simply the product and one and two so this is the result we already know if the two bases are relatively prime let's actually go ahead and prove this however suppose there is a solution so x is equivalent to b1 mod n1 but it's also equivalent to b2 mod m2 so it is a multiple of n1 plus p1 but it is also equal to a multiple of n2 plus v2 so i can set these two things equal to each other and get the following k2 n2 minus k1n1 is equal to b1 minus b2 so now what we have is a linear combination of the two bases n1 and n2 with a result of this difference b1 minus b2 therefore the greatest common factor of n1 and n2 must be a factor of that difference okay this is bay zoo's identity linear combinations of two numbers form multiples of the greatest common factor so here we have a linear combination of n1 and n2 therefore this is a multiple of the greatest common factor and that's what we have written right here but if n1 n2's greatest common factor is a factor of the difference b1 minus b2 that's the very definition that the two numbers are equivalent modulo that base so if there are solutions then this must be true now what we need to prove is that if this is true then we can find a solution THE PROOF CONTINUED so we've shown that this pair of equivalences can only have solutions if the two target values are equivalent modulo the greatest common divisor of the two bases now we have to go the other way assume that the two target values are equivalent modulo the greatest common divisor of the bases and then show that there is a solution and once we've done that show that all solutions must be equivalent modulo the lcm let's actually take care of that last part now suppose i have two different solutions so both x and x prime are equivalent to b1 mod n1 therefore they're equivalent to each other modulo and one but they're also going to be equivalent to each other modulo and two and we have this old result that if two things are equivalent modulo two different bases they must be equivalent modulo the least common multiple so if x and x prime both solve both equivalences they must be equivalent modulo the lcm of n1 and n2 so that takes care of that last bit the only thing remaining in this theorem then is assuming b1 and b2 are equivalent modulo this gcd can we find a solution x to both well we can finally finish up this proof THE PROOF COMPLETED the only thing we have left to do is assuming b1 and b2 are equivalent modulo that greatest common divisor show that the system has solutions so for convenience let's let that greatest common divisor be written as d so we get to assume that b1 and b2 are equivalent to modulo d now d was the greatest common divisor of n1 and n2 so we can write n1 is d times something and n2 is d times something and since this common factor d is in fact the largest thing that can be factored out of both the numbers that are left behind j and k must be relatively prime to one another also since b1 is equivalent to b2 modulo d then b1 is equal to a multiple of d plus b2 so suppose x is a solution to our system of equivalences it must satisfy that x is equivalent to b1 mod n1 in other words x is a multiple of n1 plus v1 but also x is equivalent to b2 mod n2 so it's a multiple of n2 plus b2 if these two things are both equal to x they must be equal to each other n1 t1 plus b1 is equal to n2 t2 plus b2 and now what we're going to do is we're going to substitute this for n1 this for n2 and this for b1 so here we have n1 t1 plus b1 equals n2 t2 plus b2 we can cancel the b2 off of both sides here great now every term has a common factor of d which is not zero so i can divide it cancel it out get jt1 minus k t 2 equals negative i once i move some things around this has solutions if and only if this target value of negative i is in fact a multiple of the gcd of j and k that's just bay zoo's identity again here i have a linear combination of j and k therefore this should be a multiple of uh the greatest common divisor and vice versa since the greatest common divisor was one this is definitely a multiple of one so since this here has solutions t1 and t2 i can take the t1 and t2 that solve this and come all the way back here and solve my original so assuming that b1 was equivalent to b2 modulo n1 and 2's greatest common divisor we were able to deduce there must be a solution and we've already shown there can be only one solution modulo the lcm of n1 and n2 so we finally finished this proof two linear equivalences have solutions a mutual solution x if and only if the difference of the two target values is a multiple of the greatest common factor of the two bases FINDING SOLUTIONS now on the previous slide we finished this lengthy proof that solutions can exist under certain circumstances but it's actually not a terribly convenient way to find solutions in general so if we want to solve this pair of equivalences and we assume that the greatest common divisor of the two bases is a factor of that difference v1 minus b2 let's keep all the notation from the proof we just finished so d is the greatest common divisor of n1 and n2 n1 is dj n2 is dk j and k must be relatively prime since b1 was equivalent to b2 mod n2 we can write b1 is equal to something plus b2 what we needed to do is having solved for j k and i now we need to solve for t1 and t2 if you can solve for t1 and t2 you can come back to what x was taken to be equal to and plug in t1 or t2 into either of these two things you should get the same number either way but solving this line right here requires using the extended euclidean algorithm first to solve since j and k are relatively prime find integers z1 and z2 so that this combination produces a 1 then multiply everything by negative i and isolate what this integer and this integer are and the extended euclidean algorithm while very fast for computers to do can be a little lengthy if you're working by hand if we're solving systems by hand and the bases are not extremely large specifically we can perform addition of elements modulo the bases in our head pretty quickly there is a faster method if you're just brute forcing it on paper because of an exam setting or something BRUTE FORCE, CLEVERLY APPLIED so we're going to do brute force but we're going to be marginally clever about it we're asked to solve this pair of equivalences first we check that this is true because if this isn't true there are no solutions anyway so suppose this condition is met solutions exist but how do we find them if we have to do it by hand if we can perform addition modulo one of the bases i'm just going to use modulo n2 then we can solve the problem by brute force fairly quickly we simply compute we need the answer to be b1 mod n1 so b one is one such number or b one plus n one or plus two n one or plus three n one and so forth so now i have a bunch of numbers that are all equivalent to b one mod n one if i simply take all of these and check what they're equal to sorry equivalent to modulo m2 eventually i'll find something that is equivalent to the desired value mod n2 so we simply keep checking b1 plus multiples of n1 modulo and 2 until we find the desired value because all of these are definitely equivalent to b1 mod n1 so once i find a multiple of n1 to add here so that this is equivalent to b2 mod n2 then we would be done let's try AN EXAMPLE to do an example of this technique find an integer x that solves equivalent to 8 mod 12 but also equivalent to 11 mod 15. so let's go ahead and solve it we check what's the greatest common factor of 12 and 15 it's 3 8 minus 11 would be negative 3 so 8 and 11 are equivalent modulo that gcd so solutions do exist to find so we're going to solve by brute force but we're not just going to guess and check i'm not just going to say uh 28 or 42 uh 17 we're not just going to randomly throw numbers out there we simply keep computing 8 plus larger and larger multiples of 12 they will all be equivalent to 8 mod 12. and all we do is we check one at a time until we are equivalent to 11 modulo 15. so 8 with no additional 12s added on is equivalent to 8 mod 15 not 11. if i add 12 to both sides i would get 20 but 20 modulo 15 is equivalent to 5. if i add 12 again i would get 17 but 17 is equivalent to 2. if i add 12 again i get 14. if i add 12 again i finally get 11. so starting with 8 which is definitely equivalent to 8 mod 12 and adding 12 which doesn't change the fact that all of these numbers are now equivalent to 8 mod 12. i just kept checking what are they mod 15 until we got the desired value of 11. so 56 is a solution to this problem as long as you're dealing with fairly small bases this is a pretty viable technique if you had extremely large bases then you would really be relying on computer assistance to solve these in a reasonable amount of time but if your unreasonable and terribly cruel professor is asking you to do some of these by hands and listing out all of the steps then hopefully you're working with pretty small numbers and this is really all the work that's required |
3003 | https://artofproblemsolving.com/wiki/index.php/Geometric_sequence?srsltid=AfmBOor8hXWazyiz7hUgfGYr7iLv9cRxN1XxzGvAf9L8ExcBn6LiEfMo | Art of Problem Solving
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Geometric sequence
In algebra, a geometric sequence, sometimes called a geometric progression, is a sequence of numbers such that the ratio between any two consecutive terms is constant. This constant is called the common ratio of the sequence.
For example, is a geometric sequence with common ratio and is a geometric sequence with common ratio ; however, and are not geometric sequences, as the ratio between consecutive terms varies.
More formally, the sequence is a geometric progression if and only if . A similar definition holds for infinite geometric sequences. It appears most frequently in its three-term form: namely, that constants , , and are in geometric progression if and only if .
Contents
1 Properties
2 Sum
2.1 Finite
2.2 Infinite
3 Problems
3.1 Introductory
3.2 Intermediate
4 See also
Properties
Because each term is a common multiple of the one before it, every term of a geometric sequence can be expressed as the sum of the first term and a multiple of the common ratio. Let be the first term, be the th term, and be the common ratio of any geometric sequence; then, .
A common lemma is that a sequence is in geometric progression if and only if is the geometric mean of and for any consecutive terms . In symbols, . This is mostly used to perform substitutions, though it occasionally serves as a definition of geometric sequences.
Sum
A geometric series is the sum of all the terms of a geometric sequence. They come in two varieties, both of which have their own formulas: finitely or infinitely many terms.
Finite
A finite geometric series with first term , common ratio not equal to one, and total terms has a value equal to .
Proof: Let the geometric series have value . Then Factoring out , mulltiplying both sides by , and using the difference of powers factorization yields Dividing both sides by yields , as desired.
Infinite
An infinite geometric series converges if and only if ; if this condition is satisfied, the series converges to .
Proof: The proof that the series convergence if and only if is an easy application of the ratio test from calculus; thus, such a proof is beyond the scope of this article. If one assumes convergence, there is an elementary proof of the formula that uses telescoping. Using the terms defined above, Multiplying both sides by and adding , we find that Thus, , and so .
Problems
Here are some problems with solutions that utilize geometric sequences and series.
Introductory
2025 AMC 8 Problem 20
Intermediate
1965 AHSME Problem 36
2005 AIME II Problem 3
2007 AIME II Problem 12
See also
Arithmetic sequence
Harmonic sequence
Sequence
Series
Retrieved from "
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Algebra
Sequences and series
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3004 | https://planetmath.org/debruijnerdhostheorem | De Bruijn–Erdős theorem
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finite plane
De Bruijn–Erdős theorem
Let be a set of ‘points’, say , and let , , … be different subsets of of which any two have exactly one point in common. Now the De Bruijn–Erdős theorem says that , and that if then (up to renumbering) at least one of the following three must be true:
•
for , and
•
for , and
•
is of the form , each contains points, and each point is contained in of the
and we recognise the last case as projective planes of order . For () the first and last case overlap, and for () the last two cases do. The second case is also known as a near-pencil. The last two cases together are examples of linear spaces.
To exclude the first two cases one usually defines projective planes to satisfy some non-triviality conditions; unfortunately that also excludes projective planes of order 0 and 1.
The formulation above was found in the literature [Cam94]. Naturally, if the are points we tend to interpret the subsets as lines. However, interpreted in this way the condition that every two share an says rather more than is needed for the structure to be a finite plane ( space) as it insists that no two lines are parallel. The absence of the dual condition, for two to share a , actually means we have too little for the structure to be a finite plane (linear space), as for two points we may not have a line through them. And indeed, while the second and third cases are finite planes without parallel lines, the first case is not a plane.
Erdős–De Bruijn theorem
A dual formulation (which we could whimsically call the Erdős–De Bruijn Theorem) remedies both under- and over-specification for a plane. Indeed, the conditions are now exactly tailored to make the structure a finite plane (with some parallel lines in the first of the three cases).
Let be a set of ‘points’, say , and let , , … be subsets of (‘lines’) such that for any two points there is a unique that contains them both. We must now be careful about the points (the former lines) being “different” (this was easier to formulate in the previous version, in which form it was a simple incidence structure): this condition now takes the form that no two points must have the same collection of lines that are incident with them. Then , and if then (up to renumbering) at least one of the following three must be true:
•
for , and
•
for , and
•
is of the form , each contains points, and each point is contained in of the
For a proof of the theorem (in the former version) see e.g. Cameron [Cam94].
References
1
Cam94 Peter J. Cameron, Combinatorics: topics, techniques, algorithms, Camb. Univ. Pr. 1994, ISBN 0 521 45761 0 pjc/comb/ pjc/comb/ (solutions, errata &c.)
| | |
--- |
| Title | De Bruijn–Erdős theorem |
| Canonical name | DeBruijnErdHosTheorem |
| Date of creation | 2013-03-22 15:11:21 |
| Last modified on | 2013-03-22 15:11:21 |
| Owner | marijke (8873) |
| Last modified by | marijke (8873) |
| Numerical id | 7 |
| Author | marijke (8873) |
| Entry type | Theorem |
| Classification | msc 05B25 |
| Related topic | FiniteProjectivePlane4 |
| Related topic | IncidenceStructures |
| Related topic | Geometry |
| Related topic | LinearSpace2 | |
3005 | https://utminers.utep.edu/xzeng/2017spring_math5330/MATH_5330_Computational_Methods_of_Linear_Algebra_files/ln10.pdf | MATH 5330: Computational Methods of Linear Algebra Lecture Note 10: Householder Transformation Xianyi Zeng Department of Mathematical Sciences, UTEP 1 The QR Decomposition Using Givens rotations allows us to write A = QE where Q is orthogonal and E is of the row echelon form. Note that the lower-triangular part of E is always zero, i.e. eij = 0 if i > j; thus this is decomposition is also known as the QR decomposition, where “R” stands for right-triangular or upper triangular. Henceforth we will write A = QR instead of A = QE. In the special case when A is square and non-singular, we immediately see that R = Q−1A is also non-singular; hence all its diagonal elements are non-zero. Furthermore, we can compute the inverse of A easily by: A−1 = R−1Qt .
The advantage of using the QR decomposition over the LU decomposition to invert a non-singular matrix is that all its components have an a priori estimate. However, the QR decomposition requires much more computational cost (∼2n3) than the Gaussian elimination (∼2n3/3). Robustness does not come free!
Givens rotations belong to one of three widely used methods to compute A = QR: • Gram Schmidt: Since every column of A is a linear combination of the columns of Q, we have col(A) ⊆col(Q); thus in the end the QR decomposition can be reduced to orthogonalization of the column vectors of A. We have already seen in the Arnoldi’s method that, this can be achieved by the Gram Schmidt process. The major problem of the Gram Schmidt process is similar to that of Gaussian elimination: The algorithm involves dividing by a number that cannot be estimated a priori, hence it can breakdown unexpected.
• Givens rotations: This method is more robust than the Gram Schmidt, and in each rotation only two adjacent rows are involved so it is more bandwidth efficient and reduce cache misses.
A major objection for using the Givens rotation is its complexity in implementation; partic-ularly people found out that the ordering of the rotations actually matter in practice , and determining the optimal order is a non-trivial problem.
• Householder transformation: This method is robust like the one using Givens rotations, easier to implement, and requires less computation. However, it is less bandwidth efficient and more difficult to parallelize than the latter.
At the end of this section, we prove that if A is square and non-singular, the QR decomposition is unique, if we require in addition that all diagonal entries of R are positive. In fact, if A=Q1R1 = Q2R2 where both Q1 and Q2 are orthogonal and both R1 and R2 are upper-triangular, we then have: Qt 2Q1 = R2R−1 1 .
1 The left hand side is orthogonal whereas the right hand side is upper triangular. It is not difficult to see that the only upper triangular matrix that is also orthogonal is diagonal matrices with diagonal elements being ±1. By the assumption that the diagonal elements of both R1 and R2 are positive, the only possibility is R2R−1 1 = I, or equivalently Q1 = Q2 and R1 = R2.
2 The Householder Transformation A major motivation for using Givens transform to construct the QR decomposition is that rotations preserve the L2-norm of vectors.
These are, however, not the only operations that have this property. For example, the reflection about any plane also preserve the L2-norm of vectors in Rn.
The method utilizing this latter property is built on the Householder transformation.
There are at least two ways to describe a Householder matrix. Algebraically, a Householder matrix differs from the identity matrix by a rank one matrix as follows: Hv = I −2vvt , (2.1) where v is a unit vector. Furthermore, Hv is symmetric and orthogonal; and we’ll check the latter: Ht vHv = (I −2vvt)(I −2vvt) = I −4vvt +4v(vtv)vt = I −4vvt +4vvt = I .
Now let u be any vector, and we compute Hvu: Hvu = (I −2vvt)u = u−2(u·v)v .
(2.2) The second term looks familiar: (u·v)v is the projection of u onto the straight line spanned by the vector v. Indeed, we see that: u = (u·v)v+[u−(u·v)v] is the orthogonal decomposition of u w.r.t. V = span(v): v·[u−(u·v)v] = v·u−(u·v)||v||2 = 0 .
Thus (2.2) gives the reflection of u w.r.t. V⊥, in the sense that the midpoint of the line connecting u and Hvu is the orthogonal projection of u onto V⊥. This is the geometrical description of a Householder transform, and a 2D example is depicted in Figure 1.
O V⊥ u v Hvu Figure 1: Reflection of u by the line V⊥in R2.
2 In 2D, V⊥itself is a straight line; in Rn, V⊥has dimension n−1 and it has normal vector v.
We call V⊥a hyperplane with unit normal vector v.
As an immediate application of the Householder transform, we can essentially map u to any other direction by choosing v carefully. In particular, we would like to find v such that Hvu=||u||e1, i.e., all the components except the first one are zero. Let α = u·v for now, then we need: u1 −2αv1 = ||u|| ; ui −2αvi = 0 , i = 2,···,n .
Hence v is given by: v1 = u1 −||u|| 2α , vi = ui 2α , i = 2,···,n .
To proceed, note that v needs to be a unit vector, we thusly have: 4α2 = (u1 −||u||)2 + n X i=2 u2 i = 2||u||2 −2u1||u|| .
Choosing 2α = q 2||u||2 −2u1||u|| we obtain the desired unit vector v: v = u1−||u|| √ 2||u||2−2u1||u|| u2 √ 2||u||2−2u1||u|| .
.
.
un √ 2||u||2−2u1||u|| .
(2.3) Finally, it is not difficult to check that: v·u = u2 1 −u1||u|| q 2||u||2 −2u1||u|| + n X i=2 u2 i q 2||u||2 −2u1||u|| = ||u||2 −u1||u|| q 2||u||2 −2u1||u|| = 2α2 2α = α .
For convenience, we shall denote this Householder matrix by H(u), i.e., I −2vvt where v is given by (2.3).
3 The QR Decomposition using Householder Transformations The idea is to use Householder transformation to construct a sequence of matrices H1, H2, ···, Hn such that HnHn−1···H1A is of the row echelon form. Again, the purpose of each Hi is to put the i-th column of A in the desired form without touching the previous i−1 columns.
The purpose of H1 is to have the first column of H1A possess at most one non-zero, which is the first component. This is clearly achieved by defining H1 as the Householder matrix that corresponds to (2.3) where u is replaced by the first column of A. Let a(0) 1 be the first column of 3 A(0) def = = A, following our previous notation we have H1 = H(a(0) 1 ): [A]1 7→[H1A]1 is given by a11 a21 a31 .
.
.
am1 7→ p a2 11 +a2 21 +···+a2 m1 0 0 .
.
.
0 .
Let A(1) = H1A; now we look at the second column. Again, there are two situations: if a(1) 11 = p a2 11 +···+a2 m1 = 0, we define H2 = H(a(1) 2 ), where a(1) 2 is the entire second column of A(1): [A(1)]2 7→[H2A(1)]2 is given by a(1) 12 a(1) 22 a(1) 32 .
.
.
a(1) m2 7→ q (a(1) 12 )2 +(a(1) 21 )2 +···+(a(1) m1)2 0 0 .
.
.
0 .
If, however, a(1) 11 is not zero, we do not want to touch the first row of A(1) and we only need to put the last (m−2) elements of the second column of A(1) into zero. To this end, we abuse the notation and denote: a(1) 2 = a(1) 22 a(1) 32 .
.
.
a(1) m2 and define: H2 = " 1 H(a(1) 2 ) # .
It is not difficult to check that H2A(1) keeps the first column of A(1) unchanged; furthermore, the first two columns of A transform as: [A]1:2 7→[H2H1A]1:2 is given by a11 a12 a21 a22 a31 a32 .
.
.
.
.
.
am1 am2 7→ a(1) 11 a(1) 12 0 q (a(1) 22 )2 +(a(1) 32 )2 +···+(a(1) m2)2 0 0 .
.
.
.
.
.
0 0 .
Now suppose we have already computed H1 , ··· , Hk. In order to constructHk+1, we first identify the vector a(k) k+1, which is composed of the last m−p elements of the (k+1)-th column of A(k) that we need to transform, and then define: Hk+1 = " Ip H(a(k) k+1) # .
4 Here Ip is the identity matrix in Rp×p.
To summarize, the Householder version of QR decomposition is given by Algorithm 3.1.
Algorithm 3.1 Orthogonal Reduction by Householder Transformation 1: Set p = 1 and Q = Im 2: for i = 1,2,···,n do 3: Compute s = a2 p+1,i +···+a2 m,i 4: if s ̸= 0 then 5: Compute β = q s+a2 pi 6: Compute γ = p 2(β2 −apiβ) 7: Compute v = [(api −β)/γ , ap+1,i/γ , ··· , ami/γ]t 8: A[p : m,i : n] ←HvA[p : m,i : n] 9: Q[p : m,p : m] ←Q[p : m,p : m]Hv 10: end if 11: if api! = 0 then 12: Set p ←p+1 13: end if 14: end for Here we use a notation M[I,J ] to denote a sub-matrix of M: I is a subset of all the row indices of M and J is a subset of all column indices of J ; and M[I,J ] is the |I|×|J | matrix composed of elements at the intersections of rows denoted by I and columns designated by J .
4 Analysis and Other Aspects First of all, we note that updating the matrices A[p : m,i : n] and Q[p : m,p : m] in Algorithm 3.1 does not require constructing Hv explicitly. Let us assume for simplicity Hv ∈Rm×m and we want to compute HvA. Note that: HvA = (I −2vvt)A = A−2v(vtA) , can be accomplished column by column. That is, for each column ai of A, we replace ai by: ai ←ai −2(ai ·v)v , which requires 4m flops and essentially no additional storage. Similarly, computing QHv leads to: QHv = Q(I −2vvt) = Q−2(Qv)vt , and this can be done row by row, with 4m flops for updating each row of Q.
Finally, we look at the complexity of Algorithm 3.1. Note that the line 3 will be executed every outer loop, unless we smartly terminate the program when p reaches r=min(m,n). But in any case, the computational cost associated with line 3 is bounded by (2m−1)n∼2mn flops. The execution of the block from line 4 to line 10, however, will leads to an increase in the value of p; hence this 5 block can be run at most r times, with the complexity bounded by 2r square root operations and the number of flops: r X p=1 [2+3+m−p+2+4(m−p+1)(n−ip +1)] where ip is the value i such that the if block with p is executed. Clearly ip ≥p, and we obtain the estimate: r X p=1 [2+3+m−p+2+4(m−p+1)(n−ip +1)] ≤ r X p=1 [m−p+7+4(m−p+1)(n−p+1)] ∼2mr(n−r)+2nr(m−r)+ 2 3r3 .
Comparing Householder transformation and Givens rotation, the former requires only nearly two thirds of the computational cost of the latter; however, because each Householder transformation work on (almost) the entire column of A simultaneously, it is less friendly to parallelization.
References M. I. Gillespie and D. O. Olesky. Ordering givens rotations for sparse QR factorization. SIAM J. Matrix Anal. A., 16(3):1024–1041, 1995.
6 |
3006 | https://saami.org/glossary/kinetic-energy-formula/ | Skip to content
KINETIC ENERGY (FORMULA)
< Back to home
Based on the general equation for kinetic energy, Ek=½mv2, which directly uses mass (m) and velocity (v), the following formula is used to calculate the kinetic energy of a projectile (expressed in foot-pounds):
E=(WV2)/(14000gc).
In this formula, W = weight of projectile, in grains; V = velocity, in feet per second; gc= gravitational constant, 32.174 ft/s2
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Allusion Definition and Examples
Apr 13, 2023
An allusion is a short reference to a person, object, place, event, or other literary work the reader is likely to be familiar with. Allusion allows an author to pack a lot of meaning and importance into a phrase or a single word. On the other hand, Allusions are only successful if the reader understands and recognizes them. When an allusion is misinterpreted or unclear, it loses its potency because it confuses the reader. Here are some Allusion examples.
For example, in his work Ragtime, E.L. Doctorow’s character “Little Boy” instructs another character who is traveling to Europe to “warn the Duke.” This is a reference to Archduke Franz Ferdinand of Austria, whose assassination set the series of events that led to World War I. Doctorow’s Allusion is beneficial for readers who notice and comprehend the connection. It emphasizes the importance of how one deed in history may have long-term global ramifications for readers.
What is Allusion?
An allusion is a brief reference to an individual, position, thing, event, or other erudite work that the anthology is likely familiar with. Allusion is an erudite system that enables a pen to army a lot of significance and meaning into a single word or expression. But references only work if the anthology recognizes them, understands them, and rightly infers and interprets them. An allusion might lose its impact by confusing the anthology if it’s unclear or misinterpreted.
Allusion Examples
Here are some different types of allusion we’ll be covering with examples:
Common Allusions in Everyday Speech
People frequently make references in regular discourse, sometimes without realizing it and sometimes without understanding what they are alluding to. These allusions are usually to popular cultures, such as movies, music, novels, famous personalities, etc. Here are some instances of popular allusions in ordinary speech, along with the original material they allude to:
To me, his grin is like kryptonite: Superman’s weakness
She had the feeling she had a golden ticket: Charlie and the Chocolate Factory
He’s young, scrappy, and eager: Hamilton
I wish I could click my heels together: The Wizard of Oz
My automobile could turn into a pumpkin if I don’t get home by midnight: Cinderella
She has a Cheshire cat grin on her face: Alice in Wonderland
His work is analogous to removing a sword from a stone: King Arthur’s Legend
Is there an Einstein among your physics students? Albert Einstein
He who shall not be named is my maths teacher: Voldemort from the Harry Potter series
I’m trying to sound like Queen B: Beyonce
Today may be the Ides of March: Shakespeare’s Julius Caesar
It would be a good idea to sit on my thinking chair right now: Blue’s Clues
I’m dressed like a caped crusader: Batman
Examples of Classical Mythology Allusions
Authors frequently use Allusion as a literary tactic to establish context for an accidental mention of something or a fleeting reference. Because most readers are familiar with ancient myths, their storylines, and characters, Roman and Greek mythology are frequently used as sources for references in literature, either indirectly or directly. Here are some examples of classical mythology allusions:
Achilles’ heel (alluding to the one weakness of Achilles)
Arrow of Love (Allusion to Cupid)
Carrying the weight of the world on your shoulders (Allusion to Atlas)
Pushing a boulder uphill every day (Allusion to Sisyphus)
Hot as Hades (king of the underworld/alluding to the god of death)
Looking like Venus (alluding to the goddess of beauty)
Herculean effort (alluding to the strength of Hercules)
Opening Pandora’s box (alluding to Pandora’s myth of letting trouble into the world)
Protean form of management (alluding to Proteus who could change his shape)
Can’t stop staring at himself (Allusion to Narcissus)
Famous Biblical Allusion Examples
Another commonly cited source for authors who use Allusion as a literary method is the Bible. Here are a few well-known examples of Biblical Allusion:
Garden (Eden, creation)
Snake (Satan, serpent)
Flood (Noah’s Ark)
Apple/fruit (temptation, tree of knowledge)
Cross (Christ, crucifixion)
Great patience (Job)
Fraternal competition (Cain and Abel)
Betrayal (Judas)
Generosity/philanthropy (Good Samaritan)
Slingshot/stone (David and Goliath)
Allusion Examples in Literature
In all kinds of writing, Allusion is a very powerful literary device. Authors can use allusions to create characters by linking them with other well-known and recognized archetypes or symbols. Literary allusions can also provide context for the reader by comparing or contrasting one work. Furthermore, by referring to the storyline and character of another work, Allusion can give exposition to a tale, allowing the reader to understand more about the story’s events or character motives.
When making allusions in a literary work, authors must strike a balance between what they choose to reference and how to weave it into their work to be clear and significant to the reader. Here are some examples of common allusions for an author to consider when employing this literary device:
Self-reference–when an author refers to another author’s work
Single connection–when an author uses Allusion to connect one work to another
Causal reference–when a reference is made but is not necessary to the plot
Corrective reference–when an author refers to another work that is opposed to their own
Apparent reference–when an author alludes to a certain source but in an unusual manner
Multiple references–when an author employs several references
As a literary device, Allusion is used to provide a context in academic work by referring to a well-known or recognizable person, location, event, or another piece of literature. Allusions aid in developing a link between an author’s invention and its identification with something else that the reader would recognize. This is useful for character development, contextualizing a tale, and building locations. The majority of literary references are not immediately explained to the reader but rely on the reader’s ability to infer.
Allusions Examples in Poetry
Illustration 1: The Bluest Eye (Toni Morrison)
Morrison references in her work the recognizable terminology of the Dick and Jane reading primer series, which was utilized in school curricula from the 1930s until the 1970s. The series used basic, repetitive verbiage and depicted a white, middle-class family consisting of a boy, girl, mother, father, dog, and cat living in a picturesque American neighborhood. In this section, Pecola’s character confronts the Dick and Jane series’ language, meaning, and aesthetic by running the words together and making them difficult to grasp for the novel’s reader.
This symbolizes Pecola’s mental breakdown and the discrepancy of pictures between her surroundings and experience as an African American girl in 1940 Ohio and the white youngsters in Dick and Jane’s world. Morrison’s literary trick of Allusion to Dick and Jane is powerful because readers widely recognized and understood it. It adds meaning to the meaninglessness Pecola sees in the primer’s words.
Illustration 2: Nothing Gold Can Stay (Robert Frost)
Frost alludes to the “fall of man” and exile from Paradise in his poem by referring to Eden and how it “sank to grief.” This reference connects the human experience to the poet’s sorrow about the natural cycle of life and death, beginning and ending. The connection to Eden also adds an interesting context to the poem. The reader can deduce that man’s fall was a natural and probably unavoidable occurrence, adding to the appreciation of life in its fleeting existence.
Illustration 3: As I Lay Dying (William Faulkner)
Faulkner’s work extensively emphasizes the reader’s capacity to detect and infer story components and character motivations. This is especially true for Addie Bundren’s character, who can only narrate one chapter before dying. In this section, Addie recalls her adulterous romance with Reverent Whitfield, which resulted in the birth of her son Jewel. In addition, this paragraph references Nathaniel Hawthorne’s novel The Scarlet Letter, in which Hester Prynne conceived a child as a result of her romance with Reverend Dimmesdale.
This metaphor by Faulkner serves to understand Addie’s character, her connection with Jewel, and her family’s fragmented structure. On the other hand, Faulkner depends on the reader to pick up on this brief connection to Hawthorne’s work, deduce its relevance and meaning, and apply it to his novel. If the reader misses or misinterprets Faulkner’s subtle Allusion, the Allusion is rendered ineffectual and meaningless.
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3008 | https://brilliant.org/courses/logical-languages/ | Sign inSign up
Learn to analyze arguments using Venn diagrams.
26 Lessons182 Exercises
Introducing Venn Diagrams
Sets
Counting in Sets
Complement
Intersection
Union
Set Operations and Counting
Empty Sets
Nonempty Sets
Counting with Unknowns
Limited Knowledge
Subsets
Determining Existence
Nested Subsets
Subsets and Venn Diagrams
Disjoint
Arguments
Validity
False Premises
Soundness
Diagramming Arguments
Arguments with Negation
Handling Existence
Arguing Existence
Arguing Existence and Negation
Analyzing Arguments
Introducing Venn Diagrams
Course description
Continue your adventures in logic by learning how to use Venn diagrams to visualize set operations. Use sets and logic to evaluate arguments for soundness and validity.
By the end of this course, you'll be able to systematically analyze a range of arguments.
Topics covered
Venn Diagrams
Euler Diagrams
Set Operations
Intersection
Union
Complement
Logical Quantifiers
Premises
Validity
Soundness
Syllogisms
Syllogism Fallacies
Prerequisites and next steps
This course is for anyone looking to sharpen their logical reasoning skills.
Prerequisites
Logic
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3009 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4894580/ | Guidelines for Expressing the Uncertainty of Measurement Results Containing Uncorrected Bias - PMC
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Learn more: PMC Disclaimer | PMC Copyright Notice
J Res Natl Inst Stand Technol
. 1997 Sep-Oct;102(5):577–585. doi: 10.6028/jres.102.039
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Guidelines for Expressing the Uncertainty of Measurement Results Containing Uncorrected Bias
Steven D Phillips
Steven D Phillips
1 National Institute of Standards and Technology, Gaithersburg, MD 20899-0001
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1, Keith R Eberhardt
Keith R Eberhardt
1 National Institute of Standards and Technology, Gaithersburg, MD 20899-0001
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1, Brian Parry
Brian Parry
2 Boeing Corporation, Seattle, WA 98124-2207
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1 National Institute of Standards and Technology, Gaithersburg, MD 20899-0001
2 Boeing Corporation, Seattle, WA 98124-2207
Accepted 1997 Apr 25.
The Journal of Research of the National Institute of Standards and Technology is a publication of the U.S. Government. The papers are in the public domain and are not subject to copyright in the United States. Articles from J Res may contain photographs or illustrations copyrighted by other commercial organizations or individuals that may not be used without obtaining prior approval from the holder of the copyright.
PMC Copyright notice
PMCID: PMC4894580 PMID: 27805145
Abstract
This paper proposes a method to extend the current ISO Guide to the Expression of Uncertainty in Measurement to include the case of known, but uncorrected, measurement bias. It is strongly recommended that measurement results be corrected for bias, however in some situations this may not be practical, hence an extension of the Guide is proposed to address this special situation. The method keeps with the spirit of the Guide in maintaining the link between uncertainty and statistical confidence. Similarly, the method maintains the transferability of one uncertainty statement to be included as a component in another uncertainty analysis. The procedure involves modifying the calculation of the expanded uncertainty, allowing it to become asymmetric about the measurement value. The method is compared to other alternative procedures, and an illustration of how it affects tolerance zones is presented.
Keywords: bias, error, expanded uncertainty, measurement, uncertainty
1. Introduction
Recently, the ISO Guide to the Expression of Uncertainty in Measurement (the Guide) , and the associated NIST adaptation , have described a unified convention for expressing measurement uncertainty. Application of the Guide has extended beyond calibration and research laboratories and into the industrial domain of manufactured products. At the factory floor level the recommended (and strongly preferred) practice of correcting for all known bias effects may not be economically possible due to such factors as limited instrumentation, operator training, and the large throughput of measurements. Since the Guide does not deal directly with the situation where a known measurement bias is present but is uncorrected, we propose a simple convention to extend the Guide’s procedures to address this special case. Uncorrected measurement bias may arise in situations where applying a correction for a known measurement bias would be costly, but increasing the measurement uncertainty to allow for the uncorrected bias would still result in an acceptable uncertainty statement. Initially, it might seem paradoxical to be aware of a measurement bias but fail to correct for it; however, such situations are rather common. For example, the user of an automated instrument may know a bias occurs under certain measurement situations, and be unable to modify the behavior of the instrument. Since “paper and pencil” corrections to each measurement value can be time consuming and error prone, particularly under high measurement throughput situations, it may be more economically reasonable to simply account for this bias by enlarging the uncertainty value that is attached to every measurement result.
A few examples of measurements that include uncorrected bias are now presented to illustrate the situation. A manufacturer of a precision positional indicator may know that all the indicators produced read approximately 0.5 % too high, with only a small variation of 0.1 % (standard deviation) between indicators. This is within the required 1 % relative uncertainty specification for the indicators and satisfies the customer’s needs. It may be expensive and difficult to adjust the manufacturing process to reduce the 0.5 % bias, or even to apply a 0.5 % correction to each unit; consequently, it is easier to subsume this bias into the uncertainty statement. Another example may be an automated instrument that is sensitive to some slowly varying parameter such as temperature, atmospheric pressure, humidity, etc. The instrument may lack a sensor to input this parameter and the user may be “locked out” of the software which records the measurement results; hence an automated correction cannot be performed to account for this systematic bias. However, the user may be able to specify the upper and lower acceptance limits, i.e., the conformance zone, for the measurement; see Fig. 1. If this bias is sufficiently small, it may be economically sensible to subsume it into the uncertainty statement. Doing so will alter the expanded uncertainty, and hence modify the conformance zone.
Fig. 1.
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The functional specification of a part and the corresponding inspection zones; U is the expanded uncertainty of the measurement. Parts are accepted if the measurement result is within the conformance zone.
The Guide primarily addresses the situation in which all known biases have been corrected, which is the recommended practice. However, Appendix F of the Guide does briefly discuss the situation of uncorrected measurement bias. It is our intention to extend this procedure and to provide examples of its implementation. Our motivation for this effort includes the observation that some industrial practitioners, in an effort to be consistent with the Guide, have included the bias as an ordinary uncertainty source which is added in the usual root-sum-of-squares (RSS) manner. This has the undesirable effect of incorrectly stating the expanded uncertainty, because the bias is added in an RSS manner and is multiplied by the coverage factor.1 Hence, the relationship between measurement confidence and the expanded uncertainty is broken, as illustrated in detail later. Since many parts are accepted or rejected on the basis that the measurement results demonstrate conformance with the part specification (see Fig. 1), it is important not to misstate the uncertainty (or confidence) associated with the measurement.
This document describes a convention to account explicitly for uncorrected measurement bias in an uncertainty statement. We believe any method to include measurement bias in the uncertainty statement should have the following desirable properties.
The final quoted uncertainty must be greater than or equal to the uncertainty that would be quoted if the bias was corrected. Underestimating the uncertainty indicates an invalid uncertainty statement. Similarly, excessively overestimating the uncertainty indicates a poorly constructed uncertainty statement.
The method must reduce to the method given by the Guide when the bias correction is applied.
For any coverage factor and any magnitude of bias, the level of confidence for the expanded uncertainty should be at least the level obtained for the case of corrected bias, e.g., if the distribution of the values that could reasonably be assigned to the measurand is Gaussian, then k = 2 should imply at least 95 % confidence.
The method should be transferable so that both the uncertainty and the bias from one result can be used as components in another uncertainty statement.
The method should be simple and inexpensive to implement.
2. Recommendations for Measurements Involving Uncorrected Bias
As described in the Guide, measurement results should be corrected for bias, and the uncertainty in the bias correction should be included as a contribution to the combined standard uncertainty. However, when correcting for the measurement bias is not practical, it still should be accounted for explicitly in the uncertainty statement. In our proposed approach, a complete uncertainty statement includes the combined standard uncertainty (computed as if the measurement result was to be corrected for the bias), an explicit statement of the (signed) bias value, and an expanded uncertainty which includes the effect of the bias term.
The usual method of using the expanded uncertainty U, for a measurement result y which has an unknown (“true”) value of the measurand Y, is to produce an uncertainty interval (with the level of confidence determined by the coverage factor) given by: y – U ≤ Y ≤ y + U. Consequently, the measurement result is often stated as: y ± U.
In the case where the result is corrected for a bias δ, a similar uncertainty interval can be constructed for the corrected measurement result y cor = (y − δ) given by: y cor − U ≤ Y ≤ y cor + U. This is equivalent to the uncertainty interval of: y – (U + δ) ≤ Y ≤ y + (U – δ). Consequently, the measurement result could be stated as: .
This can lead to the unfortunate possibility that one of the uncertainty limits may become negative, e.g., if the bias is positive and δ>U then the upper uncertainty limit will be negative. This may confuse practitioners, particularly when constructing diagrams such as Fig. 1; consequently, we propose the additional requirement that the uncertainty limits be greater than or equal to zero for all values of δ, which always maintains non-negative uncertainty limits at a cost of a somewhat wider uncertainty interval.
Hence, for a measurement result y which includes an uncorrected bias δ, the value of the measurand Y is estimated by the following uncertainty interval where U is the usual expanded uncertainty that would be calculated if the measurement had been corrected for bias; see Fig. 2. An uncertainty interval in the presence of uncorrected bias is given by:
Fig. 2.
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Uncorrected measurement result y, having bias δ (δ> 0), shown with expanded uncertainties U+ (black) and U− (dark gray) (top). The corresponding corrected measurement result y cor, is also shown together with the usual expanded uncertainty U (light gray) (bottom). The same situation with δ< 0.
Note that a large bias may result in a one sided uncertainty interval, e.g., if δ>U then U− = U + δ and U+ = 0. (One could propose a symmetric uncertainty interval, with the expanded uncertainty given by the larger of U+ or U−, but this further reduces the conformance zone with no additional economic benefit.)
When computing an uncertainty statement for cases where there are several sources of uncorrected bias, biases are algebraically added together (explicitly accounting for the sign of the bias). The resulting net bias is stated together with the combined standard uncertainty. Occasionally, the case may arise where multiple sources of uncertainty have bias and these biases are not independent. To avoid “double counting” the bias sources, the degree of overlap of the biases is estimated and this amount is subtracted from the bias summation. The uncertainty in the overlap correction is added in a RSS manner to the combined standard uncertainty. Finally, we point out that the expanded uncertainty must be re-computed if the coverage factor is changed, and in particular, that U± (k = 2) ≠ 2 × U− (k = 1).
The proposed approach is recommended for its simplicity and utilitarian value (see Fig. 3), even though it can somewhat overestimate the uncertainty. However, for a given coverage factor the corresponding level of confidence will be at least as high as would be the case if the bias had been corrected.
Fig. 3.
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An example of the functional specification of a part and the corresponding inspection zones for a measurement result with uncorrected bias; U+ and U− are the expanded uncertainties which include the effects of measurement bias (top: bias > 0, bottom: bias < 0). Parts are accepted if the measurement result is within the conformance zone.
3. Comparison With Other Methods of Combining Uncorrected Bias
We compare our proposed method of treating uncorrected bias with two other procedures which have been proposed to address this problem. The first procedure treats the uncorrected bias as another uncertainty source and simply sums it in an RSS manner with the usual combined standard uncertainty; we denote this method as RSS u c. The second method sums the bias in a RSS manner with the usual expanded uncertainty; we denote this as RSS U. In contrast our proposed method algebraically sums the signed bias with the expanded uncertainty (unless the bias is large), so we denote our method as SUM U. The three methods are shown below.
RSS u c Method:
where
RSS U Method:
where
SUM U Method:
where
and .
Figure 4 illustrates some important differences between the three methods. The three plots display the actual statistical confidence of the three methods versus the magnitude of the uncorrected bias for coverage factors of k = 1, 2, and 3. Gaussian (normal) distributions are assumed in all cases. For example, in the k = 2 case, ideally we would like to maintain a 95 % (strictly speaking, 95.44 %) confidence for all values of the uncorrected bias. Our proposed method (SUM U) maintains this confidence until the ratio of the bias to the combined standard uncertainty becomes larger than the coverage factor. For such large bias values, the SUM U method produces uncertainty intervals that are slightly conservative (larger than necessary to produce valid 95 % confidence levels.) The RSS u c method, on the other hand, can produce uncertainties that are considerably larger than necessary. For example, with k = 2 and a bias twice as large as the combined standard uncertainty (δ/u c = 2), the actual achieved confidence level of the interval is nearly 100 %, rather than the nominal 95 % (see Fig. 4.) Although this overstatement of the uncertainty is not necessarily disastrous, it can come at the significant cost of consuming most of the part tolerance zone, i.e., specification zone, as we will soon describe. In contrast, the RSS U method seriously understates the true uncertainty. For example, with a coverage factor of k = 2 and an uncorrected bias twice as large as the combined standard uncertainty (δ/u c = 2), the uncertainty interval is under-sized to the extent that the actual achieved confidence is less than 80 %, which falls significantly short of the nominal 95 % confidence level.
Fig. 4.
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Comparison of the actual achieved confidence levels resulting from three methods of incorporating uncorrected bias in uncertainty intervals. The methods are the proposed SUM U method (solid line), the RSS u c method (dotted line) and the RSS U method (dashed line). (In the case k = 1, RSS u c = RSS U.)
The three plots in Fig. 5 display the relative sizes of the expanded uncertainty interval for each of the three methods as a function of the magnitude of the uncorrected bias, for the coverage factors k = 1, 2, and 3. The scale of the ordinate on the left hand side of the plots is defined as the full width of the uncertainty interval divided by the combined standard uncertainty, and hence the ratio would be equal to 2 k (where k is the coverage factor) if the bias had been corrected. As seen in the figure, the SUM U method consistently produces the smallest expanded uncertainty interval of any of the methods for all values of bias and coverage factors.
Fig. 5.
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Comparison of uncertainty interval lengths for three methods of incorporating uncorrected bias in uncertainty intervals. The methods are the proposed SUM U method (solid line), the RSS u c method (dotted line) and the RSS U method (dashed line). The scale on the right-hand axis of the plots assumes that the value of the expanded uncertainty represents a gauging ratio of 4:1 in the case of zero bias (δ = 0). (In the case k = 1, RSS u c = RSS U.)
An example of how the size of the expanded uncertainty interval might impact the user is shown on the right hand ordinate. This axis describes the percentage of the specification zone that is consumed by the expanded uncertainty interval for the somewhat typical case where the ratio of the specification zone to expanded uncertainty interval zone (if the bias had been corrected) is 4:1. (Obviously, the right hand ordinate axis is numerically correct only for the particular example of a gauging ratio of 4:1; other ratios, while having qualitatively similar behavior, would have different percentages of the specification zone consumed by the expanded uncertainty interval.) The plot illustrates that the SUM U method consumes the smallest percentage of the specification zone compared to the other two methods. For example, in the k = 2 case, and for an uncorrected bias equal to four times the combined standard uncertainty (δ/u c = 4), the SUM U method would consume 37.5 % of the specification zone (compared to 25 % if the bias had been corrected), while the RSS U and RSS u c methods consume 56 % and over 100 %, respectively.
Figures 4 and 5 illustrate that, of the three methods described, our proposed method (SUM U) offers a significant advantage in reducing the impact on the user of the uncorrected bias when subsumed into the uncertainty statement. This method always maintains an actual level of confidence equal to or greater than the nominal confidence corresponding to the coverage factor in use. While our examples are based on the Gaussian distribution, the SUM U method retains this relationship (between k and the confidence level) for any distribution shape because the resulting interval always contains at least the same interval as would be covered by the corrected result. Furthermore, of the three methods discussed, the SUM U method minimizes the percentage of the specification zone consumed by the expanded uncertainty interval. In addition, the method avoids negative expanded uncertainties, which could be confusing to the user when determining the conformance zone (as shown in Fig. 3).
4. Examples of Uncertainty Statements Containing Uncorrected Bias
The following examples should clarify the procedure for expressing measurement uncertainty in the presence of uncorrected bias. We use the same terminology as the Guide when referring to the evaluation of the bias, i.e., Type A corresponds to any valid statistical method for treating the data and Type B corresponds to a bias evaluated by other means. For each example we give a worked numerical case to illustrate the procedure; in these examples all expanded uncertainties are evaluated with a coverage factor of k = 2. These examples are contrived to illustrate the procedure of accounting for uncorrected bias and are not designed to describe the subtleties of creating an uncertainty statement; consequently many uncertainty sources may have been omitted or simplified.
4.1 Example 1: One Type A bias
Consider a measurement result y, having a constant bias of estimated magnitude δ 1. Assume that δ 1 is assessed directly by repeated measurements of a reference standard having a combined standard uncertainty of u ref. Specifically, suppose δ 1 is evaluated as the average deviation from the reference standard’s calibrated value found from N 1 measurements. Let the experimental standard deviation of the N 1 measurements be s; the standard uncertainty of the estimated bias is then . The combined standard uncertainty of the measurement result is given below, where u 1 accounts for the combination of all other uncertainty sources not directly associated with the bias. Note that u 1 already includes the repeatability of the measurement, i.e., the standard deviation s, since this source of uncertainty is always present and is unaffected by the fixed bias. The combined standard uncertainty is the same quantity that would be determined if the measurement had been corrected for the bias. Note that the expanded uncertainty is treated asymmetrically and the results depend on the sign of the bias. In this example δ 1< 0 and ku c + δ 1> 0.
The uncertainty interval is given by y –U−≤Y≤y+U+. Equivalently the result can be stated with the expanded uncertainty as .
Numerical case: Suppose that a measuring machine designed to inspect parts of length 100 mm is checked with a reference standard having a combined standard uncertainty of u ref = 1.5 μm. A total of 15 measurements are recorded having an experimental standard deviation of 3.0 μm with the mean result 4.0 μm smaller than the calibrated value of the reference standard, i.e., the bias is negative. From previous work it is known that all other uncertainty sources combined yield a standard uncertainty of 5.0 μm, i.e., u 1 = 5.0 μm. Therefore:
and the measurement result y can be stated with the expanded uncertainty as: .
4.2 Example 2: One Type B bias
For some measurements, the bias might be estimated rather than directly measured. For example, length measurements made on the factory floor often are not corrected back to the standard temperature of 20 °C. Hence, the uncorrected thermal expansion represents a measurement bias. Suppose the factory floor temperature varies between 20 °C and 30 °C, about an estimated mean of 25 °C. The estimated magnitude of the bias is given by δ 2 (δ 2> 0) which accounts for the length deviation due to the 5 °C mean uncorrected thermal expansion. The variability of the temperature can be described by a uniform distribution of full width 10 °C, i.e., by a standard uncertainty of 2.9 °C which, when multiplied by the appropriate coefficient of thermal expansion, gives rise to the corresponding standard uncertainty u temp. The combined standard uncertainty and expanded uncertainty are given below, where u 2 would be the combined standard uncertainty for the measurement if the measurement had been corrected back to 20 °C. (The value of u 2 includes the uncertainties in the temperature measurements, the uncertainties in the thermal expansion coefficient, and other effects.) In this example δ 2> 0 and ku c – δ 2> 0.
The uncertainty interval is given by y –U−≤Y≤y+U+. Equivalently the result can be stated with the expanded uncertainty as .
Numerical case: Suppose a measuring machine that inspects parts of length 100 mm has u 2 = 7.0 μm, and the machine’s scale has a thermal expansion coefficient of 9 (μm/m)/°C, and the part under inspection has a thermal expansion coefficient of 22 (μm/m)/°C. Then the differential thermal expansion is 22 (μm/m)/°C – 9 (μm/m)/°C = 13 (μm/m)/°C, corresponding to an average bias of 13(μm/m)/°C×5 °C×0.100 m=6.5μm. The standard uncertainty associated with the variation in the temperature (modeled as a uniform distribution) is given by u temp = 13 (μm/m)/°C × 0.100 m × 2.9 °C= 3.8 μm. Therefore:
and the measurement result y can be stated with the expanded uncertainty as: .
4.3 Example 3: Combination of independent biases
An uncertainty statement consists of two uncertainty sources given by those of examples 1 and 2, which are assumed to be independent. The resulting uncertainty statement is given below. Note that δ 3 is the sum of the two biases, and that we assume δ 3> 0 and ku c–δ 3> 0; u c1 and u c2 are the combined standard uncertainties from examples 1 and 2, respectively.
The uncertainty interval is given by y –U−≤Y≤y+U+. Equivalently the result can be stated with the expanded uncertainty as .
Numerical case: Using the values given in examples 1 and 2 we find
and the measurement result y can be stated with the expanded uncertainty as: .
4.4 Example 4: Combination of independent and dependent biases
The measuring instrument described by the uncertainty statement of Example 3 is modified by an accessory that does not add variability but produces an additional bias δ. This bias is assessed by repeated measurements, i.e., found from N 4 measurements of a second (independent) reference standard (having a combined standard uncertainty of u ref2). The measurements collectively have an experimental standard deviation s (this is the same standard deviation found in example 1), and a mean value differing from the calibrated value by δ, with δ< 0 and δ<δ 1< 0. It is estimated that between 30 % to 50 % of the bias estimated by δ is already accounted for in δ 1. To avoid double counting, 0.4 δ (which is the best estimate of the overlap, i.e., the average of 30 % and 50 % = 40 % = 0.4) is subtracted from the bias summation. A standard uncertainty of , corresponding to a uniform distribution (from 0.3 δ to 0.5 δ with half width 0.1 δ), accounting for the uncertainty of the bias overlap is added in an RSS manner to the other standard uncertainties. We assume the total net bias δ 4> 0 and ku c – δ 4> 0, as shown below.
The uncertainty interval is given by y –U−≤Y≤y+U+. Equivalently the result can be stated with the expanded uncertainty as .
Numerical case: The additional bias δ = –2.0 μm is evaluated as the mean of 10 measurements using a reference standard with combined standard uncertainty u ref 2 = 1.0 μm. Using the values given in the previous examples, we find
and the measurement result y can be stated with the expanded uncertainty as: .
Acknowledgments
This work was supported in part by the U.S. Air Force Calibration Coordination Group (CCG) program and by the Advanced Technology Program (ATP). The authors thank the many individuals who reviewed this and earlier drafts of the manuscript.
Biography
About the authors: Steve Phillips is a physicist in the Precision Engineering Division of the Manufacturing Engineering Laboratory at NIST. Keith Eberhardt is a mathematical statistician in the Statistical Engineering Division of the Information Technology Laboratory at NIST. Brian Parry P. E. is a Principal Engineer with The Boeing Company. The National Institute of Standards and Technology is an agency of the Technology Administration, U.S. Department of Commerce.
Footnotes
1
An inaccurate method for calculating an expanded uncertainty can lead to the uncertainty being either overstated or understated, depending on the values of the bias, combined standard uncertainty, the coverage factor, and the shape of the distribution. When the uncertainty is overstated (i.e., is too large), the nominal confidence level claimed will be smaller than it should be to properly describe the given uncertainty. Conversely, if the uncertainty is understated (i.e., is too small), the nominal confidence level claimed will be larger than it should be.
5. References
1.International Organization for Standardization . Guide to the Expression of Uncertainty in Measurement. Geneva, Switzerland: 1993. corrected and reprinted 1995. [Google Scholar]
2.Taylor BN, Kuyatt CE. Guidelines for Evaluating and Expressing the Uncertainty of NIST Measurement Results. 1994 Edition. National Institute of Standards and Technology; Gaithersburg, MD 20899: 1994. (NIST Technical Note 1297). [Google Scholar]
3.Draft International Standard, Geometrical Product Specifications (GPS) Inspection by measurement of workpieces and measuring instruments—Part 1: Decision rules for providing conformance or non-conformance with specification. 1997 ISO/DIS 14253-1. [Google Scholar]
Articles from Journal of Research of the National Institute of Standards and Technology are provided here courtesy of National Institute of Standards and Technology
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Abstract
1. Introduction
2. Recommendations for Measurements Involving Uncorrected Bias
3. Comparison With Other Methods of Combining Uncorrected Bias
4. Examples of Uncertainty Statements Containing Uncorrected Bias
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3010 | https://par.nsf.gov/servlets/purl/10545645 | Lower Bounds for Non-adaptive Shortest Path Relaxation
David Eppstein (B)
Computer Science Department, University of California, Irvine, Irvine, CA 92697, USA
eppstein@uci.edu
Abstract. We consider single-source shortest path algorithms that per-form a sequence of relaxation steps whose ordering depends only on the input graph structure and not on its weights or the results of prior steps. Each step examines one edge of the graph, and replaces the tentative distance to the endpoint of the edge by its minimum with the tentative distance to the start of the edge, plus the edge length. As we prove, among such algorithms, the Bellman-Ford algorithm has optimal com-plexity for dense graphs and near-optimal complexity for sparse graphs, as a function of the number of edges and vertices in the given graph. Our analysis holds both for deterministic algorithms and for randomized algorithms that find shortest path distances with high probability.
1 Introduction
Dijkstra’s algorithm finds shortest paths in directed graphs when all edge weights are non-negative, but the problem becomes more difficult when negative edge weights (but not negative cycles) are allowed. In this case, despite recent breakthroughs on near-linear time bounds for graphs with small integer edge weights , the best strongly-polynomial time bound for single-source shortest paths remains that of the Bellman-Ford algorithm [4,10,18], which takes time
O(mn ) on graphs with m edges and n vertices, or O(n3) on dense graphs. Both Dijkstra’s algorithm and the Bellman-Ford algorithm (as well as an unnamed linear-time algorithm for single-source shortest paths in directed acyclic graphs) can be unified under the framework of relaxation algorithms ,also called label-correcting algorithms . These algorithms initialize tentative distances D[v] from the source vertex to each other vertex v, by setting D[s] = 0
and D[v] = +∞ for v = s. Then, they repeatedly relax the edges of the graph. This means, that for a given edge u → v, the algorithm updates D[v] to
D[u] + length( u → v). In Dijkstra’s algorithm, each edge u → v is relaxed once, in sorted order by the tentative distance D[u]. In the Bellman-Ford algorithm, an edge can be relaxed many times. The algorithm starts with the tentative dis-tance equal to the correct distance for s, but not for the other vertices. When-ever the algorithm relaxes an edge u → v in the shortest path tree, at a time when u already has the correct distance, the tentative distance to v becomes correct as well. Thus, the goal in designing the algorithm is to perform these
c©The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 P. Morin and S. Suri (Eds.): WADS 2023, LNCS 14079, pp. 416–429, 2023. _27 Lower Bounds for Non-adaptive Shortest Path Relaxation 417
distance-correcting relaxations while wasting as little effort as possible on other relaxations that do not correct any distance, and on the overhead in selecting which relaxation to perform. We would like to prove or disprove the optimality of the Bellman–Ford algo-rithm among a general class of strongly-polynomial shortest path algorithms, without restricting the types of computation such an algorithm can perform, but such a result appears to remain far out of reach. Instead, in this work we focus only on relaxation algorithms, asking: how few relaxation steps are needed? Note that, without further assumptions, a shortest path algorithm could “cheat”, computing a shortest path tree in some other way and then performing only n−1
relaxation steps in a top-down traversal of a shortest path tree. To focus purely on relaxation, and prevent such cheating, we consider non-adaptive relaxation algorithms , in which the sequence of relaxation steps is determined only by the structure of the given graph, and not on its weights nor on the outcome of earlier relaxation steps. Dijkstra’s algorithm is adaptive, but the linear-time DAG algo-rithm is non-adaptive. Another example of a non-adaptive algorithm comes from past work on the graphs in which, like DAGs, it is possible to relax every edge once in a fixed order and guarantee that all tentative distances are correct . As usually described, the Bellman-Ford algorithm is adaptive. Its typical optimiza-tions include adaptive rules that disallow repeatedly relaxing any edge u → v
unless the tentative distance to u has decreased since the previous relaxation, and that stop the entire algorithm when no more allowed relaxations can be found. However, its same asymptotic time bounds can be achieved by a non-adaptive version of the Bellman-Ford algorithm, with a round-robin relaxation sequence, one that merely repeats n − 1 rounds of relaxing all edges in the same order per round. A non-adaptive asynchronous distributed form of the Bellman-Ford algorithm is widely used in distance vector routing of internet traffic, to maintain paths of minimum hop count between major internet gateways .
1.1 Known Upper Bounds
We do not require non-adaptive relaxation algorithms to be round-robin, but we are unaware of any way to take advantage of this extra flexibility. Nevertheless, among round-robin algorithms, there is still freedom to choose the ordering of edges within each round, and this freedom can lead to improved constant factors in the number of relaxation steps performed by the Bellman-Ford algorithm. Yen described a method based on the following idea. Choose an arbitrary linear ordering for the vertices, and partition the edges into two subsets: the edges that are directed from an earlier vertex to a later vertex in the ordering, and the edges that are directed from a later vertex to an earlier vertex. Both of these two edge subsets define directed acyclic subgraphs of the given graph, with the chosen linear ordering or its reverse as a topological ordering. Use a round-robin edge ordering that first relaxes all of the edges of the first subgraph, in its topological order, and then relaxes all of the edges of the second subgraph, in its topological order. If any shortest path is divided into contiguous subpaths that lie within one of these two DAGs, then each two consecutive subpaths from the first and 418 D. Eppstein
second DAG will be relaxed in order by each round of the algorithm. In the worst case, there is a single shortest path of n − 1 edges, alternating between the two DAGs, requiring n/ 2 rounds of relaxation. For complete directed graphs, this method uses ( 12 + o(1) )n3 relaxation steps, instead of the (1 + o(1) )n3 that might be used by a less-careful round-robin method. As we showed in earlier work , an additional constant factor savings can be obtained by a randomized algorithm that selects from a random distribution of non-adaptive relaxation sequences, and that obtains a correct output with high probability rather than with certainty. To do so, use Yen’s method, but choose the vertex ordering as a uniformly random permutation of the vertices, rather than arbitrarily. In any shortest path tree, each vertex with more than one child reduces the number of steps from the source to the deepest leaf by one, reducing the number of alternations between the two DAGs. For each remaining vertex with one child in the tree, the probability that it lies between its parent and child in the randomly selected ordering is 13 , and when this happens, it does not contribute to the bound on the number of alternations. With high probability, the number of these non-contributing vertices is close to one third of the single-child vertices. Therefore, with high probability, the maximum number of alternations between the two DAGs among paths on the shortest path tree is ( 23 + o(1) )n,and an algorithm that uses this method to perform ( 13 + o(1) )n3 relaxation steps will find the correct shortest paths with high probability. The worst-case asymptotic time of these methods remains O(n3) for complete graphs, and O(mn ) for arbitrary graphs with m vertices and n edges. Both Yen’s method and the randomized permutation method can also be used in adaptive versions of the Bellman-Ford algorithm, with better constant factors and in the randomized case leading to a Las Vegas algorithm rather than a Monte Carlo algorithm, but it is their non-adaptive variants that concern us here.
1.2 New Lower Bounds
We provide the following results: – Any deterministic non-adaptive relaxation algorithm for single-source short-est paths on a complete directed graph with n vertices must use ( 16 − o(1) )n3
relaxation steps. – Any randomized non-adaptive relaxation algorithm for shortest paths on a complete directed graph with n vertices, that with high probability sets all distances correctly, must use ( 112 − o(1) )n3 relaxation steps. – For any m and n with n ≤ m ≤ 2(n
2
), there exists a directed graph on m edges and n vertices on which any deterministic or high-probability randomized non-adaptive relaxation algorithm for shortest paths must use Ω(mn/ log n)
relaxation steps. When m = Ω(n1+ ε) for some ε > 0, the lower bound improves to Ω(mn ).These lower bounds hold even on graphs for which all edges weights are zero and one, for which an adaptive algorithm, Dial’s algorithm, can find shortest paths in linear time . Lower Bounds for Non-adaptive Shortest Path Relaxation 419
1.3 Related Work
Although we are not aware of prior work in the precise model of computation that we use, variants of the Bellman-Ford algorithm have been studied and shown optimal for some other related problems: – The k-walk problem asks for a sequence of exactly k edges, starting and one vertex and ending at the other, allowing repeated edges. The Bellman-Ford algorithm can be modified to find the shortest k-walk between two vertices in time O(kn 2), non-adaptively. In any non-adaptive relaxation algorithm, the only arithmetic operations on path lengths and edge weights are addition and minimization, and these operations are performed in a fixed order. There-fore, the sequence of these operations can be expanded into a circuit, with two kinds of gates: minimization and addition. The resulting (min , +) -circuit model of computation is somewhat more general than the class of relaxation algorithms, because the sequence of operations performed in this model does not need to come from a sequence of relaxation steps. The k-walk version of the Bellman-Ford algorithm is nearly optimal in the (min , +) -circuit model: circuit size Ω(k(n − k)n) is necessary . However, this k-walk problem is different from the shortest path problem, so this bound does not directly apply to shortest paths. – Under conditional hypotheses that are standard in fine-grained complexity analysis, the O(km ) time of Bellman-Ford for finding paths of at most k
steps, for graphs of m edges, is again nearly optimal: neither the exponent of k nor the exponent of m can be reduced to a constant less than one. For large-enough k, the shortest path of at most k steps is just the usual shortest path, but this lower bound applies only for choices of k that are small enough to allow the result to differ from the shortest path . – Another related problem is the all hops shortest path problem, which asks to simultaneously compute k paths, having distinct numbers of edges from one to a given parameter k. Again, this can be done in time O(km ) by a variant of the Bellman-Ford algorithm, and it has an unconditional Ω(km ) lower bound for algorithms that access the edge weights only by path length comparisons, as Bellman-Ford does [6,11]. Because it demands multiple paths as output, this lower bound does not apply to algorithms that compute only a single shortest path. – Meyer et al. study a version of the Bellman-Ford algorithm, in which edges are relaxed in a specific (adaptive) order. They construct sparse graphs, with
O(n) edges, on which this algorithm takes Ω(n2) time, even in the average case for edge weights uniformly drawn from a unit interval. This bound applies only to this algorithm and not to other relaxation orders.
2 Deterministic Lower Bound for Complete Graphs
The simplest of our results, and the prototype for our other results, is a lower bound on the number of relaxations needed by a deterministic non-adaptive 420 D. Eppstein
relaxation algorithm, in the worst case, on a complete directed graph with n
vertices.
Theorem 1. Any deterministic non-adaptive relaxation algorithm for single-source shortest paths on a complete directed graph with n vertices must use at least ( 16 − o(1) )n3 relaxation steps. Proof. Fix the sequence σ of relaxation steps chosen by any such algorithm. We will find an assignment of weights for the complete directed graph, such that the distances obtained by the relaxation algorithm are not all correct until ( 16 −
o(1) )n3 relaxation steps have taken place. Therefore, in order for the algorithm to be correct, it must make this many steps. For the weights we choose, the shortest path tree will form a single directed path, of n − 1 edges, starting at the source vertex. In order for the relaxation algorithm to achieve correct distances to all vertices, its sequence of relaxations must include a subsequence consisting of all path edges in order. The weights of these edges are unimportant (because we are considering only non-adaptive algorithms) so we may set all path edges to have weight zero and all other edges to have weight one. To determine this path, we choose one at a time its edges in even positions: its second, fourth, sixth, etc., edge. These chosen edges include every vertex in the path, so choosing them will also determine the edges in odd positions. When choosing the ith edge (for an even number i), we make the choice greedily, to maximize the position in σ of the step that relaxes this edge and makes its endpoint have the correct distance. Let si denote this position, with s0 = 0 as a base case recording the fact that, before we have relaxed any edges, the source vertex already has the correct distance. Then the length of σ is at least equal to the telescoping sum
(s2 − s0) + ( s4 − s2) + ( s6 − s4) + · · · .
When choosing edge i, for an even position i, there are i − 1 earlier vertices, whose position in the shortest path is already determined, and n−i+1 remaining vertices. Between step si−2 and step si of the relaxation sequence σ, it must relax all n − i + 1 edges from the last endpoint of edge i − 2 to one of these remaining vertices, and all 2(n−i+1 2
) edges between pairs of the vertices that remain to be corrected. For, if it did not do so, there would be an edge that it had not relaxed, and choosing this edge next would cause si to be greater; but this would violate the greedy choice of edge i to make si as large as possible. Therefore,
si − si−2 ≥ (n − i + 1) + 2
(n − i + 1 2
)
= ( n − i + 1) 2.
Summing over all (n − 1) /2 choices of edges in even positions gives, as a lower bound on the total number of relaxation steps,
∑
i=2 ,4,6,...
si − si−2 ≥ ∑
i=2 ,4,6,...
(n − i + 1) 2 = n3 − n
6 ,
where the closed form for the summation follows easily by induction. Lower Bounds for Non-adaptive Shortest Path Relaxation 421
3 Randomized Lower Bound for Complete Graphs
It does not make much sense to consider expected time analysis for non-adaptive algorithms, because these algorithms have a fixed stopping time (determined as a function of the given graph), and we want their output to be correct with high probability rather than in any expected sense. Nevertheless, it is often easier to lower-bound the expected behavior of randomized algorithms, by using Yao’s principle , according to which the expected cost of a randomized algorithm on its worst-case input can be lower bounded by the cost of the best deterministic algorithm against any random distribution of inputs. In order to convert high-probability time bounds into expectations, we con-sider randomized non-adaptive algorithms that are guaranteed to produce the correct distances, and we define the reduced cost of such an algorithm to be the number of relaxations that it performs until all distances are correct, ignoring any remaining relaxations after that point.
Lemma 1. If a randomized non-adaptive relaxation algorithm A takes s(G)
steps on any weighted input graph G and computes all distances from the source vertex correctly with probability 1 − o(1) , then there exists a randomized non-adaptive relaxation algorithm B that is guaranteed to produce correct distances and whose expected reduced cost, on weighted graphs G with n vertices and m
edges, is at most s(G) + o(mn ).Proof. Construct algorithm B by using the relaxation sequence from algorithm
A, appending onto it the sequence of relaxations from a conventional non-adaptive deterministic Bellman-Ford algorithm. Then with probability 1 − o(1)
the relaxed cost of B counts only the relaxation sequence from algorithm A, of length s(G). With probability o(1) the relaxed cost extends into the determinis-tic Bellman-Ford part of the sequence, of length O(mn ). Because this happens with low probability, its contribution to the expected reduced cost is o(mn ).
Corollary 1. Any lower bound on expected reduced cost is also a valid lower bound, up to an additive o(mn ) term, on the number of relaxation steps for a randomized non-adaptive relaxation algorithm that produces correct distances with high probability.
With this conversion to expected values in hand, we may now formulate Yao’s principle as it applies to our problem. We need the following notation:
Definition 1. For any graph G, with a specified source vertex, let WG be the family of assignments of real weights to edges of G. Let DG be the family of probability distributions of weights in WG, and let ΣG be the class of relaxation sequences on G that are guaranteed to produce correct distances from the speci-fied source vertex. For any randomized non-adaptive relaxation algorithm A and weight vector w ∈ WG, let rG(A, w ) denote the expected reduced cost of run-ning algorithm A on G with edges weighted by w. For σ ∈ ΣG and D ∈ D G
let ρG(σ, D ) be the expected reduced cost of sequence σ on weight vectors drawn from D.422 D. Eppstein
Lemma 2 (Yao’s principle). For any graph G with specified source vertex, and any randomized non-adaptive relaxation algorithm A,
min
A
max
w∈WG
rG(A, w ) = max
D∈D G
min
σ∈ΣG
ρG(σ, D ).
Proof. This is just the minimax principle for zero-sum games, applied to a game in which one player chooses a relaxation sequence σ ∈ ΣG, the other player chooses a weight vector w ∈ WG, and the outcome of the game is the reduced cost for σ on w. According to that principle, the value of the best mixed strategy for the sequence player, against its worst-case pure strategy (the left hand side of the equality in the lemma) equals the value of the best mixed strategy for the weight player, against its worst-case pure strategy (the right hand side).
Corollary 2. For any weight distribution D ∈ D G, min σ∈ΣG ρG(σ, D ) is a valid lower bound on the expected reduced cost of any randomized non-adaptive relax-ation algorithm that is guaranteed to produce correct distances. Proof. An arbitrary algorithm A can only have a greater or equal value to the left hand side of Lemma 2, and an arbitrary weight distribution D can only have a smaller or equal value to the right hand side. So the expected reduced cost of the algorithm, on a worst-case input, can only be greater than or equal to the value given for D in the statement of the corollary.
Theorem 2. Any randomized non-adaptive relaxation algorithm for shortest paths on a complete directed graph with n vertices, that with high probability sets all distances correctly, must use at least ( 112 − o(1) )n3 relaxation steps. Proof. We apply Corollary 2 to a weight distribution D defined as follows: we choose a random permutation of the vertices of the given complete graph, start-ing with the source vertex, we make the weight of edges connecting consecutive vertices in order along this permutation zero, and we make all other weights one. Thus, each weighting of the complete graph drawn from this distribution will have a unique shortest path tree in the form of a single path, with all paths from the source vertex equally likely. For any weight vector w drawn from D, let πw
be this path. Let σ be any relaxation sequence in ΣD . As in the proof of Theorem 1, we define si (for a weight vector w to be determined) to be the step at which the second endpoint of the ith edge of πw has its shortest path distance set correctly. Let Ci denote the conditional probability distribution obtained from D by fixing the choice of the first i edges of πw. Under condition Ci, the remaining
n − i − 1 vertices remain equally likely to be permuted in any order. There are
2(n−i−12
) choices for edge i + 2 , each of which is equally likely. Therefore, the expected value of si+2 − si is greater than or equal to the average, among these edges, of their distance along sequence σ from position si. (It is greater than or equal, rather than equal, because this analysis does not take into account the requirement that edge i + 1 must be relaxed first, before we relax edge i + 2 .) Sequence σ can minimize this average if, in σ, the next 2(n−i−12
) relaxation Lower Bounds for Non-adaptive Shortest Path Relaxation 423
Fig. 1. Schematic view of the graphs used for our lower bound construction
steps after si are exactly these distinct edges. When σ packs the edges in this minimizing way, the average is 2(n−i−12
)/2; for other sequences it can only be greater. Therefore,
E[si+2 − si | Ci] ≥
(n − i − 12
)
.
Summing these expected differences, over the sequence of values si for even i,and applying Corollary 1 and Corollary 2, gives the result.
4 Lower Bounds for Incomplete Graphs
In our lower bounds for complete graphs, the edges in even and odd positions of the shortest paths perform very different functions. The edges in even positions are the ones that, at each step in the shortest path, force the relaxation sequence to have a large subsequence of relaxation steps. Intuitively, this is because there are many possible choices for the edge at the next step and all of these possibili-ties (in the deterministic bound) or many of these possibilities (in the randomized bound) must be relaxed before reaching the edge that is actually chosen. The edges in odd positions, on the other hand, do not contribute much directly to the length of the sequence of relaxation steps. Instead, they are used to connect the edges in the even positions into a single shortest path. To construct graphs that are not complete, for which we can prove analogous lower bounds, we make this dichotomy more explicit. For a chosen “capacity” parameter c, we will construct graphs that have two designated subsets of c
vertices, S and T (with the source vertex contained in subset S). We will connect the vertices in T to the vertices in S by a biregular bipartite directed graph of some degree d ≈ m/ 2c, a graph in which each vertex in T has exactly d outgoing neighbors and each vertex in S has exactly d incoming neighbors. This biregular 424 D. Eppstein
graph will perform the function of the even position edges in our complete graph lower bounds: it will have many edges to choose from, forcing any relaxation algorithm to make a long subsequence of relaxations between each two chosen edges. The detailed structure of this graph is not important for our bounds. In the other direction, from S to T , we will construct a special graph with the property that, no matter which sequence of disjoint edges we choose from the biregular graph, we can complete this sequence to a path. A schematic view of this construction is depicted in Fig. 1. We begin the more detailed description of this structure by defining the graphs we need to connect from S to T . The following definition is standard:
Definition 2. A rearrangeable non-blocking network of capacity c is a directed graph G with c vertices labeled as inputs, and another c vertices labeled as out-puts, with the following property. For all systems of pairs of inputs and outputs that include each input and output vertex at most once, there exists in G a system of vertex-disjoint paths from the input to the output of each pair.
Observation 3. A complete bipartite graph Kc,c , with its edges directed from
c input vertices to c output vertices, is a rearrangeable non-blocking network of capacity c, with 2c vertices and c2 edges. In this case, the disjoint paths realizing any system of disjoint input-output pairs is just a matching, formed by the edges from the input to the output in each pair.
Lemma 4. For any capacity c, there exist rearrangeable non-blocking network of capacity c with O(c log c) vertices and edges.
Pippenger credits the proof of Lemma 4 to Beizer , who used a recursive construction. A more recent construction of Alon and Capalbo is based on blowing up an expander graph, producing enough copies of each vertex that a system of edge-disjoint paths in the expander can be transformed into a system of vertex-disjoint paths in the non-blocking network. Their networks are non-blocking in a stronger sense (the vertex-disjoint paths can be found incrementally and efficiently), but we do not need that additional property. A simple counting argument shows that o(c log c) edges is not possible: to have enough subsets of edges to connect c! possible systems of pairs, the number of edges must be at least log 2 c!. For non-blocking networks with fewer vertices and more edges we turn to an older construction of Clos :
Lemma 5 (Clos ). Suppose that there exists a rearrangeable non-blocking network Gc of capacity c with n vertices and m edges. Then there exists a rear-rangeable non-blocking network of capacity c2 with 3cn − 2c2 vertices and 3cm
edges. Proof. Construct 3c copies of Gc, identified as c input subunits, c internal sub-units, and c output subunits. The input subunits have together c2 input vertices, which will be the inputs of the whole network. Similarly, the output subunits have together c2 output vertices, which will be the outputs of the whole net-work. Identify each output vertex of an input subunit with an input vertex of an Lower Bounds for Non-adaptive Shortest Path Relaxation 425
Fig. 2. Three rearrangeable non-blocking networks of capacity 16. Each network’s input vertices are in its left column and its output vertices are in the right column. Left: Complete bipartite graph. Center: Three-stage Clos network, with pairs of input and output vertices in consecutive stages connected by edges rather than being identified as single vertices. Right: Nine-stage network obtained by expanding each subunit of the center network into a three-stage network.
internal subunit, in such a way that each pair of these subunits has exactly one identified vertex. Similarly, identify each output vertex of an internal subunit with an input vertex of an output subunit, in such a way that each pair of these subunits has exactly one identified vertex. An example of this network, for c = 4 and Gc = K4,4, can be seen in an expanded form as the middle network of Fig. 2. For greater legibility of the figure, instead of identifying pairs of vertices between subunits, these pairs have been connected by added edges. Contracting these edges would produce the network described above. To produce vertex-disjoint paths connecting any system of disjoint pairs of inputs and outputs, consider these pairs as defining a multigraph connecting the input subunits to the output subunits of the overall network. This multigraph has maximum degree c (each input or output subunit participates in at most c pairs), and we may apply a theorem of Dénes Kőnig according to which every bipartite multigraph with maximum degree c has an edge coloring using c colors . These colors may be associated with the c internal subunits, and used to designate which internal subunit each path should pass through. Once this designation is made, each subunit has its own system of disjoint pairs of inputs and outputs through which its paths should go, and the paths through each subunit can be completed using the assumption that it is rearrangeable non-blocking.
Corollary 3. For any constant ε > 0 and any integer c ≥ 1, there exist rear-rangeable non-blocking networks of capacity c with O(c) vertices and O(c1+ ε)
edges. Proof. We prove the result by induction on the integer i = log 2 1/ε . As a base case this is true for ε = 1 (for which i = 0 ) and for arbitrary c, using the complete bipartite graph as the network. For smaller values of ε, apply the induc-426 D. Eppstein
tion hypothesis with the parameters 2ε and √c, to produce a rearrangeable non-blocking network N of capacity √c with O(√c) vertices and O(c1/2+ ε)
edges. Applying Lemma 5 to N produces a rearrangeable non-blocking network of capacity ≥ c with O(c) vertices and O(c1+ ε) edges, as desired. Deleting excess vertices to reduce the capacity to exactly c completes the induction.
Theorem 3. For any m and n with n ≤ m ≤ 2(n
2
), there exists a directed graph on m edges and n vertices on which any deterministic or high-probability random-ized non-adaptive relaxation algorithm for shortest paths must use Ω(mn/ log n)
relaxation steps. When m = Ω(n1+ ε) for some ε > 0, the lower bound improves to Ω(mn ).Proof. We construct a graph according to the construction outlined above, in which we choose a capacity c, set up two disjoint sets S and T of c vertices, connect T to S by a biregular bipartite digraph of some degree d, and connect S
to T by a rearrangeable non-blocking network of capacity c. We allocate at least
m/ 2 edges to the biregular graph, and the rest to the non-blocking network, giving d ≈ m/ 2c. For the Ω(mn/ log n) bound, we use the non-blocking network of Lemma 4, with c = Θ( n/ log n). For the Ω(mn ) bound, we use the non-blocking network of Corollary 3, with c = Θ( n). In both cases, we can choose the parameters of these networks to achieve these asymptotic bounds without exceeding the given numbers n and m of vertices and edges. We pad the resulting graph with additional vertices and edges in order to make the numbers of vertices and edges be exactly n and m, and set the weights of these padding edges to be high enough that they do not interfere with the remaining construction. Next, we choose a random distribution on weights for the resulting network so that, for every relaxation sequence σ, the expected reduced cost of σ, for weights from this distribution, matches the lower bound in the statement of the lemma. For deterministic non-adaptive relaxation algorithms, this will give the desired lower bound directly, via the simple fact that the worst case of any distribution is always at least its expectation. For randomized algorithms, the lower bound will follow using Corollary 1 and Corollary 2 to convert the lower bound on expected reduced cost into a high-probability lower bound. As in Theorem 2, the random distribution on weights that we use is deter-mined from a random distribution on paths from the source, such that the shortest path tree for the weighted graph will contain the chosen path. We can accomplish this by setting the lengths of the path edges to zero and all other edge lengths to one. Unlike in Theorem 2, these paths will not necessarily include all vertices in the graph and the shortest path tree may contain other branches. To choose a random path, we simply choose a sequence of edges in the biregular graph, one at a time, in order along the path. In each step, we choose uniformly at random among the subset of edges in the biregular graph that are disjoint from already-chosen edges. Because of the biregularity of the biregular part of our graph, each chosen edge is incident to at most 2( d − 1) other edges, and eliminates these other edges from being chosen later. At least c/ 2 choices are possible before there are no more disjoint edges, and throughout the first c/ 4Lower Bounds for Non-adaptive Shortest Path Relaxation 427
choices there will remain at least m/ 4 edges to choose from, disjoint from all previous edges. The sequence ends when there are no more such edges to choose. Once we have chosen this sequence of edges from the biregular graph, we con-struct a set of vertex-disjoint paths in the rearrangeable nonblocking network that connects them in sequence into a single path. For any given relaxation sequence σ, as in the proof of Theorem 2, let τ be the subsequence of edges in σ that belong to the biregular part of the graph, and consider a modified relaxation algorithm that, after relaxing each edge in τ ,immediately relaxes all edges of the non-blocking network. Define the reduced cost for τ to be the number of relaxation steps made from τ before all dis-tances are correct, not counting the relaxation steps in the non-blocking net-work. Clearly, this is at most equal to the reduced cost for σ, because σ might fail to relax a path in the non-blocking network when τ succeeds, causing the computation of shortest path distances using σ to fall behind that for τ . Define
ti to be the step in the relaxation sequence for τ that relaxes the ith chosen edge from the biregular graph, making the distance to its ending vertex correct. Then the expectation of ti − ti−1 (conditioned on the choice of the first i − 1
edges is at least the average, over all edges that were available to be chosen as the ith edge, of the number of steps along τ from ti−1 to the next occurrence of that edge. This expectation is minimized when the edges occurring immediately following position ti−1 in τ are exactly the next available edges, and is equal to half the number of available edges; for other possibilities for τ , the expectation can only be even larger. The expected reduced cost for τ equals the sum of these differences ti − ti−1. Since there are Ω(c) steps in which the number of available edges is Ω(m), the expected reduced cost for τ is Ω(cm ). The expected reduced cost for σ can only be larger, and plugging in the value of c (coming from our choice of which type of non-blocking network to use) gives the result.
5 Conclusions and Open Problems
We have shown that, for a wide range of choices for m and n, the Bellman-Ford algorithm is asymptotically optimal among non-adaptive relaxation algorithms. Adaptive versions of the Bellman-Ford algorithm are faster, but only by con-stant factors. Is it possible to prove that, among adaptive relaxation algorithms, Bellman-Ford is optimal? Doing so would require a careful specification of what information about the results of relaxation steps can be used in choosing how to adapt the relaxation sequence. The constant factors of 16 and 112 in our deterministic and randomized lower bounds for complete graphs are far from the constant factors of 12 and 13 in the corresponding upper bounds. Can these gaps be tightened? Is it possible to make them tight enough to distinguish deterministic and randomized complex-ity? Alternatively, is it possible to improve the deterministic methods to match the known randomized upper bound? For sparse graphs ( m = O(n)), our lower bound falls short of the Bellman-Ford upper bound by a logarithmic factor. Can the lower bound in this range be improved, or can the Bellman-Ford algorithm for sparse graphs be improved? 428 D. Eppstein
In this work, we considered the worst-case number of relaxation steps used by non-adaptive relaxation algorithms for the parameters m and n. But it is also natural to look at this complexity for individual graphs, with unknown weights. For any given graph, there is some relaxation sequence that is guaranteed to find shortest path distances for all weightings of that graph, with as few relaxation steps as possible. An algorithm of Haddad and Schäffer can find such a sequence for the special case of graphs for which it is as short as possible, one relaxation per edge. What is the complexity of finding or approximating it more generally?
Acknowledgements. This research was supported in part by NSF grant CCF-2212129.
References
Alon, N., Capalbo, M.R.: Finding disjoint paths in expanders deterministically and online. In: 48th Annual IEEE Symposium on Foundations of Computer Science (FOCS 2007), October 20–23, 2007, Providence, RI, USA, Proceedings, pp. 518– 524. IEEE Computer Society (2007). 2. Bannister, M.J., Eppstein, D.: Randomized speedup of the Bellman-Ford algo-rithm. In: Martínez, C., Hwang, H-K (eds.), Proceedings of the 9th Meeting on Analytic Algorithmics and Combinatorics, ANALCO 2012, Kyoto, Japan, Jan-uary 16, 2012, pp. 41–47. SIAM (2012). 63. Beizer, B.: The analysis and synthesis of signal switching networks. In: Proceedings of the Symposium on Mathematical Theory of Automata, New York, April 1962, pp. 563–572. Polytechnic Press of the Polytechnic Institute of Brooklyn (1962) 4. Bellman, R.: On a routing problem. Q. Appl. Math. 16 , 87–90 (1958). org/10.1090/qam/102435 5. Bernstein, A., Nanongkai, D., Wulff-Nilsen, C.: Negative-weight single-source shortest paths in near-linear time. In: 63rd IEEE Annual Symposium on Founda-tions of Computer Science, FOCS 2022, Denver, CO, USA, October 31 - Novem-ber 3, 2022, pp. 600–611. IEEE (2022). 00063 6. Cheng, G., Ansari, N.: Finding all hops shortest paths. IEEE Commun. Lett. 8(2), 122–124 (2004). 7. Clos, C.: A study of non-blocking switching networks. Bell Syst. Tech. J. 32 (2), 406–424 (1953). 8. Narsingh Deo and Chi Yin Pang: Shortest-path algorithms: taxonomy and anno-tation. Networks 14 (2), 275–323 (1984). 9. Dial, R.B.: Algorithm 360: Shortest-path forest with topological ordering [H]. Com-mun. ACM 12 (11), 632–633 (1969). 10. Ford, L.R., Jr. Fulkerson, D.R.: A shortest chain algorithm. In: Flows in Networks, pp. 130–134. Princeton University Press (1962) 11. Guérin, R., Orda, A.: Computing shortest paths for any number of hops. IEEE/ACM Trans. Networking 10 (5), 613–620 (2002). TNET.2002.803917 12. Haddad, R.W., Schäffer, A.A.: Recognizing Bellman-Ford-orderable graphs. SIAM J. Discret. Math. 1(4), 447–471 (1988). Lower Bounds for Non-adaptive Shortest Path Relaxation 429 13. Hedrick, C.: Routing Information Protocol. Request for Comments, RFC 1058. Network Working Group (1988). 14. Jukna, S., Schnitger, G.: On the optimality of Bellman-Ford-Moore shortest path algorithm. Theoret. Comput. Sci. 628 , 101–109 (2016). tcs.2016.03.014 15. Kociumaka, T., Polak, A.: Bellman-Ford is optimal for shortest hop-bounded paths. Electronic preprint arxiv:2211.07325 (2023) 16. Kőnig, D.: Über Graphen und ihre Anwendung auf Determinantentheorie und Men-genlehre. Math. Ann. 77 , 453–465 (1916). 17. Meyer, U., Negoescu, A., Weichert, V.: New bounds for old algorithms: on the average-case behavior of classic single-source shortest-paths approaches. In: Marchetti-Spaccamela, A., Segal, M. (eds.) TAPAS 2011. LNCS, vol. 6595, pp. 217–228. Springer, Heidelberg (2011). 18. Moore, E.F.: The shortest path through a maze. In: Proceedings of International Symposium on Switching Theory 1957, Part II, pp. 285–292. Harvard University Press, Cambridge, Massachusetts (1959) 19. Pippenger, N.: On rearrangeable and non-blocking switching networks. J. Comput. Syst. Sci. 17 (2), 145–162 (1978). 20. Yao, A.C.C.: Probabilistic computations: toward a unified measure of complexity. In: 18th Annual Symposium on Foundations of Computer Science, Providence, Rhode Island, USA, 31 October - 1 November 1977, pp. 222–227. IEEE Computer Society (1977). 21. Yen, J.Y.: Shortest Path Network Problems , volume 18 of Mathematical Systems in Economics . Verlag Anton Hain, Meisenheim am Glan (1975) |
3011 | https://www.doubtnut.com/qna/642521846 | A metallic rod of length l is moved perpendicular to its length with velocity v in a magnetic field →B acting perpendicular to the plane in which rod moves. Derive the expression for the inducced emf.
To derive the expression for the induced electromotive force (emf) in a metallic rod moving in a magnetic field, we can follow these steps: Step 1: Understand the Setup We have a metallic rod of length L moving with a velocity v perpendicular to its length in a magnetic field →B that is also perpendicular to the plane of motion of the rod. Step 2: Determine the Change in Area As the rod moves, the area associated with the magnetic flux changes. The change in area dA as the rod moves a small distance dx can be expressed as: dA=L⋅dx Step 3: Calculate the Change in Magnetic Flux The magnetic flux Φ through the area A is given by: Φ=→B⋅→A=B⋅A⋅cos(θ) Since the magnetic field →B is perpendicular to the area vector (which is also perpendicular to the motion of the rod), the angle θ=0∘, and thus cos(0)=1. Therefore, the change in magnetic flux dΦ due to the change in area is: dΦ=B⋅dA=B⋅(L⋅dx) Step 4: Apply Faraday's Law of Electromagnetic Induction According to Faraday's law, the induced emf E is given by the negative rate of change of magnetic flux: E=−dΦdt Substituting the expression for dΦ: E=−d(B⋅L⋅dx)dt Step 5: Relate dx to Time The term dxdt represents the velocity v of the rod. Therefore, we can rewrite the equation: E=−B⋅L⋅dxdt=−B⋅L⋅v Step 6: Final Expression for Induced EMF The magnitude of the induced emf is given by: |E|=B⋅L⋅v Thus, the expression for the induced emf in the rod moving in a magnetic field is: E=B⋅L⋅v
To derive the expression for the induced electromotive force (emf) in a metallic rod moving in a magnetic field, we can follow these steps:
Step 1: Understand the Setup
We have a metallic rod of length L moving with a velocity v perpendicular to its length in a magnetic field →B that is also perpendicular to the plane of motion of the rod.
Step 2: Determine the Change in Area
As the rod moves, the area associated with the magnetic flux changes. The change in area dA as the rod moves a small distance dx can be expressed as:
dA=L⋅dx
Step 3: Calculate the Change in Magnetic Flux
The magnetic flux Φ through the area A is given by:
Φ=→B⋅→A=B⋅A⋅cos(θ)
Since the magnetic field →B is perpendicular to the area vector (which is also perpendicular to the motion of the rod), the angle θ=0∘, and thus cos(0)=1. Therefore, the change in magnetic flux dΦ due to the change in area is:
dΦ=B⋅dA=B⋅(L⋅dx)
Step 4: Apply Faraday's Law of Electromagnetic Induction
According to Faraday's law, the induced emf E is given by the negative rate of change of magnetic flux:
E=−dΦdt
Substituting the expression for dΦ:
E=−d(B⋅L⋅dx)dt
Step 5: Relate dx to Time
The term dxdt represents the velocity v of the rod. Therefore, we can rewrite the equation:
E=−B⋅L⋅dxdt=−B⋅L⋅v
Step 6: Final Expression for Induced EMF
The magnitude of the induced emf is given by:
|E|=B⋅L⋅v
Thus, the expression for the induced emf in the rod moving in a magnetic field is:
E=B⋅L⋅v
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Contents
(Top)
1 Usage
2 Energy equivalents
3 History
3.1 Origin of the concept
3.2 Isotopic variation
3.3 Joint definition by IUPAP and IUPAC
3.4 Adoption by BIPM
3.5 Unit name
3.6 2019 revision of the SI
4 Measurement
4.1 Josef Loschmidt
4.2 Jean Perrin
4.3 Coulometry
4.4 Electron mass measurement
4.5 X-ray crystal density methods
5 See also
6 Notes
7 References
8 External links
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From Wikipedia, the free encyclopedia
Standard unit of mass for atomic-scale chemical species
Not to be confused with atomic units.
| dalton(unified atomic mass unit) |
| Unit of | mass |
| Symbol | Da or u |
| Named after | John Dalton |
| Conversions |
| 1 Da or u in ... | ... is equal to ... |
| |
| kg | 1.66053906892(52)×10−27 |
| mu | 1 |
| MeV/c2 | 931.49410372(29) |
The dalton or unified atomic mass unit (symbols: Da or u, respectively) is a unit of mass defined as 1/12 of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state and at rest. It is a non-SI unit accepted for use with SI. The word "unified" emphasizes that the definition was accepted by both IUPAP and IUPAC. The atomic mass constant, denoted mu, is an atomic-scale reference mass, defined identically, but it is not a unit of mass. Expressed in terms of ma(12C), the atomic mass of carbon-12: mu = ma(12C)/12 = 1 Da. The dalton's numerical value in terms of the fixed-h kilogram is an experimentally determined quantity that, along with its inherent uncertainty, is updated periodically. The 2022 CODATA recommended value of the atomic mass constant expressed in the SI base unit kilogram is:
mu = 1.66053906892(52)×10−27 kg.
As of June 2025[update], the value given for the dalton (1 Da = 1 u = mu) in the SI Brochure is still listed as the 2018 CODATA recommended value:
1 Da = mu = 1.66053906660(50)×10−27 kg.
This was the value used in the calculation of g/Da, the traditional definition of the Avogadro number. This resulted in the value g/Da = 6.022 140 762 081 123 . . . × 1023, which was then rounded to 9 significant figures and fixed at exactly that value for the 2019 redefinition of the mole.
The value serves as a conversion factor of mass from daltons to kilograms, which can easily be converted to grams and other metric units of mass. The 2019 revision of the SI redefined the kilogram by fixing the value of the Planck constant (h), improving the precision of the atomic mass constant expressed in SI units by anchoring it to fixed physical constants. Although the dalton remains defined via carbon-12, the revision enhances traceability and accuracy in atomic mass measurements.
The mole is a unit of amount of substance used in chemistry and physics, such that the mass of one mole of a substance expressed in grams (i.e., the molar mass in g/mol or kg/kmol) is numerically equal to the average mass of an elementary entity of the substance (atom, molecule, or formula unit) expressed in daltons. For example, the average mass of one molecule of water is about 18.0153 Da, and the mass of one mole of water is about 18.0153 g. A protein whose molecule has an average mass of 64 kDa would have a molar mass of 64 kg/mol. However, while this equality can be assumed for practical purposes, it is only approximate, because of the 2019 redefinition of the mole.
Usage
[edit]
The dalton is commonly used in physics and chemistry to express the mass of atomic-scale objects, such as atoms, molecules, and elementary particles, both for discrete instances and multiple types of ensemble averages. For example, an atom of helium-4 has a mass of 4.0026 Da. This is an intrinsic property of the isotope and all helium-4 atoms have the same mass. Acetylsalicylic acid (aspirin), C9H8O4, has an average mass of about 180.157 Da. However, there are no acetylsalicylic acid molecules with this mass. The two most common masses of individual acetylsalicylic acid molecules are 180.0423 Da, having the most common isotopes, and 181.0456 Da, in which one carbon is carbon-13.
The molecular masses of proteins, nucleic acids, and other large polymers are often expressed with the unit kilodalton (kDa) and megadalton (MDa). Titin, one of the largest known proteins, has a molecular mass of between 3 and 3.7 megadaltons. The DNA of chromosome 1 in the human genome has about 249 million base pairs, each with an average mass of about 650 Da, or 156 GDa total.
In general, the mass in daltons of an atom is numerically close but not exactly equal to the number of nucleons in its nucleus. It follows that the molar mass of a compound (grams per mole) is numerically close to the average number of nucleons contained in each molecule. By definition, the mass of an atom of carbon-12 is 12 daltons, which corresponds with the number of nucleons that it has (6 protons and 6 neutrons). However, the mass of an atomic-scale object is affected by the binding energy of the nucleons in its atomic nuclei, as well as the mass and binding energy of its electrons. Therefore, this equality holds only for the carbon-12 atom in the stated conditions, and will vary for other substances. For example, the mass of an unbound atom of the common hydrogen isotope (hydrogen-1, protium) is 1.007825032241(94) Da,[a] the mass of a proton is 1.0072764665789(83) Da, the mass of a free neutron is 1.00866491606(40) Da, and the mass of a hydrogen-2 (deuterium) atom is 2.014101778114(122) Da. In general, the difference (absolute mass excess) is less than 0.1%; exceptions include hydrogen-1 (about 0.8%), helium-3 (0.5%), lithium-6 (0.25%) and beryllium (0.14%).
The dalton differs from the unit of mass in the system of atomic units, which is the electron rest mass (me).
Energy equivalents
[edit]
The atomic mass constant can also be expressed as its energy-equivalent, muc2. The CODATA recommended values are:
muc2 = 1.49241808768(46)×10−10 J = 931.49410372(29) MeV
The mass-equivalent is commonly used in place of a unit of mass in particle physics, and these values are also important for the practical determination of relative atomic masses.
History
[edit]
Origin of the concept
[edit]
The interpretation of the law of definite proportions in terms of the atomic theory of matter implied that the masses of atoms of various elements had definite ratios that depended on the elements. While the actual masses were unknown, the relative masses could be deduced from that law. In 1803 John Dalton proposed to use the (still unknown) atomic mass of the lightest atom, hydrogen, as the natural unit of atomic mass. This was the basis of the atomic weight scale.
For technical reasons, in 1898, chemist Wilhelm Ostwald and others proposed to redefine the unit of atomic mass as 1/16 the mass of an oxygen atom. That proposal was formally adopted by the International Committee on Atomic Weights (ICAW) in 1903. That was approximately the mass of one hydrogen atom, but oxygen was more amenable to experimental determination. This suggestion was made before the discovery of isotopes in 1912. Physicist Jean Perrin had adopted the same definition in 1909 during his experiments to determine the atomic masses and the Avogadro constant. This definition remained unchanged until 1961. Perrin also defined the "mole" as an amount of a compound that contained as many molecules as 32 grams of oxygen (O2). He called that number the Avogadro number in honor of physicist Amedeo Avogadro.
Isotopic variation
[edit]
The discovery of isotopes of oxygen in 1929 required a more precise definition of the unit. Two distinct definitions came into use. Chemists choose to define the AMU as 1/16 of the average mass of an oxygen atom as found in nature; that is, the average of the masses of the known isotopes, weighted by their natural abundance. Physicists, on the other hand, defined it as 1/16 of the mass of an atom of the isotope oxygen-16 (16O).
Joint definition by IUPAP and IUPAC
[edit]
The existence of two distinct units with the same name was confusing, and the difference (about 1.000282 in relative terms) was large enough to affect high-precision measurements. Moreover, it was discovered that the isotopes of oxygen had different natural abundances in water and in air.
In April 1957 Alfred O. C. Nier suggested to Josef Mattauch that the carbon-12 be adopted as mass scale because of carbon's use as a secondary standard in mass spectrometry. Also, carbon-12 implied acceptable relative changes in the atomic weight scale, i.e., 42 parts-per-million (ppm) compared to 275 ppm for oxygen-16 which would not be acceptable to chemists.
Following the approval of the International Union of Pure and Applied Physics (IUPAP) General Assembly at Ottawa, Canada in 1960 and the International Union of Pure and Applied Chemistry (IUPAC) General Assembly at Montreal, Canada in 1961, the atomic weights were officially given on the carbon-12 scale for the first time.
The new unit was named the "unified atomic mass unit" and given a new symbol "u", to replace the old "amu" that had been used for the oxygen-based unit. However, the old symbol "amu" has sometimes been used, after 1961, to refer to the new unit, particularly in lay and preparatory contexts.
With this new definition, the standard atomic weight of carbon is about 12.011 Da, and that of oxygen is about 15.999 Da. These values, generally used in chemistry, are based on averages of many samples from Earth's crust, its atmosphere, and organic materials.
Adoption by BIPM
[edit]
The IUPAC 1961 definition of the unified atomic mass unit, with that name and symbol "u", was adopted by the International Bureau for Weights and Measures (BIPM) in 1971 as a non-SI unit accepted for use with the SI.
Unit name
[edit]
In 1993, the IUPAC proposed the shorter name "dalton" (with symbol "Da") for the unified atomic mass unit. As with other unit names such as watt and newton, "dalton" is not capitalized in English, but its symbol, "Da", is capitalized. The name was endorsed by the International Union of Pure and Applied Physics (IUPAP) in 2005.
In 2003 the name was recommended to the BIPM by the Consultative Committee for Units, part of the CIPM, as it "is shorter and works better with [SI] prefixes". In 2006, the BIPM included the dalton in its 8th edition of the SI brochure of formal definitions as a non-SI unit accepted for use with the SI. The name was also listed as an alternative to "unified atomic mass unit" by the International Organization for Standardization in 2009. It is now recommended by several scientific publishers, and some of them consider "atomic mass unit" and "amu" deprecated. In 2019, the BIPM retained the dalton in its 9th edition of the SI brochure, while dropping the unified atomic mass unit from its table of non-SI units accepted for use with the SI, but secondarily notes that the dalton (Da) and the unified atomic mass unit (u) are alternative names (and symbols) for the same unit.
2019 revision of the SI
[edit]
The definition of the dalton was not affected by the 2019 revision of the SI, that is, 1 Da in the SI is still 1/12 of the mass of a carbon-12 atom, a quantity that must be determined experimentally in terms of SI units. However, the definition of a mole was changed to be the amount of substance consisting of exactly 6.02214076×1023 entities, and the definition of the kilogram was changed as well. As a consequence, the molar mass constant (Mu) remains close to but no longer exactly 1 g/mol, meaning that the mass in grams of one mole of any substance remains nearly but no longer exactly numerically equal to the average mass of an entity in daltons, although the relative standard uncertainty of 4.5×10−10 at the time of the redefinition is insignificant for all practical purposes. Letting ent denote one entity, the smallest amount of any substance (retaining its chemical properties), one mole is an aggregate of an Avogadro number of entities, i.e. 1 mol = N0 ent ≈ (g/Da) ent. This means that the appropriate atomic-scale unit for molar mass is dalton per entity (Da/ent), equal to the constant Mu, very nearly equal to 1 g/mol. For Da/ent to be exactly equal to g/mol, the dalton would need to be redefined exactly in terms of the (fixed-h) kilogram and the Avogadro number so that 1 Da = (1/N0) g, satisfying the relationships N0 = g/Da and 1 mol = N0 ent = (g/Da) ent, thus conforming to the original mole concept—that the Avogadro number is the gram-to-dalton mass unit ratio.
Measurement
[edit]
Though relative atomic masses are defined for neutral atoms, they are measured (by mass spectrometry) for ions: hence, the measured values must be corrected for the mass of the electrons that were removed to form the ions, and also for the mass equivalent of the electron binding energy, Eb/muc2. The total binding energy of the six electrons in a carbon-12 atom is 1030.1089 eV = 1.6504163×10−16 J: Eb/muc2 = 1.1058674×10−6, or about one part in 10 million of the mass of the atom.
Before the 2019 revision of the SI, experiments were aimed to determine the value of the Avogadro constant for finding the value of the unified atomic mass unit.
Josef Loschmidt
[edit]
A reasonably accurate value of the atomic mass unit was first obtained indirectly by Josef Loschmidt in 1865, by estimating the number of particles in a given volume of gas.
Jean Perrin
[edit]
Perrin estimated the Avogadro number by a variety of methods, at the turn of the 20th century. He was awarded the 1926 Nobel Prize in Physics, largely for this work.
Coulometry
[edit]
Main article: Coulometry
The electric charge per mole of elementary charges is a constant called the Faraday constant, F, whose value had been essentially known since 1834 when Michael Faraday published his works on electrolysis. In 1910, Robert Millikan obtained the first measurement of the charge on an electron, −e. The quotient F/e provided an estimate of the Avogadro constant.
The classic experiment is that of Bower and Davis at NIST, and relies on dissolving silver metal away from the anode of an electrolysis cell, while passing a constant electric current I for a known time t. If m is the mass of silver lost from the anode and Ar the atomic weight of silver, then the Faraday constant is given by:
The NIST scientists devised a method to compensate for silver lost from the anode by mechanical causes, and conducted an isotope analysis of the silver used to determine its atomic weight. Their value for the conventional Faraday constant was F90 = 96485.39(13) C/mol, which corresponds to a value for the Avogadro constant of 6.0221449(78)×1023 mol−1: both values have a relative standard uncertainty of 1.3×10−6.
Electron mass measurement
[edit]
In practice, the atomic mass constant is determined from the electron rest mass me and the electron relative atomic mass Ar(e) (that is, the mass of electron divided by the atomic mass constant). The relative atomic mass of the electron can be measured in cyclotron experiments, while the rest mass of the electron can be derived from other physical constants.
where c is the speed of light, h is the Planck constant, α is the fine-structure constant, and R∞ is the Rydberg constant.
As may be observed from the old values (2014 CODATA) in the table below, the main limiting factor in the precision of the Avogadro constant was the uncertainty in the value of the Planck constant, as all the other constants that contribute to the calculation were known more precisely.
| Constant | Symbol | 2014 CODATA values | Relativestandarduncertainty | Correlationcoefficientwith NA |
--- ---
| Proton–electron mass ratio | mp/me | 1836.15267389(17) | 9.5×10−11 | −0.0003 |
| Molar mass constant | Mu | 1 g/mol | 0 (defined) | — |
| Rydberg constant | R∞ | 10973731.568508(65) m−1 | 5.9×10−12 | −0.0002 |
| Planck constant | h | 6.626070040(81)×10−34 J⋅s | 1.2×10−8 | −0.9993 |
| Speed of light | c | 299792458 m/s | 0 (defined) | — |
| Fine structure constant | α | 7.2973525664(17)×10−3 | 2.3×10−10 | 0.0193 |
| Avogadro constant | NA | 6.022140857(74)×1023 mol−1 | 1.2×10−8 | 1 |
The power of having defined values of universal constants as is presently the case can be understood from the table below (2018 CODATA).
| Constant | Symbol | 2018 CODATA values | Relativestandarduncertainty | Correlationcoefficientwith NA |
--- ---
| Proton–electron mass ratio | mp/me | 1836.15267343(11) | 6.0×10−11 | — |
| Molar mass constant | Mu | 0.99999999965(30) g/mol | 3.0×10−10 | — |
| Rydberg constant | R∞ | 10973731.568160(21) m−1 | 1.9×10−12 | — |
| Planck constant | h | 6.62607015×10−34 J⋅s | 0 (defined) | — |
| Speed of light | c | 299792458 m/s | 0 (defined) | — |
| Fine structure constant | α | 7.2973525693(11)×10−3 | 1.5×10−10 | — |
| Avogadro constant | NA | 6.02214076×1023 mol−1 | 0 (defined) | — |
X-ray crystal density methods
[edit]
Silicon single crystals may be produced today in commercial facilities with extremely high purity and with few lattice defects. This method defined the Avogadro constant as the ratio of the molar volume, Vm, to the atomic volume Vatom: where Vatom = Vcell/n and n is the number of atoms per unit cell of volume Vcell.
The unit cell of silicon has a cubic packing arrangement of 8 atoms, and the unit cell volume may be measured by determining a single unit cell parameter, the length a of one of the sides of the cube. The CODATA value of a for silicon is 5.431020511(89)×10−10 m.
In practice, measurements are carried out on a distance known as d220(Si), which is the distance between the planes denoted by the Miller indices {220}, and is equal to a/√8.
The isotope proportional composition of the sample used must be measured and taken into account. Silicon occurs in three stable isotopes (28Si, 29Si, 30Si), and the natural variation in their proportions is greater than other uncertainties in the measurements. The atomic weight Ar for the sample crystal can be calculated, as the standard atomic weights of the three nuclides are known with great accuracy. This, together with the measured density ρ of the sample, allows the molar volume Vm to be determined: where Mu is the molar mass constant. The CODATA value for the molar volume of silicon is 1.205883199(60)×10−5 m3⋅mol−1, with a relative standard uncertainty of 4.9×10−8.
See also
[edit]
Physics portal
Mass (mass spectrometry)
Kendrick mass
Monoisotopic mass
Mass-to-charge ratio
Notes
[edit]
^ The digits in parentheses indicate the uncertainty; see Uncertainty notation.
References
[edit]
^ a b c d e Bureau International des Poids et Mesures (2019): The International System of Units (SI), 9th edition, English version, page 146. Available at the BIPM website.
^ IUPAC, Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (2006–) "atomic mass constant". doi:10.1351/goldbook.A00497
^ a b International Standard ISO 80000-10:2019 – Quantities and units – Part 10: Atomic and nuclear physics, International Organization for Standardization, 2019
^ a b c d Holden, Norman E. (2004). "Atomic Weights and the International Committee—A Historical Review". Chemistry International. 26 (1): 4–7.
^ Taylor, Barry N. (2009). "Molar mass and related quantities in the new SI". Metrologia. 46 (3): L16 – L19. doi:10.1088/0026-1394/46/3/L01. S2CID 115540416.
^ "2022 CODATA Value: atomic mass constant". The NIST Reference on Constants, Units, and Uncertainty. NIST. May 2024. Retrieved 2024-05-18.
^ The International System of Units (9th(v3.01) ed.), International Bureau of Weights and Measures, August 2024, p. 145, ISBN 978-92-822-2272-0
^ a b Berg, Jeremy M.; Tymoczko, John L.; Stryer, Lubert (2007). "2". Biochemistry (6th ed.). Macmillan. p. 35. ISBN 978-0-7167-8724-2.
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^ Meng Wang, G. Audi, F.G. Kondev, W.J. Huang, S. Naimi, and Xing Xu (2017): "The Ame2016 atomic mass evaluation (II). Tables, graphs and references". Chinese Physics C, volume 41, issue 3, article 030003, pages 1-441. doi:10.1088/1674-1137/41/3/030003
^ "2022 CODATA Value: atomic mass constant energy equivalent". The NIST Reference on Constants, Units, and Uncertainty. NIST. May 2024. Retrieved 2024-05-18.
^ "2022 CODATA Value: atomic mass constant energy equivalent in MeV". The NIST Reference on Constants, Units, and Uncertainty. NIST. May 2024. Retrieved 2024-05-18.
^ a b c Petley, B. W. (1989). "The atomic mass unit". IEEE Trans. Instrum. Meas. 38 (2): 175–179. Bibcode:1989ITIM...38..175P. doi:10.1109/19.192268.
^ Perrin, Jean (1909). "Mouvement brownien et réalité moléculaire". Annales de Chimie et de Physique. 8e Série. 18: 1–114. Extract in English, translation by Frederick Soddy.
^ Chang, Raymond (2005). Physical Chemistry for the Biosciences. University Science Books. p. 5. ISBN 978-1-891389-33-7.
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^ Mills, Ian; Cvitaš, Tomislav; Homann, Klaus; Kallay, Nikola; Kuchitsu, Kozo (1993). Quantities, Units and Symbols in Physical Chemistry International Union of Pure and Applied Chemistry; Physical Chemistry Division (2nd ed.). International Union of Pure and Applied Chemistry and published for them by Blackwell Science Ltd. ISBN 978-0-632-03583-0.
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^ International Standard ISO 80000-1:2009 – Quantities and Units – Part 1: General. International Organization for Standardization. 2009.
^ "Instructions to Authors". AoB Plants. Oxford journals; Oxford University Press. Archived from the original on 2011-11-03. Retrieved 2010-08-22.
^ "Author guidelines". Rapid Communications in Mass Spectrometry. 64 (7). Wiley-Blackwell: 839. 2010. Bibcode:2010ApSpe..64..839.. doi:10.1366/000370210791666192.
^ International Bureau for Weights and Measures (2017): Proceedings of the 106th meeting of the International Committee for Weights and Measures (CIPM), 16-17 and 20 October 2017, page 23. Available at the BIPM website Archived 2021-02-21 at the Wayback Machine.
^ International Bureau for Weights and Measures (2018): Resolutions Adopted – 26th Conference Générale des Poids et Mesures Archived 2018-11-19 at the Wayback Machine. Available at the BIPM website.
^ Lehmann, H. P.; Fuentes-Arderiu, X.; Bertello, L. F. (2016-02-29). "Unified Atomic Mass Unit". Glossary of Terms in Quantities and Units in Clinical Chemistry. doi:10.1515/iupac.68.2930.
^ Mohr, Peter J.; Taylor, Barry N. (2005). "CODATA recommended values of the fundamental physical constants: 2002" (PDF). Reviews of Modern Physics. 77 (1): 1–107. Bibcode:2005RvMP...77....1M. doi:10.1103/RevModPhys.77.1. Archived from the original (PDF) on 2017-10-01.
^ Loschmidt, J. (1865). "Zur Grösse der Luftmoleküle". Sitzungsberichte der Kaiserlichen Akademie der Wissenschaften Wien. 52 (2): 395–413. English translation.
^ Oseen, C.W. (December 10, 1926). Presentation Speech for the 1926 Nobel Prize in Physics.
^ (1974): Introduction to the constants for nonexperts, 1900–1920 From the Encyclopaedia Britannica, 15th edition; reproduced by NIST. Accessed on 2019-07-03.
^ This account is based on the review in Mohr, Peter J.; Taylor, Barry N. (1999). "CODATA recommended values of the fundamental physical constants: 1998" (PDF). Journal of Physical and Chemical Reference Data. 28 (6): 1713–1852. Bibcode:1999JPCRD..28.1713M. doi:10.1063/1.556049. Archived from the original (PDF) on 2017-10-01.
^ Mohr, Peter J.; Taylor, Barry N. (1999). "CODATA recommended values of the fundamental physical constants: 1998" (PDF). Journal of Physical and Chemical Reference Data. 28 (6): 1713–1852. Bibcode:1999JPCRD..28.1713M. doi:10.1063/1.556049. Archived from the original (PDF) on 2017-10-01.
^ "Constants bibliography, source of the CODATA internationally recommended values". The NIST Reference on Constants, Units, and Uncertainty. Retrieved 4 August 2021.
^ "Unit Cell Formula". Mineralogy Database. 2000–2005. Retrieved 2007-12-09.
^ "2022 CODATA Value: lattice parameter of silicon". The NIST Reference on Constants, Units, and Uncertainty. NIST. May 2024. Retrieved 2024-05-18.
^ "2022 CODATA Value: molar volume of silicon". The NIST Reference on Constants, Units, and Uncertainty. NIST. May 2024. Retrieved 2024-05-18.
External links
[edit]
"Atomic weights and isotopic compositions". physics.nist.gov. Physical Reference Data. National Institute for Standards and Technology. 23 August 2009.
"Atomic mass unit". sizes.com. Archived from the original on 2008-01-15.
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Learning Inequalities =====================
Last update!
by Sayan, Jul 7, 2023, 7:32 AM
Hi AoPS,
About 12 years ago, when I was a high school student, I joined this community. It was very exciting! I was interacting with a lot of similar-minded people and we are all interested in solving Olympiad problems. I was spending hours and hours each day on this website. It was like an addiction. I was particularly very fond of inequalities. So, I decided to make this blog collecting several inequality problems that I like.
I was not an extraordinary math genius. I passed some regional math level Olympiad and that's it. Nevertheless, AoPS did help me a lot in learning and mastering the techniques. As a result, at the end of high school, I got admission to the college of my first choice. Gradually I started to spend way less time in AoPS. I only did some moderator-related activities and moved my blog to a wordpress blog where I posted a bit of undergraduate mathematics. However, I soon grew bored with that as well. Nevertheless, I was doing decently in academics, and after completing my undergrad and master's, I went on to do a PhD in Math at a very decent university.
I no longer engage in Olympiad-level math. Although I am involved in math, the situation now is quite different. I am an adult now and I have a multitude of responsibilities to consider. Mathematics has become my profession—I teach and conduct research. However, the sheer joy and satisfaction I experience when solving a math problem are as profound as they were when I was a high school student. It is a sentiment I aspire to carry with me throughout my lifetime.
Thank you and bye AoPS,
Sayan.
Edit: I still check AoPS once in a while! Feel free to leave any comment!
This post has been edited 1 time. Last edited by Sayan, Jun 16, 2025, 3:29 AM
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Moving my blog
by Sayan, Jan 7, 2016, 1:23 PM
Hi there!
I moving this blog to sayanmath.wordpress.com !
That blog is still under construction.
Some older entries of this blog will be there. New entries will mainly be on undergraduates/graduates topic.
Thank you!
Sayan
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Simulating Laplace random numbers from Normal distribution
by Sayan, May 20, 2014, 4:53 AM
In this short section we will look at a different thing from our main theme - Inequalities. I assumed here that the reader have a basic knowledge about continuous random variables. We will show how to draw random numbers from Laplace distribution from Standard normal random numbers.
Proposition: Suppose and are independent and identically distributed random variables each following . Then is a continuous random variable with the following probability density function (pdf)
Such a distribution is termed as standard Laplace distribution. It is also sometimes called as Double exponential distribution.
Suppose we can generate standard normal random variable as many as we want. Then using this proposition we can draw random numbers from Laplace distribution. Of course there are also other ways to simulate random numbers from Laplace. Before going to the proof of the proposition we need some machinery.
Lemma 1[Convolution formula]: Suppose and are independent random variables with pdf and , then is a continuous random variable with pdf given by
Proof: Define .
Fix
Enough to show:
Note that
This proves the lemma. Note that here we switched the integrals in the penultimate step. Since the function is non-negative this is justified by Fubini's Theorem. We would using this switching of integrals throughout the text.
Lemma 2: Suppose and are independent random variables with pdf and , then is a continuous random variable with pdf given by
Proof: Define .
Fix
Enough to show:
Note that
Exercise: Under same conditions of Lemma 2, prove that has pdf
Now we are all set to prove our main proposition.
Note that . Since are all independent, and have the same pdf given by
Therefore has pdf
Note that we can write
Let
Hence
Since it is well known and easy to show that Now we do the polar transformation. Let . Then it can be checked that becomes , and the trainsformation changes the range to . Hence the integral can be written as
Now let us compute the integral separately
Therefore we have
which is exactly what we want to show. This concludes the proof. It is quite fascinating to see that we started with a big multiple integral which was ultimately reduced to a mere exponential function.
References:
Multiple integral
Laplace distribution
This post has been edited 2 times. Last edited by Sayan, Jul 1, 2017, 2:04 AM
Laplace distributionsimulationprobability
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Symmetric n variable inequalities
by Sayan, Oct 10, 2013, 5:35 AM
Its been a while since I posted last time in my blog. In this section we will look up to some strong variable inequalities.
Symmetric Sums: Let be real numbers. We define symmetric sum as the coefficient of in the polynomial
and symmetric average is defined by . For example for variables we have
Newton's Inequality: Let be non-negative reals. Then for all we have
Proof: See here
Maclaurin's Inequality: Let be nonnegative reals. Let their symmetric averages be . Then we have
Proof: We will use induction and Newton's Inequality. Let us prove that holds. This inequality is equivalent to
which is obvious.
Suppose the chain of inequalities is true upto . Then by applying Newton's inequality we have
Hence
Thus by induction the proof is complete.
(own): Let denote the arithmetic mean of -th degree symmetric sum of (positive) variables. If and for and for we have the following inequalities:
Proof: To prove this we use induction on .
For , both inequalities are equivalent to which is Newton's inequality. Let us suppose both holds for all
Then for
And
Hence by induction the two inequalities are true.
Some applications of the above two inequalities:
(Lastnightstar): If s uch that t hen
Proof: We use the same notations as used in . The condition is equivalent to
Hence the inequality after homogenization becomes
which is
(Lastnightstar): If then
Proof: Using the same notations by AM-GM
Thus
Again by we have . Using this we have
which is equivalent to
which is precisely what we want to show. Hence the proof is complete.
Note that for three variables becomes
which is the well-known Vasc inequality.
Let us use the notations used by Lastnightstar from here.
Let be nonnegative reals. Let
denote the general arithmetic mean of the symmetric sum of variables.
Then Maclaurin's inequality is equivalent to and Newton's inequality is equivalent to .
Under this notations the two inequalities in become and respectively.
: show that .
Proof: The inequality is equivalent to
Which is true because by Holder:
Similarly we can prove:
: If then .
Show that for suitable values of .
Proof: The inequality is very easy. By AM-GM and cauchy schwarz we have
(Lastnightstar): Let and let . Show that
Proof: Let us prove for variables. For variables the inequality is equivalent to
which Schur's inequality of degree.
For variables,
Let us simplify the left side:
Note that
Hence
where
Now let us simplify the right side:
Note that
Therefore
Note that Since it is exactly equivalent to for variables namely , i.e., Schur's inequality of degree.
Hence we have
Show that
Proof: It can be proved similarly like the last one.
Show that
Proof: Note that by power mean inequality we have .
By Maclaurin's inequality we have and
Hence
We have
Proof: Note that
We will prove the inequality for . That inequality is equivalent to
Note that which is just equivalent to for and variables as .
Also observe that . Since by power mean inequality and by Maclaurin .
Hence we have for all . We call the left side and right side of this inequality as and .
Observe that and . Hence we have
Prove that for all .
Proof: Let us call . We will show that .
Note that .
By Maclaurin's Inequality and power mean inequality we have . And
Hence we have
Let us replace the variables, . Then the last inequality is equivalent to .
Hence it is enough to show that . We will proceed by induction.
Note that and is Schur's 3rd degree inequality. So assume that the inequality holds for all . Therefore we need to prove .
Note that for variables gives
Similarly we can obtain other inequalities by changing the roles of variables. Summing up all the inequalities we get
Again note that by -th degree Schur we have
Adding these inequalities over all possible combinations of we get
Adding (1) and (2) we get
Hence the proof is complete.
Prove that for all .
Proof: It can be proved similarly like .
Show that the following inequality holds when :
Proof: The inequality is equivalent to
For , the inequality is equivalent to which is obvious.
For , we see that
Adding these two inequalities we get the desired result.
: Let be positive reals such that . Show that
Proof: The condition is equivalent to . Note that:
which is obvious by AM-GM. Hence by using this inequality we have
By AM-GM we have
By we have . Hence we have
Therefore we get
Inspired from this, I proposed this inequality in Mathematical Reflections for .
(own) [MR: S280] be positive reals such that . Show that
Proof: It is well known and easy to prove that for all positive reals we have
and
Hence
(own): Let such that . Let and
Show that
where
Proof: gives and . Both are known results.
We use induction. Assume that it is true for all . Then for we do two cases.
1)
2) . We have and .
Multiplying we get
Hence
By AM-GM
Hence
Hence for is true.
Thus by induction for all the following is true:
The following inequality also can be used in proving .
(xzlbq): Let such that . Let . Prove that:
Proof:
which is true.
Note that if is true,
Hence is true. Thus by induction the proof is complete. .
Note: Some of the constants here are not the best one. But they are quite good ones, given the fact that we used only elementary inequalities to prove them!
To be updated soon...
This post has been edited 9 times. Last edited by Sayan, Jul 1, 2017, 2:18 AM
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No Calculus, only AM-GM!
by Sayan, Dec 21, 2012, 11:02 AM
Here I will show some problems that are normally solved by calculus, but AM-GM can be used to prove it too. Let us start with simple one:
If find the maximum of
Solution: By AM-GM we have . Hence maximum is and it occurs when
If is a real root of show that (India 2004)
Solution: First note , Since by AM-GM we have
Again,
This implies
Let be a positive real number such that . Prove that
Solution: Note that
And for the left side
Let be a real number such that . Prove that
Solution: Note that and
And
Also we can show
$a^6+1 & =a^2+3a+\frac3a$
Let , find the maximum of where are positive reals.
Solution: Let Hence we have to maximize Note that by AM-GM
$\implies a^p(1-a^2)^{\frac{q}2} \le \sqrt{\frac{p^pq^q}{(p+q)^{p+q}}$
Equality holds when
Let be a real number such that . Prove that .
Solution: First I will prove is positive, let us assume the opposite. Then,
But then, and which is a contradiction. So, is positive. Note the following:
Now since
be a real number such that prove that .
Solution: First of all notice is positive and and
Note that
[\begin{align}a^7 & =a^6-a^3+3a^2 \ & =a^5-a^2+3a-a^3+3a^2 \ & =a^4+3-a+2a^2+3a-a^3 \ & =a^3-1+\frac3a+3+2a^2+2a-a^3\ & = 2a^2+2a+\frac3a+2 \ & \ge 2a^2+2+2\sqrt6 \ & >2\sqrt3+2+2\sqrt6 > 4\sqrt{18}+2 > 10]
From here we get another sharper bound . In fact we also have . Can you prove it? We can also find another sharper upper bound only by AM-GM.
If , find the maximum of where are positive reals.
Consider two positive real numbers and such that . Find the minimum value of .
Solution: Let we have and
Note
Again, where and
So,
Equality holds when
, are positive and . Find the range of
Solution: Clearly if one of the variable is large enough the expression tends to infinity. So we are looking for the minimum. By AM-GM:
and
So,
Minimum occurs when . Range: ![Image 412: $5\sqrt5,\infty)$
Prove that the polynomial equation has no real solution. (Refinement of ISI 2012)
Solution: First observe that there is no negative solution. Since gives which is impossible for a non-negative Hence we assume is positive. By AM-GM and
Thus
This proves the result.
Let show that
Solution: Let at and maximization occur. Then, by weighted AM-GM
, and
We want to choose such that when these terms are multiplied the power of and becomes . In that case the variables in the fraction will cancel out, giving an attainable upper numerical bound. Indeed that will be the maximum value. Hence we have
and
Keeping in mind that are positive, solving we get
Hence the required maximum is
Equality holds iff
Let and . Show that
Solution: WLOG assume is between . Thus,
Note that
Again by AM-GM
[\begin{align} x^2y+yz^2+2xyz=y(x+z)^2 & =y(1-y)^2 \ & \le 4\left(\frac{\frac{1-y}2+\frac{1-y}2+y}3\right)^3=\frac4{27}]
This post has been edited 5 times. Last edited by Sayan, Jun 21, 2013, 2:58 AM
AM-GM
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a^2+b^2+c^2+abc=4 and its consequences
by Sayan, Nov 11, 2012, 3:25 AM
If are non-negative reals such that , then we have the following nice inequalities:
Proof: Let then
Proof(mudok): Two of have the same sign. Assume that . If then it is done. If then Contradiction. So
Proof:
(USAMO 2001):
(mudok):
(own) Refinement of :
(own) Refinement of :
(mudok)
This post has been edited 2 times. Last edited by Sayan, Dec 15, 2012, 1:30 PM
inequalitiesUsamo 2001
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A collection of Jack Garfunkel inequalities
by Sayan, Aug 4, 2012, 9:55 AM
Problem 1: Prove that for positive reals we have
When equality is attained?
Proof:
Setting
Let We need to prove:
Using CS we have:
We will prove:
Which is obviously true because
Equality holds for
Problem 2: If where show that
Proof:
Again,
Hence
Equality holds when
Problem 3: Given is with incenter . Lines are drawn to meet the incircle for the first time in respectively. Prove that
is not less than the perimeter of the triangle of maximum perimeter that can be inscribed in circle .
Problem 4: In a triangle whose circumcircle has unit diameter, let and denote the lengths of the median and the internal angle bisector to side respectively. Prove that
Problem 5: Given is a triangle with altitudes and medians to sides respectively. Prove that
with equality if and only if the triangle is equilateral
Problem 6: Given a triangle with sides , let denote the angle bisectors extended to the circumcircle of the triangle. Prove that
with equality attained in the equilateral triangle.
Problem 7: Consider the following three inequalities for the angles of a triangle
3rd inequality is trivially implied from 2nd one. Prove the first inequality and show that 1st one implies the second inequality.
Solution: Let us denote the triangle sides as . Note that
And it is well known that and . And hence the conclusion follows.
Problem 8: If , prove that
Problem 9: If is the incentre of triangle and lines meet the circumcircle of the triangle again in respectively, prove that
Problem 10: Given a triangle , its centroid , and the pedal triangle os its incenter . The segments meet the incircle in ; and the segments meet the incircle in . Let denote the perimeter od a triangle, then prove the following double inequality
Problem 11: If are the interior angles of a convex quadrilateral , prove that
with equality if and only if is a rectangle.
Problem 12: Let denote the lengths of the medians to sides respectively of triangle , and let denote the lengths of these medians extended to the circumcircle of the triangle. Prove that
Problem 13: A triangle hass sides and the medians of this triangle are used s sides of a new triangle. If is the inradius of this new triangle, prove or disprove that
with equality just when the original triangle is equilateral.
Problem 14: If are the angles of a triangle, show that
Problem 15: Given a triangle , let be the lengths of its internal angle bisectors and let be the lengths of these bisectors extended to the circumcircle of the triangle. Prove that
Problem 16: If are the angles of a triangle then show that
Solution: See here and here
Problem 17: Let be a triangle with sides and let be the circumradius of thetriangle formed by using as sides the medians of triangle . Prove that
Problem 18: Given is a triangle with sides and medians is the usual order, circumradius , and inradius . Prove that
(a)
(b)
(c)
(d)
Problem 19: For a triangle with circumradius and inradius , let . An inequality involving elemnts of triangle will be called strong or weak respectively according as or
(a) Prove that the following inequality is strong:
(b) Prove that the following inequality is weak
Problem 20: For a triangle with angles show that
Problem 21: Show that, for any triangle ,
Problem 22: If are angles of a triangle, show that
Problem 23: If are angles of any triangle show that
Problem 24: If triangle is acute-angled prove or disprove that
(a)
(b)
Problem 25:(Also known as Walker inequality) Let be the circumradius, inradius and semiperimeter respectively of an acute-angled triangle. Prove or disprove that
When does equality occur?
Solution: See here
Problem 26: Let and be two nonequilateral triangles such that and . Prove that
where and are the semiperimeter and inradius of triangles and , repectively.
Problem 27:(a) If , prove that
(b) Let be the inradius of a triangle and the radii of its three Malfatti circles. Deduce from (a) that
Problem 28: For , let be the center and the radius of the Malfatti circle nearest in triangle . Prove that
When does equality occur?
Problem 29: If are angles of a triangle show that
Problem 30: If are angles of a triangle show that
Problem 31: If are the angles of an acute triangle , prove that
Problem 32: In the figure, and the three circles with centres represent the malfatti configuration. Circle is externally tangent to these three circles and the sides of triangle are each tangent to and one of the smaller circles. Prove that
where stands for perimeter. Equality is attained when is equilateral.
Problem 33: Prove (or disprove) that for an acute triangle ,
where denote the medians to sides and denote the semiperimeter of .
Problem 34: If is the equilateral triangle of minimum perimeter that can be inscribed in a triangle , show that the perimeter of is at most half the perimeter of , with equality when is equilateral.
Problem 35: be a triangles with three Malfatti circles inscribed. be the centres of the three circles, and let and be the inradii of triangles and respectively. Prove that
Problem 36: If are the angles of a triangle, prove that
Problem 37: Isosceles right triangles are constructed outwardly on the sides of a triangle , with the right angles at , and triangle is drawn. Prove or disprove that
where are the angles of .
Problem 38: Prove or disprove that if are the angles of a triangle,
Problem 39: If are the angles of a triangle, prove that
Problem 40: If are the angles of a triangle, prove or disprove that
Problem 41: Find the smallest constant such that
for all positive .
Problem 42: is any point inside a triangle . Lines are drawn and angles are denoted by respectively. Prove or disprove that
Problem 43: Triangle is formed by joining the feet of the altitudes of an acute triangle . Prove that
where and are the semiperimeters and inradii of and respectively.
Problem 44: Show that
where the sums are cyclic over the angles of an acute triangle.
Problem 45: Let be a triangle with angles and sides and let be a triangle with sides
where is a positive constant. Prove that (i.e., is in a sense "more equilateral" that )
Last updated on 7.30am UTC on 31th Dec 2013
This post has been edited 13 times. Last edited by Sayan, Dec 31, 2013, 6:27 AM
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IMO 2012, geeralisation of Problem 2
by Sayan, Jul 12, 2012, 8:02 AM
First we restate the problem used in the contest.
(IMO 2012) Let be an integer, and let be positive real numbers such that . Prove that
A generalisation was proposed by Ion Bursuc
Generalisation
ionbursuc wrote:
If positive reals satisfy and ,prove that
where .
Proof 1(own)
Observe that for ,
Now note that for
So we have
Thus
Proof 2(kunny)
By taking natural logarithm, the problem can be rephrased as follows:
If, for each positive integer , holds, then prove that :
Proof
Let , from , has a local minimum
Therefore, for , we have
.
Since , yielding $\sum_{k=2}^n (\lambda_k+\mu_k)\ln (\lambda_k a_k+1)}\geq \sum_{k=2}^n {(\lambda_k+\mu_k)\ln (\lambda_k+\mu_k)-\mu_k\ln \mu_k}.$
Now note that
So we are done!
NEW Refinement of IMO 2012:
Quote:
If positive reals satisfy and ,prove that
Note that for , Hence and
Hence
This post has been edited 1 time. Last edited by Sayan, Apr 22, 2013, 6:43 AM
IMO 2012 problem 2inequalities
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Fresh inequalities
by Sayan, Jun 13, 2012, 5:47 AM
Problem 1 (proposed by Nicusorz)
Let , Prove that
let
and
$S =\sum \frac{a_i^{p+1}}{a_i^2+\prod a_i} \ge n^n \sum \frac{a_i^{p+1}}{n^na_i^2+1}& \ge n^n \frac{\left[\sum a_i^{\frac{p+1}{2}}\right]^2}{n^n \sum a_i^2+n}$
now by power mean inequality if we have
thus we have
Problem 2(proposed by oldbeginner)
For and prove that
now
note that
thus we have to prove
equivalently we have to prove
and it is true from power mean inequality.
hence the inequality is proved and equality holds when
Problem 3(proposed by old beginner)
For and prove that
thus
Problem 4 (proposed by old beginner)
For and prove that
from problem 3 we know
so we have to prove
now,
this is true because and
this is true because
combining the two we get the desired result, and equality holds when
Problem 5
Let x,y,z be positive numbers such that .Prove that:
Here i used Mixing Variables
let
WLOG, assume
so,
now,
since
so,
so it remains to prove
but,
and equality holds when
Problem 6(proposed by nicusorz)
Prove that
use of integration makes the problem easy.
Problem 7(proposed by oldbeginner)
For prove that
Solution:
Now we simplify the dinominator
now let us simplify
Thus we have
Hence it is left to prove that
it is equivalent to which is true by power mean inequality
Thus equality holds when
Problem 8(proposed by xzlbq)
Let , prove that:
Solution:
Then which is obvious
So it is enough to show But this is equivalent to
Now note that which proves the inequality
This post has been edited 8 times. Last edited by Sayan, Jun 6, 2013, 5:51 AM
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Some recent problems from AoPS
by Sayan, Apr 8, 2012, 3:12 AM
This problem is inspired by this problem
Problem 1
How many triangles with ordered integral sides can be formed which have perimeter ( is an integer)?
Here and triangles are considered as different
Solution:
First of all lets make ordering
let
To avoid floor and ceiling function, I choose
Maximum value of is and minimum value of is
Let the number of triplets be denoted as .
thus,
Thus
similarly when
when
when and so on.
Thus,
The number of isosceles triangles are
while there is one equilateral triangle.
Thus, the number of ordered sides is
lets replace by
thus we have
with similar efforts for all other leaving different residues modulo , a general result was obtained as follows:
Let denotes the number of ordered triplets of sides of triangle with perimeter , then
Some related problems are:
a) The number of ways to place 11 identical balls in three distinct boxes so that any two boxes together will contain more balls than the other one ?
b)In how many ways 101 shares can be distributed among three persons if sum of shares owned by any two of them is greater than the no. of shares owned by the third?
But if ordering does not matters what will be the number of triangles??
This gives our 1st corollary.
Problem 2
What is the number of triangles with perimeter having integral sides?
Solution:
Let denotes the number of triangles only
then when we already have
Summing up the other results we have the following:
TasK
Can you prove the first problem using generating functions?
Problem 3
Find all positive integral solution of the equation
Solution
A trivial solution is we will prove that it is the only one.
This solution is based on bunch of modulo work.
First with modulo we get
thus we get are of same parity.
Then if are even
forcing to be even
if are odd forcing to be odd.
finally by checking with modulo , we have which is possible only when is odd.
thus we get that all are odds.
now we check by modulo
we get
again, and
thus, or
now, let us assume and we check the equation by modulo
we have
again and
now we work with modulo
by fermat's theorem,
thus it is enough to check those cases where
Case 1:
then we have
, and
trying all combinations, none of them satisfy the given equation.
Case 2:
then
, and
Here also none of the combination satisfy, thus we get a contradiction, hence
now the equation is equivalent to
let us assume , now observe that
hence for all odd , hence we arrive at a contradiction. Thus , yielding
Hence the only solution is
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3014 | https://epubs.siam.org/doi/10.1137/080744037 | The Sign-Rank of AC$^0$ | SIAM Journal on Computing
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HomeSIAM Journal on ComputingVol. 39, Iss. 5 (2010)10.1137/080744037
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The Sign-Rank of AC 0
Authors: Alexander A.Razborov and Alexander A.SherstovAuthors Info & Affiliations
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The sign-rank of a matrix 𝐴=[𝐴 𝑖𝑗] with ±1 entries is the least rank of a real matrix 𝐵=[𝐵 𝑖𝑗] with 𝐴 𝑖𝑗𝐵 𝑖𝑗>0 for all 𝑖,𝑗. We obtain the first exponential lower bound on the sign-rank of a function in 𝖠 𝖢 0. Namely, let 𝑓(𝑥,𝑦)=⋀𝑖=1,…,𝑚⋁𝑗=1,…,𝑚 2(𝑥 𝑖𝑗∧𝑦 𝑖𝑗). We show that the matrix [𝑓(𝑥,𝑦)]𝑥,𝑦 has sign-rank e x p(Ω(𝑚)). This in particular implies that Σ 𝑐𝑐 2⊈𝖴 𝖯 𝖯 𝑐𝑐, which solves a longstanding open problem in communication complexity posed by Babai, Frankl, and Simon [Proceedings of the 27 th Symposium on Foundations of Computer Science (FOCS), 1986, pp. 337–347]. Our result additionally implies a lower bound in learning theory. Specifically, let 𝜙 1,…,𝜙 𝑟:{0,1}𝑛→ℝ be functions such that every DNF formula 𝑓:{0,1}𝑛→{−1,+1} of polynomial size has the representation 𝑓≡s g n(𝑎 1𝜙 1+⋯+𝑎 𝑟𝜙 𝑟) for some reals 𝑎 1,…,𝑎 𝑟. We prove that then 𝑟⩾e x p(Ω(𝑛 1/3)), which essentially matches an upper bound of e x p(˜𝑂(𝑛 1/3)), due to Klivans and Servedio [J. Comput. System Sci., 68 (2004), pp. 303–318]. Finally, our work yields the first exponential lower bound on the size of threshold-of-majority circuits computing a function in 𝖠 𝖢 0. This substantially generalizes and strengthens the results of Krause and Pudlák [Theoret. Comput. Sci., 174 (1997), pp. 137–156].
MSC codes
03D15
68Q15
68Q17
Keywords
sign-rank
communication complexity
complexity classes Σ 𝑐𝑐 2
Π 𝑐𝑐 2
and 𝖴 𝖯 𝖯 𝑐𝑐
constant-depth AND/OR/NOT circuits
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Information & Authors
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Information
Published In
SIAM Journal on Computing
Volume 39 • Issue 5 • January 2010
Pages: 1833 - 1855
DOI: 10.1137/080744037
ISSN (online): 1095-7111
Copyright
Copyright © 2010 Society for Industrial and Applied Mathematics.
History
Submitted: 16 December 2008
Accepted: 12 October 2009
Published online: 22 January 2010
MSC codes
03D15
68Q15
68Q17
Keywords
sign-rank
communication complexity
complexity classes Σ 𝑐𝑐 2
Π 𝑐𝑐 2
and 𝖴 𝖯 𝖯 𝑐𝑐
constant-depth AND/OR/NOT circuits
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Alexander A.Razborov
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Alexander A.Sherstov
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References
References
1.
M. Alekhnovich, M. Braverman, V. Feldman, A. R. Klivans, and T. Pitassi, Learnability and automatizability, in Proceedings of the 45th Symposium on Foundations of Computer Science (FOCS), 2004, pp. 621–630.
Google Scholar
2.
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HomeSIAM Journal on ComputingVol. 39, Iss. 5 (2010)10.1137/080744037
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3015 | https://study.com/skill/learn/finding-the-restriction-on-a-variable-in-a-denominator-in-a-linear-equation-explanation.html | Finding the Restriction on a Variable in a Denominator in a Linear Equation | Algebra | Study.com
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Copyright
Finding the Restriction on a Variable in a Denominator in a Linear Equation
Algebra 2 Skills Practice
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00:04 Example 1
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Briana Tiede, Amy McKenney
Instructors Briana Tiede
Briana has tutored middle school and high school math for over 15 years. She has a BA in math from Salisbury University and a master's in math from Villanova University.
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Amy McKenney
Amy has taught high school mathematics for over 14 years. She has a master's degree in education from Plymouth State University and her undergraduate degree in mathematics. She is certified to teach grades 7-12 mathematics.
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Example SolutionsPractice Questions
Finding the Restrictions on a Variable in a Denominator in a Linear Equation
Step 1: Set the linear denominator equal to zero.
Step 2: Find the solutions that make the denominator equal to zero by solving the equation created in step 1. These solutions are the restrictions on the variable, because we can never divide by 0.
Step 3: State the solutions found in step 2 as the restrictions on the variable.
Finding the Restriction on a Variable in a Denominator in a Linear Equation Vocabulary
Undefined/Division by Zero: A mathematical expression is undefined when the denominator is equal to zero. For example, {eq}\frac{4}{0}=\text{undefined} {/eq}.
Linear Denominator: A linear denominator is a denominator of a rational expression that is linear, or of the form ax + b, where a and b are constants.
Rational Function: A rational function is a function that contains the variable in the denominator of a rational expression. That is, it is a function of the form {eq}\frac{f(x)}{g(x)} {/eq}, where f(x) and g(x) are functions in x. Since we can never divide by 0, a restriction on the variable of a rational function with a linear denominator would be any value of the variable that makes the linear denominator equal to 0.
Let's use these steps and definitions to practice finding restrictions on a variable in a denominator in a linear equation with two examples.
Finding the Restriction on a Variable in a Denominator in a Linear Equation Example 1:
Determine the restrictions on the variable for the following rational function with a linear denominator:
$$f(x)=\frac{2}{x-4} $$
Step 1: Set the denominator equal to zero.
$$x-4=0 $$
Step 2: Solve for x.
$$x-4=0 $$
We solve for x by adding 4 to both sides.
$$x-4+4=0+4 $$
Because {eq}-4+4=0 {/eq}, we are left with the following when we simplify both sides of this equation:
$$x=4 $$
Step 3: State the restrictions on the variable.
The denominator is 0 when {eq}x=4 {/eq}. This is the only value that makes the equation undefined.
Therefore, the restriction on the variable in the linear denominator of the given function is {eq}x\ne 4 {/eq}.
Finding the Restriction on a Variable in a Denominator in a Linear Equation Example 2:
Determine the restrictions on the variable for the following rational function with a linear denominator:
$$f(x)=\frac{2}{5x+6} $$
Step 1: Set the denominator equal to zero.
$$5x + 6 = 0 $$
Step 2: Solve for x.
$$\begin{align} 5x + 6 &= 0\ 5x &= -6\ x &= -\frac{6}{5} \end{align} $$
Step 3: State the restrictions on the variable.
Our restriction on the variable is {eq}x \neq -\frac{6}{5} {/eq}.
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Finding the Restrictions on a Variable in a Denominator in a Linear Equation
Vocabulary
Example 1
Example 2
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Finding the Restriction on a Variable in a Denominator in a Linear Equation
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3016 | https://byjus.com/maths/area-of-isosceles-triangle/ | The area of an isosceles triangle is the amount of region enclosed by it in a two-dimensional space. The general formula for the area of triangle is equal to half the product of the base and height of the triangle. Here, a detailed explanation of the isosceles triangle area, its formula and derivation are given along with a few solved example questions to make it easier to have a deeper understanding of this concept.
Check more mathematics formulas here.
What is the Formula for Area of Isosceles Triangle?
The total area covered by an isosceles triangle is known as its area. For an isosceles triangle, the area can be easily calculated if the height (i.e. the altitude) and the base are known. Multiplying the height with the base and dividing it by 2, results in the area of the isosceles triangle.
What is an isosceles triangle?
An isosceles triangle is a triangle that has any of its two sides equal in length. This property is equivalent to two angles of the triangle being equal. An isosceles triangle has two equal sides and two equal angles. The name derives from the Greek iso (same) and Skelos (leg). An equilateral triangle is a special case of the isosceles triangle, where all three sides and angles of the triangle are equal.
An isosceles triangle has two equal side lengths and two equal angles, the corners at which these sides meet the third side is symmetrical in shape. If a perpendicular line is drawn from the point of intersection of two equal sides to the base of the unequal side, then two right-angle triangles are generated.
| |
| Table of Contents: Formula List of Formulae How to Calculate Area? Derivation of Formula Area of Right Isosceles Triangle Perimeter Formula of Area Using Trigonometry Examples Practice Questions FAQs |
Area of Isosceles Triangle Formula
The area of an isosceles triangle is given by the following formula:
| |
| Area = ½ × base × Height |
| | |
--- |
| The perimeter of the isosceles triangle | P = 2a + b |
| The altitude of the isosceles triangle | h = √(a2 − b2/4) |
List of Formulas to Find Isosceles Triangle Area
| Formulas to Find Area of Isosceles Triangle | |
--- |
| Using base and Height | A = ½ × b × h where b = base and h = height |
| Using all three sides | A = ½[√(a2 − b2 ⁄4) × b] a is the measure of equal sides b is the base of triangle |
| Using the length of 2 sides and an angle between them | A = ½ × a × b × sin(α) a is the measure of equal sides b is the base of triangle |
| Using two angles and length between them | A = [a2×sin(β)×sin(α)/ 2×sin(2π−α−β)] a is the measure of equal sides b is the base of triangle α is the measure of equal angles β is the angle opposite to the base |
| Area formula for an isosceles right triangle | A = ½ × a2 a is the measure of equal sides |
How to Calculate Area if Only Sides of an Isosceles Triangle are Known?
If the length of the equal sides and the length of the base of an isosceles triangle are known, then the height or altitude of the triangle is to be calculated using the following formula:
| |
| Altitude of an Isosceles Triangle = √(a2 − b2/4) |
| |
| Area of Isosceles Triangle Using Only Sides = ½[√(a2 − b2 /4) × b] |
b = base of the isosceles triangle
h = height of the isosceles triangle
a = length of the two equal sides
Derivation for Isosceles Triangle Area Using Heron’s Formula
The area of an isosceles triangle can be easily derived using Heron’s formula as explained below.
According to Heron’s formula,
Area = √[s(s−a)(s−b)(s−c)]
Where, s = ½(a + b + c)
Now, for an isosceles triangle,
s = ½(a + a + b)
⇒ s = ½(2a + b)
Or, s = a + (b/2)
Area = √[s(s−a)(s−b)(s−c)]
Or, Area = √[s (s−a)2 (s−b)]
⇒ Area = (s−a)× √[s (s−b)]
Substituting the value of “s”
⇒ Area = (a + b/2 − a)× √[(a + b/2) × ((a + b/2) − b)]
⇒ Area = b/2 × √[(a + b/2) × (a − b/2)]
Or, area of isosceles triangle = b/2 × √(a2 − b2/4)
Area of Isosceles Right Triangle Formula
| |
| The formula for Isosceles Right Triangle Area= ½ × a2 |
Derivation:
Area = ½ ×base × height
area = ½ × a × a = a2/2
Perimeter of Isosceles Right Triangle Formula
| |
| P = a(2+√2) |
Derivation:
The perimeter of an isosceles right triangle is the sum of all the sides of an isosceles right triangle.
Suppose the two equal sides are a. Using Pythagoras theorem the unequal side is found to be a√2.
Hence, perimeter of isosceles right triangle = a+a+a√2
= 2a+a√2
= a(2+√2)
= a(2+√2)
Area of Isosceles Triangle Using Trigonometry
Using Length of 2 Sides and Angle Between Them
A = ½ × b × c × sin(α)
Using 2 Angles and Length Between Them
A = [c2×sin(β)×sin(α)/ 2×sin(2π−α−β)]
Solved Examples
Example 1:
Find the area of an isosceles triangle given b = 12 cm and h = 17 cm?
Solution:
Base of the triangle (b) = 12 cm
Height of the triangle (h) = 17 cm
Area of Isosceles Triangle = (1/2) × b × h
= (1/2) × 12 × 17
= 6 × 17
= 102 cm2
Example 2:
Find the length of the base of an isosceles triangle whose area is 243 cm2, and the altitude of the triangle is 27 cm.
Solution:
Area of the triangle = A = 243 cm2
Height of the triangle (h) = 27 cm
The base of the triangle = b =?
Area of Isosceles Triangle = (1/2) × b × h
243 = (1/2) × b × 27
243 = (b×27)/2
b = (243×2)/27
b = 18 cm
Thus, the base of the triangle is 18 cm.
Question 3:
Find the area, altitude and perimeter of an isosceles triangle given a = 5 cm (length of two equal sides), b = 9 cm (base).
Solution:
Given, a = 5 cm
b = 9 cm
Perimeter of an isosceles triangle
= 2a + b
= 2(5) + 9 cm
= 10 + 9 cm
= 19 cm
Altitude of an isosceles triangle
h = √(a2 − b2/4)
= √(52 − 92/4)
= √(25 − 81/4) cm
= √(25–81/4) cm
= √(25−20.25) cm
= √4.75 cm
h = 2.179 cm
Area of an isosceles triangle
= (b×h)/2
= (9×2.179)/2 cm²
= 19.611/2 cm²
A = 9.81 cm²
Question 4:
Find the area, altitude and perimeter of an isosceles triangle given a = 12 cm, b = 7 cm.
Solution:
Given,
a = 12 cm
b = 7 cm
Perimeter of an isosceles triangle
= 2a + b
= 2(12) + 7 cm
= 24 + 7 cm
P = 31 cm
Altitude of an isosceles triangle
= √(a2 − b2⁄4)
= √(122−72/4) cm
= √(144−49/4) cm
= √(144−12.25) cm
= √131.75 cm
h = 11.478 cm
Area of an isosceles triangle
= (b×h)/2
= (7×11.478)/2 cm²
= 80.346/2 cm²
= 40.173 cm²
Practice Questions
Find the altitude of the triangle if the length of its base is 25 cm and the area enclosed is 375 cm2?
The length of the base of an isosceles triangle is half of its altitude. If the altitude of the triangle is 14cm, find the area enclosed by it?
Find the area of an isosceles triangle, whose length of two equal sides is 5 cm and the length of the third side is 6 cm?
Find the length of each side of a right isosceles triangle whose area is 112.5 cm2.
Frequently Asked Questions on Area of Isosceles Triangle
Q1
What is an Isosceles Triangle?
An isosceles triangle can be defined as a special type of triangle whose 2 sides are equal in measure. For an isosceles triangle, along with two sides, two angles are also equal in measure.
Q2
What does the Area of an Isosceles Triangle Mean?
The area of an isosceles triangle is defined as the amount of space occupied by the isosceles triangle in the two-dimensional plane.
Q3
What is the Formula for Area of Isosceles Triangle?
To calculate the area of an isosceles triangle, the following formula is used:
A = ½ × b × h
Q4
What is the Formula for Perimeter of Isosceles Triangle?
The formula to calculate the perimeter of an isosceles triangle is:
P = 2a + b
Test your knowledge on Area Of Isosceles Triangle
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3017 | https://medcoeckapwstorprd01.blob.core.usgovcloudapi.net/pfw-images/dbimages/Ch%2031.pdf | 451 Pediatric Care Chapter 31 Pediatric Care Introduction The military surgeon must be prepared to address the unique challenges that pediatric patients present, not only in war scenarios, but also in noncombat military operations other than war. Anatomical and Physiological Considerations Fluid, electrolyte, and nutrition.
o Maintenance fluid requirements in children may be estimated using a weight-based nomogram (Table 31-1) or a length-based method, such as the Broselow Pediatric Emergency Tape. There is an increasing realization that hyponatremia caused by hypotonic intravenous fluids is a significant cause of morbidity and mortality in the injured child. Administration of D51/2NS with 20 mEq/L of KC1 is acceptable even in infants. Monitoring serum electrolytes serves as is a useful guide in assessing the adequacy of resuscitation. Because infants and small children have limited glycogen stores, administration of glucose early in the resuscitation period (within 6 hours) is critical.
Table 31-1. Hourly Fluid Requirements for Children Weight (kg) Hourly Volume Fluid D51/2NS + 20 mEq KCl/L 11–20 kg 40 mL + 2 mL/kg over 10 kg D51/2NS + 20 mEq KCl/L >20 kg 60 mL + 1 mL/kg over 20 kg D51/2NS + 20 mEq KCl/L o Fluid resuscitation is best performed with normal (0.9%) saline at 20 mL/kg boluses. (See Evaluation and Diagnosis.) 452 Emergency War Surgery o Normal urine output -infants: 2 mL/kg/h -children: 1 mL/kg/h -adults: 0.5 mL/kg/h o Daily caloric and protein requirements may be estimated based on weight and age (Table 31-2).
Table 31-2. Daily Caloric and Protein Requirements for Children Body Weight Protein Age (yrs) (kcal/kg) (g/kg Body Weight) 0–1 90–120 2.0–3.5 1–7 75–90 2.0–2.5 7–12 60–75 2.0 12–18 30–60 1.5 >18 25–30 1.0 o Breast milk is optimal when initiating oral intake in infants, and offers many advantages over other preparations. Standard infant formulas provide 20 kcal/oz. An estimate of the amount of formula needed to provide 120 kcal/kg/d is: Infant’s weight (kg) × 22-30 = Amount (in mL) of formula needed q4h.
o If pediatric-appropriate enteral formulas are unavailable, adult preparations may be utilized beyond infancy by diluting them with free water: 100 mL/kg of free water per day for the first 10 kg, 50 mL/kg/day for the second 10 kg, and then 20 mL/kg/day for every kg over that up to 2.4 L/day.
Pulmonary.
o In contrast to adults the most common cause of cardiac arrest in children is respiratory arrest. Hypoxemia can lead to bradycardia with hypoperfusion, followed rapidly by cardiac arrest.
o Newborns are obligate nasal breathers; thus, nasal airways and nasogastric tubes should be avoided if possible in favor of oropharyngeal airways and orogastric tubes.
453 Pediatric Care o The child’s larynx is positioned more anteriorly compared to adults, and a more forward positioning of the head (“sniffing position”) will facilitate visualization of the trachea.
o The acceptable range of PaO2 (60–90 mm Hg) corresponds to an oxygen saturation of 92%–97%. Oxygen saturations in the low 90s are adequate in infants due to increased levels of fetal hemoglobin (Hgb F). Supraphysiologic concentrations of inspired oxygen are unnecessary, and are harmful, especially to preterm infants.
o Infants breathe primarily using their diaphragm; thus, increases in intraabdominal pressure or other problems that limit diaphragmatic movement may significantly impair respiration. An example of this is gastric distension. In the traumatically injured infant or child, gastric decompression is important to preserve respiratory function and prevent vagal stimulation. Children have limited functional residual capacity; therefore, maneuvers such as c-spine and backboard immobilization may impair respiration, especially in the face of gastric distension.
Cardiovascular.
o Vital signs by age group are shown in Table 31-3.
Table 31-3. Normal Vital Signs for Age Weight Respiration BP Age (kg) Rate Pulse (Systolic) Premie <3 40–60 130–150 42 ± 10 Term 3 40 120–140 60 ± 10 1–5 years ~10–20 20–30 100–130 95 ± 30 6–10 years 20–32 12–25 75–100 100 ± 15 Adolescent 50 12–18 70 120 ± 20 o Cardiac stroke volume in infants is relatively fixed. Therefore, bradycardia or relative bradycardia may be associated with a significant decrease in cardiac output (CO=HR × SV). 454 Emergency War Surgery Obtaining venous access in a hypotensive child may be extremely difficult and may delay the administration of critically needed fluids or medications. Limit peripheral IV access attempts to two within 90 seconds for the child in shock, then immediately proceed to intraosseous needle access. (See Chapter 7, Shock, Damage Control Resuscitation, and Vascular Access.) NEVER utilize the sternum as a location for IO access in small children. The humerus should be used only as a last resort. In general, the tibial plateau or distal femur are the preferred locations.
Burns.
o An infant or child’s head comprises a greater percentage of the total body surface area, with the lower extremities being a lesser percentage. The area encompassed by the palm of the patient’s hand corresponds to approximately 1% of the total body surface area, and may be useful in calculating the total burn surface area (Fig. 31-1).
Gastrointestinal.
o Esophageal reflux is physiologic in the newborn period, and decreases as a child becomes ambulatory, solid food comprises a greater part of the diet, and lower esophageal sphincter tone increases. o Upright positioning and institution of thickened feedings often improves symptomatic children.
o Children (especially infants) are predisposed to hypoglycemia due to their low hepatic glycogen stores. Full-term infants will tolerate NPO status for approximately 5 days (with an appropriate D10 solution). Premature infants will tolerate only 3 days of NPO status prior to the initiation of total parenteral nutrition. Approximately 5–7 mg/kg/min of glucose is necessary to maintain normoglycemia.
18% 9% 9% 18% Back 13% Each buttock 2.5% Legs 7% on each side 14% 14% Fig. 31-1. Body surface area percentages for infants and children.
455 Pediatric Care o A child’s GI tract is very sensitive to insults, including hypoxemia, enteric infection, electrolyte and acid-base abnormalities, and systemic illness. This may be manifest as impaired motility and feeding intolerance.
o Gastroenteritis with diarrhea, often associated with fevers, is also a very common cause of severe dehydration.
Hematology and blood volume.
o Infants have a physiologic anemia (hematocrit of 30%–33%) as production of adult hemoglobin (Hgb A) replaces fetal hemoglobin (Hgb F) during the first 3–5 months of life.
o Estimates of blood volume are as follows: Age Estimate Volume (mL/kg) Newborn 90 Infant 80 School-age child 70 Renal.
o Infants and young children have a limited ability to concentrate urine (maximum: 400–600 mOsm/L) and a fixed ability to excrete sodium, and are therefore at risk for hypernatremia if excessive sodium is administered. o Maintenance electrolyte requirements: 1 mEq/kg/day of potassium 2 mEq/kg/day of sodium 3 mEq/kg/day of chloride o Serum electrolytes should be monitored daily in children who are NPO with significant GI losses or children who are being resuscitated. GI losses such as gastric fluid output should be carefully measured and replaced with the appropriate IV fluids. Bolusing children with potassium-containing fluids can be very dangerous. It is preferable to tolerate mild to moderate hypokalemia while replacing losses very slowly (over 12 hours) while on an EKG monitor, if available. It is much safer to replace ongoing losses as they occur, preferably by the enteral route, than to use an IV bolus for measured deficits.
Thermoregulation.
o Infants and young children are predisposed to heat loss due to diminished quantities of adipose tissue, and they 456 Emergency War Surgery compensate poorly for wide fluctuations in ambient temperatures. Children have a higher ratio of body surface area to mass, and therefore are likely to experience increased insensible water loss and become dehydrated more quickly than adults when febrile.
o Reduce exposure to heat loss, and keep infants and children in a regulated warm environment. Active warming measures such as warming IV fluids and using thermal blankets are critical for infants and small children during trauma evaluations and in the operating room. Small children will rapidly become hypothermic if this is not done as part of standard practice.
Immune system.
o Premature infants have an immature immune system, causing a 60-fold increased risk of sepsis. All elective surgery in infants under 30 days of age requires 48 hours of prophylactic antibiotics (with anaerobic coverage added when appropriate) after the first week of life.
o Early signs of sepsis may be subtle in infants and may be manifested as lethargy, intolerance to feedings, fever, hypothermia, tachycardia, and irritability before a rise in white blood cell count. Leukopenia and thrombocytopenia in infants and small children may be associated with an overwhelming infectious process.
Evaluation and Diagnosis Pediatric cervical spine clearance can be performed with a physical examination in awake children without neurological deficits. If there is no midline tenderness or pain with active motion, the spine can be cleared. Obtunded children, those with focal neurological deficits, and those with tenderness should have further imaging, dictated by imaging modalities available at a given facility. Although children are at higher risk for spinal cord injury without radiographic abnormality (SCIWORA) than adults, most cases may be managed with immobilization alone. Plain films will detect most clinically significant c-spine injuries in children. Unlike in adults, a normal CT of the c-spine in small children will not rule out a ligamentous injury. Pseudosubluxation of C2 on C3 may 457 Pediatric Care be evident on x-ray. If in doubt, continue immobilization of the cervical spine. Because of the larger cranium in infants, the body should be placed on a pad in order to maintain normal anatomic alignment of the airway during backboard immobilization, and sufficient padding should be placed under the occiput to prevent pressure necrosis of the scalp. Basic ATLS guidelines should direct the initial assessment and evaluation for all children involved in traumas.
o Modified Glasgow Coma Scale scores for children < 4 years old: Verbal Response Verbal Score Appropriate words/social 5 smile/fixes/follows Cries, but consolable 4 Persistently irritable 3 Restless, agitated 2 None 1 Treatment The treatment algorithm shown here provides the proper sequence for the rapid sequence intubation of the pediatric patient (Fig. 31-2).
o Avoid succinylcholine in patients with burn or crush injuries, or at risk for increased intracranial pressure.
o Ketamine may prevent hypotension in patients not in septic shock.
Equipment and Supplies Accessory pediatric medical/surgical equipment arranged according to age and weight appears in Table 31-4.
Surgical instruments.
o If a pediatric surgical set is not immediately available, a peripheral vascular set will usually contain instruments delicate enough to accomplish most tasks in newborns.
Commonly Used Drugs and Dosages All doses are IV or IM.
Phenobarbital: 10–20 mg/kg IV at a rate not to exceed 1 mg/ kg/min (maximum dose: 40 mg/kg).
Diazepam: 0.04–0.3 mg/kg/dose.
458 Emergency War Surgery Fig. 31-2. Rapid sequence intubation for the pediatric patient.
Preoxygenate Atropine 0.02 mg/kg Sedation Midazolam Fentanyl 0.1 mg/kg (5 mg maximum) 2–5 g/kg Cricoid pressure Intubate, Check Tube Position, Release Cricoid Pressure Paralysis Succinylcholine chloride —Weight <10 kg: 2 mg/kg —Weight >10 kg: 1 mg/kg Rocuronium 1 mg/kg OR Vecuronium 0.1–0.2 mg/kg Midazolam: 0.1 mg/kg IV (maximum: 5 mg).
Atropine: 0.02 mg/kg IV.
Phenytoin: 15–20 mg/kg IV; administered at 0.5–1.5 mL/kg/ min as a loading dose, then 4–7 mg/kg/d IV for maintenance.
Mannitol: 0.25–1.0 g/kg IV.
Succinylcholine chloride: 2 mg/kg IV for <10 kg and 1 mg/ kg IV for >10 kg.
Ampicillin: 25–50 mg/kg IV q6h; 100–200 mg/kg/d divided q6h.
Gentamicin: 4.5–7.5 mg/kg IV qd [once daily dosing (ODD)]; keep doses in manual for q8h dosing.
459 Pediatric Care Table 31-4. Pediatric Resuscitation Equipment and Supplies Airway/Breathing Circulation Supplemental Equipment Age, Weight Oral Bag Laryngo- IV NG Chest Urinary (kg) O2 Mask Airway Valve scope ET Tube Stylet Suction BP Cuff Cath Tube Tube Cath C-collar Premie Premie 2.5–3.0 Premie 24 5 Fr 3 kg Newborn Infant Infant 0 Straight No cuff 6 Fr 6–8 Fr Newborn gauge 12 Fr 10–14 Fr Feeding — 0–6 mo Infant 3.0–3.5 Newborn 22 5–8 Fr 3.5 kg Newborn Small Infant 1 Straight No cuff 6 Fr 8 Fr Infant gauge 12 Fr 12–18 Fr Feeding — 6–12 mo 3.5–4.0 Infant 22 7 kg Pediatric Small Pediatric 1 Straight No cuff 6 Fr 8–10 Fr Child gauge 12 Fr 14–20 Fr 8 Fr Small 1–3 yrs 4.0–4.5 20–22 10–12 kg Pediatric Small Pediatric 1 Straight No cuff 6 Fr 10 Fr Child gauge 12 Fr 14–24 Fr 10 Fr Small 4–7 yrs 2 Straight 5.0–5.5 20 16–18 kg Pediatric Medium Pediatric or curved No cuff 14 Fr 14 Fr Child gauge 12 Fr 20–32 Fr 10–12 Fr Small 8–10 yrs Medium Pediatric 2–3 Straight 5.5–6.5 Child 18–20 24–30 kg Adult Large Adult or curved Cuffed 14 Fr 14 Fr Adult gauge 12 Fr 28–38 Fr 12 Fr Medium BP: blood pressure; Cath: catheter; C-collar: cervical collar; ET: endotracheal; Fr: French (gauge); IV: intravenous; NG: nasogastric; O2: oxygen.
460 Emergency War Surgery Metronidazole: 7.5 mg/kg IV q6h.
Acetaminophen: 15 mg/kg PO q4h.
Cefazolin: 25–100 mg/kg/d divided q6h–q8h.
Clindamycin: 15–40 mg/kg/d divided q6h–q8h.
Hypertonic saline (3%): 5–10 mL/kg.
Morphine: 0.1–0.2 mg/kg q2h–q4h PRN.
Ketamine: 0.5–1.5 mg/kg IV over 1 minute >3 months; 2–4 mg/kg IM.
Surgical Management Basics.
o As a general guideline, transverse abdominal incisions should be used in infants. This minimizes the risk of postoperative dehiscence, while still allowing adequate exposure of all areas of the abdomen except the gastro-esophageal junction.
o Absorbable sutures, such as VICRYL or PDS (2-0), should be used to close the rectus fascia, regardless of the incision. The skin can then be closed using staples or absorbable monofilament suture (eg, MONOCRYL 6-0).
o Cricothyroidotomy should not be done in children under age 10. If a surgical airway is required, a large-bore IV connected to high flow oxygen can be placed into the cricothyroid membrane or the trachea while preparations are made for an urgent tracheostomy. It is important to allow for passive expiration during this procedure because pneumothoraces can occur if high flow oxygen is administered in the face of an upper airway obstruction. Use of a needle cricothroidomy for more than 45 minutes will result in hypercarbia. Tracheostomy in small children is done as it is in adults. The presence of an esophageal tube is critical to avoid inadvertent injury to the esophagus.
o The vast majority of thoracic injuries in children can be managed with chest tube alone. Children who present in extremis due to penetrating thoracic injury should undergo a resuscitative thoracotomy similarly to adults. However, for children with blunt trauma, survival after resuscitative thoracotomy is so poor that its use in this setting cannot be justified, particularly in the austere environment.
461 Pediatric Care o When thoracotomy is required in small children, double lumen endotracheal tubes with lung isolation is usually not necessary. Placement of a bronchial blocker or standard endobronchial placement of a single lumen tube will allow for sufficient exposure in small children.
References Creamer KM, Fuenfer MM, eds. Pediatric Surgery and Medicine for Hostile En- vironments. Fort Sam Houston, TX: Borden Institute; 2016.
Tschudy MM, Arcara KM, eds. The Harriet Lane Handbook: A Manual for Pediatric House Officers. 19th ed. Philadelphia, PA: Elsevier Mosby; 2012.
For Clinical Practice Guidelines, go to 462 Emergency War Surgery |
3018 | http://www.algebra.com/algebra/homework/logarithm/logarithm.faq.question.1195515.html | SOLUTION: Solve x= log2(1/8) by rewriting in exponential form
SOLUTION: Solve x= log2(1/8) by rewriting in exponential form
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Algebra: Logarithm SectionSolvers SolversLessons LessonsAnswers archive Answers Discover more Math Algebra Inc Mathematics math Click here to see ALL problems on logarithm Question 1195515: Solve x= log2(1/8) by rewriting in exponential form Found 2 solutions by Alan3354, greenestamps: Answer byAlan3354(69443) (Show Source): You can put this solution on YOUR website! Solve by rewriting in exponential form --- x = -3 Answer bygreenestamps(13200) (Show Source): You can put this solution on YOUR website! (1) Read the given equation out loud: "x is the logarithm, base 2, of 1/8". (2) Now say out loud 100 times (okay, maybe 3 or 4): "The logarithm is the exponent". Now you know the base is 2 and the exponent is x; so the "1/8" has to be the answer: Then solve by writing 1/8 as a power of 2: The bases are the same, so the exponents are equal:
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3020 | https://mathworld.wolfram.com/IntervalArithmetic.html | TOPICS
Interval Arithmetic
Interval arithmetic is the arithmetic of quantities that lie within specified ranges (i.e., intervals) instead of having definite known values. Interval arithmetic can be especially useful when working with data that is subject to measurement errors or uncertainties. It can be considered a rigorous version of significance arithmetic (a.k.a., automatic precision control).
It is powerful enough to provide rigorous mathematical proofs (de la Llave 1991, Hutchings et al. 2000, Tucker 2002, Gutowski 2003), but rigor comes at a price. In particular, interval arithmetic can be slow, and often gives overly pessimistic results for real-world computations.
See also
Floating-Point Arithmetic, Interval, Projectively Extended Real Numbers
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References
de la Llave, R. In Computer Aided Proofs in Analysis (Ed. K. Meyer and D. Schmidt). New York: Springer-Verlag, 1991.Marlov, S. M. In Scientific Computing and Validated Numerics (Ed. G. Alefeld; A. Frommer, and B. Lang). Berlin: Akademie Verlag, 1996.Gutowski, M. W. "Power and Beauty of Interval Methods." 20 Feb 2003. M.; Morgan, F.; Ritoré; M.; and Ros, A. Electron. Res. Announc. Amer. Math. Soc. 6, 45, 2000.Jaulin, L.; Kieffer, M.; Didrit, O.; and Walter, É. Applied Interval Analysis. London: Springer-Verlag, 2003.Kearfott, B. R. Euromath Bull. 2, 95, 1996.Petkovič M. S.; and Petkovič, L. D. Complex Interval Arithmetic and Its Applications. Berlin: Wiley, 1998.Popova, E. D. and Ullrich, C. P. "Simplication of Symbolic-Numerical Interval Expressions." In Proceedings of the 1998 International Symposium on Symbolic and Algebraic Computation (Ed. O. Gloor). New York: ACM Press, pp. 207-214, 1998.Schenkel, A.; Wehr, J.; and Wittwer, P. Math. Phys. Electr. J. 6, 2000.Shokin, Y. I. In Scientific Computing and Validated Numerics (Ed. G. Alefeld; A. Frommer, and B. Lang). Berlin: Akademie Verlag, 1996.Trott, M. "Interval Arithmetic." §1.1.2 in The Mathematica GuideBook for Numerics. New York: Springer-Verlag, pp. 54-66, 2006. W. Found. Comput. Math. 2, 53, 2002.Wolfram, S. A New Kind of Science. Champaign, IL: Wolfram Media, p. 1168, 2002.
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Finding Unit Rates with Equivalent Ratios
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August 27, 2024
When comparing amounts and making judgments in daily life, unit rates are an invaluable tool. In other words, a unit rate represents the connection between two measures by indicating the amount of one quantity that is equivalent to one unit of another. For instance, understanding the price per item while purchasing groceries might help you get the greatest bargain. We frequently use equivalent ratios—ratios that represent the same connection even when the quantities differ—to quickly obtain unit rates. When you're grocery shopping, cooking, or vacation planning, you may streamline computations and draw insightful comparisons by knowing how to employ similar ratios. Let's examine the relationship between unit rates and equivalent ratios in more detail!
What is a Unit Rate?
When comparing two distinct quantities, one is stated per unit of the other, and this comparison is done using a unit rate. It indicates the ratio of one item to one unit of another. The unit pricing would be $2 per apple, for instance, if you paid $8 for four apples. Unit rates are crucial for making well-informed decisions since they are frequently utilized in daily calculations such as expenses, speeds, and pricing. They aid in streamlining comparisons and facilitating comprehension of computations.
The Importance of Equivalent Ratios
Two or more ratios that represent the same connection between items are called equivalent ratios. The ratios 1:2, 2:4, and 3:6 are comparable, for instance. Comprehending equivalent ratios is essential for determining unit rates since they make the procedure easier.
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Steps to Find Unit Rates Using Equivalent Ratios
Identify the Ratios: Decide the two amounts you wish to compare first. Let's take an example where you are examining the cost of oranges. The price-to-weight ratio is $6 for three pounds of oranges if you purchase them for that amount.
Set Up the Ratio: Put this information in the fractional form: $6. Three pounds, three pounds, six bucks. This facilitates the visualization of the two quantities' connection.
Find Equivalent Ratios: By identifying analogous ratios, you may simplify the ratio and determine the unit rate.
Interpret the Result: We may now interpret the unit rate that we have determined. In this example, comparing the price of oranges to other fruits or other quantities is made simple by the knowledge that they cost $2 per pound.
Real-Life Applications
We use unit rates in our daily lives and it's not only an academic exercise. Here are a few instances:
Shopping: Understanding the unit rate while comparing costs can help you choose which brand has the greatest value.
Traveling: Knowing miles per gallon will help you more efficiently budget for petrol expenses while you're making travel plans.
Cooking: Converting amounts is a common requirement in recipes; understanding unit rates helps make ingredient scaling easier.
Practice Makes Perfect
Practice unit rate calculations using real-world examples to gain proficiency. You can attempt the following scenarios:
At the grocery shop, compare the costs of various-sized packages of the same product.
By tracking the distance you've traveled over time, you can determine your speed.
Analyze the overall cost of the ingredients in a recipe to find the cost per serving
Using comparable ratios to find unit rates is a useful ability that may improve your comprehension of real-world scenarios. You can make better decisions whether you travel, buy, or cook if you use this strategy. So, keep in mind to simplify and determine that crucial unit rate the next time you come across a ratio.
FAQs (Frequently Asked Questions)
Q1: What is a unit rate?
Ans: A unit rate is a comparison of two different quantities expressed as a quantity per one unit of another. For example, if you pay $12 for 3 pounds of apples, the unit rate is $4 per pound.
Q2: How do I find a unit rate?
Ans: To find a unit rate, divide the first quantity by the second quantity. For instance, if a car travels 150 miles in 3 hours, divide 150 by 3 to find the unit rate of 50 miles per hour.
Q3: What are equivalent ratios?
Ans: Equivalent ratios are ratios that express the same relationship between quantities, even if the numbers are different. For example, 2:4, 3:6, and 1:2 are all equivalent ratios.
Q4: How do equivalent ratios help in finding unit rates?
Ans: Equivalent ratios allow you to simplify ratios to find unit rates easily. By finding a ratio equivalent to 1 (e.g., 3:3 or 4:4), you can determine the cost or rate per unit more clearly.
Q5: Can you give an example of finding a unit rate using equivalent ratios?
Ans: Sure! If 10 pounds of rice costs $20, you can express this as a ratio: $20 for 10 pounds. By simplifying, you can find that $20 divided by 10 equals $2 per pound, making the unit rate $2 per pound.
Q6: Why are unit rates important?
Ans: Unit rates are important because they help us compare prices, make budget decisions, and understand quantities more easily in everyday life, from grocery shopping to travel planning.
Q7: How can I practice finding unit rates?
Ans: You can practice by comparing prices at the grocery store, calculating speeds while driving, or figuring out costs per serving when cooking. Look for real-world examples where you can apply unit rates and equivalent ratios.
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3022 | https://math.stackexchange.com/questions/3520936/random-point-x-y-is-uniformly-distributed-in-the-unit-square | probability theory - Random point $(X,Y)$ is uniformly distributed in the unit square. - Mathematics Stack Exchange
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Random point (X,Y)(X,Y) is uniformly distributed in the unit square.
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Random point (X,Y)(X,Y) is uniformly distributed in the unit square. At which values for the constant "c" the events |X−Y|≥c|X−Y|≥c and X+Y≤3 c X+Y≤3 c are independent?
I found that p.d.f. of (X+Y)(X+Y) as: f(X+Y)(X+Y)=(X+Y)f(X+Y)(X+Y)=(X+Y) when 0<(X+Y)≤1 0<(X+Y)≤1 and f(X+Y)(X+Y)=1−(X+Y)f(X+Y)(X+Y)=1−(X+Y) when 1<(X+Y)≤2 1<(X+Y)≤2. Also found that p.d.f. of (X-Y) as: f(X−Y)(X−Y)=1+(X−Y)f(X−Y)(X−Y)=1+(X−Y) when −1<(X−Y)≤0−1<(X−Y)≤0 and f(X−Y)(X−Y)=1−(X−Y)f(X−Y)(X−Y)=1−(X−Y) when 0<(X−Y)≤1 0<(X−Y)≤1.
I think that to find when the events are independent I need to find when P r{|X−Y|≥c,X+Y≤3 c}=P r{|X−Y|≥c}∗P r{X+Y≤3 c}P r{|X−Y|≥c,X+Y≤3 c}=P r{|X−Y|≥c}∗P r{X+Y≤3 c} but I have no idea how to continue.
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edited Jan 24, 2020 at 13:10
Noa Even
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1 Draw unit square and find all the probabilities as the area of the corresponding parts of the square.NCh –NCh 2020-01-24 13:17:54 +00:00 Commented Jan 24, 2020 at 13:17
Thanks you. If the two events don't overlap inside the unit square would that be enough to say that they are independent?GeorgiD –GeorgiD 2020-01-24 13:30:25 +00:00 Commented Jan 24, 2020 at 13:30
These are always dependent.Math-fun –Math-fun 2020-01-24 14:01:52 +00:00 Commented Jan 24, 2020 at 14:01
@Math-fun They definitely aren't dependent if c≥1 3 c≥1 3.WoolierThanThou –WoolierThanThou 2020-01-24 14:47:00 +00:00 Commented Jan 24, 2020 at 14:47
@GeorgiD If events don't overlap and have positive probabilities, they are dependent.NCh –NCh 2020-01-24 15:26:02 +00:00 Commented Jan 24, 2020 at 15:26
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The region X+Y≤3 c X+Y≤3 c is the region below the downward sloping line. The region |X−Y|≥c|X−Y|≥c is the region outside the parallel lines. So for c≤1 3 c≤1 3 we have the situation in the first image. The shaded regions show the overlap.
As c c increases, the parallel lines move further apart and the downward sloping line moves up.
So for 1 3<c<1 2 1 3<c<1 2 we have the situation in the second image, and for c>1 2 c>1 2 we have the situation in the third image. Note that for c>2 3 c>2 3 the downward sloping line has moved out of the square, so the whole square satisfies the inequality X+Y≤3 c X+Y≤3 c.
c≤1 3 c≤1 3">
The condition for independence of events A,B A,B is p(A and B)=p(A)p(B)p(A and B)=p(A)p(B). So we have to calculate those three probabilities. Let us call |X−Y|≥c|X−Y|≥c event A A, and X+Y≤3 c X+Y≤3 c event B B. The area outside the parallel lines is just the area of a square side 1−c 1−c, so p(A)=(1−c)2 p(A)=(1−c)2. For c<1 3 c<1 3 the area below the downward sloping line is 1 2(3 c)2 1 2(3 c)2. For 1 3≤c≤2 3 1 3≤c≤2 3 it is 1−1 2(2−3 c)2=6 c−9 2 c 2−1 1−1 2(2−3 c)2=6 c−9 2 c 2−1. For c>2 3 c>2 3 it is 1. So we have p(B)p(B). If we plot the product p(A)p(B)p(A)p(B) we get:
We now calculate the area of the overlap region. For c≤1 3 c≤1 3 it is two halves of a square diagonal 2 c 2 c. So the area is 2 c 2 2 c 2. For 1 3<c<1 2 1 3<c<1 2 it is a square side 1−c 1−c less a square diagonal 2−4 c 2−4 c, giving an area of (1−c)2−1 2(2−4 c)2=6 c−7 c 2−1(1−c)2−1 2(2−4 c)2=6 c−7 c 2−1. For c>1 2 c>1 2 it is two halves of a square side 1−c 1−c, so area (1−c)2(1−c)2. So we have p(A and B)p(A and B). If we plot p(A)p(B)p(A)p(B) and p(A and B)p(A and B) together we get:
So we guess they are equal for c=1 3 c=1 3 and c≥2 3 c≥2 3. Checking we find that indeed p(A)p(B)=p(A and B)=2 9 p(A)p(B)=p(A and B)=2 9 for c=1 3 c=1 3.
For 1 3<c<1 2 1 3<c<1 2 some tiresome algebra shows there are indeed no solutions. For 1 2<c<2 3 1 2<c<2 3 we have
p(A)p(B)=(1−c)2(6 c−9 2 c 2−1)≠(1−c)2=p(A and B)p(A)p(B)=(1−c)2(6 c−9 2 c 2−1)≠(1−c)2=p(A and B)
For c>2 3 c>2 3 we have p(B)=1 p(B)=1 and so p(A)p(B)=p(A and B)p(A)p(B)=p(A and B).
So the values of c c giving independence are c=1 3 c=1 3 and 2 3≤c<1 2 3≤c<1. Note that we are told that X,Y X,Y are points in the unit square, so we must have 0≤c≤1 0≤c≤1. If c<0 c<0 then X+Y≤3 c X+Y≤3 c is impossible for both X,Y≥0 X,Y≥0; and if c>1 c>1, then |X−Y|≥c|X−Y|≥c is impossible. Alternatively, note that if p(A)=0 p(A)=0 or p(B)=0 p(B)=0, then p(A and B)=0 p(A and B)=0 also, and so A,B A,B are independent (albeit trivially).
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edited Jan 25, 2020 at 8:16
answered Jan 24, 2020 at 19:32
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Thanks a lot for your explanation GeorgiD –GeorgiD 2020-01-24 21:28:51 +00:00 Commented Jan 24, 2020 at 21:28
You forgot c≤0 c≤0.NCh –NCh 2020-01-25 03:43:51 +00:00 Commented Jan 25, 2020 at 3:43
@NCh Thanks. Fixed.almagest –almagest 2020-01-25 08:16:20 +00:00 Commented Jan 25, 2020 at 8:16
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3023 | https://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1038&context=physicsgallup | University of Nebraska - Lincoln University of Nebraska - Lincoln
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Gordon Gallup Publications Research Papers in Physics and Astronomy
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Quantum-Mechanical Analysis of a Longitudinal Stern-Gerlach Quantum-Mechanical Analysis of a Longitudinal Stern-Gerlach
Effect Effect
Gordon A. Gallup UNL , ggallup1@unl.edu
Herman Batelaan University of Nebraska - Lincoln , hbatelaan@unl.edu
Timothy J. Gay University of Nebraska - Lincoln , tgay1@unl.edu
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Quantum-Mechanical Analysis of a Longitudinal Stern-Gerlach Effect
G. A. Gallup, H. Batelaan, and T. J. Gay
Behlen Laboratory of Physics, University of Nebraska, Lincoln, Nebraska 68588-0111
(Received 28 August 2000 )We present the results of a rigorous quantum-mechanical calculation of the propagation of electrons through an inhomogeneous magnetic field with axial symmetry. A complete spin polarization of the beam is demonstrated assuming that a Landau eigenstate can be inserted into the field. This is in contrast with the semiclassical situation, where the spin splitting is blurred. DOI: 10.1103/PhysRevLett.86.4508 PACS numbers: 34.80.Nz, 03.65.Ta, 14.60.Cd The Stern-Gerlach experiment is one of the most impor-tant in the history of physics and is often used to illustrate the nature of spin in quantum mechanics . Curiously, a magnet of the type used by Stern and Gerlach does not work with beams of electrons because of the combined ef-fects of the Lorentz force and the Heisenberg uncertainty principle. This was shown first by Mott and Bohr in 1928 . Pauli subsequently made a more general argument that no device based on the concept of classical particle trajec-tories and macroscopic magnetic fields could be used to separate an electron beam by spin or to measure the elec-tron’s magnetic moment . The Bohr/Mott/Pauli argu-ment is codified in numerous textbooks . In this Letter we show that, in fact, it is possible to ob-serve spin splitting of a beam of electrons using a longitu-dinal magnetic-field configuration instead of the standard transverse geometry of Stern and Gerlach. The longitudi-nal configuration has the advantage that the electrons ex-perience only off-axis Lorentz forces that are significantly smaller than the on-axis forces in the transverse geome-try. Such an idea was first proposed by Brillouin , but was specifically rejected by Pauli. Recently, however, we discovered an error in the reasoning Pauli used against Brillouin’s idea, and analyzed a counterexample using classical particle trajectories in which spin splitting of elec-trons could be achieved equal to the blurring caused by the Lorentz forces [6–8]. Although these results are intrigu-ing, they do not take into account the wave nature of the electron, and thus do not address the central question: Can a spin separation really be expected? In this article we re-port the results of a rigorous quantum-mechanical analysis of the longitudinal Stern-Gerlach problem, which corre-sponds to physical reality. We obtain the surprising result that complete separation can be achieved, an improvement over the semiclassical situation. Complete spin splitting is thus shown to be, fundamentally, a quantum-mechanical effect. It is clear that the magnetic moment of individual elec-trons can be measured; Dehmelt and his colleagues accom-plished this with a modified Penning trap . Separation of electrons by spin does not follow from these experi-ments, however, even though Pauli used the same general argument against both. Quantum calculations have shown spin separations in some static situations [10,11]. For the beam configuration considered by Pauli et al., however, the situation is more ambiguous. Adler and Garroway and Stenholm verified Pauli’s analysis of the trans-verse-field geometry. More importantly for this discussion, Bloch and Dehmelt sketched quantum arguments for longitudinal-field geometries and suggested that com-plete isolation of the lowest energy “spin-backward” state should be possible. They did not, however, consider the nondestructive case of full transmission of both spin com-ponents in a beam. An experiment of the “Bloch-Dehmelt” type was performed by Knight and his colleagues in the mid 1960s, and observation of a low energy ,10 28 eV
tail of (presumably) spin-polarized electrons was reported . This work was never formally published, and, ap-parently, could not be reproduced . Sannikov and Conte et al. considered the problem we take up here, i.e., longitudinal spatial spin separation of a fully trans-mitted beam, but did not use electron wave packets hav-ing extended transverse dimensions. This precludes any elimination of the splitting due to blurring of the separate spin states. Our calculation begins with the full nonrelativistic Ham-iltonian for an electron in the magnetic field Br, f, z of a simple current ring. The ring lies in the x-y plane, has radius R, and is centered at the origin. Thus,
Br, f, z 2 12 r dB z
dz cos fˆx 1 sin fˆy
1 Bzˆz , (1) where
Bz B0
√
R
pR2 1 z2
!3, (2) and B0 . 0 is the field at the origin assuming rR ø
Schrödinger’s equation for this problem in cylindrical coordinates about the z axis is
H Lz 1 Hz c i ¯h ≠c ≠t , (3) 4508 0031-9007 01 86(20) 4508(4)$15.00 © 2001 The American Physical Society V OLUME 86, N UMBER 20 P H Y S I C A L R E V I E W L E T T E R S 14 M AY 2001
H Lz 2 ¯h22me
1
r≠≠r
μ
r ≠≠r
∂
1 L2
z
2m er2
1 v Lz L z 1 gS z 1 12 m evLz 2r2 ,
(4)
Hz 2 ¯h22m e
≠2
≠z2 1 g dvLdz xS x 1 yS y , (5) where vLz is the Larmor frequency jeB zj
2mec
, Lz is the op-erator for the canonical angular momentum of the electron about the z axis, Sx , Sy , and S z are the Cartesian spin op-erators, m e is the electron mass, and g 21 1 a e is the anomalous electron gyromagnetic ratio. Equations (4) and (5) bear some discussion. The fi rst two terms of (4) and the fi rst term of (5) correspond to the electron ’s kinetic en-ergy. In (4), both the fourth term and the part of the third term involving L z correspond to the classical 2 morb ? B
potential of an electron with orbital angular momentum about z. The latter part of the third term in (4) and the sec-ond term in (5) are the equivalent terms for the magnetic potential associated with electron spin. If B were uniform, the last term of (5) would vanish, and both m l and ms, the quantum numbers associated with Lz and Sz , would sepa-rately be good. Generally, for the Hamiltonian of (3), only
m j m l 1 m s is a good quantum number. We now construct complete wave packet solutions of the form
c X
n
X
ml
X
ms
Rm l
n
r, fhm s a nm l m s z, t , (6) where h is a spinor, the anm l m s z, t are functions that contain all of the explicit z and t dependence of c as the electrons move along the magnetic fi eld, and
H L Rm l
n
hm s E nm l m s Rm l
n
hm s . (7) Note that the Rm l
n
r, f have a parametric dependence upon z, but otherwise are solutions to the standard Landau problem. They span x-y space and can be written as
Rm l
n
r, f Nnjm l jpa r jm l jLjm l j
n
ar 2
3 exp 2 12 ar 2e im l f, (8)
Ljm l j
n
x 21nn! xjm l j e x d ndx n x n1jm l je2x , (9)
a az mvz
¯h , (10) and
N njm l j
s
n! apn 1 jml j! , (11) where
E nm l m s ¯hvLz 2n 1 jm l j 1 m l 1 gm s 1 1 . (12) In order to leave m l as an explicit quantum number, we have not followed the common practice of denoting 2n 1
jm l j as another single integer. Substituting (6) into (3) and using (7), we obtain a set of coupled equations for the anm l m s . The coupling terms in these equations determine how nonadiabatic the electron transmission process is, i.e., how likely it is for an electron in a given Landau uniform-fi eld eigenstate to depart from that state over the course of its passage through the mag-netic fi eld. We estimate these terms by using the physi-cal quantities considered in the semiclassical calculation of Ref. . The electrons follow a 2 m path length that has a midpoint at the center of a 2 cm radius current ring, where the fi eld magnitude B0 is 10 T. Their initial speed is taken to be 10 5 ms 28 meV . These values yield a reasonable splitting 631 mm of the spin in a semiclassi-cal model for on-axis trajectories. From the point of view of this paper, the 10 T maximum fi eld means that magnetic potential energies for the electron along the trajectory will always be ø28 meV for the lowest Landau levels. The above experimental conditions justify two ap-proximations. First, the z dependence of the Landau Hamiltonian [Eq. (4)] could cause transitions between uniform-fi eld Landau eigenstates. We fi nd that these off-diagonal coupling terms can be neglected with one ex-ception: those associated with the magnetic-fi eld gradient that arise from the second term on the right of Eq. (5), corresponding to electron spin fl ip. Second, given that the magnetic potential energies are much smaller than the electron kinetic energy, the WKB approximation is valid for the longitudinal wave function propagation. The spin-fl ip probability can be characterized by the ratio of coupling terms U:
Ç Uij U ii 2 U jj
Ç
3zR
4ae
pa0 R1 1 zR214 , (13) where i and j label states connected by the transverse spin operator. Although not negligible, this quantity is ,10 22
at all values of z, and corresponds to spin-fl ip probabilities of the same order of magnitude. Thus the electrons traverse the magnetic fi eld almost completely adiabatically; if a wave packet that corresponds to a Landau eigenfunction can be inserted into the fi eld, the probability is high that it will emerge in the same state. It is interesting to note that the small amount of spin fl ipping that does occur is inversely proportional to the electron g-factor anomaly. We now consider the transmission of the two Landau wave packets with n, ml , m s 0, 0, 612. These states are most strongly coupled to the 0, 61, 712 states, re-spectively, but this coupling is negligible, as discussed above. The packets are superpositions of plane waves such that 4509 V OLUME 86, N UMBER 20 P H Y S I C A L R E V I E W L E T T E R S 14 M AY 2001
a00 612z, t 0 a6z, 0
2p ¯h212 Z
fp z
3 exp ip z z ¯h dp z , (14) where fpz is taken to be Gaussian-like with a spatial width along z of x20 and a momentum spread of ¯hp2x0.At later times,
a6z, t 2p ¯h212 Z
fp
3 exp
μ
i
∑ pz
¯h 1 d 6E 2 Et
¯h
∏∂
dp , (15) where we have used the WKB approximation for
d6E 2pa0 RG
μ E
2 6 a e ¯hv0
∂
, (16) and
Gx px
Z 2
∑μ
1 2 1
x1 1 j 232
∂12
2 1
∏
dj .
(17) The value of x20 should be chosen to minimize the spread-ing of the wave packet along ˆz over the electron fl ight time
ly0. This condition yields
x0, min
s
¯hl m ey0
(18) and corresponds to 80 mm in the present case. The WKB phase shift can be expanded as
d 6p d0 1 d 1, 6p 2 p0 1 12d2p 2 p02
1 . . . (19) and interpreted as follows: the wave packet is displaced by a distance ¯hd1, 6 relative to its position in the absence of a magnetic potential, and spreads by an amount correspond-ing to the normal spreading of a free wave packet at time
t plus an extra amount corresponding to an additional time increment, m e ¯hd2.The results of our calculation for the “spin-forward ” and
“spin-backward ” minimum uncertainty wave packets are shown in Fig. 1. Their most striking feature is the virtually complete separation of the two packets, in marked contrast with our previous calculations . In those calculations,
“magnetic bottle ” forces associated with the mechanical orbital angular momentum of the electron “smeared ” both packets by an amount equal to their centroid splitting. In the present situation, the individual wave packet spreading is essentially that which one would observe in a fi eld-free measurement, with a very small additional spreading char-acterized by d2. The lack of spreading, when compared with the semiclassical case, results because the eigenen-ergies of the electron wave packet depend not on the me-chanical angular momentum but on the canonical angular momentum L z , which is sharp . Since the electron wave packet is characterized by a sharp eigenenergy, the only
00.2 0.4 0.6 0.8 1-0.1 -0.08 -0.06 -0.04 -0.02 00.02 Probability Density Position on z axis (mm) 0,0,-1/2 0,0,+1/2 2,0,-1/2 2,0,+1/2 3,0,-1/2 3,0,+1/2
FIG. 1. Electron probability density vs distance along the z
axis after traversal of a 2 m fl ight path. Each peak is marked with its principal n, m l , m s values. Distance indicated is the deviation from the position of the leading packet, which equals the fi eld-free position. The n 2 and n 3 peaks are contami-nant contributions caused by the insertion of the electron into the magnetic fi eld (see text). relevant “force ” acting on it is proportional to the lon-gitudinal spatial derivative of the magnetic potential (and thus ≠B≠z) and is therefore sharp as well. The complete spin splitting is thus seen to be, fundamentally, a quantum-mechanical effect. A quantitative measurement of the spin splitting can be made using the parameter F SW, where W is the full width at half maximum of one of the two spin components along the z axis, and S is the splitting distance separating the two spin component centroids. Thus F 0 in the
fi eld-free case, unity if the spins are just resolved using Rayleigh ’s criterion, and ¿1 for “complete ” splitting. F is plotted in Fig. 2 as a function of mean electron energy E0.Wave packets obeying the minimum spreading criterion have widths at the end of a given fi eld-free path length
l proportional to E2140 , whereas the spin splitting varies as E210 . Thus, even at the lowest energies we consider, magnetic splitting always dominates natural wave packet spreading. The initial packets must have the spatial dimensions and angular momentum properties of the ground Landau states in a minimum-spreading longitudinal con fi guration. This means that they must be cylindrically symmetric so that m l 0. Assuming an energy of 28 meV, minimum spreading requires a longitudinal velocity uncertainty of 1 m s and a pulse duration of 2 ps. We assume that the electron beam is de fi ned by two (or more) circular aper-tures along the symmetry axis of the current ring. The insertion of Landau eigenstates into the B-fi eld region oc-curs at a fi nal aperture in, e.g., a highly permeable bound-ary wall. It can be shown that this aperture ’s diameter should be d p20 heB , where B is the fi eld several di-ameters, d, inside the container. This value of d ensures 4510 V OLUME 86, N UMBER 20 P H Y S I C A L R E V I E W L E T T E R S 14 M AY 2001
0123456780.02 0.06 0.10 0.14 0.18 Separation Factor, F Electron Energy (eV)
FIG. 2. Spin splitting fi gure of merit F vs electron energy. When F 1, the opposite-spin wave packets just meet the Rayleigh resolution criterion. maximum overlap between the a00 612 states and an inci-dent wave whose transverse dimensions are much larger than the aperture. In this case, electrons emerging into the magnetic-fi eld region have an 82% probability of be-ing in the n 0 state. The chance of being in a state with
n , 16 is 95%. The effect of these “contaminant ” m l 0
states is shown in Fig. 1. For an apparatus whose typical dimension is 1 m, B
will be of the order of a Gauss at the entrance aperture
d 5m. The size of the aperture is large enough to prevent appreciable diffraction as the electrons enter the magnetic-fi eld region; their de Broglie wavelength at this energy is ,10 28 m. Finally we note that the magnetic
fi eld “leaks ” out into the low-fi eld region from the entrance aperture with an inverse third power spatial dependence on the beam axis . This results in negligible distortion of the electron plane waves as they approach the aperture. It appears there is no fundamental physics preventing the observation of spin splitting. We speculate that the most pernicious problem for the realization of our thought experiment would be the nonideal nature of real collimat-ing apertures, manifesting itself in spurious electric and magnetic-fi eld effects. A better experimental approach may well be one of the type discussed by Conte et al. , in which the beam to be polarized is that of a synchrotron, and the separation is effected through a series of longitu-dinal Stern-Gerlach “kicks. ”
The authors thank Emil Sidky and Marlan Scully for useful conversations. This work was supported in part by NSF Grant No. PHY-9702350, Research Corp. Grant No. RI0380, and TIAA-CREF.
W. Gerlach and O. Stern, Z. Phys. 9, 349 (1922). N. F. Mott, Proc. R. Soc. London A 124 , 425 (1929). W. Pauli, in Magnetism (Gauthier-Villars, Brussels, 1930), pp. 217 –226. See, e.g., C. Cohen-Tannoudji, B. Diu, and F. Laloe,
Quantum Mechanics (Wiley, New York, 1977), p. 969; E. Merzbacher, Quantum Mechanics (Wiley, New York, 1970), 2nd ed., p. 254; N. F. Mott and H. S. W. Massey,
The Theory of Atomic Collisions (Clarendon Press, Oxford, 1965), 3rd ed., p. 214; H. C. Ohanian, Prin-ciples of Quantum Mechanics (Prentice-Hall, Englewood Cliffs, NJ, 1990), p. 243; J. Kessler, Polarized Electrons
(Springer-Verlag, Berlin, 1985), 2nd ed., p. 2; Quantum Theory and Measurement, edited by J. A. Wheeler and W. H. Zurek (Princeton University Press, Princeton, NJ, 1983), p. 701; G. Baym, Lectures on Quantum Mechanics
(Benjamin, New York, 1969), p. 328. L. Brillouin, Proc. Natl. Acad. Sci. U.S.A. 14 , 755 (1928). H. Batelaan, T. J. Gay, and J. J. Schwendiman, Phys. Rev. Lett. 79 , 4517 (1997). G. H. Rutherford and R. Grobe, Phys. Rev. Lett. 81 , 4772 (1998). H. Batelaan and T. J. Gay, Phys. Rev. Lett. 81 , 4773 (1998). H. Dehmelt, Z. Phys. D 10 , 127 (1988). H. Martens and W. M. d. Muynck, J. Phys. A 26 , 2001 (1993). G. H. Rutherford and R. Grobe, J. Phys. A 31 , 9331 (1998). F. Adler, Helv. Phys. Acta 10 , 455 (1937). B. M. Garraway and S. Stenholm, Phys. Rev. A 60 , 63 (1999). F. Bloch, Physica 19 , 821 (1953). L. V. Knight, in Physics (Stanford University Press, Palo Alto, CA, 1965), p. 97. S. S. Sannikov, Sov. Phys. JETP 19 , 1186 (1964). M. Conte, A. Penzo, and M. Pusterla, Nuovo Cimento
108A , 127 (1995). J. D. Jackson, Classical Electrodynamic (Wiley, New York, 1975), 2nd ed., pp. 125 –127. 4511 |
3024 | https://steemit.com/mathematics/@dkmathstats/the-indicator-function | The Indicator Function — Steemit
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The Indicator Function
dkmathstats(61)in#mathematics • 8 years ago
Hi there, this post will be about the indicator function. I was introduced to this topic informally by a former theoretical physicist who later got into financial mathematics and taught us probability theory. The contents here is based from my old notes, my memory and a part from this book.
It is assumed that the reader is familiar with basic probability theory and basic set theory.
A more longer and more technical version of this post is from my website here.
The math text and images rendered with QuickLaTeX.com.
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Definition
Properties
Some basic properties of the indicator function are based from set theory. The set A' is the complement of the set A, A∩B is the intersection of the sets A and B, and A∪B is the union of the two sets A and B.
Since the indicator function is a random variable, we can find the expected value and the variance of it. The P(A) notation stands for the probability of outcomes in set A occurring.
Heaviside Step Function Example
Changing Integral Bounds Examples
Note that the integral at a single point is zero (area of a point is zero). There is no difference in using an open interval or closed interval.
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#math#probability#steemstem#functions
8 years ago in#mathematicsbydkmathstats(61)
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3025 | https://www.oxfordlearnersdictionaries.com/us/definition/english/convenience | Definition of convenience noun from the Oxford Advanced Learner's Dictionary
convenience
noun
/kənˈviːniəns/
/kənˈviːniəns/
Idioms
jump to other results
[uncountable] the quality of being useful, easy or suitable for somebody
We have provided seats for the convenience of our customers.
For (the sake of) convenience, the two groups have been treated as one in this report.
The position of the house combines quietness and convenience.
In this resort you can enjoy all the comfort and convenience of modern tourism.compare inconvenience see also flag of convenience, marriage of convenience
Extra Examples
All our holiday chalets include a microwave and food processor for extra convenience.
An order form is enclosed for your convenience.
I keep my cookbooks in the kitchen for convenience.
The system is based on administrative convenience rather than public benefit.
We leave the keys near the front door for the sake of convenience.
Oxford Collocations Dictionaryadjective
great
added
additional
…verb + convenience
offer
provideconvenience + noun
food
storepreposition
at your convenience
for (your) conveniencephrases
a marriage of convenience
at your earliest convenience
comfort and convenience
…See full entry
Take your English to the next level
The Oxford Learner’s Thesaurus explains the difference between groups of similar words. Try it for free as part of the Oxford Advanced Learner’s Dictionary app
2. [countable] something that is useful and can make things easier or quicker to do, or more comfortable
It was a great convenience to have the school so near.
The house had all the modern conveniences (= central heating, etc.) that were unusual at that time.see also public convenience
Extra Examples
They wouldn't like to live without modern conveniences such as microwaves.
It's a great convenience living next door to a post office.
It's a great convenience living near the station.
Oxford Collocations Dictionaryadjective
great
modernSee full entry
Word Originlate Middle English: from Latin convenientia, from convenient- ‘assembling, agreeing’, from the verb convenire ‘assemble, agree, fit’, from con- ‘together’ + venire ‘come’.
Idioms
at somebody’s convenience
(formal) at a time or a place that is suitable for somebody
Can you telephone me at your convenience to arrange a meeting?
at your earliest convenience
(business) as soon as possible
Please contact us at your earliest convenience.
See convenience in the Oxford Advanced American DictionarySee convenience in the Oxford Learner's Dictionary of Academic English
Check pronunciation:
convenience
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: convenience food noun
convenience store noun
public convenience noun
flag of convenience noun
marriage of convenience noun
convenience foods
convenience stores
flags of convenience
marriages of convenience
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Nearby words
convene verb
convener noun
convenience noun
convenience food noun
convenience store noun
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3026 | https://www.quora.com/Why-do-magnesium-ions-have-a-charge-of-2+ | Why do magnesium ions have a charge of 2+? - Quora
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Magnesium Ion
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Why do magnesium ions have a charge of 2+?
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Magnesium ions have a charge of 2+ because magnesium (Mg) is an alkaline earth metal that has two electrons in its outermost electron shell (the 3s subshell). The electron configuration of a neutral magnesium atom is:
Mg:1 s 2 2 s 2 2 p 6 3 s 2 Mg:1 s 2 2 s 2 2 p 6 3 s 2
To achieve a stable electron configuration, similar to that of the noble gases, magnesium tends to lose these two outer electrons during chemical reactions. By losing two electrons, magnesium becomes a cation with a charge of 2+, represented as Mg 2+Mg 2+.
The loss of two negatively charged electrons results in a net positive charge of +2, which i
Continue Reading
Magnesium ions have a charge of 2+ because magnesium (Mg) is an alkaline earth metal that has two electrons in its outermost electron shell (the 3s subshell). The electron configuration of a neutral magnesium atom is:
Mg:1 s 2 2 s 2 2 p 6 3 s 2 Mg:1 s 2 2 s 2 2 p 6 3 s 2
To achieve a stable electron configuration, similar to that of the noble gases, magnesium tends to lose these two outer electrons during chemical reactions. By losing two electrons, magnesium becomes a cation with a charge of 2+, represented as Mg 2+Mg 2+.
The loss of two negatively charged electrons results in a net positive charge of +2, which is why magnesium ions are commonly found with a 2+ charge in compounds and solutions.
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What is the charge on magnesium?
Mg forms +2 ions, but not +1 ions. Why?
Why is it that magnesium usually forms Mg2+ ions, and why can’t it form Mg1+ ions by losing only one electron instead of two at a time?
Why does magnesium lose 2 electrons?
What are the benefits of magnesium threonate?
Adnan Khan
I have done BS Chemistry 4 years program and now I'm teaching in college · Author has 61 answers and 147.7K answer views
·8y
Because magnesium is one of the group llA (Alkaline earth metal) element of the periodic table.
One thing which we must kept in mind is that the elements group numbers also represent the number of electrons in these elements VALENCE or OUTERMOST SHELL .
The valence or outermost shell electrons of an element only take part in a reaction and the core shell electrons remain intact .
One thing more the number of maximum bonds that an element form in order to form a molecule of a compound can also be easily predicted by the element group number.
Suppose Hydrogen is the group first element so it must ha
Continue Reading
Because magnesium is one of the group llA (Alkaline earth metal) element of the periodic table.
One thing which we must kept in mind is that the elements group numbers also represent the number of electrons in these elements VALENCE or OUTERMOST SHELL .
The valence or outermost shell electrons of an element only take part in a reaction and the core shell electrons remain intact .
One thing more the number of maximum bonds that an element form in order to form a molecule of a compound can also be easily predicted by the element group number.
Suppose Hydrogen is the group first element so it must have 1 electron in its outermostor valence shell it must also form only one bond with other elements and hydrogen does so in the following compounds;
The Lewis structure for the following compounds of hydrogen;
HCl hydrochloric acid
H. .Cl
HF hydroflouric acid
H. .F
CH4, HBr , HI etc
There are eight groups of the elements in the periodic table as lA llA lllA lVA VA VlA VllA and the last one VlllA
CHEMICAL BOND FORMATION
Octate Rule; This rule was put forward by LEWiS according to him.
“Atom loss, gain , share or transferre elentrons in order to be surrounded by eight electrons .”
Just like group VlllA elements such as ;
He Ne Ar Kr Xe Rn
Metals are ELECTROPOSITIVE (means nucleus lover or to get loss their electrons) in nature so they readily reacts with non-metals and oxidize (loss electrons) to be surrounded by the electrons like the group VlllA the “nobel gas” eelements.
As we know that the group llA elements are called “ ALKALINE EARTH METALS” they've 2 electrons in their outer most shell which they loss or transfer complete to other electronegative ( electron lovers) in order to achieve group VlllA elements like stable electronic configuration (means surrounded by the same number of electrons) because they're chemically stable elements of the periodic table.
Elemens of group llA are given below
llA
Berilium Be
MAGNESIUM Mg
Calcium Ca
Strontium St
Barium Ba
Radium Ra
As your desired element Magnesaium is an ALKALINE EARTH METAL and as we've discussed the chemical properties of metals earlier above so
Being the member of group llA elements it would used the have 2 electrons in its valence shell and it would loss or completely transfer these 2 electrons to the electonegative element( F Cl Br I etc) to get stability or nobel gases or GROUP VllA elements like Elctronic configuration.
Mg ———-> Mg2+ + 2e- ( electrons) Oxidation
2Cl + 2e- —————-> 2Cl- Reduction
Mg+ Cl —————→
As magnesium has lost 2 electrons so the electrons will be lesser than the number of protons present in nucleus by 2 number so it will no more be remained Neutral and will show 2 electrons scarce by +2 charge.
I hope it would be proved fruitfull remember me in prayers
Thanks for reading
Muhammad Adnan
Always stay blessed
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99 15
Asad Khan
Nothing to see here, just an A-level Chemistry student. · Author has 52 answers and 67.2K answer views
·Updated 8y
They have a charge of +2 due to multiple reasons. It's electronic configuration is of 2, 8, 2. This results in a state in which it has 2 valence electrons, which it is keen to lose. This is because it is the aim of all atoms to achieve a full electronic shell, like a noble gas, which gives it stability(remember Mg is more stable than Mg2+ because it needs energy to produce Mg2+ and we will see it in just compound materials). As a result, when the magnesium reacts with any anion or anything else, its electrons are donated to other atoms (or ions now). Because the electrons are donated to other
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They have a charge of +2 due to multiple reasons. It's electronic configuration is of 2, 8, 2. This results in a state in which it has 2 valence electrons, which it is keen to lose. This is because it is the aim of all atoms to achieve a full electronic shell, like a noble gas, which gives it stability(remember Mg is more stable than Mg2+ because it needs energy to produce Mg2+ and we will see it in just compound materials). As a result, when the magnesium reacts with any anion or anything else, its electrons are donated to other atoms (or ions now). Because the electrons are donated to other ions, it now has two protons more than electrons, which gives it a relative charge of 2 positive charges, as it now is deficient in the negative charge provided by the electrons. As a final result, there is a +2 charge on the magnesium ion, as there is a charge difference of 2 electrons' amount, but in a way that there is a greater positive charge.
Hope this helps.
Upvote ·
9 9
9 1
Yan Kee
Studied Chemistry · Author has 314 answers and 2.7M answer views
·9y
The outer shell of magnesium have two electrons it really ”wants” to “give” away the two electrons so if magnesium react with other elements magnesium “gives” away the two electron and magnesium have an ion of 2+ charge here’s an example of magnesium oxide the magnesium gives two electron to oxygen and magnesium have a charge of 2+.
Why magnesium have to “give” away the electron?
Because it it needs to “gives” away two electrons to have a stable electron configuration like neon so magnesium ion have 2+ charge to stable it.
Hope this helps:)
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The outer shell of magnesium have two electrons it really ”wants” to “give” away the two electrons so if magnesium react with other elements magnesium “gives” away the two electron and magnesium have an ion of 2+ charge here’s an example of magnesium oxide the magnesium gives two electron to oxygen and magnesium have a charge of 2+.
Why magnesium have to “give” away the electron?
Because it it needs to “gives” away two electrons to have a stable electron configuration like neon so magnesium ion have 2+ charge to stable it.
Hope this helps:)
Upvote ·
9 4
Related questions
More answers below
Why are we nearly all magnesium deficient?
Does +1 charged magnesium ions exist?
What is the formula for magnesium ion, Mg+2 or Mg2+?
What is the charge of op magnesium?
Whose atomic size of (Na+1) sodium ion and magnesium (Mg+2) ion is more? Why?
Samiul Islam
Actor (2010–present) · Author has 179 answers and 1.3M answer views
·7y
Metals react with non-metals to form salts - for example, sodium chloride. When they are molten, these compounds conduct electricity, which shows that they are made up of charged particles called ions.
Metal ions
When a metal reacts with a non-metal, each metal atom loses the electron, or electrons, from its outer shell. The atom loses negative electrons but still has the same number of positive protons, so it has an overall positive charge. It's not an atom now. Instead it is called an ion. Here are some examples of metal ions:
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How much does it cost to have an employee?
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That depends! There are many factors to consider. If you’ve decided your business needs additional support, selecting the appropriate pay rate for a new hire is only the first step in determining the funds you’ll need available to grow your team. You’ll have to figure out if you can actually afford the total cost of adding someone to your payroll.
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99 31
John Chapman
Ex Chemistry teacher and author · Author has 4K answers and 5.6M answer views
·8y
Because the outer electron shell of a magnesium atom has two electrons in it. Both are lost when it changes to an ion leaving it with 12 positively charged protons and 10 negatively charged electrons. This produces a +2 positive charge.
Upvote ·
9 8
Adi
Ph.D in Rare Earth&Organometallic Chemistry, Imperial College London (Graduated 2014) · Author has 2.2K answers and 4.7M answer views
·7y
Related
Mg forms +2 ions, but not +1 ions. Why?
You can actually form Mg1+ compounds, they are just incredibly rare compared to Mg2+ compounds. For example:
Mg2+ is more stable as Mg has two valence electrons. Losing both allows Mg to have a noble gas configuration, which is what all elements strive towards.
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You can actually form Mg1+ compounds, they are just incredibly rare compared to Mg2+ compounds. For example:
Mg2+ is more stable as Mg has two valence electrons. Losing both allows Mg to have a noble gas configuration, which is what all elements strive towards.
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There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you?
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9 3
Arjun Bhardwaj
9y
Mg (At. No. 12) is in Group 2 of the periodic table , it has 2 electrons in its valence shell hence valency is 2 .Thus, it has a tendency to loose 2 electrons and form Mg²+ to attain nearest noble gas configuration
Upvote ·
9 4
Divyam Verma
Undergrad at BITS Pilani
·6y
Originally Answered: Why is the magnesium valency +2? ·
Because it is present in II nd group in which all atoms lose two electron to reach their nearest noble gas configuration. In Magnesium’s case it is Neon
Mg: 1s2 2s2 2p6 3s2
Mg2+:1s2 2s2 2p6 = Ne
Or
As Magnesium has two electron in its outermost shell(Mg: 1s2 2s2 2p6 3s2), it may lose two electrons.
Hope it helps
-Divyam V.
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What is small business insurance?
Small business insurance is a comprehensive type of coverage designed to help protect small businesses from various risks and liabilities. It encompasses a range of policies based on the different aspects of a business’s operations, allowing owners to focus on growth and success.
The primary purpose of small business insurance is to help safeguard a business’s financial health. It acts as a safety net, helping to mitigate financial losses that could arise from the unexpected, such as property damage, lawsuits, or employee injuries. For small business owners, it’s important for recovering quickl
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Small business insurance is a comprehensive type of coverage designed to help protect small businesses from various risks and liabilities. It encompasses a range of policies based on the different aspects of a business’s operations, allowing owners to focus on growth and success.
The primary purpose of small business insurance is to help safeguard a business’s financial health. It acts as a safety net, helping to mitigate financial losses that could arise from the unexpected, such as property damage, lawsuits, or employee injuries. For small business owners, it’s important for recovering quickly and maintaining operations.
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999 555
9 1
9 3
Areeba Amjad
Studied GCE O Levels at The City School (Pakistan)
·2y
Originally Answered: Why is the magnesium valency +2? ·
Magnesium belongs to group 2 and has a proton number 12. It’s electronic configuration is (2,8,2) so it easily looses 2 electron to attain the electronic configuration of a noble gases hence after loosing 2 electons it has more proton than electon so have +2 charge or valency.
Upvote ·
Héctor López
almost PhD from University of Buenos Aires
·7y
Originally Answered: Why is the magnesium valency +2? ·
Two electrons in the outer shell (2.8.2 is it’s configuration) To get to the nearest noble gas’ configuration these 2 electrons must be lost ( to a non metal atom, say oxygen,) to be uso electrónico with neon. That is, valence 2.
Upvote ·
9 2
Danny Warren
HS Science and Math Teacher (2016–present)
·Updated 4y
Related
Why does calcium ion have a charge of 2+?
Calcium has an electron configuration of:
1s2 2s2 2p6 3s2 3p6 4s2
This means it has two valence electrons, which are the electrons in the outermost shell (in this case, the only two electrons in energy level 4).
You may know about the octet rule. In simple terms, most atoms like having eight electrons in the outermost energy level. This is what is referred to a stable valence or a stable octet.
If you
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Calcium has an electron configuration of:
1s2 2s2 2p6 3s2 3p6 4s2
This means it has two valence electrons, which are the electrons in the outermost shell (in this case, the only two electrons in energy level 4).
You may know about the octet rule. In simple terms, most atoms like having eight electrons in the outermost energy level. This is what is referred to a stable valence or a stable octet.
If you look at the picture, energy-level 1 has two electrons (which is stable), energy-level 2 has eight electrons (which is stable), energy-level 3 has eight electrons (which is stable), but energy-level 4 has only two electrons. This makes calcium reactive.
Since calcium has only two valence electrons, it will likely give up two electrons to another atom/ion. Now calcium will have eight valence electrons in energy-level 3, since energy-level 4’s electrons are gone. For calcium, losing two valence electrons (and dropping down to energy-level 3 with its eight electrons) is way easier than finding another atom to steal six electrons from to fill energy-level 4 up with eight electrons.
Since ...
Upvote ·
99 15
David Young
Former Full-time Tutor of Physics, Chemistry, Mathematics
·6y
Be a use magnesium shares group 2 with other alkaline earth metals whose atoms have 2 (valence) electrons in their outer shell. Cations from group 2 elements have a valence (charge) or +2 since the atoms give up those two electrons in a complete transfer to any of the anions which are attempting to attract (through their electronegativity) those 2 electrons.
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Related questions
What is the charge on magnesium?
Mg forms +2 ions, but not +1 ions. Why?
Why is it that magnesium usually forms Mg2+ ions, and why can’t it form Mg1+ ions by losing only one electron instead of two at a time?
Why does magnesium lose 2 electrons?
What are the benefits of magnesium threonate?
Why are we nearly all magnesium deficient?
Does +1 charged magnesium ions exist?
What is the formula for magnesium ion, Mg+2 or Mg2+?
What is the charge of op magnesium?
Whose atomic size of (Na+1) sodium ion and magnesium (Mg+2) ion is more? Why?
Why does calcium have a +2 charge while magnesium has a +1 charge?
How can we get magnesium +2 ions from magnesium oxide?
What reagent separate Barium ions with Magnesium ions?
Magnesium burns brightly in oxygen to produce magnesium oxide, which contains Mg 2+ and O 2-. The formation of both these ions from their elements is strongly endothermic . Why, therefore should magnesium combine with with oxygen?
Why can't other metal ions with 2+ charge replace magnesium in chlorophyll?
Related questions
What is the charge on magnesium?
Mg forms +2 ions, but not +1 ions. Why?
Why is it that magnesium usually forms Mg2+ ions, and why can’t it form Mg1+ ions by losing only one electron instead of two at a time?
Why does magnesium lose 2 electrons?
What are the benefits of magnesium threonate?
Why are we nearly all magnesium deficient?
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3027 | https://www.geogebra.org/m/n24fCgrm | SAS Congruence – GeoGebra
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An activity to show that triangles with two congruent sides and a congruent, included angle are congruent.
Use the given angle and line segments, equal to a respective side, angle and side of the given triangle, to create a second triangle. Use the “Angle” and “Distance or Length” tools to show that all the angles of the new triangle are equal to the corresponding angles of the given triangle.
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3028 | https://planetmath.org/lawofrareevents | law of rare events
Let X be distributed as Bin(n,p), a binomial random variable with parameters n and p. Suppose
| | | |
---
| | limn→∞np=λ, | |
where λ is a positive real constant, then X is asymptotically distributed as Poisson(λ), a Poisson distribution with parameter λ.
Basically, when the size of the population n is very large and the occurrence of certain event A is rare, where p, the probability of A is very small, the binomial random variable X can be approximated by a Poisson random variable.
Sketch of Proof. Let X∼Bin(n,p). So
| | | | | |
--- ---
| | P(X=m) | = | n!m!(n-m)!pm(1-p)n-m | |
| | | = | n!nm(n-m)!(np)mm!(1-npn)n-m | |
| | | = | n!nm(n-m)!(np)mm!(1-npn)n(1-npn)-m. | |
As n→∞,
| | | |
---
| | n!nm(n-m)!=nnn-1n⋯n-m+1n≈1, | |
| | | |
---
| | (1-npn)-m≈(1-λn)-m≈1, | |
| | | |
---
| | (1-npn)n≈(1-λn)n≈e-λ, | |
and
| | | |
---
| | (np)mm!≈λmm!. | |
Therefore,
| | | |
---
| | P(X=m)≈λmm!e-λ=Poisson(λ). | |
Example. Suppose in a given year, the number of fatal automobile accidents has a binomial distribution for a particular insuarance company with five hundred automobile insurance policies. On average, there is one policy out of the five hundred that will be involved in a fatal crash. What is the probability that there will be no fatal accidents (out of five hundred policies) in any particular year?
Solution. If X be the number of fatal accidents in a year from a population of 500 auto policies, then X∼Bin(n,p) with n=500 and p=1/500. λ=500×1/500=1 and so
| | | |
---
| | P(X=0)≈e-1≈0.368. | |
Using the binomial distribution, we have
| | | |
---
| | P(X=0)=(1-1500)500≈0.367. | |
| | |
--- |
| Title | law of rare events |
| Canonical name | LawOfRareEvents |
| Date of creation | 2013-03-22 14:39:32 |
| Last modified on | 2013-03-22 14:39:32 |
| Owner | CWoo (3771) |
| Last modified by | CWoo (3771) |
| Numerical id | 6 |
| Author | CWoo (3771) |
| Entry type | Definition |
| Classification | msc 62P05 |
| Classification | msc 60E99 |
| Classification | msc 60F99 |
| Synonym | Poisson theorem | |
3029 | https://www.ck12.org/flexi/chemistry/electron-configurations/how-many-orbitals-are-in-the-3rd-shell/ | Flexi answers - How many orbitals are in the 3rd shell? | CK-12 Foundation
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In the n = 3 shell, there are a total of 9 orbitals. This is because there are three subshells in the n = 3 shell: 3s, 3p, and 3d. The 3s subshell has 1 orbital, the 3p subshell has 3 orbitals, and the 3d subshell has 5 orbitals. So, 1 + 3 + 5 = 9 orbitals in total.
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3030 | https://www.math.cmu.edu/~af1p/Texfiles/RandomCubic2Greedy.pdf | A greedy algorithm for finding a large 2-matching on a random cubic graph Deepak Bal∗ Patrick Bennett† Tom Bohman‡ Alan Frieze§ October 26, 2017 Abstract A 2-matching of a graph G is a spanning subgraph with maximum degree two. The size of a 2-matching U is the number of edges in U and this is at least n −κ(U) where n is the number of vertices of G and κ denotes the number of components. In this paper, we analyze the performance of a greedy algorithm 2greedy for finding a large 2-matching on a random 3-regular graph. We prove that with high probability, the algorithm outputs a 2-matching U with κ(U) = ˜ Θ n1/5 .
1 Introduction In this paper we analyze the performance of a generalization of the well-known Karp-Sipser algo-rithm [14, 13, 1, 4] for finding a large matching in a sparse random graph. A 2-matching U of a graph G is a spanning subgraph with maximum degree two. Our aim is to show that w.h.p. our algorithm finds a large 2-matching in a random cubic graph. The algorithm 2greedy is described below and has been partially analyzed on the random graph Gδ≥3 n,cn, c ≥10 in Frieze . The random graph Gδ≥3 n,m is chosen uniformly at random from the collection of all graphs that have n vertices, m edges and minimum degree δ(G) ≥3. In , the 2-matching output by the algorithm is used to find a Hamilton cycle in O(n1.5+o(1)) time w.h.p. Previously, the best known result for this model was that Gδ≥3 n,cn is Hamiltonian for c ≥64 due to Bollob´ as, Cooper, Fenner and Frieze . It is conjectured that Gδ≥3 n,cn is Hamiltonian w.h.p. for all c ≥3/2.
The existence of Hamilton cycles in other random graph models with O(n) edges has also been the subject of much research. In such graphs, the requirement δ ≥3 is necessary to avoid three vertices of degree two sharing a common neighbor. This obvious obstruction occurs w.h.p. in many models with O(n) edges and δ = 2. G3-out is a random graph where each vertex chooses 3 neighbors ∗Department of Mathematical Sciences, Montclair State University, Montclair, NJ †Department of Mathematics, Western Michigan University, Kalamazoo, MI. Research supported by a grant from the Simons Foundation #426894 ‡Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh, PA. Research supported in part by NSF Grant DMS1001638 §Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh, PA. Research supported in part by NSF Grant CCF1013110 1 uniformly at random. This graph has minimum degree 3 and average degree 6. Bohman and Frieze proved that G3-out is Hamiltonian w.h.p. also by building a large 2-matching into a Hamilton cycle . Robinson and Wormald proved that r-regular graphs with r ≥3 are Hamiltonian w.h.p. using an intricate second moment approach ,. Before this result, Frieze proved Hamiltonicity of r-regular graphs w.h.p. for r ≥85 using an algorithmic approach . An algorithmic proof of Hamiltonicity for r ≥3 was given in . In the binomial random graph Gn,p with p = c/n, there is no 2-factor and therefore studying the size of the largest 2-matching is an interesting problem. In the recent paper , an explicit asymptotic formula is given for the maximum size of a 2- matching in such graphs. The paper generalizes this result to random hypergraphs.
In addition to the Hamiltonicity of Gδ≥3 n,cn for 3/2 < c < 10, the Hamiltonicity of random graphs Gn,d with O(n) edges and a fixed degree sequence d is a wide open question. One of the difficulties being that at present we do not know how to couple two graphs Gi = Gdi, i = 1, 2, where d1 ≥d2 so that G1 ⊇G2. One natural example is the Hamiltonicity of a graph chosen uniformly at random from all the collection of all graphs with n/2 vertices of degree 3 and n/2 vertices of degree 4 (this particular question was posed by Wormald). For both Gδ≥3 n,cn and graphs with a fixed degree sequence one might hope to prove Hamiltonicity by first using 2greedy to produce a large 2-matching and then using an extension rotation argument to convert this 2-matching into a Hamilton cycle. In this paper we provide evidence that the first half of this broad program is feasible by showing that 2greedy finds a very large 2-matching for the sparsest of the models with minimum degree 3, the random cubic graph itself (which is the same as Gδ≥3 n,cn for c = 3/2).
The size of a 2-matching U is the number of edges in U and this is at least n −κ(U) where κ denotes the number of components. It was shown in that w.h.p. the Karp-Sipser algorithm only leaves ˜ Θ(n1/5) vertices unmatched. Here we prove the corresponding result for 2greedy on a random cubic graph.
Theorem 1.1. Algorithm 2greedy run on a random 3-regular graph with n vertices outputs a 2-matching U with κ(U) = ˜ Θ(n1/5), w.h.p.
Here the notation f(n) = ˜ Θ(g(n)) denotes c1g(n) logd1 n ≤f(n) ≤c1g(n) logd2 n for absolute constants c1, c2, d1, d2 for n sufficientl large.
We prove Theorem 1.1 using the differential equations method for establishing dynamic concentra-tion. The remainder of the paper is organized as follows. The 2greedy algorithm is introduced in the next Section, and the random variables we track are given in Section 3. The trajectories that we expect these variables to follow are given in Section 4. A heuristic explanation of why 2greedy should produce a 2-matching with Θ(n1/5) components is also given in Section 4. In Section 5 we state and prove our dynamic concentration result. The proof of Thereom 1.1 is then completed in Sections 5, 6, and 7.
1.1 Values for the constants Throughout the proof above, we collect various constraints on the constants in (5.16), (5.23), (5.25), (5.32), (5.35), (5.36), (5.40), (5.41), (5.42), (5.43), (5.45) and (5.46).
CA = 400, Ch = 500, Cpy = 2 · 103, Cℓ= 200, Cx = 12 · 106, 2 Cζ = 65 · 106, Cα = 6 · 108, CvB = 2 · 108, CB = 108, CT = 5 · 105.
2 The Algorithm The Karp-Sipser algorithm for finding a large matching in a sparse random graph is essentially the greedy algorithm, with one slight modification that makes a big difference. While there are vertices of degree one in the graph, the algorithm adds to the matching an edge incident with such a vertex. Otherwise, the algorithm chooses a random edge to add to the matching. The idea is that no mistakes are made while pendant edges are chosen since such edges are always contained in some maximum matching. The algorithm presented in is a generalization of Karp-Sipser for 2-matchings. Our algorithm is essentially the same as that presented in applied to random cubic graphs. A few slight modifications have been made to ease the analysis and to account for the change in model.
We assume that our input (multi-)graph G = G([n], E) is generated by the configuration model of Bollob´ as . Let W = [3n] be our set of configuration points and let Wi = [3(i −1) + 1, 3i], i ∈[n], partition W into 3-sets. The function φ : W →[n] is defined by w ∈Wφ(w). Given a pairing F (i.e. a partition of W into m = 3n/2 pairs) we obtain a (multi-)graph GF with vertex set [n] and an edge (φ(u), φ(v)) for each {u, v} ∈F. Choosing a pairing F uniformly at random from all possible pairings Ωof the points of W produces a random (multi-)graph GF .
It is known that conditional on GF being simple, i.e. having no loops or multi-edges, that it is equally likely to be any (simple) cubic graph. Further, GF is simple with probability (1−o(1))e−2.
So from now on we work with G = GF .
As the algorithm progresses, it grows a 2-matching and deletes vertices and edges from the input graph G. We let Γ = (VΓ, EΓ) be the current state of G, and for each v ∈VΓ let dΓ(v) be the degree of v in Γ. Throughout the algorithm we keep track of the following: • U is the set of edges of the current 2-matching. The internal vertices and edges of the paths and cycles in U will have been deleted from Γ.
• b(v) is the 0-1 indicator for vertex v ∈[n] being adjacent to an edge of U.
• Yk = {v ∈VΓ : dΓ(v) = k, b(v) = 0}, k = 0, 1, 2, 3.
• Zk = {v ∈VΓ : dΓ(v) = k, b(v) = 1}, k = 0, 1, 2.
We refer to the sets Y3 and Z2 as Y and Z throughout. The basic idea of the algorithm is as follows. We add edges to the 2-matching one by one, which sometimes forces us to delete edges.
These deletions may put vertices in danger of having degree less than 2 in the final 2-matching.
Thus, we prioritize the edges that we add to U, so as to match the dangerous vertices first. More precisely, at each iteration of the algorithm, a vertex v is chosen and an adjacent edge is added to U. We choose v from the first non-empty set in the following list: Y1, Y2, Z1, Y, Z. As in the Karp-Sipser algorithm, taking edges adjacent to the vertices in Y1, Y2 and Z1 is not a mistake. We will prove that by proceeding in this manner, we do not create too many components.
Note that the algorithm as written below can take any cubic (multi-)graph as input. However we intend to analyze its performance on the random cubic (multi-)graph GF . An important as-pect of our analysis is that we only reveal adjacencies (pairings) of GF as the need arises in the 3 algorithm. When a vertex v is chosen and its neighbor in the configuration is exposed it is called a selection move. Call the revealed neighbor, w the selection. The edge {v, w} is removed from Γ and added to U. If the selection w is a vertex in Z, then once {v, w} is added to U, we must delete the other edge adjacent to w. Hence we reveal the other edge {w, x} in the configuration adjacent to w. Call this exposure a deletion move and the vertex x, the affected vertex.
Details of the algorithm are now given.
Algorithm 2Greedy: Initially, all vertices are in Y . Iterate the following steps as long as one of the conditions holds.
Step 1(a) Y1 ̸= ∅.
Choose a random vertex v of Y1. Suppose its unique (selected) neighbor in Γ is w. Remove {v, w} from Γ and add it to U. Set b(v) = 1 and move v to Z0.
Re-assign(w). ( This means place w in the set Zk if it now has degree k ≤1 in U, or else remove w from Γ if it has degree 2 in U).
Step 1(b) Y1 = ∅, Y2 ̸= ∅.
Choose a random vertex v of Y2. Randomly select one of the two neighbors of v in Γ and call it w.
If w = v ({v} comprises an isolated component in Γ with a loop), then remove (v, v) from Γ and move v from Y2 to Y0.
Otherwise, remove {v, w} from Γ and add it to U. Set b(v) = 1 and move it to Z1.
Re-assign(w).
Step 1(c) Y1 = Y2 = ∅, Z1 ̸= ∅.
Choose a random vertex v of Z1. v is the endpoint of a path in U. Suppose the unique (selected) neighbor of v in Γ is w. Remove {v, w} from Γ and add it to U. Remove v from Γ.
Re-assign(w).
Step 2 Y1 = Y2 = Z1 = ∅, Y ̸= ∅.
Choose a random vertex v of Y . Randomly select one of the three neighbors of v in Γ and call it w.
If w = v, then we remove loop {v, v} from Γ and move v to Y1.
Otherwise, remove {v, w} from Γ and add it to U. Set b(v) = 1 and move it to Z.
Re-assign(w).
Step 3 Y1 = Y2 = Z1 = Y = ∅, Z ̸= ∅ The remaining (multi-)graph is 2-regular since Z is the set of degree 2 vertices.
Put a maximum matching on this remaining (multi-)graph. Add the edges of this matching to U.
Step 4 Return U (the algorithm terminates here).
4 Subroutine Re-assign(w): 1. If b(w) = 0: Set b(w) = 1 and move w from Y to Z, Y2 to Z1 or Y1 to Z0 depending on the initial state of w.
2. If b(w) = 1: Remove w from Γ. If w was in Z prior to removal, then the removal of w from Γ causes an edge (w, w′), to be deleted from Γ. Move w′ to the appropriate new set. For example, if w′ were in Z, it would be moved to Z1; if w′ were in Y , it would be moved to Y2, etc.
To see that this algorithm produces a 2-matching, note first that in Steps 1 and 2, only one edge (from Γ) at a time is added to U and it is never a loop. Every vertex in Γ is adjacent to at most one edge in U. Thus the addition of such an edge can only create vertices of degree at most 2 in U. When a vertex gets degree 2 in U, it is removed from Γ, thus deleting all of its other edges.
Immediately before Step 3, the vertices of Z have degree 1 in U, thus adding a matching among these vertices will only increase their degree to at most 2 in U.
Note that the final 2-matching U may contain cycles, since in steps 1(c) and 3, we may insert an edge that closes a cycle. However this is not a problem because a 2-matching can contain cycles.
We include here cycles of length two i.e. multi-edges. Note that the expected number of multi-edges in G is O(1) and so this is not an issue.
3 The Variables In this section we will describe the variables which are tracked as the algorithm proceeds. Through-out the paper, in a slight abuse of notation, we let Y, Z, etc. refer to both the sets and the size of the set. Let M refer to the size of EΓ. We also define the variable ζ := Y1 + 2Y2 + Z1.
We sometimes consider ζ to be the set Y1 ∪Y2 ∪Z1. Note that, unlike Y , Z, etc. the size of this set is not the same as ζ, however there are Y1 + 2Y2 + Z1 half-edges (i.e. unpaired configuration points) in Γ that are adjacent to Y1 ∪Y2 ∪Z1.
If X is a variable indexed by i, we define ∆X(i) := X(i + 1) −X(i).
3.1 The sequences σ, δ We define two sequences σ, δ indexed by the step number i. σ(i) will indicate what type of vertex is selected during a selection move, and δ(i) will do the same for deletion moves.
Formally, σ is a sequence of the following symbols: Y, Z, ζ, loop, multi. We will put σ(i) = loop only when step i is of type 2 and the selection move reveals a loop. We put σ(i) = multi in the 5 following case: step i is of type 1(c), w = u ∈Z, where u is the other end of the path in U that contains v. Furthermore, the edge {v, u} is already in U (so we have revealed a multi-edge). The only way this happens is when v ∈Z1, u ∈Z, {v, u} ∈U, and the selection made at step i happens to select the vertex u. Otherwise we just put σ(i) = Y, Z, ζ according to whether the selected vertex is in Y, Z, ζ.
Note that the symbols loop, multi are for very specific events, and not just any loop or multi-edge. If step i is of type 1(b) and our selection move reveals a loop, then we put σ(i) = ζ. Also, if step i is of type 1(c) and the selection move reveals a multi-edge whose other endpoint is also in Z1 then we put σ(i) = ζ as well. Using loop, multi in this way will allow us to define variables A, B whose one step changes do not depend on whether or not ζ > 0.
δ is a sequence of the following symbols: Y, Z, ζ, ∅. We will put δ(i) = ∅when there is no deletion move at step i (i.e. when σ(i) / ∈{Z, multi}). Otherwise δ(i) just indicates the type of the affected vertex that the deletion move chooses (here we don’t make any distinctions regarding loops or multi-edges).
3.2 The variables A, B We will define the following two important variables: A := Y + ζ.
B := 2Y + Z + ζ.
A is a natural quantity to define, since Step 3 of the algorithm begins precisely when A = 0. B is also natural because it represents the number of half-edges which will (optimistically) be added to our current 2-matching before termination. We will see that A and B are also nice variables in that their 1-step changes ∆A(i), ∆B(i) do not depend on what type of step we take at step i. Here A(i), B(i), . . . , denote the values of the corresponding variables A, B, . . . , at the end of i iterations of 2greedy. We have ∆Y (i) = −1ζ(i)=0 −1σ(i)=Y − 1σ(i)=Z + 1σ(i)=multi 1δ(i)=Y .
(3.1) To justify the above equation, note that if the selection is a Y vertex then Y decreases by 1.
We may also lose a Y vertex if there is a deletion (i.e. if the selection is a Z vertex or in the event of a multi selection) and the affected vertex is in Y . Finally, we lose one more Y vertex whenever ζ = 0.
The following equations are justified similarly, by considering the effect due to selections, deletions, and whether ζ = 0 and Y ̸= 0 (we do not consider Step 3 here).
∆Z(i) = 1ζ(i)=0 + 1σ(i)=Y −1σ(i)=Z −1σ(i)=loop −1σ(i)=multi.
− 1σ(i)=Z + 1σ(i)=multi 1δ(i)=Z.
(3.2) Observe that if σ(i) = loop then we do not increase Z even though ζ(i) = 0 and so we subtract one to counter 1ζ(i)=0. In this case ζ increases and this feeds into the next equation.
6 ∆ζ(i) = −1ζ(i)>0 + 1σ(i)=loop −1σ(i)=ζ + 1σ(i)=Z + 1σ(i)=multi −1δ(i)=ζ + 1δ(i)=Z + 2 · 1δ(i)=Y .
(3.3) and note that these all depend on whether ζ = 0 (i.e. whether step i is of type 1 or 2). Now consider the identity 1δ(i)=Y + 1δ(i)=Z + 1δ(i)=ζ = 1σ(i)=Z + 1σ(i)=multi.
which states that we make a deletion move if and only if our selection move was Z or multi. Then we have that ∆A(i) = ∆Y (i) + ∆ζ(i) = −1 −1σ(i)=Y −1σ(i)=ζ + 1σ(i)=loop+1σ(i)=Z+1σ(i)=multi − 1σ(i)=Z + 1σ(i)=multi 2 · 1δ(i)=ζ.
(3.4) ∆B(i) = 2∆Y (i) + ∆Z(i) + ∆ζ(i) = −2 + 1σ(i)=loop −1δ(i)=ζ.
(3.5) which do not depend on whether ζ = 0. Note also that if we establish dynamic concentration on A, B, ζ then we implicitly establish concentration on Y, Z, M since Y = A −ζ.
(3.6) Z = B −2A + ζ.
(3.7) 2M = 3Y + 2Z + ζ = 2B −A.
(3.8) 4 The expected behavior of A, B, ζ In this section, we will non-rigorously predict the behavior of the variables and some facts about the process. Throughout the paper, unless otherwise specified, t refers to the scaled version of i, so t := i n, where t ≤n, since we add an edge to U each round and U is a 2-matching.
Heuristically, we assume there exist differentiable functions a, b such that A(i) ≈na(t), B(i) ≈ nb(t). Further, we assume that ζ stays “small”. We will prove later that these assumptions are indeed valid. We also let pz := 2Z 2M , py := 3Y 2M , pζ := ζ 2M .
where we have omitted the dependence on i for ease of notation. Note that these represent the probabilities that a selection or deletion move is Z, Y or ζ respectively. So for example E 1σ(i)=Z = pz. We can claim this because in the configuration model, we can arbitrarily change the pairing of unpaired configuration points while still being consistent with the history of the process. We are using the method of “deferred decisions”.
7 4.1 The trajectory b(t) Since B(0) = 2n, and recalling (3.5), we see that B(i) = 2n −2i + X j≤i 1σ(j)=loop −1δ(j)=ζ .
(4.1) The probability that σ(j) = loop or δ(j) = ζ on any step j should be negligible. Thus we expect B(i) ≈2n −2i = 2n(1 −t), so we will set b(t) = 2(1 −t).
(4.2) 4.2 The trajectory a(t) We derive an ODE that a should satisfy. Since py = 3(A−ζ) 2B−A ≈ 3a(t) 2b(t)−a(t) and pz = 2(B−2A+ζ) 2B−A ≈2b(t)−4a(t) 2b(t)−a(t) we should have (referring to (3.4) and ignoring all ζ, loop, and multi events since they should be negligible) a′(t) ≈E[∆A(i)] ≈−1 −py + pz ≈− 6a(t) 2b(t) −a(t).
Thus a(t) should satisfy a′ = − 6a 4 −4t −a.
(4.3) Lemma 4.1. The unique solution to (4.3) with boundary condition a(0) = 1 is a(t) = 7 + 2t −6 √ 5 + 4t cos 1 3 arccos 11 + 14t + 2t2 (5 + 4t)3/2 + π 3 .
Proof. The substitution a = (1 −t)x yields a separable ODE: (1 −t)x′ = x − 6x 4 −x or dt 1 −t = dx 3 x + 2 −2 x .
This can be solved directly and together with x(0) = 1 this gives 1 1 −t = (x + 2)3 27x2 .
After substituting back we arrive at 0 = (a + 2 −2t)3 −27a2 = a3 −(6t + 21)a2 + 12(1 −t)2a + 8(1 −t)3.
We make the substitution a = r + 7 + 2t to obtain the equation r3 −27(5 + 4t)r −54(11 + 14t + 2t2) = 0.
8 Putting p = 27(5 + 4t) and q = −54(11 + 14t + 2t2) and using the cosine formula for the solution of a cubic equation we have three roots rk(t) = 2 rp 3 cos 1 3 arccos −3q 2p r3 p + k2π 3 = 6 √ 5 + 4t cos 1 3 arccos 11 + 14t + 2t2 (5 + 4t)3/2 + k2π 3 , k = 0, 1, 2.
We can assume that 0 ≤arccos(x) ≤π.
We now have three possibilities for a, viz. ak(t) = 7 + 2t+rk(t), k = 0, 1, 2. We use the boundary condition a(0) = 1 to see which choice is correct.
Putting t = 0 gives rk(0) = 6 √ 5 cos 1 3 arccos 11 53/2 + k2π 3 , k = 0, 1, 2.
Also, using the identity cos(3θ) = 4 cos3(θ) −3 cos(θ) with θ = arccos 1 √ 5 we see that cos 3 arccos 1 √ 5 = −11 53/2 from which we deduce that 1 3 arccos 11 53/2 + π = 1 3 arccos −11 53/2 = arccos 1 √ 5 and then similarly cos 1 3 arccos 11 53/2 + 4π 3 = −1 √ 5.
It follows that r2(0) = −6 and a2(0) = 1. Furthermore, r0(0), r1(0) = 3 ± 6 √ 3 will be different from r2(0) and so a(t) = a2(t) or a(t) = 7 + 2t + 6 √ 5 + 4t cos 1 3 arccos 11 + 14t + 2t2 (5 + 4t)3/2 + 4π 3 = 7 + 2t −6 √ 5 + 4t cos 1 3 arccos 11 + 14t + 2t2 (5 + 4t)3/2 + π 3 .
(4.4) From here we can see that a(t) →0 as t →1−. (t = i/n and i ≤n and a 2-matching has at most n edges.) More precisely, lim t→1− a(t) (1 −t) 3 2 = 2 3 3 2 .
(4.5) To confirm this, we use the facts that for δ →0 arccos (1 −δ) = √ 2δ + O(δ3/2), 9 √ 1 −δ = 1 −1 2δ + O(δ2), cos π 3 + δ = 1 2 − √ 3 2 δ + O(δ2), and the fact that 11 + 14(1 −ε) + 2(1 −ε)2 (5 + 4(1 −ε))3/2 = 1 −4ε3 729 + O(ε4).
Rewriting, we see a(1 −ε) = 9−2ε −18 1−2 9ε + O(ε2) × cos 1 3 8ε3 729 1/2 + O(ε5/2) !
+ π 3 !
= 9−2ε −18 1−2 9ε + O(ε2) × 1 2 1 −2 27 2 3 1/2 ε3/2 + O(ε5/2) !
= 2 3 3/2 ε3/2 + O(ε2), which implies (4.5).
Additionally, d dt a(t) (1 −t)3/2 = − 6a 4(1 −t) −a(1 −t)−3/2 + 3 2a · (1 −t)−5/2 = a · (1 −t)−5/2 3 2 − 6(1 −t) 4(1 −t) −a < 0.
(4.6) Since a(0) = 1, for all 0 ≤t ≤1 we have 2 3 3 2 (1 −t)3/2 ≤a(t) ≤(1 −t)3/2 ≤1 −t.
(4.7) This inequality is used extensively in Section 5.
4.3 Downward drift of ζ We expect ζ to be “small”, and to heuristically justify that claim we will show that whenever ζ is positive, it is likely to decrease. In this section therefore, we are implicitly assuming that py, pz ≫pζ. Let us first note that this is non-trivial, i.e. it is possible for ζ to grow. Suppose the algorithm executes Step 1(b). So v ∈Y2 and w is one of its two neighbors. One possible scenario is that w ∈Z which means there will be a deletion move this step (case 2 of re-assign(w)). This deletion move may affect a vertex u ∈Y . So v moved from Y2 to Z1 and u moved from Y to Y2.
Thus the net change in ζ is −1 + 2 = +1. Assume that ζ(i) > 0. In the following table, we once again make use of the fact that δ(i) ̸= ∅if and only if σ(i) ∈{Z, multi}. So for example, 1δ=Y = (1σ=Z + 1σ=multi)1δ=Y .
Then from (3.3) we see that if ζ(i) > 0, 10 ∆ζ = 1 with prob. pzpy + O 1 M .
0 with prob. p2 z + O 1 M .
−1 with prob. py + O 1 M .
−2 with prob. O (pζ).
(4.8) Each entry in the above table can be understood by considering the most likely way for ∆ζ to take the appropriate value. For example, the most likely way for ∆ζ = 1 is for σ = Z and δ = Y (even though this is not the only way for ∆ζ = 1). Specifically, P [∆ζ(i) = 1] = P [σ(i) = Z, δ(i) = Y ] +P [σ(i) = multi, δ(i) = Y ] = 2Z 2M −1 3Y 2M −3 + O 1 M = pzpy + O 1 M .
Similarly, the most likely way for ∆ζ = 0 is for σ = Z and δ = Z. The only way to get ∆ζ = −1 is to have σ = Y (in which case δ = ∅), and the only ways to get ∆ζ = −2 involve σ = ζ or δ = ζ.
Therefore, roughly speaking we have E[∆ζ(i)] = pzpy −py + O (pζ) ≈2b −4a 2b −a · 3a 2b −a − 3a 2b −a = − 9a2 (2b −a)2 (4.9) and are motivated to define Φ(t) := 9a2 (2b −a)2 = Θ(1 −t) (4.10) to represent the downward drift of ζ(i) (if it is positive) at step i.
4.4 Expected behavior of ζ In the last subsection we estimated E[∆ζ(i)] when ζ > 0, using (4.8). We can also use (4.8) to estimate the variance when ζ > 0. We see that V ar[∆ζ(i)|ζ > 0] = Θ(py) = Θ (1 −t) 1 2 .
Thus, to model the behavior of ζ(i) we consider a simpler variable: a lazy random walk Xτ(k) with Xτ(0) = 0, expected 1-step change E [∆Xτ] = −(1 −τ) and V ar[∆Xτ] = (1 −τ) 1 2 . After s steps, we have E [Xτ(s)] = −(1 −τ)s and V ar[Xτ(s)] = (1 −τ) 1 2 s. There is at least constant (bounded away from 0) probability that Xτ(s) is, say, 1 standard deviation above its mean. However, the probability that Xτ(s) is too many standard deviations larger than that is negligible. In other words, it is reasonable to have a displacement as large as Xτ(s) = −(1 −τ)s + (1 −τ) 1 4 s 1 2 , but not much larger. The quantity ψ(s) := −(1 −τ)s + (1 −τ) 1 4 s 1 2 is negative for s > (1 −τ)−3 2 . Also ψ(s) is maximized when s = 1 4(1 −τ)−3 2 , where we have ψ(s) = 1 4(1 −τ)−1 2 .
Now we reconsider the variable ζ. Roughly speaking, ζ(i) behaves like the lazy random walk considered above, so long as we restrict the variable i to a short range (so that t does not change significantly), and we have ζ(i) > 0 for this range of i. We have ζ(0) = 0, and ζ has a negative drift 11 so it’s likely that ζ(j) = 0 for many j > 0. Specifically, if j is an index such that ζ(j) = 0, then we expect ζ(i) to behave like Xτ(i −j) with τ = j n, so long as i is not significantly larger than j. Thus we expect to have ζ(i) = 0 for some j ≤i ≤j + (1 −τ)−3 2 . Also, for all j ≤i ≤j + (1 −τ)−3 2 we should have ζ(i) ≤1 4(1 −τ)−1 2 . But this rough analysis does not make sense toward the end of the process: indeed, for j > n −n 3 5 (i.e. for 1 −τ < n−2 5 ), we have j + (1 −τ)−3 2 > n. However, we can still say something about what happens when j is large, since the variable s cannot be any bigger than n −j. Now for j ≥n −n 3 5 and s ≤n −j we have ψ(s) ≤(1 −τ) 1 4 s 1 2 ≤n−1/10 · n3/10 = n1/5.
Thus, we never expect ζ to be larger than n 1 5 , even towards the end of the process.
4.5 Why do we have ˜ Θ n 1 5 many components?
At any step of the algorithm, we expect the components of the 2-matching to be mostly paths (and a few cycles). We would like the algorithm to keep making the paths longer, but sometimes it isn’t possible to make a path any longer because of deletion moves. Specifically, for example, if one endpoint of a path is in Z1, and then there is a deletion move which affects that endpoint, then that end of the path will never grow. If the same thing happens to the other endpoint of the path, then the path will never get longer, and will never be connected to any of the other paths. Similarly, the number of components in the final 2-matching can be increased by a deletion move that affects a vertex in Y1 or Y2. Thus we can bound the number of components in the final 2-matching by bounding the number of steps i such that δ(i) = ζ.
Roughly, P[δ(i) = ζ] = 2Z 2M−1 · ζ 2M−3 = O 1 n min (1 −t)−3 2 , n 1 5 1−t logO(1) n . So integrating, we estimate the total number of components as O Z 1−1 n 0 min ( 1 n(1 −t)−3 2 , n 1 5 1 −t logO(1) n ) dt !
= O n 1 5 logO(1) n .
Indeed, we will see in section 7 that a matching (up to log factors) lower bound also holds. Very roughly speaking, this is because we expect there to be a positive proportion of steps i where ζ(i) is more than its expectation by a standard deviation.
We will now rigorously justify the above claims about the performance of the algorithm 2greedy.
5 The stopping time T and dynamic concentration In this section, we introduce a stopping time T, before which A and B stay close to their trajectories, and ζ does roughly what we expect it to do. We will also introduce “error” terms for both A, B and a “correction” term α for the variable A. For most of the process, α will stay smaller than the error term for A. However, toward the end of the process α will be significant. Using α in our calculations thus allows us to track the process farther. As it turns out, the variable B does not need an analogous “correction” term.
We define the following random variables which represent “actual error” in A, B: ea(i) := A(i) −na(t) −α(i).
12 eb(i) := B(i) −nb(t).
The definition of α(i) is through a recurrence – see (5.13).
We define the stopping time T as the minimum of n −CT n 7 15 log 6 5 n and the first step i such that any of the four following conditions fail: |ea(i)| ≤fa (t) , (5.1) |eb(i)| ≤fb (t) , (5.2) ζ(i) ≤fζ (t) , (5.3) and for every step j < i such that ζ is positive on steps j, . . . , i, ζ(i) ≤ζ(j) − X j≤k n−2 5 log 2 5 n.
−n 1 5 log 4 5 n log(1 −t) : otherwise.
(5.7) fζ(t) := Cζ · ( (1 −t)−1 2 log n, : 1 −t > n−2 5 log 2 5 n.
n 1 5 log 4 5 n : otherwise.
(5.8) 5.1 A useful lemma We’ll use the following simple lemma several times to estimate fractions.
Lemma 5.2. For any real numbers x, y, εx, εy, if we have x, y ̸= 0 and εx x , εy y ≤1 2, then x + εx y + εy −x y = yεx −xεy y2 + O yεxεy + xε2 y y3 !
.
13 Proof.
x + εx y + εy −x y = x y ( 1 + εx x · 1 1 + εy y −1 ) = x y ( 1 + εx x · " 1 −εy y + O ε2 y y2 !# −1 ) = x y ( εx x −εy y + O εxεy xy + ε2 y y2 !) = yεx −xεy y2 + O yεxεy + xε2 y y3 !
.
5.2 T is not triggered by A We define A+(i) := A(i) −na(t) −α(i) −fa(t) = ea(i) −fa(t) (5.9) and let the stopping time Tj := min {i(j), max (j, T)} where i(j) represents the least index i ≥j such that ea(i) is not in the critical interval [ga(t), fa(t)] (5.10) where 0 < ga < fa is an as-yet unspecified function of n, t. Our strategy is to show that w.h.p. A never goes above na + α + fa because every time ea enters the critical interval, w.h.p. it does not exit the interval at the top. The use of critical intervals in a similar context was first introduced in .
Let Fi be the natural filtration of the process (so conditioning on Fi tells us the values of all the variables, among other things).
For i < T, we have from (3.4) and (3.6), (3.7), (3.8) that E[∆A(i)|Fi] = −1 −3Y 2M − ζ 2M + 2Z 2M −2 · 2Z 2M · ζ 2M + O 1 M = −(2B −A) −3(A −ζ) −ζ + 2(B −2A + ζ) 2B −A −4ζ(B −2A + ζ) (2B −A)2 + O 1 2B −A = − 6A 2B −A + 4ζ(A + B) (2B −A)2 + O 1 2B −A + ζ2 (2B −A)2 = − 6 (na + α + ea) 2(nb + eb) −(na + α + ea) + 4ζ [(na + α + ea) + (nb + eb)] [2(nb + eb) −(na + α + ea)]2 + O 1 2B −A + ζ2 (2B −A)2 = − 6a 2b −a + 12aeb −12b (α + ea) n(2b −a)2 + 4(a + b)ζ n(2b −a)2 + O 1 n(2b −a) + α2 + f2 a + f2 b + f2 ζ n2(2b −a)2 !
. (5.11) 14 The second line above follows from substituting the values of Y, Z, M in terms of A, B, ζ. The last line above also follows from Lemma 5.2 (the fourth line has two fractions with error terms in the numerators and denominators. We apply Lemma 5.2 to these fractions, regarding ea, eb, α as error terms, to arrive at the last line, making use of (5.1)-(5.3), that ea + eb + α = o(2b −a)n and that a + b = O(2b −a)). Also note that we only apply a crude form of Lemma 5.2 to the second fraction of the fourth line, as the lemma would allow us to put fewer of the resulting terms into the big-O. Note that the lemma actually implies that the big-O term includes mixed products of terms like α·fζ for example. We have simplified by using the fact that for all real numbers x and y, |xy| ≤1 2 x2 + y2 . Note that we have not put all the occurrences of ζ into the big-O term. While we will see that the ζ term inside the big-O is negligible, the ζ term outside the big-O may become significant towards the end of the process.
We are now motivated to cancel out the ζ term in the last line by recursively defining α(0) := 0.
(5.12) α(i + 1) := α(i) + 4(a + b)ζ −12bα(i) n(2b −a)2 (5.13) ≥α(i) 1 − 12b n(2b −a)2 .
We see from (4.2) and (4.7) that a + b ≤3(1 −t) ≤2b −a ≤4(1 −t).
(5.14) So using T ≤n−CT n 7 15 log 6 5 n we have that 1 − 12b n(2b−a)2 ≥0 and hence that α(i) ≥0 throughout.
From (5.13) and the definition of fζ and ζ ≤fζ, it follows that for i ≤T, 0 ≤α(i) ≤ i X j=0 4(a + b)fζ n(2b −a)2 ≤ i X j=0 4 · 3 1 −j n fζ j n n 3 1 −j n 2 = 4Cζ 3n i X j=0 1 1 −j n · 1 −j n −1 2 log n, : 1 −j n > n−2 5 log 2 5 n n 1 5 log 4 5 n : otherwise ≤4Cζ 3 Z i+1 n τ=0 1 1 −τ · ( (1 −τ)−1 2 log n, : 1 −τ > n−2 5 log 2 5 n n 1 5 log 4 5 n : otherwise dτ ≤Cα · ( log n(1 −t)−1/2 for i ≤n −n3/5 log2/5 n, n1/5 log9/5 n for n −n3/5 log2/5 n < i ≤T, (5.15) since Cα and Cζ satisfy Cα > 8Cζ.
(5.16) Note that we can pass from the sum on the second line to the integral on the third line since the integrand is an increasing function. Also note that in evaluating the integral, the value of the antiderivative at τ = i+1 n is asymptotically the same as the value at τ = t = i n, so the last inequality holds since we chose Cα large enough with room to spare.
15 Nowlet j ≤i < Tj. Note that if this holds, then by the definition of Tj, i satisfies ea(i) ∈ [ga(t), fa(t)]. We have the supermartingale condition E[∆A+(i)|Fi] = E[∆A(i)|Fi] −a′(t) −4(a + b)ζ −12bα(i) n(2b −a)2 −1 nf′ a(t) + O 1 na′′(t) + 1 n2 f′′ a (t) (5.17) ≤− 12bga n(2b −a)2 −1 nf′ a(t) + O afb n(2b −a)2 + 1 n(2b −a) + α2 + f2 a + f2 b + f2 ζ n2(2b −a)2 + 1 na′′(t) + 1 n2 f′′ a (t) !
.
(5.18) Note that in the first line we have used (5.9) and (5.13), and in the last line we have used (5.11), (5.13), the fact that ea ≥ga, and also that a satisfies the differential equation (4.3). By taking ga = Cgfa where Cg < 1 we guarantee that the corresponding critical interval is non-empty.
We will subsequently choose Cg = 3/4, see (5.24). By (4.2), (5.6) and (5.14) we have − 12bga n(2b −a)2 −1 nf′ a = −Θ n−1/2 log1/2 n(1 −t)−1/4 .
(5.19) We then see that A+(j), . . . , A+(Tj) is a supermartingale once we prove the following claim.
Claim 5.3. − 12bga n(2b−a)2 −1 nf′ a dominates the big-O term in (5.18).
Proof. Throughout this proof we will assume that (1 −t) = Ω(n−8/15 log6/5 n), see (5.5). Now (5.14) and (5.19) take care of the second big-O term in (5.18). Also it is not hard to see by (5.6) that f′′ a (t) = O(n1/2 log1/2 n(1 −t)−5/4) = o(f2 a/(2b −a)2) and so the fifth big-O term is taken care of.
Now consider the fourth big-O term. It follows from (4.3) and (4.7) that 6 2 3 3/2 (1 −t)3/2 4(1 −t) ≤−a′(t) = 6a 4 −4t −a ≤ 6(1 −t)3/2 4(1 −t) −(1 −t) and so −2 (1 −t) 1 2 ≤a′(t) ≤− r 2 3 (1 −t) 1 2 .
(5.20) Also we have a′′(t) = −6 [a′(4 −4t −a) + a(4 + a′)] (4 −4t −a)2 and then from (4.7) and (5.20) we deduce that |a′′(t)| = O (1 −t)−1/2 .
(5.21) Thus bga (2b −a)2a′′ = Ω n−8/15 log6/5 n 1/4 n1/2 log1/2 n ≫1 and this takes care of the fourth big-O term.
16 For the first and third big-O terms we must consider cases according to the value of t. First consider the case 1−t > n−2/5 log2/5 n. By (4.2), (4.7), (5.7), and our choice of ga (again see (5.24)) we have afb = O((1 −t) log n) = o(bga) since bga (1 −t) log n = Ω n−2/5 log2/5 n 3/4 n1/2 log−1/2 n ≫1 and this deals with the first term. For the third term we see by (5.6)-(5.8) and (5.15) that α2 + f2 a + f2 b + f2 ζ = O((1 −t)−1 log2 n + (1 −t)3/2n log n). Now using the bounds on t we have nbga (1 −t)3/2n log n = Ω n−2/5 log2/5 n 1/4 n1/2 log−1/2 n ≫1 and nbga (1 −t)−1 log2 n = Ω n−2/5 log2/5 n 11/4 n3/2 log−3/2 n ≫1 and so α2 + f2 a + f2 b + f2 ζ = o(nbga) so this deals with the third term, and finishes the case 1 −t > n−2/5 log2/5 n.
The other case is Ω(n−8/15 log6/5 n) ≤1 −t ≤n−2/5 log2/5 n. The only terms that change are fb = −n1/5 log4/5 n log(1 −t) = O(n1/5 log9/5 n) by (5.7) and fζ, α = O(n1/5 log9/5 n) by (5.8) and (5.15). So afb = O(n1/5 log9/5 n(1 −t)3/2) = o(bga) and f2 ζ = O(n2/5 log8/5 n) = o(nbga) using our bounds on t.
We use the following asymmetric version of the Azuma-Hoeffding inequality (for a proof see ): Lemma 5.4. Let Xj be a supermartingale, such that −C ≤∆X(j) ≤c for all j, for c < C 10. Then for any u < cm we have Pr(Xm −X0 > u) ≤exp − u2 3cCm We have by (3.4) that −2 ≤∆A ≤0.
(5.22) For an absolute bound on a′(t) we have by (5.20) and the bounds on t that −2 ≤a′(t) ≤− 2CT 3 1 2 n−4 15 log 3 5 n.
Now by (5.9) we see ∆A+ = ∆A −a′ −∆α + O 1 nf′ a + 1 na′′ and before the stopping time T we have by (5.8) and (5.15) that ζ ≤Cζn 1 5 log 4 5 n and α ≤Cαn 1 5 log 9 5 n 17 and so we have |∆α| (5.13) = 4(a + b)ζ −12bα(i) n(2b −a)2 (4.7) ≤12(1 −t)ζ + 24(1 −t)α(i) 9n(1 −t)2 ≤12Cζn 1 5 log 4 5 n + 24Cαn 1 5 log 9 5 n 9n(1 −t) ≤3Cαn−4 5 log 9 5 n(1 −t)−1 ≤3Cα CT n−4 15 log 3 5 n so ∆α a′(t) ≤3Cα C 3 2 T .
Since Cα > 2C3/2 T (5.23) we get the bounds (using (5.22) and (5.20)) ∆A+ ≤− a′ j n + ∆α (1 + o(1)) ≤ 3Cα C 3 2 T +1 + o(1) a′ j n ≤7Cα C 3 2 T 1 −j n 1 2 and ∆A+ ≥∆A −o(1) ≥−2 + o(1) ≥−3 for the supermartingale A+(j) · · · A+(Tj). Thus, if A crosses its upper boundary na(t)+α(i)+fa(t) at the stopping time T, since ∆A+ ≤7Cα C 3 2 T and this will be the first crossing there is some step j (with T = Tj) such that A+(j) ≤ga j −1 n −fa j −1 n + 7Cα C 3 2 T and A+ (Tj) > 0.
In this case, j is intended to represent the step when ea enters the critical interval, (5.10). Our choice of constants in Section 1.1 allows us to apply Lemma 5.4 and see that the probability of the supermartingale A+ having such a large upward deviation has probability at most exp − fa j−1 n −ga j−1 n −7Cα C 3 2 T !2 3 · 3 · 7Cα C 3 2 T n 1 −j n 3 2 .
As there are O (n) supermartingales A+(j), . . . , A+(Tj), we must choose fa, ga to make the above probability o 1 n . The following choice suffices: fa(t) = CA(1 −t) 3 4 n 1 2 log 1 2 n.
ga(t) = 3 4fa(t).
(5.24) 18 since the constant CA is chosen so that 1 4CA 2 63Cα C 3 2 T > 1.
(5.25) If we define A−:= A −na −α + fa = ea + fa then we may prove that A−stays positive w.h.p. in a completely analogous fashion.
5.3 T is not triggered by condition (5.4) Referring to (4.8) and (3.6)-(3.8), we may say that if ζ(i) > 0, E[∆ζ(i)|Fi] = pzpy −py + O (pζ) = − 9A2 (2B −A)2 + O ζ 2B −A .
(5.26) Now, before T we have 9a2 (2b −a)2 − 9A2 (2B −A)2 = −9 A 2B −A − a 2b −a A 2B −A + a 2b −a = −9 2b(α + ea) −2aeb n(2b −a)2 + O α2 + f2 a + f2 b n2(2b −a)2 × 2 a 2b −a + 2b(α + ea) −2aeb n(2b −a)2 + O α2 + f2 a + f2 b n2(2b −a)2 = 36a(aeb −bα −bea) n(2b −a)3 + O " α2 + f2 a + f2 b n2(2b −a)2 · a 2b −a + b(α + ea) + aeb n(2b −a)2 + α2 + f2 a + f2 b n2(2b −a)2 + aeb −bα −bea n(2b −a)2 2# = 36a(aeb −bα −bea) n(2b −a)3 + O α2 + f2 a + f2 b n2(2b −a)2 .
(5.27) On the second line we have used Lemma 5.2 and the inequalities (5.1), (5.2) and in the last step we have cleaned up the big-O using the facts α + fa + fb n(2b −a) = o(1) and a 2b −a = O(1) which follow from (5.15), (5.6), (5.7), (4.7) and (5.14). For every step j, we re-define Tj to be the stopping time Tj := min {i(j), max(j, T)} where i(j) is the least index i ≥j such that ζ(i) = 0. Also, define a sequence ζ+ j (j) · · · ζ+ j (Tj), where ζ+ j (i) := ζ(i) + X j≤k<i Φ k n −hj i n 19 where hj is some function we will choose that will make ζ+ j (i) a supermartingale. Now for j ≤i < Tj, using (5.26) and (5.27), we have E[∆ζ+ j (i)|Fi] = − 9A2 (2B −A)2 + 9a2 (2b −a)2 −1 nh′ j(t) + O ζ 2B −A + 1 n2 h′′ j (t) (5.28) = 36a(aeb −bα −bea) n(2b −a)3 −1 nh′ j(t) + O α2 + f2 a + f2 b n2(2b −a)2 + ζ 2B −A + 1 n2 h′′ j (t) (5.29) ≤36a(afb + bfa) n(2b −a)3 −1 nh′ j(t) + O α2 + f2 a + f2 b n2(2b −a)2 + fζ n(2b −a) + 1 n2 h′′ j (t) .
(5.30) In the last line we have used (5.1), (5.2), (5.3), and the fact that α ≥0. Note that by (5.6), (5.7) we have afb = o(bfa) so 36a(afb + bfa) (2b −a)3 ≤36(1 −t) 3 2 · 2(1 −t) · CA(1 −t) 3 4 n 1 2 log 1 2 n · (1 + o(1)) 64(1 −t)3 ≤ 9 8CA + o(1) n 1 2 log 1 2 n (1 −t) 1 4 (5.31) so the choice hj(t) := Ch 1 −j n 1 4 n 1 2 log 1 2 n t −j n makes the sequence a supermartingale as long as the constant Ch is chosen so that Ch > 9 8CA.
(5.32) One can verify, as in Claim 5.3, that the big-O term is dominated by the main terms.
Since hj j n = 0, we will always have ζ+ j (j) = ζ(j).
We’ll use the following supermartingale inequality due to Freedman : Lemma 5.5. Let Xi be a supermartingale, with ∆Xi ≤C for all i, and V (i) := X k≤i V ar[∆Xk|Fk] Then P [∃i : V (i) ≤v, Xi −X0 ≥d] ≤exp − d2 2(v + Cd) .
Referring to (4.8), before T we can put V ar[∆ζ+ j (i)|Fi] = V ar[∆ζ(i)|Fi] ≤E h (∆ζ(i))2 | Fi i = 1 · pzpy + 1 · py + O(pζ) + O 1 M ≤3py (5.33) since pz ≤1 and pζ = o(py) before T. Note that before T, we have py = 3Y 2M ≤ 3A 2B −A ≤ 3[n(1 −t) 3 2 + α + fa] 4n(1 −t) −2fb −n(1 −t) 3 2 −α −fa ≤ C3/2 T + Cα CT + o(1) !
(1 −t) 1 2 .
(5.34) 20 Indeed, if t ≤1 −n−2/5 log2/5 n then by (5.15), (5.6) and (5.7) we have 3[n(1 −t) 3 2 + α + fa] 4n(1 −t) −2fb −n(1 −t) 3 2 −α −fa ≤ 3[n(1 −t) 3 2 + Cα(1 −t)−1/2 log n + CA(1 −t) 3 4 n 1 2 log 1 2 n] 4n(1 −t) −2CB(1 −t)−1 2 log n −n(1 −t) 3 2 −Cα(1 −t)−1/2 log n −CA(1 −t) 3 4 n 1 2 log 1 2 n = 3n(1 −t)3/2 + o(n(1 −t)) (4 + o(1))n(1 −t) .
Whereas if t ≥1 −n−2/5 log2/5 n then 3[n(1 −t) 3 2 + α + fa] 4n(1 −t) −2fb −n(1 −t) 3 2 −α −fa ≤ 3[n(1 −t) 3 2 + Cαn1/5 log9/5 n + CA(1 −t) 3 4 n 1 2 log 1 2 n] 4n(1 −t) + 2CBn 1 5 log 4 5 n log(1 −t) −n(1 −t) 3 2 −Cαn1/5 log9/5 n −CA(1 −t) 3 4 n 1 2 log 1 2 n ≤(3 + o(1))[n(1 −t) 3 2 + Cαn1/5 log9/5 n] (4 + o(1))n(1 −t) .
Here our choice of constants Cα, CT , Cpy are such that Cpy > C3/2 T + Cα CT .
(5.35) and so py ≤Cpy(1 −t) 1 2 . Also, note that ∆ζ+ ≤3 since ∆ζ ≤2 and Φ(t) ≤9(1−t)3 9(1−t)2 ≤1 (by (4.10), (4.7) and (5.14)).
Suppose condition (5.4) triggers the stopping time T. Then there are steps j < i = T such that ζ > 0 all the way from step j to step i, and ζ+ j (i) > ℓj(t) −hj(t). We’ll need to apply Lemma 5.5 to the supermartingale ζ+ j to show this event has low probability (guiding our choice for ℓj). Note that by (5.33), in Lemma 5.5 we can plug in the following for v: V (i) = X j≤k≤i V ar[∆ζ+ j (k)|Fk] ≤3Cpy 1 −j n 1 2 (i −j).
So the unlikely event that condition (5.4) triggers T has probability at most exp − (ℓj −hj)2 2 3Cpy 1 −j n 1 2 (i −j) + 3(ℓj −hj) by Lemma 5.5. The above bound holds for each of the O(n2) pairs j, i, but note that it is with different parameters ℓj(t) for each i (ℓj(t) depends on t and therefore on i). For the union bound 21 to work, we’d like to make the above probability o 1 n2 for each pair j, i. Towards this end we consider 2 cases.
If 1 −j n 1 2 (i −j) ≤log n, then it suffices to put ℓj −hj = Cℓlog n since C2 ℓ 6Cpy + 6Cℓ > 2.
(5.36) If 1 −j n 1 2 (i −j) > log n, then again by (5.36), it suffices to put ℓj−hj = Cℓ 1 −j n 1 4 (i −j) 1 2 log 1 2 n.
Thus we choose ℓj(t) := hj(t) + Cℓmax ( log n, 1 −j n 1 4 (i −j) 1 2 log 1 2 n ) .
(5.37) With this choice, w.h.p. T is not triggered by condition (5.4).
5.4 An upper bound on ζ In this section we’ll show that T is not triggered by condition (5.3). We will see that (5.3) actually holds deterministically, assuming T is not triggered by the other conditions Lemma 5.6. W.h.p. for all j < n −2C 2 5 x n 3 5 log 2 5 n such that ζ(j −1) = 0, we have 1. ζ(j′) = 0 for some j ≤j′ ≤j + Cx 1 −j n −3 2 log n, and 2. ζ(i) ≤40C2 ℓ 1 −j n −1 2 log n for all j ≤i ≤j′ −1.
Proof. Suppose ζ(j −1) = 0. Note that we then have ζ(j) ≤2. Recall the definition (4.10) of Φ and equations (4.3) and (4.2). Now Φ(t)/(1 −t) is decreasing since d dt Φ(t) 1 −t = Φ′(t) 1 −t + Φ(t) (1 −t)2 = 18aa′ (2b −a)2(1 −t) −18a2(2b′ −a′) (2b −a)3(1 −t) + 9a2 (2b −a)2(1 −t)2 = 18a (2b −a)2(1 −t) − 6a 2b −a + a 4 − 6a 2b−a 2b −a + 9a2 (2b −a)2(1 −t)2 = 18a (2b −a)2(1 −t) × −4a(a + b) (2b −a)2 + 9a2 (2b −a)2(1 −t)2 = 9a2 (2b −a)2(1 −t) −8a + 8b (2b −a)2 + 1 1 −t = − 9a3(8b −a) (1 −t)2(2b −a)4 ≤0, where we have used b = 2(1 −t) to get the last equation and (4.7) to justify the inequality.
22 Also, using (4.5) and the definition of b, lim t→1− Φ(t) 1 −t = 1 6.
Hence Φ(t) ≥1 6 (1 −t) for all 0 ≤t ≤1. If we substitute x = i−j n then X j≤k 1, we have 1 −j n 1 4 n 1 2 x 1 2 j log 1 2 n = C 1 2 x 1 −j n −1 2 log n > log n so we can evaluate the “max” in ℓj. Also note that the coefficient of x is dominated by −1 6n 1 −j n , so the coefficient of x is at most, say −1 7n 1 −j n . Thus (5.39) gives ζ(j + nxj) ≤2+C2 x 12 n−1 log2 n 1 −j n −3 − Cx 7 −Cℓ p Cx log n 1 −j n −1 2 .
Recall that we have assumed that j < n −2C 2 5 x n 3 5 log 2 5 n and so 1 −j n > 2C 2 5 x n−2 5 log 2 5 n.
Now Cx 7 −Cℓ √Cx log n 1 −j n −1 2 C2 x 12 n−1 log2 n 1 −j n −3 = Cx 7 −Cℓ √Cx C2 x 12 n log n 1 −j n 5 2 > Cx 7 −Cℓ √Cx Cx 12 and so ζ(j + nxj) is negative for this range of j since Cx 7 −Cℓ p Cx > Cx 12 .
(5.40) 23 Therefore, ζ must have hit 0 again before step i = j +nxj. This proves the first part of the lemma.
To prove the second part, consider (5.39) for j < i < j + nxj (i.e.
for 0 < x < xj).
If x ≤n−1 log n 1 −j n −1 2 then we can put ζ(i) ≤1 12nx2 −1 7n 1 −j n x + Cℓlog n.
This is maximised at x = 0 or when x = n−1 log n 1 −j n −1 2 and for the latter we have ζ(i) ≤1 12n−1 1 −j n −1 log2 n −1 7 1 −j n 1/2 log n + Cℓlog n ≤C−2/5 x 24 n−3/5 log8/5 n − √ 2C1/5 x 7 n−1/5 log6/5 n + Cℓlog n ≤Cℓlog n.
This deals with x ≤n−1 log n 1 −j n −1 2 .
For x larger than that, we’ll put ζ(i) ≤1 12nx2 −1 7n 1 −j n x + Cℓ 1 −j n 1 4 n 1 2 x 1 2 log 1 2 n ≤C2 x 12 n−1 log2 n 1 −j n −3 + 7C2 ℓ 4 1 −j n −1 2 log n < 40C2 ℓ 1 −j n −1 2 log n.
To justify the second line we use the fact that x < xj as well as the inequality −dx + cx1/2 ≤c2 4d for real numbers x, c, d > 0 to handle the last two terms. And for the third inequality, we use Cx ≤400C2 ℓ.
(5.41) We would also like to say something about ζ(i) for i > n −2C 2 5 x n 3 5 log 2 5 n.
Lemma 5.7. W.h.p. for all i ≤T we have ζ(i) ≤Cζn 1 5 log 4 5 n.
Proof. Suppose that at step j′ ≥n −2C 2 5 x n 3 5 log 2 5 n we have ζ(j′) = 0. It follows from Lemma 5.6 that w.h.p.
such a j′ exists.
Let i ≥j′ such that ζ(j′ −1) . . . ζ(i) are all positive.
Note that we again have the bound (5.39). But now 0 ≤x ≤n−j′ n ≤2C 2 5 x n−2 5 log 2 5 n, and (5.39) gives ζ(i) ≤ 1 3C 4 5 x + 2 3 4 CℓC 3 10 x + o(1) n 1 5 log 4 5 n. Note that the term in square brackets in (5.39) is negative here.
So in particular we can say that for i ≤T we have ζ(i) ≤fζ(t) = Cζ min n (1 −t)−1 2 log n, n 1 5 log 4 5 n o , 24 since Cζ > max 40C2 ℓ, 1 3C 4 5 x + 2 3 4 CℓC 3 10 x .
(5.42) 5.5 T is not triggered by B Recall from (4.1) that eb(i) = X j≤i 1σ(i)=loop −1δ(j)=ζ .
First we’ll bound X j≤i 1δ(j)=ζ. Define B−(i) := − X j≤i 1δ(j)=ζ + 1 2fb(t).
Then E[∆B−(i)|Fi] = − 2Z 2M −1 · ζ 2M −3 + 1 2nf′ b(t) + O 1 n2 f′′ b (t) ≥− fζ n(2b −a) + 1 2nf′ b(t) + O 1 n2 f′′ b (t) .
We have used 2Z ≤2M −5 to get the second line. This is valid as Y > 0 up until time T. Here we see that if Y = 1 we must have Z1 ≥2, unless G contains a copy of K4, which happens with probability o(1).
Note that by (4.7), 3(1 −t) ≤2b −a ≤4(1 −t), so B−will be a submartingale since CB > 4 3Cζ.
(5.43) Indeed, it follows from (5.7) and (5.8) that if 1 −t > n−2/5 log2/5 n then E[∆B−(i)|Fi] ≥−Cζ(1 −t)−1/2 log n n(2b −a) + CB(1 −t)−3/2 log n 4n + O (1 −t)−5/2 log n n2 !
≥−Cζ(1 −t)−3/2 log n 3n + CB(1 −t)−3/2 log n 4n + O (1 −t)−5/2 log n n2 !
≥0.
And if 1 −t ≤n−2/5 log2/5 n then E[∆B−(i)|Fi] ≥−Cζn1/5 log4/5 n n(2b −a) + CBn1/5 log4/5 n 2n(1 −t) + O (1 −t)−2 log4/5 n n9/5 !
≥−Cζn1/5 log4/5 n 3n(1 −t) + CBn1/5 log4/5 n 2n(1 −t) + O (1 −t)−2 log4/5 n n9/5 !
.
≥0.
25 We’ll apply Lemma 5.5 to −B−. Note that before T we can put V ar[∆B−(i)|Fi] = V ar[1δ(i)=ζ|Fi]≤E[12 δ(i)=ζ] ≤pζ ≤ fζ 4n(1 −t) −na −α −fa −fb ≤ fζ 2n(1 −t) using (4.7) and α, fa, fb = o(n(1 −t)) for t ≤T.
And therefore, referring to V (i) as in Lemma 5.5, V (i) ≤ X 0≤k≤i fζ 2n(1 −t)≤Cζ ( 2(1 −t)−1 2 log n : 1 −t > n−2 5 log 2 5 n.
n 1 5 log 9 5 n : otherwise.
So for v we will plug in v = CvB · ( (1 −t)−1 2 log n : 1 −t > n−2 5 log 2 5 n.
n 1 5 log 9 5 n : otherwise.
(5.44) which is an upper bound on V (i) as long as CvB ≥2Cζ.
(5.45) Note that ∆B− ≤1 + f′ b(t) 2n + O 1 n2 f′′ b (t) = 1 + o(1), so by Lemma 5.5 the probability that −B−(i) > 1 2fb(t) is at most exp ( − 1 4f2 b (2 + o(1)) v + 1 2fb ) ≤ exp − 1 4 C2 B(1−t)−1 log2 n (2+o(1))[CvB (1−t)−1/2 log n+ 1 2 CB(1−t)−1/2 log n] 1 −t > n−2 5 log 2 5 n.
exp − 1 4 C2 Bn2/5 log8/5 n log2(1−t) (2+o(1))[CvB n1/5 log9/5 n+ 1 2 CBn1/5 log4/5 n log(1−t)] 1 −t ≤n−2 5 log 2 5 n.
which is o 1 n since 1 4C2 B 2[CvB + 1 2CB] > 1.
(5.46) So w.h.p. for all i ≤T, we have X j<i 1δ(j)=ζ ≤fb(t).
The sum X j<i 1σ(j)=loop presents less difficulty, since w.h.p. the configuration has at most log n loops total. So we can trivially say that X j<i 1σ(j)=loop ≤fb(t) and hence w.h.p. the stopping time T is not triggered by variable B.
This completes the proof of Theorem 5.1.
26 6 Upper bound on the number of components In this section we prove the following lemma which provides the upper bound for the proof of Theorem 1.1: Lemma 6.1. W.h.p. the algorithm outputs a 2-matching with O n 1 5 log 9 5 n components.
Proof. The components of our 2-matching at any step i consist of cycles and paths (including paths of length 0). First we’ll bound the number of paths in the final 2-matching. Note that these final paths have both endpoints in Z0 (or for paths of length 0 there is only one vertex which is in Y0).
These vertices must either have been in Z1 and had a half-edge deleted to land in Z0; or been in Y2 and had a half-edge deleted to land in Y1; or been in Y and had a loop revealed. In the first two cases the endpoints were in ζ and had a half-edge deleted.
So to bound the number of these paths, we bound the sum X j 1δ(j)=ζ + 1σ(j)=loop = O(log n) + X j 1δ(j)=ζ w.h.p.. Note that in light of Section 5.5, we have the bound X j 0 and trivially when ζ = 0 in the sense that when ζ = 0, ∆ζ ≥1 with probability ≈pz =1 −o(1).
Here we need to use the fact that M = Θ(n3/5) in our range.
For any i0 < i ≤iend we have µ = E[X(i)] = −6Ln−2/5 log9/5 n 29 and σ2 = V ar[X(i)] = 2w(i) −10Ln−2/5 log9/5 n −36L2n−4/5 log18/5 n.
We will split the time range i0 to iend into d = log n many intervals of length ≈n3/5 log−2 n. Recall that i0 = n −n3/5 and for all 1 ≤ℓ≤d define iℓ= iℓ−1 + n3/5 log−2 n.
For 0 ≤ℓ< d, we define a random walk starting at iℓ+ 1 and ending at iℓ+1 and let Sℓ= k X i=iℓ+1 X(i).
We note here that these d random walks are independent, identically distributed copies of the same random walk. So for 0 ≤ℓ< d we have E [Sℓ] = E iℓ+1 X i=iℓ+1 X(i) = −6Ln1/5 log−1/5 n.
and that σ · p iℓ+1 −iℓ= Θ n1/5 log−1 n because σ2 = Θ(w(i)) = Θ(n−1/5) and iℓ+1 −iℓ= n3/5 log−2 n. Note that there exists an absolute constant c such that σ · √iℓ+1 −iℓ≤cn1/5 log−1 n for all 0 ≤ℓ< d.
Hence applying Lemma 7.2 to Piℓ+1 i=iℓ+1 X(i), we see that Piℓ+1 i=iℓ+1 X(i) + 6Ln1/5 log−1/5 n σ · √iℓ+1 −iℓ d →N(0, 1).
So there exists some constant p0 > 0 such that for each 0 ≤ℓ< d (and n sufficiently large), P Piℓ+1 i=iℓ+1 X(i) + 6Ln1/5 log−1/5 n σ · √iℓ+1 −iℓ ≥6L + 1 c ≥p0.
So we get that P ∀0 ≤ℓ< 1 2d, ζ(iℓ+1) ≤n1/5 log−1/5 n ≤P ∀0 ≤ℓ< 1 2d, iℓ+1 X i=iℓ+1 X(i) ≤n1/5 log−1/5 n ≤(1 −p0) 1 2 log n = o(1).
So we know that w.h.p. there is a point ib where ζ(ib) > n1/5 log−1/5 n and b ≤1 2d. We would like to show that after n3/5 log−3 n steps, ζ has not decreased below 1 2n1/5 log−1/5 n. To prove this, we consider the sequence Sb(k) = n1/5 log−1/5 n + k X i=ib (X(i) −E[X(i)]) .
30 which is a martingale since E[X(i)|Fi−1] = E[X(i)]. We use the language of martingales here only because Lemma 5.5 is already at hand. Let ic = ib + n3/5 log−3 n < iend. Then ic X i=ib+1 V ar[X(i)] = Θ n2/5 log−3 n .
By applying Lemma 5.5 to the negative of this martingale, we have that after n3/5 log−3 n steps, P ∃i : ib ≤i ≤ic, ζ(i) ≤1 2n1/5 log−1/5 n ≤P ∃i ≤ic : Sb(i) ≤1 2n1/5 log−1/5 n ≤exp −Ω n2/5 log−2/5 n n2/5 log−3 n + n1/5 log−1/5 n !!
≤o(1).
So we know that whp, ζ(i) ≥1 2n1/5 log−1/5 n for ib ≤i ≤ic. In this time, the algorithm is likely to delete an edge adjacent to a ζ vertex. Formally, we have that there exists some q0 such that for all ib ≤i ≤ic, pz(i)pζ(i) ≥q0 = (1 −o(1)) · ζ 2M = Ω n−2/5 log−1/5 n so that if W is a random variable representing the number of i between ib and ic when δ(i) = ζ, then W stochastically dominates Bin(n3/5 log−3 n, q0).
E[Bin(n3/5 log−3 n, q0)] = Ω n1/5 log−16/5 n , so an application of the Chernoffbound tells us that, w.h.p., W = Ω n1/5 log−16/5 n .
References J. Aronson, A.M. Frieze and B. Pittel, Maximum matchings in sparse random graphs: Karp-Sipser revisited, Random Structures and Algorithms 12 (1998) 111-178.
T. Bohman, The Triangle-Free Process, Advances in Mathematics, 221 (2009) 1653-1677.
T. Bohman and A.M. Frieze,Hamilton cycles in 3-out, Random Structures and Algorithms 35 (2009) 393-417.
T. Bohman and A.M. Frieze, Karp-Sipser on random graphs with a fixed degree sequence, Combinatorics, Probability and Computing 20 (2011) 721-742.
T. Bohman, A.M. Frieze and E. Lubetzky, A note on the random greedy triangle-packing algorithm, Journal of Combinatorics 1 (2010) 477-488.
B. Bollob´ as, A probabilistic proof of an asymptotic formula for the number of labelled regular graphs, European Journal on Combinatorics 1 (1980) 311-316.
B.Bollob´ as, C. Cooper, T.I.Fenner and A.M.Frieze, On Hamilton cycles in sparse random graphs with minimum degree at least k, Journal of Graph Theory, 34 (2000) 42-59.
31 R. Durrett, Probability: Theory and Examples. Fourth Edition. Cambridge University Press.
David A. Freedman, On Tail Probabilities for Martingales, The Annals of Probability 3 (1975), 100-118.
A.M. Frieze, On a Greedy 2-Matching Algorithm and Hamilton Cycles in Random Graphs with Minimum Degree at Least Three, Random Structures and Algorithms 45 (2014) 443-497.
A.M. Frieze, Finding Hamilton cycles in sparse random graphs, Journal of Combinatorial Theory B, 44 (1988) 230-250.
A.M. Frieze, M.R. Jerrum, M. Molloy, R. Robinson and N.C. Wormald, Generating and count-ing Hamilton cycles in random regular graphs, Journal of Algorithms 21 (1996), 176-198.
A.M. Frieze, J. Radcliffe and S. Suen, Analysis of a simple greedy matching algorithm on random cubic graphs, Combinatorics, Probability and Computing 4 (1995) 47-66.
R.M. Karp and M. Sipser, Maximum matchings in sparse random graphs, Proceedings of the 22nd Annual IEEE Symposium on Foundations of Computing (1981) 364-375.
M. Lelarge, A New Approach to the Orientation of Random Hypergraphs, Proceedings of the Twenty-Third Annual ACM-SIAM Symposium on Discrete Algorithms (2012) 251-264.
R.W. Robinson and N.C. Wormald, Almost all cubic graphs are Hamiltonian. Random Struc-tures and Algorithms 3 (1992) 117-125.
R.W. Robinson and N.C. Wormald, Almost all regular graphs are Hamiltonian. Random Struc-tures and Algorithms 5 (1994) 363-374.
J. Salez, Weighted enumeration of spanning subgraphs in locally tree-like graphs. Random Structures Algorithms 43 (2013) 377-397.
32 |
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Explore forces and motion as you push household objects up and down a ramp. Lower and raise the ramp to see how the angle of inclination affects the parallel forces. Graphs show forces, energy and work.
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Predict, qualitatively, how an external force will affect the speed and direction of an object's motion.
Explain the effects with the help of a free body diagram.
Use free body diagrams to draw position, velocity, acceleration and force graphs and vice versa.
Explain how the graphs relate to one another.
Given a scenario or a graph, sketch all four graphs.
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3032 | https://study.com/academy/lesson/list-price-vs-net-price.html | Net vs. List Price | Definition & Calculation - Lesson | Study.com
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Accounting 101: Financial Accounting
Net vs. List Price | Definition & Calculation
Contributors: Jack Woerner, Mark Koscinski
Author
Author:
Jack Woerner
BA in Political Science with Emphasis on Social Studies Education at Brevard College, 6 years experience (2 years online) teaching Economics, Personal Finance, APUS Government and more. Certified Gifted/Talented Teacher.
Instructor
Instructor:
Mark Koscinski
Mark has a doctorate from Drew University and teaches accounting classes. He is a writer, editor and has experience in public and private accounting.
Learn about the net price and the list price and how these are defined economically. Discover the mechanisms that differentiate the two values and why they exist. Updated: 11/21/2023
Table of Contents
Net Price
List Price
List Price vs. Net Price
Lesson Summary
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FAQ
What is the list price?
The list price is the price set on goods and services without adding any trade discounts, deals, or promotions. The list price can also be called the catalog price or the manufacturer's suggested retail price (MSRP).
What is net price vs list price?
Net price is the price of a product after all discounts, costs, or taxes have been calculated. List price is the catalog price set on a product without any trade discounts added.
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Net Price ---------
What does net price mean? In economics, the net price definition is the final price paid for a product or service after all costs have been paid. Costs associated with a product can be operating expenses, maintenance, storage, and taxes. The net sales price is usually what a customer pays for a product. Businesses have the challenge of pricing their items in a way that effectively brings in steady revenue but also satisfies the demand of their target customer base. Pricing strategy varies from business to business, and it changes in different economic systems. Net price fluctuates depending on economic situations as well.
Customers who shop at businesses in nations that have already applied the costs to the final price will experience the net price right away. In businesses in other nations or cultures, the costs may not be realized until the customer is ready to make the purchase. There might be a special sales tax, import tax, fee, or other costs that are not seen until the item is finally purchased or even after the item is purchased. There may be a late discount applied to an item as well that could change the net price after an item is purchased, in which case the business would need to refund the customer.
Fish market pricing drastically changes depending on economic and natural conditions. They can change too because of the amount that someone buys.
List Price ----------
The list price is the full price amount for any product without factoring in discounts, promotions, or other deals. The list price is sometimes known as catalog price or the manufacturer's suggested retail price (MSRP). Companies set the list price by calculating all the costs associated with the production of that product. Businesses then set a listing price to recoup the losses from production and operating expenses but also earn profit. Organizations must account for any promotions or discounts they may want to add when determining list price so that they can still earn a profit.
List Price vs. Net Price ------------------------
Businesses must come up with a strategy when calculating the relationship of list price vs. net price. The distinction between the two prices exists because businesses need to choose their price strategy and different economic situations may require a switch in pricing methods. Choosing list price vs. net price for a business depends on the characteristics of the customers. The price may also vary depending on the price set by a business's competition. List price can offer a better profit margin because there are no discounts or promotions added. Net price is more realistic however because competition may force a business to lower prices, or a business adds discounts to meet the needs and desires of their customers.
The calculation of each price is necessary for different economic calculations. In economics, the net price formula is the list price minus any trade discounts. Accounting adjustments and tax reporting are required in order to maintain healthy financial statements for businesses. Accounting keeps up with all inventory costs and the information used from accounting can help determine the necessity of price changes and price strategy. When a business makes any income, it is taxable. The revenue brought in by businesses relies on product sales. Reporting the accurate pricing method used helps businesses when reporting their profit or loss. Economic conditions like trade discounts and market conditions can cause large price swings. Businesses need to keep their financial information organized to prepare for these events.
Trade Discounts
Trade discounts are discounts offered by business-to-business sales or discounts offered by wholesalers to retail companies. Trade discounts are usually applied when a retail business enters into a contract with a wholesale company. This relationship can benefit both parties and can help move products quicker. Manufacturers and wholesalers can offer larger discounts if more of a product is purchased. Volume purchases can change prices based upon the quantity of the purchase. Most trade discounts increase as the sales volume increases. Therefore, the individual consumer who will often purchase a very low volume will most likely pay list price. If a discount is large enough it can impact the list price that manufacturers suggest may change. The trade discount can also affect the net price for customers causing prices to go down or to increase.
Market Conditions
Market conditions in the stock market can cause changes to net and listing pricing. Corporations that have a vast number of products in the economy often have their shares traded publicly on the stock market. A corporation's value is mostly tied up to the stock price and the number of shares that are trading on any given day. If a company experiences some negative events and the news goes public, the share price may drop causing the value of the company to decrease. In response, the company may look to offer discounts or lower the price to bring in extra revenue. Prices can change in a day or even within hours especially during difficult economic times.
Market conditions can prove to be very favorable for consumers like in this 1936 photo of surplus products. Consumers can purchase cheap produce when it goes on sale because of surplus production that businesses are wanting to get rid of before losing all their cost.
Lesson Summary --------------
Net price is the final price of a good or service after all costs, discounts, and taxes have been accounted for. Net prices often change due to market demands, economic situations, unique business practices, and different country's policies. The list price, or the MSRP, is the full price of a product without any discounts, deals, or promotions. The buyer and seller will both use the net or selling price as the basis for its accounting entries. The net price formula is the list price minus trade discounts. Trade discounts are discounts offered by wholesalers or manufacturers to retailers usually as long as they purchase a large volume of products. As sales volume increases, item costs will generally fall. Therefore, larger volume purchases often receive sales discounts. Manufacturing companies will consider market conditions and product cost when setting the list price of products. The final consumer of the good or service will be the party that will most likely pay the list price. Market conditions can change depending on the economic situation or the stock market price of a corporation.
Video Transcript
Pricing Decisions
You worked hard, and your equipment manufacturing company is finally taking off. The sales and marketing departments want to produce your first company catalog. It will be printed for mailing and will also be posted as a page on the company website. Before you send your catalog to print, you must make key pricing decisions. At what price will you try to sell your products?
List Price
Manufacturers and wholesalers preparing a sales catalog usually give each item a list price, also called the catalog price. The list price is determined by several factors, including market research to see what the market will bear and the cost of the product. The list price should cover all production and operating costs and provide an acceptable profit level for the company after allowing for anticipated trade discounts. In many, if not most, cases, the company will not be able to sell its products at list price because buyers will often demand a discount for larger sales volumes.
Discounts
There are a number of different kinds of discounts that can come into play when selling a product. Let's take a look at a few of them.
Trade Discounts
An item's intended selling price, or net price, equals the list price minus a given percentage, called a trade discount. The amount of trade discount usually depends on whether a buyer is a wholesaler, retailer, or final consumer. The larger the volume of the purchase, the larger the discount given. This is sound economics and pricing strategy. A larger volume of production will spread fixed overhead costs over more and more sales units, thereby reducing the total cost of the product. Before setting the level of trade discounts, it is important to understand how much product costs will be affected by the increase in sales volume.
Additional Sales Discounts
Sales personnel may also be given the authority to add an additional sales discount, which is a smaller trade discount to convince the customer to buy if the customer is not yet committed. Sometimes an additional smaller discount makes the sale, but this tactic should be used sparingly. Other customers may demand the same discount, thereby reducing overall profitability.
Obsolete Inventory Discounts
Suppose market conditions have changed, and competition has become fiercer in your competitive space. Perhaps there has been an inadvertent buildup in inventory. In these and similar situations, it might become necessary to use obsolete inventory discounts to reduce the sales price in order to move the product. You do not want inventory to build up and tie up critical working capital. Adding another trade discount may be prudent to clear the inventory.
Accounting
The seller and the buyer of the goods both record the sale at the net price. Neither the buyer nor the seller uses the list price in its accounting records. For example, you may set a list price for small equipment your company manufactures at $1,000 in your catalog. A wholesaler might receive a 30% volume. In this case, the purchase price would be $700.
The seller makes the following journal entry to record the sale:
Dr.Accounts receivable 700
Cr.Sales 700
| To record sales to a wholesaler at a 30% discount. |
Simultaneously, the buyer will make the following entry in its books:
Dr.Equipment 700
Cr.Accounts payable 700
| To record the purchase of equipment. |
Lesson Summary
The list price is also known as the catalog price. The actual selling price, called the net price, will be the list price less any trade discounts the buyer is entitled to take. The higher the sales volume, the higher the trade discount (or given percentage off the list price) will generally be because fixed overhead costs are spread over greater sales volume. Additional discounts may be available when salesmen need to close the deal (called an additional sales discount) or because of inventory buildup (called obsolete inventory discounts). The buyer and the seller account for the sale at the net price.
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Accounting 101: Financial Accounting
14 chapters
137 lessons
13 flashcard sets
Chapter 1
Introduction to Accounting
Purpose of Accounting | ALOE Equation & Examples 11:20 minAccounting Equation Formula & Examples 8:08 minExternal and Internal Users and Uses of Accounting 8:22 minGenerally Accepted Accounting Principles | GAAP Overview & Rules 10:44 minAccounting Principles | Definition & Importance 6:34 minPeriodic Reporting & the Time Period Principle 4:33 minAccrual vs. Cash-Basis Accounting | Definition & Examples 6:20 minEthics in Accounting | Definition, Importance & Examples 7:17 minTechnology in Accounting: The Growing Role of Technology in Accounting 6:37 min
Chapter 2
Financial Statements in Accounting
Using Accrual Accounting to Make Financial Statements More Useful 5:47 minFinancial Statement | Definition, Types & Importance 4:47 minIncome Statement Overview, Format & Examples 3:19 minBalance Sheet Definition, Format & Examples 6:15 minStatement of Cash Flows | Overview & Examples 6:54 minNotes to Financial Statements | Definition, Elements & Examples 7:52 minPreparing an Income Statement | Overview, Process & Example 6:48 minBalance Sheet | Preparation, Equation & Example 11:00 minExternal Audits of Financial Statements 5:52 min
Chapter 3
Mechanics of the Accounting Cycle
Source Documents in Accounting | Definition, Types & Examples 3:52 minAccounts in Accounting | Definition & Types 7:04 minTemporary vs. Permanent Accounts | Differences & Examples 4:57 minNominal vs. Real Account | Definition & Examples 4:55 minLedgers & Charts of Accounts | Definition, Types & Relationship 3:37 minDebit & Credit in Accounting | Meaning, Importance & Examples 5:17 minRecording Transactions in Accounting | Process & Examples 8:00 minTransaction Analysis | Definition, Types & Examples 6:25 minJournal Entries & Trial Balance | Differences & Examples 5:16 min
Chapter 4
Adjusting Accounts and Preparing Financial Statements
Account Adjustments: Types, Purpose & Their Link to Financial Statements 9:00 minAccrued Revenues & Expenses | Definition, Types & Examples 4:30 minAdjusted Trial Balance | Definition, Tables & Examples 6:43 minFinancial Statements | Preparation & Examples 7:35 minClosing Entries | Definition, Examples & Purpose 8:25 minPost-Closing Trial Balance | Definition, Purpose & Examples 3:14 min
Chapter 5
Internal Controls in Accounting
Typical Problems with Financial Information 5:24 minInternal Controls in Accounting | Definition, Purpose & Examples 5:52 minSafeguards & Controls of Banking Activities 6:38 minCash Control Definition & Examples 5:27 minCash Disbursement & Receipt | Definitions & Examples 6:30 minEarnings Management | Definition, Techniques & Consequences 10:17 minBank Reconciliation Definition, Purpose & Examples 6:47 minReconciling the Bank Account After Purchases or Sales 4:40 minTechnology's Impact on the Internal Control System 5:50 minLimitations of Internal Control in Financial Reporting 5:57 minThe Sarbanes-Oxley Act | Summary & Purpose 4:31 minAccounting Auditor Definition, Role & Responsibilities 7:04 minSecurities and Exchange Commission | SEC Overview & Purpose 4:32 min
Chapter 6
Merchandising Operations and Inventory in Accounting
Merchandising Company | Definition, Types & Examples 11:45 minItems that Make Up Merchandise Inventory 8:00 minInventory Cost | Definition & Types 6:42 minAccounting for Inventory Purchases 9:29 minPerpetual & Periodic Inventory | Definition, Uses & Differences 6:00 minRecording Purchases Using the Perpetual System 7:24 minInventory Sales & Accounting | Overview & Journal Entry 8:06 minRecording Sales Using the Perpetual System 5:59 minInventory Counting Methods, Importance & Examples 5:06 minInventory Valuation Methods: Specific Identification, FIFO, LIFO & Weighted Average 9:37 minThe Effects of Financial Statements on Costing Methods 5:17 minThe Effects of Inventory Errors 4:40 minNet Realizable Value | Uses & Formula 9:12 minThe Lower of Cost or Market | LCM Definition and Method 6:01 min
Chapter 7
Receivables in Accounting
Accounts Receivable | Definition & Examples 5:57 minWhat is Revenue Recognition? - Principles, Process & Examples 7:06 minAllowance Method | Definition, Overview & Examples 6:42 minReceivables Management | Purpose, Impact & Examples 4:06 minRecording Warranty and Service Costs 4:42 minLoan Maturity Date | Definition & Overview 6:02 minConverting Receivables to Cash Before Maturity 3:08 min
Chapter 8
Completing the Operating Cycle in Accounting
Accounting for Employee Compensation: Payroll, Leave & Bonuses 4:23 minPayroll Cost Calculations 6:37 minHow to Account for Sales, Property & Income Taxes 5:16 minAccounting for Contingencies & Environmental Liabilities 5:22 minComparing the Effects of Capitalizing & Expensing 5:09 min
Chapter 9
Long-Term Assets in Accounting
Long-Term Operating Assets: Acquisition & Uses 5:55 minPlant Assets | Definition, Characteristics & Examples 3:34 minCost Principle | Definition, Advantages & Examples 6:22 minAcquisition of Property, Plant & Equipment 4:07 minDepreciation Expense| Definition, Types & Formula 8:08 minMethods of Depreciation 6:07 minPartial-Year Depreciation | Definition, Formula & Examples 6:16 minReporting Depreciation on the Balance Sheet 4:41 minComparing Revenue Expenditures & Capital Expenditures 6:11 minHow to Account for Asset Disposal 4:51 minAccounting for Natural Resource Assets & Depletion 6:46 minIntangible Assets Definition, Types & Examples 3:56 minMeasuring Property, Plant & Equipment Asset Efficiency 3:46 minAccelerated Depreciation | Definition, Methods & Calculation 8:19 min
Chapter 10
Current and Long-Term Liabilities in Accounting
Current vs. Long-Term Liabilities | Differences & Examples 5:08 minKnown Liabilities: Definition & Types 4:59 minEstimated Liabilities | Definition, Types & Examples 5:13 minContingent Liabilities: Definition & Examples 4:40 minMeasuring Long-Term Liabilities & Annuities 6:03 minAccounting for Long-Term Liabilities 8:00 minBonds Definition, Categories & Examples 6:36 minThe Advantages of Bond Financing 5:19 minDetermining a Bond's Issuance Price 6:23 minDiscount & Premium Bonds | Definition, Advantages & Disadvantages 6:41 minHow to Record the Retirement of Bonds 5:11 minUsing Debt-Related Financial Ratios 5:14 min
Chapter 11
Reporting & Analyzing Equity in Accounting
Comparing Loans & Investments 3:26 minHow to Raise Equity Financing: Proprietorships & Partnerships 5:44 minCorporations: Characteristics & the Start Up Process 5:15 minThe Components of Stockholder Equity 5:02 minComparing Common & Preferred Stock 4:54 minIssuing a Stock & Accounting for Stock Repurchases 6:19 minItems Reported in Retained Earnings 4:04 min
Chapter 12
Statement of Cash Flows in Accounting
Disclosing Non-Cash Investing and Financing Activities 4:45 minAnalyzing Cash Flow Statement Patterns 6:07 minSources Needed to Prepare the Cash Flow Statement 4:13 minThe Six-Step Process for Preparing a Statement of Cash Flows 5:38 minUsing the Statement of Cash Flows for Decision Making 3:21 minCash Flow Statement | Direct & Indirect Methods 4:09 min
Chapter 13
Financial Statement Analysis in Accounting
Financial Statement Analysis | Definition, Purpose & Examples 6:10 minStandards for Comparison in Financial Statement Analysis 6:25 minHorizontal Analysis | Overview, Formula & Examples 4:43 minVertical Analysis Formula, Calculation & Purpose 5:36 minDefining and Applying Financial Ratio Analysis 6:19 minCommonly Used Financial Ratios 8:02 minReturn on Assets | ROA Formula, Examples & Uses 4:17 minFinancial Ratios Definition, Types & Analysis 7:32 minEfficiency Ratios: Types & Formula 6:52 minLeverage Ratios | Formula, Types & Examples 6:42 minProfitability Definition & Examples 6:23 minIssues with Financial Statement Analysis 6:29 min
Chapter 14
Studying for Accounting 101
Accounting Overview FlashcardsFinancial Statements in Accounting FlashcardsMechanics of the Accounting Cycle FlashcardsAdjusting Accounts & Preparing Financial Statements FlashcardsInternal Controls in Accounting FlashcardsAccounting for Merchandising FlashcardsReceivables in Accounting FlashcardsOperating Cycle in Accounting FlashcardsLong-Term Assets in Accounting FlashcardsCurrent & Long-Term Liabilities in Accounting FlashcardsReporting & Analyzing Equity in Accounting FlashcardsStatement of Cash Flows in Accounting FlashcardsFinancial Statement Analysis in Accounting FlashcardsAccounting 101: Financial Accounting Formulas
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3034 | https://www.scribd.com/doc/243337211/thermod04-pdf | Erf A: Thermal Conduction | PDF | Science & Mathematics | Technology & Engineering
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Erf A: Thermal Conduction
This document discusses thermal conduction and introduces key concepts. It begins by defining the error function erf(a) and complementary error function erfc(a), which are used in calculatio…
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1
CHAPTER 4 THERMAL CONDUCTION
4.0
The Error Function
Before we start this chapter, let’s just make sure that we are familiar with the
error function
erf
a
. We may need it during this chapter. Here is a graph of the gaussian function .1
2
x
e y
−
π=
4.0.1
-2.5-2-1.5-1-0.5 0 0.5 1 1.5 2 2.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 x
y
x = -a x = +a
erf a
I have chosen the coefficient
π
/1 s o th at th e a re a un de r t he cu r ve, f ro m
∞−
to
∞+
is 1. The maximum value, which occurs at
x
= 0, is 5642.0/1
=π
, and it is easy to show that the half width at half the maximum is .8326.0 2 ln
=
Also of some i nterest (though not particularly in this chapter) is the square root of the second moment of area around the
y-
axis. In a mechanical context this would be called the
radius of gyration
. In a stati stical context it would be called the
standard deviation
. Either way, its value is .7071.0 2/1
=
We shall meet the gaussian function again in Chapter 6.
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2 In the present chapter we shall need to make use of the
error function
erf
a
. This is the area under the gaussian curve from :to
a x a
+=−=
.1 erf
2
dx e a
a a x
∫
+−−
π=
4.0.2 The area outside the limits 1
±=
x
, which is the area under the two “tails” of the gaussian function, is sometimes called the
complementary error function
: .erf 1 erfc
a a
−=
4.0.3 It will be clear that erf
a
goes from 0 to 1 a s
a
goes from 0 to infinity. Note also that erfc (one standard deviation) = 0.3173 erfc (two standard deviations) = 0.0455. Here are graphs of erf
a
(continuous line) and erfc
a
(dashed line) versus
a
.
0 0.5 1 1.5 2 2.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x erf a erfc a
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3 4.1
Introduction
While the subject of thermal conduction is an important one, and obviously a proper topic in the theory of heat, it is not really part of the great logical structure of
thermodynamics
, not does it require a wide or deep knowledge of thermodynamics to understand it, at least at an introductory level. In other words, this chapter is more or less a stand-alone chapter. It is not necessary to understand earlier chapters to understand this one; nor, if your primary interest is in thermodynamics, is it necessary t o understand this chapter before proceeding to later ones. That is – if you wish
−
you can skip this chapter without compromising your understanding of any later ones. 4.2
Thermal Conductivity
Figure IV.1 shows heat flowing at a rate
dQ
/
dt
along a bar of cross-sectional area
A
of material. There is a temperature gradient along the length of the bar (which is why heat is flowing down it). At a distance
x
from the end of the bar the temperature is
T
; at a distance
x
+
δ
x
it is
T
δ
T.
Note that, if heat is flowing in the positive direction as shown,
δ
T
must be negative. That is, it is cooler towards the right hand end of the bar. The temperature gradient
dT
/
dx
is negative. Heat flows in the opposite direction to the temperature gradient. The ratio of the rate of heat flow per unit area to the negative of the temperature gradient is called the
thermal conductivity
of the material: .
dx dT KA dt dQ
−=
4.2.1 I am using the symbol
K
for thermal conductivity. Other symbols often seen are
k
or
λ
. Its SI unit is W m
−
1
K
−
1
. I have defined it in a one-dimensional situation and for an isotropic medium, in which case the heat flow is opposite to the temperature gradient. One can imagine that, in an anisotropic medium, the rate of heat flow and the temperature gradient may be different parallel to the different crystallographic axes. In that case t he heat flow and th e temperature gradient m ay not be strictl y antiparallel, and the thermal conductivity is a tensor quantity. Such a situation will not concern us in this chapter.
dQ/dt T T
δ
T A x
δ
x
FIGURE IV.1
x
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4 If, in our one-dimensional example, there is no escape of heat from the sides of the bar, then the rate of flow of heat along the bar must be the same all a long the bar, which means that the temperature gradient is uniform along the length of the wire. It may be easier to im agine no heat loss from the sides than to achieve it in practice. If the bar were situated in a vacuum, there would be no loss by conduction or conve ction, and if the bar were very shiny, there would be little loss b y radiation. Order-of-magnitude values of the thermal conductivities of common substances are Air 0.03 W m
−
1
K
−
1
Water 0.6 Glass 0.8 Fe 80 Al 240 Cu 400 It is easy to imagine how heat may be conducted along a solid, with the vibrations of the atoms at one end of the solid being transmitted to the next atoms by one atom nudging the next, and so on. However, it is evident from the table, and in any case is common knowledge, that some substances (metals) conduct heat much better than others. Indeed, among the metals, there is a close correlation between the thermal and electrical conductivities (at a given temperature). This suggests that the mechanism for thermal conductivity in metals is the same as for electrical conductivity. Heat is conducted in a metal prim arily by electrons. It would be an interesting exercise to find, from the Web or from other references, the thermal and electrical conductivities of a number of metals. It may be found that therm al conductivities,
K
, are sometimes quoted in unfamiliar “practical” units, such as BTU per hour per square foot for a temperature gradient of 1 F° per inch, and converting these to SI units of W m
−
1
K
−
1
might be a bit of a challenge. Electrical conductivities,
σ
, decrease somewhat with rising temperature (so do thermal conductivities, but rather less so), so it would be important to find them all at the same temperature. Then you could see whether the ratio
σ
/
K
is indeed the same for all metals at a given temperature. This is known as the
Wiedemann-Franz Law.
First-order theory (which we do not give here) predicts that .K W 10 44.2 3 1
1 8 2
−−
Ω×=π=σ
e k T K
4.2.2 Here
k
is Boltzmann’s constant and
e
is the electronic charge. This prediction is found to be obeyed well at room temperatures and higher, but at low temperatures the electrical conductivity increases rapidly with lowering temperature, and the ratio starts to fall well below the value predicted by equation 4.2.2, approaching zero at 0 K. The reader may be familiar with the following terms in electricity
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3035 | https://bouman.chem.georgetown.edu/S02/sol10/sol10.htm | Answer Key to Problem Set 10 - Chem 056 Spring 2005
Problem Set 10
Radioactive 222 Rn and 220 Rn constantly form from the decay of uranium and thorium in rocks and soil and, being gaseous, seep out of the ground. The radon isotopes decay fairly quickly, but their products, which are also radioactive, are then in the air and attach themselves to dust particles. Thus, airborne radioactivity can accumulate to worrisome levels in poorly ventilated basements dug in ground that is rich in uranium and thorium.
(a) Describe the composition of an atom of 222Rn and compare it with that of an atom of 220Rn.
222 Rn has 86 electrons outside the nucleus, it has 86 protons and (222-86 =136) neutrons inside the nucleus. 220 Rn has 86 electrons outside the nucleus and has 86 protons and (220-86 = 134) neutrons inside the nucleus.
(b) Although 222 Rn is a decay product of 238 U, 220 Rn comes from 232 Th. How many alpha particles are emitted in forming these radon isotopes from their uranium and thorium starting points?
238-222 = 16, 16 divided by 4 = 4;
From 238 U, four alpha particles need to be emitted to produce 222 Rn.
232-220 = 12, 12 divided by 4 = 3;
From 232 Th, three alpha particles need to be emitted to produce 220 Rn.
(c) Can alpha decay alone explain the formation of these radon isotopes? If not, state what other types of decay must occur.
238 U emitting 4 alpha particles will yield 222 Po, thus, somewhere along the line 2 beta particles need to be emitted to raise the atomic number by two. The same is true for 232 Th into 220 Rn.
(d) The half-life of 222 Rn is 3.82 days, for 220 Rn, the half-life is 54 seconds. Which isotope presents a greater health risk for exposure to a given amount for a given short period of time?
220 Rn presents a greater risk because it has a faster rate of disintegration.
The half-life of a free neutron was thought to be about 1100 seconds until a new experiment established it as 876 s. Suppose that a procedure starts with 1.00 mole of free neutrons. Calculate how much time is required for the amount of neutrons to be reduced to 0.90 mole, according to the old half-life and the new half-life.
For first-order rates:
k = ln 2/ t 1/2 = 0.693/t 1/2 and
ln A - ln A 0 = -kt
k = 0.693/1100 sec (old half-life)
k = 6.30 x 10-4 sec-1
A = 0.90, A 0= 1.00
ln (0.90) - ln (1.00) = -(6.30 x 10-4 sec-1) t
t = 167 seconds
With new half-life, t = 133 seconds.
In the early 19th century, William prout hypothesized that all atomic masses were whole-number multiples of the atomic mass of hydrogen. Discuss this view in terms of the modern understanding of the atomic theory.
Hydrogen, its most abundant isotope, 1 H, has one proton in the nucleus and one electron outside the nucleus. It is true that all the other atoms of the other elements are made up of these building blocks (protons and electrons). However, upon closer inspection, it is observed that the masses of the neutrons and the protons actually change when they combine to form a heavier nucleus, thus, the whole-number multiple theory can no longer be followed.
A radioactive decay series that begins with 237 Np ends with formation of the stable nuclide 209 Bi. How many alpha particle emissions and how many beta particle emissions are involved in the sequence of radioactive decays?
237-209 = 28, 7 alpha emissions. 7 alpha emissions, however, will reduce the atomic number from 93 to 79. (Each alpha particle consists of two protons). Thus, to have Bi as the final product, 4 beta emission should occur (resulting in an increase in the atomic number by 4).
Which of the following nuclides would you expect to be radioactive (unstable)? (a)17 O (b) 176 W (c) 108 Sn
(d) 92 Zr (e) 238 Pu
One uses the belt of stability figure to answer this question.
(a) 17 O (almost 1:1 for neutron:proton ratio, which is ideal for light elements)
should be stable.
(b) 176 W (1.35:1, for neutron:proton ratio, should be 1.5 to be in the belt of stability)
should be radioactive.
(c) 108 Sn (1.16:1, for neutron:proton ratio, should be 1.4 to be stable)
should be radioactive.
(d) 92 Zr (1.3:1, for neutron:proton ratio, should be 1.25)
should be stable enough.
(e) 238 Pu, atomic number > 84, should be radioactive.
End of Answer Key to problem set 10 |
3036 | https://en.wiktionary.org/wiki/opulence | opulence - Wiktionary, the free dictionary
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1 EnglishToggle English subsection
1.1 Etymology
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opulence
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From Wiktionary, the free dictionary
English
[edit]
Etymology
[edit]
From Frenchopulence, from Latinopulentia.
Pronunciation
[edit]
IPA(key): /ˈɒp.jʊl.əns/
Audio (Southern England):Duration: 2 seconds.0:02(file)
Rhymes: -ɒpjʊləns
Noun
[edit]
opulence (countable and uncountable, pluralopulences)
Wealth.
Abundance, bounty, profusion.
Ostentatious display of wealth and luxury; plushness. quotations▼
1721, John Gay, “A Panegyrical Epiſtle to Mr. Thomas Snow, Goldſmith, near Temple-Bar; Occaſion’d by his Buying and Selling the Third South-Sea Subſcriptions, taken in by the Directors at a Thouſand per Cent”, in Miſcellanies, volume 3, published 1733, page 239:There in full Opulence a Banker dwelt,
Who all the Joys and Pangs of Riches felt;
His Side-board glitter’d with imagin’d Plate;
And his proud Fancy held a vaſt Eſtate.
C. J. Fox: The most meritorious persons have always … been removed from opulence.
Synonyms
[edit]
See also Thesaurus:wealth
Related terms
[edit]
opulent
Translations
[edit]
show ▼±wealth
[Select preferred languages] [Clear all]
Catalan: opulènciaf
Dutch: rijkdom(nl)m, weelde(nl)
Finnish: vauraus(fi)
German: Reichtum(de)m, Wohlstand(de)m
Hindi: बहुतायत(hi)f(bahutāyat), संपन्नता(hi)f(sampannatā)
Hungarian: jólét(hu), gazdagság(hu), vagyon(hu)
Latin: opēs(la)f pl, opulentiaf
Polish: obfitość(pl)f
Romanian: opulență(ro)f
Russian: бога́тство(ru)n(bogátstvo)
Spanish: opulencia(es)f
Swedish: rikedom(sv)
Add translation: More
[x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective
Noun class: Plural class:
Transliteration: (e.g. zìmǔ for 字母)
Literal translation:
Raw page name: (e.g. 疲れる for 疲れた)
Qualifier: (e.g. literally, formally, slang)
Script code: (e.g. Cyrl for Cyrillic, Latn for Latin)
Nesting: (e.g. Serbo-Croatian/Cyrillic)
show ▼±abundance
[Select preferred languages] [Clear all]
Dutch: overvloed(nl)m
Finnish: ylenpalttisuus(fi)
Hindi: बहुतायत(hi)f(bahutāyat)
Hungarian: bőség(hu)
Russian: изоби́лие(ru)n(izobílije), пы́шность(ru)f(pýšnostʹ)(figuratively)
Swedish: överflöd(sv)n
Add translation: More
[x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective
Noun class: Plural class:
Transliteration: (e.g. zìmǔ for 字母)
Literal translation:
Raw page name: (e.g. 疲れる for 疲れた)
Qualifier: (e.g. literally, formally, slang)
Script code: (e.g. Cyrl for Cyrillic, Latn for Latin)
Nesting: (e.g. Serbo-Croatian/Cyrillic)
show ▼±ostentatious display of wealth and luxury
[Select preferred languages] [Clear all]
Armenian: please add this translation if you can
Chinese: Mandarin: please add this translation if you can
Esperanto: please add this translation if you can
Finnish: loisto(fi), mahtailu(fi), ylenpalttisuus(fi)
Hungarian: fényűzés(hu)
Russian: ро́скошь(ru)f(róskošʹ), роско́шество(ru)n(roskóšestvo)
Add translation: More
[x] masc. - [x] masc. dual - [x] masc. pl. - [x] fem. - [x] fem. dual - [x] fem. pl. - [x] common - [x] common dual - [x] common pl. - [x] neuter - [x] neuter dual - [x] neuter pl. - [x] singular - [x] dual - [x] plural - [x] imperfective - [x] perfective
Noun class: Plural class:
Transliteration: (e.g. zìmǔ for 字母)
Literal translation:
Raw page name: (e.g. 疲れる for 疲れた)
Qualifier: (e.g. literally, formally, slang)
Script code: (e.g. Cyrl for Cyrillic, Latn for Latin)
Nesting: (e.g. Serbo-Croatian/Cyrillic)
French
[edit]
French Wikipedia has an article on:
opulence
Wikipedia fr
Etymology
[edit]
Borrowed from Latinopulentia.
Pronunciation
[edit]
IPA(key): /ɔ.py.lɑ̃s/
Noun
[edit]
opulencef (pluralopulences)
opulence
Related terms
[edit]
opulent
Further reading
[edit]
“opulence”, in Trésor de la langue française informatisé[Digitized Treasury of the French Language], 2012.
Retrieved from "
Categories:
English terms derived from Proto-Indo-European
English terms derived from the Proto-Indo-European root h₃ep-
English terms derived from French
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Rhymes:English/ɒpjʊləns
Rhymes:English/ɒpjʊləns/3 syllables
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3037 | https://www.et.byu.edu/~vps/ME505/IEM/05%2005.pdf | Chapter V ODE V.5 Systems of Ordinary Differential Equations 405 V V.
.5 5 S SY YS ST TE EM MS S O OF F F FI IR RS ST T O OR RD DE ER R L LI IN NE EA AR R O OD DE Es s Objectives: After the completion of this section the student - should recall the definition of a system of linear 1st order ODEs - should be able to find the general solution of a system with constant coefficients - should be able to investigate for stability the equilibrium point of a plane system Contents: V.5.1. Definitions and Notations V.5.2. Theory of Linear Systems of ODEs V.5.3. The Fundamental Set of a Linear System with Constant Coefficients V.5.4. Autonomous Systems V.5.5. Examples V.5.6. Review Questions and Exercises V.5.7. Systems of ODEs with Maple Chapter V ODE V.5 Systems of Ordinary Differential Equations 406 V.5 SYSTEMS OF 1st ORDER LINEAR ODEs In this section we will study the theory of the systems of linear 1st order ODEs. It can be shown that such systems are equivalent to a single linear differential equation of a higher order; and for both of them the most theoretical results have a similar description. Although, we will try to avoid duplication of the theoretical justification, in a practical approach, the methods of solution for a single equation and for the systems are different. In many cases, the description of the physical model is more natural to perform with the systems of ODEs, and investigation of the physical models such as dynamic, stability etc. is more visual when it is made with the help of systems. 1. DEFINITIONS AND NOTATIONS Normal form Consider a system of n linear first order ODEs written in the normal form (solved for the derivatives of unknown functions): ( ) ( ) ( ) ( ) t f x t a x t a x t a x n n 1 1 2 12 1 11 1 + + + + = ′ ( ) ( ) ( ) ( ) t f x t a x t a x t a x n n 2 2 2 22 1 21 2 + + + + = ′ (1) ( ) ( ) ( ) ( ) t f x t a x t a x t a x n n nn n n n + + + + = ′ 2 2 1 1 where ( ) ( ) ( ) t x t x t x n , , , 2 1 are unknown functions to be determined from the system (1), and coefficients ( ) t aij , ,...
2 , 1 , = j i and functions ( ) t f i are continuous functions in D R ⊂ . Matrix form Introduce the following column vectors and a matrix with entries which are the elements of the system (1): ( ) ( ) ( ) = t x t x t x n 2 1 x ( ) ( ) ( ) ′ ′ ′ = ′ t x t x t x n 2 1 x ( ) ( ) ( ) = t f t f t f n 2 1 f ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 11 12 1n 21 22 2n n1 n2 nn a t a t a t a t a t a t a t a t a t = A Then the system (1) can be written in the compact matrix form: f Ax x + = ′ (2) This is a non-homogeneous system. Without a free term f , the system is homogeneous: ′ = x Ax (3) Solution vector The solution vector (particular solution) is any column vector ( ) t x , t D ∈ which satisfies the system (2). Initial value problem Find the solution of a system of ODEs subject to initial conditions: f Ax x + = ′ subject to ( ) 0 0 t = x x where 0 1 0 2 0 0 n x x x = x , 0 i x ∈ (4) Chapter V ODE V.5 Systems of Ordinary Differential Equations 407 Existence Theorem Theorem 1 (existence and uniqueness theorem) Let the system of linear 1st order ODEs f Ax x + = ′ be normal for t D ∈ ⊂ (means that the functions ( ) t aij and ( ) t f i are continuous on D ⊂) and let 0 i x ∈. Then there exist exactly one solution ( ) t x such that ( ) 0 1 0 2 0 0 n x x t x = x 0 t D ∈ The particular solution ( ) t x of the system (2) or (3) is a vector valued function ( ) n t : → x which can be treated as a parametric definition of the curve in the space n : ( ) ( ) ( ) ( ) 1 1 2 2 n n x x t x x t t x x t = = = = x t D ∈ (5) with the coordinates 1 2 n x ,x ,...,x . This space is called a phase space, and the solution curve defined parametrically by equation (5) is called the trajectory or the orbit in the phase space. For 2-D and 3-D cases, the traditional notations for the coordinate system and, correspondingly, for unknown functions in the system are used: ( ) ( ) ( ) ( ) 11 12 13 1 x a t x a t y a t z f t ′ = + + + ( ) ( ) ( ) ( ) 21 22 23 2 y a t x a t y a t z f t ′ = + + + (1b) ( ) ( ) ( ) ( ) 31 32 33 3 z a t x a t y a t z f t ′ = + + + with the particular solutions written as ( ) ( ) ( ) ( ) x x t t y y t z z t = = = = x t D ∈ (5b) The graph of the equation (5b) defines a trajectory in the phase space (phase plane, for 2-D case). The independent variable t D ∈ can be treated as the time (can also be negative), and for any moment of time equation (5b) defines the position of a point on the trajectory – therefore, the parametric equation (5b) can be interpreted as a motion of a material point along the trajectories defined by the linear system of ODEs. The arrows on the trajectories indicate the direction of motion with the increase of time t. The family of all trajectories of the linear system is called the phase portrait. An initial value problem defines the trajectory which goes through the prescribed point. According to the Existence Theorem, the solution of any initial value problem of the linear system is unique – it means that there is only one trajectory which goes through any point of the phase space, and that the trajectories of the linear system do not intersect. trajectory phase space phase portrait Chapter V ODE V.5 Systems of Ordinary Differential Equations 408 2. THEORY OF LINEAR SYSTEMS OF ODEs It can be shown that a linear nth order ODE can be transformed to a system of n linear 1st order ODEs, and a system of n linear 1st order ODEs can be reduced to a linear nth order ODE. Therefore, the description and the properties of the general solution of the system will be similar to those of the general solution of a linear nth order ODE. Here, we will list the major results of the theory of the systems of linear 1st order ODEs. I) Homogeneous System: ′ = x Ax (3) linear independent solutions Vectors (vector functions) ( ) ( ) ( ) 1 2 n t , t ,..., t x x x are linearly independent if their linear combination equals zero for all t D ∈ ( ) ( ) ( ) 1 1 2 2 n n c t c t ...
c t 0 + + + = x x x (6) only if all coefficients are zero 1 2 n c c ...
c 0 = = = = . If vectors ( ) ( ) ( ) 1 2 n t , t ,..., t x x x are not linearly independent, then they are linearly dependent. It means that in the linear combination at least one coefficient k c 0 ≠ can be non-zero. Wronskian The Wronskian of the set of solution vectors of the homogeneous system ( ) ( ) ( ) 1 2 n t , t ,..., t x x x is defined as a determinant of the matrix whose columns are the vectors ( ) k t x : ( ) ( ) ( ) ( ) 1 2 n W t = det t t ... t x x x (7) There is a relationship of the Wronskian (7) to the Wronskian defined in 5.3. If ( ) W t 0 ≠ at least at one point t D ∈ , then the solution vectors ( ) ( ) ( ) 1 2 n t , t ,..., t x x x are linearly independent. There always exist n linear independent solutions ( ) ( ) ( ) 1 2 n t , t ,..., t x x x of the homogeneous system (3). But any n 1 + solutions of the homogeneous system (3) are linearly dependent. Fundamental set Any set of n linearly independent solutions of the system (3) ( ) ( ) ( ) 1 2 n t , t ,..., t x x x is called a fundamental set. It is obvious that homogeneous systems always possess a zero solution ( ) t ≡ x 0 (trivial solution). But any set which includes the zero vector is linearly dependent. Therefore, the fundamental set cannot include the trivial solution. Fundamental matrix A matrix with columns which are constructed from the fundamental set is said to be the fundamental matrix: ( ) ( ) ( ) ( ) 1 2 n t = t t ... t X x x x (8) General Solution Any solution of the homogeneous system (3) can be written as a linear combination of the vectors from the fundamental set. Therefore, the general solution (complete solution, complimentary solution) of the homogeneous system is a set of all its solutions and it is given by all linear combinations of the Chapter V ODE V.5 Systems of Ordinary Differential Equations 409 vectors from the fundamental set (span of the fundamental set) and it can be defined as: ( ) ( ) ( ) ( ) c 1 1 2 2 n n t c t c t ...
c t = + + + = x x x x Xc (9) where c is a vector of arbitrary constants. Equation (9) defines a family of curves in the phase space which represents the solutions of the homogeneous system. Solution of IVP The solution of the initial value problem for a homogeneous system: ′ = x Ax subject to ( ) 0 0 t = x x (10) is given by ( ) ( ) ( ) 1 0 0 t t t − = x X X x (11) where ( ) t X is the fundamental matrix and ( ) 1 0 t − X is the inverse of the fundamental matrix at 0 t t = II) Non-Homogeneous System: ′ = + x Ax f (2) Denote by ( ) p t x any particular solution of the system (2). A particular solution can be found by the method of undetermined coefficients (similar to the case of linear ODE) or by the method of variation of parameter: ( ) ( ) ( ) ( ) 1 p t t t t dt − = ∫ x X X f (12) The general solution of the non-homogeneous system (2) is given by a sum of the general solution of the homogeneous system (complementary solution) and a particular solution: ( ) t x ( ) ( ) c p t t = + x x General Solution Using equations (9) and (12), the general solution of the non-homogeneous system can be written as: ( ) t x ( ) t = X c ( ) ( ) ( ) 1 t t t dt − + ∫ X X f (13a) Solution of IVP The solution of the IVP for the non-homogeneous equation (4) can be given by: ( ) t x ( ) ( ) 1 0 0 t t − = X X x ( ) ( ) ( ) 0 t 1 t t s s ds − + ∫ X X f (13b) In a conclusion, the general solution and the solution of IVP for homogeneous and non-homogeneous systems (9), (11), (12) and (13) can be determined if the fundamental matrix of the homogeneous system ( ) t X is known. In the next section we will consider the case of linear systems with constant coefficients (autonomous systems) for which there exist the developed methods of finding the fundamental matrix. Chapter V ODE V.5 Systems of Ordinary Differential Equations 410 3. FUNDAMENTAL SET OF LINEAR SYSTEMS WITH CONSTANT COEFFICIENTS Consider the homogeneous system of linear 1st order ODEs 1 11 1 12 2 1n n x a x a x a x ′ = + + + 2 21 1 22 2 2n n x a x a x a x ′ = + + + (14) n n1 1 n2 2 nn n x a x a x a x ′ = + + + where all coefficients ij a ∈, i, j 1,2,...
= are constants. In matrix form: ′ = x Ax 11 12 1n 21 22 2n n1 n2 nn a a a a a a a a a = A (15) The Eigenvalue Problem, matrix diagonalization, matrix exponential are among the techniques used for construction of the fundamental set for a system with constant coefficients. Eigenvalue Problem: Because the linear system is of the first order, we look for the non-trivial solution of the exponential form ( ) t t eλ = x k , 1 n k k = ≠ k 0 (16) k is a non-zero (non trivial) vector of constants, where i k and λ can be the real or the complex numbers which have to be found from satisfying equation (15). Substitute the trial form (16) into matrix equation (15): ( ) ( ) t t e e λ λ ′ = k A k t t e e λ λ λ = k Ak t t e e λ λ λ − = k Ak 0 t t e e λ λ λ − = Ik Ak 0 I is the identity matrix ( ) t eλ λ − = I A k 0 can be divided by t e 0 λ > ( ) λ − = I A k 0 ( ) λ − = A I k 0 (17) This is the homogeneous system of algebraic equations, which according to the Theorem has a non-trivial solution if the determinant of the matrix of coefficients is equal to zero. Therefore, the following condition should be satisfied: ( ) det 0 λ − = A I (18) Equation (18) is the th n order algebraic equation for constant λ : ( ) det λ − A I 11 12 1n 21 22 2n n1 n2 nn a a a 1 0 0 a a a 0 1 0 det a a a 0 0 1 λ = − 11 12 1n 21 22 2n n1 n2 nn a a a a a a det a a a λ λ λ − − = − n n 1 n n 1 1 0 c c ...
c c λ λ λ − − = + + + + Chapter V ODE V.5 Systems of Ordinary Differential Equations 411 Expansion of the determinant yields an algebraic equation with real coefficients which is called the characteristic equation. According to the Fundamental Theorem of Algebra it has n roots 1 2 n , ,..., λ λ λ which can be real or complex, distinct or repeated. These roots are called the eigenvalues. After the eigenvalues are determined, they can be substituted into equation (17) and the corresponding non-zero solutions 1 2 n , ,..., k k k of the vector equation can be found. These solutions 1 2 n , ,..., k k k are called the eigenvectors corresponding to eigenvalues 1 2 n , ,..., λ λ λ . The solution of the eigenvalue problem (17) is not unique; they can be chosen in such a way that the desired solutions (16) have only the real-valued components. Then the constructed fundamental matrix also will have only real-valued entries. Let us show how it can be done: Fundamental Matrix: Case 1: All eigenvalues 1 2 n , ,..., λ λ λ are real and distinct. Then the corresponding eigenvectors vectors 1 2 n , ,..., k k k are also real-valued and linearly independent. Therefore, the fundamental matrix can be defined as ( ) n 1 2 t t t 1 2 n t e e ... eλ λ λ = X k k k (19) Exercise: show that the Wronskian is not equal to zero for any t ∈. In general, for real distinct eigenvalues 1 2 m , ,..., λ λ λ the corresponding entries of the fundamental matrix are m 1 2 t t t 1 2 m e e ... eλ λ λ k k k (20) Case 2: Let eigenvalue λ be the repeated real root of the characteristic equation (18) of multiplicity m . Then if: a) there are m linearly independent eigenvectors 1 2 m , ,..., k k k corresponding to the eigenvalue λ . Then the fundamental matrix includes m 1 2 t t t 1 2 m e e ... eλ λ λ k k k b) there is only one linearly independent eigenvector k corresponding to the eigenvalue λ . Then the other linearly independent solutions can be constructed in the following way: find the vectors , , ,...
k p q which are a solution of the vector equations ( ) λ − = A I k 0 ( ) λ − = A I p k ( ) λ − = A I q p Then the fundamental matrix will include the vectors t t t 2 t t t e , te + e , t e + te e , ... λ λ λ λ λ λ + k k p k p q (21) It can be verified with the help of the Wronskian that these vectors are linearly indenedent. Case 3: The eigenvalue λ is complex. We know that the complex roots of the algebraic equation with real coefficients appear in conjugate pairs: 1,2 i λ α β = ± The corresponding eigenvectors are also complex conjugates 1 k and 1 k . Therefore for distinct eigenvalues 1,2 i λ α β = ± there are two linearly independent solutions ( ) i t 1e α β + k and ( ) i t 1e α β − k . But Chapter V ODE V.5 Systems of Ordinary Differential Equations 412 they are complex-valued functions which is not convenient for representation of the physical problems. It can be shown that the linear combinations of these two solutions and application of the Euler formula ( ( ) a bi a e cosb i sinb e + = + ) yield the two independent real-valued solutions which can be included in the fundamental matrix: ( ) t 1 1 2 cos t sin t eα β β = − x b b (22) ( ) t 2 2 1 cos t sin t eα β β = + x b b where vectors are ( ) 1 1 Re = b k and ( ) 2 1 Im = b k . Conclusion: The solution of the eigenvalue problem for the homogeneous linear system of 1st order ODEs with constant coefficients yields the fundamental matrix X . Matrix exponential: The system of 1st order ODEs in matrix form ′ = x Ax resembles a 1st order ODE for which it is very tempting to write the solution in the traditional exponential form t eA . But how can an exponential function with the matrix be calculated? Again, we can use the analogy with the calculus of functions of a single variable and define the matrix-valued exponential function in the form of the Taylor series: ( ) k 2 3 t 2 3 k 0 t t t e = t ....
k !
2!
3!
∞ = = + + + + ∑ A A I A A A (23) in which we know how to calculate the powers of the matrix k times k = A AA A (it can be shown that the infinite series (23) is always convergent for any t). Then the fundamental matrix of the linear system (15) can be written as t e = A X (23) Then the general solution in matrix exponential form is ( ) t t t e e e t dt − = + ∫ A A A x c f (24) The solution of the IVP can be defined by ( ) ( ) 0 0 t t t t s 0 t e e e s ds − − = + ∫ A A A x x f (25) The matrix exponential form of the fundamental matrix is not used very often for the actual solution of a linear system of ODEs. But it is very convenient for derivation and proofs of the theoretical results such as existence theorems etc. Chapter V ODE V.5 Systems of Ordinary Differential Equations 413 4. AUTONOMOUS SYSTEMS: A system of 1st order ODEs is called autonomous if it can be written in the form: ( ) 1 1 1 2 n x f x ,x ,...x ′ = ( ) 2 2 1 2 n x f x ,x ,...x ′ = (26a) ( ) n n 1 2 n x f x ,x ,...x ′ = and in a matrix form: ( ) ′ = x f x (26b The unknown functions ( ) ( ) ( ) 1 2 n x t ,x t ,...x t are functions of t, but the independent variable t does not appear explicitly in the right hand side of the system (26). Autonomous systems are not necessarily linear. plane linear Here, we will consider only plane linear autonomous systems, which for autonomous systems simplicity can be written as x ax by y cx dy ′ = + ′ = + a,b,c,d ∈ t −∞< < ∞ (27a) and in matrix form ′ = x Ax (27b) The particular solution of the plane system is a 2-dimensional vector which parametrically describes a trajectory (orbit) on the phase plane: ( ) ( ) ( ) x t t y t = x ( ) ( ) x x t y y t = = t −∞< < ∞ (28) The general solution also includes an arbitrary constant vector ( ) ( ) ( ) 1 2 x t,c t, y t,c = x c (29) It defines the family of trajectories in the phase plane (phase portrait) and describes the motion of the points along the solution curves with the change of time t . The arrows on the trajectories indicate the direction of the motion of the point with the increase of time. This mapping defined by equation (29) is called a dynamical system. The initial value problem ′ = x Ax , ( ) 0 0 = x x has a unique solution ( ) ( ) ( ) 0 0 0 x t,x t, y t,y = = x x x the trajectory which goes through the prescribed point 0 x . The right hand side of the vector equation (27b) defines the vector field in the phase plane . At any point on the plane x we can draw a vector Ax and obtain a geometrical representation of the directional field. These directional vectors are tangent to the trajectories defined by the system (27b). The directional vector field can be drawn even without solving system (27), but it provides a qualitative picture of the dynamical system. 0 x ( ) 0 t, x x Chapter V ODE V.5 Systems of Ordinary Differential Equations 414 Critical Points: The important characteristics of the autonomous systems are the critical (equilibrium, stationary) points, which can be defined as the solutions not changing in time (constant solutions): if a point is placed at the equilibrium point it will remain there forever. The critical points can be defined as the solutions of the equation: ( ) = f x 0 For the plane system, critical points are the solutions of the system of equations: ax by 0 cx dy 0 + = + = If the determinant a b det ad bc 0 c d = − ≠ there is only one critical point ( ) cr 0,0 = x (isolated critical point). If the determinant a b det ad bc 0 c d = − = ⇒ ad bc = then there are infinitely many critical points which are located on the line a y x b = − These critical points are not isolated. For a non-linear dynamical system, the situation with the critical points is more complicated. Stability of critical points: Let cr x be the isolated critical point of the autonomous system (27): ( ) cr = f x 0 1. cr x is said to be stable if for any neighborhood ( ) cr U x there exists a smaller neighborhood ( ) cr V x such that for any ( ) 0 cr V ∈ x x the trajectory ( ) ( ) ( ) ( ) 0 0 cr 0 x t,x t, U y t,y = = ∈ x x x x for all t 0 ≥ It means that the trajectory which starts in V remains completely in U . 2. cr x is said to be unstable if it is not stable. It means that it does not matter how close to the critical point the starting point 0 x will be, some trajectory will go outside of any neighborhood ( ) cr U x of the critical point. 3. cr x is said to be asymptotically stable if a) cr x is stable; b) there exists a neighborhood ( ) cr W x such that ( ) 0 cr t lim t, →∞ = x x x Here, we will investigate the stability of the plane dynamical systems (27) which can have only one isolated critical point ( ) cr 0,0 = x depending on the matrix of coefficients A . stable unstable asymptotically stable Chapter V ODE V.5 Systems of Ordinary Differential Equations 415 Phase Portrait of the Plane System: x ax by y cx dy ′ = + ′ = + a b c d = A a b det det ad bc c d = = − A Eigenvalue Problem: The characteristic equation: ( ) 2 2 a b det a d ad bc Tr det 0 c d λ λ λ λ λ λ λ − − = = − + + − = − + = − A I A A Eigenvalues: ( ) ( ) ( ) 2 1,2 a d a d 4 ad bc 2 λ + ± + − − = The form of eigenvalues depends on the expression under the square root which is called the discriminant: Discriminant: ( ) ( ) ( ) 2 2 a d 4 ad bc Tr 4 det ∆= + − − = − A A I) If 0 ∆> then the eigenvalues are real and distinct 1 2 λ λ ≠ II) If 0 ∆= then the eigenvalues are real and repeated 1 2 λ λ λ = = III) If 0 ∆< then the eigenvalues are complex conjugates 1,2 i λ α β = ± Consider the possible configurations of the plane phase portrait (for simplicity of presentation, the details of solution will be skipped; derivation of some of the results will be conducted in the examples and the exercises) : I) 0 ∆> General solution: ( ) 1 2 t t 1 1 2 2 t c e c e λ λ = + x k k a) 1 2 0, 0 λ λ > > Both eigenvalues are positive Unstable node b) 1 2 0, 0 λ λ < < Both eigenvalues are negative ( ) 1 2 1 2 t t t t 1 1 2 2 1 1 2 2 t t t t lim t lim c e c e c lime c lime λ λ λ λ →∞ →∞ →∞ →∞ = + = + = x k k k k 0 Stable node c) 1 2 0, 0 λ λ > < The eigenvalues are of the opposite sign Saddle point (unstable) Chapter V ODE V.5 Systems of Ordinary Differential Equations 416 II) 0 ∆= 1 2 λ λ λ = = a) There are two linearly independent eigenvectors 1 2 , k k General solution: ( ) ( ) t t t 1 1 2 2 1 1 2 2 t c e c e c c e λ λ λ = + = + x k k k k i) 0 λ > degenerate (proper) unstable node ii) 0 λ < degenerate (proper) stable node b) There is one linearly independent eigenvector k (find , , ...
p q ). General solution: ( ) ( ) t t t 1 2 1 2 2 c c t c e c t e c te t t λ λ λ = + + = + + x k k p k p k i) 0 λ > degenerate (improper) unstable node ii) 0 λ < degenerate (improper) stable node III) 0 ∆< 1,2 i λ α β = ± eigenvectors 1 1 2 2 1 2 i , i = + = − k b b k b b a) 0 α ≠ General Solution: ( ) ( ) t 1 1 2 2 1 2 x c cos t sin t c cos t sin t eα β β β β = − + + b b b b i) 0 α > unstable focus (spiral point) ii) 0 α < stable focus (spiral point) asymptotically stable b) 0 α = , 1,2 i λ β = ± (pure imaginary, when a d = −) General Solution: ( ) ( ) 1 1 2 2 1 2 x c cos t sin t c cos t sin t β β β β = − + + b b b b stable center (not asymptotically stable) 0 λ > 0 λ > 0 α > 0 α < 0 α = Chapter V ODE V.5 Systems of Ordinary Differential Equations 417 Classification of the critical points of the plane linear system Procedure for Solution of the Linear System of 1st Order ODEs with Constant Coefficients: 1. Write the system in the normal matrix form (2): ′ = + x Ax f 2. Solve the Eigenvalue Problem (17): ( ) λ − = A I k 0 to find eigenvalues i λ and eigenvectors i k . 3. Construct the fundamental matrix (8) ( ) t X in correspondence with equations (19-22). 4. Calculate the general solution according to the variation of parameter formula (13a): ( ) t x ( ) t = X c ( ) ( ) ( ) 1 t t t dt − + ∫ X X f 5. For solution of the IVP (4) with ( ) 0 0 t = x x , use the variation of parameter formula (13b): ( ) t x ( ) ( ) 1 0 0 t t − = X X x ( ) ( ) ( ) 0 t 1 t t s s ds − + ∫ X X f det A TrA ( ) 2 Tr det 4 = A A 0 ∆= unstable degenerate node unstable node unstable focus stable focus stable degenerate node stable node stable center saddle point Chapter V ODE V.5 Systems of Ordinary Differential Equations 418 5. EXAMPLES: 1) (reduction of the system of 1st order ODEs to a higher order ODE) Consider the system of two 1st order ODEs: x 3x 2y ′ = − y 2x y ′ = − Reduction is performed by differentiation of the equations and consecutive replacement of the unknown functions until a differential equation for a single unknown function is obtained. Consider the second equation: y 2x y ′ = − ⇒ 1 1 x y y 2 2 ′ = + (☼) Differentiate the second equation with respect to t y 2x y ′′ ′ ′ = − ⇒ 1 1 x y y 2 2 ′ ′′ ′ = + Substitute expressions for x and x′ into the first equation x 3x 2y ′ = − 1 1 1 1 y y 3 y y 2y 2 2 2 2 ′′ ′ ′ + = + − Rearrange it into the equation for the function y y 2y y 0 ′′ ′ − + = This is a single 2nd order ODE, linear homogeneous with constant coefficients, which can be solved by the standard method: auxiliary equation 2 m 2m 1 0 − + = ⇒ 1,2 m 1 = general solution t t 1 2 y c e c te = + (♂) The function ( ) x t can be found from equation (☼): 1 1 x y y 2 2 ′ = + ( ) ( ) t t t t 1 2 1 2 1 1 c e c te c e c te 2 2 = + + + ( ) ( ) t t t t 1 2 1 2 1 1 c e c te c e c te 2 2 ′ = + + + t 1 2 1 c c t e 2 = + + Therefore, the general solution of the system of equations is: ( ) t t 1 2 y t c e c te = + ( ) t 1 2 1 x t c c t e 2 = + + That can be verified by direct substitution into the original system of equations. Chapter V ODE V.5 Systems of Ordinary Differential Equations 419 2) (reduction of a higher order ODE to a system of 1st order ODEs) Consider a normal th n order linear ODE ( ) ( ) ( ) ( ) ( ) ( ) n n 1 0 1 n a x y a x y ...
a x y f x − + + + = ( ) 0 a x 0 ≠ for all x (◊) Divide the equation by ( ) 0 a x 0 ≠ : ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) n n 1 1 n 0 0 0 a x a x f x y y ...
y a x a x a x − = − − − + (◊◊) Introduce the set of new functions: 1 x y = 2 x y′ = ⇒ 1 2 x x ′ = 3 x y′′ = ⇒ 2 3 x x ′ = ( ) n 1 n x y − = ⇒ n 1 n x x − ′ = Differentiate n x ( ) n n x y ′ = With these notations, the equation (◊◊) can be rewritten as ( ) ( ) ( ) ( ) ( ) ( ) 1 n n n 1 0 0 0 a x a x f x x x ...
x a x a x a x ′ = − − − + Collect these in the normal linear system of 1st order ODEs: 1 2 x x ′ = 2 3 x x ′ = n 1 n x x − ′ = ( ) ( ) ( ) ( ) ( ) ( ) 1 n n n 1 0 0 0 a x a x f x x x ...
x a x a x a x ′ = − − − + (◊◊◊) 3) Solve the 2nd order ODE y 2y y 0 ′′ ′ − + = by reduction to a system of 1st order ODEs. Applying (◊◊◊) for the 2nd order equation, we obtain 1 2 x x ′ = ( ) ( ) ( ) ( ) ( ) ( ) 1 2 2 2 1 0 0 0 a x a x f x x x x a x a x a x ′ = − − + 2 1 2x x = − (Surprisingly, this system is not identical to the system of Example 1) In matrix form: 1 1 2 2 x x 0 1 x x 1 2 ′ = − Find the fundamental set for a system with constant coefficients. Characteristic equation (18): 2 1 det 2 1 0 1 2 λ λ λ λ − = − + = − − There is only one eigenvalue 1 λ = of multiplicity 2. Find eigenvectors by plugging 1 λ = in vector equation (17): 1 2 k 1 1 0 k 1 2 1 0 − = − − Chapter V ODE V.5 Systems of Ordinary Differential Equations 420 1 2 k 1 1 0 k 1 1 0 − = − It is a singular linear system of algebraic equations, it has only one independent solution: 1 1 = k Find the vector p by solution of the equation ( ) λ − = A I p k 1 2 p 1 1 1 p 1 1 1 − = − Then 1 2 p , p can be found from the equation 1 2 p p 1 − + = One of the solutions can be 0 1 = p Then the fundamental matrix is: t t t e te e λ λ λ = + X k k p [ ] t e t = + X k k p t 1 t e 1 t 1 = + X Then the general solution of the system is given by x = Xc or in the component form ( ) t t 1 1 2 x t c e c te = + ( ) ( ) t t 2 1 2 x t c e c t 1 e = + + This is the solution of the system of ODEs to which the ODE was reduced. Recall now that ( ) 1 x t was defined as 1 x y = , therefore, the general solution is ( ) ( ) t t 1 1 2 y t x t c e c te = = + Which coincides with the previously obtained general solution (♂). The second solution can be treated as ( ) ( ) t t 2 1 2 y x t c e c t 1 e ′ = = + + Integration of this equation will duplicate the previous result. Chapter V ODE V.5 Systems of Ordinary Differential Equations 421 4a) (Linear System of 2 equations – General Solution) Find the general solution of the system of ODEs: 3t 1 1 2 x x 2x 12e ′ = + + 2t 2 1 2 x 4x 3x 18e ′ = + + Solution: 1) Rewrite the given system in the matrix form: 3t 1 1 2t 2 2 x x 1 2 12e x x 4 3 18e ′ = + 2) Solve the eigenvalue problem: 1 2 k 1 2 1 0 0 k 4 3 0 1 0 λ − = Characteristic equation: 2 1 2 1 0 1 2 det 4 5 0 4 3 0 1 4 3 λ λ λ λ λ − − = = − − = − Eigenvalues: 1 1 λ = −, 2 5 λ = (real distinct) Eigenvectors: 1 2 k 1 2 1 0 0 k 4 3 0 1 0 λ − = 1 1 λ = − 1 2 k 1 2 1 0 0 k 4 3 0 1 0 + = 1 2 k 2 2 0 k 4 4 0 = ⇒ 1 1 1 − = k 2 5 λ = 1 2 k 1 2 5 0 0 k 4 3 0 5 0 − = 1 2 k 4 2 0 k 4 2 0 − = − ⇒ 2 1 2 = k 3) Fundamental matrix: ( ) t 5t t 5t e e t e 2e − − − = X ( ) t 5t 4t 4t 4t t 5t e e t det 2e e 3e e 2e − − − = = − − = − X 0 ≠ 4) Variation of parameter formula (13a): Inverse of the Fundamental matrix: ( ) t t 1 5t 5t 2 1 e e 3 3 t 1 1 e e 3 3 − − − − − = X ( ) ( ) t t 3t 4t 3t 1 2t 2t 3t 5t 5t 2 1 e e 12e 8e 6e 3 3 t t 1 1 18e 4e 6e e e 3 3 − − − − − − − − + = = + X f Chapter V ODE V.5 Systems of Ordinary Differential Equations 422 ( ) ( ) 4t 3t 4t 3t 1 2t 3t 2t 3t 8 e dt 6 e dt 2e 2e t t dt 2e 2e 4 e dt 6 e dt − − − − − − + − + = = − − + ∫ ∫ ∫ ∫ ∫ X f Particular solution: ( ) ( ) ( ) t 5t 4t 3t 2t 1 p t 5t 2t 3t 3t 2t e e 2e 2e 4e t t t dt e 2e 2e 2e 6e 2e − − − − − − − + − = = = − − − − ∫ x X X f Complimentary Solution c x : Complimentary Solution (solution of the homogeneous system): ( ) t 5t t 5t 1 1 2 c t 5t t 5t 2 1 2 c c e c e e e t c c e 2c e e 2e − − − − − + − = = = − + x X c General solution: ( ) t x ( ) ( ) c p t t = + x x t 5t 2t 1 2 t 3t 3t 2t 1 2 c e c e 4e c e 2c e 6e 2e − − − − + − = − + − − 4b) (Linear System of 2 equations – Initial Value Problem) Find the solution of the system of ODEs: 3t 1 1 2 x x 2x 12e ′ = + + 2t 2 1 2 x 4x 3x 18e ′ = + + subject to the initial condition: ( ) 3 0 0 = x Solution: Use the fundamental set of the previous example. 3) Fundamental matrix: ( ) t 5t t 5t e e t e 2e − − − = X 4) Variation of parameter formula (13b): ( ) t t 1 5t 5t 2 1 e e 3 3 t 1 1 e e 3 3 − − − − − = X ( ) 1 2 1 3 3 0 1 1 3 3 − − = X 2 1 2 = k saddle point (eigenvalues of opposite sign) 1 1 1 − = k Chapter V ODE V.5 Systems of Ordinary Differential Equations 423 ( ) ( ) 1 0 0 t t − X X x t 5t t 5t t 5t t 5t 2 1 3 e e 2e e 3 3 1 1 0 e 2e 2e 2e 3 3 − − − − − − + = = − + ( ) ( ) t t 3t 4t 3t 1 2t 2t 3t 5t 5t 2 1 e e 12e 8e 6e 3 3 t t 1 1 18e 4e 6e e e 3 3 − − − − − − − − + = = + X f ( ) ( ) t t 4s 3s t t 4t 3t 4t 3t t 0 0 0 0 1 2t 3t t t t 2t 3t 0 2s 3s 0 0 0 8 e ds 6 e ds 2e 2e 2e 2e s s ds 2e 2e 4 2e 2e 4 e ds 6 e ds − − − − − − − − + − + − + = = = − − + − − + ∫ ∫ ∫ ∫ ∫ X f ( ) ( ) ( ) t 5t 4t 3t 2t 5t t 1 t 5t 2t 3t 3t 2t 5t 0 e e 2e 2e 4e 4e t s s ds e 2e 2e 2e 4 6e 2e 8e − − − − − − − + − + = = − − + − − + ∫ X X f Solution of IVP: ( ) t 5t 2t 5t t 2t 5t t 5t 3t 2t 5t t 3t 2t 5t 2e e 4e 4e 2e 4e 5e t 2e 2e 6e 2e 8e 2e 6e 2e 10e − − − − + − + − + = + = − + − − + − − − + x ( ) t x Chapter V ODE V.5 Systems of Ordinary Differential Equations 424 5) (stability of autonomous system) Investigate for stability the equilibrium point and sketch the phase portrait of the following autonomous system: x 2x 3y ′ = − + y 3x 2y ′ = − − Solution: 1) Rewrite the given system in matrix form: x 2 3 x y 3 2 y ′ − = − − 2) Solve the eigenvalue problem: 1 2 k 2 3 1 0 0 k 3 2 0 1 0 λ − − = − − Characteristic equation: 2 2 3 1 0 2 3 det 4 13 0 3 2 0 1 3 2 λ λ λ λ λ − −− − = = + + = − − − −− Eigenvalues: 1 2 3i λ = −+ , 2 2 3i λ = −− (complex) Eigenvectors: 1 2 k 2 3 1 0 0 k 3 2 0 1 0 λ − − = − 1 2 3i λ = −+ 1 2 k 1 2 2 3i 0 0 k 4 3 0 2 3i 0 −+ − = −+ 1 2 k 3i 3 0 k 3 3i 0 − = − − ⇒ 1 1 i = k 1 1 1 0 i i 0 1 = = + k ⇒ 1 1 0 = b , 2 0 1 = b 3) Fundamental matrix (use equation (22)): ( ) t 2t 1 1 2 1 0 cos t sin t e cos3t sin3t e 0 1 α β β − = − = − x b b ( ) t 2t 2 2 1 0 1 cos t sin t e cos3t sin3t e 1 0 α β β − = + = + x b b ( ) ( ) ( ) ( ) ( ) 2t 2t 2t 2t cos 3t e sin 3t e t sin 3t e cos 3t e − − − − = − X 4) General solution: ( ) ( ) t = x t X c ( ) ( ) ( ) ( ) 2t 2t 1 2t 2t 2 c cos 3t e sin 3t e c sin 3t e cos 3t e − − − − = − ( ) ( ) ( ) ( ) 2t 2t 1 2 2t 2t 1 2 c cos 3t e c sin 3t e c sin 3t e c cos 3t e − − − − + = − + Chapter V ODE V.5 Systems of Ordinary Differential Equations 425 In parametric form: ( ) ( ) ( ) ( ) ( ) ( ) 2t 2t 1 2 2t 2t 1 2 x t c cos 3t e c sin 3t e y t c sin 3t e c cos 3t e − − − − = + = − + According to case III) a) ii) this is an asymptotically stable focus. Sketch the phase portrait: For the particular curve, choose 1 2 c 1, c 0 = = , then ( ) ( ) ( ) ( ) 2t 2t x t cos 3t e y t sin 3t e − − = = − A graphing calculator can be used for sketching the graph of this curve, but it is important to know how to sketch the graph just from the parametric equation – we can perform it qualitatively in the following way: For t 0 = ( ) ( ) x t 1 y t 0 = = The starting point is defined. Now let us see where the curve will go under a small increase of time t ε = ( ) ( ) x t 1 y t ε ε = − = − Then continue the curve as a shrinking spiral in the determined direction: All other trajectories will be of the same shape, covering the entire plane without intersections. Here is the graph generated by Maple: Conclusion: the equilibrium point is the asymptotically stable focus (spiral point). Chapter V ODE V.5 Systems of Ordinary Differential Equations 426 6. REVIEW QUESTIONS AND EXERCISES: 1) What is a system of differential equations? 2) What type of systems did we study in this section? 3) How many solutions of a normal system of 1st order ODEs can go through an arbitrary point of the plane? 4) How many solutions of a homogeneous system of 1st order ODEs can go through the point ( ) 0,0 ? 5) Why is uniqueness not violated for a saddle equilibrium point? 6) Why is the system of two 1st order ODEs called a dynamical system? 7) What is the stability of an equilibrium point? What does it mean? EXERCISES: 1) Reduce the following ODEs to a system of 1st order ODEs: a) 2t y 5y 3y 6 y te ′′′ ′′ ′ + + − = b) ( ) iv 2y 6 y 2y y 3y t cost ′′′ ′′ ′ − + − − = Reduce the system of 1st order ODEs to a higher order ODE: c) 1 1 2 x 4x x t ′ = + + 2 1 2 x 2x 3x 2t ′ = + − 2) Matrix exponential: a) Using the definition of matrix exponential, verify the differentiation rule: t t d e e dt = A A A b) Show that if 11 22 nn a 0 0 0 a 0 0 0 a = A then 11 22 nn a t a t t a t e 0 0 0 e 0 e 0 0 e = A c) Consider IVP: solve ′ = x Ax subject to ( ) 0 0 = x x . Show that ( ) ( ) t 1 e t 0 − = A X X , where ( ) t X is the fundamental matrix. 3) Find the general solution of the following systems and sketch the solution curves of the homogeneous part of the systems: a) 1 2 x x ′ = b) 2t 1 1 2 x 5x 3x e− ′ = − + + 2 1 x x sect ′ = − + 2 1 2 x 2x 10x 1 ′ = − + c) 1 1 2 x 4x x t ′ = + + d) 1 1 2 x 2x x sint ′ = − + + 2 1 2 x 2x 3x 2t ′ = + − 2 1 x x cost ′ = − + Chapter V ODE V.5 Systems of Ordinary Differential Equations 427 e) 1 1 2 x x x ′ = − + f) 1 1 2 3 x 3x x x ′ = − − 2 2 3 x x 4x ′ = − + 2 1 2 3 x x x x t ′ = + − + 3 1 3 x x 4x ′ = − t 3 1 2 3 x x x x 2e ′ = − + + 4) Find the solution of the following Initial Value Problems and sketch the graph of the solution: a) 2t 1 1 2 x 3x x 4e ′ = − + b) 1 1 2 x 3x 2x 2 ′ = − + 4t 2 1 2 x x 3x 4e ′ = − + + 2 1 2 x 5x 3x 3 ′ = − + subject to ( ) ( ) 1 2 x 0 1, x 0 2 = = ( ) ( ) 1 2 x 2 0, x 2 0 π π = = c) 1 1 2 3 x 3x x x ′ = − − c) 1 1 2 3 x 3x x x ′ = − − 2 1 2 3 x x x x t ′ = + − + 2 1 2 3 x x x x t ′ = + − + t 3 1 2 3 x x x x 2e ′ = − + + t 3 1 2 3 x x x x 2e ′ = − + + ( ) ( ) ( ) 1 2 3 x 0 1, x 0 0, x 0 0 = = = ( ) ( ) ( ) 1 2 3 x 0 0, x 0 2, x 0 0 = = = 5) Investigate for stability the equilibrium point and sketch the phase portrait of the following autonomous systems: a) x 2x 3y ′ = + b) x x 4y ′ = + y 3x 2y ′ = − + y x y ′ = + c) x 2x 5y ′ = + d) x x 4y ′ = − y x 2y ′ = − y x 5y ′ = + e) x 2x y ′ = + f) x 2x y ′ = + y x 2y ′ = + y x 2y ′ = −+ g) x 2x y ′ = − h) x x y α ′ = −− y x 4y ′ = + y x y ′ = − Chapter V ODE V.5 Systems of Ordinary Differential Equations 428 7. LINEAR SYSTEMS OF ODEs WITH MAPLE 1. a) Find the general solution of the homogeneous system and sketch the phase portrait: 1 2 x 2x x ′ = − 2 1 2 x 3x 2x 4t ′ = − + > with(linalg): > k:=matrix(2,1,); := k 2 0 > f:=matrix(2,1,); := f 0 4 t > C:=matrix(2,1,,[c]]); := C c1 c2 Eigenvalue Problem: > A:=matrix(2,2,); := A 2 -1 3 -2 > eigenvects(A); , [ ] , , 1 1 { } [ ] , 1 1 [ ] , , -1 1 { } [ ] , 1 3 Fundamental matrix: > X:=matrix(2,2,); := X et e ( ) −t et 3 e ( ) −t Complimentary Solution - General Solution of Homogeneous System: > Xc:=evalm(X&C); := Xc + et c1 e ( ) −t c2 + et c1 3 e ( ) −t c2 -Phase Portrait: > x(t):=exp(t)c+exp(-t)c; := ( ) x t + et c1 e ( ) −t c2 > y(t):=exp(t)c+3exp(-t)c; := ( ) y t + et c1 3 e ( ) −t c2 > p:={seq(seq(subs({c=i/22,c=j/22}, [x(t),y(t),t=-4..4]),i=-4..4),j=-4..4)}: > plot(p,x=-5..5,y=-10..10,color=black); Chapter V ODE V.5 Systems of Ordinary Differential Equations 429 b) Find the general solution of the non-homogeneous system: Particular Solution - Variation of Parameter: > Xinv:=simplify(inverse(X)); := Xinv 3 2 e ( ) −t −1 2 e ( ) −t −1 2 et 1 2 et > simplify(evalm(Xinv&f)); −2 e ( ) −t t 2 et t > map(int,%,t); + 2 e ( ) −t t 2 e ( ) −t − 2 et t 2 et > Xp:=simplify(evalm(X&%)); := Xp 4 t − 8 t 4 General Solution: > GS:=evalm(Xc+Xp); := GS + + et c1 e ( ) −t c2 4 t + + − et c1 3 e ( ) −t c2 8 t 4 Solution Curves: > xn(t):=exp(t)c+exp(-t)c+4t; := ( ) xn t + + et c1 e ( ) −t c2 4 t > yn(t):=exp(t)c+3exp(-t)c+8t-4; := ( ) yn t + + − et c1 3 e ( ) −t c2 8 t 4 > pn:={seq(seq(subs({c=i/2,c=j/2}, [xn(t),yn(t),t=-20..20]),i=-3..3),j=-2..2)}: > plot(pn,x=-8..6,y=-10..12,color=black, numpoints=500); Chapter V ODE V.5 Systems of Ordinary Differential Equations 430 c) Find the solution of the Initial Value Problem: Solution of IVP - Variation of parameter formula (13b): > Xinv0:=simplify(subs(t=0,evalm(Xinv))); := Xinv0 3 2 -1 2 -1 2 1 2 > X1:=evalm(evalm(X&Xinv0)&k); := X1 − 3 et e ( ) −t − 3 et 3 e ( ) −t > X2:=simplify(evalm(Xinv&f)); := X2 −2 e ( ) −t t 2 et t > X3:=subs(t=s,evalm(X2)); := X3 −2 e ( ) −s s 2 es s > X4:=simplify(map(int,X3,s=0..t)); := X4 + − 2 e ( ) −t t 2 e ( ) −t 2 − + 2 et t 2 et 2 > X5:=simplify(evalm(X&X4)); := X5 − + 4 t 2 et 2 e ( ) −t − − + 8 t 4 2 et 6 e ( ) −t > XS:=evalm(X1+X5); := XS + + et e ( ) −t 4 t + + − et 3 e ( ) −t 8 t 4 Graph of the solution of IVP: > u:=exp(t)+exp(-t)+4t; := u + + et e ( ) −t 4 t > v:=exp(t)+3exp(-t)+8t-4; := v + + − et 3 e ( ) −t 8 t 4 > plot([u,v,t=-3..2]); Chapter V ODE V.5 Systems of Ordinary Differential Equations 431 Application of the standard Maple procedure for solution of the system of ODEs. 2. Investigate for stability the equilibrium point and sketch the phase portrait of the following autonomous system: x 2x 3y ′ = + y 3x 2y ′ = − + > eq1:=diff(x(t),t)=-4x(t)+3y(t); := eq1 = d d t ( ) x t − + 4 ( ) x t 3 ( ) y t > eq2:=diff(y(t),t)=-2x(t)+1y(t); := eq2 = d d t ( ) y t − + 2 ( ) x t ( ) y t > Solution:=dsolve({eq1,eq2},{x(t),y(t)}); := Solution { } , = ( ) x t + _C1 e ( ) −2 t _C2 e ( ) −t = ( ) y t + 2 3 _C1 e ( ) −2 t _C2 e ( ) −t > assign(Solution): > p:={seq(seq(subs({_C1=i/2,_C2=j/2}, [x(t),y(t),t=-2..2]),i=-5..5),j=-5..5)}: > plot(p,x=-3..3,y=-3..3,color=black,scaling=constrained); The matrix of coefficients has two real distinct negative eigenvalues 1 2 2, 1 λ λ = − = −. The equilibrium point is a stable node. Chapter V ODE V.5 Systems of Ordinary Differential Equations 432 |
3038 | https://math.stackexchange.com/questions/3793432/how-do-you-find-the-number-of-contiguous-subarrays-of-size-k-in-a-given-array | combinatorics - How do you find the number of contiguous subarrays of size $k$ in a given array? - Mathematics Stack Exchange
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How do you find the number of contiguous subarrays of size k k in a given array?
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For example: Given the array [1,2,3,4,5,6,7,8,9][1,2,3,4,5,6,7,8,9] where N N is the length of the array and k k is the subarray size. Here N=9 N=9 and given k=5 k=5, we find that N−k+1 N−k+1 contiguous subarrays of size k k can be found. How can we prove N−k+1 N−k+1 as the number of contiguous subarrays of size k k? I'm sure it is intuitive, but I can't wrap my head around it.
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edited Aug 16, 2020 at 23:49
coolusernamecoolusername
asked Aug 16, 2020 at 23:35
coolusernamecoolusername
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HINT: If a subarray of length k k starts at position m m, at what position does it end?Brian M. Scott –Brian M. Scott 2020-08-16 23:39:15 +00:00 Commented Aug 16, 2020 at 23:39
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Instead of looking at the answer for a general value of k k, let's look at specific examples.
First of all, how many subarrays of length one are there? The answer to this question is n n. Why? Because we can choose any of the n n elements to be in our array.
Next, how many subarrays of length two are there? The answer to this question is n−1 n−1. Why? Because we can choose any of the n n elements except for the last element to be the "start" of the array (and the element directly after it will also be included). Note that we cannot "start" the array at the last element since there's no element to include afterward.
Continuing with the exact same reasoning, we can see that the answer for subarrays of length k k must be n−(k−1)=n−k+1 n−(k−1)=n−k+1 since we're able to "start" the array anywhere except for the last k−1 k−1 positions.
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answered Aug 16, 2020 at 23:45
Ekesh KumarEkesh Kumar
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1 I've edited the question. I'm sorry I meant contiguous subarrays, does the proof still hold?coolusername –coolusername 2020-08-16 23:57:18 +00:00 Commented Aug 16, 2020 at 23:57
Got it, thanks a lot! Is there any way to write a proof for this though?coolusername –coolusername 2020-08-21 20:46:17 +00:00 Commented Aug 21, 2020 at 20:46
No, I don't see how we could prove this using induction, please let me know if you do.coolusername –coolusername 2020-08-21 21:00:18 +00:00 Commented Aug 21, 2020 at 21:00
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Contents
1 Latest election
2 Result in history
2.1 1910 South Africa general election
2.2 1929 South Africa general election
2.3 1994 South Africa general election
3 History
3.1 Before the Union
3.2 At the Union
3.3 Enfranchisement of white women and poor whites
3.4 Segregation of black voters
3.5 Segregation of coloured voters
3.6 Republic referendum
3.7 Tricameral Parliament
3.8 End of apartheid
4 Voting districts
5 List of elections
6 See also
7 References
8 External links
Elections in South Africa
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From Wikipedia, the free encyclopedia
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| Part of a series on the |
| Politics ofSouth Africa |
| Constitution Founding provisions Bill of Rights Chapter nine institutions |
| Executive President Cyril Ramaphosa Deputy President Paul Mashatile Cabinet Departments Shadow Cabinet |
| Legislature National Council of Provinces Chairperson Refilwe Mtsweni-Tsipane National Assembly Speaker Thoko Didiza Leader of the Opposition John Hlophe |
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| Elections General 1910 1915 1920 1921 1924 1929 1933 1938 1943 1948 1953 1958 1961 1966 1970 1974 1977 1981 1984 1987 1989 1994 1999 2004 2009 2014 2019 2024 Municipal 2000 2006 2011 2016 2021 Referendums 1960 1983 1992 Political parties |
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| v t e |
Elections in South Africa include elections for the National Assembly (the lower house of parliament), the provincial legislatures, and municipal councils. Elections are held on a five-year cycle and are conducted by the Electoral Commission (IEC), which is an independent body established by the constitution. The most recent elections for the National Assembly and provincial legislatures were held in 2024, while the most recent elections for municipal councils were held in 2021.
In elections for the National Assembly every South African citizen who is 18 or older is eligible to vote, including (since the 2014 election) those resident outside South Africa. Voters must be registered with the Electoral Commission and have a identity card or book. In elections for a provincial legislature or municipal council, only those resident within the province or municipality may vote.
The electoral system for the National Assembly and the provincial legislatures is based on party-list proportional representation, which means that parties are represented in proportion to their electoral support. Since 2024 independent candidates have also been able to stand for election. For municipal councils there is a mixed-member system in which wards elect individual councillors alongside those named from party lists.
Latest election
[edit]
| Party | National ballot | Regional ballot | Totalseats | +/– |
--- ---
| Votes | % | +/– | Seats | Votes | % | +/– | Seats |
| African National Congress | 6,459,683 | 40.18 | –17.32 | 73 | 6,231,519 | 39.38 | — | 86 | 159 | –71 |
| Democratic Alliance | 3,505,735 | 21.81 | +1.04 | 42 | 3,439,272 | 21.74 | — | 45 | 87 | +3 |
| uMkhonto weSizwe | 2,344,309 | 14.58 | New | 31 | 2,237,877 | 14.14 | — | 27 | 58 | New |
| Economic Freedom Fighters | 1,529,961 | 9.52 | –1.28 | 17 | 1,556,965 | 9.84 | — | 22 | 39 | –5 |
| Inkatha Freedom Party | 618,207 | 3.85 | +0.47 | 8 | 688,570 | 4.35 | — | 9 | 17 | +3 |
| Patriotic Alliance | 330,425 | 2.06 | +2.02 | 5 | 345,880 | 2.19 | — | 4 | 9 | +9 |
| Freedom Front Plus | 218,850 | 1.36 | –1.02 | 4 | 234,477 | 1.48 | — | 2 | 6 | –4 |
| ActionSA | 192,373 | 1.20 | New | 4 | 219,477 | 1.39 | — | 2 | 6 | New |
| African Christian Democratic Party | 96,575 | 0.60 | –0.24 | 3 | 93,581 | 0.59 | — | 0 | 3 | –1 |
| United Democratic Movement | 78,448 | 0.49 | +0.04 | 2 | 85,618 | 0.54 | — | 1 | 3 | +1 |
| Rise Mzansi | 67,975 | 0.42 | New | 1 | 70,142 | 0.44 | — | 1 | 2 | New |
| Build One South Africa | 65,912 | 0.41 | New | 2 | 69,020 | 0.44 | — | 0 | 2 | New |
| African Transformation Movement | 63,554 | 0.40 | –0.04 | 2 | 66,831 | 0.42 | — | 0 | 2 | 0 |
| Al Jama-ah | 39,067 | 0.24 | +0.06 | 2 | 53,337 | 0.34 | — | 0 | 2 | +1 |
| National Coloured Congress | 37,422 | 0.23 | New | 1 | 47,178 | 0.30 | — | 1 | 2 | New |
| Pan Africanist Congress of Azania | 36,716 | 0.23 | +0.04 | 1 | 40,788 | 0.26 | — | 0 | 1 | 0 |
| United Africans Transformation | 35,679 | 0.22 | New | 1 | 32,185 | 0.20 | — | 0 | 1 | New |
| Good | 29,501 | 0.18 | –0.22 | 1 | 36,103 | 0.23 | — | 0 | 1 | –1 |
| #Hope4SA | 27,206 | 0.17 | New | 0 | 16,872 | 0.11 | — | 0 | 0 | New |
| Allied Movement for Change | 22,055 | 0.14 | New | 0 | 18,393 | 0.12 | — | 0 | 0 | New |
| United Independent Movement | 20,003 | 0.12 | New | 0 | 18,907 | 0.12 | — | 0 | 0 | New |
| African Independent Congress | 19,900 | 0.12 | –0.16 | 0 | 3,833 | 0.02 | — | 0 | 0 | –2 |
| National Freedom Party | 19,397 | 0.12 | –0.23 | 0 | 22,726 | 0.14 | — | 0 | 0 | –2 |
| Azanian People's Organisation | 19,048 | 0.12 | +0.05 | 0 | 18,741 | 0.12 | — | 0 | 0 | 0 |
| African Congress for Transformation | 18,354 | 0.11 | New | 0 | 348 | 0.00 | — | 0 | 0 | New |
| African Heart Congress | 16,306 | 0.10 | New | 0 | 3,579 | 0.02 | — | 0 | 0 | New |
| Congress of the People | 14,177 | 0.09 | –0.18 | 0 | 16,768 | 0.11 | — | 0 | 0 | –2 |
| African People's Convention | 13,195 | 0.08 | –0.03 | 0 | 14,693 | 0.09 | — | 0 | 0 | 0 |
| Africa Restoration Alliance | 11,108 | 0.07 | New | 0 | 12,651 | 0.08 | — | 0 | 0 | New |
| Forum for Service Delivery | 11,077 | 0.07 | +0.03 | 0 | 7,444 | 0.05 | — | 0 | 0 | 0 |
| Democratic Liberal Congress | 10,904 | 0.07 | +0.01 | 0 | 7,022 | 0.04 | — | 0 | 0 | 0 |
| Alliance of Citizens for Change | 9,336 | 0.06 | New | 0 | 11,217 | 0.07 | — | 0 | 0 | New |
| Action Alliance Development Party [af] | 7,802 | 0.05 | New | 0 | 4,600 | 0.03 | — | 0 | 0 | New |
| Conservatives in Action [af] | 7,424 | 0.05 | New | 0 | 1,115 | 0.01 | — | 0 | 0 | New |
| South African Royal Kingdoms Organisation [af] | 6,685 | 0.04 | New | 0 | 3,195 | 0.02 | — | 0 | 0 | New |
| Northern Cape Communities Movement [af] | 6,629 | 0.04 | New | 0 | 7,016 | 0.04 | — | 0 | 0 | New |
| People's Movement for Change | 5,539 | 0.03 | New | 0 | 7,045 | 0.04 | — | 0 | 0 | New |
| Abantu Batho Congress | 5,531 | 0.03 | New | 0 | 3,552 | 0.02 | — | 0 | 0 | New |
| Economic Liberators Forum [af] | 5,408 | 0.03 | New | 0 | 7,115 | 0.04 | — | 0 | 0 | New |
| Organic Humanity Movement | 5,241 | 0.03 | New | 0 | 6,457 | 0.04 | — | 0 | 0 | New |
| African Content Movement | 5,107 | 0.03 | 0.00 | 0 | 4,617 | 0.03 | — | 0 | 0 | 0 |
| Sizwe Ummah Nation | 5,016 | 0.03 | New | 0 | 4,869 | 0.03 | — | 0 | 0 | New |
| South African Rainbow Alliance | 4,796 | 0.03 | New | 0 | 7,645 | 0.05 | — | 0 | 0 | New |
| African People's Movement | 4,601 | 0.03 | New | 0 | 4,200 | 0.03 | — | 0 | 0 | New |
| Able Leadership [af] | 3,867 | 0.02 | New | 0 | 3,161 | 0.02 | — | 0 | 0 | New |
| Referendum Party | 3,834 | 0.02 | New | 0 | 4,206 | 0.03 | — | 0 | 0 | New |
| All Citizens Party [af] | 3,693 | 0.02 | New | 0 | 1,644 | 0.01 | — | 0 | 0 | New |
| Africa Africans Reclaim [af] | 3,371 | 0.02 | New | 0 | 2,565 | 0.02 | — | 0 | 0 | New |
| Citizans [af] | 2,992 | 0.02 | New | 0 | 4,084 | 0.03 | — | 0 | 0 | New |
| Xiluva | 2,592 | 0.02 | New | 0 | 1,167 | 0.01 | — | 0 | 0 | New |
| African Movement Congress [af] | 2,141 | 0.01 | New | 0 | 1,550 | 0.01 | — | 0 | 0 | New |
| Free Democrats | 1,992 | 0.01 | 0.00 | 0 | 2,276 | 0.01 | — | 0 | 0 | 0 |
| Independents | 19,304 | 0.12 | 0 | 0 | New |
| Total | 16,076,719 | 100.00 | – | 200 | 15,823,397 | 100.00 | – | 200 | 400 | 0 |
| Valid votes | 16,076,719 | 98.69 | 15,823,397 | 99.02 |
| Invalid/blank votes | 213,437 | 1.31 | 156,834 | 0.98 |
| Total votes | 16,290,156 | 100.00 | 15,980,231 | 100.00 |
| Registered voters/turnout | 27,782,081 | 58.64 | 27,782,081 | 57.52 |
| Source: Electoral Commission of South Africa, IOL |
Result in history
[edit]
1910 South Africa general election
[edit]
| Party | Votes | % | Seats |
--- --- |
| Unionist Party (South Africa) | 39,766 | 37.65 | 36 |
| South African Party | 30,052 | 28.45 | 66 |
| Labour Party (South Africa) | 11,549 | 10.93 | 3 |
| Socialist Party | 448 | 0.42 | 0 |
| Other | 4,245 | 4.02 | 4 |
| Independents | 19,563 | 18.52 | 12 |
| Total | 105,623 | 100.00 | 121 |
1929 South Africa general election
[edit]
| Party | Votes | % | Seats |
--- --- |
| South African Party | 159,896 | 46.50 | 61 |
| National Party (South Africa) | 141,579 | 41.17 | 78 |
| Labour Party (South Africa) | 33,919 | 9.86 | 8 |
| Independents | 8,503 | 2.47 | 1 |
| Total | 343,897 | 100.00 | 148 |
1994 South Africa general election
[edit]
| Party | Votes | % | Seats |
--- --- |
| African National Congress | 12,237,655 | 62.65 | 252 |
| National Party (South Africa) | 3,983,690 | 20.39 | 82 |
| Inkatha Freedom Party | 2,058,294 | 10.54 | 43 |
| Freedom Front Plus | 424,555 | 2.17 | 9 |
| Democratic Party (South Africa) | 338,426 | 1.73 | 7 |
| Pan Africanist Congress of Azania | 243,478 | 1.25 | 5 |
| African Christian Democratic Party | 88,104 | 0.45 | 2 |
| Africa Muslim Party | 34,466 | 0.18 | 0 |
| African Moderates Congress Party | 27,690 | 0.14 | 0 |
| Dikwankwetla Party of South Africa | 19,451 | 0.10 | 0 |
| Federal Party | 17,663 | 0.09 | 0 |
| Minority Front | 13,433 | 0.07 | 0 |
| Sport Organisation for Collective Contributions and Equal Rights | 10,575 | 0.05 | 0 |
| African Democratic Movement | 9,886 | 0.05 | 0 |
| Women's Rights Peace Party | 6,434 | 0.03 | 0 |
| Ximoko Party | 6,320 | 0.03 | 0 |
| Keep It Straight and Simple Party | 5,916 | 0.03 | 0 |
| Workers Organisation for Socialist Action | 4,169 | 0.02 | 0 |
| Luso-South African Party | 3,293 | 0.02 | 0 |
| Total | 19,533,498 | 100.00 | 400 |
History
[edit]
Before the Union
[edit]
When the British took over the Cape, first in 1795 and then more permanently in 1806, they inherited a large, thinly populated pastoral society that depended on the labour of slaves and a rural workforce of indigenous Khoekhoe whose condition was akin to serfdom. In 1806 the entire population of the colony consisted of fewer than 80,000 people: 26,768 whites, 1,200 free blacks (manumitted slaves), 29,861 slaves and 20,426 Khoikhoi . In 1807, the British government ended the slave trade and, finally, in 1833 outlawed the practice of slavery throughout the Empire. [citation needed]
Under pressure from the humanitarian lobby in the UK, acting in concert with a local missionary campaign, the government abolished the Khoekhoe's serf-like status by the declaration of Ordinance 50 of 1828. According to the government's instructions, 'all Hottentots and other free persons of colour, lawfully residing within the said Colony, are and shall be, in the most full and ample manner, entitled to all and every the rights, privileges and benefits of the law, to which any other His Majesty's subjects, lawfully residing within the said Colony, are or can be entitled.' This became known as the "Hottentots' Magna Carta". The equality of all people appeared to be assured. As a result, municipal boards set up in the colony's towns and villages from 1836 allowed any male resident who lived in a property with a yearly rent £10 or more to vote for his town board. From the very beginning, therefore, persons of color were able to participate in local elections.
This principle of a non-racial franchise was to be entrenched when the Cape was granted representative government in 1853. Part of the reason may have been give propertied Khoekhoe a greater stake in the political system. In 1850 the Khoekhoe of the Kat River Settlement had risen in rebellion. Although the uprising was defeated, it did influence the minds of colonial officials and politicians who were responsible for drawing up the 1853 constitution. The Cape's Attorney General, William Porter said: 'I would rather meet the Hottentot at the hustings voting for his representative than meet the Hottentot in the wilds with a gun on his shoulder.' In terms of the constitution of 1853, any man who owned property worth at least £25 was entitled to vote for or stand in the Cape's Parliament. By 1886 Africans made up 43 per cent of the vote in six constituencies of the Eastern Cape, and were a real political force. It was not long before white politicians began challenging the rights of Coloureds and Africans to vote. The passage of the Franchise and Ballot Act, which raised the property qualification from £25 to £75 in 1892, met with an angry response from African and Coloured voters.
These developments were not without criticism or opposition. The abolition of slavery, the declaration of Ordinance 50, and the accompanying extension of rights to the black population, were deeply resented by the white Dutch farmers of the Cape as undermining their way of life. Starting in 1834, thousands of these Boers set out on the Great Trek in the hope of leaving British control behind them. In the interior of southern Africa they would establish independent states, the South African Republic (also called the Transvaal Republic) and the Orange Free State, whose constitutions enshrined the principle of inequality between white and black in church and state.[citation needed]
After the Second Anglo-Boer War, the white peoples made peace and came together at the National Convention in October 1909. It brought together politicians from the Cape, Transvaal, Natal and the Orange River Colony, as well as Rhodesia. They aimed to draw up a constitution for the Union of South Africa, uniting the British possessions. The result was the South Africa Act.
Only white men were present at the National Convention; women and all other racial groups were excluded. This was in some ways unusual. Black men had enjoyed the vote in the Cape since the 1850s and – as long as they had sufficient property, income and education – continued to do so. By 1909 there were 14,388 Coloured and 6,633 African voters in the Cape. Between them they made up 14.8% of the electorate. In Natal African men also had the right to vote, but it was so constrained that it was almost theoretical. They had to prove they had property and that they were 'civilised' and had been so for seven years. The Governor might then grant them the vote. Indians also had to overcome obstacles designed to prevent them from being enfranchised. In the Transvaal and Orange Free State, only white men could vote.
The Cape argued that their non-racial franchise should be extended across the proposed Union. This was rejected by the Transvaal and Orange Free State. Finally, a compromise was arrived at, maintaining the Cape's existing voting system without extending it to the rest of the country, but insisting that this compromise was entrenched in the constitution. Africans and Coloured people would retain most of their voting rights in the Cape, but would not receive them in any other part of the Union.
African and Coloured politicians came together to resist these plans, and called on a former Cape Prime Minister, William Philip Schreiner, to lead a delegation to the UK to call for the Cape franchise to be implemented in the whole of South Africa. The delegation was unsuccessful in its appeal, despite receiving considerable support from the infant Labour Party and other liberal British organisations.
At the Union
[edit]
The Union of South Africa was created on 31 May 1910 by the South Africa Act 1909, an act of the British Parliament. The House of Assembly (the lower house of the newly created Parliament of South Africa) and the provincial councils were elected by first-past-the-post voting in single-member electoral divisions. The franchise in these elections was initially the same as the franchise for the lower houses in the four colonies that had formed the Union, so there were different qualifications in different provinces.
In the Transvaal and the Orange Free State, the vote was limited by law to white men aged 21 or over. In Natal the vote was limited to men over 21 who met property and literacy qualifications; in theory, this could include non-white men but in practice only very small numbers managed to qualify: in 1910 over 99% of the electorate was white. In the Cape Province the franchise was also limited to men over 21 who met property and literacy qualifications, and non-white men did qualify in significant numbers, making up approximately 15% of the electorate in 1910. The qualifications in the Cape and Natal also excluded a substantial number of poorer white men. Only white men could stand for election to the House of Assembly, even from the Cape constituencies. The franchise rights of non-white voters in the Cape (but not in Natal) were entrenched in the South Africa Act by a provision that they could only be reduced by an act of Parliament passed by a two-thirds majority of both houses of Parliament sitting in a joint session.
Enfranchisement of white women and poor whites
[edit]
In 1930 the National Party government of J. B. M. Hertzog passed the Women's Enfranchisement Act, which extended the right to vote and the right to stand for election to all white women over the age of 21. In the following year the Franchise Laws Amendment Act lifted the property and literacy requirements for white male voters in the Cape and Natal, with the result that all white citizens over 21 were enfranchised. As the exclusion of women and the literacy and property qualifications continued to apply to non-white voters, these acts had the effect of diluting their electoral power by more than doubling the size of the white electorate.
At the next following general election in 1933, Leila Reitz was elected as the first female MP, representing Parktown for the South African Party.
Segregation of black voters
[edit]
In 1936 the Hertzog government enacted the Representation of Natives Act, which removed black voters from the common voters' rolls and placed them on separate "native voters' rolls". The act was passed by the required two-thirds majority in a joint session. Black voters could no longer vote in ordinary elections for the House of Assembly or the Cape Provincial Council; instead they would separately elect three members of the assembly and two members of the council. Four senators would also be indirectly elected by chiefs, tribal councils and local councils for "native areas". The Representation of Natives Act was repealed in 1959 and consequently the seats of the "native representative members" were abolished in 1960. From this point, the only political representation of black South Africans was in the Bantustan legislatures.
Segregation of coloured voters
[edit]
Main article: Coloured vote constitutional crisis
After coming to power in 1948 the National Party engaged in a policy of removing coloured voters similarly to black voters. In 1951 Parliament passed the Separate Representation of Voters Act, which removed coloured voters from the common voters roll and instead allowed them to separately elect four MPs. The act was challenged on the basis that it had not been passed with a two-thirds majority in a joint sitting, as required by the South Africa Act, and in 1952 the Appellate Division of the Supreme Court declared it to be invalid.
A subsequent attempt by the government to circumvent the Supreme Court by creating a High Court of Parliament failed. In the election of 1953, coloured voters in the Cape cast their ballots in the same constituencies as white voters. In 1955, the government introduced a new act which reconstituted the Senate, providing the two-thirds majority necessary to validate the Separate Representation of Voters Act.
This separate representation of coloured voters in the House of Assembly was ended in 1970. Instead, all coloured adults were given the right to vote for the Coloured Persons' Representative Council, which had limited legislative powers and was permanently dissolved in 1980.
Republic referendum
[edit]
In 1960 a whites-only referendum was held to decide whether South Africa should become a republic. No changes were made to the franchise with the Republic's emergence in 1961. However, with the policy of establishing Bantustans, the remaining black representation in the Senate was completely removed.
Tricameral Parliament
[edit]
In 1983 a referendum on constitutional reform was held, as a result of which the Tricameral Parliament was formed, consisting of three separate houses to represent white, coloured and Indian South Africans. The existing House of Assembly was retained with its white electorate, while two new houses were created: the House of Representatives elected by coloured voters, and the House of Delegates elected by Indian voters. Many Indians and Coloureds rejected this powerless government as it was a strategy by the government to divide and rule over the nonwhite vote. Elections to these houses were conducted on the basis of first-past-the-post voting in single-member electoral divisions.
End of apartheid
[edit]
During the negotiations to end apartheid the Interim Constitution was enacted. It introduced universal suffrage on a non-racial basis, and replaced first-past-the-post voting with party-list proportional representation. South Africans of all races took part in the first fully democratic elections in 1994. "Universal adult suffrage, a national common voters roll, regular elections and a multi-party system of democratic government" are founding principles of the 1996 Constitution of South Africa, and the right of all citizens to vote is included in the Bill of Rights.
In the post-apartheid era, the Constitutional Court has struck down two attempts by the government to deny the vote to convicted criminals in prison. The court has also ruled that South Africans living outside the country must be allowed to vote. In 2020 in the New Nation Movement case the court ruled that the pure party-list electoral system is unconstitutional because it prevents individuals from standing as candidates without joining a political party; Parliament was given two years to adopt a new electoral system. As of the 2024 General Elections a third provincial ballot was adopted. This ballot facilitates the election of independent candidates.
Voting districts
[edit]
Each voter in South Africa is assigned to a voting district based on the voter's residence at the time that they registered to vote. Each voting district is uniquely associated with a single voting station. Voters who are outside their registered district on election day may vote at another polling station, but additional paperwork is required. Voting districts have no significance outside of the election process, and district boundaries are drawn for purposes of efficiently planning and administering elections. Urban voting districts are drawn to have a population of around 3,000 within a radius of 7.5 kilometres (4.7 mi), and rural voting districts are drawn to have a population of around 1,200 within a radius of 10 kilometres (6.2 mi).
As of 2019, there were 22,933 voting districts nationwide. The district boundaries are set by the Electoral Commission's Delimitation Directorate, and are reviewed and adjusted before each election.
List of elections
[edit]
Main article: List of elections in South Africa
Since 1910, parliamentary general elections have been held on the following dates:
15 September 1910
20 October 1915
10 March 1920
8 February 1921
19 June 1924
12 June 1929
17 May 1933
18 May 1938
7 July 1943
26 May 1948
15 April 1953
16 April 1958
18 October 1961
30 March 1966
22 April 1970
24 April 1974
30 November 1977
29 April 1981
22 and 28 August 1984 (House of Representatives and House of Delegates)
6 May 1987 (House of Assembly)
6 September 1989 (all three houses)
26–29 April 1994
2 June 1999
14 April 2004
22 April 2009
7 May 2014
8 May 2019
29 May 2024
See also
[edit]
Electoral calendar
Electoral system
References
[edit]
^ Martin Legassick, The struggle for the Eastern Cape, 1800 – 1854, KMM Review Publishing Company, Johannesburg, 2010, p. 5
^ Report of Commission of Inquiry regarding Cape Coloured Population of the Union, U.G. 54 – 1937, Government Printer, Pretoria, 1937, p. 212, para. 1034
^ MacAulay, Zachary (July 1829). "Natives of South Africa". Anti-Slavery Monthly Reporter. No. 50. Retrieved 11 May 2021.
^ Report of Commission of Inquiry regarding Cape Coloured Population of the Union, U.G. 54 – 1937, Government Printer, Pretoria, 1937, p.213, para 1037
^ Stanley Trapido, The friends of the natives: merchants, peasants and the political and ideological structure of liberalism in the Cape, 1854-1910, in Shula Marks and Anthony Atmore (eds.), Economy and society in pre-industrial South Africa, Longman, London, 1980, p. 262
^ Andre Odendaal, The Founders: The Origins of the ANC and the Struggle for Democracy in South Africa, Jacana, Auckland Park, 2012, p.96
^ Richard van der Ross, A Political and Social History of the Cape Coloured People, 1880 – 1970, Thesis submitted to the University of Cape Town, 1973, UCT Special Collections, p. 11
^ L. M. Thompson, The Unification of South Africa, 1902 – 1910, Oxford at the Clarendon Press, Oxford, 1960, p. 110
^ L. M. Thompson, The Unification of South Africa, 1902 – 1910, Oxford at the Clarendon Press, Oxford, 1960, p. 111
^ National Archive, Secret Despatch from Selborne to Crewe, 24 October 1908, PRO CO 879/106/8
^ Martin Plaut, Promise and Despair, the First Struggle for a Non-Racial South Africa, Jacana Press, 2016
^ "2024 General Elections, 29 May | South African Government".
^ a b c Electoral Commission of South Africa. "About voting districts and stations". Retrieved 24 February 2020. Each voting district is serviced by one voting station only.
^ "Can you vote at any voting station? Yes, you can, but…". 7 May 2019. Retrieved 24 February 2020.
^ "Publication of the Voting District including mobile stations". SABC News. 7 March 2019. Retrieved 24 February 2020.
External links
[edit]
"Women's suffrage". Archived from the original on 5 March 2006. Retrieved 4 March 2006.
Adam Carr's Election Archive
African Elections Database
Democratic Development in South Africa Archived 3 March 2016 at the Wayback Machine from the Dean Peter Krogh Foreign Affairs Digital Archives Archived 12 January 2012 at the Wayback Machine
| v t e Elections and referendums in South Africa |
| General elections | 1910 1915 1920 1921 1924 1929 1933 1938 1943 1948 1953 1958 1961 1966 1970 1974 1977 1981 1984 1987 1989 1994 1999 2004 2009 2014 2019 2024 Next |
| Provincial elections | 1910 1917 1935 1936 1943 1949 1954 1959 1965 1970 1974 1981 1994 1999 2004 2009 2014 2019 2024 |
| Municipal elections | 1995–96 2000 2006 2011 2016 2021 2026 |
| Referendums | 1960 1983 1992 |
| See also: Elections and referendums in the Boer republics and South African colonies Elections and referendums in South African Bantustans |
| v t e Presidential elections in South Africa |
| Ceremonial State President | 1961 1967 1968 1975 1978 1979 |
| Executive State President | 1984 1989 |
| President | 1994 1999 2004 2008 2009 2014 2018 2019 2024 |
| v t e Elections and referendums in the Boer republics and South African colonial territories |
| | | | Cape Colony | | Parliamentary elections | 1854 1858–59 1862 1866 1870 1874 1878–79 1884 1888–89 1891 1894 1898 1903 1904 1908 | |
| | | | Griqualand West | | Parliamentary elections | 1873 | |
| | | | Natalia Republic and Natal | | Parliamentary elections | 1838 1839 1840 1841 1842 1857 1859 1880 1883 1892 1893 1897 1901 1902 1906 | | Referendums | 1909 | |
| | | | Orange Free State | | Presidential elections | 1854 1855 1860 1863 1868 1873 1878 1883 1888 1893 1896 | | Parliamentary elections | 1854 1907 | |
| | | | Transvaal | | Presidential elections | 1872 1883 1888 1893 1898 | | Parliamentary elections | 1855 1856 1893 1907 | |
| See also: Elections in South Africa |
| v t e Elections and referendums in South African Bantustans |
| | | | Elections | | Bophuthatswana | 1972 1977 1982 1987 | | Ciskei | 1973 1978 1986 | | Gazankulu | 1973 1978 1983 1989 | | KwaNdebele | 1988 | | Lebowa | 1973 1978 1983 | | QwaQwa | 1975 1980 1985 1990 | | Transkei | 1963 1968 1973 1976 1981 1986 | | Venda | 1973 1978 1984 | |
| | | | | | Ciskei | | |
| See also: Elections in South Africa |
| v t e Elections in Africa |
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3040 | https://electronics.stackexchange.com/questions/376048/how-may-i-select-a-smoothing-capacitor-to-mitigate-led-flicker-without-the-use-o | pwm - How may I select a smoothing capacitor to mitigate LED flicker without the use of a scope? - Electrical Engineering Stack Exchange
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How may I select a smoothing capacitor to mitigate LED flicker without the use of a scope?
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I recently replaced a set of incandescent bulbs from the interior of my old car with some very inexpensive festoon LEDs. Unfortunately the LEDs are not 100% compatible with the lighting system as dimming is accomplished via pulsing the voltage signal. The pulsing effectively dims the incandescent bulb, but causes the LEDs to flicker/blink.
I think a quick solution to mitigate or eliminate the flicker is to solder a ceramic or film capacitor in parallel with the LED power supply terminals. However, I don't have access to a scope to measure the pulse frequency so I can't calculate an ideal capacitor value to use.
The bulbs run at 12V, draw ~250mA of current -- can someone suggest a safe range of capacitor values I can experiment with? Perhaps the frequency needs to be estimated for this question to make any sense.
I may also be able to emulate the resistance of the incandescent bulb with a resistor in series, but I don't want to lose the intensity that the LEDs are currently providing.
led
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asked May 23, 2018 at 21:00
Steve GuidiSteve Guidi
393 2 2 gold badges 5 5 silver badges 17 17 bronze badges
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A lot of multimeters provide you with Hz measurements, and its likely related to mains power, but in the end your dimmer is switching on the LEDs PSU that often in a second on and off, so better get a compatible system.PlasmaHH –PlasmaHH 2018-05-23 21:02:53 +00:00 Commented May 23, 2018 at 21:02
Caps will not be a great solution due to the low rate of pulses and high peak currents. Is this a Lada?Tony Stewart EE since 1975 –Tony Stewart EE since 1975 2018-05-23 21:11:20 +00:00 Commented May 23, 2018 at 21:11
@Tony: Not that old :). A 2000 VW.Steve Guidi –Steve Guidi 2018-05-23 21:12:51 +00:00 Commented May 23, 2018 at 21:12
you could try any capacitor, as long as its voltage rating is higher that 18V or so (an automotive power system runs higher than 12V) .... use a capacitor from a scrap PC power supply if you have one jsotola –jsotola 2018-05-24 04:52:41 +00:00 Commented May 24, 2018 at 4:52
The solution is not to buy cheap poorly designed LED light bulbs.Misunderstood –Misunderstood 2018-05-30 15:27:39 +00:00 Commented May 30, 2018 at 15:27
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There's a visual difference between flicker and blink...
Flicker (if you can see it) would be something like 40-60 Hz.
Blinking would be much lower, maybe 10 Hz or less.
The difference is important, because flicker can be solved with a cap, but blinking could indicate that there is a bit of a problem that won't be solved with a cap.
For example, the car's electronics could measure the current drawn by the bulb, then decide that the current is too low (because you installed LEDs instead of incandescents) and then decide the bulb is blown, then sleep for half a second, and then restart the cycle. That would result in blinking.
Or the driver chip in your LED bulbs could be "cheap" (ie, braindead), and being driven by PWM pulses could give it a bit of a seizure.
I've had good success with sets of lights by not replacing them all with LEDs, but leaving at least one incandescent bulb. This obviously works only in the case where you got several lights in parallel, but it seems to fool the electronics into thinking that the original bulbs are present. So try this first (ie, one LED, the rest incandescent).
However if the issue is indeed that the PWM frequency is too low, the problem with a capacitor in parallel with the LED bulb is that the discharged capacitor acts like a short and will draw a high current when the bulb turns on, which may cause problems with the switching MOSFET. In this case, a RC network to filter the voltage, may be a better option, or just use a low quality cap with enough ESR to not cause a current spike.
(for example, using a 0.1R ESR cap in parallel with the lights would cause a 12V / 0.1R = 120A current spike at turn-on... but a cap with a few ohms ESR would be okay)
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answered May 23, 2018 at 21:36
bobfluxbobflux
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I had some of the bulbs removed from the circuit(s) in the car when testing, and that's when I observed the blinking (low frequency). After I fitted all the bulbs, the dimming function worked as expected in almost all of the lighting circuits. Some combination of the extra circuit resistance or logic in the lighting controller must have been satisfied and things are pretty much normal now. Only two circuits exhibit blinking now, but it is negligible: very low luminosity and one circuit is rarely used (glove box).Steve Guidi –Steve Guidi 2018-06-01 17:19:35 +00:00 Commented Jun 1, 2018 at 17:19
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It would make more sense to bypass your current incandescent bulb dimming system, as it is looking to power a much heavier load than you are proposing.
Are their enough lights on all the time to achieve a significant power savings/ reduced load on your electrical system?
I would determine if you still want to do this considering the time and probability of laying on your back in a funny position for some time, working under the dashboard to accomplish this. Plus the cost, of course.
If you still want to do it, there are some pre-made switcher (DC-DC) boards which could be purchased inexpensively, that would be a better solution.
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answered May 23, 2018 at 21:22
Sara HeartSara Heart
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3041 | https://arxiv.org/pdf/1607.06266 | A LIOUVIIE-TYPE THEOREMS FOR SOME CLASSES
OF COMPLETE RIEMANNIAN ALMOST PRODUCT MANIFOLDS AND
FOR SPECIAL MAPPINGS OF COMPLETE RIEMANNIAN MANIFOLDS
S ERGEY S TEPANOV
Abstract. In the present paper we prove Liouville-type theorems: non-existence theorems for some complete Riemannian almost product manifolds and submersions of complete Rieman-nian manifolds which generalize similar results for compact manifolds.
Keywords : complete Riemannian manifold, two complementary orthogonal distributions, submersion, non-existence theorems.
Mathematical Subject Classification 53C15; 53C20; 53C23
Introduction
S. Bochner devised an analytic technique to obtain non-existence theorems for some geometric objects on a closed (compact, boundaryless) Riemannian manifold, under some curvature assumption (see ). Currently, there are two different points of view about classical Bochner technique ; the first one uses the Green’s divergence theorem , and the second uses the Hopf’s theorem which were obtained from the Stokes’s theorem and clas-sical maximum principle for compact Riemannian manifold, respectively. In particular, a good account of applications of the Bochner technique in differential geometry of Rie-mannian almost product manifolds and submersions may be found in . We recall here that a Riemannian almost product manifold is a Riemannian manifold ( M, g ) equipped with two complementary orthogonal distributions. For instance, the total space of any submersion of an arbitrary Riemannian manifold onto another Riemannian manifold ad-mits such a structure.
In the present paper we will use a generalized Bochner technique : our proofs will be based on generalized divergence theorems and a generalized maximal principle for com-plete, noncompact Riemannian manifolds (see ). We will prove Liouville type non-
-existence theorems for some complete, noncompact Riemannian almost product mani-folds, conformal and projective diffeomorphisms and submersions of complete, noncom-
Stepanov Sergey
Department of Mathematics, Finance University under the Government of Russian Federation,
Leningradsky Prospect, 49-55, 125468 Moscow, Russian Federation
e-mail: s.e.stepanov@mail.ru
The author was supported by RBRF grant 16-01-00053-а(Russia).
pact Riemannian manifolds which generalize similar well known results for closed mani-folds.
This paper is based on our report on the conference “Differential Geometry and its Appli-cations” (July 11-15, 2016, Brno, Czech Republic).
Three global divergence theorems
Let ( M, g ) an n-dimensional oriented Riemannian manifold ( M, g ) with volume form
=gVol d ndx ... dx gdet ∧∧1 for positively oriented local coordinates . Then we can define the divergence of the vector field X via the formula
n
xx ,..., 1
Xdiv
( ) ( ) ggX Vol dXdiv Vol did = where denotes contraction with X (see [4, p. 281-283]). Furthermore, if
X
i
ω is an ( n − 1)-form on ( M, g ), then we can write gX Vol di=
ω where X
=
ωg∗ for the Hodge star operator relative to g. Thus when
ω is an ( n − 1)-form with compact support in an orientable n-dimensional Riemannian manifold ( M, g ) without boundary, the Stokes theorem d
ω = 0 follows the classic Green divergence theorem ∫М
( )∫ =
M
0gVol dXdiv
if the vector field X has compact support in a (not necessarily oriented) Riemannian manifold ( M, g ) (see [5, p. 11]). On the other hand, there are some Lp(M, g )-extensions of the classical Green divergence theorem to complete, noncompact Riemannian manifolds without boundary. Firstly, we formulate the following (see )
Theorem 1 . Let (M, g ) be geodesically complete Riemannian manifold and X be a smooth vector field on (M, g) which satisfies the conditions ( )gMLX ,1
∈ and , then where ( g,MLXdiv 1
∈ ) ( )∫ =
Mg
Vol dXdiv 0 X is a norm of the vector field X induced by the metric g .
Later, L. Karp showed in the generalized version of Theorem 1. Namely, he has pro-vided the following theorem.
Theorem 2 . Let (M, g ) be a complete, noncompact Riemannian manifold and X be a smooth vector field on (M, g ) which satisfies the condition ( ) ( )∫∞→ rBrB grVol dXrlim 21inf = 0 (1.2)
for the geodesic ball B (r) of radius r with the center at some fixed Mx ∈ . If div X has an integral (i.e. if either (div X )+ or (div X )– is integrable ) then ( )∫ =
Mg
Vol dXdiv 0 .
In particular, from the above theorem we conclude that if outside some compact set div X
is everywhere ≥ 0 (or ≤ 0) then ( )∫ =
Mg
Vol dXdiv 0 .
In conclusion, we formulate the third generalized Green’s divergence theorem (see ; ), which can be regarded, as a consequence of the above two theorems and Yau lemma from .
Theorem 3 . Let X be a smooth vector field on a connected complete, noncompact and oriented Riemannian manifold (M, g ), such that div X ≥ 0 ( or div X ≤ 0) everywhere on
(M, g ). If the norm ( gMLX ,1
∈ ), then 0=Xdiv .
The Laplace-Beltrami operator of any is defined as MCf ∞∈ ( fgrad div f g= )Δ where
grad f is the unique vector field that satisfies ( ) ( )Xdf fgrad X, g = for all vector fields X
on M. The scalar function is said to be harmonic if it satisfies MCf ∞∈ 0=Δ f . It is well known, that if is a harmonic function on any complete Riemannian mani-fold satisfying for some
MCf ∞∈
( g,MLf p
∈ ) ∞<< p1 , then f is constant (see ).
In addition, we recall that the scalar function is called subharmonic (resp. su-perharmonic ) if
MCf ∞∈
0≥Δ f (resp. 0≤Δ f ). In particular, if ( M, g ) is compact then every harmonic (subharmonic and superharmonic) function is constant by the Hopf’s theorem. On the other hand, Yau has proved in that any subharmonic function de-fined on a complete, noncompact Riemannian manifold with
MCf 2
∈
∫ ∞<
Mg
dV df is har-monic. Then based on this statement (or on the Theorem 3) we conclude that the follow-ing lemma is true.
Lemma . If (M, g ) is a complete, noncompact Riemannian manifold, then any superhar-monic function with gradient in MCf 2
∈ ( )g,ML1 is harmonic. Proof . Let ( M, g ) be a complete Riemannian manifold and f be a scalar function such that ,MCf 2
∈ 0Δ ≤f and ( )gMLfgrad ,1
∈ . If we suppose that
ϕ = − f then the above conditions can be written in the form
ϕ ∈ C2M, Δ
ϕ ≥ 0 and ( gMLgrad ,1
∈ϕ ). In this case, from the Yau statement we conclude that Δ
ϕ = 0 and hence f = −
ϕ is a harmonic function.
Liouville-type theorems for some complete Riemannian almost product manifolds
Let ( M, g ) be an n-dimensional ( )2≥п Riemannian manifold with the Levi-Civita con-nection ∇ and TM = V ⊕ H be an orthogonal decomposition of the tangent bundle TM
into vertical V and horizontal H distributions of dimensions n – m and m, respectively. We shall use the symbols V and H to denote the orthogonal projections onto V and H ,respectively. In this case we can define a Riemannian almost product manifold (see ) as the triple ( M, g, P ) for P
= V – H , where ( M, g ) is a Riemannian manifold M and P is a (1,1)-tensor field on M satisfying PP
2
= id and g (P, P ) = g. In addition, the eigenspaces of
P corresponding to the eigenvalues 1 and – 1, at each point, determine two orthogonal complementary distributions V and H.
The second fundamental form QH and the integrability tensor F H of H are define by (see [12, p. 148])
QH ( )HX HY V HY HX ∇+∇= 21 , FH ( )HX HY V HY HX ∇−∇= 21
for any smooth vector fields X and Y on M. It is well known that QH vanishes if and only if H is a totally geodesic distribution . We recall that a distribution on a Riemannian mani-fold is totally geodesic if each geodesic which is tangent to it at point remains for its en-tire length (see [12, p. 150]). On the other hand, FH vanishes if and only if H is an inte-grable distribution. A maximal connected integral manifold of H is called a leaf of the fo-liation. The collection of leafs of H is called a foliation of M. By interchanging and we define the corresponding tensor fields QV and FV for V = H ⊥.
We define now the mixed scalar curvature of ( M, g ) as the function ( )∑ ∑ = +==
manma
E,Esec s
11mix
αα
where is the sectional curvature of the mixed plane (αEEa ,sec ) }{α
π EEspan a ,= for the local orthonormal frames { } and mEE ,..., 1 { }nm EE ,..., 1+ on TM adapted to V and H, re-spectively (see ; [13, p. 23]). It is easy to see that this expression is independent of the chosen frames. With this in hand we can now state the formula which can be found in and . Namely, the following formula:
( ) =+ HVdiv
ξ
ξ
222222mix HVHVHV
FFQQs
ξξ +−−−++= (2.1)
where VgV Qtrace =
ξ and HgH Qtrace =
ξ are the mean curvature vectors of V and H ,respectively (see ; [12, p. 149]).
Assume that V and H are totally umbilical distributions , i.e. ( ) ( ) VV V,VgтпQ
ξ⊗−= −1
and (see ; [12, p. 151]). In this case the formula (2.1) can be rewrite in the form (see ; ; )
( ) HH H,HgтQ
ξ⊗= −1
( ) =+ HVdiv
ξ
ξ 2222mix 11
HVHVтттптп
FFs
ξξ −−−− +−−− (2.2)
If in addition to the assumption above we now suppose that ( M, g ) is a connected com-plete and oriented Riemannian manifold without boundary and with nonpositive mixed scalar curvature smix , then from (3.2) we obtain the inequality ( ) 0≤+ HVdiv
ξ
ξ . If at the same time, ( g,MLHV
1
∈+ξξ ) then by Theorem 4 we conclude that ( ) 0=+ HVdiv
ξ
ξ .In this case, from (3.2) we obtain 2
H
ξ 022 === NV
ξ . It means that V and H are two integrable distributions with totally geodesic integral manifolds (totally geodesic fo-liations). We fix now a point Мх ∈ and let and be the maximal integral mani-folds of distributions through x, respectively. Then by the de Rham decomposition theo-rem (see [4, p. 187]) we conclude that if ( M, g ) is a simply connected Riemannian mani-fold then it is isometric to the direct product
1
М 2М
( )2121 gg,MM ⊕× of some Riemannian manifolds and ( for the Riemannian metric and which induced by
g on and , respectively. In addition, we recall that every simply connected mani-
( )11 g,M )22 g,M 1g 2g
1
М 2Мfold M is orientable. Summarizing, we formulate the statement which generalizes a theorem on two orthogonal complete totally umbilical distributions on compact Riemannian manifold with non positive mixed scalar curvature that has been proved in ; ; and .
Theorem 4 . Let (M, g ) be a complete, noncompact and simply connected Riemannian manifold with two orthogonal complementary totally umbilical distributions V and H
such that their mean curvature vectors V
ξ and H
ξ satisfy the condition
( g,MLHV
1
∈+ξξ ). If the mixed scalar curvature of (M, g ) is nonpositive then V
and H are integrable and (M, g ) is isometric to a direct product ( )
mix
s
2121
gg,MM ⊕× of some Riemannian manifolds ( ) and 21 g,M ( )21 g,M such that integral manifolds of V and
H correspond to the canonical foliations of the product 21 MM × .
We consider now an ( n − 1)-dimensional totally geodesic distribution H on ( M, g ). In this case the formula (2.2) can be rewrite in the form (see )
=Vdiv
ξ 2mix HFs − (2.2)
where is the vertical Ricci curvature for an orthonormal frame at a point
( ) ( 1121mix e,eRic e,esec sm
== ∑=αα )}{ ne,..., e1 Mx ∈ such that H x = { }1span e and V x = { }ne,..., e2span .Hence, an immediate consequence of (2.2) and Theorem 4 is following
Corollary 1 . Let (M, g ) be an n-dimensional complete noncompact and simply connected Riemannian manifold with (n − 1)-dimensional totally geodesic horizontal distribution H.
If the vertical Ricci curvature of (M, g ) is nonpositive and ( g,MLV
1
∈ξ ))
for the mean curvature vector of V, then H is integrable and (M, g ) is isometric to a direct product
of Riemannian manifolds ( 2121 gg,MM ⊕× ( )21 g,M and such that and integral manifolds of V and H correspond to the canonical foliations of the product .
( 22 g,M )
1dim 1 =M
21
MM ×
The integral formula (2.1) can be reformulated as follows (see )
222mix 21
422 HVVH FFPsdiv div −−∇+=+ξξ HV . (2.3) where and . If V and H are inte-grable distributions then (2.3) can be rewrite in the form
( )∑=∇=
maaHEH
E,gdiv a
1
ξξV (∑+=∇=
nmVEV
E,gdiv
1
αα
ξξαH )
2mix 21
422 Psdiv div VH ∇+=+ξξ HV . (2.4)
Suppose now that all integral manifolds of the vertical distribution V are minimal sub-manifolds of the Riemannian manifold ( M, g ) and . Then from (2.4) we obtain 0mix ≥s
2mix 41
2 Psdiv H ∇+=ξV ≥ 0. (2.5)
If at least one connected complete and oriented maximal integral manifold M
′ of V exists. We assume that M
′ equipped with the Riemannian metric g′ inherited from ( M, g) such that ( g,MLH ′′∈ 1
ξ ) for the mean curvature vector Н
ξ of H = V ⊥ which belongs to
М ′ at each point x ∈ M
′. Then by applying Theorem 3 to Н
ξ , from (2.5) we get
0=′ Hgdiv
ξ . Therefore, if all integral manifolds of V are connected complete and ori-ented minimal submanifolds of the Riemannian manifold ( M, g ) and Н
ξ is a L1-vector field for every of them, then 0=∇Р . In this case V and H are two integrable distributions with totally geodesic integral manifolds (totally geodesic foliations) on ( M, g ) (see ). If at the same time, ( M, g ) is complete, noncompact and simply connected Riemannian manifold then by the de Rham decomposition theorem (see [4, p. 187]) it is isometric to the direct product ( )2121 gg,MM ⊕× of some Riemannian manifolds ( )11 g,M and for the Riemannian metric and which induced by g on and , re-spectively. Summarizing, we formulate the statement which
( 22 g,M ))
1
g 2g 1М 2М
generalizes the main theorem of .
Theorem 5. Let (M, g ) be complete, noncompact and simply connected Riemannian manifold. If the following three conditions are satisfied:
1) (M, g ) admits an integrable distribution V such that an arbitrary integral manifold
of V which equipped with the Riemannian metric inherited from (M, g) is a connected complete and oriented minimal submanifold of (M, g);
( g,M ′′ g′2) the orthogonal complementary distribution H = V ⊥ is also integrable and its mean curvature vectors H
ξ satisfies the condition ( )g,MLH ′′∈ 1
ξ ;
3) the mixed scalar curvature , then (M, g ) is isometric to a direct product
of some Riemannian manifolds
0mix ≥s
( 2121 gg,MM ⊕× ) ( )21 g,M and such that integral manifolds of V and H correspond to the canonical foliations of the product .
( 21 g,M )
21
MM ×
Remark 1. If in addition, at least one closed integral manifold M
′ of V exists , then, by applying the classic Green divergence theorem to H
ξ , from (2.5) we get
( ) 08 2mix =∇+ ′′∫ g
М
dVol Ps
where is the volume form of gdVol ′ ( )g,M ′′ . If ( )g,M ′′ is non-oriented we can consider its orientable double cover. In this case the inequality is a condition of nonexis-tence of two orthogonal complementary foliations one of which consists of minimal sub-manifolds.
0mix >s
Applications to the theory of projective mappings of Riemannian manifolds
We recall here the definition of pregeodesic and geodesic curves . Namely, a pregeodesic
curve is a smooth curve
γ : t ∈ J ⊂ ℝ → ( )t
γ ∈ M on a Riemannian manifold ( M, g ), which becomes a geodesic curve after a change of parameter. Let us change the parame-ter along
γ so that t becomes an affine parameter . Then 0=∇ ХХ for dt dХ
γ= , and
γ
is called a geodesic curve . By analyzing of the last equation, one can conclude that either γ is an immersion, i.e., 0≠dt d
γ for all t ∈ J, or ( )t
γ is a point of M.
Let ( M, g ) and ( gM , ) be Riemannian manifolds of dimension n ≥ 2. Then a smooth map
( ) ( gMgMf ,,: → ) of Riemannian manifolds is a projective map if ( )
γf is a pre-geodesic in ( gM , ) for an arbitrary pregeodesic
γ in ( M, g ) (see ). In particular, if a projective map ( ) ( g,Mg,Mf →: ) is called totally geodesic if it maps linearly pa-rametrized geodesics of ( M, g ) to linearly parametrized geodesics of ( g,M ). An equiva-lent definition is that f is connection-preserving, or affine . The global structure of these maps is investigates in the paper .
For a projective diffeomorphism ( ) ( )g,Mg,Mf →: we have (see [21, p. 135])
( )( )ψψψ dddnRic Ric ⊗−∇−+= 1 (3.1)
where Ric and Ric denote the Ricci tensors of ( M, g ) and ( )g,M , respectively, and
( ) С
gdet gn +⎟⎟⎠⎞⎜⎜⎝⎛+= det log 121
ψ (3.2)
for some constant C. Now we can formulate the following
Theorem 6 . Let (M, g ) be a connected complete, noncompact Riemannian manifold and
( ) ( g,Mg,Mf →: ) be a projective diffeomorphism onto another Riemannian manifold
( g,M ) such that sRic trace g ≥ for the Ricci tensor Ric of ( )g,M and the scalar curva-ture s of (M, g ). If the gradient of the function ⎟⎟⎠⎞⎜⎜⎝⎛
gg
det det log has integrable norm on (M, g )
then f is affine map.
Proof. We conclude immediately from (3.1) that
( ) 211
ψψ grad sRic trace n g +−
−
=Δ (3.3)
Let sRic trace g ≥ then (3.3) shows 0≥Δ
ψ . If ( M, g ) is a complete, noncompact Rie-mannian manifold and ( )g,MLgrad 1
∈ψ then by the Yau statement (see [10, p. 660]) we conclude that 0=Δ
ψ and
ψ must be harmonic on ( M, g ). At the same time, we see from (3.3) that
ψ is constant. Then according to the formula (40.8) from [21, p. 133] we obtain 0=∇ g . Hence by , f is affine map.
Let ( M, g ) and ( gM , ) be Riemannian manifolds of dimension n and m such that n > m. A surjective map ( ) ( gMgMf ,,: → ) is a submersion if it has maximal rank m at any point
x of M, that is, each differential map of f is surjective, hence, for xf∗ Му ∈ . In this case, for an arbitrary ( )yf 1− Му ∈ is an ( n – m )-dimensional closed submanifold М ′ of (M, g) (see [22, p.11]). We call the submanifolds ( )yf 1− fibers .Putting Vx , for any ( )хfKer ∗= Mx ∈ , we obtain an integrable vertical distribution V
which corresponds to the foliation of M determined by the fibres of f, since each Vx
coincides with tangent space of ( )yfTx
1−
= ( )уf 1− at x for ( ) yxf = .
Let H be the complementary distribution of V determined by the Riemannian metric g,
i.e. H x = V x⊥ at each x ∈ M. So, at any Mx ∈ , one has the orthogonal decomposition
Tx(M) = Vx ⊕ Hx where Hx is called the horizontal space at x . Thus we have defined a Riemannian almost product structure on ( M, g ).
Consider now an n-dimensional simple connected complete Riemannian manifold ( M, g ), and suppose that a projective submersion ( ) ( )gMgMf ,,: → onto an m-dimensional (m < n) Riemannian manifold ( gМ , ) exists. Then each pregeodesic line М⊂
γ which is an integral curve of the distribution is mapped into a point ∗fKer ( )
γf in M . Note that this fact does not contradict the definition of projective submersion.
In addition, we have proved in and that ( M, g ) is isometric to a twisted product
( )22121 2 geg,MM
α
+× of some Riemannian manifolds ( )21 g,M and , and for smooth function
( 22 g,M )
→× 212 : ММ
α ℝ such that all fibres of submersion and their orthogo-nal complements correspond to the canonical foliations of (see and ). In this case, the following corollary of Theorem 4 is true.
21
MM ×
Corollary 2 . Let (M, g ) be an n-dimensional complete, noncompact and simply connected Riemannian manifold and ( ) ( )gMgMf ,,: → be a projective submersion onto another m-dimensional (m < n ) Riemannian manifold ( )gM , such that the mean curvature vector
H
ξ of the horizontal distribution ( )⊥
∗
fKer satisfies the condition ( g,MLH
1
∈ξ ). If the mixed scalar curvature of (M, g ) is nonpositive then mix s ( )⊥∗fKer is integrable and (M, g) is isometric to a direct product ( )2121 gg,MM ⊕× of some Riemannian manifolds and ( such that integral manifolds of and correspond to the canonical foliations of the product
( )21 g,M ) )21 g,M ∗fKer ( ⊥∗fKer
21
MM × .
Moreover, we have proved in that if a simple connected complete n-dimensional Riemannian manifold ( M, g ) has a nonnegative sectional curvature and admits a projec-tive submersion onto another m-dimensional ( m < n) Riemannian manifold ( )gM , , then (M, g ) is isometric to a direct product ( )2121 gg,MM ⊕× of some Riemannian manifolds and ( such that the integral manifolds of and ( )21 g,M ) )22 g,M ∗fKer ( ⊥
∗
fKer corre-spond to the canonical foliations of the product 21 MM × . We can formulate now a state-ment which will supplement this theorem. The statement is a corollary of Theorem 1 and Theorem 6.
Corollary 3 . Let (M, g ) be an n-dimensional complete, noncompact and simply connected Riemannian manifold and ( ) ( )gMgMf ,,: → be a projective submersion onto another m-dimensional (m < n ) Riemannian manifold ( )gM , with connected fibres. If the mixed scalar curvature then (M, g ) is isometric to a direct product ( )0mix ≥s 2121 gg,MM ⊕×
of some Riemannian manifolds and ( )21 g,M ( )22 g,M such that the integral manifolds of
and ∗fKer ( )⊥
∗
fKer correspond to the canonical foliations of the product 21 MM × .
Proof. Let ( M, g ) be an n-dimensional complete, noncompact and simply connected Rie-mannian manifold and ( ) ( )gMgMf ,,: → be a projective submersion onto another m-dimensional ( m < n ) Riemannian manifold ( )gM , with connected fibres . It follows from the above, the fibre for an arbitrary ( )yf 1− Му ∈ is an a ( n – m )-dimensional closed connected submanifold М ′ of ( M, g) equipped with the Riemannian metric g′ inherited from ( M, g). The mean curvature vector Н
ξ of H = ( )⊥
∗
fKer belongs to МТ х ′ at each point x ∈ M
′ then, by applying the classic Green divergence theorem 0 to ( ) =∫ ′ ′
МgH
Vol ddiv
ξV H
ξ , from (2.4) we get the following equation
( ) g
М
Vol dPs ′′∫ ∇+ 2mix 412 = 0 (3.4)
where is the volume form of gdVol ′ ( )g,M ′′ . If ( )g,M ′′ is non-oriented we can consider its orientable double cover. If the mixed scalar curvature then from (3.4) we ob-tain that
0mix ≥s
0=∇ Р at each point of M
′. At the same time, we recall that M
′ is an arbitrary fibre of the projective submersion ( ) ( )gMgMf ,,: → . Therefore, and ∗fKer ( )⊥
∗
fKer
are two integrable distributions with totally geodesic integral manifolds (totally geodesic foliations) on the complete, noncompact and simply connected Riemannian manifold ( M, g). Then by the well known de Rham decomposition theorem it is isometric to the direct product of some Riemannian manifolds ( 2121 gg,MM ⊕× ) ( )11 g,M and ( )22 g,M for the Riemannian metric and which induced by g on and , respectively. The proof of our corollary is complete.
1
g 2g 1М 2М
Using the equality (3.4) once again we get the following
Corollary 4. Let (M, g) be an n-dimensional Riemannian manifold and
( ) ( gMgMf ,,: → ) be a submersion onto another m-dimensional (m < n ) Riemannian manifold ( gM , ) with connected fibres. If the mixed scalar curvature then 0mix >s
( ) ( gMgMf ,,: → ) is not a projective submersion.
Proof . For the case we can rewrite (3.4) in the form 0mix >s
( ) g
М
Vol dPs ′′∫ ∇+ 2mix 412 > 0.
The contradiction just obtained with the classic Green divergence theorem completes the proof of our corollary.
Applications to the theory of conformal mappings of Riemannian manifolds
Let ( M, g ) and ( gM , ) be Riemannian manifolds of dimension n ≥ 2. Then a diffeomor-phism ( ) ( )g,Mg,Mf →: is called conformal if it preserves angles between any pair curves. In this case, geg
σ2
= for some scalar function
σ . In particular, if the function
σ
is constant then f is a homothetic mapping .
If MC 2
∈
σ then (see [21, p. 90])
( ) ( )( ) 22 2112
σσσ grad nnnsse −−−Δ−−= (4.1)
where s denote the scalar curvature ( )g,M . Now we can formulate the following
Theorem 6 . Let (M, g ) be an n-dimensional (n ≥ 3) complete, noncompact Riemannian manifold and ( ) ( g,Mg,Mf →: ) be a conformal diffeomorphism onto another Rieman-nian manifold ( g,M ) such that geg
σ2
= and ses
σ2−
≥ for some function MC 2
∈
σand the scalar curvatures s and s of (M, g) and ( g,M ), respectively. If
( g,MLgrad 1
∈σ ), then f is a homothetic mapping.
Proof . If ( ) ( g,Mg,Mf →: ) is a conformal diffeomorphism a connected complete non-compact and oriented Riemannian manifold (M, g ) onto another Riemannian manifold
( g,M ) such that geg
σ2
= for some function MC 2
∈
σ , then from (4.1) we obtain
( ) ( )( ) 22 2112
σσσ grad nnsesn −−−−=Δ− . (4.2)
Let ses
σ2
≤ then (2) shows 0≤Δ
σ . It means that
σ is a superharmonic function. By the condition of our theorem, the gradient of
σ has integrable norm on ( M, g ) and we obtain from (4.2) that 0=Δ
σ and
σ must be harmonic (see our Lemma). Since n ≥ 3, we see from (4.2) that
σ is constant. The proof of the theorem is complete.
Let ( M, g ) and ( gM , ) be Riemannian manifolds of dimension n and m for n > m.A submersion ( ) ( gMgMf ,,: → ))
is called a horizontal conformal if restricted to the horizontal distribution H = ( is conformal mapping.
∗
f
⊥∗
fKer
Next, we consider a horizontal conformal submersion ( ) ( gMgMf ,,: → ) for the case m< n. We note here that horizontal conformal mappings were introduced by Ishihara . From the above discussion, one can conclude that the notion of horizontally conformal mappings is a generalization of concept of Riemannian submersions. In addition, we note that a natural projection onto any factor of a double-twisted product
( )22212121 gg,MM
λλ +× of any Riemannian manifolds ( )aa g,M and smooth positive func-tions →× 21 MM:a
λ ℝ for an arbitrary a = 1, 2 is horizontal conformal submersion with umbilical fibres (see ).
Let ( ) ( gMgMf ,,: → ) be a horizontal conformal submersion and be an umbilical distribution then (2.2) can be rewrite in the form
∗
fKer
( ) =+ HVdiv
ξ
ξ 222mix 11
HVHтттптп
Fs
ξξ −−−− +−− . (4.3)
In this case, we can formulate a corollary of Theorem 4 which generalizes our theorem on the horizontal conformal submersions of compact Riemannian manifolds with non positive mixed scalar curvature that has been proved in (see also ). Corollary 5 . Let (M, g ) be an n-dimensional complete, noncompact and simply connected Riemannian manifold and ( ) ( )gMgMf ,,: → be a horizontal conformal submersion with umbilical fibres onto another m-dimensional (m < n ) Riemannian manifold ( )gM , .If the mean curvature vector V
ξ of and the mean curvature vector ∗fKer H
ξ of satisfy the condition ( ⊥∗fKer ) ( )g,MLHV
1
∈+ξξ and the mixed scalar curvature
of (M, g ) is nonpositive then
mix
s
( )⊥
∗
fKer is integrable and (M, g ) is isometric to a direct product of some Riemannian manifolds ( 2121 gg,MM ⊕× ) ( )21 g,M and ( such that integral manifolds of and
)21 g,M
∗
fKer ( )⊥
∗
fKer correspond to the canonical foliations of the product .21 MM ×
Applications to the theory of Riemannian submersions
A submersion ( ) ( gMgMf ,,: → ) is called Riemannian submersion if ( preserves the length of the horizontal vectors at each point
) хf∗
Mx ∈ (see [13, p. 3] ). In this case, the horizontal distribution H = ( is totally geodesic (see ). In the paper and in the monograph [13, p. 235] was proved the following theorem. Let
)⊥∗fKer
( ) ( gMgMf ,,: → ) be a Riemannian submersion with totally umbilical fibres. If ( M, g )is a closed and orientable manifold with nonpositive mixed sectional curvature (i.e. sec (X,Y ) ≤ 0 for every horizontal vector field X and for every vertical vector field Y), then all fibres are totally geodesic and horizontal distribution H = ( )⊥∗fKer is integrable, and the mixed sectional curvature is equals to zero. We present a generalization of this theorem. The following result is deduced immediately from Corollary 6.
Corollary 6 . Let (M, g ) be an n-dimensional complete, noncompact and simply connected Riemannian manifold and ( gM , ) be another m-dimensional (m < n ) Riemannian mani-fold and ( ) ( gMgMf ,,: → ) be a Riemannian submersion with totally umbilical fibres . If the mixed scalar curvature is nonpositive and the mean curvature vector mix s V
ξ of fibres satisfies the condition ( g,MLV
1
∈ξ ) ))
, then the horizontal distribution ( is inte-grable and the Riemannian manifold (M, g) is isometric to a direct product
of some Riemannian manifolds
⊥∗
fKer
( 2121 gg,MM ⊕× ( )21 g,M and ( such that the )22 g,Mintegral manifolds of and ∗fKer ( )⊥
∗
fKer correspond to the canonical foliations of the product .21 MM ×
We know from that there are no Riemannian submersions from closed Riemannian manifolds with positive Ricci curvature to Riemannian manifolds with nonpositive Ricci curvature. The following statement is a direct consequence of Corollary 3 and comple-ments the above vanishing theorem.
Corollary 7 . Let (M, g ) be an n-dimensional complete, noncompact and simply connected Riemannian manifold and ( ) ( )gMgMf ,,: → be a Riemannian submersion onto an (n –
1) -dimensional Riemannian manifold ( )gM , . If the vertical Ricci curvature of (M, g ) is nonpositive and the mean curvature vector V
ξ of fibres satisfies the condition
( g,MLV
1
∈ξ ), then the horizontal distribution ( )⊥∗fKer is integrable and the Rieman-nian manifold (M, g ) is isometric to direct product ( )2121 gg,MM ⊕× of some Rieman-nian manifolds ( ) and ( such that 21 g,M )22 g,M 11 =Mdim and the integral manifolds of and ( correspond to the canonical foliations of the product ∗fKer )⊥∗fKer 21 MM × .
Applications to the theory of harmonic submersions of Riemannian manifolds
A smooth mapping ( ) ( )gMgMf ,,: → is said to be harmonic if f provides an extremum of the energy functional ( ) ∫Ω∗Ω = gdffE Vol 2 for each relatively closed open subset
M⊂Ω with respect to the variations of f that are compactly supported in Ω .
If ( M, g ) is an n-dimensional Riemannian manifold and ( ) ( )gMgMf ,,: → is a harmonic submersion onto another m-dimensional (m < n ) Riemannian manifold ( )gM ,
then each its fibre ( M
′, g
′) is an ( n – m )-dimensional closed imbedded minimal submani-fold of ( M, g ) (see ). Then from the above arguments and Theorem 4 we conclude that the following corollary is true.
Corollary 9 . Let (M, g ) be an n-dimensional complete, noncompact and simply connected Riemannian manifold and ( ) ( )gMgMf ,,: → be a harmonic submersion onto another m-dimensional (m < n ) Riemannian manifold ( )gM , with connected fibres. If the hori-zontal distribution ( )⊥∗fKer is integrable and the mixed scalar curvature is non-negative, then (M, g ) is isometric to a direct product
mix
s
( )2121 gg,MM ⊕× of some Rie-mannian manifolds ( )21 g,M and ( )22 g,M such that the integral manifolds of and
correspond to the canonical foliations of the product .
∗
fKer
( ⊥∗fKer ) 21 MM ×
Using the Remark 1 we get the following
Corollary 8. Let (M, g) be an n-dimensional Riemannian manifold and
( ) ( gMgMf ,,: → ) be a submersion onto another m-dimensional (m < n ) Riemannian manifold ( gM , ) )with connected fibres. If the horizontal distribution ( is inte-grable and the mixed scalar curvature is positive, then
⊥∗
fKer
mix
s ( ) ( gMgMf ,,: → ) is not harmonic.
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3042 | https://www.youtube.com/watch?v=0nN3H7w2LnI | Art of Problem Solving: Simon's Favorite Factoring Trick
Art of Problem Solving
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Art of Problem Solving: Simon's Favorite Factoring Trick with English Subtitles Created By Philjae Chang, KGSEA MATH CIRCLE Now you know I like to show off some of my best stuff in these videos. But in this one, I'm gonna show you Simon's favorite factoring trick, which I learned from Simon himself. You don't, you don't know Simon? Ah you're missing out. Simon is wicked smart, plays a mean cello, just got his Ph.D from Stanford, which means he probably knows more math than I do.
Simon is wicked smart, plays a mean cello, just got his Ph.D from Stanford, which means he probably knows more math than I do. Shhh... What? Oh yeah, you are asking why don't we have Simon doing this video. I'll tell you why, cause Simon's thousand miles away. That's what I call job security. Keeping Simon thousand miles away. Now, I'm gonna teach you his favorite factoring trick with this problem right here. We are trying to find all pairs of positive integers that satisfy this equation. And this bit about, trying to only find integers that satisfy this equation, very important, even has a special name. We call equations like this Diophantine equations. This is the big fancy word that says we're only looking for integer solutions. Now, let's see here, what do you gonna use to solve this? Of course, we're gonna use Simon's favorite factoring trick. We're gonna try to factor. Now the bit about integers is a clue that we're gonna try to factor. We'll see a little bit later why. But of course, we're doing a video on Simon's favorite factoring trick, we're not gonna break out Simon's favorite guessing and checking trick. Cause guessing and checking on this is gonna be a real pain. Where can we factor... Well, I can't factor the whole left side, but I can factor an "m" out of those first two terms. I have m times (n+3) minus 8n equals 59. And now looks like we're stuck, can't factor anything else right here. So now I'm gonna break out one of my favorite strategies, which is wishful thinking. Now, I would like to be able to factor more over here. And if this n were to magically turn into an (n+3), then I could factor out (n+3) from the two terms. So I wish, I wish really hard to turn that into an (n+3), and I see how to make my wishes come true. I can get an (n+3) out there, if just I have a -8 times 3 to factor the -8 out of. Now if I subtract 8 times 3 on the left side, can't play favorites, gotta subract it on the right side as well. And now over here, I can factor out that -8. Check out what we have, we factor out the -8, and we've made our wishes come true. We've got that -8 times (n+3). Over here, 39 minus, "59" minus 24 is 35. And now, we can factor again. And this, this step right here, where we tapped on that little -8 times 3. That's Simon's favorite factoring trick. Throwing a lean, little extra term there on both sides to let us factor the left hand side. Simon is wicked smart. So now we have, this(m-8) times this(n+3) equals 35. Now here's where the bit about integers, this "diophantine" bit is so important. Because m and n are integers, (m-8) and (n+3) are integers. Now there are only so many pairs of integers that multiply together to give us 35. And here, we're restricting even more, we want positive integers. So if m and n are positive where we know at least n+3 has to be positive. So we need m-8 to be positive as well to get a positive 35. So we need both of these to be positive. So now we just have to think about the ways, in which we can write 35 as a product two positive integers. Well, there aren't a whole lot of those. There's 1 times 35. There's 5 times 7. 7 times 5. And 35 times 1. Now we'll look at each one of these possibilities and we'll figure out what values of m and n go with each. So for the first one here, 1 and 35. Well if m-8 is 1, that means m is 9. n+3 is 35, that tells us n is 32. Come down to here. m-8 is 5, that gives us 13 for m. And then, n+3 is 7, that gives us 4 for n. And then on to this one. m-8 is 7, that gives us 15 for m. n+3 is 5, that gives us just 2 for n. And finally, this last one, m-8 is 35, that gives us 43 for m. But n+3 equals 1, that makes n=-2. Read the question, positive integers. So here are the three possibilities, you would've had a hard time trying to find these just by guessing and checking. Checking, checking is a good idea. Let's check these real quick. Stick in (9,32) in there. 9 times 32. Well, we're gonna have -8 times 32 over there, when we put n=32 in there. So 9 times 32 minus 8 times 32, that's just gonna leave 32, plus 3 times 9 is 27. 32 + 27 = 59 Try this one, 13 times 4 gives us 52. Plus 3 times 13 is 39. 52 plus 39 gives us 91, minus 8 times 4. 91 - 32 = 59 Put in these two, (15,2). 15 times 2 is 30. Plus 3 times 15, that's another 45. That brings us up to 75. 75 minus 8 times 2. 75 - 16 =59. All three work. These are correct. Now, when I'm breaking out Simon's favorite factoring trick, I don't always go through all these steps. I jump straight into my wishful thinking. Here's what it looks like. I'll look at this problem and I see a product of variables, and I see each of variables in a linear term. And I think of Simon, makes Simon happy, too. But, I think of Simon when I see this and I think, okay I'm gonna try to make that left hand side factored. And when I get it factored, it's gonna look something like this. Two factors equal some number. And, we're gonna have "m" over here. And an "n" over here. Cause when I multiply these two out, these are both gonna be binomials. So I'm gonna get m times n, that's gonna give me that mn right there. And I need a plus 3m right there. So need to multiply this m times a (+)3. Plus 3. There we go. And I need a minus 8n that come out when we multiply this out. So I need a -8 right here. -8 times n, that's gonna give me the -8n. Course when I multiply all this stuff out, I'm also gonna get this -8 times a (+)3. That's gonna giv me a -24. I don't have a -24 up there. So I'll put one in. And that's what gets me to think, subtract 24 from both sides. Now, this has taken our steps little bit out of order, but we get to the same equation we have before. And now, we can solve the problem in the same ways as before. So, moral of the story here is, you see integers, you're trying to solve a diophantine equation, think factoring. And when you see product of variables, and an each of those variables in a linear term, you think of Simon. Art of Problem Solving: Simon's Favorite Factoring Trick with English Subtitles Created By Philjae Chang, KGSEA MATH CIRCLE |
3043 | https://www.cuemath.com/questions/what-is-8-as-a-decimal/ | What is 8% as a decimal?
A decimalnumber can be defined as a number whose whole number part and the fractional part is separated by a decimal point.
Answer: 8% as a decimal is 0.08
Let us proceed step by step to determine the value of 8% as a decimal.
Explanation:
In order to find out 8 percent as a decimal, we divide 8 by 100.
We can perform the following steps to determine the value of 8% as a decimal.
Step 1: Change the given percentage into a fraction, so we can write 8% as 8 / 100.
Step 2: After that change the obtained fraction into a decimal by dividing the numerator by the denominator.
Hence, 8 ÷ 100 = 0.08.
You can also verify the result by using Cuemath's percent to decimal calculator.
Therefore, 8% as a decimal is 0.08 |
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Reproductive Fitness
In subject area:Agricultural and Biological Sciences
Reproductive fitness is defined as an individual's success in passing on genes to the next generation, which can be enhanced through the allocation of resources to either mating effort or parenting effort.
AI generated definition based on: Neuroscience & Biobehavioral Reviews, 2019
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Chapter
Attributes of successful biopesticides
2022, Plant Nematode BiopesticidesAnwar L. Bilgrami, Anish Khan
High reproductive potentials
Reproductive success is a measure of fitness, which determines the ability of organisms to carry their progeny. There is a significant relationship between fecundity and fertility as has been explained by Bradshaw and McMahon (2008). Subsequently, the presence of too many organisms overcrowds and depletes resources such as space, breeding sites, food, etc. Such a condition appears to be unprofitable for individuals wasting their energy associated with fertility. In some plant nematode biopesticidal organisms such as predatory mononchs and, to some extent, in predatory dorylaims, overcrowding of conspecific species reduced fertility and fecundity. This is, however, not the case with other plant nematode biopesticidal agents, e.g., diplogasterid predators, which are more fertile and reproduce a large number of offspring without any shortcomings. Biphasic mode (switching over to bacteria in absence of prey nematodes) of feeding is the reason why density does not compromise their survival and resources.
One of the beneficial traits of the biopesticidal organism is high reproductive potential to keep a high population threshold in the field, but there must be a judicious balance between reproduction and fertility. How the fecundity of biological control agents contributes to their success as plant nematode biopesticides must be seen in the light of their biology, feeding, and available resources. Fecundity is a reproductive capacity that a female individual can have over its lifetime, whereas fertility refers to its current reproductive performance. Both fecundity and fertility are important from a biological control point of view. While selecting a biological control agent for nematode management, factors such as fecundity, fertility, age, body size, population density, mate choice, and environmental variability must always be considered. Also, the plant nematode biopesticidal agents must show reproductive fitness in terms of fecundity and survival. Reproductive fitness can be improved by selecting preferred food, environment, ecology, and soil type.
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Book2022, Plant Nematode BiopesticidesAnwar L. Bilgrami, Anish Khan
Chapter
Weed Population Genetics
2018, Fundamentals of Weed Science (Fifth Edition)Michael J. Christoffers
5.1Fitness and Components of Fitness
Fitness is the ability of a plant to pass its alleles on to progeny. In population genetics, fitness is the main phenotypic characteristic of interest because it represents the final and ultimate measure of a plant's reproductive success. In understanding how and why weed populations evolve owing to phenotypic interactions with the environment, fitness is almost always a main player. A plant's level of fitness is determined by various genetic and environmental factors throughout its life cycle; three of the main components of fitness are survival, fecundity, and mating ability. Survival is the ability of a plant to reach reproductive maturity, and fecundity is its capacity to produce gametes. Mating ability is the likelihood that a plant's gametes will be involved in a fertilization event that results in offspring.
Directly determining weed fitness is often not a major goal of weed science research. For example, weed control measures are usually evaluated by their effectiveness in reducing weed populations or weed competition with crops. Herbicide-resistant weeds are often evaluated based on biomass after treatment compared with wild-type, susceptible plants. Although survival is indeed a component of fitness, and competitiveness or biomass may be related to reproductive ability, these measures are not complete and direct estimates of fitness. However, various studies have identified cases of weed genetic variation with clear implications for fitness. For example, the psbA gene codes for the D1 protein of PSII, and a missense mutation substituting glycine for serine at amino acid 264 of D1 is known to confer resistance to triazines and some other PSII-inhibiting herbicides in various weeds. However, this mutation also causes reduced fitness compared with wild-type plants grown in the absence of herbicide (reviewed by Holt and Thill, 1994). This pleiotropic reduction in fitness (fitness cost) is likely the result of reduced PSII efficiency owing to the mutated D1 protein. Triazine-resistant weeds also sometimes show reduced fitness compared with wild-type plants when treated with alternative herbicides, such as bentazon (Gadamski et al., 2000). There is also increasing evidence that other forms of herbicide resistance often, but not always, have a fitness cost (reviewed by Vila-Aiub et al., 2009).
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Book2018, Fundamentals of Weed Science (Fifth Edition)Michael J. Christoffers
Chapter
Plant Physiology
2008, Encyclopedia of EcologyU. Lüttge
General Importance of Physiological Processes for Fitness at the Community Level
Fitness is frequently defined and quantified as the number of seeds produced (‘Darwinian fitness’). However, this may cause difficulties, because it is too limited a view. This is readily recognized when we consider, for example, K- and r-strategies of plants with small and high seed numbers, respectively, or clonal growth, where particular plant species may dominate entire ecosystems without any generative propagation. Germination of seeds and establishment and survival of seedlings are essential. Thus, we see that the entire complement of physiological processes contributes to fitness, for example, in addition to the capacity of photosynthesis treated above in some detail as a special case story, and similarly important functions such as dormancy of seeds with germination at the ecologically appropriate time and the physiology of development, competition, defense, and mutualism as alluded to above.
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Reference work2008, Encyclopedia of EcologyU. Lüttge
Review article
Plant science and agricultural productivity: Why are we hitting the yield ceiling?
2013, Plant ScienceStève de Bossoreille de Ribou, ... Ioan Negrutiu
2.3Female reproductive fitness
Female reproductive fitness has historically been considered a major yield trait, with seed size and the number of seeds as the two quantitative trait components of female productivity in plants. Artificial selection and cropping constraints are expected to decrease the variability of seed size. A few large-effect QTLs are accounting for more than 20% of the variance in seed size in several crops and refs therein). Seed number is a plastic trait responding more readily to changes in the availability of resources. A trade-off between seed number and seed size can occur depending on resources (also see Section 6.3).
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Journal2013, Plant ScienceStève de Bossoreille de Ribou, ... Ioan Negrutiu
Chapter
Human Reproductive Ecology
2015, Basics in Human EvolutionClaudia R. Valeggia, Alejandra Núñez-de la Mora
Challenges and Future Directions of the Field of Human Reproductive Ecology
The main goal of HRE has been to understand how human reproduction responds to different environments and the consequences of variation in response in terms of health, reproductive success, and ultimately, reproductive fitness—in other words, to establish whether the flexibility and responsiveness of human reproductive function are adaptive.
As a collective, the findings of the past two decades provide a very persuasive argument in favor of the notion that much observed phenotypic variation in human reproductive function can be conceptualized as the product of natural selection rather than of developmental constraints or pathology (Ellison and Jasienska, 2007). One of the main challenges for future research is to demonstrate an actual relative advantage of such variation in terms of reproductive fitness using real data from contemporary populations. Some existing long-term epidemiological and clinical studies that include biomarker measurements provide an extraordinary opportunity for this (Stearns et al., 2010b).
Another aspect of reproductive variation that is instrumental to understanding the underlying evolutionary processes is its genetic basis. However, reproductive ecologists have so far done comparatively little work in this respect (Jasienska et al., 2006a; Iversen et al., 2012); much still needs to be learned about genetic variation, genotype–phenotype, and gene–environment interactions in relation to variation in reproductive function (Jasienska, 2008), a fact that identifies this area as ripe for future research. In this context, it is becoming increasingly clear that early developmental ecologies, likely acting through epigenetic processes, may program the reproductive trajectories followed later in life (Nuñez-de la Mora et al., 2007; Nuñez-De La Mora et al., 2008).
The trade-off concept, that individual differential allocation of finite energy among different life functions underlies variation in life history strategies, is central to the theoretical framework guiding much of HRE research. At a population level, trade-offs between reproductive effort and some other life history traits have been well described (Lummaa, 2010). At an individual level, however, the demonstration of trade-offs between physiological vital functions has been less straightforward (Clancy et al., 2013). The advent of new reliable methods for the simultaneous quantification of biomarkers associated with different vital functions (O’Connor et al., 2003; McDade et al., 2007; Valeggia, 2007; Salvante et al., 2012) is likely to remove some existing methodological constraints as well as boost research on trade-offs and on the dynamics of interactions between life functions. The development of suitable statistical techniques to manage the resulting data will need to follow suit (Stearns et al., 2010a).
Finally, perhaps one of the greatest challenges for HRE and many other disciplines relates to the biological and social consequences of the unprecedented change that most populations in the contemporary world are experiencing. The rapid economic, ecological, and cultural changes are generating new parameters and selective pressures, and as a result different trade-offs that can already be quantified as shifts in some human life history variables (Kirk et al., 2001; Byars et al., 2010; Courtiol et al., 2013). Understanding the variations in physiology, behavior, and life history patterns as humans experience these new biocultural environments is likely to engage human reproductive ecologists for decades to come.
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Book2015, Basics in Human EvolutionClaudia R. Valeggia, Alejandra Núñez-de la Mora
Chapter
Breeding Oilseed Crops for Climate Change
2016, Breeding Oilseed Crops for Sustainable ProductionAbdullah A. Jaradat
Reproductive Fitness in Oilseed Crops under Climate Change
Biological and agronomic characteristics of reproductive fitness in oilseed crops are critical for genetic manipulation and to increase seed and oil yields. Integrated high-throughput plant-phenotyping (HTPP) and genome-wide association analyses in oilseed crops have the potential of accelerating the rate of trait discovery and improving phenotypic predictions of yield sink size (e.g., seed or fruit) and reproductive resilience under GCC stresses. Especially when high heritability is associated with moderate to high genetic advance for the duration of the reproductive phase, then there will be a reasonable possibility for improvement of traits related to reproductive fitness (Jain, 2014). Several reproductive fitness traits are usually strongly correlated and QTL colocalized across environments (e.g., flowering time, dry rosette weight, seed yield, harvest index, ratio of reproductive to vegetative organs, seed C% and N%, and 1000-seed weight). These positive correlations are consistent with life history theory in that genotypic variation in resource acquisition masks tradeoffs (Chardon et al., 2014). Constitutive overexpression of a newly discovered Gγ protein (AGG3; Choudhury et al., 2014) in Camelina sativa resulted in faster vegetative and reproductive growth, along with increased PS efficiency. Several components of reproductive fitness (e.g., seed size, seed mass, and seed number per plant) increased by 15–40%, and higher oil yield per plant was achieved when AGG3 was expressed constitutively or specifically in seed tissue.
The risk of high-temperature impact on oilseed crop production is increasing with increased global warming. Impaired fertility and yield loss due to heat stress are widely reported in oilseed crops (Yu et al., 2014). Adverse climatic conditions and environmental stresses particularly impact male reproductive development in oilseed crops. When exposed to stress, male gametophytic organs often show abnormalities at the morphological, structural, and metabolic levels; such changes usually lead to meiotic defects or premature spore abortion and male reproductive sterility. Pollen viability, flower number, bud number, seed and pod number, as components of the reproductive system, are used to identify genotypes with high-temperature tolerance. The number and length of buds emerged on main stem, and pod number produced under high temperatures, might provide a useful preliminary screening criterion for high-temperature tolerance in oilseed crops such as B. napus and identify genetic resources (e.g., B. rapa) to provide genes for heat stress tolerance (Annisa et al., 2013).
Depending on the type of stress involved (e.g., heat, cold, or drought) and the duration of exposure, the underlying cellular defects are highly variable and involve a wide range of aberrations (De Storme and Geelen, 2014). Interactions between leaf senescence and resource allocation to seeds, hence reproductive fitness, were revealed by QTL metaanalysis of traits related to senescence, resource allocation, and seed yield in Arabidopsis thaliana. Sequential and monocarpic senescence observed at vegetative and reproductive stages, respectively facilitate N remobilization and control the duration of C fixation. However, genetic and environmental factors control N and C resource allocation to seeds (Chardon et al., 2014). Heat-responsive genes, with organ-specific expression, have been identified through transcriptional profiling at the seed-filling stage in siliqua wall and seed of B. napus (Yu et al., 2014). The molecular bases of response to heat during late reproductive stages provided valuable information and identified gene resources for the genetic improvement of heat tolerance in B. napus. When reproductive tolerance to high temperature was increased by 2°C (Singh et al., 2012), reproductive efficiency was improved, especially after the plants were subjected to drought stress. Large genotypic variation for heat tolerance was found at the reproductive stage in several oilseed accessions (spring-type B. rapa accessions and one B. juncea accession). Heat tolerance was determined by the ability of accessions to set seed equally at normal (i.e., control) and high-temperature treatments (Annisa et al., 2013).
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Book2016, Breeding Oilseed Crops for Sustainable ProductionAbdullah A. Jaradat
Review article
Reproductive value and the evolution of altruism
2022, Trends in Ecology & EvolutionAntónio M.M. Rodrigues, Andy Gardner
Reproductive value
Natural selection arises when individuals that differ genetically also differ in their reproductive success. However, individuals may vary systematically in reproductive success for reasons other than the genes they carry, and this variation can have important consequences for natural selection. For example, if males and females were always equal from the perspective of natural selection, parents would simply maximize their number of offspring, and the sex ratio of the brood would be an irrelevant detail. However, males and females are not necessarily equal: if, for any reason, there is a biased sex ratio, then newborns of the rarer sex will, on average, leave more descendants than newborns of the more common sex, under diplodiploid inheritance . In panmictic populations this leads natural selection to favor parents who invest more heavily in offspring of the rarer sex, which tends to neutralize any population bias in investment into one sex over the other [1–4]. The rarer-sex effect (see Glossary) highlights a more general principle of how natural selection acts in relation to class-structured populations.
To solve the problem of class structure, Fisher developed the concept of individual reproductive value, which measures how well individuals of a particular age, sex, or any other type of class transmit copies of their genes to future generations. The reproductive value of an individual is given by the combined reproductive value of all her offspring, in which the contribution of each offspring is weighted by the genetic share of the individual in that offspring (Box 1) [1,5]. Put another way, reproductive value measures the relative probability that a gene picked at random from the distant future traces its origin to the focal individual in the present generation, and it is this quantity – rather than number of offspring or grandoffspring per se – that defines the proper measure of Darwinian fitness. Fisher [1,6] introduced the concept in relation to age-structured populations, and used data from the 1911 Australian census to show how female reproductive value increased during childhood as the individual approached reproductive maturity, peaked shortly thereafter, and then declined towards zero at older ages. From the standpoint of natural selection, the life of an individual is most valuable when her reproductive value peaks.
Box 1
Reproductive value and Hamilton’s rule
The total reproductive value of a class of individuals is the probability that a gene picked at random from the distant future traces its ancestry back to this class in the present generation. Accordingly, class reproductive value (c) describes the relative importance of this segment of the population from the perspective of natural selection, and hence coincides with the concept of force of selection in relation to a particular class [1,7,8]. This can be calculated directly from the 'gene flow' between the different classes using mathematics analogous to the way in which Google calculates the PageRank of webpages [10,11]. For example, in haplodiploid species with non-overlapping generations, daughters inherit half of their genes from their mothers (i.e., 1 − ϕ = ½), and half from their fathers (i.e., ϕ = ½), whereas sons inherit all of their genes from their mothers (i.e., μ = 1, 1 − μ = 0). This implies that the class reproductive value of females satisfies cf = (1 − ϕ)cf + μcm = ½cf + cm , and the class reproductive value of males satisfies cm = ϕcf + (1 − μ)cm = ½ cf (Figure I). Using matrix notation, we can rewrite these equations as:
[I]
or c = cP, where P is the gene-flow matrix and is the dominant left-eigenvector of P. Following the convention that class reproductive values sum to unity, this yields cf = 2/3 and cm = 1/3. Under neutrality, the reproductive value of a class is shared equally among all the individuals of that class, such that individual reproductive value (v) satisfies cf = vfuf for females and cm = vmum for males, where uf and um are the equilibrium frequencies of the two classes.
Reproductive value provides one of the three measures of value that feature in the general version of Hamilton’s rule of inclusive-fitness theory, the other two being relatedness and fitness effect [12–15]. To illustrate, consider an altruistic behavior in which a daughter forgoes her own clutch of C offspring to help her mother raise B extra offspring. Hamilton’s rule is given by:
[II]
where rF→D is the relatedness between mother and daughter, rF→S is the relatedness between mother and son, rF→F is the relatedness between sisters, rF→M is the relatedness between sister and brother, and z is the sex ratio. The selective advantage of altruism increases as the life-for-life relatedness valuation of the siblings of the focal female, namely RS = (1 − z)vfrF→F + zvmrF→M increases relative to that of her own offspring, in other words RO = (1 − z)vfrF→D + zvmrF→S.
Reproductive value considerations are crucial when individuals must decide how to behave towards their relatives [13,16,17]. Inclusive fitness theory shows that an individual should value each of her social partners according to how well they transmit copies of her genes. This is not only a matter of the extent to which they carry copies of her genes (relatedness) but also concerns the ability of each social partner to transmit copies of their own genes to future generations (reproductive value). Hamilton was the first to integrate the concept of reproductive value into the theory of inclusive fitness by defining life-for-life relatedness as relatedness multiplied by the ratio of the recipient and actor reproductive values, in the context of understanding the evolution of social behavior in insects (Box 1). Despite these early insights, the crucial role of reproductive value in driving the evolution of altruism has been overshadowed by research on relatedness. To remedy this we provide an overview of ways in which reproductive value may modulate patterns of altruism – in relation to individual differences in age, sex and general condition – across the natural world (Figure 1 for illustrative examples).
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Journal2022, Trends in Ecology & EvolutionAntónio M.M. Rodrigues, Andy Gardner
Chapter
Human Reproductive Ecology
2015, Basics in Human EvolutionClaudia R. Valeggia, Alejandra Núñez-de la Mora
Abstract
Human reproductive ecology (HRE) studies human reproduction in the context of local ecological variables. The perspective of HRE is evolutionary, and its central goal is to provide models for explaining variation in reproductive patterns as adaptations to the environment. Reproductive function shows significant variation between and within individuals and populations. Human reproductive function has been shown to vary in response to ecological variables such as nutrition, physical activity, acute immunological challenges, psychological stress, and social context. HRE allows us to explore what the consequences of that variation are in terms of health, fertility, and ultimately, reproductive fitness.
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Book2015, Basics in Human EvolutionClaudia R. Valeggia, Alejandra Núñez-de la Mora
Review article
Is Female Health Cyclical? Evolutionary Perspectives on Menstruation
2018, Trends in Ecology & EvolutionAlexandra Alvergne, Vedrana Högqvist Tabor
Most biomedical research on female health focuses on reproductive health, which, as pointed out by the social sciences, reduces female health to women’s reproductive function . Recent health policy reports thus call for a research program that moves beyond the study of reproduction to tackle the future challenges for women’s health: cardiovascular diseases, cancers, autoimmune diseases, diabetes, and mental health . Arguably, the global women’s health agenda goes much beyond reproductive concerns, but we contend that it is incorrect to conceptualize women’s health as being separate from women’s evolved reproductive system. From an evolutionary perspective, the regulation of the reproductive function is viewed as the very foundation of health. This is because reproductive features have a great impact on reproductive fitness and are thus particularly malleable to evolution by natural selection. This evolutionary force is constrained by resource allocation trade-offs and thus energetic investment in reproductive function imposes constraints on the potential for the body to be ‘healthy’ (i.e., to invest in immune-competence). The core insights to be taken from an evolutionary approach are that the body is best viewed as an evolved system of interconnected functions (growth, immunity, reproduction), and from the perspective of natural selection, health is only a means to the end of reproduction . While it is well known that the neuro-endocrine and immune systems are intimately connected , the biomedical sciences have yet to conceptualize reproductive health as foundational for nonreproductive health.
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Journal2018, Trends in Ecology & EvolutionAlexandra Alvergne, Vedrana Högqvist Tabor
Review article
Group selection and kin selection: formally equivalent approaches
2011, Trends in Ecology & EvolutionJames A.R. Marshall
In , a model for the evolution of eusociality is presented entirely in terms of personal fitness, along with the claim that this means inclusive fitness is not involved. In fact, Hamilton himself proposed precisely the same approach to modelling social evolution as part of inclusive fitness theory, under the name of ‘neighbour-modulated fitness’ . This approach was used in early models of interactions between relatives and in generalising Hamilton's rule, where it was shown to be equivalent to the inclusive fitness accounting approach [30,33]. Today, the term ‘personal fitness’ is generally used, and it is recognised as being equivalent to inclusive fitness , as well as being increasingly popular among theorists owing to the frequent difficulty of identifying all the consequences of a social behaviour (as illustrated in the main text) . Thus models using personal fitness, such as that in , are readily interpreted in inclusive fitness terms (in fact, the authors themselves concede this equivalence in the supplementary information for their paper ). Given the equivalence of the approaches, one might ask why the inclusive fitness concept is needed at all, and might claim that social evolution can be explained entirely in terms of classical Darwinian fitness, extended to include the effects of the behaviours of others on the fitness of an individual . The answer is that, although mathematically the two are equivalent, in evolutionary terms they are very different. Hamilton introduced the concept of inclusive fitness as a new quantity that selection should maximise, as a replacement for classical fitness [1,2]. Formal justification for the concept of natural selection leading to the maximisation of inclusive fitness has also been developed [88,89]. Informally, the reason inclusive fitness, rather than personal fitness, is maximised by natural selection is because an individual actor only has control over their own inclusive fitness, but does not have full control over their personal fitness. This is because the inclusive fitness is calculated in terms of the effects of the behaviours of an individual on their own direct fitness, and on that of their genetic relatives (see discussion in main text) and, because the behaviour of an actor is under their own control, so is their inclusive fitness. By contrast, only part of the personal fitness of an actor is due to their own behaviour and, hence, under their own control, with the remainder the result of social partners over whom they might have no influence. Despite the aforementioned mathematical equivalence, because only inclusive fitness is under the control of an individual, it is this quantity that they should ‘act as if to maximise’ [88,90].
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Journal2011, Trends in Ecology & EvolutionJames A.R. Marshall
Related terms:
Hatchery
Inbreeding
Life History
Natural Selection
Genetic Divergence
Mouse
Receptor
Allele
Population Size
Evolution
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3045 | https://www.spanishunicorn.com/ejercicios-de-gramatica-preterito-imperfecto/ | Spanish UnicornEjercicios de español
A2 . A2-B1 . B1-B2 . Ejercicios de gramática . Pasados
Pretérito Imperfecto -17.
On 05/12/2017 by SpanishUnicorn
Ejercicios de gramática – Pretérito Imperfecto
¿Ya completaste todos los ejercicios de gramática? ¿Los de pretérito indefinido? ¿Sí? Entonces podemos continuar con algo nuevo: Ejercicios de gramática – Pretérito Imperfecto
NIVELES A2 – B1
La conjugación no es nada difícil.
¿Listos? ¿Preparado?
¡Empecemos!
¿Qué es el Pretérito Imperfecto?
Español:El pretérito imperfecto es un tiempo verbal del pasado que usamos para hablar de acciones habituales, descripciones o situaciones en progreso en el pasado. No indica cuándo empieza ni cuándo termina la acción, sino que nos muestra el “fondo” de la historia.
English:The pretérito imperfecto is a past tense used to talk about habitual actions, descriptions, or ongoing situations in the past. It doesn’t show when the action started or ended; instead, it sets the background of the story.
Usos principales del Pretérito Imperfecto
1. Descripciones en el pasado
Español: Usamos el imperfecto para describir personas, lugares o cosas en el pasado.
English: We use the imperfect to describe people, places, or things in the past.
Ejemplos:
Mi abuela era muy amable. → My grandmother was very kind.
La casa tenía un jardín enorme. → The house had a huge garden.
El cielo estaba gris y hacía frío. → The sky was gray and it was cold.
2. Acciones habituales o repetitivas
Español: Para hablar de cosas que pasaban una y otra vez en el pasado.
English: To talk about things that used to happen repeatedly in the past.
Ejemplos:
Los domingos siempre íbamos al parque. → On Sundays we always went to the park.
Cuando era niño, jugaba con mis primos todos los días. → When I was a child, I used to play with my cousins every day.
Ella cocinaba sopa cada viernes. → She used to cook soup every Friday.
3. Acciones en progreso en el pasado
Español: Cuando una acción estaba en desarrollo en un momento del pasado.
English: When an action was ongoing at a certain moment in the past.
Ejemplos:
Yo leía un libro mientras ella cocinaba. → I was reading a book while she was cooking.
Los niños dormían cuando llegamos a casa. → The children were sleeping when we arrived home.
Él estudiaba en la biblioteca a las cinco de la tarde. → He was studying in the library at five in the afternoon.
4. Contexto o ambiente de una historia
Español: Para situar la escena, el clima, la hora, los sentimientos…
English: To set the scene, the weather, the time, or the feelings.
Ejemplos:
Era una tarde de verano y hacía mucho calor. → It was a summer afternoon and it was very hot.
Eran las diez de la noche y la ciudad estaba vacía. → It was ten at night and the city was empty.
Todos estaban nerviosos antes del examen. → Everyone was nervous before the exam.
2. Marcadores temporales del Pretérito Imperfecto:
entonces
de pequeño
cuando + imperfecto
todos los días
a menudo
frecuentemente
con frecuencia
muchas veces
algunas veces
siempre
casi siempre
Ejemplos con marcadores temporales del Pretérito Imperfecto
1. entonces (then, at that time)
Entonces vivíamos en un piso muy pequeño. → Back then we lived in a very small flat.
No trabajaba, porque entonces estudiaba medicina. → I wasn’t working, because at that time I was studying medicine.
Entonces no había internet en las casas. → Back then there was no internet at home.
2. de pequeño (as a child)
De pequeño iba al colegio en bicicleta. → As a child I used to go to school by bike.
De pequeño tenía miedo de la oscuridad. → As a child I was afraid of the dark.
De pequeño jugaba mucho en la calle. → As a child I used to play a lot in the street.
3. cuando + imperfecto (when + imperfect)
Cuando era joven, salía mucho con mis amigos. → When I was young, I used to go out a lot with my friends.
Cuando tenía tiempo libre, siempre pintaba cuadros. → When I had free time, I always painted pictures.
Cuando vivíamos en Madrid, íbamos al Retiro todos los domingos. → When we lived in Madrid, we went to El Retiro every Sunday.
4. todos los días (every day)
Todos los días leía el periódico en el bar. → Every day I read the newspaper in the bar.
Todos los días mis padres me llevaban al colegio. → Every day my parents took me to school.
Todos los días desayunábamos churros con chocolate. → Every day we had churros with hot chocolate for breakfast.
5. a menudo (often)
A menudo visitábamos a mis abuelos los sábados. → We often visited my grandparents on Saturdays.
A menudo jugaba al tenis con mis amigos. → I often played tennis with my friends.
A menudo escuchaba música en mi habitación. → I often listened to music in my room.
6. frecuentemente (frequently)
Frecuentemente iba al cine con mi hermano. → I frequently went to the cinema with my brother.
Frecuentemente estudiaba hasta muy tarde. → I frequently studied until very late.
Frecuentemente llovía en mi ciudad. → It frequently rained in my city.
7. con frecuencia (frequently/often)
Con frecuenciahablábamos de política en casa. → We often talked about politics at home.
Con frecuencia me visitaban mis primos. → My cousins often visited me.
Con frecuencia me quedaba dormido en clase. → I often fell asleep in class.
8. muchas veces (many times)
Muchas veces viajábamos a la playa en verano. → Many times we used to travel to the beach in summer.
Muchas veces veía películas en la tele. → Many times I watched movies on TV.
Muchas veces cocinaba pasta porque era barato. → Many times I cooked pasta because it was cheap.
9. algunas veces (sometimes)
Algunas veces salíamos a cenar fuera. → Sometimes we went out for dinner.
Algunas veces me quedaba en casa leyendo. → Sometimes I stayed at home reading.
Algunas veces ayudaba a mi padre en el trabajo. → Sometimes I helped my father at work.
10. siempre (always)
Siempre llevaba sombrero en invierno. → He always wore a hat in winter.
Siempre hablábamos en voz baja en la biblioteca. → We always spoke quietly in the library.
Siempre se olvidaba de sus llaves. → He always forgot his keys.
11. casi siempre (almost always)
Casi siempre estudiaba en la cafetería. → I almost always studied in the café.
Casi siempre llegaba tarde a clase. → He almost always arrived late to class.
Casi siempre comíamos en casa de mis abuelos los domingos. → We almost always had lunch at my grandparents’ house on Sundays.
3. Conjugación – pretérito imperfecto:
No es tan difícil, ¿verdad?
Por suerte solo hay tres verbos irregulares: SER, IR, VER.
Primero vamos a practicar la conjugación de los verbos irregulares:
Ejercicios de conjugación – pretérito imperfecto
Verbs regulares
Verbos irregulares – Pretérito imperfecto
Compra aquí los ejercicios
Ejercicios de gramática – Pretérito Imperfecto
Ejercicio 8.
Ejercicio 9.
Ejercicio 10.
Compra aquí los +200 ejercicios
Ejercicio 11: Crucigrama de Pretérito Imperfecto
Ejercicio 12: Descripción de imágenes
¿Qué hacían?
¿Dónde?
¿Qué llevaban?
¿Cómo vivían?
¿Cuándo…?
Ejercicios relacionados
Ejercicios de conjugación: pretérito imperfecto
Imperfecto o indefinido – ejercicios de gramática
Test de pretérito imperfecto – 30 ejercicios
Ejercicios de gramática: imperfecto
Indefinido o imperfecto – ejercicios y conjugación
Comprensión auditiva
Descarga: Ejercicios de gramática – Pretérito Imperfecto
ConjugacionDownload
imperfecto-ejercicios-3Descarga
Mi tío (SER) ………………. una persona muy seria.
Con mis amigos (IR, nosotros) ………………. al cine si (TENER) ………………. dinero.
Siempre (HABLAR, él) ………………. de política.
Nuestra casa (ESTAR) ………………. al lado del mercado donde (HABER) ………………. mucha gente.
Ahora no fumo, pero antes (FUMAR, yo) ………………. mucho.
Mientras (COMER, ellos) comían (VER, ellos) ………………. la tele.
(SER) ………………. un día frío de invierno. Las estrellas (BRILLAR) ………………. en el cielo.
(IR, vosotros) ………………. a la escuela de lunes a viernes.
Pepe (TENER) ………………. un perro agresivo que (LLAMARSE) ………………. Rufo.
Antes (VER, tú) ………………. programas muy interesantes.
Ejercicios de escucha Vocabulario español Ejercicios de gramática Ejercicios de Pretérito Perfecto Ejercicios de Verbos Reflexivos Ejercicios de Verbos Irregulares en Presente El Imperativo en Español Ejercicios de Subjuntivo Sobre nosotros Contacto Política de Cookies, Política de Privacidad
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3046 | https://www.johndcook.com/blog/2020/02/28/sum-of-squared-digits/ | Sum of squared digits
Posted on by John
Take a positive integer x, square each of its digits, and sum. Now do the same to the result, over and over. What happens?
To find out, let’s write a little Python code that sums the squares of the digits.
```
def G(x):
return sum(int(d)2 for d in str(x))
```
This function turns a number into a string, and iterates over the characters in the string, turning each one back into an integer and squaring it.
Now let’s plot the trajectories of the iterations of G.
```
def iter(x, n):
for _ in range(n):
x = G(x)
return x
for x in range(1, 40):
y = [iter(x, n) for n in range(1, 12)]
plt.plot(y)
```
This produces the following plot.
For every starting value, the sequence of iterations either gets stuck on 1 or it goes in the cycle 4, 16, 37, 58, 89, 145, 42, 20, 4, … . This is a theorem of A. Porges published in 1945 .
To see how the sequences eventually hit 1 or 4, let’s modify our iteration function to stop at 4 rather than cycling.
```
def iter4(x, n):
for _ in range(n):
if x != 4:
x = G(x)
return x
for x in range(1, 40):
y = [iter4(x, n) for n in range(1, 16)]
plt.plot(y)
```
This produces the following plot.
Update: Here’s a better, or at least complementary, way to look at the iterations. Now the horizontal axis represents the starting points x and the points stacked vertically over x are the iterates of G starting at x.
```
def orbit(x):
pts = set()
while x not in pts:
pts.add(x)
x = G(x)
return pts
for x in range(1, 81):
for y in orbit(x):
plt.plot(x, y, "bo", markersize=2)
plt.xlabel("$x$")
plt.ylabel("Iterations of $x$")
plt.savefig("porges3.png")
```
Porges, A set of eight numbers, Amer. Math. Monthly, 52(1945) 379-382.
One thought on “Sum of squared digits”
Jonathan
When I was playing around with this my visualization (done with pen and paper instead of code) was a directed graph, with edges from x to G(x). I don’t think that plt has anything for this, but I’ll bet there’s a Javascript library out there that does. Time to update my skill set!
Oh, and the next fun question that arises is “What happens when you change the base?”. Since I was pen-and-papering it I never got beyond doing it in base 8 (I believe there were two different [non-trivial] cycles a number could fall into). But with code one could explore a whole range of bases.
Comments are closed. |
3047 | https://brilliant.org/wiki/recurrence-relations/ | Recurrence Relations | Brilliant Math & Science Wiki
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Andy Hayes, Eddie The Head, Happy Melodies, and
6 others
Jaydee Lucero
U Z
Satyabrata Dash
Bob Kadylo
Zandra Vinegar
Jimin Khim
contributed
A recurrence relation is an equation that uses recursion to relate terms in a sequence or elements in an array. It is a way to define a sequence or array in terms of itself. Recurrence relations have applications in many areas of mathematics:
number theory - the Fibonacci sequence
combinatorics - distribution of objects into bins
calculus - Euler's method
and many more.
Recurrence relations are used when an exhaustive approach to problem solving is simply too arduous to be practical. Although it is not ideal to compute the terms in a sequence one at a time by using previous terms, this approach can be much more efficient than the alternative of exhaustive casework.
Sometimes, a recurrence relation can be "solved" by defining the terms of a sequence in terms of its index rather than previous terms in the sequence. This gives a closed form expression for each term in the sequence and eliminates the need for an iterative process to solve for terms in the sequence. There are several ways to accomplish this:
solving linear recurrence relations
solving recurrence relations with generating functions
solving recurrence relations with the substitution method
solving recurrence relations with the method of summation factors.
Even if a solution of this form is not possible, a recurrence relation is still useful, as it can be used to develop computer algorithms. When terms in a sequence are stored, dynamic programming allows one to compute new terms in a sequence efficiently. Recurrence relations are also applicable for recursive backtracking, in which recursion is used to optimize algorithms.
Contents
Setting up a Recurrence Relation
Applications of Recurrence Relations
Using Generating Functions to Solve Recurrence Relations
Method of Summation Factors
Setting up a Recurrence Relation
Recurrence relations are used to reduce complicated problems to an iterative process based on simpler versions of the problem. An example problem in which this approach can be used is the Tower of Hanoi puzzle.
The Tower of Hanoi puzzle consists of three vertical pegs and several disks of various sizes. Each disk has a hole in its center for the pegs to go through.
A Tower of Hanoi puzzle with 8 disks
The rules of the puzzle are as follows:
The puzzle begins with all disks placed on one of the pegs. They are placed in order of largest to smallest, bottom to top.
The goal of the puzzle is to move all of the disks onto another peg.
Only one disk may be moved at a time, and disks are always placed onto pegs.
Disks may only be moved onto an empty peg or onto a larger disk.
Let T n T_n T n be defined as the minimum number of moves needed to solve a puzzle with n n n disks.
Image Credit: Ævar Arnfjörð Bjarmason, Wikimedia Commons
It's not immediately clear that a recursion solution will work for this problem. However, there are a couple things about this problem that make it a good candidate to solve with a recurrence relation.
Identifying a candidate problem to solve with a recurrence relation:
The problem can be reduced to simpler cases. The Tower of Hanoi puzzle can be simplified by removing some of the disks.
There is a numerical value, n n n, to identify each case. For the Tower of Hanoi puzzle, this numerical value is the number of disks.
The problem increases in complexity as the numerical identifier increases. As the number of disks in the Tower of Hanoi problem increases, it becomes more difficult to find a solution.
The goal for this exercise will be to develop a recurrence relation for T n T_n T n. Rather than try to tackle a complicated puzzle, like the one with 8 disks pictured above, it is often best to start off with the most simple version of the problem possible.
Step 1: Define a Base Case
The most simple version of the Tower of Hanoi puzzle would contain only one disk.
In terms of the recurrence relation, n=1 n=1 n=1.
T 1=1 T_1=1 T 1=1, because it would only take 1 1 1 move to move all the disks to another peg.
The base case is often trivially simple. However, it will be necessary for developing more complicated cases. Even though the goal of this exercise is to come up with an equation for T n T_n T n, it's not immediately clear what that equation will look like. It's necessary to do casework to get an understanding of how the problem works. After a couple of cases have been completed, one starts to develop an understanding of how to set up the recurrence relation.
Step 2: Develop More Complicated Cases
Below is the solution to a Tower of Hanoi puzzle with n=2 n=2 n=2.
It can be seen from above that T 2=3 T_2=3 T 2=3.
Below is a solution to a Tower of Hanoi puzzle with n=3 n=3 n=3.
It can be seen from above that T 3=7 T_3=7 T 3=7.
You can continue developing more complicated cases as needed. The goal of this process is to understand how the problem works, and to begin to think about how to set up the recurrence relation.
Although two cases were analyzed in the example above, it may sometimes be necessary to develop more cases in order to gain a better understanding of the problem. With the two cases above, one can probably begin to see how the cases are related to each other.
Step 3: Write the Recurrence Relation
Think about how the cases are related to each other.
With n=1 n=1 n=1, it was a simple matter to move the disk once, and then the puzzle was complete.
With n=2 n=2 n=2, the smaller disk had to be moved before the larger disk could be moved. Then, the smaller disk was placed on the larger disk to complete the puzzle.
With n=3 n=3 n=3, the smaller disks had to be moved before the largest disk could be moved. Then, the smaller disks were placed on the larger disk to complete the puzzle.
One can begin to see a pattern in how these solutions are structured. You move the smaller disks, then you move the largest disk, then you move the smaller disks back onto the largest disk to complete the puzzle.
In terms of n n n,
Do T n−1 T_{n-1}T n−1 moves to get the smaller disks off the largest disk.
Do 1 1 1 move to move the largest disk.
Do T n−1 T_{n-1}T n−1 moves to get the smaller disks back onto the largest disk.
In total, the number of moves for n n n disks is
T n=2 T n−1+1.T_n=2T_{n-1}+1.T n=2 T n−1+1.
One can confirm the recurrence relation written above by matching it to the known values of T n T_n T n:
T 1=1 T 2=2 T 1+1=2(1)+1=3 T 3=2 T 2+1=2(3)+1=7.\begin{aligned} T_1&=1\ T_2&=2T_1+1=2(1)+1=3\ T_3&=2T_2+1=2(3)+1=7. \end{aligned}T 1T 2T 3=1=2 T 1+1=2(1)+1=3=2 T 2+1=2(3)+1=7.
The recurrence relation for the Tower of Hanoi puzzle is an example of a linear recurrence relation. It can be put into a closed form solution using the techniques discussed in the given link.
G n=3 G n−1−1 G_n=3G_{n-1}-1 G n=3 G n−1−1 G n=3 G n−1−G n−2 G_n=3G_{n-1}-G_{n-2}G n=3 G n−1−G n−2G n=3 G n−1 G_n=3G_{n-1}G n=3 G n−1G n=G n−1+G n−2 G_n=G_{n-1}+G_{n-2}G n=G n−1+G n−2
Reveal the answer
A string of n n n characters is made from the characters A\text{A}A, B\text{B}B, and C\text{C}C.
Let G n G_n G n be the number of such strings of n n n characters which contain no instance of AB\text{AB}AB (in that order).
Write a recurrence relation for G n G_n G n.
The correct answer is: G n=3 G n−1−G n−2 G_n=3G_{n-1}-G_{n-2}G n=3 G n−1−G n−2
Applications of Recurrence Relations
Number Theory
The Fibonacci Sequence
Harmonic Numbers
Pell Numbers
Pell's Equation
Partition of an Integer
Combinatorics
Distributions of Distinct Objects into Identical Bins
Distributions of Identical Objects into Identical Bins
Pascal's Triangle
Binomial Coefficient
Derangements
Calculus
Arithmetic Progressions
Geometric Progressions
Euler's Method
Computer Science
Recursive Backtracking
Dynamic Programming
Memoization
Using Generating Functions to Solve Recurrence Relations
Main Article: Using Generating Functions to Solve Recurrence Relations
One method to solve recurrence relations is to use a generating function.
A generating function is a power series whose coefficients correspond to terms in a sequence of numbers.
What is the generating function of the Fibonacci sequence?
The generating function of the Fibonacci sequence is
f(x)=∑n=0∞F n x n=1+x+2 x 2+3 x 3+5 x 4+⋯,f(x)=\sum_{n=0}^\infty F_n x^n = 1 + x + 2x^2 + 3x^3 + 5x^4+\cdots, f(x)=n=0∑∞F nx n=1+x+2 x 2+3 x 3+5 x 4+⋯,
where F n F_n F n is the n th n^{\text{th}}n th Fibonacci number. Using techniques demonstrated here, it can be shown that this generating function has a closed form expression
f(x)=∑n=0∞F n x n=1 1−x−x 2.□f(x)=\sum_{n=0}^\infty F_n x^n =\frac{1}{1 - x - x^2}.\ _\square f(x)=n=0∑∞F nx n=1−x−x 2 1.□
Use the generating function of T n T_n T n to write a closed form expression for T n T_n T n.
Recall the recurrence relation for T n T_n T n:
T n+1=2 T n+1.T_{n+1} = 2T_{n}+1.T n+1=2 T n+1.
It has already been established that T 1=1 T_1=1 T 1=1, but a value for T 0 T_0 T 0 is also required. Let T 0=0 T_0 = 0 T 0=0, and note that the recurrence relation still holds.
Define the generating function:
T(x)=∑n=0∞T n x n=T 0+T 1 x+T 2 x 2+⋯.\begin{aligned} T(x) &= \sum\limits_{n = 0}^{\infty} T_n x^n \ \ &= T_0 + T_1 x + T_2 x^2 + \cdots. \end{aligned}T(x)=n=0∑∞T nx n=T 0+T 1x+T 2x 2+⋯.
Recall that T 0=0 T_0=0 T 0=0. Divide both sides of the equation by x x x, which gives
T(x)x=T 1+T 2 x+T 3 x 2+⋯=∑n=0∞T n+1 x n.\begin{aligned} \frac{T(x)}{x} &= T_1 + T_2 x + T_3 x^2 + \cdots \ \ &= \sum\limits_{n=0}^{\infty} T_{n+1} x^n. \end{aligned}x T(x)=T 1+T 2x+T 3x 2+⋯=n=0∑∞T n+1x n.
Now substitute the recurrence relation for T n+1 T_{n+1}T n+1 into the equation:
T(x)x=∑n=0∞(2 T n+1)x n=2 T(x)+∑n=0∞x n.\begin{aligned} \frac{T(x)}{x} &= \sum\limits_{n = 0}^{\infty} (2T_{n}+1 )x^n \ \ &= 2T(x) + \sum\limits_{n = 0}^{\infty} x^n. \end{aligned}x T(x)=n=0∑∞(2 T n+1)x n=2 T(x)+n=0∑∞x n.
Provided that ∣x∣<1|x|<1∣x∣<1, the series above converges as an infinite geometric progression:
∑n=0∞x n=1 1−x,∣x∣<1.\sum\limits_{n=0}^{\infty}x^n=\frac{1}{1-x},\ \, |x|<1.n=0∑∞x n=1−x 1,∣x∣<1.
Substituting this expression gives
T(x)x=2 T(x)+1 1−x,∣x∣<1.\frac{T(x)}{x} = 2T(x) + \frac{1}{1-x},\ \, |x|<1.x T(x)=2 T(x)+1−x 1,∣x∣<1.
Solving for T(x)T(x)T(x),
T(x)=x(1−x)(1−2 x),∣x∣<1.T(x) = \frac{x}{(1-x)(1-2x)},\ \, |x|<1.T(x)=(1−x)(1−2 x)x,∣x∣<1.
Now that a closed form expression for the generating function has been found, the goal shifts to finding a closed-form expression for T n T_n T n.
As the generating function is equal to a rational expression with binomial factors in the denominator, a good starting point would be to do partial fraction decomposition:
T(x)=x(1−x)(1−2 x)=x(2 1−2 x−1 1−x).\begin{aligned} T(x)&=\frac{x}{(1-x)(1-2x)} \ \ &= x\left( \frac{2}{1-2x} - \frac{1}{1-x}\right). \end{aligned}T(x)=(1−x)(1−2 x)x=x(1−2 x 2−1−x 1).
Recall infinite geometric progressions. For the next step, the two rational expressions are put back into power series form using the identities:
2 1−2 x=∑n=0∞2 n+1 x n 1 1−x=∑n=0∞x n.\begin{aligned} \frac{2}{1-2x}&=\sum\limits_{n=0}^{\infty}{2^{n+1}x^n} \ \ \frac{1}{1-x}&=\sum\limits_{n=0}^{\infty}x^n. \end{aligned}1−2 x 21−x 1=n=0∑∞2 n+1 x n=n=0∑∞x n.
Substituting these identities gives
T(x)=x(∑n=0∞2 n+1 x n−∑n=0∞x n)=∑n=0∞(2 n+1−1)x n+1=∑n=1∞(2 n−1)x n.\begin{aligned} T(x) &= x\left(\sum\limits_{n=0}^{\infty}{2^{n+1}x^n}-\sum\limits_{n=0}^{\infty}x^n \right) \ \ &=\sum\limits_{n=0}^{\infty}{(2^{n+1}-1)x^{n+1}} \ \ &=\sum\limits_{n=1}^{\infty}{(2^{n}-1)x^{n}}. \end{aligned}T(x)=x(n=0∑∞2 n+1 x n−n=0∑∞x n)=n=0∑∞(2 n+1−1)x n+1=n=1∑∞(2 n−1)x n.
From here, it is clear that the coefficient of the generating function (((the series T n)T_n)T n) is given by
T n=2 n−1.□\boxed{T_n = 2^{n}-1}.\ _\square T n=2 n−1.□
Method of Summation Factors
Main Article: Method of Summation Factors
There is another way of solving recurrence relations of the form A a n=B a n−1+C Aa_n = Ba_{n-1} + C A a n=B a n−1+C, where A A A, B B B and C C C are functions of n n n, known as the method of summation factors.
This method can be used in the Tower of Hanoi problem, resolved once more in the example below.
Show that the closed form of the recurrence relation T 0=0 T_0 = 0 T 0=0, T n=2 T n−1+1 T_n = 2T_{n-1} + 1 T n=2 T n−1+1 is T n=2 n−1 T_n = 2^n - 1 T n=2 n−1.
The summation factor s n s_n s n with A(n)=1 A(n) = 1 A(n)=1 and B(n)=2 B(n) = 2 B(n)=2 is
s n=1⋅1⋅⋯⋅1 2⋅2⋅⋯⋅2=1 2 n−1,s_n = \frac{1 \cdot 1 \cdot \cdots \cdot 1}{2 \cdot 2 \cdot \cdots \cdot 2} = \frac{1}{2^{n-1}},s n=2⋅2⋅⋯⋅2 1⋅1⋅⋯⋅1=2 n−1 1,
where the 1's and 2's occur n−1 n-1 n−1 times each. Multiplying this to the recurrence relation gives
1 2 n−1 T n=1 2 n−2 T n−1+1 2 n−1.\frac{1}{2^{n-1}} T_n = \frac{1}{2^{n-2}}T_{n-1} + \frac{1}{2^{n-1}}.2 n−1 1T n=2 n−2 1T n−1+2 n−1 1.
Let b n=1 2 n−1 T n b_n = \dfrac{1}{2^{n-1}} T_n b n=2 n−1 1T n. Then b n−1=1 2 n−2 T n−1 b_{n - 1} = \dfrac{1}{2^{n-2}}T_{n-1}b n−1=2 n−2 1T n−1 and b 0=1 2 0−1 T 0=0 b_0 = \dfrac{1}{2^{0-1}} T_0 = 0 b 0=2 0−1 1T 0=0. So,
b n=b n−1+1 2 n−1.b_n = b_{n - 1} + \frac{1}{2^{n-1}}.b n=b n−1+2 n−1 1.
It follows that the closed form of b n b_n b n is
b n=0+∑k=1 n 1 2 k−1=1⋅(1 2)n−1 1 2−1=2−1 2 n−1.b_n = 0 + \sum_{k = 1}^n \frac{1}{2^{k-1}} = \frac{1 \cdot \left(\frac12\right)^n - 1}{\frac12 - 1} = 2 - \frac{1}{2^{n - 1}}.b n=0+k=1∑n2 k−1 1=2 1−1 1⋅(2 1)n−1=2−2 n−1 1.
Note that the summation above is a finite geometric series. But since b n=1 2 n−1 T n b_n = \dfrac{1}{2^{n-1}} T_n b n=2 n−1 1T n, the closed form of T n T_n T n is therefore
T n=2 n−1(2−1 2 n−1)=2 n−1.□T_n = 2^{n - 1} \left( 2 - \frac{1}{2^{n - 1}} \right) = \boxed{2^n - 1}.\ _\square T n=2 n−1(2−2 n−1 1)=2 n−1.□
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3048 | https://courses.lumenlearning.com/odessa-collegealgebra/chapter/graphing-parabolas-with-vertices-at-the-origin/ | Graphing Parabolas with Vertices at the Origin | College Algebra
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College Algebra
The Parabola
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Graphing Parabolas with Vertices at the Origin
In The Ellipse, we saw that an ellipse is formed when a plane cuts through a right circular cone. If the plane is parallel to the edge of the cone, an unbounded curve is formed. This curve is a parabola.
Figure 2. Parabola
Like the ellipse and hyperbola, the parabola can also be defined by a set of points in the coordinate plane. A parabola is the set of all points (x,y)(x,y) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix.
We previously learned about a parabola’s vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola. See Figure 3. Notice that the axis of symmetry passes through the focus and vertex and is perpendicular to the directrix. The vertex is the midpoint between the directrix and the focus.
The line segment that passes through the focus and is parallel to the directrix is called the latus rectum. The endpoints of the latus rectum lie on the curve. By definition, the distance d d from the focus to any point P P on the parabola is equal to the distance from P P to the directrix.
Figure 3. Key features of the parabola
To work with parabolas in the coordinate plane, we consider two cases: those with a vertex at the origin and those with a vertex at a point other than the origin. We begin with the former.
Figure 4
Let (x,y)(x,y) be a point on the parabola with vertex (0,0)(0,0), focus (0,p)(0,p), and directrix y=−p y=−p as shown in Figure 4. The distance d d from point (x,y)(x,y) to point (x,−p)(x,−p)on the directrix is the difference of the y-values: d=y+p d=y+p. The distance from the focus (0,p)(0,p) to the point (x,y)(x,y) is also equal to d d and can be expressed using the distance formula.
d=√(x−0)2+(y−p)2=√x 2+(y−p)2 d=(x−0)2+(y−p)2=x 2+(y−p)2
Set the two expressions for d d equal to each other and solve for y y to derive the equation of the parabola. We do this because the distance from (x,y)(x,y) to (0,p)(0,p) equals the distance from (x,y)(x,y) to (x,−p)(x,−p).
√x 2+(y−p)2=y+p x 2+(y−p)2=y+p
We then square both sides of the equation, expand the squared terms, and simplify by combining like terms.
x 2+(y−p)2=(y+p)2 x 2+y 2−2 p y+p 2=y 2+2 p y+p 2 x 2−2 p y=2 p y x 2=4 p y x 2+(y−p)2=(y+p)2 x 2+y 2−2 p y+p 2=y 2+2 p y+p 2 x 2−2 p y=2 p y x 2=4 p y
The equations of parabolas with vertex (0,0)(0,0) are y 2=4 p x y 2=4 p x when the x-axis is the axis of symmetry and x 2=4 p y x 2=4 p y when the y-axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features.
A General Note: Standard Forms of Parabolas with Vertex (0, 0)
The table below and Figure 5 summarize the standard features of parabolas with a vertex at the origin.
Axis of SymmetryEquationFocusDirectrixEndpoints of Latus Rectum
x-axis y 2=4 p x y 2=4 p x(p,0)(p,0)x=−p x=−p(p,±2 p)(p,±2 p)
y-axis x 2=4 p y x 2=4 p y(0,p)(0,p)y=−p y=−p(±2 p,p)(±2 p,p)
Figure 5. (a) When p>0 p>0 and the axis of symmetry is the x-axis, the parabola opens right. (b) When p<0 p<0 and the axis of symmetry is the x-axis, the parabola opens left. (c) When p<0 p<0 and the axis of symmetry is the y-axis, the parabola opens up. (d) When p<0 p<0 and the axis of symmetry is the y-axis, the parabola opens down.
The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and latus rectum. When given a standard equation for a parabola centered at the origin, we can easily identify the key features to graph the parabola.
A line is said to be tangent to a curve if it intersects the curve at exactly one point. If we sketch lines tangent to the parabola at the endpoints of the latus rectum, these lines intersect on the axis of symmetry, as shown in Figure 6.
Figure 6
How To: Given a standard form equation for a parabola centered at (0, 0), sketch the graph.
Determine which of the standard forms applies to the given equation: y 2=4 p x y 2=4 p x or x 2=4 p y x 2=4 p y.
Use the standard form identified in Step 1 to determine the axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum.
If the equation is in the form y 2=4 p x y 2=4 p x, then
the axis of symmetry is the x-axis, y=0 y=0
set 4 p 4 p equal to the coefficient of x in the given equation to solve for p p. If p>0 p>0, the parabola opens right. If p<0 p<0, the parabola opens left.
use p p to find the coordinates of the focus, (p,0)(p,0)
use p p to find the equation of the directrix, x=−p x=−p
use p p to find the endpoints of the latus rectum, (p,±2 p)(p,±2 p). Alternately, substitute x=p x=p into the original equation.
If the equation is in the form x 2=4 p y x 2=4 p y, then
the axis of symmetry is the y-axis, x=0 x=0
set 4 p 4 p equal to the coefficient of y in the given equation to solve for p p. If p>0 p>0, the parabola opens up. If p<0 p<0, the parabola opens down.
use p p to find the coordinates of the focus, (0,p)(0,p)
use p p to find equation of the directrix, y=−p y=−p
use p p to find the endpoints of the latus rectum, (±2 p,p)(±2 p,p)
Plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.
Example 1: Graphing a Parabola with Vertex (0, 0) and the x-axis as the Axis of Symmetry
Graph y 2=24 x y 2=24 x. Identify and label the focus, directrix, and endpoints of the latus rectum.
Solution
The standard form that applies to the given equation is y 2=4 p x y 2=4 p x. Thus, the axis of symmetry is the x-axis. It follows that:
24=4 p 24=4 p, so p=6 p=6. Since p>0 p>0, the parabola opens right the coordinates of the focus are (p,0)=(6,0)(p,0)=(6,0)
the equation of the directrix is x=−p=−6 x=−p=−6
the endpoints of the latus rectum have the same x-coordinate at the focus. To find the endpoints, substitute x=6 x=6 into the original equation: (6,±12)(6,±12)
Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.
Figure 7
Try It 1
Graph y 2=−16 x y 2=−16 x. Identify and label the focus, directrix, and endpoints of the latus rectum.
Solution
Example 2: Graphing a Parabola with Vertex (0, 0) and the y-axis as the Axis of Symmetry
Graph x 2=−6 y x 2=−6 y. Identify and label the focus, directrix, and endpoints of the latus rectum.
Solution
The standard form that applies to the given equation is x 2=4 p y x 2=4 p y. Thus, the axis of symmetry is the y-axis. It follows that:
−6=4 p−6=4 p, so p=−3 2 p=−3 2. Since p<0 p<0, the parabola opens down.
the coordinates of the focus are (0,p)=(0,−3 2)(0,p)=(0,−3 2)
the equation of the directrix is y=−p=3 2 y=−p=3 2
the endpoints of the latus rectum can be found by substituting y=3 2 y=3 2 into the original equation, (±3,−3 2)(±3,−3 2)
Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.
Figure 8
Try It 2
Graph x 2=8 y x 2=8 y. Identify and label the focus, directrix, and endpoints of the latus rectum.
Solution
Writing Equations of Parabolas in Standard Form
In the previous examples, we used the standard form equation of a parabola to calculate the locations of its key features. We can also use the calculations in reverse to write an equation for a parabola when given its key features.
How To: Given its focus and directrix, write the equation for a parabola in standard form.
Determine whether the axis of symmetry is the x– or y-axis.
If the given coordinates of the focus have the form (p,0)(p,0), then the axis of symmetry is the x-axis. Use the standard form y 2=4 p x y 2=4 p x.
If the given coordinates of the focus have the form (0,p)(0,p), then the axis of symmetry is the y-axis. Use the standard form x 2=4 p y x 2=4 p y.
Multiply 4 p 4 p.
Substitute the value from Step 2 into the equation determined in Step 1.
Example 3: Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix
What is the equation for the parabola with focus(−1 2,0)(−1 2,0) and directrixx=1 2?x=1 2?
Solution
The focus has the form (p,0)(p,0), so the equation will have the form y 2=4 p x y 2=4 p x.
Multiplying 4 p 4 p, we have 4 p=4(−1 2)=−2 4 p=4(−1 2)=−2.Substituting for 4 p 4 p, we have y 2=4 p x=−2 x y 2=4 p x=−2 x.
Therefore, the equation for the parabola is y 2=−2 x y 2=−2 x.
Try It 3
What is the equation for the parabola with focus (0,7 2)(0,7 2) and directrix y=−7 2?y=−7 2?
Solution
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3049 | https://brilliant.org/wiki/riemann-zeta-function/ | Riemann Zeta Function
Andrew Ellinor,
Aareyan Manzoor,
Aditya Kumar,
and
Rasel Parvej
Kishore S. Shenoy
Yao Liu
Mayank Chaturvedi
A Former Brilliant Member
Alex G
Rahil Sehgal
Worranat Pakornrat
Sravanth C.
Chew-Seong Cheong
Ralph James
Jimin Khim
Calvin Lin
contributed
Contents
Definition
Euler Product Representation
Integral Representation
Functional Equations
Zeta Function over Even and Negative Integers
Relation to Prime Zeta
Relation to Prime Counting Function π(x)
Riemann Zeta as a Special Case of Other Known Functions
Values of Riemann Zeta Function
The Riemann zeta function for s∈C with Re(s)>1 is defined as
ζ(s)=n=1∑∞ns1.
It is then defined by analytical continuation to a meromorphic function on the whole C by a functional equation.
Euler Product Representation
The Riemann zeta function for s∈C with Re(s)>1 can be written as
ζ(s)=p prime∏1−p−s1.
Dirichlet series is used here.
Proof 1:
We know that
n=1∑∞nsf(n)=p=prime∏1−f(p)p−s1
such that the summation converges and f(n) is a completely multiplicative function. Since 1(n)=1 is completely multiplicative, we haven=1∑∞ns1=p=prime∏1−p−s1. □
Proof 2:
Since ns1 is multiplicative, we take the product of the sum over all prime powers:
n=1∑∞ns1=p=prime∏k=0∑∞(pk)s1.
The LHS's sum is just the geometric progression sum. We have it equal to
n=1∑∞ns1=p=prime∏1−p−s1. □
Proof 3:
(The addition of this proof in this section is intended to simplify the discussion of Euler's proof on the infinitude of number of primes. This is actually a detailed form of Proof 2.)
This proof will deal only with real number s>1. We'll also ignore technicalities about convergence. But the absolute convergence (not just conditional convergence) of the sum and product can be used to justify our manipulations.
Recall the formula for the sum of a geometric series
1+r+r2+r3+⋯=1−r1=(1−r)−1,
where ∣r∣<1. Letting r=p−s, where p is a prime, we obtain
1+ps1+p2s1+p3s1+⋯=(1−p−s)−1.
Now, consider the product
(1−2−s)−1(1−3−s)−1=(1+2s1+4s1+8s1+⋯)(1+3s1+9s1+27s1+⋯).
When we expand this, we obtain
1+2s1+4s1+8s1+⋯+3s1+2s3s1+4s3s1+⋯+9s1+2s9s1+4s9s1+⋯=1+2s1+3s1+4s1+6s1+8s1+9s1+12s1+⋯=n∈S(2,3)∑ns1,
where S(p,q,⋯) denotes the set of all such integers (including 1=p0q0⋯) whose prime factorizations involve only primes in {p,q,⋯}, which is a set of primes. Note that each integer from S(2,3) occurs exactly once in the sum. For example, 12∈S(2,3), and the term 12s1 comes from 4s1×3s1. The uniqueness of factorization says that this is the only way it can occur.
When we multiply by (1−5−s)−1, we get
(1−2−s)−1(1−3−s)−1(1−5−s)−1=n∈S(2,3,5)∑ns1.
In general, if pm is the mth prime, we have
(1−2−s)−1(1−3−s)−1(1−5−s)−1⋯(1−pm−s)−1=n∈S(2,3,5,…,pm)∑ns1.
Let m→∞. The left side converges to the product over all primes. Since every positive integer has a prime factorization, each positive integer n lies in S(2,3,5,⋯,pm) for a sufficiently large integer m. Therefore, the right side converges to the sum over all positive integers n. This gives the identity in the theorem. That is,
n=1∑∞ns1=p=prime∏1−p−s1. □
An interesting result that comes from this is the fact that there are infinite prime numbers.
As at s≤1 it diverges, the product must be taken over infinite numbers, and hence there are infinite primes.
p=prime∏(1−p−11+1+p−11−1)
Find the value of the closed form of the above product.
The correct answer is: 2.5
p prime∏p2+1p2
If the product above is equal to BπA for positive integers A and B, find 2A+B.
The correct answer is: 19
Integral Representation
The zeta function can be represented as
Γ(s)ζ(s)=∫0∞ex−1xs−1dx.
Consider the gamma function
Γ(s)=∫0∞xs−1e−xdx.
We substitute x=nu and thus dx=ndu:
Γ(s)⇒Γ(s)ns1=∫0∞ns−1us−1e−nundu=∫0∞nsus−1e−nudu=∫0∞us−1e−nudu.
We take the sum over all the positive integers as n:
Γ(s)n=1∑∞ns1=n=1∑∞∫0∞us−1e−nudu=∫0∞us−1n=1∑∞e−nudu.
We apply the formula for the sum of a geometric progression:
Γ(s)ζ(s)=∫0∞eu−1us−1du. □
Functional Equations
π−2sΓ(2s)ζ(s)=π−21−sΓ(21−s)ζ(1−s)
Consider
Γ(2s)=∫0∞x2s−1e−xdx.
Let x=πn2u→dx=πn2du→∣0∞=∣0∞. So
π−2sΓ(2s)n−s=∫0∞x2s−1e−πn2xdx.
Summing over all positive integers,
π−2sΓ(2s)n=1∑∞n−s=n=1∑∞∫0∞x2s−1e−πn2xdx=∫0∞x2s−1n=1∑∞e−πn2xdx.
We recall the Jacobi theta function:
θ(x)ψ(x)θ(x)xθ(x)=n=−∞∑∞e−πn2x=2n=1∑∞e−πn2x+1=n=1∑∞e−πn2x=2ψ(x)+1=θ(x−1).
So, let
ξ(s)=π−2sΓ(2s)ζ(s)=∫0∞x2s−1ψ(x)dx.
Now split up in two parts
ξ(s)=∫01x2s−1ψ(x)dx+∫1∞x2s−1ψ(x)dx.
Consider
xθ(x)=θ(x−1)→x(2ψ(x)+1)=2ψ(x−1)+1→ψ(x)=x1ψ(x1)+2x1−21.
Putting this in the first integral,
∫01x2s−1ψ(x)dx=∫01x2s−1(x1ψ(x1)+2x1−21)dx=∫01x2s−23ψ(x1)+21(x2s−23−x2s−1)dx.
We can easily integrate the terms independent of ψ to get the integral to be
∫01x2s−23ψ(x1)dx+s(s−1)1.
Change x with x1 to get
∫1∞x−2s−21ψ(x)dx+s(s−1)1.
So,
ξ(s)=∫01x2s−1ψ(x)dx+∫1∞x2s−1ψ(x)dx=∫1∞(x2(1−s)−1+x2s−1)ψ(x)dx+s(s−1)1.
Note that
ξ(s)=ξ(1−s).
So
π−2sΓ(2s)ζ(s)=π−21−sΓ(21−s)ζ(1−s). □
ζ(s)=2sπs−1sin(2πs)Γ(1−s)ζ(1−s)
Consider
π−2sΓ(2s)ζ(s)⇒π−2sΓ(2s)Γ(2s+1)ζ(s)=π−21−sΓ(21−s)ζ(1−s)=π−21−sΓ(2s+1)Γ(1−21+s)ζ(1−s).
Using Lagrange duplication formula and Euler reflection formula,
π−2s2s−1πΓ(s)ζ(s)⇒ζ(1−s)=π−21−scos(2πs)πζ(1−s)=21−sπ−scos(2πs)Γ(s)ζ(s).
Replacing s with 1−s gives
ζ(s)=2sπs−1sin(2πs)Γ(1−s)ζ(1−s). □
Zeta Function over Even and Negative Integers
For any positive even integer 2n,
ζ(2n)=2(2n)!(−1)n+1β2n(2π)2n,
where βn denotes the nth Bernoulli number.
Consider
πssin(πs)=n=1∏∞(1−n2s2).
Taking log and d.w.r.s
πscot(πs)=1−2n=1∑∞1−n2s2n2s2=1−2n=1∑∞n2s2k=0∑∞(n2s2)k=1−2n=1∑∞k=1∑∞(n2s2)k.
Interchanging the summations,
πscot(πs)=1−2k=1∑∞s2kn=1∑∞n−2k=1−2k=1∑∞ζ(2k)s2k.
Now consider
πscot(πs)=iπs+e2iπs−12πis=iπs+n=0∑∞n!βn(2iπs)n.
The Taylor series of es−1s is used. Since β2n+1=0 for 2n+1 being an odd positive integer >1, β0=1,β1=−0.5:
πscot(πs)=1−2n=1∑∞2(2n!)−β2n(2iπs)2n.
Since in a power series the coefficients must be the same, we have
1−2k=1∑∞ζ(2k)s2k⇒ζ(2n)=1−2n=1∑∞2(2n!)−β2n(2iπs)2n=2(2n)!(−1)n+1β2n(2π)2n. □
ζ(−n)=−n+1βn+1,
where −n is a negative integer.
Consider
ζ(s)=2sπs−1sin(2πs)Γ(1−s)ζ(1−s).
First notice if s is −2n for n positive integer, the sin is zero and hence the LHS is zero. So
ζ(−2n)=0=β2n+1=−2n+1β2n+1∀n∈N.
This is also known as the trivial zeroes of the zeta function.
Now replace s with −2n+1, then the sin becomes (−1)n and
ζ(−2n+1)=2−2n+1π−2nΓ(2n)ζ(2n)=2−2n+1π−2n(2n−1)!(−1)n2(2n)!(−1)n+1β2n(2π)2n=−2nβ2n=−2n−1+1β2n−1+1.
So, we have
ζ(−n)=−n+1βn+1. □
Relation to Prime Zeta
ln(ζ(s))=n=1∑∞nP(sn)
View Prime Zeta Function.
We recall the Euler product definition:
ζ(s)=p=prime∏1−p−s1.
Taking log of both sides,
ln(ζ(s))=p=prime∑(−ln(1−p−s)).
Using the common Taylor series of log, we have
ln(ζ(s))=p=prime∑n=1∑∞np−sn=n=1∑∞n1p=prime∑p−sn.
We recall
P(x)=p=prime∑p−x,
so the summation is just
ln(ζ(s))=n=1∑∞nP(sn). □
Relation to Prime Counting Function π(x)
Riemann Zeta as a Special Case of Other Known Functions
Special Case of Hurwitz Zeta Function
Hurwitz zeta function is defined as
ζ(a,b)=n=0∑∞(b+n)a1.
If we take b=1, we'll get
ζ(a,1)=n=0∑∞(1+n)a1.
Now on rearranging the limits, we get
ζ(a,1)=n=1∑∞(n)a1=ζ(a).
Hence, we can conclude that the Riemann zeta function is the special case of the Hurwitz zeta function.
Therefore, ζ(a,1)=ζ(a).
Special Case of Polygamma Function
Polygamma function is defined as
ψa(x)=dxa+1da+1(lnΓ(x)).
It is also represented as
ψa(x)=n=0∑∞(x+n)a(−1)aa!=ζ(a,x).
Now for x=1 we get
ψa(1)=n=0∑∞(1+n)a(−1)nn!=ζ(a).
Hence, we can conclude that the Riemann zeta function is the special case of Polygamma function.
Therefore, ψa(1)=ζ(a).
Special Case of Dirichlet Series
View Dirichlet Series.
We define a Dirichlet series as
n=1∑∞nsf(n),
where f is an arithmetic function. At the special case of f(n)=1(n)=1, it is the zeta function. Also, many other Dirichlet series give result in terms of the zeta function, as you can see in the wiki or these notes.
These relations are very useful while solving problems.
Values of Riemann Zeta Function
Lemma: k=1∑∞ζ(2k)x2k=21(1−πxcot(πx))
Proof:
We have
f(x)=k=1∑∞ζ(2k)x2k=k=1∑∞i=1∑∞i2kx2k=i=1∑∞1−i2x2i2x2=i=1∑∞i2−x2x2=−2xi=1∑∞(x−i1+x+i1)=−2x(πcot(πx)−x1)=21(1−πxcot(πx)).(1)
Now, we deviate from what we have obtained at (1). W'll evaluate the series expansion of xcot(x).
Consider the series xcot(x)=∑k=0∞akx2k. Here an is an unknown real valued function:
cos(x)=xsin(x)k=0∑∞akx2k.
On using the Taylor series of expansion of cos(x) and sin(x), we get
n=0∑∞(−1)n(2n)!x2n=n=0∑∞(−1)n(2n+1)!x2nk=0∑∞akx2k=n=0∑∞(k=0∑n(−1)k(2k+1)!an−k)x2n.(2)
On comparing coefficients in (2), we get
an=(2n)!(−1)n−i=1∑n(−1)i(2i+1)!an−i=(2n+1)!(−1)n2n−i=1∑n−1(−1)i(2i+1)!an−i.(3)
From (1), we can deduce that an=−2π2nζ(2n).
Hence (3) becomes
ζ(2n)=(2n+1)!(−1)n−1π2nn+k=1∑n−1(2k+1)!(−1)k−1π2kζ(2n−2k).
This equation is true for n>1 with n being an integer.
From the recursive relation we get
ζ(2)=3!π2=6π2
ζ(4)=−5!π42+3!π2ζ(2)=90π4
ζ(6)=7!π63−5!π4ζ(2)+3!π2ζ(4)=945π6
ζ(8)=−9!π84+7!π6ζ(2)−5!π4ζ(4)+3!π2ζ(6)=9450π8.
Cite as:
Riemann Zeta Function.
Brilliant.org.
Retrieved
03:16, September 28, 2025,
from |
3050 | https://www.ncbi.nlm.nih.gov/books/NBK559161/ | Liver Transplantation - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
StatPearls [Internet].
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Liver Transplantation
Yara Dababneh; Omar Y. Mousa.
Author Information and Affiliations
Authors
Yara Dababneh 1; Omar Y. Mousa 2.
Affiliations
1 The Hashemite University
2 Mayo Clinic
Last Update: June 13, 2025.
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Continuing Education Activity
Cirrhosis and decompensated liver disease were the ninth leading cause of death for men in the United States in 2016, highlighting the critical need for advanced treatment strategies. Liver transplantation (LT) is a lifesaving and proven intervention that offers patients with acute or chronic end-stage liver disease a chance at restored health, improved lifestyle, and an extended lifespan, often by 15 years. As a last resort when all medical therapies have failed, LT has become a cornerstone in the management of a wide range of liver diseases, including decompensated cirrhosis, hepatocellular carcinoma, metabolic disorders, and acute liver failure. Advances in surgical techniques, immunosuppression, and long-term care, combined with the growing use of living donors and split grafts, have significantly improved survival rates and patient quality of life. Yet, LT remains a complex, high-stakes procedure with risks such as graft rejection, infection, long-term metabolic complications, and malignancies, demanding careful patient selection, rigorous surgical precision, and expert postoperative management.
This comprehensive educational activity offers an in-depth, evidence-based review of liver transplantation, guiding learners through updated anatomy, physiology, clinical indications, contraindications, surgical approaches, and the evolving landscape of transplant medicine. As the burden of liver disease shifts—with rising rates of metabolic dysfunction-associated steatohepatitis and alcohol-related liver disease—this course equips clinicians with the tools to navigate nuanced patient selection, including model for end-stage liver disease (MELD)-based prioritization, MELD-exception scenarios, and updated scoring systems like MELD-sodium (Na) and MELD 3.0. Participants gain expertise in perioperative and postoperative management, complication recognition, and outcome optimization, with special attention to transplant indications in autoimmune hepatitis, cholangiopathies, hepatocellular carcinoma, and metabolic diseases. By completing this course, learners enhance their clinical decision-making, strengthen interprofessional communication, and improve care coordination alongside hepatologists, surgeons, psychiatrists, nutritionists, social workers, nurses, pharmacists, and advanced practitioners—ultimately elevating patient-centered care, safety, and transplant outcomes.
Objectives:
Identify clinical scenarios and disease states that warrant evaluation for liver transplantation, including acute liver failure, acute-on-chronic liver failure, cirrhosis with high model for end-stage liver disease or hepatocellular carcinoma, and metabolic liver diseases.
Assess liver transplant candidates for liver health using validated prognostic scoring systems such as the model for end-stage liver disease (MELD), MELD-sodium (Na), and the newer MELD 3.0, including MELD exception criteria.
Differentiate surgical approaches to liver transplantation and recognize common early and late postoperative complications.
Collaborate with interprofessional healthcare team members—including hepatologists, surgeons, psychiatrists, nutritionists, nurses, and social workers—to optimize outcomes for patients undergoing liver transplant.
Access free multiple choice questions on this topic.
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Introduction
Cirrhosis and decompensated liver disease were the ninth leading cause of death for men in 2016 in the United States.Liver transplantation (LT) is a lifesaving gift and procedure, as well as a proven intervention in caring for individuals with acute and chronic end-stage liver disease.This procedure restores normal health, lifestyle, and extends lifespan by 15 years.The advent of liver transplantation has provided a critical safety net for treating advanced liver diseases when all other medical interventions have been exhausted. According to the Scientific Registry of Transplant Recipients data, overall patient survival is excellent, reaching 90% at 1 year following deceased donor LT and 77% at 5 years.Since LT was first attempted in 1963, there have been continuous advances and major improvements in the surgical technique, type of organ donation, with the expansion of the organ donation pool, and a significant focus on the quality of life for both recipients and donors.Still, there are important challenges, including the shortage of donor organs, the selection of LT candidates, and organ allocation.
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Anatomy and Physiology
The liver, the largest organ in the human body, lies beneath the eighth through twelfth ribs on the right side.Anatomically, it consists of 4 lobes: the right and left lobes (divided by the falciform ligament), the quadrate lobe, and the caudate lobe. However, these are not true functional lobes; functionally, the liver is divided into right and left lobes by the Cantlie line, which runs through the gallbladder bed and the notch of the inferior vena cava. Each lobe is subdivided into 2 segments, and these are further divided into subsegments based on hepatic arterial and portal venous blood supply, biliary drainage, and hepatic venous outflow. The segments are numbered 1 to 8 in the Couinaud classification, with the caudate lobe designated as segment 1 and the rest numbered clockwise.
The liver’s unique dual blood supply comes from both systemic (hepatic artery) and portal (portal vein) circulations. The liver’s functional unit, the hepatocyte, is organized into zones: zone I (periportal) receives the highest oxygenated blood and focuses on oxidative metabolism; zone II (midzonal) lies between zones I and III; zone III (pericentral) is the farthest from the blood supply and specializes in drug and toxin detoxification. Importantly, bile and blood flow in opposite directions, with bile exiting the liver and blood entering to nourish it. Typically, the liver’s volume comprises about 60% from the right lobe and 40% from the left, with the left lateral segment alone accounting for roughly 20%.
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Indications
LT is indicated in cases of acute or chronic end-stage liver disease where medical therapy has failed. Patients who experience hepatic decompensation, such as hepatic encephalopathy, variceal hemorrhage, or ascites, should first undergo medical management. If they are potential LT candidates, a thorough transplant evaluation should be initiated. Notably, up to 80% of LTs are performed for decompensated cirrhosis. Patients with cirrhosis are typically classified using the Child-Turcotte-Pugh score, which integrates both biochemical markers (serum albumin, serum bilirubin, and International Normalized Ratio [INR]) and clinical findings (ascites and encephalopathy) to estimate prognosis and guide management.
Model for End-Stage Liver Disease
The model for end-stage liver disease (MELD) score was originally developed to predict survival following the transjugular intrahepatic portosystemic shunt procedure. Still, it was later found to be an effective predictor of survival in patients with cirrhosis and has become widely accepted as a key tool for prioritizing organ allocation for LT. The MELD score estimates 3-month mortality by calculating serum creatinine, bilirubin, and INR levels. For pediatric patients, a modified version—the pediatric end-stage liver disease score—replaces creatinine with factors like age, albumin, and failure to thrive. In 2016, the Organ Procurement and Transplantation Network updated MELD to include serum sodium (eg, MELD-Na), as hyponatremia is a common and important marker of cirrhosis severity.
A strong indication for LT evaluation arises when individuals with cirrhosis develop complications such as ascites, variceal bleeding, hepatic encephalopathy, or hepatocellular dysfunction, typically reflected by a MELD score 15 or greater. However, MELD has known limitations, as it does not account for complications like refractory ascites, recurrent gastrointestinal bleeding, chronic encephalopathy, hepatopulmonary syndrome, portopulmonary hypertension, or severe pruritus—all of which significantly increase mortality risk. To address these gaps, MELD exception points are granted for conditions such as hepatocellular carcinoma, Budd-Chiari syndrome, familial amyloidotic polyneuropathy, cystic fibrosis, hereditary hemorrhagic telangiectasia, polycystic liver disease, primary hyperoxaluria, recurrent cholangitis, rare metabolic diseases, and certain malignancies, including cholangiocarcinoma, to ensure fair access to transplantation. Additionally, newer scoring systems—including MELD-Na, integrated MELD, delta MELD, and the recently developed MELD 3.0 (which incorporates sex, serum albumin, and reweighted variables)—aim to improve prognostic accuracy, particularly for women and patients with lower MELD scores.
Specific Indications for LT
Acute liver failure
Patients with acute liver failure (ALF) rapidly deteriorate and develop severe liver dysfunction, elevated bilirubin, aminotransferases, encephalopathy, and coagulopathy (above 1.5). Acetaminophen is responsible for almost half the cases of ALF in the United States.ALF is considered a strong high (1a) indication for LT as it supersedes all other etiologies of chronic liver disease and takes precedence on the United Network for Organ Sharing waiting list.To achieve priority on the waiting list as an (1a) ALF case, the following criteria must be met:
Intensive care unit admission
On ventilatory support
Requiring hemodialysis
Elevated INR above 2
Development of hepatic encephalopathy within 8 weeks of the onset of symptoms
Compared to LT due to chronic disease, the 1-year survival in LT due to ALF is worse, but with higher survival rates following the first year.
Acute-on-chronic liver failure
This condition has a very high mortality without transplantation,but outcomes are influenced by multiple factors related to the recipient, intensive care unit (ICU) course, and donor characteristics. Advanced age, particularly in those older than 60, and comorbidities such as diabetes, cardiovascular disease, and a high age-adjusted Charlson Comorbidity Index (≥6) are associated with poor posttransplant survival. Frailty and sarcopenia also impact prognosis, with severe frailty being a contraindication. ICU-related factors, including respiratory failure (PaO 2/FiO 2 ≤200), mechanical ventilation, renal replacement therapy, and vasopressor use, predict worse outcomes, although stabilization before LT improves survival. The timing of LT is critical, with study results suggesting optimal outcomes when performed within 7 to 15 days of ICU admission. Pretransplant infection with multidrug-resistant organisms, active sepsis, and profound leukopenia (<500/mm³) are absolute contraindications. Several risk models, such as the transplantation for acute-on-chronic liver failure (ACLF) grade 3 model and the Sundaram ACLF-LT-mortality model, have been proposed to predict mortality after LT in ACLF; however, clinical judgment remains paramount in selecting candidates for transplantation.
Metabolic dysfunction-associated steatohepatitis
Metabolic dysfunction-associated steatohepatitis (MASH), previously known as nonalcoholic steatohepatitis (NASH),is considered among the top indications for LT.MASH is included in the spectrum of nonalcoholic fatty liver disease, ranging from steatosis alone to NASH with accompanying cirrhosis. These liver diseases are linked to metabolic syndrome with increased body mass index and obesity.Because,until now, there was no effective treatment for MASH or fibrosis (recently, drugs have shown fibrosis reversal in MASH),LT has been rising due to NASH.Also, it has been noted that patients diagnosed with MASH with and without cirrhosis have an increased risk of developing hepatocellular cancer (HCC). Still, this condition should not be considered an indication for LT without cirrhosis and a high MELD score.
Alcohol-related liver disease
This condition is currently the most common indication for LT.Patients with alcohol use disorder should be referred for psychosocial and psychiatric support before LT to ensure at least 6 months of abstinence and prevent relapses, especially since these relapses result in delisting the patient from the waiting list.In cases of acute alcoholic hepatitis that do not respond to medical therapy,LT may be required while less than 6 months of abstinence is achieved.
Hepatitis B
This infection previously resulted in increasing numbers of chronic liver disease, but with the use of hepatitis B immunoglobulins and the introduction of antivirals, hepatitis B has resulted in decreased rates of LT.Additionally, treating and controlling the infection is crucial to prevent reinfection after transplantation. Hepatitis B can be complicated by HCC, which serves as an important indication for LT.
Chronic hepatitis C
Cirrhosis due to chronic hepatitis C infection was the most common indication for LT until 2015.Since 2016, hepatitis C infection has become the third most common indication, surpassed by alcohol-related liver disease and NASH. To prevent reinfection and subsequent graft failure after LT, eradication of chronic hepatitis C infection before transplantation was historically critical.However, the emergence of novel direct-acting antiviral agents over the past decade has transformed management, now enabling the effective treatment of chronic hepatitis C infection even after LT.
Autoimmune hepatitis
This infection can lead to liver cirrhosis and failure, even with long-term corticosteroids and immunosuppression therapy. LT is indicated in ALF secondary to autoimmune hepatitis (AIH) or cases of chronic decompensated cirrhosis due to AIH.Poor outcomes and the need for LT can be predicted by the following observations: young age, a MELD score higher than 12, multiple relapses, and a delayed downward slope of aminotransferase after treatment.
Primary biliary cirrhosis
Patients with decompensated cirrhosis or severe pruritus refractory to other medical interventions require LT. Over the years, the need for LT has decreased with the use of ursodeoxycholic acid to treat primary biliary cirrhosis, which slows disease progression.
Primary sclerosing cholangitis
Since there is no effective medical therapy for PSC, LT is considered an effective treatment modality among patients with the decompensated disease or those who develop perihilar cholangiocarcinoma (within certain criteria) or recurrent bouts of bacterial cholangitis.PSC is associated with inflammatory bowel disease; therefore, frequent colonoscopy is necessary to screen for colorectal cancer before and after LT.
HCC
Patients with HCC must meet the Milan criteria to be eligible for LT. This criterion includes:
One tumor less than 5 cm in diameter, or 3 tumors each having a diameter less than 3 cm, that must be confirmed by computed tomography (CT) or magnetic resonance imaging
Without any metastasis documented by chest CT and bone scan
Absence of major vessel involvement
An unresectable tumor
Patients with HCC usually have a normal liver function, and their MELD score is usually normal or low. Therefore, they undergo an MELD exception to obtain a score that allows them to be prioritized on the waiting list for LT.
Cholangiocarcinoma
LT is considered in the management of patients with early-stage cholangiocarcinoma with nonresectable perihilar lesions (<3 cm in diameter) or underlying parenchymal liver disease such as primary sclerosing cholangitis with cirrhosis. In addition to the tumor criteria, LT should be performed in combination with neoadjuvant chemotherapy regimens to ensure higher survival rates than those achieved previously without neoadjuvant therapy.Patients are granted a MELD exception if they meet the eligibility criteria on the United Network for Organ Sharing waiting list.
Metabolic liver diseases
Wilson disease
LT is indicated in patients with ALF due to Wilson disease or in cases of decompensated cirrhosis that have failed all medical therapies.LT in Wilson disease has great outcomes, even in cases with metabolic complications like renal failure, which resolves after LT.Parents of patients who are heterozygous can serve as living donors for transplantation with successful results.Other cirrhosis-related complications, such as hepatopulmonary syndrome and portopulmonary hypertension, are also indications for LT.
Alpha-1 antitrypsin deficiency
This disease is usually diagnosed in adults without any previous history of liver disease. LT is considered the main and only treatment modality for decompensated liver disease secondary to alpha-1 antitrypsin deficiency.There is no risk of recurrence due to the expression of the donor's alpha-1 antitrypsin gene after LT. Patients should undergo screening for lung disease by chest imaging and pulmonary function testing.
Hereditary hemochromatosis
LT is indicated in decompensated cirrhosis or in patients with HCC who have hereditary hemochromatosis (HH). Cirrhosis, due to HH, accounts for the highest risk of developing HCC among all other causes of cirrhosis.The use of iron reduction therapy through phlebotomy before LT has resulted in improved outcomes posttransplantation.
Familial amyloid polyneuropathy
LT has proven effective in familial amyloid polyneuropathy due to a mutation in the transthyretin gene.LT is most effective in patients younger than 50 because LT neither improves nor reverses neurological symptoms, but rather prevents disease progression.Also, patients with hereditary renal amyloidosis inherited as an autosomal dominant gene usually benefit from liver and renal transplantation. LT doesn't seem to affect the ocular or cardiac effects of amyloidosis.
Primary hyperoxaluria type I
This is an autosomal recessive defect that results in a deficiency of alanine: glyoxylate aminotransferase in the liver, impairing the metabolism of glyoxylate and leading to the overproduction of oxalate.This disease usually leads to end-stage renal disease at 20 to 40 years. LT is effective in curing the disease.Both liver and renal transplants are reported to be successful in patients with end-stage renal disease and liver failure, and improve cardiomyopathy in these patients.
Other metabolic liver diseases that have indications for LT include:
Cystic fibrosis
Glycogen storage diseases
Graft failure
This is a significant indication for liver retransplantation. Hyperacute rejection, which causes hepatic artery thrombosis and graft failure, typically occurs immediately after surgery. Although retransplantation can be done, the outcomes are worse than those of the primary transplant.
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Contraindications
While the indications for LT are becoming broader, the contraindications are becoming fewer due to advancements in LT techniques. Still,LT has some absolute and relative contraindications.
Absolute Contraindications
MELD score of less than 15 (without MELD exception)
Advanced cardiac or pulmonary disease
Active alcohol or illicit substance use
HCC or perihilar cholangiocarcinoma with metastatic spread
Untreated septic shock or sepsis
An anatomic abnormality that precludes LT
Intrahepatic cholangiocarcinoma
Extrahepatic malignancy, unless the patient is tumor-free for more than 2 years, with a low probability for recurrence
Hemangiosarcoma
Fulminant hepatic failure leading to sustained intracranial pressure above 50 mm Hg or cerebral perfusion pressure less than 40 mm Hg
Lack of psychosocial support and severe psychological disease
Severe pulmonary hypertension
Although acquired immunodeficiency syndrome used to be considered an absolute contraindication for LT, recent centers are selecting patients with human immunodeficiency virus alone as candidates for LT.
Relative Contraindications
General debility
Persistent noncompliance
Advanced age
Extensive previous abdominal surgery
Extensive portal or mesenteric thrombosis
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Personnel
To deliver the highest quality of patient care in LT, a coordinated interprofessional team approach is essential. In the pretransplant phase, evaluation by a hepatologist, transplant surgeon, and transplant nurse coordinator is critical to assess the patient’s medical status, review necessary vaccinations, medications, and required lifestyle modifications, explain the surgical options, and provide a detailed overview of the posttransplant phase—including immunosuppression, potential complications, and expected outcomes. Psychiatric assessment by a transplant psychiatrist is important to address any issues of alcohol or substance abuse and to ensure the patient has appropriate insight and preparedness for the procedure and its consequences. Social workers play a key role in evaluating the patient’s social support system, especially to ensure adequate posttransplant home care and adjustment. Additionally, a specialized administrative team works to secure insurance coverage for the transplant surgery and long-term immunosuppressive medications. Nutritionists are essential both before and after transplant to optimize nutritional status and guide dietary changes, particularly in patients managing chronic comorbidities such as diabetes, hypertension, and hyperlipidemia.
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Preparation
The preoperative evaluation must address numerous important aspects and health concerns in LT candidates. This evaluation should include a detailed and comprehensive history and physical examination, as well as laboratory tests and imaging studies, to conduct a full systematic review of the patient and manage them accordingly. An extensive review is available at the American Association for the Study of Liver Disease's Practice Guidelines website. Below is a summary of those guidelines:
Obesity
Patients should be evaluated for increased body mass index (BMI) as it increases the perioperative risks and reduces survival in patients with LT.Patients with a BMI of 30 kg/m² or higher should be referred to a dietitian. A BMI above 40 kg/m² is considered a relative contraindication for an LT.
Coronary artery disease
Perioperative cardiac risk assessment is crucial. All patients should undergo an electrocardiogram, as well as cardiac stress testing (especially if the patient is 50 or older), which can be either physical or chemical stress testing. If patients were found to have stenosis, coronary revascularization should be done before LT.
Age
Although the prognosis of transplant in patients older than 70 is not as good as in younger patients, older age is not a contraindication to LT in patients without or with controlled comorbidities.Recently, it has been demonstrated that carefully selected older adult candidates can benefit from LT, thereby restoring their expected lifespan.
Pulmonary hypertension
When the mean pulmonary artery pressure (MPAP) is equal to or above 25 mm Hg and is associated with portal hypertension (HTN), the condition is referred to as portopulmonary HTN. Moderate to severe portopulmonary HTN is associated with a higher mortality rate after LT; the mortality rate can reach 100% if MPAP is above 50 mm Hg.Pulmonary HTN is diagnosed by echocardiography, and if severe, right heart catheterization is a gold standard test to confirm the diagnosis. Vasodilators treat this condition, and LT is indicated in patients who respond to vasodilator therapy, reducing MPAP to less than 35 mm Hg and pulmonary vascular resistance less than 400 dynes/s/cm.
Hepatopulmonary syndrome
In this syndrome, patients with chronic liver disease experience shortness of breath and hypoxemia, especially those with portal HTN. This is due to microvascular dilation of the pulmonary vessels, resulting in an intrapulmonary shunt.Patients should be screened with pulse oximetry prior to LT.Affected patients may require a longer recovery period and long-term supplemental oxygen posttransplant, depending on the severity of hepatopulmonary syndrome.Hepatorenal syndrome is not a contraindication for LT.
Renal dysfunction
Patients with renal disease must be diagnosed before LT, as renal dysfunction significantly increases mortality.Simultaneous liver and kidney transplants are indicated if a patient has a glomerular filtration rate less than 30 mL/min, indicating chronic kidney disease or acute kidney failure that requires dialysis for more than 6 weeks. This is also indicated in case severe glomerulosclerosis is present.Hepatorenal syndrome is not a contraindication for LT and is sometimes the treatment of choice. Some forms of kidney disease, such as immunoglobulin A nephropathy, may improve after LT.
The United Network for Organ Sharing renal safety net ensures priority access to kidney transplantation for LT recipients who develop end-stage renal disease within 12 months post-LT, promoting equitable organ allocation. This policy is particularly relevant for patients with hepatorenal syndrome, who frequently experience severe pretransplant renal dysfunction, with some progressing to persistent kidney failure despite hepatic recovery. By facilitating timely kidney transplantation for those ineligible for simultaneous liver-kidney transplantation, the safety net aims to mitigate long-term dialysis dependence and improve posttransplant outcomes.
Cigarette smoking
Smoking increases mortality among LT recipients due to cardiac disease and also increases the risk of hepatic artery thrombosis.Smoking should be prohibited, and many hospitals consider smoking cessation a requirement to be listed for LT.
Extrahepatic malignancy
Patients should undergo all age-appropriate screening before undergoing LT. If they have any increased risk factors for a specific cancer, they should undergo further testing for the specific cancer type.Any patient with a previously diagnosed malignancy should be treated and cured before undergoing LT.
Infectious disease
All infectious diseases should be treated effectively before LT. Screening serologies in the blood should include viral infections such as hepatitis A and B (to ensure immunity through vaccination and absence of active infection), cytomegalovirus, Epstein-Barr virus, bacterial infections (including tuberculosis and syphilis), and fungal infections (such as Strongyloides and Coccidioidomycosis). All live attenuated viral vaccines should be administered before LT as they are contraindicated once immunosuppression is initiated after transplantation.
Nutrition
A nutritionist should evaluate patients before undergoing LT, as it is essential to address all nutritional deficiencies related to chronic liver disease and fat malabsorption. Adequate dietary control, particularly regarding other comorbidities such as diabetes mellitus, HTN, and hyperlipidemia, should be emphasized.Sarcopenia is a serious problem in cirrhosis and can be assessed by measuring the skeletal muscle index at the L3 level using CT scan images.The thickness of the psoas muscle also has a prognostic role. Assessment of frailty using simple tools, such as the Liver Frailty Index, is recommended before transplantation, followed by prehabilitation measures. Evaluation of osteoporosis using bone densitometry is also essential and should be performed for all patients.
Bone disease
Densitometry, vitamin D, and calcium levels should be obtained before LT in all candidates. Osteoporosis is relatively common in all patients with chronic liver disease due to malabsorption of vitamin D and in cases of autoimmune hepatitis due to corticosteroid use.
Human immunodeficiency virus
Patients affected by the condition can be considered candidates for LT only if their CD4 counts are above 100 µL and their viral load is undetectable before the procedure. Human immunodeficiency virus is not considered a contraindication to LT due to the advent of effective antiretroviral therapy.
Psychological evaluation
LT candidates must be evaluated for any psychiatric disorders that might affect their prognosis, compliance with medication, and medical directives.Their social support systems and caregiver availability should also be assessed, especially in patients with encephalopathy.For example, patients who have depressive symptoms, mainly in the immediate postoperative period, usually have a poor outcome after LT.Substance use should also be carefully evaluated.
Portal vein thrombosis
This is not an absolute contraindication to transplantation. However, a dynamic 3-phase CT scan should be performed pretransplant in all candidates to assess the extent and severity of thrombosis. Extensive portal vein thrombosis (PVT) involving the entire splenoportal axis may pose significant surgical challenges and could serve as a contraindication to LT. Careful preoperative evaluation is essential to determine the feasibility and potential need for surgical or interventional strategies to manage PVT in transplant candidates.
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Technique or Treatment
Any LT procedure consists of 2 main components: the donor and the recipient. Recipient operations involve complete removal of the patient’s native liver following dissection of the hepatic ligamentous attachments and hilar structures, with the inferior vena cava (IVC) encircled to secure blood control. Donors may be either deceased or living, as detailed below:
Living donor liver transplantation
Initially reserved for pediatric cases, living donor liver transplant is now also performed in adults due to the rising demand and shortage of deceased donor organs. Unlike whole grafts from deceased donors, living donor grafts are partial and involve smaller hepatic arteries, hepatic veins, and portal veins, requiring careful dissection and precise implantation. Adequate space is created by incising along the hepatic vein’s sides to enable proper arterial, portal, and biliary reconstruction.The anastomoses are performed sequentially: hepatic vein (ensuring sufficient length), portal vein, hepatic artery (challenging due to short tributaries), and finally, duct-to-duct bile duct anastomosis. Living donor grafts may include the left lateral sector (approximately 20% of liver volume), the left lobe (approximately 40%), or the right lobe (approximately 60%). In some cases, dual grafts (2 left lobes from 2 donors) are implanted into 1 recipient. Donors undergoing hepatectomy typically receive a right subcostal incision extending into the midline, sparing dissection of the rectus muscles on both sides. For right lobe donations, the left lobe should be secured to the anterior abdominal wall before wound closure.
Deceased donor liver transplantation
Whole LT is the most common approach. Brain-dead donor transplantation uses grafts from donors with confirmed brain death, while donation after cardiac death involves graft procurement after irreversible cardiac arrest. The donor’s liver is usually prepared separately. Once the recipient’s surgical field is ready, the donor liver is brought in and anastomosed, beginning with the suprahepatic IVC, followed by the infrahepatic IVC, and then the portal vein. Once these connections are complete, clamps are removed, allowing the portal vein to initiate blood inflow and perfuse the liver. The hepatic arteries are connected near the gastroduodenal artery anastomoses, and finally, the bile duct is reconstructed.
Domino LT is primarily used in familial amyloid polyneuropathy (FAP), where the native liver (which functions normally but carries the genetic defect) is transplanted into another recipient, typically older than 55, to reduce the risk of disease transmission. This requires precise bench reconstruction of suprahepatic veins, as the IVC is preserved in the FAP donor.
Partial LT is used when a full graft is unavailable or to address certain metabolic deficiencies, ensuring a graft-to-recipient weight ratio of at least 0.8% to achieve adequate posttransplant function.
Auxiliary LT involves implanting a partial graft alongside the native liver to provide temporary support in ALF or to correct metabolic conditions, often in younger patients with viral or autoimmune ALF. This can be performed orthotopically or heterotopically, depending on the clinical context.
Split LT enables a single donor liver to be divided between 2 recipients. When shared between an adult and a child, the right lobe (including segment IV) is allocated to the adult, while the left lateral segment (segments II and III) is given to the pediatric recipient. In cases where 2 adults receive a split graft, the liver is divided into 2 hemilivers to ensure sufficient mass for each. The first attempt at split grafting was reported in 2003, where a deceased donor liver was divided, resulting in the right lobe (without the middle hepatic vein) functioning as an allograft, similar to the modified right lobe technique in living donor LT, while the left part retained the IVC and common hepatic artery.
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Complications
Complications that occur after LT are classified as either early or late.
Early Complications
These include:
Primary nonfunction of the liver allograft
Hepatic artery thrombosis
Acute cell rejection
Biliary complications
Infection
In the first postoperative week, liver enzymes typically trend downward as the graft recovers and begins to regenerate. However, close monitoring is essential, as several serious complications can arise during this critical period. The most serious early complication is primary nonfunction of the liver allograft, which presents immediately after transplantation. Clinically, this is characterized by the absence of bile production or the production of clear bile, accompanied by worsening liver enzymes and rising bilirubin levels. Without urgent retransplantation, the patient’s survival is at extreme risk.
During the first 48 to 72 hours posttransplant, it is common to observe elevated liver enzymes due to graft injury from cold and warm ischemia during organ removal and implantation. However, it is vital to exclude hepatic artery thrombosis in this setting, as this complication often occurs early but can also develop later.Hepatic artery thrombosis may present asymptomatically or with fever and rising liver enzymes, and if untreated, can progress to hepatic ischemia, necrosis, and ischemic cholangiopathy. Severe cases, especially those occurring within the first week, often necessitate urgent retransplantation. Diagnosis is made using Doppler ultrasound.
Acute cellular rejection occurs in up to 50% of LT recipients, typically within the first 2 months after transplantation. Most cases respond well to corticosteroid therapy. For corticosteroid-resistant rejection, antithymocyte globulin is used. A definitive diagnosis of acute cellular rejection is made through liver biopsy, and despite the high incidence, long-term outcomes with appropriate treatment are generally favorable. Biliary complications are also common, with the biliary anastomosis representing the most frequent site of biliary strictures. Management includes endoscopic dilation, stenting, or, in rare cases, surgical revision. Nonanastomotic (ischemic) strictures may develop as a result of hepatic artery thrombosis, ABO blood type incompatibility, prolonged graft ischemia time (either warm or cold), or when using grafts donated after cardiac death.
Finally, infection is a significant early risk due to the required immunosuppression after LT. Opportunistic infections are common and include cytomegalovirus, the most frequent viral infection; Candida species, the most common fungal pathogens; and other serious infections such as Pneumocystis jirovecii, Aspergillus, Nocardia, and Cryptococcus. Additionally, immunosuppressive agents such as tacrolimus and cyclosporine can cause neurologic and renal impairment, as well as the development of hyperglycemia.
Late Complications
The long-term toxic effects of immunosuppressive medications predominantly drive late complications following LT. Late complications are also a result of recurrent disease posttransplant and de novo malignancies.
The most common late complications include:
Chronic kidney disease
HTN
Diabetes mellitus
Dyslipidemia
Calcineurin inhibitors, when combined with preexisting chronic kidney disease and HTN, significantly contribute to the development of posttransplant renal failure. Management strategies involve strict blood pressure control and either dose reduction or discontinuation of calcineurin inhibitors.Cardiovascular disease risk is markedly elevated in LT recipients due to immunosuppression-related increases in diabetes mellitus, HTN, obesity, and dyslipidemia. When combined with preexisting high-risk lifestyle factors, this leads to a substantial rise in atherosclerotic disease burden.
Bone health complications are also notable, particularly the increased risk of osteoporosis. Long-term corticosteroid use, along with malnutrition and vitamin D deficiency associated with chronic liver disease, exacerbate bone loss. However, this complication has become less frequent due to successful interventions with bisphosphonate therapy and efforts to reduce corticosteroid dosing.
Neurologic impairments are another recognized late complication, most commonly manifesting as tremors, but also including insomnia and paresthesias, primarily as a result of calcineurin inhibitor toxicity. Recurrent liver diseases can occur after transplantation. These include the recurrence of hepatitis C virus and hepatitis B virus infections, both of which can now be effectively managed posttransplant. Additionally, other chronic liver diseases may recur, including nonalcoholic steatohepatitis, primary biliary cholangitis, primary sclerosing cholangitis, autoimmune hepatitis, and hepatocellular carcinoma.
Malignancies that arise de novo are a major long-term cause of mortality in LT recipients. Several factors increase the risk of posttransplant malignancy, including chronic immunosuppression, viral infections, alcohol use, cigarette smoking, and older recipient age. The most frequently observed malignancies in this population include skin cancers, posttransplant lymphoproliferative disease, and anogenital cancers such as cervical, vulvar, and anal cancers.
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Clinical Significance
LT is 1 of the most transformative and lifesaving procedures in modern medicine, offering definitive treatment for both acute and chronic end-stage liver disease when medical therapies fail. Transplantation has profoundly reshaped the prognosis of patients with chronic decompensated cirrhosis, due to causes such as alcohol-related liver disease, hepatitis C infection, nonalcoholic steatohepatitis, and autoimmune hepatitis (AIH), as well as acute liver failure from drug-induced injury, viral hepatitis, AIH, or indeterminate causes.Without transplantation, these conditions carry extremely poor prognoses, with mortality rates exceeding 80% in many advanced cases. LT dramatically changes this outlook, providing the possibility of long-term survival, with current 1-year survival rates above 85% and 10-year survival approaching 60% to 70% at experienced transplant centers.
Beyond extending life,LT restores key physiologic and functional capacities, leading to dramatic improvements in quality of life. Following LT, patients often experience resolution of hepatic encephalopathy, ascites, variceal bleeding, and jaundice, along with restoration of appetite, muscle mass, and physical capacity. Many return to work, school, and social engagement, and in children, transplantation supports normal growth, neurodevelopment, and long-term health.
Importantly, advances in living donor LT (LDLT) have further expanded access and improved outcomes. When LDLT was first introduced, its results were inferior to those of deceased donor LT. However, LDLT has since been proven to significantly reduce mortality by shortening waiting list times, particularly in both pediatric and adult populations. East Asian countries, where deceased donor availability remains limited, have played a pivotal role in refining LDLT by developing advanced surgical techniques, improving graft selection, and optimizing portal and hepatic blood flow. These innovations have revolutionized LT worldwide, leading to comparable survival rates between deceased donor LT and LDLT and markedly improving the quality of life of patients with chronic liver disease or acute liver failure, even among those with high MELD scores above 30.
LT has also driven broader innovations in surgery and ethics, including the development of split LT, where 1 donor liver is divided between 2 recipients, and domino transplantation, where livers with metabolic disease are used in select older recipients. Additionally, expanding the donor pool through the use of donation after circulatory death and marginal grafts has improved equitable access to transplantation. However, these advances raise complex ethical questions about consent, allocation, and distributive justice.
Beyond surgery, LT has propelled advances in transplant immunology. The development of calcineurin inhibitors, mechanistic (or mammalian) target of rapamycin inhibitors, and monoclonal antibodies has improved both acute and chronic rejection rates; at the same time, efforts to achieve immune tolerance and implement minimization protocols continue to evolve. The lessons learned from LT immunology extend beyond the liver, benefiting the fields of kidney, heart, and lung transplant as well.
From a public health perspective,LT reduces healthcare system burdens by decreasing hospitalizations for complications of decompensated cirrhosis, lowering the frequency of ICU stays, and minimizing the costs associated with end-stage care. Additionally, recipients who regain health are often able to return to work and participate in their communities, contributing economically and socially. However, the long-term costs of immunosuppressive therapy, cancer surveillance, cardiovascular risk management, and recurrent disease monitoring remain significant and require ongoing attention.
Despite its remarkable success,LT faces major challenges. Organ shortages persist worldwide, leading to high waitlist mortality. Disparities in access to transplantation disproportionately affect minority and socioeconomically disadvantaged populations, and long-term complications—such as chronic kidney disease, cardiovascular disease, metabolic syndrome, and de novo malignancies—remain common. Additionally, recurrent liver diseases such as hepatitis C, hepatitis B, nonalcoholic steatohepatitis, primary biliary cholangitis, primary sclerosing cholangitis, autoimmune hepatitis, and hepatocellular carcinoma can reemerge after transplantation, necessitating vigilant posttransplant monitoring and management.
In conclusion, LT is a life-saving, quality-of-life-transforming intervention that not only improves individual outcomes but also drives innovations in surgery, immunology, and public health. The evolution of techniques such as LDLT, along with advancements in immunosuppression and innovations in graft preservation, will continue to shape the future of LT and extend its life-changing benefits to an even broader patient population.
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Enhancing Healthcare Team Outcomes
Effective liver transplantation care requires the seamless integration of skills, strategies, and interprofessional communication among clinicians, nurses, pharmacists, dietitians, social workers, and other healthcare professionals. Clinicians and transplant surgeons lead the complex diagnostic evaluation, surgical planning, and perioperative management, while hepatologists and advanced practitioners, such as nurse practitioners or physician assistants, coordinate the management of chronic liver disease, immunosuppression titration, and long-term follow-up. Nurses play a crucial role in patient education, bedside monitoring, and the early recognition of postoperative complications, thereby ensuring patient safety through vigilant assessment and timely communication. Pharmacists contribute expertise in immunosuppressive drug selection, drug interaction management, and patient education on adherence, which is crucial to preventing rejection and minimizing adverse effects. Dietitians ensure patients maintain appropriate nutritional status before and after transplant, while social workers coordinate psychosocial support, insurance navigation, and home care planning to promote sustained recovery.
To enhance patient-centered care, the liver transplant team must employ clear, structured communication strategies such as multidisciplinary rounds, shared electronic medical records, standardized handoff protocols, and checklists. Regular team meetings facilitate the alignment of care plans, discussion of patient progress, and identification of barriers to adherence or recovery, thereby promoting collaborative problem-solving and effective patient care. Prioritizing patient and family involvement in decision-making ensures that care aligns with individual values, preferences, and goals. Effective coordination reduces medical errors, improves transplant outcomes, minimizes hospital readmissions, and strengthens team performance, ultimately delivering safer, more efficient, and more compassionate care for this vulnerable patient population.
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Disclosure:Yara Dababneh declares no relevant financial relationships with ineligible companies.
Disclosure:Omar Mousa declares no relevant financial relationships with ineligible companies.
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3051 | https://pdfs.semanticscholar.org/2611/af34bab123dfb2ab3be1c16c573dad565af2.pdf | Review Ascorbate is a multifunctional micronutrient whose synthesis is lacking in primates Junichi Fujii Department of Biochemistry and Molecular Biology, Graduate School of Medical Science, Yamagata University, Yamagata 990-9585, Japan (Received 6 November, 2020; Accepted 7 December, 2020; Published online 25 March, 2021) Ascorbate (vitamin C) is an essential micronutrient in primates, and exhibits multiple physiological functions. In addition to antioxidative action, ascorbate provides reducing power to α-ketoglutarate-dependent non-heme iron dioxygenases, such as prolyl hydroxylases. Demethylation of histones and DNA with the aid of ascorbate results in the reactivation of epigenetically silenced genes. Ascorbate and its oxidized form, dehydroascorbate, have attracted interest in terms of their roles in cancer therapy. The last step in the biosynthesis of ascorbate is catalyzed by L-gulono-γ-lactone oxidase whose gene Gulo is commonly mutated in all animals that do not synthesize ascorbate. One common explanation for this deficiency is based on the increased availability of ascorbate from foods.
In fact, pathways for ascorbate synthesis and the detoxification of xenobiotics by glucuronate conjugation share the metabolic processes up to UDP-glucuronate, which prompts another hypothesis, namely, that ascorbate-incompetent animals might have developed stronger detoxification systems in return for their lack of ability to produce ascorbate, which would allow them to cope with their situation. Here, we overview recent advances in ascorbate research and propose that an enhanced glucuronate conjugation reaction may have applied positive selection pressure on ascorbate-incompetent animals, thus allowing them to dominate the animal kingdom.
Key Words: ascorbate, vitamin C, glucuronate conjugation, detoxification V itamins are nutrients that are, in the original meaning, essential micronutrients that cannot be synthesized by an organism. About a century ago, Szent-Györgyi(1) discovered that high levels of ascorbate (Asc) are stored in the adrenal gland, and this stimulated research interest in this compound.
The inability to synthesize a vitamin is generally due to a lack genes responsible for their syntheses that developed during the evolutional process, however, applying this scenario to Asc is not quite correct. In fact, most organisms, including mammals, are able to synthesize Asc from glucose as the primary carbohydrate.
On the contrary, the ancestors of primates and some other species once had the ability to synthesize Asc but have lost this ability during their evolution.(2,3) Humans need to ingest massive amounts of Asc compared to other vitamins for health maintenance; the recommended daily requirement for Asc is 75–110 mg/day for adult humans.(4) A high excretion rate from the body through the kidneys; a half-life being approximately one day, is the main contributor to this massive requirement. Regarding physiological roles, antioxidation and electron donation in the synthesis of biological compounds, such as collagen and catecholamines, are well established functions of Asc.(5,6) Recent advances in this area have revealed that the contribution of Asc can be extended to the oxygen-dependent regulation of gene expression via the hypoxia-inducible factor (HIF)-1α(7) and the epigenetic regulation of genes.(8,9) In the meantime, since the proposal by Linus Pauling on the use of Asc for cancer therapy, the benefits of high doses of Asc in the treatment of cancer has been a subject of considerable debate.(10) Accumulating evidence, based on basic and clinical studies, are now confirming that Asc can, in fact, be useful in cancer prevention and treatment.(11) Phylogenetically, vertebrates higher than fish possess the Gulo encoding l-gulono-γ-lactone oxidase (Gulo), which catalyzes the last step in Asc synthesis.(12) A mutation in Gulo occurred about 63,000,000 years ago, which makes primates unable to synthesize Asc.(13) In fact, mutations have been observed in Gulo in all incompetent animal species, which include guinea pigs, bats, passeriformes birds, and teleost fish.(3,14,15) These animals can ingest sufficient amounts of Asc from foods in ordinary life, so that Asc production may not be essential for their health.
However, the question remains about what has allowed these incompetent animals in which Gulo was mutated to dominate during their evolution.
In this review article, we briefly overview the multiple functions of Asc with specific emphasis on recent advances in our understanding of the beneficial roles of Asc in our health and propose a mechanism regarding selection pressure that may have allowed animals that are unable to produce Asc continue to dominate.
Anti-oxidative and Redox Reaction of Asc The chemistry of ascrobic acid has been extensively overviewed,(16,17) and this is not the subject of this article.
Accordingly, only minimal information concerning the structure and the chemical nature of Asc that are required for our understanding of its physiologica funcitons, are provided here.
While ascorbic acid is present mainly in three states: the fully oxidized form (dehydroascorbate; DHA), the ascorbyl radical and the fully reduced monoanion form, Asc is the dominant form at physiological pH and is a potent reducing agent (Fig. 1).
Asc readily undergoes a one-electron oxidation to form an ascorbyl radical and then DHA. In other words, due to its high electron-donating ability, Asc is involved in a variety of reactions that require reducing equivalents in both enzymatic and non-enzymatic reactions. The resulting oxidized form, DHA is either recycled by several processes that are described below or is converted into 2,3-diketo-l-gulonic acid, which is then further metabolized via l-xylose, l-threonic acid plus oxalic acid, or l-xylonic acid and l-lyxonic acid.(6) Asc concentrations in the plasma/extracellular fluid and cells/tissues are 0.04–0.08 mM and 0.3–10 mM, respectively.(10,18) A short half-life of To whom correspondence should be addressed.
E-mail: jfujii@med.id.yamagata-u.ac.jp He received “The SFRR Japan Prize” in 2020 in recognition of his outstanding work.
doi: 10.3164/jcbn.20-181 ©2021 JCBN J. Clin. Biochem. Nutr. | July 2021 | vol. 69 | no. 1 | 1–15 approximately one day in mammals is mainly due to the high excretion rate and insufficient reuptake in the kidney. Due to less abundance of DHA and high level of reductive recycling, the issue of how much of the DHA-derived metabolites are subsequently metabolized remain ambiguous.
Anti-oxidation is a major function of Asc and has been extensively reviewed.(17,19,20) Asc is a safe and relatively cheap antioxidative compound and is generally used as a food additive in commercial food products. The antioxidative capacity of Asc is largely attributed to its ability to donate an electron and be converted into a radical species, typically a hydroxyl radical, a vitamin E (α-tocopherol) radical and a thyl radical.(19) While Asc is a highly water-soluble molecule and can function to suppress hydrophilic radical species, vitamin E provides protection against lipid peroxidation by donating electrons to lipid peroxyl radicals (LOO•). Experimental data indicate that Asc is able to recycle vitamin E radicals back to vitamin E, and hence, also functions to protect lipids against peroxidation. Because of this, both Asc and vitamin E are often discussed together as nutritional antioxidants.(16,21) Because lipid peroxidation products trigger ferroptosis, a newly identified type of iron-dependent non-apoptotic cell death,(22,23) Asc may also be involved in the suppression of ferroptosis via the reductive recycling of vitamin E, a potent anti-ferroptotic compound. In the meantime, Asc is not a very effective scavenger of superoxide, a primary oxygen radical, according to the rate constant for this reaction (2.7 × 105 M−1s−1)(24) which is about half that for the spontaneous dismutation of superoxides. However, because of the abundance of Asc in tissues, this allows the reaction between superoxide and Asc to proceed sufficiently rapidly to suppress superoxide to permissible levels in vivo.(25) Because superoxide is the oxygen radical that is first produced under a variety of reactions with oxygen consumption, its elimination would be of importance for suppressing subsequent radical chain reactions and hence provide relief from oxidative stress.
Asc is also closely associated with iron homeostasis and function. In cooperation with ferrireductase including duodenal cytochrome B, ferric iron (Fe3+) is reduced to ferrous iron (Fe2+) by Asc and is then taken up by apical membranes in the intestinal lumen via the divalent metal ion transporter DMT1.(26) Hence, an Asc insufficiency decreases the intestinal absorption of iron and causes iron-deficient anemia.(27) Asc stimulates the synthesis of the iron-storage protein ferritin, inhibits its lysosomal degradation, and decreases the efflux of iron from cells.(28) Moreover, Asc-dependent enzymatic reactions are dependent on iron in most cases and copper occasionally.(5) It is noteworthy, however, that Asc in pharmacologic concentrations is converted to excessive levels of ascorbyl radicals and is simultaneously involved in production of hydrogen peroxide in a free-iron dependent manner in extracellular fluid.(29) Thus, Asc may also act as a pro-oxidant and exert cytotoxicity, depending on the redox environment, which appears to be an underlying mechanism for the anti-cancer action of Asc.(30) Outline of Asc Biosynthesis Asc is synthesized dominantly in the liver from glucose as a primary carbohydrate via several processes,(31,32) and the pathway cross-talks with other glucose metabolic pathways (Fig. 2).
Accordingly nutritional conditions can have a significant effect on Asc synthesis and can have an influence on the Asc levels in the body. Upon feeding, cells take up glucose from the blood via a glucose transporter (GLUT) and phosphorylate glucose to glucose-6-phosphate (glucose-6-P) via the action of hexokinase/ glucokinase. Glucose-6-P is then either degraded by glycolysis, converted to other carbohydrates via the pentose phosphate pathway, or isomerized to glucose-1-phosphate (glucose-1-P) by phosphoglucomutase. Glucose-1-P uridylyltransferase, also referred to as UDP-glucose pyrophosphorylase, then converts glucose-1-P into UDP-glucose which serves as the communal O OH O− OH O H H CH2OH O OH O O O H H CH2OH .
− O OH O O O H H CH2OH O R HO O R .O LOOH LOO･ Ascorbate (Asc) Dihydroascorbate (DHA) Ascorbyl radical +e−, H+ Tocopherol Tocopherol radical −e− , −H+ Lipid peroxyl radical Lipid −e− +e−, H+ Fig. 1. Interconversion of Asc, ascorbyl radical and DHA and their roles in the vitamin E-mediated elimination of lipid radical. Three major forms of ascorbic acid and their redox-mediated interconversion are depicted on the top. Vitamin E (tocopherol) donates electron to lipid hydroxyl radical (LOO•) and is converted to tocopherol radical. Asc may donate an electron to a tocopherol radical for recycling and is converted to an ascorbyl radical, which is then oxidized to DHA. Asc collectively can protect lipids from oxidation reactions as well. Schemes show conceptual reactions and are not stoichiometric. R; alkyl side chain.
2 doi: 10.3164/jcbn.20-181 ©2021 JCBN precursor for glycogen synthesis and Asc synthesis.(32) UDP-glucose is oxidized to UDP-glucuronate, which then flows to the Asc synthesis or is used for glucuronate conjugation for purposes of detoxification.(33) In the biosynthesis of Asc, d-glucuronate is produced from UDP-glucuronate via either d-glucuronide-1-phosphate (d-glucuronide-1-P) or d-β-glucuronide. The resulting d-glucuronate is reduced by members of the aldo-keto reductase family (Akr)-1(34) to l-gulonate, which is then converted to either l-gulono-γ-lactone for Asc synthesis or 3-keto-l-gulonate leading to the pentose phosphate pathway.
The conversion of l-gulonate to l-gulono-γ-lactone is catalyzed by gluconolactonase (Rgn). l-Gulono-γ-lactone is transported into the endoplasmic reticulum (ER) lumen and eventually oxidized to Asc by the catalytic action of Gulo.(32) Glycogen is an energy storage form of carbohydrate, and glycogenolysis, upon stimulation with a hyperglycemic hormone such as glucagon, provides glucose to the body via the blood. In this response, glucose-1-P is released by glycogen phosphorylase, isomerized to glucose-6-P by phosphoglucomutase, and finally dephosphorylated to glucose by glucose 6-phosphatase (G6Pase) in the lumen of the ER (Fig. 3). Glucagon, the levels of which are increased in blood under hypoglycemic conditions and the principle hormone for maintaining blood glucose levels, stimulates glycogenolysis but does not activate Asc synthesis.(35) There is a large body of evidence to support the conclusion that Asc is also synthesized upon activation of glycogenolysis in competent animals such as rodents.(2,32) For example, Asc synthesis is reportedly stimulated under glutathione depletion by either inhibiting glutathione synthesis(36) or activating glutathione conjugation in rodents,(35,37) although blood glucose levels remain unchanged by these treatments.(37,38) For this purpose, glucose-1-P is converted into UDP-glucose only for Asc synthesis. Thus, there appears to be mechanisms that preferentially force glycogen-derived glucose-1-P to flow toward Asc production. These phenomena imply the existence of a line of communication between the glutathione system and the Asc system, but the molecular basis for this crosstalk is not clearly understood.(32) We will revisit this issue in a later section and discuss it from evolutional aspect of Asc synthesis.
It is also of note that oxygen consumption by Gulo produces hydrogen peroxide as a byproduct, which may be reduced to water during the oxidative folding of secretory proteins.
Although an excessive production of hydrogen peroxide may trigger oxidative stress and eventually ER stress partly due to stimulating the misfolding of proteins,(39) a causal connection between the hydrogen peroxide released by the Gulo reaction and ER stress is obscure. Upon the dysfunction of Gulo in Asc-incompetent animals, upstream intermediary compounds of l-gulono-γ-lactone tend to flow toward other metabolic pathways and may be metabolized in different ways.
Asc Transport and the Reductive Recycling of DHA The resulting Asc is released into the blood and delivered to organs. Asc is taken up by cells via a member of the Na+-dependent vitamin C transporter (SVCT) family that is driven by the Na+ gradient (Fig. 4).(40,41) On the other hand, DHA is taken up via facilitative glucose transporters (42) that include GLUT1, GLUT3, and GLUT4(43,44) as well as others.(45) Given presence of abundant levels of Na+ in the extracellular fluid and that Asc is the major form in the blood and extracellular milieu,(46) SVCTs dominantly contribute to the uptake of Asc by cells and generate large concentration gradients.(47) Consequently, Asc levels are maintained at high levels in cells, 0.5 to 10 mM compared to 50 to 80 μM in blood plasma.(48) The distribution of the two isoforms of the Asc transporters is discrete: SVCT1 (also referred to as solute carrier family 23 member 1; Slc23a1) in the intestine, kidney and liver cells, whereas it is SVCT2 (Slc23a2) in the brain, eye and other organs. SVCT1 which is expressed in the intestine and kidney, is involved in the absorption of Asc from nutritional sources and reabsorption from filtrate, D-Glucuronate Glucose-1-P UDP-Glucose UDP-Glucuronate β-D-Glucuronide 3-Keto-L-gulonate L-Gulonate Asc L-Gulono-γ-lactone Gulo Glycogen Glucuronate conjugation ER Pentose phosphate pathway ER Glucose Glucose-6-P Glucose Blood Glucose Glycolysis D-Glucuronide-1-P Akr1 Rgn Fig. 2. Pathways for Asc synthesis and other related carbohydrate metabolism pathways from glucose. Solid lines indicate carbon flow to Asc synthesis. Dotted lines indicate flows that simultaneously or occasionally procced depending on the physiological conditions of the animals. Gulo in the ER is absent in incompetent animals of the Asc synthesis.
J. Fujii J. Clin. Biochem. Nutr. | July 2021 | vol. 69 | no. 1 | 3 ©2021 JCBN ER Glucose-1-P G6PT Glucose-6-P Glucose Glycogen G6Pase UDP-Glucose GT L-gulono-γ-lactone Asc Glucose-6-P Pi Blood glucose L-gulono-γ-lactone Asc declined GSH Xenobiotics Glucagon ?
H2O2 O2 Gulo Fig. 3. Roles of ER in the production of glucose via glucose 6-P and Asc via UDP-glucose from glycogen in rodents. Glucose 1-P from glycogen is the common precursor for plasma glucose and Asc. Glucagon stimulates glycogenolysis but is not involved in UDP-glucose formation. In the meantime, xenobiotics and glutathione consumption stimulate the production of UDP-glucose and the consequent Asc synthesis, but not blood glucose. Dephosphorylation of glucose 6-P by glucose 6-phosphatase (G6Pase) releases glucose in the ER. Gulo oxidizes L-gulono-γ-lactone to Asc by means of the oxidation of molecular oxygen, leading to the production of hydrogen peroxide.
SVCT DHA Na+ Asc Na+ Asc DHA DHA ER GLUT Glc Glc e− Asc Asc synthesis Gulo Asc Asc Reductase GSH Asc Asc GLUT1 Fig. 4. Transport system for Asc and DHA and the reductive recycling of DHA. Asc is transported via the SVCT which is driven by a Na+ electrochemical gradient. Some types of GLUT are responsible for DHA uptake as well as glucose (Glc) uptake. DHA is reduced to Asc by either enzymatic reactions or an electron donation from glutathione. DHA is partly transported into mitochondria via GLUT and reduced to Asc by electrons from the respiratory chain.
4 doi: 10.3164/jcbn.20-181 ©2021 JCBN respectively. SVCT2 is expressed in the brain and accounts for the uptake of Asc from the cerebrospinal fluid,(49) and Asc has significant roles in neuronal functions, can be therapeutic target of neurodegenerative diseases such as Alzheimer’s disease, Parkinson’s disease and Huntington’s disease.(50) The birth of a mouse with a deficiency in SVCT2 is accompanied by respiratory failure and intraparenchymal brain hemorrhage and dies within a few minute after birth.(45) An SVCT2 deficiency also impairs adrenal chromaffin cell function and decreases the tissue adrenalin and noradrenalin and plasma levels of corticosterone, although these appear to not be the cause for lethality.(51) Regarding cellular organelles, next to the ER, mitochondria are rich in Asc. SVCT2 is reportedly present in the mitochondrial membrane and appears to be responsible for the mitochondrial transport of Asc.(52) Genetic ablation demonstrates the roles of SVCT1 in renal Asc reabsorption and perinatal survival.(53) Although SVCT1 is generally believed to be responsible for the intestinal absorption of Asc, its transport across the intestines is normal in SVCT1-knockout (KO) mice,(53) indicating the presence of an alternative transport mechanism. Functional analyses of the glucose transporters GLUT2 and GLUT5-12 that are expressed in Xenopus oocytes indicate that GLUT2 and GLUT8 are able to take up DHA.(54) Hence, DHA is taken up from intestinal fluids via GLUT2 and GLUT8, and after reduction to Asc, is recruited through the liver. Asc is abundantly present in the central nervous system. SVCT2 is highly expressed and appears to function in the cortex and cerebellum of the fetal rat brain.(55) However, the blood-brain barrier may interfere with the absorption of hydrophilic compounds, including Asc. GLUT1, which is expressed on endothelial cells at the blood-brain barrier, appears to transport DHA into the brain and is responsible for supplying Asc to neurons.(56) Asc has multiple roles in the brain, so please refer to recent review articles(18,50) concerning details regarding the regulation of Asc homeostasis and functions.
Because the mouse can synthesize Asc, the ablation of these transporter genes under a genetic background with defected Asc synthesis may cause the production of different phenotypes.
DHA is partly reduced by glutathione,(20) and at a more accelerated rate, by that utilize an electron from glutathione or NADH/NADPH. NADH-cytochrome b5 reductase is one such enzyme that is responsible for the reduction of DHA,(57,58) but there appears to be more enzymes involved in its reductive recycling.(59,60) For example, protein disulfide isomerase (PDI) catalyzes the reduction of DHA in conjunction with disulfide bond formation in proteins within the ER, which, as viewed from the opposite side, implies that DHA plays a role in oxidative protein folding.(61) DHA is incorporated into mitochondria via GLUT1 which is localized on the inner mitochondrial membrane(62) and is then reduced to Asc by an electron from either glutathione(63) or the electron transport complex III.(64) Upon reduction, Asc may be transported out of mitochondria via SVCT2.
Asc as a Cofactor for Non-heme Iron Dioxygenases In addition to antioxidation, another well-recognized action of Asc is to donate electrons for enzymatic processes that require reducing equivalents for the production of a variety of compounds, such as collagen, monoamines, and steroid hormones.(5,65) Asc is a cofactor for α-ketoglutarate (αKG)-dependent non-heme iron dioxygenases that include prolyl 4-hydroxylase for collagen synthesis and a unique prolyl hydroxylase for HIF-1α.(7,66) The prolyl hydroxylases utilize Asc, molecular oxygen and αKG as co-substrates for proline hydroxylation and result in the formation of DHA, carbon dioxide and succinate (Fig. 5A).
Collagens are the most abundant proteins in the body, constituting ~30% of the total protein mass in the human body, and the proline residues are highly hydroxylated (~100 in each molecule).(67) Asc is a well-established electron donor for the prolyl 4-hydrolase reaction that is required for the maturation of collagen. Scurvy is a disease that is accompanied by aberrant collagen synthesis and is caused by a Asc insufficiency.(68) On the contrary, the excessive production of collagen, notably by myofibroblasts, may cause fibrosis under pathological conditions and result in tissue remodeling.(69) Vertebrates produce three types of collagen prolyl 4-hydroxylase that function in the ER where disulfide bond-mediated oxidative protein folding occurs (Fig. 5B). Prolyl 4-hydroxylases consist of a unique catalytic α-subunit, which is responsible for the actual proline hydroxylation, and a common β-subunit, which is identical to PDI.(66) PDI alone functions as an independent enzyme that plays roles in the oxidative folding of general secretory proteins in the ER as well as in the restoration of erroneous disulfide bonds.(70) Hence, prolyl 4-hydroxylases may be the self-contained enzyme that participates in the maturation of collagen. The requirement of PDI as a subunit for prolyl 4-hydroxylases may be associated with the lethality in PDI-KO mice due to the inability to synthesize collagen.(71) While Asc supports collagen synthesis by donating an electron for proline hydroxylation, DHA may be involved in the oxidative folding of collagen and other secretory proteins by accepting an electron via PDI.(61) However, because the rate of PDI-medicated reduction of DHA is not very rapid, (72) the extent of the contribution of DHA to oxidative protein folding is obscure.
Animal Models with Defected Asc Synthesis About 60 years after the discovery of Asc by Szent-Györgyi,(1) genetic defects in the human gene encoding Gulo were characterized.(13) Among rodents, the guinea pig also carries an evolutionary defect in the Gulo gene(73) and has been utilized as a model animal for Asc function in vivo.(74) A natural mutant the Wistar rat strain, osteogenic disorder Shionogi (ODS), also carries a mutation in Gulo and develops scurvy,(75) despite it being less popular as a pathological model compared to the guinea pig. Since the mouse is the most popular laboratory animal, a huge amount of data on gene function and metabolism has accumulated. Thus, the mouse with a genetic ablation of the specific gene for Asc synthesis would be more advantageous as a pathological model animal.
The Gulo-KO mouse, which exhibits pathological conditions similar to scurvy in humans, was established by the homologous recombination of the gene(76) and has been employed successfully to study the physiological action of Asc.(77) The penultimate reaction in the Asc synthesis, the dehydration of l-gulonate to l-gulono-γ-lactone, is catalyzed by Rgn. The gene product of Rgn was found to be identical to the senescence marker protein 30.(78) The preceding reaction of Rgn is the NADPH-dependent reduction of d-glucuronic acid to l-gulonate.(79) This process is catalyzed by two closely related enzymes, namely, aldehyde reductase (Akr1a) and aldose reductase (Akr1b) in the mouse, which are members of the aldo-keto reductase superfamily.(34,80) The contribution of Akr1a and Akr1b in the mouse is 85–90% and 10–15%, respectively. Even in the presence of ~15% Asc for the wild-type (WT) mouse, the Akr1a-KO mouse starts dying after 20 weeks and does not survive beyond one year, whereas Asc supplementation extends their life-span to a considerable extent.(81) On the other hand, the Akr1b-KO mouse shows no such phenotypic abnormality.(34) These genetically modified mice have been extensively employed in studies of the roles of Asc from the standpoints of physiology and pathophysiology.(77,82) While the oxidative conversion of l-gulono-γ-lactone to Asc appears to be the sole reaction catalyzed by Gulo, additional functions may reside in other gene products that are involved in J. Fujii J. Clin. Biochem. Nutr. | July 2021 | vol. 69 | no. 1 | 5 ©2021 JCBN the synthesis of Asc. Notably in the case of Akr1a/1b, multiple carbonyl compounds can serve as substrates for these enzymes,(83) and hence, an Akr1a-deficiency may result in the development of novel phenotypes beyond those expected from the Asc depletion, as has been observed in some reports.(84,85) Moreover, such a genetic deficiency may also change the upstream metabolic flow and result in the development of phenotypes that are not directly associated with Asc functions. To attribute the corresponding phenotype to an Asc deficiency, a simple but reliable criterion is based on observations of whether ot not Asc supplementation could rescue the animals. EDTA, a chelator of divalent cations is sometimes included in supplemental Asc solutions in order to suppress the spontaneous oxidation of Asc. However, because the inclusion of EDTA may cause deficits in metal absorption by metal chelation, notably iron, in the intestine and cause adverse effect in animals, researchers must take this into account in investigations of Asc-involved biological actions. Considering the sufficient absorption of DHA by intestinal cells via GLUT, (53,54) in our studies, we do not include a metal chelator in the Asc solution but, instead, change the Asc-containing water every 2–3 days to avoid the formation of such artifacts.
The genetically modified mice are particularly advantageous in investigating the physiological consequences of an Asc deficiency. For example, when brain slices of the Rgn-KO mouse are subjected to the hypoxia-reperfusion treatment, Asc effectively suppresses superoxide production as judged by measuring the fluorescence of lucigenin.(86) Asc deprivation causes pulmonary emphysema with decreased collagen synthesis in the Rgn-KO mouse.(87) A double deficiency of Rgn and SOD1, which encodes the superoxide dismutase isoform which is abundantly present in cytosol and to a more limiter extent in mitochondria, shows premature death and aberrant lipid metabolism in the absence of Asc supplementation.(88) We recently found that a double deficiency of Akr1a and SOD1 causes inevitable death in the mice within several weeks after the Asc supplementation is ceased.(25) Among the organs examined, oxidative lung injury appears to be the primary cause for the death of the double KO mouse. These observations strongly support the requisite role of Asc in eliminating superoxide from the body, notably in the lung that is continuously exposed to hyperoxygenic conditions compared to other organs.
To the contrary, an Asc deficiency due to genetic ablation does not always validate the conclusion that Asc is required for enzymatic reactions that have been demonstrated in in vitro ER nascent preprocollagen mRNA Scurvy Fibrosis -SH -SH procollagen mature collagen DHA Proline hydroxylation Thiol oxidation OH OH OH OH OH S S S S OH OH OH O2 H2O H2O2 Normal Asc DHA Asc DHA Asc OH O N CH2 C H2C CH2 C H O N CH C H2C CH2 C H Proline Hydroxyproline O2+αKG CO2+Suc Prolyl hydroxylase Iron A B Fig. 5. Proline hydroxylation and the maturation of collagen in the ER. (A) Proly hydroxylase catalyzes the oxidative conversion of proline to hydroxyproline by employing Asc, molecular oxygen and α-ketoglutarate (αKG) and results in the production of DHA, carbon dioxide (CO2) and succinate (Suc). (B) Oxidative folding and the prolyl hydroxylation of collagen proceed in the ER lumen. Procollagen and other secretory proteins undergo oxidative folding by means of ERO1 and other thiol oxidases. The resulting procollagen is hydroxylated at proline residues by prolyl hydroxylase coupled with Asc oxidation at multiple positions (~100 proline residues in a molecule) and then secreted out of the cell. DHA may also be involved in thiol oxidation. Both an Asc insufficiency and abundance may cause pathological conditions, such as scurvy and fibrosis, respectively.
6 doi: 10.3164/jcbn.20-181 ©2021 JCBN assays. For example, normal collagen status is observed in the Gul-KO mouse despite the complete absence of Asc and eventual death.(89) In the Rgn-KO mouse, the collagen content is in the skin is not affected, but the levels in the lung are decreased.(90) When the Akr1a-KO mouse is subjected to unilateral ureteral obstruction, a pathological model for kidney fibrosis, collagen production is elevated in both Akr1a-KO and WT mice, but no evident difference is observed between the two genotypic mice.(81) In other instances, while the levels of noradrenaline and adrenaline are low in adrenal glands of Rgn-KO mice,(91) these levels are not changed in the brain of the Akr1a-KO mouse.(92,93) Instead, the levels of adrenaline and noradrenaline are somewhat elevated in the Gulo-KO mice compared to WT mice in a water restraint stress model.(94) These observations imply that Asc is not the sole electron donor for these enzymatic systems, but that other redox-active compounds may compensate for the Asc-insufficient situation. Studies employing Asc-deficient animals, hence, demonstrate that Asc is actually involved in multiple reactions, but, in some instances, redox compounds other than Asc appears to be involved. Thiol compounds, more specifically, glutathione, are promising alternatives to Asc due to their abundance and electron donating ability.
Epigenetic Action of Asc Reactive oxygen species (ROS) cause base modifications in DNA and lead to genetic mutations, which may ultimately induce cancer development. 8-Oxo-7,8-dihydroguanine (8-oxoG), is the most frequently modified base, and consequently causes transversion of the GC pair to a TA pair. Hence, the DNA repair glycosylase OGG1 that excises 8-oxoG plays a pivotal role in the suppression of ROS-induced mutation and cancer.(95) While the protective effects of Asc against oxidative damage appear to largely dependent on its antioxidant ability,(96) recent studies have revealed that Asc regulates gene expression via a novel mechanism. The methylation of DNA and histones contributes to the epigenetic regulation of gene expression. DNA methylation at the 5-position of cytosine (5mC) is catalyzed by DNA methyltransferases (DNMTs). While DNMT3 family enzymes are responsible for de novo methylation and newly establishing the 5mC in DNA, DNMT1 is involved in the maintenance of 5mC in the daughter strand during DNA replication.(97) In these cytosine methylation reactions, S-adenosylmethionine, which is produced in the process of methionine metabolism, serves as a methyl group donor. Cytosine methylation catalyzed by DNMTs tends to cause chromatin compaction, leading to heterochromatin formation and consequently the silencing of some genes.
In the meantime, histones undergo a variety of modifications, which includes methylation of lysine and arginine residues, acetylation, and ubiquitination, and also affect gene expression epigenetically.(98) Concerning Asc functions, the methylation status in lysine residues is a subject of interest. Histone H3-specific methyltransferases were initially found to methylate lysine 9.(99) Thereafter, several specific histone lysine methyltransferases and histone lysine demethylases were identified.(98) While lysine acetylation is largely correlated with the transcriptional activation of genes, the methylation of lysine results in either transcriptional activation or suppression, depending on the position and degree of methylation in the histones.
Because Asc modulates gene expression epigenetically, it has attracted the attention of researchers in the fields of stem cell and cancer research. For example, Asc in fact enhances the generation of induced pluripotent stem cells (iPSCs) from both humans and mice.(100) Asc is a cofactor for Fe2+- and αKG-dependent dioxygenases that involve the demethylation of DNA and histones (Fig. 6).(9,101,102) The ten-eleven translocation (TET) methylcytosine dioxygenase is the enzyme that catalyzes the conversion of 5mC into 5-hydroxymethylcytosine(103) and then 5-carboxylcytosine, which JHDM TET RNA polymerase mRNA Asc 3Asc 3DHA DHA Base excision DNA repair + Transcriptionally inactive Transcriptionally active 5-Methyl cytosine Trimethyl lysine CH3 CH3 CH3 CH3 NH3+ N+ NH2 N N O Histone Iron Iron DNA K K H NH2 N N O H H Fig. 6. Epigenetic action of Asc towards gene activation. Electron donation from Asc to TET reaction supports the conversion of 5-methyl cytosine to 5-hydroxyl cytosine and then 5-carboxylcytosine, which is consequently replaced to cytosine by the base excision DNA repair system. Jumonji C (JmjC)-domain-containing histone demethylases (JHDMs) removes methyl groups from lysines in histones. Demethylation of DNA and histones may be able to activate consequent gene expression. Only concept of the reactions are presented schematically.
J. Fujii J. Clin. Biochem. Nutr. | July 2021 | vol. 69 | no. 1 | 7 ©2021 JCBN is consequently replaced with an unmodified cytosine via a base excision DNA repair system.(104) Asc was first found to induce TET-dependent DNA demethylation in embryonic stem (ES) cells.(105) The TET-catalyzed reaction is accompanied by the conversion of Fe2+ to Fe3+. The electron-donating nature of Asc supports the reductive recycling of Fe3+ to Fe2+.
Consequently, genes that have been epigenetically silenced by the methylation of cytosine undergo reactivation by Asc-supported TET reactions.(8) The Jumonji C-domain-containing histone demethylase 1 (JHDM1) is the first identified histone demethylase that specifically demethylates histone H3 at lysine 36 (H3K36) and is conserved from yeast to humans.(106) JHDM1a/1b has been demonstrated to be the Asc-dependent H3K36 demethylase in mouse embryonic fibroblasts during reprogramming into iPSCs.(107) As the end of this reaction, trimethyl lysine is converted back to lysine in conjunction with the oxidation of three Asc molecules to DHA. Currently, the family contains approximately 20 proteins that catalytically remove methyl groups from methyl lysine with the aid of Asc.(108) Through the epigenetic remodeling of chromatin, an Asc deficiency would be expected to affect a variety of physiological processes, including development, differentiation, and aging and be consequently involved in the development of pathological conditions in animals.
Reactions of Asc in Prevention and Treatment of Cancer An early trial with high-dose Asc in the treatment of caner patiens resulted in beneficial action,(109,110) but controversial results were reported in later studies.(111,112) Thus, the use of Asc in the treatment of tumor growth has been a subject of debate since the hypothesis was first announced by Linus Pauling.(11) Asc, in fact, exerts opposing effects, either carcinogenic or tumoricidal, in cancer, which largely depends on the redox status of the microenvironment. Iron can contribute to carcinogenic processes in several ways.(113) Fe3+ bound to ferritin is not toxic in the bound form, however, due to its electron-donating ability, Asc causes the free form of Fe3+ to be converted to Fe2+. Fe2+ supplies electrons for the reductive conversion of hydrogen peroxide to hydroxyl radicals, which are highly reactive, oxidize a variety of molecules including DNA and results in the mutation of genes. Thus, Asc, in conjunction with free iron, may result in elevated ROS levels and exhibit carcinogenicity. By contrast, Asc eliminates ROS as an antioxidant and hence prevents their mutagenic action.
Recent studies have unveiled the anti-tumorigenic action of Asc in which serveral mechanisms are involved. The results of animal experiments indicate that Asc is beneficial in suppressing carcer development; the growth and metastasis of transplanted tumor cells observed in Gulo-KO mice were decreased by supplementation of Asc at physiological concentrations.(114–116) Asc appears to suppress chemical carcinogenesis as well. The administration of N-nitrosodiethylamine causes the complete death in Rgn-KO mice within 4 months and therefore provides no information regarding tumorigenesis.(117) Although Akr1a-KO mice are also mostly dead within a half-year, a few surviving mice developed extreme hepatocellular carcinomas, and Asc-supplementation markedly suppressed hepatocellular carcinoma development.(118) Because methylation of cytosine in DNA and lysine residues in histones epigenetically suppresses the expression of genes that include tumor suppressor genes, their Asc-mediated demethylation may lead to their reactivation.(9,98) Thus, the epigenetic actions of Asc would very likely contribute to the anticarcinogenic action.
Electron donation from Asc to a specific prolyl hydroxylase is responsible for the regulation of HIF-1α, which is a transcription factor and the master regulator of genes whose expression increases oxygen availability.(119) HIF-1α levels are increased in many solid tumors and reportedly support their development because the inner mass of solid tumor tends to be hypoxic.
The constitutive activation of HIF-1α, hence, induces genes that increase the supply of oxygen and energy production, leading to the survival and proliferation of tumor cells. Asc is utilized for the hydroxylation of specific prolyl residues in HIF-1α, leading to the generation of a binding site for the von Hippel-Lidau (pVHL) tumor suppressor protein, which is made up of a ubiquitin ligase complex (Fig. 7). After polyubiquitination, HIF-1α undergoes degradation by proteasomes. As a result, Asc in the presence of oxygen may restrict the proliferation and survival of tumor cells.(7,120) There is an emerging body of evidence to indicate that pharmacologically high-doses of Asc acts as a potential therapeutic agent for some types of cancers.(120–122) Because the plasma Asc level is strictly controlled; ≈80 μM,(123) the oral administration of a high dose of Asc does not reach effective levels. To administer a high dose of Asc for therapeutic purposes, intravenous administration is required; e.g.
the intravenous administration of 10 g of Asc to cancer patients results in 1–5 mM concentrations in the plasma.(124,125) Upon Asc treatment, mitochondrial oxidative metabolism is stimulated and this increases the production of superoxide and hydrogen peroxide, which then disrupts iron metabolism in cancer cells.(126) The resulting labile iron together with hydrogen peroxide sensitizes cancer cells, notably non-small-cell lung cancer and glioblastoma, to pharmacological Asc doses in a redox-active manner. The synergistic action of Asc and iron coordinately trigger lipid peroxidation, which results in ferroptosis in cancer cells.(30,127) In the meantime, colorectal cancer cells bearing mutations in Kras and Braf are selectively killed by the action of a high-dose of DHA.(128) The strategy for killing cancer cells by DHA is different from that for Asc. In this anti-tumor action, glutathione is oxidized in the process of the reduction of DHA to Asc, which results in an elevation in ROS levels.
Oxidative DNA damage then stimulates the poly-(ADP-ribose) polymerase-mediated consumption of NAD+, leading to energy depletion and the eventual death of cancer cells. In fact, clinical trials using cancer patients with a fasting-mimicking diet and vitamin C appears to be promising therapeutics for the treatment of colorectal cancer and possibly other tumors with KRAS mutations.(129) Thus, Asc at physiologic concentrations prevents tumorigenesis, whereas both Asc and DHA at pharmacological doses exert tumoricidal actions.
Evolutional Aspect of Asc Synthesis Animals inherently do not possess some genes that are responsible for certain types of metabolism, which leads them to be incapable of synthesizing the corresponding compounds, e.g., essential amino acids and most vitamins. Contrary to these, Asc-incompetent animals had originally possessed functional Gulo but have lost its function due to mutations in the course of evolution.(13) Thus, scenario dealing with Asc synthesis is not comparable to those of essential amino acids or other vitamins. If sufficient Asc is not ingested, our bodies suffer disadvantages due to the inability to produce Asc. Nevertheless, Asc-incompetent animals, including primates, the guinea pig, the bat and passeriformes birds have actually survived.(3,14) In fact, these Asc-incompetent animals have even prospered. The teleost fish, which is the most abundant group in the modern fish family, also cannot synthesize Asc. Different from other Asc-incompetent animals, however, no relic of Gulo has been found in the teleost fish genome by searches of gene data bases.(15) Because the ancestral species can synthesize Asc, it is postulated that Gulo was completely eliminated during the 8 doi: 10.3164/jcbn.20-181 ©2021 JCBN course of evolution of the teleost fish.(3) It is of note that Asc incompetence in these animals is caused solely by mutations in Gulo, which can be attributed to independent events during the evolutionary process.
The reason why Asc-incompetent animals are no longer able to synthesize Asc but remain prosperous is not understood and this constitutes an interesting issue from the standpoint of evolution and nutritional biochemistry as well. The most common explanation for the inability to synthesize Asc in primates is as follows; primates can climb trees and therefore gain access to sufficient amounts of Asc-rich foods, notably fruits, and hence do not need to synthesize Asc by themselves.(3) This notion may sound reasonable if the animals do not develop serious damage to their health caused by an Asc insufficiency.
In fact, however, Asc exerts a variety of important roles in animal physiology, and an Asc insufficiency causes an unhealthy situation, and in severe cases, fatal conditions such as scurvy.(68) Most genetic mutations, as are found in hereditary diseases, are disadvantageous to the host animals, and hence, tend to be eliminated from the gene pool by death or failure in reproduction.
When the mutation is neutral, meaning the mutation does not influence the survival of animals, both the original and mutated genes can be maintained in the gene pool of the animal species because there is no selection pressure on them. However, no active Gulo remains among primates, which suggests that either Gulo itself is disadvantageous or the mutated Gulo is more advantageous for these animals. The question therefore arises as to what mechanism has actually made the incompetent animals lose their ability to synthesize such an important nutrient by assuming the risk of a deficiency disorder?
Gulo is an ER-resident enzyme that converts L-gulono-γ-lactone to Asc by an oxidation reaction in conjunction with the reduction of molecular oxygen to hydrogen peroxide. Because even a very high dose of Asc is not toxic to animals in the general sense,(123) the presence of excessive levels of Asc does not appear to be harmful to animals. Hydrogen peroxide, a byproduct of the Gulo reaction, is released into the ER lumen, and once excessive levels are reached, it may exert unfavorable events due to its oxidizing power.(130) If Gulo-derived hydrogen peroxide is toxic enough, Asc-competent animals would be more vulnerable to the ER stress compared to the incompetent animals. In fact, the ER is the most hydrogen peroxide-rich organelle and this situation appears to be caused by the action of endoplasmic reticulum oxidoreductin 1 (Ero1), which is involved in oxidative protein folding in the ER and produces large amounts of hydrogen peroxide during the course of the enzymatic reaction.(131,132) Several peroxidases are localized in the ER and utilize hydrogen peroxide for thiol oxidation during oxidative protein folding in the ER.(70) Peroxiredoxin 4 (Prdx4) is one such thiol oxidase that oxidizes thiols in the PDI family or newly synthesized proteins to disulfides.(133) The genetic ablation of Prdx4 alone shows no obvious changes in the mouse,(134) but a mouse with a double deficiency of Prdx4 and Ero1 develops atypical scurvy due to Asc consumption and insufficient collagen production.(135) Thus, an Asc insufficiency rather than excessive hydrogen peroxide appears to cause deleterious effects in the ER. This notion is also supported by the finding that an insufficient supply of oxygen decreases the production of hydrogen peroxide and the capacity for oxidative protein folding, which consequently leads to ER stress.(31) Moreover, no reports have yet shown that Gulo-derived hydrogen peroxide alone induces ER damage, although excessive ROS, which is caused by a double deficiency of Prdx4 and SOD1, induces ER stress and hepatic damage in mice.(136) Thus far, there is no robust Prolyl hydroxylase OH OH pVHL E2 Ubiquitin OH OH Gene induction VEGF GLUT Erythropoietin etc… DHA Asc HIF-1α Transcription HIF-1α/β Hypoxia Responsive gene Proteasome Hydroxylated HIF-1α Nucleus Normoxia O2 + αKG CO2 + Suc Iron HIF-1β Fig. 7. Roles of Asc in controlling HIF-1α action. Under normoxic conditions, prolyl hydroxylase catalyzes the hydroxylation of specific proline residues in HIF-1α, which leads to polyubiquitination by the ubiquitin ligase activity of pVHL, followed by degradation via proteasomes. Under hypoxic conditions, the short oxygen supply suppresses proly hydroxylation, which results in the stabilization of HIF-1α. After translocation to the nucleus, HIF-1α dimerizes with HIF-1β, which is constitutively present largely in the nucleus, and together with other transcriptional factors, activates transcription of corresponding genes, such as erythropoietin, vascular epidermal growth factor (VEGF), and GLUT.
J. Fujii J. Clin. Biochem. Nutr. | July 2021 | vol. 69 | no. 1 | 9 ©2021 JCBN body of evidence to show that the Gulo-catalyzed reaction is a disadvantage in competent animals. Given that Gulo-deficient, Asc-incompetent animals are rather prosperous among close animal groups in the animal kingdom now, it is rational to conclude that the Asc-incompetent animals have gained an advantage in gaining prosperity in return for the loss for Asc synthesis.
Enhanced Detoxification via Glucuronate Conjugation Reactions May be Used to Rationalize the Loss of Asc Synthesis The question still remains as to what function is gained in return for the loss of Asc synthesis and has allowed incompetent animals to become dominant during the course of evolution. Instead of focusing on the Gulo-catalyzed reaction itself, a re-examination of the upstream metabolic pathway of the Asc synthesis may provide a clue in this issue.
The Asc-synthesizing pathway constitutes one of the glucose metabolism networks and the processes up to UDP-glucuronate is shared with that of glucuronate conjugation (Fig. 8). UDP-glucuronate either proceeds to form l-gulonate for Asc synthesis or becomes the substrate for conjugation reactions of xenobiotics, which is performed by UDP-glucuronosyltransferase (UGT).(33,137) Sharing the processes between the Asc synthesis and glucuronate conjugation may explain why xenobiotics stimulate Asc synthesis. UGT itself or a closely related enzyme may be responsible for the following removal of UDP from UDP-glucuronate.(59) If UGT is identical to the enzyme, it can be regarded as the gate-keeping enzyme that dictates which pathways UDP-glucuronate takes, either glucuronate conjugation or Asc synthesis, depending on the presence or absence of xenobiotics. Upon the mutation in Gulo, however, UDP-glucuronate dominantly flows to the glucuronate conjugation reaction. Consequently, a loss of the Asc synthesis would result in enhanced detoxification by means of stimulated glucuronate conjugation.
While three major conjugation reactions, glucuronate conjugation, glutathione conjugation and sulfate conjugation, are involved in detoxification of xenobiotic compounds, which are mostly hydrophobic in nature, the inhibition of glutathione synthesis by buthionine sulfoximine(36) or glutathione consumption by stimulated glutathione conjugation(37) reportedly increases Asc-synthesis in rodents. In the case of the xenobiotic-mediated glutathione consumption, the accelerated conversion of UDP-glucuronate to glucuronate is a likely cause for the elevated Asc synthesis.(35) Hence, two detoxification pathways, glutathione conjugation and glucuronate conjugation, appear to be coordinately functioning from the standpoint of the detoxification of xenobiotics. Although, no metabolic pathways UDP-Glucose Asc L-Gulono-γ-lactone Gulo Xenobiotic (X) GSH X-SG Detoxification Glutathione conjugation Glucuronate conjugation Antioxidation Glucose-1-P GSH consumption UDP-Glucuronic acid (GluA) X-GluA Oxidative damage GST UGT Fig. 8. Schematic representation of and crosstalk between two detoxification pathways, glutathione conjugation and glucuronate conjugation in the liver. While glutathione conjugation is catalyzed by glutathione S-transferases (GST), glucuronate conjugation is catalyzed by UGT. UGT may also catalyze the hydrolysis of UDP-glucuronate, resulting in the formation of D-glucuronate. Pathways for glucuronate conjugation and Asc synthesis share the metabolic processes up to UDP-glucuronate. Ablation of enzymes responsible for the downstream reaction causes the accumulation of intermediary compounds that include UDP-glucuronate. As a result, the glucuronate conjugation reaction is enhanced in animals that have a defect in Asc synthesis and preserve glutathione for antioxidation.
10 doi: 10.3164/jcbn.20-181 ©2021 JCBN are directly shared between the two conjugation reactions, these pathways are in fact under common genetic regulation.
Especially Nrf2 is a redox-sensitive transcriptional regulatory factor that activates antioxidative genes, including those for glutathione synthesis, and also genes for detoxification, which includes UGT.(138) Because glutathione is also a pivotal electron donor for glutathione peroxidase reactions in antioxidation, it is conceivable that a preferential metabolic flow to glucuronate conjugation would save glutathione by providing relief for the glutathione conjugation reaction, which would lead to an augmented antioxidative capacity. Fruits are generally rich in Asc but also sometimes contain toxic compounds, notably in immature stages.(139) Considering that ancestral primates obtain food that could contain relatively high levels of toxic compounds, individuals that have acquired strong detoxification systems, by means of glucuronate conjugation, and an antioxidation system, by means of glutathione, in return for waiving Asc synthesis would have an advantage in terms of survival.
This hypothetical adaptation may rationalize prosperity of the Asc-incompetent animals. However, is there evidence for it? Our recent studies on drug-induced hepatic injuries caused by acetaminophen (APAP) may provide a clue to this issue.(140–142) APAP is a popular antifebrile and analgesic agent and is generally regarded as a safe medicine. Upon an overdose, however, APAP induces liver injury, which is a leading cause of drug-induced acute liver failure in several countries.(143) Numerous studies have been carried out and the rsults show that ingested APAP is mostly excreted via either glucuronate conjugation or sulfate conjugation(144,145) but is also partly oxidized to N-acetyl-p-benzoquinone imine (NAPQI) by cytochrome P450 (CYP) enzymes.(146) The resulting NAPQI then undergoes glutathione conjugation, which leads to the consumption of glutathione and consequent hepatic damage. The administration of N-acetyl cysteine, a precursor for cysteine and eventually glutathione, is the established treatment for an APAP overdose,(143) implying the importance of glutathione maintenance in liver homeostasis.
We employed Akrla-KO mice to examine the protective effects of Asc on APAP-induced hepatic injury. While the Akrla-KO mice are more vulnerable to high doses of APAP than the WT mice, Asc supplementation renders the Akrla-KO mouse to be markedly resistant to APAP hepatotoxicity.(140) These results alone can be simply explained by assuming that Asc exerted a hepatoprotective action via antioxidation against elevated ROS, which are elevated by the CYP reaction and the consumption of glutathione due to glutathione conjugation of NAPQI. Our most recent study, however, had some unexplainable results based on the simple antioxidant action of Asc.(142) Under conditions of an iron load, the hepatotoxicity of APAP was comparable between WT and the Akr1a-KO mice. To our surprise, however, Asc supplementation dramatically ameliorated the APAP-induced hepatic injury in the iron-loaded Akr1a-KO mice but had no effect on the iron-loaded WT mice. Because WT mice possesses both endogenously produced Asc and supplemented Asc, the Asc contents are actually higher in the WT mice than in the Akr1a-KO mouse, and hence, the results cannot be rationalized by Asc function alone. In the meantime, our metabolomic study on the Akr1a-KO mice indicated that the Akr1a-KO mice had an excessive accumulation of glucuronate (41-fold in the Akr1a-KO vs the WT mice) and its oxidized metabolite saccharate (102-fold in the Akr1a-KO vs the WT mice) in the liver.(80) These collective results imply that elevated oxidative stress due to an Asc deficiency in the Akr1a-KO mice makes them vulnerable to APAP, but the suppression of ROS production by Asc supplementation reinforces hepatic resistance, which is accomplished by efficient detoxification via enhanced glucuronate conjugation.
It should also be noted that the levels of glucuronate-conjugated APAP are actually the highest in the guinea pig among rodents that were administered APAP.(147) Guinea pig is consistently less sensitive to APAP hepatotoxicity. These circumstantial evidences support our hypothesis on the prosperity of Asc-incompetent animals; the restriction in the carbohydrate flow to Asc synthesis enhances the detoxification of APAP by stimulated glucuronate conjugation, and at the same time, preserves glutathione, which collectively leads to the KO mouse being more resistant to APAP. Thus, the preferential carbohydrate flow to glucuronate conjugation reaction due to the loss of Asc synthesis appears to be advantageous to Gulo-deficient animals.
This hypothesis explains the phenomena observed in the Asc-incompetent animals and genetically modified mice but is largely based on circumstantial evidences at this moment. Thus, further experimentation will be required to verify this hypothetical mechanism for the loss of Asc synthesis.
Brief Summary Asc is an essential micronutrient the synthesis of which primates have lost a long time ago. Because Asc exerts pivotal actions in a variety of metabolic processes and gene expression as well as functioning as an antioxidant in the body, a deficiency causes scurvy and is associated with cancer and many other diseases due to dysfunctions in physiological processes. The issue of why primates have lost synthesizing such an important molecule remains a mystery. Herein, we propose a hypothetical mechanism, which is different from the ordinary explanation; i.e., the loss of Asc synthesis due to the mutation in Gulo enhances detoxification reactions by glucuronate conjugation, which is accomplished by a preferential carbohydrate flow.
Acknowledgments This study was supported, in part, by grants from the Japan Society for the Promotion and Science (JSPS) and the National Research Foundation (NRF) of Republic of Korea under the Japan-Korea Basic Scientific Cooperation Program.
Abbreviations Akr aldo-keto reductase Akr1a aldehyde reductase Akr1b aldose reductase APAP acetaminophen Asc ascorbate CYP cytochrome P450 d-glucuronide-1-P d-glucuronide-1-phosphate DHA dehydroascorbate DNMT DNA methyltransferase ER endoplasmic reticulum Ero1 endoplasmic reticulum oxidoreductin 1 Fe3+ ferric iron Fe2+ ferrous iron Glc glucose glucose-6-P glucose-6-phosphate glucose-1-P glucose-1-phosphate GLUT glucose transporter G6Pase glucose 6-phosphatase Gulo l-gulono-γ-lactone oxidase HIF hypoxia-inducible factor H3K36 histone H3 at lysine 36 iPSC induced pluripotent stem cell JHDM Jumonji C-domain-containing histone demethylase αKG α-ketoglutarate KO knockout J. Fujii J. Clin. Biochem. Nutr. | July 2021 | vol. 69 | no. 1 | 11 ©2021 JCBN 5mC DNA methylation at the 5-position of cytosine NAPQI N-acetyl-p-benzoquinone imine 8-oxoG 8-oxo-7,8-dihydroguanine PDI protein disulfide isomerase Prdx4 peroxiredoxin 4 Rgn gluconolactonase ROS reactive oxygen species SVCT Na+-dependent vitamin C transporter TET ten-eleven translocation UGT UDP-glucuronosyltransferase WT wild-type Conflict of Interest No potential conflicts of interest were disclosed.
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Contents
1 History
1.1 China
1.2 Japan
1.3 India
1.4 Middle East, North Africa, Muslim Iberia
1.5 Latin Europe
1.6 Europe after 15th century
2 Some famous magic squares
2.1 Luo Shu magic square
2.2 Magic square in Parshavnath temple
2.3 Albrecht Dürer's magic square
2.4 Sagrada Família magic square
3 Properties of magic squares
3.1 Magic constant
3.2 Magic square of order 1 is trivial
3.3 Magic square of order 2 cannot be constructed
3.4 Center of mass
3.5 Moment of inertia
3.6 Birkhoff–von Neumann decomposition
4 Classification of magic squares
5 Enumeration of magic squares
6 Transformations that preserve the magic property
6.1 For any magic square
6.2 For associative magic squares
6.3 For pan-diagonal magic squares
6.4 For bordered magic squares
6.5 For composite magic squares
7 Special methods of construction
7.1 A method for constructing a magic square of order 3
7.2 A method for constructing a magic square of odd order
7.3 A method of constructing a magic square of doubly even order
7.3.1 Generic pattern
7.3.2 A construction of a magic square of order 4
7.3.3 An extension of the above example for Orders 8 and 12
8 Method of superposition
8.1 Euler's method
8.2 Narayana-De la Hire's method for even orders
9 Method of borders
9.1 Bordering method for order 3
9.2 Bordering method for order 5
9.3 Continuous enumeration methods
10 Method of composition
10.1 For squares of order m × n where m, n > 2
10.2 For squares of doubly even order
10.3 Medjig-method for squares of even order 2n, where n > 2
11 Solving partially completed magic squares
12 Variations of the magic square
12.1 Extra constraints
12.2 Multiplicative magic squares
12.3 Multiplicative magic squares of complex numbers
12.4 Additive-multiplicative magic and semimagic squares
12.5 Geometric magic squares
12.6 Area magic squares
12.7 Other magic shapes
13 Related problems
13.1 n-Queens problem
14 Magic squares in occultism
15 Magic squares in popular culture
16 See also
17 Notes
18 References
19 Further reading
Magic square
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From Wikipedia, the free encyclopedia
Square of numbers with equal row, column and diagonal totals
This article is about mathematical arrays. For the association football formation, see Formation (association football) § 4–2–2–2 (magic rectangle).
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| This article may be too long to read and navigate comfortably. Consider splitting content into sub-articles, condensing it, or adding subheadings. Please discuss this issue on the article's talk page. (June 2023) |
In mathematics, especially historical and recreational mathematics, a square array of numbers, usually positive integers, is called a magic square if the sums of the numbers in each row, each column, and both main diagonals are the same. The order of the magic square is the number of integers along one side (n), and the constant sum is called the magic constant. If the array includes just the positive integers , the magic square is said to be normal. Some authors take magic square to mean normal magic square.
Magic squares that include repeated entries do not fall under this definition and are referred to as trivial. Some well-known examples, including the Sagrada Família magic square are trivial in this sense. When all the rows and columns but not both diagonals sum to the magic constant, this gives a semimagic square (sometimes called orthomagic square).
The mathematical study of magic squares typically deals with its construction, classification, and enumeration. Although completely general methods for producing all the magic squares of all orders do not exist, historically three general techniques have been discovered: by bordering, by making composite magic squares, and by adding two preliminary squares. There are also more specific strategies like the continuous enumeration method that reproduces specific patterns. Magic squares are generally classified according to their order n as: odd if n is odd, evenly even (also referred to as "doubly even") if n is a multiple of 4, oddly even (also known as "singly even") if n is any other even number. This classification is based on different techniques required to construct odd, evenly even, and oddly even squares. Beside this, depending on further properties, magic squares are also classified as associative magic squares, pandiagonal magic squares, most-perfect magic squares, and so on. More challengingly, attempts have also been made to classify all the magic squares of a given order as transformations of a smaller set of squares. Except for n ≤ 5, the enumeration of higher-order magic squares is still an open challenge. The enumeration of most-perfect magic squares of any order was only accomplished in the late 20th century.
Magic squares have a long history, dating back to at least 190 BCE in China. At various times they have acquired occult or mythical significance, and have appeared as symbols in works of art. In modern times they have been generalized a number of ways, including using extra or different constraints, multiplying instead of adding cells, using alternate shapes or more than two dimensions, and replacing numbers with shapes and addition with geometric operations.
History
[edit]
The third-order magic square was known to Chinese mathematicians as early as 190 BCE, and explicitly given by the first century of the common era. The first dateable instance of the fourth-order magic square occurred in 587 CE in India. Specimens of magic squares of order 3 to 9 appear in an encyclopedia from Baghdad c. 983, the Encyclopedia of the Brethren of Purity (Rasa'il Ikhwan al-Safa). By the end of the 12th century, the general methods for constructing magic squares were well established. Around this time, some of these squares were increasingly used in conjunction with magic letters, as in Shams Al-ma'arif, for occult purposes. In India, all the fourth-order pandiagonal magic squares were enumerated by Narayana in 1356. Magic squares were made known to Europe through translation of Arabic sources as occult objects during the Renaissance, and the general theory had to be re-discovered independent of prior developments in China, India, and Middle East. Also notable are the ancient cultures with a tradition of mathematics and numerology that did not discover the magic squares: Greeks, Babylonians, Egyptians, and Pre-Columbian Americans.
China
[edit]
While ancient references to the pattern of even and odd numbers in the 3×3 magic square appear in the I Ching, the first unequivocal instance of this magic square appears in the chapter called Mingtang (Bright Hall) of a 1st-century book Da Dai Liji (Record of Rites by the Elder Dai), which purported to describe ancient Chinese rites of the Zhou dynasty. These numbers also occur in a possibly earlier mathematical text called Shushu jiyi (Memoir on Some Traditions of Mathematical Art), said to be written in 190 BCE. This is the earliest appearance of a magic square on record; and it was mainly used for divination and astrology. The 3×3 magic square was referred to as the "Nine Halls" by earlier Chinese mathematicians. The identification of the 3×3 magic square to the legendary Luoshu chart was only made in the 12th century, after which it was referred to as the Luoshu square. The oldest surviving Chinese treatise that displays magic squares of order larger than 3 is Yang Hui's Xugu zheqi suanfa (Continuation of Ancient Mathematical Methods for Elucidating the Strange) written in 1275. The contents of Yang Hui's treatise were collected from older works, both native and foreign; and he only explains the construction of third and fourth-order magic squares, while merely passing on the finished diagrams of larger squares. He gives a magic square of order 3, two squares for each order of 4 to 8, one of order nine, and one semi-magic square of order 10. He also gives six magic circles of varying complexity.
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | --- | 4 | 9 | 2 | | 3 | 5 | 7 | | 8 | 1 | 6 | | | | | | | --- --- | | 2 | 16 | 13 | 3 | | 11 | 5 | 8 | 10 | | 7 | 9 | 12 | 6 | | 14 | 4 | 1 | 15 | | | | | | | | --- --- | 1 | 23 | 16 | 4 | 21 | | 15 | 14 | 7 | 18 | 11 | | 24 | 17 | 13 | 9 | 2 | | 20 | 8 | 19 | 12 | 6 | | 5 | 3 | 10 | 22 | 25 | | | | | | | | | --- --- --- | | 13 | 22 | 18 | 27 | 11 | 20 | | 31 | 4 | 36 | 9 | 29 | 2 | | 12 | 21 | 14 | 23 | 16 | 25 | | 30 | 3 | 5 | 32 | 34 | 7 | | 17 | 26 | 10 | 19 | 15 | 24 | | 8 | 35 | 28 | 1 | 6 | 33 | |
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | | | | | | --- --- --- | 46 | 8 | 16 | 20 | 29 | 7 | 49 | | 3 | 40 | 35 | 36 | 18 | 41 | 2 | | 44 | 12 | 33 | 23 | 19 | 38 | 6 | | 28 | 26 | 11 | 25 | 39 | 24 | 22 | | 5 | 37 | 31 | 27 | 17 | 13 | 45 | | 48 | 9 | 15 | 14 | 32 | 10 | 47 | | 1 | 43 | 34 | 30 | 21 | 42 | 4 | | | | | | | | | | | --- --- --- --- | | 61 | 3 | 2 | 64 | 57 | 7 | 6 | 60 | | 12 | 54 | 55 | 9 | 16 | 50 | 51 | 13 | | 20 | 46 | 47 | 17 | 24 | 42 | 43 | 21 | | 37 | 27 | 26 | 40 | 33 | 31 | 30 | 36 | | 29 | 35 | 34 | 32 | 25 | 39 | 38 | 28 | | 44 | 22 | 23 | 41 | 48 | 18 | 19 | 45 | | 52 | 14 | 15 | 49 | 56 | 10 | 11 | 53 | | 5 | 59 | 58 | 8 | 1 | 63 | 62 | 4 | | | | | | | | | | | | --- --- --- --- | 31 | 76 | 13 | 36 | 81 | 18 | 29 | 74 | 11 | | 22 | 40 | 58 | 27 | 45 | 63 | 20 | 38 | 56 | | 67 | 4 | 49 | 72 | 9 | 54 | 65 | 2 | 47 | | 30 | 75 | 12 | 32 | 77 | 14 | 34 | 79 | 16 | | 21 | 39 | 57 | 23 | 41 | 59 | 25 | 43 | 61 | | 66 | 3 | 48 | 68 | 5 | 50 | 70 | 7 | 52 | | 35 | 80 | 17 | 28 | 73 | 10 | 33 | 78 | 15 | | 26 | 44 | 62 | 19 | 37 | 55 | 24 | 42 | 60 | | 71 | 8 | 53 | 64 | 1 | 46 | 69 | 6 | 51 | |
The above magic squares of orders 3 to 9 are taken from Yang Hui's treatise, in which the Luo Shu principle is clearly evident. The order 5 square is a bordered magic square, with central 3×3 square formed according to Luo Shu principle. The order 9 square is a composite magic square, in which the nine 3×3 sub squares are also magic. After Yang Hui, magic squares frequently occur in Chinese mathematics such as in Ding Yidong's Dayan suoyin (c. 1300), Cheng Dawei's Suanfa tongzong (1593), Fang Zhongtong's Shuduyan (1661) which contains magic circles, cubes and spheres, Zhang Chao's Xinzhai zazu (c. 1650), who published China's first magic square of order ten, and lastly Bao Qishou's Binaishanfang ji (c. 1880), who gave various three dimensional magic configurations. However, despite being the first to discover the magic squares and getting a head start by several centuries, the Chinese development of the magic squares are much inferior compared to the Indian, Middle Eastern, or European developments. The high point of Chinese mathematics that deals with the magic squares seems to be contained in the work of Yang Hui; but even as a collection of older methods, this work is much more primitive, lacking general methods for constructing magic squares of any order, compared to a similar collection written around the same time by the Byzantine scholar Manuel Moschopoulos. This is possibly because of the Chinese scholars' enthralment with the Lo Shu principle, which they tried to adapt to solve higher squares; and after Yang Hui and the fall of Yuan dynasty, their systematic purging of the foreign influences in Chinese mathematics.
Japan
[edit]
Japan and China have similar mathematical traditions and have repeatedly influenced each other in the history of magic squares. The Japanese interest in magic squares began after the dissemination of Chinese works—Yang Hui's Suanfa and Cheng Dawei's Suanfa tongzong—in the 17th century, and as a result, almost all the wasans devoted their time to its study.
In the 1660 edition of Ketsugi-sho, Isomura Kittoku gave both odd and even ordered bordered magic squares as well as magic circles; while the 1684 edition of the same book contained a large section on magic squares, demonstrating that he had a general method for constructing bordered magic squares. In Jinko-ki (1665) by Muramatsu Kudayu Mosei, both magic squares and magic circles are displayed. The largest square Mosei constructs is of 19th order. Various magic squares and magic circles were also published by Nozawa Teicho in Dokai-sho (1666), Sato Seiko in Kongenki (1666), and Hosino Sanenobu in Ko-ko-gen Sho (1673). One of Seki Takakazu's Seven Books (Hojin Yensan) (1683) is devoted completely to magic squares and circles. This is the first Japanese book to give a general treatment of magic squares in which the algorithms for constructing odd, singly even and doubly even bordered magic squares are clearly described. In 1694 and 1695, Yueki Ando gave different methods to create the magic squares and displayed squares of order 3 to 30. A fourth-order magic cube was constructed by Yoshizane Tanaka (1651–1719) in Rakusho-kikan (1683). The study of magic squares was continued by Seki's pupils, notably by Katahiro Takebe, whose squares were displayed in the fourth volume of Ichigen Kappo by Shukei Irie, Yoshisuke Matsunaga in Hojin-Shin-jutsu, Yoshihiro Kurushima in Kyushi Iko who rediscovered a method to produce the odd squares given by Agrippa, and Naonobu Ajima. Thus by the beginning of the 18th century, the Japanese mathematicians were in possession of methods to construct magic squares of arbitrary order. After this, attempts at enumerating the magic squares was initiated by Nushizumi Yamaji.
India
[edit]
The 3×3 magic square first appears in India in Gargasamhita by Garga, who recommends its use to pacify the nine planets (navagraha). The oldest version of this text dates from 100 CE, but the passage on planets could not have been written earlier than 400 CE. The first dateable instance of 3×3 magic square in India occur in a medical text Siddhayog (c. 966 CE) by Vrnda, which was prescribed to women in labor in order to have easy delivery.
The oldest dateable fourth order magic square in the world is found in an encyclopaedic work written by Varahamihira around 587 CE called Brhat Samhita. The magic square is constructed for the purpose of making perfumes using 4 substances selected from 16 different substances. Each cell of the square represents a particular ingredient, while the number in the cell represents the proportion of the associated ingredient, such that the mixture of any four combination of ingredients along the columns, rows, diagonals, and so on, gives the total volume of the mixture to be 18. Although the book is mostly about divination, the magic square is given as a matter of combinatorial design, and no magical properties are attributed to it. The special features of this magic square were commented on by Bhattotpala (c. 900 CE)
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | --- --- | | 2 | 3 | 5 | 8 | | 5 | 8 | 2 | 3 | | 4 | 1 | 7 | 6 | | 7 | 6 | 4 | 1 | | | | | | | --- --- | | 10 | 3 | 13 | 8 | | 5 | 16 | 2 | 11 | | 4 | 9 | 7 | 14 | | 15 | 6 | 12 | 1 | |
The square of Varahamihira as given above has sum of 18. Here the numbers 1 to 8 appear twice in the square. It is a pan-diagonal magic square. Four different magic squares can be obtained by adding 8 to one of the two sets of 1 to 8 sequence. The sequence is selected such that the number 8 is added exactly twice in each row, each column and each of the main diagonals. One of the possible magic squares shown in the right side. This magic square is remarkable in that it is a 90 degree rotation of a magic square that appears in the 13th century Islamic world as one of the most popular magic squares.
The construction of 4th-order magic square is detailed in a work titled Kaksaputa, composed by the alchemist Nagarjuna around 10th century CE. All of the squares given by Nagarjuna are 4×4 magic squares, and one of them is called Nagarjuniya after him. Nagarjuna gave a method of constructing 4×4 magic square using a primary skeleton square, given an odd or even magic sum. The Nagarjuniya square is given below, and has the sum total of 100.
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | --- --- | | 30 | 16 | 18 | 36 | | 10 | 44 | 22 | 24 | | 32 | 14 | 20 | 34 | | 28 | 26 | 40 | 6 | | | | | | | --- --- | | 7 | 1 | 4 | 6 | | 2 | 8 | 5 | 3 | | 5 | 3 | 2 | 8 | | 4 | 6 | 7 | 1 | |
The Nagarjuniya square is a pan-diagonal magic square. It is made up of two arithmetic progressions starting from 6 and 16 with eight terms each, with a common difference between successive terms as 4. When these two progressions are reduced to the normal progression of 1 to 8, the adjacent square is obtained.
Around 12th-century, a 4×4 magic square was inscribed on the wall of Parshvanath temple in Khajuraho, India. Several Jain hymns teach how to make magic squares, although they are undateable.
As far as is known, the first systematic study of magic squares in India was conducted by Thakkar Pheru, a Jain scholar, in his Ganitasara Kaumudi (c. 1315). This work contains a small section on magic squares which consists of nine verses. Here he gives a square of order four, and alludes to its rearrangement; classifies magic squares into three (odd, evenly even, and oddly even) according to its order; gives a square of order six; and prescribes one method each for constructing even and odd squares. For the even squares, Pheru divides the square into component squares of order four, and puts the numbers into cells according to the pattern of a standard square of order four. For odd squares, Pheru gives the method using horse move or knight's move. Although algorithmically different, it gives the same square as the De la Loubere's method.
The next comprehensive work on magic squares was taken up by Narayana Pandit, who in the fourteenth chapter of his Ganita Kaumudi (1356) gives general methods for their construction, along with the principles governing such constructions. It consists of 55 verses for rules and 17 verses for examples. Narayana gives a method to construct all the pan-magic squares of fourth order using knight's move; enumerates the number of pan-diagonal magic squares of order four, 384, including every variation made by rotation and reflection; three general methods for squares having any order and constant sum when a standard square of the same order is known; two methods each for constructing evenly even, oddly even, and odd squares when the sum is given. While Narayana describes one older method for each species of square, he claims the method of superposition for evenly even and odd squares and a method of interchange for oddly even squares to be his own invention. The superposition method was later re-discovered by De la Hire in Europe. In the last section, he conceives of other figures, such as circles, rectangles, and hexagons, in which the numbers may be arranged to possess properties similar to those of magic squares. Below are some of the magic squares constructed by Narayana:
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | | --- | 8 | 1 | 6 | | 3 | 5 | 7 | | 4 | 9 | 2 | | | | | | | --- --- | | 1 | 14 | 4 | 15 | | 8 | 11 | 5 | 10 | | 13 | 2 | 16 | 3 | | 12 | 7 | 9 | 6 | | | | | | | | --- --- | 16 | 14 | 7 | 30 | 23 | | 24 | 17 | 10 | 8 | 31 | | 32 | 25 | 18 | 11 | 4 | | 5 | 28 | 26 | 19 | 12 | | 13 | 6 | 29 | 22 | 20 | | | | | | | | | --- --- --- | | 1 | 35 | 4 | 33 | 32 | 6 | | 25 | 11 | 9 | 28 | 8 | 30 | | 24 | 14 | 18 | 16 | 17 | 22 | | 13 | 23 | 19 | 21 | 20 | 15 | | 12 | 26 | 27 | 10 | 29 | 7 | | 36 | 2 | 34 | 3 | 5 | 31 | | | | | | | | | | --- --- --- | 35 | 26 | 17 | 1 | 62 | 53 | 44 | | 46 | 37 | 21 | 12 | 3 | 64 | 55 | | 57 | 41 | 32 | 23 | 14 | 5 | 66 | | 61 | 52 | 43 | 34 | 25 | 16 | 7 | | 2 | 63 | 54 | 45 | 36 | 27 | 11 | | 13 | 4 | 65 | 56 | 47 | 31 | 22 | | 24 | 15 | 6 | 67 | 51 | 42 | 33 | | | | | | | | | | | --- --- --- --- | | 60 | 53 | 44 | 37 | 4 | 13 | 20 | 29 | | 3 | 14 | 19 | 30 | 59 | 54 | 43 | 38 | | 58 | 55 | 42 | 39 | 2 | 15 | 18 | 31 | | 1 | 16 | 17 | 32 | 57 | 56 | 41 | 40 | | 61 | 52 | 45 | 36 | 5 | 12 | 21 | 28 | | 6 | 11 | 22 | 27 | 62 | 51 | 46 | 35 | | 63 | 50 | 47 | 34 | 7 | 10 | 23 | 26 | | 8 | 9 | 24 | 25 | 64 | 49 | 48 | 33 | |
The order 8 square is interesting in itself since it is an instance of the most-perfect magic square. Incidentally, Narayana states that the purpose of studying magic squares is to construct yantra, to destroy the ego of bad mathematicians, and for the pleasure of good mathematicians. The subject of magic squares is referred to as bhadraganita and Narayana states that it was first taught to men by god Shiva.
Middle East, North Africa, Muslim Iberia
[edit]
Although the early history of magic squares in Persia and Arabia is not known, it has been suggested that they were known in pre-Islamic times. It is clear, however, that the study of magic squares was common in medieval Islam, and it was thought to have begun after the introduction of chess into the region. The first dateable appearance of a magic square of order 3 occurs in Jābir ibn Hayyān's (fl. c. 721 – c. 815) Kitab al-mawazin al-Saghir (The Small Book of Balances) where the magic square and its related numerology is associated with alchemy. While it is known that treatises on magic squares were written in the 9th century, the earliest extant treaties date from the 10th-century: one by Abu'l-Wafa al-Buzjani (c. 998) and another by Ali b. Ahmad al-Antaki (c. 987). These early treatises were purely mathematical, and the Arabic designation for magic squares used is wafq al-a'dad, which translates as harmonious disposition of the numbers. By the end of 10th century, the two treatises by Buzjani and Antaki makes it clear that the Middle Eastern mathematicians had understood how to construct bordered squares of any order as well as simple magic squares of small orders (n ≤ 6) which were used to make composite magic squares. A specimen of magic squares of orders 3 to 9 devised by Middle Eastern mathematicians appear in an encyclopedia from Baghdad c. 983, the Rasa'il Ikhwan al-Safa (the Encyclopedia of the Brethren of Purity). The squares of order 3 to 7 from Rasa'il are given below:
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | --- | 2 | 7 | 6 | | 9 | 5 | 1 | | 4 | 3 | 8 | | | | | | | --- --- | | 4 | 14 | 15 | 1 | | 9 | 7 | 6 | 12 | | 5 | 11 | 10 | 8 | | 16 | 2 | 3 | 13 | | | | | | | | --- --- | 21 | 3 | 4 | 12 | 25 | | 15 | 17 | 6 | 19 | 8 | | 10 | 24 | 13 | 2 | 16 | | 18 | 7 | 20 | 9 | 11 | | 1 | 14 | 22 | 23 | 5 | | | | | | | | | --- --- --- | | 11 | 22 | 32 | 5 | 23 | 18 | | 25 | 16 | 7 | 30 | 13 | 20 | | 27 | 6 | 35 | 36 | 4 | 3 | | 10 | 31 | 1 | 2 | 33 | 34 | | 14 | 19 | 8 | 29 | 26 | 15 | | 24 | 17 | 28 | 9 | 12 | 21 | | | | | | | | | | --- --- --- | 47 | 11 | 8 | 9 | 6 | 45 | 49 | | 4 | 37 | 20 | 17 | 16 | 35 | 46 | | 2 | 18 | 26 | 21 | 28 | 32 | 48 | | 43 | 19 | 27 | 25 | 23 | 31 | 7 | | 38 | 36 | 22 | 29 | 24 | 14 | 12 | | 40 | 15 | 30 | 33 | 34 | 13 | 10 | | 1 | 39 | 42 | 41 | 44 | 5 | 3 | |
The 11th century saw the finding of several ways to construct simple magic squares for odd and evenly even orders; the more difficult case of oddly even case (n = 4k + 2) was solved by Ibn al-Haytham with k even (c. 1040), and completely by the beginning of 12th century, if not already in the latter half of the 11th century. Around the same time, pandiagonal squares were being constructed. Treaties on magic squares were numerous in the 11th and 12th century. These later developments tended to be improvements on or simplifications of existing methods. From the 13th century onwards, magic squares were increasingly put to occult purposes. However, much of these later texts written for occult purposes merely depict certain magic squares and mention their attributes, without describing their principle of construction, with only some authors keeping the general theory alive. One such occultist was the Algerian Ahmad al-Buni (c. 1225), who gave general methods on constructing bordered magic squares; some others were the 17th century Egyptian Shabramallisi and the 18th century Nigerian al-Kishnawi.
The magic square of order three was described as a child-bearing charm since its first literary appearances in the alchemical works of Jābir ibn Hayyān (fl. c. 721 – c. 815) and al-Ghazālī (1058–1111) and it was preserved in the tradition of the planetary tables. The earliest occurrence of the association of seven magic squares to the virtues of the seven heavenly bodies appear in Andalusian scholar Ibn Zarkali's (known as Azarquiel in Europe) (1029–1087) Kitāb tadbīrāt al-kawākib (Book on the Influences of the Planets). A century later, the Algerian scholar Ahmad al-Buni attributed mystical properties to magic squares in his highly influential book Shams al-Ma'arif (The Book of the Sun of Gnosis and the Subtleties of Elevated Things), which also describes their construction. This tradition about a series of magic squares from order three to nine, which are associated with the seven planets, survives in Greek, Arabic, and Latin versions. There are also references to the use of magic squares in astrological calculations, a practice that seems to have originated with the Arabs.
Latin Europe
[edit]
Unlike in Persia and Arabia, better documentation exists of how the magic squares were transmitted to Europe. Around 1315, influenced by Arab sources, the Greek Byzantine scholar Manuel Moschopoulos wrote a mathematical treatise on the subject of magic squares, leaving out the mysticism of his Middle Eastern predecessors, where he gave two methods for odd squares and two methods for evenly even squares. Moschopoulos was essentially unknown to the Latin Europe until the late 17th century, when Philippe de la Hire rediscovered his treatise in the Royal Library of Paris. However, he was not the first European to have written on magic squares; and the magic squares were disseminated to rest of Europe through Spain and Italy as occult objects. The early occult treaties that displayed the squares did not describe how they were constructed. Thus the entire theory had to be rediscovered.
Magic squares had first appeared in Europe in Kitāb tadbīrāt al-kawākib (Book on the Influences of the Planets) written by Ibn Zarkali of Toledo, Al-Andalus, as planetary squares by 11th century. The magic square of three was discussed in numerological manner in early 12th century by Jewish scholar Abraham ibn Ezra of Toledo, which influenced later Kabbalists. Ibn Zarkali's work was translated as Libro de Astromagia in the 1280s, due to Alfonso X of Castille. In the Alfonsine text, magic squares of different orders are assigned to the respective planets, as in the Islamic literature; unfortunately, of all the squares discussed, the Mars magic square of order five is the only square exhibited in the manuscript.
Magic squares surface again in Florence, Italy in the 14th century. A 6×6 and a 9×9 square are exhibited in a manuscript of the Trattato d'Abbaco (Treatise of the Abacus) by Paolo Dagomari. It is interesting to observe that Paolo Dagomari, like Pacioli after him, refers to the squares as a useful basis for inventing mathematical questions and games, and does not mention any magical use. Incidentally, though, he also refers to them as being respectively the Sun's and the Moon's squares, and mentions that they enter astrological calculations that are not better specified. As said, the same point of view seems to motivate the fellow Florentine Luca Pacioli, who describes 3×3 to 9×9 squares in his work De Viribus Quantitatis by the end of 15th century.
Europe after 15th century
[edit]
The planetary squares had disseminated into northern Europe by the end of the 15th century. For instance, the Cracow manuscript of Picatrix from Poland displays magic squares of orders 3 to 9. The same set of squares as in the Cracow manuscript later appears in the writings of Paracelsus in Archidoxa Magica (1567), although in highly garbled form. In 1514 Albrecht Dürer immortalized a 4×4 square in his famous engraving Melencolia I. Paracelsus' contemporary Heinrich Cornelius Agrippa von Nettesheim published his famous three volume book De occulta philosophia in 1531, where he devoted Chapter 22 of Book II to the planetary squares shown below. The same set of squares given by Agrippa reappear in 1539 in Practica Arithmetice by Girolamo Cardano, where he explains the construction of the odd ordered squares using "diamond method", which was later reproduced by Bachet. The tradition of planetary squares was continued into the 17th century by Athanasius Kircher in Oedipi Aegyptici (1653). In Germany, mathematical treaties concerning magic squares were written in 1544 by Michael Stifel in Arithmetica Integra, who rediscovered the bordered squares, and Adam Riese, who rediscovered the continuous numbering method to construct odd ordered squares published by Agrippa. However, due to the religious upheavals of that time, these works were unknown to the rest of Europe.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| Saturn=15 | 4 | 9 | 2 | | 3 | 5 | 7 | | 8 | 1 | 6 | | Jupiter=34 | 4 | 14 | 15 | 1 | | 9 | 7 | 6 | 12 | | 5 | 11 | 10 | 8 | | 16 | 2 | 3 | 13 | | Mars=65 | 11 | 24 | 7 | 20 | 3 | | 4 | 12 | 25 | 8 | 16 | | 17 | 5 | 13 | 21 | 9 | | 10 | 18 | 1 | 14 | 22 | | 23 | 6 | 19 | 2 | 15 | | Sol=111 | 6 | 32 | 3 | 34 | 35 | 1 | | 7 | 11 | 27 | 28 | 8 | 30 | | 19 | 14 | 16 | 15 | 23 | 24 | | 18 | 20 | 22 | 21 | 17 | 13 | | 25 | 29 | 10 | 9 | 26 | 12 | | 36 | 5 | 33 | 4 | 2 | 31 | |
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| Venus=175 | 22 | 47 | 16 | 41 | 10 | 35 | 4 | | 5 | 23 | 48 | 17 | 42 | 11 | 29 | | 30 | 6 | 24 | 49 | 18 | 36 | 12 | | 13 | 31 | 7 | 25 | 43 | 19 | 37 | | 38 | 14 | 32 | 1 | 26 | 44 | 20 | | 21 | 39 | 8 | 33 | 2 | 27 | 45 | | 46 | 15 | 40 | 9 | 34 | 3 | 28 | | Mercury=260 | 8 | 58 | 59 | 5 | 4 | 62 | 63 | 1 | | 49 | 15 | 14 | 52 | 53 | 11 | 10 | 56 | | 41 | 23 | 22 | 44 | 45 | 19 | 18 | 48 | | 32 | 34 | 35 | 29 | 28 | 38 | 39 | 25 | | 40 | 26 | 27 | 37 | 36 | 30 | 31 | 33 | | 17 | 47 | 46 | 20 | 21 | 43 | 42 | 24 | | 9 | 55 | 54 | 12 | 13 | 51 | 50 | 16 | | 64 | 2 | 3 | 61 | 60 | 6 | 7 | 57 | | Luna=369 | 37 | 78 | 29 | 70 | 21 | 62 | 13 | 54 | 5 | | 6 | 38 | 79 | 30 | 71 | 22 | 63 | 14 | 46 | | 47 | 7 | 39 | 80 | 31 | 72 | 23 | 55 | 15 | | 16 | 48 | 8 | 40 | 81 | 32 | 64 | 24 | 56 | | 57 | 17 | 49 | 9 | 41 | 73 | 33 | 65 | 25 | | 26 | 58 | 18 | 50 | 1 | 42 | 74 | 34 | 66 | | 67 | 27 | 59 | 10 | 51 | 2 | 43 | 75 | 35 | | 36 | 68 | 19 | 60 | 11 | 52 | 3 | 44 | 76 | | 77 | 28 | 69 | 20 | 61 | 12 | 53 | 4 | 45 | |
In 1624 France, Claude Gaspard Bachet described the "diamond method" for constructing Agrippa's odd ordered squares in his book Problèmes Plaisants. During 1640 Bernard Frenicle de Bessy and Pierre Fermat exchanged letters on magic squares and cubes, and in one of the letters Fermat boasts of being able to construct 1,004,144,995,344 magic squares of order 8 by his method. An early account on the construction of bordered squares was given by Antoine Arnauld in his Nouveaux éléments de géométrie (1667). In the two treatise Des quarrez ou tables magiques and Table générale des quarrez magiques de quatre de côté, published posthumously in 1693, twenty years after his death, Bernard Frenicle de Bessy demonstrated that there were exactly 880 distinct magic squares of order four. Frenicle gave methods to construct magic square of any odd and even order, where the even ordered squares were constructed using borders. He also showed that interchanging rows and columns of a magic square produced new magic squares. In 1691, Simon de la Loubère described the Indian continuous method of constructing odd ordered magic squares in his book Du Royaume de Siam, which he had learned while returning from a diplomatic mission to Siam, which was faster than Bachet's method. In an attempt to explain its working, de la Loubere used the primary numbers and root numbers, and rediscovered the method of adding two preliminary squares. This method was further investigated by Abbe Poignard in Traité des quarrés sublimes (1704), by Philippe de La Hire in Mémoires de l'Académie des Sciences for the Royal Academy (1705), and by Joseph Sauveur in Construction des quarrés magiques (1710). Concentric bordered squares were also studied by De la Hire in 1705, while Sauveur introduced magic cubes and lettered squares, which was taken up later by Euler in 1776, who is often credited for devising them. In 1750 d'Ons-le-Bray rediscovered the method of constructing doubly even and singly even squares using bordering technique; while in 1767 Benjamin Franklin published a semi-magic square which had the properties of eponymous Franklin square. By this time the earlier mysticism attached to the magic squares had completely vanished, and the subject was treated as a part of recreational mathematics.
In the 19th century, Bernard Violle gave a comprehensive treatment of magic squares in his three volume Traité complet des carrés magiques (1837–1838), which also described magic cubes, parallelograms, parallelopipeds, and circles. Pandiagonal squares were extensively studied by Andrew Hollingworth Frost, who learned it while in the town of Nasik, India, (thus calling them Nasik squares) in a series of articles: On the knight's path (1877), On the General Properties of Nasik Squares (1878), On the General Properties of Nasik Cubes (1878), On the construction of Nasik Squares of any order (1896). He showed that it is impossible to have normal singly-even pandiagonal magic square. Frederick A.P. Barnard constructed inlaid magic squares and other three dimensional magic figures like magic spheres and magic cylinders in Theory of magic squares and of magic cubes (1888). In 1897, Emroy McClintock published On the most perfect form of magic squares, coining the words pandiagonal square and most perfect square, which had previously been referred to as perfect, or diabolic, or Nasik.
Some famous magic squares
[edit]
Luo Shu magic square
[edit]
Main article: Lo Shu Square
Legends dating from as early as 650 BCE tell the story of the Lo Shu (洛書) or "scroll of the river Lo". According to the legend, there was at one time in ancient China a huge flood. While the great king Yu was trying to channel the water out to sea, a turtle emerged from it with a curious pattern on its shell: a 3×3 grid in which circular dots of numbers were arranged, such that the sum of the numbers in each row, column and diagonal was the same: 15. According to the legend, thereafter people were able to use this pattern in a certain way to control the river and protect themselves from floods.[citation needed] The Lo Shu Square, as the magic square on the turtle shell is called, is the unique normal magic square of order three in which 1 is at the bottom and 2 is in the upper right corner. Every normal magic square of order three is obtained from the Lo Shu by rotation or reflection.
Magic square in Parshavnath temple
[edit]
There is a well-known 12th-century 4×4 normal magic square inscribed on the wall of the Parshvanath temple in Khajuraho, India.
| | | | |
--- --- |
| 7 | 12 | 1 | 14 |
| 2 | 13 | 8 | 11 |
| 16 | 3 | 10 | 5 |
| 9 | 6 | 15 | 4 |
This is known as the Chautisa Yantra (Chautisa, 34; Yantra, lit. "device"), since its magic sum is 34. It is one of the three 4×4 pandiagonal magic squares and is also an instance of the most-perfect magic square. The study of this square led to the appreciation of pandiagonal squares by European mathematicians in the late 19th century. Pandiagonal squares were referred to as Nasik squares or Jain squares in older English literature.
Albrecht Dürer's magic square
[edit]
The order four normal magic square Albrecht Dürer immortalized in his 1514 engraving Melencolia I, referred to above, is believed to be the first seen in European art. The square associated with Jupiter appears as a talisman used to drive away melancholy. It is very similar to Yang Hui's square, which was created in China about 250 years before Dürer's time. As with every order 4 normal magic square, the magic sum is 34. But in the Durer square this sum is also found in each of the quadrants, in the center four squares, and in the corner squares (of the 4×4 as well as the four contained 3×3 grids). This sum can also be found in the four outer numbers clockwise from the corners (3+8+14+9) and likewise the four counter-clockwise (the locations of four queens in the two solutions of the 4 queens puzzle), the two sets of four symmetrical numbers (2+8+9+15 and 3+5+12+14), the sum of the middle two entries of the two outer columns and rows (5+9+8+12 and 3+2+15+14), and in four kite or cross shaped quartets (3+5+11+15, 2+10+8+14, 3+9+7+15, and 2+6+12+14). The two numbers in the middle of the bottom row give the date of the engraving: 1514. The numbers 1 and 4 at either side of the date correspond respectively to the letters "A" and "D", the initials of the artist.
| | | | |
--- --- |
| 16 | 3 | 2 | 13 |
| 5 | 10 | 11 | 8 |
| 9 | 6 | 7 | 12 |
| 4 | 15 | 14 | 1 |
Dürer's magic square can also be extended to a magic cube.
See also: Albrecht Dürer's magic square
Sagrada Família magic square
[edit]
The Passion façade of the Sagrada Família church in Barcelona, conceptualized by Antoni Gaudí and designed by sculptor Josep Subirachs, features a trivial order 4 magic square: The magic constant of the square is 33, the age of Jesus at the time of the Passion. Structurally, it is very similar to the Melancholia magic square, but it has had the numbers in four of the cells reduced by 1.
| | | | |
--- --- |
| 1 | 14 | 14 | 4 |
| 11 | 7 | 6 | 9 |
| 8 | 10 | 10 | 5 |
| 13 | 2 | 3 | 15 |
Trivial squares such as this one are not generally mathematically interesting and only have historical significance. Lee Sallows has pointed out that, due to Subirachs's ignorance of magic square theory, the renowned sculptor made a needless blunder, and supports this assertion by giving several examples of non-trivial 4×4 magic squares showing the desired magic constant of 33.
Similarly to Dürer's magic square, the Sagrada Familia's magic square can also be extended to a magic cube.
Properties of magic squares
[edit]
Magic constant
[edit]
Main article: Magic constant
The constant that is the sum of any row, or column, or diagonal is called the magic constant or magic sum, M. Every normal magic square has a constant dependent on the order n, calculated by the formula . This can be demonstrated by noting that the sum of is . Since the sum of each row is , the sum of rows is , which when divided by the order n yields the magic constant as . For normal magic squares of orders n = 3, 4, 5, 6, 7, and 8, the magic constants are, respectively: 15, 34, 65, 111, 175, and 260 (sequence A006003 in the OEIS).
Magic square of order 1 is trivial
[edit]
The 1×1 magic square, with only one cell containing the number 1, is called trivial, because it is typically not under consideration when discussing magic squares; but it is indeed a magic square by definition, if a single cell is regarded as a square of order one.
Magic square of order 2 cannot be constructed
[edit]
Normal magic squares of all sizes can be constructed except 2×2 (that is, where order n = 2).
Center of mass
[edit]
If the numbers in the magic square are seen as masses located in various cells, then the center of mass of a magic square coincides with its geometric center.
Moment of inertia
[edit]
The moment of inertia of a magic square has been defined as the sum over all cells of the number in the cell times the squared distance from the center of the cell to the center of the square; here the unit of measurement is the width of one cell. (Thus for example a corner cell of a 3×3 square has a distance of a non-corner edge cell has a distance of 1, and the center cell has a distance of 0.) Then all magic squares of a given order have the same moment of inertia as each other. For the order-3 case the moment of inertia is always 60, while for the order-4 case the moment of inertia is always 340. In general, for the n×n case the moment of inertia is
Birkhoff–von Neumann decomposition
[edit]
Dividing each number of the magic square by the magic constant will yield a doubly stochastic matrix, whose row sums and column sums equal to unity. However, unlike the doubly stochastic matrix, the diagonal sums of such matrices will also equal to unity. Thus, such matrices constitute a subset of doubly stochastic matrix. The Birkhoff–von Neumann theorem states that for any doubly stochastic matrix , there exists real numbers , where and permutation matrices such that
This representation may not be unique in general. By Marcus-Ree theorem, however, there need not be more than terms in any decomposition. Clearly, this decomposition carries over to magic squares as well, since a magic square can be recovered from a doubly stochastic matrix by multiplying it by the magic constant.
Classification of magic squares
[edit]
While the classification of magic squares can be done in many ways, some useful categories are given below. An n×n square array of integers 1, 2, ..., n2 is called:
Semi-magic square when its rows and columns sum to give the magic constant.
Simple magic square when its rows, columns, and two diagonals sum to give magic constant and no more. They are also known as ordinary magic squares or normal magic squares.
Self-complementary magic square when it is a magic square which when complemented (i.e. each number subtracted from n2 + 1) will give a rotated or reflected version of the original magic square.
Associative magic square when it is a magic square with a further property that every number added to the number equidistant, in a straight line, from the center gives n2 + 1. They are also called symmetric magic squares. Associative magic squares do not exist for squares of singly even order. All associative magic square are self-complementary magic squares as well.
Pandiagonal magic square when it is a magic square with a further property that the broken diagonals sum to the magic constant. They are also called panmagic squares, perfect squares, diabolic squares, Jain squares, or Nasik squares. Panmagic squares do not exist for singly even orders. However, singly even non-normal squares can be panmagic.
Ultra magic square when it is both associative and pandiagonal magic square. Ultra magic square exist only for orders n ≥ 5.
Bordered magic square when it is a magic square and it remains magic when the rows and columns on the outer edge are removed. They are also called concentric bordered magic squares if removing a border of a square successively gives another smaller bordered magic square. Bordered magic square do not exist for order 4.
Composite magic square when it is a magic square that is created by "multiplying" (in some sense) smaller magic squares, such that the order of the composite magic square is a multiple of the order of the smaller squares. Such squares can usually be partitioned into smaller non-overlapping magic sub-squares.
Inlaid magic square when it is a magic square inside which a magic sub-square is embedded, regardless of construction technique. The embedded magic sub-squares are themselves referred to as inlays.
Most-perfect magic square when it is a pandiagonal magic square with two further properties (i) each 2×2 subsquare add to 1/k of the magic constant where n = 4k, and (ii) all pairs of integers distant n/2 along any diagonal (major or broken) are complementary (i.e. they sum to n2 + 1). The first property is referred to as compactness, while the second property is referred to as completeness. Most-perfect magic squares exist only for squares of doubly even order. All the pandiagonal squares of order 4 are also most perfect.
Franklin magic square when it is a doubly even magic square with three further properties (i) every bent diagonal adds to the magic constant, (ii) every half row and half column starting at an outside edge adds to half the magic constant, and (iii) the square is compact.
Multimagic square when it is a magic square that remains magic even if all its numbers are replaced by their k-th power for 1 ≤ k ≤ P. They are also known as P-multimagic square or satanic squares. They are also referred to as bimagic squares, trimagic squares, tetramagic squares, and pentamagic squares when the value of P is 2, 3, 4, and 5 respectively.
Enumeration of magic squares
[edit]
Unsolved problem in mathematics
How many magic tori and magic squares of order n are there for and , respectively?
More unsolved problems in mathematics
Low-order squares
There is only one (trivial) magic square of order 1 and no magic square of order 2. As mentioned above, the set of normal squares of order three constitutes a single equivalence class-all equivalent to the Lo Shu square. Thus there is basically just one normal magic square of order 3.
The number of different n × n magic squares for n from 1 to 6, not counting rotations and reflections is:
: 1, 0, 1, 880, 275305224, 17753889197660635632. (sequence A006052 in the OEIS)
Magic tori
Cross-referenced to the above sequence, a new classification enumerates the magic tori that display these magic squares. The number of magic tori of order n from 1 to 5, is:
: 1, 0, 1, 255, 251449712 (sequence A270876 in the OEIS).
Higher-order squares and tori
The number of distinct normal magic squares rapidly increases for higher orders.
The 880 magic squares of order 4 are displayed on 255 magic tori of order 4 and the 275,305,224 squares of order 5 are displayed on 251,449,712 magic tori of order 5. The numbers of magic tori and distinct normal squares are not yet known for orders beyond 5 and 6, respectively.[citation needed]
Algorithms tend to only generate magic squares of a certain type or classification, making counting all possible magic squares quite difficult. Since traditional counting methods have proven unsuccessful, statistical analysis using the Monte Carlo method has been applied. The basic principle applied to magic squares is to randomly generate n × n matrices of elements 1 to n2 and check if the result is a magic square. The probability that a randomly generated matrix of numbers is a magic square is then used to approximate the number of magic squares.
More intricate versions of the Monte Carlo method, such as the exchange Monte Carlo, and Monte Carlo backtracking have produced even more accurate estimations. Using these methods it has been shown that the probability of magic squares decreases rapidly as n increases. Using fitting functions give the curves seen to the right.
Transformations that preserve the magic property
[edit]
For any magic square
[edit]
A magic square remains magic when its numbers are multiplied by any constant.
A magic square remains magic when a constant is added or subtracted to its numbers, or if its numbers are subtracted from a constant. In particular, if every element in a normal magic square is subtracted from n2 + 1, the resulting square is the complement of the original square. In the example below, elements of 4×4 square on the left is subtracted from 17 to obtain the complement of the square on the right.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | --- --- | | 10 | 3 | 13 | 8 | | 5 | 16 | 2 | 11 | | 4 | 9 | 7 | 14 | | 15 | 6 | 12 | 1 | | | | | | | --- --- | | 7 | 14 | 4 | 9 | | 12 | 1 | 15 | 6 | | 13 | 8 | 10 | 3 | | 2 | 11 | 5 | 16 | |
The numbers of a magic square can be substituted with corresponding numbers from a set of s arithmetic progressions with the same common difference among r terms, such that r × s = n2, and whose initial terms are also in arithmetic progression, to obtain a non-normal magic square. Here either s or r should be a multiple of n. Let us have s arithmetic progressions given by
: where a is the initial term, c is the common difference of the arithmetic progressions, and d is the common difference among the initial terms of each progression. The new magic constant will be
: If s = r = n, then we have the simplification
: If we further have a = c = 1 and d = n, we obtain the usual M = n(n2+1)/2. For given M we can find the required a, c, and d by solving the linear Diophantine equation. In the examples below, we have order 4 normal magic square on the left most side. The second square is a corresponding non-normal magic square with r = 8, s = 2, a = 1, c = 1, and d = 10 such that the new magic constant is M = 38. The third square is an order 5 normal magic square, which is a 90 degree clockwise rotated version of the square generated by De la Loubere method. On the right most side is a corresponding non-normal magic square with a = 4, c = 1, and d = 6 such that the new magic constant is M = 90.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | --- --- | | 1 | 15 | 14 | 4 | | 12 | 6 | 7 | 9 | | 8 | 10 | 11 | 5 | | 13 | 3 | 2 | 16 | | | | | | | --- --- | | 1 | 17 | 16 | 4 | | 14 | 6 | 7 | 11 | | 8 | 12 | 13 | 5 | | 15 | 3 | 2 | 18 | | | | | | | | --- --- | 11 | 10 | 4 | 23 | 17 | | 18 | 12 | 6 | 5 | 24 | | 25 | 19 | 13 | 7 | 1 | | 2 | 21 | 20 | 14 | 8 | | 9 | 3 | 22 | 16 | 15 | | | | | | | | --- --- | 16 | 14 | 7 | 30 | 23 | | 24 | 17 | 10 | 8 | 31 | | 32 | 25 | 18 | 11 | 4 | | 5 | 28 | 26 | 19 | 12 | | 13 | 6 | 29 | 22 | 20 | |
Any magic square can be rotated and reflected to produce 8 trivially distinct squares. In magic square theory, all of these are generally deemed equivalent and the eight such squares are said to make up a single equivalence class. In discussing magic squares, equivalent squares are usually not considered as distinct. The 8 equivalent squares are given for the 3×3 magic square below:
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | --- | 8 | 1 | 6 | | 3 | 5 | 7 | | 4 | 9 | 2 | | | | | | --- | 6 | 1 | 8 | | 7 | 5 | 3 | | 2 | 9 | 4 | | | | | | --- | 2 | 7 | 6 | | 9 | 5 | 1 | | 4 | 3 | 8 | | | | | | --- | 4 | 3 | 8 | | 9 | 5 | 1 | | 2 | 7 | 6 | |
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | --- | 2 | 9 | 4 | | 7 | 5 | 3 | | 6 | 1 | 8 | | | | | | --- | 4 | 9 | 2 | | 3 | 5 | 7 | | 8 | 1 | 6 | | | | | | --- | 8 | 3 | 4 | | 1 | 5 | 9 | | 6 | 7 | 2 | | | | | | --- | 6 | 7 | 2 | | 1 | 5 | 9 | | 8 | 3 | 4 | |
Given any magic square, another magic square of the same order can be formed by interchanging the row and the column which intersect in a cell on a diagonal with the row and the column which intersect in the complementary cell (i.e. cell symmetrically opposite from the center) of the same diagonal. For an even square, there are n/2 pairs of rows and columns that can be interchanged; thus we can obtain 2n/2 equivalent magic squares by combining such interchanges. For odd square, there are (n−1)/2 pairs of rows and columns that can be interchanged; and 2(n−1)/2 equivalent magic squares obtained by combining such interchanges. Interchanging all the rows and columns rotates the square by 180 degree. In the example using a 4×4 magic square, the left square is the original square, while the right square is the new square obtained by interchanging the 1st and 4th rows and columns.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | --- --- | | 1 | 15 | 14 | 4 | | 12 | 6 | 7 | 9 | | 8 | 10 | 11 | 5 | | 13 | 3 | 2 | 16 | | | | | | | --- --- | | 16 | 3 | 2 | 13 | | 9 | 6 | 7 | 12 | | 5 | 10 | 11 | 8 | | 4 | 15 | 14 | 1 | |
Given any magic square, another magic square of the same order can be formed by interchanging two rows on one side of the center line, and then interchanging the corresponding two rows on the other side of the center line; then interchanging like columns. For an even square, since there are n/2 same sided rows and columns, there are n(n−2)/8 pairs of such rows and columns that can be interchanged. Thus we can obtain 2n(n−2)/8 equivalent magic squares by combining such interchanges. For odd square, since there are (n−1)/2 same sided rows and columns, there are (n−1)(n−3)/8 pairs of such rows and columns that can be interchanged. Thus, there are 2(n−1)(n−3)/8 equivalent magic squares obtained by combining such interchanges. Interchanging every possible pairs of rows and columns rotates each quadrant of the square by 180 degree. In the example using a 4×4 magic square, the left square is the original square, while the right square is the new square obtained by this transformation. In the middle square, row 1 has been interchanged with row 2; and row 3 and 4 has been interchanged. The final square on the right is obtained by interchanging columns 1 and 2, and columns 3 and 4 of the middle square. In this particular example, this transform amounts to rotating the quadrants by 180 degree. The middle square is also a magic square, since the original square is an associative magic square.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | | | --- --- | | 1 | 15 | 14 | 4 | | 12 | 6 | 7 | 9 | | 8 | 10 | 11 | 5 | | 13 | 3 | 2 | 16 | | | | | | | --- --- | | 12 | 6 | 7 | 9 | | 1 | 15 | 14 | 4 | | 13 | 3 | 2 | 16 | | 8 | 10 | 11 | 5 | | | | | | | --- --- | | 6 | 12 | 9 | 7 | | 15 | 1 | 4 | 14 | | 3 | 13 | 16 | 2 | | 10 | 8 | 5 | 11 | |
A magic square remains magic when any of its non-central rows x and y are interchanged, along with the interchange of their complementary rows n − x + 1 and n − y + 1; and then interchanging like columns. This is a generalization of the above two transforms. When y = n − x + 1, this transform reduces to the first of the above two transforms. When x and y are on the same side of the center line, this transform reduces to the second of the above two transforms. In the example below, the original square is on the left side, while the final square on the right. The middle square has been obtained by interchanging rows 1 and 3, and rows 2 and 4 of the original square. The final square on the right is obtained by interchanging columns 1 and 3, and columns 2 and 4 of the middle square. In this example, this transform amounts to interchanging the quadrants diagonally. Since the original square is associative, the middle square also happens to be magic.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | | | --- --- | | 1 | 15 | 14 | 4 | | 12 | 6 | 7 | 9 | | 8 | 10 | 11 | 5 | | 13 | 3 | 2 | 16 | | | | | | | --- --- | | 8 | 10 | 11 | 5 | | 13 | 3 | 2 | 16 | | 1 | 15 | 14 | 4 | | 12 | 6 | 7 | 9 | | | | | | | --- --- | | 11 | 5 | 8 | 10 | | 2 | 16 | 13 | 3 | | 14 | 4 | 1 | 15 | | 7 | 9 | 12 | 6 | |
A magic square remains magic when its quadrants are diagonally interchanged because this is another symmetric permutation of the form described above. For even-order , permute the rows and columns by permutation where for , and for . For odd-order , permute rows and columns by permutation where for , and for . For odd ordered square, the halves of the central row and column are also interchanged. Examples for order 4 and 5 magic squares are given below:
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | --- --- | | 1 | 15 | 14 | 4 | | 12 | 6 | 7 | 9 | | 8 | 10 | 11 | 5 | | 13 | 3 | 2 | 16 | | | | | | | --- --- | | 11 | 5 | 8 | 10 | | 2 | 16 | 13 | 3 | | 14 | 4 | 1 | 15 | | 7 | 9 | 12 | 6 | | | | | | | | --- --- | 17 | 24 | 1 | 8 | 15 | | 23 | 5 | 7 | 14 | 16 | | 4 | 6 | 13 | 20 | 22 | | 10 | 12 | 19 | 21 | 3 | | 11 | 18 | 25 | 2 | 9 | | | | | | | | --- --- | 21 | 3 | 19 | 10 | 12 | | 2 | 9 | 25 | 11 | 18 | | 20 | 22 | 13 | 4 | 6 | | 8 | 15 | 1 | 17 | 24 | | 14 | 16 | 7 | 23 | 5 | |
For associative magic squares
[edit]
An associative magic square remains associative when two rows or columns equidistant from the center are interchanged. For an even square, there are n/2 pairs of rows or columns that can be interchanged; thus {{{1}}} equivalent magic squares by combining such interchanges can be obtained. For odd square, there are (n − 1)/2 pairs of rows or columns that can be interchanged; and 2n−1 equivalent magic squares obtained by combining such interchanges. Interchanging all the rows flips the square vertically (i.e. reflected along the horizontal axis), while interchanging all the columns flips the square horizontally (i.e. reflected along the vertical axis). In the example below, a 4×4 associative magic square on the left is transformed into a square on the right by interchanging the second and third row, yielding the famous Durer's magic square.
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | --- --- | | 16 | 3 | 2 | 13 | | 9 | 6 | 7 | 12 | | 5 | 10 | 11 | 8 | | 4 | 15 | 14 | 1 | | | | | | | --- --- | | 16 | 3 | 2 | 13 | | 5 | 10 | 11 | 8 | | 9 | 6 | 7 | 12 | | 4 | 15 | 14 | 1 | |
An associative magic square remains associative when two same sided rows (or columns) are interchanged along with corresponding other sided rows (or columns). For an even square, since there are n/2 same sided rows (or columns), there are n(n − 2)/8 pairs of such rows (or columns) that can be interchanged. Thus, 2n(n − 2)/8 × 2n(n − 2)/8 = 2n(n − 2)/4 equivalent magic squares can be obtained by combining such interchanges. For odd square, since there are (n − 1)/2 same sided rows or columns, there are (n − 1)(n − 3)/8 pairs of such rows or columns that can be interchanged. Thus, there are 2(n − 1)(n − 3)/8 × 2(n − 1)(n − 3)/8 = 2(n − 1)(n − 3)/4 equivalent magic squares obtained by combining such interchanges. Interchanging all the same sided rows flips each quadrants of the square vertically, while interchanging all the same sided columns flips each quadrant of the square horizontally. In the example below, the original square is on the left, whose rows 1 and 2 are interchanged with each other, along with rows 3 and 4, to obtain the transformed square on the right.
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | --- --- | | 1 | 15 | 14 | 4 | | 12 | 6 | 7 | 9 | | 8 | 10 | 11 | 5 | | 13 | 3 | 2 | 16 | | | | | | | --- --- | | 12 | 6 | 7 | 9 | | 1 | 15 | 14 | 4 | | 13 | 3 | 2 | 16 | | 8 | 10 | 11 | 5 | |
For pan-diagonal magic squares
[edit]
A pan-diagonal magic square remains a pan-diagonal magic square under cyclic shifting of rows or of columns or both. This allows us to position a given number in any one of the n2 cells of an n order square. Thus, for a given pan-magic square, there are n2 equivalent pan-magic squares. In the example below, the original square on the left is transformed by shifting the first row to the bottom to obtain a new pan-magic square in the middle. Next, the 1st and 2nd column of the middle pan-magic square is circularly shifted to the right to obtain a new pan-magic square on the right.
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | | | --- --- | | 10 | 3 | 13 | 8 | | 5 | 16 | 2 | 11 | | 4 | 9 | 7 | 14 | | 15 | 6 | 12 | 1 | | | | | | | --- --- | | 5 | 16 | 2 | 11 | | 4 | 9 | 7 | 14 | | 15 | 6 | 12 | 1 | | 10 | 3 | 13 | 8 | | | | | | | --- --- | | 2 | 11 | 5 | 16 | | 7 | 14 | 4 | 9 | | 12 | 1 | 15 | 6 | | 13 | 8 | 10 | 3 | |
For bordered magic squares
[edit]
A bordered magic square remains a bordered magic square after permuting the border cells in the rows or columns, together with their corresponding complementary terms, keeping the corner cells fixed. Since the cells in each row and column of every concentric border can be permuted independently, when the order n ≥ 5 is odd, there are equivalent bordered squares. When n ≥ 6 is even, there are equivalent bordered squares. In the example below, a square of order 5 is given whose border row has been permuted and (3!)2 = 36 such equivalent squares can be obtained.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | | --- --- | 1 | 23 | 16 | 4 | 21 | | 15 | 14 | 7 | 18 | 11 | | 24 | 17 | 13 | 9 | 2 | | 20 | 8 | 19 | 12 | 6 | | 5 | 3 | 10 | 22 | 25 | | | | | | | | --- --- | 1 | 16 | 23 | 4 | 21 | | 15 | 14 | 7 | 18 | 11 | | 24 | 17 | 13 | 9 | 2 | | 20 | 8 | 19 | 12 | 6 | | 5 | 10 | 3 | 22 | 25 | |
A bordered magic square remains a bordered magic square after each of its concentric borders are independently rotated or reflected with respect to the central core magic square. If there are b borders, then this transform will yield 8b equivalent squares. In the example below of the 5×5 magic square, the border has been rotated 90 degrees anti-clockwise.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | | --- --- | 1 | 23 | 16 | 4 | 21 | | 15 | 14 | 7 | 18 | 11 | | 24 | 17 | 13 | 9 | 2 | | 20 | 8 | 19 | 12 | 6 | | 5 | 3 | 10 | 22 | 25 | | | | | | | | --- --- | 21 | 11 | 2 | 6 | 25 | | 4 | 14 | 7 | 18 | 22 | | 16 | 17 | 13 | 9 | 10 | | 23 | 8 | 19 | 12 | 3 | | 1 | 15 | 24 | 20 | 5 | |
For composite magic squares
[edit]
A composite magic square remains a composite magic square when the embedded magic squares undergo transformations that do not disturb the magic property (e.g. rotation, reflection, shifting of rows and columns, and so on).
Special methods of construction
[edit]
Over the millennia, many ways to construct magic squares have been discovered. These methods can be classified as general methods and special methods, in the sense that general methods allow us to construct more than a single magic square of a given order, whereas special methods allow us to construct just one magic square of a given order. Special methods are specific algorithms whereas general methods may require some trial-and-error.
Special methods are the most simple ways to construct magic squares. They follow certain algorithms which generate regular patterns of numbers in a square. The correctness of these special methods can be proved using one of the general methods given in later sections. After a magic square has been constructed using a special method, the transformations described in the previous section can be applied to yield further magic squares. Special methods are usually referred to using the name of the author(s) (if known) who described the method, for e.g. De la Loubere's method, Starchey's method, Bachet's method, etc.
Magic squares are believed to exist for all orders, except for order 2. Magic squares can be classified according to their order as odd, doubly even (n divisible by four), and singly even (n even, but not divisible by four). This classification is based on the fact that entirely different techniques need to be employed to construct these different species of squares. Odd and doubly even magic squares are easy to generate; the construction of singly even magic squares is more difficult but several methods exist, including John Horton Conway's LUX method for magic squares and the Strachey method for magic squares.
A method for constructing a magic square of order 3
[edit]
In the 19th century, Édouard Lucas devised the general formula for order 3 magic squares. Consider the following table made up of positive integers a, b and c:
| | | |
---
| c − b | c + (a + b) | c − a |
| c − (a − b) | c | c + (a − b) |
| c + a | c − (a + b) | c + b |
These nine numbers will be distinct positive integers forming a magic square with the magic constant 3c so long as 0 < a < b < c − a and b ≠ 2a. Moreover, every 3×3 magic square of distinct positive integers is of this form.
In 1997 Lee Sallows discovered that leaving aside rotations and reflections, then every distinct parallelogram drawn on the Argand diagram defines a unique 3×3 magic square, and vice versa, a result that had never previously been noted.
A method for constructing a magic square of odd order
[edit]
See also: Siamese method
A method for constructing magic squares of odd order was published by the French diplomat de la Loubère in his book, A new historical relation of the kingdom of Siam (Du Royaume de Siam, 1693), in the chapter entitled The problem of the magical square according to the Indians. The method operates as follows:
The method prescribes starting in the central column of the first row with the number 1. After that, the fundamental movement for filling the squares is diagonally up and right, one step at a time. If a square is filled with a multiple of the order n, one moves vertically down one square instead, then continues as before. When an "up and to the right" move would leave the square, it is wrapped around to the last row or first column, respectively.
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--- --- --- --- --- --- --- --- --- --- --- --- --- |
| step 1 | 1 | | | | | | | | | step 2 | 1 | | | | | | 2 | | step 3 | 1 | | | 3 | | | 2 | | step 4 | 1 | | | 3 | | | 4 | 2 | |
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| step 5 | 1 | | | 3 | 5 | | | 4 | 2 | | step 6 | 1 | 6 | | 3 | 5 | | | 4 | 2 | | step 7 | 1 | 6 | | 3 | 5 | 7 | | 4 | 2 | | step 8 | 8 | 1 | 6 | | 3 | 5 | 7 | | 4 | 2 | | step 9 | 8 | 1 | 6 | | 3 | 5 | 7 | | 4 | 9 | 2 | |
Starting from other squares rather than the central column of the first row is possible, but then only the row and column sums will be identical and result in a magic sum, whereas the diagonal sums will differ. The result will thus be a semimagic square and not a true magic square. Moving in directions other than north east can also result in magic squares.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| Order 3 | 8 | 1 | 6 | | 3 | 5 | 7 | | 4 | 9 | 2 | | Order 5 | 17 | 24 | 1 | 8 | 15 | | 23 | 5 | 7 | 14 | 16 | | 4 | 6 | 13 | 20 | 22 | | 10 | 12 | 19 | 21 | 3 | | 11 | 18 | 25 | 2 | 9 | | Order 9 | 47 | 58 | 69 | 80 | 1 | 12 | 23 | 34 | 45 | | 57 | 68 | 79 | 9 | 11 | 22 | 33 | 44 | 46 | | 67 | 78 | 8 | 10 | 21 | 32 | 43 | 54 | 56 | | 77 | 7 | 18 | 20 | 31 | 42 | 53 | 55 | 66 | | 6 | 17 | 19 | 30 | 41 | 52 | 63 | 65 | 76 | | 16 | 27 | 29 | 40 | 51 | 62 | 64 | 75 | 5 | | 26 | 28 | 39 | 50 | 61 | 72 | 74 | 4 | 15 | | 36 | 38 | 49 | 60 | 71 | 73 | 3 | 14 | 25 | | 37 | 48 | 59 | 70 | 81 | 2 | 13 | 24 | 35 | |
A method of constructing a magic square of doubly even order
[edit]
Doubly even means that n is an even multiple of an even integer; or 4p (e.g. 4, 8, 12), where p is an integer.
Generic pattern
[edit]
All the numbers are written in order from left to right across each row in turn, starting from the top left hand corner. Numbers are then either retained in the same place or interchanged with their diametrically opposite numbers in a certain regular pattern. In the magic square of order four, the numbers in the four central squares and one square at each corner are retained in the same place and the others are interchanged with their diametrically opposite numbers.
A construction of a magic square of order 4
[edit]
Starting from top left, go left to right through each row of the square, counting each cell from 1 to 16 and filling the cells along the diagonals with its corresponding number. Once the bottom right cell is reached, continue by going right to left, starting from the bottom right of the table through each row, and fill in the non-diagonal cells counting up from 1 to 16 with its corresponding number. As shown below:
| | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| M = Order 4 | 1 | 4 | | 6 | 7 | | | 10 | 11 | | | 13 | 16 | | M = Order 4 | 1 | 15 | 14 | 4 | | 12 | 6 | 7 | 9 | | 8 | 10 | 11 | 5 | | 13 | 3 | 2 | 16 | |
An extension of the above example for Orders 8 and 12
[edit]
First generate a pattern table, where a '1' indicates selecting from the square where the numbers are written in order 1 to n2 (left-to-right, top-to-bottom), and a '0' indicates selecting from the square where the numbers are written in reverse order n2 to 1. For M = 4, the pattern table is as shown below (third matrix from left). With the unaltered cells (cells with '1') shaded, a criss-cross pattern is obtained.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| M = Order 4 | 1 | 2 | 3 | 4 | | 5 | 6 | 7 | 8 | | 9 | 10 | 11 | 12 | | 13 | 14 | 15 | 16 | | M = Order 4 | 16 | 15 | 14 | 13 | | 12 | 11 | 10 | 9 | | 8 | 7 | 6 | 5 | | 4 | 3 | 2 | 1 | | M = Order 4 | 1 | 0 | 0 | 1 | | 0 | 1 | 1 | 0 | | 0 | 1 | 1 | 0 | | 1 | 0 | 0 | 1 | | M = Order 4 | 1 | 15 | 14 | 4 | | 12 | 6 | 7 | 9 | | 8 | 10 | 11 | 5 | | 13 | 3 | 2 | 16 | |
The patterns are a) there are equal number of '1's and '0's in each row and column; b) each row and each column are "palindromic"; c) the left- and right-halves are mirror images; and d) the top- and bottom-halves are mirror images (c and d imply b). The pattern table can be denoted using hexadecimals as (9, 6, 6, 9) for simplicity (1-nibble per row, 4 rows). The simplest method of generating the required pattern for higher ordered doubly even squares is to copy the generic pattern for the fourth-order square in each four-by-four sub-squares.
For M = 8, possible choices for the pattern are (99, 66, 66, 99, 99, 66, 66, 99); (3C, 3C, C3, C3, C3, C3, 3C, 3C); (A5, 5A, A5, 5A, 5A, A5, 5A, A5) (2-nibbles per row, 8 rows).
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| M = Order 8 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | | M = Order 8 | 1 | 4 | 5 | 8 | | 10 | 11 | 14 | 15 | | | 18 | 19 | 22 | 23 | | | 25 | 28 | 29 | 32 | | 33 | 36 | 37 | 40 | | 42 | 43 | 46 | 47 | | | 50 | 51 | 54 | 55 | | | 57 | 60 | 61 | 64 | | M = Order 8 | 1 | 63 | 62 | 4 | 5 | 59 | 58 | 8 | | 56 | 10 | 11 | 53 | 52 | 14 | 15 | 49 | | 48 | 18 | 19 | 45 | 44 | 22 | 23 | 41 | | 25 | 39 | 38 | 28 | 29 | 35 | 34 | 32 | | 33 | 31 | 30 | 36 | 37 | 27 | 26 | 40 | | 24 | 42 | 43 | 21 | 20 | 46 | 47 | 17 | | 16 | 50 | 51 | 13 | 12 | 54 | 55 | 9 | | 57 | 7 | 6 | 60 | 61 | 3 | 2 | 64 | |
For M = 12, the pattern table (E07, E07, E07, 1F8, 1F8, 1F8, 1F8, 1F8, 1F8, E07, E07, E07) yields a magic square (3-nibbles per row, 12 rows.) It is possible to count the number of choices one has based on the pattern table, taking rotational symmetries into account.
Method of superposition
[edit]
The earliest discovery of the superposition method was made by the Indian mathematician Narayana in the 14th century. The same method was later re-discovered and studied in early 18th century Europe by de la Loubere, Poignard, de La Hire, and Sauveur; and the method is usually referred to as de la Hire's method. Although Euler's work on magic square was unoriginal, he famously conjectured the impossibility of constructing the evenly odd ordered mutually orthogonal Graeco-Latin squares. This conjecture was disproved in the mid 20th century. For clarity of exposition, two important variations of this method can be distinguished.
Euler's method
[edit]
This method consists in constructing two preliminary squares, which when added together gives the magic square. As a running example, a 3×3 magic square is considered. Each number of the 3×3 natural square by a pair of numbers can be labeled as
| | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- |
| | | | | --- | 1 | 2 | 3 | | 4 | 5 | 6 | | 7 | 8 | 9 | | | | | | --- | αa | αb | αc | | βa | βb | βc | | γa | γb | γc | |
where every pair of Greek and Latin alphabets, e.g. αa, are meant to be added together, i.e. αa = α + a. Here, (α, β, γ) = (0, 3, 6) and (a, b, c) = (1, 2, 3). The numbers 0, 3, and 6 are referred to as the root numbers while the numbers 1, 2, and 3 are referred to as the primary numbers. An important general constraint here is
a Greek letter is paired with a Latin letter only once.
Thus, the original square can now be split into two simpler squares:
| | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- |
| | | | | --- | α | α | α | | β | β | β | | γ | γ | γ | | | | | | --- | a | b | c | | a | b | c | | a | b | c | |
The lettered squares are referred to as Greek square or Latin square if they are filled with Greek or Latin letters, respectively. A magic square can be constructed by ensuring that the Greek and Latin squares are magic squares too. The converse of this statement is also often, but not always (e.g. bordered magic squares), true: A magic square can be decomposed into a Greek and a Latin square, which are themselves magic squares. Thus the method is useful for both synthesis as well as analysis of a magic square. Lastly, by examining the pattern in which the numbers are laid out in the finished square, it is often possible to come up with a faster algorithm to construct higher order squares that replicate the given pattern, without the necessity of creating the preliminary Greek and Latin squares.
During the construction of the 3×3 magic square, the Greek and Latin squares with just three unique terms are much easier to deal with than the original square with nine different terms. The row sum and the column sum of the Greek square will be the same, α + β + γ, if
each letter appears exactly once in a given column or a row.
This can be achieved by cyclic permutation of α, β, and γ. Satisfaction of these two conditions ensures that the resulting square is a semi-magic square; and such Greek and Latin squares are said to be mutually orthogonal to each other. For a given order n, there are at most n − 1 squares in a set of mutually orthogonal squares, not counting the variations due to permutation of the symbols. This upper bound is exact when n is a prime number.
In order to construct a magic square, we should also ensure that the diagonals sum to magic constant. For this, we have a third condition:
either all the letters should appear exactly once in both the diagonals; or in case of odd ordered squares, one of the diagonals should consist entirely of the middle term, while the other diagonal should have all the letters exactly once.
The mutually orthogonal Greek and Latin squares that satisfy the first part of the third condition (that all letters appear in both the diagonals) are said to be mutually orthogonal doubly diagonal Graeco-Latin squares.
Odd squares: For the 3×3 odd square, since α, β, and γ are in arithmetic progression, their sum is equal to the product of the square's order and the middle term, i.e. α + β + γ = 3 β. Thus, the diagonal sums will be equal if we have βs in the main diagonal and α, β, γ in the skew diagonal. Similarly, for the Latin square. The resulting Greek and Latin squares and their combination will be as below. The Latin square is just a 90 degree anti-clockwise rotation of the Greek square (or equivalently, flipping about the vertical axis) with the corresponding letters interchanged. Substituting the values of the Greek and Latin letters will give the 3×3 magic square.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | --- | β | α | γ | | γ | β | α | | α | γ | β | | | | | | --- | c | a | b | | a | b | c | | b | c | a | | | | | | --- | βc | αa | γb | | γa | βb | αc | | αb | γc | βa | | | | | | --- | 6 | 1 | 8 | | 7 | 5 | 3 | | 2 | 9 | 4 | |
For the odd squares, this method explains why the Siamese method (method of De la Loubere) and its variants work. This basic method can be used to construct odd ordered magic squares of higher orders. To summarise:
For odd ordered squares, to construct Greek square, place the middle term along the main diagonal, and place the rest of the terms along the skew diagonal. The remaining empty cells are filled by diagonal moves. The Latin square can be constructed by rotating or flipping the Greek square, and replacing the corresponding alphabets. The magic square is obtained by adding the Greek and Latin squares.
A peculiarity of the construction method given above for the odd magic squares is that the middle number (n2 + 1)/2 will always appear at the center cell of the magic square. Since there are (n − 1)! ways to arrange the skew diagonal terms, we can obtain (n − 1)! Greek squares this way; same with the Latin squares. Also, since each Greek square can be paired with (n − 1)! Latin squares, and since for each of Greek square the middle term may be arbitrarily placed in the main diagonal or the skew diagonal (and correspondingly along the skew diagonal or the main diagonal for the Latin squares), we can construct a total of 2 × (n − 1)! × (n − 1)! magic squares using this method. For n = 3, 5, and 7, this will give 8, 1152, and 1,036,800 different magic squares, respectively. Dividing by 8 to neglect equivalent squares due to rotation and reflections, we obtain 1, 144, and 129,600 essentially different magic squares, respectively.
As another example, the construction of 5×5 magic square is given. Numbers are directly written in place of alphabets. The numbered squares are referred to as primary square or root square if they are filled with primary numbers or root numbers, respectively. The numbers are placed about the skew diagonal in the root square such that the middle column of the resulting root square has 0, 5, 10, 15, 20 (from bottom to top). The primary square is obtained by rotating the root square counter-clockwise by 90 degrees, and replacing the numbers. The resulting square is an associative magic square, in which every pair of numbers symmetrically opposite to the center sum up to the same value, 26. For e.g., 16+10, 3+23, 6+20, etc. In the finished square, 1 is placed at center cell of bottom row, and successive numbers are placed via elongated knight's move (two cells right, two cells down), or equivalently, bishop's move (two cells diagonally down right). When a collision occurs, the break move is to move one cell up. All the odd numbers occur inside the central diamond formed by 1, 5, 25 and 21, while the even numbers are placed at the corners. The occurrence of the even numbers can be deduced by copying the square to the adjacent sides. The even numbers from four adjacent squares will form a cross.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | --- | | 10 | 5 | | 10 | 20 | | | 10 | | | 0 | 10 | | | 15 | 10 | | | | | | --- | 10 | 0 | 5 | | 10 | 20 | 0 | | 0 | 10 | | | 0 | 10 | | | 15 | 0 | 10 | | | | | | | | --- --- | 10 | 15 | 20 | 0 | 5 | | 5 | 10 | 15 | 20 | 0 | | 0 | 5 | 10 | 15 | 20 | | 20 | 0 | 5 | 10 | 15 | | 15 | 20 | 0 | 5 | 10 | | | | | | | | --- --- | 2 | 1 | 5 | 4 | 3 | | 1 | 5 | 4 | 3 | 2 | | 5 | 4 | 3 | 2 | 1 | | 4 | 3 | 2 | 1 | 5 | | 3 | 2 | 1 | 5 | 4 | | | | | | | | --- --- | 12 | 16 | 25 | 4 | 8 | | 6 | 15 | 19 | 23 | 2 | | 5 | 9 | 13 | 17 | 21 | | 24 | 3 | 7 | 11 | 20 | 24 | | 18 | 22 | 1 | 10 | 14 | 18 | 22 | | 4 | 8 | 12 | 16 | | 2 | 6 | |
A variation of the above example, where the skew diagonal sequence is taken in different order, is given below. The resulting magic square is the flipped version of the famous Agrippa's Mars magic square. It is an associative magic square and is the same as that produced by Moschopoulos's method. Here the resulting square starts with 1 placed in the cell which is to the right of the centre cell, and proceeds as De la Loubere's method, with downwards-right move. When a collision occurs, the break move is to shift two cells to the right.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | --- | | 10 | 20 | | 10 | 15 | | | 10 | | | 5 | 10 | | | 0 | 10 | | | | | | | | --- --- | 10 | 0 | 15 | 5 | 20 | | 20 | 10 | 0 | 15 | 5 | | 5 | 20 | 10 | 0 | 15 | | 15 | 5 | 20 | 10 | 0 | | 0 | 15 | 5 | 20 | 10 | | | | | | | | --- --- | 1 | 4 | 2 | 5 | 3 | | 4 | 2 | 5 | 3 | 1 | | 2 | 5 | 3 | 1 | 4 | | 5 | 3 | 1 | 4 | 2 | | 3 | 1 | 4 | 2 | 5 | | | | | | | | --- --- | 11 | 4 | 17 | 10 | 23 | | 24 | 12 | 5 | 18 | 6 | | 7 | 25 | 13 | 1 | 19 | | 20 | 8 | 21 | 14 | 2 | | 3 | 16 | 9 | 22 | 15 | |
In the previous examples, for the Greek square, the second row can be obtained from the first row by circularly shifting it to the right by one cell. Similarly, the third row is a circularly shifted version of the second row by one cell to the right; and so on. Likewise, the rows of the Latin square is circularly shifted to the left by one cell. The row shifts for the Greek and Latin squares are in mutually opposite direction. It is possible to circularly shift the rows by more than one cell to create the Greek and Latin square.
For odd ordered squares, whose order is not divisible by three, we can create the Greek squares by shifting a row by two places to the left or to the right to form the next row. The Latin square is made by flipping the Greek square along the main diagonal and interchanging the corresponding letters. This gives us a Latin square whose rows are created by shifting the row in the direction opposite to that of the Greek square. A Greek square and a Latin square should be paired such that their row shifts are in mutually opposite direction. The magic square is obtained by adding the Greek and Latin squares. When the order also happens to be a prime number, this method always creates pandiagonal magic square.
This essentially re-creates the knight's move. All the letters will appear in both the diagonals, ensuring correct diagonal sum. Since there are n! permutations of the Greek letters by which we can create the first row of the Greek square, there are thus n! Greek squares that can be created by shifting the first row in one direction. Likewise, there are n! such Latin squares created by shifting the first row in the opposite direction. Since a Greek square can be combined with any Latin square with opposite row shifts, there are n! × n! such combinations. Lastly, since the Greek square can be created by shifting the rows either to the left or to the right, there are a total of 2 × n! × n! magic squares that can be formed by this method. For n = 5 and 7, since they are prime numbers, this method creates 28,800 and 50,803,200 pandiagonal magic squares. Dividing by 8 to neglect equivalent squares due to rotation and reflections, we obtain 3,600 and 6,350,400 equivalent squares. Further dividing by n2 to neglect equivalent panmagic squares due to cyclic shifting of rows or columns, we obtain 144 and 129,600 essentially different panmagic squares. For order 5 squares, these are the only panmagic square there are. The condition that the square's order not be divisible by 3 means that we cannot construct squares of orders 9, 15, 21, 27, and so on, by this method.
In the example below, the square has been constructed such that 1 is at the center cell. In the finished square, the numbers can be continuously enumerated by the knight's move (two cells up, one cell right). When collision occurs, the break move is to move one cell up, one cell left. The resulting square is a pandiagonal magic square. This square also has a further diabolical property that any five cells in quincunx pattern formed by any odd sub-square, including wrap around, sum to the magic constant, 65. For e.g., 13+7+1+20+24, 23+1+9+15+17, 13+21+10+19+2 etc. Also the four corners of any 5×5 square and the central cell, as well as the middle cells of each side together with the central cell, including wrap around, give the magic sum: 13+10+19+22+1 and 20+24+12+8+1. Lastly the four rhomboids that form elongated crosses also give the magic sum: 23+1+9+24+8, 15+1+17+20+12, 14+1+18+13+19, 7+1+25+22+10. Such squares with 1 at the center cell are also called God's magic squares in Islamic amulet design, where the center cell is either left blank or filled with God's name.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | | --- --- | 10 | 15 | 20 | 0 | 5 | | 0 | 5 | 10 | 15 | 20 | | 15 | 20 | 0 | 5 | 10 | | 5 | 10 | 15 | 20 | 0 | | 20 | 0 | 5 | 10 | 15 | | | | | | | | --- --- | 3 | 1 | 4 | 2 | 5 | | 4 | 2 | 5 | 3 | 1 | | 5 | 3 | 1 | 4 | 2 | | 1 | 4 | 2 | 5 | 3 | | 2 | 5 | 3 | 1 | 4 | | | | | | | | --- --- | 13 | 16 | 24 | 2 | 10 | | 4 | 7 | 15 | 18 | 21 | | 20 | 23 | 1 | 9 | 12 | | 6 | 14 | 17 | 25 | 3 | | 22 | 5 | 8 | 11 | 19 | |
We can also combine the Greek and Latin squares constructed by different methods. In the example below, the primary square is made using knight's move. We have re-created the magic square obtained by De la Loubere's method. As before, we can form 8 × (n − 1)! × n! magic squares by this combination. For n = 5 and 7, this will create 23,040 and 29,030,400 magic squares. After dividing by 8 in order to neglect equivalent squares due to rotation and reflection, we get 2,880 and 3,628,800 squares.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | | --- --- | 15 | 20 | 0 | 5 | 10 | | 20 | 0 | 5 | 10 | 15 | | 0 | 5 | 10 | 15 | 20 | | 5 | 10 | 15 | 20 | 0 | | 10 | 15 | 20 | 0 | 5 | | | | | | | | --- --- | 2 | 4 | 1 | 3 | 5 | | 3 | 5 | 2 | 4 | 1 | | 4 | 1 | 3 | 5 | 2 | | 5 | 2 | 4 | 1 | 3 | | 1 | 3 | 5 | 2 | 4 | | | | | | | | --- --- | 17 | 24 | 1 | 8 | 15 | | 23 | 5 | 7 | 14 | 16 | | 4 | 6 | 13 | 20 | 22 | | 10 | 12 | 19 | 21 | 3 | | 11 | 18 | 25 | 2 | 9 | |
For order 5 squares, these three methods give a complete census of the number of magic squares that can be constructed by the method of superposition. Neglecting the rotation and reflections, the total number of magic squares of order 5 produced by the superposition method is 144 + 3,600 + 2,880 = 6,624.
Even squares: We can also construct even ordered squares in this fashion. Since there is no middle term among the Greek and Latin alphabets for even ordered squares, in addition to the first two constraint, for the diagonal sums to yield the magic constant, all the letters in the alphabet should appear in the main diagonal and in the skew diagonal.
An example of a 4×4 square is given below. For the given diagonal and skew diagonal in the Greek square, the rest of the cells can be filled using the condition that each letter appear only once in a row and a column.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | --- | | α | δ | | δ | α | | | γ | β | | | β | γ | | | | | | | --- --- | | α | β | γ | δ | | γ | δ | α | β | | δ | γ | β | α | | β | α | δ | γ | | | | | | | --- --- | | a | b | c | d | | d | c | b | a | | b | a | d | c | | c | d | a | b | | | | | | | --- --- | | αa | βb | γc | δd | | γd | δc | αb | βa | | δb | γa | βd | αc | | βc | αd | δa | γb | | | | | | | --- --- | | 1 | 6 | 11 | 16 | | 12 | 15 | 2 | 5 | | 14 | 9 | 8 | 3 | | 7 | 4 | 13 | 10 | |
Using these two Graeco-Latin squares, we can construct 2 × 4! × 4! = 1,152 magic squares. Dividing by 8 to eliminate equivalent squares due to rotation and reflections, we get 144 essentially different magic squares of order 4. These are the only magic squares constructible by the Euler method, since there are only two mutually orthogonal doubly diagonal Graeco-Latin squares of order 4.
Similarly, an 8×8 magic square can be constructed as below. Here the order of appearance of the numbers is not important; however the quadrants imitate the layout pattern of the 4×4 Graeco-Latin squares.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | | | | | | | --- --- --- --- | | 0 | 8 | 16 | 24 | 32 | 40 | 48 | 56 | | 24 | 16 | 8 | 0 | 56 | 48 | 40 | 32 | | 48 | 56 | 32 | 40 | 16 | 24 | 0 | 8 | | 40 | 32 | 56 | 48 | 8 | 0 | 24 | 16 | | 56 | 48 | 40 | 32 | 24 | 16 | 8 | 0 | | 32 | 40 | 48 | 56 | 0 | 8 | 16 | 24 | | 8 | 0 | 24 | 16 | 40 | 32 | 56 | 48 | | 16 | 24 | 0 | 8 | 48 | 56 | 32 | 40 | | | | | | | | | | | --- --- --- --- | | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | | 3 | 4 | 1 | 2 | 7 | 8 | 5 | 6 | | 5 | 6 | 7 | 8 | 1 | 2 | 3 | 4 | | 7 | 8 | 5 | 6 | 3 | 4 | 1 | 2 | | 4 | 3 | 2 | 1 | 8 | 7 | 6 | 5 | | 2 | 1 | 4 | 3 | 6 | 5 | 8 | 7 | | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | | 6 | 5 | 8 | 7 | 2 | 1 | 4 | 3 | | | | | | | | | | | --- --- --- --- | | 1 | 10 | 19 | 28 | 37 | 46 | 55 | 64 | | 27 | 20 | 9 | 2 | 63 | 56 | 45 | 38 | | 53 | 62 | 39 | 48 | 17 | 26 | 3 | 12 | | 47 | 40 | 61 | 54 | 11 | 4 | 25 | 18 | | 60 | 51 | 42 | 33 | 32 | 23 | 14 | 5 | | 34 | 41 | 52 | 59 | 6 | 13 | 24 | 31 | | 16 | 7 | 30 | 21 | 44 | 35 | 58 | 49 | | 22 | 29 | 8 | 15 | 50 | 57 | 36 | 43 | |
Euler's method has given rise to the study of Graeco-Latin squares. Euler's method for constructing magic squares is valid for any order except 2 and 6.
Variations: Magic squares constructed from mutually orthogonal doubly diagonal Graeco-Latin squares are interesting in themselves since the magic property emerges from the relative position of the alphabets in the square, and not due to any arithmetic property of the value assigned to them. This means that we can assign any value to the alphabets of such squares and still obtain a magic square. This is the basis for constructing squares that display some information (e.g. birthdays, years, etc.) in the square and for creating "reversible squares". For example, we can display the number π ≈ 3.141592 at the bottom row of a 4×4 magic square using the Graeco-Latin square given above by assigning (α, β, γ, δ) = (10, 0, 90, 15) and (a, b, c, d) = (0, 2, 3, 4). We will obtain the following non-normal magic square with the magic sum 124:
| | | | |
--- --- |
| 10 | 2 | 93 | 19 |
| 94 | 18 | 12 | 0 |
| 17 | 90 | 4 | 13 |
| 3 | 14 | 15 | 92 |
Narayana-De la Hire's method for even orders
[edit]
Narayana-De la Hire's method for odd square is the same as that of Euler's. However, for even squares, we drop the second requirement that each Greek and Latin letter appear only once in a given row or column. This allows us to take advantage of the fact that the sum of an arithmetic progression with an even number of terms is equal to the sum of two opposite symmetric terms multiplied by half the total number of terms. Thus, when constructing the Greek or Latin squares,
for even ordered squares, a letter can appear n/2 times in a column but only once in a row, or vice versa.
As a running example, if we take a 4×4 square, where the Greek and Latin terms have the values (α, β, γ, δ) = (0, 4, 8, 12) and (a, b, c, d) = (1, 2, 3, 4), respectively, then we have α + β + γ + δ = 2 (α + δ) = 2 (β + γ). Similarly, a + b + c + d = 2 (a + d) = 2 (b + c). This means that the complementary pair α and δ (or β and γ) can appear twice in a column (or a row) and still give the desired magic sum. Thus, we can construct:
For even ordered squares, the Greek magic square is made by first placing the Greek alphabets along the main diagonal in some order. The skew diagonal is then filled in the same order or by picking the terms that are complementary to the terms in the main diagonal. Finally, the remaining cells are filled column wise. Given a column, we use the complementary terms in the diagonal cells intersected by that column, making sure that they appear only once in a given row but n/2 times in the given column. The Latin square is obtained by flipping or rotating the Greek square and interchanging the corresponding alphabets. The final magic square is obtained by adding the Greek and Latin squares.
In the example given below, the main diagonal (from top left to bottom right) is filled with sequence ordered as α, β, γ, δ, while the skew diagonal (from bottom left to top right) filled in the same order. The remaining cells are then filled column wise such that the complementary letters appears only once within a row, but twice within a column. In the first column, since α appears on the 1st and 4th row, the remaining cells are filled with its complementary term δ. Similarly, the empty cells in the 2nd column are filled with γ; in 3rd column β; and 4th column α. Each Greek letter appears only once along the rows, but twice along the columns. As such, the row sums are α + β + γ + δ while the column sums are either 2 (α + δ) or 2 (β + γ). Likewise for the Latin square, which is obtained by flipping the Greek square along the main diagonal and interchanging the corresponding letters.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | --- | | α | δ | | β | γ | | | β | γ | | | α | δ | | | | | | | --- --- | | α | γ | β | δ | | δ | β | γ | α | | δ | β | γ | α | | α | γ | β | δ | | | | | | | --- --- | | a | d | d | a | | c | b | b | c | | b | c | c | b | | d | a | a | d | | | | | | | --- --- | | αa | γd | βd | δa | | δc | βb | γb | αc | | δb | βc | γc | αb | | αd | γa | βa | δd | | | | | | | --- --- | | 1 | 12 | 8 | 13 | | 15 | 6 | 10 | 3 | | 14 | 7 | 11 | 2 | | 4 | 9 | 5 | 16 | |
The above example explains why the "criss-cross" method for doubly even magic square works. Another possible 4×4 magic square, which is also pan-diagonal as well as most-perfect, is constructed below using the same rule. However, the diagonal sequence is chosen such that all four letters α, β, γ, δ appear inside the central 2×2 sub-square. Remaining cells are filled column wise such that each letter appears only once within a row. In the 1st column, the empty cells need to be filled with one of the letters selected from the complementary pair α and δ. Given the 1st column, the entry in the 2nd row can only be δ since α is already there in the 2nd row; while, in the 3rd row the entry can only be α since δ is already present in the 3rd row. We proceed similarly until all cells are filled. The Latin square given below has been obtained by flipping the Greek square along the main diagonal and replacing the Greek alphabets with corresponding Latin alphabets.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | --- | | α | γ | | γ | α | | | β | δ | | | δ | β | | | | | | | --- --- | | α | β | δ | γ | | δ | γ | α | β | | α | β | δ | γ | | δ | γ | α | β | | | | | | | --- --- | | a | d | a | d | | b | c | b | c | | d | a | d | a | | c | b | c | b | | | | | | | --- --- | | αa | βd | δa | γd | | δb | γc | αb | βc | | αd | βa | δd | γa | | δc | γb | αc | βb | | | | | | | --- --- | | 1 | 8 | 13 | 12 | | 14 | 11 | 2 | 7 | | 4 | 5 | 16 | 9 | | 15 | 10 | 3 | 6 | |
We can use this approach to construct singly even magic squares as well. However, we have to be more careful in this case since the criteria of pairing the Greek and Latin alphabets uniquely is not automatically satisfied. Violation of this condition leads to some missing numbers in the final square, while duplicating others. Thus, here is an important proviso:
For singly even squares, in the Greek square, check the cells of the columns which is vertically paired to its complement. In such a case, the corresponding cell of the Latin square must contain the same letter as its horizontally paired cell.
Below is a construction of a 6×6 magic square, where the numbers are directly given, rather than the alphabets. The second square is constructed by flipping the first square along the main diagonal. Here in the first column of the root square the 3rd cell is paired with its complement in the 4th cells. Thus, in the primary square, the numbers in the 1st and 6th cell of the 3rd row are same. Likewise, with other columns and rows. In this example the flipped version of the root square satisfies this proviso.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | | | | | --- --- --- | | 0 | 24 | 18 | 12 | 6 | 30 | | 30 | 6 | 12 | 18 | 24 | 0 | | 0 | 24 | 12 | 18 | 6 | 30 | | 30 | 24 | 12 | 18 | 6 | 0 | | 30 | 6 | 18 | 12 | 24 | 0 | | 0 | 6 | 18 | 12 | 24 | 30 | | | | | | | | | --- --- --- | | 1 | 6 | 1 | 6 | 6 | 1 | | 5 | 2 | 5 | 5 | 2 | 2 | | 4 | 3 | 3 | 3 | 4 | 4 | | 3 | 4 | 4 | 4 | 3 | 3 | | 2 | 5 | 2 | 2 | 5 | 5 | | 6 | 1 | 6 | 1 | 1 | 6 | | | | | | | | | --- --- --- | | 1 | 30 | 19 | 18 | 12 | 31 | | 35 | 8 | 17 | 23 | 26 | 2 | | 4 | 27 | 15 | 21 | 10 | 34 | | 33 | 28 | 16 | 22 | 9 | 3 | | 32 | 11 | 20 | 14 | 29 | 5 | | 6 | 7 | 24 | 13 | 25 | 36 | |
As another example of a 6×6 magic square constructed this way is given below. Here the diagonal entries are arranged differently. The primary square is constructed by flipping the root square about the main diagonal. In the second square the proviso for singly even square is not satisfied, leading to a non-normal magic square (third square) where the numbers 3, 13, 24, and 34 are duplicated while missing the numbers 4, 18, 19, and 33.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | | | | | --- --- --- | | 6 | 30 | 12 | 18 | 0 | 24 | | 24 | 0 | 12 | 18 | 30 | 6 | | 24 | 0 | 18 | 12 | 30 | 6 | | 6 | 30 | 18 | 12 | 0 | 24 | | 24 | 0 | 18 | 12 | 30 | 6 | | 6 | 30 | 12 | 18 | 0 | 24 | | | | | | | | | --- --- --- | | 2 | 5 | 5 | 2 | 5 | 2 | | 6 | 1 | 1 | 6 | 1 | 6 | | 3 | 3 | 4 | 4 | 4 | 3 | | 4 | 4 | 3 | 3 | 3 | 4 | | 1 | 6 | 6 | 1 | 6 | 1 | | 5 | 2 | 2 | 5 | 2 | 5 | | | | | | | | | --- --- --- | | 8 | 35 | 17 | 20 | 5 | 26 | | 30 | 1 | 13 | 24 | 31 | 12 | | 27 | 3 | 22 | 16 | 34 | 9 | | 10 | 34 | 21 | 15 | 3 | 28 | | 25 | 6 | 24 | 13 | 36 | 7 | | 11 | 32 | 14 | 23 | 2 | 29 | |
The last condition is a bit arbitrary and may not always need to be invoked, as in this example, where in the root square each cell is vertically paired with its complement:
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | | | | | --- --- --- | | 6 | 30 | 12 | 24 | 18 | 0 | | 6 | 0 | 18 | 24 | 12 | 30 | | 24 | 0 | 12 | 6 | 18 | 30 | | 6 | 30 | 18 | 24 | 12 | 0 | | 24 | 30 | 12 | 6 | 18 | 0 | | 24 | 0 | 18 | 6 | 12 | 30 | | | | | | | | | --- --- --- | | 2 | 2 | 5 | 2 | 5 | 5 | | 6 | 1 | 1 | 6 | 6 | 1 | | 3 | 4 | 3 | 4 | 3 | 4 | | 5 | 5 | 2 | 5 | 2 | 2 | | 4 | 3 | 4 | 3 | 4 | 3 | | 1 | 6 | 6 | 1 | 1 | 6 | | | | | | | | | --- --- --- | | 8 | 32 | 17 | 26 | 23 | 5 | | 12 | 1 | 19 | 30 | 18 | 31 | | 27 | 4 | 15 | 10 | 21 | 34 | | 11 | 35 | 20 | 29 | 14 | 2 | | 28 | 33 | 16 | 9 | 22 | 3 | | 25 | 6 | 24 | 7 | 13 | 36 | |
As one more example, we have generated an 8×8 magic square. Unlike the criss-cross pattern of the earlier section for evenly even square, here we have a checkered pattern for the altered and unaltered cells. Also, in each quadrant the odd and even numbers appear in alternating columns.
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | | | | | | | --- --- --- --- | | 0 | 48 | 16 | 32 | 24 | 40 | 8 | 56 | | 56 | 8 | 40 | 24 | 32 | 16 | 48 | 0 | | 0 | 48 | 16 | 32 | 24 | 40 | 8 | 56 | | 56 | 8 | 40 | 24 | 32 | 16 | 48 | 0 | | 56 | 8 | 40 | 24 | 32 | 16 | 48 | 0 | | 0 | 48 | 16 | 32 | 24 | 40 | 8 | 56 | | 56 | 8 | 40 | 24 | 32 | 16 | 48 | 0 | | 0 | 48 | 16 | 32 | 24 | 40 | 8 | 56 | | | | | | | | | | | --- --- --- --- | | 1 | 8 | 1 | 8 | 8 | 1 | 8 | 1 | | 7 | 2 | 7 | 2 | 2 | 7 | 2 | 7 | | 3 | 6 | 3 | 6 | 6 | 3 | 6 | 3 | | 5 | 4 | 5 | 4 | 4 | 5 | 4 | 5 | | 4 | 5 | 4 | 5 | 5 | 4 | 5 | 4 | | 6 | 3 | 6 | 3 | 3 | 6 | 3 | 6 | | 2 | 7 | 2 | 7 | 7 | 2 | 7 | 2 | | 8 | 1 | 8 | 1 | 1 | 8 | 1 | 8 | | | | | | | | | | | --- --- --- --- | | 1 | 56 | 17 | 40 | 32 | 41 | 16 | 57 | | 63 | 10 | 47 | 26 | 34 | 23 | 50 | 7 | | 3 | 54 | 19 | 38 | 30 | 43 | 14 | 59 | | 61 | 12 | 45 | 28 | 36 | 21 | 42 | 5 | | 60 | 13 | 44 | 29 | 37 | 20 | 53 | 4 | | 6 | 51 | 22 | 35 | 27 | 46 | 11 | 62 | | 58 | 15 | 42 | 31 | 39 | 18 | 55 | 2 | | 8 | 47 | 24 | 33 | 25 | 48 | 9 | 64 | |
Variations: A number of variations of the basic idea are possible: a complementary pair can appear n/2 times or less in a column. That is, a column of a Greek square can be constructed using more than one complementary pair. This method allows us to imbue the magic square with far richer properties. The idea can also be extended to the diagonals too. An example of an 8×8 magic square is given below. In the finished square each of four quadrants are pan-magic squares as well, each quadrant with same magic constant 130.
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | | | | | | | --- --- --- --- | | 0 | 48 | 56 | 8 | 16 | 32 | 40 | 24 | | 56 | 8 | 0 | 48 | 40 | 24 | 16 | 32 | | 0 | 48 | 56 | 8 | 16 | 32 | 40 | 24 | | 56 | 8 | 0 | 48 | 40 | 24 | 16 | 32 | | 48 | 0 | 8 | 56 | 32 | 16 | 24 | 40 | | 8 | 56 | 48 | 0 | 24 | 40 | 32 | 16 | | 48 | 0 | 8 | 56 | 32 | 16 | 24 | 40 | | 8 | 56 | 48 | 0 | 24 | 40 | 32 | 16 | | | | | | | | | | | --- --- --- --- | | 1 | 8 | 1 | 8 | 7 | 2 | 7 | 2 | | 7 | 2 | 7 | 2 | 1 | 8 | 1 | 8 | | 8 | 1 | 8 | 1 | 2 | 7 | 2 | 7 | | 2 | 7 | 2 | 7 | 8 | 1 | 8 | 1 | | 3 | 6 | 3 | 6 | 5 | 4 | 5 | 4 | | 5 | 4 | 5 | 4 | 3 | 6 | 3 | 6 | | 6 | 3 | 6 | 3 | 4 | 5 | 4 | 5 | | 4 | 5 | 4 | 5 | 6 | 3 | 6 | 3 | | | | | | | | | | | --- --- --- --- | | 1 | 56 | 57 | 16 | 23 | 34 | 47 | 26 | | 63 | 10 | 7 | 50 | 41 | 32 | 17 | 40 | | 8 | 49 | 64 | 9 | 18 | 39 | 42 | 31 | | 58 | 15 | 2 | 55 | 48 | 25 | 24 | 33 | | 51 | 6 | 11 | 62 | 37 | 20 | 29 | 44 | | 13 | 60 | 53 | 4 | 27 | 46 | 35 | 22 | | 54 | 3 | 14 | 59 | 36 | 21 | 28 | 45 | | 12 | 61 | 52 | 5 | 30 | 43 | 38 | 19 | |
Method of borders
[edit]
Bordering method for order 3
[edit]
In this method, the objective is to wrap a border around a smaller magic square which serves as a core. Consider the 3×3 square for example. Subtracting the middle number 5 from each number 1, 2, ..., 9, we obtain 0, ±1, ±2, ±3, and ±4, which we will, for lack of better words, following S. Harry White, refer to as bone numbers. The magic constant of a magic square, which we will refer to as the skeleton square, made by these bone numbers will be zero since adding all the rows of a magic square will give nM = Σ k = 0; thus M = 0.
It is not difficult to argue that the middle number should be placed at the center cell: let x be the number placed in the middle cell, then the sum of the middle column, middle row, and the two diagonals give Σ k + 3 x = 4 M. SinceΣ k = 3 M, we have x = M / 3. Here M = 0, so x = 0.
Putting the middle number 0 in the center cell, we want to construct a border such that the resulting square is magic. Let the border be given by:
| | | |
---
| u | a | v |
| b∗ | 0 | b |
| v∗ | a∗ | u∗ |
Since the sum of each row, column, and diagonals must be a constant (which is zero), we have
: a + a∗ = 0,
: b + b∗ = 0,
: u + u∗ = 0,
: v + v∗ = 0.
Now, if we have chosen a, b, u, and v, then we have a∗ = −a, b∗ = −b, u∗ = −u, and v∗ = −v. This means that if we assign a given number to a variable, say a = 1, then its complement will be assigned to a∗, i.e. a∗ = −1. Thus out of eight unknown variables, it is sufficient to specify the value of only four variables. We will consider a, b, u, and v as independent variables, while a∗, b∗, u∗, and v∗ as dependent variables. This allows us to consider a bone number ±x as a single number regardless of sign because (1) its assignment to a given variable, say a, will automatically imply that the same number of opposite sign will be shared with its complement a∗, and (2) two independent variables, say a and b, cannot be assigned the same bone number. But how should we choose a, b, u, and v? We have the sum of the top row and the sum of the right column as
: u + a + v = 0,
: v + b + u∗ = 0.
Since 0 is an even number, there are only two ways that the sum of three integers will yield an even number: 1) if all three were even, or 2) if two were odd and one was even. Since in our choice of numbers we only have two even non-zero number (±2 and ±4), the first statement is false. Hence, it must be the case that the second statement is true: that two of the numbers are odd and one even.
The only way that both the above two equations can satisfy this parity condition simultaneously, and still be consistent with the set of numbers we have, is when u and v are odd. For on the contrary, if we had assumed u and a to be odd and v to be even in the first equation, then u∗ = −u will be odd in the second equation, making b odd as well, in order to satisfy the parity condition. But this requires three odd numbers (u, a, and b), contradicting the fact that we only have two odd numbers (±1 and ±3) which we can use. This proves that the odd bone numbers occupy the corners cells. When converted to normal numbers by adding 5, this implies that the corners of a 3×3 magic square are all occupied by even numbers.
Thus, taking u = 1 and v = 3, we have a = −4 and b = −2. Hence, the finished skeleton square will be as in the left. Adding 5 to each number, we get the finished magic square.
| | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- |
| | | | | --- | 1 | −4 | 3 | | 2 | 0 | −2 | | −3 | 4 | −1 | | | | | | --- | 6 | 1 | 8 | | 7 | 5 | 3 | | 2 | 9 | 4 | |
Similar argument can be used to construct larger squares. Since there does not exist a 2×2 magic square around which we can wrap a border to construct a 4×4 magic square, the next smallest order for which we can construct bordered square is the order 5.
Bordering method for order 5
[edit]
Consider the fifth-order square. For this, we have a 3×3 magic core, around which we will wrap a magic border. The bone numbers to be used will be ±5, ±6, ±7, ±8, ±9, ±10, ±11, and ±12. Disregarding the signs, we have 8 bone numbers, 4 of which are even and 4 of which are odd. In general, for a square of any order n, there will be 4(n − 1) border cells, which are to be filled using 2(n − 1) bone numbers. Let the magic border be given as
| | | | | |
--- ---
| u | a | b | c | v |
| d∗ | d |
| e∗ | e |
| f∗ | f |
| v∗ | a∗ | b∗ | c∗ | u∗ |
As before, we should
place a bone number and its complement opposite to each other, so that the magic sum will be zero.
It is sufficient to determine the numbers u, v, a, b, c, d, e, f to describe the magic border. As before, we have the two constraint equations for the top row and right column:
: u + a + b + c + v = 0
: v + d + e + f + u = 0.
Multiple solutions are possible. The standard procedure is to
first try to determine the corner cells, after which we will try to determine the rest of the border.
There are 28 ways of choosing two numbers from the set of 8 bone numbers for the corner cells u and v. However, not all pairs are admissible. Among the 28 pairs, 16 pairs are made of an even and an odd number, 6 pairs have both as even numbers, while 6 pairs have them both as odd numbers.
We can prove that the corner cells u and v cannot have an even and an odd number. This is because if this were so, then the sums u + v and v + u∗ will be odd, and since 0 is an even number, the sums a + b + c and d + e + f should be odd as well. The only way that the sum of three integers will result in an odd number is when 1) two of them are even and one is odd, or 2) when all three are odd. Since the corner cells are assumed to be odd and even, neither of these two statements are compatible with the fact that we only have 3 even and 3 odd bone numbers at our disposal. This proves that u and v cannot have different parity. This eliminates 16 possibilities.
Using similar type reasoning we can also draw some conclusions about the sets {a, b, c} and {d, e, f}. If u and v are both even, then both the sets should have two odd numbers and one even number. If u and v are both odd, then one of the sets should have three even numbers while the other set should have one even number and two odd numbers.
As a running example, consider the case when both u and v are even. The 6 possible pairs are: (6, 8), (6, 10), (6, 12), (8, 10), (8, 12), and (10, 12). Since the sums u + v and v + u∗ are even, the sums a + b + c and d + e + f should be even as well. The only way that the sum of three integers will result in an even number is when 1) two of them are odd and one is even, or 2) when all three are even. The fact that the two corner cells are even means that we have only 2 even numbers at our disposal. Thus, the second statement is not compatible with this fact. Hence, it must be the case that the first statement is true: two of the three numbers should be odd, while one be even.
Now let a, b, d, e be odd numbers while c and f be even numbers. Given the odd bone numbers at our disposal: ±5, ±7, ±9, and ±11, their differences range from D = {±2, ±4, ±6} while their sums range from S = {±12, ±14, ±16, ±18, ±20}. It is also useful to have a table of their sum and differences for later reference. Now, given the corner cells (u, v), we can check its admissibility by checking if the sums u + v + c and v + u∗ + f fall within the set D or S. The admissibility of the corner numbers is a necessary but not a sufficient condition for the solution to exist.
For example, if we consider the pair (u, v) = (8, 12), then u + v = 20 and v + u = 6; and we will have ±6 and ±10 even bone numbers at our disposal. Taking c = ±6, we have the sum u + v + c to be 26 and 14, depending on the sign of ±6 taken, both of which do not fall within the sets D or S. Likewise, taking c = ±10, we have the sum u + v + c to be 30 and 10, both of which again do not fall within the sets D or S. Thus, the pair (8, 12) is not admissible. By similar process of reasoning, we can also rule out the pair (6, 12).
As another example, if we consider the pair (u, v) = (10, 12), then u + v = 22 and v + u∗ = 2; and we will have ± 6 and ± 8 even bone numbers at our disposal. Taking c = ±6, we have the sum u + v + c to be 28 and 16. While 28 does not fall within the sets D or S, 16 falls in set S. By inspection, we find that if (a, b) = (−7, − 9), then a + b = −16; and it will satisfy the first constraint equation. Also, taking f = ± 8, we have the sum v + u∗ + f to be 10 and -6. While 10 does not fall within the sets D or S, −6 falls in set D. Since −7 and −9 have already been assigned to a and b, clearly (d, e) = (-5, 11) so that d + e = 6; and it will satisfy the second constraint equation.
Likewise, taking c = ±8, we have the sum u + v + c to be 30 and 14. While 30 does not fall within the sets D or S, 14 falls in set S. By inspection, we find that if (a, b) = (−5, −9), then a + b = −14. Also, taking f = ± 6, we have the sum v + u∗ + f to be 8 and -4. While 8 does not fall within the sets D or S, −4 falls in set D. Clearly, (d, e) = (−7, 11) so that d + e = 4, and the second constraint equation will be satisfied.
Hence the corner pair (u, v) = (10, 12) is admissible; and it admits two solutions: and . The finished skeleton squares are given below. The magic square is obtained by adding 13 to each cells.
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | | --- --- | 10 | -7 | -9 | -6 | 12 | | 5 | -5 | | -11 | 11 | | 8 | -8 | | -12 | 7 | 9 | 6 | -10 | | | | | | | | --- --- | 23 | 6 | 4 | 7 | 25 | | 18 | 8 | | 2 | 24 | | 21 | 5 | | 1 | 20 | 22 | 19 | 3 | | | | | | | | --- --- | 10 | -5 | -9 | -8 | 12 | | 7 | -7 | | -11 | 11 | | 6 | -6 | | -12 | 5 | 9 | 8 | -10 | | | | | | | | --- --- | 23 | 8 | 4 | 5 | 25 | | 20 | 6 | | 2 | 24 | | 19 | 7 | | 1 | 18 | 22 | 21 | 3 | |
Using similar process of reasoning, we can construct the following table for the values of u, v, a, b, c, d, e, f expressed as bone numbers as given below. There are only 6 possible choices for the corner cells, which leads to 10 possible border solutions.
| u, v | a, b, c | d, e, f |
---
| 12, 10 | -6, -7, -9 | -11, 5, 8 |
| 12, 10 | -5, -8, -9 | -11, 6, 7 |
| 11, 5 | 6, -10, -12 | -9, 7, 8 |
| 10, 6 | 5, -9, -12 | -11, 7, 8 |
| 10, 6 | 7, -11, -12 | -9, 5, 8 |
| 9, 7 | 5, -10, -11 | -12, 6, 8 |
| 9, 7 | 6, -10, -12 | -11, 5, 8 |
| 8, 6 | 7, -10, -11 | -12, 5, 9 |
| 8, 6 | 9, -11, -12 | -10, 5, 7 |
| 7, 5 | 9, -10, -11 | -12, 6, 8 |
Given this group of 10 borders, we can construct 10×8×(3!)2 = 2880 essentially different bordered magic squares. Here the bone numbers ±5, ..., ±12 were consecutive. More bordered squares can be constructed if the numbers are not consecutive. If non-consecutive bone numbers were also used, then there are a total of 605 magic borders. Thus, the total number of order 5 essentially different bordered magic squares (with consecutive and non-consecutive numbers) is 174,240. See history. The number of fifth-order magic squares constructible via the bordering method is about 26 times larger than via the superposition method.
Continuous enumeration methods
[edit]
Exhaustive enumeration of all the borders of a magic square of a given order, as done previously, is very tedious. As such a structured solution is often desirable, which allows us to construct a border for a square of any order. Below we give three algorithms for constructing border for odd, doubly even, and singly even squares. These continuous enumeration algorithms were discovered in 10th century by Arab scholars; and their earliest surviving exposition comes from the two treatises by al-Buzjani and al-Antaki, although they themselves were not the discoverers. Since then many more such algorithms have been discovered.
Odd-ordered squares: The following is the algorithm given by al-Buzjani to construct a border for odd squares. A peculiarity of this method is that for order n square, the two adjacent corners are numbers and .
Starting from the cell above the lower left corner, we put the numbers alternately in left column and bottom row until we arrive at the middle cell. The next number is written in the middle cell of the bottom row just reached, after which we fill the cell in the upper left corner, then the middle cell of the right column, then the upper right corner. After this, starting from the cell above middle cell of the right column already filled, we resume the alternate placement of the numbers in the right column and the top row. Once half of the border cells are filled, the other half are filled by numbers complementary to opposite cells. The subsequent inner borders is filled in the same manner, until the square of order 3 is filled.
Below is an example for 9th-order square.
| | | | | | | | | |
--- --- --- ---
| 8 | 80 | 78 | 76 | 75 | 12 | 14 | 16 | 10 |
| 67 | 22 | 64 | 62 | 61 | 26 | 28 | 24 | 15 |
| 69 | 55 | 32 | 52 | 51 | 36 | 34 | 27 | 13 |
| 71 | 57 | 47 | 38 | 45 | 40 | 35 | 25 | 11 |
| 73 | 59 | 49 | 43 | 41 | 39 | 33 | 23 | 9 |
| 5 | 19 | 29 | 42 | 37 | 44 | 53 | 63 | 77 |
| 3 | 17 | 48 | 30 | 31 | 46 | 50 | 65 | 79 |
| 1 | 58 | 18 | 20 | 21 | 56 | 54 | 60 | 81 |
| 72 | 2 | 4 | 6 | 7 | 70 | 68 | 66 | 74 |
Doubly even order: The following is the method given by al-Antaki. Consider an empty border of order n = 4k with k ≥ 3. The peculiarity of this algorithm is that the adjacent corner cells are occupied by numbers n and .
Starting at the upper left corner cell, we put the successive numbers by groups of four, the first one next to the corner, the second and the third on the bottom, and the fourth at the top, and so on until there remains in the top row (excluding the corners) six empty cells. We then write the next two numbers above and the next four below. We then fill the upper corners, first left then right. We place the next number below the upper right corner in the right column, the next number on the other side in the left column. We then resume placing groups of four consecutive numbers in the two columns as before. Once half of the border cells are filled, the other half are filled by numbers complementary to opposite cells.
The example below gives the border for order 16 square.
| | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- |
| 15 | 1 | 255 | 254 | 4 | 5 | 251 | 250 | 8 | 9 | 10 | 246 | 245 | 244 | 243 | 16 |
| 240 | 17 |
| 18 | 239 |
| 19 | 238 |
| 237 | 20 |
| 236 | 21 |
| 22 | 235 |
| 23 | 234 |
| 233 | 24 |
| 232 | 25 |
| 26 | 231 |
| 27 | 230 |
| 229 | 28 |
| 228 | 29 |
| 30 | 227 |
| 241 | 256 | 2 | 3 | 253 | 252 | 6 | 7 | 249 | 248 | 247 | 11 | 12 | 13 | 14 | 242 |
For order 8 square, we just begin directly with the six cells.
| | | | | | | | |
--- --- --- --- |
| 7 | 1 | 2 | 62 | 61 | 60 | 59 | 8 |
| 56 | 9 |
| 10 | 55 |
| 11 | 54 |
| 53 | 12 |
| 52 | 13 |
| 14 | 51 |
| 57 | 64 | 63 | 3 | 4 | 5 | 6 | 58 |
Singly even order: For singly even order, we have the algorithm given by al-Antaki. Here the corner cells are occupied by n and n − 1. Below is an example of 10th-order square.
Start by placing 1 at the bottom row next to the left corner cell, then place 2 in the top row. After this, place 3 at the bottom row and turn around the border in anti-clockwise direction placing the next numbers, until n − 2 is reached on the right column. The next two numbers are placed in the upper corners (n − 1 in upper left corner and n in upper right corner). Then, the next two numbers are placed on the left column, then we resume the cyclic placement of the numbers until half of all the border cells are filled. Once half of the border cells are filled, the other half are filled by numbers complementary to opposite cells.
| | | | | | | | | | |
--- --- --- --- --- |
| 9 | 100 | 2 | 98 | 5 | 94 | 88 | 15 | 84 | 10 |
| 83 | 18 |
| 16 | 85 |
| 87 | 14 |
| 12 | 89 |
| 11 | 90 |
| 93 | 8 |
| 6 | 95 |
| 97 | 4 |
| 91 | 1 | 99 | 3 | 96 | 7 | 13 | 86 | 17 | 92 |
Method of composition
[edit]
For squares of order m × n where m, n > 2
[edit]
This is a method reminiscent of the Kronecker product of two matrices, that builds an nm × nm magic square from an n × n magic square and an m × m magic square. The "product" of two magic squares creates a magic square of higher order than the two multiplicands. Let the two magic squares be of orders m and n. The final square will be of order m × n. Divide the square of order m × n into m × m sub-squares, such that there are a total of n2 such sub-squares. In the square of order n, reduce by 1 the value of all the numbers. Multiply these reduced values by m2, and place the results in the corresponding sub-squares of the m × n whole square. The squares of order m are added n2 times to the sub-squares of the final square. The peculiarity of this construction method is that each magic subsquare will have different magic sums. The square made of such magic sums from each magic subsquare will again be a magic square. The smallest composite magic square of order 9, composed of two order 3 squares is given below.
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| Order 3 | 8 | 1 | 6 | | 3 | 5 | 7 | | 4 | 9 | 2 | | Order 3×3 | 63 | 63 | 63 | 0 | 0 | 0 | 45 | 45 | 45 | | 63 | 63 | 63 | 0 | 0 | 0 | 45 | 45 | 45 | | 63 | 63 | 63 | 0 | 0 | 0 | 45 | 45 | 45 | | 18 | 18 | 18 | 36 | 36 | 36 | 54 | 54 | 54 | | 18 | 18 | 18 | 36 | 36 | 36 | 54 | 54 | 54 | | 18 | 18 | 18 | 36 | 36 | 36 | 54 | 54 | 54 | | 27 | 27 | 27 | 72 | 72 | 72 | 9 | 9 | 9 | | 27 | 27 | 27 | 72 | 72 | 72 | 9 | 9 | 9 | | 27 | 27 | 27 | 72 | 72 | 72 | 9 | 9 | 9 | | Order 3×3 | 71 | 64 | 69 | 8 | 1 | 6 | 53 | 46 | 51 | | 66 | 68 | 70 | 3 | 5 | 7 | 48 | 50 | 52 | | 67 | 72 | 65 | 4 | 9 | 2 | 49 | 54 | 47 | | 26 | 19 | 24 | 44 | 37 | 42 | 62 | 55 | 60 | | 21 | 23 | 25 | 39 | 41 | 43 | 57 | 59 | 61 | | 22 | 27 | 20 | 40 | 45 | 38 | 58 | 63 | 56 | | 35 | 28 | 33 | 80 | 73 | 78 | 17 | 10 | 15 | | 30 | 32 | 34 | 75 | 77 | 79 | 12 | 14 | 16 | | 31 | 36 | 29 | 76 | 81 | 74 | 13 | 18 | 11 | |
Since each of the 3×3 sub-squares can be independently rotated and reflected into 8 different squares, from this single 9×9 composite square we can derive 89 = 134,217,728 essentially different 9×9 composite squares. Plenty more composite magic squares can also be derived if we select non-consecutive numbers in the magic sub-squares, like in Yang Hui's version of the 9×9 composite magic square. The next smallest composite magic squares of order 12, composed of magic squares of order 3 and 4 are given below.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| Order 3 | 2 | 9 | 4 | | 7 | 5 | 3 | | 6 | 1 | 8 | | Order 4 | 1 | 14 | 11 | 8 | | 12 | 7 | 2 | 13 | | 6 | 9 | 16 | 3 | | 15 | 4 | 5 | 10 | | Order 3 × 4 | 2 | 9 | 4 | 119 | 126 | 121 | 92 | 99 | 94 | 65 | 72 | 67 | | 7 | 5 | 3 | 124 | 122 | 120 | 97 | 95 | 93 | 70 | 68 | 66 | | 6 | 1 | 8 | 123 | 118 | 125 | 96 | 91 | 98 | 69 | 64 | 71 | | 101 | 108 | 103 | 56 | 63 | 58 | 11 | 18 | 13 | 110 | 117 | 112 | | 106 | 104 | 102 | 61 | 59 | 57 | 16 | 14 | 12 | 115 | 113 | 111 | | 105 | 100 | 107 | 60 | 55 | 62 | 15 | 10 | 17 | 114 | 109 | 116 | | 47 | 54 | 49 | 74 | 81 | 76 | 137 | 144 | 139 | 20 | 27 | 22 | | 52 | 50 | 48 | 79 | 77 | 75 | 142 | 140 | 138 | 25 | 23 | 21 | | 51 | 46 | 53 | 78 | 73 | 80 | 141 | 136 | 143 | 24 | 19 | 26 | | 128 | 135 | 130 | 29 | 36 | 31 | 38 | 45 | 40 | 83 | 90 | 85 | | 133 | 131 | 129 | 34 | 32 | 30 | 43 | 41 | 39 | 88 | 86 | 84 | | 132 | 127 | 134 | 33 | 28 | 35 | 42 | 37 | 44 | 87 | 82 | 89 | | Order 4 × 3 | 17 | 30 | 27 | 24 | 129 | 142 | 139 | 136 | 49 | 62 | 59 | 56 | | 28 | 23 | 18 | 29 | 140 | 135 | 130 | 141 | 60 | 55 | 50 | 61 | | 22 | 25 | 32 | 19 | 134 | 137 | 144 | 131 | 54 | 57 | 64 | 51 | | 31 | 20 | 21 | 26 | 143 | 132 | 133 | 138 | 63 | 52 | 53 | 58 | | 97 | 110 | 107 | 104 | 65 | 78 | 75 | 72 | 33 | 46 | 43 | 40 | | 108 | 103 | 98 | 109 | 76 | 71 | 66 | 77 | 44 | 39 | 34 | 45 | | 102 | 105 | 112 | 99 | 70 | 73 | 80 | 67 | 38 | 41 | 48 | 35 | | 111 | 100 | 101 | 106 | 79 | 68 | 69 | 74 | 47 | 36 | 37 | 42 | | 81 | 94 | 91 | 88 | 1 | 14 | 11 | 8 | 113 | 126 | 123 | 120 | | 92 | 87 | 82 | 93 | 12 | 7 | 2 | 13 | 124 | 119 | 114 | 125 | | 86 | 89 | 96 | 83 | 6 | 9 | 16 | 3 | 118 | 121 | 128 | 115 | | 95 | 84 | 85 | 90 | 15 | 4 | 5 | 10 | 127 | 116 | 117 | 122 | |
For the base squares, there is only one essentially different 3rd order square, while there 880 essentially different 4th-order squares that we can choose from. Each pairing can produce two different composite squares. Since each magic sub-squares in each composite square can be expressed in 8 different forms due to rotations and reflections, there can be 1×880×89 + 880×1×816 ≈ 2.476×1017 essentially different 12×12 composite magic squares created this way, with consecutive numbers in each sub-square. In general, if there are cm and cn essentially different magic squares of order m and n, then we can form cm × cn × ( 8m2 + 8n2) composite squares of order mn, provided m ≠ n. If m = n, then we can form (cm)2 × 8m2 composite squares of order m2.
For squares of doubly even order
[edit]
When the squares are of doubly even order, we can construct a composite magic square in a manner more elegant than the above process, in the sense that every magic subsquare will have the same magic constant. Let n be the order of the main square and m the order of the equal subsquares. The subsquares are filled one by one, in any order, with a continuous sequence of m2/2 smaller numbers (i.e. numbers less than or equal to n2/2) together with their complements to n2 + 1. Each subsquare as a whole will yield the same magic sum. The advantage of this type of composite square is that each subsquare is filled in the same way and their arrangement is arbitrary. Thus, the knowledge of a single construction of even order will suffice to fill the whole square. Furthermore, if the subsquares are filled in the natural sequence, then the resulting square will be pandiagonal. The magic sum of the subsquares is related to the magic sum of the whole square by where n = km.
In the examples below, we have divided the order 12 square into nine subsquares of order 4 filled each with eight smaller numbers and, in the corresponding bishop's cells (two cells diagonally across, including wrap arounds, in the 4×4 subsquare), their complements to n2 + 1 = 145. Each subsquare is pandiagonal with magic constant 290; while the whole square on the left is also pandiagonal with magic constant 870.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | | | | | | | | | --- --- --- --- --- --- | | 1 | 142 | 139 | 8 | 9 | 134 | 131 | 16 | 17 | 126 | 123 | 24 | | 140 | 7 | 2 | 141 | 132 | 15 | 10 | 133 | 124 | 23 | 18 | 125 | | 6 | 137 | 144 | 3 | 14 | 129 | 136 | 11 | 22 | 121 | 128 | 19 | | 143 | 4 | 5 | 138 | 135 | 12 | 13 | 130 | 127 | 20 | 21 | 122 | | 25 | 118 | 115 | 32 | 33 | 110 | 107 | 40 | 41 | 102 | 99 | 48 | | 116 | 31 | 26 | 117 | 108 | 39 | 34 | 109 | 100 | 47 | 42 | 101 | | 30 | 113 | 120 | 27 | 38 | 105 | 112 | 35 | 46 | 97 | 104 | 43 | | 119 | 28 | 29 | 114 | 111 | 36 | 37 | 106 | 103 | 44 | 45 | 98 | | 49 | 94 | 91 | 56 | 57 | 86 | 83 | 64 | 65 | 78 | 75 | 72 | | 92 | 55 | 50 | 93 | 84 | 63 | 58 | 85 | 76 | 71 | 66 | 77 | | 54 | 89 | 96 | 51 | 62 | 81 | 88 | 59 | 70 | 73 | 80 | 67 | | 95 | 52 | 53 | 90 | 87 | 60 | 61 | 82 | 79 | 68 | 69 | 74 | | | | | | | | | | | | | | | --- --- --- --- --- --- | | 69 | 74 | 79 | 68 | 29 | 114 | 119 | 28 | 61 | 82 | 87 | 60 | | 75 | 72 | 65 | 78 | 115 | 32 | 25 | 118 | 83 | 64 | 57 | 86 | | 66 | 77 | 76 | 71 | 26 | 117 | 116 | 31 | 58 | 85 | 84 | 63 | | 80 | 67 | 70 | 73 | 120 | 27 | 30 | 113 | 88 | 59 | 62 | 81 | | 21 | 122 | 127 | 20 | 53 | 90 | 95 | 52 | 13 | 130 | 135 | 12 | | 123 | 24 | 17 | 126 | 91 | 56 | 49 | 94 | 131 | 16 | 9 | 134 | | 18 | 125 | 124 | 23 | 50 | 93 | 92 | 55 | 10 | 133 | 132 | 15 | | 128 | 19 | 22 | 121 | 96 | 51 | 54 | 89 | 136 | 11 | 14 | 129 | | 45 | 98 | 103 | 44 | 5 | 138 | 143 | 4 | 37 | 106 | 111 | 36 | | 99 | 48 | 41 | 102 | 139 | 8 | 1 | 142 | 107 | 40 | 33 | 110 | | 42 | 101 | 100 | 47 | 2 | 141 | 140 | 7 | 34 | 109 | 108 | 39 | | 104 | 43 | 46 | 97 | 144 | 3 | 6 | 137 | 112 | 35 | 38 | 105 | |
In another example below, we have divided the order 12 square into four order 6 squares. Each of the order 6 squares are filled with eighteen small numbers and their complements using bordering technique given by al-Antaki. If we remove the shaded borders of the order 6 subsquares and form an order 8 square, then this order 8 square is again a magic square. In its full generality, we can take any m2/2 smaller numbers together with their complements to n2 + 1 to fill the subsquares, not necessarily in continuous sequence.
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--- --- --- --- --- --- |
| 60 | 82 | 88 | 56 | 90 | 59 | 24 | 118 | 124 | 20 | 126 | 23 |
| 64 | 69 | 74 | 79 | 68 | 81 | 28 | 33 | 110 | 115 | 32 | 117 |
| 83 | 75 | 72 | 65 | 78 | 62 | 119 | 111 | 36 | 29 | 114 | 26 |
| 84 | 66 | 77 | 76 | 71 | 61 | 120 | 30 | 113 | 112 | 35 | 25 |
| 58 | 80 | 67 | 70 | 73 | 87 | 22 | 116 | 31 | 34 | 109 | 123 |
| 86 | 63 | 57 | 89 | 55 | 85 | 122 | 27 | 21 | 125 | 19 | 121 |
| 6 | 136 | 142 | 2 | 144 | 5 | 42 | 100 | 106 | 38 | 108 | 41 |
| 10 | 15 | 128 | 133 | 14 | 135 | 46 | 51 | 92 | 97 | 50 | 99 |
| 137 | 129 | 18 | 11 | 132 | 8 | 101 | 93 | 54 | 47 | 96 | 44 |
| 138 | 12 | 131 | 130 | 17 | 7 | 102 | 48 | 95 | 94 | 53 | 43 |
| 4 | 134 | 13 | 16 | 127 | 141 | 40 | 98 | 49 | 52 | 91 | 105 |
| 140 | 9 | 3 | 143 | 1 | 139 | 104 | 45 | 39 | 107 | 37 | 103 |
Medjig-method for squares of even order 2n, where n > 2
[edit]
In this method a magic square is "multiplied" with a medjig square to create a larger magic square. The namesake of this method derives from mathematical game called medjig created by Willem Barink in 2006, although the method itself is much older.[citation needed] An early instance of a magic square constructed using this method occurs in Yang Hui's text for order 6 magic square.[citation needed] The LUX method to construct singly even magic squares is a special case of the medjig method, where only 3 out of 24 patterns are used to construct the medjig square.[citation needed]
The pieces of the medjig puzzle are 2×2 squares on which the numbers 0, 1, 2 and 3 are placed. There are three basic patterns by which the numbers 0, 1, 2 and 3 can be placed in a 2×2 square, where 0 is at the top left corner:
| | | | | | | | | | | | | | | |
--- --- --- --- --- --- ---
| | | | --- | | 0 | 1 | | 2 | 3 | | | | | --- | | 0 | 1 | | 3 | 2 | | | | | --- | | 0 | 2 | | 3 | 1 | |
Each pattern can be reflected and rotated to obtain 8 equivalent patterns, giving us a total of 3×8 = 24 patterns. The aim of the puzzle is to take n2 medjig pieces and arrange them in an n × n medjig square in such a way that each row, column, along with the two long diagonals, formed by the medjig square sums to 3n, the magic constant of the medjig square. An n × n medjig square can create a 2n × 2n magic square where n > 2.
Given an n×n medjig square and an n×n magic square base, a magic square of order 2n×2n can be constructed as follows:
Each cell of an n×n magic square is associated with a corresponding 2×2 subsquare of the medjig square
Fill each 2×2 subsquares of the medjig square with the four numbers from 1 to 4n2 that equal the original number modulo n2, i.e. x+n2y where x is the corresponding number from the magic square and y is a number from 0 to 3 in the 2×2 subsquares.
Assuming that we have an initial magic square base, the challenge lies in constructing a medjig square. For reference, the sums of each medjig piece along the rows, columns and diagonals, denoted in italics, are:
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| | | | | --- | 1 | 0 | 1 | | 5 | 2 | 3 | | 3 | 2 | 4 | 3 | | | | | | --- | 1 | 0 | 1 | | 5 | 3 | 2 | | 4 | 3 | 3 | 2 | | | | | | --- | 2 | 0 | 2 | | 4 | 3 | 1 | | 5 | 3 | 3 | 1 | |
Doubly even squares: The smallest even ordered medjig square is of order 2 with magic constant 6. While it is possible to construct a 2×2 medjig square, we cannot construct a 4×4 magic square from it since 2×2 magic squares required to "multiply" it does not exist. Nevertheless, it is worth constructing these 2×2 medjig squares. The magic constant 6 can be partitioned into two parts in three ways as 6 = 5 + 1 = 4 + 2 = 3 + 3. There exist 96 such 2×2 medjig squares.[citation needed] In the examples below, each 2×2 medjig square is made by combining different orientations of a single medjig piece.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| Medjig 2×2 | 0 | 1 | 3 | 2 | | 2 | 3 | 1 | 0 | | 3 | 2 | 0 | 1 | | 1 | 0 | 2 | 3 | | Medjig 2×2 | 0 | 1 | 2 | 3 | | 3 | 2 | 1 | 0 | | 1 | 0 | 3 | 2 | | 2 | 3 | 0 | 1 | | Medjig 2×2 | 0 | 2 | 3 | 1 | | 3 | 1 | 0 | 2 | | 0 | 2 | 3 | 1 | | 3 | 1 | 0 | 2 | |
We can use the 2×2 medjig squares to construct larger even ordered medjig squares. One possible approach is to simply combine the 2×2 medjig squares together. Another possibility is to wrap a smaller medjig square core with a medjig border. The pieces of a 2×2 medjig square can form the corner pieces of the border. Yet another possibility is to append a row and a column to an odd ordered medjig square. An example of an 8×8 magic square is constructed below by combining four copies of the left most 2×2 medjig square given above:
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| Order 4 | 1 | 14 | 4 | 15 | | 8 | 11 | 5 | 10 | | 13 | 2 | 16 | 3 | | 12 | 7 | 9 | 6 | | Medjig 4 × 4 | 0 | 1 | 3 | 2 | 0 | 1 | 3 | 2 | | 2 | 3 | 1 | 0 | 2 | 3 | 1 | 0 | | 3 | 2 | 0 | 1 | 3 | 2 | 0 | 1 | | 1 | 0 | 2 | 3 | 1 | 0 | 2 | 3 | | 0 | 1 | 3 | 2 | 0 | 1 | 3 | 2 | | 2 | 3 | 1 | 0 | 2 | 3 | 1 | 0 | | 3 | 2 | 0 | 1 | 3 | 2 | 0 | 1 | | 1 | 0 | 2 | 3 | 1 | 0 | 2 | 3 | | Order 8 | 1 | 17 | 62 | 46 | 4 | 20 | 63 | 47 | | 33 | 49 | 30 | 14 | 36 | 52 | 31 | 15 | | 56 | 40 | 11 | 27 | 53 | 37 | 10 | 26 | | 24 | 8 | 43 | 59 | 21 | 5 | 42 | 58 | | 13 | 29 | 50 | 34 | 16 | 32 | 51 | 35 | | 45 | 61 | 18 | 2 | 48 | 64 | 19 | 3 | | 60 | 44 | 7 | 23 | 57 | 41 | 6 | 22 | | 28 | 12 | 39 | 55 | 25 | 9 | 38 | 54 | |
The next example is constructed by bordering a 2×2 medjig square core.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| Order 4 | 1 | 14 | 4 | 15 | | 8 | 11 | 5 | 10 | | 13 | 2 | 16 | 3 | | 12 | 7 | 9 | 6 | | Medjig 4 × 4 | 0 | 1 | 0 | 1 | 2 | 3 | 3 | 2 | | 2 | 3 | 3 | 2 | 1 | 0 | 1 | 0 | | 0 | 3 | 0 | 2 | 3 | 1 | 0 | 3 | | 1 | 2 | 3 | 1 | 0 | 2 | 1 | 2 | | 2 | 1 | 0 | 2 | 3 | 1 | 2 | 1 | | 3 | 0 | 3 | 1 | 0 | 2 | 3 | 0 | | 3 | 2 | 0 | 1 | 2 | 3 | 0 | 1 | | 1 | 0 | 3 | 2 | 1 | 0 | 2 | 3 | | Order 8 | 1 | 17 | 14 | 30 | 36 | 52 | 63 | 47 | | 33 | 49 | 62 | 46 | 20 | 4 | 31 | 15 | | 8 | 56 | 11 | 43 | 53 | 21 | 10 | 58 | | 24 | 40 | 59 | 27 | 5 | 37 | 26 | 42 | | 45 | 29 | 2 | 34 | 64 | 32 | 35 | 19 | | 61 | 13 | 50 | 18 | 16 | 48 | 51 | 3 | | 60 | 44 | 7 | 23 | 41 | 57 | 6 | 22 | | 28 | 12 | 55 | 39 | 25 | 9 | 38 | 54 | |
Singly even squares: Medjig square of order 1 does not exist. As such, the smallest odd ordered medjig square is of order 3, with magic constant 9. There are only 7 ways of partitioning the integer 9, our magic constant, into three parts.[citation needed] If these three parts correspond to three of the medjig pieces in a row, column or diagonal, then the relevant partitions for us are:
: 9 = 1 + 3 + 5 = 1 + 4 + 4 = 2 + 3 + 4 = 2 + 2 + 5 = 3 + 3 + 3.
A 3×3 medjig square can be constructed with some trial-and-error, as in the left most square below. Another approach is to add a row and a column to a 2×2 medjig square. In the middle square below, a left column and bottom row has been added, creating an L-shaped medjig border, to a 2×2 medjig square given previously. The right most square below is essentially same as the middle square, except that the row and column has been added in the middle to form a cross while the pieces of 2×2 medjig square are placed at the corners.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| Medjig 3 × 3 | 2 | 3 | 0 | 2 | 0 | 2 | | 1 | 0 | 3 | 1 | 3 | 1 | | 3 | 1 | 1 | 2 | 2 | 0 | | 0 | 2 | 0 | 3 | 3 | 1 | | 3 | 2 | 2 | 0 | 0 | 2 | | 0 | 1 | 3 | 1 | 1 | 3 | | Medjig 3 × 3 | 0 | 3 | 0 | 1 | 3 | 2 | | 2 | 1 | 2 | 3 | 1 | 0 | | 3 | 0 | 3 | 2 | 0 | 1 | | 2 | 1 | 1 | 0 | 2 | 3 | | 0 | 1 | 3 | 1 | 3 | 1 | | 2 | 3 | 0 | 2 | 0 | 2 | | Medjig 3 × 3 | 0 | 1 | 0 | 3 | 3 | 2 | | 2 | 3 | 2 | 1 | 1 | 0 | | 3 | 1 | 0 | 1 | 3 | 1 | | 0 | 2 | 2 | 3 | 0 | 2 | | 3 | 2 | 3 | 0 | 0 | 1 | | 1 | 0 | 2 | 1 | 2 | 3 | |
Once a 3×3 medjig square has been constructed, it can be converted into a 6×6 magic square. For example, using the left most 3×3 medjig square given above:
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| Order 3 | 8 | 1 | 6 | | 3 | 5 | 7 | | 4 | 9 | 2 | | Medjig 3 × 3 | 2 | 3 | 0 | 2 | 0 | 2 | | 1 | 0 | 3 | 1 | 3 | 1 | | 3 | 1 | 1 | 2 | 2 | 0 | | 0 | 2 | 0 | 3 | 3 | 1 | | 3 | 2 | 2 | 0 | 0 | 2 | | 0 | 1 | 3 | 1 | 1 | 3 | | Order 6 | 26 | 35 | 1 | 19 | 6 | 24 | | 17 | 8 | 28 | 10 | 33 | 15 | | 30 | 12 | 14 | 23 | 25 | 7 | | 3 | 21 | 5 | 32 | 34 | 16 | | 31 | 22 | 27 | 9 | 2 | 20 | | 4 | 13 | 36 | 18 | 11 | 29 | |
There are 1,740,800 such 3×3 medjig squares. An easy approach to construct higher order odd medjig square is by wrapping a smaller odd ordered medjig square with a medjig border, just as with even ordered medjig squares. Another approach is to append a row and a column to an even ordered medjig square. Approaches such as the LUX method can also be used. In the example below, a 5×5 medjig square is created by wrapping a medjig border around a 3×3 medjig square given previously:
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| Order 5 | 17 | 24 | 1 | 8 | 15 | | 23 | 5 | 7 | 14 | 16 | | 4 | 6 | 13 | 20 | 22 | | 10 | 12 | 19 | 21 | 3 | | 11 | 18 | 25 | 2 | 9 | | Medjig 5 × 5 | 0 | 1 | 3 | 1 | 0 | 1 | 3 | 1 | 3 | 2 | | 2 | 3 | 0 | 2 | 2 | 3 | 0 | 2 | 1 | 0 | | 3 | 0 | 2 | 3 | 0 | 2 | 0 | 2 | 1 | 2 | | 1 | 2 | 1 | 0 | 3 | 1 | 3 | 1 | 3 | 0 | | 0 | 2 | 3 | 1 | 1 | 2 | 2 | 0 | 1 | 3 | | 1 | 3 | 0 | 2 | 0 | 3 | 3 | 1 | 0 | 2 | | 3 | 0 | 3 | 2 | 2 | 0 | 0 | 2 | 1 | 2 | | 1 | 2 | 0 | 1 | 3 | 1 | 1 | 3 | 3 | 0 | | 3 | 2 | 1 | 3 | 1 | 0 | 1 | 3 | 0 | 1 | | 1 | 0 | 2 | 0 | 3 | 2 | 2 | 0 | 2 | 3 | | Order 10 | 17 | 42 | 99 | 49 | 1 | 26 | 83 | 33 | 90 | 65 | | 67 | 92 | 24 | 74 | 51 | 76 | 8 | 58 | 40 | 15 | | 98 | 23 | 55 | 80 | 7 | 57 | 14 | 64 | 41 | 66 | | 48 | 73 | 30 | 5 | 82 | 32 | 89 | 39 | 91 | 16 | | 4 | 54 | 81 | 31 | 38 | 63 | 70 | 20 | 47 | 97 | | 29 | 79 | 6 | 56 | 13 | 88 | 95 | 45 | 22 | 72 | | 85 | 10 | 87 | 62 | 69 | 19 | 21 | 71 | 28 | 53 | | 35 | 60 | 12 | 37 | 94 | 44 | 46 | 96 | 78 | 3 | | 86 | 61 | 43 | 93 | 100 | 25 | 27 | 77 | 9 | 34 | | 36 | 11 | 68 | 18 | 95 | 75 | 52 | 2 | 59 | 84 | |
Solving partially completed magic squares
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Solving partially completed magic squares is a popular mathematical pastime. The techniques needed are similar to those used in Sudoku or KenKen puzzles, and involve deducing the values of unfilled squares using logic and permutation group theory (Sudoku grids are not magic squares but are based on a related idea called Graeco-Latin squares).
Variations of the magic square
[edit]
Extra constraints
[edit]
Certain extra restrictions can be imposed on magic squares.
If raising each number to the nth power yields another magic square, the result is a bimagic (n = 2), a trimagic (n = 3), or, in general, a multimagic square.
A magic square in which the number of letters in the name of each number in the square generates another magic square is called an alphamagic square.
There are magic squares consisting entirely of primes. Rudolf Ondrejka (1928–2001) discovered the following 3×3 magic square of primes, in this case nine Chen primes:
| | | |
---
| 17 | 89 | 71 |
| 113 | 59 | 5 |
| 47 | 29 | 101 |
The Green–Tao theorem implies that there are arbitrarily large magic squares consisting of primes.
The following "reversible magic square" has a magic constant of 264 both upside down and right way up:
| | | | |
--- --- |
| 96 | 11 | 89 | 68 |
| 88 | 69 | 91 | 16 |
| 61 | 86 | 18 | 99 |
| 19 | 98 | 66 | 81 |
When the extra constraint is to display some date, especially a birth date, then such magic squares are called birthday magic square. An early instance of such birthday magic square was created by Srinivasa Ramanujan. He created a 4×4 square in which he entered his date of birth in DD–MM–CC–YY format in the top row and the magic happened with additions and subtractions of numbers in squares. Not only do the rows, columns, and diagonals add up to the same number, but the four corners, the four middle squares (17, 9, 24, 89), the first and last rows two middle numbers (12, 18, 86, 23), and the first and last columns two middle numbers (88, 10, 25, 16) all add up to the sum of 139.
Multiplicative magic squares
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Instead of adding the numbers in each row, column and diagonal, one can apply some other operation. For example, a multiplicative magic square has a constant product of numbers. A multiplicative magic square can be derived from an additive magic square by raising 2 (or any other integer) to the power of each element, because the logarithm of the product of 2 numbers is the sum of logarithm of each. Alternatively, if any 3 numbers in a line are 2a, 2b and 2c, their product is 2a+b+c, which is constant if a+b+c is constant, as they would be if a, b and c were taken from ordinary (additive) magic square. For example, the original Lo-Shu magic square becomes:
M = 32768
| 16 | 512 | 4 |
| 8 | 32 | 128 |
| 256 | 2 | 64 |
Other examples of multiplicative magic squares include:
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| M = 216 | 2 | 9 | 12 | | 36 | 6 | 1 | | 3 | 4 | 18 | | M = 6720 | 1 | 6 | 20 | 56 | | 40 | 28 | 2 | 3 | | 14 | 5 | 24 | 4 | | 12 | 8 | 7 | 10 | | M = 6,227,020,800 | 27 | 50 | 66 | 84 | 13 | 2 | 32 | | 24 | 52 | 3 | 40 | 54 | 70 | 11 | | 56 | 9 | 20 | 44 | 36 | 65 | 6 | | 55 | 72 | 91 | 1 | 16 | 36 | 30 | | 4 | 24 | 45 | 60 | 77 | 12 | 26 | | 10 | 22 | 48 | 39 | 5 | 48 | 63 | | 78 | 7 | 8 | 18 | 40 | 33 | 60 | |
Multiplicative magic squares of complex numbers
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Still using Ali Skalli's non iterative method, it is possible to produce an infinity of multiplicative magic squares of complex numbers belonging to set. On the example below, the real and imaginary parts are integer numbers, but they can also belong to the entire set of real numbers . The product is: −352,507,340,640 − 400,599,719,520 i.
Skalli multiplicative 7×7 of complex numbers
| 21 | +14i | −70 | +30i | −93 | −9i | −105 | −217i | 16 | +50i | 4 | −14i | 14 | −8i |
| 63 | −35i | 28 | +114i | −14i | 2 | +6i | 3 | −11i | 211 | +357i | −123 | −87i |
| 31 | −15i | 13 | −13i | −103 | +69i | −261 | −213i | 49 | −49i | −46 | +2i | −6 | +2i |
| 102 | −84i | −28 | −14i | 43 | +247i | −10 | −2i | 5 | +9i | 31 | −27i | −77 | +91i |
| −22 | −6i | 7 | +7i | 8 | +14i | 50 | +20i | −525 | −492i | −28 | −42i | −73 | +17i |
| 54 | +68i | 138 | −165i | −56 | −98i | −63 | +35i | 4 | −8i | 2 | −4i | 70 | −53i |
| 24 | +22i | −46 | −16i | 6 | −4i | 17 | +20i | 110 | +160i | 84 | −189i | 42 | −14i |
Additive-multiplicative magic and semimagic squares
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Additive-multiplicative magic squares and semimagic squares satisfy properties of both ordinary and multiplicative magic squares and semimagic squares, respectively.
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| First knownadditive-multiplicative magic square 8×8 found by W. W. Horner in 1955 Sum = 840 Product = 2058068231856000 | 162 | 207 | 51 | 26 | 133 | 120 | 116 | 25 | | 105 | 152 | 100 | 29 | 138 | 243 | 39 | 34 | | 92 | 27 | 91 | 136 | 45 | 38 | 150 | 261 | | 57 | 30 | 174 | 225 | 108 | 23 | 119 | 104 | | 58 | 75 | 171 | 90 | 17 | 52 | 216 | 161 | | 13 | 68 | 184 | 189 | 50 | 87 | 135 | 114 | | 200 | 203 | 15 | 76 | 117 | 102 | 46 | 81 | | 153 | 78 | 54 | 69 | 232 | 175 | 19 | 60 | | Smallest known additive-multiplicative semimagic square 4×4 found by L. Morgenstern in 2007 Sum = 247 Product = 3369600 | 156 | 18 | 48 | 25 | | 30 | 144 | 60 | 13 | | 16 | 20 | 130 | 81 | | 45 | 65 | 9 | 128 | |
It is unknown if any additive-multiplicative magic squares smaller than 7×7 exist, but it has been proven that no 3×3 or 4×4 additive-multiplicative magic squares and no 3×3 additive-multiplicative semimagic squares exist.
Smallest known additive-multiplicative magic square 7×7 found by Sébastien Miquel(Sébastien Miquel) in August 2016 Sum = 465 Product = 150885504000
| 126 | 66 | 50 | 90 | 48 | 1 | 84 |
| 20 | 70 | 16 | 54 | 189 | 110 | 6 |
| 100 | 2 | 22 | 98 | 36 | 72 | 135 |
| 96 | 60 | 81 | 4 | 10 | 49 | 165 |
| 3 | 63 | 30 | 176 | 120 | 45 | 28 |
| 99 | 180 | 14 | 25 | 7 | 108 | 32 |
| 21 | 24 | 252 | 18 | 55 | 80 | 15 |
Geometric magic squares
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Magic squares may be constructed which contain geometric shapes instead of numbers. Such squares, known as geometric magic squares, were invented and named by Lee Sallows in 2001.
In the example shown the shapes appearing are two dimensional. It is clear that all magic squares are geometric, in that the numbers that appear in numerical magic squares can be interpreted as a shorthand notation which indicates the lengths of straight line segments that are the geometric 'shapes' occurring in the square. That is, numerical magic squares are that special case of a geometric magic square using one dimensional shapes.
Area magic squares
[edit]
In 2017, following initial ideas of William Walkington and Inder Taneja, the first linear area magic square (L-AMS) was constructed by Walter Trump.
Other magic shapes
[edit]
Other two dimensional shapes than squares can be considered. The general case is to consider a design with N parts to be magic if the N parts are labeled with the numbers 1 through N and a number of identical sub-designs give the same sum. Examples include magic circles, magic rectangles, magic triangles magic stars, magic hexagons, magic diamonds. Going up in dimension results in magic spheres, magic cylinders, magic cubes, magic parallelepiped, magic solids, and other magic hypercubes.
Possible magic shapes are constrained by the number of equal-sized, equal-sum subsets of the chosen set of labels. For example, if one proposes to form a magic shape labeling the parts with {1, 2, 3, 4}, the sub-designs will have to be labeled with {1,4} and {2,3}.
Related problems
[edit]
n-Queens problem
[edit]
In 1992, Demirörs, Rafraf, and Tanik published a method for converting some magic squares into n-queens solutions, and vice versa.
Magic squares in occultism
[edit]
Magic squares of order 3 through 9, assigned to the seven planets, and described as means to attract the influence of planets and their angels (or demons) during magical practices, can be found in several manuscripts all around Europe starting at least since the 15th century. Among the best known, the Liber de Angelis, a magical handbook written around 1440, is included in Cambridge Univ. Lib. MS Dd.xi.45. The text of the Liber de Angelis is very close to that of De septem quadraturis planetarum seu quadrati magici, another handbook of planetary image magic contained in the Codex 793 of the Biblioteka Jagiellońska (Ms BJ 793). The magical operations involve engraving the appropriate square on a plate made with the metal assigned to the corresponding planet, as well as performing a variety of rituals. For instance, the 3×3 square, that belongs to Saturn, has to be inscribed on a lead plate. It will, in particular, help women during a difficult childbirth.
In about 1510 Heinrich Cornelius Agrippa wrote De Occulta Philosophia, drawing on the Hermetic and magical works of Marsilio Ficino and Pico della Mirandola. In its 1531 edition, he expounded on the magical virtues of the seven magical squares of orders 3 to 9, each associated with one of the astrological planets, much in the same way as the older texts did. This book was very influential throughout Europe until the Counter-Reformation, and Agrippa's magic squares, sometimes called kameas, continue to be used within modern ceremonial magic in much the same way as he first prescribed.
The most common use for these kameas is to provide a pattern upon which to construct the sigils of spirits, angels or demons; the letters of the entity's name are converted into numbers, and lines are traced through the pattern that these successive numbers make on the kamea. In a magical context, the term magic square is also applied to a variety of word squares or number squares found in magical grimoires, including some that do not follow any obvious pattern, and even those with differing numbers of rows and columns. They are generally intended for use as talismans. For instance the following squares are: The Sator square, one of the most famous magic squares found in a number of grimoires including the Key of Solomon; a square "to overcome envy", from The Book of Power; and two squares from The Book of the Sacred Magic of Abramelin the Mage, the first to cause the illusion of a superb palace to appear, and the second to be worn on the head of a child during an angelic invocation:
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | | | | | --- --- | S | A | T | O | R | | A | R | E | P | O | | T | E | N | E | T | | O | P | E | R | A | | R | O | T | A | S | | | | | | | --- --- | | 6 | 66 | 848 | 938 | | 8 | 11 | 544 | 839 | | 1 | 11 | 383 | 839 | | 2 | 73 | 774 | 447 | | | | | | | | --- --- | H | E | S | E | B | | E | Q | A | L | | | S | | | E | G | | | B | | | | | | | | --- --- | | A | D | A | M | | D | A | R | A | | A | R | A | D | | M | A | D | A | |
Magic squares in popular culture
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In Goethe's Faust, the witch's spell used to make a youth elixir for Faust, the Hexen-Einmal-Eins [de], has been interpreted as a construction of a magic square.
American composer John Cage used magic squares as an early technique for introducing chance operations into musical composition, prior to his use of the I Ching in 1951.
The English composer Peter Maxwell Davies has used magic squares to structure many of his compositions. For example, his 1975 Ave Maris Stella uses the 9×9 magic square of Moon while his 1977 A Mirror of Whitening Light uses the 8×8 magic square of Mercury to create the entire set of notes and durations for the piece. His other works that employ magic squares include The Lighthouse (1979), Resurrection (1987), Strathclyde Concerto No. 3 for Horn and Trumpet (1989), as well as many of his symphonies. According to Davies' own account:
A magic square in a musical composition is not a block of numbers – it is a generating principle, to be learned and known intimately, perceived inwardly as a multi-dimensional projection into that vast (chaotic!) area of the internal ear – the space/time crucible – where music is conceived. ... Projected onto the page, a magic square is a dead, black conglomeration of digits; tune in, and one hears a powerful, orbiting dynamo of musical images, glowing with numen and lumen.
Magic squares, including Benjamin Franklin's, appear as clues to the mystery in Katherine Neville's novels The Eight and The Fire.
Magic squares play a role in Steve Martin's 2003 novel The Pleasure of My Company.
Dürer's magic square and his Melencolia I both also played large roles in Dan Brown's 2009 novel, The Lost Symbol.
In the 2011 Korean television drama Deep Rooted Tree, King Sejong is shown attempting to construct a 33×33 magic square using lunch boxes. He ultimately discovers the "pyramid method" and completes the magic square with the help of an army of court attendants. This inspires him to create a more just form of government ruled by reason and words rather than military might.
On October 9, 2014, the post office of Macao in the People's Republic of China issued a series of stamps based on magic squares. The figure below shows the stamps featuring the nine magic squares chosen to be in this collection.
The metallic artifact at the center of The X-Files episode "Biogenesis" is alleged by Chuck Burks to be a magic square.
Mathematician Matt Parker attempted to create a 3×3 magic square using square numbers in a YouTube video on the Numberphile channel. His failed attempt is known as the Parker square.
The first season Stargate Atlantis episode "Brotherhood" involves completing a magic square as part of a puzzle guarding a powerful Ancient artefact.
Magic Squares are also featured in the 2019 Spanish film Vivir dos veces.
See also
[edit]
Antimagic square
Arithmetic sequence
Associative magic square
Combinatorial design
Freudenthal magic square
Hexagonal tortoise problem
John R. Hendricks
Latin square
Magic circle
Magic cube
Magic cube classes
Magic polygon
Magic series
Square matrices
Most-perfect magic square
Nasik magic hypercube
Prime reciprocal magic square
Room square
Sigil (magic)
Sriramachakra
Sudoku
Unsolved problems in mathematics
Vedic square
Notes
[edit]
^ Miller, Jeff (September 3, 2016). "Earlier Known Uses of Some of the Words of Mathematics (M)".
^ Schwartzman, Steven (1994). The Words of Mathematics: An Etymological Dictionary of Mathematical Terms Used in English. MAA. p. 130.
^ Wolfram MathWorld: Magic Square Weisstein, Eric W.
^ The most famous Arabic book on magic, named "Shams Al-ma'arif (Arabic: كتاب شمس المعارف), for Ahmed bin Ali Al-boni, who died about 1225 (622 AH). Reprinted in Beirut in 1985
^ a b c d e Yoke, Ho Peng (2008). "Magic Squares in China". Encyclopaedia of the History of Science, Technology, and Medicine in Non-Western Cultures. Encyopaedia of the History of Science, Technology, and Medicine in Non-Western Cultures (2 ed.). Springer. pp. 1252–1259. doi:10.1007/978-1-4020-4425-0_9350. ISBN 978-1-4020-4559-2.
^ Andrews, William Symes (1917). Magic Squares and Cubes (2nd ed.). Open Court Publishing Company. p. 122.
^ a b c d e f g h i Cammann, Schuyler (April 1960). "The Evolution of Magic Squares in China" (PDF). Journal of the American Oriental Society. 80 (2): 116–124. doi:10.2307/595587. JSTOR 595587.
^ a b c d e Swetz, Frank J. (2008). The Legacy of the Luoshu (2nd ed.). A.K. Peters/CRC Press.
^ O'Connor, J.J.; Robertson, E.F. "Yang Hui". MacTutor History of Mathematics Archive. Retrieved 15 March 2018.
^ The Influence of Chinese Mathematical Arts on Seki Kowa by Shigeru Jochi, MA, School of Oriental and African Studies, University of London, 1993
^ Smith, David Eugene; Mikami, Yoshio (1914). A history of Japanese mathematics. Open Court Publishing Company. p. 69–75. Isomura Kittoku.
^ Smith, David Eugene; Mikami, Yoshio (1914). A history of Japanese mathematics. Open Court Publishing Company. p. 79–80. Isomura Kittoku.
^ Smith, David Eugene; Mikami, Yoshio (1914). A history of Japanese mathematics. Open Court Publishing Company. p. 116–122. Isomura Kittoku.
^ Smith, David Eugene; Mikami, Yoshio (1914). A history of Japanese mathematics. Open Court Publishing Company. p. 178. Isomura Kittoku.
^ Michiwaki, Yoshimasa (2008). "Magic Squares in Japanese Mathematics". Encyclopaedia of the History of Science, Technology, and Medicine in Non-Western Cultures. Encyopaedia of the History of Science, Technology, and Medicine in Non-Western Cultures (2 ed.). Springer. pp. 1252–1259. doi:10.1007/978-1-4020-4425-0_9154. ISBN 978-1-4020-4559-2.
^ a b Mikami, Yoshio (1917). Magic squares in Japanese mathematics (in Japanese). Tokyo: Imperial Academy of Science.
^ a b c d e f g Hayashi, Takao (2008). "Magic Squares in Indian Mathematics". Encyclopaedia of the History of Science, Technology, and Medicine in Non-Western Cultures (2 ed.). Springer. pp. 1252–1259. doi:10.1007/978-1-4020-4425-0_9778. ISBN 978-1-4020-4559-2.
^ a b c d e Datta, Bibhutibhusan; Singh, Awadhesh Narayan (1992). "Magic Squares in India" (PDF). Indian Journal of History of Science. 27 (1): 51–120. Archived from the original (PDF) on 2018-01-17. Retrieved 2018-01-16.
^ Hayashi, Takao (1987). "Varahamihira's Pandiagonal Magic Square of the Order Four" (PDF). Historia Mathematica. 14 (2): 159–166. doi:10.1016/0315-0860(87)90019-X.
^ J. P. Hogendijk, A. I. Sabra, The Enterprise of Science in Islam: New Perspectives, Published by MIT Press, 2003, ISBN 0-262-19482-1, p. xv.
^ Helaine Selin, Ubiratan D'Ambrosio, Mathematics Across Cultures: The History of Non-Western Mathematics, Published by Springer, 2001, ISBN 1-4020-0260-2, p. 160.
^ a b c d e f Sesiano, Jacques (November 2003). "Construction of magic squares using the knight's move in Islamic mathematics" (PDF). Archive for History of Exact Sciences. 58 (1): 1–20. doi:10.1007/s00407-003-0071-4. S2CID 123219466.
^ a b Sesiano, Jacques (1997). "Magic squares in Islamic mathematics". Encyclopaedia of the History of Science, Technology, and Medicine in Non-Western Cultures. pp. 1259–1260.
^ a b c d e f g Sesiano, Jacques (2007). Magic squares in the tenth century: Two Arabic treatises by Antaki and Buzjani. Springer.
^ Sesiano, J., Abūal-Wafā\rasp's treatise on magic squares (French), Z. Gesch. Arab.-Islam. Wiss. 12 (1998), 121–244.
^ a b c Cammann, Schuyler (February 1969). "Islamic and Indian Magic Squares, Part I". History of Religions. 8 (3): 181–209. doi:10.1086/462584. S2CID 162095408.
^ Sesiano, Jacques (2004). "Quelques methodes arabes de construction des carres magiques impairs (some Arabic construction methods of odd magical squares)". Bulletin de la Société Vaudoise des Sciences Naturelles (in French). 83 (1): 51–76.
^ Peter, J. Barta, The Seal-Ring of Proportion and the magic rings (2016), pp. 6–9.
^ a b Needham, Joseph (1987). Theoretical Influences of China on Arabic Alchemy. UC Biblioteca Geral 1.
^ Jābir ibn Hayyān, Book of the Scales. French translation in: Marcelin Berthelot (1827–1907), Histoire de sciences. La chimie au moyen ge, Tom. III: L'alchimie arabe. Paris, 1893. [rprt.. Osnabruck: O. Zeller, 1967], pp. 139–162, in particular: pp. 150–151
^ al-Ghazālī, Deliverance From Error (al-munqidh min al-ḍalāl ) ch. 145. Arabic: al-Munkidh min al-dalal. ed. J. Saliba – K. Ayyad. Damascus: Maktab al-Nashr al-'Arabi, 1934, p. 79. English tr.: Richard Joseph McCarthy, Freedom and Fulfillment: An annotated translation of al-Ghazali's al-Munkidh min al-Dalal and other relevant works of al-Ghazali. Boston, Twayer, 1980. He refers a book titled 'The Marvels of Special Properties' as his source. This square was named in the Orient as the Seal of Ghazali after him.
^ a b c d Comes, Rosa (2016). "The Transmission of Azarquiel's Magic Squares in Latin Europe". In Wallis, Faith; Wisnovsky, Robert (eds.). Medieval Textual Cultures: Agents of Transmission, Translation and Transformation. Judaism, Christianity, and Islam – Tension, Transmission, Transformation. Vol. 6. Walter de Gruyter GmbH & Co KG. pp. 159–198. ISBN 978-3-11-046730-7.
^ The Latin version is Liber de septem figuris septem planetarum figurarum Geberi regis Indorum. This treatise is the identified source of Dürer and Heinrich Cornelius Agrippa von Nettesheim. Cf. Peter, J. Barta, The Seal-Ring of Proportion and the magic rings (2016), pp. 8–9, n. 10
^ Sesiano, Jacques (2004). Les carrés magiques dans les pays islamiques (in French). PPUR presses polytechniques.
^ Schimmel, Annemarie (1993). The mystery of numbers. New York: Oxford University Press.
^ "The Magic Squares of Manuel Moschopoulos - Introduction | Mathematical Association of America". www.maa.org. Archived from the original on 2019-12-05. Retrieved 2018-01-18.
^ a b c d Cammann, Schuyler (May 1969). "Islamic and Indian Magic Squares, part II". History of Religions. 8 (4): 271–299. doi:10.1086/462589. JSTOR 1062018. S2CID 224806255.
^ presently in the Biblioteca Vaticana (cod. Reg. Lat. 1283a)
^ See Alfonso X el Sabio, Astromagia (Ms. Reg. lat. 1283a), a cura di A.D'Agostino, Napoli, Liguori, 1992
^ Mars magic square appears in figure 1 of "Saturn and Melancholy: Studies in the History of Natural Philosophy, Religion, and Art" by Raymond Klibansky, Erwin Panofsky and Fritz Saxl, Basic Books (1964)
^ The squares can be seen on folios 20 and 21 of MS. 2433, at the Biblioteca Universitaria of Bologna. They also appear on folio 69rv of Plimpton 167, a manuscript copy of the Trattato dell'Abbaco from the 15th century in the Library of Columbia University.
^ In a 1981 article ("Zur Frühgeschichte der magischen Quadrate in Westeuropa" i.e. "Prehistory of Magic Squares in Western Europe", Sudhoffs Archiv Kiel (1981) vol. 65, pp. 313–338) German scholar Menso Folkerts lists several manuscripts in which the "Trattato d'Abbaco" by Dagomari contains the two magic square. Folkerts quotes a 1923 article by Amedeo Agostini in the Bollettino dell'Unione Matematica Italiana: "A. Agostini in der Handschrift Bologna, Biblioteca Universitaria, Ms. 2433, f. 20v–21r; siehe Bollettino della Unione Matematica Italiana 2 (1923), 77f. Agostini bemerkte nicht, dass die Quadrate zur Abhandlung des Paolo dell'Abbaco gehören und auch in anderen Handschriften dieses Werks vorkommen, z. B. New York, Columbia University, Plimpton 167, f. 69rv; Paris, BN, ital. 946, f. 37v–38r; Florenz, Bibl. Naz., II. IX. 57, f. 86r, und Targioni 9, f. 77r; Florenz, Bibl. Riccard., Ms. 1169, f. 94–95."
^ This manuscript text (circa 1496–1508) is also at the Biblioteca Universitaria in Bologna. It can be seen in full at the address Archived 2012-03-01 at the Wayback Machine
^ Pacioli states: A lastronomia summamente hanno mostrato li supremi di quella commo Ptolomeo, al bumasar ali, al fragano, Geber et gli altri tutti La forza et virtu de numeri eserli necessaria (Masters of astronomy, such as Ptolemy, Albumasar, Alfraganus, Jabir and all the others, have shown that the force and the virtue of numbers are necessary to that science) and then goes on to describe the seven planetary squares, with no mention of magical applications.
^ a b c Muurinen, Ismo (2020). Fermat, magic squares and the idea of self-supporting blocks (PDF) (MSc). University of Helsinki.
^ Chabert, Jean-Luc, ed. (1999). "2.4 Arnauld's Borders Method". A History of Algorithms: From the Pebble to the Microchip. Springer. pp. 70–80. doi:10.1007/978-3-642-18192-4. ISBN 978-3-540-63369-3.
^ O'Connor, J.J.; Robertson, E.F. "Benjamin Franklin". MacTutor History of Mathematics Archive. Retrieved 15 December 2018.
^ a b c Rouse Ball, W.W. "Magic Squares". Mathematical Recreations and Essays (4 ed.). London: Mac Millan and Co., Limited. pp. 122–142.
^ Andrews, William Symes (1917). Magic Squares and Cubes (2nd ed.). Open Court Publishing Company. pp. 124–126.
^ "Virtual Home of Paul Muljadi". Archived from the original on 2005-11-09. Retrieved 2005-03-18.
^ "Magic cube with Dürer's square" Ali Skalli's magic squares and magic cubes
^ "The magic square on the Passion façade: keys to understanding it". 7 February 2018.
^ Letters: The Mathematical Intelligencer; 2003; 25; 4: pp. 6–7.
^ Skalli, Ali (14 October 2009). "Magic cube with Gaudi's square". Archived from the original on 15 December 2021.
^ Adler, Allan; Alejandre, Suzanne. "Why there are no 2x2 magic squares". mathforum.org. Archived from the original on 2018-03-02.
^ a b Loly, Peter (March 2004) [1 August 2016]. "The invariance of the moment of inertia of magic squares" (PDF). Mathematical Gazette. 88 (511): 151–153. CiteSeerX 10.1.1.552.7296. doi:10.1017/S002555720017456X. S2CID 125989925. Archived from the original (PDF) on 14 November 2017. Retrieved 5 June 2017.
^ Marcus, M.; Ree, R. (1959). "Diagonals of doubly stochastic matrices". The Quarterly Journal of Mathematics. 10 (1): 296–302. doi:10.1093/qmath/10.1.296.
^ How many magic squares are there? by Walter Trump, Nürnberg, January 11, 2001
^ a b Anything but square: from magic squares to Sudoku by Hardeep Aiden, Plus Magazine, March 1, 2006
^ Kitajima, Akimasa; Kikuchi, Macoto; Altmann, Eduardo G. (14 May 2015). "Numerous but Rare: An Exploration of Magic Squares". PLOS ONE. 10 (5) e0125062. Bibcode:2015PLoSO..1025062K. doi:10.1371/journal.pone.0125062. PMC 4431883. PMID 25973764.
^ a b c d e f Kraitchik, Maurice (1953). "Magic Squares". Mathematical Recreations (2nd ed.). New York: Dover Publications, Inc. pp. 142–192. ISBN 9780486201634.{{cite book}}: CS1 maint: ignored ISBN errors (link)
^ a b Sallows, Lee (Fall 1997) [9 January 2009]. "The lost theorem". The Mathematical Intelligencer. 19 (4): 51–54. doi:10.1007/BF03024415. S2CID 122385051.
^ a b White, S. Harry. "Associative Magic Squares". budshaw.ca.
^ a b Hawley, Del (2011). "Magic Squares II". nrich.maths.org. University of Cambridge.
^ Mathematical Circles Squared By Phillip E. Johnson, Howard Whitley Eves, p. 22
^ The Canterbury Puzzles and Other Curious Problems, Henry Ernest Dudeney, 1907
^ Bordered Square Numbers, S. Harry White, 2009
^ Some illustrated comments on 5×5 golden magic matrices and on 5×5 Stifelsche Quadrate, George P. H. Styan, 2014.
^ Hartley, M. "Making Big Magic Squares".
^ 2N Composite Squares, S. Harry White, 2009
^ Karl Fulves, Self-working Number Magic (Dover Magic Books)
^ Stifel, Michael (1544), Arithmetica integra (in Latin), pp. 29–30.
^ "8x8 multiplicative magic square of complex numbers" Ali Skalli's magic squares and magic cubes
^ "Multimagie.com – Additive-Multiplicative magic squares, 8th and 9th-order". Retrieved 26 August 2015.
^ "Multimagie.com – Smallest additive-multiplicative magic square". Retrieved 16 January 2024.
^ Magic squares are given a whole new dimension, The Observer, April 3, 2011
^ Les carrés magiques géométriques by Jean-Paul Delahaye, Pour La Science No. 428, June 2013
^ "Area Magic Squares". Futility Closet. 2017-01-19. Retrieved 2017-06-12.
^ a b Magic Designs, Robert B. Ely III, Journal of Recreational Mathematics volume 1 number 1, January 1968
^ "MathWorld News: There Are No Magic Knight's Tours on the Chessboard".
^ Mayhematics, "12×12 Magic Knight's Tours"
^ Demirörs, O.; Rafraf, N.; Tanik, M. M. "Obtaining n-queens solutions from magic squares and constructing magic squares from n-queens solutions". Journal of Recreational Mathematics. 24 (272–280): 1992.
^ See Juris Lidaka, The Book of Angels, Rings, Characters and Images of the Planets in Conjuring Spirits, C. Fangier ed. (Pennsylvania State University Press, 1994)
^ Benedek Láng, Demons in Krakow, and Image Magic in a Magical Handbook, in Christian Demonology and Popular Mythology, Gábor Klaniczay and Éva Pócs eds. (Central European University Press, 2006)
^ According to the correspondence principle, each of the seven planets is associated to a given metal: lead to Saturn, iron to Mars, gold to the Sun, etc.
^ Drury, Nevill (1992). Dictionary of Mysticism and the Esoteric Traditions. Bridport, Dorset: Prism Press. ISBN 978-1-85327-075-8.
^ "The Book of Power: Cabbalistic Secrets of Master Aptolcater, Mage of Adrianople", transl. 1724. In Shah, Idries (1957). The Secret Lore of Magic. London: Frederick Muller Ltd.
^ Holger Vietor: Das Hexen-Einmaleins – der Weg zur Entschlüsselung. In: Goethe-Jahrbuch 122. Wallstein Verlag, Göttingen 2005, ISBN 3-8353-2195-1, S. 325–327 (German).
^ Norbert Herrmann: Mathematik und Gott und die Welt. 3-te Auflage, Springer, Berlin/ Heidelberg 2018, ISBN 978-3-662-56387-8, S. 27–31 (German).
^ Kostelanetz 2003, 68.
^ Gareth E. Roberts (March 23, 2015). "Composing with Numbers: Sir Peter Maxwell Davies and Magic Squares" (PDF). Retrieved December 25, 2018.
^ a b Roberts, Gareth E. (2016). "8 Mathematical Modern Music". From Music to Mathematics: Exploring the Connections. JHU Press. ISBN 978-1-4214-1918-3.
^ Macau Post Office web site Archived 2014-11-11 at the Wayback Machine
^ Macau's magic square stamps just made philately even more nerdy The Guardian Science, November 3, 2014
^ Michelle Erica Green (June 15, 1997). "Biogenesis on The X-Files". littlereview.com. The Little Review. Retrieved March 25, 2017. Moreover, it's a magic square, a pattern in which God supposedly instructed the early Hebrews to gain power from names or their numeric equivalents.
^ Zack Handlen (November 17, 2012). "The X-Files: "Biogenesis" / Millennium: "Goodbye To All That"". The A.V. Club. The Onion, Inc. Retrieved March 25, 2017. I love when they bring the nerdy FBI guy in to explain the concept of "the magic square", which he does by telling us that magic squares have been around for a while, and then nothing else. Unless I missed something, all I have at this point is that magic squares are squares that people once thought were magic.
References
[edit]
Weisstein, Eric W. "Magic Square". MathWorld.
Magic Squares at Convergence
Wikisource has the text of the 1911 Encyclopædia Britannica article "Magic Square".
John Lee Fults, Magic Squares. (La Salle, Illinois: Open Court, 1974).
Cliff Pickover, The Zen of Magic Squares, Circles, and Stars (Princeton, New Jersey: Princeton University Press)
Leonhard Euler, On magic squares
Leonhard Euler, Investigations on new type of magic square
William H. Benson and Oswald Jacoby, "New Recreations with Magic Squares". (New York: Dover, 1976).
Kostelanetz, Richard. 2003. Conversing with Cage, 2nd edition. New York: Routledge. ISBN 0-415-93792-2.
Further reading
[edit]
Andrews, W.S. (1917). Magic Squares and Cubes (2nd ed.). Open Court Publishing. p. 428.
Block, Seymour (2009). Before Sudoku: The World of Magic Squares. Oxford University Press. ISBN 978-0-19-536790-4.
Schinz, Alfred (1996). The Magic Square: Cities in Ancient China. Edition Axel Menges. p. 428. ISBN 978-3-930698-02-8.
McCranie, Judson (November 1988). "Magic Squares of All Orders". Mathematics Teacher. 81 (8): 674–78. doi:10.5951/MT.81.8.0674.
Ollerenshaw, Kathleen; Bree, David (October 1998). Most perfect pandiagonal magic squares: their construction and enumeration. The Institute of Mathematics and its Applications. ISBN 978-0-905091-06-8.
Benjamin, Arthur T.; Brown, Ethan J. (November 2014). "Challenging Magic Squares for Magicians" (PDF). The College Mathematics Journal. 45 (2): 92–100. doi:10.4169/college.math.j.45.2.092. S2CID 125255312.
| v t e Magic polygons |
| Types | Magic circle Magic hexagon Magic hexagram Magic square Magic star Magic triangle |
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| Higher dimensional shapes | Magic cube + classes Magic hypercube Magic hyperbeam |
| Classification | Associative magic square Pandiagonal magic square Multimagic square |
| Related concepts | Latin square Word square Number Scrabble Eight queens puzzle Magic constant Magic graph Magic series |
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3053 | https://byjus.com/wavelength-to-frequency-formula/ | The wavelength of any sinusoidal wave is defined as the spatial period of the wave, that is, the distance over the shape of the wave repeats itself. The wavelength is denoted by a Greek letter lambda (λ) and is calculated in the units of length or metre.
Frequency is defined as the number of time a recurring event occurs in one second. For a sinusoidal wave, we define frequency as the number of cycles or crest or trough completed in one second. Frequency is denoted by f or ν and is calculated in the units of Hertz.
As we know, for a sinusoidal wave moving at a fixed speed, the wavelength of the wave is inversely proportional to its frequency. Calculate the wavelength to frequencyusing BYJU’S online calculator.
The Formula for Wavelength to Frequency
The wavelength to frequency formula is given by
Speed = Frequency x Wavelength
Wavelength = (Speed of the wave)/(Frequency of the wave)
As mentioned above, all the quantities are represented by a symbol. The symbolic representation of the formula given above can be seen as:
C = f × λ
In terms of wavelength to frequency, the formula is given as:
| |
| |
Where,
λ is the wavelength of the wave under consideration expressed in the units of a metre,
C is the speed of the wave in the given medium, expressed in terms of m/s
f is the frequency of the wave expressed in terms of Hertz.
Solved Examples
Example 1 In an experiment, the wavelength of a photon particle was observed to be 500 nm. What can be said about the frequency of the wave?
Solution: Given, the wavelength of the photon particle = 500 nm.
In order to calculate the frequency of the photon particle, we use the formula given above.
| |
| |
As we know, the speed of light is c = 3×108
Substituting the known values in the equation above, we get that,
f=6×10-4Hz
The frequency of the wave is equal to 6×10-4Hz.
Example 2 For a light ray having a wavelength equal to 200 nm, calculate the frequency.
Solution: Given, the wavelength of the light ray = 200 nm. In order to calculate the frequency of the light ray, we use the formula given above.
| |
| |
As we know, the speed of light is c = 3×108.
Substituting the known values in the equation above, we get that,
f=1.5×1014Hz
The frequency of the wave is equal to 1.5×1014Hz.
Stay tuned with BYJU’S to learn more about other Physics related articles.
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3054 | https://www.cs.fsu.edu/~myers/cgs4406/notes/overload.html | Operator Overloads
Operator overloading
Fundamentals:
There are many operators available that work on built-in types, like int and double.
Operator overloading means the process of creating new versions of these operators for use with user-defined types.
It is not as difficult as it sounds. Some points to note:
An operator in C++ is just a special kind of function. Overloading an operator simply involves writing a function.
C++ already does some operator overloading implicitly on built-in types. Consider the fact that the + operator already works for ints, floats, doubles, and chars. There is really a different version of the + operator for each type.
Operator overloading is done for the purpose of using familiar operator notation on new user-defined types.
Some rules and facts:
Overloading an operator cannot change its precedence.
Overloading an operator cannot change its associativity.
Overloading an operator cannot change its "arity" (i.e. number of operands)
It is not possible to create new operators -- only new versions of existing ones.
Operator meaning on the built-in types cannot be changed.
Friend:
Operator overloads can be written as member functions of a class or as friend functions. Remember that the keyword friend is used to give something outside a class (like another class or a function) access to all of the members of the class.
A simple arithmetic operator overload:
To see how operator overloading works, consider the following example.
We should have no trouble conceptually with the normal uses of the arithmetic operators:
int x = 3, y = 6, z;
float a = 3.4, b = 2.1, c;
z = x + y;
c = a / b;
However, consider the idea of doing arithmetic on our Fraction objects. How can we accomplish it? Would the following make sense to the compiler?
Fraction n1, n2, n3;
n3 = n1 + n2; // is this legal?
It should be clear that this would not make sense to the compiler. How would the computer automatically know about getting the common denominator, for instance? Fraction is a user-defined type -- the language designers didn't know we were going to build it!
So, how do we add Fraction objects? One way is to write a function, let's call it Sum, that takes two Fraction objects as parameters, and returns a Fraction as a result:
Fraction Sum(Fraction f1, Fraction f2);
However, the + notation would certainly be more convenient! We can write the same function (just like the one above) using an operator overload instead.
Format: Since an operator is a function, we declare and define one just like we would a function. A declaration of an operator overload requires a name (which is the operator symbol), a return type, and a parameter list. In addition, we need the keyword operator.
ReturnType operator OperatorName(ParameterList)
Example of the Fraction class operator overload declaration for +
Fraction operator+(Fraction f1, Fraction f2);
Notice that this function is not a member function of the Fraction class. Instead, two Fraction objects are passed in as parameters. If we want this function to have access to the private data, we can declare this function to be a friend of the Fraction class:
friend Fraction operator+(Fraction f1, Fraction f2);
Note that this is not something that should be done with all functions. It is a generally accepted practice for operator overloads, since this function is intended for use with this class and it allows faster performance.
Here is how this function might be defined. Notice that we must tell the computer how to DO the addition -- i.e. finding the common denominator and adding the adjusted numerators. (If we wrote the Sum function listed above, it would have a very similar definition.)
Fraction operator+(Fraction f1, Fraction f2)
{
Fraction r; // declare a Fraction to hold the result
// load result Fraction with sum of adjusted numerators
r.numerator = (f1.numeratorf2.denominator)
+ (f2.numeratorf1.denominator);
// load result with the common denominator
r.denominator = f1.denominator f2.denominator;
return r; // return the result Fraction
}
Once this operator overload is defined, then the following is legitimate:
Fraction n1, n2, n3;
n3 = n1 + n2; // now it is legal!
Click here to see the Fraction class with the + operator added in.
Operator overloads as member functions:
Some operator overloads can be defined as member functions or as outside (and possibly friend) functions. Others must be one or the other. If a binary operator is defined as an outside function, the function prototype will have two parameters. If a binary operator is defined as a member function, the prototype will list one parameter (the second operand) - the first operand will be the object itself.
Example of the Fraction class + overload as a member function.
Fraction Fraction::operator+(Fraction f2)
{
Fraction r; // result
r.numerator = (numerator f2.denominator)
+ (f2.numerator denominator);
r.denominator = (denominator f2.denominator);
return r;
}
Notice that there is no dot-operator on one of the sets of member data variables. These variables are part of the object we are in (as a member function). The second object was passed in as a parameter. In the following call, the + operator will run as a member function of object n1, and object n2 will be passed in as a parameter.
Fraction n1, n2, n3;
n3 = n1 + n2;
Click here to see the Fraction class with the + operator as a member function.
Other examples of operator overloads (prototypes only):
// multiplication overload for Fractions
friend Fraction operator(Fraction f1, Fraction f2);
// less-than comparison overload for Fractions
friend bool operator<(Fraction f1, Fraction f2);
// addition operator to add a Fraction and an integer
friend Fraction operator+(Fraction f, int n);
// same as above, but this one allows the int to come first in the call
friend Fraction operator+(int n, Fraction f);
Overloading the insertion << and extraction >> operators:
As with other operators, the << and >> operators are defined for the basic types. If you build your own class, don't expect << to automatically work with your new types of objects! If you want it to work, you have to teach the computer how to do such output. Consider the following: Fraction f;
cout << f; // how would the machine know how to do this?
We have no reason to expect the second line to work! The insertion operator << is only pre-defined for built-in types. The iostream.h library doesn't know about the Fraction type.
The << operator is a binary operator (2 parameters, left side and right side). The first parameter is always an ostream object (we've mostly used cout, so far). Because of this, it cannot be defined as a member function (it would have to be a member of the ostream class, which we cannot change). The << and >> operators should always be defined as outside functions (usually friend functions). The second parameter is whatever new type it is being overloaded to print:
friend ostream& operator << (ostream& s, Fraction f);
This declaration has all of the usual parts for defining a function. The name is operator<< (the keyword operator and the operator symbol). The return type is ostream&. The parameters are (ostream& s, Fraction f). When defining overloads of << and >> , always pass the stream parameters by reference. A better way to write this operator is:
friend ostream& operator << (ostream& s, const Fraction& f);
Notice that the first one passes the Fraction by value (and makes a copy). The second passes by reference (avoiding the overhead of a copy). It is declared as a const because the Fraction does not need to change if we are just doing output.
Here is the corresponding prototype for extraction >>
friend istream& operator >> (istream& s, Fraction& f);
Notice that the Fraction parameter for >> is also a reference parameter. This is because we are getting input into the object, so we need to work on the original, not a copy.
Remember the Show() function of the Fraction class?
void Fraction::Show()
{
cout << numerator << '/' << denominator;
}
Here is how the << operator would be defined for Fraction. Notice how similar it is to the Show() function.
ostream& operator << (ostream& s, const Fraction& f)
{
s << f.numerator << '/' << f.denominator;
return s;
}
Note the differences between this and the Show() function.
In this function, we must have a return statement, because we have a return type (as opposed to "void" in the original Show function). We return the ostream itself.
Note also that we must use "s" (not "cout") in the function body. This is the formal parameter, and a nickname for whatever was passed in (which could be cout, but also could be a different ostream).
Last, notice that this is not a member function of the Fraction class, but rather, a friend function. So, we can access the private data, but we must do it through the object:
Once this is defined, we can use a Fraction object in a cout statement:
Fraction f1;
So now, instead of:
cout << "Fraction f1 is ";
f1.Show();
cout << '\n';
We can write:
cout << "Fraction f1 is " << f1 << '\n';
Click here to see the Fraction class with the << overload used instead of Show().
Now, what would the definition of the >> overload look like for Fraction? Try it!
Phonebook database example with << and >> overloads in the Entry class. |
3055 | https://www.chemicalaid.com/tools/sigfigscalculator.php?expression=0.00210&hl=en | Significant Figures in 0.00210
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Significant Figures in 0.00210
0.00210
How many significant figures does 0.00210 have?
0.00210 has 3 significant figures and 5 decimals. 0.00210 rounded to 2 sig figs is 0.0021, and to 1 sig figs is 0.002. To count the number of sig figs in 0.00210, ignore the leading insignificant digits (0.00) to get the significant digits (210).
| Result | 0.00210 |
| Sig Figs | 3 (0.00210) |
| Decimals | 5 (0.00210) |
| Scientific Notation | 2.10 × 10-3 |
| E-Notation | 2.10e-3 |
| Words | zero point zero zero two one |
0.002 Rounded to Fewer Sig Figs
| 2 | 0.0021 | 2.1 × 10-3 |
| 1 | 0.002 | 2 × 10-3 |
+-×÷
+-×÷^ (Exponentiation)× 10 nlog 10(n)log e(n)exact(n)
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The following sig fig rules are used:
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To count trailing zeros, add a decimal point at the end (e.g. 1000.) or use scientific notation (e.g. 1.000 × 10^3 or 1.000e3).
Zeros have all their digits counted as significant (e.g. 0 = 1, 0.00 = 3).
Rounds when required, after parentheses, and on the final step.
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((4.790 × 10^-2) - (6.10 × 10^-5)) ÷ (3.045 × 10^2)
24.2 + 17.56
13.4 + 0.79
124 + 11.78
400 + 12.4
13.63 - 4.95 × 0.29
1.024e9
1300000.80000
1497142857
0.056 × 1.136
12.56 + 3.01
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3056 | https://blogs.sas.com/content/iml/2015/06/24/sensitivity-newtons-method.html | # Blogs
# Blogs
The sensitivity of Newton's method to an initial guess
2
By
Rick Wicklin on The DO Loop
In my article about finding an initial guess for root-finding algorithms, I stated that Newton's root-finding method
"might not converge or might converge to a root that is far away from the root that you wanted to find." A reader wanted more information about that statement.
I have previously shown how to implement Newton's method in SAS.
The behavior of Newton's method depends on the initial guess. If you provide a guess that is sufficiently close to a simple root, Newton's method will converge quadratically to the nearby root. However, if your guess is near a critical point of the function, Newton's method will produce a "next guess" that is far away from the initial guess. Further iterations might converge to an arbitrary root, might endlessly cycle in a periodic or aperiodic manner, or might diverge to infinity.
The dynamics of Newton iteration can be quite complex. If you owned a PC in the 80's and early 90's, you might have spent countless hours computing Mandelbrot sets and Julia sets. You might have seen pictures like the one at the beginning of this article, which show the domains of attraction for Newton's iteration for a cubic polynomial. In the picture, each point in the complex plane is colored according to which root Newton's method converges to when it begins at that point. The points that eventually converge to a root are the Fatou set, whereas the points that do not converge form the Julia set.
The sensitivity of Newton's method to an initial guess Click To Tweet
The sensitivity of Newton's method
You can perform the same kind of computer experiments for Newton's method applied to a real function. (You can
download the SAS/IML programs that I used to create the graphs in this article.) Consider the polynomial
f(x) = x5
– 2 x4
– 10 x3
+ 20 x2
+ 9 x
– 18.
The roots of the polynomial are {-3, -1, 1, 2, 3}. The polynomial has critical points (where the derivative vanishes) near -2.3, -0.2, 1.5, and 2.6. Recall that Newton's method involves iteration of the rational function N(x) = x – f(x)/f'(x), which has singularities at the critical points of f.
You can ask the following question: For each point, x, to which root does Newton's method converge when x is the initial guess? You can also keep track of how many iterations it takes for Newton's method to converge to a root.
If you apply Newton's method to 250 initial conditions on the interval [-4, 4], you get the results that are summarized in the needle plot to the left. (Click to enlarge.) The color of the needle at x indicates the root to which x converges under Newton's method. The height of the needle indicates the number of iterations required to converge.
You can see that initial guesses that are close to a root converge to the nearby root in five or fewer iterations. Near the critical points, Newton's method requires more iterations to converge, often more than 10 and sometimes more than 20 iterations. The regions near the critical points
are interlaced bands of color, which indicates that the dynamics of Newton's method is complicated in those regions. A small change in the initial guess can result in a big difference in the Newton iterations.
The dynamics near 0 seem interesting, so let's zoom into that region. In particular, repeat the previous computation for 250 initial conditions on the interval [-0.5, 0.5]. The needle plot to the left reveals additional details. All roots can be reached from initial guesses in this region, even though the nearest roots are at x = -1 and x = 1 (roots #2 and #3). Again, there are regions for which many iterations are required and there is an interlacing of colors, which indicates that newton's method is sensitive to the initial guess.
You can zoom again into a multicolored region and repeat the computations on a new interval. The behavior continues ad infinitum. You can find two initial guesses that differ by an arbitrarily small amount, yet their iteration under Newton's method eventually diverge and result in different roots. This is known as sensitive dependence on initial conditions, or, more poetically, as the butterfly effect.
Conclusions
Newton's method is one of my favorite root-finding techniques. However, it is important to understand that the famous quadratic convergence of Newton's method applies to initial guesses that are close to a root. For an arbitrary initial guess, Newton's method can be result in divergence, periodic orbits, or convergence to a far-away root.
Fractals and chaos are fun topics to explore, but not when you need to find a root as part of your work. Therefore I recommend using a pre-search method, as described in
my previous article, to obtain a good initial guess.
Further details
You can
download the SAS program that I used to generate the images in this article. The needle plots were created by using the NEEDLE statement in PROC SGPLOT.
The image at the top of this article is from the Wikipedia article on Newton fractals, which are
the fractals formed by Newton iterations of a complex function.
For a mathematical treatment of some of the complexities of Newton's method, see
Walsh (1995), The College Mathematics Journal, 26(1).
Tags Math Numerical Analysis
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About Author
Rick Wicklin
Distinguished Researcher in Computational Statistics
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Rick Wicklin, PhD, is a distinguished researcher in computational statistics at SAS and is a principal developer of SAS/IML software. His areas of expertise include computational statistics, simulation, statistical graphics, and modern methods in statistical data analysis. Rick is author of the books Statistical Programming with SAS/IML Software and Simulating Data with SAS.
2 Comments
Anders Sköllermo on
I used Newton's method several times earlier (40 years ago). Actually it was a Fortran program from IMSL (and also from NAG) that used a combination of bisection and Newtons's method,
to make sure that a solution should be found. They worked Very well.
There are however problems that are more difficult to solve. E.g. how to find the biggest solution OR how to find all solutions.
Consider (x-3)(x-2)(x-1). Obviously the zeroes are 1, 2 and 3.
But (xx+1)(x-3) only has one zero.
/ Br Anders
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3057 | https://www-math.umd.edu/research/colloquium/88-math/undergraduate/311-textbook-list.html | Department of Mathematics - Textbook List
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Textbook List
The textbooks below are always or sometimes used in the courses indicated. In some cases, different texts may be used in different sections of the same course. Textbooks in all courses numbered 300 and above are chosen by the individual instructor. Check the course listings inTestudoto determine texts offered in specific sections.Please note that for courses numbered 246 and below, all sections use the same textbooks. Testudo has been known to incorrectly say "No books listed yet" for some sections; if you see this, please check another section of the same course. Also, some textbooks listed in Testudo as "Recommended" are in fact alternative forms of the required texts.
Some textbooks may be available only from the University Book Center, but most can be bought less expensively from online retailers. You are encouraged to explore this option, with the following caveats. You must order books with plenty of time to spare, because shipping or inventory delays may occur, especially near the start of the semester. Also, you must make sure you are buying the correct edition of a book. Delays or other problems with online orders are not grounds for being excused from class assignments.
MATH 007 (Formerly Math 010/011)
Course Name:Algebra for MATH 110
Text(s):
Intermediate Algebra Review, A Small Group Approach, by Debra Gray Franklin. Custom Published by Thomson, 2002. ISBN: 0759336245 (Current)(Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Finite Mathematics for the Managerial, Life, and Social Sciences by Soo. T. Tan. Published by Cengage Learning. ISBN: 9781285464657 (11th Edition) (Sp17) (F17) (Sp18) (F18) (Sp19)
Finite Mathematics for the Managerial, Life, and Social Sciences, by Soo T. Tan. Published by Cengage Learning. ISBN: 9781337405782 (12th Edition) (Current)(F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
MATH 010 (No Longer being taught - combine with Math 007 Spring 2017)
Course Name:Algebra for MATH 110
Text(s):
Intermediate Algebra Review, a Small Group Approach, by Debra Gray Franklin. Custom Published by Thomson, 2002.ISBN: 0759336245 (F06)(Sp07)(F07)(Sp08) (F08)(Sp09)(F09) (Sp10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16)
Finite Mathematics, by Waner and Costenoble. Published by Cengage Learning. ISBN: 9781305016248 (New Custom Edition (Bundel Set) - Fall 2013) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16)
Finite Mathematics, by Waner and Costenoble. Published by Thomson Learning, 2004.ISBN: 111629447 (New Custom Edition - Fall 2010) (F06)(Sp07)(F07)(Sp08) (F08)(Sp09) (F09) (Sp10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp13)
MATH 011 (No longer being taught - combine with Math 007 Spring 2017)
Course Name:Algebra for MATH 111
Text(s):
Intermediate Algebra Review, a Small Group Approach, by Debra Gray Franklin. Custom Published by Thomson, 2002.ISBN: 0759336245 (F06) (Sp07)(F07)(Sp08) (F08)(Sp09) (F09)(Sp 10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp12) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16)
Introduction to Probability by S.T. Tan. Custom Published by Cengage Learning. ISBN: 9781305014879 (New Custom Edition - Fall 2014) (F14) (Sp15) (F15) (Sp16) (F16)
Introduction to Probability, by S.T. Tan, Custom Published by Cengage Learning. ISBN: 9781285877914 (New Custom Edition - Fall 2013) (F13) (Sp14)
Introduction to Probability, by S.T. Tan, Custom Published by Cengage Learning.ISBN: 1133397514 (New Custom Edition Fall 2011) (F06)(Sp07) (F07)(Sp08) (F08)(Sp09)(F09)(Sp10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp13)
MATH 013
Course Name:Algebra for MATH 113
Text(s):
Intermediate Algebra Review, a Small Group Approach by Debra Gray Franklin. Custom Published by Thomson, 2002.ISBN: 0759336245 (Current)(F06) (Sp07)(F07) (Sp08) (F08) (Sp09)(F09) (Sp10)(F10)(Sp11) (F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
OpenStax, Algebra and Trigonometry (Free Downloadable Desk Copy on online) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Algebra and Trigonometry with MYMATHLAB, by Robert Blitzer. Published by Prentice Hall, ISBN: 9780134758848 6TH New VALUEPACK (Sp18) (F18) (Sp19)
Algebra and Trigonometry with MYMATHLAB, by Robert Blitzer. Published by Prentice Hall, 2016. ISBN: 9781323305584 5TH VALUEPACK (Sp16) (F16) (Sp17) (F17)
College Algebra with Trigonometry, by Robert Blitzer. Published by Prentice Hall. ISBN: 9781269326759 (New Custom Edition (Bundle Set) - Fall 2013) (F13) (Sp14) (F14) (Sp15) (F15)
College Algebra with Trigonometry, by Robert Blitzer. Published by Prentice Hall. ISBN: 1256710784 (New Custom 2nd Edition)(F12) (Sp13)
College Algebra by Robert Blitzer. Published by Prentice Hall.ISBN:05558470114 (New Custom Bundle Set) (F06)(Sp07)(F07)(Sp08) (F08)(Sp09)(F09)(Sp10)(F10)(Sp11)(F11)(Sp12)
Supplement: Resource Manual, MATH 113, College Algebra with Applications, 3rd Edition, by J. Stone. Published by Kendall Hunt. ISBN: 9780757541582 (F06) (Sp07) (F07) (Sp08) (F08) (Sp09) (F09) (Sp10) (F10) (Sp11)
MATH 015
Course Name:Algebra for MATH 115
Text(s):
Intermediate Algebra Review, a Small Group Approach, by Debra Gray Franklin. Custom Published by Thomson, 2002.ISBN: 0759336245 (Current)(F06)(Sp07)(F07)(Sp08) (F08)(Sp09)(F09)(Sp10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Pre-Calculus: Mathematics for Calculus, by J. Stewart, L. Redlin and S. Watson. Published by Brooks/Cole.ISBN: 9780357758830(New Custom 8th Edition Fall 2023)(Current)(F23) (Sp24) (F24) (Sp25)(F25)
Pre-Calculus: Mathematics for Calculus, by J. Stewart, L. Redlin and S. Watson. Published by Brooks/Cole.ISBN: 9781305701618 (New Custom 7th Edition Fall 2015) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21)(F21) (Sp22) (F22) (Sp23)
Pre-Calculus: Mathematics for Calculus, by J. Stewart, L. Redlin and S. Watson. Published by Brooks/Cole.ISBN: 9781133393849 (New Custom Bundle Set Fall 2011) (F06) (Sp07) (F07)(Sp08) (F08)(Sp09)(F09)(Sp10)(F10) (Sp11)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15)
MATH 107 (Formerly Math 110/111)
Course Name: Introduction to Math Modeling and Probability
Text(s):
Finite Mathematics for the Managerial, Life, and Social Sciences by Soo. T. Tan. Published by Cengage Learning. ISBN: 9781337405782 (New 12th Edition) (Current)(F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Finite Mathematics for the Managerial, Life, and Social Sciences by Soo. T. Tan. Published by Cengage Learning. ISBN: 9781285464657 (11th Edition) (Sp17) (F17) (Sp18) (F18) (Sp19)
MATH 110 (No longer being taught - combine with Math 107 Spring 2017)
Course Name:Elementary Mathematical Models
Text(s):
Finite Mathematics, by Waner and Costenoble. Published by Thomson Learning, 2004.ISBN: 111629447(New Custom Edition - Fall 2010) (F06)(Sp07) (F07)(Sp08)(Sum08)(F08)(Sp09) (Sum09)(F09)(Sp10)(F10)(Sp11) (F11)(Sp12) (F12) (Sp13)
Finite Mathematics, by Waner and Costenoble. Published by Cengage Learning. ISBN: 9781305016248. (New Custom Editon (Bundle Set) - Fall 2013) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16)
MATH 111 (No longer being taught - combine with Math 107 Spring 2017)
Course Name:Introduction to Probability
Text(s):
Introduction to Probability by S.T. Tan. Custom Published by Cengage Learning. ISBN: 9781305014879 (New Custom Edition - Fall 2014) (F14) (Sp15) (F15) (Sp16) (F16)
Introduction to Probability, by S.T. Tan, Custom Published by Cengage Learning. ISBN: 9781285877914 (New Custom Edition - Fall 2013) (F13) (Sp14) (F14)
Finite Mathematics Student Solution Manual, by S.T. Tan. Published by Cengage Learning.ISBN: 1133363075(New Custom Edition 2011 (F09)(Sp10)(F10) (Sp11)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16)
Introduction to Probability, by S.T. Tan. Published by Cengag Learning. ISBN: 9781285325231 (New Custom Edition Fall 2012) (F12) (Sp13)
Introduction to Probability, by S.T. Tan. Custom Published by Cengage Learning. ISBN: 0495214442(F06)(Sp07) (F07)(Sp08) (Sum08) (Sp09) (Sum09)
Introduction to Probability, by S.T. Tan. Custom Published by Cengage Learning.ISBN: 1133397514 (New Custom Edition Fall 2011) (F09) (Sp10) (F10) (Sp11)(F11)(Sp12)
MATH 112 (No longer being taught - combine with Math 113 Fall 2013)
Course Name:College Algebra with Applications and Trigonometry
Text(s):
College Algebra with Trigonometry, by Robert Blitzer. Published by Prentice Hall. ISBN: 1256710784 (New Custom 2nd Edition) (F12) (Sp13)
Algebra & Trigonometry,(New Custom Bundle Set Edition)by Robert Blitzer. Published by Prentice Hall.ISBN: 0558216447 (F06) (F07) (F08)(Sp09)(F09) (Sp10) (F10)(Sp11)(F11)(Sp12)
Supplement: Resource Manual, MATH 113, College Algebra with Applications, 3rd Edition, by J. Stone. Published by Kendall Hunt.ISBN: 9780757541582 (F06) (F07)(F08)(Sp09)(F09)(Sp10)(F10)(Sp11)
MATH 113
Course Name:College Algebra with Applications
Text(s):
OpenStax, Algebra and Trigonometry (Free Downloadable Desk Copy online) (Current)(F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Algebra and Trigonometry with MYMATHLAB, by Robert Blitzer. Published by Prentice Hall, ISBN: 9780134758848 6TH New VALUEPACK (Sp18) (F18) (Sp19)
Algebra and Trigonometry with MYMATHLAB, by Robert Blitzer. Published by Prentice Hall, 2016. ISBN: 9781323305584 5TH VALUEPACK (Sp16) (F16) (Sp17) (F17)
College Algebra with Trigonometry, by Robert Blitzer. Published by Prentice Hall. ISBN: 9781269326759 (New Custom Edition (Bundle Set) - Fall 2013) (F13) (Sp14) (F14) (Sp15) (F15)
College Algebra with Trigonometry, by Robert Blitzer. Published by Prentice Hall. ISBN: 1256720784 (New Custom 2nd Edition) (F12) (Sp13)
College Algebra, 4th Edition, by Robert Blitzer. Published by Prentice Hall. ISBN: 978-0536502094(Bundle Set)(F06)(Sp07)(F07)(Sp08) (Sum08) (F08)(Sp09)(Sum09)
College Algebra,by Robert Blitzer. Published by Prentice Hall.ISBN: 0558470114 (New Custom Bundle Set Edition) (F09)(Sp10) (F10)(Sp11)(F11)(Sp12)
Supplement: Resource Manual, MATH 113, College Algebra with Applications, 3rd Edition, by J. Stone. Published by Kendall Hunt. ISBN: 9780757541582(F06) (Sp07)(F07)(Sp08) (Sum08)(F08)(Sp09) (Sum09)(F09)(Sp10) (F10) (Sp11)
MATH 115
Course Name:Pre-calculus
Text(s):
Pre-Calculus: Mathematics for Calculus, by J. Stewart, L. Redlin and S. Watson. Published by Brooks/Cole.ISBN: 9780357758830(New Custom 8th Edition Fall 2023)(Current) (F23) (Sp24) (F24) (Sp25)(F25)
Pre-Calculus: Mathematics for Calculus, by J. Stewart, L. Redlin and S. Watson. Published by Brooks/Cole.ISBN: 9781305701618 (New Custom 7th Edition Fall 2015) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23)
Pre-Calculus: Mathematics for Calculus,by J. Stewart, L. Redlin and S. Watson. Published by Brooks/Cole.ISBN: 9781133393849 (New Custom Edition (Bundle Set) Fall 2011) (F06)(Sp07)(F07)(Sp08)(Sum08)(F08) (Sp09)(Sum09)(F09)(Sp10) (F10)(Sp11)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15)
MATH 120 (Formerly Math 220)
Course Name: Elementary Calculus I
Text(s):
Calculus and its Applications, Books a la Carte with Custom MyMathLab, by Goldstein, D. Lay and D. Schneider.Published by Prentice Hall.ISBN: 9780137590612(New 15th Edition Bundle Set - Spring 2024)(Current) (Sp24) (F24) (Sp25)(F25)
Calculus and its Applications, Books a la Carte with Custom MyMathLab, by Goldstein, D. Lay and D. Schneider. Published by Prentice Hall.ISBN: 9780134768687 (New 14th Edition Bundle Set - Fall 2017) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23)
120 Coursepack by Washington.Published by Ming Press. ISBN: 2818440181672(Current) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25) (F25)
Calculus and its Applications, Books a la Carte with Custom MyMathLab, by Goldstein, D. Lay and D. Schneider. Published by Prentice Hall.ISBN: 9781323322741 (New 13th Edition Bundle Set - Fall 2015) (Sp16) (F16)
(Sp17)
MATH 121 (Formerly Math 221)
Course Name: Elementary Calculus II
Text(s):
Calculus and its Applications, Books a la Carte with Custom MyMathLab, by Goldstein, D. Lay and D. Schneider.Published by Prentice Hall.ISBN: 9780137590612(New 15th Edition Bundle Set - Spring 2024)(Current)(Sp25)(F25)
Calculus and its Applications, Books a la Carte with Custom MyMathLab, by Goldstein, D. Lay and D. Schneider. Published by Prentice Hall.ISBN: 9780134768687 (New 14th Edition Bundle Set - Fall 2017) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)
Calculus and its Applications, Books a la Carte with Custom MyMathLab, by Goldstein, D. Lay and D. Schneider. Published by Prentice Hall.ISBN: 9781323322741 (New 13th Edition Bundle Set - Fall 2015) (Sp16) (F16)
(Sp17) (F17)
MATH 130 (Math 130 being renamed to Math 135 Fall 2018)
Course Name:Calculus for Life Science I(NEW COURSE THAT STARTED SPRING 2008)
Text(s):
Biocalculus: Calculus, Probability, and Statistics for the Life Sciences, by J. Stewart, T. Day. Published by Cengage Learning. New 1st Edition - Fall 2015, ISBN: 9781305607866 (F15) (Sp16) (F16) (Sp17) (F17) (Sp18)
Calculus with Applications for the Life Sciences, by R. Greenwell, N. Ritchey, M. Lial. Published by Addison Wesley. New 2nd Edition, ISBN: 9780321964038 (F14) (Sp15)
Calculus with Applications for the Life Sciences, by R. Greenwell, N. Ritchey, M. Lial. Published by Addison Wesley.ISBN: 9780201745825(Sp09)(F09)(Sp10) (F10) (Sp11)(F11) (Sp12) (F12) (Sp13) (F13) (Sp14)
Student Solutions Manual for Calculus for the Life Sciences -(OPTIONAL)by R. Greenwell, N. Ritchey, M. Lial. Published by Addison Wesley.ISBN: 9780201770162 (Sp09)(F09) (Sp10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14)
Modeling the Dynamics of Life, by Frederick R. Adler. Published by Cengage Learning. ISBN: 0534404863(Sp08) (F08)
MATH 131 (Math 131 being renamed to Math 135 Fall 2018)
Course Name:Calculus for Life Science II(NEW COURSE THAT STARTED FALL 2008)
Text(s):
Biocalculus: Calculus, Probability, and Statistics for the Life Sciences, by J. Stewart, T. Day. Published by Cengage Learning. New 1st Edition - Fall 2015, ISBN: 9781305607866 (Sp16) (F16) (Sp17) (F17) (Sp18)
Calculus with Applications for the Life Sciences, by R. Greenwell, N. Ritchey, M. Lial. Published by Addison Wesley. New 2nd Edition, ISBN: 9780321964038 (Sp15) (F15)
Calculus with Applications for the Life Sciences, by R. Greenwell, N. Ritchey, M. Lial. Published by Addison Wesley.ISBN: 9780201745825(Sp09)(F09) p10)(F10)(Sp11)(F11) (Sp12) (F12) (Sp13) (F13) (Sp14) (F14)
Solutions Manual for Calculus for the Life Sciences -(OPTIONAL)by R. Greenwell, N. Ritchey, M. Lial. Published by Addison Wesley.ISBN: 0201770172 (Sp09)(F09) (Sp10)(F10)(Sp11) (F11) (Sp12) (F12) (Sp13) (F13) (Sp14) (F14)
Modeling the Dynamics of Life, by Frederick R. Adler. Published by Cengage Learning. ISBN: 0534404863(F08) (Sp09)
MATH 135 (NEW COURSE THAT STARTS THIS FALL 2018)
Course Name: Discrete Mathematics for Life Sciences
Text(s):
Mathematics For The Life Sciences, by Erin N. Bodine. Published by Princeton University Press. ISBN: 9780691150727 (Current) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
MATH 136 (NEW COURSE THAT STARTS THIS SPRING 2019)
Course Name: Discrete Mathematics for Life Sciences
Text(s):
Mathematics For The Life Sciences, by Erin N. Bodine. Published by Princeton University Press. ISBN: 9780691150727 (Current) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
MATH 140
Course Name:Calculus I
Text(s):
Calculus, by R. Ellis and D. Gulick. Published by Cengage Learning.ISBN: 9781133436751 (New Custom Edition Fall 2011) (Current)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Calculus Student Solutions Manual, by R. Ellis and D. Gulick. Published by Cengage Learning.ISBN:9780759331778(OPTIONAL) (Current)
(F06)(Sp07)(F07)(Sp08) (Sum08) (F08)(Sp09)(Sum09) (F09) (Sp10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Calculus, 6th Edition, by R. Ellis and D. Gulick. Published by Thomson Learning. ISBN: 978-0-7-5931379-8(F06)(Sp07)(F07)(Sp08) (Sum08) (F08)(Sp09)(Sum09) (F09) (Sp10)(F10)(Sp11)
MATH 140H
Course Name:Calculus I
Text(s):
Calculus, by R. Ellis and D. Gulick. Published by Cengage Learning.ISBN: 9781133436751 (New Custom Edition Fall 2011) (Current)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Calculus Student Solutions Manual, by R. Ellis and D. Gulick. Published by Cengage Learning.ISBN:9780759331778(OPTIONAL)
(F06)(Sp07)(F07)(Sp08) (Sum08) (F08)(Sp09)(Sum09) (F09) (Sp10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Calculus, 6th Edition, by R. Ellis and D. Gulick. Published by Thomson Learning. ISBN: 978-0-7-5931379-8(F06)(Sp07)(F07)(Sp08) (F08) (Sp09)(F09)(Sp10)(F10) (Sp11)
MATH 141
Course Name:Calculus II
Text(s):
Calculus, by R. Ellis and D. Gulick. Published by Cengage Learning.ISBN: 9781133436751 (New Custom Edition Fall 2011) (Current) (Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Calculus Student Solutions Manual, by R. Ellis and D. Gulick. Published by Cengage Learning.ISBN:9780759331778(OPTIONAL) (Current)
(F06)(Sp07)(F07)(Sp08) (Sum08) (F08)(Sp09)(Sum09) (F09) (Sp10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Calculus, 6th Edition, by R. Ellis and D. Gulick. Published by Thomson Learning. ISBN: 978-0-7-5931379-8(F06)(Sp07)(F07)(Sp08) (Sum08)
(F08)(Sp09)(Sum09) (F09)(Sp10)(F10)(Sp11) (F11)
MATH 141H
Course Name:Calculus II
Text(s):
Calculus, by R. Ellis and D. Gulick. Published by Cengage Learning.ISBN: 9781133436751 (New Custom Edition Fall 2011) (Current) (Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Calculus Student Solutions Manual, by R. Ellis and D. Gulick. Published by Cengage Learning.ISBN:9780759331778(OPTIONAL) (Current)
(F06)(Sp07)(F07)(Sp08) (Sum08) (F08)(Sp09)(Sum09) (F09) (Sp10)(F10)(Sp11) (F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Calculus, 6th Edition, by R. Ellis and D. Gulick. Published by Thomson Learning. ISBN: 978-0-7-5931379-8(F06)(Sp07)(F07)(Sp08) (F08) (Sp09)(F09)(Sp10) (F10)(Sp11) (F11)
MATH 212
Course Name:Elements of Numbers and Operations
Text(s):
Mathematics for Elemetarty Teachers/w Activities, by Beckman. Published by Prentice Hall. ISBN: 9780136937654 (New 6th Edition) Current(Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Mathematics for Elemetarty Teachers/w Activities, by Beckman. Published by Prentice Hall. ISBN: 9780134392790 (New 5th Edition Bundle Set) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21)
Mathematics for Elemetarty Teachers/w Activities, by Beckman. Published by Prentice Hall. ISBN: 9780321825728 (New 4th Edition Bundle Set) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17)
Mathematics for Elementary School Teachers,(New Custom 3rd Edition)by Beckmann. Published by Addison Wesley.ISBN: 0321654277 (New Custom 3rd Edition Bundle Set) (F10) (Sp11)(F11)(Sp12) (F12) (Sp13)
Mathematics for Elementary School Teachers, (Bundle Set) by Beckmann, Published by Addison Wesley. ISBN: 0321447174 (F06)(Sp07) (F07)(Sp08) (Sum08)(F08)(Sp09)(Sum09) (F09)(Sp10)
MATH 213
Course Name:Elements of Geometry and Measurement
Text(s):
Mathematics for Elemetarty Teachers/w Activities, by Beckman. Published by Prentice Hall. ISBN: 9780136937654 (New 6th Edition) (Current)(F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Mathematics for Elemetarty Teachers/w Activities, by Beckman. Published by Prentice Hall. ISBN: 9780134392790 (New 5th Edition Bundle Set) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22)
Mathematics for Elemetarty Teachers/w Activities, by Beckman. Published by Prentice Hall. ISBN: 9780321825728 (New 4th Edition Bundle Set) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17)
Mathematics for Elementary School Teachers, by Beckmann. Published by Addison Wesley.ISBN: 0321646967 (New Custom 3rd Edition) (Current) (Sp11)(F11)(Sp12) (F12) (Sp13) (F13)
Mathematics for Elementary School Teachers, (Bundle Set) by Beckmann, Published by Addison Wesley. ISBN: 0321447174(F06) (Sp07)(F07) (Sp08) (Sum08)(F08)(Sp09)(Sum09) (F09)(Sp10)(F10)
MATH 214
Course Name:Elements of Probability and Statistics
Text(s):
Intro Stats, by Deveaux, Velleman & Bock.Published by Prentice Hall. ISBN: 9780321825278(New 4th Edition) (Current) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Intro Stats, by Deveaux, Velleman & Bock. (3rd Edition) Published by Addison Wesley.ISBN: 9780321500458 (New Custom Ed.) (F08) (Sp09)(F09)(Sp10)(F10)(Sp11) (F11)(Sp12) (F12) (Sp13) (F13)
Mathematics for Elementary Teachers Activities Manual, by Beckmann. Published by Addison Wesley. (New Custom 3rd Edition) ISBN: 0321449762(Sp11) (F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14)
Mathematics for Elementary Teachers Activities Manual, (New Custom Edition) by Beckmann. Published by Addison Wesley. ISBN: 0321449762 (F06)(Sp07)(F07)(Sp08) (F08)(Sp09)(F09)(Sp10)(F10)
Statistical Reasoning for Everyday Life, 2nd Ed., by Bennett, Briggs, Triola. Published by Addison Wesley. ISBN: 0201771284(F06) (Sp07)(F07)(Sp08)
Student Solutions Manual for Statistical Reasoning, etc. ISBN: 020183846X (F06)(Sp07) (F07) (Sp08)
MATH 220 (Math 220 being renamed to Math 120 Spring 2017)
Course Name:Elementary Calculus I
Text(s):
Calculus and its Applications, Books a la Carte with Custom MyMathLab, by Goldstein, D. Lay and D. Schneider. Published by Prentice Hall.ISBN: 9781323322741 (New 13th Edition Bundle Set - Fall 2015) (Sp16) (F16)
Calculus and its Applications w/Access (LL), by Goldstein, D. Lay and D. Schneider. Published by Prentice Hall.ISBN: 9780321921796 (New 13th Edition Bundle Set - Fall 2015) (F15)
Calculus and its Applications, by Goldstein, D. Lay and D. Schneider. Published by Prentice Hall.ISBN: 9780321936141 (New 13th Edition Bundle Set) (Sp14) (F14) (Sp15)
Calculus and its Applications,(New 12th Edition Bundle Set)by Goldstein, D. Lay and D. Schneider. Published by Prentice Hall.ISBN: 0321643658 (F09) (Sp10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp13) (F13)
Calculus and its Applications, 11th Edition, by Goldstein, D. Lay, and D. Schneider. Published by Prentice Hall, 2004. ISBN: 0131746251 (Current)(F06)(Sp07)(F07)(Sp08) (Sum08)(F08) (Sp09) (Sum09)
Study Guide with Selected Applications and Visual Calculus, 11th Edition, by D. Lay and D. Schneider. Published by Prentice-Hall, 2004. (bundle with textbook) (Current)(F06)(Sp07)(F07) (Sp08) (Sum08)(F08)(Sp09) (Sum09)
MATH 221 (Math 221 being renamed to Math 121 Spring 2017)
Course Name:Elementary Calculus II
Text(s):
Calculus and its Applications, Books a la Carte with Custom MyMathLab, by Goldstein, D. Lay and D. Schneider. Published by Prentice Hall.ISBN: 9781323322741 (New 13th Edition Bundle Set - Fall 2015) (Sp16) (F16)
Calculus and its Applications, by Goldstein, D. Lay and D. Schneider. Published by Prentice Hall. ISBN: 9780321936141 (New 13th Edition Bundle Set)(Sp14) (F14) (Sp15) (F15)
Calculus and its Applications,(New 12th Edition Bundle Set)by Goldstein, D. Lay and D. Schneider. Published by Prentice Hall. ISBN: 0321643658 (Sp10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp13) (F13)
Calculus and its Applications, 11th Edition, by Goldstein, D. Lay, and D. Schneider. Published by Prentice-Hall, 2004. ISBN: 0131746251 (Sp07)(F07) (Sp08) (Sum08)(F08)(Sp09)(Sum09) (F09)
Study Guide with Selected Applications and Visual Calculus, 11th Edition, by D. Lay and D. Schneider. Published by Prentice-Hall, 2004. (bundle with textbook) (Sp07) (F07) (Sp08) (Sum08) (F08) (Sp09) (Sum09) (F09)
MATH 240
Course Name:Introduction to Linear Algebra
Text(s):
Linear Algebra and its Applications/w Study Guide, by David Lay. Published by Addison Wesley.ISBN: 9780135851258(New 6th Edition Bundle Set - Fall 2020) Current) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Linear Algebra and its Applications/w Study Guide, by David Lay. Published by Addison Wesley.ISBN: 9780134279190 (New 5th Edition Bundle Set - Fall 2015) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21)
Linear Algebra and its Applications/w Study Guide, by David Lay. Published by Addison Wesley.ISBN: 9780134279190 (New 4th Edition Bundle Set - Fall 2014) Current) (F14) (Sp15) (F15) (Sp16)
Linear Algebra and its Applications, 3rd Edition, by David Lay. Published by Addison Wesley. ISBN: 0321287134 (F06) (Sp07) (F07) (Sp08) (Sum08) (F08) (Sp09) (Sum09) (F09) (Sp10) (F10) (Sp11)
Study Guide - Linear Algebra and Its Applications, 3rd Edition by David Lay. Published by Addison Wesley. ISBN: 0321280660 (F06) (Sp07)(F07)(Sp08) (Sum08) (F08)(Sp09)(Sum09) (F09)(Sp10)(F10)(Sp11)
Linear Algebra and its Applications/w Study Guide, by David Lay. Published by Addison Wesley.ISBN: 9780321791542 (New 4th Edition Bundle Set - Fall 2011) (F11)(Sp12) (F12) (Sp13) (F13) (Sp14)
MATH 241
Course Name:Calculus III
Text(s):
Calculus, by R. Ellis and D. Gulick. Published by CENGAGE Learning.ISBN: 9781133436751 (New Custom Edition Fall 2011 but you can still use the 6th Edition for Math 241/241H) (Current) (F06) (Sp07) (F07) (Sp08) (Sum08) (F08) (Sp09) (Sum09) (F09) (Sp10) (F10) (Sp11) (F11) (Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Calculus Student Solutions Manual, by R. Ellis and D. Gulick. Published by Cengage Learning.ISBN:9780759331778(OPTIONAL) (Current)
(F06)(Sp07)(F07)(Sp08) (Sum08) (F08)(Sp09)(Sum09) (F09) (Sp10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Multivariable Calculus with MATLAB by Lipsman. Published by Springer. ISBN: 9783319650692 (OPTIONAL) (Current)(F21) (Sp22)
A Guide to MATLAB: For Beginners and Experienced Users, (New 3rd Edition)by B. Hunt, R. Lipsman, and J. Rosenberg. Published by Cambridge Univ. Press. ISBN: 9781107662223 (OPTIONAL) (Current) (F15) (F16) (F17) (F18) (F19) (F20) (F21) (Sp22) (F22) (F23)
A MATLAB: Companion for Multivariable Calculus, by Jeffery Cooper. Pulished by Harcourt/Academic Press.ISBN: 9780121876258(OPTIONAL) (Current)(Sp05)(Sp07) (Sp08) (Sp09) (Sp10) (Sp11)(Sp12) (Sp13) (Sp14) (Sp15) (Sp16) (Sp17) (Sp18) (F18) (Sp19) (Sp20) (Sp21) (Sp22) (Sp23) (Sp24) (Sp25)
Calculus, 6th Edition, by R. Ellis and D. Gulick. Published by Thomson Publishing.ISBN: 9780759313798 (F06)(Sp07)(F07)(Sp08) (Sum08)
(F08)(Sp09)(Sum09) (F09)(Sp10) (F10)(Sp11) (F11)(Sp12) (Sp20)
A Guide to MATLAB: For Beginners and Experienced Users, (New 2nd Edition)by B. Hunt, R. Lipsman, and J. Rosenberg. Published by Cambridge Univ. Press. ISBN: 9780521615655(F06) (F07)(F08) (F09) (F10) (F11) (F12) (F13) (F14) (F19)
MATH 241H
Course Name:Calculus III
Text(s):
Calculus, by R. Ellis and D. Gulick. Published by Cengage Learning.ISBN: 9781133436751 (New Custom Edition Fall 2011 but you can still use the 6th Edition for Math 241/241H)(Current) (F06) (Sp07) (F07) (Sp08) (Sum08) (F08) (Sp09) (Sum09) (F09) (Sp10) (F10) (Sp11) (F11) (Sp11) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Calculus Student Solutions Manual, by R. Ellis and D. Gulick. Published by Cengage Learning.ISBN:9780759331778 (OPTIONAL)(Current)
(F06)(Sp07)(F07)(Sp08) (Sum08) (F08)(Sp09)(Sum09) (F09) (Sp10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp17) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Multivariable Calculus with MATLAB by Lipsman. Published by Springer. ISBN: 9783319650692 (OPTIONAL) (Current)(F21)
A Guide to MATLAB: For Beginners and Experienced Users, by B. Hunt, R. Lipsman, and J. Rosenberg. (New 3rd Edition) Published by Cambridge Univ. Press. ISBN: 9781107662223 (Current) (F15) (F16) (F17) (F19) (F21) (F22) (F23)
A MATLAB: Companion for Multivariable Calculus, by Jeffery Cooper. Pulished by Harcourt/Academic Press.ISBN: 9780121876258 (Current) (Sp05)(Sp07) (Sp08) (Sp09)(Sp10)(Sp11)(Sp12) (Sp13) (Sp14) (Sp15) (Sp16) (Sp17) (Sp18) (F18) (Sp19) (Sp20) (Sp21) (Sp24) (Sp25)
Div,Grad, and Curl and All That,(An informal text on Vector Calculus), 4th Edition, by H.M. Schey. Published by W.W. Norton. ISBN: 0393925161(F05)
Calculus, 6th Edition, by R. Ellis and D. Gulick. Published by Thomson Publishing.ISBN: 9780759313798 (F06) (Sp07) (F07)(Sp08) (F08) (Sp09)(F09)(Sp10)(F10)(Sp11) (F11)(Sp12)
A Guide to MATLAB: For Beginners and Experienced Users, by B. Hunt, R. Lipsman, and J. Rosenberg. (New 2nd Edition) Published by Cambridge Univ. Press. ISBN: 9780521615655 (F06) (F07) (F08) (F09)(F10) (F11) (F13) (F14)
MATH 243
Course Name: Introduction to Linear Algebra and Differential Equations
Text(s):
Differential Equations and Linear Algebra by C. Henry Edwards, David E. Penney, David T. Calvis. Published by Prentice Hall. ISBN: 9780134497181 (Current)(Sp24) (F24) (Sp25)(F25)
MATH 246
Course Name:Differential Equations for Scientists and Engineers
Text(s):
Differential Equations with MATLAB, by Hunt, Lipsman, Osborn and Rosenberg. Published by J. Wiley. ISBN: 9781118376805 (New 3rd Edition - Fall 2012) (Current) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Differential Equations with MATLAB, by Hunt, Lipsman, Osborn and Rosenberg. Published by J. Wiley. ISBN: 9781119231141 ( 3rd Edition Online- UPDATED copy) - (Optional) (Current) (F20) (Sp21) (F21) (Sp22) (F22) (F23)
Elementary Differential Equations by Boyce and Diprima. (New Custom Edition Textbook Bundle with MATLAB - Fall 2012) Published by Wiley. ISBN: 9781118443798 (F12) (Sp13)
Elementary Differential Equations,(New Custom 9th Edition)by Boyce and DiPrima. Published by Wiley.ISBN: 9780470541074 (F09) (Sp10)(F10) (Sp11) (F11)(Sp12) (F12) (Sp13)
Differential Equations with MATLAB, 2nd Edition, by Hunt, Lipsman, Osborn and Rosenberg. Published by J. Wiley. (bundle with textbook) (Current)(F06) (Sp07)(F07)(Sp08)(Sum08)(F08)(Sp09)(Sum09) (F09)(Sp10)(F10)(Sp11)(F11)(Sp12)
Elementary Differential Equations, 8th Edition, by Boyce and DiPrima. Published by Wiley. ISBN: 0471729574 (F06) (Sp07) (F07) (Sp08) (Sum08)(F08) (Sp09)(Sum09)
MATH 246H
Course Name:Differential Equations for Scientists and Engineers
Text(s):
Differential Equations with MATLAB, by Hunt, Lipsman, Osborn and Rosenberg. Published by Wiley. ISBN: 9781118376805(New 3rd Edition - Fall 2012) (Current) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23) (Sp24) (F24) (Sp25)(F25)
Differential Equations with MATLAB, by Hunt, Lipsman, Osborn and Rosenberg. Published by J. Wiley. ISBN: 9781119231141 (New 3rd Edition Online- UPDATED copy) - (Optional)(Current) (F20) (Sp21) (F21) (Sp22) (F22) (Sp23) (F23)
Elementary Differential Equations by Boyce and Diprima. (New Custom Edition Textbook bundle with MATLAB - Fall 2012). Published by Wiley. ISBN: 9781118443798 (F12) (Sp13)
Elementary Differential Equations,(New Custom 9th Edition)by Boyce and DiPrima. Published by Wiley.ISBN: 9780470541074 (F09)(Sp10) (F10) (Sp11)(F11)(Sp12) (F12) (Sp13)
Differential Equations with MATLAB, 2nd Edition, by Hunt, Lipsman, Osborn and Rosenberg. Published by J. Wiley.(bundle with textbook) (F06) (Sp07) (F07) (Sp08) (F08)(Sp09) (Sum09) (F09) (Sp10) (F10) (Sp11)(F11)(Sp12)
Elementary Differential Equations, 8th Edition, by Boyce and DiPrima. ISBN: 0471729574 (F06) (Sp07)(F07)(Sp08) (F08) (Sp09)
MATH 274
Course Name: History of Mathematics
Text(s):
A History of Mathematics - (New 3rd Edition) by V.J. Katz. Published by Addison Wesley.ISBN: 9780134689524 (Current) (Sp07) (Sp09) (F11) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (Sp22) (F23)
Heritage of Thales, by Anglin. Published by Springer-Velarg.ISBN: 9780387945446 (Sp10)(Sp12) (F12) (Sp13) (F13) (F14) (F15)
The Emergence of Number - (2nd Edition) by J.N. Crossley. Published by World Scientific. ISBN: 9971504146 (F10) (Sp11)
Math Through the Ages: etc., by W.P. Berlinghoff, F.Q. Gouvea. Published by Oxton House Pub. ISBN:978-1881929215(Sp09)
MATH 307
Course Name:A Condensed Introduction to Analysis
Text(s):
Mathematical Proofs:A Transition to Advanced Mathematics, by Chartrand, Polimeni, Zhang. Published by Addison Wesley. ISBN: 0321390539 (W10)
Advanced Calculus, by P.M. Fitzpatrick. Published by AMS. ISBN: 0534376037(Note: this textbook is used in MATH410) (W10)
MATH 310
Course Name: Introduction to Analysis
Text(s):
Transition to Advanced Mathematics, by D. Smith, M. Eggen, R. St. Andre, New 8th Edition. Published by Cengage Learning. ISBN: 9781285463261 (Current)(Sp18) (F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F23) (Sp24) (F24) (Sp25)(F25)
How to Prove It: A Structured Approach by Daniel J. Velleman. Published by Cambridge University Press. ISBN: 9780521861243 (Sp24)
Professor is using ONLINE Publicly. (Current) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (F23)
Mathematical Proofs:A Transition to Advanced Mathematics, by Chartrand, Polimeni, Zhang. Published by Prentice Hall.ISBN: 9780321390530 (New 3rd Edition) (Sp09) (Sp10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17)
Advanced Calculus, by P.M. Fitzpatrick. Published by AMS.ISBN: 9780821847916 (Optional) (2nd Edition) (Sp09) (Sp10)(F10)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17)
Professor is using his own book posted ONLINE. (Sp13) (F13) (Sp14) (F14) (Sp15) (F15)
Infinite Sequences and Series, by Konrad Knopp. Published by Dover. ISBN: 04866015366 (Sp09) (Sum09) (Sp10)
Mathematical Thinking-Problem Solving & Proofs, 2nd Edition, by DAngelo, West. Published by Prentice Hall. ISBN: 0130144126(F06) (Sp07)(F07)(Sp08) (Sum08) (F08) (F09)
Analysis with an Introduction to Proof, 4th Edition, by Steven Lay. Published by Prentice Hall. ISBN: 9780131481015 (F06) (F16)
Logic: Techniques of Formal Reasoning, 2nd Edition, by Kalish, Montague, Mar. Published by Wadsworth. ISBN: 0155511818(F05
Transition to Advanced Mathematics, 7th Edition. Published by Cengage Learning. ISBN: 9780495562023 (F16)
MATH 312 (New Course Spring 2012)
Course Name:Math Reasoning & Proof for Pre-Service Middle School Teachers
Text(s):
Teachers Notes - (Current) (Sp17) (Sp22)
Teaching Fractions & Ratios for Understanding: Essential Content Knowledge & Instructional Strategies for Teachers by Susan J. Lamon. (New 3rd Edition) Published by Lawrence Erlbaum Assoc., Inc.ISBN: 9780415886123 (Sp12) (Sp13)
MATH 314 (New Course Spring 2011)
Course Name:Intro to Probability, Data, Analysis and Statistics for Preservice Middle School Teachers
Text(s):
Developing Essential Understanding of Statistics for Teaching Mathematics in Grades 6-8, by Kader. Published by NCTM. ISBN: 9780873536721 (F14) (Sp16)
Teacher Notes - (F15) (F18) (Sp20) (Sp22) (F22) (Sp24)
Intro Stats, by Deveaux, Velleman & Bock. Published by Addison Wesley. ISBN: 9780321699121 (New 4th Edition - Fall 2013) (F13)
Intro Stats, by Deveaux, Velleman & Bock. (3rd Edition) Published by Addison Wesley. ISBN: 9780321699121 (New Custom Edition) (Sp11) (Sp13)
MATH 315 (New Course Spring 2010)
Course Name:Algebra for Preservice Middle School Teachers
Text(s):
Teacher Notes - (F15) (Sp17) (F18) (Sp19) (F22)
Algebra: Form + Function, McCallum. Published by Wiley. ISBN: 9780470521434 (F14)
Algebra for Elementary and Middle School Teachers: An Inquiry Approach, (2nd Edition) by S. Stump, K. Roebuck and J. Bishop. Published by Pearson Custom Publishing. (2nd Edition) ISBN: 9780558387778 (F10) (F11) (F12) (F13) (Sp15)
Algebra for Elementary and Middle School Teachers: An Inquiry Approach, 1st Edition by S. Stump, K. Roebuck and J. Bishop. Published by Pearson Custom Publishing. ISBN: 053656762X (Sp10)
MATH 340
Course Name:Multivariable Calculus, Linear Algebra & Differential Equations I: Honors.
Text(s):
No Textbook is needed (Current) (F22)(F25)
Calculus and Analysis in Euclidean Space, by Jerry Shurman. Published by Springer International Publishing. ISBN: 9783319493121 (F17)
Advanced Calculus of Several Variables, by Edwards. Published by Dover. ISBN: 9780486683362 (Current) (F11) (F16) (F17) (F18) (F19) (F23)
A Visual Introduction to Differential Forms and Calculus on Manifolds by J.P. Fortney. Published by Springer International Publishing. ISBN: 9783319969916 (OPTIONAL) (F19)
Advanced Calculus by Shlomo Sternberg, Lynn Loomis (Revised Edition). Published by World Scientific Publishing Company. ISBN: 9789814583930 (F16)
Vector Calculus, by Marsden. Published by Freeman, W. H & Company. ISBN: 9781429215084 (F14) (F15)
Vector Calculus, Linear Algebra and Differential Forms, by Hubbard and Hubbard. Published by Prentice Hall. (New 4th Edition - Fall 2013) ISBN: 9780971576650 (F13)
Multivariable Mathematics, by R. Williamson, H. Trotter. (4th Edition) Published by Prentice Hall.ISBN: 9780130672766(F09) (F10) (F12)
Student Solution Manual, by R. Williamson, H. Trotter. (4th Edition) Published by Prentice Hall. (Optional)ISBN: 9780130461926 (F09) (F10) (F12)
Matrices and Linear Transformations, 2nd Edition, by Charles G. Cullen. Published by Dover. ISBN: 0486663280 (F06)(F07) (F08)
Vector Calculus, 3rd Edition, by Susan J. Colley. Published by Prentice Hall. ISBN: 0131858742(F06)(F07) (F08)
Vector Calculus, Linear Algebra and Differential Forms, by Hubbard and Hubbard. Published by Prentice Hall. ISBN: 0130414085(F05)
MATH 341
Course Name:Multivariable Calculus, Linear Algebra & Differential Equations II: Honors.
Text(s):
No Textbook is needed (Current) (Sp20) (Sp22)
Linear Algebra and its Applications/w Study Guide, by David Lay. Published by Addison Wesley.ISBN: 9780321982384 (Sp18)
Introduction to Differential Equations, by Taylor. Published by AMS. ISBN: 9780821852712 (Sp16)
Differential Equations Dynamical System and An Introduction to Chaos, by M. Hirsch. Published by Elsevier. ISBN: 9780123820105 (Sp16) (Sp17)
Differential Equations and their Applications (An Introduction to Applied Math), by Martin Braun. Published by Springer-Verlag.ISBN: 9780387978949 (4th Edition) (Sp06) (Sp07) (Sp08) (Sp09)(Sp12) (Sp14) (Sp15) (Sp19)
Advanced Calculus of Several Variables, by Edwards. Published by Dover. ISBN: 9780486683362
Multivariable Mathematics, 1st Edition by R. Williamson, H. Trotter. (4th Edition) Published by Prentice Hall.ISBN: 9780130672766 (Sp10) (Sp11) (Sp13)
Student Solution Manual, by R. Williamson, H. Trotter. (4th Edition) Published by Prentice Hall. (Optional) ISBN: 9780130461926 (Sp10) (Sp11) (Sp13)
Matrices and Linear Transformation, by C.G.Cullen. Published by Dover. ISBN: 9780486663289 (Sp06) (Sp07) (Sp09) (Sp15)
Vector Calculus, by S. Colley. Published by Prentice Hall. ISBN: 0131858742 (Sp07) (Sp09)
Vector Calculus, Linear Algebra and Differential Forms, by Hubbard and Hubbard. Published by Prentice Hall. ISBN: 0130414085
Differential Equations with MATLAB, by Coombes, Hunt,Lipsman,Osborn,Stuck. Published by Wiley. ISBN: 0471718122 (Sp04) (Sp08)
MATH 350 and 351
Course Name:Honors Analysis I and II(COURSES THAT ARE NOT BEING OFFERED ANY MORE)
Text(s):
Advanced Calculus, 3rd Edition, by R.C. Buck.
Elementary Classical Analysis, 2nd Edition, by J.E. Marsden and M.J. Hoffman.
Principles of Mathematical Analysis, 3rd Edition, by W. Rudin.
MATH 400
Course Name:Vectors and Matrices
Text(s):
Linear Algebra and its Applications, 3rd Edition, by David Lay. Published by Addision Wesley. ISBN: 0321287134 (Sp05) (Sp07)
Student Study Guide, by D. Lay. Published by Addison Wesley.ISBN: 0321280660 (Sp07)
MATH 401
Course Name:Applications of Linear Algebra
Text(s):
Linear Algebra and its Applications/w Study Guide, by David Lay. Published by Addison Wesley.ISBN: 9780135851258 (New 6th Edition - Fall 2020) (Current) (F20) (Sp20) (Sp21) (F21) (Sp22) (F23) (Sp24) (F24) (Sp25)
Linear Algebra and its Applications/w Study Guide, by David Lay. Published by Addison Wesley.ISBN: 9780321982384 (New 5th Edition - Fall 2015) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F20)
Professor is using his online notes. (Current) (Sp20) (F21) (Sp22) (F22) (F25)
Applied Linear Algebra, 2nd Edition, by P.J. Olver & C. Shakiban. Published by Prentice Hall.ISBN: 9783319910406 (Current) (F06) (F08) (Sp10)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (F16) (Sp17) (F17) (Sp18) (F18) (F23) (Sp25)
Linear Algebra and its Applications/w Study Guide, by David Lay. Published by Addison Wesley.ISBN: 9780321791542 (New 4th Edition Bundle Set - Fall 2011) (F11) (F12) (F13) (Sp14) (F14) (Sp15)
Linear Algebra and its Applications, by G. Strang. Published by Thomson Publishing.ISBN: 9780030105678 (4th Edition) (F06) (Sp07) (Sp09) (Sum09) (F09) (Sp11)(Sp12) (Sp13) (F13) (Sp14) (F14) (F15) (Sp16) (F22)
Linear Algebra and its Applications, 3rd Edition, by D. Lay. Published by Addison Wesley. ISBN: 0321287134 (F06) (Sp07) (F07) (Sp08) (Sum08) (F08) (Sp09) (F09) (Sp10) (F10) (Sp11)
Student Study Guide, 3rd Edition, by D. Lay. Published by Addison Wesley. ISBN: 0321280660 (F06) (Sp07)(F07)(Sp08) (Sum08) (F08) (Sp09) (F09) (Sp10) (F10) (Sp11)
MATH 402
Course Name:Algebraic Structures
Text(s):
Contemporary Abstract Algebra, (New 10th Edition) by J. Gallian. Published by Cengage Learning. ISBN: 9780367651787 (Current) (F24) (Sp25)
A First Course in Abstract Algebra,by Fraleigh. Published by Cengage Learning. ISBN: 9780201763904 (7th Edition)(Sp20)
Comtemporary Abstract Algebra, (New 9th Edition) by J. Gallian. Published by Cengage Learning. ISBN: 9781305657960 (Sp17) (F18) (F19) (F21)
Professor is using her notes and FREE DOWNLOADABLE TEXTBOOK: (F14) (Sp16) (F17)
Abstract Algebra: Theory and Applications, by T. W. Judson. Published by Orthogonal Publishing. ISBN: 9781133599708 (Sp14)
Contemporary Abstract Algebra, (New 8th Edition) by J. Gallian. Published by Cengage Learning. ISBN: 9781133599708 (Current) (Sp13) (F13) (Sp15) (F15) (F22)
Contemporary Abstract Algebra,(New 7th Edition), by J. Gallian. Published by Cengage Learning.ISBN: 9780547165097 (F06) (Sp07)(F07) (Sp08) (F08)(Sp09) (F09) (Sp10) (F10)(Sp11)(F11)(Sp12) (F12)
Professor is using his online notes. (Current) (F16) (Sp18) (Sp19) (Sp22)
MATH 403
Course Name:Abstract Algebra
Text(s):
Abstract Algeba, (3rd Edition) by David S. Dummit, Richard M. Foote. Published by Wiley. ISBN: 9780471433347 (Current) (Sp25)
Contemporary Abstract Algebra, (New 10th Edition), by J. Gallian.Published by Cengage Learning. ISBN: 9780367651787 (Current) (Sp24) (F24) (Sp25)
Contemporary Abstract Algebra, (New 9th Edition), by J. Gallian. Published by Cengage Learning. ISBN: 9781305657960 (Current) (F17) (Sp18) (Sp19) (F19) (Sp20) (Sp21) (F21) (Sp22) (F22) (Sp24)
Visual Group Theory by Nathan Carter. Published by MAA. ISBN: 9780883857571 (OPTIONAL) (F19)
First Course in Abstract Algebra, by Rotman. Published by Prentice Hall. ISBN: 9780131862678 (Sp15) (F15) (Sp19)
A First Course in Abstract Algebra,by Fraleigh. Published by Cengage Learning. ISBN: 9780201763904 (7th Edition) (Sp14) (F14) (F18) (F22) (F23) (Sp24)
Contemporary Abstract Algebra, (New 8th Edition) by J. Gallian. Published by Cengage Learning. ISBN: 9781133599708 (Sp13) (F13) (Sp14) (Sp15) (F15)
Abstract Algebra, 3rd Edition, by I.N. Herstein. Published by Wiley.ISBN: 9780471368793 (Current) (F06) (Sp08) (F08) (F10)(Sp12) (F12) (Sp13) (F13) (Sp16) (F16) (Sp17) (F18) (Sp20) (Sp22) (F23)
Modern Algebra with Applications, by W. Gilbert, W. Nicholson. Published by Wiley. ISNB: 9780471414513 (2nd Edition) (F13)
Contemporary Abstract Algebra, (New 7th Edition), by J. Gallian. Published by Cengage Learning. ISBN: 0618514716(Sp06) (Sp07) (Sp09) (F09) (Sp10) (Sp11) (F11)(Sp12)
Topics In Algebra, 2nd Edition, by I. Herstein. Published by Wiley. ISBN: 0471010901 (F07) (F11)
MATH 404
Course Name:Field Theory
Text(s):
Galois Theory, by I. Stewart, New 5th Edition. Published by Chapman + Hall/CRC.ISBN: 9781032101590 (Sp24)
Galois Theory, by I. Stewart. Published by Chapman + Hall/CRC.ISBN: 978148224520 (New 4th Edition) 2016) (Sp16) (Sp18) (Sp19)
Galois Theory, by I. Stewart. Published by Chapman + Hall/CRC.ISBN: 9781584883937 (3rd Ed.) (Sp07) (Sp08) (Sp09) (Sp10) (Sp11) (Sp13) (Sp14)
Fields & Galois Theory by Howie. Published by Springer-Verlag.ISBN: 9781852339869 (Current) (Sp12) (Sp20) (Sp22) (Sp25)
Contemporary Abstract Algebra, 6th Edition, by J. Gallian. Published by Houghton Mifflin. ISBN: 0618514716(Sp05)
MATH 405
Course Name:Linear Algebra
Text(s):
Linear Algebra Done Right, (4th Edition), by Sheldon Axler. Published by Springer-Verlag. ISBN: 9783031410260 (Current)(F25)
Linear Algebra, 3rd Edition, by Lang. Published by Springer-Verlag. ISBN: 9781441930811(Current) (Sp22)
Linear Algebra Done Right, (3rd Edition), by Sheldon Axler. Published by Springer-Verlag. ISBN: 9783319110790 (Current) (Sp19) (Sp20) (F21) (F22) (F25)
Finite Dimensional Vector Spaces, by P.R. Halmos (2nd Edition). Published by Dover. ISBN: 9780486814865 (F17)
Linear Algebra by Curtis. Published by Springer-Verlag. ISBN: 9781461270195 (F17) (Sp24)
Linear Algebra by Curtis. Published by Springer-Verlag. ISBN: 9780387909929 (F14) (Sp16) (F16) (F19) (Sp20)
Linear Algebra, by Hoffman & Kunze. Published by Prentice Hall. ISBN: 9780135367971 (2nd Edition) (Current) (F06) (Sp07)(F07)(Sp08) (F08) (F12) (Sp13) (F13) (Sp14) (Sp15) (F15) (Sp18) (F21) (Sp22)
Linear Algebra by Schaum. Published by MCG. ISBN: 9780071794565 (Sp13)
Introduction to Linear Algebra, by Y. Katznelson and Y. Katznelson. Published by AMS.ISBN: 9780821844199 (Sp12)
Introduction to Linear Algebra, (2nd Edition), by S. Lang. Published by Springer-Verlag. ISBN: 0387962050 (F11)
Finite Dimensional Vector Spaces, by P.R. Halmos. Published by Springer-Verlag. ISBN: 9780387900933 (Sp10)
Linear Algebra, 3rd Edition, by Lang. Published by Springer-Verlag. ISBN: 9780387964126 (Current) (Sp05) (F08) (Sp09) (F09) (F10) (Sp11) (F18) (Sp19) (F23)
MATH 406
Course Name:Introduction to Number Theory
Text(s):
An Introduction to the Theory of Numbers by Ivan Niven. Published by Wiley. ISBN: 9780471625469 (F21)
Elementary Number Theory, by Strayer. Published by Waveland. ISBN: 9781577662242 (F16)
Intro to Number Theory w/Cryptography, (New 2nd - Fall 2013) by Kraft and Washington. Published by Taylor & Francis.ISBN: 9781138063471 (Current) (F13) (Sp14) (F14) (Sp15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F21) (Sp22) (F23) (Sp24) (F24) (Sp25)
Elementary Number Theory and its Applications, by Rosen. Published by Addison Wesley.ISBN: 9780321500311 (Current) (New 6th Edition) (F06) (Sp07) (F07) (Sp08)(F08) (Sp09) (Sum09) (F09) (Sp10) (Sp11)(F11) (Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F17) (Sp18) (F18) (Sp19) (Sp20) (F21) (Sp22) (Sp24) (Sp25)
Fundamentals of Number Theory,1st Edition, by W. LeVeque. Published by Dover. ISBN: 9780486689067 (Sp11) (F11)
Elementary Number Theory,(2nd Edition), by Underwood Dudley. Published by Dover. ISBN:97804886469317 (Current)(F09) (F10) (Sp11)(F11) (Sp12) (F12) (Sp13) (F14) (F15) (F16) (Sp17) (Sp18) (Sp22) (F23) (Sp24)(F25)
The Higher Arithmetic, by H. Davenport. Published by Addison Wesley. ISBN: 9780521722360 (F09) (Sp20)
A Concise Introduction to the Theory of Numbers, by Alan Baker. Published by Cambridge University Press. ISBN: 9780521243834 (OPTIONAL) (Sp20)
Topics in Number Theory, Vols. I and II. by W. J. LeVeque. Published by Dover. ISBN: 0486425398 (Sp09)
(Sp10) A Friendly Introduction to Number Theory, 4th Edition, by J. Silverman. Published by Prentice Hall.ISBN: 9780321816191 (Sp18)
An Illustrated Theory of Numbers, by Martin H. Weissman. Published by AMS.ISBN: 9781470463717 (Current) (Sp18) (F22) (F23) (Sp24)(F25)
A Friendly Introduction to Number Theory, 3rd Edition, by J. Silverman. Published by Prentice Hall.ISBN: 9780321816191 (F06) (F16)
Elementary Number Theory, by D. M. Burton. Published by McGraw-Hill. ISBN: 9780073383149 (Sp07) (F07) (Sp08) (Sum08) (F08) (Sp15)
MATH 410
Course Name:Advanced Calculus I
Text(s):
Advanced Calculus, by P.M. Fitzpatrick. Published by AMS.ISBN: 9780821847916 (2nd Edition Revised)(Current)(F06) (Sp07)(F07)(Sp08) (Sum08)(F08) (Sp09) (Sum09) (F09)(Sp10)(F10)(Sp11)(F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F21) (Sp22) (F22) (F23) (Sp24) (F24) (Sp25)(F25)
Working Analysis, 1st Edition, by J. Cooper. Published by Academic Press. ISBN: 0121876047(F06) (Sp07)(F07)
MATH 411
Course Name:Advanced Calculus II
Text(s):
Advanced Calculus, by P.M. Fitzpatrick. Published by AMS.ISBN: 9780821847916 (2nd Edition Revised)(Current)(F06) (Sp07) (F07)(Sp08) (Sum08)(F08)(Sp09)(Sum09) (F09)(Sp10)(F10) (F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F21) (Sp22) (F22) (F23) (Sp24) (F24) (Sp25)(F25)
MATH 412
Course Name:Advanced Calculus with Applications
Text(s):
Working Analysis, 1st Editon, by Jeffery Cooper. Published by Addison Wesley. ISBN: 0121876047 (Sp05) (Sp07) (Sp08) (Sp09) (Sp10) (Sp11)
Professor is using his notes(Sp04)
MATH 414
Course Name:Ordinary Differential Equations
Text(s):
The Qualitative Theory of Ordinary Differential Equations: An Introduction, 1st Edition, by Brauer-Nohel. Published by Dover. ISBN: 0486658465(F06) (F08)
MATH 416
Course Name:Applied Harmonic Analysis: Intro. to Signal Processing
Text(s):
First Course in Wavelets w/Fourier Anal by Boggess. Published by Wiley. ISBN: 9780470431177 (F15)
Mathematics for Multimedia, (New Revised Edition) by M.V. Wickerhauser. Published by Springer-Verlag.ISBN: 9780817648794 (Sp10) (Sp11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (Sp16) (F16) (Sp17) (Sp18) (Sp20)
Mathematics for Multimedia, 1st Edition, by Mladen V. Wickerhauser. Published by Academic Press. ISBN: 0127484515(Sp08)
Professor is using his notes(Sp19) (F19) (Sp22) (F22) (F23) (Sp24)
MATH/AMSC 420
Course Name:Mathematical Modeling
Text(s):
Professor is using his notes (Sp19) (Sp20) (Sp24)
Topics in Mathematical Modeling, by K. Tung. Published by Princeton University Press. ISBN: 069116423 (OPTIONAL) (Sp11)
Guide to Mathematical Modeling, New 2ND Edition by D. Edwards and M. Hamson. Published by CRC Press, 1990. ISBN: 9780831133375 (OPTIONAL) (Sp03) (Sp11)
Concepts of Mathematical Modeling, by W. Meyer. Published by Dover. ISBN: 0486435156 (OPTIONAL) (Sp11)
MATH 424
Course Name:Introduction to the Mathematics of Finance
Text(s):
Professor is using notes no textbook (Current) (F25)
Mathematics for Finance: An Introduction to Financial Engineering by Marek Capiński, Tomasz Zastawniak. Published by Springer-Verlag. ISBN: 978-0857290816 (Current) (Sp25)
An Elementary Intro. to Math. Finance:Options & Other Topics, (3rd Edition) by Sheldon Ross. Published by Cambridge Univ. Press. ISBN: 9780521192538 (Current) (F06) (F07) (F08) (F09) (F10) (F11) (F12) (F13) (F14) (F15) (F16) (F17) (F18) (Sp19) (F19) (Sp20) (F21) (Sp22) (F22) (F23)
Financial Modeling Using Excell & VBA, by C. Sengupta. Published by Wiley.ISBN: 0471267686(F06)
Advanced Modeling in Finance Using Excell & VBA, by M. Jackson, M. Staunton. Published by Wiley.ISBN: 0471499226(F06)
Numerical Methods in Finance: A Matlab-Based Intro, by P. Brandimarte. Published by Wiley-Interscience. ISBN: 0471396869(F05)
MATH 430
Course Name:Euclidean and Non-Euclidean Geometries
Text(s):
Geometry: Euclid and Beyond, by Hartshorne. (Optional) Published by Springer-Verlag. ISBN: 9780387986500 (F11) (F13) (Sp14)
The Four Pillars of Geometry, by J. Stillwell. Published by Springer-Verlag. ISBN: 9780387255309 (Current)(F10) (F11) (F13) (Sp14) (F14) (Sp16) (Sp17) (Sp19) (Sp20) (F20) (Sp22) (F23) (Sp25)(F25)
Foundations of Geometry, by G.A. Venema. (2nd Edition) Published by Prentice Hall.ISBN: 9780136020585 (Current) (F06) (Sp07) (F07) (Sp08) (F08) (Sp09) (F09) (Sp10) (F10) (Sp11)(Sp12) (F12) (Sp13) (Sp15) (F15) (F16) (Sp18) (F18) (F19) (Sp20) (F21) (F22)
Experiencing Geometry: etc, 3rd Edition, by D. Henderson & D. Taimina. Published by Prentice Hall. ISBN: 0131437488 (Sp06)
Roads to Geometry, 3rd Edition, by E. Wallace & S. West. Published by Prentice Hall. ISBN: 0130413968(Sp05)
Advanced Euclidean Geometry, by Alfred Posamentier. Published by Key Curriculum Press. ISBN: 1930190859(F04)
Geometry from Euclid to Knots, by S. Stahl. Published by Prentice Hall. ISBN:0130329274(F04)
Modern Geometry, by D. Thomas. Published by Brooks/Cole.ISBN: 0534666914(Sp03)
MATH 431
Course Name:Geometry for Computer Graphics
Text(s):
Perspective and Projective Geometry by Crannell, Frantz, Futamura. Published by Springer-Verlag. ISBN: 978-0691196565 (Current) (F23)
Geometric Algebra: An Algebraic System for Computer Games and Animation, by John A. Vince. Published by Springer-Verlag. ISBN: 9781848823785 (F16)
Mathematics for Computer Graphics by J. Vince. Published by Springer-Verlag. ISBN: 9781447162896 (F15) (F16)
Applied Geometry for Computer Graphics and CAD, by D. Marsh. Published by Springer-Verlag. ISBN: 9781852338015 (F09) (F10) (F12) (F13) (F14) (F16)
3-D Computer Graphics: A Math Introduction with OpenGL, 1st Ed., by Samuel R. Buss. Published by Cambridge University Press. ISBN: 0521821037(F08)
Geometric Tools for Computer Graphics, by P. Schneider & D. Eberly. Published by Academic Press. ISBN: 1558605947(F08)
NOTES by G. Helzer. (Sp03)
NOTES by Professor. (F06) (F18) (F21) (F22)
MATH 432
Course Name:Introduction to Topology
Text(s):
Topology, 2nd Edition, Classic Version by J. Munkres. (Previously published by Prentice Hall.) ISBN:9780134689517 (Current) (F06) (F07) (F08) (F09) (F10) (F11) (F12) (F13) (F14) (F15) (F16) (F17) (F18) (F19) (F21) (F23) (F24)
Topology of Surfaces, by L.C. Kinsey. Published by Springer. ISBN: 0387941029(F04)
MATH 436
Course Name:Differential Geometry of Curves and Surfaces I
Text(s):
Elementary Differential Geometry, by Andrew Pressley. Published by Springer-Verlag. ISBN: 9781848828902 (Current) (F06) (F07) (F09) (F10) (F11) (F12) (F14) (F15) (F16) (F18) (F19) (F21) (F22)
Online notes at www.math.uga.edu/~shifrin/ShifrinDiffGeo.pdf(F06)
Diff. Geometry of Curves & Surfaces, by Manfredo Do Carmo. Published by Prentice Hall.ISBN: 0132125897(F05)
MATH 437
Course Name:Differential Forms and their Applications
Text(s):
Differential Topology by Guillemin and Pollack. Published by American Mathematical Society. ISBN: 9780821851937 (REPRINT) (Current) (Sp25)
Vector Analysis, by Klaus Janich. Published by Springer-Verlag. ISBN: 9781441931443 (Sp16) (Sp20)
Geometric Approach to Differential Forms, by Bachman. Published by Springer-Verlag. ISBN: 9780817683030 (2nd Edition) (Sp14) (Sp15)
Calculus on Manifolds: A Modern Approach To Classical Theorems of Advanced Calculus, by M. Spivak. Published by Westview Press. ISBN: 9780805390216 (Sp11) (Sp13) (Sp17)
Differential Forms and their Applications, by Manfredo Do Carmo. Published by Springer-Verlag.ISBN: 9783540576181 (Current) (Sp04)(Sp12) (Sp18) (Sp22) (Sp24)
Differential Forms and Connections, 1st Edition, by R.W.R. Darling. Published by Cambridge University. ISBN: 9780521468008 (Sp04) (Sp07) (Sp19)
MATH 445
Course Name:Elementary Mathematics Logic
Text(s):
Mathematical Intro to Logic by Enderton. Published by Academic Press. ISBN: 9780122384523 (F14)
What Is Mathematical Logic, by J.N. Crossley, et al. Published by Dover. ISBN: 0486264041 (F09)
A First Course in Logic, by Shawn Hedman. Published by Oxford University Press. ISBN: 01985229813(F08)
The Development of Mathematical Logic, 2nd Edition, by P.H. Nidditch. Published by St. Augustine Press, Inc. ISBN: 1855065827(F04)
Professor will follow ONLINE Notes: (F12) (F15) (F16) (F19) (F21) (F22) (F23)
MATH 446
Course Name:Axiomatic Set Theory
Text(s):
The Joy of Sets: Fundamentals of Contemprary Set Theory, by Keith Devlin. Published by Springer-Verlag. ISBN: 9781461269410 (Sp04) (Sp14) (Sp15)
Notes on Set Theory, by Y. Moschovakis. Published by Springer-Verlag.ISBN: 9780387287232 (Sp07) (Sp08) (Sp10) (Sp11)(Sp12)
MATH 447
Course Name:Introduction to Mathematical Logic(COURSE IS NOT BEING OFFERED ANY MORE)
Text(s):
A Mathematical Introduction to Logic, by H.B. Enderton.
Computability and Logic, by G. Boolos and R. Jeffrey.
MATH 450/CMSC 450
Course Name:Logic for Computer Science
Text(s):
Logic for Mathematics and Computer Science, by Stanley N. Burris. Published by Prentice Hall. ISBN: 013285742(F06)
MATH 452/AMSC 452
Course Name:Introduction to Dynamics and Chaos
Text(s):
NONLINEAR DYNAMICS AND CHAOS, by Steven H. Strogatz. Published by CRC Press. ISBN:978-0367026509 (Current) (Sp25)
A First Course in Discrete Dynamical Systems, by Richard A. Holmgren. Published by Springer. ISBN: 978-0387947808 (Current) (F22) (F23)
An Introduction to Chaotic Dynamical Systems, by DeVaney. Published by Perseus D. ISBN: 9780813340852 (F16)
Encounters with Chaos +Fractals by Denny Gulick. Published by Taylor. ISBN: 9781584885177 (F15)
First Course in Chaotic Dynamical System by DeVaney. Published by Westview. ISBN: 9780201554069 (Current) (F14) (F17) (F18) (F19) (F21) (Sp22)
Encounters with Chaos, by D. Gulick. Published by McGraw Hill. ISBN: 0070252033 (Sp09) (F11) (Printed at Bel-Jean and sold in Maryland Book Exchange)
Notes by Professor. (Printed at Bel-Jean and sold in Maryland Book Exchange) (F07)
Chaos and Intro. to Dynamical Systems, 1st Edition, by Alligood, Sauer & Yorke. Published by Springer-Verlag.ISBN: 9780387946771 (F06) (Sp10)
An Introduction to Dynamical Systems, 3rd Edition, by K. Alligood, T. Sauer, J. Yorke. Published by Springer-Verlag. ISBN: 9780387946771 (Sp05) (Sp13) (F16)
MATH 456/CMSC 456
Course Name:Cryptology
Text(s):
Introduction to Cryptography with Coding Theory, by W. Trappe & L. Washington. (3rd Edition) Published by Prentice Hall.ISBN: 9780136731542 (Current) (F23) (Sp24) (F24) (Sp25)
Introduction to Modern Cryptography, (3rd Edition) by J. Katz and Y. Lindell. Published by Chapman Hall/CRC. ISBN: 9780815354369 (Current) (Sp22) (F22) (F23) (Sp24) (F24) (Sp25)
Introduction to Cryptography with Coding Theory, by W. Trappe & L. Washington. (3rd Edition) Published by Prentice Hall.ISBN: 9780134859064 (F21) (Sp22) (F22)
Introduction to Modern Cryptography, by J. Katz and Y. Lindell. Published by Chapman Hall/CRC. (2nd Edition) ISBN: 9781466570269 (Sp16) (F16) (Sp19) (Sp20)
Introduction to Cryptography with Coding Theory, by W. Trappe & L. Washington. (2nd Edition) Published by Prentice Hall.ISBN: 9780131862395 (F06) (Sp07) (Sp08) (Sp09) (Sp10) (Sp11)(Sp12) (Sp13) (Sp14) (Sp15) (F15) (Sp17) (Sp18) (F18) (SP19) (F19)
Introduction to Modern Cryptography, by J. Katz and Y. Lindell. Published by Chapman Hall/CRC. ISBN: 9781584885511 (F09) (F10) (F11) (F12) (F13) (F14)
MATH 461
Course Name:Linear Algebra for Scientists and Engineers
Text(s):
Linear Algebra and its Applications,(New 6th - Spring 2021), by David Lay. Published by Pearson.ISBN: 9780135851258 (F21) (Sp22) (F22) (F23) (Sp24) (F24) (Sp25)(F25)
A First Course in Linear Algebra an Open Text by Lyryx Learning. (Free Downloadable for Students) (F19) (Sp20)
Linear Algebra and its Applications,(New 5th - Fall 2014), by David Lay. Published by Addison Wesley.ISBN: 9780321982384 (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19)
Linear Algebra with Applications, 3rd Edition, by D. Lay. Published by Addison Wesley. ISBN: 0321287134 (Sp05) (F07) (Sp08) (Sum08) (Sp09) (Sum09) (Sp10) (F10) (Sp11)
Introduction to Linear Algebra, (New 4th Edition), by Gilbert Strang. Published by Wellesley Cambridge Press. ISBN: 9780980232714 (F06) (Sp07) (F08) (F09)
Linear Algebra and its Applications/w Study Guide,(New 4th Edition Bundle Set - Fall 2011), by David Lay. Published by Addison Wesley.ISBN: 9780321791542 (F11)(Sp12) (F12) (Sp13) (F13) (Sp14)
MATH 462
Course Name:Partial Differential Equations for Scientists and Engineers
Text(s):
Partial Differential Equations, by Rustum Choksi. Published by AMS. ISBN:978-1470464912 (Current) (F25)
Professor will be using Canvas Lecture Notes (Sp20)
Introduction to Partial Differential Equations, by Walter Strauss. (2nd Edition) Published by Wiley. ISBN: 9780470054567 (Current) (F06) (Sp07) (F07) (Sp08) (Sum08) (F08) (Sp09) (Sum09) (F09) (Sp10) (F10) (Sp11) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (F21) (F22) (F23) (Sp24) (F24) (Sp25)
First Course in Partial Differential Equations, by Weinberger. Published by Dover. ISBN: 9780486686400 (F11) (F13)
Partial Differential Equations, by Evans. Published by AMS. (Optional) ISBN: 9780821849743 (F11)
Partial Differential Equations of Mathematical Physics, by S.L. Sobolev. Published by Dover. (Optional) ISBN: 9780486659640 (F11)
Introduction to Partial Differential Equations with MATLAB, by J. Cooper. Published by Springer.ISBN: 9780817639679 (Sp07) (Sp10)(Sp12)
MATH 463
Course Name:Complex Variables for Scientists and Engineers
Text(s):
Complex Variables for Scientists and Engineers, 2nd Edition by John D. Paliouras, Douglas S. Meadows. Published by Prentice Hall. ISBN: 9780486493473 (F18) (Sp18) (Sp19)
Fund. of Complex Analysis for Math. Science & Enginering, 3rd Edition, by Saff and Snider. Published by Prentice Hall.ISBN: 9780139078743 (F06) (F07) (F08) (F10) (Sp11)(Sp12) (Sp15) (Sp16) (F16) (F17) (Sp20)
Complex Variables and Applications, (New 9th Edition - Spring 2014), by Churchill/Brown. Published by McGraw Hill.ISBN: 9780073383170 (Current)(F06) (Sp07)(F07) (Sp08) (Sum08) (F08)(Sp09) (F09) (Sp10) (F10)(Sp11) (F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (F16) (Sp17) (F17) (Sp18) (Sp19) (F19) (Sp20) (F21) (Sp22) (F22) (F23) (Sp24)(F25)
Professor will use his notes. (Sp08) (F15) (Sp16)
MATH 464
Course Name:Transform Methods for Scientists and Engineers
Text(s):
Professor will using his own PDF notes (Sp20) (Sp24) (Sp25)(F25)
Analysis, 2nd Edition. Published by AMS. ISBN: 9780821827833 (F18)
Harmonic Analysis and Applications, 1st Edition by J.J. Benedetto. (Optional) Published by CRC.ISBN: 9780849378799 (Current) (F06) (Sp07)(F07)(Sp12) (F13) (Sp14) (F18) (F23)
A First Course in Fourier Analysis,(2nd Edition)by D.W. Kammler. Published by Prentice Hall.ISBN: 9780521709798 (Current) (F04) (Sp08) (F08) (Sp09)(F09)(Sp10)(F10) (Sp11)(F11) (Sp12) (F12) (Sp13) (F13) (Sp15) (F15) (F16) (Sp17) (Sp18) (F18) (Sp19) (F19) (F21) (F22)
Introduction to the Mathematics of Medical Imaging by C. Epstein. Published by SIAM. ISBN: 9780898716429 (Sp09) (F09)
MATH 470
Course Name:Mathematics for Secondary Education
Text(s):
Mathematics for Human Flourishing by Francis Su. Published by Yale University Press. ISBN: 9780300258516 (Current) (Sp22)(Sp24) (Sp25)
Professor will use his notes. (F16) (Sp20)
Math. for High Sch Teachers-An Advanced Perspective, by Usiskin,Peressini,Marchisotto & Stanley. Published by Prentice Hall.ISBN: 9780130449412 (F06) (F07) (F08) (F09) (F10)(F11) (F12) (F13) (F14)
MATH 475/CMSC 475
Course Name:Combinatorics and Graph Theory
Text(s):
Combinatorics and Graph Theory, by John Harris, J. Hirst, M. Mossinghoff. Published by Springer. ISBN: 9780387797106 (F21)
First Course in Graph Theory, by Gary Chartrand. Published by Dover. ISBN: 9780486483689 (OPTIONAL) (Sp16) (Sp17) (Sp18) (F18) (Sp20)
Introduction to Enumerative Combinatorics, 2nd Edition by Miklos Bona. Published by Taylor & Francis
ISBN: 9781482249095 (Current) (OPTIONAL) (Sp16) (Sp17) (Sp18) (F18) (Sp20) (Sp22) (F22) (F23)
Introduction to Graph Theory by Douglas West. Published by Prentice Hall. ISBN: 9780130144003 (OPTIONAL) (Sp16) (Sp17)
Introductory Combinatorics, by R. Brualdi. Published by Prentice Hall.ISBN: 9780136020400 (5th Edition) (Sp06) (Sp10)(Sp12) (Sp13) (Sp14) (Sp15)
Applied Combinatorics, 4th Edition, by Alan Tucker. Published by Wiley. ISBN: 047143809X (Sp06) (Sp09)
Introductory Combinatorics, 3rd Edition, by Kenneth Bogart. Published by CENGAGE Learning. ISBN:0121108309 (Sp08)
MATH 489
Course Name:Research Interactions in Mathematics
Text(s):
None
STAT 100
Course Name:Elementary Statistics and Probability
Text(s):
MYSTATLAB Student Access Code Card by Pearson. Published by Pearson. ISBN: 9780321694645 (Required) (F15) (Sp16) (F16) (Sp17) (Students please go to class first and see if you need to purchase these material for your class)
Intro Stats-W/CD+ACCESS CODE (Looseleaf), 4th by DeVeaux. Published by Prentice Hall. ISBN: 9780321891242 (Optional) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17)
(Students please go to class first and see if you need to purchase these material for your class)
Intro Stats-W/DVD+ACCESS CODE, 4th by DeVeaux. Published by Prentice Hall. ISBN: 9780321869852 (Optional) (Sp15) (F15) (Sp16) (F16) (Sp17)
(Students please go to class first and see if you need to purchase these material for your class) (Current)(F18) (Sp19) (F19) (Sp20) (F20) (Sp21) (F21) (Sp22) (F22) (F23) (Sp24) (F24) (Sp25) (F24) (Sp25)(F25)
Statistics: Principles and Methods, by R. A. Johnson & G. K. Bhattacharyya. Published by J. Wiley. ISBN: 9780470560044 (Bundle Set) (New 6th Edition 2010) (F10) (Sp11) (F11) (Sp12) (F12) (Sp13) (F13) (Sp14) (F14)
MINITAB Release 14 (Student Edition).(On Campus Now: Math Bldg (Room 0203), PG-2 and Across Campus in WAM Labs) (F06) (Sp07)(F07)(Sp08) (Sum08)(F08)(Sp09)(Sum09) (F09)(Sp10)(F10)(Sp11)(F11)(Sp12)
Statistics: Principles and Methods, 5th Edition, by R.A. Johnson and G.K. Bhattacharyya. Published by J. Wiley, 1996. ISBN: 0470054611 (Bundle Set) (F06) (Sp07) (F07)(Sp08) (Sum08)(F08) (Sp09)(Sum09) (F09)(Sp10)
STAT 400
Course Name:Applied Probability and Statistics I
Text(s):
STAT 400 + ENHANCED WEBASSIGN(IP),(New Custom 9th Edition - Fall 2015) by J. Devore. Published by Cengage Learning. ISBN: 9781305764477 (Current) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F21) (Sp22) (F22) (F23) (Sp24) (F24) (Sp25)(F25)
Probability and Statistics for Engineering and the Sciences, by J. Devore. Published by Cengage Learning Publishing.ISBN: 9781305763029 (New Custom 9th Edition - Fall 2015) (F15)
Probability and Statistics for Engineering and the Sciences, by J. Devore. Published by Cengage Learning Publishing.ISBN: 9781285105123(New Custom 8th Edition - Fall 2011) (F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15)
Probability and Statistics for Engineering and the Sciences, NEW CUSTOM 7TH EDITION, by J. Devore. Published by Cengage Learning Publishing. ISBN: 1111521387 (F10) (Sp11)
Probability and Statistics for Engineering and the Sciences, 7th Edition,by J. Devore. Published by Thomson Publishing. ISBN: 0495557447(F06) (Sp07) (F07)(Sp08) (Sum08) (F08)(Sp09) (Sum09) (F09)(Sp10)
STAT 401
Course Name:Applied Probability and Statistics II
Text(s):
Probability and Statistics for Engineering and the Sciences,(New 9th Edition - Fall 2015)by J. Devore. Published by Cengage Learning Publishing.ISBN: 9781305251809 (Current)(F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F21) (Sp22) (F22) (F23) (Sp24) (F24) (Sp25)(F25)
Probability and Statistics for Engineering and the Sciences, by J. Devore. Published by Cengage Learning Publishing.ISBN: 9780538733526 (New 8th Edition - Fall 2011) (F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15)
Probability and Statistics for Engineering and the Sciences, 7th Edition, by J. Devore. Published by Thomson Publishing. ISBN: 0495557447 (F06) (Sp07) (F07) (Sp08) (F08) (Sp09) (Sp10) (F10) (Sp11)
STAT 405
Course Name:Stochastic Models for Queues and Networks
Text(s):
Probabilistic Stochastic Processes and Queuing Theory, by Nelson. Published by Springer Verlag
STAT 410
Course Name:Introduction to Probability Theory
Text(s):
Introduction to Probability, by David F. Anderson, Benedek Valko. Published by Cambridge University Press. ISBN: 9781108415859 (Current)(Sp24)(F25)
A First Course in Probability, by Sheldon Ross. Published by Prentice Hall.ISBN: 9780134753119 (Current) (New 10th Edition - Spring 2019) (Sp19) (F19) (Sp20) (F21) (Sp22) (F22) (Sp24) (F24) (Sp25)(F25)
A First Course in Probability, by Sheldon Ross. Published by Prentice Hall.ISBN: 9780321794772 (New 9th Edition - Fall 2013) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19)
A First Course in Probability,8th Edition, by Sheldon Ross. Published by Prentice Hall.ISBN: 978-0136033134(F06) (Sp07)(F07)(Sp08) (F08)(Sp09) (F09)(Sp10) (F10) (Sp11)(F11)(Sp12) (F12) (Sp13)
STAT 420/SURV 420
Course Name:Introduction to Statistics
Text(s):
Statistical Inference, 2nd Edition, by George Casella, Roger Berger. Published by Cengenge Learning. ISBN: 9788131503942 (Current) (F22)
Introduction to Mathematical Statistics, (New 8th Edition ) by R. Hogg & A.T. Craig. Published by Prentice Hall.ISBN: 9780134686998 (Current)(F21) (Sp22) (F22) (Sp24) (F24) (Sp25)(F25)
Introduction to Mathematical Statistics, (New 7th Edition - Fall 2013) by R. Hogg & A.T. Craig. Published by Prentice Hall.ISBN: 9780321795434 (Sp06) (Sp07) (Sp08) (Sp09) (Sp10) (Sp11)(Sp12) (Sp13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20)
STAT 426
Course Name:Introduction to Data Science and Machine Learning
Text(s):
An Introduction to Statistical Learning: with Applications in R(2nd Edition online) by G. James, D. Witten, T. Hastie, R. Tibshirani. Published by Springer-Verlag. ISBN: 9781071614204 (Current) (F22) (Sp25)
Professor using his notes (F18) (F19) (Sp20) (F21) (Sp24)(F25)
STAT 430
Course Name:Introduction to Statistical Computation with SAS
Text(s):
Applied Statistics and the SAS Programming Language,(5th Edition)by R.P. Cody and J.K. Smith. Published by Prentice-Hall.ISBN: 9780131465329 (Current)(F06) (F07) (Sp08) (F08) (Sp09)(F09)(Sp10) (F10) (Sp11) (F11)(Sp12) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19) (Sp20) (F21) (Sp22) (F22) (Sp24) (F24) (Sp25)(F25)
Applied Regression Analysis, by N. Draper and H. Smith. Published by Wiley.ISBN: 9780471170822 (OPTIONAL)(F09)(Sp12) (F16)
The Little SAS Book: A Primer, by L. Delwiche & S. Slaughter. Published by SAS Publishing.ISBN: 9781612903439 (OPTIONAL)(Current) (Sp12) (F16) (Sp17) (F17) (SP18) (F18) (Sp19) (F19) (Sp20) (F21) (Sp22) (F22) (Sp24) (F24) (Sp25)(F25)
Professor using his notes(F25)
STAT 440/SURV 440
Course Name:Sampling Theory
Text(s):
Sampling, by Thompson. (3rd Edition) Published by Wiley. ISBN: 9780470402313 (Current) (F13) (Sp17) (F17) (Sp18) (F18) (Sp19) (Sp20) (F21) (Sp22) (F22) (Sp24) (F24) (Sp25)(F25)
Sampling Statistics, by Fuller. Published by Wiley. ISBN: 97804700454602 (F10) (F15) (Sp16)
Model Assisted Survey Sampling, by C. Sarndal, B. Swensson, J. Wretman. Published by Springer-Verlag. ISBN: 0387406204(F06)(F07)(F08)
Sampling: Design and Analysis, (2nd Edition) by S. L. Lohr. Published by Thomson Publishing. ISBN: 9780495105275 (Current) (F05) (F07) (F09) (F12) (F16) (F19) (F22)
STAT 450
Course Name:Regression and Variance Analysis(COURSE IS NOT BEING OFFERED ANY MORE)
Text(s):
Applied Regression Analysis, 3rd Edition, by Draper and Smith. Published by Wiley and Sons.
STAT 464
Course Name:Introduction to Biostatistics
Text(s):
Fundamentals of Biostatistics, by Bernard Rosner.(New 8th Edition - Spring 2020)Published by Cengage Learning. ISBN: 9781305268920 (Current) (Sp20) (F21) (Sp22) (F22) (Sp24) (F24) (Sp25)(F25)
Fundamentals of Biostatistics (with data disk), by Bernard Rosner. (New 7th Edition) Published by Cengage Learning. ISBN: 9780538733496 (F10) (F11) (F12) (Sp13) (F13) (Sp14) (F14) (Sp15) (F15) (Sp16) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (F19)
Fundamentals of Biostatistics (with data disk), 6th Edition, by Bernard Rosner. Published by Thomson Publishing. ISBN: 9780534418205 (F06) (F07) (F08) (F09)
STAT 470
Course Name:Actuarial Mathematics
Text(s):
Actuarial Mathematics for Life Contingent Risks, by Dickson. (2nd Edition) Published by Cambridge University Press.ISBN: 9781107044074 (Sp14) (F14) (Sp15) (Sp16) (Sp17) (Sp18) (F19) (F21)
Theory of Interest and Life Contingencies, etc., 3rd Edition, by M.M. Parmenter. Published by ACTEX Pub.ISBN: 9781566983334(Current) (Sp05) (Sp08) (Sp09) (Sp11)(Sp12) (Sp13) (F15) (F16) (F17) (Sp19) (Sp20) (F21) (Sp22) (F22) (Sp24) (F24)(Sp25)
Life Insurance Mathematics, 3rd Edition, by Hans Gerber. Published by Springer-Verlag. ISBN: 354062242X (Sp06)
Fundamentals of Acturial Math, by S. David Promislow. Published by Wiley. ISBN: 9780470016893 (Sp08)
AMSC 420 SeeMATH 420
AMSC 452 SeeMATH 452
AMSC 460/CMSC 460
Course Name:Computational Methods
Text(s):
Professor is using notes:(F22)
Numerical Analysis by R. Burden, J. Faires, A. Burden. Published by Cengage Learning. ISBN: 9781305253667 (Current) (F19) (Sp20) (F21) (Sp22) (Sp24) (F24) (Sp25)
First Course In Numerical Methods, by Uri M. Ascher, Chen Greif. Published by SIAM. ISBN: 9780898719970 (Sp19) (F19) (F21)
Introduction to Numerical Analysis, by Suli. (1st Edition) Published by Cambridge University Press. ISBN: 9780521007948 (Sp11) (F11) (F12) (Sp13) (F13) (Sp14) (Sp15) (F15) (Sp17) (Sp18)
Introduction to Scientific Computing, 2nd Edition, by C.F. van Loan, Published by Prentice Hall.ISBN: 9780139491573 (Sp19)
Numerical Methods: Design, Analysis and Computer Implementation of Algorithms by A. Greenbaum & T. Chartier. Published by Princeton University Press. ISBN: 9780691151229 (F12) (F14)
(Sp05) (Sp08) (Sp11)(Sp12)
Numerical Computing with Matlab, by Cleve B. Moler. Published by SIAM.ISBN: 9780898716603 (Sp05) (Sp08) (Sum08) (Sp09) (Sum09) (F09)(Sp12) (F12) (F14) (F16) (Sp17) (F17) (Sp18) (F18) (Sp19) (Sp20)
Numerical Mathematics & Computing, by W. Cheney & D. Kincaid. Published by Cengage Learning. ISBN: 978-0495114758 (Sp09) (F09)
AMSC 466/CMSC 466
Course Name:Introduction to Numerical Analysis
Text(s):
Numerical Analysis by Timothy Sauer. 3rd Edition.Published by Pearson.ISBN:9780134696454 ( Current)(Sp25)(F25)
Professor is using notes: (F22)
Numerical Analysis by W. Gautschi. Published by Springer-Verlag. ISBN: 9780817682583 (F19)
Numerical Mathematics, by Quarteroni. Published by Springer-Verlag. ISBN: 9783540346586 (OPTIONAL) (F16)
Numerical Computing with Matlab, by Cleve B. Moler, Published by SIAM.ISBN: 9780898716603 (Revised Edition) (Sp15) (F15)
An Introduction to Numerical Analysis, (1st Edition) by E. Suli & D. Mayers. Published by Cambridge University Press.ISBN: 9780521007948 (OPTIONAL) (F05) (F10) (F11)(Sp12) (F12) (Sp14) (Sp15) (F15) (F16) (Sp17) (F18) (Sp19) (F19) (Sp24) (F24) (Sp25)
Afternotes on Numerical Analysis, by G.W. Stewart. Published by SIAM.ISBN: 97808988713626 (Sp08) (Sp09) (F09) (Sp11) (Sp13) (F13) (F16) (F17) (Sp18) (F21)
Elementary Numerical Analysis, 3rd Edition, by Atkinson-Han. Published by Wiley. ISBN: 0471433373 (F04) (F11)
Introduction to Numerical Analysis, 2nd Edition, by K.E. Atkinson. Published by J. Wiley. ISBN: 0471624896 (Current) (F06) (Sp08) (F09) (Sp11)
Fundamentals of Numerical Computing, by Shampine, Allen Pruess. Published by Wiley. ISBN: 0471163635 (F07)
Numerical Methods, by G. Dahlquist, A. Bjorck. Published by Dover. ISBN: 0486428079 (F06)
AMSC 477
Course Name:Optimization
Text(s):
Optimization in Operations Research, by R. Rardin. Prentice Hall, 1998.
Archives: F2011-S2012F2012-S2013F2013-S2014F2014-S2015F2015-S2016F2016-S2017F2017-S2018F2018-S2019F2019-S2020F2020-S2021F2021-S2022F2022-S2023F2023-S2024F2024-S2025
Magic+Math conference activitiesWhen: Wed, December 31, 1969 - 7:00pm
Where:
The Unitary Dual#### Speaker: Jeffrey Adams (University of Maryland) -
When: Wed, September 17, 2025 - 3:15pm
Where: Kirwan Hall 3206
View Abstract ### View Abstract
Abstract: Classifying the irreducible unitary representations of a Lie group (the Unitary Dual) is a major problem with a long history. It is well known that the answer is complicated. The Atlas of Lie Groups and Representations project was started in 2002 with the goal of understanding the unitary dual using computational tools. In this talk I will describe a major result of this project: a description of the unitary dual. The answer is in terms of an algorithm which we have implemented in the atlas software, and recently used to compute the answer for E8. I will also discuss parallel efforts to understand the unitary dual conceptually, based on recent progress on Arthur's conjectures. These conjectures by Jim Arthur were first announced at a conference at the University of Maryland in 1983. This is joint work of the atlas project, whose members are listed at www.liegroups.org.
Common divisors of binomial coefficients and invariable generation of finite simple groups#### Speaker: John Shareshian (Washington University ) -
When: Wed, September 24, 2025 - 3:15pm
Where: Kirwan Hall 3206
View Abstract ### View Abstract
Abstract: I will discuss joint work with Russ Woodroofe and joint work with Bob Guralnick and Woodroofe. It follows quickly from a theorem of Kummer that, given an integer n>1, the greatest common divisor of the nontrivial binomial coefficeints {{n} \choose {k}}, k=1,2,...,n-1, is larger than 1 if and only if n is a prime power. We considered the problem of finding the smallest number of primes such that each of these binomial coefficients is divisible by at least one of the chosen primes.
This problem is closely related to the problem of invariable generation of alternating groups by Sylow subgroups - if there exist primes p and r such that A_n is generated by subgroups P and R whenever P is a Sylow p-subgroup and R is a Sylow r-subgroup, then one of p,r divides every nontrivial binomial coefficient.
This leads to similar questions: given a simple group G, can one find a pair (x,y) of elements of G such that g^{-1}xg and h^{-1}yh generate G for all g,h? What if we put restrictions on the orders of x and y?
I will discuss our progress on these problems and some motivation for our work.
Artificial Intelligence: Thinking, fast and slow#### Speaker: Jim Yorke (University of Maryland) -
When: Wed, October 1, 2025 - 3:15pm
Where: Kirwan Hall 3206
View Abstract ### View Abstract
Abstract: I compare the ways humans and artificial intelligences think, building on Daniel Kahneman’s Systems 1 and 2 (Nobel Prize in Economics, 2002). System 1 is fast subconscious thought like recognizing a face, or walking, or other every-day tasks. System 2 is slow conscious analysis, like figuring out how I cross a mud puddle without soaking my shoes or searching through an audience to see if I know anyone. System 2 uses System 1 repeatedly. Talking and writing use System 2. I add a third component that I call System 3 — learning, improving both intuition (System 1) and analysis (System 2). All 3 components have explicit analogues in AIs. We can separate the roles of System 1 and 2 by comparing fast chess with slow chess. Fast chess allows 3 seconds per move which is essentially pure System 1 thought. Slow chess allows about 3 minutes per move, sixty times slower. Based on international chess ratings, I find that the best players at fast chess are the best at slow. Superb slow System 2 thinking is achieved only by chaining together superb intuitive System 1 insights. I give examples from mathematics and physics requiring System 1 intuition, including some from my new paper ``Tactics in Proofs", written with Boris Hasselblatt. I think AI needs the most improvement in its training System 2 for carrying out complex tasks, while humans may benefit greatly from improvement in the training of System 1, intuition.
Skipper (TBA)#### Speaker: Rachel Skipper (University of Utah) -
When: Wed, October 8, 2025 - 3:15pm
Where: Kirwan Hall 3206
View Abstract ### View Abstract
Abstract: TBA
Landscapes of harmonic maps (cancelled)#### Speaker: Antoine Song (Caltech) -
When: Wed, December 3, 2025 - 3:15pm
Where: Kirwan Hall 3206
View Abstract ### View Abstract
Abstract: Harmonic maps from a surface to a target manifold are nonlinear analogue of harmonic functions. They form a fundamental class of objects in differential geometry, but most of the time, they are very hard to describe explicitly. In recent years, people have started to study their shape under "typical", "large" or "random" constraints. In this talk, I will give a biased survey of the developments in this field, which connect geometric analysis to dynamical systems and random matrix theory.
Ioana (TBA)#### Speaker: Adrian Ioana (University of California, San Diego) -
When: Wed, February 4, 2026 - 3:15pm
Where: Kirwan Hall 3206
View Abstract ### View Abstract
Abstract: TBA
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Right derivative increasing implies convex
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Asked 4 years, 6 months ago
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I want to prove the following: If we have a function f f (not necessarily differentiable), but with right and left derivative f′±(x)≡lim y→x±f(y)−f(x)y−x
f′±(x)≡lim y→x±f(y)−f(x)y−x
existing at all x x in R R with f′+(x)≤f′+(y)f′+(x)≤f′+(y)∀x≤y∀x≤y, then f f is convex.
This is easily proven if f f is differentiable using the mean value theorem, but nothing like the same strategy (as far as I can see) works for this example. Any help would be massively appreciated!
calculus
derivatives
convex-analysis
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asked Mar 12, 2021 at 10:40
qp212223qp212223
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One can show that f(y)−f(x)y−x≤f′+(y)≤f(z)−f(y)z−y
f(y)−f(x)y−x≤f′+(y)≤f(z)−f(y)z−y
for x<y<z x<y<z, which implies that f f is convex.
For the right inequality, fix y y and consider the function h(z)=f(z)−f(y)−f′+(y)(z−y)
h(z)=f(z)−f(y)−f′+(y)(z−y)
for z≥y. Then h has a right derivative with h′+(z)=f′+(z)−f′+(y)≥0. It follows that h is non-decreasing, compare f:(0,1)→R countinous with non-negative right-hand derivative is non-decreasing. So h(z)≥h(y)=0, and that implies the desired estimate.
The proof of the left inequality works similarly.
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answered Mar 12, 2021 at 11:09
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1 This is just brilliant! Thank you.qp212223 –qp212223 2021-03-12 23:54:46 +00:00 Commented Mar 12, 2021 at 23:54
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Using mean value inequality
A version of the mean value inequality is:
Let f:[a,b]→R be a continuous, right (or left) differentiable map with m≤f′+(x)≤M for all x∈[a,b]. Then
m(y−x)≤f(y)−f(x)≤M(y−x) for all x,y∈[a,b].
You can see here for proof. See third approach where you'll realize that the proof can be adapted without major change for right differentiable maps only.
Based on this version of the mean value inequality, the result you're looking for follows in a straightforward way. And by the way, the mean value inequality is interesting by itself!
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edited Mar 12, 2021 at 13:07
answered Mar 12, 2021 at 11:23
mathcounterexamples.netmathcounterexamples.net
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5f:(0,1)→R countinous with non-negative right-hand derivative is non-decreasing
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14A real differentiable function is convex if and only if its derivative is monotonically increasing
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4if f convex, f has bounded first derivative iff f uniformly continuous
1Inequality between function and its derivative
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0The right derivative of a convex function f is right continuous ⟺f is differentiable
8The right derivative, f′+, of a convex function f is continuous ⟺f is differentiable.
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nature
eye
clinical study
article
Clinical Study
Published: 12 June 2009
Ocular manifestations of congenital toxoplasmosis
J Melamed1,2,
G U Eckert2,
V S Spadoni2,
E G Lago3&
…
F Uberti4
Show authors
Eyevolume 24,pages 528–534 (2010)Cite this article
9613 Accesses
59 Citations
3 Altmetric
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Abstract
Purpose
To evaluate the ocular manifestations of congenital toxoplasmosis at the first ophthalmological examination of children up to the age of 12 months.
Methods
Cross-sectional study of 44 children with a confirmed diagnosis of congenital toxoplasmosis. In all patients, complete ophthalmological examinations were performed under sedation. The patients underwent biomicroscopy of the anterior segment, skiascopy under cyclopegia, and indirect binocular ophthalmoscopy with maximum mydriasis.
Results
The mean age of patients was 4.2 months. Of the 44 children evaluated, 31 (70.4%) presented ocular involvement and 29 (65.9%) of them had retinochoroiditis lesions. The retinochoroiditis lesions were bilateral in 22 (75.8%) patients and unilateral in 7 (24.2%). The retinochoroiditis lesions were active in 8 (15.7%) eyes and had healed in 43 (84.3%). Most of the lesions were concentrated in the papillomacular area (76.3%). Other associated ocular alterations were present in 22 children, the most prevalent being cataract, microphthalmia, and strabismus.
Conclusion
Ocular involvement in congenital toxoplasmosis might be much more frequent and occurs earlier than previously described.
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Acknowledgements
This study was designed by JM, GUE, and VSS. This study was conducted by JM, VSS, EGL, and FU. The collection, management, analysis, and interpretation of data was done by JM, GUE, and EGL. The preparation, review, or approval of the paper was done by JM, GUE, VSS, EGL, FU.
Author information
Authors and Affiliations
Department of Ophthalmology and Otorhinolaryngology, Faculdade de Medicina da Universidade Federal do Rio Grande do Sul (UFRGS), Porto Alegre, Rio Grande do Sul, Brazil
J Melamed
Uveitis Section, Ophthalmology Service, Hospital de Clínicas de Porto Alegre (HCPA), Porto Alegre, Brazil
J Melamed,G U Eckert&V S Spadoni
Department of Pediatrics, Pontifícia Universidade Católica do Rio Grande do Sul, Porto Alegre, Rio Grande do Sul, Brazil
E G Lago
Department of Pediatrics, Faculdade de Medicina da Universidade Federal do Rio Grande do Sul, Porto Alegre, Rio Grande do Sul, Brazil
F Uberti
Authors
1. J MelamedView author publications Search author on:PubMedGoogle Scholar
2. G U EckertView author publications Search author on:PubMedGoogle Scholar
3. V S SpadoniView author publications Search author on:PubMedGoogle Scholar
4. E G LagoView author publications Search author on:PubMedGoogle Scholar
5. F UbertiView author publications Search author on:PubMedGoogle Scholar
Corresponding author
Correspondence to J Melamed.
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Melamed, J., Eckert, G., Spadoni, V. et al. Ocular manifestations of congenital toxoplasmosis. Eye24, 528–534 (2010).
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Received: 29 December 2008
Revised: 11 May 2009
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Published: 12 June 2009
Issue Date: April 2010
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Keywords
Toxoplasma gondii
congenital toxoplasmosis
ocular manifestations
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Remington JS, McLeod R, Thulliez P, Desmonts G . Toxoplasmosis. In: Remington JS, Klein JO, Wilson CB, Baker CJ (eds). Infectious Diseases of the Fetus and Newborn Infant. 6th ed. Elsevier-Saunders: Philadelphia, 2006, pp 948–1091.
Google Scholar
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3061 | https://brainly.com/question/22022607 | [FREE] Simplify the radical expression: -4\sqrt{160}. A. -4\sqrt{80} B. -4\sqrt{16} C. -16\sqrt{10} D. - brainly.com
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Simplify the radical expression: −4 160.
A. −4 80
B. −4 16
C. −16 10
D. 10
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c
Explanation
160 = 16 10 = 4 4 2 5 = 2 2 2 2 2 5
−4 160=−4 2∗2∗2∗2∗2∗5=−4∗2∗2 2∗5=−16 10
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The radical expression −4 160 simplifies to −16 10. The correct answer is option C. This simplification involves breaking down the number 160 into factors and extracting the square root of perfect squares.
Explanation
To simplify the radical expression −4 160, we start by breaking down the number inside the square root.
Factor 160 into its prime factors:
160=16×10=4×4×10=4×4×2×5.
Now we can simplify −4 160:
−4 160=−4 16×10
−4 16×10
−4×4 10=−16 10
Therefore, the expression simplifies to −16 10.
Thus, the correct answer is option C: −16 10.
Examples & Evidence
For example, simplifying −4 36 would yield −4×6=−24. This same method can be applied to other square roots as seen in the process above.
This solution is based on the properties of square roots and the ability to factor numbers into their prime components, supported by basic algebra.
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3062 | https://www.youtube.com/watch?v=J_iTmLNtudY | 5.3 Limiting Reactant Calculations | High School Chemistry
Chad's Prep
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Posted: 28 Oct 2020
Chad introduces limiting reactant calculations using a non-chemistry example involving bicycles. He then shows how to perform calculations involving mole-to-mole ratios from a balanced chemical equation. He concludes the lesson by working a complete limiting reactant calculation including all of the following:
1) How to determine the limiting reactant
2) How to calculate the theoretical yield
3) How to determine the amount of reagent in excess
4) How to perform percent yield calculations involving actual and theoretical yields
I've created a high school chemistry page that organizes all my videos by chapter - just an easier way for you to watch my YouTube videos. Check it out at
00:00 Lesson Introduction
00:48 Non-Chemistry Related Introduction to Limiting Reactant Calculations
04:01 Introduction to Calculations with Mole-to-Mole Ratios
10:12 Limiting Reactant Calculations (including Theoretical Yield and Reactant in Excess)
20:28 Percent Yield Calculations
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Transcript:
Lesson Introduction limiting reagent calculations this is kind of the meat and potatoes of the stoichiometry chapter here it's also kind of like the iconic high school chemistry lesson the one that kind of is most memorable for students coming away from their high school chemistry experience now we'll deal with limiting reagents or reactants we'll deal with the reagent in excess we'll deal with a theoretical yield so we'll also deal with like percent yield calculations all of these will be encompassed in this lesson now this is your first time joining me my name is chad and welcome to chad's prep where my goal is to make science both understandable and hopefully even enjoyable now this is my high school chemistry playlist i'll be posting these lessons weekly throughout the 2020-21 school year so if you don't want to miss one subscribe to the channel click the bell notifications you'll be notified every time i post a new lesson Non-Chemistry Related Introduction to Limiting Reactant Calculations all right so if you look at what is really at the basis of these limiting reagent calculations it's actually not difficult it's not difficult conceptually it's something you can see in your everyday life but the moment we start applying it to like molecules and compounds and things of a sort uh it gets complicated in a hurry and it's things that are a little bit abstract that you're not used to seeing and you're not used to visualizing atoms and molecules in your head and stuff so before we apply it there we're going to apply it to a little more tangible example here and we're going to talk about uh building bicycles so i i'm going to quit my job and i'm going to go to local junkyards and i'm going to collect bicycle tires and bicycle frames and i'm going to make recycled bicycles that's what i'm gonna do and every recycled bicycle requires two tires and one frame and so i've written it kind of like a chemical equation here so a couple questions i might ask you is well if i've got 10 frames how many tires do i need to convert all of the frames into bicycles okay and probably doing this in your head you didn't need this chemical equation to see that but i can see from the chemical equation that every one frame requires two tires to make the bicycle and so in this case if i've got 10 frames that 1 to 2 ratio tells me i'm going to need 20 tires so the way we'd actually formally do the dimensional analysis here is we'd put frames on bottom we'd put tires on top and we just use the coefficients as the conversion every one frame requires two tires just like so we put frames on bottom so they'd cancel tires on top because that's what we're trying to get to and what you probably already did in your head is just said i need 20 tires for 10 frames chad it's a bicycle sweet and i'm hoping it's that simple because once we make this something not involving a tangible everyday you know uh set of of objects it's going to be a little more challenging in chemical terms but hopefully you'll realize it's exactly analogous to this now what if i told you that i went to a bunch of local junkyards and my haul today came back with in my trailer was 20 frames and 30 tires and the question is how many bicycles can i actually assemble well in this case 20 frames that's enough to make 20 bikes okay but 30 tires that's only enough to make 15 bicycles and i can see that i'm going to run out of tires before i run out of frames and the one you run out of first we call that the limiting reactant or limiting reagent same diff totally synonymous and so in this case it's not the one you necessarily have less of because i can see i have less frames so it's the one that runs out first and the idea the reason it's not the one you run out you know that you have less of is because they don't always react in a one to one ratio it also depends on what ratio they actually react with so in this case again 20 frames that's enough to make 20 bikes because it's a one to one ratio but 30 tires that's only enough to make 15 bicycles because of the two to one ratio and so as a result we're gonna run out of tires first that's gonna be our limiting reagent or limiting reactant as the case may be and we're only gonna be able to make 15 bicycles now i'm not showing the formal calculation here because it's probably something that you can pretty much see in your head but just keep in mind because what we do with chemicals and compounds here in a little bit really will be totally analogous to this with one extra little wrinkle or larger wrinkle that will make it a little more challenging but at its heart the calculations will be very very similar all right so the Introduction to Calculations with Mole-to-Mole Ratios reaction we're going to deal with here is the creation of ammonia nh3 is the formula for ammonia and here we see that one mole of n2 reacts with three moles of h2 to produce two moles of nh3 so here we've got a balanced chemical reaction and my first question for you is how many moles of nitrogen do i have if you said 1 you're wrong my next question for you is how many moles of hydrogen h2 do i have if you said three moles you're also wrong and if i asked you how much money do i have in my wallet first correct answer you've had you said i don't know chad how would i know that's exactly the right answer it was also the same answer to the first two questions how should you know how much hydrogen i have or how much nitrogen i have what you've like been to my house and checked out you know i've got like tanks lying on the side of my house and you've measured how much nitrogen gas or hydrogen gas is in those tanks or something so the point is this when you see a balanced chemical reaction it never ever ever tells you what you have it only tells you the ratio in which things react for every mole of nitrogen you do have three times as many mole of hydrogen molecules will be required and it'll replace it will produce twice as many moles of ammonia so it's always just a ratio in which things react doesn't have anything to do with how much you actually have how much of anything you start with is extra information that's provided in addition to the balanced chemical equation so first question here in this section says how many moles of hydrogen gas are required for six moles of nitrogen gas to react completely so we're told that i have six moles of nitrogen gas i could have had any amount i've got six moles of nitrogen gas and the question is how many moles of hydrogen gas do i need for it to react completely and once again i can see that one mole of nitrogen requires three moles of hydrogen to react completely three times as many well three times as many would be 18 and again maybe you did that in your head but formally the process which you're going to want to get down because if the numbers aren't things you can do in your head and stuff or things that you can't visualize quite so easily as this then you're going to want to make sure you have that process down quickly so in this case i want to convert from moles of nitrogen to moles of hydrogen and this is a mole to mole ratio if you notice and we said a couple lessons ago that there's two places you get a multiple ratio one is from a chemical formula but the other that we use in the limiting range context here is the coefficients in a balanced reaction and the moles of hydrogen that's where the three is and the moles of nitrogen is the unwritten one it's a three to one ratio and so in this case we can see just as we already predicted that there are six moles of nitrogen are going to acquire 18 moles of hydrogen in order to react completely all right let's take a look at the second question here so how many holes of hydrogen gas required for 56 grams of nitrogen to react completely and so this is one step more challenging because instead of giving you the moles of nitrogen like we did from the get-go in the last question i'm starting you off with grams and the problem is that again the coefficients those aren't a gram to gram ratio those are a mole to mole ratio so if i give you the grams the very first thing you should do is get it into moles because the ratios you've got are all mole mole ratios not gram to gram all right so in this case 56 grams first thing we'll do is convert it to moles and the relationship between grams and moles is always the molar mass right off the periodic table so in this case we'll put grams of nitrogen on bottom moles of nitrogen on top and your molar mass is always for one mole and if you look at the periodic table be careful this is diatomic it's not just n which would be 14 grams it's n2 which is going to be 28 grams per mole of n2 so and i rounded out to a whole number because nitrogen is really close to 14 so i round it to 14 and 2 times 14 is 28 and 56 divided by 28 we see that we've got exactly 2 moles of n2 and once i see that i've got 2 moles of n2 a 1 to 3 ratio 2 moles of n2 would require 6 moles of hydrogen to react completely so but rather than stopping and calculating and then converting it further you can actually make it part of one big calculation and so right now we've got moles of nitrogen but we really want to convert this into moles of hydrogen so we'll put moles of nitrogen on bottom so that they cancel we'll put moles of hydrogen on top and once again that's our mole to mole ratio that three to one ratio right from the coefficients in the balanced reaction and 56 over 28 is two moles of nitrogen and that three to one ratio tells me that it's going to require six moles of hydrogen cool this is the wrinkle that can potentially make this just a little bit tougher because everything in the balanced chemical reaction it's a multiple ratio and as long as i'm giving you moles and asking about moles this is no more difficult than the bicycle example we did earlier but because we'll commonly give you instead of the moles we'll give you the grams of a substance and instead of asking you about the moles of a substance we might even ask you about grams as well and so we might have had one more step at the end here if i wanted to know how many grams of hydrogen i needed well then i'd have to convert this to grams so and it turns out it'd be 12 grams because one mole of h2 weighs 2 grams so then 6 moles would weigh 12 grams it would have been one more step in the calculation and it would have been a classic what we call grams to moles to moles to grams type calculation all right so let's make this a little bit harder now now we've kind of got how this works but essentially you can be asked anything about any ratio i can relate any two species in the reaction to each other so one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia and so once again let's say i told you had 10 moles of n2 how many moles of hydrogen would you need for it to react completely 30 1 to 3 ratio is a 10 to 30 ratio how many moles of ammonia could that 10 moles of n2 produce well based on the 1 2 ratio the maximum it could boost would be 20 moles 10 to 20 is the same as a one to two ratio again you can compare any two species and use the coefficients from the balanced equation as your multiple ratio Limiting Reactant Calculations (including Theoretical Yield and Reactant in Excess) all right so now we're going to do limiting reagent calculations proper and your limiting range of calculations we're going to give you an amount of more than one reactant in this case both reactants in this reaction and you're going to figure out who the limiting reactant is and what's your theoretical yield and things of this sort so in this case i'm giving you that you're combining 112 grams of nitrogen and 30 grams of hydrogen together and they're gonna react to completion and the first question is who's the limiting reagent okay well you figure that out it's based on this one to three ratio so but that's a one to three ratio in moles and you've got grams and so the first thing you want to do is convert these both into moles to see which one you don't have enough of to match that one to three ratio that the reaction calls for and so first thing we'll do here is we'll take our grams and nitrogen bottom to cancel convert that to moles of nitrogen on top and one mole weighs 28 grams molar mass right off the periodic table same thing for hydrogen we'll put grams of hydrogen on bottom moles of hydrogen on top and one mole of h2 and one of the most common mistakes students make when with a diatomic is they get that molar mass and they just use it right off periodic table instead of doubling it so this wouldn't be one gram it's going to be two grams for h2 those will cancel and we'll get moles in both cases so the top one 112 divided by 28 turns out is exactly equal to 4 moles of nitrogen so and 30 divided by 2 is equal to 15 moles of hydrogen so things are looking a little bit different here so a lot of students start this off they're like oh my gosh we have way more grams of nitrogen than grams of hydrogen so the hydrogen's definitely got to be the limiting reactant well it doesn't work that way so one by the time you convert them to moles they got very different molar masses and it's not about who has more grams it's partly about who's got more moles but also what ratio do they react and all those come into play and so don't just look at whichever one you have the most grams of from the get-go and be like he's not the limiting reactant it's the one we have less of because it doesn't work that way necessarily it could it could not and if you've got a good chemistry teacher who wants to make sure you really know your stuff they're going to give you an example just like i have here where if you make that assumption you're going to be wrong so be very careful here now how do we actually figure out who's the limiting reagent here now that we know how many moles of each of these we actually have well there's a few different ways so my favorite is this though is to look at your balanced reaction and i'll kind of explain it before we just kind of make it part of a process here so in this case every time we do this reaction it takes one mole of nitrogen well i've got four moles of nitrogen so my question for you is how many times can we actually do the reaction well four times every time i do it it takes one mole of nitrogen i've got four moles so i can do it four times cool now i've got 15 moles of of hydrogen gas here so my question for you is then if every time i do the reaction as written it requires three moles of hydrogen well then how many times can i do the reaction based on that amount of hydrogen well in this case five times so and so i have enough nitrogen to the reaction four times i have enough hydrogen to do the reaction five times so the question is can i do it four times or can i do it five times well i can only do it four times so it's kind of like if i own a construction company and i tell you that i have enough bricks to make a thousand homes with my construction company but i have enough cement to make two homes well can i make a thousand homes or can i make two homes i can only make two homes and i'm out of cement i might have way more bricks and we'd call that the reagent in excess but once i've made two homes i'm out of cement that's my limiting reactant i'm out i'm done i can't make any more houses until i get some more cement cool works the same way here i have enough n2 to do the action four times i have enough h due to the reaction five times well then i can only do it four times and n2 is my limiting reagent and the way i actually calculate this is part of a process and again this is not the only way a lot of people would look at this and like it's a one to three ratio so look at your one to three ratio and you'd see that you're a little light a one to three ratio would be a four to twelve ratio and you've got more than enough hydrogen and that's a valid way there's nothing wrong with that so but students mix that up sometimes and my favorite way that students mess up the least is just simply take the number of moles of each of these and divide them by their coefficient so i'm going to take this one and divide by one this one divide by three just their coefficient from the balanced reaction whichever number comes out smaller is your limiting reactant 4 divided by 1 is 4. 15 divided by 3 is 5 and so this is my limiting reactant and notice when you divide by the coefficient it effectively is telling you how many times you can do the reaction so now we know the nitrogen is the limiting reactant the answer to that first question cool and i just wrote that out real quick because that's part of my process but i don't actually use that as part of any future calculation if i'm going to continue on from here we're going to find out that these are the numbers i'm still going to work with not the numbers after i divided by the coefficient all right so part b and this says how much nh3 is formed okay if we look at how much nh3 is formed so you always want to start with the amount of product you're forming and technically this could be called the theoretical yield by the way from your limiting reactant though that's where you always start so there's my limiting reactant that's where we start and so in this case what you want to do is convert that to the moles of whatever you're converting to in this case i want moles of nh3 then you'll convert to moles of nh3 and you're converting from moles of n2 so that has to go on bottom and notice that's a mole to mole ratio and again in this context your multiple ratio always comes from the coefficients in the balanced reaction well the coefficient for ammonia is two the coefficient for n2 is one and we can see that if we stopped here that we would be able to make eight moles of ammonia now unfortunately my question on the handout only said how much nh3 is formed now if this were a multiple choice question on a multi-choice test you'd look and you look at all the answer choices and you'd see what units they're in are they in moles great then you'd stop here and it'd be 8 moles but what if all the answer choices were in grams well then you have to take this a step further because my question only says how much is formed i wasn't specific but you're more likely to see a question that either says how many moles are formed or how many grams of ammonia nh3 are formed and so in this case i want to take this all the way to grams so but i just wanted to i put it generic here on the sheet so that you'd realize there's two different ways it could be asked but here we're going to take it all the way to grams and so we're not done yet how do i get from moles of ammonia to grams well again connection between moles and grams is always your molar mass and so right off the periodic table nitrogen weighs 14 each of the three hydrogens weighs one and one mole therefore weighs 17 grams rounding the veneers whole number since they're all close to a whole number and so now we're going to have 4 times 2 is 8 and 8 times 17. and again you can use your calculator be a great time to do so or you can do it in your head so 8 times 17 8 times 10 is 80. 8 times 7 is 56 and 80 plus 56 is 136. and once again when you when you're calculating out how much product you can make we call that your theoretical yield now we call it theoretical yield for a reason so this is the greatest amount of that product you could make so but it turns out you almost never get that maximum possible amount so usually you'll find that when you start mixing reactants together that there are side reactions that can happen and so instead of doing this reaction maybe these combine to form some other product as well and because that will use up some of your reactants in making something other than your desired product well then you're not going to get up to 136 grams now and so theoretically that's the most i could make if everything goes perfect so no side reactions there's something else we'll talk about way later in this course called equilibrium which comes into play but this is the amount you get theoretically if everything goes perfect and that's why we call the theoretical yield there's also something called your actual yield which is if you actually go in the lab and you know mix 112 grams of n2 and 30 grams of h2 and then measure the mass of ammonia that's actually produced you actually did the reaction and that's how much actually got made so and we can compare that actual yield to that theoretical yield and see kind of how that goes but we'll get there in a minute because we'll be asking that shortly third question for this set is how much of the reagent in excess is left over and this is the hardest question in this section and i say it's the hardest it's not you know all that hard in any other context outside of chemistry but it will be hard here so if we take a look at this so the reagent in excess the one that's going to be left over is this guy and if you want to figure out how much of this is left over you start in the same place you started out for your theoretical yield so you start off with your limiting reagent and we figured out right here that we got four moles of that limiting reagent and so from here you start with that four moles and instead of figuring out how much product you're making you're going to take that four moles of n2 and figure out how much h2 gets used up at the same time well again based on that one to three ratio and being that the numbers are nice four moles of n2 would use up three times as many moles of hydrogen based on that one to three ratio and that's 12 moles of h2 and so in this case we can see that 12 moles of h2 get used so we started with 15 moles of h2 in our container 12 of them get used and the question is how much is left over and you just take a number you start with subtract off the number that got used and you can see that you've got three moles of h2 left over or we might say in excess so it just doesn't react with anything because by the time you know you've used up this 12 moles of h2 there's no n2 left in the container cool again that's the hardest of the calculations on here but it really just has come down to this process so again start with your limiting reactant calculate out how much of that reagent gets used and then just subtract it from the amount you started with Percent Yield Calculations all right last question this section starts off with the same setup we just had 112 grams of n2 reacts with 30.0 grams of h2 so and it produces like it actually produces you went in the lab and actually did this it produces 102 grams of ammonia and the question is what is the percent yield so and i want you to think about this in one other context real quick let's say i gave you money i know you like this analogy so i'm going to give you enough money to go to the store and buy me 10 oranges i need 10 oranges so i give you enough money to get 10 oranges and you come back and you give me eight oranges and i'm like hey i gave you enough money to get 10 oranges and you're like i'm like where's my other two and you're like that's the tax for me going and shopping for you chad i'm like okay fair enough so my question though is what is the percent yield of oranges well i gave you enough to get ten you only came back with eight eight out of ten would be an eighty percent yield same thing here theoretically we could get as much as a hundred 136 grams of ammonia nh3 but in this case we only actually got 102 grams and so the percent yield is the actual over theoretical times a hundred that is your percent yield calculation here and so in our case we actually got 102 grams of ammonia theoretically we could have got a maximum of 136 grams and this comes out to a 75 percent yield cool these are all the questions relating to limiting reagents that you might see now one thing you might notice i used really nice numbers i used numbers where we could do a lot of the math in our heads so and things of this sort and this is a great great place to start but now you're going to want to get some practice and you definitely want to get some practice that's going to deal with not so nice numbers because instead of starting with like 112 grams of n2 i could have been like we start with 1.376 times 10 to the negative 4 grams of n2 and so you want to learn this process with nice numbers but then you definitely want to work a number of examples with not so nice numbers to make sure that you really really really know this process and have it down pat now if you've benefited from this lesson consider giving me a like and a share pretty much a couple of the best things you can do to support the channel and if you've got questions feel free to leave them in the comments section below if you're looking for the study guide that goes with this lesson or if you're looking for a bunch of practice problems on limiting reagent calculations check out my premium course on chadsprep.com |
3063 | https://www.arxiv.org/pdf/2411.09926 | Completing partial k-star designs
Ajani De Vas Gunasekara ∗ Daniel Horsley †
Abstract
A k-star is a complete bipartite graph K1,k . A partial k-star design of order n is a pair (V, A) where V is a set of n vertices and A is a set of edge-disjoint k-stars whose vertex sets are subsets of V . If each edge of the complete graph with vertex set V is in some star in A, then ( V, A) is a (complete) k-star design . We say that ( V, A) is completable if there is a k-star design ( V, B) such that A ⊆ B . In this paper we determine, for all k and
n, the minimum number of stars in an uncompletable partial k-star design of order n.
1 Introduction
For many classes of combinatorial design it is the case that all “small” partial designs can be completed to a full design, but that this is not the case for larger partial designs. For any such class of designs it is natural to try to determine the size threshold at which uncompletable partial designs first appear. Latin squares give one famous example of this. Smetaniuk and Anderson and Hilton independently proved a conjecture of Evans that, for all n ⩾ 2, the smallest number of filled cells in an uncompletable n×n latin square is exactly n. Investigations in this spirit have been undertaken for many objects including block designs , Hadamard Matrices , latin hypercubes , and matchings in graphs [17, Chapter 6]. In this paper we give a solution to the problem for k-star designs. It seems this specific problem for k-star designs has not previously been considered in the literature, but related problems concerning embeddings of k-star designs have been addressed in [7, 10, 11, 12]. Essentially these papers show that every partial k-star design can be completed after some number of new vertices are added to it. For general k, the upper bound on this number of new vertices was successively improved in , and . The special case of k = 3 was dealt with in . Let k ⩾ 2 be an integer. A k-star is a copy of K1,k , the complete bipartite graph with parts of size 1 and k. A k-star decomposition of a graph G is a set of k-stars whose edge sets partition
E(G). For a set V , let KV denote the complete graph with vertex set V . A partial k-star design of order n is a pair ( V, A) where V is a set of n vertices and A is a k-star decomposition of some subgraph G of KV . If G = KV , then ( V, A) is a k-star design . We say a positive integer n
is k-admissible if n
2
≡ 0 (mod k). Tarsi and Yamamoto et al. independently proved that, for n ⩾ 2, a k-star design of order n exists if and only if n ⩾ 2k and n is k-admissible. A completion of a partial k-star design ( V, A) is a k-star design ( V, B) such that A ⊆ B . Our main result is as follows.
∗School of Arts and Sciences, The University of Notre Dame Australia, NSW 2007, Australia ajani.de.vas.gunasekara@nd.edu.au
†School of Mathematics, Monash University, Victoria 3800, Australia danhorsley@gmail.com
1
arXiv:2411.09926v2 [math.CO] 18 Aug 2025
Theorem 1. Let k ⩾ 2 be an integer. For each k-admissible integer n such that n ⩾ 2k, any partial k-star design of order n with at most u(n, k ) stars has a completion, where
u(n, k ) =
(2⌊n−2
k
⌋ − 1 if n̸ ≡ 1 (mod k)
2( n−1)
k
− 2 if n ≡ 1 (mod k).Furthermore, for each k-admissible integer n > 1, there is a partial k-star design of order n
with u(n, k ) + 1 stars that has no completion.
In Section 2 we will exhibit uncompletable partial k-star designs to prove the ‘furthermore’ claim in Theorem 1. Some preliminaries and our overall approach to finding completions of partial k-star designs are given in Section 3. Then we will find the required completions for partial k-star designs of k-admissible orders less than or equal to 3 k + 1. For orders larger than 3k + 1, our approach relies on defining a certain function from the vertices of the partial design to the nonenegative integers. We define these functions in Section 5 and show that they possess certain desirable properties. In Section 6 we then use these functions to complete the proof of Theorem 1. A possible direction for future work is discussed in Section 7
2 Uncompletable partial designs
In this section we will exhibit the uncompletable partial k-star designs we require for Theorem 1. We begin by introducing some more definitions and notation that we will use throughout the paper. In a k-star with k ⩾ 2, the vertex of degree k is the centre and each other vertex is a leaf .Let G be a graph. We denote the complement of G by G. For a subset S of V (G), we denote the subgraph of G induced by S by G[S]. For a given k-star decomposition D of G, we can define a function c : V (G) → Z⩾0 called the central function , where c(x) is the number of
k-stars of D whose centre is x for each x ∈ V (G). The leftover L of a partial k-star design (V, A) is the graph with vertex set V in which each edge of KV is present if and only if it is not an edge of a star in A. Of course, a partial k-star design is completable if and only if there is a k-star decomposition of its leftover. Note that L is the graph with vertex set V whose edges are exactly the edges in k-stars in A.As we mentioned above, the orders for which a k-star design exists have been completely determined.
Theorem 2 ([14, 16]) . A k-star design of order n exists if and only if n is k-admissible and
n = 1 or n ⩾ 2k.
So, of course, any partial k-star design whose order is not k-admissible or is in {2, . . . , 2k−1}
must be uncompletable. In Lemma 3(i) and (ii) below, we establish the tightness of the bound in Theorem 1.
Lemma 3. Let k ⩾ 2 be an integer.
(i) For all k-admissible integers n > 1 such that n̸ ≡ 1 (mod k) there is a partial k-star design of order n with 2⌊n−2
k
⌋ stars that is not completable.
(ii) For all k-admissible integers n > 1 such that n ≡ 1 (mod k) there is a partial k-star design of order n with 2n−2
k
− 1 stars that is not completable.
2Proof . We first prove (i). Let ( V, A) be a partial k-star design of order n with exactly 2 ⌊n−2
k
⌋
stars and central function c such that there exist distinct vertices x1, x 2 ∈ V for which c(x1) =
c(x2) = ⌊n−2
k
⌋ and x1 and x2 are adjacent in the leftover L of ( V, A). Since n̸ ≡ 1 (mod k), we have n−kk ⩽ ⌊n−2
k
⌋ ⩽ n−2
k
and hence 1 ⩽ deg L(xi) ⩽ k − 1 for each i ∈ { 1, 2}. So, in a k-star decomposition of L, no star can be centred at x1 or x2 and hence no star can contain the edge
x1x2. So ( V, A) cannot be completed. We now prove (ii). Let ( V, A) be a partial k-star design of order n with exactly 2n−2
k
− 1stars and central function c such that there exist distinct vertices x1, x 2, x 3 ∈ V for which
c(x1) = c(x2) = n−k−1
k
, c(x3) = 1, x1 and x2 are leaves of the star centred at x3, and x1 and
x2 are adjacent in the leftover L of ( V, A). For each i ∈ { 1, 2}, this implies deg L(xi) = k − 1because xi is a leaf of one star and centre of n−k−1
k
stars. So, in a k-star decomposition of L, no star can be centred at x1 or x2 and hence no star can contain the edge x1x2. So ( V, A) cannot be completed.
3 Strategy and preliminary results
Given Lemma 3, to prove Theorem 1 it remains to show that, for any k ⩾ 2 and k-admissible integer n, every partial k-star design with at most u(n, k ) stars has a completion. We begin by observing that we can always add a star to a partial k-star design of order n with fewer than
u(n, k ) stars.
Lemma 4. Let k and n be integers such that k ⩾ 2 and n ⩾ 2k, and let (V, A′) be a partial
k-star design of order n with fewer than u(n, k ) stars. The leftover L′ of (V, A′) contains a
k-star.
Proof . Note that k(u(n, k )−1) < 2n−2k and hence |E(L′)| > n
2
−ku (n, k ) > 12 n(n−5)+2 k.Hence there is a vertex y of L′ such that deg L′ (y) > n − 5 + 4kn ⩾ 2k − 3 ⩾ k − 1where the second inequality follows because n ⩾ 2k. So, since deg L′ (y) is an integer, it is at least k.Given any partial k-star design ( V, A′) of order n with fewer than u(n, k ) stars, we can use Lemma 4 to add stars to produce a partial k-star design ( V, A) with exactly u(n, k ) stars. Any completion of ( V, A) is also a completion of ( V, A′). Thus, to prove Theorem 1 it in fact suffices to find completions for just those partial k-star designs of k-admissible orders n ⩾ 2k that have exactly u(n, k ) stars. We will now dispense with the case of partial 2-star decompositions. The graphs admitting a 2-star decomposition have been completely characterised.
Theorem 5 () . A 2-star decomposition of a connected graph G exists if and only if |E(G)| ≡
0 (mod 2) .
The following lemma completes the proof of Theorem 1 for k = 2. Note that u(n, 2) = n − 3for all integers n > 1.
Lemma 6. For all 2-admissible integers n > 1, any partial 2-star design of order n with n − 3
stars is completable.
3Proof . Note that n ⩾ 4 since n is 2-admissible. Let ( V, A) be a partial 2-star design of order n such that |A| = n − 3 and let L be the leftover of ( V, A). Since n is 2-admissible,
|E(L)| ≡ 0 (mod 2). We will show that L has only one nontrivial component and hence complete the proof by Theorem 5. Suppose for a contradiction that L has at least two nontrivial components and let w be the number of vertices in a smallest nontrivial component. Then 2 ⩽ w ⩽ n
2
and |E(L)| ⩾ w(n − w) ⩾ 2( n − 2). This contradicts the fact that |E(L)| = 2( n − 3) since |A| = n − 3. So, for the remainder of the paper we will concentrate on completing partial k-star designs where k ⩾ 3. Throughout we will often write a k-admissible integer n as ak + b where a is a nonnegative integer and b ∈ { 1, . . . , k }. For each k ⩾ 3 we have that ak + 2 is not k-admissible, and hence that b̸ = 2. We use this fact tacitly in the remainder of the paper. When n is written in this fashion we have the following equivalent definition of u(n, k ), which is often more convenient.
u(ak + b, k ) =
(2a − 2 if b = 1 2a − 1 if b ∈ { 3, . . . , k }. (1) We will take advantage of one further reduction of the problem. Call a partial k-star design (V, A) of order n reducible if n ≡ 1 (mod k), |A| = u(n, k ), and there is a vertex which is the centre of at least one star in A and is not a leaf of any star in A. The following lemma shows that it suffices to find completions only for non-reducible partial k-star designs.
Lemma 7. Let k ⩾ 2 be an integer. Suppose that, for each k-admissible integer n ⩾ 2k, every non-reducible partial k-star design of order n with at most u(n, k ) stars is completable. Then, for each k-admissible integer n ⩾ 2k, every partial k-star design of order n with at most u(n, k )
stars is completable.
Proof . Let ( V, A) be a reducible partial k-star design and say |V | = ak + 1 where a ⩾ 2 is an integer. Then A has u(ak + 1 , k ) = 2 a − 2 stars. Let L be the leftover of ( V, A) and let
x be a vertex which is the centre of at least one star in A and is not a leaf of any star in A.Then deg L(x) ≡ 0 (mod k). Let Ax be the set of stars in A that are centred at x and let A′
x
be a set of 1
k
deg L(x) edge-disjoint stars in L, all centred on x. Then ( V \ { x}, A \ A x) is a partial k-star design of order ak with at most u(ak, k ) = 2 a − 3 stars. Now ( V \ { x}, A \ A x) is non-reducible since its order is ak and so, by our hypotheses, it has a completion ( V \ { x}, A′). Then ( V, A′ ∪ A x ∪ A ′
x
) is a completion of ( V, A). Throughout the paper, for any function f that assigns real values to vertices in a set V
and for any subset T of V , we abbreviate P
x∈T
f (x) to f (T ). Let k ⩾ 2 be an integer. For any graph G such that |E(G)| ≡ 0 (mod k) we define a k-precentral function for G to be a function p : V (G) → Z⩾0 such that p(V (G)) = 1
k
|E(G)|. We say a k-precentral function for G
is realisable if it is the central function of some k-star decomposition D of G.Our approach to finding the completions necessary to prove Theorem 1 is based on a key fact which we state as Lemma 8 below. This fact is an easy consequence of a result of Tarsi [15, Theorem 2] and can also be obtained as a specialisation of results in or concerning star decompositions of multigraphs.
Lemma 8. Let p be a k-precentral function for a graph G. Then p is realisable if and only if, for every nonempty proper subset T of V (G), we have ∆+
T
⩾ ∆−
T
where ∆+
T
is the number of edges of G that are incident with at least one vertex in T and ∆−
T
= kp (T ).
4We use the notation ∆ +
T
and ∆ −
T
extensively in the remainder of the paper. We also define ∆T = ∆ +
T
− ∆−
T
so that ∆ T is nonnegative if and only if ∆ +
T
⩾ ∆−
T
. Note that all of this notation is implicitly dependent on a k-precentral function p that will always be clear from context. Through the rest of the paper our approach to showing that a partial k-star design (V, A) has a completion will be to define a k-precentral function p for its leftover L and then use Lemma 8 to show that p is realisable.
4 Small orders
In this section we complete the proof of Theorem 1 for n ∈ { 2k, . . . , 3k + 1 }. Lemmas 9, 10 and 11 below will, respectively, handle the cases where n ∈ { 2k, 2k + 1 }, where 2 k + 1 < n ⩽ 3k,and where n = 3 k + 1.
Lemma 9. Let k ⩾ 3 be an integer and let n be a k-admissible integer such that n ∈ { 2k, 2k+1 }.Any non-reducible partial k-star design of order n with u(n, k ) stars is completable.
Proof . Let ( V, A) be a partial k-star design of order n with u(n, k ) stars. If n = 2 k + 1, then
|A| = u(2 k + 1 , k ) = 2 and it follows that ( V, A) must be reducible. So, in fact, n = 2 k and
|A| = u(2 k, k ) = 1. By Theorem 2, there is a k-star design of order n. By relabelling the vertices of such a design, we can obtain a k-star design ( V, B) such that A ⊆ B .
Lemma 10. Let k ⩾ 3 be an integer and let n be a k-admissible integer such that 2k + 1 <n ⩽ 3k. Any partial k-star design of order n with three stars is completable.
Proof . Let b be the integer such that n = 2 k + b and note that b ∈ { 3, . . . , k } since n is
k-admissible. Let ( V, A) be a partial k-star design of order n with |A| = 3, let L be the leftover of ( V, A), and let c be the central function of A. Let C = {x ∈ V : c(x) ⩾ 1} and note that
|C| ∈ { 2, 3}. Let h = 0 if |C| = 3 and h = 1 if |C| = 2. Let S be a subset of V \ C such that
|S| = b + h + b(b−1) 2k − 4 and deg L(y) ⩾ deg L(x) for each y ∈ S and x ∈ V \ (C ∪ S). Note that S
is well-defined because b + h + b(b−1) 2k − 4 ∈ { 0, . . . , n − 3} since 3 ⩽ b ⩽ k and n is k-admissible. We define a k-precentral function p by setting p(x) = 2 − c(x) for each x ∈ C, p(x) = 2 for each
x ∈ S, and p(x) = 1 for each x ∈ V \ (C ∪ S). Note that p is indeed a k-precentral function because p(C) = |C| − h, p(V \ C) = 2 k + b − | C| + |S| and hence
p(V ) = 2 k + b − h + |S| = 1
k
2k + b
2
− 3 = |E(L)|
k
.
Let T be a nonempty proper subset of V and let t = |T |. By Lemma 8 it suffices to show that ∆T ⩾ 0. We consider two cases depending on the value of t.
Case 1. Suppose t ⩾ b. Then, since p(T ) ⩽ t + |S|, we have ∆+
T
⩾ t
2
+ t(2 k + b − t) − 3k, ∆−
T
⩽ k(t + |S|) = k(t + b + h − 4) + b
2
and hence ∆ T ⩾ 12 (t + 1 − b)(2 k + b − t) − hk . Noting that h ⩽ 1 and b ⩽ t ⩽ 2k + b − 1, we have ∆ T ⩾ 0.
Case 2. Suppose t ⩽ b − 1. There are 3 − c(T ) stars in A centred on vertices in V \ T and any such star can have at most t leaves in T . Also, for each x ∈ T , we have p(x) ⩽ 2 − c(x). So, ∆+
T
⩾ t
2
+ t(2 k + b − t) − kc (T ) − (3 − c(T )) t, ∆−
T
⩽ k(2 t − c(T )) 5and hence ∆T ⩾ 12 t(2 b + 2 c(T ) − 7 − t). (2) We can assume that the right side of (2) is negative for otherwise ∆ T ⩾ 0 and we are done. In particular, we have t ⩾ 2b − 6 since c(T ) ⩾ 0. Since t ⩽ b − 1, this implies
b ∈ { 3, 4, 5}. Because 2 k + b is k-admissible we have b(b − 1) ≡ 0 (mod 2 k) and hence (b, k ) ∈ { (3 , 3) , (4 , 6) , (5 , 5) , (5 , 10) }. The following table details, in each of these subcases, the value of |S| and the possible values of ( c(T ), t ) that make the right side of (2) negative.
b k |S| possible ( c(T ), t )3 3 h (0 , 1), (0 , 2), (1 , 2) 4 6 1 + h (0 , 2), (0 , 3) 5 5 3 + h (0 , 4) 5 10 2 + h (0 , 4) We claim there must be at least |S| vertices in V \ C whose degree in L is at most 1. In each subcase this follows from the pigeonhole principle using |V \ C| = 2 k + b − 3 + h,
P
x∈V\C
deg L(x) ⩽ 3k, the value of |S| given in the table above and h ⩽ 1. Thus, by our definition of S we have deg L(x) ⩽ 1 for each x ∈ S. Further, deg L(x) ⩽ 3 for each x ∈ V \ C.So, if c(T ) = 0, then ∆+
T
⩾ t
2
+ t(2 k + b − t) − | S ∩ T | − 3( t − | S ∩ T |), ∆−
T
= k(t + |S ∩ T |)and hence ∆T ⩾ 12 t(2 k + 2 b − t − 7) − (k − 2) |S ∩ T | ⩾ 12 t(2 b − t − 3) ⩾ 12 t(b − 2) > 0where the second inequality follows using |S ∩ T | ⩽ t and the third follows since t ⩽ b − 1. So we may assume c(T )̸ = 0 and hence, from the table, that ( b, k, c (T ), t ) = (3 , 3, 1, 2). Then deg L(x) ⩽ 5 for the unique x ∈ C ∩ T , deg L(x) ⩽ 1 for each x ∈ S, and deg L(x) ⩽ 3 for each
x ∈ T \ (C ∪ S). So ∆+
T
⩾ 15 − 5 − | S ∩ T | − 3(1 − | S ∩ T |), ∆−
T
= 3(2 + |S ∩ T |)and hence ∆ T ⩾ 1 − | S ∩ T | ⩾ 0 since |S ∩ T | ⩽ 1. Finally in this section we deal with the case n = 3 k + 1. Note that 3 k + 1 is k-admissible if and only if k is odd.
Lemma 11. Let k ⩾ 3 be an odd integer. Any non-reducible partial k-star design of order
3k + 1 with four stars is completable.
Proof . Let ( V, A) be a non-reducible partial k-star design of order 3 k +1 with |A| = 4, let L be the leftover of ( V, A), let c be the central function of A, and let C = {x ∈ V : c(x) ⩾ 1}. Since (V, A) is non-reducible, |E(L[C]) | ⩾ |C| and hence |C| ∈ { 3, 4}. Let h = 0 if |C| = 4 and h = 1 if |C| = 3. Let S be a subset of V \ C such that |S| = h + 12 (3 k − 7) and deg L(y) ⩾ deg L(x) for each y ∈ S and x ∈ V (S∪C). Note that S is well-defined because h+ 12 (3 k−7) ∈ { 0, . . . , 3k−3}
since k is odd. We define a k-precentral function p by setting p(x) = 2 − c(x) for each x ∈ C,6p(x) = 2 for each x ∈ S, and p(x) = 1 for each x ∈ V \ (S ∪ C). Note that p is indeed a
k-precentral function because p(C) = |C| − h, p(V \ C) = 3 k + 1 − | C| + |S| and hence
p(V ) = 3 k + 1 − h + |S| = 32 (3 k + 1) − 4 = 1
k
|E(L)|.
Let T be a nonempty proper subset of V and let t = |T |. By Lemma 8 it suffices to show that ∆T ⩾ 0. We consider two cases depending on the value of t.
Case 1. Suppose t ⩾ k + 1. Then, since p(T ) ⩽ t + |S|, we have ∆+
T
⩾ t
2
+ t(3 k + 1 − t) − 4k, ∆−
T
⩽ k(t + |S|) = 12 k(3 k + 2 t + 2 h − 7) and hence ∆ T ⩾ 12 (t − k)(3 k + 1 − t) − hk . Thus, noting that h ⩽ 1 and k + 1 ⩽ t ⩽ 3k, we have ∆ T ⩾ 0.
Case 2. Suppose t ⩽ k. There are 4 − c(T ) stars in A centred on vertices in V \ T and any such star can have at most t leaves in T . Also, for each x ∈ T , we have p(x) ⩽ 2 − c(x). So ∆+
T
⩾ t
2
+ t(3 k + 1 − t) − kc (T ) − (4 − c(T )) t, ∆−
T
⩽ k(2 t − c(T )) and hence ∆T ⩾ 12 t(2 k + 2 c(T ) − 7 − t). (3) We can assume that the right side of (3) is negative for otherwise ∆ T ⩾ 0 and we are done. In particular, we have t ⩾ 2k − 6 since c(T ) ⩾ 0. Since t ⩽ k and k is odd, this implies k ∈ { 3, 5}.The following table details, in both of these subcases, the value of |S| and the possible values of ( c(T ), t ) that make the right side of (3) negative.
k |S| possible ( c(T ), t )3 1 + h (0 , 1), (0 , 2), (0 , 3), (1 , 2), (1 , 3) 5 4 + h (0 , 4), (0 , 5) We claim there must be at least |S| vertices in V \ C whose degree in L is at most 1. In both subcases this follows from the pigeonhole principle using |V \C| = 3 k−3+ h, P
x∈V\C
deg L(x) = 4k − | E(L[C]) | ⩽ 4k − 4 + h (recall |E(L[C]) | ⩾ |C|), the value of |S| given in the table above, and h ⩽ 1. Thus, by our definition of S, we have deg L(x) ⩽ 1 for each x ∈ S. Further, deg L(x) ⩽ 4 for each x ∈ V \ C. So, if c(T ) = 0, then ∆+
T
⩾ t
2
+ t(3 k + 1 − t) − | S ∩ T | − 4( t − | S ∩ T |), ∆−
T
= k(t + |S ∩ T |)and hence ∆T ⩾ 12 t(4 k − t − 7) − (k − 3) |S ∩ T | ⩾ 12 t(2 k − t − 1) ⩾ 12 t(k − 1) > 0where the second inequality follows using |S ∩ T | ⩽ t and the third follows since t ⩽ k. So we may assume c(T )̸ = 0 and hence, from the table, that ( k, c (T )) = (3 , 1) and t ∈ { 2, 3}. Then deg L(x) ⩽ 6 for the unique x ∈ C ∩ T , deg L(x) ⩽ 1 for each x ∈ S, and deg L(x) ⩽ 4 for each
x ∈ T \ (C ∪ S). So ∆+
T
⩾ t
2
+ t(10 − t) − 6 − | S ∩ T | − 4( t − 1 − | S ∩ T |), ∆−
T
= 3( t + |S ∩ T |)7and hence ∆ T ⩾ 12 (t − 1)(4 − t) > 0 since t ∈ { 2, 3}.
5 Suitable precentral functions
For the leftovers of partial k-star designs of orders greater than 3 k + 1, we will define the k-precentral functions that we aim to realise in a more elaborate way. This section is devoted to explaining how these k-precentral functions are defined and proving that they possess certain useful properties. First, however, we require a lemma which tells us that, in the leftover of a partial k-star design of order n with u(n, k ) stars, not too many vertices can have small degree.
Lemma 12. Let k ⩾ 3 and n > 3k + 1 be integers such that n is k-admissible, let (V, A) be a non-reducible partial k-star design of order n with at most u(n, k ) stars, and let L be the leftover of (V, A). Then
(i) at most one vertex of L has degree at most k,
(ii) if two adjacent vertices of L have degree less than 2k, then every other vertex has degree at least 2k.
Proof . Let u = u(n, k ). Let n = ak + b where a ⩾ 3 is an integer and b ∈ { 1} ∪ { 3, . . . , k },and let c be the central function of A. We first prove (i). Let y and z be distinct vertices in
V . We can suppose without loss of generality that c(y) ⩽ c(z) and that, if c(y) = c(z), then no star centred at z has y as a leaf. It suffices to show that deg L(y) ⩾ k + 1. Let ℓ = 1 if a star centred at z has y as a leaf and ℓ = 0 otherwise. Now u − c(y) − c(z) stars in A are centred on vertices in V \ { y, z } and hence deg L(y) ⩽ kc (y) + u − c(y) − c(z) + ℓ ⩽ u + ( k − 2) c(y) (4) where the inequality follows because, from our supposition without loss of generality, either
c(z) ⩾ c(y) + 1 or c(y) = c(z) and ℓ = 0. If b = 1, then u = 2 a − 2, at least three vertices must have stars centred at them because ( V, A) is not reducible, and hence c(y) ⩽ a − 2. If
b ∈ { 3, . . . , k }, then u = 2 a − 1 and hence c(y) ⩽ a − 1. Thus, from (4), deg L(y) ⩽
(k(a − 2) + 2 if b = 1
k(a − 1) + 1 if b ∈ { 3, . . . , k }.Now deg L(y) = ak + b − 1 − deg L(y). So deg L(y) ⩾ 2k − 2 ⩾ k + 1 if b = 1 and deg L(y) ⩾
k + b − 2 ⩾ k + 1 if b ∈ { 3, . . . , k }.Now we prove (ii). Let S be a subset of V (L) such that |S| = 3 and L[S] is nonempty. To complete the proof it suffices to show that some vertex in S has degree at least 2 k. For each
x ∈ S, let ℓx ∈ { 0, 1, 2} denote the number of stars that are centred at a vertex in S and have x
as a leaf. Let y be a vertex in S such that c(y) ⩽ c(x) for each x ∈ S and ℓy ⩽ ℓz for each z ∈ S′
where S′ = {x ∈ S : c(x) = c(y)}. We will show that deg L(y) ⩾ 2k. Let m = c(S) − 3c(y) and notice that, by the definition of y, m is nonnegative, |S′| = 3 if m = 0, and |S′| = 2 if m = 1. Hence, because L[S] is nonempty, ( m, ℓ y) /∈ { (0 , 1) , (0 , 2) , (1 , 2) } by the definition of y and so 8m ⩾ ℓy. Thus, deg L(y) ⩾ ak + b − 1 − ℓy − kc (y) − u − c(S) = ak + b − 1 − ℓy − (k − 3) c(y) − u + m
⩾ ak + b − 1 − (k − 3) u
3
− u. (5) where the equality follows because c(S) = 3 c(y) + m, and the final inequality uses m ⩾ ℓy and
c(y) ⩽ ⌊u
3
⌋. Using ⌊u
3
⌋ ⩽ u
3
and (1), we have deg L(y) ⩾ 13 (a + 2) k if b = 1 and deg L(y) ⩾
13
(a + 1) k + b − 1 if b ∈ { 3, . . . , k }. Hence deg L(y) ⩾ 2k when a ⩾ 5. For values of a less than 5, we consider the lower bound given by (5) in a number of cases. case u lower bound on deg L(y)
a = 3, b ∈ { 3, . . . , k } 5 2k + b − 3
a = 4, b = 1 6 2ka = 4, b ∈ { 3, . . . , k } 7 2k + b − 2So deg L(y) ⩾ 2k in each case. We introduce some notation and definitions relating to k-precentral functions. Let G be a graph and let p be a k-precentral function for G. We say that p is proportional if p(x) ∈{⌊ 12k deg G(x)⌋, ⌈ 12k deg G(x)⌉} for each x ∈ V (G). Note that we can always find a proportional
k-precentral function p for G by first choosing a set S ⊆ V (G) such that 1
k
|E(G)| = |S| +
P
x∈V(G)
⌊ 12k deg G(x)⌋ and then setting p(x) = ⌊ 12k deg G(x)⌋ for x ∈ V (G) \ S and p(x) =
⌊ 12k deg G(x)⌋ + 1 for x ∈ S. We define p∗(x) = p(x) − 12k deg G(x) for each x ∈ V (G). Intuitively, we can think of p∗(x) as the ‘rounding’ that has been applied to 12k deg G(x) to obtain p(x). The definitions of k-precentral function and proportional immediately give some basic properties of
p∗ which we use often and encapsulate in the following lemma.
Lemma 13. Let G be a graph and let p be a k-precentral function for G.
(i) p∗(V (G)) = 0 .
(ii) If p is proportional then, for each x ∈ V (G), p∗(x) ∈ { −2k+1 2k , −2k+2 2k , . . . , 2k−12k }.
We further define a proportional k-central function p for G as minimal if, among all the proportional k-central functions for G, p has a minimum value of P
x∈V(G)
|p∗(x)|. The following simple lemma will be useful.
Lemma 14. If G is a graph and m is a minimal k-precentral function on G, then m∗(y) −
m∗(z) ⩽ 1 for all y, z ∈ V (G).
Proof . Suppose otherwise that there are vertices y and z of G such that m∗(y) − m∗(z) > 1. By Lemma 13(ii) it must be the case that m∗(y) is positive and hence m(y) ⩾ 1, and that
m∗(z) is negative. Define k-precentral function p by p(y) = m(y) − 1, p(z) = m(z) + 1 and
p(x) = m(x) for each x ∈ V (G) \ { y, z }. Then p∗(y) = m∗(y) − 1, p∗(z) = m∗(z) + 1 and
p∗(x) = m∗(x) for each x ∈ V (G) \ { y, z }. So p is proportional because −1 < p ∗(x) < 1 for each x ∈ V (G). Furthermore,
|p∗(y)| + |p∗(z)| = 1 − m∗(y) + m∗(z) + 1 = 2 − m∗(y) − m∗(z) < 1whereas |m∗(y)| + |m∗(z)| = m∗(y) − m∗(z) > 1. Thus P
x∈V(G)
|p∗(x)| < P
x∈V(G)
|m∗(x)|,contradicting the fact that m is minimal. 9Sometimes a minimal k-precentral function can have an obvious ‘flaw’ that prevents it from being realisable. Let m be a minimal k-precentral function for a graph G with vertex set V .We say a vertex y ∈ V is bad under m if deg G(y) < k and m(y) = 1, and we say an edge
y1y2 ∈ E(G) is bad under m if m(y1) = m(y2) = 0. When one of these flaws arises, we can repair it by modifying the function slightly. We say a k-precentral function s for G is suitable
if one of the following holds. (i) s = m for a minimal k-precentral function m under which no vertex or edge is bad. (ii) s is obtained from a minimal k-precentral function m for G under which y ∈ V is bad by setting s(y) = m(y) − 1 = 0, s(z) = m(z) + 1 for some vertex z such that m∗(z) = min {m∗(x) : x ∈ V }, and s(x) = m(x) for each x ∈ V \ { y, z }. In this case we say m is
obtained from s by repairing vertex y. Note z̸ = y by Lemma 13(i) since m∗(y) is positive. (iii) s is obtained from a minimal k-precentral function m for G under which an edge y1y2 ∈
E(G) with deg G(y1) ⩽ deg G(y2) is bad by setting s(y2) = m(y2) + 1 = 1, s(z) = m(z) − 1for some vertex z ∈ V such that m∗(z) = max {m∗(x) : x ∈ V }, and s(x) = m(x) for each
x ∈ V \ { y2, z }. In this case we say m is obtained from s by repairing edge y1y2 at y2. Note that deg G(y2) ⩾ 1, so m∗(y2) is negative and hence m∗(z) is positive by Lemma 13(i) and
m(z) ⩾ 1. From this definition it is clear that, for a given graph G with |E(G)| ≡ 0 (mod k), we can always obtain a suitable k-precentral function for G by taking a minimal k-precentral function for G and, if necessary, modifying it appropriately. However, it seems possible that a suitable
k-precentral function could still be bad at some vertex or edge. The next lemma shows that this cannot happen and also that suitable k-precentral functions are always proportional.
Lemma 15. Let k ⩾ 3 and n > 3k + 1 be integers, and let L be the leftover of a partial k-star design (V, A) of order n with u(n, k ) stars. If s is a suitable k-precentral function on L, then s
is proportional and no vertex or edge of L is bad under s.
Proof . If s = m for a minimal k-precentral function m for L under which no vertex or edge is bad, then the result holds trivially, so we may suppose otherwise. We will make repeated use of the fact that a proportional k-precentral function does not assign 0 to any vertex of degree at least 2 k.Suppose that s is obtained from a minimal k-precentral function m for L by repairing vertex
y. Then s(y) = m(y)−1 = 0, s(z) = m(z)+1 for some vertex z such that m∗(z) = min {m∗(x) :
x ∈ V }, and s(x) = m(x) for each x ∈ V \ { y, z }. It can be seen that s is proportional because
m is proportional, m∗(y) > 0 and, by Lemma 13(i), m∗(z) < 0. By Lemma 12(i), for each
x ∈ V \ { y} we have that deg L(x) ⩾ k + 1 and hence s is not bad on any vertex. We claim that s(x) ⩾ 1 for each x ∈ V \ { y} and hence that s is not bad on any edge. If deg L(x) ⩾ 2k,this follows because s is proportional and, if k + 1 ⩽ deg L(x) < 2k, then s(x) ⩾ m(x) = 1 since otherwise we would have m∗(x) < −12 and m∗(y) > 12 contradicting Lemma 14. Suppose that s is obtained from a minimal k-precentral function m for L by repairing edge
y1y2 at y2. Then s(y2) = m(y2) + 1 = 1, s(z) = m(z) − 1 for some vertex z ∈ V such that
m∗(z) = max {m∗(x) : x ∈ V }, and s(x) = m(x) for each x ∈ V \ { y2, z }. It can be seen that s
is proportional because m is proportional, m∗(y2) < 0 and, by Lemma 13(i), m∗(z) > 0. Now deg L(y1) ⩽ deg L(y2) < 2k and hence, by Lemma 12, deg L(y2) ⩾ k + 1 and deg L(x) ⩾ 2k for 10 each x ∈ V (L) \ { y1, y 2}. Noting s(y1) = m(y1) = 0, it follows that s is not bad at any vertex of L. Further, since s is proportional, s(x) ⩾ 1 for each x ∈ V \ { y1} and so s is not bad on any edge of L.Finally in this section, we establish upper bounds on the sum of the values of a suitable
k-precentral function over a subset of the vertices in a graph. This will be vital to showing that our suitable k-precentral functions are in fact realisable.
Lemma 16. Let p be a k-precentral function for a n-vertex graph G on vertex set V , let T be a nonempty proper subset of V , and let t = |T |.
(i) If p is minimal, then p∗(T ) ⩽ t(n−t)
n
.
(ii) If p is suitable, then
p∗(T ) ⩽
(t(2 n−2t−1) 2( n−1) if t < n
2(2 t−1)( n−t)2( n−1)
if t ⩾ n
2
.
Proof . Let V = V (G) and t = |T |. We first prove (i). Suppose for a contradiction that p is minimal and p∗(T ) > t(n−t)
n
. By Lemma 13(i), we also have p∗(V \ T ) < −t(n−t)
n
. But then there must exist y ∈ T with p∗(y) > n−tn and z ∈ V \ T with p∗(z) < − tn , contradicting Lemma 14. We now prove (ii). We do not retain any notation from our proof of (i). Suppose that p
is suitable. If p is minimal, then the result follows by (i), noting that t(n−t)
n
< t(2 n−2t−1) 2( n−1) when 0 < t < n
2
and t(n−t)
n
⩽ (2 t−1)( n−t)2( n−1) when n
2
⩽ t < n . (To see this, note that t(2 n−2t−1) 2( n−1) − t(n−t)
n
=
t(n−2t)2n(n−1)
and (2 t−1)( n−t)2( n−1) − t(n−t)
n
= (n−t)(2 t−n)2n(n−1) .) So we may assume that p is not minimal. Thus, by the definition of suitable, we must be in one of the following two cases.
Case 1. Suppose that p is obtained from a minimal k-precentral function m for G by repairing vertex y. Then deg G(y) ⩽ k − 1, p(y) = m(y) − 1 = 0, p(z) = m(z) + 1 for some vertex z such that m∗(z) = min {m∗(x) : x ∈ V }, and m(x) = p(x) for each x ∈ V (G) \ { y, z }. Let ℓ = m∗(z)and note that −1 < ℓ < 0 by Lemma 13(i) since m∗(y) is positive and that p∗(z) = ℓ + 1. Since m is minimal, by Lemma 14 we have ℓ ⩽ m∗(x) ⩽ ℓ + 1 for each x ∈ V . Thus
p∗(x) ⩽ ℓ + 1 for each x ∈ V and, in particular,
p∗(T ) ⩽ t(ℓ + 1) . (6) Further, p∗(y) = m∗(y) − 1 ⩽ ℓ and p∗(x) ⩾ ℓ for each x ∈ V \ { y}. Thus, p∗(V \ T ) ⩾
(n − t − 1) ℓ + p∗(y) and hence, using Lemma 13(i),
p∗(T ) ⩽ −(n − t − 1) ℓ − p∗(y). (7) Now,
p∗(T ) ⩽ t(n − t − 1 − p∗(y)) (n − 1) < t(2 n − 2t − 1) 2( n − 1)
where the first inequality follows from (6) if ℓ ⩽ − 1
n−1
(t + p∗(y)) and from (7) otherwise, and the second follows using p∗(y) = − 12k deg G(y) > −12 . This completes the proof because
t(2 n−2t−1) 2( n−1)
⩽ (2 t−1)( n−t)2( n−1) when t ⩾ n
2
.
Case 2. Suppose that p is obtained from a minimal k-precentral function m for G by repairing edge y1y2 at y2. Then p(y2) = m(y2) + 1 = 1, p(z) = m(z) − 1 for some vertex z ∈ V
such that m∗(z) = max {m∗(x) : x ∈ V }, and p(x) = m(x) for each x ∈ V \ { y2, z }. Note 11 deg L(y1) ⩽ deg L(y2) and hence, by Lemma 12(i), deg L(y2) ⩾ k + 1. Let g = m∗(z) and note that 0 < g < 1 by Lemma 13 since m∗(y2) is negative and that p∗(z) = g − 1. Since m is minimal, by Lemma 14 we have g − 1 ⩽ m∗(x) ⩽ g for each x ∈ V . Thus
p∗(x) ⩾ g − 1 for each x ∈ V , so p∗(V \ T ) ⩾ (n − t)( g − 1) and, using Lemma 13(i),
p∗(T ) ⩽ (n − t)(1 − g). (8) Further, p∗(y2) = m∗(y2) + 1 ⩾ g and p∗(x) ⩽ g for each x ∈ V \ { y2}. Thus,
p∗(T ) ⩽ (t − 1) g + p∗(y2). (9) Now,
p∗(T ) ⩽ (n − t)( t − 1 + p∗(y2)) (n − 1) < (2 t − 1)( n − t)2( n − 1)
where the first inequality follows from (8) if g ⩾ 1
n−1
(n − t − p∗(y2)) and from (9) otherwise, and the second follows using p∗(y2) = 1 − 12k deg G(y2) < 12 . This completes the proof because
(2 t−1)( n−t)2( n−1)
< t(2 n−2t−1) 2( n−1) when t < n
2
.
6 Realising suitable precentral functions
In this section we complete the proof of Theorem 1 by showing that a suitable k-precentral function on the leftover L of one of the relevant partial designs is realisable. In view of Lemma 8, we do this by taking a nonempty proper subset T of V (L) and showing that ∆ T ⩾ 0 by dividing into cases according to the size of T . Our next lemma covers the case when T is neither very large nor very small. We then deal with very large T in Lemma 18 and very small T in Lemma 19.
Lemma 17. Let k ⩾ 3 and n > 3k + 1 be integers such that n is k-admissible. Let (V, A) be a non-reducible partial k-star design of order n with u(n, k ) stars, let L be the leftover of (V, A),and let s be a suitable k-precentral function for L. If T is a subset of V (L) with 5 ⩽ |T | ⩽ n−5,then ∆T ⩾ 0.
Proof . Let u = u(n, k ), let t = |T |, let e be the number of edges incident with two vertices in T ,and let c be the number of edges incident with exactly one vertex in T . Then we have ∆ +
T
= e+c
and P
x∈T
deg L(x) = 2 e + c. Using these, together with the fact that s(x) = 12k deg L(x) + s∗(x)for each x ∈ V , we obtain ∆T = e + c − ks (T ) = e + c − 12
X
x∈T
deg L(x) + ks ∗(T )
!
= c
2
− ks ∗(T ). (10) Also,
c ⩾ t(n − t) − ku ⩾ t(n − t) − (2 n − k − 5) (11) where for the final inequality we used the definition of u and, in the case n ≡ 1 (mod k), k ⩾ 3. We consider two cases according to the value of t.
Case 1. Suppose that t ⩾ n
2
. Then s∗(T ) ⩽ (2 t−1)( n−t)2( n−1) by Lemma 16(ii). Using this and (11) 12 in (10), we obtain ∆T ⩾ c
2
− k(2 t − 1)( n − t)2( n − 1) ⩾ t(n − t) + 5 − 2n
2
− k
2
(2 t − 1)( n − t)
n−1
− 1
. (12) The coefficient of k in (12) is nonpositive since (2 t − 1)( n − t) ⩾ n − 1 using n
2
⩽ t ⩽ n − 5 and
n ⩾ 11. So we can substitute k < n−13 into (12) to obtain ∆T > (t − 4)( n − t − 5) 6 − 1 ⩾ −1.
Thus ∆ T is nonnegative, since it is an integer.
Case 2. Suppose that t < n
2
. Then s∗(T ) ⩽ t(2 n−2t−1) 2( n−1) by Lemma 16(ii). Using this and (11) in (10), we obtain ∆T ⩾ c
2
− tk (2 n − 2t − 1) 2( n − 1) ⩾ t(n − t) + 5 − 2n
2
− k
2
t(2 n − 2t − 1)
n−1
− 1
. (13) The coefficient of k in (13) is nonpositive since t(2 n − 2t − 1) ⩾ n − 1 using 5 ⩽ t < n
2
and
n ⩾ 11. So we can substitute k < n−13 into (13) and simplify to obtain ∆T > (t − 5)( n − t − 4) 6 − 1 ⩾ −1.
Thus ∆ T is nonnegative, since it is an integer.
Lemma 18. Let k ⩾ 3 and n > 3k + 1 be integers. Let (V, A) be a partial k-star design of order n with u(n, k ) stars, let L be its leftover, and let s be a suitable k-precentral function for
L. If T is a proper subset of V with |T | ⩾ n − 4, then ∆T ⩾ 0.
Proof . Let e = |E(L[V \ T ]) | and note e ⩽ 6 since |V \ T | ⩽ 4. We have ∆ +
T
= |E(L)| − e.Since ks (V ) = |E(L)|, we have ∆ −
T
= |E(L)| − ks (V \ T ). Thus ∆ T = ks (V \ T ) − e. This is obviously nonnegative if e = 0. If e ∈ { 1, 2, 3}, then s(V \ T ) ⩾ 1 because no edge in L[V \ T ]is bad under s, and hence ∆ T is nonnegative. If e ∈ { 4, 5, 6}, then L[V \ T ] must contain two nonadjacent edges and hence s(V \ T ) ⩾ 2 because no edge in L[V \ T ] is bad under s. So again ∆T is nonnegative.
Lemma 19. Let k ⩾ 3 and n > 3k + 1 be integers. Let (V, A) be a non-reducible partial k-star design of order n with u(n, k ) stars, let L be its leftover, and let s be a suitable k-precentral function for L. If T is a nonempty subset of V with |T | ⩽ 4, then ∆T ⩾ 0.
Proof . Let u = u(n, k ) and let c be the central function of A. Say n = ak + b where a ⩾ 3 and
b ∈ { 1} ∪ { 3, . . . , k }. Note that ( a, b )̸ = (3 , 1) since n > 3k + 1. Let t = |T | and e = |E(L[T ]) |.Note that a star centred at a vertex in V \ T contributes at most k toward P
x∈T
deg L(x) and a star centred at a vertex in T contributes k + i, where i is the number of leaves it has in T .Thus P
x∈T
deg L(x) ⩽ ku + t
2
− e and we have P
x∈T
deg L(x) ⩾ t(n − 1) − ku − t
2
+ e. So ∆+
T
⩾ t(n − 1) − ku − t
2
and hence ∆ T ⩾ t(n − 1) − ku − t
2
− ks (T ). We may assume that the right hand side of this inequality is negative for otherwise we are finished and hence, using
n = ak + b and (1) we have
t
2
(k(a(t − 2) + 2 − s(T )) if b = 1,
k(a(t − 2) + 1 − s(T )) + t(b − 1) if b ⩾ 3. (14) 13 For each x ∈ V , let δ(x) = deg L(x) − ks (x). Since s is proportional and no vertex is bad under s we have
δ(x) ⩾
deg L(x) if deg L(x) ∈ { 0, . . . , k − 1}
deg L(x) − k if deg L(x) ∈ { k, . . . , 2k}
deg L(x) − 2k if deg L(x) ∈ { 2k + 1 , . . . , 4k}
k + 1 if deg L(x) ⩾ 4k + 1. (15) In particular, note that δ(x) ⩾ 0, that δ(x) = 0 if and only if (deg L(x), s (x)) ∈ { (0 , 0) , (k, 1) },and that δ(x) = 1 if and only if (deg L(x), s (x)) ∈ { (1 , 0) , (k + 1 , 1) , (2 k + 1 , 2) }. Also, since ∆+
T
= ( P
x∈T
deg L(x)) − e, we have ∆ T = δ(T ) − e. So to complete the proof it suffices to assume δ(T ) ⩽ e − 1 and obtain a contradiction. We consider cases according to the value of t.
Case 1. Suppose that t ∈ { 1, 2}. If t = 1, then we immediately have the contradiction
δ(T ) ⩽ e − 1 = −1. If t = 2, then it must be that e = 1 and δ(x) = 0 for each x ∈ T , but this is impossible by Lemmas 12(i) and (15).
Case 2. Suppose that t = 3. Then e ∈ { 1, 2, 3}. In fact, by Lemmas 12(i) and (15), it must be that e = 3, (deg L(z), s (z)) ∈ { (0 , 0) , (k, 1) } for some z ∈ T , and (deg L(x), s (x)) ∈{(k + 1 , 1) , (2 k + 1 , 2) } for each x ∈ T \ { z}. So s(T ) ⩽ 5. Substituting this and t = 3 into (14), we have that k(a − 3) < 3 if b = 1 and k(a − 4) + 3 b < 6 if b ⩾ 3. The former immediately contradicts a ⩾ 4 and hence, from the latter, we must have a = 3 and 3 ⩽ b ⩽ 13 (k + 5). So
u = 5 and, since e = 3 and deg L(z) ∈ { 2k + b − 1, 3k + b − 1}, it must be the case that c(z) ⩾ 2and z is a leaf of at least two stars centred at vertices in V \ T . So, for some y ∈ T \ { z}, we have c(y) = 0 and that y is a leaf of at most three stars (note that y is not a leaf of any star centred at z because e = 3). This contradicts deg L(y) ∈ { k + b − 2, 2k + b − 2}.
Case 3. Suppose that t = 4. Then 0 ⩽ e ⩽ 6 By Lemma 12(i) we have δ(y) = 0 for at most one vertex y ∈ T . Hence it cannot be that δ(x) ⩾ 4 for any x ∈ T since δ(T ) ⩽ e − 1 ⩽ 5. So we may assume, for each x ∈ T , that δ(x) ⩽ 3 and hence that deg L(x) ⩽ 3k by (15). Since s is proportional, we have s(x) ⩽ 2 for each x ∈ T . Hence s(T ) ⩽ 8. Substituting t = 4 into (14), we have that k(2 a + 2 − s(T )) < 6 if b = 1 and k(2 a + 1 − s(T )) + 4 b < 10 if b ⩾ 3. In the former case we have an immediate contradiction to a ⩾ 4 and s(T ) ⩽ 8, and hence, from the latter, we must have a = 3, s(T ) = 8 and 3 ⩽ b ⩽ 14 (k + 9). So s(x) = 2 for each x ∈ T . Thus, for each x ∈ T , we have deg L(x) = 2 k + δ(x) and hence s∗(x) = 1 − 12k δ(x) or, equivalently,
δ(x) = 2 k − 2ks ∗(x). So, by Lemma 16,
δ(T ) = 8 k − 2k X
x∈T
s∗(x) ⩾ 8k − 2k · 4(2 n − 9) 2( n − 1) = 28 kn − 1 > 7 > e
where the second last inequality follows because n < 4k + 1 since a = 3. Theorem 1 now follows by combining the results we have obtained.
Proof of Theorem 1 . Fix n and k and let u = u(n, k ). By Lemma 3, there is an uncom-pletable partial k-star design of order n with u + 1 stars. So it suffices to show that any partial
k-star design of order n with at most u stars has a completion. In fact, by Lemma 7 it suffices to show that any non-reducible partial k-star design of order n with at most u stars has a completion. 14 Let ( V, A′) be a non-reducible partial k-star design of order n with at most u stars and let L′ be the leftover of ( V, A′). If necessary, use Lemma 4 to add stars to ( V, A′) to form a partial k-star design ( V, A) of order n with exactly u stars. It suffices to show that ( V, A) has a completion or, equivalently, to show there is k-star decomposition of its leftover L.If k = 2, then ( V, A) has a completion by Lemma 6. If k ⩾ 3 and n ⩽ 3k + 1, then ( V, A) is completable by Lemma 9, Lemma 10 or Lemma 11. So we may assume k ⩾ 3 and n > 3k + 1. Let L be the leftover of ( V, A), let s be a suitable k-precentral function for L, and let T be a nonempty proper subset of V . By Lemma 8, it suffices to show that ∆ T ⩾ 0. This follows from Lemmas 17, 18 and 19.
7 Conclusion
Our work above exactly determines the smallest number of stars in an uncompletable partial
k-star design for all k and n. One natural extension would be to investigate the same problem for k-star designs of index λ, that is, k-star decompositions of λ-fold complete multigraphs. Since Lemma 8 can be generalised to multigraphs, many of the techniques we use here for finding completions could also be applied for designs of higher indices. Uncompletable partial designs in this setting could be obtained by taking λ-fold copies of the examples we use to prove Lemma 3, but it is very possible that smaller uncompletable examples exist. One could also consider analagous problems for other classes of graph design. A natural target would be k-cycle systems.
Acknowledgments. The second author was supported by Australian Research Council grants DP220102212 and DP240101048. In the early stages of this work the first author was supported by a Postgraduate Publication Award from Monash University.
References
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9 (1980), 119–130. A. De Vas Gunasekara and D. Horsley, An Evans-style result for block designs, SIAM J. Discrete Math. 36 (2022), 47–63. 15 A. De Vas Gunasekara and D. Horsley, Smaller embeddings of partial k-star decomposi-tions, Electron. J. Combin. 30 (2023), Paper No. 1.19, 20 pp. T. Evans, Embedding incomplete Latin squares, Amer. Math. Monthly 67 (1960), 958–961. D. G. Hoffman, The real truth about star designs, Discrete Math. 284 (2004), 177–180. D. G. Hoffman and D. Roberts, Embedding partial k-star designs, J. Combin. Des. 22
(2014), 161–170. M. Noble and S. Nochumson, Embedding partial 3-star designs, Electron. J. Graph Theory Appl. 12 (2024), 289–295. M. Noble and S. N. Richardson, Balls, bins, and embeddings of partial k-star designs,
Discrete Math. 342 (2019), 111600, 4 pp. B. Smetaniuk, A new construction on Latin squares. I. A proof of the Evans conjecture,
Ars Combin. 11 (1981), 155–172. M. Tarsi, Decomposition of complete multigraphs into stars, Discrete Math. 26 (1979), 273–278. M. Tarsi, On the decomposition of a graph into stars, Discrete Math. 36 (1981), 299–304. S. Yamamoto, H. Ikeda, S. Shige-eda, K. Ushio and N. Hamada, On claw-decomposition of complete graphs and complete bigraphs, Hiroshima math. J. (1975), 33–42. Q.R. Yu and G. Liu, Graph factors and matching extensions, Springer (2010). 16 |
3064 | https://par.nsf.gov/servlets/purl/10354401 | New Bounds on the Size of Binary Codes with Large Minimum Distance
James (Chin-Jen) Pang, Hessam Mahdavifar, and S. Sandeep Pradhan
Department of Electrical Engineering and Computer Science, University of Michgan, Ann Arbor, MI 48109, USA Email: cjpang, hessam, pradhanv@umich.edu
Abstract —Let A(n, d ) denote the maximum number of code-words in a binary code of length n and minimum Hamming dis-tance d. Deriving upper and lower bounds on A(n, d ) has been a subject for extensive research in coding theory. In this paper, we examine upper and lower bounds on A(n, d ) in the high-minimum distance regime, in particular, when d = n/ 2 − Θ( √n).We will first provide a lower bound based on a cyclic construc-tion for codes of length n = 2 m − 1 and show that A(n, d =
n/ 2 − 2c−1√n) ⩾ nc, where c is an integer with 1 ⩽ c ⩽ m/ 2 − 1.With a Fourier-analytic view of Delsarte’s linear program, novel upper bounds on A(n, n/ 2 − √n) and A(n, n/ 2 − 2 √n) are ob-tained, and, to the best of the authors’ knowledge, are the first upper bounds scaling polynomially in n for the regime with d =
n/ 2 − Θ( √n).
I. I NTRODUCTION
Low-capacity scenarios are of increasingly greater impor-tance with the advent of the Internet of Things (IoT) and the next generation of mobile networks. In general, IoT devices op-erate under extreme power constraints and often need to com-municate at very low signal-to-noise ratio (SNR), e.g., −13
dB or 0.03 bits per transmission (in terms of channel capac-ity) in Narrow-Band-IoT protocols . In the standard, legacy turbo codes or convolutional codes at moderate rates together with many repetitions are adopted. This implies effective code rates as low as 1.6 × 10 −4 are supported in such protocols. It is expected, however, that repeating a moderate-rate code to enable low-rate communication will result in rate loss and suboptimal performance. As a result, studying ultra-low-rate error-correcting codes for reliable communications in such low-capacity regimes becomes necessary –. In this paper we focus on minimum distance properties of codes in the ultra-low-rate regime, which can be also described as the large minimum distance regime, to be specified later. Let C be a binary (n, M, d ) code of length n, size M , min-imum distance d = ( n − j)/2. The dimension of C is given by k = log 2 M , the rate by R = k/n , and relative distance by
δ = d/n . Given positive integers n and d, A(n, d ) denote the maximal value of M such that a (n, M, d ) code exists. A(n, d )
is a fundamental quantity in coding theory subject to extensive studies in the past. Lower bounds on A(n, d ) are usually ob-tained by constructions. For a survey on the known bounds with finite n and d, the reader is referred to and the websites , . For asymptotic lower bounds and a survey of prior asymp-totic results the reader is referred to , . In this paper,
This work was supported by the National Science Foundation under grants CCF–2132815 and CCF–1909771.
we focus on studying bounds on A(n, d ) in the high-minimum distance regime, in particular, when d = n/ 2 − Θ( √n).For j = n − 2d ⩽ 0, provided that a sufficient number of Hadamard matrices exist, a widely accepted conjecture, Plotkin and Levenshtein (see [12, Chapter 2, Theorem 8]) have es-sentially settled the problem and showed that A(2 d, d ) = 4 d,
A(n, d ) = 2 ⌊d/ (2 d − n)⌋ for even d > n/ 2, and A(n, d ) = 2
j d+1 2d+1 −n
k
for odd d > (n − 1) /2.In what follows, we consider the scenario with j > 0. When
j scales linearly with n, asymptotic results can be found in , . In particular, the conjecture is that there does not exist any binary code exceeding the Gilbert-Varshamov lower bound (Theorem 1). There are very few studies in the literature targeting the regime where j is sub-linear in n. In 1973, McEliece (see [12, Chapter 17, Theorem 38]), using the linear programming ap-proach, established the following bound that is valid for j =
o(√n):
A(n, d ) ⩽ n(j + 2) . (1) For j ≈ n1/3, codes have been constructed to meet McEliece’s upper bound, hence showing the tightness of this bound in this regime. A few improvements , have been derived in the literature in the regime j = o(n1/3). How-ever, to the best of the authors’ knowledge, no explicit upper bounds (or lower bounds with general enough parameters) on
A(n, d ) are derived in the regime j = Θ( √n). In this pa-per, we attempt to answer the following question: If the term
j = n − 2d scales as j = Θ( √n), what is the best size M
one can achieve? The rest of this paper is organized as follows. In Section II we review some well-known asymptotic bounds on A(n, d ) and examine their scaling behaviour when j = Θ( √n). In Section III-A, a BCH-like cyclic code construction is given to estab-lish a non-trivial new lower bound. In Section III-B, we review an alternative proof of a well-known first linear programming bound on A(n, d ) (formally decribed in Theorem 7) through a covering argument using Fourier analysis on the group Fn
2
.Then two upper bounds on A(n, d ) with d ⩾ n/ 2 − √n and
d ⩾ n/ 2 − 2 √n, that are strictly tighter than all prior results, are derived. Finally, the paper is concluded in Section IV. II. P RELIMINARIES
Let H2(·) denotes the binary entropy function. Let n ⩾ r be positive integers. Let Br (0, n ) ∈ { 0, 1}n denote the Hamming ball of radius r centered at 0 = (0 , 0, . . . , 0) , and its volume by
2022 IEEE International Symposium on Information Theory (ISIT)
978-1-6654-2159-1/22/$31.00 ©2022 IEEE
1963
2022 IEEE International Symposium on Information Theory (ISIT) | 978-1-6654-2159-1/22/$31.00 ©2022 IEEE | DOI: 10.1109/ISIT50566.2022.9834432
Authorized licensed use limited to: University of Michigan Library. Downloaded on September 06,2022 at 15:30:21 UTC from IEEE Xplore. Restrictions apply.
V ol (r, n )def
= |Br (0, n )| = Pri=0
ni
. We recall the following bounds 1) V ol (r, n ) ⩽ 2H2(r/n )n; and 2) V ol (r, n ) ⩾ 2H2(r/n )n−o(n) for sufficiently large n.
A. Known Bounds on Code Sizes
The following bounds on the size of binary codes can be found in standard coding theory textbooks, e.g. , . Bounds for the regime j = Θ( √n) are derived and given following the general bounds, e.g. inequalities (3), (5), (7), (8), (9), and (11). When the scaling behaviour of j matters, we choose j = 2 a√n, i.e., d = n/ 2 − a√n, for ease of comparison between bounds, as in (3), (9), and (11).
Theorem 1 (Gilbert-Varshamov or GV lower bound, ):
Let positive integers n and d ⩽ n/ 2 be given. Then
A(n, d ) ⩾ 2n
V ol (d − 1, n ) . (2) Asymptotically, suppose 0 ⩽ δ < 1/2, then there exists an infinite sequence of (n, M, d ) binary linear codes with d/n > δ
and rate R = k/n satisfying R ⩾ 1 − H2(d/n ). To evaluate Theorem 1 when j = Θ( √n), consider j = 2 a√n. Central limit theorem, coupled with the Berry–Esseen theorem, pro-vides an upper bound
V ol (d − 1, n ) ⩽ 2n Q(2 a) + O(1 /√n) ,
where Q(·) denotes the tail distribution function of the stan-dard normal distribution. Hence (2) becomes
A(n, n/ 2 − a√n) ⩾ Q(2 a) + O(1 /√n)−1 , (3) which is loose compared with the Plokin-Levenshtein bound
A(2 d, d ) = 4 d.
Theorem 2 (Hamming Bound): For every (n, M, d ) code
C ⊂ { 0, 1}n,
M ⩽ 2n/V ol (e, n ), (4) where e = ⌊(d − 1) /2⌋.In the asymptotics, Theorem 2 bounds the rate from above, in terms of the relative distance δ, by R ⩽ 1 − H2(δ/ 2) . For
j = Θ( √n), the term e = n/ 4 − Θ( √n), and V ol (e, n ) ⩾
2H2(1 /4) n−o(n). Hence Theorem 2 becomes
M ⩽ 2(1 −H2(1 /4)) n+o(n) ⩽ 20.189 n, (5) for all sufficiently large n.
Theorem 3 (Singleton Bound): The following holds for any code C ⊂ { 0, 1}n with distance d and dimension k.
k ⩽ n − d + 1 . (6) Under the regime j = Θ( √n), Theorem 3 becomes
M ⩽ 2n/ 2+ o(n), (7) which is weak compared to (5).
Theorem 4 (Plotkin Bound, ): The following holds for any code C ⊂ { 0, 1}n with distance d
1) If d = n/ 2, |C| ⩽ 2n.2) If d > n/ 2, |C| < 2
l d
2d−n
m
.One may use a combinatorial argument and Theorem 4 to de-rive the following corollary.
Corollary 5: If a (n, M, d ) binary code C has distance d < n/ 2, then the size M ⩽ d · 2n−2d+2 .Using Corollary 5, one may bound the size of any code with
d = ( n − j)/2 < n/ 2 by
M ⩽ d · 2j+2 < 2n · 2j . (8) When j scales as j = Θ( √n), the size M is bounded from above by a polynomial scaling sub-exponentially in n. In par-ticular, set j = 2 a√n, i.e. d = n/ 2 − a√n, (8) becomes
M ⩽ 2n · 22a√n. (9)
Theorem 6 (Elias-Bassalygo Bound): For sufficiently large
n, every code C ⊂ { 0, 1}n with relative distance δ and rate R
satisfies the following:
R ⩽ 1 − H2(J2(δ)) + o(1) , (10) where J2(δ)def
= 12 (1 − √1 − 2δ).Assuming d = n/ 2 − a√n, one may adopt steps similar to the proof of Theorem 6 as in [13, p.147] to show an upper bound:
M ⩽ n3 · 2 a
ln 2
√n+O(1)
. (11) The last upper bound we introduce is known as the first linear programming bound or the MRRW bound on binary er-ror correcting codes, or, alternatively, on optimal packing of Hamming balls in a Hamming cube. The bound was originally proved by McEliece, Rodemich, Rumsey, and Welch , fol-lowing Delsarte’s linear programming approach . Delsarte viewed the distance distribution of a code C of length n as an
(n + 1) -dimensional vector a = ( a0, a 1, . . . , a n), where ai is given by the number of pairs of codewords at distant i, and discovered a system of linear inequalities satisfied by every a
associated with a length-n, minimumm distance d code. The coefficients of the linear constraints can be viewed as values of a family of orthogonal polynomials called the Krawchouk polynomials. Based on the duality theorem of linear program-ming, one may find a feasible solution to the dual program, and view the obtained linear program as an extremal problem. (See [12, Chapter 17] for details.) Good feasible solutions of the dual program were constructed in . The resulting bound is the best known asymptotic upper bound on the cardinality of a code with a given minimal distance scaling linearly in n,for a significant range of the relative distance.
Theorem 7 (MRRW Bound, ): For sufficiently large n,every code C ⊂ { 0, 1}n with relative distance δ and rate R
satisfies the following:
R ⩽ H2
12 − pδ(1 − δ)
. (12)
Remark 1: Another bound, known as the second linear pro-gramming bound , is also given in in the form
R ⩽ min
0⩽u⩽1−2δ
1 + g(u2) − g(u2 + 2 δu + 2 δ), (13) where the function g(x)def
= H2((1 − √1 − x)/2) . For 0.273 ⩽
δ ⩽ 0.5, the bound (13) simplifies to that of (12). For δ <
0.273 , the inequality (13) is strictly tighter than (12).
2022 IEEE International Symposium on Information Theory (ISIT)
1964 Authorized licensed use limited to: University of Michigan Library. Downloaded on September 06,2022 at 15:30:21 UTC from IEEE Xplore. Restrictions apply. Plugging in δ = d/n into (12), we have the following bound:
M ⩽ 2nH 2
1/2−
√d/n (1 −d/n )
+o(n)
. (14) Note that, due to the o(n) term, the bound (14) is not tighter than (8) when j = Θ( √n). This appears to the contrary of the fact the MRRW bound is tighter than all the other bounds for relative distance δ > 0.273 . However, a tailored treatment of the proof technique may lead to a nontrivial bound as in the derivation of (11) from Theorem 6. In Section III-B, one such bound is given through an alternative proof of the Theorem 7 by working with the maximal eigenfunctions of Hamming balls. III. M AIN RESULTS
A. Cyclic Code with High Minimum Distance
Let n = 2 m − 1, and m ∈ N be an even integer with m ⩾ 4.Let c be an integer with 1 ⩽ c ⩽ m/ 2 − 1. We construct a binary cyclic code C of length n with high minimum distance as follows.
Theorem 8: There exists a binary cyclic code C of length
n, dimension cm , and minimum distance
d ⩾ 2m−1 − 2m/ 2+ c−1 ⩾ n
2 − 2c−1√n. (15)
Proof: Consider the finite field F = F2m and a subfield
K = F2 < F . Let α be a primitive root of unity in F , and set
αi = α1+2 m/ 2+ i
for i = 1 , 2, . . . , c . Consider the binary cyclic code with the generator polynomial
g(x) = xn − 1
Qci=1 Mαi (x) ,
where Mβ (·) is the minimal polynomial of β over K. Note that the αi’s belong to different conjugacy classes, i.e,
Ai
def
=
n
α2j
i
|j = 0 , 1, 2, . . . , m − 1
o
=
n
α2j +2 m/ 2+ i+j
|j = 0 , 1, 2, . . . , m − 1
o
are disjoint subsets of F { 0}, and |Ai| = m for each i. This is ensured by the particular choice of αi’s. More specifically, let
Pi = 2j + 2 m/ 2+ i+j mod 2m − 1|j = 0 , 1, 2, . . . , m − 1
be the set of the powers over α for elements in Ai. Each
Pi is a cyclotomic coset mod 2 in F and the length-m bi-nary representation for each p, p ′ in Pi are cyclic shifts of each other. Let pi = 1 + 2 m/ 2+ i be the coset representa-tive of Pi. The claim on the size of |Ai| holds by noting that |Ai| = |Pi| = m. To claim that Ai’s are disjoint, it suf-fice to show that the cyclotomic cosets Pi’s are disjoint. First note that for two cyclotomic cosets Pi and Pk, they are ei-ther disjoint or identical. Assume for some i̸ = k, cosets Pi
and Pk are identical. Then pi = 1 + 2 m/ 2+ i is an element in
Pk, that is, there is a p′ = 2 l + 2 m/ 2+ k+l ∈ Pk for which
pi = p′ modulo 2m − 1. As both pi and p′ are sums of two powers of 2, we note that neither m|l and m|(l + k − i), nor
m|(l − m/ 2 − i) and m|(m/ 2 + k − l), can happen. Hence
pi /∈ Pk, and thus Pi and Pk are disjoint. Thus the degree of the polynomial g(x) is n − cm .For the minimum distance, let t = 2 m−1 + 2 m/ 2+ c−1 + 1 .We show next that for j = t, t + 1 , . . . , 2m − 1, αj is a root for the generator polynomial g(x). In other words, Ai ∩ { t, t +1, . . . , 2m − 1} = ∅, for i = 1 , 2, . . . , c . This is by noting that the powers of the elements in Ai, after taking modulo 2m − 1,can be written as the sum of two powers of two, i.e., 2l + 2 j ,where the difference between l and j is at least m/ 2 − c, and that such a number does not belong to {t, t + 1 , . . . , 2m − 1}.Hence, the minimum distance of the code d is at least 2m −
t + 1 = 2 m−1 − 2m/ 2+ c−1 by BCH bound. Note that the parameters of the codes constructed in Theo-rem 8 are, in a sense, sitting between those of the first order and the second order Reed-Muller (RM) codes of length n =2m. More specifically, RM (m, 1) has minimum distance equal to n/ 2 and dimension equal to m + 1 , while RM (m, 2) has minimum distance n/ 4, and dimension 1 + m + m
2
. Hence, there is a wide gap in the minimum distance between the first order and the second order RM codes and, intuitively speaking the BCH-like codes constructed in Theorem 8 can be used to fill this gap. In particular, the parameters of the constructed code in the extreme cases of c, i.e., the cases of c = 1 and
c = m/ 2 − 1, are close to those of RM (m, 1) and RM (m, 2) ,respectively.
B. Improved Code Size Upper Bound
We will follow the covering argument by Navon and Samorodnitsky and show two upper bounds on the size of any code C with length n and minimum distance d, the first holds for any d ⩾ n/ 2 − √n, and the second for any
d ⩾ n/ 2 − 2 √n. The viewpoint presented in , provid-ing an alternative proof to Theorem 7, is different from that in , which relies on analytical properties of the Krawchouk polynomials, and instead adopts Fourier analysis on the group
Fn
2
as their main tool. In particular, the authors of exploit the expediency of working with the maximal eigenfunctions of Hamming balls. One key finding was that, given any real-valued func-tion f on {0, 1}n with a small support B ⊂ { 0, 1}n, such that the adjacency matrix of the Hamming cube acts on f by multiplying it pointwise by a large factor, the cardinality of error-correcting codes with minimum distance d can be upper bounded by n |B|. The applicability will depend on the value of the multiplying factor. By proposing functions f supported on Hamming balls B = Br (0, n ) of different radii r, one may derive a lower bound of the multiplying factor, formally called the maximal eigenvalue of adjacency matrix of the sub-graph incduced by B. This made possible a simple proof of the first linear programming bound. Let us now state the definition of the maximal eigenvalue of a graph. Let G = ( V, E ) be a (finite, undirected, simple) graph. Let AG = ( Aij ) denote the |V | × | V | adjacency matrix of G,defined by Aij = 1 if (i, j ) ∈ E and Aij = 0 otherwise for ver-tices i, j ∈ V . Note that AG is symmetric, so its eigenvalues are real, and can be ordered as λ1 ⩾ λ2 ⩾ . . . ⩾ λn. For any func-tion f on Fn
2
, the function Af sums at each point of {0, 1}n
the values of f at its neighbours. That is, the value taken by the function Af at a vertex x ∈ Fn
2
, denoted by (Af )( x) or
Af (x), is given by Af (x) = P
y∈Fn
2:wH(x,y )=1
f (y). When the graph is a subset of the cube, B ⊆ { 0, 1}n, set
λB
def
= max
⟨Af, f ⟩⟨f, f ⟩ f : Fn
2
→ R, supp (f ) ⊆ B
. (16)
2022 IEEE International Symposium on Information Theory (ISIT)
1965 Authorized licensed use limited to: University of Michigan Library. Downloaded on September 06,2022 at 15:30:21 UTC from IEEE Xplore. Restrictions apply. That is, λB is the maximal eigenvalue of adjacency matrix of the subgraph of {0, 1}n induced by B.Two lemmas were shown in to show (12).
Lemma 9 ( Prop 1.1): Let C be a code with block length n and minimal distance d. Let B be a subset of {0, 1}n
with λB ⩾ n − 2d + 1 . Then |C| = M ⩽ n |B|.
Lemma 10 ( Lemma 1.4): Let B = Br (0, n ) ⊆{0, 1}n. The maximal eigenvalue associated with B is
λB ⩾ 2pr(n − r) − o(n).To prove (12), we note that Lemma 10 implies that a ra-dius r∗ = n/ 2 − pd(n − d) + o(n) exists such that λB(r∗) ⩾
n − 2d + 1 . Lemma 9 in turn shows that any code of length n
and minimal distance d has at most n |B(r∗)| = nV ol (r∗, n )
codewords. The cardinality of a Hamming ball of radius r is
V ol (r, n ) = 2 H2(r/n )n+o(n). Equation (14) follows the above argument, hence yielding equation (12). We note that the above argument can not be used directly to show an upper bound when d = n/ 2 − Θ( √n). In particular, the o(n) term in Lemma 10 renders the search for a meaningful
r∗ impossible, as we would ideally require a subset B with λB
close to n − 2d + 1 = Θ( √n).First we provide a proposition in place of Lemma 10 when the radius of Hamming ball does not scale linearly in n.
Proposition 11: Let B = B3(0, n ) ⊆ { 0, 1}n be the Ham-ming ball of radius 3. The maximal eigenvalue associated with
B is λB ⩾ p3 + √6√n ≈ 2.334 √n.
Proof: Recall the definition of the maximal eigenvalue in (16). We prove the proposition by constructing a func-tion f with support in B, and for which ⟨Af, f ⟩/⟨f, f ⟩ =p3 + √6√n. The function f will be symmetric, namely its value at a point will depend only on the Hamming weight of the point. With a slight abuse of notation, such a function is fully defined by its values f (0) , f (1) , . . . , f (n) at Hamming weights 0, 1, . . . , n .Set f (0) = 1 , f (j) = 0 for j ⩾ 4, and let λf (i) = Af (i) =
if (i−1)+( n−i)f (i+1) for i = 0 , 1, 2 (assuming f (−1) = 0 ), where λ = t√n. We have
f (1) = λf (0)
n = t
√n , f (2) = λf (1) − 1f (0)
n − 1 = t2 − 1
n − 1 ,f (3) = λf (2) − 2f (1)
n − 2 = 1
n − 2
t2 − 1
n − 1 t√n − 2 t
√n
.
We may use the values f (i) and calculate
⟨Af, f ⟩ = 2 t√n + t(t2 − 1) 2 n√nn − 1 ,
⟨f, f ⟩ = 1 + t2 + ( t2 − 1) 2/2 + t2(t2 − 3) 2/6 + o(1) .
We are now ready to optimize the value
⟨Af, f ⟩⟨f, f ⟩ =
2t + t(t2 − 1) 2
(t6 − 3t4 + 9 t2 + 9) /6 + o(1)
√n (17) over t > 0. Taking t = p3 + √6, the square bracket term in (17) becomes p3 + √6 + o(1) .In order to provide a bound as tight as possible, we improve upon Lemma 9 and show the following proposition.
Proposition 12: Let C be a code with block length n and minimal distance d. Let B be a subset of {0, 1}n with λB >n − 2d. Then |C| = M ⩽ nλB −(n−2d) |B|.The proof can be shown using a similar argument as in the proof of Lemma 9 in , and is provided in Appendix-B for reference. With Proposition 11 and Proposition 12, we are ready to state the upper bound on A(n, n/ 2 − √n).
Theorem 13: If a (n, M, d ) binary code C has minimum distance d ⩾ n/ 2 − √n, then M ⩽ √n
√3+ √6−2 V ol (3 , n ) =
O(n3.5).
Proof: Let B = B3(0, n ) be the radius-3 Hamming ball. The maximal eigenvalue λB ⩾ p3 + √6√n according to Pro-position 11. Since n − 2d ⩽ 2√n < λ B , the cardinality of C
can be upper bounded using Proposition 12 as
M ⩽ nλB − (n − 2d) |B| ⩽
√n
p3 + √6 − 2 V ol (3 , n ).
Remark 2: We note that the argument above can upper bound the size as M = O(n3.5) as long as (n − 2d)/√n is strictly smaller than p3 + √6. That is, for any d ⩾ n/ 2 − t√n, for some constant t < p3 + √6/2 ≈ 1.167 , we have A(n, d ) =
O(n3.5).Using similar technique as in the proof of Proposition 11, one may show lower bounds of the maximal eigenvalues asso-ciated with Hamming balls of different radii. For example, we can list bounds of λBr /√n for 2 ⩽ r ⩽ 8, r ̸ = 3 :
λB2 /√n ⩾ √3, λB4 /√n ⩾ (5 + √10) 0.5 ≈ 2.857 ,λB5 /√n ⩾ 3.324 , λB6 /√n ⩾ 3.75 ,λB7 /√n ⩾ 4.14 , λB8 /√n ⩾ 4.51 .
This implies A(n, n/ 2 − t1
√n) = O(n2.5), A(n, n/ 2 −
t2
√n) = O(n4.5), and so on, for all t1 > √3/2 ≈ 0.866 and
t2 > 1.428 . In particular, we have the following bound
Theorem 14: If a (n, M, d ) binary code C has minimum dis-tance d ⩾ n/ 2 − 2 √n, then M ⩽ √n
4.14 −4
V ol (7 , n ) = O(n7.5).
Proof: The theorem can be proved using a similar argu-ment as Theorem 13 by taking B = B7(0, n ).In general, by finding the values of, or the lower bounds thereof, the maximal eigenvalues λBr for r = 1 , 2, . . . , one may obtain a sequence of upper bounds on A(n, d = n/ 2 −
t√n) for various t.IV. C ONCLUDING REMARKS
In this paper, we study the asymptotic performance bounds of the cardinality of codes with minimum distance d = n/ 2 −
Θ( √n). The codes in this regime have vanishing rate and thus renders ineffective most bounds that dictate the tradeoff be-tween code rate and relative distance. We obtain a sequence of lower bounds based on a cyclic code construction, and two upper bound for d ⩾ n/ 2 − √n and d ⩾ n/ 2 − 2 √n, respec-tively. The proposed cyclic linear code is the first construction ob-served in the specified distance regime, and allows one to con-struct codes of sizes polynomial in n. The proof of the upper
2022 IEEE International Symposium on Information Theory (ISIT)
1966 Authorized licensed use limited to: University of Michigan Library. Downloaded on September 06,2022 at 15:30:21 UTC from IEEE Xplore. Restrictions apply. bound makes extensive use of Fourier analysis on the Hamming cube as a group, and the calculation of the maximal eigenvalue associated with Hamming balls of small radii. An interesting problem for future work is to study the rela-tionship between the maximal eigenvalue λB associated with
B = Br (0, n ) and the radius r, for r ≪ n. The solution can be used to provide a sequence of upper bounds on A(n, d ) for the regime d = n/ 2 − Θ( √n).APPENDIX
A. Harmonic Analysis
We compile in this section harmonic analysis preliminaries as in , . See for a more detailed treatment. Here we list several necessary definitions and simple facts. Consider the abelian group structure Fn
2
= ( Z/2Z)n on the hypercube {0, 1}n. The characters of the abelian group Fn
2
are {χz }z∈Fn
2
, where χz : {0, 1}n → {− 1, 1} is given by
χz (x) = ( −1) ⟨x,z ⟩ and ⟨x, z ⟩ = Pni=1 xizi.Consider the R-vector space L(Fn
2
) = {f : Fn
2
→ R} en-dowed with the inner product ⟨· , ·⟩ , associated with the uniform distribution on {0, 1}n:
⟨f, g ⟩ = EUn f g = 12n
X
x∈Fn
2
f (x)g(x). (18) The set of 2n characters {χz }z∈Fn
2
form an orthonormal ba-sis in the space L(Fn
2
), equipped with uniform probability dis-tribution. That is, for each z, z ′ ∈ { 0, 1}n, ⟨χz , χ z′ ⟩ = δz,z ′ ,
where δ is the Kronecker delta function. The Fourier trans-form of a function f ∈ L (Fn
2
) is the function ˆf ∈ L (Fn
2
) given by the coefficients of the unique expansion of f in terms of the characters:
f (x) = X
z
ˆf (z)χz (x) or equivalently, ˆf (z) = ⟨f, χ z ⟩. (19) One may show that ˆˆf = 2 nf , and E f = ˆf (0) . If f, g ∈L(Fn
2
), we have Parseval’s identity: ⟨f, g ⟩ = P
z
ˆf (z)ˆ g(z) = 2n
D ˆf , ˆg
E
. A special case of the above equality is the following equality: E f 2 = P
z
ˆf (z)2
= ˆf 22 .
The convolution of f and g is defined by (f ∗ g)( x) =
Ey f (y)g(x + y). The convolution transforms to dot product:
[f ∗ g = ˆf · ˆg. The convolution operator is commutative and associative. One may also show that for arbitrary functions
f, g, h ∈ L (Fn
2
), the following equality holds:
⟨f ∗ g, h ⟩ = ⟨f, g ∗ h⟩. (20) In this section and in Appendix-B, let L(x) = 2 n for x ∈{0, 1}n with wH (x) = 1 and L(x) = 0 otherwise. For any
f ∈ L (Fn
2
) holds Af = f ∗ L because for x ∈ Fn
2
, Af (x) = P
y:dH(x,y )=1
f (y) = P
y:wH(y)=1
f (x + y) = Ey L(y)f (x +
y) = ( f ∗ L)( x). The Fourier transform of L is the function ˆL
given by ˆL(z) = ⟨L, χ z ⟩ = P
x:wH(x)=1
(−1) ⟨x,z ⟩ = n − 2 ·
wH (z).For C ⊂ Fn
2
, let 1C ∈ L (Fn
2
) be the indicator function of
C. It can be shown that a code C has minimum distance d if and only if (1 C ∗ 1C )( x) = 0 for all 0 < w H (x) < d .
B. Proof of Proposition 12
Let fB be an eigenfunction supported on B corresponding to its maximal eigenvalue λB . That is λB = ⟨Af B , f B ⟩/⟨fB , f B ⟩.It is known that the maximum can be attained with an non-negative function fB , and further we have Af B ⩾ λB fB (see [23, p.13-15 and appendix C]) for details). We write f = fB
and λ = λB interchangeably, and denote the Hamming weight of x ∈ Fn
2
by |x| = wH (x), in this proof. As f is supported on B, Cauchy-Schwarz inequality yields the following:
E2 f = ⟨f, 1B ⟩2 ⩽ E f 2 · E(1 B )2 = E f 2 · | B| /2n. (21) Let ϕ ∈ L (Fn
2
) be a function such that ( ˆϕ)2 = [ϕ ∗ ϕ =1C ∗ 1C . Equivalently, ϕ ∗ ϕ = 2 n \1C ∗ 1C = 2 n c1C
2
. Therefore we have
ϕ ∗ ϕ ⩾ 0 and E(ϕ2)
E2(ϕ) = (ϕ ∗ ϕ)(0) ˆϕ2(0) = |C| . (22) Now let F = ϕ ∗ f . We estimate the product ⟨AF, F ⟩ in two ways. First,
⟨AF, F ⟩ = ⟨(ϕ ∗ f ) ∗ L, ϕ ∗ f ⟩ = ⟨ϕ ∗ ϕ ∗ f, f ∗ L⟩
= ⟨ϕ ∗ ϕ ∗ f, Af ⟩ ⩾ ⟨ϕ ∗ ϕ ∗ f, λf ⟩
= λ⟨ϕ ∗ f, ϕ ∗ f ⟩ = λ⟨F, F ⟩ = λ E F 2.
Second, by Parseval’s identity,
⟨AF, F ⟩ = 2 nDdAF , ˆF
E
= 2 nD ˆL · ˆF , ˆF
E
= X
z
(n − 2 |z|) ˆF 2(z).
Since ˆF = ˆϕ · ˆf and ( ˆϕ)2(z) = (1 C ∗ 1C )( z), ˆF (z) = 0 for all
0 < |z| < d . We can estimate ⟨AF, F ⟩ by
X
z
(n − 2 |z|) ˆF 2(z) = n ˆF 2(0) + X
z:|z|⩾d
(n − 2 |z|) ˆF 2(z)
⩽ n ˆF 2(0) + ( n − 2d) X
z
ˆF 2(z) = n E2 F + ( n − 2d) E F 2.
Combining the two estimates, we have the following inequal-ity: n E2 F ⩾ (λ − (n − 2d)) E F 2. Since
E2 F = E2(ϕ ∗ f ) = [ [ϕ ∗ f (0)] 2 = [ ˆϕ(0) ˆf (0)] 2 = E2 ϕ E2 f,
E F 2 = ⟨F, F ⟩ = ⟨ϕ ∗ f, ϕ ∗ f ⟩ = ⟨ϕ ∗ ϕ, f ∗ f ⟩
⩾ 1/2n(ϕ ∗ ϕ)(0)( f ∗ f )(0) = 1 /2n E ϕ2 E f 2,
as ϕ ∗ ϕ = 2 n · c1C
2
⩾ 0, we now have
n E2 ϕ E2 f ⩾ (λ − (n − 2d)) 12n E ϕ2 E f 2. (23) Leveraging equations (21), (22), and (23), the size of any code
C with minimum distance d is
|C| = E ϕ2
E2 ϕ ⩽ nλ − (n − 2d) · 2n E2 f
E f 2 ⩽ nλ − (n − 2d) |B| .
2022 IEEE International Symposium on Information Theory (ISIT)
1967 Authorized licensed use limited to: University of Michigan Library. Downloaded on September 06,2022 at 15:30:21 UTC from IEEE Xplore. Restrictions apply. REFERENCES R. Ratasuk, N. Mangalvedhe, Y. Zhang, M. Robert, and J.-P. Koskinen, “Overview of narrowband IoT in LTE Rel-13,” in 2016 IEEE conference on standards for communications and networking (CSCN) . IEEE, 2016, pp. 1–7. M. Fereydounian, M. V. Jamali, H. Hassani, and H. Mahdavifar, “Chan-nel coding at low capacity,” in 2019 IEEE Information Theory Workshop (ITW) . IEEE, 2019, pp. 1–5. F. Abbasi, H. Mahdavifar, and E. Viterbo, “Hybrid non-binary repeated polar codes for low-SNR regime,” in 2021 IEEE International Sympo-sium on Information Theory (ISIT) . IEEE, 2021, pp. 1742–1747. I. Dumer and N. Gharavi, “Codes for high-noise memoryless channels,” in 2020 International Symposium on Information Theory and Its Appli-cations (ISITA) . IEEE, 2020, pp. 101–105. F. Abbasi, H. Mahdavifar, and E. Viterbo, “Polar coded repetition for low-capacity channels,” in 2020 IEEE Information Theory Workshop (ITW) .IEEE, 2021, pp. 1–5. I. Dumer and N. Gharavi, “Combined polar-LDPC design for channels with high noise,” in 2021 IEEE Information Theory Workshop (ITW) .IEEE, 2021, pp. 1–6. S. Litsyn, “An update table of the best binary codes known,” Handbook of Coding Theory , 1998. E. Agrell, “Bounds for unrestricted binary codes.” [Online]. Available: A. Brouwer. [Online]. Available: binary-1.html T. Jiang and A. Vardy, “Asymptotic improvement of the Gilbert-Varshamov bound on the size of binary codes,” IEEE Transactions on Information Theory , vol. 50, no. 8, pp. 1655–1664, 2004. P. Gaborit and G. Zemor, “Asymptotic improvement of the gilbert– varshamov bound for linear codes,” IEEE Transactions on Information Theory , vol. 54, no. 9, pp. 3865–3872, 2008. F. J. MacWilliams and N. J. A. Sloane, The theory of error correcting codes . Elsevier, 1977, vol. 16. V. Guruswami, A. Rudra, and M. Sudan, “Essential coding theory.” V. M. Sidel’nikov, “On mutual correlation of sequences,” in Doklady Akademii Nauk , vol. 196, no. 3. Russian Academy of Sciences, 1971, pp. 531–534. A. Tietäväinen, “Bounds for binary codes just outside the Plotkin range,”
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The Clinical Performance of UHMWPE in Elbow Replacements
12.2.3 Muscular Anatomy
Even though a number of muscles crossing the elbow are involved with motions of the wrist and fingers, relatively few of them function to move the elbow joint. The biceps brachii, brachialis, and brachioradialis are the primary elbow flexors. The tricpes brachii is the primary elbow extensor. Although the anconeus is thought tocontribute to elbow extension, it is also likely to provide dynamic constraint to varus and posterolateral instability of the elbow . Pronation is powered by the pronator teres, pronator quadratus, and brachioradialis (from supinated position). Supination is the result of biceps brachii, supinator, and brachioradialis (from pronated position) (Figure 12.5). It is important to note that the muscles crossing the elbow can act as dynamic stabilizers, compressing the joint surfaces and enhancing the geometric stability of its conforming surfaces.
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Book2016, UHMWPE Biomaterials Handbook (Third Edition)Judd S. Day PhD
Chapter
FORCE ANALYSIS OF ELBOW FLEXORS
1978, Proceedings of the Sixth New England Bioengineering ConferenceYoungil Youm Ph.D., ... Bruce L. Sprague M.D.
RESULTS
The anatomical study of muscle cross sectional areas showed that the average cross sectional areas of the biceps and brachialis are 1.62 and 1.82 times larger than that of the brachioradialis.
In individual muscle contribution analysis, the brachialis functions as the most powerful muscle among the flexors (Fig. 2).
The brachioradialis has the largest moment arm during elbow flexion (Fig. 3).
Both the individual muscle force and moment arm study showed that the optimal range of minimizing muscle effort and maximizing output is within 60°-80°.
The muscle contraction of the brachioradialis is the largest among the flexors, and this muscle has the largest muscle contractile velocity for the given time. Therefore, this muscle involves in rapid movement of elbow motion (Fig. 4). In the EMG study of elbow quick flexion the brachioradialis became very active, and during slow flexion of the elbow this muscle was relatively quiescent while the biceps and the brachialis showed considerable activity.
The force distribution analysis using average muscle cross sectional area is illustrated in Fig. 5. The brachialis acts as a powerful flexor during elbow flexion, as shown in the individual muscle study.
The result of this study is briefly tabulated as follows:
| Muscle | Force Contribution | Muscle Volume | Cross Sec. Area | Moment Arm | Muscle Contraction |
--- --- --- |
| Biceps | Medium | High | Medium | Medium | Medium |
| Brachialis | High | Medium | High | Low | Low |
| Brachioradialis | Low | Low | Low | High | High |
As shown in the table, every flexor plays an equally important and unique role during flexion motion.
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Book1978, Proceedings of the Sixth New England Bioengineering ConferenceYoungil Youm Ph.D., ... Bruce L. Sprague M.D.
Review article
Control Issues of Functional Electrical Stimulation: Current and Future Systems
2003, Medical Engineering & PhysicsMirjana B Popović
In some cases subjects with spinal cord lesion at C5 have their primary wrist movers (Extensor Carpi Radialis Brevis m. and Longus m. (ECRB and ECRL) and the Flexor Carpi Radialis m. (FCR) denervated). Surgical intervention (i.e., tendon transfer of the Brachioradialis m. into the ECRB) followed by electrical stimulation of the transferred muscle can restore limited voluntary control of wrist extension. As a result, wrist extension becomes sufficiently strong to overcome wrist flexion due to the stimulation of the finger and thumb flexor muscles . This procedure required analysis methods to control the position of the wrist by means of FES. Lemay et al. investigated one open loop and two closed-loop controllers. The open-loop control system used coactivation (i.e., increased stiffness) to stabilize position against disturbances in the flexion or extension direction. The open-loop controller was investigated because it was the simplest one for application since it does not require any sensors. The two closed-loop controllers, on the other hand, did not use any coactivation. One of the closed-loop methods used position and moment feedback to regulate the stiffness of the wrist with the equilibrium point at neutral . The control system was synthesized by applying methods developed by Crago et al. and Lan et al. . The conclusions from the study were that the stiffness closed-loop controller works best among the three systems tested. It required less stimulation than the open-loop controller and provided better wrist posture.
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Journal2003, Medical Engineering & PhysicsMirjana B Popović
Chapter
Neural Interface
2018, Advanced Rehabilitative TechnologyQingsong Ai, ... Sheng Quan Xie
Experimental Scheme and Result Analysis
Four healthy subjects participated in the study, one female and three males within an age range of 22–28 years old. Subjects had no history of neuromuscular disorders. All subjects read and signed the informed consent.
In the experiment, sEMG signals were collected from four muscles, that is, brachioradialis (BR), palmaris longus (PL), flexor carpi radialis (FCR), and flexor carpi ulnaris (FCU). sEMG electrodes (Biometrics, UK) were placed on the belly of each muscle. A reference electrode was placed on top of the lateral epicondyle. sEMG signals were amplified with a gain of 5000, bandpass-filtered between 10 and 500 Hz, and analog/digital sampled with 1000 Hz. The force signals were collected by the MTA400 three-axis force sensor (FUTEK, USA). Because the muscle force is not easy to obtain directly, the force between the wrist and the force sensor cannot be used as the equivalent.
The wrist is a very important joint of the human body, and the movements of the wrist can be divided into three DoFs, namely flexion (FX)/extension (EX), internal rotation (IR)/external rotation (ER), and radial deviation (RD)/ulnar deviation (UD). Because the torque of the internal and external rotation is not easy to be measured, the other two DoFs of the wrist are chosen, as shown in Fig. 8.7.
To ensure the unicity of action, the participant was asked to put the arm on a platform and tried to move the wrist in one DoF without other movements during one test. The initial action is to keep the angle between hand and arm as 180 degrees without deviation movement. Before data collection, the most voluntary contraction (MVC) experiment is needed to obtain the normalized reference for sEMG and force signals. To reduce the impact of muscle fatigue, the participant was given a rest of 3–5 min after each set of actions.
Based on the accurate recognition of the motions and the prediction of forces, the proportional myoelectric control method based on force prediction is used to control a rehabilitation robot. During the control, the robot movement speed is a function of force, namely, the subjects’ movement intentions can be reflected by the magnitude of the force.
Before the control experiment, it is necessary to determine the proportionality coefficient by experiment, in which the subject slowly moves toward a certain direction, while constantly changing the size of proportionality coefficient until the subject can freely control the rehabilitation robot.
The six DoFs robot platform in our laboratory can complete the movement in three directions (X, Y, Z), and we select the Y and Z directions to simulate the FX/EX DoF and RD/UD DoF, respectively. The subject performs the movement according to the target position displayed on a PC, and the corresponding EMG signals are collected for analysis at the same time.
Considering the safety and stability, the range of motion of the robot platform is set to be ± 10 cm in FX and EX directions, 4 cm in RD direction, and − 10 cm in UD direction. To reflect the continuity of the control strategy, five target positions are set in the FX, EX, and UD directions, and two target positions are set in the RD, a target position is set every other 2 cm. The design of the experiment scheme is shown in Fig. 8.8.
The trajectory tracking result of the flexion direction is ploted as shown in Fig. 8.9, where the solid blue line is the trajectory when the rehabilitation robot is moving at uniform speed, and the dotted red line is the actual trajectory of the robot in the proportional control. The result shows that the speed of the rehabilitation robot is constantly changing, and it is consistent with the trend of predicted force in the course of a target task movement. At the same time, the relative error between the actual position and the target position of the robot is calculated during a specific movement period, and the calculation result is 5.716%, which indicates the proposed proportional myoelectric control can achieve continuous control of the rehabilitation robot according to the human movement intention.
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Book2018, Advanced Rehabilitative TechnologyQingsong Ai, ... Sheng Quan Xie
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Anatomy of Muscle
2014, Reference Module in Biomedical SciencesP.D. Wilson
Muscles that Move the Forearm (Antebrachium)
The forearm includes two large bones: the ulna and radius. Two opposing sets of muscles move the forearm in two different directions at the elbow (Figures 11(a) and 11(b)). The biceps brachii, brachialis, and brachioradialis flex the elbow. The biceps brachii is also a powerful supinator when the limb is pronated. Extension of the elbow is accomplished by the large triceps brachii and the smaller anconeus muscles.
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Biomechanics of joints
2008, Joint Replacement TechnologyGR JOHNSON
Muscles and forces
The muscles acting across the elbow joint (brachialis, biceps brachii, brachioradialis and triceps) all produce flexion or extension moments. Pronation is produced by forearm muscles (pronator teres, pronator quadratus and flexor carpi radialis). Supination is achieved by a combination of supinator (in forearm) and biceps brachii which, because of its attachment to the ulna, provides a strong supination moment. The muscles acting at the elbow are listed in Table 1.6.
Table 1.6. Actions of major muscles at the elbow
| Direction | Muscle |
--- |
| Elbow flexion | Brachialis |
| | Biceps brachii |
| | Brachioradialis |
| Elbow extension | Triceps brachii |
| Forearm pronation | Pronator teres |
| | Pronator quadratus |
| | Flexor carpi radialis |
| Forearm supination | Supinator |
| | Biceps brachii |
(from Palastanga et al., 2006)
Copyright © 2006
The contact forces at the individual joints have been predicted using modelling approaches. Chadwick and Nicol (2000) have calculated for a range of tasks predicting loads of 1600N (2.3 BW approx.) in the humero-ulnar joint and 800 N (1.1 BW approx) in the humero-radial joint. In earlier studies of patients with rheumatoid arthritis (Amis et al., 1979), forces in the humero-ulnar joint of up to 0.65 kN in isometric extension and humero-coronoid forces of 1.49 kN have been described during isometric flexion. The corresponding forces in the humero-radial joint were 1.44 kN and 1.41 kN respectively.
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Anatomy of Muscle
2014, Reference Module in Biomedical SciencesP.D. Wilson
Muscles of the Pectoral Girdle and the Upper Extremity
For convenience, the muscles of the pectoral (i.e., shoulder) girdle and the upper extremity are divided here into six groups. These include muscles that move the shoulder, arm, forearm, wrist, and fingers.
Muscles that Move the Shoulder
The scapula, a triangular bone, articulates solidly only with the small clavicle. Because the scapula is not held firmly in place, it is free to move in several directions. The muscles that move the scapulae are extrinsic muscles – they attach from the neck and/or shoulder to the limb (Figures 11(a) and 11(b)). The trapezius is a large flat muscle that covers much of the upper back. Its fibers extend in several directions, and it elevates, retracts, and rotates the scapula freely. The rhomboideus and the levator scapulae muscles also retract and elevate the scapula. Acting antagonistically to these muscles, the pectoralis minor and the serratus anterior protract the scapula forward, as when a person reaches for something. The subclavius muscle attaches to and acts indirectly through the clavicle to pull the scapula toward the sternum.
Muscles that Move the Arm (Brachium)
Two groups of muscles act upon the shoulder joint to move the humerus: they are a combination of both extrinsic and intrinsic muscles. One group inserts close to the shoulder joint, produces small movements, and is used primarily in maintaining the articulation of the humerus with the glenoid fossa and in rotating the humerus. This group includes the supraspinatus, infraspinatus, subscapularis, and teres minor (Figure 11(c)). The term ‘rotator cuff’ is often applied to an area where tendons of these muscles that rotate the humerus fuse with tissues of the shoulder joint. This area is subject to trauma and pain in athletes who make extensive use of their arms (e.g., baseball pitchers and tennis players).
The second group of humeral muscles includes the deltoideus, pectoralis major, latissimus dorsi, teres major, and coracobrachialis muscles (Figure 11(b)). The deltoideus raises and abducts the arm and contributes to movements of the humerus associated with such activities as climbing. The other four muscles oppose the deltoideus by lowering and adducting the humerus. The intrinsic muscles crossing the shoulder joint are necessary to stabilize the shoulder, giving the deltoideus power in abduction (Rosse and Gaddum-Rosse, 1997).
Muscles that Move the Forearm (Antebrachium)
The forearm includes two large bones: the ulna and radius. Two opposing sets of muscles move the forearm in two different directions at the elbow (Figures 11(a) and 11(b)). The biceps brachii, brachialis, and brachioradialis flex the elbow. The biceps brachii is also a powerful supinator when the limb is pronated. Extension of the elbow is accomplished by the large triceps brachii and the smaller anconeus muscles.
Forearm Pronators and Flexors of the Wrist and Fingers
The anterior surface of the forearm contains three layers of muscles, most of which act to flex the wrist and the fingers (Figure 12). The superficial layer includes the pronator teres, which rotates the radius, turning the palm down (the act of pronation), and three flexors of the wrist: the flexor carpi ulnaris, flexor carpi radialis, and palmaris longus muscles.
The flexor digitorum superficialis muscle is the middle muscle on the anterior side of the forearm. Its tendon of insertion divides at the wrist and attaches to the middle phalanges of fingers 2–5. Along with the flexor digitorum profundus, which lies beneath it, the flexor digitorum superficialis flexes fingers 2–5, as when making a fist. The thumb has its own flexors.
The deep layer of forearm muscles includes the flexor digitorum profundus, pronator quadratus, and flexor pollicis longus. The flexor digitorum profundus inserts on the distal phalanges of fingers 2–5 and assists its superficial counterpart in flexing the fingers. The pronator quadratus pronates the hand, and the flexor pollicis longus flexes the thumb into the palm.
Forearm Supinator and Extensors of the Wrist and Fingers
The posterior aspect of the forearm has two layers of muscles concerned with extending the wrist and the fingers. A superficial layer consists of five muscles. Three of these are wrist extensors: the extensor carpi radialis longus, extensor carpi radialis brevis, and extensor carpi ulnaris. The remaining two also extend the fingers: the extensor digitorum communis, whose tendons are seen fanning over the back of the hand and the knuckles, and the extensor digiti minimi (Figure 13(a)). The deep layer primarily moves the fingers and includes the abductor pollicis longus, extensor pollicis longus, extensor pollicis brevis, and extensor indicis. Also in the deep layer is the supinator muscle, which rotates the palms upward, as when receiving a bowl of soup (Figure 13(b)).
Intrinsic Muscles of the Hand
In addition to tendons from forearm muscles, the hand is operated by three groups of muscles on the palmar surface. These muscles permit a versatile range of movements of the thumb and the fingers. The first group, which moves the thumb, forms a fleshy pad, the thenar eminence, at the base of the thumb. On the opposite side of the hand is the narrower hypothenar eminence, which extends from the base of the little finger (Figure 14). It consists of four muscles, three of which move the little finger. Between these two eminences is a V-shaped depression, with the three sets of midpalmer muscles that move fingers 2–5 (Figure 15; Gaudin, 1997).
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Hand grasp and reach in spinal cord injury
2015, Implantable Neuroprostheses for Restoring FunctionK.L. Kilgore
9.5.3 Clinical results of the IST-12 system
A clinical feasibility study of the implanted 12-channel myoelectrically controlled neuroprosthesis (IST-12) was conducted by the Cleveland FES Center. The IST-12 system was implanted in 15 arms in 12 C5–C6 cervical SCI subjects. All subjects were able to successfully use the MES from the ECRL (C6) or brachioradialis (C5) for proportional control of grasp opening and closing. Subjects also demonstrated the ability to generate MES from trapezius, platysma, deltoid, and biceps muscles to generate logic control. The use of myoelectric control in neuroprostheses allows considerable flexibility in the control algorithms, enabling them to be tailored to each individual subject. The elimination of the need for an externally mounted control source is extremely desirable and makes system use much simpler. The three subjects implemented with bilateral systems have demonstrated the ability to control each arm independently.
MES were successfully used to obtain a viable control signal during stimulation for all 30 recording electrodes (15 arms). MES could be recorded from implanted recording electrodes during stimulation of nearby muscles. For example, MES could be recorded from the ECRL muscle during stimulation of the extensor digitorum communis muscle. The MES recording was stable long-term, with no indication of loss of signal over time. These results demonstrate that the electrodes do not undergo continued encapsulation that would insulate the electrode from the muscle and reduce signal quality.
Study results to date indicate that every subject improved significantly in pinch force strength. In all cases, the presurgery pinch force was achieved by passive finger and thumb tone augmented with wrist extension. For most subjects, presurgery pinch force is useful only for acquiring light objects such as a piece of paper. With the neuroprosthesis turned on, pinch force typically doubled or tripled and could be used to perform a variety of tasks such as holding a fork for eating or a pen for writing.
Use of the neuroprosthesis allowed every subject to increase the number of objects they could manipulate in the grasp release test (GRT) (Mulcahey, Smith, & Betz, 2004; Smith, Mulcahey, & Betz, 1996; Wuolle et al., 1994). Every subject/arm could complete at least five of the six tasks using the neuroprosthesis, whereas before surgery only one subject could manipulate as many as four objects, and most subjects could only manipulate the two lightest objects. On average, subjects improved in the ability to manipulate three objects (range, 2–6).
The ability to perform ADL (Bryden, & Bezruczko, 2011; Bryden, Kilgore, Keith, & Peckham, 2008) was assessed as shown in Figure 9.5. Every subject demonstrated improvement in at least two activities, with one subject demonstrating improvement in 11 of 12 activities tested and two subjects demonstrating improvement in all nine activities tested. Improvement in these activities generally indicates that subjects can complete the task more independently with the neuroprosthesis than they can when the neuroprosthesis is turned off, although improvements in the quality of performance, ease of performance, and time to complete the task are also possible. All 15 arms in the 12 subjects showed improved function in eating with a fork and 13 of 15 showed improvement in writing with a pen. These tasks have been shown to be some of the most common tasks for which subjects use their neuroprosthesis in the home environment. Subjects with bilateral systems are able to perform activities such as using a fork and knife to cut food, using two hands to screw and unscrew a lid on a jar, and brushing hair while blow-drying. Other tasks in which subjects showed improvement included using an electric shaver, using a mobile phone, and swinging a tennis racquet.
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ELECTROMYOGRAPHY
1973, Physiology and Biochemistry (Second Edition)JOHN V. BASMAJIAN
C Upper Limb
In the upper limb, the classic work of Inman et al. (1944) on the shoulder girdle has been followed by a long series of scattered EMG studies. Bearn's (1961) finding, that the activity in the upper fibers of trapezius falls off after a minute or two to disappear completely while the person is upright, is especially significant. It is also rather surprising to find that serratus anterior has only slight activity in an upright posture.
At the shoulder joint, the downward dislocation of the shoulder is resisted by the superior capsule of the shoulder joint and supraspinatus (Basmajian and Bazant, 1959). This finding has also emphasized that muscles which cross a joint longitudinally are not necessarily active when there is distraction on the joint. Other work on the elbow joint has confirmed this finding. The general principle seems to be that capsules and capsular ligaments are sufficient to prevent distraction except where excessive forces are applied.
When muscle is a contributing factor, it often is part of a locking mechanism rather than a source of transarticular forces. On the other hand, during movement such as flexion and extension of a joint, certain muscles are extremely important as a transarticular component to prevent distraction of a joint. Thus, brachioradialis shows little if any activity in maintaining flexed postures even against added loads, but it is very active in either flexion or extension of the elbow. This is its shunt muscle function, i.e., it acts chiefly during rapid movement along the long axis of the moving bone to provide centripetal force. The whole question of spurt and shunt muscles has been thoroughly discussed elsewhere. Of course such muscles are not confined to the upper limb and have widespread significance in the economy of the body. (See MacConaill and Basmajian, 1969, for a fuller discussion.)
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Hand grasp and reach in spinal cord injury
2015, Implantable Neuroprostheses for Restoring FunctionK.L. Kilgore
9.3 Operating principles
Early designers of upper extremity systems recognized that control of patterned movements was the only practical approach to achieving real-time function (Peckham, Marsolais et al., 1980; Peckham, Mortimer et al., 1980). Specifically, rather than providing the user with control over individual muscles, the user was provided control over a conceptual pattern, that is, hand opening and hand closing. Direct proportional control of grasp opening and closing allows the user to control the grip strength and allows manipulation of objects in the grasp. Therefore, all neuroprosthetic systems use this type of control whenever possible. Switches, or other on-off signals, referred to as logic signals, can be used to perform system activities such as turning the stimulation on and off or initiating a lock function. The lock function allows the user to deliver a specific command that disconnects the stimulated grasp from the proportional control. This allows the user to hold an object for prolonged periods without having to maintain the proportional control at maximum output, and is a critical feature of functional neuroprosthetic systems. The system is unlocked through delivery of another specific command signal (Peckham et al., 1983).
The control algorithm developed for the IST-12 system used a common controller template that could be customized for each user (Kilgore et al., 2008). Typically, generation of two successive high-level bursts of muscle activity in the shoulder or neck was used to turn the system on. The user selected the desired grasp pattern using bursts of activity from the same muscle. Once the pattern was selected, the user gained direct proportional control of the degree of hand opening and closing through the MES level from the voluntary forearm musculature (either brachioradialis or extensor carpi radialis longus (ECRL)). Strong contraction of this muscle resulted in hand closing, whereas relaxation of the muscle resulted in hand opening. If the user desired to hold an object for a long period of time, he or she could initiate a lock command that disengaged the grasp stimulation from control by the forearm MES. The lock command was initiated by holding the MES above a high threshold for 2 s. Once the hand was locked, it remained locked until an unlock command was given. The unlock command could consist of two quick bursts of activity from the forearm (referred to as a double-click) or a quick burst of activity from the shoulder. The significant advantage of myoelectric control, in addition to eliminating the need for external switches, was that all control signals were derived ipsilaterally, so the opposite arm was free to be used as a helper hand. Therefore, myoelectric control was ideally suited for bilateral implementation.
In addition to proportional control of hand function, the user could independently activate functions such as elbow extension, forearm pronation, or shoulder stabilization by producing a specific pattern of myoelectric activity in the logic control muscles. The system could be tailored to the needs and physiology of each user based on the concept of using myoelectric activity from one or more voluntary muscles to control the electrical stimulation of one or more paralyzed muscles to produce functional movements. In general, reaching functions enhanced with electrical stimulation were either on or off and the user contracted the voluntary antagonist(s) against the stimulated musculature to modulate position as needed (Bryden, Memberg, & Crago, 2000; Grill & Peckham, 1998; Lemay & Crago, 1996).
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Related terms:
Thermal Stress
Feasibility Study
Proportional Control
Joints (Structural Components)
Control Algorithm
Axis Accelerometer
Elbow Flexor
Joint Reaction Force
Loop Controller
Reaction Force
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3066 | https://math.stackexchange.com/questions/2254824/examples-of-transforming-into-weierstrass-form | reference request - Examples of transforming into Weierstrass form - Mathematics Stack Exchange
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Examples of transforming into Weierstrass form
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Currently working through Silverman and Tate's Rational Points on Elliptic Curves.
Right now I'm at Weierstrass form, and I'm interested in seeing some examples of curves being transformed into the Weierstrass form, showing most substitutions, calculations, and maybe some information about what each substitution "does" to the curve, that is, how does it change the curve but maintain its algebraic structure?
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asked Apr 27, 2017 at 14:39
AirdishAirdish
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What do you mean by "calculations"? Do you want to see an example of how the transformation is found?Álvaro Lozano-Robledo –Álvaro Lozano-Robledo 2017-04-28 15:08:05 +00:00 Commented Apr 28, 2017 at 15:08
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The following is an excerpt from this book (Example 2.2.3):
Let d∈Z d∈Z, d≠0 d≠0 and let E E be the elliptic curve given by the cubic equation
X 3+Y 3=d Z 3 X 3+Y 3=d Z 3
with O=[1,−1,0]O=[1,−1,0]. The reader should verify that E E is a smooth curve. We wish to find a Weierstrass equation for E E. Note that if we change X=U+V X=U+V, Y=−V Y=−V, Z=W Z=W, then we obtain a new equation
U 3+3 U 2 V+3 U V 2=d W 3.U 3+3 U 2 V+3 U V 2=d W 3.
Since this equation is quadratic in V V, and cubic in W W, with no other cubic monomials that involve W W, the variable W W will end up playing the role of x x, and the variable V V will play the role of y y in our Weierstrass model. Next, we change variables to obtain a coefficient of 1 1 in front of V 2 V 2 and W 3 W 3. If we multiply the previous equation through by d 2 d 2, we obtain
d 2 U 3+3 d 2 U 2 V+3 d 2 U V 2=d 3 W 3,d 2 U 3+3 d 2 U 2 V+3 d 2 U V 2=d 3 W 3,
and now we change variables x=3 d W x=3 d W, y=9 d V y=9 d V, and z=U z=U. Then, we obtain
d 2 z+d y z 3+y 2 z 27=x 3 27,d 2 z+d y z 3+y 2 z 27=x 3 27,
or, equivalently, y 2 z+9 d y z=x 3−27 d 2 z y 2 z+9 d y z=x 3−27 d 2 z, which is a Weierstrass equation. Thus, [x,y,z]=[3 d W,9 d V,U]=[3 d Z,−9 d Y,X+Y][x,y,z]=[3 d W,9 d V,U]=[3 d Z,−9 d Y,X+Y] and we have found a change of variables ψ:E→E ˆ ψ:E→E^ given by
ψ([X,Y,Z])=[3 d Z,−9 d Y,X+Y]ψ([X,Y,Z])=[3 d Z,−9 d Y,X+Y]
such that the image lands on the curve in Weierstrass equation E ˆ:y 2 z+9 d y z=x 3−27 d 2 z E^:y 2 z+9 d y z=x 3−27 d 2 z. The map ψ ψ is invertible; the inverse map ψ−1:E ˆ→E ψ−1:E^→E is
ψ−1([x,y,z])=[9 d z+y 9 d,−y 9 d,x 3 d].ψ−1([x,y,z])=[9 d z+y 9 d,−y 9 d,x 3 d].
In affine coordinates, the change of variables is going from X 3+Y 3=d X 3+Y 3=d to the curve y 2+9 d y=x 3−27 d 2 y 2+9 d y=x 3−27 d 2 via the maps:
ψ(X,Y)ψ−1(x,y)==(3 d X+Y,−9 d Y X+Y),(9 d+y 3 x,−y 3 x).ψ(X,Y)=(3 d X+Y,−9 d Y X+Y),ψ−1(x,y)=(9 d+y 3 x,−y 3 x).
We leave it as an exercise for the reader to verify that the model can be further simplified to the form y 2=x 3−432 d 2 y 2=x 3−432 d 2.
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answered Apr 28, 2017 at 15:15
Álvaro Lozano-RobledoÁlvaro Lozano-Robledo
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Are the powers of z correct?p6majo –p6majo 2025-06-04 20:27:00 +00:00 Commented Jun 4 at 20:27
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18References for elliptic curves
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3067 | https://www.constructedadventures.com/how-to-build-a-treasure-hunt/how-to-write-a-riddle-thats-meant-to-be-solved?srsltid=AfmBOopczv_7SzTcH9T3fpMyJ6dwrSLmcIucLz35Um91LKfDbBPk9GJT | How to write a riddle that’s meant to be solved — Constructed Adventures
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How to write a riddle that’s meant to be solved
Sep 7
Written By The Wizard
Riddles are a nearly universal form of puzzle, appearing in every language and culture throughout all of recorded history. Why are riddles so popular? For puzzle designers, they are often simple to construct, yet their difficulty can range from trivial to fiendish. For solvers, riddles offer the satisfaction of an epiphany. If you're planning a puzzle adventure, riddles are a fundamental tool, able to fit into almost any puzzle shaped hole. But for as flexible riddles can be, they can also be a dead stop to the fun if the players end up stumped. Whether you're writing a riddle or solving one, a good grasp of riddle construction will help you be your most clever.
The Anatomy of a Riddle
While riddles can appear in a wide variety of formats, almost every riddle can be broken down into a few common components:
The Premise - These are the details and information players are provided with. These details can be explicitly stated, or obscured through elaborate presentations or trickier logic.
The Structure - This is the pattern that the information fits into. Without a structure, a riddle is just a reading comprehension question.
The Ask - This is what the players are challenged to answer. Sometimes the ask can be implied by the structure, especially in traditional formats where players can be expected to recognize a shared structure.
Last Thing First
The best way to come up with a riddle is to start with the answer. This is especially true in a puzzle adventure, where the answer usually helps players find the next step in the adventure. Once you know what you want them to come away with, you can think about ways to obfuscate that information through logic, poetic description, and procedural manipulation.
A Premise Makes a Promise
The details in a premise are anything the player can trust to be true. Betraying this trust is the fastest way to write an unsatisfying riddle. But writing a good premise takes more than just a collection of details. Let's look at an example.
_A flat approximation
of the rivers and the ground,
a projection with directions,
to help you get around.
What is it?_
In this easy example, the first four lines comprise the premise. It's important to note that while there are only 4 lines to the premise, at least 5 details are divulged:
The answer is something flat
The answer approximates rivers and ground
The answer is a projection
The answer contains directions
The directions assist with navigation.
As a general rule, a riddle with more details in its premise is easier to solve than one with fewer. Each detail is like an answer in the game 20 Questions, it narrows down the scope of possible solutions. They are used by players to test the answers they come up with, knowing any answer which violates a detail in the premise should be rejected.
A good premise will have details that aren't just accurate, but evocative of the answer. Each detail should lead the player toward the correct solution, both logically and thematically. When this is done well, the solution will be "self-proving" to the player, meaning that when they come up with it, they know it's the right answer because it fits the whole theme. To give a counterexample, here's the same riddle above written with less evocative details
_It can fit under a door
It rolls or maybe folds.
It helps you get to somewhere warm,
when you're somewhere cold._
While the answer is the same, this premise provides less thematic direction for the player to follow. We don't often think of the answer, which you can find below, as something that fits under a door, even though technically it could. And while you might use a one to "get somewhere warm", that's an unusual use case. This version isn't just harder, it's less fun.
Structure Sets the Stage
The structure of a riddle is the patterns it relies on to communicate the shape of the answer. It’s how you let your players know what to expect, and what to look for. In the above example, the structure is simple and traditional. There are four obscured descriptions of an object, and a challenge to identify it. Many riddles rely on a simple structure like this, with the more difficult versions offering fewer and more obscure details.
Nearly every reading person has encountered a riddle in this structure, so most people don’t need to be told “Each line describes properties of a specific thing. That thing may be a particular object, a category of objects, or even an esoteric idea or concept. Any thing that fits with all the of the descriptions is a good guess, but there’s one particular solution that fits best. The thing would be recognizable to any general person in your circumstances.”
Those implicit rules are part of the structure, hiding behind the words “What is it?” The structure lets the player know that “map” is a good guess, and “my friend Tom who shouts directions at me while I’m driving” isn’t. When Bilbo challenged Gollum with the riddle “What’s in my pocket?” Gollum’s rage comes from the violation of the implicit structure of the game they were playing. As a puzzle adventure designer, you want to avoid provoking rage from your players whenever possible.
Other riddle structures include “computational” or “procedural” riddles, where the player must follow a set of directions or keep track of a changing detail.
_Every dwarf of 16 years is sent into the mine
with two friends, and two strangers following behind
The strangers are both older, but neither older than 19
the friends are both within a year of the leader in between.
Every dwarf responsible to bring back shiny gems
one for every year of age, plus 4 more for both friends.
With every member born in a different numbered year,
when they return from mining how many gemstones will appear?_
In this example, the structure requires the player to plot the age range of the dwarves and add some numbers together to compute the solution. The structure assures players that if they come up with the right numbers and the right order of operations, they will arrive at the correct solution.
One wonderful thing about procedural riddles for puzzle adventures is that the numbers or details can refer to real world objects around your players. A detail like “The number of slides at this park multiplied by the number of trees” compels players to look around the park you’ve brought them to, which can be a great way to introduce the hook for the next puzzle.
Resist the temptation to use “gotcha” reasoning to solve a procedural puzzle. If you structure suggests that doing the calculations will get the answer, but it turns out the answer relies on tricky wording that doesn’t require any calculation, your players will feel almost cheated. In the context of a story or maybe a tabletop role-playing game, there’s room for this kind of trickery, but in a general puzzle adventure it’s probably a step too far.
Be Careful with Conundrums
Another structure you might encounter is a conundrum, which is the specific term for a riddle that relies on a play on words or a pun to function. Solving a conundrum is extraordinarily satisfying for a player up to the task, and you’ll see this type of riddle often as a way to test a hero’s wit in classic stories. Here’s an example:
What is shared by a tree and a dog?
The answer, which you can find below, is being used in two different ways at once. This property lets the player know they’ve found the answer. An answer like “shadow” may be technically accurate, but it doesn’t have that punning resonance that distinguishes it as the right answer. These riddles are especially popular in fiction, as they are compact and clever, and any hero solving one appears quite insightful. But they are notorious for interrupting a puzzle adventure when the players miss the key pun and instead get stuck on a wrong answer because they don’t know what’s missing.
If you include conundrum style puzzles, it’s wise to ensure that other thematic elements of the puzzle are pointing players in the right direction. You can go as far as to give a full example riddle with the answer already provided to teach your players the format, or you can make another puzzle mechanism that produces the answer through non-riddle means (like a cipher).
Just Ask
The ask is the way you tell your players what kind of answer you’re expecting. Your structure will help, especially if you’ve used an evocative premise, but it may be necessary to specify that you’re looking for a two word phrase, a famous character, or a type of food. Even a small change from “What is it?” to “Who am I?” at the end of a riddle can go a long way to getting your players pointed in the right direction.
In a challenging puzzle hunt for hardcore puzzlers, it’s common to skip the ask. The simple fact that there is a premise and a pattern is enough for some puzzlers to dive in and find an answer. But for a puzzle adventure, where the goal is to complete the challenge, it’s completely fine to be explicit with your ask. That can be as direct as “The answer is a valid English word of 8 letters”, or it can be more integrated like “Answer: _ _ _ _ _ _ _ _”. If nothing else, incorporating a few hint words in bold throughout the instructions text will give players some assurance that they understand what’s being asked of them.
Verification By Design
It’s important to consider your riddle solution, not just in terms of theme and challenge, but also in the way it communicates the next step to your players, and how they know when they have the right answer. Take this example:
_In a park east of Main
with a tree that has a name
find the shortest towering pine
underneath a secret find_
It may seem easy to write something like this (tailored to your play area), thinking “there’s only really one named tree on the east side of town, they’ll figure it out”. But think about what’s involved for a player to test their solution. And consider what happens when they guess wrong. If they don’t know of the named tree you had in mind, they could spend hours running around parks trying to find one. And every time they don’t find the right one, they are going to question if they’ve done something wrong, or worse, if you have.
In general, it’s probably not wise to have a riddle answer that is a location. It’s better if the riddle provides a code, password, or key that players can test easily where they are, and then for that prize to direct them to the location you want to send them too. Locations are better encoded into puzzle formats with clear, discrete solutions.
Another novice mistake is to rely on specific data from something that is onerous to search. “Riker’s Rank” is pretty easy to find on any Star Trek fan wiki. But “What Riker named his bug” might send your players through dozens of episode synopses or video essays looking for the one where Riker names his insectoid son Jean-Luc. Respect your players effort by avoiding difficult searches and allow them to easily test answers.
But Shouldn’t It Be Harder?
In a survey of people who would partake in a puzzle adventure asking how skilled they are at puzzling, the people who mark “Very Skilled” include both a group of people who participate in competitive puzzle events monthly, and a group of people that enjoyed a Sudoku book once in college. Remember that your ultimate goal when building a puzzle adventure is building a fun experience where your players overcome the challenges and win the day. You don’t win anything for stumping your players. It’s very easy to write a very hard riddle, and your players will only appreciate the high difficulty if they solve it on their own.
At the same time, a riddle on a piece of paper is not so exciting without a lot of thematic support. Writing out your riddles with a calligraphy pen on a parchment paper is definitely a nice touch, but it can only get you so far. To flesh out a riddle without just adding obscurity or difficulty, you can break the riddle up into pieces that the players must earn. A particularly useful version is a symbol cipher, like this:
In a ¤ or on a §
Turn it ¶, take a look.
What do you see?
For this example, you’d include other puzzles that taught what the symbols mean. As players discover that the symbols mean, in order, “body”, “book”, and “over”, the riddle’s solution comes into view. This is especially effective if the encoded riddle is provided early on with a promise that it will “make sense later”. That feeling when a player realizes they’ve just learned how to decipher a code they’ve held on to all day is very rewarding.
A well dressed riddle can even serve to let players know they are on the right track. When they put the pieces of a jigsaw puzzle together and discover a riddle written across it, they know they’ve found the next step of the adventure. A riddle that slowly constructs itself out of pieces as they progress can turn into a very exciting payoff when they finally have what they need to crack it.
The Difference Between Stuck and Stumped
When a player can’t progress in the Adventure because of a riddle, they are either stuck or stumped. A player is stuck when they don’t know the answer, but have an idea of things to try. They are stumped when they are out of ideas to try. A stuck player can be encouraged with a reminder, “Don’t forget about that cipher you learned earlier” or “Have you looked at all the maps carefully?” These hints feel encouraging because they let the player know they have what they need to make progress.
A stumped player needs a more direct hint “What kind of trees grow in the north?” or “Why is the second line phrased so oddly?” They need you to do a solving step for them to get them going again. And once they know that’s an option, you become a walking hint dispenser every time they get challenged, which is less fun for everyone.
How do you keep players from getting stumped? Try to minimize the amount of external knowledge your puzzles require. For example,
Aggassi’s necklace, when Dolly was born.
This (hard) riddle requires the player to know who Aggassi is, why they have a “necklace”, who Dolly is and when she was born. If you’re a fan of tennis and science headlines, this might not feel very challenging. But what if the only tennis players you knew were the Williams sisters? You’d need, at least, a web search to find out that Aggassi played tennis at the olympic level and that Dolly is a famous farm animal.
You might think “Ah, but my players have phones, they have internet, they can just look things up.” But then you’re creating a whole extra puzzle layer, perhaps unintentionally, where you are relying on your players to search for the right terms. It only takes one wrong search to send them off on a wild goose chase and derail your puzzle flow. And that’s all assuming that they even have reliable internet, that their phone is charged, and that the information you found online is still available when they go looking.
Of course, every riddle in this article relies on external knowledge to some degree. Determining what is “in scope” for your players can be very tricky. I like to use the “Dictionary vs Wikipedia” guideline. If the information could be found in a dictionary, and it’s not terribly obscure, it’s probably fair game for a general puzzle. If the information needed would be better found on Wikipedia, it’s probably not safe to rely on for a puzzle. In other words, players should be looking up words and meanings, not contexts and concepts. Not only does this help you keep your players on track, but you avoid the feeling that “the internet answered the riddle”.
If you’re writing a harder riddle, intending to send the players searching for external knowledge, it’s good to give them some guide as to where to look. For example, if your riddle involves world record trivia, it’s wise to specify whether you’re referring to the records kept by Guinness or the ones kept by the Book of Alternative Records. If your riddle involves identifying fonts by name, let your players know whether they should be looking at Google Fonts or the font list in MS Word. Knowing where to go looking shouldn’t be a part of the puzzle they have to solve for.
How to Hint
At a technical level, a puzzle is solvable when the players have what they need to discover the solution. But at a fun level, it takes a little more than that. People don’t like asking for hints, but they do like earning clues. You can use simple stunt challenges like “Carry this ice-cube on your forehead from here to the tree and back with no hands to earn a clue”, or you can use more elaborate constructions where the players discover hints in the play area around the riddle.
If players find your riddle next to, say, an Easter egg with the words “open me” written on it, that egg can contain a very basic clue. From there, additional clues can be hidden around in other eggs. Players can decide on their own how much energy they want to put into finding all the eggs and how much they want to focus on the riddle. Most likely, they won’t need to find all of the eggs, but even if they do you’ve built a path for them that requires no direct intervention. There’s a very exciting feeling when a player realizes they’ve earned “just enough” to crack a riddle without finding all the clues, it feels like finding a shortcut.
Bailing Out
Let’s consider the worst case situation: Your players have found all of your clues, they are reading everything clearly, and they are still stumped. They know they’ve been beaten by the puzzle, you know they’ve hit a wall, and the whole adventure is about to collapse in on itself. What should you do?
You could scramble, you could invent new ways to solve the puzzle on the spot and try to improvise your way to making them feel like they did something to earn progress. But this might feel like pandering, or even worse, like unpreparedness. No one wants to solve the “baby version” of the puzzle after being stumped, and trying to shoehorn in some plot reason to deliver the players the answer runs the risk of trivializing their work.
Instead, remember that the focus is the adventure, not the individual puzzles. In the interest of keeping the fun going, it’s OK to step in and bail them out. It doesn’t feel great, but it cuts the suffering off before it becomes a show stopper. When it comes to this, being simple, honest, and direct is the best approach. “I don’t want this to get in the way of the rest of the fun. Would you like to skip this puzzle for now?”
Don’t do a breakdown on the spot, just give them whatever information the riddle was meant to reveal. If they still want to revisit the riddle after the adventure is over, that’s the right time to walk them through it and discuss where it didn’t make sense. A post-event conversation is a great way to show that you care about your players and want to give them good experiences, even if one riddle fell short.
Putting it Together
So, to make a fun, solvable riddle, you need to consider:
How your players recognize your structure
How your players interpret your premise
How your players identify your ask
How your players seek additional information
And what you will do when they are stumped
You should be prepared to be tested on any assumptions you make. If you can point to a specific element in your adventure for each of those considerations, you’ve got more than a riddle, you have a plan for how to make a riddle work in your adventure.
Of course, if you’re putting a lot of work into a puzzle Adventure, playtesting is crucial. Getting external brains to work through your riddles is a really great way to find out when they are unexpectedly hard. It’s even better if you can get your riddle playtested by a few different people with different skill levels. You can always post on the Constructed Adventures playtesting thread on our Discord Channel.
Finally, once you’ve written and tested and polished your riddle, trust your work. Players rely on your confidence as a puzzle designer to feel safe investing their time in your adventures. If you undercut yourself by tinkering with the riddle or jumping in the first time they get stuck to walk them through it, your players may lose their trust that the adventure is ready for them to solve.
With careful planning, a good sense of what your players’ perspectives, and a bit of cleverness, you’ll can master the art of riddle writing. Use the styles and formats that appeal to you, write about the things that are interesting to you, and keep your focus on delivering that “Aha!” moment, and you’ll be delighting players time and time again.
The Answers
In case you got stumped, the answers are, in order: a map (twice), 93 gems, bark, spine, a gold medal.
The Wizard
Previous Previous Structures to follow when building a puzzly Adventure -----------------------------------------------------Next Next Common codes found in puzzles -----------------------------
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3068 | https://pubs.acs.org/doi/10.1021/j100795a003 | THE FORMATION OF BENZENE IN THE RADIOLYSIS OF ACETYLENE1 | The Journal of Physical Chemistry
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article January 1, 1963
THE FORMATION OF BENZENE IN THE RADIOLYSIS OF ACETYLENE 1
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Gilbert J. Mains
H. Niki
M. H. J. Wijnen
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Mariia A. Lukianova, Vladimir I. Feldman. Radiation-induced closure of the second aromatic ring: Possible way to PAH starting from a styrene-acetylene complex. Radiation Physics and Chemistry2022, 191, 109847.
Congcong Li, Jianwei Xie, Jinli Zhang, Bin Dai. Nitrogen-Modified Activated Carbon Supported Cu(II)Cu(I)/NAC Catalysts for Gas–Solid Acetylene Dimerization. Catalysis Letters2021, 151 (10) , 2990-2995.
Mariia A. Lukianova, Vladimir I. Feldman. Direct evidence for a single-step radiation-induced assembling of benzene ring from acetylene trimer at cryogenic temperatures. Radiation Physics and Chemistry2021, 183, 109417.
T. H. Fleisch, G. J. Mains. An XPS study of the UV reduction and photochromism of MoO3 and WO3. The Journal of Chemical Physics1982, 76 (2) , 780-786.
Robert W. Schmieder. Agglomeration of polyacetylenic particulates and its relation to soot formation. Symposium (International) on Combustion1982, 19 (1) , 1403-1412.
Jane Huey Hong, Ralph S. Becker. Hydrogen atom initiated chemistry. Journal of Molecular Evolution1979, 13 (1) , 15-26.
Hiroo Ogura. Pyrolysis of Acetylene behind Shock Waves. Bulletin of the Chemical Society of Japan1977, 50 (5) , 1044-1050.
Masao Tsukada, Takefumi Oka, Shoji Shida. PHOTOCHEMICAL AND RADIATION-INDUCED REACTIONS OF ACETYLENE AND HYDROGEN SULFIDE MIXTURE. SYNTHESIS OF THIOPHENE. Chemistry Letters1972, 1 (6) , 437-440.
K.J. Laidler, L.F. Loucks. Chapter 1 The Decomposition and Isomerization of Hydrocarbons. 1972, 1-148.
K.G. McLaren. Radiolysis of acetylene-nitrous oxide mixtures and the effect of carbon dioxide. International Journal for Radiation Physics and Chemistry1970, 2 (3) , 123-130.
Shoji Shida, Masao Tsukada, Noboru Fujisaki, Takefumi Oka. Mechanism of Benzene Formation in the Radiation-Induced and Photochemical Reactions of Acetylene. Bulletin of the Chemical Society of Japan1970, 43 (10) , 3314-3314.
Shoji Shida, Masao Tsukada. Mercury-Photosensitized Reaction of Acetylene. Bulletin of the Chemical Society of Japan1970, 43 (9) , 2740-2745.
Günther Bünau. Edelgasreaktionen in der Strahlenchemie. , 347-393.
J. V. Michael, H. Niki. Reaction of Hydrogen Atoms with Acetylene. The Journal of Chemical Physics1967, 46 (12) , 4969-4975.
Masao Tsukuda, Shoji Shida. Mechanism of the Benzene Formation in the Mercury-Photosensitized Reaction of Acetylene. The Journal of Chemical Physics1966, 44 (8) , 3133-3134.
Günther v. Bünau. Edelgasreaktionen in der Strahlenchemie. 1965, 347-393.
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The Journal of Physical Chemistry
Cite this: J. Phys. Chem. 1963, 67, 1, 11–16
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3069 | https://byjus.com/jee/born-haber-cycle/ | The Born Haber process, more commonly referred to as the Born Haber cycle, is a method that allows us to observe and analyse energies in a reaction. It mainly helps in describing the formation of ionic compounds from different elements. The methodology further enables us to understand the overall reaction process through a series of steps.
The Born Haber cycle was introduced in the year 1919 by German scientists Fritz Haber and Max Born. The Born Haber cycle is mainly used to calculate the lattice energy. It also involves several steps or processes, such as electron affinity, ionization energy, sublimation energy, the heat of formation and dissociation energy.
Considerations
The reaction of electropositive metals with electronegative nonmetals produces ionic solids. Alkali and alkaline earth metals react with chalcogen or halogen family elements to form compounds, which are crystalline ionic solids. Ionic compounds being stabilized by the electrostatic force of attraction between positive and negative charges are expected to have similar physical properties.
But physical properties like stability and the water solubility of these ionic compounds differ much. The difference is attributed to the difference in an enthalpy called ‘Lattice energy’ between the ionic solids.
Lattice energy is the energy that keeps together the cations and anions of the compound in fixed positions in a crystalline solid state. Lattice energy can be defined as either energy released when gaseous ions form one mole of a solid ionic compound or as the energy required to convert one mole ionic solid into its gaseous ions. There is no way to measure this lattice energy experimentally. Hess’s law of heat summation is the only indirect way of estimating the lattice energy.
Application of Hess’s law of heat summation to the formation of solid ionic compounds involves enthalpy of all processes that are necessary for the formation of the solid ionic compound from the elemental state of the constituent atoms in a form cycle such that the total energy on summation is zero.
What Is the Born Haber Cycle?
The Born Haber cycle is a cycle of enthalpy change of process that leads to the formation of a solid crystalline ionic compound from the elemental atoms in their standard state and of the enthalpy of formation of the solid compound such that the net enthalpy becomes zero.
Born Haber Cycle Examples
Example 1: Born Haber cycle of sodium chloride NaCl (or any AB-type Mono-valent ionic solid).
The heat of the formation of sodium chloride (ΔHf0) from the sodium metal and chlorine gas can be experimentally measured.
Na (s) + (1/2)Cl2 (g) → NaCl(s) ΔHf0 = -411kJ/mol
The formation of ionic solid sodium chloride from solid sodium metal and gaseous chlorine is not a single-step process but goes through several processes. Heat changes of all the processes except the lattice energy can be experimentally measured.
The processes or steps in the formation of sodium chloride are as follows:
1. Solid sodium atom sublimes to gaseous atom by absorbing heat energy (∆Hsub).
Na (s) → Na (g),
Sublimation energy ΔHsub = + 107kJ/mol
2. Gaseous sodium atom absorbs the ionization energy to release one electron and forms a gaseous sodium ion.
Na(g) → Na+(g) + 1e– ,
Ionization energy ∆HIE = +502kJ/mol
3. Diatomic gaseous chlorine breaks into two individual atoms by absorbing bond energy, such that each chlorine atom absorbs half of the bond energy of the chlorine molecule.
Cl2(g) → 2Cl(g) (1/2)
Bond dissociation energy of chlorine = (1/2) ∆Hdiss= (1/2) 242 = +121kJ/mol
4. Chlorine atom accepts an electron to form a chloride ion and releases energy equivalent to electron affinity.
Cl(g) + 1e– → Cl– (g)
Electron affinity = ∆HEA = -355kJ/mol
5. Gaseous sodium ion and gaseous chloride ion combine to form a solid sodium chloride molecule and releases energy equivalent to lattice energy.
Na+(g) + Cl– (g) → Na+ Cl– (s)
Lattice energy = ∆HLE = U = ?
The summation of enthalpy of all the processes (from step 1 to step 5) give the net enthalpy of the formation of solid crystalline sodium chloride from sodium and chlorine in their standard conditions of solid and gas, respectively. This should be equal to the experimentally measured enthalpy of the formation of solid sodium chloride.
The enthalpies are represented as a cycle in the figure.
So, ΔHf0 = ΔHsub + ∆HIE + (1/2) ∆Hdis + ∆HEA + U or ΔHf0 – (ΔHsub + ∆HIE + (1/2) ∆Hdis + ∆HEA + U)=0
411 + 107 + 502 +121 -355 +U = 0
Here, except lattice energy, all other enthalpies can be experimentally measured.
The lattice energy of the sodium chloride solid = U = ΔHf0 – (ΔHsub + ∆HIE + ½ ∆Hdis + ∆HEA).
= -411 -107 -502 -121 +355
= – 786kJ/mol
Example 2: Lattice energy of magnesium oxide (or any AB-type Divalent ionic solid).
The heat of the formation of magnesium oxide (ΔHf0) from the magnesium metal and oxygen gas can be experimentally measured.
Mg (s) + (1/2) O2 (g) → MgO(s)
ΔHf0 = -602kJ/mol
The processes or steps in the formation of magnesium oxide are as follows:
1. Solid magnesium atom sublimes to a gaseous atom by absorbing heat energy (∆Hsub).
Mg (s) → Mg (g), Sublimation energy ΔHsub = + 136kJ/mol
2. Gaseous magnesium atom releases two electrons in two steps with corresponding ionization energies.
Mg(g) → Mg+(g) + 1e–, ionization energy ∆H1IE= +738kJ/mol
Mg+(g) → Mg2+(g) + 1e–, ionization energy ∆H2IE= +1450kJ/mol
So energy of ionization = ∆HIE = 738 + 1450 = 2188kJ/mol.
3. Diatomic oxygen breaks into two individual atoms by absorbing bond energy, such that each chlorine atom absorbs half of the bond energy of the chlorine molecule.
O2(g) → 2O(g) (1/2) bond dissociation energy of oxygen = (1/2) ∆Hdiss= (1/2) 498 = +249kJ/mol
4. Oxygen atom accepts two electrons to form an oxide ion and releases energy equivalent to two-electron affinities.
O(g) + 1e– → O– (g) Electron affinity = ∆H1EA = -142kJ/mol
O–(g) + 1e– → O2- (g) Electron affinity = ∆H2EA = +798kJ/mol
The total energy released as electron affinity by the oxygen atom is = ∆HEA = +656kJ/mol
5. Gaseous magnesium ion and gaseous oxide ion combine to form a solid magnesium oxide molecule and releases energy equivalent to lattice energy.
Mg2+(g) + O2- (g) → Mg2+O2- (s) Lattice energy = ∆HLE = U = ?
The summation of enthalpy of all the processes from the starting step to the final step gives the net enthalpy of the formation of solid crystalline magnesium oxide from magnesium and oxygen in their standard conditions of solid and gas, respectively. This should be equal to the experimentally measured enthalpy of the formation of solid magnesium oxide.
The enthalpies are represented as a cycle in the figure.
So, ΔHf0 = ΔHsub + ∆HIE + (1/2) ∆Hdis + ∆HEA + U or ΔHf0 – (ΔHsub + ∆HIE + (1/2) ∆Hdis + ∆HEA + U) = 0
602 + 136 + 2188 + 249 +656 + U = 0
Here, except lattice energy, all other enthalpies can be experimentally measured.
The lattice energy of the magnesium oxide solid = U = ΔHf0 – (ΔHsub + ∆HIE + (1/2) ∆Hdis + ∆HEA).
= -602 – 136 – 2188 – 249 -656 = -3831kJ/mol
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3070 | https://math.stackexchange.com/questions/3551881/do-the-elements-of-a-set-have-to-be-unique | elementary set theory - Do the elements of a set have to be unique? - Mathematics Stack Exchange
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Do the elements of a set have to be unique?
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Does the mathematical definition of a set specify/imply that its elements be unique?
For context, this question has arisen in my mind from my experience using the Python programming language where one of the properties of the 'Set' data structure is that its elements are necessarily unique. Prior to this I had done a maths degree, but don't recall this being highlighted when learning about [mathematical] sets; certainly not with the emphasis it is given in Python.
Moreover, I feel like I came across questions or problems where sets would sometimes have repeated elements (perhaps problems in probability or combinatorics), but I may be mis-remembering this or it may have simply been an abuse of the notation. But an example might be:
What is the probability that the sum of two numbers, one each drawn randomly from the sets A={1,2,2,3,3,3}A={1,2,2,3,3,3} and B={1,2,3,4}B={1,2,3,4} is at least 6?
(where the desired answer is 8/24, rather than 3/12)
The Set Theory Wikipedia page does not use the term "unique" or "distinct" in reference to set elements. I came across this Stack Overflow question, but it's obviously geared heavily towards programming, so it's hard to know if that answers are really about the mathematical concept rather than programming data structures. It does mention that a set where repeated elements are allowed is called a Multiset, and in making this distinction, the Wikipedia page for Multiset does assert that a set is only allowed a single instance of an element. But it goes on to say that the term Multiset was only coined in the 1970s, so I'm left wondering what Mathematicians did before then if they wanted collections of objects with duplicates?
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asked Feb 18, 2020 at 22:52
TimTim
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Yes, because there is no way for you to know how many of a given element are in a set, just whether an element is in a set or not. Hence, regardless or not of whether the element is unique, to you, it will seem unique. –Rushabh Mehta Commented Feb 18, 2020 at 22:54
To allow for objects with duplicates, we make functions from some indexing set into a set of elements from which we want to build our set. For example, If I want to represent your A A, I would represent it as a function A:{1,2,3,4,5,6}→{1,2,3}A:{1,2,3,4,5,6}→{1,2,3} defined by 1↦1;2,3↦2;4,5,6↦3 1↦1;2,3↦2;4,5,6↦3 –Rushabh Mehta Commented Feb 18, 2020 at 22:56
We don't really use collections with duplicates much, anyways. Functions are far more versatile. –Rushabh Mehta Commented Feb 18, 2020 at 22:59
Possible duplicate of math.stackexchange.com/questions/934378/… –lhf Commented Feb 18, 2020 at 23:01
I think the real issue here is that of notation. As I understand it, in very fundamental set theory, you wouldn't write out things in lists in braces like A={1,2,2,3,3,3}A={1,2,2,3,3,3}. Instead, you might say that A A is the set for which 1∈A,2∈A,3∈A 1∈A,2∈A,3∈A and x∈A⟹x=1 x∈A⟹x=1, x=2 x=2, or x=3 x=3. We've said that A A has as its elements just 1 1, 2 2, and 3 3. But there's no basic logical/ set-theoretic language we could use to specify 'how many times' 1 1 or 2 2 or 3 3 appears in A A. In fact, at such a basic level the idea of 'how many times' hasn't itself been defined yet. –CardioidAss22 Commented Feb 19, 2020 at 0:38
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Does the mathematical definition of a set specify/imply that its elements be unique?
Yes.
The Set Theory Wikipedia page does not use the term "unique" or "distinct" in reference to set elements.
No, but it does say this:
"Set theory begins with a fundamental binary relation between an object o and a set A. If o is a member (or element) of A, the notation o ∈ A is used."
I interpret this as: To be an object in the universe is to be unique. For every _thing_, it is the only thing that is itself and either an object o o is a member of a set A A.... or it isn't.
Now a set has nothing about how order the things or how you pick them out or list them. And if o o is in the set A A, it doesn't matter if when asked to describe the elements of A A I mention o o first, or last, or 53 53 rd and if I say "A A has o o and it has t t and s s and it has o o, did I mention o o already, and it has q q and z z and o o and m m and o o and o o and, gee I'm mentioning o o a lot, and ..." The fact remains either o o is in the set or not. Those are the only options.
So if a set is {1,2,3,4}{1,2,3,4} that doesn't mean we can't list it as {4,3,4,1,2,4}{4,3,4,1,2,4}. In fact consider Q={a b|a,b∈Z;b≠0}Q={a b|a,b∈Z;b≠0}. That's perfectly valid but inefficient. Notice we have include then element 3 4 3 4 when we consider 3 4∈Q 3 4∈Q as 3,4∈Z 3,4∈Z. But we considered it a _second_ time when we considered −3,−4∈Z−3,−4∈Z and 3 4=−3−4 3 4=−3−4. And we considered it a _third_ time when we considered 51,68∈Z 51,68∈Z.
....
As to consider sets as lists with multiple _listings_ of elements or as listings where order _does_ matter.... well, that is why we have such concepts as _multisets_ or _sequences_. Even _functions_ is an extension of the concept.
As for a probability problem as you suggest. I imagine must would state it as something like "What is the probability that the sum of two numbers, one each drawn randomly from the _collections_ A={1,2,2,3,3,3} and B={1,2,3,4} is at least 6?" Technically we'd say A A is a multiset, not a set.
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answered Feb 18, 2020 at 23:35
fleabloodfleablood
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My source is Set Theory and Logic by Robert R. Stoll. The set theory developed by Stoll is what he calls _Intuitive Set Theory_. Other authors use the designation _Naive Set Theory_ for this foundational field of mathematics. So-called _Axiomatic Set Theory_ is a refinement developed in an effort to remove such problems as Russell's Paradox.
Stoll uses Cantor's original definitions in his development, so I will treat them as mathematically definitive.
Let us consider Cantor's concept of the term set and then analyze briefly its constituent parts. According to his definition, a setS S is any collection of definite, distinguishable objects of our intuition or of our intellect to be conceived as a whole. The objects are called the elements or members of S.…S.… (we shall use both terms synonymously).
The essential point of Cantor's concept is that a collection of objects is to be regarded as a single entity (to be conceived as a whole). The transfer of attention from individual objects to collections of individual objects as entities is commonplace, as evidenced by the presence in our language of such words as "bunch," "covey," "pride," and "flock."
……
With regard to the objects which may be allowed in a set, the phrase "objects of our intuition or of our intellect" gives considerable freedom.
……
The remaining key words in Cantor's concept of a set are "distinguishable" and "definite." The intended meaning of the former, as he used it, was this: With regard to any pair of objects qualified to appear as elements of a particular set, one must be able to determine whether they are different or the same. The attribute "definite" is interpreted as meaning that if given a set and an object, it is possible to determine whether the object is, or is not, a member of the set. The implication is that a set is completely determined by its members.
Consider the following natural language definitions:
element (3)noun - an artifact that is one of the individual parts of which a composite entity is made up; especially a part that can be separated from or attached to a system
member noun - anything that belongs to a set or class
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answered Jan 9 at 12:44
Steven Thomas HattonSteven Thomas Hatton
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1. Sets don't allow repeats and don't impose an ordering. Example: {1,2,1}={1,2}={2,1}{1,2,1}={1,2}={2,1}. This is the default. Your math teachers surely mentioned this a few times over the course of any mathematical undergraduate degree.
2. Multisets do allow repeats but don't impose an ordering. Example: {1,2,1}={2,1,1}≠{2,2,1}{1,2,1}={2,1,1}≠{2,2,1}. When you use them, you must always clearly say so to avoid ambiguities with the much, much more common set notation.
3. Sequences do allow repeats and do impose an ordering. Example: (1,2,3,2,1)≠(1,1,2,2,3)(1,2,3,2,1)≠(1,1,2,2,3). These are essentially Python lists or tuples. Sometimes they're notated as {a i}{a i}, which looks like a set but is actually a sequence. It's generally obvious from context what is meant.
4. Total orders don't allow repeats and do impose an ordering. Example: 1<2<3 1<2<3.
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answered Jan 9 at 13:51
Joshua P. SwansonJoshua P. Swanson
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Why post a new answer to an old question with an accepted answer you add nothing to? –Ethan Bolker Commented Jan 9 at 15:35
It had been bumped by someone else before me. I found the existing accepted answer a little unfocused. I also mentioned total orders, which was "new". I found your comment unlikely to be constructive. –Joshua P. Swanson Commented Jan 9 at 21:50
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3071 | https://pmc.ncbi.nlm.nih.gov/articles/PMC2643781/ | Herpes Simplex Virus Replication: Roles of Viral Proteins and Nucleoporins in Capsid-Nucleus Attachment - PMC
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Herpes Simplex Virus Replication: Roles of Viral Proteins and Nucleoporins in Capsid-Nucleus Attachment▿
Anna Maria Copeland
Anna Maria Copeland
1 Department of Microbiology and Cancer Center, University of Virginia Health System, Charlottesville, Virginia 22908
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1 Department of Microbiology and Cancer Center, University of Virginia Health System, Charlottesville, Virginia 22908
Corresponding author. Mailing address: Department of Microbiology, Box 800734, University of Virginia Health System, 1300 Jefferson Park Ave., Charlottesville, VA 22908. Phone: (434) 924-2504. Fax: (434) 982-1071. E-mail: jcb2g@virginia.edu
Received 2008 May 30; Accepted 2008 Dec 3; Issue date 2009 Feb.
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PMCID: PMC2643781 PMID: 19073727
Abstract
Replication of herpes simplex virus type 1 (HSV-1) involves a step in which a parental capsid docks onto a host nuclear pore complex (NPC). The viral genome then translocates through the nuclear pore into the nucleoplasm, where it is transcribed and replicated to propagate infection. We investigated the roles of viral and cellular proteins in the process of capsid-nucleus attachment. Vero cells were preloaded with antibodies specific for proteins of interest and infected with HSV-1 containing a green fluorescent protein-labeled capsid, and capsids bound to the nuclear surface were quantified by fluorescence microscopy. Results showed that nuclear capsid attachment was attenuated by antibodies specific for the viral tegument protein VP1/2 (UL36 gene) but not by similar antibodies specific for UL37 (a tegument protein), the major capsid protein (VP5), or VP23 (a minor capsid protein). Similar studies with antibodies specific for nucleoporins demonstrated attenuation by antibodies specific for Nup358 but not Nup214. The role of nucleoporins was further investigated with the use of small interfering RNA (siRNA). Capsid attachment to the nucleus was attenuated in cells treated with siRNA specific for either Nup214 or Nup358 but not TPR. The results are interpreted to suggest that VP1/2 is involved in specific attachment to the NPC and/or in migration of capsids to the nuclear surface. Capsids are suggested to attach to the NPC by way of the complex of Nup358 and Nup214, with high-resolution immunofluorescence studies favoring binding to Nup358.
Herpes simplex virus type 1 (HSV-1) virions consist of four prominent structures: the viral membrane, the tegument, the DNA-containing capsid, and the DNA itself. The membrane is a host-derived lipid bilayer, typically spherical in shape, with the viral glycoproteins embedded in it. Among the glycoproteins are those that bind to host receptors and initiate fusion between the cell and viral membranes, releasing the DNA-containing capsid and tegument into the cytoplasm of the host cell.
Upon entry into the cytoplasm, the capsid is transported to the nucleus by way of its interactions with the minus-end-directed microtubule motor protein dynein (14, 43). During entry and transit to the nucleus, much of the tegument dissociates from the capsid (19, 32, 33, 43), although at least two tegument proteins, VP1/2 and UL37, remain associated (19, 32). Such partially tegumented capsids bind to the host nuclear pore complex (NPC), where the viral DNA is released, a process termed uncoating. The HSV-1 capsid remains on the cytoplasmic side of the NPC, while the DNA enters the nucleus by translocating through the pore (4, 27, 45).
NPCs are large multiprotein complexes that mediate transport into and out of the nucleus (2). The vertebrate nuclear pore is a 125-MDa complex (40) that traverses both the inner and outer nuclear membranes. Each pore is composed of more than 30 different proteins (11), called nucleoporins, that are arranged with eightfold symmetry (16). NPCs can be thought of in terms of three structural regions: the nuclear basket, the central core, and the cytoplasmic filaments. The basket resides inside the nucleus, the central core is in the plane of the nuclear envelope, and the cytoplasmic filaments project into the cytoplasm (15, 44). The filaments are composed primarily of Nup358 (12, 47, 50), with Nup214 and Nup88 existing as a complex on the cytoplasmic face of the NPC (3, 28, 39). By projecting into the cytoplasm, the filaments are in a position to interact with HSV-1 capsids.
Capsid binding to the NPC occurs with a distinctive orientation. The capsid binds with a vertex facing the pore channel at a characteristic distance (40 to 50 nm) above the pore (A.M.C., unpublished observations) (18, 38). This distance is consistent with a possible interaction with the 35- to 50-nm-long cytoplasmic filaments (5, 17, 30). Additionally, interactions have been observed between nucleus-bound HSV-1 capsids and filaments emanating from the nucleus (43).
Nuclear capsid binding and genome uncoating are two processes that have remained poorly understood. The cellular factors importin-β and Ran-GTP are essential for binding (38), but neither the nucleoporins nor the viral proteins involved have been identified. VP1/2 has recently been shown to play a role in uncoating, long implied by the ts B7 mutant (26). ts B7 HSV-1 has a defect that allows nuclear capsid binding but not uncoating at the nonpermissive temperature (4, 45). The mutation maps to UL36, the gene encoding VP1/2 (4). Following nuclear capsid binding, VP1/2 is cleaved to allow uncoating to proceed (26). VP1/2's role in nuclear binding and the possible involvement of other herpesvirus proteins are not well understood.
Here we describe the results of studies in which we tested the involvement of herpesvirus capsid and tegument proteins and nucleoporins in the early steps in infection. Blocking experiments were performed with the use of herpesvirus and nucleoporin antibodies introduced into cells by syringe loading. Additionally, microscopy and small interfering RNA (siRNA) were used to address the involvement of nucleoporins. Specifically, Nup358 was our primary candidate because of its localization to the distal end of the cytoplasmic filaments (47). On the viral side, VP1/2 was of particular interest because of its role in uncoating, previous implications of VP1/2's involvement in nuclear binding, and its purported localization to the capsid vertices (8, 36, 38, 51).
Our results show that VP1/2 plays an important role early in infection prior to uncoating and that Nup358 is important for HSV-1 nuclear capsid binding.
MATERIALS AND METHODS
Cells, viruses, and virus concentration.
All experiments were performed with Vero cells and with the K26GFP (13) strain of HSV-1 (a gift of Prashant Desai). K26GFP is a KOS-derived mutant in which the gene encoding UL35 is fused to the green fluorescent protein (GFP) gene. Capsids appear green when viewed by epifluorescence due to incorporation of GFP-tagged UL35 into capsids.
Vero cells were maintained in 150-cm 2 flasks at 37°C with 7.5% CO 2 in minimal essential medium (MEM) with 10% fetal bovine serum (FBS), l-glutamine, and antibiotics. For large-scale virus purification, Vero cells were passed to six 850-cm 2 polystyrene roller bottles. Prior to infection, cells were washed one time with 30 ml per roller bottle MEM with 1% FBS, l-glutamine, and antibiotics. HSV-1 K26GFP was added at a multiplicity of infection of 3 and allowed to attach for 1 h in a low medium volume (23 ml per bottle MEM, 1% FBS with L-glutamine and antibiotics). Medium volumes were then increased to 56 ml per bottle, and cells were incubated for 24 h at 37°C. Virus was then concentrated. For this, medium was collected and spun in a clinical centrifuge to pellet cells and cell debris. The pellets were discarded, and the supernatant was saved. The supernatant was centrifuged for 1 h at 23,000 rpm at 4°C in a Beckman SW28 rotor to produce virus-containing pellets. Pellets were resuspended in 1× phosphate-buffered saline (PBS), aliquoted, and frozen.
Syringe loading.
This method was adapted from that of Clark and McNeil (7). Syringe loading is a method that allows the introduction of antibodies directly into the cytoplasm of cells by shearing the cell membrane in an antibody-rich solution. Small membrane tears allow the antibody to enter the cytoplasm. Tears then reseal, leaving a living cell with antibody in the cytoplasm. For syringe loading experiments, one 150-cm 2 flask of Vero cells (5 × 10 7 cells) was trypsinized. One-fourth of the cells were washed, suspended in 1.5 ml serum-free MEM, and allowed to rest for 20 min at 37°C with 7.5% CO 2. For each sample, 66 μl of the cell suspension was mixed with Pluronic-68 (Sigma-Aldrich) to a concentration of 2% (wt/vol) and antibody at a concentration between 0.2 and 1 mg/ml. The samples were taken up and down through a 27-gauge needle for a total of 50 strokes. The cells were then pelleted at 2,500 rpm for 5 min and resuspended in MEM (supplemented with l-glutamine and antibiotics) with 10% FBS. Cells were then plated in eight-well chambered cover glasses (growth area, 0.7 cm 2 per chamber) (Nunc Lab-Tek II chambered cover glasses).
Quantification of nuclear capsids.
To assess nuclear capsid binding, multiple confocal images were taken of infected cells. Images were central nuclear sections. Each green dot present on the nuclear annulus was counted and scored as representing one capsid. To validate the method, a time course was quantified. At early time points (0 and 1 h postinfection), very few capsids were present on the nuclear surface (Fig. 1A). At 2 h postinfection, a moderate increase was observed, and a sharp increase in nuclear capsids was seen by 3 h postinfection (Fig. 1A). Additionally, various virus dilutions were tested in infection. In these experiments, green dots on the nuclear surface increased with increasing multiplicities of infection (MOIs) (Fig. 1B).
FIG. 1.
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Nuclear capsid quantification. (A) Graph of nuclear capsid binding over time. (B) Graph of nuclear capsid binding increase with increased MOI.
Microscopy and image processing.
All fluorescence microscopy was done with a Nikon Eclipse TE2000-E fluorescence microscope (Improvision Open Lab software) or a Zeiss LSM 510 confocal microscope. Prior to grouping images, a calibration image was used to find the optimal contrast for each individual color. All experimental images were then adjusted to those calibration image settings. After images were compiled and grouped, the contrast was enhanced on two of the grouped figures using Adobe's Photoshop software.
siRNA.
Vero cells were plated in 96-well plates (0.32-cm 2 growth area per well) in antibiotic-free MEM with 10% fetal calf serum at 60% confluence 24 h prior to transfection. A lipid transfection reagent was used to deliver targeted RNA duplexes and nontargeted control RNA duplexes at a final concentration of 100 nm. More specifically, pools of targeted double-stranded RNA duplexes were obtained from Dharmacon (ON-TARGETplus SMARTpools). Four sequence targets were used per gene (Nup358 sense sequences, GCGAAGUGAUGAUAUGUUU, CAAACCACGUUAUUACUAA, CAGAACAACUUGCUAUUAG, and GAAGGAAUGUUCAUCAGGA; TPR sense sequences, GAAGAAGUGCGUAAGAAUA, UCAGUUGACUCCAGGAAUA, UCAAGGAGGUUUAGGAAUG, and GGCAUACACUUACUAGAAA; and Nup214 sense sequences, UCAAAUACCUCUAACCUAU, GCAACACCCUCCACUAAAG, GAAAUCGUGACUCUGGUUA, and CCACAAGCCUAACUAGUAC). Sequence targets were derived from human sequences because of a lack of green monkey (Vero cell) sequence availability. Each duplex pool was diluted to 10 μM in 1× buffer (10 mM KCl, 6 mM HEPES [pH 7.5], 0.2 mM MgCl 2). DharmaFECT lipid reagent 1 was diluted 1:8 in serum-free, antibiotic-free MEM. siRNAs were further diluted in serum-free, antibiotic-free MEM (7.4 μl 10 μM siRNA with 105 μl MEM). This siRNA solution was incubated for 30 min at room temperature with lipid and serum-free MEM (100 μl siRNA, 10 μl lipid, 90 μl MEM). After incubation, 450 μl MEM with 10% fetal calf serum was added to each sample. One hundred microliters of this 100 nM solution was added per well and allowed to incubate at 37°C with 7.5% CO 2 for 24 h. After 24 h, transfections were repeated. Forty-eight hours post-initial transfection, cells were trypsinized and replated into eight-well chambered cover glasses to a confluence of 60%. At 72 h post-initial transfection, cells were infected with the K26GFP strain of HSV-1 at an MOI of 300. For infection, cells were brought to room temperature. Concentrated virus was added and allowed to attach for 1 h at 4°C in MEM with 1% fetal calf serum. The medium was then aspirated to remove any unattached virus. The medium was replaced with warm MEM with 1% fetal calf serum, and cells were incubated at 37°C with 7.5% CO 2 for 3 h. Cells were then washed one time with 1× PBS and fixed by submersion in 4% paraformaldehyde for 5 min. Cells were then washed three times with 1× PBS before being immunostained (described below) and viewed in the fluorescence microscope.
Immunofluorescence.
Postfixation, cells were permeabilized by one of two methods: (i) submersion in ice-cold methanol for 4 min or (ii) submersion in digitonin (40 μg/ml in PBS) for 10 min at room temperature. Methanol-permeabilized cells were then washed once in 1× PBS, and digitonin-permeabilized cells were washed three times in 1× PBS, followed by blocking with 5% goat serum in 1× PBS at room temperature for 1 h. Primary antibody was added at the optimized dilution for 1 h at room temperature. Cells were washed three times in 1× PBS. Secondary antibodies (Alexa-conjugated goat antimouse or goat antirabbit [Molecular Probes]) were added at a dilution of 1:2,000 for 1 h at room temperature. Cells were washed three times in 1× PBS, and a 4′,6′-diamidino-2-phenylindole stain (0.5 μg/ml in 150 mM Tris-Cl [pH 7.5]) was added for 5 min. Cells were washed three times in 1× PBS and viewed by fluorescence microscopy.
Native labeling.
Cells were grown in eight-well chambered cover glasses to 75% confluence. Cells were washed one time in 1× PBS and permeabilized by the addition of digitonin (40 μg/ml in PBS) for 10 min at room temperature. Cells were washed three times in 1× PBS and labeled as described above. After a final wash following secondary antibody removal, cells were fixed by submersion in 4% paraformaldehyde for 5 min.
Colocalization analysis.
Confocal images were opened in the ImageJ software program (National Institutes of Health) and run through the Colocalization Finder plugin (C. Laummonerie and J. Metterer, 2006) to generate images in which colocalized pixels appear white. Images were then opened in Adobe Photoshop. The numbers of white pixels and green pixels present on the nuclear surface were determined. The number of white pixels for each nucleus was divided by the total white and green pixels for the nucleus and expressed as a percentage. A total of 50 nuclei per sample were analyzed.
Antibodies.
Anti-CD3 antibody was used as a control in syringe loading experiments. Anti-CD3 was purified from the hybridoma cell line CRL-1975 (29) (provided by Timothy P. Bender) and conjugated to Alexa-594 succinimidyl ester (Molecular Probes) to produce a control antibody that emits red fluorescence. The labeling instructions provided with the amine-reactive probe were followed. For Western blots, importin-β was used as a loading control and probed with monoclonal anti-importin-β (31H4; Sigma-Aldrich). For syringe loading experiments and immunofluorescence experiments, the following antibodies were used: anti-VP5 (6F) (34), anti-VP23 (1D2) (37), UL37 rabbit antiserum (42) (provided by Frank Jenkins), VP1/2 rabbit antiserum (48) (Schipke & Sodeik, personal communication) (provided by Beate Sodeik and Ari Helenius), Nup358 rabbit antiserum (49) (provided by Elias Coutavas and Günter Blobel), and Nup214 QE5 (39) (provided by Bryce Paschal). For syringe loading, monoclonal antibodies were used at a concentration of 1 mg/ml and serum was used undiluted, with the exception of Nup358 serum, which was diluted 1:100. For immunofluorescence, antibodies were used at the following dilutions: VP5, 1:250; VP23, 1:500; VP1/2, 1:250; UL37, 1:300; Nup358, 1:250; Nup214, 1:250; and TPR (1A8, Novus Biologicals), 1:600. For Western blots, antibodies were used at the following dilutions: Nup358, 1:1,000; Nup214, 1:500; TPR, 1:1,250; importin-β, 1:4,000.
RESULTS
Antibodies specific for VP1/2 and Nup358 reduce HSV-1 nuclear capsid binding.
To test the involvement of specific herpesvirus proteins and nucleoporins in the process of capsid-nucleus attachment, antibodies specific for selected proteins were introduced directly into the cytoplasm of cells by syringe loading (see Methods). The cells were subsequently infected with HSV-1, and nuclear binding was assessed to determine if the presence of antibodies affected the capsid's ability to interact with the host nuclear pore. It was reasoned that antibodies would physically block capsid-nucleus interactions if they were bound to sites (either on NPCs or capsids) important for nuclear binding. We introduced herpesvirus-specific antibodies, control antibodies, and nucleoporin-specific antibodies to test the ability of each to perturb capsid-nucleus attachment. Antibody uptake was confirmed by coloading with an Alexa 594-conjugated control antibody. Cells that exhibited a red fluorescence signal in their cytoplasm were considered effectively loaded (Fig. 2A). Antibodies were introduced 3 to 4 h prior to infection with K26GFP HSV-1. After infection for 3 h, nuclear capsid binding was assessed by fluorescence microscopy and counts of capsids bound to the surface of the nucleus (visualized as green punctum present on the nuclear surface) (Fig. 2A). A reduction in the number of nuclear capsids was interpreted as antibody interference with the capsid's ability to interact with the nuclear pore, indicating a role for the protein in the binding process.
FIG. 2.
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Inhibition of nuclear capsid binding in syringe-loaded Vero cells. (A) Fluorescence micrographs of syringe-loaded Vero cells infected with HSV-1 K26GFP. Note that accumulations of capsids are present at the nuclear surface of cells loaded with the following antibodies: control, VP23, VP5, UL37, and Nup214 (arrows). (B) Graph showing the number of nucleus-bound capsids as a percentage of control in cells syringe-loaded with herpesvirus antibodies (anti-VP5, anti-VP23, anti-VP1/2, and anti-UL37). (C) Graph of nuclear capsid binding inhibition in cells loaded with nucleoporin antibodies (anti-Nup358 and anti-Nup214). In panels B and C, bars represent means of duplicate or triplicate (VP1/2 and Nup358) experiments with the standard error shown (, P = 0.0041; , P< 0.0001). An average of between 40 and 50 cells were analyzed per sample per experiment. Cells were infected 3 to 4 h after antibody loading. Nuclear binding was assessed 3 h postinfection.
The results show a binding reduction with anti-VP1/2, one of four herpesvirus antibodies tested (Fig. 2A and B). The presence of antibodies to VP5, the triplex component VP23, and UL37 in the cytoplasm of infected cells did not affect the ability of capsids to accumulate at the nuclear surface (Fig. 2A and B). The presence of antibodies to VP1/2 in the cytoplasm of infected cells led to a 51% reduction in nuclear capsid binding compared to results for controls (Fig. 2B). The ability of VP1/2 antibodies to reduce nuclear capsid binding is interpreted as evidence that VP1/2 is involved early in infection.
Since herpesvirus capsids bind on the cytoplasmic side of the nuclear pore, similar experiments were performed with antibodies specific for the cytoplasmic nucleoporins. Like capsid antibodies, nucleoporin-specific antibodies were introduced into the cytoplasm of cells prior to infection with the K26GFP strain of HSV-1. The presence of antibodies to the cytoplasmic filament protein Nup358 in the cytoplasm of infected cells coincided with a 48% reduction in nuclear capsid binding (Fig. 2A and C). A reduction was not seen with antibodies specific for the protein Nup214 (Fig. 2A and C). This result supports our hypothesis that capsids interact with the nuclear pore by way of the cytoplasmic filaments, specifically the filament protein Nup358.
Herpesvirus antibodies recognize target antigens in infected cells.
To verify that herpesvirus antibodies were capable of binding to their target antigen in the context of intact capsids, immunofluorescence experiments were performed on cells infected with K26GFP HSV-1. In the case of VP5, VP23, and VP1/2, the antibody signal was coincident with the GFP capsid signal (Fig. 3A) in fixed cells. In unfixed cells, VP1/2 and UL37 showed staining coincident with the GFP capsid signal (Fig. 3B). Costaining was observed with virions at the cell surface and also with capsids at the nuclear surface (Fig. 3A and B). In unfixed cells, background levels were too high to allow analysis in cells stained with VP5 and VP23 antibody. The coincidence of the two fluorescent signals indicates that all four antibodies recognize capsid-associated antigen.
FIG. 3.
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Immunofluorescent labeling of herpesvirus proteins and nucleoporins. (A and B) Fluorescence micrographs of cells infected with K26GFP and immunolabeled for VP5, VP23, VP1/2, and UL37. Vero cells were infected with HSV-1 for 3 h, immunolabeled with antibodies against HSV-1 proteins, and visualized with Alexa 594-conjugated secondary antibodies. Capsids were detected by GFP signal. Note that GFP and protein labels occur coincidently at the cell surface (arrowheads) and at the nuclear surface (arrows) for VP5, VP23, VP1/2, and UL37. Scale bar, 10 μm. Cells in panel A were fixed prior to labeling, while cells in panel B were fixed after labeling. DAPI, 4′,6′-diamidino-2-phenylindole. (C) Unfixed cells were immunolabeled with nucleoporin antibodies. Both Nup358 and Nup214 antibodies exhibited a nuclear staining pattern consistent with interaction with their target antigen. Scale bar, 10 μm.
Native labeling with nucleoporin antibodies.
Native labeling experiments were performed to verify that nucleoporin antibodies bound to their target antigen under conditions similar to those in syringe-loaded cells. For both Nup358 and Nup214, the antibody was introduced into the cytoplasm of unfixed cells. For each antibody, the resulting stain was seen exclusively at the nuclear surface (Fig. 3C). This is consistent with the expected staining pattern and thus indicates that Nup358 and Nup214 antibodies bind native antigen when introduced into the cytoplasm of cells.
Specific depletion of Nup358, Nup214, and TPR by siRNA.
To further investigate the roles of nucleoporins in capsid-nucleus attachment, siRNAs were used to selectively remove nucleoporins from cells. Three nucleoporins were removed: TPR, Nup214, and Nup358. As a nuclear basket protein, TPR is not physically present on the cytoplasmic side of the nuclear pore, where herpesvirus capsids bind; therefore, its removal was not expected to influence nuclear capsid binding (9, 23). Protein reduction was assessed by Western blotting and immunofluorescence (Fig. 4). The greatest decreases in protein levels were observed at 72 h posttransfection (Fig. 4B to D), so this time was adopted for further experiments. Due to conflicting reports in the literature concerning the effect of silencing of Nup214 on the nuclear localization of Nup358 (6, 25), we performed double immunofluorescence labeling experiments to determine the effect of removal of Nup214 on Nup358 localization in our system.
FIG. 4.
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Nucleoporin removal by siRNA treatment. (A) Fluorescence micrographs of immunolabeled Vero cells transfected with siRNAs specific for one of three nucleoporins: cytoplasmic filament protein Nup358, cytoplasmic facing protein Nup214, or nuclear basket protein TPR. Panels show nucleoporin signal (red) in siRNA-treated cells relative to signal in untreated cells. Treatment is indicated in the bottom left corner of each image, and staining is indicated at the top of each image column. (B to D) Western blots with importin-β serving as a loading control. Samples were taken 72 and 96 h posttransfection. “s” indicates cells transfected once. “d” indicates cells transfected twice, at 0 and 24 h post-initial transfection. (B) Western blot of Vero cells transfected with Nup358 siRNA alone and Nup358 siRNA plus Nup214 siRNA. (C) Western blot of Vero cells transfected with TPR siRNA. (D) Western blot of Vero cells transfected with Nup214 siRNA.
We found that in cells treated with Nup214 siRNAs, removal of Nup214 resulted in a concurrent loss of Nup358 from the nuclear surface (Fig. 5) but not from the cell. With Nup214 siRNA treatment, cells maintained normal levels of Nup358 (Fig. 6B) as determined by Western blotting (data not shown), but localization to the nuclear surface was reduced. Nuclear loss of Nup358 was not observed in cells treated with either nonspecific control siRNAs or TPR siRNAs (data not shown). Treatment of cells with Nup358 siRNAs did not appear to reduce Nup214 nuclear localization (Fig. 5). These observations were consistent with a previous report that Nup214 plays a role in anchoring Nup358 to the nuclear surface (6); however, there exists some debate about this interaction (25, 47) (see the review in reference 31 for a discussion of relevant literature). We therefore cannot dismiss the possibility that off-target effects from Nup214 siRNA treatments are responsible for the Nup358 mislocalization.
FIG. 5.
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Nup214 and Nup358 localization in siRNA-treated Vero cells. Fluorescence micrographs show nuclear localization patterns of Nup214 and Nup358 with three different treatments. Vero cells were transfected with siRNA specific for Nup214 or Nup358 or were given no treatment. Seventy-two hours after transfection, cells were fixed and immunostained for Nup214 and Nup358. Treatment is indicated to the left of the image row. Label is indicated at the top of the column of images. Note that untreated cells exhibit nuclear staining for both Nup358 and Nup214. Nuclear Nup358 and Nup214 staining are greatly reduced in Nup214 siRNA-treated cells. Nuclear Nup358 but not Nup214 is reduced in cells treated with Nup358 siRNA. DAPI, 4′,6′-diamidino-2-phenylindole. Scale bar, 10 μm.
FIG. 6.
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Nuclear capsid binding in siRNA-treated cells. (A) Fluorescence micrographs of immunolabeled and 4′,6′-diamidino-2-phenylindole-stained siRNA-treated Vero cells infected with HSV-1 K26GFP. The siRNA treatment is indicated in the top left corner of each image. The cellular protein stained red is indicated in the bottom left. Capsids appear green. White arrows indicate accumulations of capsids at the nuclear surface 3 h postinfection. Scale bar, 10 μm. (B) Graph of Nup358 protein concentration (dark gray) and nuclear capsid binding (light gray) with standard error shown (, P = 0.0025; , P< 0.0001; , P = 0.002). Note that nuclear capsid binding reduction corresponds to Nup358 protein concentration reduction except in the case of Nup214 siRNA, in which capsid binding is reduced in the absence of Nup358 protein reduction.
Nuclear capsid binding is reduced in cells depleted of Nup358 or Nup214.
We infected siRNA-transfected cells 72 h after transfection, the time point of maximum protein reduction. As in syringe loading experiments, nuclear capsid binding was assessed by counts of GFP-labeled capsids present on the nuclear surface. Images of infected cells showed that untreated cells or cells treated with nontargeted siRNA or control TPR siRNA had accumulations of GFP-labeled capsids at the nuclear surface (Fig. 6A). Such accumulations were less often observed in cells treated with Nup358 siRNA, Nup214 siRNA, or both Nup358 and Nup214 siRNA (Fig. 6A). Counts from confocal images are summarized in Fig. 6B. The results showed that removal of the nuclear basket protein TPR failed to reduce the nuclear binding of capsids (Fig. 6B). In contrast, removal of Nup214, Nup358, or both Nup214 and Nup358 reduced nuclear capsid binding to 54%, 54%, and 61% of binding in control samples, respectively (Fig. 6B). For Nup358 siRNA-treated cells, nuclear capsid reduction corresponded to Nup358 protein concentration reductions to 60% of the control level (Fig. 6B). Our results suggest Nup358 is an important cellular factor involved in anchoring herpesvirus capsids to host nuclei. While Nup358 appears to be essential, these results are less revealing about Nup214's role in the binding process, since its removal could not be uncoupled from the removal of Nup358.
Nup358 resides exterior to Nup214 on the nuclear pore and colocalizes with HSV-1 capsids.
To better understand the relative locations of Nup358, Nup214, and herpesvirus capsids, immunolabeled cells were examined by fluorescence microscopy. Immunogold studies have previously shown that Nup358 resides exterior to Nup214 on the nuclear pore (47). In light of this, single Nup214 and Nup358 stains were performed on cells infected with K26GFP HSV-1 to see if a difference could be observed in the degree of colocalization between GFP HSV-1 capsids and either Nup214 or Nup358. Capsids were found to colocalize with Nup358 (Fig. 7A, panels b and d) and to a lesser degree with Nup214 staining (Fig. 7A, panels a and c). The degree of colocalization was quantified and is expressed here as the percentage of the nuclear GFP capsid signal that overlapped with the red nucleoporin stain (Fig. 7B). The above results support the idea that the capsid attaches to Nup358.
FIG. 7.
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Colocalization of herpesvirus capsids with Nup358 and Nup214. (A) Immunofluorescence of HSV-1 K26GFP-infected Vero cells. Cells were infected for 3 h and then fixed and stained for nucleoporins (red Nup214 in panels a and c or red Nup358 in panels b and d). High magnification of nuclear capsids is shown in panels c and d. A greater degree of colocalization was seen between capsids and Nup358 (panels b and d) than between capsid and Nup214 (panels a and c). Scale bar, 5 μm (panels a to d). (B) Graph of percent colocalization between capsids and nucleoporins. Each bar represents the number of green nuclear pixels overlapping with red nuclear pixels divided by the total number of green nuclear pixels, expressed as a percentage.
DISCUSSION
Virus replication can begin only after the viral genome has been delivered to a site capable of supporting the production of viral gene products and replication of the viral genome. In the case of most DNA viruses, that means delivery of the viral DNA into the host cell nucleus. This presents DNA viruses with an interesting set of problems. The virus must sequester the genome from the cellular environment with a protective protein coat or capsid during trafficking to the nucleus while retaining the ability to release the genome from its protective location only at the appropriate time to allow uptake into the nucleus. The signals and mechanisms viruses use to coordinate these essential steps are not well understood.
Some viruses employ a stepwise disassembly of the protective layers, initiated upon entry into the cell and completed at the host nucleus. Adenoviruses use this strategy. First, the fiber protein is removed when the virus is internalized. As the partially uncoated capsid is released from the endosome, a viral protease is activated, leading to further disassembly (10, 21, 22). Upon reaching the nuclear pore, the capsid interacts with Nup214, anchoring it to the pore (46). It is hypothesized that a loss of capsid integrity precedes DNA release. Following a loss of capsid integrity, the DNA translocates through the NPC as a complex with viral proteins (20, 41).
Herpesvirus virions also employ a stepwise disassembly; however, unlike the case with adenoviruses, herpesvirus capsids remain intact through the process. The viral envelope is removed from the capsid when it fuses with the cell membrane during entry. Much of the tegument also dissociates from the capsid upon entry and during transit to the nucleus (18, 19, 32, 33, 43). We have described tegument that is released from the capsid as loosely associated tegument and tegument that remains capsid bound (VP1/2 and UL37) as tightly associated (Fig. 8). Upon reaching the nucleus, HSV-1 capsids become anchored to the NPC. Nucleus-bound capsids have the interesting quality of being bound above the NPC with a vertex facing the pore. Our studies support the hypothesis that capsids attach to the NPC by way of an interaction with Nup358. Once bound to the NPC, a protease of unknown origin cleaves the tightly associated tegument, inducing a change that releases the viral DNA (26). HSV-1 genome uncoating resembles uncoating exhibited by double-stranded DNA bacteriophage, with the genome being extruded through the opening of a channel at a unique portal-containing vertex (35).
FIG. 8.
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Model of capsid binding at the nucleus. The model depicts the fate of HSV-1 capsids in newly infected cells. After fusion at the plasma membrane, the capsid and tegument enter the cytoplasm. Upon entry, loosely associated tegument is separated from the capsid, while a subset of tightly associated tegument proteins (VP1/2, UL37) remain attached as the capsid transits to the nucleus. Once at the nucleus, a vertex resident protein, possibly VP1/2, interacts with the pore to anchor the capsid with its distinctive vertex-to-pore orientation. Two nuclear pore proteins are highlighted in this model: Nup358 and Nup214. Herpesvirus capsids are shown bound to Nup358, the putative cytoplasmic filament protein.
Here we sought to develop a system that would allow us to identify viral and cellular proteins involved in nuclear capsid binding in the context of a live infection. In syringe loading experiments, the presence of antibodies to VP1/2 but not other herpesvirus proteins in the cytoplasm of infected cells led to a decrease in the number of capsids bound to the nuclear surface (Fig. 2B). The ability of VP1/2 antibodies to reduce nuclear capsid binding is interpreted as evidence that VP1/2 is involved early in infection.
This experiment was designed to test the involvement of herpesvirus proteins in capsid nucleus attachment. However, two nonexclusive hypotheses could explain our results. VP1/2 antibodies may reduce nuclear capsid binding by attaching to capsids, physically interfering with the capsids' ability to interact with the nucleus, or VP1/2 antibodies may attach to capsids and impair the capsids' ability to travel to the nucleus, perhaps by perturbing the capsids' interaction with microtubule motor proteins.
Both interpretations are consistent with the observation that VP1/2 remains capsid associated in transit to the nucleus (19, 32). VP1/2 likely resides at the capsid vertices. Its location on the capsid depends on its interaction with the minor capsid protein UL25, which is found at the capsid vertices (8, 36, 51). A role for VP1/2 in nuclear binding could therefore explain the capsid's ability to orient itself at the nuclear pore with a vertex facing the center of the pore (18, 38).
Previous studies with HSV-1 provide further support for the hypothesis that VP1/2 is involved in contacting nuclear pores. Ojala and colleagues showed that proteolytic digestion of tegumented capsids reduced nuclear capsid binding in vitro. In the experiment, the following tegument proteins were digested: VP1/2, VP13/14, VP16, and VP22 (38). VP1/2 is likely the only one of the digested tegument proteins present on incoming capsids that reach the NPC.
Alternatively, VP1/2 may function in transporting capsids to the nucleus. This interpretation would be consistent with in vitro studies with HSV-1 showing that inner tegument proteins promote motility along microtubules (48).
Experiments were also done to investigate the involvement of nucleoporins in capsid-nucleus attachment. In syringe loading experiments, Nup358 antibodies alone were found to reduce nuclear capsid binding (Fig. 2C). The inhibition of nuclear binding upon addition of Nup358 antibodies indicates that Nup358 may play an important role in anchoring herpesvirus capsids to nuclei. The capsids' distance from the pore when bound (40 to 50 nm) (18, 38) is consistent with an interaction with Nup358, which extends out from the pore as eight filaments (35 to 50 nm in length) (5, 17, 30). Additionally, fluorescence microscopy studies showed that fluorescently labeled capsids colocalize with Nup358 to a greater degree than with Nup214 (Fig. 7). Nup358's role as the nuclear receptor for herpesvirus capsids was further supported by the reduction in nuclear binding seen when Nup358 was removed by siRNA treatment of cells (Fig. 6A and B).
In light of published literature (6), it was not entirely surprising that nuclear binding was also reduced in cells treated with Nup214 siRNA (Fig. 6A and B). Nup214's localization to the cytoplasmic entrance of the NPC channel and not to the cytoplasmic filaments (47) makes it improbable that Nup214 plays a direct role in capsid-nucleus attachment. Rather, the reduction of nuclear capsid binding in cells treated with Nup214 siRNA is more likely a result of the loss of nuclear Nup358 in these cells (confirmed by fluorescence microscopy [Fig. 5]).
Binding of herpesvirus capsids to host nuclei is a required step in herpesvirus infections. It is the first step in the complex and poorly understood process of genome uncoating. Our results support a model (Fig. 8) in which capsids attach to the nuclear surface by way of an interaction with Nup358. The previous report citing the involvement of importin-β in nuclear capsid binding points to an interaction between capsids and Nup358 that may be indirect (38). It is tempting to imagine that herpesviruses hijack the importin-β nuclear import pathway for delivery of genome-containing capsids to the nucleus. It was recently demonstrated that the large tegument protein VP1/2, shown here to play an important role early in infection, possibly at the stage of nuclear capsid binding, contains a potent nuclear localization sequence (1). One can imagine importin-β recognizing and binding to incoming capsids via VP1/2's nuclear localization sequence or another region on the capsid and thus carrying the capsid cargo to the nucleus. Further, Nup358 has recently been shown to play an important role in importin α/β-dependent nuclear import (24). Once at the nucleus, capsid-bound importin-β could interact with Nup358 to anchor capsids to the NPC. Regardless of the involvement of other cellular factors, Nup358 appears to play an essential role in nuclear capsid binding.
Acknowledgments
We thank the following generous contributors of reagents: Bryce Paschal, Timothy P. Bender, Günter Blobel and Elias Coutavas, Ari Helenius, and Frank Jenkins. We also thank Dean Kedes, Laura Adang, and Rebecca Mingo for many useful discussions of experimental results and Laura Adang, Fred Homa, and Lucy Pemberton for critical review of the manuscript.
This work was supported by NRSA fellowship 1 F31 NS055455-01A1 and NIH award AI041644.
Footnotes
▿
Published ahead of print on 10 December 2008.
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3072 | https://jason-shi-f9dm.squarespace.com/s/2019Geometry_A-SOLS.pdf | Geometry A Solutions 1. A right cone in xyz-space has its apex at (0, 0, 0), and the endpoints of a diameter on its base are (12, 13, −9) and (12, −5, 15). The volume of the cone can be expressed as aπ. What is a?
Proposed by Nathan Bergman and Jacob Wachsprass.
Answer: 975 Solution: The center of the base is the midpoint of the diameter (12, 4, 3). Then the height of the cone is 13 and the radius of the cone is 15, so the volume is 975π.
2. Let ∆ABC be a triangle with circumcenter O and orthocenter H. Let D be a point on the circumcircle of ABC such that AD ⊥BC. Suppose that AB = 6, DB = 2, and the ratio area(∆ABC) area(∆HBC) = 5. Then, if OA is the length of the circumradius, then OA2 can be written in the form m n , where m, n are relatively prime nonnegative integers. Compute m + n.
Note: The circumradius is the radius of the circumcircle.
Proposed by Oliver Thakar.
Answer: 29 Solution: The key observation here is that ∆BDC and ∆BHC are in fact congruent. Then, AB×AC HB×HC = AB×AC DB×DC = 5.
Because ABCD is an orthogonal cyclic quadrilateral, we get the following relations: AB2 + DC2 = 4OA2 and AC2 + DB2 = 4OA2 In combination with AB × AC = 5DB × DC, we can solve for OA. We know that 3 5AC = DC. Thus, AB2 + DC2 = AC2 + DB2 reduces to 32 + 9 25AC2 = AC2, so that AC2 = 50. That makes 4OA2 = 50 + 4, so OA2 = 27 2 , or OA = 3 q 3 2.
3. Suppose we choose two real numbers x, y ∈[0, 1] uniformly at random. Let p be the probability that the circle with center (x, y) and radius |x −y| lies entirely within the unit square [0, 1] × [0, 1]. Then p can be written in the form m n , where m and n are relatively prime nonnegative integers. Compute m2 + n2.
Proposed by Sam Mathers.
Answer: 10 Solution: The key observation here is that ∆BDC and ∆BHC are in fact congruent. Then, AB×AC HB×HC = AB×AC DB×DC = 5.
First, suppose x > y, then we have the conditions x −y < y and x −y < 1 −x. The point of intersection of these two inequalities is when x = 2y and y = 2x −1 so x = 2 3 and y = 1 3.
Thus, the acceptable region for (x, y) is within the triangle with vertices (0, 0), ( 2 3, 1 3), and (1, 1). This has area 1 6. Multiplying this by two since we also have the case y > x, we get an area of 1 3 so the answer is 10.
4. Let BC = 6, BX = 3, CX = 5, and let F be the midpoint of BC.
Let AX ⊥BC and AF = √ 247. If AC is of the form √ b and AB is of the form √c where b and c are nonnegative integers, find 2c + 3b.
Proposed by Alan Chung. Solution by Aleksa Milojevi´ c.
Answer: 1288 Solution: Consider the circles C1 with center B and radius BX = 3 and C2 with center C and radius CX = 5. Their radical axis is a line through X perpendicular to BC, i.e. the line 1 AX. As A is on this radical axis, we get that its power w.r.t. circles C1 and C2 is the same: AB2 −BX2 = AC2 −CX2.
On the other hand, by the median formula in triangle ABC we have that AF 2 = 1 2BA2 + 1 2CA2 −1 4BC2. The two equations we have got are enough to find AB2 and AC2. Namely: AB2 + AC2 = 2(AF 2 + 1 4BC2) = 2 · 256 = 512 AB2 −AC2 = BX2 −CX2 = −16 By solving this system we get AB2 = c = 248 and AC2 = b = 264. Therefore, 2c + 3b = 1288.
5. Let Γ be a circle with center A, radius 1 and diameter BX. Let Ωbe a circle with center C, radius 1 and diameter DY , where X and Y are on the same side of AC. Γ meets Ωat two points, one of which is Z. The lines tangent to Γ and Ωthat pass through Z cut out a sector of the plane containing no part of either circle and with angle 60◦. If ∠XY C = ∠CAB and ∠XCD = 90◦, then the length of XY can be written in the form √a+ √ b c for integers a, b, c where gcd(a, b, c) = 1. Find a + b + c.
Proposed by Zachary Stier.
Answer: 16 Solution: Let the circles have radii a, c and let the angle at Z be θ. We first compute AC.
∠AZC = θ+2(90◦−θ) = 180◦−θ so the Law of Cosines gives b = AC = √ a2 + c2 + 2ab cos θ.
The given angle conditions make AXY C cyclic, and the right angle makes XY the diameter (of length d) of the circumcircle. Using Ptolemy and Pythagoras, we get the equation d3 −d(a2 + b2+c2)−2abc = 0; plugging in: d3−5d−2 √ 3 = 0. Taking d = d √ 3 gives 3d3−5d−2 = 0 which has root −1, so the original has root − √ 3; factoring gives the only positive value, d = √ 3+ √ 11 2 for a final answer 16 .
6. Let two ants stand on the perimeter of a regular 2019-gon of unit side length. One of them stands on a vertex and the other one is on the midpoint of the opposite side. They start walking along the perimeter at the same speed counterclockwise. The locus of their midpoints traces out a figure P in the plane with N corners. Let the area enclosed by convex hull of P be A B sinm( π 4038) tan( π 2019) , where A and B are coprime positive integers, and m is the smallest possible positive integer such that this formula holds. Find A + B + m + N.
Proposed by Jackson Danger Blitz.
Answer: 6065 .
Solution: The area of regular a n-gon with circumradius R is nR2 sin π n cos π n . The cir-cumradius of the starting n-gon is R0 = 1 2 sin( π n).
The locus consists of a 2019-star, which has 2019 isosceles triangles affixed to a smaller 2019-gon, for 4038 total corners.
The convex hull of this star is a 2019-gon. We can find it’s circumradius by noting that its vertices are precisely the points which are midpoints of a segment connecting a vertex of the 2019-gon and the midpoint of the opposite side.
Then a simple calucation gives that R = R0(1−cos( π 2019)) 2 . Plugging this into the formula we get that the area of the convex hull is 2019 4 sin4( π 4038) tan( π 2019) . Furthermore, sin π 4038 is not rational when taken to the powers 1, 2, 3. Hence the A = 2019, B = 4, m = 4, N = 4038, for a final answer of 6065.
2 Note: We also accepted the interpretation of a corner to not include the self-intersections.
This gives us that N = 2019. Following the above logic yields the answer of 4046, which we also accepted.
7. Let ABCD be a trapezoid such that AB ∥CD and let P = AC ∩BD, AB = 21, CD = 7, AD = 13, [ABCD] = 168. Let the line parallel to AB trough P intersect circumcircle of BCP in X. Circumcircles of BCP and APD intersect at P, Y . Let XY ∩BC = Z. If ∠ADC is obtuse, then BZ = a b , where a, b are coprime positive integers. Compute a + b.
Proposed by Aleksa Milojevi´ c. Solution by Igor Medvedev.
Answer: 17 Solution: The heigh of the trapezoid is 168 14 = 12.
By using Pythagoras theorem we find BC = 15. Now we claim BZ = CZ. With this BZ = 15 2 , so the answer would be 17.
Let M be the point of intersection of circumcircle of ∆APD and XP other than P.
We have ∠DY M = ∠DPM = ∠BPX = ∠BY X, and similarly ∠MY A = ∠MPA = ∠XPC = ∠XY C. Furthermore ∠ADP = ∠APY = ∠PCB + ∠CBP −∠Y PB = ∠Y CB + ∠CBY − ∠Y PB = ∠CBY . Hence triangles ∆Y DA and ∆Y BC are similar. The unique spiral similarity between these two triangles has center Y , call it Φ : ∆Y BC 7→∆Y DA.
Let N be the point of intersection of AD and Y M. Then Φ(Z) = N, because ∠BY Z = ∠BY X = ∠DY M = ∠DY N. This means that Y Z Y X = Y N Y M . This implies that NZ||MX.
From this we get CZ ZB = DN NA . On the other hand, Φ(Z) = N, so BZ ZC = DN NA , hence BZ ZC = CZ ZB , hence CZ = BZ.
8. Let γ and Γ be two circles such that γ is internally tangent to Γ at a point X. Let P be a point on the common tangent of γ and Γ and Y be the point on γ other than X such that PY is tangent to γ at Y . Let PY intersect Γ at A and B, such that A is in between P and B and let the tangents to Γ at A and B intersect at C. CX intersects Γ again at Z and ZY intersects Γ again at Q. If AQ = 6, AB = 10 and AX XB = 1 4. The length of QZ = p q √r, where p and q are coprime positive integers, and r is square free positive integer. Find p + q + r.
Proposed by Mel Shu. Solution by Igor Medvedev.
Answer: 28 Solution: P lies on the polar of C so CZ must be the polar of P so PZ is a tangent to Γ. In particular, PZ = PY so there exists a circle γ′ tangent to PY and PZ at Y and Z respectively.
Then the homothety centered at Z which takes γ′ to Γ takes Y to Q, so Q is the midpoint of arc AQB. It follows that AQ = BQ. Similarly XY is a bisector of angle ∠AXB.
Then AX XB = BY Y A by the angle bisector theorem.
Hence BY = 2, AY = 8.
By Stewart’s theorem applied to ABQ we have QY 2 = AQ2 −BY · Y A = 20. Then by power of a point, Y Z = 16 √ 20. Then QZ = 18 5 √ 5.
3 |
3073 | https://pmc.ncbi.nlm.nih.gov/articles/PMC7164101/ | The association of viral proteins with host cell dynein components during virus infection - PMC
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FEBS J
. 2011 Aug 8;278(17):2997–3011. doi: 10.1111/j.1742-4658.2011.08252.x
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The association of viral proteins with host cell dynein components during virus infection
Javier Merino‐Gracia
Javier Merino‐Gracia
1 Departamento de Bioquímica y Biología Molecular I, Universidad Complutense, Madrid, Spain
Find articles by Javier Merino‐Gracia
1, María F García‐Mayoral
María F García‐Mayoral
2 Departamento de Química‐Física Biológica, Instituto de Química‐Física Rocasolano, Madrid, Spain
Find articles by María F García‐Mayoral
2, Ignacio Rodríguez‐Crespo
Ignacio Rodríguez‐Crespo
1 Departamento de Bioquímica y Biología Molecular I, Universidad Complutense, Madrid, Spain
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1
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1 Departamento de Bioquímica y Biología Molecular I, Universidad Complutense, Madrid, Spain
2 Departamento de Química‐Física Biológica, Instituto de Química‐Física Rocasolano, Madrid, Spain
✉
I. Rodriguez‐Crespo, Departamento de Bioquímica y Biología Molecular I, Universidad Complutense, 28040 Madrid, Spain Fax: +34 91 394 4159 Tel: +34 91 394 4137 E‐mail: nacho@bbm1.ucm.es
Received 2011 Feb 8; Revised 2011 Jul 8; Accepted 2011 Jul 13; Issue date 2011 Sep.
© 2011 The Authors Journal compilation © 2011 FEBS
This article is being made freely available through PubMed Central as part of the COVID-19 public health emergency response. It can be used for unrestricted research re-use and analysis in any form or by any means with acknowledgement of the original source, for the duration of the public health emergency.
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PMCID: PMC7164101 PMID: 21777384
Abstract
After fusion with the cellular plasma membrane or endosomal membranes, viral particles are generally too large to diffuse freely within the crowded cytoplasm environment. Thus, they will never reach the cell nucleus or the perinuclear areas where replication or reverse transcription usually takes place. It has been proposed that many unrelated viruses are transported along microtubules in a retrograde manner using the cellular dynein machinery or, at least, some dynein components. A putative employment of the dynein motor in a dynein‐mediated transport has been suggested from experiments in which viral capsid proteins were used as bait in yeast two‐hybrid screens using libraries composed of cellular proteins and dynein‐associated chains were retrieved as virus‐interacting proteins. In most cases DYNLL1, DYNLT1 or DYNLRB1 were identified as the dynein chains that interact with viral proteins. The importance of these dynein–virus interactions has been supported, in principle, by the observation that in some cases the dynein‐interacting motifs of viral proteins altered by site‐directed mutagenesis result in non‐infective virions. Furthermore, overexpression of p50 dynamitin, which blocks the dynein–dynactin interaction, or incubation of infected cells with peptides that compete with viral polypeptides for dynein binding have been shown to alter the viral retrograde transport. Still, it remains to be proved that dynein light chains can bind simultaneously to incoming virions and to the dynein motor for retrograde transport to take place. In this review, we will analyse the association of viral proteins with dynein polypeptides and its implications for viral infection.
Keywords: dynein, DYNLL1, DYNLT1, infection, retrograde transport, virus
After fusion with the cellular plasma membrane or endosomal membranes, viral particles are generally too large to diffuse freely within the crowded cytoplasm environment. It has been proposed that many unrelated viruses are transported along microtubules in a retrograde manner using the cellular dynein machinery or, at least, some dynein components, in most cases the light chains DYNLL1, DYNLT1 or DYNLRB1.
Abbreviations
ASFV
African swine fever virus
DYNC1H
dynein heavy chain
DYNC1I
intermediate chain
DYNC1LI
light intermediate chain
DYNLL1
dynein light chain LC8
DYNLRB1
dynein light chain roadblock
DYNLT1
dynein light chain Tctex
GFP
green fluorescent protein
HHV
human herpes virus
HIV
human immunodeficiency virus
HSV
herpes simplex virus
MTOC
microtubule organizing centre
PV
papillomavirus
RV
rabies virus
siRNA
small interfering RNA
Introduction
The function of eukaryotic cells relies on the transport of macromolecules and small organelles throughout the cytoplasm. Microtubules are polar cytoskeletal filaments assembled from thousands of α/β tubulin heterodimers which are nucleated and organized by the perinuclear microtubule organizing centre (MTOC). Whereas kinesin and dynein motors use microtubules to move cargo throughout the cytoplasm, myosin motors interact with actin filaments [1, 2, 3]. Cytoplasmic dynein, frequently in cooperation with its cofactor dynactin, is a minus‐end‐directed microtubule associated motor responsible for retrograde transport (towards the nucleus) in eukaryotic cells [4, 5, 6]. This molecular motor is a large multiprotein complex of approximately 1.2 MDa that contains heavy chains, intermediate chains, light intermediate chains and light chains. Cytoplasmic dynein plays critical roles in a variety of eukaryotic cellular functions, including Golgi maintenance, nuclear migration, retrograde axonal transport and organelle positioning. In addition, cytoplasmic dynein is involved in numerous aspects of mitosis, such as spindle formation and organization, spindle orientation and mitotic checkpoint regulation [7, 8].
Cargoes transported by dynein are linked to the motor via the tail, which consists of an N‐terminal section of one or more heavy chains and a number of associated polypeptides [4, 5, 9, 10]. In general, the cytoplasmic dynein complex is resolved on SDS/polyacrylamide gels into subunit polypeptides of ∼ 530 (dynein heavy chains, DYNC1H), ∼ 74 (intermediate chains, DYNC1I), ∼ 53–59 (light intermediate chains, DYNC1LI) and ∼ 10–14 kD (light chains). Cytoplasmic dynein heavy chain, with ∼ 4650 amino acids in humans, is among the largest polypeptides found in mammalian cells. Dimeric dyneins have a conserved tail structure in which the heavy chains dimerize through protein–protein interactions mediated by amino acids 300–1140 in the tail region (Fig.1) . In addition, dynein heavy chain residues 446–701 and 649–800 are involved in intermediate chain and light intermediate chain binding respectively . In the case of cytoplasmic dynein, intermediate chain is also able to form homodimers through interactions mediated by residues 151–250 . Furthermore, in mammals, the 74 kDa cytoplasmic intermediate chain is homologous to the 67 kDa axonemal intermediate chain . Three highly conserved light chains, also shared by axonemal dyneins, Tctex1 (DYNLT1), LC8 (DYNLL1) and roadblock (LC7 or DYNLRB1), bind to distinct regions of the intermediate chains, always as homodimers [14, 15]. DYNLL, a ubiquitous molecule, is the most highly conserved among light chains and, interestingly, in spite of lacking sequence homology displays a three‐dimensional fold almost identical to that of DYNLT [16, 17, 18]. Both DYNLT and DYNLL form homodimers with α‐helices flanking a shared central β‐sheet, with the peptides from interacting partners lengthening the preformed β‐strand. Indeed, according to the recently published crystal structures [19, 20] the two short, consecutive protein stretches of dynein intermediate chain that bind to DYNLT and DYNLL adopt an extended conformation (Fig.1). However, DYNLRB, also a ubiquitous component of cytoplasmic dyneins, is structurally different from Tctex1 and LC8, with a dimer interface that includes the coiled‐coiled pairing of two α‐helices . Unlike when binding to DYNLT or DYNLL, two sequential helical segments of dynein intermediate chain associate to DYNLRB (Fig.1). In addition to these three light chain homodimers, there is also a dimer of DYNC1LI bound to the tail of cytoplasmic dyneins. Consistent with a role in cargo binding, DYNC1LI has been shown to bind to the core centrosomal protein pericentrin . Thus, since dynein light and light intermediate chains appear to associate with cargo, it has been suggested that viruses might target these cellular components during infection processes.
Figure 1.
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General architecture of the cytoplasmic dynein motor complex. The positions of DYNLL (dynein light chain LC8, green), DYNLT (dynein light chain Tctex, yellow), DYNLRB (dynein light chain roadblock, dark blue), DYNC1I (dynein intermediate chain, violet; the carboxy‐terminal WD40 repeat (β‐propeller) is depicted as a heptagon), DYNC1L (dynein light intermediate chain, light blue) and DYNC1H (dynein heavy chain, red) are shown. The crystallographic structures of the DYNLT1 homodimer (yellow) and DYNLL1 homodimer (green) binding to adjacent sequences from the dynein intermediate chain (violet) are shown at the bottom (PDB accession number In addition, the crystallographic structure of the DYNLRB homodimer (PDB accession number depicted in blue is shown on top. Note that the crystal structure of the motor domain of the dynein heavy chain has also been recently reported .
In human cells, the DYNLT1 and DYNLL1 homodimers bind to LGMAKITQVDF and KETQTP motifs respectively, both contiguously positioned in the intermediate chain sequence in which they adopt an extended conformation (Fig.1) [19, 20, 23, 24]. It is generally assumed that most of the cellular and viral polypeptides that also bind to DYNLT1 or DYNLL1 have protein stretches with similarities to these intermediate chain sequences and bind in a similar fashion [25, 26, 27, 28].
Following cell entry, several viruses exploit the cellular cytoplasmic transport mechanisms to allow them to travel long distances through the cytoplasm and reach their site of replication [29, 30, 31, 32] (Fig.2). Viruses require active transport along microtubules since diffusion of particles larger than 50 nm in diameter is restricted by the structural organization of the cytoplasm . Experiments in which microtubule‐depolymerizing agents such as colchicine, nocodazole or vinblastine were used have shown that the integrity of the microtubules is essential for virus infection. Many pathogens causing widespread illness including herpes simplex virus (HSV) [34, 35], adenovirus [35, 36], hepatitis B virus , human cytomegalovirus , human immunodeficiency virus (HIV) , African swine fever virus (ASFV) , parvovirus , influenza virus , papillomavirus (PV) and rabies virus (RV) rely on microtubules for efficient nuclear targeting and successful infection.
Figure 2.
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Viral retrograde transport model. Both the entry of the viral particle through the endosome pathway (A, C) and the direct fusion of the viral envelope to the plasma membrane (B) lead to the retrograde transport along microtubules using the cytoplasmic the dynein motor. Viral capsid proteins might associate to dynein directly (A) or a cellular receptor might bind simultaneously to a viral protein and the dynein motor (C). After reaching the MTOC at the microtubular minus end viruses are uncoated and directed to the sites of replication, production and assembly of the new viral proteins (D) such as the nucleus or the viral factories. The newly assembled particles might become transported to the cell periphery by the anterograde transport machinery.
Labelling of unenveloped viruses, such as SV40 or adenoviruses , with fluorescent dyes has proved extremely useful to show their retrograde transport. Likewise, to initiate infection, herpes virus must attach to cell surface receptors, fuse its envelope to the plasma membrane and allow the de‐enveloped capsid to be transported to the nuclear pores [47, 48, 49]. More recently, video microscopy using green fluorescent protein (GFP) tagged viral proteins has also demonstrated the retrograde advance of viruses towards the cell nucleus. In particular, retrograde transport of HSV in axons has been visualized using time‐lapse fluorescent microscopy . Similarly, GFP‐tagged poliovirus receptor, which associates to dynein light chain DYNLT1, colocalizes with Alexa fluor 555‐labelled poliovirus and both undergo retrograde transport along microtubules of cultured motor neurons . Likewise, GFP‐tagged ASFV protein p54 was found associated to microtubules during infection and nocodazole treatment abrogated this association .
Remarkably, when virus proteins were used as baits and screened against a library of cellular proteins, several dynein polypeptides, mostly dynein light chains DYNLL1 (LC8), DYNLT1 (Tctex) and DYNLRB1 (roadblock), were retrieved as interacting partners.
On the basis of these results, the hijack of the dynein motor by numerous viruses has been proposed as a common mechanism for virus delivery near the cell nucleus replication site [29, 30, 31, 32]. In this review we will analyse in detail this association and its biological significance.
Herpesviruses
Alphaherpesviruses such as HSV‐1 are unique parasites of the vertebrate peripheral nervous system. Primary infection usually occurs at an epithelial surface, after which the virus invades the termini of sensory and autonomic neurons that innerve the infected tissue. HSV‐1 binds to cell surface receptors, then loses its envelope after cell fusion and subsequently virion components, including the tegument and capsid layers together with the double‐stranded DNA, are transported in a retrograde manner along axons towards the cell bodies of these neurons [50, 53, 54]. Since simple diffusion does not allow the viral components to travel long distances, cellular microtubule‐based motors must be involved in alphaherpesvirus transport [31, 55]. In fact, the dynein motor is known to co‐localize with inbound cytosolic capsids of HSV‐1 [34, 56], virus infection can be blocked by over‐expression of the dynactin subunit p50 (dynamitin) and proteins from both the capsid and the tegument of HSV‐1 are known to associate with dynein protein members. Likewise, immunofluorescence studies of pig monocytes infected with pseudorabies virus, also a member of the Alphaherpesvirinae family, showed a clear co‐localization with dynein .
Protein UL34 of HSV‐1 is an integral membrane protein that is targeted to the inner nuclear membrane and only transiently associated with viral particles during the passage through the nuclear envelope during egress of newly assembled capsids from the nucleus into the cytosol . Surprisingly, it has been reported that UL34 is associated with the microtubular network and binds to the dynein intermediate chain (DYNC1I1) . The importance of this interaction is hard to evaluate in vivo, since UL34 from HSV‐1 and the homologous genes in all other herpesviruses encode for an essential protein .
Using a yeast two‐hybrid screen, fifteen HSV‐1 proteins were confronted with dynein light chains DYNLL1, DYNLT1 and DYNLT3, and a strong interaction could be detected between both UL35(VP26) and UL46(VP11/12) and the dynein light chain DYNLT1 and its homolog DYNLT3 . In fact recombinant capsids of HSV‐1 were microinjected into the cytoplasm, and those decorated with VP26 showed a stronger tendency to accumulate at the nuclear envelope . Besides a role of VP26 in recruiting the dynein motor, these data may also suggest that VP26 is somehow involved in binding to the nuclear pores. Nevertheless, additional HSV‐1 proteins must associate with the dynein motor, since capsids from HSV‐1 mutants that lack VP26 can still bind to purified dynein [62, 63]. In addition, studies with HSV‐1 lacking VP26 have shown no significant effect on dynein‐dependent retrograde viral transport in cell cultures . Recent results seem to indicate that both dynein (retrograde movement motor) and kinesin (anterograde movement motor) can bind isolated capsids of HSV‐1 in vitro , hence raising the possibility that this association might be important for certain steps of the viral life cycle or that these two cellular motors must somehow be coordinated for viral infection to succeed . Likewise, it has been reported that VP26 of pseudorabies virus is also not required for intracellular transport . Finally, HSV‐1 mutants lacking VP26 are not impaired in animal experiments that rely on axonal transport .
Moreover, HSV‐1 UL9, a protein that binds to the viral origin of replication, displays a consensus DYNLL1 binding motif (746‐KSTQT‐750) that is functional when tested as an isolated dodecapeptide .
The Betaherpesvirus human herpes virus 7 (HHV‐7) is known to infect CD4+ T lymphocytes and epithelial cells of salivary glands. Its protein U19, probably a transactivator according to its similarity to human cytomegalovirus UL38, displays a consensus sequence for DYNLL1 binding (RSTQT) repeated in tandem at its carboxy terminus. Using a pepscan technique, these sequences were found to efficiently associate with DYNLL1 . Interestingly, no similarity exists between HHV‐7 and HHV‐6 at this region, thus indicating that the association to this dynein light chain might be restricted to the HHV‐7 isolate.
Rabies virus
RV belongs to the lyssavirus genus of the Rhabdoviridae family, with rabies and Mokola virus as reference strains. They are small, enveloped, single‐stranded negative‐sense RNA viruses, whose genome is tightly encapsidated into a ribonucleoprotein complex with the viral proteins of the nucleocapsid, the RNA polymerase and the phosphoprotein P as non‐catalytic cofactor. RV is a highly neurotrophic virus that enters the organism by bites or injuries in the skin and muscle, where it replicates. It then enters the neuronal endings of peripheral nerves, such as neuromuscular junctions, to reach the central nervous system and causes lethal encephalitis in animals and humans. After receptor binding, RV enters its host cells through the endosomal pathway via a low‐pH‐induced membrane fusion process catalysed by the glycoprotein G, a major determinant for RV neuropathogenicity . The retrograde transport along microtubules has been shown recently by using fluorescently labelled RV glycoprotein. Incubation of in vitro differentiated NS20Y neuroblastoma cells with fluorescently labelled virus clearly showed the transport in the retrograde direction over long distances in neurites . Subsequently, all transcription and replication events take place in the cytoplasm, inside a specialized virus factory referred to as the Negri body .
Experiments performed simultaneously by two independent groups used the rabies and Mokola virus phosphoproteins as baits in yeast two‐hybrid screens and retrieved DYNLL1 as an interacting partner using PC12 cells and human brain libraries [70, 71]. Fine mapping of the DYNLL1 binding site within the P phosphoprotein of these two lyssaviruses revealed the presence of a KSTQT motif in rabies and KSIQI motif in Mokola that constituted the interacting polypeptide stretch. The P phosphoprotein of rhabdovirus is a cofactor of the RNA polymerase complex and, in fact, facilitates the binding of the polymerase to the N‐RNA complex . Using the SAD‐D29 low‐virulent strain of RV, deletion of the DYNLL1 binding motif in the P phosphoprotein resulted in a remarkable viral attenuation after intramuscular but not after intracranial inoculation . Unfortunately, no difference could be observed between wild‐type and recombinant viruses in which the DYNLL1 binding sequence had been deleted in the P phosphoprotein when non‐attenuated viral strains were used. Other authors have concluded that the deletion of the DYNLL1 binding site in phosphoprotein P did not produce a biologically important impairment of viral transport in the nervous system .
Some recent data seem to indicate that mutations in the DYNLL1 binding site within the P phosphoprotein of RV significantly attenuated viral transcription and replication in the central nervous system, hence showing that DYNLL1 binding to the viral protein has a more crucial role in viral polymerase activity than in the intracellular transport of the virus . This is in agreement with the nuclear staining of the P phosphoprotein where it co‐localizes with promyelocytic leukaemia protein . In this context, it is interesting to note that the DYNLL1 binding motif of the phosphoprotein of RV when fused to a reporter protein is not able, by itself, to promote active import into the cell nucleus although it can facilitate nuclear protein import when appended to proteins with nuclear localization sequences .
African swine fever virus
ASFV, the only member of the family Asfarviridae, is a large double‐stranded DNA virus that codes for approximately 150 proteins. ASFV enters the cell by dynamin‐ and clathrin‐dependent endocytosis, and its infectivity depends on the acidification of the endosome . Elegant studies by Alonso and co‐workers have shown that ASFV p54, a major protein of virion membranes, associates with DYNLL1, which allows the transport of the virus to the MTOC in the cell perinuclear area . Fine mapping in yeast two‐hybrid assays, site‐directed mutagenesis and the presence of a polypeptide stretch in p54 with the sequence TASQT that closely resembles the DYNLL1 consensus binding motifs [25, 27] led to the identification of the DYNLL1 binding region in p54 [78, 79]. In fact, small peptide inhibitors that display this binding sequence together with an internalization sequence can disrupt the interaction between p54 and DYNLL1 altering both infectivity and the viral egress . Likewise, the inhibition of the dynein–dynactin complex formation by the overexpression of p50 dynamitin blocks ASFV transport in infected cells . Thus, the interaction of ASFV p54 with DYNLL1 is required for efficient infectivity, virus replication and viral production yields.
Papillomavirus
PVs are small non‐enveloped double‐stranded DNA viruses that infect the stratified epithelia of skin and mucous membranes. The icosaedric capsid contains 360 copies of the major capsid protein L1 and up to 72 molecules of the minor capsid protein L2 . Whereas the L2 protein is required for egress of the viral genome from endosomes, L1 does not appear to exit the endosomal compartment . Based on immunofluorescence and co‐immunoprecipitation experiments, the L2 protein was found attached to microtubules after uncoating of incoming human PV pseudovirions . Then, the minor capsid protein L2 accompanies the viral DNA to the nucleus and subsequently to the subnuclear promyelocytic leukemia protein bodies . Since L2 and the viral genome co‐localize in the nucleus at promyelocytic leukemia protein bodies, it has been suggested that they are associated in the nucleus forming a complex . Site‐directed mutagenesis and deletion studies showed that the carboxy‐terminal region of L2 somehow associates with the dynein motor . This observation is in agreement with previous experiments that had shown that bovine PV binds to microtubules and becomes transported along them, and suggested the possibility that dynein is involved in this process .
Recently, a yeast two‐hybrid screen using PV L2 as bait against a human cDNA library retrieved dynein light chain DYNLT1 as a tight binder. In addition, in vitro binding studies and cotransfection experiments in HeLa cells proved that L2 was also able to bind to its homologue DYNLT3 . Subsequent studies have shown that depletion of DYNLT1 or DYNLT3 using small interfering RNA (siRNA) treatment inhibited human PV‐16 infection, whereas infection was increased after overexpression of these dynein light chains .
It must also be noted that human PV has another polypeptide, termed E4 (also known as E1^E4), that is expressed from an E1^E4 spliced mRNA prior to the assembly of infectious virions and accumulates to very high levels in cells supporting productive infection . Several PV types, such as 08, 47 or 21, display a polypeptide stretch with the sequence KQTQT that conforms a consensus binding sequence for DYNLL1. In vitro binding assays have shown that this sequence does, indeed, bind to DYNLL1 tightly , although no studies have been performed yet to demonstrate that this interaction occurs during viral infection.
Poliovirus
Poliovirus is an enteric virus that rarely causes disease in humans. Nevertheless, in the pre‐vaccine era ∼ 1% of infected individuals developed paralytic poliomyelitis due to viral invasion of the central nervous system and destruction of motor neurons. To gain access and sustain infection in neurons, a neurotropic virus such as poliovirus must be able to efficiently traffic in axons, which can be up to 1 m long. CD155, the human poliovirus receptor, is a member of the immunoglobulin superfamily, with three linked extracellular Ig‐like domains followed by a membrane‐spanning domain and a short cytoplasmic domain. Intramuscularly inoculated poliovirus is known to become incorporated into neural cells after binding to the first Ig‐like domain of CD155 followed by endocytosis [86, 87]. Then, the cytoplasmic domain of CD155 is known to associate to the dynein light chain DYNLT1 [86, 88] and subsequently the endosomes, together with the CD155‐bound poliovirus, undergo retrograde transport along microtubules through the axon to the neural‐cell body, where the uncoating and replication of poliovirus occur [51, 87].
Alternative splicing generates two membrane‐bound CD155 isoforms: CD155α and CD155δ. Yeast two‐hybrid analyses have identified the 50‐residue cytoplasmic domain of CD155α and the 25‐residue cytoplasmic domain of CD155δ as DYNLT1 binding partners [87, 88]. Subsequent studies have revealed that a basic motif adjacent to the transmembrane domain is required for efficient binding. In addition, purified recombinant DYNLT1 binds to the cytoplasmic domain when fused to glutathione S‐transferase in vitro .
Retrovirus
Spumaviruses, also known as foamy viruses, target the microtubule organizing centre prior to nuclear translocation. Hence, centrosomal targeting of incoming viral proteins and subsequent viral replication can be inhibited by nocodazole treatment . The efficiency of MTOC targeting was analysed by using various GFP‐tagged Gag mutant constructs of human foamy virus transfected in cultured cells, and a region located around amino acids 150–180 was found necessary for this subcellular localization. In this regard, a Leu171Gly Gag mutant displayed drastically reduced infectivity of the proviral clone . Interestingly, when COS6 cells were transfected with wild‐type Gag, but not with its Leu171Gly mutant, dynein light chain DYNLL1 could be co‐immunoprecipitated. However, the direct interaction between DYNLL1 and human foamy virus Gag protein has not been unequivocally proved .
HIV enters the cells following virus binding to CD4 and co‐receptors and the fusion of the viral membrane with the plasma membrane of the cell. During passage through the cytosol, the viral RNA genome is reverse transcribed into DNA within a structure named the reverse transcription complex that, eventually, must be imported into the nucleus, where the HIV genome is integrated into a chromosome. Initial reports concluded that depolymerization of cell microtubules with nocodazole had little effect on virus infection whereas actin depolymerization had a profound effect on infection . Subsequent studies used a GFP‐tagged Vpr incorporated into virions and in vivo fluorescence to show a microtubule‐dependent transport towards the MTOC positioned in perinuclear areas . Interestingly, the individual treatment of infected cells with either nocodazole or the F‐actin inhibitor latranculin B did not impede the movement of GFP‐labelled particles, whereas the simultaneous treatment with both compounds led to a cessation of movement. This suggests that HIV movement inside the cell depends on both actin and the microtubule network. Furthermore, the direct implication of dynein in HIV movement was further shown when infected cells were injected with anti‐dynein antibodies and viral migration along microtubule networks decreased significantly . Nonetheless, in the case of HIV and its binding to the dynein motor the exact viral protein and its dynein partner remain to be identified.
By means of broad yeast two‐hybrid screens, HIV integrase was found to bind to the yeast dynein light chain Dyn2p, the orthologue of mammalian DYNLL1. When analysed inside yeast, HIV integrase associates to the microtubular network and accumulates at the spindle pole body, the yeast equivalent of mammalian perinuclear MTOC . In fact, nocodazole treatment of transfected yeast or transfection of the GFP–integrin construct in a Δdyn 2 mutant strain resulted in the aberrant localization of HIV integrase. Nevertheless, it is not known if HIV integrase binds to DYNLL1 in mammalian cells or if this association is required for efficient virus infection.
Interestingly, the matrix protein of Mason–Pfizer monkey virus, the archetypal D‐type retrovirus, binds directly to dynein light chain DYNLT1, according to yeast two‐hybrid assays, in vitro association of recombinant proteins and in cell immunoprecipitation assays . Indeed, this association might be responsible for the retrograde transport of Gag‐synthesizing polysomes alongside microtubules or perhaps for other steps of the viral cycle. It is not known, nonetheless, if other retroviral matrix proteins associate to DYNLT1 as well.
Finally, the dynein motor has also been involved in the regulation of viral Gag and viral genomic RNA egress on endosomal membranes. In this regard, following transcription and nuclear export, the viral genomic RNA might transit towards the MTOC where it interacts with Gag proteins in a dynein‐mediated process .
Adenovirus
Adenoviruses are 90–100 nm diameter non‐enveloped dsDNA viruses that exit to the cytosol soon after receptor‐mediated endocytosis. Early studies revealed that adenoviruses associate to the dynein motor [36, 95, 96] and microinjection of function‐blocking anti‐dynein but not anti‐kinesin antibodies abolished the viral nuclear localization, consistent with a net minus‐end‐directed motility . It was then suggested that the subcellular transport of adenoviruses is the result of the equilibrium between dynein (retrograde movement) and kinesin (anterograde movement) forces [36, 46, 98].
Type 2 adenovirus E3 protein, a polypeptide involved in the downregulation of the host’s immune response, binds to a small GTPase (RRAG) that, in turn, is associated with the dynein light chain DYNLT1 . Since this viral polypeptide is not a structural component of the virion, the biological significance of the interaction remains unclear.
Dynein has been implicated in the transport of naked viral capsids from endosomes to the nuclear periphery after virus uncoating in the endosomes . However, incubation of HeLa cells with recombinant adenovirus penton base protein clearly shows that a significant population of the viral protein traffics in a retrograde manner towards the cell nucleus. Cell treatment with nocodazole or transfection with p50/dynamitin abrogates retrograde transport by at least ∼ 50% .
Recent analysis has revealed that the viral capsid hexon subunit interacts directly with the dynein intermediate chain . Using immunoprecipitation studies as well as antibody microinjection experiments the adenovirus hexon binding site was selectively localized to a single site within the intermediate chain and no significant interactions were observed with any of the three dynein light chains DYNLL1, DYNLT1 or DYNLRB1 .
Other viruses
Several other virus proteins have been reported to interact with the dynein motor. For instance, the E protein of severe acute respiratory syndrome coronavirus, a small integral membrane protein of 76 amino acids, associates, directly or indirectly, with dynein heavy chain when overexpressed in Vero cells . In addition, its non‐structural protein 3 is also known to bind to multiple cellular proteins in infected Vero cells, including the dynein heavy chain, although it is not known if this binding is direct or mediated by other dynein light chains .
Yeast two‐hybrid screening has also revealed that virion protein 35 of ebolavirus binds to DYNLL1 through the consensus binding sequence SQTQT . Interestingly, VP35 inhibits type I interferon production, thereby suppressing host innate immunity, an activity analogous to that of the P protein from RV, which also binds to DYNLL1.
In the case of canine parvovirus the role of dynein in viral infection has been inferred not only from the observation that intact microtubules are required for the traffic of viral particles towards the nucleus but also by the fact that microinjection of anti‐dynein antibodies reduced the nuclear accumulation of viral capsids and immunoprecipitation of dynein in infected cells co‐immunoprecipitated viral capsid proteins [41, 104].
A very recent paper described a silencing screen using siRNAs targeted against 5516 different cellular genes, with each gene being covered by three independent siRNAs. When Borna disease virus was used to infect an oligodendroglial cell line, silencing of dynein light chain DYNLRB1 significantly blocked virus infection. Although the exact viral protein that associated to DYNLRB1 has yet to be identified, this elegant approach has revealed the implication of this dynein light chain in viral infection .
The dynein motor is also involved in influenza virus infection. Surprisingly, endosomal acidification of this pathogen occurs in perinuclear areas after a dynein‐mediated retrograde transport has taken place, as demonstrated by experiments using anti‐dynein antibody injection .
Finally, using a pepscan technique and after screening multiple viral polypeptides with putative DYNLL1 binding sequences, polypeptides from Amsacta moorei entomopoxvirus, the polymerase from Vaccinia virus or Yam mosaic potyvirus polyprotein were shown to bind to this dynein light chain Nevertheless, additional studies are clearly needed in order to ascertain if these interactions do, indeed, take place during the virus infective cycle.
Selected interactions between viral polypeptides and dynein chain proteins are summarized in Table 1.
Table 1.
Selected viral proteins involved in a direct interaction with dynein polypeptides. The most recent dynein nomenclature is used: DYNLL (dynein light chain LC8), DYNLT (dynein light chain Tctex), DYNLRB (dynein light chain roadblock), DYNC1LI (dynein light intermediate chain) and DYNC1I (dynein intermediate chain).
| Virus | Family | Protein that binds to a dynein polypeptide | Dynein protein | Reference |
:--- :---
| Herpes simplex | Alphaherpesvirinae | Viral UL34 | DYNC1I1a | |
| Viral UL9 (helicase) | DYNLL1 | |
| Viral UL35 (VP26) | DYNLT1 and DYNLT3 | |
| Herpesvirus 7 | Betaherpesvirinae | Viral U19 | DYNLL1 | |
| African swine fever | Asfarviridae | Viral p54 | DYNLL1 | |
| Mokola | Rhabdoviridae | Viral phosphoprotein (P) | DYNLL1 | |
| Rabies | Rhabdoviridae | Viral phosphoprotein (P) | DYNLL1 | [70, 71] |
| Papillomavirus | Papillomaviridae | Viral minor capsid protein L2 | DYNLT1 and DYNLT3 | [43, 82] |
| Viral probable protein E4 | DYNLL1 | |
| Borna disease | Bornaviridae | Probably viral G surface glycoprotein | DYNLRB1 | |
| Poliovirus | Picornaviridae | Cellular CD155 receptor | DYNLT1 | [86, 88] |
| Human immunodeficiency | Retroviridae | Viral integrase | Dyn2p (yeast orthologue of DYNLL1) | |
| Mason–Pfizer monkey | Retroviridae | Viral matrix | DYNLT1 | |
| Adenovirus | Adenoviridae | Viral capsid hexon | DYNC1LI2 DYNC1LI1 | |
| Ebolavirus | Filoviridae | Viral phosphoprotein (VP35) | DYNLL1 | |
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The dimer–dimer hypothesis
Both DYNLL1 and DYNLT1 are protein members of the dynein motor in which they bind contiguously to the dynein intermediate chain [19, 20, 24]. However, only about 40% of total DYNLL1 associates to the dynein intermediate chain in a microtubule pellet of rat brain . Likewise, a significant fraction of DYNLT1 is not associated to microtubules in fibroblasts, as shown by sequential immunoprecipitation .
Although they share no significant sequence similarity, dynein light chains DYNLL1 and DYNLT1 display a very similar three‐dimensional structure and adopt identical ‘geometric specificity’ upon binding to protein ligands [19, 20]. Both of these small proteins are homodimers and structurally consist of two α‐helices followed by five β‐strands, with the second β‐strand being swapped between protomers. The proteins that bind to either DYNLL1 or DYNLT1 do so through polypeptide stretches that adopt an extended β‐strand conformation that inserts into the ligand binding grooves. The consensus protein sequence necessary for binding to dynein light chain DYNLT1 is not well known. However, numerous atomic coordinates are available for dynein light chain DYNLL1 in association with protein partners, and in all cases GIQVD or KXTQT motifs, or variations of them, are inserted into the DYNLL1 binding grove [24, 25, 27, 108]. Using yeast two‐hybrid and mutagenesis experiments, the binding region to DYNLL1 has been narrowed to TASQT for ASFV p54 and KSTQT and KSIQI in the case of the P protein of rabies and Mokola viruses respectively [70, 71] and SQTQT in the case of protein VP35 of ebolavirus . The presence of these sequences that closely resemble the KXTQT motif in various viral proteins suggests that virus association also occurs with the viral protein adopting an extended antiparallel β strand that fits into the DYNLL1 groove and extends the pre‐existing β‐sheet. The atomic coordinates of the modeled ASFV p54–DYNLL1 complex clearly indicate that this is indeed the case . In fact, RV P protein and the pro‐apoptotic Bcl‐2 family member Bim display an identical DYNLL1 binding sequence (DKSTQT) and the published NMR structure of the DYNLL1–Bim complex also shows the β‐sheet augmentation mode of binding .
Dynein light chains have been proposed to mediate cargo binding for their cellular transport. DYNLL1 is a bivalent molecule and many of its interacting partners are dimeric (or oligomeric) proteins . However, linking cargo molecules to dynein is not easily reconciled with binding data . In fact, DYNLL1 binding affinity for a dimeric partner is significantly higher compared with the same partner as a monomer [19, 20]. This observation has led to the hypothesis that two identical polypeptide segments from a dimeric partner occupy both of the binding grooves of DYNLL1 and DYNLT1. If this is the case, it is hard to accept that one DYNLL1 binding site is occupied by a viral protein whereas the other is occupied by the dynein intermediate chain as would be required for retrograde transport. Although this is theoretically possible, binding of a viral polypeptide to either DYNLL1 or DYNLT1 would require the displacement of the dynein intermediate chain from one of the binding sites, which is a thermodynamically unfavourable process. In this regard, it must be mentioned that, for instance, in the case of DYNLT1, peptides from the intermediate chain compete with the G protein β subunit , an indication that both peptides bind at the same location and therefore DYNLT1 cannot be simultaneously binding to both proteins. Therefore, if the DYNLL1 dimer (and by analogy the DYNLT1 dimer) binds to either two chains of the dynein intermediate chain or two chains of putative cargo proteins at the same location, how can viruses be transported in a retrograde manner towards the minus end of microtubules associated to the dynein motor?
It is conceivable that viral polypeptides, either as part of the virion or detached after uncoating, associate to dynein light chains using both binding sites simultaneously when these light chains are not part of the dynein motor [Fig.3A, binding modes (a) and (b)]. Hence, binding to dynein light chains might promote dimerization of viral polypeptides through the binding to intrinsically disordered regions, a function that has recently been assigned to DYNLL1 [111, 112]. This would mean that the interaction of viral polypeptides with dynein light chains is not responsible for the association with microtubules and might be responsible for other processes during the infective cycle. Moreover, this would rationalize the fact of why several viral proteins that are neither envelope glycoproteins nor belong to the virion capsid associate to dynein light chains.
Figure 3.
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(A) Model for the interaction of a generic viral capsid with cytoplasmic, non‐microtubule‐associated DYNLL. It is then conceivable that the DYNLL homodimer might bind simultaneously to two viral polypeptides when part of the viral capsid (a) or when soluble after viral disassembly (b). This is in agreement with the modeled solution structure of the complex of DYNLL1 with p54 of ASFV . (B) Three hypotheses for the association of viral proteins to the dynein molecular motor. One viral polypeptide displaces a dynein intermediate chain from one binding side of the DYNLL homodimer (a). Two DYNLL homodimers associate to one dynein intermediate chain and to one viral polypeptide simultaneously (b). Binding of the viral polypeptides displaces the dynein intermediate chains from the DYNLL binding grooves but DYNLL remains part of the dynein motor through the binding to the dynein heavy chain (red) (c).
If we focus on DYNLL1 only (a similar case could be put forward for viral polypeptides that associate to DYNLT1) three different situations might explain the proposed dynein motor–virus association responsible for the aforementioned retrograde transport.
1
It is conceivable that viral proteins could interact with DYNLL1 with one binding site occupied by the dynein intermediate chain and the opposite site of the same homodimer occupied by the viral protein (Fig.3B‐a).
2
However, since binding of DYNLL1 to dimeric partners is energetically favourable over monomeric partners, it is then conceivable that viral proteins might adopt a conformation that facilitates the binding of two dynein light chain dimers with viral polypeptides alternating with dynein intermediate chains in each groove of the homodimer (Fig.3B‐b).
3
Alternatively, virus polypeptides might bind simultaneously to the two equivalent sites within DYNLL1, hence displacing the dynein intermediate chain, but with the light chains still being part of the dynein motor through interactions with other dynein proteins (Fig.3B‐c). This might occur if the dynein heavy chain (coloured in red) could associate to DYNLL1 through a different surface, such as its α‐helices.
Virus, dynein light chains and apoptosis
Bim and Bmf are two pro‐apoptotic BH3‐only proteins that signal to the cell death machinery by sensing cellular damage. In healthy cells both Bim and Bmf are sequestered away from the sites where pro‐survival Bcl‐2 family members reside (fundamentally the endoplasmic reticulum and mitochondria membranes) through interaction with dynein light chain proteins. In these cells the light chain component of the myosin V motor complex, DYNLL2, binds to the polypeptide stretch DKATQTL present in Bmf , whereas the equivalent component of cytoplasmic dynein, DYNLL1, binds to the polypeptide stretch DKSTQTP present in the Bim isoforms BimL and BimEL . In response to apoptotic stimuli that impact upon the motor complexes, Bim or Bmf in complex with their respective light chains are released into the cytoplasm where they can interact with pro‐survival Bcl‐2 proteins via their BH3 domains.
Since virus infection is frequently associated with cell apoptosis [115, 116] it has been suggested that viral polypeptides and Bim or Bmf compete for the binding of dynein light chains DYNLL1 and DYNLL2. Thus, binding of a viral polypeptide to DYNLL1 or DYNLL2 might release Bim or Bmf which, in turn, would translocate to the mitochondria and initiate apoptosis. In this regard, transfection of Vero cells with ASFV p54 but not with a mutant protein that cannot bind to DYNLL1 triggers the release of microtubule‐associated Bim and the concomitant caspase‐9 and caspase‐3 activation . Hence, apoptosis induced by p54 results from the direct competition between Bim and p54 for their binding to DYNLL1, which suggests that virus–dynein interactions might be important not only in retrograde transport. Likewise, HIV Tat, which binds to microtubules through residues 35–50, induces apoptosis in a Bim‐dependent manner .
Conclusions
In the past few years, numerous studies of virus–host interactions have revealed the role of the dynein motor and the integrity of microtubules in virus infection. The development of microscopy techniques has also enabled the retrograde movement of viruses in the cellular cytoplasm to be tracked. In addition, the in vitro binding and transport assays using complete viral capsids and intact microtubule motors may be instrumental in further characterizing potential functions of the interactions between dynein light chains and viral proteins. Moreover, several viral polypeptides are known to associate to dynein light chains, although many of them do not belong to capsid proteins. However, dynein light chains appear also as homodimers in the cellular cytoplasm, without being part of the dynein motor. Therefore, it remains to be unambiguously established if the interaction of viral polypeptides with dynein light chains links the virion to the dynein motor when bound to microtubules and if this interaction is responsible for the observed viral retrograde transport. Hence, co‐immunoprecipitation of dynein intermediate or heavy chains bound to incoming virions would be an adequate experiment to prove that the interaction of certain viral polypeptides with dynein light chains does indeed link virions to microtubules via the dynein motor. Since DYNLL1 (and probably DYNLT1) work as dimerization clamps that bind to intrinsically disordered regions of proteins it is also conceivable that the dimerization of viral polypeptides might be important for certain viral infective processes distinct from retrograde transport. This might be the case of the interaction of DYNLL1 with the phosphoproteins of RV, Mokola virus and ebolavirus. On the other hand, a conundrum therefore exists in that certain envelope or capsid viral proteins bind to dynein light chains at the same site used by dynein intermediate chain. Thus, whether the binding of viral proteins to dynein light chains displaces dynein intermediate chain and how it is performed remains to be established.
Acknowledgements
This work was supported by grants from the Ministerio de Ciencia e Innovación BFU2009‐10442 and BQU2008‐0080, and by the Consolider‐Ingenio Centrosome 3D CSD2006‐00023. We are especially grateful to Dr Douglas Laurents (Intituto de Química‐Física Rocasolano) for extensive English editing and helpful suggestions.
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3074 | https://www.ajog.org/article/S0002-9378(23)00242-9/fulltext | The physiology and pharmacology of oxytocin in labor and in the peripartum period - American Journal of Obstetrics & Gynecology
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Expert ReviewVolume 230, Issue 3, SupplementS740-S758 March 2024 Open access
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The physiology and pharmacology of oxytocin in labor and in the peripartum period
Kerstin Uvnäs-Moberg, MD, PhD
Kerstin Uvnäs-Moberg, MD, PhD
Correspondence
Corresponding author: Kerstin Uvnäs-Moberg, MD, PhD.
kerstinuvnasmoberg@gmail.com
Affiliations
Department of Animal Environment and Health, Swedish University of Agriculture, Uppsala, Sweden
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kerstinuvnasmoberg@gmail.com
Affiliations & Notes Article Info
Department of Animal Environment and Health, Swedish University of Agriculture, Uppsala, Sweden
Publication History:
Received October 12, 2022; Revised April 5, 2023; Accepted April 9, 2023; Published online July 13, 2023
Footnotes:
Kerstin Uvnäs Moberg owns shares in a company called Oxagon AB, in which the effects of locally applied oxytocin on vaginal atrophy in menopausal women is studied.
DOI: 10.1016/j.ajog.2023.04.011 External LinkAlso available on ScienceDirect External Link
Copyright: © 2023 The Author. Published by Elsevier Inc.
User License: Creative Commons Attribution (CC BY 4.0) | Elsevier's open access license policy
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Abstract
Key words
Introduction
Oxytocin
The oxytocin receptor
Oxytocin during pregnancy, labor, and delivery
Methodological considerations regarding measurements of oxytocin levels
Pregnancy
Labor and delivery
Modulation of oxytocin release during labor
Intravenous administration of synthetic oxytocin for induction and augmentation of labor
Social support and activation of the oxytocinergic system
Conclusions
Supplementary Data (2)
References
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Abstract
Key words
Introduction
Oxytocin
The oxytocin receptor
Oxytocin during pregnancy, labor, and delivery
Methodological considerations regarding measurements of oxytocin levels
Pregnancy
Labor and delivery
Modulation of oxytocin release during labor
Intravenous administration of synthetic oxytocin for induction and augmentation of labor
Social support and activation of the oxytocinergic system
Conclusions
Supplementary Data (2)
References
Article metrics
Related Articles
Abstract
Oxytocin is a reproductive hormone implicated in the process of parturition and widely used during labor. Oxytocin is produced within the supraoptic nucleus and paraventricular nucleus of the hypothalamus and released from the posterior pituitary lobe into the circulation. Oxytocin is released in pulses with increasing frequency and amplitude in the first and second stages of labor, with a few pulses released in the third stage of labor. During labor, the fetus exerts pressure on the cervix of the uterus, which activates a feedforward reflex—the Ferguson reflex—which releases oxytocin. When myometrial contractions activate sympathetic nerves, it decreases oxytocin release. When oxytocin binds to specific myometrial oxytocin receptors, it induces myometrial contractions. High levels of circulating estrogen at term make the receptors more sensitive. In addition, oxytocin stimulates prostaglandin synthesis and release in the decidua and chorioamniotic membranes by activating a specific type of oxytocin receptor. Prostaglandins contribute to cervical ripening and uterine contractility in labor. The oxytocin system in the brain has been implicated in decreasing maternal levels of fear, pain, and stress, and oxytocin release and function during labor are stimulated by a social support. Moreover, studies suggest, but have not yet proven, that labor may be associated with long-term, behavioral and physiological adaptations in the mother and infant, possibly involving epigenetic modulation of oxytocin production and release and the oxytocin receptor. In addition, infusions of synthetic oxytocin are used to induce and augment labor. Oxytocin may be administered according to different dose regimens at increasing rates from 1 to 3 mIU/min to a maximal rate of 36 mIU/min at 15- to 40-minute intervals. The total amount of synthetic oxytocin given during labor can be 5 to 10 IU, but lower and higher amounts of oxytocin may also be given. High-dose infusions of oxytocin may shorten the duration of labor by up to 2 hours compared with no infusion of oxytocin; however, it does not lower the frequency of cesarean delivery.
When synthetic oxytocin is administered, the plasma concentration of oxytocin increases in a dose-dependent way: at infusion rates of 20 to 30 mIU/min, plasma oxytocin concentration increases approximately 2- to 3-fold above the basal level. Synthetic oxytocin administered at recommended dose levels is not likely to cross the placenta or maternal blood-brain barrier. Synthetic oxytocin should be administered with caution as high levels may induce tachystole and uterine overstimulation, with potentially negative consequences for the fetus and possibly the mother. Of note, 5 to 10 IU of synthetic oxytocin is often routinely given as an intravenous or intramuscular bolus administration after delivery to induce uterine contractility, which, in turn, induces uterine separation of the placenta and prevents postpartum hemorrhage. Furthermore, it promotes the expulsion of the placenta.
Key words
augmentation
birth
epigenetic changes
Ferguson reflex
first stage
induction
infusion of oxytocin
labor
myometrial contractions
oxytocin
plasma levels
pregnancy
prostaglandins
pulsatile secretion
receptor
second stage
synthetic oxytocin
third stage
Introduction
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Figure 1 Chemical structure of oxytocin, vasopressin and oxytocin metabolites
Show full caption Figure viewer
A, The figure shows the chemical structure of the oxytocin and vasopressin molecules. Oxytocin consists of a cyclic structure with 6 amino acids (cysteine, tyrosine, isoleucine, glutamine, asparagine, and cysteine) and a linear structure with 3 amino acids (proline, leucine, and glycine). Note that the 2 cysteines are linked together by a disulfide bond and that the glycine at the carboxy or C-terminal end of the molecule is amidated. In vasopressin, the amino acids isoleucine and leucine are exchanged by phenylalanine and arginine, respectively. B, The figure shows the main metabolic pathways by which oxytocin is degraded. As shown by the left pathway, one or several amino acids can be split off from the C-terminal end of the molecule. As shown by the right pathway, the cyclic structure of the oxytocin molecule can be opened up, for example, between the amino acids cysteine and tyrosine, to create linear forms of the oxytocin molecule. The linear oxytocin molecule can be further degraded to shorter fragments by amino acids being split off from both the amino and C-terminal ends of the molecule. In addition, a cysteine molecule may be split off from the linear molecule.
C-terminal, carboxy-terminal.
Uvnäs-Moberg. Physiology and pharmacology of oxytocin. Am J Obstet Gynecol 2024.
Oxytocin is produced in neurons of the supraoptic nucleus (SON) and paraventricular nucleus (PVN) of the hypothalamus and released into the circulation from the posterior pituitary lobe. It induces contractions of the uterus and the myoepithelial cells in the mammary gland during birth and lactation. Furthermore, these same neurons from the SON and PVN project to important regulatory areas within the brain. A plethora of centrally induced, oxytocin-linked, behavioral and physiological effects are induced concerning labor8
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Fritz Fuchs and Anne-Riita Fuchs contributed substantially to our understanding of the role of oxytocin in parturition. Their results help connect recently obtained knowledge regarding oxytocin-linked molecular mechanisms to physiological processes.
Synthetic oxytocin has been used for more than 60 years to induce and augment labor, to decrease the frequency of postpartum hemorrhage (PPH), and sometimes to stimulate milk ejection, such as when an infant is premature.24
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Currently, many birthing women receive infusions of synthetic oxytocin to induce or augment their labor. In addition, hospitals often give mothers a bolus administration of oxytocin after delivery.
This study aimed to review the function of oxytocin during labor, the differences between oxytocin in the plasma and oxytocin’s functions as a neurohormone, and the use of synthetic oxytocin in inducing and augmenting labor.
Oxytocin
Oxytocin is produced at multiple sites and induces many different effects via several mechanisms of action. Because oxytocin receptors are widely distributed, it makes more sense to refer to an oxytocin system rather than to oxytocin.10
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Burbach, J.P.H. ∙ Young, L.J. ∙ Russell, J.A.
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and is transported via axons to the posterior pituitary lobe, from where it is released into the circulation to act as a classical hormone during labor, delivery, and breastfeeding.8–10
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Figure 2 Release of oxytocin from the supraoptic and paraventricular nuclei
Show full caption Figure viewer
Oxytocin from magnocellular neurons from the SON and PVN is transported via axons of the oxytocin neurons to the posterior pituitary, where it is released into the circulation. Oxytocin from the magnocellular neurons of the SON and PVN is released from dendrites of the oxytocin neurons into the surrounding brain tissue, where it reaches neighboring areas in the brain by diffusion. Oxytocin is released into many areas of the brain from axon collaterals emanating from the axons of the magnocellular neurons from the SON and PVN, projecting to the posterior pituitary. The axon collaterals project to several brain regions, including the median eminence and anterior pituitary, amygdala, and cortex. Oxytocin is released from parvocellular neurons from the PVN into many regulatory areas in the brain. The figure illustrates how oxytocin released from the same oxytocin neurons may simultaneously be released into the circulation to stimulate uterine contractions and into the brain to influence brain function.
PVN, paraventricular nucleus; SON, supraoptic nucleus.
Uvnäs-Moberg. Physiology and pharmacology of oxytocin. Am J Obstet Gynecol 2024.
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The oxytocin and vasopressin genes are located on the human chromosome 20p13. The structure of the oxytocin and vasopressin genes and the posttranslational modifications of the 2 peptides are very similar, supporting the assumption that they have developed from a common ancestral gene.15
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Burbach, J.P.H. ∙ Young, L.J. ∙ Russell, J.A.
Oxytocin synthesis, secretion and reproduction
Knobil and Neill’s physiology of reproduction
Elsevier, 2006; 3055-3128
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The function of the oxytocin gene can be enhanced via effects in the promoter region of the gene via estrogen-alpha and estrogen-beta receptors, thyroid alpha receptors, and retinoic acid alpha and beta receptors.15
15.
Burbach, J.P.H. ∙ Young, L.J. ∙ Russell, J.A.
Oxytocin synthesis, secretion and reproduction
Knobil and Neill’s physiology of reproduction
Elsevier, 2006; 3055-3128
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The oxytocin receptor
Localization
Oxytocin receptors, which mediate the effects of oxytocin, are present in the myometrium and myoepithelial cells of the mammary glands and in many peripheral tissues, such as the heart, blood vessels, kidney, ovary, testis, and thymus; adipose tissues; deciduae; and chorioamniotic membranes.10
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In addition, oxytocin receptors are found in many areas of the brain, including the olfactory system, limbic system, hypothalamus, basal ganglia, brain stem, spinal cord, and some cortical regions.17
17.
Jurek, B. ∙ Neumann, I.D.
The oxytocin receptor: from intracellular signaling to behavior
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Structure of the oxytocin receptor
The oxytocin receptor, such as the closely related vasopressin receptor, belongs to the G-protein–type receptor, which transfers information from the cell membrane into the cell via G proteins. Only 1 oxytocin receptor has been demonstrated, whereas 3 different vasopressin receptors exist (V1a, V1b, and V2 receptors). The oxytocin receptor contains 389 amino acids and is a 7-transmembrane helix receptor with 3 extracellular and 3 intracellular loops. The extracellular loops are of great importance for the binding of oxytocin to its receptor (Figure 3).16
16.
Gimpl, G. ∙ Fahrenholz, F.
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Figure 3 Illustration of the binding between oxytocin and its receptor
Show full caption Figure viewer
The figure shows the structure of the oxytocin molecule and oxytocin receptor and a 3-dimensional illustration of the oxytocin receptor within the cell membrane. The oxytocin receptor contains 389 amino acids and is a 7-transmembrane helix receptor (TMH1-7) with 3 extracellular (EL 2-3, EL 4-5, and EL 6-7) and 3 intracellular loops. The extracellular loops are of great importance for binding of oxytocin to its receptor. The figure is redrawn from a freely available figure published in a paper by Jurek and Neumann.17
17.
Jurek, B. ∙ Neumann, I.D.
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The oxytocin and oxytocin receptor complex
The oxytocin receptor is coupled to a trimeric complex of G proteins, consisting of 1 G alpha unit and 1 beta or gamma unit. Several different subforms of the proteins exist, which contribute to different compositions of the G-protein complex. These structural differences, in part, explain why different effects may be induced when oxytocin binds to its receptor.15
15.
Burbach, J.P.H. ∙ Young, L.J. ∙ Russell, J.A.
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Molecular pathways activated in response to binding between oxytocin and its receptor
When oxytocin binds to its receptor, multiple intracellular effects are induced, the effect spectrum, in part, being dependent on the type of cell that is activated. Most experimental studies have been performed on myometrial cells. In this model, when oxytocin binds to its receptor, the beta or gamma unit is separated from the G-alpha protein. This, in turn, leads to the stimulation of phospholipase-C. This enzyme catalyzes phosphatidylinositol 4,5-bisphosphate breakdown to diacylglycerol and inositol 1,4,5, triphosphate and triggers calcium (Ca++) release from the smooth endoplasmic reticulum. This Ca++ activates calmodulin, leading to the phosphorylation of myosin light-chain kinase that phosphorylates myosin light chains, which, in turn, facilitates actin-myosin interaction and contraction. During labor, the frequency of Ca++ oscillations in the myometrial cells is linked to the amount of oxytocin-induced contractions (Figure 4).
Figure 4 Intracellular pathways activated when oxytocin binds to its receptor
Show full caption Figure viewer
Stimulation of the oxytocin receptor leads to stimulation of PLC. This enzyme catalyzes PIP2 breakdown to DAG and IP3 and triggers calcium (Ca++) release from the smooth endoplasmic reticulum. Ca++ activates CaM, leading to phosphorylation of MLCK that phosphorylates myosin light chains, which, in turn, facilitates actin-myosin interaction and contraction. In addition, the MAPK cascade is activated, which stimulates prostaglandin production and muscle contraction.
CaM, calmodulin; DAG, diacylglycerol; IP3, inositol 1,4,5, triphosphate; MAPK, mitogen-activated protein kinase; MLCK, myosin light-chain kinase; PIP2, phosphatidylinositol 4,5-bisphosphate; PLC, phospholipase-C.
Uvnäs-Moberg. Physiology and pharmacology of oxytocin. Am J Obstet Gynecol 2024.
In contrast, in myometrial cells isolated from pregnant rats, oxytocin receptor binding stimulates adenylyl cyclase 2, which suppresses Ca++ and reduces contractions. These data demonstrate an inhibition of the oxytocin-induced contractile effects on myometrial cells during pregnancy, contributing to uterine quiescence during this period.15–17
15.
Burbach, J.P.H. ∙ Young, L.J. ∙ Russell, J.A.
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In addition, intracellular pathways other than those leading to myometrial contractions may be activated in response to binding between oxytocin and its receptor. One of these pathways involves mitogen-activated protein kinase (MAPK) MEK 1-2 and ERK 1-2. By activation of this pathway, oxytocin has been shown to stimulate the growth of cardiomyocytes, stimulate osteoblasts, induce maturation of human mesenchymal stem cells, reverse osteoporosis, and improve muscle regeneration by enhancing muscle stem cell regeneration.44
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The synthesis of prostaglandins in the decidua and chorioamniotic membranes involves an oxytocin-induced activation of the MAPK/ERK pathway.15–17
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Burbach, J.P.H. ∙ Young, L.J. ∙ Russell, J.A.
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During labor, the oxytocin receptor in the myometrium gives rise to myometrial contractions, whereas the oxytocin receptors in the decidua and chorioamniotic membranes are linked with the synthesis and secretion of prostaglandins and inflammatory substances, such as leukotrienes and arachidonic acid.11–14
11.
Chibbar, R. ∙ Miller, F.D. ∙ Mitchell, B.F.
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Although both these effects are mediated by oxytocin, they are linked to specific receptor populations, which activate different intracellular signaling pathways, as described above.15–17
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The oxytocin receptor gene
There is only 1 gene coding for the expression of the oxytocin receptor, which is located on chromosome 3p25(A) in humans. The oxytocin receptor gene consists of 4 exons and 3 introns. Estrogen can bind to the promoter region of the oxytocin receptor gene via both estrogen alpha and estrogen beta receptors to activate transcription of the oxytocin receptor. Progesterone inhibits the transcription and translation of the oxytocin receptor gene, thereby leading to a relative quiescence of uterine smooth muscles and a low frequency of contractions during pregnancy.15–17
15.
Burbach, J.P.H. ∙ Young, L.J. ∙ Russell, J.A.
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Factors that influence the function and availability of the oxytocin receptor
High- and low-affinity state of the oxytocin receptor
The oxytocin receptor may exist in 2 different affinity states: low affinity (Kd of >100 nM) and high affinity (Kd of <1 to 5 nM), which optimizes the physiological activity of oxytocin. The low-affinity state can transfer into the high-affinity state and vice versa; the presence of Mg++ or Mn++ transfers the low-affinity state into the high-affinity state. Cholesterol stabilizes the oxytocin receptor in a high-affinity state.16
16.
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Effects of estrogen
The amount of oxytocin receptors in the myometrium gradually increases during pregnancy in response to the high levels of estrogens, and at term, there is a 100-fold increase in the concentration of oxytocin receptors, showing that the synthesis and number of oxytocin receptors are enhanced.15
15.
Burbach, J.P.H. ∙ Young, L.J. ∙ Russell, J.A.
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Fuchs, A.R. ∙ Periyasamy, S. ∙ Alexandrova, M. ...
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Endocrinology. 1983; 113:742-749
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In addition, estrogen increases the concentration of oxytocin receptors in the decidua and chorioamniotic membranes.11
11.
Chibbar, R. ∙ Miller, F.D. ∙ Mitchell, B.F.
Synthesis of oxytocin in amnion, chorion, and decidua may influence the timing of human parturition
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Chibbar, R. ∙ Wong, S. ∙ Miller, F.D. ...
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Oxytocin receptor internalization
When oxytocin binds to its receptor, it also activates a G-protein–coupled receptor kinase, allowing binding to beta-arrestin, which leads to oxytocin receptor internalization or desensitization. Beta-arrestin uncouples the oxytocin receptor from its G proteins, and in doing so, it allows the receptor to be internalized or incorporated into the cell.15–17
15.
Burbach, J.P.H. ∙ Young, L.J. ∙ Russell, J.A.
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Receptor desensitization
The availability of high levels of ligands over longer periods can result in receptor down-regulation and thereby insensitivity to the ligand. There is limited evidence for this type of down-regulation happening with oxytocin receptors. However, chronic exposure to high amounts of intranasally administered oxytocin in mice decreased the number of oxytocin receptors.15–17
15.
Burbach, J.P.H. ∙ Young, L.J. ∙ Russell, J.A.
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A desensitization of oxytocin receptors, which reduces the function of the receptors, hypothetically might occur after augmentation of labor by infusions of synthetic oxytocin.64–67
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Methylation of the oxytocin receptor gene
The production of oxytocin receptors depends on the level of methylation of the DNA in the promoter region of the receptor as DNA methylation in this region reduces the accessibility of the oxytocin receptor for transcriptional factors. The higher the levels of methylation, the lower the levels of oxytocin receptor messenger RNA (mRNA) and oxytocin receptor production. This mechanism has achieved much attention lately, and studies are being performed to determine whether DNA methylation of oxytocin or the oxytocin receptor occurs after delivery in human mothers and their neonates.15–17
15.
Burbach, J.P.H. ∙ Young, L.J. ∙ Russell, J.A.
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Oxytocin and its receptors increase during pregnancy in preparation for labor and delivery. During labor, peak levels of oxytocin are released into the circulation, which induces uterine contractions to allow the birth of the fetus. Immediately after delivery, peak levels of circulating oxytocin stimulate uterine contractions to expel the placenta and lower the risk of PPH. Moreover, oxytocin is released in response to breastfeeding to contract the myoepithelial cells of the mammary gland to promote milk ejection.8–10
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In addition to these classical contractile effects of circulating oxytocin, oxytocin released from neurons within the brain exerts important adaptative, physiological, and psychological effects. These effects have been reviewed below.
Methodological considerations regarding measurements of oxytocin levels
The gold standard for the analysis of oxytocin concentrations is the immunologic technique radioimmunoassay (RIA), which is a very specific and sensitive technique for oxytocin levels and gives rise to reproducible measurements. RIA can detect oxytocin levels within the picomolar range with great accuracy. The antibodies used in high-quality RIAs are well characterized and are often specific for the entire oxytocin molecule and do not detect linear fragments of oxytocin or cyclic fragments in which amino acids have been split off from the C-terminal end of the molecule (Figure 1, B). The different processes involved in RIA are well described, and plasma samples are always extracted before analysis to diminish unspecific binding.65–67
65.
Robinson, C. ∙ Schumann, R. ∙ Zhang, P. ...
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Most high-quality studies during pregnancy and labor have used RIA. More recently, a simpler immunologic technique without radioactivity, enzyme-linked immunosorbent assay (ELISA), has been developed and used for the analysis of oxytocin concentrations. If the plasma is extracted before analysis, and if the antibodies are well characterized and specific for the cyclic oxytocin molecule, high-quality results can be obtained with ELISA. Without extraction, very high oxytocin levels, and sometimes different effect patterns of oxytocin, are obtained with ELISA, which may be caused by the detection of unspecific material in the samples. Furthermore, if the antibodies are less well characterized, which is sometimes the case with ELISA, measurements may include the detection of fragments or metabolites of oxytocin, for example, linear fragments of oxytocin, which are not detected by RIA (Figure 1, B).51
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Collection and handling of samples
Oxytocin is released in short pulses with different intervals during labor, and therefore, it is necessary to collect multiple repeated blood samples at short time intervals to be able to detect the pulses. In addition, the speed by which blood samples are collected is crucial and must be standardized. Studies in which RIAs have been used most often include data based on multiple, serial blood samples, which will allow visualization of the oxytocin pulses. In contrast, studies using ELISA are generally based on only a few blood samples, reducing the value of the studies.8
8.
Uvnäs-Moberg, K. ∙ Ekström-Bergström, A. ∙ Berg, M. ...
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In addition, blood samples must be kept cold, and plasma must be frozen immediately after collection. Enzyme inhibitors, such as aprotinin (Trasylol), must be added to the samples to avoid the degradation of oxytocin in the tubes.
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Although RIA is a highly sensitive and specific method for the detection of oxytocin levels, it is an indirect method in the sense that the oxytocin values obtained are based on changes in the balance within an equilibrium reaction, involving binding between antibodies to oxytocin and iodinated oxytocin or oxytocin from the samples. Methods that directly quantify the number of oxytocin molecules, such as mass spectrometry, are being developed.
Pregnancy
Oxytocin concentrations
Most studies show a gradual 2- to 4-fold increase in basal oxytocin levels during pregnancy.8
8.
Uvnäs-Moberg, K. ∙ Ekström-Bergström, A. ∙ Berg, M. ...
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The oxytocin-producing neurons in the SON and PVN have estrogen receptors, such as estrogen-beta receptors, through which estrogen activation promotes the synthesis and release of oxytocin.15–17
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As circulating estrogen levels rise substantially during pregnancy, the increased concentrations of oxytocin may at least, in part, be secondary to the increased levels of estrogen. Some studies indicate that elevated oxytocin levels during pregnancy, in addition to the nonapeptide oxytocin, involve the release of an elongated oxytocin molecule.50
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Toward term, oxytocin pulses appear in the circulation, and uterine contractions start to appear.8
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Data obtained from rodents suggest that a central opioidergic mechanism, possibly mediated via mu-opioid receptors, actively inhibits peak levels of oxytocin during pregnancy. Allopregnanolone, a neuroactive metabolite of progesterone present in increasing amounts in the brain during pregnancy, mediates this inhibitory effect on the oxytocin neurons by stimulating the inhibitory opioidergic mechanism via gamma-aminobutyric acid A receptors. As demonstrated in animal experiments, this opioidergic inhibition is withdrawn as the production of progesterone decreases toward the end of pregnancy, allowing peak levels of oxytocin to occur.72–76
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This endogenous inhibitory mechanism is intended to reduce the risk of premature labor. In addition, the limited amounts of oxytocin in the hypothalamus and posterior pituitary are stored until labor starts, thus enabling the massive release of oxytocin that is necessary during labor and delivery.15
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Burbach, J.P.H. ∙ Young, L.J. ∙ Russell, J.A.
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Whether this opioidergic mechanism is present in humans or whether another mechanism mediates the inhibition of oxytocin neurons during pregnancy is not known. Moreover, progesterone levels do not decline in humans and nonhuman primates toward the end of pregnancy. In vivo studies of oxytocin neurons during pregnancy in women may be performed in the future using modern molecular imaging techniques.
The number of oxytocin receptors in the uterus increases substantially from the beginning of pregnancy to reach maximal levels at term. A 100-fold increase in the number of oxytocin receptors has been found in the human myometrium concerning the onset of labor.60
60.
Fuchs, A.R. ∙ Fuchs, F. ∙ Husslein, P. ...
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This increase in receptor density is likely to a large extent caused by the gradual increase of the availability of estrogens in the circulation during pregnancy, which leads to an activation of the oxytocin receptor gene.8
8.
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The increased amount of oxytocin receptors and function will allow small increases in circulating oxytocin levels to induce contractions of the myometrium at term, which further reduces the need for oxytocin during labor, which is important given the limited amounts of available oxytocin.
Labor and delivery
Plasma concentrations of oxytocin during labor and the postpartum period
Labor is associated with short-lasting pulses of oxytocin.8
8.
Uvnäs-Moberg, K. ∙ Ekström-Bergström, A. ∙ Berg, M. ...
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78.
Fuchs, A.R. ∙ Romero, R. ∙ Keefe, D. ...
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82.
Leake, R.D. ∙ Weitzman, R.E. ∙ Glatz, T.H. ...
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Goodfellow, C.F. ∙ Hull, M.G. ∙ Swaab, D.F. ...
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which occur more frequently as labor progresses to reach a maximal frequency of 3 pulses per 10 minutes at the end of the second stage of labor.77
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Fuchs, A.R. ∙ Goeschen, K. ∙ Husslein, P. ...
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Am J Obstet Gynecol. 1983; 147:497-502
Abstract
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Figure 5 demonstrates the presence of occasional oxytocin pulses at term pregnancy (Figure 5, A), an increasing frequency of pulses during the first (Figure 5, B) and second stages of labor, and a few pulses during the third stage of labor (Figure 5, C). The figures are based on data from Fuchs et al.78
78.
Fuchs, A.R. ∙ Romero, R. ∙ Keefe, D. ...
Oxytocin secretion and human parturition: pulse frequency and duration increase during spontaneous labor in women
Am J Obstet Gynecol. 1991; 165:1515-1523
Abstract
Full Text (PDF)
Scopus (174)
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Figure 5 Plasma oxytocin levels at term and during labor
Show full caption Figure viewer
The figure shows the pattern of peak levels of oxytocin in plasma at term pregnancy (A), first stage of labor (B), and second and third stages of labor (C). Samples were collected at 1-minute intervals from an indwelling catheter in the right arm. Data plots were chosen to show the highest and lowest pulse frequencies and amplitudes in each group. Each identified pulse is denoted with a cross. The figures are reproduced with permission from Fuchs et al,78
78.
Fuchs, A.R. ∙ Romero, R. ∙ Keefe, D. ...
Oxytocin secretion and human parturition: pulse frequency and duration increase during spontaneous labor in women
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with the courtesy of Dr Roberto Romero, who is a coauthor of the article.
Uvnäs-Moberg. Physiology and pharmacology of oxytocin. Am J Obstet Gynecol 2024.
In addition, the amount of oxytocin released with each pulse increases over time as labor progresses.78
78.
Fuchs, A.R. ∙ Romero, R. ∙ Keefe, D. ...
Oxytocin secretion and human parturition: pulse frequency and duration increase during spontaneous labor in women
Am J Obstet Gynecol. 1991; 165:1515-1523
Abstract
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As a consequence of the more frequent pulses, average plasma oxytocin concentrations double during labor compared with oxytocin levels at term.77
77.
Fuchs, A.R. ∙ Goeschen, K. ∙ Husslein, P. ...
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Am J Obstet Gynecol. 1983; 147:497-502
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Full Text (PDF)
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A very pronounced peak level of oxytocin often occurs concerning the delivery of the fetus.81–85
81.
Vasicka, A. ∙ Kumaresan, P. ∙ Han, G.S. ...
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Am J Obstet Gynecol. 1978; 130:263-273
Abstract
Full Text (PDF)
PubMed
Google Scholar
82.
Leake, R.D. ∙ Weitzman, R.E. ∙ Glatz, T.H. ...
Plasma oxytocin concentrations in men, nonpregnant women, and pregnant women before and during spontaneous labor
J Clin Endocrinol Metab. 1981; 53:730-733
Crossref
PubMed
Google Scholar
83.
Goodfellow, C.F. ∙ Hull, M.G. ∙ Swaab, D.F. ...
Oxytocin deficiency at delivery with epidural analgesia
Br J Obstet Gynaecol. 1983; 90:214-219
Crossref
PubMed
Google Scholar
84.
Thornton, S. ∙ Davison, J.M. ∙ Baylis, P.H.
Plasma oxytocin during the first and second stages of spontaneous human labour
Acta Endocrinol. 1992; 126:425-429
PubMed
Google Scholar
85.
Husslein, P. ∙ Fuchs, A.R. ∙ Fuchs, F.
[Oxytocin- and prostaglandin plasma concentrations before and after spontaneous labor: evidence of involvement of prostaglandins in the mechanism of placental separation]
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Some articles show decreased oxytocin levels after delivery, indicating a fall from higher levels during labor. Several studies show that there is no direct correlation in time between uterine contractions and peak levels of oxytocin during spontaneous labor.8
8.
Uvnäs-Moberg, K. ∙ Ekström-Bergström, A. ∙ Berg, M. ...
Maternal plasma levels of oxytocin during physiological childbirth - a systematic review with implications for uterine contractions and central actions of oxytocin
BMC Pregnancy Childbirth. 2019; 19:285
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79.
Gibbens, G.L. ∙ Chard, T.
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Am J Obstet Gynecol. 1976; 126:243-246
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Google Scholar
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82.
Leake, R.D. ∙ Weitzman, R.E. ∙ Glatz, T.H. ...
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Husslein, P. ∙ Fuchs, A.R. ∙ Fuchs, F.
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Kumaresan, P. ∙ Anandarangam, P.B. ∙ Dianzon, W. ...
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88.
Otsuki, Y. ∙ Yamaji, K. ∙ Fujita, M. ...
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Electrophysiological aspects of the oxytocin pulses
The oxytocin released into the circulation derives from oxytocinergic neurons that project to the posterior pituitary. The oxytocin-producing cells in the SON and PVN have been shown to get closer to each other because of retraction of the surrounding glial cells in response to the strong stimulation of the oxytocin-producing cells, such as during birth and breastfeeding.89
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As a consequence, the electrical action potential of the individual magnocellular oxytocin-secreting cells starts to synchronize. When all the neurons burst in synchrony, an oxytocin pulse is released into the circulation.91
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Examples of short-lasting pulses of oxytocin in response to nipple stimulation in women close to term are illustrated in Figure 6. To demonstrate peak levels of oxytocin, 5 blood samples were collected at 15-second intervals in association with 5 uterine contractions. The peak shaped patterns of oxytocin among individuals showed different appearances, whereas the oxytocin pattern tended to be more similar concerning the contractions in the same individual. Furthermore, in some individuals, peak levels of oxytocin were only observed in 1 of 5 samples collected at 15-second intervals, illustrating how difficult it might be to record oxytocin pulses.93
93.
Christensson, K. ∙ Nilsson, B.A. ∙ Stock, S. ...
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Figure 6 Plasma oxytocin levels in response to nipple stimulation
Show full caption Figure viewer
The figure shows pulses of oxytocin in plasma and uterine contractions in response to nipple stimulation in 6 women in late pregnancy (38–39 weeks of gestation) not in labor. Of note, 5 samples were collected at 15-second intervals from an indwelling catheter in the right arm during individual contractions. Plasma levels of oxytocin were measured with radioimmunoassay. Note that the peak levels of oxytocin correspond in time with uterine contractions, that their appearance varies between different women, and that the peak level sometimes can be observed in only 1 sample. In (A) data from 4 women are shown and in (B), 2 women with particularly high oxytocin levels are shown. The figures are redrawn based on data published by Christensson et al.93
93.
Christensson, K. ∙ Nilsson, B.A. ∙ Stock, S. ...
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Oxytocin-mediated stimulation of prostaglandin synthesis and release in labor
Prostaglandins play an important role in labor as they affect cervical ripening and myometrial contractility. As described previously, oxytocin promotes the production of prostaglandins60
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Fuchs, A.R. ∙ Fuchs, F. ∙ Husslein, P. ...
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The oxytocin released within these areas activates specific types of oxytocin receptors, within the same tissues, which are linked to the synthesis and release of prostaglandins, via a local, paracrine mechanism. The expression of both the tissue-specific oxytocin release and oxytocin receptors is promoted by estrogen.11
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Whether the activity of the autonomic nervous system plays a role in human labor and delivery is not known. The autonomic innervation of the uterus, both the sympathetic and parasympathetic branches, undergoes degenerative changes during pregnancy, as demonstrated both in animal studies and in women.112
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Figure 7 Nervous connections between the uterus and the brain
Show full caption Figure viewer
The Ferguson reflexes or parasympathetic sensory nerves are activated by the pressure exerted by the head of the fetus on the cervix and lead to an increase of oxytocin levels in the circulation and the brain. Sympathetic afferents from the myometrium are activated by myometrial contractions which decrease the release of oxytocin.
CRH, corticotropin-releasing factor.
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These data demonstrate that the rate of oxytocin infusion and the total amount of synthetic oxytocin administered during labor vary substantially between countries in Europe. More synthetic oxytocin might be given in other countries (eg, in the United States).
The sensitivity to oxytocin varies among pregnant women. In 1 study, adequate myometrial contractions were obtained by infusion rates of synthetic oxytocin varying between 0.9 and 5 mIU/minute. The reason behind this variation is not known, but the different sensitivity of the oxytocin receptors may include factors, such as different exposures or effects of increased estrogen levels during pregnancy, different body mass indices, and age of the woman.61
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and similar-sized dose-dependent increments of oxytocin concentrations were observed. As measured by RIA, the plasma oxytocin concentration rose by 1 to 2 pg/mL in response to an increase of the infusion rate by 1 mIU/min.140
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In all of these dose-response studies, oxytocin levels were measured using RIA. No dose-response study has been performed in which oxytocin levels were measured using ELISA. Of note, 1 study that measured both the average rise of oxytocin concentrations occurring during spontaneous, physiological labor and the increase of oxytocin concentrations in response to infusions of synthetic oxytocin found that infusion rates up to 10 mIU/min (0.6 IU/h) raised oxytocin concentrations to levels normally observed during the second stage of labor (from basal concentrations of 20 to 40 pg/mL). However, at higher infusion rates, a further 2- to 3-fold increase of oxytocin concentrations above those recorded at an infusion rate of 10 mIU/minute was observed. This means that plasma oxytocin concentrations rarely exceed 100 to 150 pg/mL during infusion of synthetic oxytocin during labor if recommended dose regimens are used.77
77.
Fuchs, A.R. ∙ Goeschen, K. ∙ Husslein, P. ...
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Steady-state oxytocin levels were obtained within 40 minutes of infusion. As the half-life of oxytocin in plasma is relatively short (3–6 minutes),54
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basal oxytocin concentrations are soon reached after infusions of synthetic oxytocin have been stopped. Moreover, it should be pointed out that individual oxytocin pulses during labor can reach quite substantial concentrations for a short period (Figure 4).
Hyperstimulation
The amount of oxytocin administered is crucial, as too high levels of oxytocin may induce hyperstimulation of the myometrium, including excessive frequency of contractions. This may harm the fetus as the placental and fetal blood flow may become compromised with consequent fetal hypoxia.147–150
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Recent studies show that if oxytocin infusions are performed within the recommended dose levels, they are safe. A recent multicenter, randomized study evaluated the effect of discontinuing oxytocin infusion when reaching the active phase of labor. This intervention, which is associated with a decrease in the amount of administered oxytocin, did not improve neonatal morbidity or increase the rate of other neonatal or maternal negative outcomes.151
151.
Girault, A. ∙ Sentilhes, L. ∙ Desbriere, R. ...
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In another study, the risk of adverse outcomes in response to infusion rates higher than 20 mIU/minute were compared with those induced by infusion of oxytocin at lower rates. In this double-blind randomized controlled trial, no significant difference in maternal or perinatal adverse outcomes was found when the effects of the 2 dose levels were compared.152
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Whether circulating oxytocin from the maternal circulation passes the placenta and maternal blood-brain barrier has become a matter of debate. It has been suggested that maternal circulating oxytocin levels become extremely high in response to infusions of synthetic oxytocin and that oxytocin under these circumstances would pass through the placenta and the maternal blood-brain barrier and negatively affect the fetus and the maternal brain.153
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However, there is little scientific support for a transfer of oxytocin over the placenta if oxytocin is infused during labor according to recommendations. As stated above quite small amounts (5–10 IU) of synthetic oxytocin are infused, often for a period of several hours during labor, and therefore maternal plasma oxytocin concentration are only moderately elevated. In clinical studies in which both maternal and oxytocin fetal oxytocin concentrations were measured after birth, there is no evidence of any passage of oxytocin from the mother to the fetus.155
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Overall, these data indicate that infusions of synthetic oxytocin within the recommended dose range do not influence the fetus or the mother by passing through the placenta or blood-brain barrier during labor. However, the dose of oxytocin administered is crucial. If higher amounts of oxytocin are given, oxytocin may pass the placenta and maternal blood-brain barrier.
In addition, some animal data suggest the presence of a local transport system for oxytocin from blood to areas in the brain, such as the amygdala, the receptor for advanced glycation end products (RAGE) system. In animal experiments, the RAGE system has been shown to promote maternal interaction with her pups, possibly because of the transfer of oxytocin from the circulation to the brain. Whether a similar effect occurs during physiological and induced labor in women is not yet known.157
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Postpartum administration of oxytocin
An intramuscular or intravenous administration of 5 to 10 IU of oxytocin is often given routinely at the time of the expulsion of the posterior arm of the fetus to induce uterine contractility, to promote placental separation, and to decrease the risk of PPH.158–160
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Cardiovascular events have been reported after bolus administration of oxytocin. For example, a fall in blood pressure, transient tachycardia, and chest pain were reported after intravenous administration of 10 IU of synthetic oxytocin to women who have received epidural analgesia during cesarean delivery. In addition, the ST segment of the electrocardiogram was depressed, and troponin levels were increased in some of these women, indicating a possible occurrence of ischemic tissue damage in the heart. The cardiovascular effects disappeared or were reduced when the bolus dose of oxytocin was reduced to ≤5 IU or the route of administration of oxytocin was changed from intravenous to intramuscular.162–165
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The cardiac pain may be caused by a cardiac vasoconstrictor effect as a consequence of the hypotension and tachycardia caused by the administration of oxytocin. Moreover, the contractile effects on cardiac blood vessels may be caused by a direct effect of oxytocin on cardiac oxytocin receptors, or possibly on vasopressin receptors, as oxytocin in very high levels may bind to and cross-react with vasopressin receptors.15
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The levels of oxytocin receptor mRNA and protein synthesis have been shown to be reduced in these situations showing that there is a down-regulation of the production of the oxytocin receptor. It has been suggested that this reduction of the function of oxytocin receptors is due to a direct inhibitory effect by circulating oxytocin concentrations on the oxytocin receptors, caused by the high levels of oxytocin induced by infusions of synthetic oxytocin. In vitro experiments support this assumption. When oxytocin receptors are exposed to nanomolar or micromolar concentrations of oxytocin for 5 to 10 hours, a decrease in oxytocin receptor mRNA has been demonstrated.64–67
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However, the very high concentrations of oxytocin and the long exposure time needed in these experiments for down-regulation of oxytocin receptor mRNA suggest that the results do not represent physiological conditions, as plasma oxytocin concentrations in response to infusions of synthetic oxytocin are within the picomolar range. The down-regulation or desensitization of the oxytocin receptors may be due to another mechanism, for example, related to a physiological feedback inhibition of the function of the oxytocin receptors.
Social support and activation of the oxytocinergic system
Social support is a well-known concept within the field of psychology. The presence of a supportive person decreases stress levels, including cortisol levels and inflammatory markers, and perceived anxiety in individuals of different age and sex groups, suggesting a down-regulation of the stress system.170
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A similar calming effect may be induced by the presence of a supportive person during labor and delivery. The presence of a supportive person during labor shortens its duration, decreases the rate of cesarean delivery, and reduces the use of medical interventions, such as epidural analgesia and infusions of synthetic oxytocin.1
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In addition, studies show that mothers participating in midwife-led continuity models of care during pregnancy were more likely to have a spontaneous vaginal delivery and less likely to have regional analgesia, instrumental vaginal delivery, preterm birth (<37 weeks of gestation), and fetal loss.174
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In animals, environmental cues that signal calmness and safety have been shown to activate specific nerve fibers in the cortex, which, via the amygdala-hippocampal pathway, innervate the parvocellular oxytocinergic neurons of the PVN.17
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In addition, touch or sensory stimulation of the skin may contribute to the positive effects exerted by a supportive person on the progress or experience of labor. Several studies show that oxytocin can be released and that oxytocin mRNA in the PVN can be increased in response to touch via stimulation of cutaneous sensory nerves (Figure 8). Such stimulation does not induce a pulsatile release of oxytocin but may or may not be associated with increased levels of circulating oxytocin2
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Figure 8 Oxytocin release in response to activation of cutaneous sensory nerves
Show full caption Figure viewer
The cutaneous sensory nerves respond to touch, stroking, light pressure, and warmth and may correspond to unmyelinated C-tactile fibers and to thicker myelinated nerve fibers. In addition, analogous mental triggers mediating information about safety, familiarity, support, warmth, and empathy may hypothetically trigger oxytocin release. This model demonstrates how the release of oxytocin from parvocellular neurons in the SON and PVN during labor can be reinforced by activation of cutaneous sensory nerves.
PVN, paraventricular nucleus; SON, supraoptic nucleus.
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Single-nucleotide polymorphisms
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This is a rapidly growing field of research, and more findings are to be expected soon.
Conclusions
Oxytocin is a system involving hormonal, neurosignaling, and paracrine effects. It has multiple important effects during labor, and several oxytocin-linked mechanisms cooperate to promote the birth of the neonate. Oxytocin stimulates uterine contractions by activating oxytocin receptors in the myometrium. Moreover, it stimulates the secretion of prostaglandins via a local, paracrine effect in the decidua and amnion. In the future, oxytocin may be shown to induce effects in the maternal brain to facilitate labor and delivery; to decrease levels of pain, fear, and stress;and to adapt the mother for motherhood.
Infusions of synthetic oxytocin may be administered to shorten labor or as a bolus after delivery to decrease the risk of PPH. The plasma concentrations of oxytocin are only moderately elevated in response to infusions of synthetic oxytocin and are not likely to be associated with a passage through the placenta and maternal blood-brain barrier if given within recommended dose regimens. Social support may be used in the future to support mothers during labor and delivery.
Glossary
Allopregnanolone: A neuroactive metabolite of progesterone
Amino-terminal: The end of a peptide containing an amino group
Aprotinin (Trasylol): A substance that inhibits the activity of degrading enzymes
Beta-arrestin: A molecule that helps move oxytocin receptors from the surface into the cells
Calmodulin: A substance involved in the intracellular signaling after oxytocin has bound to its receptor
Carboxy-terminal (C-terminal): The end of a peptide containing a carboxyl group
Enzyme-linked immunosorbent assay (ELISA): An immunologic method used for the determination of oxytocin levels
G-protein–coupled receptor: A group of receptors that transfer information from the cell membrane into the cell via G proteins
Hydrophilic: A substance that can be dissolved in water
Isotocin: An oxytocinlike substance in fish
Magnocellular neurons: Large oxytocin producing neurons in the SON and PVN
Mesotocin: An oxytocinlike substance in birds
Methylation: Methylation of DNA in the promoter region of genes reduces the synthesis of proteins
Mitogen-activated protein kinase (MAPK): A substance involved in the intracellular signaling leading to production of prostaglandins after oxytocin has bound to its receptor
Myosin light-chain kinase: A substance involved in the intracellular signaling after oxytocin has bound to its receptor
Neurophysin: A peptide that is synthesized together with oxytocin but cleft off before release into the circulation
Nucleus solitarius (NTS): Important integrative center for autonomic nervous function in the brain stem
Oxytocinase: An enzyme that degrades oxytocin
Paraventricular nucleus (PVN): One of the major sites of oxytocin neurons in the hypothalamus
Parvocellular neurons: Small oxytocin-producing neurons in the PVN
Phospholipase-C: A substance involved in the intracellular signaling after oxytocin has bound to its receptor
Placental leucin-amino-peptidase: An oxytocin degrading enzyme, which is produced in the placenta
Radioimmunoassay (RIA): A method used for the determination of oxytocin levels, involving radioactivity
Receptor for advanced glycation end products (RAGE): A receptor that may actively transport oxytocin over the blood-brain barrier
Supraoptic nucleus (SON): One of the major sites of oxytocin neurons in the hypothalamus
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Periodontal Anatomy – Alveolar bone
The alveolar process, which is also called the alveolar bone, is a thickened ridge of bone containing the sockets of the teeth or dental alveoli. It is located on the jaw bones which hold the teeth. The tooth-bearing bones are called the maxilla and the mandible. The curved part of each alveolar process, which is located on the jaw, is the alveolar arch.
Structure
On the maxilla, the upper fixed bone of the jaw, the alveolar process is a ridge on the inferior surface. On the mandible, the lower jaw, it is a ridge on the superior surface. This structure makes up the thickest part of the maxillae.
The alveolar process also contains an area of compact bone, which is adjacent to the periodontal ligament (PDL). This is called the lamina dura when it is being viewed on radiographs. This portion is attached to the cementum of the roots via the periodontal ligament. It is also uniformly lighter. The integrity of the lamina dura is vital in the process of studying radiographs for pathological lesions.
The alveolar process has a supporting bone. Both contain the same components which includes fibers, cells, intercellular substances, nerves, blood vessels, and lymphatics.
The alveolar process is the lining of the alveolus or tooth socket. While the alveolar process is made up of compact bone, it may also be called the cribriform plate. This is because it contains holes where Volkmann canals pass from the alveolar bone and into the PDL. The alveolar bone proper may also be called bundle bone. This is because the Sharpey fibers, which are part of the fibers in the PDL, insert at this location. Similar to the fibers of the cemental surface, Sharpey fibers in the alveolar bone proper are inserted at 90 degrees. They are fewer in number, but thicker in diameter compared to those which are present in the cementum. Just as they are in the cellular cementum, the Sharpey fibers located in the bone are typically mineralized only partially at their periphery.
The alveolar crest is the most cervical rim included in the alveolar bone proper. In healthy cases, the alveolar crest is slightly apical to the cementoenamel junction (CEJ) by about 1.5-2 mm. The alveolar crests of the neighboring teeth should also be uniform in their height along the jaw.
The supporting alveolar bone contains both trabecular and cortical bone. The trabecular bone contains cancellous bone which is located between the alveolar bone proper and the cortical bone plates. The alveolar bone located between two neighboring teeth is the interdental septum. The cortical bone, or cortical plates, contains plates of compact bone on the alveolar bone’s facial and lingual surfaces. These cortical plates are typically about 1.5-3 mm thick over the posterior teeth. However, this thickness can drastically vary around the anterior teeth.
Composition - Inorganic Material
By weight, the alveolar bone is comprised of 67% inorganic material. The inorganic material mainly contains calcium and phosphate. The mineral content is primarily in the form of calcium hydroxyapatite crystals.
Composition - Organic Material
The remaining 33% of the alveolar bone is comprised of organic material. The organic material contains both collagen and non-collagenous material. The cellular component of the bone contains osteoblasts, osteocytes and osteoclasts. |
3076 | https://www.osmosis.org/blog/usmle-step-1-question-of-the-day-convex-lesion | USMLE® Step 1 Question of the Day: Convex lesion - Osmosis Blog
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USMLE® Step 1 Question of the Day: Convex lesion
Allopathic Medicine (MD), Exam Prep, Questions of the Day, USMLE Step 1, USMLE® Step 1
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Jan 17, 2024
— Updated:
August 27, 2025
—
Osmosis Team
In This Article
Toggle
The correct answer to today’s USMLE® Step 1 Question is…
C.Extra-axial injury; supratentorial; secondary brain injury
Incorrect answer explanations
A. Intra-axial injury; supratentorial; primary brain injury
B. Extra-axial injury; supratentorial; primary brain injury
D. Extra-axial injury; infratentorial; secondary brain injury
Main Explanation
Major Takeaway
This week, we are sharing another USMLE® Step 1-style practice question to test your knowledge of medical topics. Today’s case focuses on a 25-year-old man who suffered head trauma in a bar altercation, resulting in loss of consciousness. Despite initial refusal for medical evaluation, he later exhibits alarming symptoms, including vomiting and sudden unconsciousness. Explore the diagnostic journey as a non-contrast CT reveals a convex lesion with inferolateral pupil dilation, shedding light on the distinct injury patterns and potential complications in this critical emergency department scenario. Test your diagnostic skills with this USMLE Step 1 question.
A 25-year-old man is brought to the emergency department following an altercation at a bar where he was hit over the left side of his head with a bottle. The patient subsequently lost consciousness and spontaneously awakened minutes later. He initially refused medical evaluation, but a few hours later, he began vomiting and agreed to be taken to the hospital. On arrival, the patient suddenly loses consciousness. Temperature is 36.0°C (96.8°F), pulse is 48/min, respirations are 8/min and irregular, and blood pressure is 144/40 mmHg. On physical examination, the patient does not respond to painful stimuli. The left pupil is dilated and inferolaterally abducted. A non-contrast CT is obtained and demonstrates a convex lesion with mass effect and downward displacement of the inferior temporal lobe. This patient’s clinical presentation is best defined by which of the following injury patterns?
A. Intra-axial injury; supratentorial; primary brain injury
B. Extra-axial injury; supratentorial; primary brain injury
C. Extra-axial injury; supratentorial; secondary brain injury
D. Extra-axial injury; infratentorial; secondary brain injury
Scroll down for the correct answer!
The correct answer to today’s USMLE® Step 1 Question is…
C.Extra-axial injury; supratentorial; secondary brain injury
Before we get to the Main Explanation, let’s look at the incorrect answer explanations. Skip to the bottom if you want to see the correct answer right away!
Incorrect answer explanations
A. Intra-axial injury; supratentorial; primary brain injury
Incorrect:Diffuse axonal injury (DAI) is an example of this injury pattern. Patients with DAI often present with deep coma without elevated intracranial pressure. DAI occurs as a result of traumatic shearing forces applied to the brain when the head is rapidly accelerated and/or decelerated affecting the gray-white junction. Another example of intra-axial injury is a focal cerebral contusion. This patient more likely has an extra-axial injury.
B. Extra-axial injury; supratentorial; primary brain injury
Incorrect:Examples of this injury pattern include epidural, subdural, and subarachnoid hematomas without signs of brain herniation. This patient has imaging and exam findings indicative of herniation, a secondary brain injury.
D. Extra-axial injury; infratentorial; secondary brain injury
Incorrect:An example of this injury pattern includes downward cerebellar/tonsillar herniation or upward cerebellar/transtentorial herniation from a tumor or hematoma secondary to trauma. This patient’s presentation is consistent with supratentorial herniation.
Main Explanation
This patient presents with blunt head trauma and classic features of an epidural hematoma including a “lucid interval,” a lens shaped lesion, and signs of impending herniation. The physical examination demonstrates a blown, dilated pupil, indicating uncal herniation secondary to extrinsic compression of cranial nerve III. Lastly, the CT results confirm the diagnosis of an epidural hematoma, which is an extra-axial bleed, with secondary brain herniation.Traumatic brain injury (TBI) can be classified as primary (e.g. epidural hematoma) or secondary (e.g. brain herniation), the latter of which occurs due to mass effect and increased intracranial pressure. Another way to classify herniation is with respect to the tentorium: supratentorial herniation (cingulate, a.k.a. subfalcine), central transtentorial herniation, or infratentorial herniation (downward/cerebellar tonsillar herniation or upward cerebellar/transtentorial herniation). Finally, TBI that occurs within the brain parenchyma is classified as intra-axial (e.g. diffuse axonal injury) or extra-axial (e.g., epidural, subdural).
Major Takeaway
Pyruvate dehydrogenase complex(PDC)deficiencyis anX-linked recessive disorder characterized by hypotonia, poor feeding, developmental and intellectual delay, and seizures. Laboratory testing will demonstrateelevated lactate and alaninelevels. Treatment involves adhering to aketogenic diet.
References
Gupta, N., Rutledge, C. (2019) Pyruvate dehydrogenase complex deficiency: An unusual cause of recurrent lactic acidosis in a paediatric critical care unit. The Journal of Critical Care Medicine. 5(2), 71-75. Doi: 10.2478/jccm-2019-0012.
Haddad, A., Mohiuddin, S.S. (2020) “Biochemistry, citric acid cycle”. StatPearls [Internet]. Web Address:
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As the title indicates, I'd like to prove the following:
If f:R→R is a continuous function on [a,b], then f attains its maximum.
Now, I do have a working proof: [a,b] is a connected, compact space, which means that because f is continuous, f([a,b]) is compact and connected as well. Therefore, f([a,b]) is a closed interval, which means it has both a minimum and, as desired, a maximum.
What I would like, however, is a proof that doesn't require such general or sophisticated framework. In particular, I'd like to know if there's a proof that is understandable to somebody beginning calculus, one that (at the very least) doesn't invoke compactness. Any comments, hints, or solutions are welcome and apreciated.
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edited Jul 5, 2013 at 10:50
Vishal Gupta
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asked Jul 5, 2013 at 10:13
Ben GrossmannBen Grossmann
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It seems that I've made the mistake of not checking wikipedia first. This page has a nice proof along the same lines of some of these answerers: en.wikipedia.org/wiki/…
– Ben Grossmann
Commented
Jul 7, 2013 at 13:11
Note, this problem is badly formed - you define the domain of f as R and then ask about a maximum. What you want is the maximum on [a,b]
– Thomas Andrews
Commented
Nov 29, 2013 at 20:38
Is their anything around for if a function is continuous on [a,b] attains a min value on [a,b]. I know how to prove it using sequences and compact sets. I need a third way to go about it. Any suggestions?
– OLE
Commented
Nov 27, 2016 at 6:20
@EduardoO. are you asking specifically about attaining the minimum as opposed to the maximum? Then of course, all the same arguments work. If you'd prefer not to slightly rephrase one of the arguments below, it suffices to show that the function g(x)=−f(x) attains a maximum, then use one of the arguments directly.
– Ben Grossmann
Commented
Nov 27, 2016 at 6:24
@Omnomnomnom To be honest some of the arguments below get me lost as some of things I have yet to learn or are completely new to me. I need a proof that I can interpret to show the minimum, yes opposed to the maximum. Thanks for any guidance you can provide
– OLE
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Nov 27, 2016 at 6:30
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Here’s a sketch of one possible argument. Let u=supx∈[a,b]f(x). (Note that I allow the possibility that u=∞.) There is a sequence ⟨xn:n∈N⟩ in [a,b] such that for each n∈N, u−f(xn)<12n if u∈R and f(xn)>n if u=∞. Extract a monotone subsequence ⟨xnk:k∈N⟩. Being a monotone, bounded sequence, ⟨xnk:k∈N⟩ converges to some y. (Note that you have to use the completeness of R in some way, and this is the most elementary that occurs to me.) Moreover, y∈[a,b], and f is continuous, so f(y)=limk→∞f(xnk)=u. (Note that this shows that in fact u∈R.)
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edited Jul 10, 2023 at 19:29
answered Jul 5, 2013 at 10:31
Brian M. ScottBrian M. Scott
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I really like this proof! I think you do need to show that u is finite before extracting a suitable sequence, otherwise you can't use convergence. Perhaps that needs its own proof, then.
– Ben Grossmann
Commented
Jul 5, 2013 at 11:02
2
Of course, this is just the compactness proof in disguise. It contains a proof that [a,b] is sequentially compact.
– Chris Eagle
Commented
Jul 7, 2013 at 9:28
@user14111 good catch, I think you're right about that. As for my earlier comment: if f is unbounded on [a,b], then by the construction described f would map a convergent sequence to an unbounded sequence, which is a contradiction of continuity, so boundedness doesn't require its own proof after all.
– Ben Grossmann
Commented
Jul 7, 2013 at 13:09
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@BenGrossmann: (Oops! It’s only taken me ten years to notice these comments.) Any sequence in R has a monotone subsequence, and ⟨xn:n∈N⟩ lies in [a,b], any subsequence is bounded irrespective of whether u∈R. That u∈R then follows from the fact that it’s f(y) for some y∈[a,b].
– Brian M. Scott
Commented
Jul 10, 2023 at 19:29
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Yes, quite a few such proofs exist. You first prove that a continuous function is bounded, and apply the Bolzano-Weierstrass theorem. Here is such a proof from my lecture notes in a first course in analysis:
By the earlier result, f is bounded. Consider the image of [a,b] under f:
A=f([a,b])={f(x):x∈[a,b]}.
f bounded means that A is bounded as a subset of R. We also have that A≠∅. We will prove the existence of x2, and the existence of x1 is proved similarly.
Since A is bounded and non-empty, there is a supremum: M=supA. Given any positive integer n, M−1/n cannot be an upper bound for A. Then there exists some xn∈[a,b] such that
M−1n<f(xn)≤M.(∗)
By Bolzano-Weierstrass, xn has a convergent subsequence xnj→x. Since a≤xnj≤b, a≤x≤b. By the continuity of f, f(xnj)→f(x). But by (∗), f(xnj)→M. By the uniqueness of the limit, f(x)=M. Now set x2=x. □
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edited Jul 4, 2023 at 13:41
Shuhul Mujoo
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This theorem ultimately depends on the completeness of real number system and can't be proven without using any result equivalent to the completeness of real number system. If we assume that every continuous function on a closed interval is bounded then we can provide a very simple proof of the current problem being discussed here.
Under this assumption we have the existence of M=sup{f(x)∣x∈[a,b]}. If f(x)≠M for all x∈[a,b] then the function g(x)=1/(M−f(x)) is continuous in [a,b] and hence bounded in [a,b]. On the other hand by definition of M we can find x∈[a,b] such that M−f(x) can be made arbitrarily small. Thus g(x) can be made arbitrarily large. This contradiction shows that we must have f(x)=M for some x∈[a,b].
The result that every continuous function is bounded on a closed interval is itself another property of continuous functions which can't be proved without using completeness of real number system.
I have presented various proofs of these properties of continuous function here.
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answered Jul 7, 2013 at 9:25
Paramanand Singh♦Paramanand Singh
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Thank you for your answer, that's a neat twist! I looked through your linked proofs, which do an impressive job of explaining the topological notion of compactness on R. However, I think the better way to approach the proof that "continuous function on a closed interval is bounded" is to use the fact that sequential continuity is equivalent to continuity for metric spaces, and proceed via the B-W theorem (which itself has a neat little proof here en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem#Proof )
– Ben Grossmann
Commented
Jul 7, 2013 at 13:23
Omnomnomnom, actually the kind of proofs which I have put in my blogs are somewhat easier to understand but limited in their generality (especially they will need lot of development if we want to go for Rn). Whereas the proofs based on BW and sequential continuity are the ones which can be applied in a very general context. It also depends on the intended audience and my blog is targeted at people who probably never had undergraduate / graduate education in maths (like myself), but who find maths exciting.
– Paramanand Singh
♦
Commented
Jul 8, 2013 at 4:17
I think sequential continuity is a lot easier to understand than the Heine Borel principle. In this case, it means we don't need to worry about compactness.
– Ben Grossmann
Commented
Jul 8, 2013 at 4:47
I agree the proofs based on Heine Borel are on a higher level. Even I had a very difficult time understanding Heine Borel when I first studied from Hardy's Pure Mathematics.
– Paramanand Singh
♦
Commented
Jul 8, 2013 at 5:08
It seems I didn't give your link a good enough look before making my judgment. I think your direct use of Dedekind's Theorem is far more intuitive that the notion of sequential compactness.
– Ben Grossmann
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Jul 8, 2013 at 12:07
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I don't know how you can avoid compactness entirely. Here's one that uses sequential compactness in the form of the Bolzano-Weierstrass theorem, "every bounded sequence has a convergent subsequence". (To justify that theorem to beginners, you could take it as an axiom that every bounded monotonic sequence converges, and show them the proof that every sequence has a monotonic subsequence.) For some reason that escapes me, this proof is often done in two stages, first proving that the function is bounded and then showing that the it attains a maximum.
Given a continuous real-valued function f on [a,b], we will show that
the set Y=f([a,b]) has a greatest element.
For each positive integer n, define a finite set Qn={p/q:p,q integers, 0<q≤n,|p|≤n}.
Choose yn∈Y so as to maximize the number of elements in the set {r∈Qn:yn>r}, and choose xn∈[a,b] with f(xn)=yn.
The sequence {xn} has a subsequence converging to a point c∈[a,b]. Since f is continuous, the corresponding subsequence of {yn} converges to f(c). We will show that f(c) is the greatest element of Y.
Assume for a contradiction that f(c)r whenever r∈Qn. Since r∈Qn for all
sufficiently large n, we have yn>r>f(c) for all sufficiently
large n. But this is absurd, since {yn} has a subsequence converging
to f(c).
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edited Jul 5, 2013 at 10:40
answered Jul 5, 2013 at 10:33
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I found the following proof by Takeshi Saito in "How to Learn Mathematics(New Edition)" edited by Kunihiko Kodaira.
Let X be a non-empty compact space.
Let f(x) be a real valued continuous function on X.
Assume that f(x) doesn't attain a maximum value on X.
Then for any x∈X, there exists t∈X such that f(x)<f(t).
Let Ut:={x∈X|f(x)<f(t)} for t∈X.
f(x) is continuous on X, so Ut is an open set of X.
f(x) doesn't attain a maximum value on X, so (Ut)t∈X is an open cover of X.
X is compact, so there exists a finite subcover Ut1,⋯,Utn.
Let k∈{1,⋯,n} be an integer such that f(tk)=max{f(t1),⋯,f(tn)}.
tk∈X. So there exists an integer l such that tk∈Utl.
Then, f(tk)<f(tl)≤f(tk). This is a contradiction.
So, f(x) attains a maximum value on X.
We now prove that [0,1] is compact.
The discrete space {0,1} is finite. So {0,1} is compact. So, {0,1}N is compact.
Let f:{0,1}N→R be a function such that f({an}):=∑∞n=1an2n.
Then, f is continuous, so f({0,1}N)=[0,1] is compact.
Let a,b∈R such that a≤b.
Then, [a,b]≠∅.
And it is easy to show that [a,b] is a compact space:
Let g:[0,1]→[a,b] be a function such that g(x)=(b−a)x+a.
Since g is continuous, g([0,1])=[a,b] is compact.
So, if f(x) is a real valued continuous function on [a,b], then f(x) attains a maximum value on [a,b].
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edited Jun 29, 2020 at 8:43
answered Feb 2, 2019 at 9:51
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Theorem. Let f:[a,b]→R be continuous. Then f([a,b]) is bounded.
Proof (indirectly) avoiding compactness: Since f is continuous at a, there exists δ>0 such that for any x∈[a,a+δ), we have f(a)−1<f(x)<f(a)+1. Hence f is bounded on [a,a+δ/2]. Write
A={x∈[a,b]:f is bounded on [a,x]}.
Then a+δ/2∈A, and b is an upper bound of A. So, let c=sup(A). Now, a+δ/2≤c≤b. Continuity of f at c implies that there exists δ1>0 such that for any x∈(c−δ1,c+δ1)∩[a,b], we have f(c)−1<f(x)<f(c)+1. That is, f is bounded on (c−δ1,c+δ1)∩[a,b]. Let d∈(c−δ1,c). Then f is bounded on [a,d]. It follows that
f is bounded on [a,d]∪((c−δ1,c+δ1)∩[a,b]).
Since [a,c]⊆[a,d]∪((c−δ1,c+δ1)∩[a,b]), we have c∈A.
Further, if c<b, then choose δ2=min{δ1,(b−c)/2}. Then [a,c+δ2]⊆[a,d]∪((c−δ1,c+δ1)∩[a,b]). It follows that f is bounded on [a,c+δ2]. So, c+δ2∈A. This contradicts the fact that c=sup(A). Therefore, c=b.
Hence c=b and c∈A. That is, b∈A. Therefore, f is bonded on [a,b]. QED
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answered Dec 31, 2018 at 10:00
ArvindArvind
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Here is a sketch...
The following is a special instance of Cantor's intersection theorem and it can be proved using an infimum \ supremum argument:
Proposition: For a,b∈R with a<b let there be given a decreasing nested sequence (Ck)k≥0 of non-empty closed subsets of the closed interval [a,b], i.e.
C0⊃C1⊃⋯⊃Cn⊃Cn+1⊃⋯,
Then it follows that
⋂k=0∞Ck≠∅.
Using this it can be easily proven that any continuous real-valued function s on [a,b] must be bounded above (set Ck to the inverse image of [k,+∞) under s and use a proof by contradiction). Moreover, s must be bounded below (if not, −s would not be bounded above).
Once we know that continuous functions are bounded, we can easily show that they attain both a maximum and minimum; see Paramanand Singh's answer. For completeness we use the reasoning here.
Let f:[a,b]→R be a continuous function.
Let
α=infimum(f<[a,b]>)
and
β=supremum(f<[a,b]>)
If f never attains the supremum, then the function
g(x)=−1f(x)−β
defined on [a,b] is not bounded above, a contradiction.
If f never attains the infimum , then the function
h(x)=1f(x)−α
defined on [a,b] is not bounded above, a contradiction..
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answered Jul 12, 2023 at 1:57
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9
9.1
9.1.1
9.1.1.1
9.1.1.1 - Minitab: Confidence Interval for 2 Proportions
Minitab can be used to construct a confidence interval for the difference between two proportions using the normal approximation method. Note that the confidence intervals given in the Minitab output assume that n p≥10 and n(1−p)≥10 for both groups. If this assumption is not true, you should use bootstrapping methods in StatKey.
Minitab®–Constructing a Confidence Interval with Raw Data
Let's estimate the difference between the proportion of females who have tried weed and the proportion of males who have tried weed.
Open Minitab file: class_survey.mpx
Select Stat > Basic Statistics > 2 Proportions
Select Both samples are in one column from the dropdown
Double click the variable Try Weed in the Samples box
Double click the variable Biological Sex in the Sample IDs box
Keep the default Options
Click OK
This should result in the following output:
Method
Event: Try_Wee = Yes
p 1: proportion where Try_Weed = Yes and Biological Sex = Female
p 2: proportion where Try-Weed = Yes and Biological Sex = Male
Difference: p 1-p 2
Descriptive Statistics: Try Weed
| Biological Sex | N | Event | Sample p |
--- --- |
| Female | 127 | 56 | 0.440945 |
| Male | 99 | 62 | 0.626263 |
Estimation for Difference
| Difference | 95% CI for Difference |
--- |
| -0.185318 | (-0.313920, -0.056716) |
CI based on normal approximation
Test
| Null hypothesis | H 0: p 1−p 2=0 |
| Alternative hypothesis | H 1: p 1−p 2≠0 |
| Method | Z-Value | P-Value |
---
| | | |
| Normal approximation | -2.82 | 0.005 |
| Fisher's exact | | 0.007 |
Minitab®–Constructing a Confidence Interval with Summarized Data
Let's estimate the difference between the proportion of Penn State World Campus graduate students who have children to the proportion of Penn State University Park graduate students who have children.In our representative sample there were 120 World Campus graduate students; 92 had children. There were 160 University Park graduate students; 23 had children.
Open Minitab
Select Stat > Basic Statistics > Two-Sample Proportion
Select Summarized data in the dropdown
For Sample 1 next to Number of events enter 92 and next to Number of trials enter 120
For Sample 2 next to Number of events enter 23 and next to Number of trials enter 160
Keep the default Options
Click OK
This should result in the following output:
Method
p 1: proportion where Sample 1 = Event
p 2: proportion where Sample 2 = Event
Difference: p 1-p 2
Descriptive Statistics
| Sample | N | Event | Sample p |
--- --- |
| Sample 1 | 120 | 92 | 0.766667 |
| Sample 2 | 160 | 23 | 0.143750 |
Estimation for Difference
| Difference | 95% CI for Difference |
--- |
| 0.622917 | (0.529740, 0.716093) |
CI based on normal approximation
Test| Null hypothesis | H 0: p 1−p 2=0 |
| Alternative hypothesis | H 1: p 1−p 2≠0 |
| Method | Z-Value | P-Value |
---
| Normal approximation | 13.10 | 0.000 |
| Fisher's exact | | 0.000 |
«Previous 9.1.1 - Confidence Intervals
Next 9.1.2 - Hypothesis Testing»
Lessons
Welcome to STAT 200!
0: Prerequisite Skills
0.1 - Review of Algebra
0.1.1 - Order of Operations
0.1.2 - Summations
0.1.3 - Basic Linear Equations
0.2 - Introduction to Minitab
0.3 - Word's Equation Editor
0.4 - Canvas' Equation Editor
1: Collecting Data
1.1 - Cases & Variables
1.1.1 - Categorical & Quantitative Variables
1.1.2 - Explanatory & Response Variables
1.2 - Samples & Populations
1.2.1 - Sampling Bias
1.2.2 - Sampling Methods
1.2.2.1 - Minitab: Simple Random Sampling
1.3 - Other Sources of Bias
1.4 - Research Study Design
1.4.1 - Confounding Variables
1.4.2 - Causal Conclusions
1.4.3 - Independent and Paired Samples
1.4.4 - Control and Placebo Groups
1.4.5 - Blinding
1.5 - Lesson 1 Summary
2: Describing Data, Part 1
2.1 - Categorical Variables
2.1.1 - One Categorical Variable
2.1.1.1 - Risk and Odds
2.1.1.2 - Visual Representations
2.1.1.2.1 - Minitab: Frequency Tables
2.1.1.2.2 - Minitab: Pie Charts
2.1.1.2.3 - Minitab: Bar Charts
2.1.2 - Two Categorical Variables
2.1.2.1 - Minitab: Two-Way Contingency Table
2.1.2.2 - Minitab: Clustered Bar Chart
2.1.2.3 - Minitab: Stacked Bar Chart
2.1.3 - Probability Rules
2.1.3.1 - Range of Probabilities
2.1.3.2 - Combinations of Events
2.1.3.2.1 - Disjoint & Independent Events
2.1.3.2.2 - Intersections
2.1.3.2.3 - Unions
2.1.3.2.4 - Complements
2.1.3.2.5 - Conditional Probability
2.1.3.2.5.1 - Advanced Conditional Probability Applications
2.2 - One Quantitative Variable
2.2.1 - Graphs: Dotplots and Histograms
2.2.2 - Outliers
2.2.3 - Shape
2.2.4 - Measures of Central Tendency
2.2.4.1 - Skewness & Central Tendency
2.2.5 - Measures of Spread
2.2.6 - Minitab: Central Tendency & Variability
2.2.7 - The Empirical Rule
2.2.8 - z-scores
2.2.9 - Percentiles
2.2.10 - Five Number Summary
2.3 - Lesson 2 Summary
3: Describing Data, Part 2
3.1 - Single Boxplot
3.2 - Identifying Outliers: IQR Method
3.3 - One Quantitative and One Categorical Variable
3.4 - Two Quantitative Variables
3.4.1 - Scatterplots
3.4.1.1 - Minitab: Simple Scatterplot
3.4.2 - Correlation
3.4.2.1 - Formulas for Computing Pearson's r
3.4.2.2 - Example of Computing r by Hand (Optional)
3.4.2.3 - Minitab: Compute Pearson's r
3.4.3 - Simple Linear Regression
3.4.3.1 - Minitab: SLR
3.4.3.2 - Example: Interpreting Output
3.5 - Relations between Multiple Variables
3.5.1 - Scatterplot with Groups
3.5.2 - Bubble Plots
3.5.3 - Time Series Plot
3.6 - Lesson 3 Summary
4: Confidence Intervals
4.1 - Sampling Distributions
4.1.1 - StatKey Examples
4.1.1.1 - NFL Salaries (One Mean)
4.1.1.2 - Coin Flipping (One Proportion)
4.1.2 - Copying Data into StatKey
4.1.3 - Impact of Sample Size
4.2 - Introduction to Confidence Intervals
4.2.1 - Interpreting Confidence Intervals
4.2.2 - Applying Confidence Intervals
4.3 - Introduction to Bootstrapping
4.3.1 - Example: Bootstrap Distribution for Proportion of Peanuts
4.3.2 - Example: Bootstrap Distribution for Difference in Mean Exercise
4.4 - Bootstrap Confidence Interval
4.4.1 - StatKey: Standard Error Method
4.4.1.1 - Example: Proportion of Lactose Intolerant German Adults
4.4.1.2 - Example: Difference in Mean Commute Times
4.4.2 - StatKey: Percentile Method
4.4.2.1 - Example: Correlation Between Quiz & Exam Scores
4.4.2.2 - Example: Difference in Dieting by Biological Sex
4.4.2.3 - Example: One sample mean sodium content
4.5 - Paired Samples
4.6 - Impact of Sample Size on Confidence Intervals
4.7 - Lesson 4 Summary
5: Hypothesis Testing, Part 1
5.1 - Introduction to Hypothesis Testing
5.2 - Writing Hypotheses
5.2.1 - Examples
5.3 - Randomization Procedures
5.3.1 - StatKey Randomization Methods (Optional)
5.4 - p-values
5.5 - Randomization Test Examples in StatKey
5.5.1 - Single Proportion Example: PA Residency
5.5.2 - Paired Means Example: Age
5.5.3 - Difference in Means Example: Exercise by Biological Sex
5.5.4 - Correlation Example: Quiz & Exam Scores
5.6 - Lesson 5 Summary
6: Hypothesis Testing, Part 2
6.1 - Type I and Type II Errors
6.2 - Significance Levels
6.3 - Issues with Multiple Testing
6.4 - Practical Significance
6.5 - Power
6.6 - Confidence Intervals & Hypothesis Testing
6.7 - Lesson 6 Summary
7: Normal Distributions
7.1 - Standard Normal Distribution
7.2 - Minitab: Finding Proportions Under a Normal Distribution
7.2.1 - Proportion 'Less Than'
7.2.1.1 - Example: P(Z<-1)
7.2.1.2 - Example: P(SATM<540)
7.2.2 - Proportion 'Greater Than'
7.2.2.1 - Example: P(Z>0.5)
7.2.3 - Proportion 'In between'
7.2.3.1 - Example: Proportion Between z -2 and +2
7.2.4 - Proportion 'More Extreme Than'
7.3 - Minitab: Finding Values Given Proportions
7.3.1 - Top X%
7.3.2 - Bottom X%
7.3.3 - Middle X%
7.4 - Central Limit Theorem
7.4.1 - Hypothesis Testing
7.4.1.1 - Video Example: Mean Body Temperature
7.4.1.2 - Video Example: Correlation Between Printer Price and PPM
7.4.1.3 - Example: Proportion NFL Coin Toss Wins
7.4.1.4 - Example: Proportion of Women Students
7.4.1.5 - Example: Mean Quiz Score
7.4.1.6 - Example: Difference in Mean Commute Times
7.4.2 - Confidence Intervals
7.4.2.1 - Video Example: 98% CI for Mean Atlanta Commute Time
7.4.2.2 - Video Example: 90% CI for the Correlation between Height and Weight
7.4.2.3 - Example: 99% CI for Proportion of Women Students
7.5 - Lesson 7 Summary
8: Inference for One Sample
8.1 - One Sample Proportion
8.1.1 - Confidence Intervals
8.1.1.1 - Normal Approximation Formulas
8.1.1.1.1 - Video Example: PA Residency
8.1.1.1.2 - Video Example: Dog Ownership
8.1.1.1.3 - Video Example: Books
8.1.1.1.4 - Example: Retirement
8.1.1.2 - Minitab: Confidence Interval for a Proportion
8.1.1.2.1 - Example with Summarized Data
8.1.1.2.2 - Example with Summarized Data
8.1.1.3 - Computing Necessary Sample Size
8.1.2 - Hypothesis Testing
8.1.2.1 - Normal Approximation Method Formulas
8.1.2.1.1 - Video Example: Male Babies
8.1.2.1.2 - Example: Handedness
8.1.2.1.3 - Example: Ice Cream
8.1.2.1.4 - Example: Overweight Citizens
8.1.2.2 - Minitab: Hypothesis Tests for One Proportion
8.1.2.2.1 - Minitab: 1 Proportion z Test, Raw Data
8.1.2.2.2 - Minitab: 1 Sample Proportion z test, Summary Data
8.1.2.2.2.1 - Minitab Example: Normal Approx. Method
8.2 - One Sample Mean
8.2.1 - t Distribution
8.2.2 - Confidence Intervals
8.2.2.1 - Formulas
8.2.2.1.1 - Example: MLB Age
8.2.2.1.2- Example: Sleep Deprivation
8.2.2.1.3 - Example: Milk
8.2.2.2 - Minitab: Confidence Interval of a Mean
8.2.2.2.1 - Example: Age of Pitchers (Summarized Data)
8.2.2.2.2 - Example: Coffee Sales (Data in Column)
8.2.2.3 - Computing Necessary Sample Size
8.2.2.3.1 - Example: Estimating IQ
8.2.2.3.2 - Video Example: Age
8.2.2.3.3 - Video Example: Cookie Weights
8.2.3 - Hypothesis Testing
8.2.3.1 - One Sample Mean t Test, Formulas
8.2.3.1.1 - Video Example: Book Costs
8.2.3.1.2 : Example: Pulse Rate
8.2.3.1.3 - Example: Coffee
8.2.3.1.4 - Example: Transportation Costs
8.2.3.2 - Minitab: One Sample Mean t Tests
8.2.3.2.1 - Minitab: 1 Sample Mean t Test, Raw Data
8.2.3.2.2 - Minitab: 1 Sample Mean t Test, Summarized Data
8.2.3.3 - One Sample Mean z Test (Optional)
8.3 - Paired Means
8.3.1 - Confidence Intervals
8.3.1.1. - Example: Change in Knowledge
8.3.1.2 - Video Example: Difference in Exam Scores
8.3.2 - Hypothesis Testing
8.3.2.1 - Example: Quiz Scores
8.3.3 - Minitab: Paired Means Test
8.3.3.1 - Example: SAT Scores
8.3.3.2 - Example: Marriage Age (Summarized Data)
8.4 - Lesson 8 Summary
9: Inference for Two Samples
9.1 - Two Independent Proportions
9.1.1 - Confidence Intervals
9.1.1.1 - Minitab: Confidence Interval for 2 Proportions
9.1.2 - Hypothesis Testing
9.1.2.1 - Normal Approximation Method Formulas
9.1.2.1.1 – Example: Ice Cream
9.1.2.1.2 – Example: Same Sex Marriage
9.1.2.2 - Minitab: Difference Between 2 Independent Proportions
9.1.2.2.1 - Example: Dating
9.2 - Two Independent Means
9.2.1 - Confidence Intervals
9.2.1.1 - Minitab: Confidence Interval Between 2 Independent Means
9.2.1.1.1 - Video Example: Mean Difference in Exam Scores, Summarized Data
9.2.2 - Hypothesis Testing
9.2.2.1 - Minitab: Independent Means t Test
9.2.2.1.1 - Example: Summarized Data
9.2.2.1.3 - Example: Height by Sex
9.3 - Lesson 9 Summary
10: One-Way ANOVA
10.1 - Introduction to the F Distribution
10.2 - Hypothesis Testing
10.3 - Pairwise Comparisons
10.4 - Minitab: One-Way ANOVA
10.5 - Example: SAT-Math Scores by Award Preference
10.6 - Example: Exam Grade by Professor
10.7 - Lesson 10 Summary
11: Chi-Square Tests
11.1 - Reviews
11.1.1 - Frequency Table
11.1.2 - Two-Way Contingency Table
11.1.3 - Probability Distribution Plots
11.1.4 - Conditional Probabilities and Independence
11.2 - Goodness of Fit Test
11.2.1 - Five Step Hypothesis Testing Procedure
11.2.1.1 - Video: Cupcakes (Equal Proportions)
11.2.1.2- Cards (Equal Proportions)
11.2.1.3 - Roulette Wheel (Different Proportions)
11.2.2 - Minitab: Goodness-of-Fit Test
11.2.2.1 - Example: Summarized Data, Equal Proportions
11.2.2.2 - Example: Summarized Data, Different Proportions
11.3 - Chi-Square Test of Independence
11.3.1 - Example: Gender and Online Learning
11.3.2 - Minitab: Test of Independence
11.3.2.1 - Example: Raw Data
11.3.2.2 - Example: Summarized Data
11.3.3 - Relative Risk
11.4 - Lesson 11 Summary
12: Correlation & Simple Linear Regression
12.1 - Review: Scatterplots
12.2 - Correlation
12.2.1 - Hypothesis Testing
12.2.1.1 - Example: Quiz & Exam Scores
12.2.1.2 - Example: Age & Height
12.2.1.3 - Example: Temperature & Coffee Sales
12.2.2 - Correlation Matrix
12.2.2.1 - Example: Student Survey
12.2.2.2 - Example: Body Correlation Matrix
12.3 - Simple Linear Regression
12.3.1 - Formulas
12.3.2 - Assumptions
12.3.3 - Minitab - Simple Linear Regression
12.3.4 - Hypothesis Testing for Slope
12.3.4.1 - Example: Quiz and exam scores
12.3.4.2 - Example: Business Decisions
12.3.5 - Confidence Interval for Slope
12.3.5.1 - Example: Quiz and exam scores
12.4 - Coefficient of Determination
12.5 - Cautions
12.6 - Correlation & Regression Example
12.7 - Lesson 12 Summary
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News>Bath Sheets vs Towels: Which Is Right for You? Full Comparison and Guide
Bath Sheets vs Towels: Which Is Right for You? Full Comparison and Guide
August 13, 2025 Author: Miriam Splett
When stepping out of a warm shower or a relaxing bath, the last thing you want is to be met with a towel that’s too small, rough, or slow to dry. That’s where the decision between a bath towel and a bath sheet becomes important. But what exactly is a bath sheet, how big is it compared to a regular bath towel, and why are more people switching to them - especially when made fromluxurious bamboo?
This guide breaks down everything you need to know, from size and comfort to drying power and which one is best suited to your lifestyle. Plus, we’ll share why our Luxury Bamboo Towel Set is a customer favorite and what makes it perfect for your skin, your bathroom, and the planet.
What Is a Bath Sheet?
A bath sheet is essentially a larger, more luxurious version of a standard bath towel. It’s designed to wrap comfortably around your entire body - perfect if you enjoy that spa-like feeling after a bath or need a little more coverage.
Average Bath Sheet Size:
Typically, a bath sheet measures around 90 x 160 cm, while a standard bath towel is around 70 x 140 cm.
With more surface area, a bath sheet offers:
Greater absorbency
Full body coverage
A plush, wrap-around feel
What's the Difference Between a Bath Sheet and a Bath Towel?
Here’s a quick breakdown:
| Feature | Bath Towel | Bath Sheet |
---
| Size | Smaller (approx. 70x140 cm) | Larger (approx. 90x160 cm) |
| Absorbency | Ideal for quick drying | More absorbent due to larger surface |
| Best For | Everyday use, smaller bathrooms | Lounging, full wrap-around comfort |
| Feeling | Functional | Luxurious and indulgent |
When Should You Choose a Bath Sheet?
A bath sheet is perfect if:
You prefer more coverage post-shower
You want a towel that feels spa-like and indulgent
You need higher absorbency
You enjoy lounging in your towel before getting dressed
However, if you have limited bathroom space or prefer a lighter, more compact option, bath towels might suit you better.
Introducing Our Luxury Bamboo Towel Set
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✅ Beautiful and Functional
With a soft white shade and generous size, these towels elevate your bathroom aesthetic while performing brilliantly.
Benefits of Bamboo Towels Over Cotton
If you’re used to cotton towels, switching to bamboo towels will feel like an upgrade in every way:
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Softer after every wash
More sustainable and eco-conscious
Care Tips for Bamboo Bath Sheets and Towels
To keep your towels looking and feeling their best:
Wash at 30-40°C using a gentle detergent
Avoid fabric softeners which can reduce absorbency
Tumble dry low or air dry
Skip bleach and harsh chemicals
The better you treat your bamboo towels, the longer they’ll pamper you in return.
Final Thoughts: Should You Upgrade to a Bath Sheet?
If comfort, absorbency, and luxury matter to you, a bath sheet is the clear winner. And when it’s made from a bamboo blend, you get the added benefit of sustainability, breathability, and spa-level softness.
Our Luxury Bamboo Towel Set doesn’t just help you dry off - it elevates your entire self-care routine. Whether you're winding down after a long day or starting your morning fresh, wrapping yourself in a bamboo bath towel makes every moment softer.
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A REVIEW OF SOME BASIC CONCEPTS AND RESULTS FROM THEORETICAL LINEAR ALGEBRA
2.3.7 Orthogonal Projection
Let S be a subspace of ℂn. Then an n × n matrix P having the properties: (i) R(P) = S, (ii) P = P (P is Hermitian), (iii) P2 = P (P is idempotent) is called the orthogonal projection onto S or simply the projection matrix. We denote the orthogonal projection P onto S by Ps. The orthogonal projection onto a subspace is unique.
Let V = (v1,…, vk), where {v1,…, vk} is an orthonormal basis for a subspace S. Then,
is the unique orthogonal projection onto S. Note that V is not unique, but Ps is.
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Chapter
Elements of Descriptive Geometry
1.7 The Orthogonal Projection
1.7.1 Definition
Parallel projections are used for oblique axonometric projections in Engineering Graphics, Architecture, and, to a greater extent, in general graphics. In Descriptive Geometry, however, we go one step farther and we assume that the projection rays are perpendicular to the image plane. One simple example is shown in Fig. 1.18. We obtain thus the orthogonal projection. This is a particular case of the parallel projection; therefore, it inherits all the properties of the parallel projection. In addition, the assumption of projection rays perpendicular to the image plane yields a new property that is described in the next section. Fig. 1.19 shows a more interesting example of orthogonal projection; the reader is invited to compare it with Fig. 1.15.
In technical drawing, when using several orthogonal projections to define one object, we talk about orthographic projections. Standard definitions related to orthographic representations can be read in ISO 5456-2:1996(E).
1.7.2 The Projection of a Right Angle
Fig. 1.20 shows two line segments, and , and two image planes, and , perpendicular one to another. As shown in Section 1.8, such projection planes are used in Descriptive Geometry. In Fig. 1.20 the angle is right and the side is horizontal, i.e. parallel to the projection plane . The figure shows also the horizontal and the frontal projections of this angle. By horizontal projection we mean the projection on the horizontal plane , and by frontal projection, the projection on the frontal plane . The following theorem states an important property of orthogonal projections.
If one of the sides of a right angle is parallel to one of the projection planes, the (orthogonal) projection of the angle on that plane is also a right angle.
It is possible to prove this theorem by considering the scalar products of the vectors and and of their horizontal projections. Let us give, however, an experimental proof of the right-angle theorem. We invite the reader to take a cardboard box and a square set (triangle). First, locate the square set in the corner of the box, and orient it parallel to the bottom, as in Fig. 1.21. In the electronic version of book the square set is shown in yellow, and its horizontal projection in a darker, brown hue. This projection is right-angled.
Let now the side b of the square set remain horizontal, while we rotate the side a as in the left-hand side of Fig. 1.22. The square set still fits well in the corner of the box and the horizontal projection remains right-angled. Alternatively, keep the side a horizontal and rotate the side b as in the right-hand side of Fig. 1.22. The square set again fits well in the corner of the box and the horizontal projection remains right-angled. Finally, try to rotate both sides a and b of the square set so that none of them is horizontal, or, in other words, none of them is parallel to the bottom. It is impossible to do this without changing the angle of the box corner. In fact it must be increased. This means that the horizontal projection is no more right-angled.
Is this theorem valid only for orthogonal projections? Doesn't it hold also for an oblique parallel projection? The answer is that in the general case it does not hold. Without loss of generality, we invite the reader to check the case of a parallel, oblique projection in which the projecting rays are parallel to the frontal projection plane, , but make the angle α with the horizontal projection plane, .
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Book2019, Geometry for Naval ArchitectsAdrian Biran
Chapter
Online Learning: the Stochastic Gradient Descent Family of Algorithms
2020, Machine Learning (Second Edition)Sergios Theodoridis
Orthogonal Projections
Projections and projection matrices/operators play a crucial part in machine learning, signal processing, and optimization in general; after all, a projection corresponds to a minimization task when the loss is interpreted as a “distance.” Let A be an , matrix with column vectors, , and x an l-dimensional vector. The orthogonal projection of x on the subspace spanned by the columns of A (assumed to be linearly independent) is given by (Appendix A)
(5.65)
where in complex spaces the transpose operation is replaced by the Hermitian one. One can easily check that is orthogonal to and
When A has orthonormal columns, we obtain the (familiar from geometry) expansion
Thus, the factor , for the general case, accounts for the lack of orthonormality of the columns of A. The matrix is known as the respective projection matrix and as the projection matrix on the respective orthogonal complement space.
The simplest case occurs when ; then the projection of x onto is equal to
and the corresponding projection matrices are given by
Fig. 5.14 illustrates the geometry.
Let us apply the previously reported linear algebra results in the case of the APA of (5.63) and (5.64), and rewrite them as
The first term on the right-hand side is the projection . This is the most natural. By the definition of the respective affine set, as the intersection of the hyperplanes
each vector is orthogonal to the respective hyperplane (Fig. 5.13) (Problem 5.12). Hence, projecting on the intersection of all these hyperplanes is equivalent to projecting on an affine set, which is orthogonal to all . Note that the matrix is the projection matrix that projects on the subspace spanned by . The second term accounts for the fact that the affine set on which we project does not include the origin, but it is translated to another point in the space, whose direction is determined by . Fig. 5.15 illustrates the case for and . Because does not lie on the line (plane) , whose direction is defined by , we know from geometry (and it is easily checked) that its distance from this line is . Also, lies on the negative side of the straight line, so that . Hence, taking into account the directions of the involved vectors, it turns out that
Thus, for this specific case, the correction
coincides with the recursion of the APA.
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Book2020, Machine Learning (Second Edition)Sergios Theodoridis
Chapter
INTRODUCTION
2005, The Geometrical Tolerancing Desk ReferencePaul Green
Orthographic Representation
Technical drawings usually consist of various two dimensional views to define an object, this is known as orthogonal projection. The two orthogonal projection methods used internationally are first angle projection and third angle projection. Third angle projection (figure 1.2) is used mainly in The United States and Canada whilst first angle projection (figure 1.1) is used mainly throughout Europe and the rest of the world. Both first and third angle projection have equal status and are approved internationally. In ISO Standard ISO1101: 2004 all figures have been drawn in first angle projection with dimensions and tolerances in millimetres. As this book is based on this standard first angle projection has been used throughout the book.
The identifying symbol for first angle projection or for third angle projection should be shown in the title block of your drawing.
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Book2005, The Geometrical Tolerancing Desk ReferencePaul Green
Chapter
Elements of Descriptive Geometry
2019, Geometry for Naval ArchitectsAdrian Biran
1.7.1 Definition
Parallel projections are used for oblique axonometric projections in Engineering Graphics, Architecture, and, to a greater extent, in general graphics. In Descriptive Geometry, however, we go one step farther and we assume that the projection rays are perpendicular to the image plane. One simple example is shown in Fig. 1.18. We obtain thus the orthogonal projection. This is a particular case of the parallel projection; therefore, it inherits all the properties of the parallel projection. In addition, the assumption of projection rays perpendicular to the image plane yields a new property that is described in the next section. Fig. 1.19 shows a more interesting example of orthogonal projection; the reader is invited to compare it with Fig. 1.15.
In technical drawing, when using several orthogonal projections to define one object, we talk about orthographic projections. Standard definitions related to orthographic representations can be read in ISO 5456-2:1996(E).
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Book2019, Geometry for Naval ArchitectsAdrian Biran
Chapter
Geometrical Transformations
2019, Geometry for Naval ArchitectsAdrian Biran
8.4.4 The Orthographic Projection as Limit of Perspective Projection
In Section 1.6 we have shown that the parallel projection is obtained when ‘the projection centre is sent to infinity’. In Section 1.7 we went one step farther and introduced the orthogonal projection assuming ‘that the projection rays are perpendicular to the image plane’. In this section we prove algebraically that the orthogonal projection can be derived, indeed, from the perspective projection by sending the projection centre (point O in Fig. 8.10) to infinity. We return to Eqs (8.24) remarking that we cannot use anymore the coordinates z because they go to infinity. Therefore, we move the origin of coordinates to the image plane and define the horizontal coordinate by
Using L'Hôpital's rule we obtain from Eqs (8.24)
(8.32)
The image is projected at . Finally, the orthogonal projection is obtained carrying on the transformation
(8.33)
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Book2019, Geometry for Naval ArchitectsAdrian Biran
Chapter
Elements of Descriptive Geometry
2019, Geometry for Naval ArchitectsAdrian Biran
1.17 What Is Axonometry
The preceding sections covered the basics of object representation by orthogonal projections on two planes that are perpendicular one to the other. For this method we use the term orthographic projections. The projection planes are also planes of coordinates. In three-dimensional space we have three axes of coordinates and each one is perpendicular to the other two. In the sketch, in the horizontal plane, , we see the projections of the x and y axes as perpendicular one to the other, while in the frontal plane, , the projections of the x and z axes are perpendicular one to the other. As the orthogonal projection is a special case of the parallel projection, it inherits the properties of the latter. In particular, distances and angles parallel to a projection plane are projected on that plane in true size. It is possible to measure directly such distances in any direction. To fully benefit of this property objects should be placed with the main faces (or features) parallel to projection planes. In general, drawing and measuring in orthographic projections present no special difficulties. On the other hand, to perceive the shape of a three-dimensional object one has to look at two projections and synthesize mentally the complete image. This operation requires certain spatial abilities and some training. Moreover, excepting certain simple objects, it is not possible to show all dimensions in one projection. In general, the dimensions must be distributed over at least two projections. For simple objects it is possible to avoid the above drawbacks by projecting on a single plane that makes an angle with two or all three coordinate planes. For reasons explained below we call this method axonometry. The axonometry can be also used for showing pictorially the main features of complex objects.
Oblique axonometry uses parallel projection and the projection rays are oblique to the projection plane. Two popular variants are the cavalier and the cabinet projections. Such projections are used in architecture, and in catalogues of machine details. In this book we consider only orthogonal axonometric projections in which the projection rays are perpendicular to the projection plane. In general, the projections of the coordinate axes are not perpendicular one to another, and each axis is projected at its particular scale. Therefore, it is possible to measure only in directions parallel with the projected axes, hence the name axonometry, a term composed of two Greek words that mean ‘axes’ and ‘measuring’.
In Section 1.5 we have written that for reconstituting a three-dimensional object one needs two different images. This observation is valid also for axonometric projections. An axonometric projection of an object and of a system of coordinates is ambiguous as to the position of the object with respect to the axes of the coordinates. To solve the ambiguity one has to add a projection of the object on one of the coordinate planes. Obviously, providing a stereo pair of axonometric images would also solve the problem.
1.17.1 The Law of Scales
In an axonometric projection the axes of coordinates are projected at scales that may differ from one axis to another. We are going to show that these scales are not independent, but fulfill a simple relationship. To be more specific, the sum of squares of the scales equals 2. In Fig. 1.67 the projection plane is inclined with respect to all coordinate planes; it is identified by its traces h, f, and s. The traces h and f intercept the x-axis in the point , the traces h and s intercept the y-axis in the point , and the traces f and s intercept the z-axis in the point . The orthogonal projection of the origin of coordinates on the oblique projection plane is .
We take apart the triangle and show it in Fig. 1.68. As is perpendicular to the projection plane, it is perpendicular to all lines belonging to that plane and in particular to . Let the notation of angles be , . The segment is projected as . This means that distances measured parallel to the x-axis are projected at the scale
(1.12)
Similarly, in the triangle the segment is perpendicular to . Let the notation of angles be , . The segment is projected as . This means that distances measured parallel to the y-axis are projected at the scale
(1.13)
Finally, in the triangle the segment is perpendicular to . Let the notation of angles be , . The segment is projected as . This means that distances measured parallel to the z-axis are projected at the scale
(1.14)
Squaring both sides of Eqs (1.12) to (1.14), and considering the trigonometric relationships in the three triangles, we obtain
(1.15)
Adding these equations side by side yields
(1.16)
The quantities between parentheses are the direction cosines of the normal and their sum equals 1. Indeed, the projections of this normal on the three axes are , , . The length of can be calculated from its projections by Pythagoras' theorem in three dimensions
which gives
Substituting this value into Eq. (1.16) we obtain the fundamental law of axonometry
(1.17)
1.17.2 Isometry
When the isometric projection is displayed by means of specialized software it is easy to choose any position of the projection plane and let the computer draw the projection as required by Eq. (1.17). Doing the same in manual work could be very difficult. Therefore, there is an interest in simplified procedures. The first idea that may cross our mind is to let all three scales be equal, that is
(1.18)
Intuition may tell us that Eq. (1.18) is fulfilled when, in Fig. 1.67,
(1.19)
We can prove Eq. (1.19) deriving from Fig. 1.68 the equality
Squaring both sides and after some substitutions we get
In the same way we write
Dividing the above equations side by side and extracting the square roots yields
Taking into account the equality of scales, that is Eq. (1.18), we obtain Eq. (1.19). Then, the right-angled triangles , , and are congruent and . As shown in Fig. 1.70, the triangle of traces, , is equilateral and its angles are all equal to . The projected origin is the centre of the circumscribed circle and the projected axes bisect the angles of the triangle into angles of . The angles between the projected axes are all equal to , as drawn in Fig. 1.69. We draw these axes at from a horizontal. As a first and classical example, Fig. 1.71 shows the isometric cube. The circles inscribed in the faces of the cube are projected as ellipses. This cube appears in the German standard DIN 5, Part 1 (1970).
In manual work it is impractical to multiply all dimensions by the scale 0.82. Therefore, various standards allow to draw at the scale 1:1. Comparing such a drawing with the corresponding orthographic projections causes a sensation of enlargement. Some CAD programs that can display isometric projections let the user choose between the 1:1 and the scales.
Example 1.1 Isometric drawings of piping systems
An outstanding application of the isometric projection is in the drawing of piping systems. The advantages are obvious even for such a simple example as that shown in Fig. 1.72. This drawing was carried on with the help of a CAD program, using an isometric grid. Pipes are represented as single lines following the axes of the actual pipes. This single view is sufficient for understanding how the pipes are connected and for writing all dimensions. Fig. 1.73 shows the same system in orthographic projections. To completely define the system we used a front view, a top view, and two sections. To understand the connections between the four pipes that run in parallel with the x-axis, one has to see all four projections and mentally synthesize the knowledge.
Marine systems consist of dozens of pipes, valves, and other fittings. Therefore, one can appreciate that it would be practically impossible to design and build such systems without isometric drawings. A detailed example of a piping system, and valuable explanations are included in the German standard DIN 5, Part 1 (1970).
1.17.3 An Ambiguity of the Isometric Projection
Fig. 1.74 shows in isometric projection a system of coordinates and a point, P. This projection is not sufficient to define the position of the point P with respect to the system of coordinates. As seen in Fig. 1.75, to the given projection of the point P we can associate as horizontal projection any point lying on the vertical through P, below the intersection with the y-axis. Three examples are , and The frontal projections, , and can be obtained by an obvious construction and thus we completely define the points , and that all can correspond to the given isometric projection P.
To conclude the discussion we mention that for solids that extend in one direction much more than in the other two, the isometric projection may look unnatural. The cause is that from our visual experience we expect parallel lines to converge, unless they are parallel to the image plane. In such cases it is recommended to use the dimetric projection in which only two scales are equal. Good definitions and illustrations can be seen, for example, in the German standard DIN 5, Part 2 (1970). When all three scales are different, we speak about a trimetric projection. These matters are detailed in some textbooks on engineering drawing (for example Chirone and Tornincasa, 2005), or in specialized books, such as the superbly illustrated one of Pavanelli et al. (2003).
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8.4 Perspective Projections
8.4.1 The Projection Matrix
In Chapter 1, Appendix 1.A, we have shown that orthographic projections can be obtained multiplying the array of coordinates by a projection matrix. In this section we explain how to use matrices for central projections. This section should be regarded as a continuation of Section 1.4. We use the term perspective projection as an alternative for the term central projection. To derive a transformation matrix refer to Fig. 8.10. The projection centre is the point O. We place in this point the origin of the world coordinates. The z-axis is horizontal. The line Oz is the optical axis. The object to be projected is the point P; its coordinates are x, y, z. The image plane is situated at the distance f from O. In the case of a camera the distance f is the focal length. The optical axis pierces the image plane in the point called principal point or image centre. The central, or perspective projection of P is , the point in which the view line pierces the image plane. The coordinates of in the image plane are X, Y. is the orthogonal projection of P on the plane xz and the orthogonal projection of on the optical axis. From similar triangles we derive
(8.23)
From these equations we obtain the coordinates of in the image plane
(8.24)
Obviously . Arranged in matrix form Eqs (8.24) become
(8.25)
The third component is f as the projected point lies in the image plane. An example of function that calculates the perspective projection of a given point is shown below.
function Q = PerspPoint( P, f)
%PERSPPOINT Yields the central projection of point P.
% Input arguments:
% P, column vector of the homogeneous coordinates of the point P
% f, distance between centre of projection and image plane,
% for example the focal length of a camera.
% create projection matrix
PM = [ f 0 0 0
0 f 0 0
0 0 f 0
0 0 1 0 ];
% calculate homogeneous coordinates of the projected point
P1 = PMP;
% convert to Euclidean coordinates in the image plane
Q = P1(1:2)/P(3);
end
The following MATLAB script calculates the perspective projection of a unit cube centred about the optical axis.
%PERSPCUBE1 Perspective view of unit cube centered about the optical axis.
% define frame
plot([ -0.05 0.05 ], [ 0 0 ], 'k-', [ 0 0 ], [ -0.05, 0.05 ], 'k-')
axis equal, axis off
hold on
% define cube vertices
P1 = [ -0.5; -0.5; 2; 1 ]; P2 = [ 0.5; -0.5; 2; 1 ];
P3 = [ 0.5; 0.5; 2; 1 ]; P4 = [ -0.5; 0.5; 2; 1 ];
d = [ 0; 0; 1; 0 ]; % cube depth
P5 = P1 + d; P6 = P2 + d; P7 = P3 + d; P8 = P4 + d;
f = 0.08;
Q1 = PerspPoint(P1, f); Q2 = PerspPoint(P2, f);
Q3 = PerspPoint(P3, f); Q4 = PerspPoint(P4, f);
Ff = [ Q1 Q2 Q3 Q4 Q1 ] % front face projection in homogeneous coordinates
Ffe = Ff(1:2, 1:5) % projection of front face in Euclidean coordinates
pline(Ffe, 1.5, 'k', '-')
Q5 = PerspPoint(P5, f); Q6 = PerspPoint(P6, f);
Q7 = PerspPoint(P7, f); Q8 = PerspPoint(P8, f);
Fb = [ Q5 Q6 Q7 Q8 Q5 ] % back face projection in homogeneous coordinates
Fbe = Fb(1:2, 1:5) % projection of back face in Euclidean coordinates
pline(Fbe, 1.5, 'k', '-')
pline([ Q2(1:2) Q6(1:2) ], 1.5,'k', '-')
pline([ Q3(1:2) Q7(1:2) ], 1.5,'k', '-')
pline([ Q4(1:2) Q8(1:2) ], 1.5,'k', '-')
pline([ Q1(1:2) Q5(1:2) ], 1.5,'k', '-')
hold off
The result is shown in Fig. 8.11. The vanishing point coincides with the origin of the coordinates x, y. To obtain a projection with two vanishing points we rotate the cube around the vertical axis of coordinates. To exemplify this add the following lines to the script PerspCube1.
% create matrix of rotation by -30 deg around the y-axis
TR = [ cosd(-30) 0 sind(-30) 0
0 1 0 0
-sind(-30) 0 cosd(-30) 0
0 0 0 1 ];
% rotate
P1 = TRP1; P2 = TRP2; P3 = TRP3;
P4 = TRP4; P5 = TRP5; P6 = TRP6;
P7 = TRP7; P8 = TRP8;
8.4.2 Ideal and Vanishing Points
We consider a line passing through a point P and having the direction defined by a vector v. The parametric equation of this line, in the system of coordinates assumed in Fig. 8.10, is
(8.26)
The equation of a line parallel to this and passing through a point is
(8.27)
As scaling the homogeneous coordinates of a point by the same factor does not change the corresponding Euclidean coordinates of that point we divide Eq. (8.26) by t and get
(8.28)
The point at infinity of the line is
(8.29)
In a similar way we obtain the same point at infinity for the line . We conclude that
1.
: the lines and intersect in the point defined by Eq. (8.29);
2.
: the coordinates of the point at infinity of a set of parallel line depend only on the direction of those lines. Actually, these coordinates represent the common direction.
As we made no restriction on the direction of the lines, conclusion 1 can be generalized to say that all lines that have the same direction also have the same point at infinity. We go now a step ahead and calculate the perspective projection of the point at infinity found above. As scaling all homogeneous coordinates by the same factor does not change the corresponding Euclidean coordinates, we use Eq. (8.25) multiplied by z and write
(8.30)
Dividing all components by we get
(8.31)
Analyzing this result we reach algebraically conclusions that in Chapter 1 were presented geometrically. These conclusions are listed below.
VANISHING POINTS.
: If the ideal point is projected at a finite distance and is the vanishing point for all the lines with the direction defined by the vector v. The condition means that the lines are not parallel to the image plane.
LINES PARALLEL TO THE IMAGE PLANE.
: The condition means lines parallel to the image point. Eq. (8.31) shows that there is no visible vanishing point. Lines parallel to the image plane are projected as parallel lines.
LINES PERPENDICULAR TO THE IMAGE PLANE.
: For lines perpendicular to the image point . The vanishing point is the image centre, the point in which the optical axis pierces the image plane.
8.4.3 The Vanishing Line
In projective geometry it is taught that all the points at infinity of a plane lie on a straight line called the line at infinity of that plane. Correspondingly, in a perspective projection the vanishing points of a plane lie on the vanishing line of that plane. To give an example we show in Fig. 8.13 the perspective view of a hexagon that lies in a horizontal plane below the centre of projection. Starting from the vertex closest to the centre of projection, the first edge is parallel to the fourth, the second to the fifth, and the third to the sixth. The parallel edges are projected as lines that intersect in three vanishing points that are collinear. In perspective drawing the common line is called horizon line. This property is used in the drawing of perspective views. In a three-point perspective we have a vanishing point that does not lie on the horizon line. An example is shown in Fig. 1.14.
8.4.4 The Orthographic Projection as Limit of Perspective Projection
In Section 1.6 we have shown that the parallel projection is obtained when ‘the projection centre is sent to infinity’. In Section 1.7 we went one step farther and introduced the orthogonal projection assuming ‘that the projection rays are perpendicular to the image plane’. In this section we prove algebraically that the orthogonal projection can be derived, indeed, from the perspective projection by sending the projection centre (point O in Fig. 8.10) to infinity. We return to Eqs (8.24) remarking that we cannot use anymore the coordinates z because they go to infinity. Therefore, we move the origin of coordinates to the image plane and define the horizontal coordinate by
Using L'Hôpital's rule we obtain from Eqs (8.24)
(8.32)
The image is projected at . Finally, the orthogonal projection is obtained carrying on the transformation
(8.33)
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Elements of Descriptive Geometry
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1.7.2 The Projection of a Right Angle
Fig. 1.20 shows two line segments, and , and two image planes, and , perpendicular one to another. As shown in Section 1.8, such projection planes are used in Descriptive Geometry. In Fig. 1.20 the angle is right and the side is horizontal, i.e. parallel to the projection plane . The figure shows also the horizontal and the frontal projections of this angle. By horizontal projection we mean the projection on the horizontal plane , and by frontal projection, the projection on the frontal plane . The following theorem states an important property of orthogonal projections.
If one of the sides of a right angle is parallel to one of the projection planes, the (orthogonal) projection of the angle on that plane is also a right angle.
It is possible to prove this theorem by considering the scalar products of the vectors and and of their horizontal projections. Let us give, however, an experimental proof of the right-angle theorem. We invite the reader to take a cardboard box and a square set (triangle). First, locate the square set in the corner of the box, and orient it parallel to the bottom, as in Fig. 1.21. In the electronic version of book the square set is shown in yellow, and its horizontal projection in a darker, brown hue. This projection is right-angled.
Let now the side b of the square set remain horizontal, while we rotate the side a as in the left-hand side of Fig. 1.22. The square set still fits well in the corner of the box and the horizontal projection remains right-angled. Alternatively, keep the side a horizontal and rotate the side b as in the right-hand side of Fig. 1.22. The square set again fits well in the corner of the box and the horizontal projection remains right-angled. Finally, try to rotate both sides a and b of the square set so that none of them is horizontal, or, in other words, none of them is parallel to the bottom. It is impossible to do this without changing the angle of the box corner. In fact it must be increased. This means that the horizontal projection is no more right-angled.
Is this theorem valid only for orthogonal projections? Doesn't it hold also for an oblique parallel projection? The answer is that in the general case it does not hold. Without loss of generality, we invite the reader to check the case of a parallel, oblique projection in which the projecting rays are parallel to the frontal projection plane, , but make the angle α with the horizontal projection plane, .
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Topics of Functional Analysis
2008, Advanced Mathematical Tools for Automatic Control Engineers: Deterministic Techniques, Volume 1Alexander S. Poznyak
18.5.4 Orthogonal projection operators in Hilbert spaces
Let M be a subspace of a Hilbert space .
Definition 18.22
The operator , acting in such that
(18.60)
is called the orthogonal projection operator to the subspace M.
Lemma 18.10
The element y = Px is unique and (x − y, z) = 0 for any z ∈ M.
Proof
See subsection 18.3.2.
The following evident properties of the projection operator hold.
Proposition 18.2
1.
: x ∈ M if and only if Px = x.
2.
: Let M⊥be the orthogonal complement to M, that is,
(18.61)
Then any can be represented as where y ∈ M and Z ⊥ M. Then the operator , defining the orthogonal projection any point from to , has the following representation:
(18.62)
3.
: if and only if Px = 0.
4.
: P is a linear operator, i.e., for any realandwe have
(18.63)
5.
: (18.64)
Indeed, which implies and thus . On the other hand, if with . Then . The inequalities and give (18.64).
6.
: (18.65)
since for any x ∈ M we have P2(Px) = Px.
7.
: P is self-adjoint, that is,
(18.66)
8.
: For any
(18.67)
which implies
(18.68)
9.
: if and only if x ∈ M.
10.
: For any
(18.69)
which follows from (18.67), the Cauchy-Bounyakovsk-Schwarz inequality and (18.64).
11.
: Let and . Then A is obligatory an orthogonal projection operator to some subspace . Indeed, since it follows that and .
The following lemma can be easily verified.
Lemma 18.11
Let P1be the orthogonal projector to a subspace M1and P2be the orthogonal projector to a subspace M2. Then the following four statements are equivalent:
1.
2.
3.
: for any
4.
: for any
Corollary 18.9
1.
: if and only if P1P2= 0
2.
: P1P2 is a projector if and only if P1P2 = P2P1.
3.
: Let Pi (1 = 1,…,N) be a projection operator. Then is a projection operator too if and only if
4.
: P1 − P2is a projection operator if and only if P1P2 = P2, or, equivalently, when
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Related terms:
Nuclear Magnetic Resonance
Principal Components
Least Square
Singular Value Decomposition
Eigenvalue
Eigenvector
Component Analysis
Projection Operator
Random Variable ξ
Singular Vector
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3081 | https://en.khanacademy.org/math/probability/binomial-probability-a2/binomial-probability-a2i/v/free-throw-binomial-probability-distribution | Free throw binomial probability distribution (video) | Khan Academy
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AP®︎/College Statistics
Course: AP®︎/College Statistics>Unit 8
Lesson 5: Introduction to the binomial distribution
Binomial variables
Recognizing binomial variables
10% Rule of assuming "independence" between trials
Identifying binomial variables
Binomial probability example
Generalizing k scores in n attempts
Free throw binomial probability distribution
Graphing basketball binomial distribution
Binompdf and binomcdf functions
Binomial probability (basic)
Binomial probability formula
Calculating binomial probability
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AP®︎/College Statistics>
Random variables and probability distributions>
Introduction to the binomial distribution
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Free throw binomial probability distribution
AP.STATS: UNC‑3 (EU), UNC‑3.B (LO), UNC‑3.B.1 (EK)
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About About this video Transcript
Sal uses the binomial distribution to calculate the probability of making different number of free throws.
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tanwarkml 10 years ago Posted 10 years ago. Direct link to tanwarkml's post “In the first video (no. o...” more In the first video (no. of heads) we only take combinatorics and not the individual probability (p(H) = 0.5 p(T) =0.5). In this video, we take both P(Score)= 0.7 P(MIss)=0.3 and the combinatorics i.e. the different ways in which you could score and miss. What am I missing?
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robshowsides 10 years ago Posted 10 years ago. Direct link to robshowsides's post “Good question. When the ...” more Good question. When the two events have equal probability, we can just use combinatorics, because every outcome is equally likely. Say we flip a coin six times. Then p(HHHTTT) = p(HHHHHH) = p(TTHHTT) = ... etc = (0.5)^6 = 1/(2^6) = 1/64
But when the two possibilities have different probabilities, we need to use those probabilities. So if p(S)= 0.7 and p(M)=0.3, then p(SSSSSS) = (0.7)^6, p(SMSMSM) = (0.7)^3 (0.3)^3, p(SSMMSS) = (0.7)^4 (0.3)^2. So there are still 64 different outcomes of 6 shots, like flipping 6 coins, but they have seven different groups with different probabilities that can be collected -- groups with 0,1,2,3,4,5, or 6 scores out of 6 shots. So there are 6C4 different ways he could have 4 scores and 2 misses, all with the same probability. If you DID use the probabilities for H and T, you would still get the correct answer, of course.
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Tyler 6 years ago Posted 6 years ago. Direct link to Tyler's post “Doesn't each attempt affe...” more Doesn't each attempt affect the probability of the next? Since the probability is just an average or something like that. I feel like the 70% & 30% should change each attempt. (since this isn't something like a coin toss)
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tfeng1822 2 years ago Posted 2 years ago. Direct link to tfeng1822's post “If that were the case, it...” more If that were the case, it would not be a binomial probability, and the point of this video would be thrown out the window. I think it is reasonable to assume that the 70% and 30% should be constant, because how do you calculate a change in chance of attempt?
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Cecilie 10 years ago Posted 10 years ago. Direct link to Cecilie's post “But how can I calculate 1...” more But how can I calculate 120c3? It´s easy when it´s only 6c0,1,2,3 and so on, but what do I do when the number is 120c3? My problem is this: 120c30,05^30.95117...
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robshowsides 10 years ago Posted 10 years ago. Direct link to robshowsides's post “120C3 = (120!)/(117! 3!...” more 120C3 = (120!)/(117! 3!), right? Think about what that would look like if you wrote it out:
(120119118117116115...321)/((117116115...321)(321)), right? Can you see that lots and lots of terms will cancel? Not much will be left after you cancel...
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rana 2 years ago Posted 2 years ago. Direct link to rana's post “I couldnt get what he mea...” more I couldnt get what he means at 7:15 ;(
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daniella 2 years ago Posted 2 years ago. Direct link to daniella's post “At 7:15, they're emphasiz...” more At 7:15 , they're emphasizing that the probability distribution for the number of successful free throws in 6 attempts is not symmetric, unlike some other distributions. The lack of symmetry arises from the fact that the probability of making a free throw is not 0.5 (as it would be in a fair coin flip scenario), but rather 0.7 in this example. The speaker is highlighting that although there isn't a symmetrical shape to the distribution, there is still a pattern in the coefficients of the binomial distribution.
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Ariel Olshevsky 10 years ago Posted 10 years ago. Direct link to Ariel Olshevsky's post “so when you take only 6 f...” more so when you take only 6 free throws you can calculate the probability of some number of "success" with that formula, but when you got lets say 300 free throws and you want to know whats the probability of scoring AT LEAST 20 of them how would you do that? don't tell me I need to solve the equation for these big numbers....
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Nicholas Mihm 4 years ago Posted 4 years ago. Direct link to Nicholas Mihm's post “Why did we add in the (6/...” more Why did we add in the (6/choose #)? In the last videos like binomial probability example, we just multiplied 0.7^2 and 0.3^1. But now we are multiplying the (6/choose #) in as well. Can someone explain to me why this is?
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daniella 2 years ago Posted 2 years ago. Direct link to daniella's post “The inclusion of n choose...” more The inclusion of n choose k (often denoted as (nCk)) in the binomial probability calculation represents the number of ways you can choose k successes out of n attempts. It's a fundamental part of binomial probability calculations because it accounts for the different possible arrangements of successes within the n trials. Multiplying by (nCk) ensures that each combination of successes is appropriately weighted in the overall probability calculation.
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Aarnav Sharma 5 years ago Posted 5 years ago. Direct link to Aarnav Sharma's post “Why does the probability ...” more Why does the probability go up, but at 6 choose 6 it goes down?
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daniella 2 years ago Posted 2 years ago. Direct link to daniella's post “The probability goes up a...” more The probability goes up as the number of successful attempts increases because you are considering scenarios where more free throws are made. However, at 6 choose 6, you are calculating the probability of making all 6 free throws, which is the highest possible success scenario in this case. The probability may seem to decrease here due to rounding or computational precision, but conceptually, it represents the scenario where all attempts are successful.
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azkkanoor156 5 years ago Posted 5 years ago. Direct link to azkkanoor156's post “AT 7.40 - how is 6 choose...” more AT 7.40 - how is 6 choose 2 is 15?
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Jimmy 2 years ago Posted 2 years ago. Direct link to Jimmy's post “Using the combinations fo...” more Using the combinations formula you get 65/2=15
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Sara García 5 months ago Posted 5 months ago. Direct link to Sara García's post “A las 5:31, ¿cómo es la l...” more A las 5:31 , ¿cómo es la luna lo suficientemente grande como para bloquear el sol? ¿No es el sol mucho más grande que la luna?
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Steve Logan 9 years ago Posted 9 years ago. Direct link to Steve Logan's post “i don't quite understand ...” more i don't quite understand why the closest we came to the shooting percentage of 70% is 32.4%?
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Video transcript
[Voiceover] Now that we've spent a couple of videos exploring a scenario where I'm taking multiple free throws and figuring out the probability of making K of the scores and six attempts or in N attempts. Let's actually define a random variable using this scenario and see if we can construct it's probability distribution and we'll actually see that it's a binomial distribution. So, let's define the random variable X. So, let's say that X is equal to the number, the number of made shots, number of made free throws when taking, when taking six free throws. So, it's how many of the six do you make? And we're going to assume what we assumed in the first video in this series of this, these making free throws. So, we're gonna assume the seventy percent free throw probability right over here. So, assuming assumptions, assuming seventy percent free throw, free throw percentage. All right. So, let's figure out the probabilities of the different values that X could actually take on. So, let's see, what is the probability, what is the probability that X is equal to zero? That even though you have a seventy percent free throw percentage that you make none of the shots and actually you could calculate this through probably some common sense without using all of these fancy things but just to make things consistent, I'm gonna write it out. So, this is going to be, this is going to be, it's going to be equal to six choose zero times zero point seven to the zeroth power times zero point three to the sixth power, and this right over here is gonna end up being one. This over here's going to end up being one. And so, you're just gonna be left with zero point three to the sixth power and I calculated it ahead of time. So, if we round to the nearest, if we do, if we round our percentages to the nearest tenth, this is going to give you approximately, approximately, well, if we round the decimal to the nearest, to the nearest thousandth you're gonna get something like that which is approximately equal to zero point one percent chance of you missing all of them. So, roughly I'm speaking, roughly here a one in a thousand, one in a thousand chance of that happening, of missing all six free throws. Now, let's keep going, this is fun. So, what is the probability that our random variable is equal to 1? Well, this is going to be six choose one times zero point seven to the first power times zero point three to the six minus first power. So, it's a fifth power. And I calculated this and this is approximately zero point zero one or we could say one percent. So, it's still a fairly low probability. Ten times more likely than this roughly but still a fairly low probability. Let's keep going. So, the probability that x is equal to two, well, that's what our first video was essentially. So, this is going to be six choose two times zero point seven squared times zero, zero point three to the fourth power, and we saw that this is approximately going to be zero, zero point zero six, or we could say six percent, and obviously you could type these things in a calculator and get a much more precise answer but just for the sake of just getting a sense of what these probabilities all look like, that's why I'm giving these rough estimates. Kind of, I guess you could say, to the closest, maybe, tenth of a, tenth of a percent. And actually and if you round to the closest tenth of a percent you actually actually get to six point oh percent and this is one point, one point oh percent 'cause this we actually went to a tenth of a percent here but let's keep going. We're obviously going to have to do a few more of these. So, let me just make sure I have enough real estate. All right, so, the probability that our random variable is equal to three is going to be six choosethree and I'm sure you could probably fill this out on your own but I'm going to do it. Zero point seven to the third power times zero point three to the six minus three which is a third power which is approximately equal to, well, it's going to be zero point one eight five or eighteen point five, eighteen point five percent. So, yeah, that's definitely within the realm of possibility. I mean, all of these are in the realm of possibility but it's starting to be a non-insignificant probability. So, now, let's do the probability that our random variable is equal to four. Well, it's going to be six choose four times zero point seven to the fourth power times zero point three to the six minus four or second power which is equal to, this is going to get equal to or approximately, 'cause I am, I am taking away a little bit of the precision when I write these things down. Zero point three two four. So, approximately thirty two point four percent chance of making exactly four out of the six free throws. All right, two more to go. Let's see, I have not used purple as yet. So, the probability that a random variable is equal to five it's gonna be six choose five or zero, at times I should say, zero point seven to the fifth power times zero point three to the first power and that is going to be roughly, roughly zero point three zero three which is thirty point three percent. That's interesting, one more left. So, the probability that I make all of them, of all six, is going to be equal to, is equal to six choose six and zero point seven to the sixth power times zero point three to the zeroth power which is, this right over here is going to be one, this is going to be one, so it's really just zero point seven to the sixth, to the sixth power and this is approximately zero point one one eight. I calculated that ahead of time, which is eleven point eight. Eleven point eight percent. And so, there's something interesting that's going on here. The first time we looked binomial distribution we said, "Hey, there's a symmetry as we kind of "got to some type of a peak and went down, "but I don't see that symmetry here." And the reason you're not seeing that symmetry is that you are more likely to make a free throw than not. It was you have a seventy percent free throw probability. This is no longer just flipping a fair coin. Where you will see the symmetry is in these coefficients. If you calculate these coefficients, six choose zero is one. Six choose six is one. You would see that six choose one is six and six choose five is six. You'd see six choose two is fifteen and six choose four is also fifteen, and then six choose three is twenty. So, you definitely see, you definitely see the symmetry in the coefficients but then these things are weighted by the fact that you're more likely to make something than miss something. If these were both point five then you would also see the symmetry right over here. And you can plot this to essentially visualize what the probability distribution looks like for this example and I encourage you to do that. To take these different cases, just like we did in that first example with the fair coin, and plot these. But this essentially does give you the probability distribution for, for the random variable in question. This is, I just wrote it out instead of just visualizing it but it says, "Okay, well, what's the... "These are the different values "that this random variable can take on." It can't, it can't take on negative one or it can't be fifteen point five or pi or one million. These are the only seven values that this random variable can take on and I've just given you the probabilities, or I guess you could say the rough probabilities of the random variable taking on each of those, each of those seven values.
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3082 | https://web.physics.wustl.edu/classinfo/322/Pdf/errorProp.pdf | Consider a box having length L, width W and height H. How is the change of volume (V) of the box related to the changes in length, height and width? Since V = LHW, if only the length varies the change in volume is ∆V = ∆LHW, and L∆HW if only the height changes and LH∆W if only the width changes. If all three dimensions change, then the change in volume is simply the sum of these individual changes: ∆V = ∆LHW + L∆HW + LH∆W (2) The variation of a function of several variables due to the change in only one of the variables is how we define partial derivatives. The first term in the equation above is simply the partial derivative of V with respect to L times the change in length L. We can write the total differential change in volume in terms of partial derivatives as: dV = δV δL dL + δV δH dH + δV δW dV (3) Now suppose g(x,y,z) is a quantity that is a function of the measured variables x,y,z. Then g itself is uncertain due to the uncertainties of each of these measured values. A measure of the scatter of the individual measured values of each variable about their mean is provided by the average square deviation of g and is given by the sum of the square deviations divided by N. In fact, this becomes the commonly accepted definition of the square of the standard deviation„ when N is replaced by N-1.
σ2 g = 1 N −1 $ −i (gi −g)2 (4) (where gi is an individual value and g is the average value) This definition keeps σ meaningful for small populations (when N = 1, or a small number).
Statistically, 68% of the measurements will fall within 1σ and 95% within 2σ. The variation (deviation from the mean) of the ith value of g is related to the variations of the measured variables by: (gi −g) = (xi −x)δg δx + (yi −y)δg δy + (zi −z)δg δz (5) Therefore, the square of the standard deviation is given by: σ2 g = ( 1 N −1) $ i [(xi −x)δg δx + (yi −y)δg δy + (zi −z)δg δz]2 (6) or after expansion by: σ2 g = ( 1 N −1)[ $ i (xi −x)2(δg δx)2 + $ i (yi −y)2(δg δy)2 + $ i (zi −z)2(δg δz)2] (7) 3 The cross terms (second line in Eq. 7) vanish in the summation if the variables are linearly independent. That is, if the variation in x, for instance, is independent of the variations in y and z.
Note also that: σ2 x = 1 N −1 $ i (xi −x)2 ; σ2 y = 1 N −1 $ i (yi −y)2 ; σ2 z = 1 N −1 $ i (zi −z)2 (8) Substituting the expressions from Eq. 8 into Eq. 7 yields: σ2 g = σ2 x(δg δx)2 + σ2 y(δg δy)2 + σ2 z(δg δz)2 (9) which relates the standard deviation of the computed function σg to the standard deviations of the measured quantities. Since the standard deviation σ is related to the standard deviation of the mean (or standard error) σm by the relationship σm = σ/ √ N, we can find the expected uncertainty of the computed quantity g from the uncertainties of the measured quantities x,y,z. This quantity is also known as the standard deviation of the mean or the standard error of the mean. Now lets see how this works for the different functional forms.
2.1 Addition and Subtraction First we consider a quantity g(x,y,z) that is a function of three independent parameters consisting of sums and differences: g(x, y, z) = ax + by + cz or g(x, y, z) = ax −by −cz (10) The partial derivatives are δg δx = a, δg δy = ±b δg δz = ±c (11) The sign depends upon whether the terms are added or subtracted. We can use the expression derived above (Eq. 9) to find σg as a function of σx, σy and σz σ2 g = σ2 x(δg δx)2 + σ2 y(δg δy)2 + σ2 z(δg δz)2 = a2σ2 x + b2σ2 y + c2σ2 z (12) Note here that the minus signs vanish because of the squares. The square of the standard error is σ2 m = σ2 g N = a2σ2 x + b2σ2 y + c2σ2 z N = a2σ2 mx + b2σ2 my + c2σ2 mz (13) Thus, the uncertainty for measurements that have a functional dependence that involves sums and differences is σm = " a2σ2 mx + b2σ2 my + c2σ2 mz = % 1 N (a2σ2 x + b2σ2 y + c2σ2 z) (14) Rule for Addition and Subtraction The overall uncertainty is equal to the square root of the sum of the squares of the uncertainties of each of the individual terms.
4 2.2 Multiplication and Division Next we consider a quantity g(x,y) that is a function of two independent parameters consisting of a single multiplication or division g(x, y) = ±axy or g(x, y) = ±ax/y (15) For the case of multiplication we have δg δx = ±ay and δg δy = ±ax and σ2 g = σ2 x(δg δx)2 + σ2 y(δg δy)2 = a2y2σ2 x + a2x2σ2 y and for division δg δx = ±a y and δg δy = ±ax y2 and σ2 g = σ2 x(δg δx)2 + σ2 y(δg δy)2 = a2 y2σ2 x + a2x2 y4 σ2 y Dividing by g2 (in each case) results in the following expression σ2 g g2 = σ2 x x2 + σ2 y y2 for both multiplication and division. Now, recalling that the standard error is σm = σ/ √ N, and that σm/g is the fractional error of g, we have σm g = % (σmx x )2 + (σmy y )2 for both multiplication and division.
Rule for Multiplication and Division The Fractional Error of the quantity (fractional overall uncertainty) is equal to the square root of the sum of the squares of the individual fractional errors (note that σmx/x is the fractional error of x, etc.).
5 2.3 Powers At first glance one may think that powers are just products and we proceed as described above for multiplication. For instance, the function, g(x,y) = Cxy2 = Cxyy is a constant C times the product of three variables x, y and y, but the last two are obviously not independent variables. Therefore, the treatment above is no longer valid and we must develop the proper expression for variables raised to some power.
Consider a function of a single variable given by g(x) = ax±b which has the following partial derivative δg δx = ±abx±(b−1) Since σ2 g = σ2 x(δg δx)2 or σg = σx δg δx = σxabx±(b−1) (16) we can obtain the fractional uncertainty by dividing Eq. 16 by g = ax±b σg g = bσx x Recalling that the standard error is defined as σm = σ N we obtain the fractional uncertainty in terms of the standard error σm g = bσmx x Rule for Powers For measurements that have the functional form , the fractional error on g(x,y,z) = Cxpyqzr , the fractional error on g is given by σm g = % (pσmx x )2 + (qσmy y )2 + (rσmz z )2 3 Least Squares Fitting In many cases the measurement of the function f(x), f(x,y), etc. is performed by varying one of the independent variables over a large range. For example, consider the function y = A + Bx. Instead of making numerous measurements of y for one particular value of x, we can make a series of N measurements of yi , one for each of several values of the quantity x = xi, where i runs from 1 to N.
Therefore we have N sets of data points represented as (xi, yi). Given this situation, what will be the best estimate for the values A and B and the error in the various quantities y, A and B? One powerful approach to determine these quantities is that of least-squares fitting and is discussed below.
6 xi yi x2 i xiyi yi −A −Bxi # xi # yi # x2 i # xiyi #(yi −A −Bxi)2 Table 1: Least-squares analysis for straight line fitting 3.3 Uncertainties in A, B and y The standard deviation for the various quantities can be evaluated as under: σ2 y = 1 N −2 N $ i=1 (yi −A −Bxi)2 (23) The denominator of N-2 comes from the fact that there are only N -2 independent variable in the N measurements, as we have determined A and B.
σ2 A = σ2 y # x2 i ∆; and σ2 B = Nσ2 y ∆ (24) where ∆is the denominator of equation 22.
4 Data Rejection Sometimes when we make a series of measurements of a specific quantity one of the measured values disagrees strikingly with all the other measured values. When this happens the experimenter is presented with the situation where he/she must decide whether the anomalous measurement resulted from some mistake (glitch in the measurement system) and should be rejected or was a bona fide measurement that should be kept. If careful records were kept sometimes we can establish a definite cause for the anomalous measurement and therefore justifiably reject the measurement.
If an external cause can not be found for the anomalous result, then the truly honest course of action is to repeat the measurement many times. If the anomaly shows up again then hopefully the cause may be found. Either as a glitch in the measurement system or as a real physical effect. If the anomaly does not recur, then due to the increased number of measurements made there will be no significant difference in our final answer whether we include the anomaly or not.
If it is impossible to re take the measurements then the experimenter must decide whether or not to reject the anomaly by examining the measured data and the properties of a Gaussian distribution.
The rejection of data is a subjective controversial question, on which experts disagree. The experimenter who rejects data may reasonably be accused of fixing his/her data. The situation is made worse by the possibility that the anomalous result may reflect some important physical effect.
One criterion for rejecting suspect data is Chauvenet’s criterion.
8 4.1 Chauvenet’s Criterion for Data Rejection Suppose we make N measurements x1, x2, ..... xNof the same quantity x.
1. Using all the values of the N measurements made calculate the mean (x) and standard deviation (σx). If one of the measurements (call it xsuspect) greatly differs from x and looks suspicious, then calculate tsuspect = xsuspect −x σx the number of standard deviations by which xsuspect differs from x.
2. 2. We next find the probability P(outside tsuspect σx) that a legitimate measurement will differ from x by or tsuspect or more standard deviations.
P(outside tsuspectσx) = 1 −P(within tsuspectσx) 3. Finally, we multiply by N, the total number of measurements, to arrive at n(worse than xsuspect) = NP(outside tsuspectσx) This n is the number of measurements expected to be at least as bad as xsuspect.
If n is less than 1 2, then xsuspect fails Chauvenet’s criterion and is rejected.
9 5 Reporting data and error 5.1 Significant figures (digits) 5.2 How to estimate error Errors when Reading Scales The most common measurements in the lab are done with devices that have a marked scale. Let's look at an example. We will measure the length of the pendulum from the pivot point to the visible end of the mass. The situation is schematically shown in the figure below (numbers are in centimeters) Even using this idealized, zoomed-in picture, we cannot tell for sure whether the length to the end of the mass is 128.89 cm or 128.88 cm. However, it is certainly closer to 128.9 cm than to 128.8 cm or 129.0 cm. Thus we can state with absolute confidence that the length L is We call the first term, 128.9 cm, the "central value" and the second term, 0.1 cm, the "error" or "uncertainty". If pressed, we could get a little bit better precision from the picture. However, in a real situation, the precision of 0.1 cm for measurements done with the centimeter ruler is as good as you can get. In this course we will report error in measurements to be ± the last readable digit. 5.3 How to report error Number of Significant Digits As a general rule you should round the error to just one significant digit for your final answer and keep two significant digits for intermediate results. So for a time measurement T, we would write Now, what about the number of significant digits for the central value of T itself? Suppose the calculator gives us the value T = 3.57361382 sec instead of 3.6 sec. It is ridiculous to write our final answer for T as Since our error already affects the number 5 in the tenth place of 3.57361382, we should round our value to that decimal place. So the correct answer is This principle is true in general. We should round our central value to the rightmost decimal place at which our error applies. Thus if we have two decimal places in our error we should round our central value to the hundredths decimal place. 5.3 Maintaining Significant Digits in Calculations Once the number of significant figures various values have been determined, the issue then becomes dealing with significant figures when these values are used in calculations. When combining values with different degrees of precision, the precision of the final answer can be no greater than the least precise measurement. However, it is a good idea to keep one more digit than is significant during the calculation to reduce rounding errors. In the end, however, the answer must be expressed with the proper number of significant figures. When adding and subtracting, round the final result to have the same precision (same number of decimal places) as the least precise initial value, regardless of the significant figures of any one term. For example, 98.112 + 2.3 = 100.412 but this value must be rounded to 100.4 (the precision of the least precise term). When multiplying, dividing, or taking roots, the result should have the same number of significant figures as the least precise number in the calculation. For example, (3.69)x(2.3059) = 8.5088, which should be rounded to 8.51 (three significant figures like 3.69). Logarithms and Antilogithms When calculating the logarithm of a number, retain in the mantissa (the number to the right of the decimal point in the logarithm) the same number of significant figures as there are in the number whose logarithm is being found. For example, When calculating the antilogarithm of a number, the resulting value should have the same number of significant figures as the mantissa in the logarithm. For example, If a calculation involves a combination of mathematical operations, perform the calculation using more figures than will be significant to arrive at a value. Then, go back and look at the individual steps of the calculation and determine how many significant figure could carry through to the final result based on the above conversions. For Example, Addition and Subtraction Multiplication, Division, and Roots log(3.000x10 4 ) = 4.477121, which should be rounded to 4.477 log(3x10 4 ) = 4.477121, but this value should be rounded to 4 antilog(0.301) = 1.9998, which should be rounded to 2.00, antilog(0.30) = 1.9998, which should be rounded to 2.0 Multiple Mathematical Operations X = ((5.254+0.0016)/34.6) -2.23110 -3 Calculate the value of X using more digits than will be significant. In this case X = 0.1496649538 Then, go back and look at each piece of the equation to determine the significant figures. 5.254 + 0.0016 = 5.256 (since the sum is limited to the thousandths place by 5.254) 5.256 / 34.6 = 0.152 (since the quotient is limited to 3 significant figures by 34.6) 0.152 - 0.002231 = 0.149 (since the difference is limited to the thousandths place by 0.152) The value initially obtained for X (0.1496649538) should be rounded to have 3 significant digits Therefore, the final answer is 0.150 or 1.50x10 -1 . The Rules of Rounding When a value contains too many significant figures, it must be rounded off. There are two methods that are commonly used to minimize the error introduced into a value to rounding. Method 1: This method involves underestimating the value when rounding the five digits 0, 1, 2, 3, and 4, and overestimating the value when rounding the five digits 5, 6, 7, 8, and 9. With this approach, if the value of the digit(s) to the right of the last significant figures smaller than 5, drop this digit and leave the remaining number unchanged. Thus, 2.794 becomes 2.79. If the value of the digit(s) to the right of the last significant digit is 5 or larger, drop this digit and add 1 to the preceding digit. Thus, 2.795 becomes 2.80. Method 2: This method takes into account that zero doesn't really require rounding and when rounding 5, this value is exactly centered between the underestimated value if it is rounded down and the overestimated value if it is rounded up. Therefore, five should be rounded up half of the time and down half of the time. Since it would difficult to keep track of this when performing numerous measurements or calculations, 5 is rounded down when the preceding significant digit is even and 5 is rounded up when the preceding significant digit is odd. Values less than 5 are rounded down and values greater than 5 are rounded up. For example, 2.785 would be rounded down to 2.78 and 2.775 would be rounded up to 2.78. |
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Geometric inequality
A geometric inequality is an inequality involving various measures (angles, lengths, areas, etc.) in geometry.
Contents
1 Triangle Inequality
2 Pythagorean Inequality
3 Isoperimetric Inequality
4 Trigonometric Inequalities
5 Euler's inequality
6 Ptolemy's inequality
7 Erdos-Mordell inequality
Triangle Inequality
The Triangle Inequality says that the sum of the lengths of any two sides of a nondegenerate triangle is greater than the length of the third side. This inequality is particularly useful and shows up frequently on Intermediate level geometry problems. It also provides the basis for the definition of a metric space in analysis.
Pythagorean Inequality
The Pythagorean Inequality is a generalization of the Pythagorean Theorem. The Theorem states that in a right triangle with sides of length we have . The Inequality extends this to obtuse and acute triangles. The inequality says:
For an acute triangle with sides of length , . For an obtuse triangle with sides , .
This inequality is a direct result of the Law of Cosines, although it is also possible to prove without using trigonometry.
Isoperimetric Inequality
The Isoperimetric Inequality states that if a figure in the plane has area and perimeter, then . This means that given a perimeter for a plane figure, the circle has the largest area. Conversely, of all plane figures with area , the circle has the least perimeter.
Trigonometric Inequalities
In , .
Proof: is a concave function from . Therefore we may use Jensen's inequality:
Alternatively, we may use a method that can be called "perturbation". If we let all the angles be equal, we prove that if we make one angle greater and the other one smaller, we will decrease the total value of the expression. To prove this, all we need to show is if , then . This inequality reduces to , which is equivalent to . Since this is always true for , this inequality is true. Therefore, the maximum value of this expression is when , which gives us the value .
Similarly, in , .
Euler's inequality
Euler's inequality states that with equality when is equailateral, where and denote the circumradius and inradius of triangle , respectively.
Proof: The distance from the circumcenter and incenter of a triangle can be expressed as , meaning or equivalently with equality if and only if the incenter equals the circumcenter, namely the triangle is equilateral.
Ptolemy's inequality
Ptolemy's inequality states that for any quadrilateral , with equality when quadrilateral is cyclic.
First Proof: Let P be the point such that . By SAS we also have that . By the triangle inequality, . calculating the lengths, we obtain an equivalent statement: . Multiplying by we get the desired result with equality when P is on DC. This happens when . But so , or quadrilateral is cyclic.
Second Proof (using inversion): Let the inversion map B,C and D to B',C' and D' respectively. We then have By the triangle inequality, we have By multiplying on both sides we get the desired result with equality when is collinear, implying either ABCD is cyclic or collinear.
Erdos-Mordell inequality
The Erdős–Mordell inequality states that if lies in then where are the foot of the altitudes from to and , respectively.
Proof: First, we prove a lemma.
Mordell's Lemma:
Proof of Lemma: Let and be the projections of and onto line Note that is cyclic with diameter By the Law of Sines, Since is cyclic, we have that and are supplementary. Since is a line, This means that Similarly, So the problem is reduced to proving that but this is obvious by the Pythagorean Theorem.
Now the rest of the problem is straightforward. We know that Adding these cyclically implies By AM-GM, with equality when ABC is equilateral and P is the center of it. This article is a stub. Help us out by expanding it.
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3084 | http://nlaresourcecenter.lipidjournal.com/Content/PDFs/6-Debarber-Tuberous.pdf | Official Publication of the National Lipid Association 13 Some inherited lipid disorders can be recognized by the deposition of cholesterol and other lipids in lesions on the body termed xanthomas. Lipid deposition can occur in cutaneous and subcutaneous structures, including tendons, with the pattern of deposition characteristic of the underlying genetic disorder. Tuberous xanthomas are nodules frequently localized to the extensor surfaces of elbows, knees, knuckles, and buttocks. Tendon xanthomas are subcutaneous nodules found in fascia, ligaments, and tendons, which can occur particularly in the Achilles tendon and extensor tendons of the hands. (See Figure 1) The focus of this article is to highlight less common genetic disorders that cause tuberous and tendon xanthomas to promote better diagnosis of these disorders.
The most common hyperlipidemia underlying tuberous and tendon xanthomas is familial hypercholestrolemia (FH), caused by defects in the genes encoding the low-density lipoprotein (LDL) receptor, apolipoprotein B (apoB)-100, PCSK9, and LDLRAP1,1-7 although a number of other genetic disorders can cause these types of xanthoma. (See Table 1) Tuberous xanthomas also can occur in type III hyperlipidemia (familial dysbetalipoproteinemia) and sitosterolemia (phytosterolemia). Tendon xanthomas can occur in sitosterolemia and cerebrotendinous xanthomatosis (CTX). Outside of xanthomas in children with Alagille syndrome, finding xanthomas in children is extremely suggestive of homozygous familial hypercholestrolemia, sitosterolemia or CTX. For CTX especially, the appearance of xanthomas often occurs early in the course of disease and prompt diagnosis and early treatment is crucial in preventing disease progression and development of severe neurological disease.
Sitosterolemia (along with autosomal recessive FH because of mutations in LDL receptor adaptor protein 1 [LDLRAP1] and ARH) should be considered as the underlying cause of xanthomas when hypercholesterolemia is present and in the absence of a family history of FH. CTX should be considered as the underlying cause of xanthomas in the absence of marked hypercholesterolemia Guest Editorial: Tuberous and Tendon Xanthomas: Don’t Overlook Sitosterolemia or Cerebrotendinous Xanthomatosis P. BARTON DUELL, MD, FNLA President, Pacific Lipid Association Knight Cardiovascular Institute Associate Professor, Division of Endocrinology, Diabetes, and Clinical Nutrition Oregon Health & Science University Portland, OR ANDREA E. DEBARBER, PhD Supervisor, Sterol Analysis Laboratory Associate Director, Bioanalytical Shared Resource Facility Research Assistant Professor, Physiology & Pharmacology Department Oregon Health & Science University Portland, OR Discuss this article at www.lipid.org/lipidspin 14 LipidSpin Disorder: heterozygous familial hypercholestrolemia (FH), homozygous FH.1-7 Types of xanthoma: Tuberous, tendinous and xanthelasma (additional types of xanthoma may be formed, particularly in homozygous FH).
Genotype: Defects in LDL receptor gene, in apoB-100 gene (specifically LDL receptor-binding domain), proprotein convertase subtilisin kexin type 9 (PCSK9) and LDL receptor adaptor protein 1 (LDLRAP1). The latter condition is transmitted as a recessive trait and in its homozygous form is known as autosomal recessive hypercholesterolemia (ARH). FH is the most common disorder with an estimated prevalence of 1 in 200-300 ndividuals affected; approximately 1% of families with premature coronary heart disease (CHD) have heterozygous FH. Phenotype: The disorder is characterized by significant risk of coronary atherosclerosis and for homozygous FH development of CHD and aortic stenosis prior to age 20 years. Biochemically, plasma low-density lipoprotein cholesterol (LDL-C) is elevated (> 190 mg/dL is suggestive of heterozygous FH and >450-500 mg/dL homozygous FH, although there may be considerable overlap between the LDL-C concentrations in heterozygous and homozygous FH). Familial defective apoB-100 and PCKS9 gain of function mutations cannot be clinically distinguished from classical FH, although the elevation in LDL-C is less severe. Treatment: A diet low in cholesterol and saturated fat, statins, ezetimibe, anion exchange resins, and niacin. Two anti-PCSK9 monoclonal antibody therapies, alirocumab and evolocumab, are efficacious to lower LDL-C, with evolocumab approved by the U.S. Food and Drug Administration (FDA) for treatment of mild homozygous FH. Treatment for homozygous FH also can include high-dose statins, ezetimibe, niacin, bile acid sequestrants, mipomersen (an antisense therapeutic for apoB), lomitapide, evolocumab, and low-density lipoprotein (LDL) apheresis. Historically, ileal bypass and liver transplantation have also been used to treat this condition.
Disorder: type III hyperlipidemia (familial dysbetalioproteinemia).
Types of xanthoma: Tuberous xanthomas may be present, tuboeruptive xanthomas normally are observed, xanthomas striatum palmaris is a characteristic cutaneous feature (although their absence does not rule out this disorder).
Genotype: Apolipoprotein E2/2 (apoE2/2) occasionally apoE deficiency or hepatic lipase deficiency.
Phenotype: The disorder is defined by accumulation of chylomicron remnants and very low-density-lipoprotein (VLDL) remnants, premature CHD, other vascular disease, gout, and diabetes. Biochemically, elevated plasma total cholesterol and triglycerides are observed (both in the range 250-500 mg/dL), non-high-density lipoprotein cholesterol (non-HDL-C) (total cholesterol-HDL cholesterol) is elevated (>200 mg/dL), LDL-C is decreased and HDL-C is normal (with apoE deficiency, HDL cholesterol may be elevated). E2/E2 homozygosity only causes type III hyperlipidemia when overproduction of VLDL also is present (e.g. hypothyroid, diabetes, alcohol excess, hormonal changes). The abnormal genotype is necessary, but nor sufficient to cause this condition.
Treatment: Regular exercise and weight loss (if overweight); a diet low in cholesterol, saturated fat and sugar; as well as treatment with fibric acid derivatives, statins, niacin, and fish oil. Treatment of the underlying “second hit” or cause of VLDL overproduction is necessary to correct the lipid disorder. Treatment often leads to regression of xanthomas.
Disorder: sitosterolemia (phytosterolemia).
Types of xanthoma: Tuberous and tendinous.
Genotype: Defects in ATPbinding cassette sub-family G members 5 (ABCG5) and 8 (ABCG8) transporter genes.8 Phenotype: The disorder is characterized by premature CHD and abnormal hematologic findings may be present (hemolytic anemia, abnormally shaped erythrocytes and large platelets.8,9 Biochemically, LDL-C may be normal to elevated with plant sterols, such as β-sitosterol and campesterol, markedly elevated8,9 (for typical plasma sitosterol concentrations see Table 2). Phytosterols and free cholesterol are absorbed by the Niemann-Pick C1-Like 1 protein expressed on enterocytes. Phytosterols are then normally excreted back into the intestinal lumen by the ABCG5/8 transporter. Sitosterolemia occurs when this transporter is defective leading to pathologic absorption of high levels of phytosterols Treatment: A diet low in shellfish sterols, plant sterols (avoidance of vegetable oils, margarine, nuts, seeds, avocados, chocolate, and shellfish) and cholesterol, along with ezetimibe treatment can lower plant sterols by 10 to 50% and stabilize xanthomas.8,9 Bile acid sequestrants such as cholestryramine may be useful.
Disorder: cerebrotendinous xanthomatosis (CTX).
Types of xanthoma: Tendinous.
Genotype: Defects in CYP27A1 gene that encodes sterol 27-hydroxylase enzyme.10 Phenotype: The disorder is characterized by recurrent diarrhea in infancy and childhood, juvenile cataracts, xanthomas, and development of neurological disease (cerebellar ataxia, cognitive decline, and dementia).10,11 Biochemically, plasma cholesterol is only modestly elevated, with cholestanol markedly elevated (for typical plasma concentrations see Table 2). Synthesis of chenodeoxycholic acid is blocked in this disorder, leading to pathological induction of the bile acid pathway and production of excessive cholestanol.10 Treatment: In adults, 250 mg TID/day oral chenodeoxycholic acid (weight-based dosing is used in children) to normalize bile acid synthesis and cholestanol production.10,11 Statin treatment may be useful. It can take up to six months for cholestanol to return to normal levels. Treatment normally stabilizes xanthomas.
Table 1. Genetic disorders that can cause tuberous and tendon xanthomas.
Official Publication of the National Lipid Association 15 and when persistent diarrhea, juvenile cataracts, and cognitive impairment may be present. Biochemical diagnosis of sitosterolemia and CTX cannot be accomplished using standard laboratory methods of cholesterol measurement. Specialized analyses of cholestanol and plant sterols are required, typically using gas chromatography mass spectrometry (GC-MS). In sitosterolemia concentrations of the plant sterol sitosterol can be as high as 10-65 mg/ dL.8 Typical plant sterol concentrations in healthy individuals are 100-fold lower than cholesterol (for sitosterol 0.21 ± 0.7 mg/dL 8; see also Table 2). A plasma sitosterol concentration > 1mg/dL is considered diagnostic for sitosterolemia. A diagnostic threshold of 1 mg/dL helps avoid false positive results, although ezetimibe treatment may result in a false negative result. Identification of pathogenic mutations in ABCG5 and ABCG8 by molecular genetic testing also can be useful to confirm sitosterolemia. In untreated CTX patients, concentrations of cholestanol can range from 0.84 mg/dL to 6.6 mg/ dL.12 The range of the mean cholestanol concentration in healthy individuals is around 0.13 mg/dL.12 (See Table 2) A plasma cholestanol concentration > 1mg/ dL is considered diagnostic for CTX. Because cholestanol can be elevated in liver disease and concentrations of up to 0.7 and 1.1 mg/dL have been reported in healthy individuals,13,14 a diagnostic threshold of 1mg/dL helps avoid false positive results. Bile acid intermediates that accumulate in CTX also may be useful for biochemical diagnosis of this disorder.12 Identification of pathogenic mutations in CYP27A1 by molecular genetic testing also can be useful in confirming CTX. The plasma concentration ranges for sitosterol and cholestanol in unaffected and affected individuals as historically determined by the Oregon Health and Science University (OHSU) Sterol Analysis Laboratory are provided in Table 2.
Sitosterolemia and CTX should not be overlooked when evaluating the cause of tuberous and tendon xanthomas, because correct diagnosis and appropriate treatment of these rare underlying genetic disorders are critical to ensure the best outcome for patients affected by these disorders. n Disclosure statement: Dr. DeBarber received a research grant from Retrophin Inc. to perform a screening pilot study to screen for cerebrotendinous xanthomatosis in newborns. Dr. Duell has received research grants from Genzyme, Regeneron, Retrophin, and Amgen. He has received honoraria from Genzyme, Sanofi, Regeneron, Retrophin, Lilly, and Kaneka. References are listed on page 35.
Figure 1. Left panel: Thickening of the Achilles tendon in cerebrotendinous xanthomatosis (CTX). Middle panel: Knee tendon xanthomata in CTX. Right panel: Hand tendon xanthomata in CTX. Similar tendon xanthomas occur in patients affected with familial hypercholesterolemia (FH) and sitosterolemia. Photographs obtained by W. Connor and printed with permission.
Disorder Total Cholesterol (mg/dL) Cholestanol (mg/dL) Sitosterol (mg/dL) Unaffected individuals (n=180) 150 ± 51.7 [52.8-459] 0.237 ± 0.226 [0.065-0.794] 0.273 ± 0.203 [0.015-0.719] CTX-affected untreated (n=6) 185 ± 26.2 [154-231] 2.29 ± 1.50 [0.839-5.52] 0.353 ± 0.154 [0.053-0.480] CTX-affected treated longer than 6 months (n=40 samples from n=7 individuals)ǂ 162 ± 34.6 [95.2-242] 0.338 ± 0.186 [0.076-0.874] 0.267 ± 0.127 [0.067-0.589] Sitosterolemia-affected+ (n=2 samples from one individual)ǂ 130, 185 0.781, 1.24 10.6, 9.33 Table 2. Concentrations of plasma total cholesterol, cholestanol and sitosterol in CTX and sitosterolemia measured by gas chromatography mass spectrometry (GC-MS).
The mean concentration ± S.D. and [range of results] are given.
+Treatment status unknown.
Samples are analyzed for the same patients at least annually for therapeutic monitoring.
Historical data from samples submitted for biochemical CTX and sitosterolemia diagnosis to the Sterol Analysis Laboratory, OHSU. This is one of only a few laboratories in the U.S. that performs biochemical testing for CTX and sitosterolemia. For questions regarding obtaining testing please email Dr. DeBarber at debarber@ohsu.edu. |
3085 | https://www.jpeds.com/article/S0022-3476(20)31089-1/fulltext | Peritonsillar Abscess in an Infant - The Journal of Pediatrics
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Insights and ImagesVolume 227p320-321 December 2020
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Peritonsillar Abscess in an Infant
Bruna B.C.Olson, MD
Bruna B.C.Olson, MD
Affiliations
Department of Pediatrics
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∙ Mark R.Zonfrillo, MD, MSCE
Mark R.Zonfrillo, MD, MSCE
Affiliations
Department of Pediatrics
Department of Emergency Medicine
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∙ Lauren M.Allister, MD
Lauren M.Allister, MD
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Department of Pediatrics
Department of Emergency Medicine, Alpert Medical School, Brown University, Providence, RI
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Affiliations & Notes Article Info
1 Department of Pediatrics
2 Department of Pediatrics
3 Department of Emergency Medicine
4 Department of Pediatrics
5 Department of Emergency Medicine, Alpert Medical School, Brown University, Providence, RI
Publication History:
Received August 15, 2020; Accepted August 20, 2020; Published online August 21, 2020
DOI: 10.1016/j.jpeds.2020.08.053 External LinkAlso available on ScienceDirect External Link
Copyright: © 2020 Elsevier Inc. All rights reserved.
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An 8-month-old girl presented to the emergency department (ED) with a 3-day history of fever, decreased oral intake, and limited neck and jaw movement. Her examination was notable for right-sided palatal edema, a left-shifted uvula, and an obstructed view of the oropharynx. A computed tomography (CT) scan of the neck demonstrated a 4-cm right peritonsillar abscess with significant mass effect on the airway (Figure 1). She was managed with noninvasive airway support and intravenous steroids plus intravenous ampicillin/sulbactam. She was taken to the operating room by the otolaryngology service, who drained 12 mL of purulent fluid from the abscess. She rapidly improved and was discharged from the hospital on day 4 with a course of oral amoxicillin/clavulanic acid.
Figure viewer
Figure 1 Axial computed tomography of the neck showing a 4‐cm right peritonsillar abscess (black arrow) with significant mass effect and compression of the airway (white arrow)
Peritonsillar abscess is a suppurative tissue infection that occurs in the palatine tonsils. The presentation includes fever, dysphagia, drooling, and trismus. It is a rare condition in children younger than 5 years of age, with a mean age of onset of 12 years and two-thirds of the cases occurring in children aged >10 years.1
1.
Hsiao, H. ∙ Huang, Y. ∙ Hsia, S. ...
Clinical features of peritonsillar abscess in children
Pediatr Neonatol. 2012; 53:366-370
Full Text
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Scopus (28)
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Sepsis, jugular vein thrombosis, and airway obstruction are potential serious complications.2
2.
Bochner, R. ∙ Gangar, M. ∙ Belamarich, P.
A clinical approach to tonsillitis, tonsillar hypertrophy, and peritonsillar and retropharyngeal abscesses
Pediatr Rev. 2017; 38:81-92
Crossref
Scopus (19)
PubMed
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Smaller infant airways may be more susceptible to compromise, as evidenced in this patient, and thus more difficult to support (noninvasively or invasively) in critical scenarios.
As in this case, fever and decreased oral intake may be the only initial complaints in an infant with a peritonsillar abscess. These common pediatric symptoms are nonspecific, making a peritonsillar abscess difficult to diagnose, especially if clear visualization of the oropharynx is not performed or if these symptoms are attributed to another more common infectious etiology in this age group, such as Coxsackievirus or primary herpetic gingivostomatitis. Careful physical examination, including good visualization of the oropharynx, is key to suspecting the diagnosis of a peritonsillar abscess in an infant. Recognizing the vulnerability of the infant airway in the setting of a deep space neck infection is paramount to supporting the airway and preparing for possible decompensation. Neck imaging and subsequent medical and surgical management are crucial to the prompt and effective treatment of this uncommon but potentially life-threatening infection in infancy.
References
1.
Hsiao, H. ∙ Huang, Y. ∙ Hsia, S. ...
Clinical features of peritonsillar abscess in children
Pediatr Neonatol. 2012; 53:366-370
Full Text
Full Text (PDF)
Scopus (28)
PubMed
Google Scholar
2.
Bochner, R. ∙ Gangar, M. ∙ Belamarich, P.
A clinical approach to tonsillitis, tonsillar hypertrophy, and peritonsillar and retropharyngeal abscesses
Pediatr Rev. 2017; 38:81-92
Crossref
Scopus (19)
PubMed
Google Scholar
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3086 | https://www.frontiersin.org/journals/public-health/articles/10.3389/fpubh.2021.559595/full | Your new experience awaits. Try the new design now and help us make it even better
REVIEW article
Front. Public Health, 15 February 2021
Sec. Clinical Diabetes
Volume 9 - 2021 |
Susceptibility for Some Infectious Diseases in Patients With Diabetes: The Key Role of Glycemia
Jesús Chávez-Reyes1Carlos E. Escárcega-González2,3Erika Chavira-Suárez4Angel León-Buitimea2,3Priscila Vázquez-León1José R. Morones-Ramírez2,3Carlos M. Villalón5Andrés Quintanar-Stephano1Bruno A. Marichal-Cancino1
1Departamento de Fisiología y Farmacología, Centro de Ciencias Básicas, Universidad Autónoma de Aguascalientes, Aguascalientes, Mexico
2Facultad de Ciencias Químicas, Universidad Autónoma de Nuevo León, Nuevo León, Mexico
3Centro de Investigación en Biotecnología y Nanotecnología, Facultad de Ciencias Químicas, Universidad Autónoma de Nuevo León, Nuevo León, Mexico
4Unidad de Vinculación Científica de la Facultad de Medicina, Universidad Nacional Autónoma de México en el Instituto Nacional de Medicina Genómica, Mexico City, Mexico
5Departamento de Farmacobiología, Cinvestav-Coapa, Mexico City, Mexico
Uncontrolled diabetes results in several metabolic alterations including hyperglycemia. Indeed, several preclinical and clinical studies have suggested that this condition may induce susceptibility and the development of more aggressive infectious diseases, especially those caused by some bacteria (including Chlamydophila pneumoniae, Haemophilus influenzae, and Streptococcus pneumoniae, among others) and viruses [such as coronavirus 2 (CoV2), Influenza A virus, Hepatitis B, etc.]. Although the precise mechanisms that link glycemia to the exacerbated infections remain elusive, hyperglycemia is known to induce a wide array of changes in the immune system activity, including alterations in: (i) the microenvironment of immune cells (e.g., pH, blood viscosity and other biochemical parameters); (ii) the supply of energy to infectious bacteria; (iii) the inflammatory response; and (iv) oxidative stress as a result of bacterial proliferative metabolism. Consistent with this evidence, some bacterial infections are typical (and/or have a worse prognosis) in patients with hypercaloric diets and a stressful lifestyle (conditions that promote hyperglycemic episodes). On this basis, the present review is particularly focused on: (i) the role of diabetes in the development of some bacterial and viral infections by analyzing preclinical and clinical findings; (ii) discussing the possible mechanisms by which hyperglycemia may increase the susceptibility for developing infections; and (iii) further understanding the impact of hyperglycemia on the immune system.
Introduction
Diabetes mellitus is a chronic and complex illness characterized by several metabolic alterations including dyslipidemia and hyperglycemia, among others (1). According to the American Diabetes Association (A.D.A.), diabetes mellitus (DM) can be classified into the following categories: (i) type 1 diabetes mellitus (T1DM), characterized by the loss of pancreatic β-cells induced by an autoimmune response; (ii) type 2 diabetes mellitus (T2DM), identified by the gradual loss of insulin secretion and/or the development of insulin resistance; (iii) gestational DM, developed in some pregnant women; and (iv) other types of DM that are due to miscellaneous causes (2, 3). Interestingly, patients with uncontrolled DM (regardless of type) have alterations in healing latency and susceptibility for developing some emerging infectious (mainly bacterial) diseases. In addition, compared to non-diabetic normoglycemic patients, DM patients are at higher risk for developing the current severe acute respiratory syndrome coronavirus 2 (SARS-COV2) caused by the coronavirus 2 (CoV2) that has shocked the world economy and created a global health pandemic emergency named COVID-19 (4–7). Moreover, the restoration of normoglycemia seems to be related to a better prognosis for bacterial infections (5); whereas in COVID-19 diabetic patients, no obvious conclusions have been reached about the impact of normoglycemic treatments on the development and outcome of this particular disease (8).
People with metabolic impairments (i.e., fasting hyperglycemia, postprandial hyperglycemia and DM) show greater ranges of glucose levels (2). Indeed, fasting hyperglycemia (when food has not been taken for at least 8 h) is a metabolic disorder characterized by levels of plasma glucose above 110 mg/dL, a condition commonly observed in diabetic patients (9). Fasting hyperglycemia (from now on simply referred to as hyperglycemia) has been involved in deleterious effects such as tissue damage associated with oxidative stress and immunological impairments (10), which increase the susceptibility to acquire bacterial infections and COVID-19 (8, 11–17) (see below).
Remarkably, the effects of hypoglycemia induced by anorexia on the clinical outcome of infected patients have been discussed with no consensus (18). Moreover, Wang et al. (19) have suggested that: (i) the pathogenic nature (i.e., bacterial, viral, etc.) and infection profile may be key factors for the prognosis and clinical outcome in a preclinical model of bacterial infection; (ii) glucose plays a key role in the outcome of infected animals; and (iii) survival of animals under bacterial sepsis (with Listeria monocytogenes) was dramatically decreased when they were gavage-fed. In contrast, if these animals received glucose (i.p.), all animals died (19). As anorexia is an important stereotypic behavior of the sickness response, it could be an adaptive strategy for combating some infectious illnesses. In this sense, it has also been reported that bacteria from other groups (e.g., Salmonella) may induce inhibition of anorexia via Salmonella leucine rich repeat protein (SlrP) which inhibits interleukin-1β (IL-1β). This effect seems to maintain the conditions for increasing the opportunity for Salmonella to infect other hosts (20). Although these experiments were carried out in preclinical models, the results suggest that glycemia is so important that hosts and infectious agents have developed adaptive strategies to control glucose levels during the progression of an infectious process (11). Moreover, the control of glucose levels may determine the infection course and/or the recovery times (Figure 1); therefore, the understanding of the mechanisms involved on the hijacking of glucose control during infections may have an enormous medical utility.
FIGURE 1
Figure 1. Illustration of a hypothetical outcome in an experimental model of bacterial pneumonia in normal conditions vs. during hyperglycemia. Under normoglycemia conditions the immune response handles successfully bacterial infections. Nevertheless, hyperglycemia impairs the immune response by inducing several glucose-related factors, including those mentioned above. These scenarios could determine the outcome during bacterial infections.
The relationship between hyperglycemia and susceptibility to infections has been described extensively in diabetic patients. Nevertheless, very few reports have analyzed the appropriate management of glycemia according to the infection type, immunological responsiveness, and clinical variables (i.e., patient age, time or period elapsed with diabetes, etc.). To focus on these aspects, the present review has considered information on: (i) the effect of glycemia on infection outcome and immune cells physiology; (ii) the biochemical alterations in cell physiology during diabetes and/or hyperglycemia; and (iii) the impact of pharmaceutical care interventions for glycemia control on some of the most frequent emerging infectious diseases.
Methods and Inclusion Criteria
To consider the relevant literature in this theoretical review, we searched for studies published in various databases such as Science direct, Pubmed central and Google Scholar. These databases included any combination of the main key terms “bacterial infections,” “COVID-19,” “diabetes,” “influenza A virus,” “hepatitis B and C viruses,” “human immunodeficiency virus,” and “hyperglycemia” among themselves and with important topics such as: “rheological properties of blood,” “biochemical alterations of diabetic immune cells,” “immune response on hyperglycemic environment,” “hypoglycemic drugs,” “bacterial infection outcome,” and “comorbidities with COVID-19.” Around 600 articles published from 1966 up to 2020 were perused, and only 260 of those articles with experimental and/or theoretical information which related hyperglycemia and/or diabetes to bacterial infections and/or COVID-19 and some other viruses were included in this review.
The Association Between Hyperglycemia and Common Infectious (Mainly Bacterial) Diseases in Diabetes
Effect of Hyperglycemia on the Immune Response
In general, alterations in the immune system during hyperglycemia seem to be associated with mechanisms that include lower secretion of inflammatory cytokines, depression in neutrophils and T cells function, as well as decreases in humoral immunity (21, 22). Moreover, it is documented that hyperglycemia may delay the recuperation of tissues (e.g., via changes in the secretion of growth factors and collagenase levels) (12, 23); this, in turn, may lead to increased susceptibility of these tissues to develop secondary emerging infections (mainly bacterial). Other alterations in the immune response induced by hyperglycemia can be explained by biochemical and/or cellular events, such as: (i) creation of advanced glycation end-products (AGEs), which reduce the expression of myeloid cells surface proteins known as class I major histocompatibility complex (24); (ii) decreased migration of polymorphonuclear leukocytes, chemotaxis, and/or phagocytic activity (25); (iii) inhibition on G6PD (see below) (26); and (iv) increased apoptosis of polymorphonuclear leukocytes and reduced transmigration through the endothelium (27). Clearly, the control of glycemia may be mandatory for dealing with emerging infections (mainly bacterial), in view that: (i) some bacteria grow better in a high glucose environment (28); and (ii) a hyperglycemic state seems to negatively affect the body's ability to respond to antimicrobial therapy (29).
Common infections related to T1DM and T2DM are those of the respiratory and urinary tracts. Indeed, it has been reported that patients with T2DM have alterations in chemotaxis, phagocytosis, antigen presentation and proliferation/function of T cells in response to Mycobacterium tuberculosis, which facilitates infection and its symptomatic progression (30). Certainly, the impaired chemotaxis of leukocytes does not depend on the type of diabetes mellitus (31). Other tissues/organs that are also commonly compromised in diabetic patients include the skin, bone marrow, gastrointestinal tract and liver, among others (21, 22, 32). This susceptibility for developing infections may lead to complications in the management of diabetic patients, such as post-operative infections, sepsis, chronic periodontitis, emphysematous cholecystitis, emphysematous pyelonephritis, malignant external otitis, rhinocerebral mucormycosis, gangrenous cholecystitis, and others (21, 22, 32, 33).
It is noteworthy that foot infections are highly common in patients with diabetes, which usually start after a foot wound that eventually leads to ulceration. In this respect, neuropathy seems to be an important component of foot ulceration which, in turn, increases the risk of amputation (34). The wound is predisposed to a loss of sensitivity because of the damage in neuron fibers by pathophysiological mechanisms not fully understood (34–36). It has been suggested that a vascular endothelium damage produced by inflammation and oxidative stress (36) may produce alterations in the microcirculation and, finally, nerve damage (37).
In many cases, these infections cause ischemia at the wound site, ultimately leading to amputations (38). Moreover: (i) immunological disturbances in neutrophil functions such as chemotaxis, phagocytosis and intracellular killing may contribute to exacerbate infections (12, 31, 38–41); and (ii) AGEs may influence the appearance of a chronic immune imbalance by activating pro-inflammatory cells which, in turn, would lead to a chronic subclinical inflammation that hinders the correct function of the immune system to fight infections and to deal with wound healing (42).
Thus, the typical hyperglycemia present in patients with diabetes could be related to an increased risk of different types of infections. Interestingly, cancer patients treated with glucocorticoids showed increased infection rates (43). Indeed, glucocorticoids are direct immunosuppressors that may increase hyperglycemia by hepatic gluconeogenesis and inhibition of glucose intake (43, 44). All these lines of evidence, strongly suggest that hyperglycemia may induce an adequate environment for several infectious pathogens; and hence, a suitable glycemic control would decrease the rate of infection risk (45–49).
Finally, it is logical to assume that the changes in the immune system produced by hyperglycemia as an occasional (transient) event (e.g., stress) should be quite different from the changes induced by a chronic hyperglycemia. Nevertheless, in any hyperglycemic condition (i.e., transient or chronic), patients may be susceptible to some of the same clinical complications, including poor wound healing and an increased rate of infection (50). In fact, acute glucose elevation in critically injured trauma patients may be predictive of infections (51); whereas hyperglycemia at admission (with no indication about the cause) is a predictor of infections in critically ill trauma patients (52). Clearly, chronic hyperglycemia involves compensatory mechanisms (not discussed here) that are absent when it is due to an occasional event; in both cases, normalization of the glucose levels seems to be a useful practice to improve the nosocomial outcome (50, 53, 54).
Stress-Induced Hyperglycemia and Infections
Besides diabetes, another condition that commonly predisposes to hyperglycemia is stress. The stress-induced hyperglycemia (SIH) generally refers to a metabolic condition with a transient hyperglycemia associated with clinical illness (55). The SIH is a common problem in patients admitted to intensive care units (ICU) (50), even in the absence of pre-existing diabetes (55), and it is defined as an increase above 200 mg/dL of blood glucose (52, 56, 57).
The SIH is especially dangerous in chronic critical illnesses (58), as the organs' functions become aberrant increasing the risk of death. In less severe cases, SIH seems to affect the normal immune response, since hyperglycemia (as discussed above) is associated with an increased risk to infections (50). Hyperglycemia is certainly related to a higher risk of infections and sepsis in patients of ICU (59), an increased risk of complications in patients who underwent orthopedic trauma surgery (60), and surgical site infections in non-diabetic orthopedic trauma patients (61) (see Table 1).
TABLE 1
Table 1. Main complications during bacterial infections in diabetic patients.
A stress response is associated with an increased pro-inflammatory response characterized by the release of several cytokines, including tumor necrosis factor-alpha (TNF-α), IL-1 and IL-6, which are related to insulin resistance (59). Indeed: (i) TNF-α inhibits tyrosine kinases and decreases tyrosine phosphorylation of the insulin receptor (92); (ii) IL-1 suppresses glucose transporter-4 (GLUT-4) translocation by a decreased activation of the phosphoinositide-3-kinase (PI3K) mechanism (93); and (iii) IL-6 increases the release of adrenocorticotrophic hormone (94), with all the scenarios resulting in insulin resistance.
Additionally, there may be an integration of a vicious cycle when taking a hypercaloric fatty diet, which can also induce an increase in catecholamines' release (95), in combination with lifestyle factors leading to chronic stress (resulting in a plasma increase in catecholamines and cortisol). In this scenario, catecholamines via β-adrenoceptors expressed in adipocytes, liver, skeletal and smooth muscle cells may increase the metabolism of glycogen and triglycerides for increasing blood glucose, fatty acids, glycerol, and other local vascular actions (96–98). Accordingly, this SIH may lead to immunosuppression (see Figure 2).
FIGURE 2
Figure 2. Development of a vicious cycle induced by a stressful lifestyle. Stressors promote the behavior of hypercaloric food binge which, in turn, increases the release of catecholamines and cortisol (immunosuppressants). Subsequently, catecholamines induce glycolysis, reinforce the maladaptive eating patterns, and negatively modulate immune cell activity. On the other hand, cortisol induces chronic immunosuppression.
The contribution of SIH in the diabetic patient is a complex issue that seems to worsen the glycemic status of patients with type 2 diabetes; whereas the autonomic damage induced by neuropathy in type 1 diabetes leads to contradictory and non-conclusive data (99). Patients with necrotizing fasciitis (and diabetes) developed adverse outcomes when SIH generated glycemic gaps with increases in glucose >146 mg/dL (86). Thus, synergy of diabetes and SIH is a phenomenon of high medical impact (both in infectious and non-infectious clinical admission) that remains to be completely characterized (53, 100–104).
Infectious Diseases in Patients With Diabetes: Exacerbation and Susceptibility to SARS-CoV-2 and Other Viral Diseases
Diabetes is a condition that may potentiate infectious diseases and predispose patients to acquiring more severe diseases. To support this notion, a recent matched cohort study analyzed the incidence infection rate from 306,011 patients (102,493 patients with type 1 and 2 diabetes), and reported that patients with diabetes (especially type 1) are more susceptible to developing severe infectious diseases (105). In addition, patients with diabetes are more vulnerable to fungal, viral, and bacterial infections than the non-diabetic population, exhibiting a worse prognosis once the infection is installed (106).
According to epidemiological studies, the most common infectious diseases of hospitalization in children, adolescents and adults with diabetes are lower tract respiratory infections (pneumonia, among others), diabetic foot infection, skin and soft tissue abscesses, and urinary tract infections (107, 108). The respiratory tract infections are the primary comorbidity associated with severe or lethal infections that increases hospitalizations in individuals with diabetes (109, 110).
Pneumonia is the hospitalization leading cause of severe lower respiratory tract infections, and it is an important risk factor for infectious illnesses in diabetic patients (111). The main fungal and bacterial pathogens associated with pneumonia infections are Mycoplasma pneumoniae, Chlamydophila pneumoniae, Legionella pneumophila, methicillin-resistant Staphylococcus aureus, Haemophilus influenzae, and Streptococcus pneumoniae (109). This diversity of pneumonia development pathogens may denote a complex biological interaction between wild microorganisms residing in the human body, the host immunophysiology, and the pathogenic pneumonia specificity.
On the other hand, hormones like cortisol, glucagon and catecholamines released during certain conditions, such as trauma, infection and surgery (57) increase gluconeogenesis and decrease peripheral glucose uptake (52). Interestingly, the association between sympathetic hyperactivity (e.g., induced by chronic stress), hyperglycemia, hypothermia and immunosuppression of the acquired immunity seems to be mainly mediated by activation of α-adrenoceptors (112, 113).
Emerging global health studies have reported that other respiratory tract infections with high mortality rate in patients with diabetes, besides pneumonia (114), are those promoted by viral agents. These include the influenza viruses, the Middle East respiratory syndrome coronavirus (MERS-CoV), the severe acute respiratory syndrome coronavirus (SARS-CoV) and, most recently, the SARS-CoV-2 (see below), the last viral infection outbreak across the globe (7, 17, 115, 116).
SARS-CoV-2 and Other Viral Diseases: Impact of Hyperglycemia
Patients with hyperglycemia have been reported to be susceptible to develop a severe form of COVID-19, which is a risk factor for fatality (6, 17). Newsworthy, diabetes provides a ~3-fold higher risk of fatality as compared to the non-diabetic population among the COVID-19 sufferers (7). Moreover, diabetes increased the length of hospital stays for COVID-19 patients from 9.8 days in non-diabetic patients to 14.4 days in diabetic patients in a retrospective cross-sectional study that was conducted in England (14). In this regard, it has been described that a proper control of glycemia by antidiabetic drugs can be beneficial in reducing the risk of death in diabetic patients with COVID-19 (16). Indeed, it was inferred that DPP-4 inhibitors might be beneficial to prevent or treat COVID-19 disease (117). Although this certainly opens a new field of interest in the treatment of SARS-CoV-2 pneumonia, further studies and research are required on this topic.
Bacterial infections are frequently identified after typical viral respiratory infections and they are important causes of morbidity and mortality. In patients with COVID-19, bacterial comorbidity has been reported to be low (i.e., an overall proportion of 6.9 %) in a recent metanalysis reported by Langford et al. (118). Notoriously, the comorbidity was slightly higher in critical patients (i.e., 8.1%) (118).
DM, hypertension, cardiovascular diseases, and obesity are the top four comorbidities worldwide associated to critically ill patients with COVID-19 and mortality (15, 119–121). Indeed, 5 to 10% of patients with SARS-CoV-2 pneumonia require intensive care unit (ICU) admission and mechanical ventilation. Patients requiring invasive mechanical ventilation are strongly related to poor outcome with high mortality rate in Chinese and American populations (122, 123).
Unfortunately, studies on the unfavorable outcomes and mortality rate related to pathogenic co-infections that worsen respiratory tract function in people with diabetes and COVID-19 infection are limited (124). However, bacterial, and viral pathogenic co-infections have been studied in patients with SARS-CoV-2 pneumonia requiring ICU admission. These studies showed that methicillin-sensitive Staphylococcus aureus, Haemophilus influenzae, Streptococcus pneumoniae, Enterobacteriaceae, Pseudomonas aeruginosa, Moraxella catarrhalis, and Acinetobacter baumannii were the 28% of bacterial strains isolated by experimental laboratory procedures (cultures or PCRs assays) (125). Importantly, no viral co-infection has been detected in the critically ill COVID-19 patients, supporting the idea that respiratory infections are often depending on combinatorial factors associated to geography, season, human physiology, and behavior, as well as pathogenic interactions. Therefore, it is mandatory to determine the different biological mechanisms used for each viral infection and/or co-infection of pathogens that aggravate states of health or disease to pursue appropriate treatments.
Admittedly, pathogenesis of COVID-19 viremia remains unclear. However, some lines of evidence suggest that high levels of systemic glucose increase glucose concentration in the epithelial secretion of the respiratory tract, disrupting the orchestration of the innate and humoral immunological response. This includes, in particular, hyperglycemia-induced changes in coagulation, worsening of endothelial function, and reproduction of inflammatory cytokines (126).
Data about hospitalization for infectious diseases in diabetic and non-diabetic subjects have been associated with various hyperglycemic conditions on admission, increasing poor outcomes and mortality rates. Moreover, hyperglycemia on admission was clearly associated with undiagnosed DM, strongly suggesting that an optimal glycemic control that reduces glycemic fluctuations during hospitalization should be a beneficial clinical practice for viremia control (127).
The American Diabetes Association (A.D.A.) recommends a blood glucose level of 140 to 180 mg /dL (7.8 to 10.0 mmol/L) for most critically ill patients and patients who are not in good health. Glycemic control during clinical procedures could be accompanied by insulin therapy if the hyperglycemia persists starting at a threshold ≥ 180 mg/dL (128). Thus, patients with COVID-19 and with/without DM should have a well-controlled blood glucose (129).
Influenza A Virus
The influenza A virus (IAV) induces a self-limited infection in most patients, which is characterized by several symptoms such as myalgia, fever, and dry cough (130). Nevertheless, patients with diabetes experience a more severe type of this disease (131) that is represented by a triple risk of hospitalization and double risk of fatality compared with non-diabetic sufferers (132). Despite the fact that IAV infects upon 15% of the world's population every year (133), the full mechanisms underlying its pathogenesis, especially on patients with diabetes, remain thus far inconclusive. In this sense, Hulme et al. (132) reported that the IAV infection in hyperglycemic conditions increases the endothelial damage leading to a pronounced inflammatory response; this explains, at least in part, the severity of the symptoms in patients with diabetes (132). In support of this notion, Kohio and Adamson (134) reported an enhanced IAV replication rate in pulmonary epithelial cells under elevated glucose concentrations in vitro.
Admittedly, the specific mechanisms that underline susceptibility factors for viral infections development in patients with diabetes (Table 2) are poorly understood. However, an experimental study (132) that used an in vitro and in vivo model of pulmonary epithelial-endothelial cells exposed to a high glucose concentration (12 mM) demonstrated an increased barrier damage after co-cultured cells were infected with IAV; this, in turn, augmented pulmonary edema associated with a pro-inflammatory response (132). Thus, controlling hyperglycemia seems to be important for hospitalized patients with severe viral infections and diabetes (144).
TABLE 2
Table 2. Main complications during viral infections in patients with diabetes.
Hepatitis B and C Viruses
The Hepatitis C virus (HCV) and Hepatitis B virus (HBV) are known causes of hepatic decompensation, liver cirrhosis, and hepatocellular carcinoma (HCC), being two major public health problems worldwide (145–147). The evidence for a link between HCV and DM has been proposed several decades ago (148). In this sense, based on the meta-analysis of Gou and colleagues (145), patients with T2DM are more prone to HCV infection (~3.5-fold increase) compared with the risk in the non-diabetic group. In the case of HBV, the diabetic condition predisposes to acquiring the infection (147, 149). Moreover, there is a high association between diabetes and the higher risk for a worse outcome of HCV and HBV infection (139, 141, 150, 151). HCV patients with diabetes have a higher incidence of HCC compared to non-diabetic HCV patients (radio 1.73) (152). Interestingly, the use of several hypoglycemic drugs improves the prognosis for this type of cancer (153, 154).
Human Immunodeficiency Virus
Human immunodeficiency virus (HIV) increases the risk for developing T2DM (155, 156). Likewise, patients with HIV are prone to diabetes in younger people and in the absence of obesity (157, 158). In this sense, several hypotheses have been proposed to understand the mechanisms for this link, including the effects of antiretroviral drugs (ARVD), lipodystrophy, co-infections, and autoimmunity (156). The use of ARVD in patients with HIV, which include atazanavir, darunavir, and saquinavir, interfere with the GLUT-4 dynamics by increasing insulin resistance and reducing insulin secretion (159). On the other hand, it has been recognized that the HIV infection and/or its treatment can induce lipodystrophy (i.e., an abnormal distribution of fat in the body); this raises the levels of TNF-α which, in turn, contributes to increasing insulin resistance and finally triggering diabetes (160). The third hypothesis to understand the relation HIV-diabetes includes the co-participation of HCV; in this sense, the increased intrahepatic TNF-α may be a trigger to develop diabetes (161). Finally, the autoimmune hypothesis explains that some HIV-patients may undergo beta cell destruction, developing the autoimmune diabetes observed in some HIV-infected patients (162).
Physicochemical Changes During Hyperglycemia: Effects on the Immune System
Rheological Properties and Blood Viscosity
Rheological properties of blood may impact function, metabolism, motility and even the latency for clearing toxins of blood cells (163, 164). Changes in rheological conditions have been reported during diabetes and hyperglycemia, which may alter red blood cells physiology and the local microcirculation (163, 165). Indeed, some of the blood rheological properties that have been reported to be disturbed during hyperglycemia and/or diabetes include: (i) an increment in serum osmolarity (166); (ii) erythrocyte deformation that is produced by glycosylation of membrane proteins (167, 168); (iii) changes in pH (169); and (iv) an increase in blood viscosity (164, 165). All these alterations may impair the immune system activity and could explain the impact that glycemia has on the clinical outcome (Figure 1).
Furthermore, increased blood viscosity may lead to hemoconcentration and vasodilatation that increases edema (164). In close connection with this response, coagulation directly affects blood viscosity, increasing the risk for developing microangiopathy (168). In fact, anomalous erythrocyte deformability and platelet aggregation impair microcirculation, which leads to hypoxia in hyperglycemia and diabetes (170, 171). In this sense, a decrease in oxygen supply could impair the immune response because in those cells oxygen is essential for destroying infectious microorganisms (172). As a result, oxygen supplementation: (i) avoids surgical infections during the perioperative period (173, 174); and (ii) can be used to prevent infections and promote wound healing (175).
To round off and complete the above rheological scenario, it is to be noted that the concentration of fibrinogen and globulins are also important factors involved in blood viscosity (168). In fact, an increase in plasma fibrinogen in diabetic patients is a determining factor for blood viscosity (176). This, in turn, will alter oxygen supply resulting in an impaired immune response.
pH
Any change in pH may be detrimental for the proper functioning of the whole body (169), including the diabetic sufferers. In this sense, diabetic ketoacidosis (DKA) is a common hyperglycemic condition that affects both T1DM and T2DM patients, resulting in a decreased venous blood pH (below 7.3) (177). DKA results from an altered metabolism of glucose mainly produced by a decreased or abolished production of insulin (178). This, in turn, promotes the metabolism of triglycerides into glycerol and fatty acids, with the latter being further oxidized to ketone bodies, mainly acetoacetate and β-hydroxybutyrate (178, 179). Ketone bodies are weak acids that weigh down blood buffering capacity (carried out by bicarbonate anion), altering pH and resulting in a metabolic acidosis (177, 180).
As the most severe complication of DM, patients with DKA have more difficulty to handle infections (179, 181). Admittedly, it is not clear whether DM may increase the susceptibility for all infections; however, many of them (mainly the bacterial ones) are more severe, frequent and/or typical of diabetic patients (21). For instance, some of the most common infections in these patients are pneumonia and urinary tract infections (179, 182), as well as other infections difficult to manage, such as mucormycosis (183, 184), aspergillosis (185), tuberculous meningitis (186), and pulmonary coccidioidomycosis (187).
Several reports have shown the role of pH in the immune response. For example, with a pH below 6.5: (i) the mobility of polymorphonuclear leukocytes was impaired (188), which could result in delayed migration of leukocytes; (ii) chemotaxis was inhibited (188, 189); and (iii) the production of superoxide anion was decreased in neutrophils (190), resulting in an impaired “respiratory burst” (191). However, phagocytosis in bovine neutrophils was hardly affected when they were challenged with Staphylococcus aureus at acidic pH (192). Moreover, Loeffler et al. (193) reported an inhibition in lymphocytes proliferation induced by interleukin-2 (IL-2) at acidic pH. Nevertheless, only some functions seem to be affected in lymphocytes at an acidic pH, namely, at pH 6.7 (as compared with pH 7.1) an increase in lymphocytes mobilization was reported (194, 195). A possible explanation for this finding is that every cell type and specific functions are differentially altered by pH gradients. Obviously, further studies are required to understand the molecular mechanisms underlying each cellular type and the corresponding physiological phenomena.
Other important alterations induced by hyperglycemia in the circulatory system are related to a miss-functionality of the enzymatic machinery of blood cells, including Na+/K+-ATPase activity and glucose-6-phosphate dehydrogenase (G6PD) (see below).
Alterations in Na+/K+-ATPase Activity
Na+/K+-ATPase is a transmembrane protein responsible for maintaining intracellular Na+/K+ balance by generating the gradients of Na+ and K+ (196). This enzyme is expressed ubiquitously in almost all cell types, regulating a plethora of functions such as the reabsorption of glucose and amino acids (which depends on a Na+ gradient) in distal convoluted tubule, motility in sperm cells, action potentials in synaptic neurons, etc. (197). In erythrocytes, this enzyme is involved in maintaining their volume and water homeostasis (198); while in lymphocytes, their proliferation induced by a variety of stimulus is dependent on Na+/K+-ATPase activity (199). Interestingly, Na+/K+-ATPase activity is decreased in the erythrocytes from T2DM sufferers (198, 200), but its expression remains unaltered (201). These findings suggest that the activity of Na+/K+-ATPase may be used as a potential biomarker for detecting early phases of T2DM (202). Within this context, one theory that explains the effects of hyperglycemia on Na+/K+-ATPase is by glycosylation, which induces the impairment of the ATPase activity in erythrocytes (202). In fact, this enzyme has several glycosylation sites located at β-subunits, some of them related to protein maturation (203) and other functional processes (197). These lines of evidence show the importance of glycosylation in Na+/K+-ATPase activity.
On the other hand, Na+/K+-ATPase partake in the functionality of immune cells (199, 204, 205). Indeed, proliferation of lymphocytes is dependent on Na+/K+-ATPase activity (199) and the expression of nuclear factor of activated T cells transcription complex (NFAT) of thymocytes (206); this factor is essential for the production of Interleukin-2 (207), a cytokine produced by lymphocytes during a microbial infection (208). Hence, immunologic and hematologic deficiencies in diabetic patients are related to multiple alterations, which may include aberrant activity of the Na+/K+-ATPase.
In agreement with the above findings, a reasonable possibility to explain the alterations in immune system activity during diabetes is that the Na+/K+-ATPase activity could be equally decreased in both lymphocytes and erythrocytes (since these cell types are in the same environment) (201). Besides this, protein glycosylation can occur by enzymatic, but also by non-enzymatic ways; in this respect, glucose is chemically attached to proteins by Schiff base and Amadori product adducts, resulting in a variety of biological effects, including inactivation of enzymes (209), such as Na+/K+-ATPase. It has even been reported that a deficiency in glucose-6-phosphate dehydrogenase, an enzyme altered in diabetes, increases protein glycosylation (210), supporting the idea previously proposed (see below).
Glucose-6-Phosphate Dehydrogenase
Glucose-6-phosphate dehydrogenase (G6PD) is an enzyme expressed ubiquitously in all mammalian tissues. It plays an important role in the pentose pathway catalyzing the first reaction in this metabolic route, which is necessary to convert glucose into pentose sugars (211). This pathway produces nicotinamide adenine dinucleotide phosphate hydrogen (NADPH), an antioxidant molecule that catalyzes the reaction to regenerate reduced glutathione (212).
Many studies have reported the importance of G6PD in antioxidant defense against toxicity of reactive oxygen species (ROS) (211, 213). Interestingly, a relationship is established between diabetes and a decrease in G6PD activity in a variety of cells from rats (212) and humans (214). Additionally, this enzyme plays an important role against infections (213, 215) and in T cell proliferation (216). In keeping with this view, a deficiency of this enzyme in leukocytes is related to serious infectious diseases, such as chronic granulomatous disease (172, 217).
Admittedly, the specific molecular mechanisms that explain the effects of chronic hyperglycemia on G6PD activity in immune cells remain uncertain. For example, Xu et al. (26) showed evidence of inhibition of this enzyme via phosphorylation by protein kinase A in kidney cortex of diabetic rats pretreated with streptozotocin. Similar results were observed in aortic endothelial cells cultured under hyperglycemic conditions (218). Another possibility to explain the effect of glucose on G6PD activity is via protein glycosylation produced by a high glucose concentration (219).
In summary, the above physicochemical alterations resulting from hyperglycemia impair the immune response, predisposing diabetic subjects to acquire infections as well as exacerbates them.
Potential Benefits of Hypoglycemic Drugs on the Outcome of Clinical Infections
Hypoglycemic Drugs and Their Clinical Effects on Bacterial Infections
An uncontrolled blood glucose level is associated with an increase in microvascular and macrovascular complications in diabetic patients (220). Likewise, a hyperglycemic state results in multiple consequences, including osmotic diuresis, fluid/electrolyte imbalance, poor wound healing, impaired immune response, and increased susceptibility to infections, among others (22, 221). Accordingly, these pathophysiological conditions have led to the implementation of therapeutic strategies for a tight glycemic control in patients with T2DM, resulting in the development of the so-called glucose-lowering drugs (i.e., Oral Antidiabetic Drugs; OADs).
Several lines of evidence have shown that the use of OADs to maintain tight blood glucose concentrations between 80 and 110 mg/dl decreases infection-related complications and mortality (see Table 3). For example, metformin, which is the first-line pharmacological agent for T2DM treatment (233), reduced airway glucose permeability and prevented the higher load of Staphylococcus aureus (S. aureus) induced by hyperglycemia (224). Similarly, metformin pre-treatment inhibited the glucose-induced growth of Pseudomonas aeruginosa, increased transepithelial electrical resistance (TEER) and decreased glucose flux in an epithelial cell culture model (234). In this sense, mutants of genes affecting glucose uptake of P. aeruginosa decreased the bacterial loads on streptozotocin-induced hyperglycemic mice compared to control.
TABLE 3
Table 3. Pharmacodynamics of some hypoglycemic drugs and their reported effects on infectious processes.
Interestingly: (i) metformin pre-treatment of hyperglycemic animals reduced both airway glucose and bacterial load (234); (ii) the incidence of tuberculosis has been related to abnormal glucose levels, whereas metformin is a protective agent in the treatment of tuberculosis in diabetic patients (235); (iii) metformin treatment was also associated with an increased risk of bacterial pneumonia in patients with chronic obstructive pulmonary disease from a nationwide cohort study (Taiwan) (236); and (iv) pneumonia is a swelling disease usually caused by a bacterial infection commonly associated with diabetic patients (237).
Consistent with the above findings, diabetic patients with community-acquired pneumonia (CAP) developed worse results and longer hospital stays in comparison to patients with CAP without diabetes (238); accordingly, it is important to discuss the relationship between the use of OADs and pneumonia. Indeed, these data support airway glucose as a critical determinant of increased bacterial load during diabetes (225).
Moreover, Mendy et al. (223) analyzed data from the National Health and Nutrition Examination Survey during 1988–1994 and 1999–2010 for participants aged 40 years or older who had diabetes and were followed up for mortality through 2011. Their results showed that metformin was associated with a decreased risk for chronic lower respiratory diseases (CLRD) mortality in the overall population (HR: 0.39, 95% CI: 0.15–0.99) and among participants with baseline CLRD (HR: 0.30, 95% CI: 0.10–0.93) (223).
Likewise, Pan et al. (226) investigated the effect of metformin vs. sulfonylureas on tuberculosis risk in patients with T2DM. The study demonstrated that patients with T2DM treated with metformin in the initial 2 years, had a significant reduced risk of tuberculosis as compared to those receiving sulfonylureas as initial treatment (226).
Furthermore, Shih et al. (227) reported the relationship between the use of AODs and the risk of hospitalization for sepsis. The authors found that the use of metformin was associated with ~20% reduced risk of sepsis as compared with non-use. In contrast, meglitinides and sulfonylureas were associated with increased risk of future sepsis events, but this association was not evident among recent and current sulfonylurea users. Moreover, the DPP-4 inhibitors and thiazolidinediones on sepsis were neutral, nevertheless, the occurrence of sepsis in current thiazolidinediones users was reduced (227).
On the other hand, some studies have shown that pretreatment with dapagliflozin, a sodium-glucose co-transporter 2 inhibitor, reduced blood and bronchoalveolar lavage glucose concentrations and P. aeruginosa CFU in leptin receptor-deficient (db/db) mice, as compared to those seen in wild type (WT) mice (239).
In summary, the available evidence thus far has established the increased susceptibility to certain types of infections related to hyperglycemia in T2DM. Clearly, further studies on the mechanisms regulating OADs and bacterial action on specific tissues/organs are required. Such studies could yield potential alternatives to prevent/suppress hyperglycemia and bacterial infections.
Moreover, the risk for developing infections is increased in hyperglycemic environments, where there is a lower production of interleukins, a reduced chemotaxis and phagocytic activity, and a gastrointestinal dysmotility (22). The use of specific OADs such as metformin is associated with reduced hospital-treated infections, septicemia prognosis, and some kinds of respiratory illnesses (224, 240). Indeed, another study in diabetic patients has shown a reduction in autoimmune diseases by an acute intervention with OADs, such as DPP-4 in combination with other hypoglycemic drugs (241). One mechanism that may improve those immune response effects is through GLP-1 action that induces insulin secretion and inhibits glucagon secretion, ameliorating the glycemic variability (242).
To conclude this section, it is to be noted that experimental anorexia seems to play an important protective role in supporting the recuperation of bacterial infections (19). Moreover, mortality in critical illnesses (e.g., sepsis, severe burning, etc.) may increase via alterations in immune cell activity that, in turn, may be mediated by the release of stress hormones (cortisol, catecholamines, etc.) and the hyperglycemia that these hormones induce (57). As hyperglycemia impairs periphery glucose usage, administration of insulin improves cellular uptake and attenuates the inflammatory response (57).
A higher risk of CAP was found with other OADs, except with dipeptidyl-peptidase 4 (DPP-4) inhibitors (237). Indeed, in a retrospective cohort and a meta-analysis study, DPP-4 inhibitors failed to increase the risk of pneumonia during diabetes (243). These controversial data about the use OADs and the outcome of bacterial infections in diabetic patients point out the necessity for more detailed analyses and clinical observations.
In view that hyperglycemia may be a determinant factor in the outcome for bacterial infections, any effort for controlling the increases glucose levels is valuable. Another interesting approach may be the supplementation with calcium and vitamin D because it decreases insulin resistance and hyperglycemia (244); nevertheless, some strategies must be considered to ponder the risks and benefits.
Further Considerations
Hypoglycemia occurs when there exists a lack of adequate food intake, excessive exercise, a stressful experience, excessive alcohol consumption, concurrent infections, severe digestive and urologic diseases, and/or after taking antidiabetic medications (245). This suggests that hypoglycemia is an endocrine alarming signal that is triggered to level the required concentrations of blood glucose in the body.
Considering that fasting plasma glucose is normally maintained between 70 and 99 mg/dL (2), a biomarker associated with high blood glucose levels is HbA1c, whose normal range is between 4 and 5.7% in healthy people. Less than 7% of HbA1c is found in controlled people with diabetes, and above 8% is found in people with uncontrolled diabetes (246). Low or high levels of HbA1c have been related to severe hypoglycemic episodes with a glucose-lowering regimen in patients with diabetes (247).
Glycemic control and reduction of hyperglycemia or hypoglycemia events are the main challenges in the clinical experience to achieve decreases of blood glucose variability (248). Indeed, levels of glucose and its constant fluctuations are good indicators of organ dysfunctions such as those associated with infections (249). Patients with diabetes often suffer from chronic low-grade infections such as periodontitis and foot ulceration. Surgery-site infections and susceptibility to septic shock increase with pre and post-operative glucose levels and their variability (250, 251), suggesting that glucose monitoring is one of the key elements in hyperglycemia and hypoglycemia management diseases where the immune system is compromised.
Glucose variability is currently considered more deleterious than chronic hyperglycemia in the development of diabetes-related complications (252). However, some studies suggest that an intensive glucose control does not improve some of the diabetes-associated complications such as cardiovascular failures, raising the mortality rate (253). Furthermore, a tight glucose control induces hypoglycemic episodes and the increased response of the immune system, impacting on coagulant factors, pro-inflammatory cytokines, proatherogenic cell adhesion molecules, and nitric oxide-mediated vasodilatation. Innate immunity response is activated nearly after acute or chronic infections are experienced by diabetes sufferers. For this, the study of the suppression of innate immune system is a key factor, since it exacerbates the inflammatory response after an acute hypoglycemia episode, inducing prothrombotic changes and increasing platelet reactivity (254).
Another consideration is that under normal and pathophysiological metabolic functions, individuals course with glucose swings during the day (255), correlating them with the gastric emptying rate and postprandial glucose levels. Glycemic fluctuations are limited by low glucose levels that slow the gastric emptying or by high glucose that accelerate it (256). However, a hypoglycemic state promotes reverse effects; hence, the gastric emptying is accelerated and the absorption speed of nutrients is increased to reach the physiological glycemic levels, suggesting that gastrointestinal motility and gastric motor function are important factors to consider for a therapy of glycemic control (257).
During physiological gastric emptying, carbohydrates and proteins are evacuated faster than lipids for their caloric content. The evacuation of these macronutrients begins at 20 to 40 min after food intake and when they reach the intestine, incretin hormones are secreted to blood. Glucagon-like peptide-1 (GLP-1) is an incretin hormone that stimulates insulin secretion, reduces glucagon secretion, and delays gastric emptying in a glucose-dependent manner (256, 258).
Significantly, glucose-lowering therapies through the use of diverse drug classes have been reported as an important source of heart failure risk, particularly with differential effects on insulin (259). Consequently, older patients are the most affected population, especially if a diminished food intake, excessive alcohol use, combination of non-prescribed medications, concomitant infections, and diabetic complications are also taken into account (260). Because of this, an intervention with a forced hypoglycemia should be considered with caution according to disease timing, age, nutritional behaviors, type of medications and concomitant infections.
General Conclusion
Hyperglycemia clearly induces physiological and immunological disorders in body tissues/organs that may predispose and exacerbate some infectious diseases. Therefore, the control of glucose levels could be an alternative tool to contribute to the fight against infections not only in diabetic patients, but also in other conditions that induce hyperglycemia, such as SIH. In addition, several studies have shown the potential benefits of controlling it (e.g., pharmacological approaches), opening a new option to improve the outcome of some infections (bacterial and viral). It is worth noting that the authors of this review agree that glycemic control is necessary as part of good intervention strategies to treat current and emerging infectious diseases. Admittedly, the clinical evidence for reducing glycemic exposure requires more supportive data, specifically for hypoglycemia as a tool to fight infections in humans. Notwithstanding, this review summarizes enough preclinical evidence to increase our chances of beating infections by focusing on the key role of glycemic control.
Author Contributions
JC-R, CE-G, and BM-C developed the central idea of this article and wrote the manuscript. BM-C proposed the central idea, made the graphs, and obtained funding. PV-L, EC-S, AL-B, and AQ-S provided original ideas, developed some sections, and reviewed the manuscript. CMV and JM-R discussed the central ideas, reviewed, edited, and corrected the manuscript. All authors contributed to the article and approved the submitted version.
Funding
BM-C was supported by Dirección General de Investigación y Posgrado from Autonomous University of Aguascalientes by the research grant: PIBB19-1. CMV was financially supported by the SEP-Cinvestav Research Support Fund (grant No. 50). JM-R was supported financially by PAICYT 2019–2020 and 2020–2021, Science Grants from the Universidad Autónoma de Nuevo León and by the Fronteras de la Ciencia CONACyT grant 1502. AL-B was a recipient of a Beca de Posdoctorado Nacional 2018–2020.
Conflict of Interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
The handling editor and reviewer AV declared a shared affiliation with one of the authors EC-S at time of review.
Acknowledgments
We thank Verónica Rodríguez Gallegos for her support in the elaboration of graphics.
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Keywords: infections, diabetes, immune system, hyperglycemia, COVID-19
Citation: Chávez-Reyes J, Escárcega-González CE, Chavira-Suárez E, León-Buitimea A, Vázquez-León P, Morones-Ramírez JR, Villalón CM, Quintanar-Stephano A and Marichal-Cancino BA (2021) Susceptibility for Some Infectious Diseases in Patients With Diabetes: The Key Role of Glycemia. Front. Public Health 9:559595. doi: 10.3389/fpubh.2021.559595
Received: 27 May 2020; Accepted: 12 January 2021;
Published: 16 February 2021.
Edited by:
Marcia Hiriart, National Autonomous University of Mexico, Mexico
Reviewed by:
Hidetaka Hamasaki, Hamasaki Clinic, Japan
Antonio R. Villa, Universidad Nacional Autónoma de México, Mexico
Copyright © 2021 Chávez-Reyes, Escárcega-González, Chavira-Suárez, León-Buitimea, Vázquez-León, Morones-Ramírez, Villalón, Quintanar-Stephano and Marichal-Cancino. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
Correspondence: Bruno A. Marichal-Cancino, bruno.marichal@edu.uaa.mx |
3087 | https://stackoverflow.com/questions/62910392/c-program-to-find-out-both-real-and-complex-roots-for-quadratic-equation-ax2 | math - C program to find out both real and complex roots for quadratic equation (ax^2 + bx + c = 0) - Stack Overflow
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C program to find out both real and complex roots for quadratic equation (ax^2 + bx + c = 0)
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Asked 5 years, 2 months ago
Modified5 years, 2 months ago
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This is the code I tried to get real and complex value for this equation, ax^2 + bx + c = 0.
```c
include
include
int main() {
double a, b, c, d, x1, x2, i = sqrt(-1);
scanf("%lf %lf %lf", &a, &b, &c);
d = (b b) - (4 a c);
x1 = (-b + sqrt(d)) / (a + a);
x2 = (-b - sqrt(d)) / (a + a);
if (d < 0) {
printf("Complex solution\n");
printf("R1 = %.2lf\n", x1);
printf("R2 = %.2lf\n", x2);
} else
if (a == 0)
printf("Value of a must be non zero\n");
else {
printf("R1 = %.2lf\n", x1);
printf("R2 = %.2lf\n", x2);
}
return 0;
}
```
And I want to get an answer with i (eg: -.4 + 4i).
c
math
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edited Jul 15, 2020 at 8:53
chqrlie
150k 12 12 gold badges 143 143 silver badges 226 226 bronze badges
asked Jul 15, 2020 at 8:12
MariaMaria
67 1 1 gold badge 3 3 silver badges 7 7 bronze badges
5
What is your question? Please edit your post and add a question that can be answered.anatolyg –anatolyg 2020-07-15 08:18:12 +00:00 Commented Jul 15, 2020 at 8:18
can you share some inputs and outputs you applied and got from your code?walid barakat –walid barakat 2020-07-15 08:22:45 +00:00 Commented Jul 15, 2020 at 8:22
If d is less than zero, then sqrt(d) will have already failed before you make the d < 0 check.Adrian Mole –Adrian Mole 2020-07-15 08:23:06 +00:00 Commented Jul 15, 2020 at 8:23
1 Even before, initializing i = sqrt(-1); will fail.Pierre François –Pierre François 2020-07-15 08:29:10 +00:00 Commented Jul 15, 2020 at 8:29
1 @PierreFrançois: i = sqrt(-1); will produce a NaN value, unlikely to cause an exception or any such failure. If the compiler is smart, it will not even compute i as it is never used and the initializer is a pure function.chqrlie –chqrlie 2020-07-15 09:06:21 +00:00 Commented Jul 15, 2020 at 9:06
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Use <complex.h>
```c
include
include
include
include
int cprint(const char pre, complex double x, const char post) {
int n = printf("%s", pre);
n += printf("%.2f", creal(x));
if (cimag(x)) n += printf(" %+.2fi", cimag(x));
n += printf("%s", post);
return n;
}
int main(void) {
double a, b, c, d;
complex double x1, x2;
if (scanf("%lf%lf%lf", &a, &b, &c) != 3) {
printf("Bad input\n");
exit(EXIT_FAILURE);
}
if (a == 0) {
printf("Value of a must be non zero\n");
exit(EXIT_FAILURE);
}
d = (b b) - (4 a c);
x1 = (-b + csqrt(d)) / (a + a); // using csqrt()
x2 = (-b - csqrt(d)) / (a + a); // complex square root
if (d < 0) {
printf("Complex solution\n");
}
cprint("R1 = ", x1, "\n");
cprint("R2 = ", x2, "\n");
return 0;
}
```
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edited Jul 15, 2020 at 8:55
answered Jul 15, 2020 at 8:28
pmgpmg
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Maria
MariaOver a year ago
Thanks for your approach, Sir. But I am a beginner. Most of the functions used in your code are unknown to me.
2020-07-15T14:44:09.797Z+00:00
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pmg
pmgOver a year ago
The only new function I've used is csqrt(): it calculates the square root of complex numbers, returning a complex number. Other than that I've used complex numbers (and <complex.h>) which you can think of as an (integrated) array of 2 double values, one for the real part and another for the imaginary part.
2020-07-15T15:01:28.473Z+00:00
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Maria
MariaOver a year ago
Okay. Now I got it.
2020-07-17T06:46:11.773Z+00:00
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You can calculate the complex part yourself:
```c
include
include
int main()
{
double a,b,c;
scanf("%lf %lf %lf",&a,&b,&c);
if (a==0)
{
printf("Value of a must be non zero\n");
return 0;
}
double d = (bb)-(4ac);
if (d<0){
double imag = sqrt(-d)/(a+a);
double real = -b/(a+a);
printf("Complex solution\n");
printf("R1 = %.2f + i %.2f\n",real, imag);
printf("R2 = %.2f - i %.2f\n",real, imag);
}
else
{
double x1 = (-b + sqrt(d))/(a+a);
double x2 = (-b - sqrt(d))/(a+a);
printf("R1 = %.2f\n",x1);
printf("R2 = %.2f\n",x2);
}
return 0;
}
```
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edited Jul 15, 2020 at 9:41
answered Jul 15, 2020 at 8:31
mchmch
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chqrlie
chqrlieOver a year ago
The OP seems to prefer i instead of j for the imaginary root of -1
2020-07-15T09:00:04.173Z+00:00
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Your program has multiple issues:
sqrt() is a real function, -1 is outside its domain.
you can use complex numbers if the compiler supports them by defining x1 and x2 as complex double x1, x2;. But it is just as easy to compute the complex components yourself.
you should check the return value of scanf() to detect invalid input.
if a == 0 the equation becomes a first degree equation, that may or may not have solutions.
Here is a modified version:
```c
include
include
int main() {
double a, b, c, d;
if (scanf("%lf %lf %lf", &a, &b, &c) != 3) {
printf("Invalid input\n");
return 1;
}
if (a == 0) {
printf("Value of a must be non zero\n");
if (b == 0) {
if (c == 0) {
printf("Equality is true for all values of x\n");
} else {
printf("No solution\n");
}
} else {
double x = (-c) / b;
printf("Single solution\n");
printf("R1 = %.2f\n", x);
}
return 0;
}
d = (b b) - (4 a c);
if (d < 0) {
double x = (-b) / (a + a);
double y = fabs(sqrt(-d) / (a + a));
printf("Complex solutions\n");
printf("R1 = %.2f - %.2fi\n", x, y);
printf("R2 = %.2f + %.2fi\n", x, y);
} else
if (d == 0) {
double x = (-b) / (a + a);
printf("Single real solution\n");
printf("R1 = %.2f\n", x);
} else {
double x1 = (-b + sqrt(d)) / (a + a);
double x2 = (-b - sqrt(d)) / (a + a);
printf("Real solutions\n");
printf("R1 = %.2f\n", x1);
printf("R2 = %.2f\n", x2);
}
return 0;
}
```
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edited Jul 17, 2020 at 17:14
answered Jul 15, 2020 at 9:23
chqrliechqrlie
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When you declare a variable as double, you declare a real number, not a complex one. For declaring and using complex numbers, use the library complex.h as described in How to work with complex numbers in C?.
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answered Jul 15, 2020 at 8:27
Pierre FrançoisPierre François
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Maria
MariaOver a year ago
Thanks a lot for your approach, Sir. As I am beginner. So, complex.h library function is unknown to me.
2020-07-15T14:49:37.103Z+00:00
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well, let's talk a little bit about quadratic equation :
The term b2-4ac is known as the discriminant of a quadratic equation. The discriminant tells the nature of the roots.
If discriminant is greater than 0, the roots are real and different.
If discriminant is equal to 0, the roots are real and equal.
If discriminant is less than 0, the roots are complex and different.
i guess if we follow the math, you can write a smooth code for that.
but for a small hint, you can find more explanation about quadratic equation and how to implement a C program to solve it here : :) .
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edited Jul 15, 2020 at 8:46
answered Jul 15, 2020 at 8:36
walid barakatwalid barakat
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3088 | https://www.academia.edu/86656172/The_Art_of_Electronics_Paul_Horowitz_and_Winfield_Hill_R | Academia.edu no longer supports Internet Explorer.
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The Art of Electronics - Paul Horowitz & Winfield Hill R
Alberto Albuquerque
2015
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This volume is intended as an electronic circuit design textbook and reference book; it begins at a level suitable for those with no previous exposure to electronics and carries the reader through to a reasonable degree of proficiency in electronic circuit design.
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How to find a point in the middle of N points? [closed]
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Asked 9 years, 9 months ago
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Let's say I have N points defined by their coordinates (X, Y). How could I find coordinates of another point X, which is in the middle of that N points (i.e. distance between X and each point from N is more or less equal). Is there any algorithm to do it?
javascript
algorithm
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edited Dec 10, 2015 at 20:22
LA_LA_
asked Dec 10, 2015 at 20:09
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20.5k 61 61 gold badges 181 181 silver badges 319 319 bronze badges
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Can the downvoter explain ?user1196549 –user1196549 2015-12-10 20:11:59 +00:00 Commented Dec 10, 2015 at 20:11
2 Is this similar to your question?gtgaxiola –gtgaxiola 2015-12-10 20:12:23 +00:00 Commented Dec 10, 2015 at 20:12
2 What does this have to do with programming and or javascript? @YvesDaoust, I suspect downvoted because "any algorithm to do it" screams "off topic for is there a library..[etc]".Sterling Archer –Sterling Archer 2015-12-10 20:13:33 +00:00 Commented Dec 10, 2015 at 20:13
Seems here is a Java Solution for finding Centroidgtgaxiola –gtgaxiola 2015-12-10 20:14:28 +00:00 Commented Dec 10, 2015 at 20:14
2 @MarcB OP may still be below high school age or not have received formal education. That's no reason to discourage him.Zirak –Zirak 2015-12-10 20:16:29 +00:00 Commented Dec 10, 2015 at 20:16
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You probably mis-stated the question by thinking that a point "in the middle" is equidistant to all others.
As you can't achieve that, a better approach is to minimize the total Euclidean distance of the center to the N points.
It turns out that this problem has no easy analytical solution, as you need to minimize
javascript
D = Σ √(X - Xc)² + (Y - Yc)²
Deriving on Xc, you get
javascript
D'x = Σ (X - Xc) / √(X - Xc)² + (Y - Yc)² = 0
a complex nonlinear expression. This problem is known as the "geometric median" and you'll find more info here (in particular, the Weiszfeld's algorithm).
On the opposite, minimizing the sum of the squared distances
javascript
D = Σ (X - Xc)² + (Y - Yc)²
is trivial as
javascript
D'x = 2 Σ (X - Xc) = 0
which yields the centroid.
javascript
Xc = Σ X / N, Yc = Σ Y / N.
You can also think of the medoid of your points, which is the point from the given set that minimizes the sum of the distances to the others. It takes N(N-1)/2 distance computations to find it.
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edited Dec 10, 2015 at 20:39
answered Dec 10, 2015 at 20:24
user1196549 user1196549
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There may not be such a point. You can average the Xs and Ys and get a point that's in the middle (a.k.a "centroid").
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answered Dec 10, 2015 at 20:15
Michael Bar-SinaiMichael Bar-Sinai
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For 2 points, this "middle" exists. For more, we can't be sure.
You can calculate the centroid with the average x and y of the points.
```javascript
if (points.length > 0) {
var x_acc = 0;
var y_acc = 0;
for (var i = 0; i < points.length; i++) {
x_acc += points[i].x;
y_acc += points[i].y;
}
var centroid_x = x_acc / points.length;
var centroid_y = y_acc / points.length;
}
```
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edited Oct 17, 2018 at 19:55
merlin
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answered Dec 10, 2015 at 20:22
carlos pradocarlos prado
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3090 | https://math.stackexchange.com/questions/4281655/how-to-find-the-range-of-a-function | Stack Exchange Network
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How to find the range of a function?
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Asked
Modified 3 years, 11 months ago
Viewed 27k times
4
$\begingroup$
I need to find the range of $\frac{x^2+6}{2x+1}$
I know that $x$ cannot be $-1/2$. I graphed the function on Desmos and I can see that there is a vertical asymptote at $x=-1/2$ However, I'm having trouble finding the range of this function?
I can understand what the range is by looking at the graph however I don't know how to find this algebraically. Is there a method that I can use every time to get the range?
Also I know there is an oblique asymptote at $1/2x-1/4$. How can I use this to help me?
The range according to the graph is $(-\infty,-3)\cup(2,\infty)$
calculus
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edited Oct 20, 2021 at 1:05
user8290579user8290579
asked Oct 20, 2021 at 0:54
user8290579user8290579
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$\endgroup$
4
$\begingroup$ Helpful here $\endgroup$
user10101
– user10101
2021-10-20 00:58:39 +00:00
Commented Oct 20, 2021 at 0:58
$\begingroup$ Have you calculated the derivative? $\endgroup$
N. F. Taussig
– N. F. Taussig
2021-10-20 01:17:29 +00:00
Commented Oct 20, 2021 at 1:17
1
$\begingroup$ i haven't learned about the derivative $\endgroup$
user8290579
– user8290579
2021-10-20 01:31:20 +00:00
Commented Oct 20, 2021 at 1:31
$\begingroup$ @user8290579 Your calculated range is incorrect. The range will be $(-\infty, -3]\cup[2, \infty)$... $\endgroup$
An Alien
– An Alien
2021-10-20 01:47:54 +00:00
Commented Oct 20, 2021 at 1:47
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Here's a method without calculus.
Let $f(x)=y=\frac{x^2+6}{2x+1}\Rightarrow 2xy+y=x^2+6\Rightarrow x^2-2xy+6-y=0\Rightarrow x=\frac{2y±\sqrt{4y^2-24+4y}}{2}$ Denominator can never be $0$ in this case. So the domain depends only upon $4y^2-24+4y\Rightarrow 4y^2+4y-24¥0\Rightarrow 4(y-2)(y+3)¥0\Rightarrow (y-2)(y+3)¥0\Rightarrow y\in(- \infty, -3]\cup[2,\infty).$
Hence the range of the $f(x)$ is $(- \infty, -3]\cup[2,\infty)$
I guess you know solving inequalities...
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answered Oct 20, 2021 at 1:36
An AlienAn Alien
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To find the range of a function, my first instinct is to check whether the graph has an inverse. We can do this informally using the Horizontal Line Test. If no Horizontal Line intersects the function more than once, then the function has an inverse. With that, we can find $f(y) = x$. The "domain" of $f(y)$ would be the range of the function $f(x)$.
Else, we can use the derivatives of the function to help us find the maximum and minimum which gives us the lower and upper bound of the range of $f(x)$.
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answered Oct 20, 2021 at 1:34
user3497user3497
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Using the resource shared by @user10101, we have that:
$$y=\dfrac{x^2+6}{2x+1}\Rightarrow y(2x+1)=x^2+6\Rightarrow 0=x^2-2xy-y+6$$
Then we can use the quadratic formula, which yields: $$x=\dfrac{2y\pm\sqrt{4y^2+4y-24}}{2}=y\pm\sqrt{y^2+y-6}=y\pm\sqrt{(y+3)(y-2)}$$
Since the term inside the square root has to be nonnegative, we have that if $y\geq2$ then the term is always positive. On the other hand, if $y<2$ then $(y-2)$ is negative, so $(y+3)$ has to be negative to get a positive sign, which happens when $y\leq-3$. That gives you the range you obtained.
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edited Oct 20, 2021 at 1:44
answered Oct 20, 2021 at 1:39
Schach21Schach21
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$\begingroup$ Term inside square root need not to be positive... It needs to be non-negative $\endgroup$
An Alien
– An Alien
2021-10-20 01:43:09 +00:00
Commented Oct 20, 2021 at 1:43
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3091 | https://math.tools/numbers/prime-factors/3240 | MATH MathTOOLS
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3092 | https://www.dictionary.com/e/widow-vs-widower/ | Widow vs. Widower: What’s the Difference?
⚡ Quick summary
A widow is a woman who has lost a spouse by death and has not remarried. A widower is a man who has lost a spouse by death and has not remarried.
The words widow and widower are both used to describe a person who has remained unmarried after their spouse passes away. What do these two similar words mean, and why do we use two different words to mean almost the same thing?
In this article, we will define widow and widower, explain where the word widower came from, and give a few alternatives that people may want to use in place of these words.
When to use widow or widower
A widow is a woman whose spouse has died and who hasn’t remarried.
A widower is a man whose spouse has died and who hasn’t remarried.
In Old English, the feminine form widuwe was used to mean “a woman whose husband had died,” and the masculine form widuwa was used to mean “a man whose wife had died,” though it was used much less frequently. In Middle English, the word widewer or wedewer was used to refer to a man whose wife had died. This word emerged by combining the more common feminine Old English form with the suffix –er, which generally meant “a man related to something.” The –er suffix was most often used to indicate professions, such as in the Middle English synger (singer) and hatter.
Go Behind The Words!
Today and throughout history, the word widow (and its older forms) has been much more commonly used than the word widower for many reasons. Health data shows that women are more likely to live longer than men. Also, men are significantly more likely to remarry than women. And historically it was much more common for younger women to marry older men than for younger men to marry older women. Many other historical factors, such as poor sanitation during childbirth and the death of soldiers during war, also contribute to these usage differences.
Although both the words widow and widower are still commonly used in formal and informal contexts, some people may avoid using these terms because they limit a person’s identity to their association with their late spouse or misgender the person. People who feel this way may avoid using either widow or widower or avoid mentioning a person’s marital status at all.
Talking about loss can be hard without the right words. Learn what "grieve" and "bereave" mean.
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3093 | https://projecteuclid.org/journals/probability-surveys/volume-8/issue-none/A-basic-theory-of-Benfords-Law/10.1214/11-PS175.pdf | Probability Surveys Vol. 8 (2011) 1–126 ISSN: 1549-5787 DOI: 10.1214/11-PS175 A basic theory of Benford’s Law∗ Arno Berger†,‡ Mathematical and Statistical Sciences University of Alberta Edmonton, Alberta T6G 2G1, Canada e-mail: aberger@math.ualberta.ca and Theodore P. Hill‡ School of Mathematics Georgia Institute of Technology Atlanta, GA 30332–0160, USA e-mail: hill@math.gatech.edu Abstract: Drawing from a large, diverse body of work, this survey presents a comprehensive and unified introduction to the mathematics underlying the prevalent logarithmic distribution of significant digits and significands, often referred to as Benford’s Law (BL) or, in a special case, as the First Digit Law. The invariance properties that characterize BL are developed in detail. Special attention is given to the emergence of BL in a wide variety of deterministic and random processes. Though mainly expository in nature, the article also provides strengthened versions of, and simplified proofs for, many key results in the literature. Numerous intriguing problems for future research arise naturally.
AMS 2000 subject classifications: Primary 60-01, 11K06, 37M10, 39A60; secondary 37A45, 60F15, 62E10.
Keywords and phrases: Benford’s Law, significant digits, uniform distri-bution mod 1, scale-invariance, base-invariance, sum-invariance, shadowing, difference equation, random probability measure, mixture of distributions.
Received June 2011.
Contents 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2 2 Significant digits and the significand . . . . . . . . . . . . . . . . . . .
6 2.1 Significant digits . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 2.2 The significand . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 2.3 The significand σ-algebra . . . . . . . . . . . . . . . . . . . . . .
8 3 The Benford property . . . . . . . . . . . . . . . . . . . . . . . . . . .
16 3.1 Benford sequences . . . . . . . . . . . . . . . . . . . . . . . . . .
16 ∗This is an original survey paper.
†Supported by an Nserc Discovery Grant.
‡The authors are grateful to S. Evans for pointing them towards a proof of Theorem 6.6.
1 2 A. Berger and T.P. Hill 3.2 Benford functions . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 3.3 Benford distributions and random variables . . . . . . . . . . . .
22 4 Characterizations of Benford’s Law . . . . . . . . . . . . . . . . . . . .
27 4.1 The uniform distribution characterization . . . . . . . . . . . . .
27 4.2 The scale-invariance characterization . . . . . . . . . . . . . . . .
40 4.3 The base-invariance characterization . . . . . . . . . . . . . . . .
52 4.4 The sum-invariance characterization . . . . . . . . . . . . . . . .
59 5 Benford’s Law for deterministic processes . . . . . . . . . . . . . . . .
64 5.1 One-dimensional discrete-time processes . . . . . . . . . . . . . .
64 5.2 Multi-dimensional discrete-time processes . . . . . . . . . . . . .
85 5.3 Differential equations . . . . . . . . . . . . . . . . . . . . . . . . .
95 6 Benford’s Law for random processes . . . . . . . . . . . . . . . . . . . 101 6.1 Independent random variables . . . . . . . . . . . . . . . . . . . . 101 6.2 Mixtures of distributions . . . . . . . . . . . . . . . . . . . . . . . 111 6.3 Random maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 List of symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 1. Introduction Benford’s Law, or BL for short, is the observation that in many collections of numbers, be they e.g. mathematical tables, real-life data, or combinations thereof, the leading significant digits are not uniformly distributed, as might be expected, but are heavily skewed toward the smaller digits. More specifically, BL says that the significant digits in many datasets follow a very particular logarithmic distribution. In its most common formulation, namely the special case of first significant decimal (i.e. base-10) digits, BL is also known as the First-Digit Phenomenon and reads Prob(D1 = d1) = log10 1 + d−1 1 for all d1 = 1, 2, . . ., 9 ; (1.1) here D1 denotes the first significant decimal digit, e.g.
D1( √ 2) = D1(1.414 . . .) = 1 , D1(π−1) = D1(0.3183 . . .) = 3 , D1(eπ) = D1(23.14 . . .) = 2 .
Thus, for example, (1.1) asserts that Prob (D1 =1) = log10 2 = 0.3010 . . . , Prob(D1 =2) = log10 3 2 = 0.1760 . . . , hence the two smallest digits occur with a combined probability close to 50 percent, whereas the two largest digits together have a probability of less than 10 percent, Prob (D1 =8) = log10 9 8 = 0.05115 . . . , Prob(D1 =9) = log10 10 9 = 0.04575 . . . .
A basic theory of Benford’s Law 3 A crucial part of the content of (1.1), of course, is an appropriate formulation or interpretation of “Prob”. In practice, this can take several forms. For sequences of real numbers (xn), for example, Prob usually refers to the proportion (or relative frequency) of times n for which an event such as D1 = 1 occurs. Thus Prob(D1 =1) is the limiting proportion, as N →∞, of times n ≤N that the first significant digit of xn equals 1. Implicit in this usage of Prob is the assumption that all limiting proportions of interest actually exist. Similarly, for real-valued functions f : [0, +∞) →R, Prob (D1 = 1) refers to the limiting proportion, as T →∞, of the total length of time t < T for which the first significant digit of f(t) is 1. For a random variable or probability distribution, on the other hand, Prob simply denotes the underlying probability, e.g. if X a random variable then Prob (D1(X) = 1) is the probability that the first significant digit of X equals 1. Finite datasets of real numbers can also be dealt with this way, with Prob being the empirical distribution of the dataset.
All of these approaches to (1.1) will be studied in detail in subsequent chap-ters. Fig 1 illustrates several of the possible settings, including simple sequences such as the Fibonacci numbers (Fn) = (1, 1, 2, 3, 5, 8, 13, . . .), and real-life data from [Ben] as well as recent census statistics; in addition, it previews some of the many scenarios, also to be discussed later, that lead to exact conformance with BL. In Fig 1 and throughout, #A denotes the cardinality (number of elements) of the finite set A.
In a form more complete than (1.1), BL is a statement about the joint dis-tribution of all decimal digits: For every positive integer m, Prob (D1, D2, . . . , Dm) = (d1, d2, . . . , dm) = log10 1 + Xm j=1 10m−jdj −1 (1.2) holds for all m-tuples (d1, d2, . . . , dm), where d1 is an integer in {1, 2, . . ., 9} and for j ≥2, dj is an integer in {0, 1, . . ., 9}; here D2, D3, D4 etc. represent the second, third, forth etc. significant decimal digit, e.g.
D2( √ 2) = 4 , D3(π−1) = 8 , D4(eπ) = 4 .
Thus, for example, (1.2) implies that Prob (D1, D2, D3) = (3, 1, 4) = log10 315 314 = 0.001380 . . . .
A perhaps surprising corollary of the general form of BL is that the significant digits are dependent, and not independent as one might expect [Hi2]. Indeed, from (1.2) it follows for instance that the (unconditional) probability that the second digit equals 1 is Prob(D2 = 1) = X9 j=1 log10 1 + 1 10j + 1 = log10 6029312 4638501 = 0.1138 . . . , whereas, given that the first digit equals 1, the (conditional) probability that the second digit equals 1 as well is Prob(D2 = 1|D1 = 1) = log10 12 −log10 11 log10 2 = 0.1255 . . . .
4 A. Berger and T.P. Hill 1 1 1 1 1 1 1 · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · d d d d d d d 9 9 9 9 9 9 9 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.3 0.3 0.3 0.3 0.3 0.3 0.3 ρ ρN ρN ρX ρX ρX ρX Original data from [Ben] N →∞ see Chapters 3 and 5 mixture of many datasets see Chapter 6 products of i.i.d. variables see Chapter 6 X uniform on [0, 1) X exponential EX = 1 (#X = 20, 229) U.S. 1990 Census data (#X = 3, 141) exact BL Example: (xn) = (Fn) Fn+2 = Fn+1 + Fn (n ∈N), F1 = F2 = 1 N = 10 N = 102 ρ(d) = log10(1 + d−1) Finite dataset X ρX(d) := #{x ∈X : D1(x) = d} #X Random variable X ρX (d) := P(D1(X) = d) Sequence (xn) ρN (d) := #{1 ≤n ≤N : D1(xn) = d} N Fig 1. Different interpretations of (1.1) for sequences, datasets, and random variables, re-spectively, and scenarios that may lead to exact conformance with BL.
This dependence among significant digits decreases rapidly, in fact exponentially, as the distance between the digits increases. For example, it follows easily from (1.2) that Prob (Dm = 1|D1 = 1) = Prob (Dm = 1) + O(10−m) as m →∞.
(Here and throughout, the order symbol O is used as usual: If (an) and (bn) are sequences of real numbers then an = O(bn) as n →∞simply means that |an| ≤c|bn| for all n, with some constant c > 0.) A basic theory of Benford’s Law 5 A related consequence of (1.2) is that the distribution of the m-th significant digit approaches the uniform distribution on {0, 1, . . ., 9} exponentially fast also, e.g.
Prob(Dm = 1) = 1 10 + 63 20 ln 1010−m + O(10−2m) as m →∞.
Apparently first discovered by polymath S. Newcomb [Ne] in the 1880’s, (1.1) and (1.2) were rediscovered by physicist F. Benford [Ben] and, Newcomb’s ar-ticle having been forgotten at the time, came to be known as Benford’s Law.
Today, BL appears in a broad spectrum of mathematics, ranging from differen-tial equations to number theory to statistics. Simultaneously, the applications of BL are mushrooming — from diagnostic tests for mathematical models in biology and finance to fraud detection. For instance, the U.S. Internal Revenue Service uses BL to ferret out suspicious tax returns, political scientists use it to identify voter fraud, and engineers to detect altered digital images. As R. Raimi already observed some 35 years ago [Ra1, p.512], “This particular logarithmic distribution of the first digits, while not universal, is so common and yet so surprising at first glance that it has given rise to a varied literature, among the authors of which are mathematicians, statisticians, economists, engineers, physicists, and amateurs.” At the time of writing, the online database [BH2] contains more than 600 articles on the subject.
It is the purpose of this article to explain the basic terminology, mathematical concepts and results concerning BL in an elementary and accessible manner.
Having read this survey, the reader will find it all the more enjoyable to browse the multifarious literature where a wide range of extensions and refinements as well as applications are discussed.
Note. Throughout this overview of the basic theory of BL, attention will more or less exclusively be restricted to significant decimal (i.e. base-10) digits. From now on, therefore, log x will always denote the logarithm base 10 of x, while ln x is the natural logarithm of x. For convenience, the convention log 0 := 0 will be adopted. All results stated here only with respect to base 10 carry over easily to arbitrary integer bases b ≥2, and the interested reader may find some pertinent details e.g. in [BBH]. The general form of (1.2) with respect to any such base b is Prob D(b) 1 , D(b) 2 , . . . , D(b) m = (d1, d2, . . . , dm) = logb 1+ Xm j=1bm−jdj −1 , (1.3) where logb denotes the base-b logarithm and D(b) 1 , D(b) 2 , D(b) 3 etc. are, respec-tively, the first, second, third etc. significant digits base b; in particular, there-fore, d1 is an integer in {1, 2, . . ., b −1}, and for j ≥2, dj is an integer in {0, 1, . . ., b −1}. Note that in the case m = 1 and b = 2, (1.3) reduces to Prob D(2) 1 = 1 = 1, which trivially is true because the first significant digit base 2 of every non-zero number equals 1.
♣ 6 A. Berger and T.P. Hill 2. Significant digits and the significand Benford’s Law is a statement about the statistical distribution of significant (decimal) digits or, equivalently, about significands viz. fraction parts in floating-point arithmetic. Thus a natural starting point for any study of BL is the formal definition of significant digits and the significand (function).
2.1. Significant digits Definition 2.1. For every non-zero real number x, the first significant decimal digit of x, denoted by D1(x), is the unique integer j ∈{1, 2, . . . , 9} satisfying 10kj ≤|x| < 10k(j + 1) for some (necessarily unique) k ∈Z.
Similarly, for every m ≥2, m ∈N, the m-th significant decimal digit of x, denoted by Dm(x), is defined inductively as the unique integer j ∈{0, 1, . . ., 9} such that 10k Xm−1 i=1 Di(x)10m−i + j ≤|x| < 10k Xm−1 i=1 Di(x)10m−i + j + 1 for some (necessarily unique) k ∈Z; for convenience, Dm(0) := 0 for all m ∈N.
Note that, by definition, the first significant digit D1(x) of x ̸= 0 is never zero, whereas the second, third, etc. significant digits may be any integers in {0, 1, . . ., 9}.
Example 2.2.
D1( √ 2) = D1(− √ 2) = D1(10 √ 2) = 1 , D2( √ 2) = 4 , D3( √ 2) = 1 ; D1(π−1) = D1(10π−1) = 3 , D2(π−1) = 1 , D3(π−1) = 8 .
2.2. The significand The significand of a real number is its coefficient when it is expressed in floating-point (“scientific notation”) form, more precisely Definition 2.3. The (decimal) significand function S : R →[1, 10) is defined as follows: If x ̸= 0 then S(x) = t, where t is the unique number in [1, 10) with |x| = 10kt for some (necessarily unique) k ∈Z; if x = 0 then, for convenience, S(0) := 0.
Observe that, for all x ∈R, S(10kx) = S(x) for every k ∈Z , and also S S(x) = S(x). Explicitly, S is given by S(x) = 10log |x|−⌊log |x|⌋ for all x ̸= 0 ; here ⌊t⌋denotes, for any real number t, the largest integer less than or equal to t. (The function t 7→⌊t⌋is often referred to as the “floor function”.) A basic theory of Benford’s Law 7 1 1 1 2 3 10 10 −1 −1 −2 −3 −10 S(x) log S(x) x log |x| Fig 2. Graphing the (decimal) significand function S.
Note. The original word used in American English to describe the coefficient of floating-point numbers in computer hardware seems to have been mantissa, and this usage remains common in computing and among computer scientists.
However, this use of the word mantissa is discouraged by the IEEE floating-point standard committee and by some professionals such as W. Kahan and D. Knuth because it conflicts with the pre-existing usage of mantissa for the fractional part of a logarithm. In accordance with the IEEE standard, only the term significand will be used henceforth. (With the significand as in Definition 2.3, the (traditional) mantissa would simply be log S.) The reader should also note that in some places in the literature, the significand is taken to have values in [0.1, 1) rather than in [1, 10).
♣ Example 2.4.
S( √ 2) = S(10 √ 2) = √ 2 = 1.414 . . . , S(π−1) = S(10π−1) = 10π−1 = 3.183 . . . .
The significand uniquely determines the significant digits, and vice versa.
This relationship is recorded in the following proposition which immediately follows from Definitions 2.1 and 2.3.
Proposition 2.5. For every real number x: (i) S(x) = P m∈N 101−mDm(x); (ii) Dm(x) = ⌊10m−1S(x)⌋−10⌊10m−2S(x)⌋for every m ∈N.
Thus, Proposition 2.5(i) expresses the significand of a number as an explicit function of the significant digits of that number, and (ii) expresses the significant digits as a function of the significand.
It is important to note that the definition of significand and significant digits per se does not involve any decimal expansion of x. However, it is clear from Proposition 2.5(i) that the significant digits provide a decimal expansion of S(x), and that a sequence (dm) in {0, 1, . . ., 9} is the sequence of significant digits of some positive real number if and only if d1 ̸= 0 and dm ̸= 9 for infinitely many m.
8 A. Berger and T.P. Hill Example 2.6. It follows from Proposition 2.5, together with Examples 2.2 and 2.4, that S( √ 2) = D1( √ 2) + 10−1D2( √ 2) + 10−2D3( √ 2) + . . . = 1.414 . . . = √ 2 , as well as D1( √ 2) = ⌊ √ 2⌋= 1 , D2( √ 2) = ⌊10 √ 2⌋−10⌊ √ 2⌋= 4 , D3( √ 2) = ⌊100 √ 2⌋−10⌊10 √ 2⌋= 1 , etc .
As the significant digits determine the significand, and are in turn determined by it, the informal version (1.2) of BL in the Introduction has an immediate and very concise counterpart in terms of the significand function, namely Prob (S ≤t) = log t for all 1 ≤t < 10 .
(2.1) (Recall that log denotes the base-10 logarithm throughout.) As noted earlier, the formal versions of (1.2) and (2.1) will be developed in detail below.
2.3. The significand σ-algebra The informal statements (1.1), (1.2) and (2.1) of BL involve probabilities. Hence to formulate mathematically precise versions of these statements, it is necessary to re-formulate them in the setting of rigorous probability theory.
The fundamental concept of standard modern probability theory is that of a probability space (Ω, A, P); here Ω, A and P are, respectively, a non-empty set, a σ-algebra on Ω, and a probability measure on (Ω, A). Recall that a σ-algebra A on Ωis simply a family of subsets of Ωsuch that ∅∈A, and A is closed under taking complements and countable unions, that is, A ∈A = ⇒Ac := {ω ∈Ω: ω ̸∈A} ∈A , as well as An ∈A for all n ∈N = ⇒ [ n∈N An ∈A .
Given any collection E of subsets of Ω, there exists a (unique) smallest σ-algebra on Ωcontaining E, referred to as the σ-algebra generated by E and denoted by σ(E). Perhaps the most important example is the so-called Borel σ-algebra B on R: By definition, B is the σ-algebra generated by all intervals. If C ⊂R then B(C) is understood to be the σ-algebra C ∩B := {C ∩B : B ∈B} on C; for brevity, write B[a, b) instead of B [a, b) and B+ instead of B(R+), where R+ = {t ∈R : t > 0}.
In general, given any function f : Ω→R, recall that, for every C ⊂R, the set f −1(C) ⊂Ω, called the pre-image of C under f, is defined as f −1(C) = {ω ∈Ω: f(ω) ∈C} .
A basic theory of Benford’s Law 9 The σ-algebra on Ωgenerated by E = {f −1(I) : I ⊂R an interval} is also referred to as the σ-algebra generated by f; it will be denoted by σ(f).
Thus σ(f) is the smallest σ-algebra on Ωthat contains all sets of the form {ω ∈Ω: a ≤f(ω) ≤b}, for every a, b ∈R. It is easy to check that in fact σ(f) = {f −1(B) : B ∈B}. Similarly, a whole family F of functions f : Ω→R may be considered, and σ(F) := σ [ f∈F σ(f) = σ f −1(I) : I ⊂R an interval, f ∈F is then simply the smallest σ-algebra on Ωcontaining all sets {ω ∈Ω: a ≤ f(ω) ≤b} for all a, b ∈R and all f ∈F.
In probability theory, the elements of a σ-algebra A on Ωare often referred to as events, and functions f : Ω→R with σ(f) ⊂A are called random variables.
Probability textbooks typically use symbols X, Y etc., rather than f, g etc., to denote random variables, and this practice will be adhered to here also. Thus, for example, for a Bernoulli random variable X on (R, B) taking only the values 0 and 1, σ(X) is the sub-σ-algebra of B given by σ(X) = ∅, {0}, {1}, {0, 1}, R, R{0}, R{1}, R{0, 1} ; here, and throughout, A\B = A ∩Bc is the set of all elements of A that are not in B.
As the third ingredient in the concept of a probability space, a probability measure on (Ω, A) is a function P : A →[0, 1] such that P(∅) = 0, P(Ω) = 1, and P [ n∈N An = X n∈N P(An) holds whenever the sets An ∈A are disjoint. The obvious probabilistic interpre-tation of P is that, for every A ∈A, the number P(A) ∈[0, 1] is the probability that the event {ω ∈A} occurs. Two of the most important examples of proba-bility measures are the discrete uniform distribution on a non-empty finite set A, where the probability of any set B ⊂A is simply #(B ∩A) #A , and its continuous counterpart the uniform distribution λa,b with a < b, more technically referred to as (normalized) Lebesgue measure on [a, b), or more pre-cisely on [a, b), B[a, b) , given by λa,b [c, d] := d −c b −a for every [c, d] ⊂[a, b) .
(2.2) In advanced analysis courses, it is shown that (2.2) does indeed entail a unique, consistent definition of λa,b(B) for every B ∈B[a, b); in particular λa,b [a, b) = 10 A. Berger and T.P. Hill 1. Another example of a probability measure, on any (Ω, A), is the Dirac mea-sure (or point mass) concentrated at some ω ∈Ω, symbolized by δω. In this case, δω(A) = 1 if ω ∈A, and δω(A) = 0 otherwise. Throughout, unspecified probability measures on (Ω, A) with Ω⊂R and A ⊂B will typically be denoted by capital Roman letters P, Q etc.
In view of the above, the key step in formulating BL precisely is identifying the appropriate probability space, and hence in particular the correct σ-algebra.
As it turns out, in the significant digit framework there is only one natural candidate which, although different from B, is nevertheless both intuitive and easy to describe.
Definition 2.7. The significand σ-algebra S is the σ-algebra on R+ generated by the significand function S, i.e. S = R+ ∩σ(S).
The importance of the σ-algebra S comes from the fact that for every event A ∈S and every x > 0, knowing S(x) is enough to decide whether x ∈A or x ̸∈A. Worded slightly more formally, this observation reads as follows.
Lemma 2.8. For every function f : R+ →R the following statements are equivalent: (i) f can be described completely in terms of S, that is, f(x) = ϕ S(x) holds for all x ∈R+, with some function ϕ : [1, 10) →R satisfying σ(ϕ) ⊂ B[1, 10).
(ii) σ(f) ⊂S.
Proof. First assume (i) and let I ⊂R be any interval. Then B = ϕ−1(I) ∈B and f −1(I) = S−1 ϕ−1(I) = S−1(B) ∈S, showing that σ(f) ⊂S.
Conversely, if σ(f) ⊂S then f(10x) = f(x) holds for all x > 0. Indeed, assuming by way of contradiction that, say, f(x0) < f(10x0) for some x0 > 0, let A := f −1 f(x0) −1, f(x0) + f(10x0) 2 ∈σ(f) ⊂S and note that x0 ∈A while 10x0 ̸∈A. Since A = S−1(B) for some B ∈B, this leads to the contradiction that S(x0) ∈B and S(x0) = S(10x0) ̸∈B. Hence f(10x) = f(x) for all x > 0, and by induction also f(10kx) = f(x) for all k ∈Z.
Given x ∈[1, 10), pick any y > 0 with S(y) = x and define ϕ(x) := f(y). Since any two choices of y differ by a factor 10k for some k ∈Z, ϕ : [1, 10) →R is well-defined, and ϕ S(y) = f(y) holds for all y > 0. Moreover, for any interval I ⊂R and x > 0, ϕ(x) ∈I holds if and only if x ∈S k∈Z 10kf −1(I). By assumption, the latter set belongs to S, which in turn shows that σ(ϕ) ⊂B[1, 10).
Informally put, Lemma 2.8 states that the significand σ-algebra S is the family of all events A ⊂R+ that can be described completely in terms of their significands, or equivalently (by Theorem 2.9 below) in terms of their significant digits. For example, the set A1 of positive numbers whose first significant digit is 1 and whose third significant digit is not 7, i.e.
A1 = {x > 0 : D1(x) = 1, D3(x) ̸= 7} , A basic theory of Benford’s Law 11 belongs to S, as does the set A2 of all x > 0 whose significant digits are all 5 or 6, i.e.
A2 = {x > 0 : Dm(x) ∈{5, 6} for all m ∈N} , or the set A3 of numbers whose significand is rational, A3 = {x > 0 : S(x) ∈Q} .
On the other hand, the interval [1, 2], for instance, does not belong to S. This follows from the next theorem which provides a useful characterization of the significand sets, i.e. the members of the family S. For its formulation, for every t ∈R and every set C ⊂R, let tC := {tc : c ∈C}.
Theorem 2.9 ([Hi2]). For every A ∈S, A = [ k∈Z 10kS(A) (2.3) holds, where S(A) = {S(x) : x ∈A} ⊂[1, 10). Moreover, S = R+ ∩σ(D1, D2, D3, . . .) = n[ k∈Z 10kB : B ∈B[1, 10) o .
(2.4) Proof. By definition, S = R+ ∩σ(S) = R+ ∩{S−1(B) : B ∈B} = R+ ∩{S−1(B) : B ∈B[1, 10)} .
Thus, given any A ∈S, there exists a set B ∈B[1, 10) with A = R+ ∩S−1(B) = S k∈Z 10kB. Since S(A) = B, it follows that (2.3) holds for all A ∈S.
To prove (2.4), first observe that by Proposition 2.5(i) the significand func-tion S is completely determined by the significant digits D1, D2, D3, . . . , so σ(S) ⊂σ(D1, D2, D3, . . .) and hence S ⊂R+∩σ(D1, D2, D3, . . .). Conversely, ac-cording to Proposition 2.5(ii), every Dm is determined by S, thus σ(Dm) ⊂σ(S) for all m ∈N, showing that σ(D1, D2, D3, . . .) ⊂σ(S) as well. To verify the re-maining equality in (2.4), note that for every A ∈S, S(A) ∈B[1, 10) and hence A = S k∈Z 10kB for B = S(A), by (2.3). Conversely, every set of the form S k∈Z 10kB = R+ ∩S−1(B) with B ∈B[1, 10) obviously belongs to S.
Note that for every A ∈S there is a unique B ∈B[1, 10) such that A = S k∈Z 10kB, and (2.3) shows that in fact B = S(A).
Example 2.10. The set A4 of positive numbers with A4 = {10k : k ∈Z} = {. . . , 0.01, 0.1, 1, 10, 100, . . .} belongs to S. This can be seen either by observing that A4 is the set of positive reals with significand exactly equal to 1, i.e. A4 = R+ ∩S−1({1}), or by noting that A4 = {x > 0 : D1(x) = 1, Dm(x) = 0 for all m ≥2}, or by using (2.4) and the fact that A4 = S k∈Z 10k{1} and {1} ∈B[1, 10).
12 A. Berger and T.P. Hill Example 2.11. The singleton set {1} and the interval [1, 2] do not belong to S, since the number 1 cannot be distinguished from the number 10, for instance, using only significant digits. Nor can the interval [1, 2] be distinguished from [10, 20]. Formally, neither of these sets is of the form S k∈Z 10kB for any B ∈ B[1, 10).
Although the significand function and σ-algebra above were defined in the setting of real numbers, the same concepts carry over immediately to the most fundamental setting of all, the set of positive integers. In this case, the induced σ-algebra is interesting in its own right.
Example 2.12. The restriction SN of S to subsets of N, i.e. SN = {N∩A : A ∈S} is a σ-algebra on N. A characterization of SN analogous to that of S given in Theorem 2.9 is as follows: Denote by N✚ 10 the set of all positive integers not divisible by 10, i.e. N✚ 10 = N\10N. Then SN = n A ⊂N : A = [ l∈N0 10lB for some B ⊂N✚ 10 o .
A typical member of SN is {271, 2710, 3141, 27100, 31410, 271000, 314100, . . .} .
Note that for instance the set {31410, 314100, 3141000, . . .} does not belong to SN since 31410 is indistinguishable from 3141 in terms of significant digits, so if the former number were to belong to A ∈SN then the latter would too. Note also that the corresponding significand function on N still only takes values in [1, 10), as before, but may never be an irrational number. In fact, the possible values of S on N are even more restricted: S(n) = t for some n ∈N if and only if t ∈[1, 10) and 10lt ∈N for some integer l ≥0.
The next lemma establishes some basic closure properties of the significand σ-algebra that will be essential later in studying characteristic aspects of BL such as scale- and base-invariance. To concisely formulate these properties, for every C ⊂R+ and n ∈N, let C1/n := {t > 0 : tn ∈C}.
Lemma 2.13. The following properties hold for the significand σ-algebra S: (i) S is self-similar with respect to multiplication by integer powers of 10, i.e.
10kA = A for every A ∈S and k ∈Z .
(ii) S is closed under multiplication by a scalar, i.e.
αA ∈S for every A ∈S and α > 0 .
(iii) S is closed under integral roots, i.e.
A1/n ∈S for every A ∈S and n ∈N .
A basic theory of Benford’s Law 13 Informally, property (i) says that every significand set remains unchanged when multiplied by an integer power of 10 — reflecting the simple fact that shifting the decimal point keeps all the significant digits, and hence the set itself, unchanged; (ii) asserts that if every element of a set expressible solely in terms of significant digits is multiplied by a positive constant, then the new set is also expressible by significant digits; correspondingly, (iii) states that the collection of square (cubic, fourth etc.) roots of the elements of every significand set is also expressible in terms of its significant digits alone.
Proof. (i) This is obvious from (2.3) since S(10kA) = S(A) for every k.
(ii) Given A ∈S, by (2.4) there exists B ∈B[1, 10) such that A = S k∈Z 10kB.
In view of (i), assume without loss of generality that 1 < α < 10. Then αA = [ k∈Z 10kαB = [ k∈Z 10k αB∩[α, 10) ∪ α 10B∩[1, α) = [ k∈Z 10kC , with C = αB ∩[α, 10) ∪ α 10B ∩[1, α) ∈B[1, 10), showing that αA ∈S.
(iii) Since intervals of the form [1, t] generate B[1, 10), i.e. since B[1, 10) = σ {[1, t] : 1 < t < 10} , it is enough to verify the claim for the special case A = S k∈Z 10k[1, 10s] for every 0 < s < 1. In this case A1/n = [ k∈Z10k/n[1, 10s/n] = [ k∈Z10k [n−1 j=0 [10j/n, 10(j+s)/n] = [ k∈Z10kC , with C = Sn−1 j=0 [10j/n, 10(j+s)/n] ∈B[1, 10). Hence A1/n ∈S.
Remark. Note that S is not closed under taking integer powers: If A ∈S and n ∈N, then An ∈S if and only if S(A)n = B ∪10B ∪. . . ∪10n−1B for some B ∈B[1, 10) .
For example, consider A5 = [ k∈Z 10k{1, √ 10} ∈S , for which S(A5)2 = {1, 10} = {1} ∪10{1} and hence A2 5 ∈S, whereas choosing A6 = [ k∈Z 10k{2, √ 10} leads to S(A6)2 = {4, 10}, and correspondingly A2 6 ̸∈S.
♣ Since, by Theorem 2.9, the significand σ-algebra S is the same as the sig-nificant digit σ-algebra σ(D1, D2, D3, . . .), the closure properties established in Lemma 2.13 carry over to sets determined by significant digits. The next exam-ple illustrates closure under multiplication by a scalar and integral roots.
Example 2.14. Let A7 be the set of positive real numbers with first significant digit 1, i.e.
A7 = {x > 0 : D1(x) = 1} = {x > 0 : 1 ≤S(x) < 2} = [ k∈Z 10k[1, 2) .
14 A. Berger and T.P. Hill 0 0 0 0 .1 .01 .2 .2 .4 1 1 2 2 4 4 10 10 20 20 20 1 √ 2 1 √ 10 1 √ 5 √ 10 2 √ 5 10 10 √ 2 R+ R+ R+ R+ A7 = {D1 = 1} = {1 ≤S < 2} ∈S 2A7 ∈S √A7 ∈S A2 7 ̸∈S Fig 3. The σ-algebra S is closed under multiplication by a scalar and integral roots but not under integer powers (bottom), see Example 2.14.
Then 2A7 = x > 0 : D1(x) ∈{2, 3} = {x > 0 : 2 ≤S(x) < 3} = [ k∈Z10k[2, 4) ∈S , and also A1/2 7 = x > 0 : S(x)∈[1, √ 2)∪[ √ 10, √ 20) = [ k∈Z10k [1, √ 2)∪[ √ 10, 2 √ 5) ∈S , whereas on the other hand clearly A2 7 = [ k∈Z102k[1, 4) ̸∈S , since e.g. [1, 4) ⊂A2 7 but [10, 40) ̸⊂A2 7; see Fig 3.
Example 2.15. Recall the significand σ-algebra SN on the positive integers defined in Example 2.12. Unlike its continuous counterpart S, the family SN is not even closed under multiplication by a positive integer, since for example A8 = N ∩{x > 0 : S(x) = 2} = {2, 20, 200, . . .} ∈SN , A basic theory of Benford’s Law 15 but 5A8 = {10, 100, 1000, . . .} ̸∈SN .
Of course, this does not rule out that some events determined by significant digits, i.e. some members of SN, still belong to SN after multiplication by an integer. For example, if A9 = {n ∈N : D1(n) = 1} = {1, 10, 11, . . ., 19, 100, 101, . . .} ∈SN then 3A9 = {3, 30, 33, . . ., 57, 300, 303, . . .} ∈SN .
It is easy to see that, more generally, SN is closed under multiplication by m ∈N precisely if gcd (m, 10) = 1, that is, whenever m and 10 have no non-trivial common factor. Moreover, like S, the σ-algebra SN is closed under integral roots: If A = S l∈N0 10l b A with b A ⊂N✚ 10 then A1/n = S l∈N0 10l b A1/n ∈SN. With A9 from above, for instance, A1/2 9 = n ∈N : S(n) ∈[1, √ 2) ∪[ √ 10, √ 20) = {1, 4, 10, 11, 12, 13, 14, 32, 33, . . ., 44, 100, 101, . . .} ∈SN .
Thus many of the conclusions drawn later for positive real numbers carry over to positive integers in a straightforward way.
The next lemma provides a very convenient framework for studying probabil-ities on the significand σ-algebra by translating them into probability measures on the classical space of Borel subsets of [0, 1), that is, on [0, 1), B[0, 1) . For a proper formulation, observe that for every function f : Ω→R with A ⊃σ(f) and every probability measure P on (Ω, A), f and P together induce a probability measure f∗P on (R, B) in a natural way, namely by setting f∗P(B) = P f −1(B) for all B ∈B .
(2.5) Other symbols commonly used in textbooks to denote f∗P include P ◦f −1 and Pf. In the case of a linear function f, i.e. for f(t) ≡αt with some α ∈R, instead of f∗P simply write α∗P. The special case of interest for significands is (Ω, A) = (R+, S) and f = log S.
Lemma 2.16. The function ℓ: R+ →[0, 1) defined by ℓ(x) = log S(x) es-tablishes a one-to-one and onto correspondence (measure isomorphism) between probability measures on (R+, S) and on [0, 1), B[0, 1) , respectively.
Proof. From ℓ−1 [a, b] = S−1 [10a, 10b] for all 0 ≤a < b < 1, it follows that σ(ℓ) = R+ ∩σ(S) = S, and hence ℓ∗P according to (2.5) is a well-defined probability measure on [0, 1), B[0, 1) .
Conversely, given any probability measure P on [0, 1), B[0, 1) and any A in S, let B ∈B[0, 1) be the unique set for which A = S k∈Z 10k10B, where 10B = {10s : s ∈B}, and define PP (A) := P(B) .
16 A. Berger and T.P. Hill It is readily confirmed that ℓ(A) = B, ℓ−1(B) = A, and PP is a well-defined probability measure on (R+, S). Moreover ℓ∗PP (B) = PP ℓ−1(B) = PP (A) = P(B) for all B ∈B[0, 1) , showing that ℓ∗PP = P, and hence every probability measure on [0, 1), B[0, 1) is of the form ℓ∗P with the appropriate P. On the other hand, Pℓ∗P(A) = ℓ∗P(B) = P ℓ−1(B) = P(A) for all A ∈S , and hence the correspondence P 7→ℓ∗P is one-to-one as well. Overall P ↔ℓ∗P is bijective.
From the proof of Lemma 2.16 it is clear that a bijective correspondence between probability measures on (R+, S) and on [0, 1), B[0, 1) , respectively, could have been established in many other ways as well, e.g. by using the func-tion e ℓ(x) = 1 9(S(x) −1) instead of ℓ. The special role of ℓaccording to that lemma only becomes apparent through its relation to BL. To see this, denote by B the (unique) probability measure on (R+, S) with B {x > 0 : S(x) ≤t} = B [ k∈Z 10k[1, t] = log t for all 1 ≤t < 10 .
In view of (2.1), the probability measure B on (R+, S) is the most natural for-malization of BL. On the other hand, it will become clear in subsequent chapters that on [0, 1), B[0, 1) the uniform distribution λ0,1 has many special properties and hence plays a very distinguished role. The relevance of the specific choice for ℓin Lemma 2.16, therefore, is that ℓ∗B = λ0,1. The reader will learn shortly why, for a deeper understanding of BL, the latter relation is very beneficial indeed.
3. The Benford property In order to translate the informal versions (1.1), (1.2) and (2.1) of BL into more precise statements about various types of mathematical objects, it is necessary to specify exactly what the Benford property means for any one of these objects.
For the purpose of the present chapter, the objects of interest fall into three categories: sequences of real numbers, real-valued functions defined on [0, +∞); and probability distributions and random variables. (Recall also Fig 1.) 3.1. Benford sequences A sequence (xn) = (x1, x2, x3, . . .) of real numbers is a (base-10) Benford se-quence, or simply Benford, if, as N →∞, the limiting proportion of indices n ≤N for which xn has first significant digit d1 exists and equals log(1 + d−1 1 ) for all d1 ∈{1, 2, . . ., 9}, and similarly for the limiting proportions of the occur-rences of all other finite blocks of initial significant digits. The formal definition is as follows.
A basic theory of Benford’s Law 17 Definition 3.1. A sequence (xn) of real numbers is a Benford sequence, or Benford for short, if limN→∞ #{1 ≤n ≤N : S(xn) ≤t} N = log t for all t ∈[1, 10) , or equivalently, if for all m ∈N, all d1 ∈{1, 2, . . ., 9} and all dj ∈{0, 1, . . ., 9}, j ≥2, limN→∞ # 1 ≤n ≤N : Dj(xn) = dj for j = 1, 2, . . . , m N = log 1+ Xm j=1 10m−jdj −1 .
As will be shown below, the sequence of powers of 2, namely (2n)=(2, 4, 8, . . .) is Benford. However, it is not Benford base 2 since the second significant digit base 2 of 2n is 0 for every n, whereas the generalized version (1.3) of BL requires that 0 < Prob D(2) 2 = 0 = 1 −Prob D(2) 2 = 1 = log2 3 −1 < 1. Similarly, (3n), the sequence of powers of 3 is Benford, and so is the sequence of factorials (n!) as well as the sequence (Fn) of Fibonacci numbers. Simple examples of sequences that are not Benford are the positive integers (n), the powers of 10 and the sequence of logarithms (log n).
The notion of Benford sequence according to Definition 3.1 offers a natural interpretation of Prob in the informal expressions (1.1)–(1.3): A sequence (xn) is Benford if, when one of the first N entries in (xn) is chosen (uniformly) at random, the probability that this entry’s first significant digit is d approaches the Benford probability log(1 + d−1) as N →∞, for every d ∈{1, 2, . . ., 9}, and similarly for all other blocks of significant digits.
Example 3.2. Two specific sequences of positive integers will be used repeat-edly to illustrate key concepts concerning BL: the Fibonacci numbers and the prime numbers. Both sequences play prominent roles in many areas of mathe-matics.
(i) As will be seen in Example 4.12, the sequences of Fibonacci numbers (Fn) = (1, 1, 2, 3, 5, 8, 13, . . .), where every entry is simply the sum of its two predecessors, and F1 = F2 = 1, is Benford. Already the first N = 102 elements of the sequence conform very well to the first-digit version (1.1) of BL, with Prob being interpreted as relative frequency, see Fig 4. The conformance gets even better if the first N = 104 elements are considered, see Fig 5.
(ii) In Example 4.11(v), it will become apparent that the sequence of prime numbers (pn) = (2, 3, 5, 7, 11, 13, 17, . . .) is not Benford. Fig 4 shows how, ac-cordingly, the first hundred prime numbers do not conform well to (1.1). More-over, the conformance gets even worse if the first ten thousand primes are con-sidered (Fig 5).
18 A. Berger and T.P. Hill 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1346269 2178309 3524578 5702887 9227465 14930352 24157817 39088169 63245986 165580141 165580141 267914296 433494437 701408733 1134903170 1836311903 2971215073 4807526976 7778742049 12586269025 20365011074 32951280099 53316291173 86267571272 139583862445 225851433717 365435296162 591286729879 956722026041 1548008755920 2504730781961 4052739537881 6557470319842 10610209857723 17167680177565 27777890035288 44945570212853 72723460248141 117669030460994 190392490709135 308061521170129 498454011879264 806515533049393 1304969544928657 2111485077978050 3416454622906707 5527939700884757 8944394323791464 14472334024676221 23416728348467685 37889062373143906 61305790721611591 99194853094755497 160500643816367088 259695496911122585 420196140727489673 679891637638612258 1100087778366101931 1779979416004714189 2880067194370816120 4660046610375530309 7540113804746346429 12200160415121876738 19740274219868223167 31940434634990099905 51680708854858323072 83621143489848422977 135301852344706746049 218922995834555169026 354224848179261915075 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 1 2 3 4 5 6 7 8 9 30 18 13 9 8 6 5 7 4 25 19 19 20 8 2 4 2 1 30.10 17.60 12.49 9.691 7.918 6.694 5.799 5.115 4.575 Fibonacci numbers F1, F2, . . . , F100 Prime numbers p1, p2, . . . , p100 #{D1 =d} Fibonacci Prime 102·log(1+d−1) d Fig 4. The first one-hundred Fibonacci numbers conform to the first digit law (1.1) quite well (top and bottom), while the first one-hundred prime numbers clearly do not (center and bottom).
Remark. Based on discrete density and summability definitions, many alterna-tive notions of Benford sequences have been proposed, utilizing e.g. reiteration of Ces` aro averaging [Fl], and logarithmic density methods. The reader is referred to [Ra1, Ra2] for an excellent summary of those approaches. Those methods, however, do not offer as natural an interpretation of “Prob” as Definition 3.1.
On this, Raimi [Ra1, p.529] remarks that “[t]he discrete summability schemes [. . . ] can also be tortured into probability interpretations, though none of the authors [. . . ] (except Diaconis) does so”.
A basic theory of Benford’s Law 19 1 2 3 4 5 6 7 8 9 30 18 13 9 8 6 5 7 4 3011 1762 1250 968 792 668 580 513 456 25 19 19 20 8 2 4 2 1 1601 1129 1097 1069 1055 1013 1027 1003 1006 103 ·R 18.84 0.1574(!) 103.0 140.9 Fibonacci Prime d #{D1 =d} = N ·ρN(d) N =102 N =102 N =104 N =104 104 ·log(1+d−1) 3010. 1760. 1249. 969.1 791.8 669.4 579.9 511.5 457.5 Fig 5. Increasing the sample size from N = 102 to N = 104 entails an even better conformance with (1.1) for the Fibonacci numbers, as measured by means of the quantity R = max9 d=1 ρN(d) −log(1 + d−1) . For the primes, on the other hand, the rather poor con-formance does not improve at all.
Only the notion according to Definition 3.1 will be used henceforth. How-ever, to get an impression how alternative concepts may relate to Definition 3.1 analytically, denote, for any set C ⊂R, by 1C the indicator function of C, i.e.
1C : R →{0, 1} with 1C(t) = 1 if t ∈C , 0 otherwise.
Given any sequence (xn) and any number t ∈[1, 10), consider the sequence 1[1,t)(S(xn)) . Clearly, since the latter contains only zeros and ones, it will usually not converge. It may, however, converge in some averaged sense, to a limit that may depend on t. Specifically, (xn) is Benford if and only if limN→∞ PN n=1 1[1,t) S(xn) N = log t for all t ∈[1, 10) .
(3.1) Instead of (3.1), one could more generally consider the convergence of XN n=1 an1[1,t) S(xn) XN n=1 an , (3.2) where the an can be virtually any non-negative numbers with PN n=1 an →+∞ as N →∞. With this, (3.1) corresponds to the special case an = 1 for all n. Another popular choice in (3.2), related to the number-theoretic concept of logarithmic (or analytic) density [Se], is an = n−1 for all n, in which case (ln N)−1 PN n=1 an →1. Utilizing the latter, a sequence (xn) of real numbers 20 A. Berger and T.P. Hill might be (and has been, see [Ra1]) called weakly Benford if limN→∞ 1 ln N XN n=1 1[1,t) S(xn) n = log t for all t ∈[1, 10) .
It is easy to check that every Benford sequence is weakly Benford. To see that the converse does not hold in general, take for instance (xn) = (n). A short calculation confirms that, for every t ∈[1, 10), lim infN→∞ PN n=1 1[1,t) S(n) N = t −1 9 , whereas lim supN→∞ PN n=1 1[1,t) S(n) N = 10 9 · t −1 t , showing that (n) is not Benford. (Recall that the limit inferior and limit superior of a sequence (an), denoted by lim infn→∞an and lim supn→∞an, are defined, respectively, as the smallest and largest accumulation value of (an).) On the other hand, (n) turns out to be weakly Benford: Indeed, given N, let LN := ⌊log N⌋. For any t ∈[1, 10), it follows from the elementary estimate 1 ln 10LN+1 XLN−1 i=0 X⌊10it⌋ j=10i 1 j ≤ 1 ln N XN n=1 1[1,t) S(n) n ≤ 1 ln⌊10LNt⌋ XLN i=0 X⌊10it⌋ j=10i 1 j , together with X⌊10it⌋ j=10i 1 j = 10−i X⌊10it⌋−10i j=0 1 1 + 10−ij → Z t−1 0 dτ 1 + τ = ln t , as i →∞, as well as limL→∞ ln 10L+1 L = limL→∞ ln⌊10Lt⌋ L = ln 10 and the Cauchy Limit Theorem that limN→∞ 1 ln N XN n=1 1[1,t) S(n) n = ln t ln 10 = log t , i.e., (n) is weakly Benford. In a similar manner, the sequence (pn) can be shown to be weakly Benford without being Benford, see [GG, Wh].
♣ 3.2. Benford functions BL also appears frequently in real-valued functions such as e.g. those arising as solutions of initial value problems for differential equations (see Section 5.3 below). Thus, the starting point is to define what it means for a function to follow BL.
A basic theory of Benford’s Law 21 Recall that a function f : R →R is (Borel) measurable if f −1(I) is a Borel set, i.e. f −1(I) ∈B, for every interval I ⊂R. With the terminology introduced in Section 2.3, this is equivalent to saying that σ(f) ⊂B. Slightly more gener-ally, for any set Ωand any σ-algebra A on Ω, a function f : Ω→R is (Borel) measurable if σ(f) ⊂A. The collection of Borel measurable functions f : R →R contains all functions of practical interest. For example, every piecewise contin-uous function (meaning that f has at most countably many discontinuities) is measurable. Thus every polynomial, trigonometric and exponential function is measurable, and so is every probability density function of any relevance. In fact, it is a difficult exercise to produce a function that is not measurable, or a set C ⊂R that is not a member of B, and this can be done only in a non-constructive way. For all practical purposes, therefore, the reader may simply read “set” for “Borel set”, and “function” for “Borel measurable function”.
Recall that given a set Ωand a σ-algebra A on Ω, a measure µ on (Ω, A) is a function µ : A →[0, +∞] that has all the properties of a probability measure, except that µ(A) may also be bigger than 1, and even infinite. By far the most important example is the so-called Lebesgue measure on (R, B), denoted by λ here and throughout. The basic, and in fact defining property of λ is that λ [a, b] = b −a for every interval [a, b] ⊂R. The relation between the measure λ and the probability measures λa,b considered earlier is such that, for instance, λ(B) = limN→∞2Nλ−N,N B ∩[−N, N] for every B ∈B .
It is customary to also use the symbol λ, often without a subscript etc., to denote the restriction of Lebesgue measure to C, B(C) with the Borel set C being clear from the context.
In analogy to the terminology for sequences, a function f is a (base-10) Ben-ford function, or simply Benford, if the limiting proportion of the time τ < T that the first digit of f(τ) equals d1 is exactly log(1 + d−1 1 ), and similarly for the other significant digits, and in fact the significand. The formal definition is as follows.
Definition 3.3. A (Borel measurable) function f : [0, +∞) →R is Benford if limT →+∞ λ τ ∈[0, T ) : S f(τ) ≤t T = log t for all t ∈[1, 10) , or equivalently, if for all m ∈N, all d1 ∈{1, 2, . . ., 9} and all dj ∈{0, 1, . . ., 9}, j ≥2, limT →+∞ λ τ ∈[0, T ) : Dj f(τ) = dj for j = 1, 2, . . . , m T = log 1+ Xm j=110m−jdj −1 .
Directly analogous to the probabilistic interpretation of a Benford sequence, the definition of a Benford function given in Definition 3.3 also offers a natural probabilistic interpretation: A function f : [0, +∞) →R is Benford if, when a 22 A. Berger and T.P. Hill time τ is chosen (uniformly) at random in [0, T ), the probability that the first digit of f(τ) is d approaches log(1+d−1) as T →+∞, for every d ∈{1, 2, . . ., 9}, and similarly for all other blocks of significant digits.
As will be seen in Example 4.5 below, the function f(t) = eαt is Benford whenever α ̸= 0, but f(t) = t and f(t) = sin2 t, for instance, are not.
3.3. Benford distributions and random variables BL appears prominently in a wide variety of statistics and probability settings, such as e.g. in products of independent, identically distributed random variables, mixtures of random samples, and stochastic models like geometric Brownian motion that are of great importance for the stochastic modelling of real-world processes. This section lays the foundations for analyzing the Benford prop-erty for probability distributions and random variables. The term independent, identically distributed will henceforth be abbreviated i.i.d., in accordance with standard stochastic terminology.
Recall from Section 2.3 that a probability space is a triple (Ω, A, P) where Ω is a set, often referred to as the set of outcomes, A is a σ-algebra (the family of events), and P is a probability measure. A (real-valued) random variable X on (Ω, A, P) is simply a Borel measurable function X : Ω→R, and its distribution PX is the probability measure on (R, B) defined by PX (−∞, t] = P(X ≤t) for all t ∈R .
Thus with the notation introduced in (2.5), simply PX = X∗P. The expectation, or expected (or mean) value of X is EX = Z Ω X dP = Z R t dPX(t) , provided that this integral exists. More generally, for every measurable function g : R →R, the expectation of the random variable g(X) is Eg(X) = Z Ω g(X) dP = Z R g(t) dPX(t) .
In particular, if EX exists, then var X := E(X −EX)2 is the variance of X.
Any probability measure on (R, B) will be referred to as a Borel probability measure on R. Again, since all subsets of R of any practical interest are Borel sets, the specifier “Borel” will be suppressed unless there is a potential for confu-sion, i.e., the reader may read “probability measure on R” for “Borel probability measure on R”. Any probability measure P on R is uniquely determined by its distribution function FP , defined as FP (t) = P (−∞, t] for all t ∈R .
It is easy to check that the function FP is right-continuous and non-decreasing, with limt→−∞FP (t) = 0 and limt→+∞FP (t) = 1. For the sake of notational A basic theory of Benford’s Law 23 simplicity, write FX instead of FPX for every random variable X. The probability measure P, or any random variable X with PX = P, is continuous (or atomless) if P({t}) = 0 for every t ∈R, or equivalently if the function FP is continuous.
It is absolutely continuous (a.c.) if, for any B ∈B, P(B) = 0 holds whenever λ(B) = 0. By the Radon–Nikodym Theorem, this is equivalent to P having a density, i.e. to the existence of a measurable function fP : R →[0, +∞) such that P [a, b] = Z b a fP (t) dt for all [a, b] ⊂R .
(3.3) Again, for simplicity write fX instead of fPX for every a.c. random variable X.
Note that (3.3) implies R +∞ −∞fP (t) dt = 1. Every a.c. probability measure on (R, B) is continuous but not vice versa, see e.g. [CT]. Given any probability P on (R, B), denote | · |∗P simply by |P|, that is, |P|(B) = P {t ∈R : |t| ∈B} for all B ∈B .
Clearly, |P| is concentrated on [0, +∞), i.e. |P| [0, +∞) = 1, and F|P |(t) = ( 0 if t < 0 , FP (t) −FP (−t) + P({−t}) if t ≥0 ; in particular, therefore, if P is continuous or a.c. then so is |P|, its density in the latter case being fP (t) + fP (−t) · 1[0,+∞), where fP is the density of P.
Definition 3.4. A Borel probability measure P on R is Benford if P {x ∈R : S(x) ≤t} = S∗P {0} ∪[1, t] = log t for all t ∈[1, 10) .
A random variable X on a probability space (Ω, A, P) is Benford if PX is Ben-ford, i.e. if P S(X) ≤t = PX {x ∈R : S(x) ≤t} = log t for all t ∈[1, 10) , or equivalently, if for all m ∈N, all d1 ∈{1, 2, . . ., 9} and all dj ∈{0, 1, . . ., 9}, j ≥2, P Dj(X) = dj for j = 1, 2, . . . , m = log 1 + Xm j=1 10m−jdj −1 .
Example 3.5. If X is a Benford random variable on some probability space (Ω, A, P), then from (1.1) and the numerical values given in Chapter 1, P(D1(X) = 1) = P(1 ≤S(X) < 2) = log 2 = 0.3010 . . . , P(D1(X) = 9) = log 10 9 = 0.04575 . . . , P D1(X), D2(X), D3(X) = (3, 1, 4) = log 315 314 = 0.001380 . . . .
24 A. Berger and T.P. Hill 1 1 1 1 1 1 1 0 0 0 0 0 0 10 10 0.5 0.5 0.2 1 ln 10 t t s s log t s P(S(X) < t) fS(X) flog S(X) P(log S(X) < s) 1[0,1) 1[1,10) t ln 10 Fig 6. The distribution functions (top) and densities of S(X) and log S(X), respectively, for a Benford random variable X.
As the following example shows, there are many probability measures on the positive real numbers, and correspondingly many positive random variables that are Benford.
Example 3.6. For every integer k, the probability measure Pk with density fk(x) = 1 x ln 10 on [10k, 10k+1) is Benford, and so is e.g. 1 2(Pk +Pk+1). In fact, ev-ery convex combination of the (Pk)k∈Z, i.e. every probability measure P k∈Z qkPk with 0 ≤qk ≤1 for all k and P k∈Z qk = 1, is Benford.
As will be seen in Example 6.4 below, if U is a random variable uniformly distributed on [0, 1), then the random variable X = 10U is Benford, but the random variable Xlog 2 = 2U is not.
Definition 3.7. The Benford distribution B is the unique probability measure on (R+, S) with B(S ≤t) = B [ k∈Z 10k[1, t] = log t for all t ∈[1, 10) , or equivalently, for all m ∈N, all d1 ∈{1, 2, . . ., 9} and all dj ∈{0, 1, . . ., 9}, j ≥2, B Dj = dj for j = 1, 2, . . . , m = log 1 + Xm j=1 10m−jdj −1 .
The combination of Definitions 3.4 and 3.7 gives A basic theory of Benford’s Law 25 Proposition 3.8. A Borel probability measure P on R is Benford if and only if |P|(A) = B(A) for all A ∈S .
In particular, if P(R+) = 1 then P is Benford precisely if P(A) = B(A) for all A ∈S.
Note that the Benford distribution B is a probability distribution on the significant digits, or the significand, of the underlying data, and not on the raw data themselves. That is, B is a probability measure on the family of sets defined by the base-10 significand, i.e. on (R+, S), but not on the bigger (R+, B+) or the still bigger (R, B). For example, the probability B({1}) is not defined, simply because the set {1} cannot be defined in terms of significant digits or significands alone, and hence does not belong to the domain of B.
Example 3.9. In the framework of the Examples 2.12 and 2.15, it is tempting to call a probability P on (N, SN) a Benford distribution on N if P {n ∈N : S(n) ≤t} = log t for all t ∈[1, 10) .
However, no such probability exists! To see this, for every n ∈N✚ 10 let An = S l∈N0 10l{n} ∈SN and note that N equals the disjoint union of the sets An, and S(An) = {10⟨log n⟩}; here ⟨log n⟩∈[0, 1) denotes the fractional part of log n, that is, ⟨log n⟩= log n −⌊log n⌋. With qn := P(An) therefore P n∈N ✚ 10 qn = 1 and S∗P = P n∈N ✚ 10 qnδ10⟨log n⟩. Since the set of discontinuities of t 7→FS∗P (t) is 10⟨log n⟩: qn ̸= 0 ̸= ∅, it is impossible to have FS∗P (t) = log t for all t ∈[1, 10). Note that, as a consequence, a Borel probability measure P on R concentrated on N, i.e. with P(N) = 1, cannot be Benford.
On the other hand, given ε > 0 it is not hard to find a probability Pε on (N, SN) with Pε {n ∈N : S(n) ≤t} −log t < ε for all t ∈[1, 10) .
(3.4) For a concrete example, for any N ∈N consider the probability measure QN := c−1 N X10N+1−1 j=10N j−1δj , where cN = P10N+1−1 j=10N j−1. Note that QN may be thought of as a discrete approximation of the Benford probability PN in Example 3.6. From S∗QN = c−1 N X10N+1−1 j=10N j−1δS(j) = c−1 N X10N+1−10N j=1 1 10N + j −1δ1+10−N (j−1) , together with the elementary estimate ln M+1 L < PM j=L j−1 < ln M L−1, valid for all L, M ∈N with 2 ≤L < M, it is straightforward to deduce that, for all 1 ≤t < 10, S∗QN [1, t] −log t < −log(1 −10−N) = 10−N ln 10 + O(10−2N) as N →∞.
26 A. Berger and T.P. Hill Thus (3.4) is guaranteed by taking Pε = QN with N sufficiently large. A short calculation confirms that it suffices to choose N > 1 + | log ε|.
Example 3.10. (i) If X is distributed according to U(0, 1), the uniform dis-tribution on [0, 1), i.e. PX = λ0,1, then for every 1 ≤t < 10, P S(X) ≤t = λ0,1 [ k∈Z 10k[1, t] = X n∈N 10−n(t −1) = t −1 9 ̸≡log t , showing that S(X) is uniform on [1, 10), and hence λ0,1 is not Benford.
(ii) If X is distributed according to exp(1), the exponential distribution with mean 1, whose distribution function is given by Fexp(1)(t) = max(0, 1 −e−t), then P(D1(X) = 1) = P X ∈ [ k∈Z 10k[1, 2) = X k∈Z e−10k −e−2·10k > e−1/10 −e−2/10 + e−1 −e−2 + e−10 −e−20 = 0.3186 . . . > log 2 , and hence exp(1) is not Benford either. (See [EL, LSE, MN] for a detailed analysis of the exponential distribution’s relation to BL.) (iii) Let X be distributed according to the Beta 1 2, 1 2 - or arcsin-distribution, meaning that P(X ≤s) = 2 π arcsin √s for all 0 ≤s < 1. It follows that, for every 1 ≤t < 10, FS(X)(t) = P(S(X) ≤t) = P X ∈ [ n∈N 10−n[1, t] = 2 π X∞ n=1 arcsin(10−n/2√ t) −arcsin(10−n/2) = 2 π X∞ l=0 (2l)!
22l(l!)2(2l + 1) · tl+1/2 −1 10l+1/2 −1 , and hence in particular FS(X)( √ 10) = 2 π X∞ l=0 (2l)!
22l(l!)2(2l + 1) · 1 10l/2+1/4 + 1 < 2 π X∞ l=0 (2l)!
22l(l!)2(2l + 1)10−(l/2+1/4) = 2 π arcsin(10−1/4) = 0.3801 . . . < 2 5 , which in turn shows that X is not Benford, as FB( √ 10) = 1 2. Alternatively, FS(X) is easily seen to be strictly convex on [1, 10) and therefore FS(X)(t) ≡log t cannot possibly hold.
A basic theory of Benford’s Law 27 4. Characterizations of Benford’s Law The purpose of this chapter is to establish and illustrate four useful charac-terizations of the Benford property in the context of sequences, functions, dis-tributions and random variables, respectively. These characterizations will be instrumental in demonstrating that certain datasets are, or are not, Benford, and helpful for predicting which empirical data are likely to follow BL closely.
4.1. The uniform distribution characterization The uniform distribution characterization is undoubtedly the most basic and powerful of all characterizations, mainly because the mathematical theory of uniform distribution mod 1 is very well developed, see e.g. [DT, KN] for author-itative surveys of the theory.
Here and throughout, denote by ⟨t⟩the fractional part of any real number t, that is ⟨t⟩= t−⌊t⌋. For example, ⟨π⟩= ⟨3.1415 . . .⟩= 0.1415 . . . = π −3. Recall that λ0,1 denotes Lebesgue measure on [0, 1), B[0, 1) .
Definition 4.1. A sequence (xn) of real numbers is uniformly distributed mod-ulo 1, abbreviated henceforth as u.d. mod 1, if limN→∞ #{1 ≤n ≤N : ⟨xn⟩≤s} N = s for all s ∈[0, 1) ; a (Borel measurable) function f : [0, +∞) →R is u.d. mod 1 if limT →+∞ λ{τ ∈[0, T ) : ⟨f(τ)⟩≤s} T = s for all s ∈[0, 1) ; a random variable X on a probability space (Ω, A, P) is u.d. mod 1 if P(⟨X⟩≤s) = s for all s ∈[0, 1) ; and a probability measure P on (R, B) is u.d. mod 1 if P({x : ⟨x⟩≤s}) = P [ k∈Z[k, k + s] = s for all s ∈[0, 1) .
The next simple theorem (cf. [Di]) is one of the main tools in the theory of BL because it allows application of the powerful theory of uniform distribution mod 1. (Recall the convention log 0 := 0.) Theorem 4.2 (Uniform distribution characterization). A sequence of real num-bers (respectively, a Borel measurable function, a random variable, a Borel prob-ability measure) is Benford if and only if the decimal logarithm of its absolute value is uniformly distributed modulo 1.
28 A. Berger and T.P. Hill Proof. Let X be a random variable and, without loss of generality, assume that P(X = 0) = 0. Then, for all s ∈[0, 1), P(⟨log |X|⟩≤s) = P log |X| ∈ [ k∈Z[k, k + s] = P |X| ∈ [ k∈Z[10k, 10k+s] = P(S(X) ≤10s) .
Hence, by Definitions 3.4 and 4.1, X is Benford if and only if P(S(X) ≤10s) = log 10s = s for all s ∈[0, 1), i.e., if and only if log |X| is u.d. mod 1.
The proofs for sequences, functions, and probability distributions are com-pletely analogous.
Next, several tools from the basic theory of uniform distribution mod 1 will be recorded that will be useful, via Theorem 4.2, in establishing the Benford property for many sequences, functions, and random variables.
Lemma 4.3.
(i) The sequence (xn) is u.d. mod 1 if and only if the sequence (kxn +b) is u.d. mod 1 for every k ∈Z{0} and every b ∈R. Also, (xn) is u.d. mod 1 if and only if (yn) is u.d. mod 1 whenever limn→∞|yn−xn| = 0.
(ii) The function f is u.d. mod 1 if and only if t 7→kf(t) + b is u.d. mod 1 for every non-zero integer k and every b ∈R.
(iii) The random variable X is u.d. mod 1 if and only if kX + b is u.d. mod 1 for every non-zero integer k and every b ∈R.
Proof. (i) The “if” part is obvious with k = 1, b = 0. For the “only if” part, assume that (xn) is u.d. mod 1. Note first that limN→∞ #{1 ≤n ≤N : ⟨xn⟩∈C} N = λ0,1(C) holds whenever C is a finite union of intervals. Let k ∈Z be non-zero and observe that, for any 0 < s < 1, x : ⟨kx⟩≤s = n x : ⟨x⟩∈Sk−1 j=0 h j k, j+s k io if k > 0 , n x : ⟨x⟩∈S|k|−1 j=0 h j+1−s |k| , j+1 |k| io if k < 0 .
Consequently, limN→∞ #{1 ≤n ≤N : ⟨kxn⟩≤s} N = λ0,1 Sk−1 j=0 h j k, j+s k i if k > 0 , λ0,1 S|k|−1 j=0 h j+1−s |k| , j+1 |k| i if k < 0 , = ( k · s k if k > 0 |k| · s |k| if k < 0 = s , A basic theory of Benford’s Law 29 showing that (kxn) is u.d. mod 1. Similarly, note that, for any b, s ∈(0, 1), x : ⟨x + b⟩≤s = ( {x : ⟨x⟩∈[0, s −b] ∪[1 −b, 1)} if s ≥b , {x : ⟨x⟩∈[1 −b, 1 + s −b]} if s < b .
Thus, assuming without loss of generality that 0 < b < 1, limN→∞ #{1 ≤n ≤N : ⟨xn + b⟩≤s} N = ( λ0,1 [0, s −b] ∪[1 −b, 1) if s ≥b λ0,1 [1 −b, 1 + s −b] if s < b = s , and hence (xn + b) is also u.d. mod 1. The second assertion is clear from the definition.
The proofs of (ii) and (iii) are completely analogous.
Example 4.4. (i) The sequence (nπ) = (π, 2π, 3π, . . .) is u.d. mod 1, by Weyl’s Equidistribution Theorem, see Proposition 4.8(i) below. Similarly, the sequence (xn) = (n √ 2) is u.d. mod 1, whereas (xn √ 2) = (2n) = (2, 4, 6, . . .) clearly is not, as ⟨2n⟩= 0 for all n. Thus the requirement in Lemma 4.3(i) that k be an integer cannot be dropped.
For an analogous example using random variables, let X be uniform on [0, 2), that is PX = λ0,2. Then X is u.d. mod 1, but X √ 2 is not because P ⟨X √ 2⟩≤s = 3 2 √ 2s if s ∈[0, 2 √ 2 −2) , 1 √ 2s + √ 2−1 √ 2 if s ∈[2 √ 2 −2, 1) .
(ii) The sequence (log n) is not u.d. mod 1. A straightforward calculation shows that N −1#{1 ≤n ≤N : ⟨log n⟩≤s} N∈N has, for every s ∈[0, 1), 1 9(10s −1) and 10 9 (1 −10−s) as its limit inferior and limit superior, respectively.
Example 4.5. (i) The function f(t) = at + b with real a, b is u.d. mod 1 if and only if a ̸= 0. Clearly, if a = 0 then f is constant and hence not u.d. mod 1. On the other hand, if a > 0 then ⟨aτ + b⟩≤s if and only if τ ∈ k−b a , k−b+s a for some k ∈Z. Note that each of the intervals k−b a , k−b+s a has the same length s a. Thus, given T > 0 and s ∈[0, 1), s a(⌊aT ⌋−2) ≤λ {τ ∈[0, T ) : ⟨aτ + b⟩≤s} ≤s a(⌊aT ⌋+ 2) , and since limT →+∞ s aT (⌊aT ⌋± 2) = s, the function f is u.d. mod 1. The argu-ment for the case a < 0 is similar.
As a consequence, although the function f(t) = αt is not Benford for any α, the function f(t) = eαt is Benford whenever α ̸= 0, via Theorem 4.2, since log f(t) = αt/ ln 10 is u.d. mod 1, see Fig 7.
30 A. Berger and T.P. Hill (ii) The function f(t) = log |at+b| is not u.d. mod 1 for any a, b ∈R. Indeed, if a = 0 then f is constant and hence not u.d. mod 1. On the other hand, for a ̸= 0 essentially the same calculation as in Example 4.4(ii) above shows that, for every s ∈[0, 1), lim infT →+∞ λ({τ ∈[0, T ) : ⟨log |aτ + b|⟩≤s}) T = 1 9(10s −1) , and lim supT →+∞ λ({τ ∈[0, T ) : ⟨log |aτ + b|⟩≤s}) T = 10 9 (1 −10−s) .
Again, this implies that f(t) = at + b is not Benford for any a, b.
Similarly, f(t) = −log(1 + t2) is not u.d. mod 1, and hence f(t) = (1 + t2)−1 is not Benford, see Fig 7.
(iii) The function f(t) = et is u.d. mod 1. To see this, let T > 0 and N := ⌊eT ⌋, and recall that t −1 2t2 ≤ln(1 + t) ≤t for all t ≥0. Given 0 ≤s < 1, it follows from λ ({τ ∈[0, T ) : ⟨eτ⟩≤s}) = XN−1 n=1 ln 1 + s n + (T −ln N) that s PN−1 n=1 n−1 −1 2s2 PN−1 n=1 n−2 ln(N + 1) ≤λ ({τ ∈[0, T ) : ⟨eτ⟩≤s}) T ≤s PN−1 n=1 n−1 + ln(1 + N −1) ln N , and hence indeed limT →+∞T −1λ ({τ ∈[0, T ) : ⟨eτ⟩≤s}) = s.
As a consequence, the super-exponential function f(t) = eeαt is also Benford if α ̸= 0.
(iv) For the function f(t) = sin2 t, it is straightforward to check that, given any 0 ≤s < 1, limT →+∞ λ({τ ∈[0, T ) : ⟨sin2 τ⟩≤s}) T = 2 π arcsin √s .
Thus, asymptotically ⟨f⟩is not uniform on [0, 1) but rather arcsin-distributed, see Example 3.10(iii).
(v) For the function f(t) = log(sin2 t), it follows from (iv) that the asymptotic distribution of ⟨f⟩has the density d ds 2 π X∞ n=1 arcsin 10(s−n)/2 −arcsin10−n/2 = ln 10 π X∞ n=1 1 √ 10n−s −1 > ln 10 π · 10s/2 101/2 −1 , A basic theory of Benford’s Law 31 1 1 1 0 0 0 0 0 0 10 10 10 10 10 10 t t t f1(t) = e−t S(f1) f2(t) = (1 + t2)−1 S(f2) f3(t) = sin2 t S(f3) Fig 7. While the function f1 is Benford, the functions f2, f3 are not, see Example 4.5.
for 0 ≤s < 1. Thus clearly f is not u.d. mod 1, showing that t 7→sin2 t is not Benford, see Fig 7.
Example 4.6. (i) If the random variable X is uniformly distributed on [0, 2) then it is clearly u.d. mod 1. However, if X is uniform on, say, [0, π) then X is not u.d. mod 1.
(ii) No exponential random variable is u.d. mod 1. Specifically, let X be an exponential random variable with mean σ, i.e.
FX(t) = max(0, 1 −e−t/σ) , t ∈R .
Hence var X = σ2. For every l ≥0, P(l ≤X < l + 1 2) = FX(l + 1 2) −FX(l) > FX(l + 1) −FX(l + 1 2) = P(l + 1 2 ≤X < l + 1) , and since P∞ l=0 P(l ≤X < l + 1) = 1, this implies that P(⟨X⟩< 1 2) = X∞ l=0 P(l ≤X < l + 1 2) > 1 2 , showing that X is not u.d. mod 1. To obtain more explicit information, observe that, for every 0 ≤s < 1, F⟨X⟩(s) = P(⟨X⟩≤s) = X∞ l=0 FX(l + s) −FX(l) = 1 −e−s/σ 1 −e−1/σ , 32 A. Berger and T.P. Hill from which it follows via a straightforward calculation that max0≤s<1 F⟨X⟩(s) −s = 1 e1/σ −1 −σ + σ ln(σe1/σ −σ) =: Rii(σ) .
Note that Rii(1) = ln(e −1) −e−2 e−1 = 0.1233 . . . < 1 8. Moreover, Rii(σ) = 1 8σ + O(σ−2) as σ →+∞, which shows that even though X is not u.d. mod 1, the deviation of ⟨X⟩from uniform is small for large σ. As a consequence, 10X resembles a Benford random variable ever more closely as σ →+∞.
(iii) If X is a normal random variable then X is not u.d. mod 1, and neither is |X| or max(0, X). While this is easily checked by a direct calculation as in (ii), it is again illuminating to obtain more quantitative information. To this end, assume that X is a normal variable with mean 0 and variance σ2. By means of Fourier series [Pi], it can be shown that, for every 0 ≤s < 1, F⟨X⟩(s) −s = X∞ n=1 sin(2πns) πn e−2σ2π2n2 .
From this, it follows that Riii(σ) := max0≤s<1 F⟨X⟩(s) −s ≤1 π X∞ n=1 n−1e−2σ2π2n2 , and hence in particular Riii(σ) = e−2σ2π2 π + O(e−8σ2π2) as σ →+∞, showing that Riii(σ), the deviation of ⟨X⟩from uniformity, goes to zero very rapidly as σ →+∞. Already for σ = 1 one finds that Riii(1) < 8.516 · 10−10.
Thus even though a standard normal random variable X is not u.d. mod 1, the distribution of ⟨X⟩is extremely close to uniform. Consequently, a log-normal random variable with large variance is practically indistinguishable from a Ben-ford random variable.
Corollary 4.7.
(i) A sequence (xn) is Benford if and only if, for all α ∈R and k ∈Z with αk ̸= 0, the sequence (αxk n) is also Benford.
(ii) A function f : [0, +∞) →R is Benford if and only if 1/f is Benford.
(iii) A random variable X is Benford if and only if 1/X is Benford.
The next two statements, recorded here for ease of reference, list several key tools concerning uniform distribution mod 1, which via Theorem 4.2 will be used to determine Benford properties of sequences, functions, and random variables.
Conclusion (i) in Proposition 4.8 is Weyl’s classical uniform distribution result [KN, Thm.3.3], conclusion (ii) is an immediate consequence of Weyl’s criterion [KN, Thm.2.1], conclusion (iii) is [Ber2, Lem.2.8], and conclusion (iv) is [BBH, Lem.2.4.(i)].
A basic theory of Benford’s Law 33 Proposition 4.8. Let (xn) be a sequence of real numbers.
(i) If limn→∞(xn+1 −xn) = θ for some irrational θ, then (xn) is u.d. mod 1.
(ii) If (xn) is periodic, i.e. xn+p = xn for some p ∈N and all n, then (nθ+xn) is u.d. mod 1 if and only if θ is irrational.
(iii) The sequence (xn) is u.d. mod 1 if and only if (xn + α log n) is u.d. mod 1 for all α ∈R.
(iv) If (xn) is u.d. mod 1 and non-decreasing, then (xn/ log n) is unbounded.
The converse of (i) is not true in general: (xn) may be u.d. mod 1 even if (xn+1 −xn) has a rational limit. Also, in (ii) the sequence (nθ) cannot be replaced by an arbitrary uniformly distributed sequence (θn), i.e. (θn +xn) may not be u.d. mod 1 even though (θn) is u.d. mod 1 and (xn) is periodic.
Another very useful result is Koksma’s metric theorem [KN, Thm.4.3]. For its formulation, recall that a property of real numbers is said to hold for almost every (a.e.) x ∈[a, b) if there exists a set N ∈B[a, b) with λa,b(N) = 0 such that the property holds for every x ̸∈N. The probabilistic interpretation of a given property of real numbers holding for a.e. x is that this property holds almost surely (a.s.), i.e. with probability one, for every random variable that has a density (i.e., is absolutely continuous).
Proposition 4.9. Let fn be continuously differentiable on [a, b] for all n ∈N.
If f ′ m −f ′ n is monotone and |f ′ m(x) −f ′ n(x)| ≥α > 0 for all m ̸= n, where α does not depend on x, m and n, then fn(x) is u.d. mod 1 for almost every x ∈[a, b].
Theorem 4.10 ([BHKR]). If a, b, α, β are real numbers with a ̸= 0 and |α| > |β| then (αna + βnb) is Benford if and only if log |α| is irrational.
Proof. Since a ̸= 0 and |α| > |β|, limn→∞ βnb αna = 0, and therefore log |αna + βnb| −log |αna| = log 1 + βnb αna →0 , showing that (log |αna+βnb|) is u.d. mod 1 if and only if (log |αna|) = (log |a|+ n log |α|) is. According to Proposition 4.8(i), this is the case whenever log |α| is irrational. On the other hand, if log |α| is rational then ⟨log |a|+n log |α|⟩attains only finitely many values and hence (log |a| + n log |α|) is not u.d. mod 1. An application of Theorem 4.2 therefore completes the proof.
Example 4.11. (i) By Theorem 4.10 the sequence (2n) is Benford since log 2 is irrational, but (10n) is not Benford since log 10 = 1 ∈Q. Similarly, (0.2n), (3n), (0.3n), 0.01 · 0.2n + 0.2 · 0.01n are Benford, whereas (0.1n), √ 10 n , 0.1 · 0.02n + 0.02 · 0.1n are not.
(ii) The sequence 0.2n + (−0.2)n is not Benford, since all odd terms are zero, but 0.2n + (−0.2)n + 0.03n is Benford — although this does not follow directly from Theorem 4.10.
34 A. Berger and T.P. Hill (iii) By Proposition 4.9, the sequence (x, 2x, 3x, . . .) = (nx) is u.d. mod 1 for almost every real x, but clearly not for every x, as for example x = 1 shows.
Consequently, by Theorem 4.2, (10nx) is Benford for almost all real x, but not e.g. for x = 1 or, more generally, whenever x is rational.
(iv) By Proposition 4.8(iv) or Example 4.4(ii), the sequence (log n) is not u.d. mod 1, so the sequence (n) of positive integers is not Benford, and neither is (αn) for any α ∈R, see also Fig 8.
(v) Consider the sequence (pn) of prime numbers. By the Prime Number Theorem, pn = O(n log n) as n →∞. Hence it follows from Proposition 4.8(iv) that (pn) is not Benford, see Fig 8 Example 4.12. Consider the sequence (Fn) = (1, 1, 2, 3, 5, 8, 13, . . .) of Fi-bonacci numbers, defined inductively as Fn+2 = Fn+1 + Fn for all n ∈N, with F1 = F2 = 1. It is well known (and easy to check) that Fn = 1 √ 5 1 + √ 5 2 !
n − 1 − √ 5 2 !
n!
= 1 √ 5 ϕn −(−ϕ−1)n for all n ∈N , where ϕ = 1 2(1 + √ 5) ≈1.618. Since ϕ > 1 and log ϕ is irrational, (Fn) is Benford, by Theorem 4.10, see also Fig 8. Sequences such as (Fn) which are generated by linear recurrence relations will be studied in detail in Section 5.2.
Theorem 4.13. Let X, Y be random variables. Then: (i) If X is u.d. mod 1 and Y is independent of X, then X + Y is u.d. mod 1.
(ii) If ⟨X⟩and ⟨X + α⟩have the same distribution for some irrational α then X is u.d. mod 1.
(iii) If (Xn) is an i.i.d. sequence of random variables and X1 is not purely atomic (i.e. P(X1 ∈C) < 1 for every countable set C ⊂R), then limn→∞P DXn j=1 Xj E ≤s = s for every 0 ≤s < 1 , (4.1) that is, Pn j=1 Xj →U(0, 1) in distribution as n →∞.
Proof. The proof is most transparently done by means of some elementary Fourier analysis. To this end, for any random variable Z with values in [0, 1), or equivalently for the associated probability measure PZ on [0, 1), B[0, 1) , let c PZ(k) = E(e2πıkZ) = Z 1 0 e2πıksdPZ(s) = Z 1 0 cos(2πks) dPZ(s) + ı Z 1 0 sin(2πks) dPZ(s) , k ∈Z .
The bi-infinite sequence c PZ(k) k∈Z, referred to as the Fourier (or Fourier– Stieltjes) coefficients of Z or PZ, is a bounded sequence of complex numbers, A basic theory of Benford’s Law 35 (Fn) = (1, 1, 2, 3, 5, 8, 13, . . . ) S(Fn) = (1, 1, 2, 3, 5, 8, 1.3, . . . ) (2n) = (2, 4, 6, 8, 10, 12, 14, . . . ) S(2n) = (2, 4, 6, 8, 1, 1.2, 1.4, . . . ) (xn) = (⌊10(n+2)/5⌋) = (3, 6, 10, 15, 25, 39, 63, . . . ) S(xn) = (3, 6, 1, 1.5, 2.5, 3.9, 6.3, . . . ) (pn) = (2, 3, 5, 7, 11, 13, 17, . . .) S(pn) = (2, 3, 5, 7, 1.1, 1.3, 1.7, . . .) ρN (1) = #{1 ≤n ≤N : D1 = 1} N 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 0.1 0.1 0.1 0.2 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.4 0.5 0.5 0.5 0.6 0.8 1.0 log10N log10N log10N log10N ρN (1) ρN (1) ρN (1) Fig 8. For a Benford sequence, limN→∞ρN(1) = log 2. Thus if ρN(1) N∈N does not con-verge (center) or has a different limit (bottom), then the sequence in question is not Benford, see also Example 4.11.
with |c PZ(k)| ≤1 for all k ∈Z, and c PZ(0) = 1. The three single most important properties of Fourier coefficients are that c PZ(k) k∈Z uniquely determines PZ, i.e. PZ1 = PZ2 whenever d PZ1(k) = d PZ2(k) for all k ∈Z; that \ P⟨Z1+Z2⟩(k) = d PZ1(k) · d PZ2(k) for all k, provided that Z1 and Z2 are independent; and that Zn →Z in distribution if and only if limn→∞d PZn(k) = c PZ(k) for every k, see e.g. [CT] for an authoritative discussion of this material. Also note that the sequence of Fourier coefficients is extremely simple if Z is uniform, i.e. for Z = U(0, 1), namely \ PU(0,1)(k) = d λ0,1(k) = 1 if k = 0 , 0 otherwise.
36 A. Berger and T.P. Hill With these preparations, the proof of the theorem is very short indeed.
(i) Since [ P⟨X⟩(k) = 0 for all k ̸= 0, \ P⟨X+Y ⟩(k) = c PX(k) · c PY (k) = 0 , which in turn shows that ⟨X + Y ⟩= U(0, 1), i.e. X + Y is u.d. mod 1.
(ii) Note that if Z = α with probability one then c PZ(k) = e2πıkα for every k ∈Z. Consequently, if ⟨X⟩and ⟨X + α⟩have the same distribution then [ P⟨X⟩(k) = \ P⟨X+α⟩(k) = e2πıkα[ P⟨X⟩(k) for every k ∈Z. If α is irrational then e2πıkα ̸= 1 for all k ̸= 0, implying that [ P⟨X⟩(k) = 0. Thus [ P⟨X⟩= d λ0,1 and hence P⟨X⟩= λ0,1, i.e. ⟨X⟩= U(0, 1).
(iii) Assume that X1, X2, . . . are independent and all have the same distribu-tion. Then, for every k ∈Z and n ∈N, \ P⟨X1+...+Xn⟩(k) = \ P⟨X1⟩(k) n .
Recall that |\ P⟨X1⟩(k)| ≤1. Thus \ P⟨X1+...+Xn⟩(k) →0 as n →∞, and hence ⟨X1 + . . . + Xn⟩→U(0, 1) in distribution, unless |\ P⟨X1⟩(k0)| = 1 for some non-zero integer k0. In the latter case, let \ P⟨X1⟩(k0) = e2πıθ with the appropriate θ ∈[0, 1). It follows from 0 = 1 −e−2πıθ\ P⟨X1⟩(k0) = 1 − \ P⟨X1−θ/k0⟩(k0) = Z 1 0 1 −cos(2πk0s) dP⟨X1−θ/k0⟩(s) ≥0 , that cos(2πk0⟨X1 −θ/k0⟩) = cos 2π(k0X −θ) = 1 with probability one. Hence P(k0X1 ∈θ + Z) = 1, and X1 is purely atomic. (In fact, X1 is concentrated on a lattice {a + k/|k0| : k ∈Z} with the appropriate a > 0.) Example 4.14. (i) Let (Xn) be an i.i.d. sequence of Cauchy random variables, i.e.
fX1(t) = 1 π(1 + t2) , t ∈R .
It is well known, or readily checked by a direct calculation, that 1 n Pn j=1 Xj is again Cauchy. Thus f⟨Pn j=1 Xj⟩(s) = 1 π X k∈Z n n2 + (s + k)2 , 0 ≤s < 1 , from which it follows that, uniformly in s, f⟨Pn j=1 Xj⟩(s) = 1 πn X k∈Z 1 1 + (s + k)/n 2 →1 π Z +∞ −∞ dt 1 + t2 = 1 as n →∞.
A basic theory of Benford’s Law 37 As asserted by Theorem 4.13, therefore, for every 0 ≤s < 1, limn→∞P DXn j=1 Xj E ≤s = limn→∞ Z s 0 f⟨Pn j=1 Xj⟩(σ) dσ = Z s 0 1 dσ = s .
(ii) Consider an i.i.d. sequence (Xn) where P(X1 = 0) = P(X1 = √ 2) = 1 2.
In this case d PX1(k) = 1 2 1 + e2πık √ 2 = eπık √ 2 cos(πk √ 2) , k ∈Z .
Note that |d PX1(k)| = | cos(πk √ 2)| < 1 for all k ̸= 0. Hence \ P⟨Pn j=1 Xj⟩(k) = \ P⟨X1⟩(k)n →0 as n →∞, which in turn shows that (4.1) holds, even though X1 is purely atomic.
On the other hand, if P(X1 = 0) = P(X1 = 1 2) = 1 2 then X1 is also purely atomic, but P Xn j=1 Xj = 1 2l = 2−n n l for all n ∈N, l = 0, 1, . . . , n , and consequently, for every n, P DXn j=1 Xj E = 0 = Xn l=0, l even 2−n n l = 1 2 , showing that (4.1) does not hold in this case. Correspondingly, d PX1(k) = 1 2 1 + (−1)k , and so d PX1(k) = 1 whenever k is even.
A careful inspection of the above proof shows that, in the setting of Theorem 4.13(iii), (4.1) holds if and only if P(X1 ∈a + 1 mZ) < 1 for every a ∈R and m ∈N. While the “if” part has been proved above, for the “only if” part simply note that if P(X1 ∈a + 1 mZ) = 1 for some a ∈R and m ∈N then ⟨X1 + . . . + Xn⟩is, for every n ∈N and possibly up to a rotation, concentrated on the set {0, 1 m, . . . , m−1 m } = ⟨1 mZ⟩and hence does not converge in distribution to U(0, 1).
None of the familiar classical probability distributions or random variables, such as e.g. normal, uniform, exponential, beta, binomial, or gamma distribu-tions are Benford. Specifically, no uniform distribution is even close to BL, no matter how large its range or how it is centered. This statement can be quanti-fied explicitly as follows.
Proposition 4.15 ([Ber5, BH3]). For every uniformly distributed random vari-able X, max0≤s<1 F⟨log X⟩(s) −s ≥−9 + ln 10 + 9 ln 9 −9 ln ln 10 18 ln 10 = 0.1334 . . . , and this bound is sharp.
38 A. Berger and T.P. Hill Similarly, all exponential and normal random variables are uniformly bound-ed away from BL, as is explained in detail in [BH3]. However, as the following example shows, some distributions do come fairly close to being Benford.
Example 4.16. (i) Let X be exponential with mean 1, that is FX(t) = max(0, 1 −e−t) , t ∈R .
An explicit calculation shows that, for every 1 ≤t < 10, P(S(X) ≤t) = X k∈Z FX(10kt) −FX(10k) = X k∈Z e−10k −e−10kt .
Since P(S(X) ≤t) ̸≡log t, the random variable X is not Benford. Numerically, one finds max1≤t<10 |P(S(X) < t) −log t| < 3.054 · 10−2, see also Fig 9. Thus even though X is not exactly Benford, it is close to being Benford in the sense that |P(S(X) ≤t) −log t| is small for all t ∈[1, 10).
(ii) Let X be standard normal. Then, for every t ∈[1, 10), P(S(X) ≤t) = X k∈Z Φ(10kt) −Φ(10k) , where Φ is the distribution function of X, that is Φ(t) = FX(t) = P(X ≤t) = 1 √ 2π Z t −∞ e−1 2 τ 2dτ , t ∈R .
Numerically, one finds max1≤t≤10 |P(S(X) < t) −log t| < 6.052 · 10−2. Though larger than in the exponential case, the deviation of X from BL is still rather small.
0.04 0.04 −0.04 −0.04 10 10 1 1 t t X standard normal X exponential, EX = 1 FS(X)(t) −log t FS(X)(t) −log t ∥FS(X)(t)−log t∥∞≈6.052·10−2 ∥FS(X)(t)−log t∥∞≈3.054·10−2 Fig 9. For standard exponential (left) and normal random variables X, the distribution of S(X) deviates from BL only slightly. Note, however, that non-standard normal variables can be far from BL, e.g., if EX = 75 and var X = 1 then D1(X) = 7 with very high probability.
A basic theory of Benford’s Law 39 The next result says that every random variable X with a density is asymp-totically uniformly distributed on lattices of intervals as the size of the intervals goes to zero. Equivalently, ⟨nX⟩is asymptotically uniform, as n →∞. This result has been the basis for several recent fallacious arguments claiming that if a random variable X has a density with a very large “spread” then log X must also have a density with large spread and thus, by the theorem, must be close to u.d. mod 1, implying in turn that X must be close to Benford (cf. [Fel, Few]).
The error in those arguments is that, regardless of which notion of “spread” is used, the variable X may have large spread and at the same time the variable log X may have small spread; for details, the reader is referred to [BH3].
Theorem 4.17. If X has a density then limn→∞P(⟨nX⟩≤s) = s for all 0 ≤s < 1 , (4.2) that is, ⟨nX⟩→U(0, 1) in distribution as n →∞.
Proof. Since ⟨nX⟩= n⟨X⟩ , it can be assumed that X only takes values in [0, 1). Let f be the density of X, i.e. f : [0, 1] →R is a non-negative measurable function with P(X ≤s) = R s 0 f(σ) dσ for all s ∈[0, 1). From P(⟨nX⟩≤s) = P X ∈ [n−1 l=0 l n, l + s n = Xn−1 l=0 Z (l+s)/n l/n f(σ) dσ = Z s 0 1 n Xn−1 l=0 f l + σ n dσ , it follows that the density of ⟨nX⟩is given by f⟨nX⟩(s) = 1 n Xn−1 l=0 f l + s n , 0 ≤s < 1 .
Note that if f is continuous, or merely Riemann integrable, then, as n →∞, f⟨nX⟩(s) → Z 1 0 f(σ) dσ = 1 for all s ∈[0, 1) .
In general, given any ε > 0 there exists a continuous density gε such that R 1 0 |f(σ) −gε(σ)| dσ < ε and hence Z 1 0 |f⟨nX⟩(σ) −1| dσ ≤ Z 1 0 1 n Xn−1 l=0 f l + σ n −1 n Xn−1 l=0 gε l + σ n dσ + Z 1 0 1 n Xn−1 l=0 gε l + σ n −1 dσ ≤ Z 1 0 |f(σ) −gε(σ)| dσ + Z 1 0 1 n Xn−1 l=0 gε l + σ n − Z 1 0 g(τ) dτ dσ , which in turn shows that lim supn→∞ Z 1 0 |f⟨nX⟩(σ) −1| dσ ≤ε , 40 A. Berger and T.P. Hill and since ε > 0 was arbitrary, R 1 0 |f⟨nX⟩(σ) −1| dσ →0 as n →∞. From this, the claim follows immediately because, for every 0 ≤s < 1, P(⟨nX⟩≤s) −s = Z s 0 (f⟨nX⟩(σ) −1) dσ ≤ Z 1 0 |f⟨nX⟩(σ) −1| dσ →0 .
Remark. If X does not have a density, then (4.2) may not hold. Trivially, if X is an integer with probability one then P(⟨nX⟩≤s) = 1 for every n and 0 ≤s < 1. Hence (4.2) fails. For a simple continuous example, let X be uniformly distributed on the classical Cantor middle thirds set. In more probabilistic terms, X = 2 P∞ j=1 3−jXj where the Xj are i.i.d. with P(X1 = 0) = P(X1 = 1) = 1 2.
Then PX ̸= λ0,1 but ⟨3X⟩has the same distribution as X, and so has ⟨3nX⟩ for every n ∈N. Thus (4.2) fails again.
In fact, using the Fourier analysis tools introduced in the proof of Theorem 4.13, together with the observation that \ P⟨nX⟩(k) = [ P⟨X⟩(nk) for all n ∈N, k ∈Z , it is clear that (4.2) holds if and only if X has the property that [ P⟨X⟩(k) →0 as |k| →∞, i.e. precisely if P⟨X⟩is a so-called Rajchman probability. As Theorem 4.17 shows, a probability on [0, 1) is Rajchman whenever it is a.c. (In advanced calculus, this fact is usually referred to as the Riemann–Lebesgue Lemma.) The converse is not true, i.e., there exist Rajchman probabilities on [0, 1) that are not a.c., see [Ly].
♣ 4.2. The scale-invariance characterization One popular hypothesis often related to BL is that of scale-invariance. Infor-mally put, scale-invariance captures the intuitively attractive notion that any universal law should be independent of units. For instance, if a sufficiently large aggregation of data is converted from meters to feet, US$ to e etc., then while the individual numbers change, the statements about the overall distribution of significant digits should not be affected by this change. R. Pinkham [Pi] credits R. Hamming with the idea of scale-invariance, and attempts to prove that the Benford distribution is the only scale-invariant distribution. Pinkham’s argument has subsequently been used by numerous authors to explain the ap-pearance of BL in many real-life data, by arguing that the data in question should be invariant under changes of scale and thus must be Benford.
Although this scale-invariance conclusion is correct in the proper setting, see Theorem 4.20 below, Pinkham’s argument contains a fatal error. As Knuth [Kn] observes, the error is Pinkham’s implicit assumption that there is a scale-invariant Borel probability measure on R+, when in fact such a probability measure does not exist, cf. [Ra1]. Indeed, the only real-valued random variable X that is scale-invariant, i.e., X and αX have the same distribution for all A basic theory of Benford’s Law 41 scaling factors α > 0, is the random variable that is constant equal to zero, that is P(X = 0) = 1. Clearly, any such random variable is scale-invariant since X = αX with probability one. To see that this is the only scale-invariant random variable, suppose that P(|X| > c) = δ > 0 for some c > 0. Then P(|αX| > c) = P(|X| > c/α) ց 0 as α ց 0, so for sufficiently small positive α, P(|αX| > c) < δ = P(|X| > c), contradicting scale-invariance. Thus no non-zero random variable is scale-invariant. Note, however, that the measure on (R+, B+) defined as µ [a, b] := Z b a dt t = log b a for all [a, b] ⊂R+ , is scale invariant because, for every α > 0, α∗µ [a, b] = Z b/α a/α dt t = log b a = µ [a, b] .
Obviously, µ is not finite, i.e. µ(R+) = +∞, but is still σ-finite. (Generally, a measure µ on (Ω, A) is σ-finite if Ω= S n∈N An for some sequence (An) in A, and µ(An) < +∞for all n.) In a similar spirit, a sequence (xn) of real numbers may be called scale-invariant if limN→∞ #{1 ≤n ≤N : αxn ∈[a, b]} N = limN→∞ #{1 ≤n ≤N : xn ∈[a, b]} N holds for all α > 0 and [a, b] ⊂R. For example, the sequence 2, 2−1, 2, 3, 2−1, 3−1, 2, 3, 4, 2−1, 3−1, 4−1, . . . , 2, 3, . . ., n, 2−1, 3−1, . . . , n−1, 2 . . .
is scale-invariant. As above, it is not hard to see that limN→∞ #{1 ≤n ≤N : xn ∈[a, b]} N = 0 for all [a, b] ⊂R{0} , holds whenever (xn) is scale-invariant. Most elements of a scale-invariant se-quence of real numbers, therefore, are very close to either 0 or ±∞.
While a positive random variable X cannot be scale-invariant, as shown above, it may nevertheless have scale-invariant significant digits. For this, how-ever, X has to be Benford. In fact, Theorem 4.20 below shows that being Benford is (not only necessary but) also sufficient for X to have scale-invariant significant digits. The result will first be stated in terms of probability distributions.
Definition 4.18. Let A ⊃S be a σ-algebra on R+. A probability measure P on (R+, A) has scale-invariant significant digits if P(αA) = P(A) for all α > 0 and A ∈S , or equivalently if for all m ∈N, all d1 ∈{1, 2, . . ., 9} and all dj ∈{0, 1, . . ., 9}, j ≥2, P x:Dj(αx) = dj for j = 1, 2, . . .m =P x:Dj(x) = dj for j = 1, 2, . . . , m holds for every α > 0.
42 A. Berger and T.P. Hill Example 4.19. (i) The Benford probability measure B on (R+, S) has scale-invariant significant digits. This follows from Theorem 4.20 below but can also be seen from a direct calculation. Indeed, if A = S k∈Z 10k[a, b] with 1 ≤a < b < 10, then, given any α > 0, αA = [ k∈Z 10k+log α[a, b] = [ k∈Z 10k+⟨log α⟩[a, b] = [ k∈Z 10kB , where the set B is given by B = 10⟨log α⟩a, 10⟨log α⟩b if 0 ≤⟨log α⟩< 1 −log b , 1, 10⟨log α⟩−1b ∪ 10⟨log α⟩a, 10 if 1 −log b ≤⟨log α⟩< 1 −log a , 10⟨log α⟩−1a, 10⟨log α⟩−1b if 1 −log a ≤⟨log α⟩< 1 .
From this, it follows that B(αA) = log 10⟨log α⟩b −log 10⟨log α⟩a log 10⟨log α⟩−1b + 1 −log 10⟨log α⟩a log 10⟨log α⟩−1b −log 10⟨log α⟩−1a = log b −log a = B(A) , showing that B has scale-invariant digits.
(ii) The Dirac probability measure δ1 concentrated at the constant 1 does not have scale-invariant significant digits, since δ2 = 2∗δ1 yet δ1(D1 = 1) = 1 ̸= 0 = δ2(D1 = 1).
(iii) The uniform distribution on [0, 1) does not have scale-invariant digits, since if X is distributed according to λ0,1 then, for example P(D1(X) = 1) = 1 9 < 11 27 = P D1 3 2X = 1 .
As mentioned earlier, the Benford distribution is the only probability measure (on the significand σ-algebra) having scale-invariant significant digits.
Theorem 4.20 (Scale-invariance characterization [Hi1]). A probability measure P on (R+, A) with A ⊃S has scale-invariant significant digits if and only if P(A) = B(A) for every A ∈S, i.e., if and only if P is Benford.
Proof. Fix any probability measure P on (R+, A), denote by P0 its restriction to (R+, S), and let Q := ℓ∗P0 with ℓgiven by Lemma 2.16. According to Lemma 2.16, Q is a probability measure on [0, 1), B[0, 1) . Moreover, under the corre-spondence established by ℓ, P0(αA) = P0(A) for all α > 0, A ∈S (4.3) A basic theory of Benford’s Law 43 A = {D1 = 2} log A = {2 ≤S < 3} 3A log 3A S log S B(3A) = B(A) λ0,1(log 3A) = λ0,1(log A) 1 2 3 4 5 6 7 8 9 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Fig 10. Visualizing the scale-invariant significant digits of BL.
is equivalent to Q(⟨t + B⟩) = Q(B) for all t ∈R, B ∈B[0, 1) , (4.4) where ⟨t + B⟩= {⟨t + x⟩: x ∈B}. Pick a random variable X such that the distribution of X is given by Q. With this, (4.4) simply means that, for every t ∈R, the distributions of ⟨X⟩and ⟨t+X⟩coincide. By Theorem 4.13(i) and (ii) this is the case if and only if X is u.d. mod 1, i.e. Q = λ0,1. (For the “if” part, note that a constant random variable is independent from every other random variable.) Hence (4.3) is equivalent to P0 = (ℓ−1)∗λ0,1 = B.
Example 4.21. For every integer k, let qk > 0 and fk(t) = 1 t ln 10 if 10k ≤t < 10k+1 , 0 otherwise.
If P k∈Z qk = 1 then, according to Example 3.6, P k∈Z qkfk is the density of a Benford probability measure P on (R+, B+). By Theorem 4.20, P has scale-invariant significant digits. Note that, in full agreement with earlier observations, P is not scale-invariant, as for instance qk = P [10k, 10k+1) = P 10k−l[10l, 10l+1) = P [10l, 10l+1) = ql cannot possibly hold for all pairs (k, l) of integers.
In analogy to Definition 4.18, a sequence (xn) of real numbers is said to have scale-invariant significant digits if limN→∞ #{1 ≤n ≤N : S(αxn) < t} N = limN→∞ #{1 ≤n ≤N : S(xn) < t} N for all α > 0, t ∈[1, 10) .
(4.5) 44 A. Berger and T.P. Hill Implicit in (4.5) is the assumption that the limits on either side exit for all t.
A similar definition can be considered for real-valued functions. To formulate an analog of Theorem 4.20 using this terminology, recall that a set A ⊂N has density ρ ∈[0, 1] if the limit limN→∞#{1 ≤n ≤N : n ∈A}/N exists and equals ρ. For example, ρ({n : n even }) = 1 2 and ρ({n : n prime }) = 0, whereas {n : D1(n) = 1} does not have a density.
Theorem 4.22.
(i) For any sequence (xn) of real numbers, let {n : xn ̸= 0} = {n1 < n2 < . . .}. Then (xn) has scale-invariant significant digits if and only if {n : xn ̸= 0} has a density and either ρ({n : xn = 0}) = 1 or else (xnj)j∈N is Benford. In particular, if ρ({n : xn = 0}) = 0 then the sequence (xn) has scale-invariant significant digits if and only if it is Benford.
(ii) A (Borel measurable) function f : [0, +∞) →R with λ {t ≥0 : f(t) = 0} < +∞has scale-invariant significant digits if and only if it is Benford.
Moreover, f is Benford precisely if αf is Benford for every α ̸= 0.
Proof. (i) Assume first that (xn) has scale-invariant significant digits. According to (4.5), G(s) := limN→∞ #{1 ≤n ≤N : S(xn) < 10s} N exists for every 0 ≤s < 1. In particular, {n : xn = 0} has a density G(0). For G(0) = 1 there is nothing else to show. Thus, assume G(0) < 1 from now on, and define a non-decreasing function H : [0, 1) →R as H(s) = G(s) −G(0) 1 −G(0) , 0 ≤s < 1 .
Note that H(s) = limN→∞ #{1 ≤n ≤N : S(xn) < 10s, xn ̸= 0} #{1 ≤n ≤N : xn ̸= 0} = limN→∞ #{1 ≤j ≤N : S(xnj) < 10s} N , so H takes into account only the non-zero entries in (xn). Define h : R →R as h(s) = H(⟨s⟩) −⟨s⟩ for all s ∈R .
Clearly, h is 1-periodic, with h(0) = 0 and |h(s)| ≤1 for all s ∈R. In terms of the function H, the invariance property (4.5) simply reads H(s) = ( H(1 + s −⟨log α⟩) −H(1 −⟨log α⟩) if s < ⟨log α⟩, 1 −H(1 −⟨log α⟩) + H(s −⟨log α⟩) if s ≥⟨log α⟩, provided that log α ̸∈Z. In terms of h, this is equivalent to h(s) = h(1 + s −⟨log α⟩) −h(1 −⟨log α⟩) for all s ∈R, α > 0 .
(4.6) A basic theory of Benford’s Law 45 As a consequence, s 7→h(1 + s −⟨log α⟩) −h(s) is constant for every α > 0.
Since the function h is bounded and 1-periodic, it can be represented (at least in the L2-sense) by a Fourier series h(s) = X k∈Z cke2πıks , from which it follows that h(1 + s −⟨log α⟩) −h(s) = X k∈Z ck e2πık(1+s−⟨log α⟩) −e2πıks = X k∈Z ck e−2πık⟨log α⟩−1 e2πıks .
Pick α > 0 such that ⟨log α⟩is irrational, e.g. α = 2. Then e−2πık⟨log α⟩̸= 1 whenever k ̸= 0, which in turn implies that ck = 0 for all k ̸= 0, i.e. h is constant almost everywhere. Thus H(s) = s + c0 for a.e. s ∈[0, 1), and in fact H(s) ≡s because H is non-decreasing with H(0) = 0. Overall, therefore, limN→∞ #{1 ≤j ≤N : S(xnj) < 10s} N = s for all s ∈[0, 1) , showing that (xnj) is Benford.
Conversely, if ρ({n : xn = 0}) = 1 then (4.5) holds with both sides being equal to 1 for all t ∈[1, 10). Assume, therefore, that ρ({n : xn = 0}) < 1 and (xnj) is Benford. By the above, h(s) ≡0, so (4.6) and hence also (4.5) hold, i.e., (xn) has scale-invariant significant digits.
The proof of (ii) is completely analogous, utilizing G(s) := limT →+∞ λ τ ∈[0, T ) : S f(τ) < 10s T , 0 ≤s < 1 .
Note that the assumption λ {t ≥0 : f(t) = 0} < +∞implies G(0) = 0.
Example 4.23. Let (xn) equal either the sequence of Fibonacci or prime num-bers. In both cases, xn ̸= 0 for all n, and hence by Theorem 4.22(i) (xn) has scale-invariant significant digits if and only if it is Benford. Thus (Fn) does have scale-invariant significant digits, and (pn) does not. These facts are illus-trated empirically in Fig 11 to 13 which show the relevant data for, respectively, the first 102 (Fig 11 and 12) and 104 (Fig 13) entries of either sequence, and compare them with the respective expected values for BL.
The next example is an elegant and entertaining application of the ideas underlying Theorems 4.20 and 4.22 to the mathematical theory of games. The game may be easily understood by a schoolchild, yet it has proven a challenge for game theorists not familiar with BL.
46 A. Berger and T.P. Hill 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1346269 2178309 3524578 5702887 9227465 14930352 24157817 39088169 63245986 165580141 165580141 267914296 433494437 701408733 1134903170 1836311903 2971215073 4807526976 7778742049 12586269025 20365011074 32951280099 53316291173 86267571272 139583862445 225851433717 365435296162 591286729879 956722026041 1548008755920 2504730781961 4052739537881 6557470319842 10610209857723 17167680177565 27777890035288 44945570212853 72723460248141 117669030460994 190392490709135 308061521170129 498454011879264 806515533049393 1304969544928657 2111485077978050 3416454622906707 5527939700884757 8944394323791464 14472334024676221 23416728348467685 37889062373143906 61305790721611591 99194853094755497 160500643816367088 259695496911122585 420196140727489673 679891637638612258 1100087778366101931 1779979416004714189 2880067194370816120 4660046610375530309 7540113804746346429 12200160415121876738 19740274219868223167 31940434634990099905 51680708854858323072 83621143489848422977 135301852344706746049 218922995834555169026 354224848179261915075 2 2 4 6 10 16 26 42 68 110 178 288 466 754 1220 1974 3194 5168 8362 13530 21892 35422 57314 92736 150050 242786 392836 635622 1028458 1664080 2692538 4356618 7049156 11405774 18454930 29860704 48315634 78176338 126491972 204668310 331160282 535828592 866988874 1402817466 2269806340 3672623806 5942430146 9615053952 15557484098 25172538050 40730022148 65902560198 106632582346 172535142544 279167724890 451702867434 730870592324 1182573459758 1913444052082 3096017511840 5009461563922 8105479075762 13114940639684 21220419715446 34335360355130 55555780070576 89891140425706 145446920496282 235338060921988 380784981418270 616123042340258 996908023758528 1613031066098786 2609939089857314 4222970155956100 6832909245813414 11055879401769514 17888788647582928 28944668049352442 46833456696935370 75778124746287812 122611581443223182 198389706189510994 321001287632734176 519390993822245170 840392281454979346 1359783275277224516 2200175556732203862 3559958832009428378 5760134388741632240 9320093220751060618 15080227609492692858 24400320830243753476 39480548439736446334 63880869269980199810 103361417709716646144 167242286979696845954 270603704689413492098 437845991669110338052 708449696358523830150 7 7 14 21 35 56 91 147 238 385 623 1008 1631 2639 4270 6909 11179 18088 29267 47355 76622 123977 200599 324576 525175 849751 1374926 2224677 3599603 5824280 9423883 15248163 24672046 39920209 64592255 104512464 169104719 273617183 442721902 716339085 1159060987 1875400072 3034461059 4909861131 7944322190 12854183321 20798505511 33652688832 54451194343 88103883175 142555077518 230658960693 373214038211 603872998904 977087037115 1580960036019 2558047073134 4139007109153 6697054182287 10836061291440 17533115473727 28369176765167 45902292238894 74271469004061 120173761242955 194445230247016 314618991489971 509064221736987 823683213226958 1332747434963945 2156430648190903 3489178083154848 5645608731345751 9134786814500599 14780395545846350 23915182360346949 38695577906193299 62610760266540248 101306338172733547 163917098439273795 265223436612007342 429140535051281137 694363971663288479 1123504506714569616 1817868478377858095 2941372985092427711 4759241463470285806 7700614448562713517 12459855912032999323 20160470360595712840 32620326272628712163 52780796633224425003 85401122905853137166 138181919539077562169 223583042444930699335 361764961984008261504 585348004428938960839 947112966412947222343 1532460970841886183182 2479573937254833405525 1 2 3 4 5 6 7 8 9 30 18 13 9 8 6 5 7 4 30 19 11 10 8 7 6 5 4 29 19 13 8 8 7 5 4 5 30.10 17.60 12.49 9.691 7.918 6.694 5.799 5.115 4.575 original data rescaled by 2 rescaled by 7 (Fn)100 n=1 (2Fn)100 n=1 (7Fn)100 n=1 #{D1 =d} 103 ·R 18.84 14.93 16.91 102·log(1 + d−1) d Fig 11. Illustrating the (approximate) scale-invariance of the first one-hundred Fibonacci numbers, cf. Fig 5.
A basic theory of Benford’s Law 47 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 4 6 10 14 22 26 34 38 46 58 62 74 82 86 94 106 118 122 134 142 146 158 166 178 194 202 206 214 218 226 254 262 274 278 298 302 314 326 334 346 358 362 382 386 394 398 422 446 454 458 466 478 482 502 514 526 538 542 554 562 566 586 614 622 626 634 662 674 694 698 706 718 734 746 758 766 778 794 802 818 838 842 862 866 878 886 898 914 922 926 934 958 974 982 998 1006 1018 1042 1046 1082 14 21 35 49 77 91 119 133 161 203 217 259 287 301 329 371 413 427 469 497 511 553 581 623 679 707 721 749 763 791 889 917 959 973 1043 1057 1099 1141 1169 1211 1253 1267 1337 1351 1379 1393 1477 1561 1589 1603 1631 1673 1687 1757 1799 1841 1883 1897 1939 1967 1981 2051 2149 2177 2191 2219 2317 2359 2429 2443 2471 2513 2569 2611 2653 2681 2723 2779 2807 2863 2933 2947 3017 3031 3073 3101 3143 3199 3227 3241 3269 3353 3409 3437 3493 3521 3563 3647 3661 3787 1 2 3 4 5 6 7 8 9 25 19 19 20 8 2 4 2 1 17 12 13 9 10 10 9 11 9 31 26 22 5 3 2 6 1 4 30.10 17.60 12.49 9.691 7.918 6.694 5.799 5.115 4.575 original data rescaled by 2 rescaled by 7 (pn)100 n=1 (2pn)100 n=1 (7pn)100 n=1 #{D1 =d} 103·R 103.0 131.0 95.06 102·log(1 + d−1) d Fig 12. Illustrating the lack of scale-invariance for the first one-hundred prime numbers.
Example 4.24 ([Mo]). Consider a two-person game where Player A and Player B each independently choose a (real) number greater than or equal to 1, and Player A wins if the product of their two numbers starts with a 1, 2, or 3; otherwise, Player B wins. Using the tools presented in this section, it may easily be seen that there is a strategy for Player A to choose her numbers so that she wins with probability at least log 4 ∼ = 60.2%, no matter what strategy Player B uses. Conversely, there is a strategy for Player B so that Player A will win no more than log 4 of the time, no matter what strategy Player A uses.
The idea is simple, using the scale-invariance property of BL discussed above.
If Player A chooses her number X randomly according to BL, then since BL is scale-invariant, it follows from Theorem 4.13(i) and Example 4.19(i) that X · y is still Benford no matter what number y Player B chooses, so Player A will win with the probability that a Benford random variable has first significant digit less than 4, i.e. with probability exactly log 4. Conversely, if Player B chooses his number Y according to BL then, using scale-invariance again, x · Y is Benford, so Player A will again win with the probability exactly log 4. In fact, as will now be shown, BL is the only optimal strategy for each player.
48 A. Berger and T.P. Hill 1 2 3 4 5 6 7 8 9 104·log(1+d−1) 3010.
1760.
1249.
969.1 791.8 669.4 579.9 511.5 457.5 3011 1762 1250 968 792 668 580 513 456 3009 1763 1248 970 792 670 580 511 457 3009 1762 1249 969 791 668 583 511 458 1601 1129 1097 1069 1055 1013 1027 1003 1006 1653 1572 1504 1469 1445 1434 584 174 165 5104 1016 585 573 556 556 541 543 526 103·R 0.1574 0.2087 0.3080 140.9 209.3 135.7 Fibonacci Prime d original original ×2 ×2 ×7 ×7 Fig 13. When the sample size is increased from N = 102 to N = 104 the Fibonacci numbers are even closer to scale-invariance. For the primes, this is not the case at all, see also Fig 5.
To prepare for the formal argument, model the strategy of Player A, i.e. the way this player chooses a number, by a probability measure P on (R+, B+).
For example, if Player A chooses the same number a all the time, then P = δa.
(Game theorists refer to this as a pure strategy.) Similarly, Q represents the strategy of Player B. Denote by M+ the set of all probability measures on (R+, B+) and, given P, Q ∈M+, let p(P, Q) ∈[0, 1] be the probability that Player A wins, i.e., the product of the chosen numbers begins with 1, 2, or 3, assuming Players A and B choose their numbers independently and according to the strategies P and Q, respectively. It is natural for Player A to try to maximize infQ∈M+ p(P, Q), whereas Player B aims at minimizing supP ∈M+ p(P, Q). Which strategies should the players choose, and what probabilities of winning/losing are achievable/unavoidable?
In view of the informal discussion above, it may not come as a complete surprise that these questions ultimately have very simple answers. A little pre-paratory work is required though. To this end, for every 0 ≤s < 1 and P ∈M+, let GP (s) := P {x > 0 : S(x) ≤10s} , and note that s 7→GP (s) is non-decreasing, right-continuous, with GP (0) ≥0 as well as lims↑1 GP (s) = 1. (With the terminology and notation introduced in Sections 2.3 and 3.3 simply GP (s) = FS∗P (10s).) Extend GP to a (non-decreasing, right-continuous) function GP : R →R by setting GP (s) := GP (⟨s⟩) + ⌊s⌋ for all s ∈R , A basic theory of Benford’s Law 49 and let gP (s) := GP (s) −s. Since gP (s + 1) = GP (s + 1) −(s + 1) = GP (⟨s⟩) −⟨s⟩= gP (s) , the function gP is 1-periodic with gP (0) = 0. Also, gP is Riemann integrable, and |gP (s)| ≤1 for all s ∈R. With these preliminary definitions, observe now that, given any a > 0, p(P, δa) = ( GP (log 4 −⟨log a⟩) + 1 −GP (1 −⟨log a⟩) if ⟨log a⟩< log 4 GP (1 + log 4 −⟨log a⟩) −GP (1 −⟨log a⟩) if ⟨log a⟩≥log 4 = gP (1 + log 4 −⟨log a⟩) −gP (1 −⟨log a⟩) + log 4 = log 4 + hP (⟨log a⟩) , where the 1-periodic, Riemann integrable function hP : R →R is given by hP (s) = gP (1 + log 4 −s) −gP (1 −s) , s ∈R .
From R 1 0 hP (s) ds = 0, it follows that cP := infs∈R hP (s) ≤0. Consequently, if cP < 0 then infQ∈M+ p(P, Q) ≤infa>0 p(P, δa) = log 4 + cP < log 4 .
On the other hand, if cP = 0 then necessarily hP (s) = 0 for a.e. s and hence, as gP is right-continuous, gP (−s + log 4) = gP (−s) for all s ∈R .
This in turn implies that gP (⟨n log 4⟩) = gP (0) for all n ∈N. Recall now that gP has at most countably many discontinuities and that (⟨n log 4⟩) is u.d. mod 1 and hence dense in the interval [0, 1). Thus, if 0 < s0 < 1 is a point of continuity of gP , then choosing a sequence 1 ≤n1 < n2 < . . . with limj→∞⟨nj log 4⟩= s0 shows that gP (s0) = limj→∞gP (⟨nj log 4⟩) = gP (0) .
With the possible exception of at most countably many s therefore, GP (s) = s + gP (0) whenever 0 ≤s < 1. But since s 7→GP (s) is non-decreasing with GP (s) ≥0 and lims↑1 GP (s) = 1, gP (0) = 0 and GP (s) = s must in fact hold for all s, i.e.
P {x > 0 : S(x) ≤10s} ≡s .
In other words, P is Benford. Overall therefore infQ∈M+ p(P, Q) ≤log 4 = 0.6020 . . . , with equality holding if and only if P is Benford. Thus the unique optimal strategy for Player A is to choose her numbers according to BL.
A completely analogous argument shows that supP ∈M+ p(P, Q) ≥log 4 , 50 A. Berger and T.P. Hill with equality holding if and only if Q is Benford. Hence the unique optimal strategy for Player B to minimize the probability of loosing is also to choose numbers obeying BL. Overall, supP ∈M+ infQ∈M+ p(P, Q) = log 4 = infQ∈M+ supP ∈M+ p(P, Q) holds, and the value (expected gain) of one game for Player A is given by log 4 −(1 −log 4) = 0.2041 . . . > 1 5.
If both players are required to choose positive integers then their strategies are probabilities on (N, N ∩B). Denote by MN the set of all such probabilities.
Since {⟨log n⟩: n ∈N} is dense in [0, 1), the above argument shows that infQ∈MN p(P, Q) < log 4 for every P ∈MN, and similarly supP ∈MN p(P, Q) > log 4 for every Q ∈MN. On the other hand, given ε > 0, it is not hard to find Pε, Qε ∈MN such that log 4 −ε < infQ∈MN p(Pε, Q) < log 4 < supP ∈MN p(P, Qε) < log 4 + ε .
Indeed, it is enough to choose Pε, Qε such that these probabilities approximate BL sufficiently well. (Recall Example 3.9 which also showed that no P ∈MN is Benford.) When played with positive integers only, therefore, the game has no optimal strategy for either player, but there are ε-optimal strategies for every ε > 0, and supP ∈MN infQ∈MN p(P, Q) = log 4 = infQ∈MN supP ∈MN p(P, Q) still holds.
Theorem 4.20 showed that for a probability measure P on (R+, B+) to have scale-invariant significant digits it is necessary (and sufficient) that P be Ben-ford. In fact, as noted in [Sm], this conclusion already follows from a much weaker assumption: It is enough to require that the probability of a single significant digit remain unchanged under scaling.
Theorem 4.25. For every random variable X with P(X = 0) = 0 the following statements are equivalent: (i) X is Benford.
(ii) There exists a number d ∈{1, 2, . . ., 9} such that P(D1(αX) = d) = P(D1(X) = d) for all α > 0 .
In particular, (ii) implies that P(D1(X) = d) = log(1 + d−1).
A basic theory of Benford’s Law 51 Proof. Assume first that X is Benford. By Theorem 4.20, X has scale-invariant significant digits. Thus for every α > 0, P(D1(αX) = d) = log(1 + d−1) = P(D1(X) = d) for all d = 1, 2, . . ., 9 .
Conversely, assume that (ii) holds. Similarly as in the proof of Theorem 4.22(i), for every 0 ≤s < 1 let GX(s) := P(S(X) < 10s) .
Hence GX is non-decreasing and left-continuous, with GX(0) = 0, and P(D1(X) = d) = GX log(1 + d) −GX(log d) .
Extend GX to a (non-decreasing, left-continuous) function GX : R →R by setting GX(s) := GX(⟨s⟩) + ⌊s⌋, and let gX(s) := GX(s) −s. Hence gX is 1-periodic, Riemann-integrable, with gX(0) = 0 and |gX(s)| ≤1. Specifically, P(D1(X) = d) = gX log(1 + d) −gX(log d) + log(1 + d−1) , and essentially the same calculation as in Example 4.24 shows that P(D1(αX) = d) = gX(log(1 + d) −⟨log α⟩) −gX(log d −⟨log α⟩) + log(1 + d−1) .
With the 1-periodic, Riemann-integrable hX : R →R given by hX(s) = gX(log(1 + d) −s) −gX(log d −s) , the assumption that P(D1(αX) = d) = P(D1(X) = d) for all α > 0 simply means that hX(s) ≡hX(0), i.e., hX is constant, and so is the function s 7→ gX(log(1 + d) −s) −gX(log d −s). The same Fourier series argument as in the proof of Theorem 4.22 now applies: From gX(s) = X k∈Z cke2πıks , it follows that gX(log(1 + d) −s) −gX(log d −s) = X k∈Z ck e2πık log(1+d)−e2πık log d e2πıks = X k∈Z cke2πık log d e2πık log(1+d−1)−1 e2πıks , and since log(1 + d−1) is irrational for every d ∈N, necessarily ck = 0 for all k ̸= 0, i.e., gX is constant almost everywhere, and GX(s) = s + c0 for a.e.
s ∈[0, 1). As GX is non-decreasing with GX(0) = 0, overall, GX(s) ≡s, which in turn shows that X is Benford.
Remark. A close inspection of the above proof shows that Theorem 4.25 can still be strengthened in different ways. On the one hand, other significant digits can 52 A. Berger and T.P. Hill be considered. For example, the theorem (and its proof also) remain virtually unchanged if in (ii) it is assumed that, for some m ≥2 and some d ∈{0, 1, . . ., 9}, P(Dm(αX) = d) = P(Dm(X) = d) for all α > 0 .
On the other hand, it is enough to assume in (ii) that, for some d ∈{1, 2, . . ., 9}, P(D1(αnX) = d) = P(D1(X) = d) for all n ∈N , with the sequence (αn) of positive numbers being such that {⟨log αn⟩: n ∈N} is dense in [0, 1). Possible choices for such a sequence include (2n), (n2), and the sequence of prime numbers. For example, therefore, X is Benford if and only if P(D1(2nX) = 1) = P(D1(X) = 1) for all n ∈N .
♣ Example 4.26 ([Sm]). (“Ones-scaling-test”) In view of the last remark, to in-formally test whether a sample of data comes from a Benford distribution, sim-ply compare the proportion of the sample that has first significant digit 1 with the proportion after the data has been re-scaled, i.e. multiplied by α, α2, α3, . . ., where log α is irrational, e.g. α = 2. In fact, it is enough to consider only re-scalings by αn2, n = 1, 2, 3, . . .. On the other hand, note that merely assuming P(D1(2X) = d) = P(D1(X) = d) for all d = 1, 2, . . . , 9 , (4.7) is not sufficient to guarantee that X is Benford. Indeed, (4.7) holds for instance if X attains each of the four values 1, 2, 4, 8 with equal probability 1 4.
4.3. The base-invariance characterization One possible drawback to the hypothesis of scale-invariance in some tables is the special role played by the constant 1. For example, consider two physical laws, namely Newton’s lex secunda F = ma and Einstein’s famous E = mc2.
Both laws involve universal constants. In Newton’s law, the constant is usually made equal to 1 by the choice of units of measurement, and this 1 is then not recorded in most tables of universal constants. On the other hand, the speed of light c in Einstein’s equation is typically recorded as a fundamental constant. If a “complete” list of universal physical constants also included the 1s, it seems plausible that this special constant might occur with strictly positive frequency.
But that would clearly violate scale-invariance, since then the constant 2, and in fact every other constant as well would occur with this same positive probability, which is impossible.
Instead, suppose it is assumed that any reasonable universal significant-digit law should have base-invariant significant digits, that is, the law should be equally valid when rewritten in terms of bases other than 10. In fact, all of the classical arguments supporting BL carry over mutatis mutandis [Ra1] to other bases. As will be seen shortly, a hypothesis of base-invariant significant digits characterizes mixtures of BL and a Dirac probability measure concentrated on the special constant 1 which may occur with positive probability.
A basic theory of Benford’s Law 53 Just as the only scale-invariant real-valued random variable is 0 with prob-ability one, the only positive random variable X that is base-invariant, i.e.
X = 10Y with some random variable Y for which Y, 2Y, 3Y, . . . all have the same distribution, is the random variable which almost surely equals 1, that is, P(X = 1) = 1. This follows from the fact that all nY have the same distribution for n = 1, 2, 3 . . ., and hence P(Y = 0) = 1, as shown in the previous section.
On the other hand, a positive random variable (or sequence, function, dis-tribution) can have base-invariant significant digits. The idea behind base-invariance of significant digits is simply this: A base-10 significand event A corresponds to the base-100 event A1/2, since the new base b = 100 is the square of the original base b = 10. As a concrete example, denote by A the set of positive reals with first significant digit 1, i.e.
A = {x > 0 : D1(x) = 1} = {x > 0 : S(x) ∈[1, 2)} .
It is easy to see that A1/2 is the set A1/2 = {x > 0 : S(x) ∈[1, √ 2) ∪[ √ 10, √ 20)} .
Consider now the base-100 significand function S100, i.e., for any x ̸= 0, S100(x) is the unique number in [1, 100) such that |x| = 100kS100(x) for some, necessarily unique k ∈Z. (To emphasize that the usual significand function S is taken relative to base 10, it will be denoted S10 throughout this section.) Clearly, A = {x > 0 : S100(x) ∈[1, 2) ∪[10, 20)} .
Hence, letting a = log 2, x > 0 : Sb(x) ∈[1, ba/2) ∪[b1/2, b(1+a)/2) = ( A1/2 if b = 10 , A if b = 100 .
Thus, if a distribution P on the significand σ-algebra S has base-invariant sig-nificant digits, then P(A) and P(A1/2) should be the same, and similarly for other integral roots (corresponding to other integral powers of the original base b = 10). Thus P(A) = P(A1/n) should hold for all n. (Recall from Lemma 2.13(iii) that A1/n ∈S for all A ∈S and n ∈N, so those probabilities are well-defined.) This motivates the following definition.
Definition 4.27. Let A ⊃S be a σ-algebra on R+. A probability measure P on (R+, A) has base-invariant significant digits if P(A) = P(A1/n) holds for all A ∈S and n ∈N.
Example 4.28. (i) Recall that δa denotes the Dirac measure concentrated at the point a, that is, δa(A) = 1 if a ∈A, and δa(A) = 0 if a ̸∈A. The probability measure δ1 clearly has base-invariant significant digits since 1 ∈A if and only if 1 ∈A1/n. Similarly, δ10k has base-invariant significant digits for every k ∈Z.
On the other hand, δ2 does not have base-invariant significant digits since, with A = {x > 0 : S10(x) ∈[1, 3)}, δ2(A) = 1 yet δ2(A1/2) = 0.
54 A. Berger and T.P. Hill A = {D1 = 2} log A = {2 ≤S < 3} √ A log √ A S log S B( √ A) = B(A) λ0,1(log √ A) = λ0,1(log A) 1 2 3 4 5 6 7 8 9 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Fig 14. Visualizing the base-invariant significant digits of BL.
(ii) It is easy to see that the Benford distribution B has base-invariant sig-nificant digits. Indeed, for any 0 ≤s < 1, let A = {x > 0 : S10(x) ∈[1, 10s)} = [ k∈Z 10k[1, 10s) ∈S .
Then, as seen in the proof of Lemma 2.13(iii), A1/n = [ k∈Z 10k [n−1 j=0 [10j/n, 10(j+s)/n) and therefore B(A1/n) = Xn−1 j=0 log 10(j+s)/n −log 10j/n = Xn−1 j=0 j + s n −j n = s = B(A) .
(iii) The uniform distribution λ0,1 on [0, 1) does not have base-invariant significant digits. For instance, again taking A = {x > 0 : D1(x) = 1} leads to λ0,1(A1/2) = X n∈N 10−n( √ 2 −1 + √ 20 − √ 10) = 1 9 + ( √ 5 −1)(2 − √ 2) 9 > 1 9 = λ0,1(A) .
(iv) The probability measure 1 2δ1 + 1 2B has base-invariant significant digits since both δ1 and B do.
Example 4.29. Completely analogously to the case of scale-invariance, it is possible to introduce a notion of a sequence or function having base-invariant significant digits and to formulate an analoge of Theorem 4.22 in the context of Theorem 4.30 below. With this, the sequence (Fn) has base-invariant significant digits, whereas the sequence (pn) does not. As in Example 4.23, this is illustrated empirically in Fig 15 to 17.
A basic theory of Benford’s Law 55 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1346269 2178309 3524578 5702887 9227465 14930352 24157817 39088169 63245986 165580141 165580141 267914296 433494437 701408733 1134903170 1836311903 2971215073 4807526976 7778742049 12586269025 20365011074 32951280099 53316291173 86267571272 139583862445 225851433717 365435296162 591286729879 956722026041 1548008755920 2504730781961 4052739537881 6557470319842 10610209857723 17167680177565 27777890035288 44945570212853 72723460248141 117669030460994 190392490709135 308061521170129 498454011879264 806515533049393 1304969544928657 2111485077978050 3416454622906707 5527939700884757 8944394323791464 14472334024676221 23416728348467685 37889062373143906 61305790721611591 99194853094755497 160500643816367088 259695496911122585 420196140727489673 679891637638612258 1100087778366101931 1779979416004714189 2880067194370816120 4660046610375530309 7540113804746346429 12200160415121876738 19740274219868223167 31940434634990099905 51680708854858323072 83621143489848422977 135301852344706746049 218922995834555169026 354224848179261915075 1.000 1.000 4.000 9.000 2.500 6.400 1.690 4.410 1.156 3.025 7.921 2.073 5.428 1.421 3.721 9.741 2.550 6.677 1.748 4.576 1.198 3.136 8.212 2.149 5.628 1.473 3.858 1.010 2.644 6.922 1.812 4.745 1.242 3.252 8.514 2.229 5.836 1.527 4.000 1.047 2.741 7.177 1.879 4.919 1.288 3.372 8.828 2.311 6.050 1.584 4.147 1.085 2.842 7.442 1.948 5.100 1.335 3.496 9.153 2.396 6.273 1.642 4.300 1.125 2.947 7.716 2.020 5.288 1.384 3.624 9.490 2.484 6.504 1.702 4.458 1.167 3.055 8.000 2.094 5.483 1.435 3.758 9.839 2.576 6.744 1.765 4.622 1.210 3.168 8.294 2.171 5.685 1.488 3.896 1.020 2.670 6.992 1.830 4.792 1.254 1.000 1.000 1.280 2.187 7.812 2.097 6.274 1.801 5.252 1.522 4.423 1.283 3.728 1.082 3.142 9.124 2.649 7.692 2.233 6.484 1.882 5.466 1.587 4.608 1.337 3.884 1.127 3.274 9.508 2.760 8.015 2.327 6.756 1.961 5.696 1.653 4.801 1.394 4.047 1.175 3.412 9.907 2.876 8.352 2.424 7.040 2.044 5.935 1.723 5.003 1.452 4.217 1.224 3.555 1.032 2.997 8.703 2.526 7.336 2.130 6.184 1.795 5.213 1.513 4.395 1.276 3.705 1.075 3.123 9.068 2.633 7.644 2.219 6.444 1.871 5.432 1.577 4.579 1.329 3.860 1.120 3.254 9.449 2.743 7.966 2.312 6.715 1.949 5.661 1.643 4.772 1.385 4.023 1.168 3.391 9.846 2.858 8.300 2.410 6.997 1 2 3 4 5 6 7 8 9 30 18 13 9 8 6 5 7 4 31 17 12 11 7 8 4 5 5 31 18 11 9 8 7 6 4 6 30.10 17.60 12.49 9.691 7.918 6.694 5.799 5.115 4.575 original data base 7→base2 base 7→base7 (Fn)100 n=1 (F 2 n)100 n=1 (F 7 n)100 n=1 103 ·R 18.84 17.99 14.93 102·log(1 + d−1) d #{D1 =d} Fig 15. Illustrating the (approximate) base-invariance of the first one-hundred Fibonacci num-bers. (In the two middle tables, the values of S(F 2 n) and S(F 7 n), respectively, are shown to four correct digits.) 56 A. Berger and T.P. Hill 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 4.000 9.000 2.500 4.900 1.210 1.690 2.890 3.610 5.290 8.410 9.610 1.369 1.681 1.849 2.209 2.809 3.481 3.721 4.489 5.041 5.329 6.241 6.889 7.921 9.409 1.020 1.060 1.144 1.188 1.276 1.612 1.716 1.876 1.932 2.220 2.280 2.464 2.656 2.788 2.992 3.204 3.276 3.648 3.724 3.880 3.960 4.452 4.972 5.152 5.244 5.428 5.712 5.808 6.300 6.604 6.916 7.236 7.344 7.672 7.896 8.008 8.584 9.424 9.672 9.796 1.004 1.095 1.135 1.204 1.218 1.246 1.288 1.346 1.391 1.436 1.466 1.513 1.576 1.608 1.672 1.755 1.772 1.857 1.874 1.927 1.962 2.016 2.088 2.125 2.143 2.180 2.294 2.371 2.410 2.490 2.530 2.590 2.714 2.735 2.926 1.280 2.187 7.812 8.235 1.948 6.274 4.103 8.938 3.404 1.724 2.751 9.493 1.947 2.718 5.066 1.174 2.488 3.142 6.060 9.095 1.104 1.920 2.713 4.423 8.079 1.072 1.229 1.605 1.828 2.352 5.328 6.620 9.058 1.002 1.630 1.789 2.351 3.057 3.622 4.637 5.888 6.364 9.273 9.974 1.151 1.235 1.861 2.742 3.105 3.302 3.728 4.454 4.721 6.276 7.405 8.703 1.019 1.073 1.251 1.383 1.453 1.853 2.570 2.813 2.943 3.216 4.353 4.936 6.057 6.306 6.830 7.685 8.967 1.004 1.123 1.208 1.347 1.554 1.667 1.914 2.267 2.344 2.762 2.853 3.142 3.348 3.678 4.163 4.424 4.561 4.844 5.785 6.496 6.879 7.703 8.146 8.851 1.041 1.070 1.356 1 2 3 4 5 6 7 8 9 25 19 19 20 8 2 4 2 1 35 24 9 5 8 5 5 3 6 d 33 15 11 11 4 10 4 7 5 30.10 17.60 12.49 9.691 7.918 6.694 5.799 5.115 4.575 original data base 7→base2 base 7→base7 (pn)100 n=1 (p2 n)100 n=1 (p7 n)100 n=1 #{D1 =d} 103·R 103.0 63.90 39.18 102 ·log(1 + d−1) Fig 16. Illustrating the lack of base-invariance for the first one-hundred prime numbers. (In the two middle tables, the values of S(p2 n) and S(p7 n), respectively, are shown to four correct digits.) The next theorem is the main result for base-invariant significant digits. It shows that convex combinations as in Example 4.28(iv) are the only probability distributions with base-invariant significant digits. To put the argument in per-spective, recall that the proof of the scale-invariance theorem (Theorem 4.20) ultimately depended on Theorem 4.13(i,ii) which in turn was proved analyti-cally using Fourier analysis. The situation here is similar: An analytical result (Lemma 4.32 below) identifies all probability measures on [0, 1), B[0, 1) that are invariant under every map x 7→⟨nx⟩on [0, 1). Once this tool is available, it is straightforward to prove Theorem 4.30 (Base-invariance characterization [Hi1]). A probability measure P on (R+, A) with A ⊃S has base-invariant significant digits if and only if, for some q ∈[0, 1], P(A) = qδ1(A) + (1 −q)B(A) for every A ∈S .
(4.8) A basic theory of Benford’s Law 57 1 2 3 4 5 6 7 8 9 104·log(1+d−1) 3010.
1760.
1249.
969.1 791.8 669.4 579.9 511.5 457.5 3011 1762 1250 968 792 668 580 513 456 3012 1760 1248 971 791 672 577 513 456 3011 1762 1248 969 791 671 579 511 458 1601 1129 1097 1069 1055 1013 1027 1003 1006 3012 1626 1200 987 798 716 609 536 516 2340 1437 1195 1036 944 844 775 745 684 103·R 0.1574 0.2919 0.1532 140.9 67.02 36.85 Fibonacci Prime d original original b 7→b2 b 7→b2 b 7→b7 b 7→b7 Fig 17. Increasing the sample size from N = 102 to N = 104 makes the Fibonacci numbers’ leading digits even more closely base-invariant. As in the case of scale-invariance, this is not at all true for the primes, cf. Fig 13.
Corollary 4.31. A continuous probability measure P on R+ has base-invariant significant digits if and only if P(A) = B(A) for all A ∈S, i.e., if and only if P is Benford.
Recall that λ0,1 denotes Lebesgue measure on [0, 1), B[0, 1) . For each n ∈N, denote the map x 7→⟨nx⟩of [0, 1) into itself by Tn. Generally, if T : [0, 1) →R is measurable, and T [0, 1) ⊂[0, 1), a probability measure P on [0, 1), B[0, 1) is said to be T -invariant, or T is P-preserving, if T∗P = P. Which probability measures are Tn-invariant for all n ∈N? A complete answer to this question is provided by Lemma 4.32. A probability measure P on [0, 1), B[0, 1) is Tn-invariant for all n ∈N if and only if P = qδ0 + (1 −q)λ0,1 for some q ∈[0, 1].
Proof. From the proof of Theorem 4.13 recall the definition of the Fourier coef-ficients of P, b P(k) = Z 1 0 e2πıksdP(s) , k ∈Z , and observe that d TnP(k) = b P(nk) for all k ∈Z, n ∈N .
Assume first that P = qδ0 + (1 −q)λ0,1 for some q ∈[0, 1]. From b δ0(k) ≡1 and 58 A. Berger and T.P. Hill d λ0,1(k) = 0 for all k ̸= 0, it follows that b P(k) = 1 if k = 0 , q if k ̸= 0 .
For every n ∈N and k ∈Z{0}, therefore, d TnP(k) = q, and clearly d TnP(0) = 1.
Thus d TnP = b P and since the Fourier coefficients of P determine P uniquely, Tn∗P = P for all n ∈N.
Conversely, assume that P is Tn-invariant for all n ∈N. In this case, b P(n) = d TnP(1) = b P(1), and similarly b P(−n) = d TnP(−1) = b P(−1). Since generally b P(−k) = b P(k), there exists q ∈C such that b P(k) = q if k > 0 , 1 if k = 0 , q if k < 0 .
Also, observe that for every t ∈R limn→∞ 1 n Xn j=1 e2πıtj = ( 1 if t ∈Z , 0 if t ̸∈Z .
Using this and the Dominated Convergence Theorem, it follows from P({0}) = Z 1 0 limn→∞ 1 n Xn j=1 e2πısjdP(s) = limn→∞ 1 n Xn j=1 b P(j) = q , that q is real, and in fact q ∈[0, 1]. Hence the Fourier coefficients of P are exactly the same as those of qδ0 + (1 −q)λ0,1. By uniqueness, therefore, P = qδ0 + (1 −q)λ0,1.
Remark. Note that P is Tmn-invariant if it is both Tm- and Tn-invariant. Thus, in Lemma 4.32 it is enough to require that P be Tn-invariant whenever n is a prime number.
It is natural to ask how small the set M of natural numbers n can be chosen for which Tn-invariance really has to be required in Lemma 4.32. By the observation just made, it can be assumed that M is closed under multiplication, hence a (multiplicative) semi-group. If M is lacunary, i.e. M ⊂{pm : m ∈N} for some p ∈N, then probability measures P satisfying Tn∗P = P for all n ∈M exist in abundance, and hence an analogue of Lemma 4.32 cannot hold. If, on the other hand, M is not lacunary, then it is not known in general whether an appropriate analogue of Lemma 4.32 may hold. For example, if M = {2m13m2 : m1, m2 ∈N0} then the probability measure P = 1 4 P4 j=1 δj/5 is Tn-invariant for every n ∈M, but it is a famous open question of H. Furstenberg [Ei] whether any continuous probability measure with this property exists — except, of course, for P = λ0,1.
♣ A basic theory of Benford’s Law 59 Proof of Theorem 4.30. As in the proof of Theorem 4.20, fix a probability mea-sure P on (R+, A), denote by P0 its restriction to (R+, S), and let Q = ℓ∗P0.
Observe that P0 has base-invariant significant digits if and only if Q is Tn-invariant for all n ∈N. Indeed, with 0 ≤s < 1 and A = {x > 0 : S10(x) < 10s}, Tn∗Q [0, s) = Q [n−1 j=0 h j n, j + s n = P0 [ k∈Z 10k [n−1 j=0 [10j/n, 10(j+s)/n) = P0(A1/n) and hence Tn∗Q = Q for all n precisely if P0 has base-invariant significant digits.
In this case, by Lemma 4.32, Q = qδ0 + (1 −q)λ0,1 for some q ∈[0, 1], which in turn implies that P0(A) = qδ1(A) + (1 −q)B(A) for every A ∈S.
Corollary 4.33. If a probability measure on R+ has scale-invariant significant digits then it also has base-invariant significant digits.
4.4. The sum-invariance characterization No finite data set can obey BL exactly, since the Benford probabilities of sets with m given significant digits become arbitrarily small as m goes to infinity, and no discrete probability measure with finitely many atoms can take arbitrarily small positive values. But, as first observed by M. Nigrini [Ni], if a table of real data approximately follows BL, then the sum of the significands of all entries in the table with first significant digit 1 is very close to the sum of the significands of all entries with first significant digit 2, and to the sum of the significands of entries with the other possible first significant digits as well. This clearly implies that the table must contain more entries starting with 1 than with 2, more entries starting with 2 than with 3, and so forth. Similarly, the sums of significands of entries with D1 = d1, . . . , Dm = dm are approximately equal for all tuples (d1, . . . , dm) of a fixed length m. In fact, even the sum-invariance of first or first and second digits yields a distribution close to BL, see Fig 18 and 19. Nigrini conjectured, and partially proved, that this sum-invariance property also characterizes BL. Note that it is the significands of the data, rather than the data themselves, that are summed up. Simply summing up the raw data will not lead to any meaningful conclusion, as the resulting sums may be dominated by a few very large numbers. It is only through considering significands that the magnitude of the individual numbers becomes irrelevant.
To motivate a precise definition of sum-invariance, note that if (xn) is Benford then the set {xn : n ∈N} is necessarily infinite, and consequently, for every d ∈{1, 2, . . ., 9}, the sum P n:D1(xn)=d S(xn) is infinite as well. To compare such sums, it is natural to normalise them by considering limiting averages. To this end, for every m ∈N, d1 ∈{1, 2, . . ., 9} and dj ∈{0, 1, . . ., 9}, j ≥2, define Sd1,...,dm(x) := S(x) if D1(x), . . . , Dm(x) = (d1, . . . , dm) , 0 otherwise.
60 A. Berger and T.P. Hill d Nd = #{xn=d} dNd Nd/N log(1+d−1) 1 2520 2520 0.3535 0.3010 2 1260 2520 0.1767 0.1761 3 840 2520 0.1178 0.1240 4 630 2520 0.0884 0.0969 5 504 2520 0.0707 0.0792 6 620 2520 0.0589 0.0669 7 360 2520 0.0505 0.0580 8 315 2520 0.0442 0.0512 9 280 2520 0.0393 0.0458 N = P d Nd = 7129 Fig 18. A (hypothetical) sample x1, x2, . . . xN containing N = 7129 numbers from {1, 2, . . . , 9} and showing exact sum-invariance for the first digit. Note that the relative frequencies Nd/N are quite close to the Benford probabilities log(1 + d−1).
(d1, d2) Nd1,d2 = #{xn =10d1 + d2} Nd1,d2/N P d2 Nd1,d2/N log(1+d−1 1 ) .
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0.30607 0.30102 0.04510 0.04575 (1, 0) 6972037522971247716453380893531230355680 0.04258 (1, 1) 6338215929973861560412164448664754868800 0.03871 (1, 2) 5810031269142706430377817411276025296400 0.03549 (1, 3) 5363105786900959781887216071947100273600 0.03276 (1, 4) 4980026802122319797466700638236593111200 0.03042 (1, 5) 4648025015314165144302253929020820237120 0.02839 (1, 6) 4357523451857029822783363058457018972300 0.02661 (1, 7) 4101198542924263362619635819724253150400 0.02505 (1, 8) 3873354179428470953585211607517350197600 0.02366 (1, 9) 3669493433142761956028095207121700187200 0.02241 (2, 0) 3486018761485623858226690446765615177840 0.02129 (8, 9) 783375002581039069264424819497891051200 0.00478 (9, 0) 774670835885694190717042321503470039520 0.00473 (9, 1) 766157969557279968841030867421014324800 0.00468 (9, 2) 757830165540353012657976184079481560400 0.00463 (9, 3) 749681454082929861984234504680777457600 0.00458 (9, 4) 741706119465026352814189456758641527200 0.00453 (9, 5) 733898686628552391205619041424340037440 0.00448 (9, 6) 726253908642838303797227176409503162050 0.00444 (9, 7) 718766754945489455304472257065075294400 0.00439 (9, 8) 711432400303188542495242948319513301600 0.00434 (9, 9) 704246214441540173379129383184972763200 0.00430 N = P d1,d2 Nd1,d2 = 163731975056100444033114230488313094880847 ≈1.637·1041 (10d1 + d2)Nd1,d2 ≡69720375229712477164533808935312303556800 ≈6.972·1040 Fig 19. An (even more hypothetical) sample x1, x2, . . . xN containing N ≈1.637 · 1041 num-bers from {10, 11, . . . , 99} and showing exact sum-invariance for the first two digits. When compared with the values in Fig 18, the relative frequencies P d2 Nd1,d2/N of the first digits are even closer to the Benford values log(1 + d−1 1 ).
A basic theory of Benford’s Law 61 1 2 3 4 5 6 7 8 9 42.71 43.82 44.75 40.35 43.28 38.67 37.10 59.21 38.58 37.67 47.68 65.92 89.59 42.17 12.80 29.30 17.20 9.700 Fibonacci Prime d N = 102 Exact sum-invariance: 102·ESd = 100 ln 10 ≈43.43 for d = 1, 2, . . . , 9 Fig 20. Except for d = 8, the value of P D1=d S does not vary much with d for the first one-hundred Fibonacci numbers, but it varies wildly for the first one-hundred primes.
Definition 4.34. A sequence (xn) of real numbers has sum-invariant significant digits if, for every m ∈N, the limit limN→∞ PN n=1 Sd1,...,dm(xn) N exists and is independent of d1, . . . , dm.
In particular, therefore, if (xn) has sum-invariant significant digits then there exists c > 0 such that limN→∞ PN n=1 Sd1(xn) N = c for all d1 = 1, 2, . . . , 9.
As will follow from Theorem 4.37 below, the sequence (2n) and the Fibonacci sequence (Fn) have sum-invariant significant digits. Clearly, (10n) does not have sum-invariant significant digits since all the first digits are 1, i.e. for all N, PN n=1 Sd1(10n) N = ( 1 if d1 = 1 , 0 if d1 ≥2.
Not too surprisingly, the sequence (pn) of prime numbers does not have sum-invariant significant digits either, see Fig 20.
The definitions of sum-invariance of significant digits for functions, distribu-tions and random variables are similar, and it is in the context of distributions and random variables that the sum-invariance characterization of BL will be established. Informally, a probability distribution has sum-invariant significant digits if in a collection of numbers with that distribution, the sums of (the sig-nificands of) all entries with first significant digit 1 is the same as each of the sums of all entries with the other first significant digits; and the sum of all the entries with, say, first two significant digits 1 and 3, respectively, is the same as the sum of all entries with any other combination of first two significant digits, etc; and similarly for all other finite initial sequences of significant digits. In complete analogy to Definition 4.34, this is put more formally by Definition 4.35. A random variable X has sum-invariant significant digits if, for every m ∈N, the value of ESd1,...,dm(X) is independent of d1, . . . , dm.
62 A. Berger and T.P. Hill Example 4.36. (i) If X is uniformly distributed on [0, 1), then X does not have sum-invariant significant digits. This follows from Theorem 4.37 below but can also be seen by a simple direct calculation. Indeed, for every d1 ∈{1, 2, . . ., 9}, ESd1(X) = X n∈N 10n Z 10−n(d1+1) 10−nd1 t dt = 2d1 + 1 18 , which obviously depends on d1.
(ii) Similarly, if P(X = 1) = 1 then X does not have sum-invariant significant digits, as ESd1(X) = ( 1 if d1 = 1 , 0 if d1 ≥2 .
(iii) Assume that X is Benford. For every m ∈N, d1 ∈{1, 2, . . ., 9} and dj ∈{0, 1, . . ., 9}, j ≥2, ESd1,...,dm(X) = Z d1+10−1d2+...+101−m(dm+1) d1+10−1d2+...+101−mdm t · 1 t ln 10 dt = 101−m ln 10 .
Thus X has sum-invariant significant digits. Note, however, that even in this example the higher moments of Sd1,...,dm(X) generally depend on d1, . . . , dm, as for instance ESd1(X)2 = 2d1 + 1 2 ln 10 , d1 = 1, 2, . . . , 9 .
This example shows that it would be too restrictive to require in Definition 4.35 that the distribution of the random variable Sd1,...,dm(X), rather than its expectation, be independent of d1, . . . , dm.
According to Example 4.36(iii) every Benford random variable has sum-invariant significant digits. As hinted at earlier, the converse is also true, i.e., sum-invariant significant digits characterize BL.
Theorem 4.37 (Sum-invariance characterization [Al]). A random variable X with P(X = 0) = 0 has sum-invariant significant digits if and only if it is Benford.
Proof. The “if”-part has been verified in Example 4.36(iii). To prove the “only if”-part, assume that X has sum-invariant significant digits. For every m ∈N, d1 ∈{1, 2, . . ., 9} and dj ∈{0, 1, . . ., 9}, j ≥2, let Jd1,...,dm := d1+10−1d2 + . . . + 101−mdm, d1 + 10−1d2 + . . . + 101−m(dm + 1) = 1 ≤x < 10 : D1(x), D2(x), . . . , Dm(x) = (d1, d2, . . . dm) .
With this, Sd1,...,dm(X) = S(X)1Jd1,...,dm S(X) , A basic theory of Benford’s Law 63 and by assumption ESd1,...,dm(X) is independent of d1, . . . , dm. Note that each of the 9 · 10m−1 intervals Jd1,...,dm has the same length λ(Jd1,...,dm) = 101−m.
Consequently, ESd1,...,dm(X) = 1 9 · 10m−1 ES(X) = λ(Jd1,...,dm) 9 ES(X) , and since the family Jd1,...,dm : m ∈N, d1 ∈{1, 2, . . ., 9} and dj ∈{0, 1, . . ., 9}, j ≥2 generates B[1, 10), E S(X)1[a,b) S(X) = b −a 9 ES(X) (4.9) holds for every 1 ≤a < b < 10. Given any 1 < t < 10, consider the sequence of functions (fn), where fn : R →R is given by fn = Xn j=1 n n + (t −1)j 1 1+(t−1) j−1 n ,1+(t−1) j n .
Note that fn(τ) ↑ 1[1,t)(τ) τ as n →∞, uniformly in τ. Hence by the Monotone Convergence Theorem and (4.9), P(1 ≤S(X) < t) = E1[1,t) S(X) = E S(X) 1 S(X)1[1,t) S(X) = limn→∞E S(X)fn S(X) = limn→∞ Xn j=1 n n + (t −1)j E S(X)1 1+(t−1) j−1 n ,1+(t−1) j n S(X) = limn→∞ Xn j=1 n n + (t −1)j · t −1 9n ES(X) = ES(X) 9 limn→∞ t −1 n Xn j=1 1 1 + (t −1)j/n = ES(X) 9 Z 1 0 t −1 1 + (t −1)σ dσ = ES(X) 9 ln t .
From P(1 ≤S(X) < 10) = P(X ̸= 0) = 1, it follows that ES(X) = 9 ln 10 and hence P(S(X) < t) = ln t ln 10 = log t for all t ∈[1, 10) , i.e., X is Benford.
64 A. Berger and T.P. Hill Remarks. (i) As shown by Example 4.36(iii) and Theorem 4.37, a random vari-able X has sum-invariant significant digits if and only if ESd1,...,dm(X) = 101−m ln 10 P(X ̸= 0) holds for all m ∈N, d1 ∈{1, 2, . . ., 9} and dj ∈{0, 1, . . ., 9}, j ≥2.
(ii) Theorem 4.37 provides another informal test for goodness-of-fit to BL: Simply calculate the differences between the sums of the significands of the data corresponding to the same initial sequence of significant digits, see [Ni].
♣ 5. Benford’s Law for deterministic processes The goal of this chapter is to present the basic theory of BL in the context of deterministic processes, such as iterates of maps, powers of matrices, and solutions of differential equations. Except for somewhat artificial examples, pro-cesses with linear growth are not Benford, and among the others, there is a clear distinction between those with exponential growth or decay, and those with super-exponential growth or decay. In the exponential case, processes typically are Benford for all starting points in a region, but are not Benford with respect to other bases. In contrast, super-exponential processes typically are Benford for all bases, but have small sets (of measure zero) of exceptional points whose orbits or trajectories are not Benford.
5.1. One-dimensional discrete-time processes This section presents some elementary facts about BL for one-dimensional dis-crete-time processes. The focus is first on processes with exponential growth or decay, then on processes with doubly-exponential or more general growth or decay. Finally, some possible applications such as Newton’s method, and extensions to nonautonomous and chaotic systems are discussed briefly.
Processes with exponential growth or decay Many classical integer sequences exhibiting exponential growth are known to be Benford.
Example 5.1. (i) Recall from Examples 4.11(i) and 4.12 that (2n) and the Fibonacci sequence (Fn) are Benford. Similarly, (n!) is Benford [BBH, Di], see also Fig 21.
(ii) Recall from the remark on p.18 that (n) is not Benford, but weakly Benford in the sense explained there, and the same is true for the sequence of prime numbers.
A basic theory of Benford’s Law 65 (2n) (n!) (Fn) 103·log(1+d−1) d #{D1 = d} 1 301 293 301 301.0 2 176 176 177 176.0 3 125 124 125 124.9 4 97 102 96 96.91 5 79 69 80 79.18 6 69 87 67 66.94 7 56 51 56 57.99 8 52 51 53 51.15 9 45 47 45 45.75 Fig 21. Empirical frequencies of D1 for the first 103 terms of the sequences (2n), (n!) and the Fibonacci numbers (Fn), as compared with the Benford probabilities.
Let T : C →C be a measurable map that maps C ⊂R into itself, and for every n ∈N denote by T n the n-fold iterate of T , i.e. T 1 := T and T n+1 := T n ◦T ; also let T 0 be the identity map idC on C, that is, T 0(x) = x for all x ∈C. The orbit of x0 ∈C is the sequence OT (x0) := T n−1(x0) n∈N = x0, T (x0), T 2(x0), . . .
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Note that this interpretation of the orbit as a sequence differs from terminology sometimes used in dynamical systems theory (e.g. [KH]) according to which the orbit of x0 is the mere set {T n−1(x0) : n ∈N}.
Example 5.2. (i) If T (x) = 2x then OT (x0) = (x0, 2x0, 22x0, . . .) = (2n−1x0) for all x0. Hence limn→∞|xn| = +∞whenever x0 ̸= 0.
(ii) If T (x) = x2 then OT (x0) = (x0, x2 0, x22 0 , . . .) = x2n−1 0 for all x0. Here xn approaches 0 or +∞depending on whether |x0| < 1 or |x0| > 1. Moreover, OT (±1) = (±1, 1, 1, . . .).
(iii) If T (x) = 1 + x2 then OT (x0) = (x0, 1 + x2 0, 2 + 2x2 0 + x4 0, . . .). Since xn ≥n for all x0 and n ∈N, limn→∞xn = +∞for every x0.
Recall from Example 4.11(i) that (2n) is Benford, and in fact (2nx0) is Ben-ford for every x0 ̸= 0, by Theorem 4.22. In other words, Example 5.2(i) says that with T (x) = 2x, the orbit OT (x0) is Benford whenever x0 ̸= 0. The goal of the present sub-section is to extend this observation to a much wider class of maps T . The main result (Theorem 5.8) rests upon three simple lemmas.
Lemma 5.3. Let T (x) = ax with a ∈R. Then OT (x0) is Benford for every x0 ̸= 0 or for no x0 at all, depending on whether log |a| is irrational or rational, respectively.
Proof. By Theorem 4.10, OT (x0) = (an−1x0) is Benford for every x0 ̸= 0 or none, depending on whether log |a| is irrational or not.
Example 5.4. (i) Let T (x) = 4x. Since log 4 is irrational, OT (x0) = (4n−1x0) is Benford for every x0 ̸= 0; in particular OT (4) = (4n) is Benford. Note, however, that (4n) is not base-2 Benford since log2 4 = 2 is rational, and correspond-ingly the second binary digit of 4n is identically equal to zero, whereas for a 66 A. Berger and T.P. Hill d · · · · · · 9 1 1 1 1 1 0 0 log10 5 log10 2 2 5 0.1 0.1 0.2 0.3 10 10 S ◦T (x) 0 histograms for N = 103 iterations compared to BL 2x x/5 y + log 2 y −log 5 x = 10y, y = log x log S ◦T (10y) y x 1 Fig 22. With T(x) = 2x, OT (x0) is Benford for all x0 ̸= 0.
2-Benford sequence the second binary digit is zero only with a relative frequency of log2(3/2) ≈0.5850.
(ii) Since log π is irrational, every orbit of T (x) = πx is Benford, unless x0 = 0. Here OT (x0) is actually base-b Benford for every b ∈N{1}.
Clearly, the simple proof of Lemma 5.3 works only for maps that are exactly linear. The same argument would for instance not work for T (x) = 2x + e−x even though T (x) ≈2x for large x. To establish the Benford behavior of maps like this, a simple version of shadowing will be used. While the argument em-ployed here is elementary, note that in dynamical systems theory, shadowing is a powerful and sophisticated tool, see e.g. [Pa].
To explain the basic idea, fix T as above, i.e. let T (x) = 2x + e−x and note first that T (x) ≥max(0, x + 1) for all x, and hence limn→∞T n(x0) = +∞for every x0. While no explicit analytical expression is available for T n(x0), it is certainly plausible to expect that, for large n, the orbit OT (x0) should resemble an orbit of the linear map x 7→2x. Fortunately, this is easily made rigorous. To this end, note that T n(x0) = 2nx0 + Xn j=1 2n−je−T j−1(x0) holds for every n ∈N and x0 ∈R. Since T n(x0) ≥0 for all n and x0, the number x0 := x0 + X∞ j=1 2−je−T j−1(x0) > x0 + e−x0 2 A basic theory of Benford’s Law 67 is well-defined and positive, and a short calculation using the fact that T n(x) ≥ x + n confirms that T n(x0) −2nx0 = X∞ j=n+1 2n−je−T j−1(x0) ≤ X∞ j=1 2−je−(x0+j+n−1) = e1−n−x0 2e −1 , (5.1) and hence |T n(x0) −2nx0| →0 exponentially fast as n →∞. As will be seen shortly, this implies that OT (x0) is Benford for all x0 ∈R. Note also that even if |T n(x0) −2ny| were merely required to remain bounded as n →∞, the only choice for y would still be y = x0. Moreover, x0 depends continuously on x0. As the following lemma shows, these observations hold in greater generality.
Lemma 5.5 (Shadowing Lemma). Let T : R →R be a map, and β a real number with |β| > 1. If supx∈R |T (x) −βx| < +∞then there exists, for every x ∈R, one and only one point x such that the sequence (T n(x)−βnx) is bounded.
Proof. Let ∆(x) := T (x) −βx and note that D := supx∈R |∆(x)| < +∞by assumption. With this, for all x ∈R and n ∈N0, T n(x) = βnx + Xn j=1 βn−j∆◦T j−1(x) .
Using this expression, together with the well-defined number x := x + X∞ j=1 β−j∆◦T j−1(x) , it follows that |T n(x) −βnx| = X∞ j=n+1 βn−j∆◦T j−1(x) ≤ X∞ j=1 |β|−j|∆◦T j+n−1(x)| ≤ D |β| −1 , and hence (T n(x) −βnx) is bounded. Moreover, the identity T n(x) −βny = T n(x) −βnx −βn(y −x) shows that (T n(x) −βny) is bounded only if y = x.
Remarks. (i) From the proof of Lemma 5.5 it can be seen that the map h : x 7→x is continuous whenever T is continuous. In general, h need not be one-to-one.
For example, h(x) = 0 for every x for which OT (x) is bounded. Also note that if lim|x|→+∞|∆(x)| = 0 then lim|x|→+∞|h(x) −x| = 0 as well. This is often the case in applications and may be used to improve the bounds on |T n(x) −βnx|.
For example, for the map T (x) = 2x+e−x considered above, the rough estimate T n(x0) ≥2nx0 −e−x0 2e −1 68 A. Berger and T.P. Hill obtained from (5.1) can be substituted into (5.1) again, leading to the much more accurate |T n(x0) −2nx0| = O e−2nx0 as n →∞.
(ii) Stronger, quantitative versions of the Shadowing Lemma have been estab-lished. They are very useful for an analysis of BL in more complicated systems, see e.g. [BBH] or [Ber3].
♣ Example 5.6. (i) Let T (x) = 2x+1. For this simple map, T n can be computed explicitly, and it is illuminating to compare the explicit findings with Lemma 5.5. From T n(x) = 2nx + 2n −1 , it is clear that (T n(x) −2nx) is unbounded for every x ∈R. However, using x := x + 1, one obtains T n(x) −2nx ≡−1 , and hence (T n(x) −2nx) is bounded.
(ii) Strictly speaking, the map T (x) = 2x+e−x studied above does not meet the assumptions of Lemma 5.5, as ∆(x) = e−x is not bounded for x →−∞. The conclusion of the lemma, however, does hold nevertheless because ∆is bounded on R+ and T maps R into R+. Put differently, x is well-defined for every x ∈R.
(iii) Let T (x) = 2x −e−x. Note that T has a unique fixed point x∗, i.e.
T (x∗) = x∗; numerically, x∗≈0.5671. Lemma 5.5 applies to T for x > x∗. To see this formally, replace T (x) by x∗+ 2(x−x∗) whenever x ≤x∗and note that this modification of T does not affect OT (x0) for x0 ≥x∗. Thus for every x ≥x∗ there exists an x such that (T n(x) −2nx) is bounded. Lemma 5.7 below implies that OT (x0) is Benford for all x0 > x∗. Clearly, OT (x∗) = (x∗, x∗, x∗, . . .) is not Benford. If x0 < x∗then T n(x0) →−∞super-exponentially fast. The Benford properties of OT (x0) in this case will be analyzed in the next sub-section.
The next lemma enables application of Lemma 5.5 to establish the Benford property for orbits of a wide class of maps.
Lemma 5.7.
(i) Assume that (an) and (bn) are sequences of real numbers with |an| →+∞and supn∈N |an −bn| < +∞. Then (bn) is Benford if and only if (an) is Benford.
(ii) Suppose that the measurable functions f, g : [0, +∞) →R are such that |f(t)| →+∞as t →+∞, and supt≥0 |f(t) −g(t)| < +∞. Then f is Benford if and only if g is Benford.
Proof. To prove (i), let c := supn∈N |an −bn| + 1. By discarding finitely many terms if necessary, it can be assumed that |an|, |bn| ≥2c for all n. From A basic theory of Benford’s Law 69 −log 1 + c |an| −c ≤log |bn| |bn| + c ≤log |bn| |an| ≤log |an| + c |an| ≤log 1 + c |an| −c , it follows that log |bn| −log |an| = log |bn| |an| ≤log 1 + c |an| −c →0 as n →∞.
Lemma 4.3(i) now shows that (log |bn|) is u.d. mod 1 if and only (log |an|) is.
The proof of (ii) is completely analogous.
Lemmas 5.5 and 5.7 can now easily be combined to produce the desired gen-eral result. The theorem is formulated for orbits converging to zero. As explained in the subsequent Example 5.9, a reciprocal version holds for orbits converging to ±∞.
Theorem 5.8 ([BBH]). Let T : R →R be a C2-map with T (0) = 0. Assume that 0 < |T ′(0)| < 1. Then OT (x0) is Benford for all x0 ̸= 0 sufficiently close to 0 if and only if log |T ′(0)| is irrational. If log |T ′(0)| is rational then OT (x0) is not Benford for any x0 sufficiently close to 0.
Proof. Let α := T ′(0) and observe that there exists a continuous function f : R →R such that T (x) = αx 1 −xf(x) . In particular, T (x) ̸= 0 for all x ̸= 0 sufficiently close to 0. Define e T(x) := T (x−1)−1 = x2 α x −f(x−1) , and note that e T(x) −α−1x = x α · f(x−1) x −f(x−1) = f(x−1) α + f(x−1)2 α x −f(x−1) .
From this it is clear that sup|x|≥ξ | e T(x) −α−1x| is finite, provided that ξ is sufficiently large. Hence Lemma 5.5 shows that for every x with |x| sufficiently large, | e T n(x) −α−nx| is bounded with an appropriate x ̸= 0. Lemma 5.7 implies that O e T (x0) is Benford if and only if (α1−nx0) is, which in turn is the case precisely if log |α| is irrational. The result then follows from noting that, for all x0 ̸= 0 with |x0| sufficiently small, OT (x0) = e T n−1(x−1 0 )−1 n∈N, and Corollary 4.7(i) which shows that (x−1 n ) is Benford whenever (xn) is.
Example 5.9. (i) For T (x) = 1 2x + 1 4x2, the orbit OT (x0) is Benford for every x0 ̸= 0 sufficiently close to 0. A simple graphical analysis shows that limn→∞T n(x) = 0 if and only if −4 < x < 2. Thus for every x0 ∈(−4, 2){0}, 70 A. Berger and T.P. Hill OT (x0) is Benford. Clearly, OT (−4) = (−4, 2, 2, . . .) and OT (2) = (2, 2, 2, . . .) are not Benford. For x0 < −4 or x0 > 2, one might try to mimic the proof of Theorem 5.8 and consider e T(x) := T (x−1)−1 = 4x2 1 + 2x near x = 0. Note that indeed e T is a smooth (C∞) map near x = 0, and e T(0) = 0.
However, e T ′(0) = 0 as well, and Theorem 5.8 does not apply. It will follow from the main result of the next subsection (Theorem 5.12) that for almost every point x0 ∈R[−4, 2] the orbit OT (x0) is Benford. However, R[−4, 2] also contains a large set of exceptional points, i.e. points whose orbit is not Benford.
(ii) To see that Theorem 5.8 applies to the map T (x) = 2x + e−x considered in Example 5.6(ii), let e T(x) := T (x−2)−1/2 = x √ 2 + x2e−1/x2 , x ̸= 0 .
With e T(0) := 0, the map e T : R →R is smooth, and e T ′(0) = 1 √ 2. Moreover, limn→∞e T n(x) = 0 for every x ∈R. By Theorem 5.8, O e T (x0) is Benford for every x0 ̸= 0, and hence OT (x0) is Benford for every x0 ̸= 0 as well, because T n(x) = e T n(|x|−1/2)−2 for all n.
(iii) As in (ii), Theorem 5.8 applies to the map T (x) = 10x + e2−x. Note that again limn→∞T n(x) = +∞for every x ∈R, but since log 10 is rational, no T -orbit is Benford. In fact, it is not hard to see that for every m ∈N and x ∈R, the sequence of m-th significant digits of T n(x), i.e.
Dm(T n(x)) n∈N is eventually constant.
Remark. Theorem 5.8 remains essentially unchanged if the case |T ′(0)| = 1 is also allowed, the conclusion being that in this case OT (x0) is not Benford for any x near 0. However, this extension requires the explicit assumption that x = 0 be attracting, see [Ber4]. (If |T ′(0)| < 1 then x = 0 is automatically attracting.) For a simple example, consider the smooth map T (x) = √ 1 + x2. While limn→∞T n(x) = +∞for every x ∈R, it follows from the explicit formula T n(x) = √ n + x2 that OT (x0) is not Benford, as (log p n + x2 0) is not u.d. mod 1, by Proposition 4.8(iv). The extended version of Theorem 5.8 just mentioned easily leads to the same conclusion because e T(x) := T (x−1)−1 = x √ x2 + 1 is smooth, with e T(0) = 0 and e T ′(0) = 1, and x = 0 is an attracting fixed point for e T.
To see that the situation can be more complicated if |T ′(0)| = 1 yet x = 0 is not attracting, fix α > 1 and consider the map Tα(x) = αx −(α −1)x 1 + x2 , A basic theory of Benford’s Law 71 for which Tα(0) = 0, T ′ α(0) = 1, and x = 0 is repelling. As far as the dynamics near x = 0 is concerned, all maps Tα are the same. However, e Tα(x) = Tα(x−1)−1 = x 1 + x2 α + x2 is smooth with e T ′ α(0) = α−1. Hence it is clear from Theorem 5.8 that OTα(x0) is Benford for all x0 ̸= 0 or for none, depending on whether log α is irrational or not.
♣ Processes with super-exponential growth or decay As was seen in the previous subsection, for the maps T : x 7→α(x + e−x) with α > 1, either all orbits are Benford (if log α is irrational) or else none are (if log α is rational). This all-or-nothing behavior is linked to the exponential growth of orbits since, by the Shadowing Lemma 5.5, T n(x) = αnx + O(e−n) as n →∞.
For an altogether different scenario, consider the smooth map T : x 7→ p 30 + 12x2 + x4 .
As before, limn→∞T n(x) = +∞for every x ∈R. However, it follows from T (x)2 + 6 = (x2 + 6)2 that T n(x) = p (x2 + 6)2n −6 = (x2 + 6)2n−1 + O 6−2n−1 as n →∞, showing that every T -orbit grows at a doubly-exponential rate. Is OT (x0) Ben-ford for some or even all x0 ∈R? The main result of this subsection, Theorem 5.12 below, shows that indeed OT (x0) is Benford for most x0. While it is dif-ficult to explicitly produce even a single x0 with this property, it is very easy to see that OT (x0) cannot be Benford for every x0. Indeed, taking for example x0 = 2, one obtains OT (2) = (2, √ 94, √ 9994, √ 999994, . . .) , and it is clear that D1 T n(2) = 9 for every n ∈N. Hence OT (2) is not Benford.
For another example, choose x0 = √ 104/3 −6 = 3.943 . . . for which the sequence of first significant digits is eventually 2-periodic, D1(T n−1(x0)) = (3, 2, 4, 2, 4, 2, 4, . . .) .
As also shown by Theorem 5.12, for maps like T there are always many excep-tional points.
The following is an analog of Lemma 5.3 in the doubly-exponential setting.
Recall that a statement holds for almost every x if there is a set of Lebesgue measure zero that contains all x for which the statement does not hold.
72 A. Berger and T.P. Hill Lemma 5.10. Let T (x) = αxβ for some α > 0 and β > 1. Then OT (x0) is Benford for almost every x0 > 0, but there also exist uncountably many exceptional points, i.e. x0 > 0 for which OT (x0) is not Benford.
Proof. Note first that letting e T(x) = cT (c−1x) for any c > 0 implies OT (x) = c−1O e T (cx), and with c = α(β−1)−1 one finds e T(x) = xβ. Without loss of general-ity, it can therefore be assumed that α = 1, i.e. T (x) = xβ. Define R : R →R as R(y) = log T (10y) = βy. Since x 7→log x establishes a bijective correspondence between both the points and the nullsets in R+ and R, respectively, all that has to be shown is that OR(y) is u.d. mod 1 for a.e. y ∈R, but also that OR(y) fails to be u.d. mod 1 for at least uncountably many y. To see the former, let fn(y) = Rn(y) = βny. Clearly, f ′ n(y)−f ′ m(y) = βn−m(βm −1) is monotone, and |f ′ n −f ′ m| ≥β −1 > 0 whenever m ̸= n. By Proposition 4.9, therefore, OR(y) is u.d. mod 1 for a.e. y ∈R.
The statement concerning exceptional points will be proved here only under the additional assumption that β is an integer, see [Ber4] for the remaining cases. Given an integer β ≥2, let (ηn) be any sequence of 0s and 1s such that ηnηn+1 = 0 for all n ∈N, that is, (ηn) does not contain two consecutive 1s.
With this, consider y0 := X∞ j=1 ηjβ−j and observe that, for every n ∈N, 0 ≤⟨βny0⟩= X∞ j=n+1 ηjβn−j ≤1 β + 1 β2(β −1) < 1 , from which it is clear that (βny0) is not u.d. mod 1. The proof is completed by noting that there are uncountably many different sequences (ηn), and each sequence defines a different point y0.
Example 5.11. Let T (x) = x2. By Lemma 5.10, OT (x0) is Benford for almost every but not for every x0 ∈R, as for instance T n(x) = x2n always has first significant digit D1 = 1 if x = 10k for some k ∈Z.
To study maps like T (x) = √ 30 + 12x2 + x4 mentioned above, Lemma 5.10 has to be extended. Note that e T(x) = T (x−1)−1 = x2 √ 1 + 12x2 + 30x4 , so e T(x) ≈x2 near x = 0. Again the technique of shadowing can be applied to relate the dynamics of e T to the one of x 7→x2 covered by Lemma 5.10. The following is an analog of Theorem 5.8 for the case when T is dominated by power-like terms.
Theorem 5.12 ([BBH]). Let T be a smooth map with T (0) = 0, and assume that T ′(0) = 0 but T (p)(0) ̸= 0 for some p ∈N{1}. Then OT (x0) is Benford for almost every x0 sufficiently close to 0, but there are also uncountably many exceptional points.
A basic theory of Benford’s Law 73 d · · · · · · 9 1 1 1 1 1 0.1 0.1 0.2 0.3 10 10 S ◦T (x) 0 0 0 histograms for N = 103 iterations compared to BL 1 2 √ 10 2x x2 2y y + log 2 x = 10y, y = log x log S ◦T (10y) y x 1 Fig 23. With T(x) = x2, OT (x0) is Benford for almost every, but not every x0 ∈R.
Proof. Without loss of generality, assume that p = min{j ∈N : T (j)(0) ̸= 0}.
The map T can be written in the form T (x) = αxp 1 + f(x) where f is a C∞-function with f(0) = 0, and α ̸= 0. As in the proof of Lemma 5.10, it may be assumed that α = 1. Let R(y) = −log T (10−y) = py −log 1 + f(10−y) , so that OT (x0) is Benford if and only if OR(−log x0) is u.d. mod 1. As the proof of Lemma 5.10 has shown, (pny) is u.d. mod 1 for a.e. y ∈R. Moreover, Lemma 5.5 applies to R, and it can be checked by term-by-term differentiation that the shadowing map h : y 7→y = y − X∞ j=1 p−j log 1 + f 10−Rj(y) is a C∞-diffeomorphism on [y0, +∞) for y0 sufficiently large. For a.e. sufficiently large y, therefore, OR(y) is u.d. mod 1. As explained earlier, this means that OT (x0) is Benford for a.e. x0 sufficiently close to 0. The existence of exceptional points follows similarly as in the proof of Lemma 5.10.
Example 5.13. (i) Consider the map T (x) = 1 2(x2 + x4) and note that limn→∞T n(x) = 0 if and only if |x| < 1. Theorem 5.12 shows that OT (x0) is Benford for a.e. x0 ∈(−1, 1). If |x| > 1 then limn→∞T n(x) = +∞, and the reciprocal version of Theorem 5.12 applies to e T (x) := T (x−1)−1 = 2x4 1 + x2 near x = 0. Overall, therefore, OT (x0) is Benford for a.e. x0 ∈R.
74 A. Berger and T.P. Hill (ii) For T (x) = √ 30 + 12x2 + x4, Theorem 5.12 applied to e T(x) := T (x−1)−1 = x2 √ 1 + 12x2 + 30x4 shows that OT (x0) is Benford for a.e. x0 ∈R.
(iii) Let T (x) = 1 + x2. Again Theorem 5.12 applied to e T(x) = T (x−1)−1 = x2 1 + x2 , shows that OT (x0) is Benford for a.e. x0 ∈R. As also asserted by that theorem, there are many exceptional points as well. For example, it can be shown that with x0 = limn→∞ r . . .
qp 102n −1 −1 . . . = 9.949 . . . , the first significant digit of T n−1(x0) always equals 9, i.e. D1 T n−1(x0) = 9 for all n ∈N. (In fact, x0 is the only point with this property, see [BBH] for details.) Remarks. (i) Note that while in Lemma 5.3 and Theorem 5.8 OT (x0) is Benford either for all x0 or for none at all, Lemma 5.10 and Theorem 5.12 guarantee the coexistence of many x0 for which OT (x0) is Benford and many exceptional points. The latter form an uncountable set of Lebesgue measure zero. From a measure-theoretic point of view, therefore, exceptional points are extremely rare.
It can be shown, however, that the points x0 for which OT (x0) is Benford form a set of first category, i.e. a countable union of nowhere dense sets. In particular, the exceptional points are dense in a neighbourhood of x = 0. (Recall that a set M is dense in C ⊂R if, given any c ∈C and ε > 0, there exists an m ∈M with |m−c| < ε.) Thus from a topological point of view, most points are exceptional.
This discrepancy between the measure-theoretic and the topological point of view is not uncommon in ergodic theory and may explain why it is difficult to explicitly find even a single point x0 for which OT (x0) is Benford for, say, T (x) = 1 + x2 — despite the fact that Theorem 5.12 guarantees the existence of such points in abundance.
(ii) Theorem 5.12 covers for instance all polynomial or rational functions of degree at least two, for |x| sufficiently large. An example not covered by that theorem is T (x) = ex or, more precisely, its reciprocal e T(x) = e−1/x. In this case, OT (x0) grows even faster than doubly-exponential. Theorem 5.21 below shows that nevertheless OT (x0) is Benford for a.e. x0 ∈R. Again, there is also a (measure-theoretically small yet topologically large) set of exceptional points.
(iii) In the context of Lemma 5.10 and Theorem 5.12, and in view of (i), many interesting questions may be asked. For instance, OT (x0) is Benford for a.e. x0 ∈R if T (x) = x2. What if x0 = 2, i.e., is OT (2) = (22n−1) Benford?
More generally, let T be any polynomial with integer coefficients and degree at A basic theory of Benford’s Law 75 least two. Then OT (x0) is Benford for almost all sufficiently large |x0|. Is OT (k) Benford for some, or even many integers k? In the case of T (x) = x2, this is equivalent to asking whether (2n log |k|) is u.d. mod 1 or, in number-theoretic terminology, whether log |k| is 2-normal. At present, 2-normality of common mathematical constants such as log 2, π or e is a well-known open problem, considered to be exceedingly difficult. Similarly, one may ask whether (F2n) is Benford. Again, this may be a very hard problem, contrasting the simple fact that (F|P (n)|) is Benford whenever P is a non-constant polynomial with integer coefficients.
♣ To conclude the present section on one-dimensional processes, a few possible applications and extensions of the results above will be discussed. The presenta-tion is very brief and mostly based on examples; for any details, the interested reader may wish to consult the references mentioned in the text.
An application: Newton’s method and related algorithms In scientific calculations using digital computers and floating point arithmetic, roundofferrors are inevitable, and as Knuth points out in his classic text The Art of Computer Programming [Kn, pp.253–255] In order to analyze the average behavior of floating-point arithmetic algorithms (and in particular to determine their average running time), we need some statis-tical information that allows us to determine how often various cases arise . . . [If, for example, the] leading digits tend to be small [that] makes the most obvious techniques of “average error” estimation for floating-point calculations invalid.
The relative error due to rounding is usually . . . more than expected.
Thus for the problem of finding numerically the root of a function by means of Newton’s Method (NM), it is important to study the distribution of significant digits (or significands) of the approximations generated by the method. As will be seen shortly, the differences between successive Newton approximations, and the differences between the successive approximations and the unknown root often exhibit exactly the type of non-uniformity of significant digits alluded to by Knuth — they typically follow BL.
Throughout this subsection, let f : I →R be a differentiable function defined on some open interval I ⊂R, and denote by Nf the map associated with f by NM, that is Nf(x) := x −f(x) f ′(x) for all x ∈I with f ′(x) ̸= 0.
For Nf to be defined wherever f is, set Nf(x) := x if f ′(x) = 0. Using NM for finding roots of f (i.e. real numbers x∗with f(x∗) = 0) amounts to picking an initial point x0 ∈I and iterating Nf. Henceforth, (xn) will denote the sequence of iterates of Nf starting at x0, that is (xn) = ONf (x0).
Clearly, if (xn) converges to x∗, say, and if Nf is continuous at x∗, then Nf(x∗) = x∗, so x∗is a fixed point of Nf, and f(x∗) = 0. (Note that according 76 A. Berger and T.P. Hill to the definition of Nf used here, Nf(x∗) = x∗could also mean that f ′(x∗) = 0.
If, however, f ′(x∗) = 0 yet f(x∗) ̸= 0 then Nf is not continuous at x∗unless f is constant.) It is this correspondence between the roots of f and the fixed points of Nf that makes NM work locally. Often, every fixed point x∗of Nf is attracting, i.e. limn→∞N n f (x0) = x∗for all x0 sufficiently close to x∗. (Observe that if f is linear near x∗, i.e. f(x) = c(x−x∗) for some c ̸= 0, then Nf(x) = x∗ for all x near x∗.) To formulate a result about BL for NM, it will be assumed that f : I →R is real-analytic. Recall that this means that f can, in a neighbourhood of every point of I, be represented by its Taylor series. Although real-analyticity is a strong assumption indeed, the class of real-analytic functions covers most prac-tically relevant cases, including all polynomials, and all rational, exponential, and trigonometric functions, and compositions thereof.
If f : I →R is real-analytic and x∗∈I a root of f, i.e. if f(x∗) = 0, then f(x) = (x −x∗)mg(x) for some m ∈N and some real-analytic g : I →R with g(x∗) ̸= 0. The number m is the multiplicity of the root x∗; if m = 1 then x∗is referred to as a simple root. The following theorem becomes plausible upon observing that f(x) = (x −x∗)mg(x) implies that Nf is real-analytic in a neighbourhood of x∗, and N ′ f(x) = f(x)f ′′(x) f ′(x)2 = m(m −1)g(x)2 + 2m(x −x∗)g′(x)g(x) + (x −x∗)2g′′(x)g(x) m2g(x)2 + 2m(x −x∗)g′(x)g(x) + (x −x∗)2g′(x)2 , so that in particular N ′ f(x∗) = 1 −m−1.
Theorem 5.14 ([BH1]). Let f : I →R be real-analytic with f(x∗) = 0, and assume that f is not linear.
(i) If x∗is a simple root, then (xn −x∗) and (xn+1 −xn) are both Benford for (Lebesgue) almost every, but not every x0 in a neighbourhood of x∗.
(ii) If x∗is a root of multiplicity at least two, then (xn −x∗) and (xn+1 −xn) are Benford for all x0 ̸= x∗sufficiently close to x∗.
The full proof of Theorem 5.14 can be found in [BH1]. It uses the following lemma which may be of independent interest for studying BL in other numerical approximation procedures. Part (i) is an analog of Lemma 5.7, and (ii) and (iii) follow directly from Theorem 5.12 and 5.8, respectively.
Lemma 5.15. Let T : I →I be C∞with T (y∗) = y∗for some y∗∈I.
(i) If T ′(y∗) ̸= 1, then for all y0 such that limn→∞T n(y0) = y∗, the sequence (T n(y0) −y∗) is Benford precisely when T n+1(y0) −T n(y0) is Benford.
(ii) If T ′(y∗) = 0 but T (p)(y∗) ̸= 0 for some p ∈N{1}, then (T n(y0) −y∗) is Benford for (Lebesgue) almost every, but not every y0 in a neighbourhood of y∗.
A basic theory of Benford’s Law 77 (iii) If 0 < |T ′(y∗)| < 1, then (T n(y0) −y∗) is Benford for all y0 ̸= y∗suffi-ciently close to y∗precisely when log |T ′(y∗)| is irrational.
Example 5.16. (i) Let f(x) = x/(1 −x) for x < 1. Then f has a simple root at x∗= 0, and Nf(x) = x2. By Theorem 5.14(i), the sequences (xn) and (xn+1 −xn) are both Benford sequences for (Lebesgue) almost every x0 in a neighbourhood of 0.
(ii) Let f(x) = x2. Then f has a double root at x∗= 0 and Nf(x) = x/2, so by Theorem 5.14(ii), the sequence of iterates (xn) of Nf as well as (xn+1−xn) are both Benford for all starting points x0 ̸= 0. (They are not, however, 2-Benford.) Utilizing Lemma 5.15, an analog of Theorem 5.14 can be established for other root-finding algorithms as well.
Example 5.17. Let f(x) = x + x3 and consider the successive approximations (yn) generated by the Jacobi-Steffensen method, yn+1 = yn − f(yn)2 f(yn) −f yn −f(yn) , n ∈N0 .
For almost every, but not every y0 near 0, (yn) is Benford. This follows from Lemma 5.15(ii), since yn = Jn f (y0) with the Jacobi-Steffensen transformation Jn f (x) = −y5 1 −y2 1 + y2 −y4 + y6 , and Jf(y) ≈−y5 near y = 0. Alternatively, Jf = N ˜ f with the real-analytic function ˜ f(x) = (x + x3)e 1 4 x4−x2, so Theorem 5.14(i) applies directly as well.
If f fails to be real-analytic, then Nf may not be well-behaved analytically.
For instance, Nf may have discontinuities even if f is C∞. Pathologies like this can cause NM to fail for a variety of reasons, of which the reader can gain an impression from [BH1, Sec.4]. Even if Nf is smooth, (xn) may not be Benford.
Example 5.18. Let f be the C∞-function f(x) = e−1/x2 if x ̸= 0 , 0 if x = 0 , for which Nf(x) = x(1 −1 2x2) is C∞as well. Note that limn→∞N n f (x) = 0 if and only if |x| < 2. In this case, however, ONf (x) is not Benford. This follows from the extended version of Theorem 5.8 mentioned in the remark on p.70 but can also be seen directly. Indeed, let T (x) = x 1+|x| and note that N ′ f(x) > 0, T ′(x) > 0 and |T (x)| ≤|Nf(x)| holds whenever |x| ≤1 2. From this it follows that |N n f (x)| ≥|T n(x)| = |x| 1 + n|x| for all n ∈N , 78 A. Berger and T.P. Hill and consequently (log |N n f (x)|) is not u.d. mod 1 by Proposition 4.8(iv), i.e., ONf (x) is not Benford. On the other hand, if |x0| > 2 then limn→∞|N n f (x0)| = +∞, and Theorem 5.12, applied to e T(x) := Nf(x−1)−1 = − 2x3 1 −2x2 near x = 0, shows that ONf (x0) is Benford for almost every, but not every x0 in this case.
Theorem 5.14 has important practical implications for estimating roots of a function via NM using floating-point arithmetic. One type of error in scientific computations is overflow (or underflow), which occurs when the running compu-tations exceed the largest (or smallest, in absolute value) floating-point number allowed by the computer. Feldstein and Turner [FT, p.241] show that under “the assumption of the logarithmic distribution of numbers [i.e. BL] floating-point addition and subtraction can result in overflow and underflow with alarming frequency . . . ” Together with Theorem 5.14, this suggests that special attention should be given to overflow and underflow errors in any computer algorithm used to estimate roots by means of NM.
Another important type of error in scientific computing arises due to round-off. In estimating a root from its Newton approximations, for example, a rule for stopping the algorithm must be specified, such as “stop when n = 106” or “stop when the differences between successive approximations are less than 10−6”.
Every stopping rule will result in some round-offerror, and Theorem 5.14 shows that this difference is generally Benford. In fact, justified by heuristics and by the extensive empirical evidence of BL in other numerical procedures, analysis of roundofferrors has often been carried out under the hypothesis of a statistical logarithmic distribution of significant digits or significands [BB]. Therefore, as Knuth points out, a naive assumption of uniformly distributed significant digits in the calculations tends to underestimate the average relative roundofferror in cases where the actual statistical distribution is skewed toward smaller leading significant digits, as is the case for BL. To obtain a rough idea of the magnitude of this underestimate when the true statistical distribution is BL, let X denote the absolute round-offerror at the time of stopping the algorithm, and let Y denote the fraction part of the approximation at the time of stopping. Then the relative error is X/Y , and assuming that X and Y are independent random vari-ables, the average (i.e., expected) relative error is simply EX ·E(1/Y ). As shown in [BH1], the assumption that Y is uniform while its true distribution is BL leads to an average underestimation of the relative error by more than one third.
The relevance of BL for scientific computing does not end here. For example, Hamming gives “a number of applications to hardware, software, and general computing which show that this distribution is not merely an amusing curiosity” [Ha, p.1609], and Schatte analyzes the speed of multiplication and division in digital computers when the statistical distribution of floating-point numbers is logarithmic and proves that, for design of computers, “[t]he base b = 8 is optimal with respect to [minimizing expected] storage use” [Scha1, p.453].
A basic theory of Benford’s Law 79 Extension I: Time-dependent systems So far, the sequences considered in this chapter have been generated by the iter-ation of a single map T or, in dynamical systems terminology, by an autonomous dynamical system. Autonomous systems constitute a classical and well-studied field. Beyond this field there has been, in the recent past, an increased interest in systems that are nonautonomous, i.e. explicitly time-dependent in one way or the other. This development is motivated and driven by important practical applications as well as pure mathematical questions. In this context, it is inter-esting to study how the results discussed previously extend to systems with the map T explicitly depending on n. In full generality, this is a very wide topic with many open problems, both conceptual and computational. Only a small number of pertinent results (without proofs) and examples will be mentioned here, and the interested reader is referred e.g. to [Ber4] for a fuller account and references as well as to [KM, LS] for an intriguing specific problem.
Throughout, let (Tn) be a sequence of maps that map R or parts thereof into itself, and for every n ∈N denote by T n the n-fold composition T n := Tn◦. . .◦T1; also let T 0 be the identity map on R. Given x0, it makes sense to consider the sequence OT (x0) := T n−1(x0) n∈N = x0, T1(x0), T2 T1(x0) , . . .
. As in the autonomous case (which corresponds to Tn being independent of n) the sequence OT (x0) is referred to as the (nonautonomous) orbit of x0.
The following is a nonautonomous variant of Theorem 5.8. A proof (of a substantially more general version) can be found in [BBH]. It relies heavily on a nonautonomous version of the Shadowing Lemma.
Theorem 5.19 ([BBH]). Let Tj : R →R be C2-maps with Tj(0) = 0 and T ′ j(0) ̸= 0 for all j ∈N, and set αj := T ′ j(0). Assume that supj max|x|≤1 |T ′′ j (x)| and P∞ n=1 Qn j=1 |αj| are both finite. If limj→∞log |αj| exists and is irrational, then OT (x0) is Benford for all x0 ̸= 0 sufficiently close to 0.
Example 5.20. (i) Let Rj(x) = (2 + j−1)x for j = 1, 2, . . .. It is easy to see that all assumptions of Theorem 5.19 are met for Tj(x) = Rj(x−1)−1 = j 2j + 1x with limj→∞log |αj| = −log 2. Hence OR(x0) is Benford for all x0 ̸= 0.
(ii) Let Tj(x) = Fj+1/Fjx for all j ∈N, where Fj denotes the j-th Fibonacci number. Since limj→∞log(Fj+1/Fj) = log 1+ √ 5 2 is irrational, and by taking reciprocals as in (i), Theorem 5.19 shows that OT (x0) is Benford for all x0 ̸= 0.
In particular, OT (F1) = (Fn) is Benford, as was already seen in Example 4.12.
Note that the same argument would not work to show that (n!) is Benford.
(iii) Consider the family of linear maps Tj(x) = 10−1+√j+1−√jx for j = 1, 2, . . .. Here Qn j=1 αj = 10−n+√n+1−1, so P+∞ n=1 Qn j=1 |αj| < +∞. However, 80 A. Berger and T.P. Hill Theorem 5.19 does not apply since limj→∞log |αj| = −1 is rational. Neverthe-less, as (√n) is u.d. mod 1 by [KN, Ex.3.9] and log |T n(x)| = −n + √ n + 1 −1 + log |x| , the sequence OT (x0) is Benford for every x0 ̸= 0.
In situations where most of the maps Tj are power-like or even more strongly expanding, the following generalization of Lemma 5.10 may be useful. (In its fully developed form, the result also extends Theorem 5.12, see [BBH, Thm.5.5] and [Ber3, Thm.3.7].) Again the reader is referred to [Ber4] for a proof.
Theorem 5.21 ([Ber4]). Assume the maps Tj : R+ →R+ satisfy, for some ξ > 0 and all j ∈N, the following conditions: (i) x 7→ln Tj(ex) is convex on [ξ, +∞); (ii) xT ′ j(x)/Tj(x) ≥βj > 0 for all x ≥ξ.
If lim infj→∞βj > 1 then OT (x0) is Benford for almost every sufficiently large x0, but there are also uncountably many exceptional points.
Example 5.22. (i) To see that Theorem 5.21 does indeed generalize Lemma 5.10, let Tj(x) = αxβ for all j ∈N. Then x 7→ln Tj(ex) = βx + ln α clearly is convex, and xT ′ j(x)/Tj(x) = β > 1 for all x > 0.
(ii) As mentioned already in (ii) of the remark on p.74, Theorem 5.21 also shows that OT (x0) with T (x) = ex is Benford for almost every, but not every x0 ∈R, as x 7→ln T (ex) = ex is convex, and xT ′(x)/T (x) = x as well as T 3(x) > e holds for all x ∈R. Similarly, the theorem applies to T (x) = 1 + x2.
(iii) For a truly nonautonomous example consider Tj(x) = x2 if j is even , 2x if j is odd , or Tj(x) = (j + 1)x .
In both cases, OT (x0) is Benford for almost every, but not every x0 ∈R.
(iv) Finally, it is important to note that Theorem 5.21 may fail if one of its hypotheses is violated even for a single j. For example Tj(x) = 10 if j = 1 , x2 if j ≥2 , satisfies (i) and (ii) for all j > 1, but does not satisfy assumption (ii) for j = 1.
Clearly, OT (x0) is not Benford for any x0 ∈R, since D1 T n(x0) = 1 for all n ∈N.
Using slightly more sophisticated tools, Theorem 5.21 can be extended so as to provide the following corollary for polynomial maps.
A basic theory of Benford’s Law 81 Corollary 5.23. Let the maps Tj be polynomials, Tj(x) = xnj + aj,nj−1xnj−1 + . . . + aj,1x + aj,0 , with nj ∈N{1} and aj,l ∈R for all j ∈N, 0 ≤l < nj. If supj∈N maxnj−1 l=0 |aj,l| < +∞then OT (x0) is Benford for almost every x0 ∈R[−ξ, ξ] with some ξ ≥0.
However, R[−ξ, ξ] also contains an uncountable dense set of exceptional points.
Example 5.24. Let Tj(x) = xj −1 for all j −1 . Then even though (Tj) do not satisfy the hypothesis (i) of Theorem 5.21, by Corollary 5.23, the orbit OT (x0) = (x0, x0 −1, x2 0 −2x0, . . .) is Benford for almost all |x0| ≥3, but that region also contains uncountably many points for which OT (x0) is not Benford.
Extension II: Chaotic dynamical systems The dynamical scenarios studied so far for their conformance with BL have all been very simple indeed: In Theorems 5.8, 5.12 and 5.19 limn→∞T n(x) = 0 holds automatically for all relevant initial values x, whereas limn→∞T n(x) = +∞in Theorem 5.21. While this dynamical simplicity does not necessarily force the behavior of S T n(x) to be equally simple (recall e.g. Example 5.13(iii)), it makes one wonder what might be observed under more general circumstances.
The present subsection presents two simple examples in this regard. Among other things, they illustrate that, as a rule, Benford sequences may be rare in more general dynamical systems.
Example 5.25. Consider the tent-map T : R →R given by T (x) = 1−|2x−1|.
Using Theorem 5.8, it is not hard to see that OT (x0) is Benford whenever x0 lies outside [0, 1]. Clearly, OT (0) = (0, 0, 0, . . .) and OT (1) = (1, 0, 0, . . .) are not Benford. As far as BL is concerned, therefore, it remains to analyze OT (x0) for 0 < x0 < 1. Define two maps τL, τR : [0, 1] →[0, 1] as τL(x) = x 2 , τR(x) = 1 −x 2 .
Then T ◦τL(x) = T ◦τR(x) = x for all x ∈[0, 1], and τL, τR can be used for a symbolic description of the dynamics of T . To this end, recall that the set Σ of all sequences consisting of the two symbols L and R, that is Σ = {L, R}N, is a compact metric space when endowed with the metric d(ω, e ω) := 2−min{n : ωn̸=e ωn} if ω ̸= e ω , 0 if ω = e ω .
Moreover, the (left) shift map σ on Σ, given by σ(ω) = (ωn+1) is a continuous map. With these ingredients, define a map h : Σ →[0, 1] as h(ω) := limn→∞τω1 ◦τω2 ◦. . . ◦τωn 1 2 .
It is easy to see that h is well defined, continuous and onto, and h ◦σ(ω) = T ◦h(ω) for all ω ∈Σ. In particular, therefore, T n−1 ◦h(ω) ∈Iωn holds for all 82 A. Berger and T.P. Hill ω ∈Σ and n ∈N, where IL = τL([0, 1]) = [0, 1 2] and IR = τR([0, 1]) = [ 1 2, 1].
Thus it is reasonable to think of ω as the “symbolic itinerary” of h(ω) under the iteration of T . (Note that h is not one-to-one, however #h−1({x}) = 1 unless x is a dyadic rational, i.e. unless 2lx is an integer for some l ∈N0.) By means of this symbolic coding, some dynamical properties of T are very easy to understand. For example, the set of x0 for which OT (x0) is periodic is dense in [0, 1]. To see this simply observe that h(ω) is periodic (under T ) whenever ω ∈Σ is periodic (under σ), and periodic sequences are dense in Σ. On the other hand, T is topologically transitive. Informally, this means that there is no non-trivial way of breaking the dynamics of T on [0, 1] into non-interacting pieces. In the present example, this property (defined and studied thoroughly e.g. in [KH]) simply means that OT (x0) is dense for at least one, but in fact many x0 ∈[0, 1]. Overall, therefore, the map T : [0, 1] →[0, 1] is chaotic in the sense of [Ber1, Def.2.21]. In particular, it exhibits the hallmark property of chaos, namely sensitive dependence on initial conditions. The latter means that, for every 0 < x < 1 and every ε > 0, a point x can be found such that |x −x| < ε yet lim supn→∞|T n(x) −T n(x)| ≥1 2 .
This follows e.g. from [Ber1, Thm.2.18] but can also be seen directly by noticing that T n is piecewise linear with slope 2n.
While the above analysis clearly reveals the complexity of the dynamics of T on [0, 1], the reader may still wonder how all this is related to BL. Is OT (x0) Benford for many, or even most x0 ∈[0, 1]? The chaotic nature of T suggests a negative answer. For a more definitive understanding, note that, for every 0 < a < 1, T∗λ0,1 [0, a] = λ0,1 0, τL(a) ∪ τR(a), 1 = a = λ0,1 [0, a] , showing that T∗λ0,1 = λ0,1, i.e. T preserves λ0,1. In fact, T is known to even be ergodic with respect to λ0,1. As a consequence of the BirkhoffErgodic Theorem, OT (x0) is distributed according to λ0,1 for Lebesgue almost every x0 ∈[0, 1].
By Example 3.10(i), for every such x0 the sequence S(T n(x0)) is uniformly distributed on [1, 10). Thus for a.e. x0 ∈[0, 1], the orbit OT (x0) is not Benford.
It remains to investigate whether OT (x0) is Benford for any x0 ∈[0, 1] at all. To this end first note that while OT (x0) is guaranteed to be uniformly distributed for a.e. x0 ∈[0, 1], there are plenty of exceptions. In fact, given any sequence ω ∈Σ whose asymptotic relative frequencies limN→∞ #{1 ≤n ≤N : ωn = L} N and limN→∞ #{1 ≤n ≤N : ωn = R} N do not both equal 1 2, or perhaps do not even exist at all, the orbit of h(ω) is not uniformly distributed. For instance, if ω = L, L, . . ., L | {z } N times , R, R, R, . . .
A basic theory of Benford’s Law 83 for some N ∈N0, then h(ω) = 21−N 3 , and T n h(ω) = 2 3 for all n ≥N. In view of this abundance of exceptional points, one may hope to identify some x0 ∈[0, 1] for which OT (x0) is Benford. Using the symbolic encoding of the dynamics, this can indeed be done as follows: Observe that T (x) = 2x whenever x ≤1 2, i.e. whenever x ∈IL, in which case log S T (x) = ⟨log 2 + log S(x)⟩.
Thus if T n(x0) ∈IL held for all n, then OT (x0) would be Benford. This is impossible since T n(x0) ∈IL for all n implies that x0 = 0, and x0 is a fixed point for T . However, since being Benford is an asymptotic property of OT (x0), it is enough for T n(x0) ∈IL to hold for most n and along arbitrarily long sequences. Concretely, let ω∗= L, L, . . . , L | {z } N1 times , R, L, L, . . ., L | {z } N2 times , R, L, L, . . ., L | {z } N3 times , R, L, . . .
, (5.2) where (Nn) is any sequence in N with Nn →∞, and set x∗= h(ω∗). According to (5.2), the orbit OT (x∗) stays in IL for the first N1 steps, then makes a one-step excursion to IR, then remains in IL for N2 steps, etc. It follows from [Ber4, Lem.2.7(i)], but can also be verified directly, that OT (x∗) is Benford. For a concrete example, choose e.g. Nn ≡2n, then ω∗= L, L, R, L, L, L, L, R, L, L, L, L, L, L, R, L, . . .
as well as x∗= h(ω∗) = X∞ n=1 21+2n−n2(−1)n+1 = 0.2422 . . . , and OT (x∗) is Benford. Notice finally that (5.2) provides uncountably many different points x∗, and hence the set {x0 ∈[0, 1] : OT (x0) is Benford } is uncountable; as initial segments of ω∗do not matter, this set is also dense in [0, 1]. To put this fact into perspective, note that with the points x∗constructed above, OT (x∗) is actually also Benford base b whenever b is not a power of 2, i.e. whenever b ̸∈{2n : n ∈N}. On the other hand, OT (x0) is not Benford base 2, 4, 8 etc. for any x0 ∈R, see [Ber4, Ex.2.11].
Example 5.26. The family of quadratic polynomials Qρ : x 7→ρx(1 −x), with ρ ∈R, often referred to as the logistic family, plays a pivotal role in dynamical systems theory, see e.g. [Ber1, KH]. Arguably the most prominent member of this family is the map Q4 which has many features in common with the tent map T from the previous example. Unlike the latter, however, Q4 is smooth, and it is this smoothness which makes the dynamics of Q4, or generally the logistic family, a much richer yet also more subtle topic.
84 A. Berger and T.P. Hill To understand the dynamics of Q4 with regards to BL, note first that near x = 0, Q4(x−1)−1 = − x2 4(1 −x) = −x2 4 + O(x3) .
Hence Theorem 5.12 applies, showing that OQ4(x0) is Benford for almost ev-ery, but not every x0 ∈R[0, 1]. As in Example 5.25, it remains to study the dynamics within the interval [0, 1]. A similar symbolic coding can be applied to demonstrate that on this interval Q4 is, in any respect, as chaotic as the tent map T . This is somewhat hard to do directly, but it becomes very simple upon introducing the homeomorphism H : [0, 1] →[0, 1] with H(x) = sin2( 1 2πx) and noting that, for all x ∈[0, 1], Q4 ◦H(x) = sin2(πx) = H ◦T (x) .
(5.3) Thus Q4 and T differ only by a change of coordinates, and all topological prop-erties of T (such as e.g. the existence of a dense set of periodic orbits, and topological transitivity) carry over to Q4. Together with T∗λ0,1 = λ0,1 it follows from (5.3) that Q4∗(H∗λ0,1) = (Q4 ◦H)∗λ0,1 = (H ◦T )∗λ0,1 = H∗(T∗λ0,1) = H∗λ0,1 , hence Q4 preserves the probability measure H∗λ0,1, and is in fact ergodic with respect to it. Note that d dxH∗λ0,1 [0, x] = d dx λ0,1 [0, 2 π arcsin √x] = 1 π p x(1 −x) , 0 < x < 1 , showing that H∗λ0,1 is simply the arcsin- or Beta( 1 2, 1 2)-distribution, and there-fore H∗λ0,1(B) = 0 if and only if λ0,1(B) = 0. Again, the BirkhoffErgodic Theorem implies that OQ4(x0) is, for almost every x0 ∈[0, 1], distributed ac-cording to H∗λ0,1, and consequently not Benford, see Example 3.10(iii). As in Example 5.25, one may wonder whether OQ4(x0) is Benford for any x0 ∈[0, 1] at all. Essentially the same argument shows that the answer is, again, positive.
With ω∗as in (5.2), the orbit of H ◦h(ω∗) spends most of its time arbitrarily close to the (unstable) fixed point at x = 0, and log S Q4(x) = ⟨log 4 + log S(x) + log(1 −x)⟩≈⟨log 4 + log S(x)⟩ whenever x > 0 is very small. A careful analysis in the spirit of Lemma 4.3(i) then shows that OQ4 H ◦h(ω∗) is indeed Benford. As in the previous example, it follows that {x0 ∈[0, 1] : OQ4(x0) is Benford } is uncountable and dense in [0, 1].
A basic theory of Benford’s Law 85 5.2. Multi-dimensional discrete-time processes The purpose of this section is to extend the basic results of the previous section to multi-dimensional systems, notably to linear, as well as some non-linear re-currence relations. Recall from Example 4.12 that the Fibonacci sequence (Fn) is Benford. Hence the linear recurrence relation xn+1 = xn + xn−1 generates a Benford sequence when started from x0 = x1 = 1. As will be seen shortly, many, but not all linear recurrence relations generate Benford sequences.
Under a BL perspective, an obvious difficulty when dealing with multi-di-mensional systems is the potential for more or less cyclic behavior, either of the orbits themselves or of their significands.
Example 5.27. (i) Let the sequence (xn) be defined recursively as xn+1 = xn −xn−1 , n = 1, 2, . . . , (5.4) with given x0, x1 ∈R. By using the matrix 0 1 −1 1 associated with (5.4), it is straightforward to derive an explicit representation for (xn), xn = x0 cos 1 3πn + 2x1 −x0 √ 3 sin 1 3πn , n = 0, 1, . . . .
From this it is clear that xn+6 = xn for all n, i.e., (xn) is 6-periodic. This oscilla-tory behavior of (xn) corresponds to the fact that the roots of the characteristic equation λ2 = λ −1 associated with (5.4) are λ = e±ıπ/3 and hence lie on the unit circle. For no choice of x0, x1, therefore, is (xn) Benford.
(ii) Consider the linear 3-step recursion xn+1 = 2xn + 10xn−1 −20xn−2 , n = 2, 3, . . . .
Again it is easy to confirm that, for any x0, x1, x2 ∈R, the value of xn is given explicitly by xn = c12n + c210n/2 + c3(−1)n10n/2 , where c1 = 10x0 −x2 6 , c2,3 = x2 −4x0 12 ± x2 + 3x1 −10x0 6 √ 10 .
Clearly, limn→∞|xn| = +∞unless x0 = x1 = x2 = 0, so unlike in (i) the sequence (xn) is not bounded or oscillatory. However, if |c2| ̸= |c3| then log |xn| = n 2 + log c110−n( 1 2 −log 2) + c2 + (−1)nc3 ≈n 2 + log |c2 + (−1)nc3| , showing that S(xn) is asymptotically 2-periodic and hence (xn) is not Benford.
Similarly, if |c2| = |c3| ̸= 0 then S(xn) is convergent along even (if c2 = c3) or odd (if c2 = −c3) indices n, and again (xn) is not Benford. Only if c2 = c3 = 0 yet c1 ̸= 0, or equivalently if 1 4x2 = 1 2x1 = x0 ̸= 0 is (xn) Benford. Obviously, 86 A. Berger and T.P. Hill the oscillatory behavior of S(xn) in this example is due to the characteristic equation λ3 = 2λ2 + 10λ −20 having two roots with the same modulus but opposite signs, namely λ = − √ 10 and λ = √ 10.
(iii) Let γ = cos(π log 2) ≈0.5852 and consider the sequence (xn) defined recursively as xn+1 = 4γxn −4xn−1 , n = 1, 2 . . . , (5.5) with given x0, x1 ∈R. As before, an explicit formula for xn is easily derived as xn = 2nx0 cos(πn log 2) + 2n−1 x1 −2γx0 p 1 −γ2 sin(πn log 2) .
Although somewhat oscillatory, the sequence (xn) is clearly unbounded. As will be shown now, however, it is not Benford. While the argument is es-sentially the same for any (x0, x1) ̸= (0, 0), for convenience let x0 = 0 and x1 = 2 sin(π log 2) ≈1.622, so that log |xn| = log 2n| sin(πn log 2)| = n log 2 + log | sin(πn log 2)| , n = 1, 2, . . . .
With the (measurable) map T : [0, 1) →[0, 1) defined as T (s) = ⟨s + log | sin(πs)|⟩, 0 ≤s < 1 , therefore simply ⟨log |xn|⟩= T (⟨n log 2⟩). Recall that (n log 2) is u.d. mod 1, and hence (⟨log |xn|⟩) is distributed according to the probability measure T∗λ0,1.
Consequently, (xn) is Benford if and only if T∗λ0,1 equals λ0,1. The latter, how-ever, is not the case. While this is clear intuitively, an easy way to see this formally is to observe that T is piecewise smooth and has a unique local max-imum at some 0 < s0 < 1. (Concretely, s0 = 1 −1 π arctan π ln 10 ≈0.7013 and T (s0) ≈0.6080.) Thus if T∗λ0,1 = λ0,1, then for all sufficiently small ε > 0, T (s0) −T (s0 −ε) ε = λ0,1 T (s0 −ε), T (s0) ε = T∗λ0,1 T (s0 −ε), T (s0) ε ≥λ0,1 [s0 −ε, s0) ε = 1 , which is impossible since T ′(s0) = 0. Hence (xn) is not Benford. The reason for this can be seen in the fact that, while log |λ| = log 2 is irrational for the characteristic roots λ = 2e±ıπ log 2 associated with (5.5), there obviously is a rational dependence between the two real numbers log |λ| and 1 2π arg λ, namely log |λ| −2( 1 2π arg λ) = 0.
The above recurrence relations are linear and have constant coefficients.
Hence they can be rewritten and analyzed using matrix-vector notation. For instance, in Example 5.27(i) xn xn+1 = 0 1 −1 1 xn−1 xn , A basic theory of Benford’s Law 87 so that, with A = 0 1 −1 1 ∈R2×2, the sequence (xn) is simply given by xn = 1 0 An x0 x1 , n = 0, 1, . . . .
It is natural, therefore, to study the Benford property of more general sequences (x⊤Any) for any A ∈Rd×d and x, y ∈Rd. Linear recurrence relations like the ones in Example 5.27 are then merely special cases. As suggested by that example, in order to guarantee the Benford property for (x⊤Any), conditions have to be imposed on A so as to rule out cyclic behavior of orbits or their significands. To prepare for these conditions, denote the real part, imaginary part, complex conjugate, and modulus (absolute value) of z ∈C by ℜz, ℑz, z, and |z|, respectively. For z ̸= 0, define arg z as the unique number in [−π, π) that satisfies z = |z|eı arg z; for notational convenience, let arg 0 := 0. Recall that real or complex numbers z1, z2, . . . , zn are rationally independent (or Q-independent) if Pn j=1 qjzj = 0 with q1, q2, . . . , qn ∈Q implies that qj = 0 for all j = 1, 2, . . . , n. A set Z ⊂C is rationally independent if every of its finite subsets is, and rationally dependent otherwise.
Let Z ⊂C be any set such that all elements of Z have the same modulus ζ, i.e., Z is contained in the periphery of a circle with radius ζ centered at the origin of the complex plain. Call the set Z resonant if either #(Z ∩R) = 2 or the numbers 1, log ζ and the elements of 1 2π arg Z are rationally dependent, where 1 2π arg Z = 1 2π arg z : z ∈Z {−1 2, 0}.
Given A ∈Rd×d, recall that the spectrum σ(A) ⊂C of A is simply the set of all eigenvalues of A. Denote by σ(A)+ the “upper half” of the spectrum, i.e., let σ(A)+ = {λ ∈σ(A) : ℑλ ≥0}. Usage of σ(A)+ refers to the fact that non-real eigenvalues of real matrices always occur in conjugate pairs, and hence σ(A)+ only contains one of the conjugates.
With the above preparations, what will shortly turn out to be an appropriate condition on A reads as follows.
Definition 5.28. A matrix A ∈Rd×d is Benford regular (base 10) if σ(A)+ contains no resonant set.
Note that in the simplest case, i.e. for d = 1, the matrix A = [a] is Benford regular if and only if log |a| is irrational. Hence Benford regularity may be con-sidered a generalization of this irrationality property. Also note that A is regular (invertible) whenever it is Benford regular.
Example 5.29. None of the matrices associated with the recurrence rela-tions in Example 5.27 is Benford regular. Indeed, in (i), A = 0 1 −1 1 , hence σ(A)+ = {eıπ/3}, and clearly log |eıπ/3| = 0 is rational. Similarly, in (ii), A = 0 1 0 0 0 1 −10 10 2 , and σ(A)+ = {− √ 10, 2, √ 10} contains the resonant set {− √ 10, √ 10}. Finally, for (iii), A = 0 1 −4 4γ , and σ(A)+ = {2eıπ log 2} is res-onant.
88 A. Berger and T.P. Hill Example 5.30. Let A = 1 −1 1 1 ∈R2×2, with characteristic polynomial pA(λ) = λ2 −2λ+2, and hence σ(A)+ = { √ 2eıπ/4}. As 1, log √ 2 and 1 2π · π 4 = 1 8 are rationally dependent, the matrix A is not Benford regular.
Example 5.31. Consider A = 0 1 1 1 ∈R2×2. The characteristic polynomial of A is pA(λ) = λ2 −λ −1, and so, with ϕ = 1 2(1 + √ 5), the eigenvalues of A are ϕ and −ϕ−1. Since pA is irreducible and has two roots of different absolute value, it follows that log ϕ is irrational (in fact, even transcendental). Thus A is Benford regular.
With the one-dimensional result (Lemma 5.3), as well as Example 5.27 and Definition 5.28 in mind, it seems realistic to hope that iterating (i.e. taking powers of) any matrix A ∈Rd×d produces many Benford sequences, provided that A is Benford regular. This is indeed the case. To concisely formulate the pertinent result, call a sequence (zn) of complex numbers terminating if zn = 0 for all sufficiently large n.
Theorem 5.32 ([Ber2]). Assume that A ∈Rd×d is Benford regular. Then, for every x, y ∈Rd, the sequence (x⊤Any) is either Benford or terminating. Also, (∥Anx∥) is Benford for every x ̸= 0.
The proof of Theorem 5.32 will make use of the following variant of [Ber2, Lem.2.9].
Proposition 5.33. Assume that the real numbers 1, ρ0, ρ1, . . . , ρm are Q-inde-pendent. Let (zn) be a convergent sequence in C, and at least one of the numbers c1, c2, . . . , cm ∈C non-zero. Then (xn) given by xn = nρ0 + log ℜ c1e2πınρ1 + . . . + cme2πınρm + zn is u.d. mod 1.
Proof of Theorem 5.32. Given A ∈Rd×d, let σ(A)+ = {λ1, λ2, . . . , λs}, where s ≤d and, without loss of generality, |λ1| ≥|λ2| ≥. . . ≥|λs|. Fix x, y ∈Rd and recall that there exist (possibly non-real) polynomials p1, p2, . . . , ps of degrees at most d −1 such that x⊤Any = ℜ p1(n)λn 1 + . . . + ps(n)λn s , n = 0, 1, . . . .
(5.6) (This follows e.g. from the Jordan Normal Form Theorem.) If (x⊤Any) is not terminating, then it can be assumed that p1 ̸= 0. (Otherwise relabel the pj and λj, and reduce s accordingly.) Now distinguish two cases.
Case 1: |λ1| > |λ2| In this case, λ1 is a dominant eigenvalue. Denote by k the degree of p1 and let c := limn→∞n−kp1(n). Note that c is a non-zero number that is real whenever A basic theory of Benford’s Law 89 λ1 is real. From |x⊤Any| = |λ1|nnk ℜ n−kp1(n) λ1 |λ1| n + n−kp2(n) λ2 |λ1| n + . . .
+ n−kps(n) λs |λ1| n = |λ1|nnk|ℜ(ceın arg λ1 + zn)| , with zn = (n−kp1(n) −c) λ1 |λ1| n + Ps j=2 n−kpj(n) λj |λ1| n →0, it follows that log |x⊤Any| = n log |λ1| + k log n + log |ℜ(ce2πın arg λ1 2π + zn)| .
In view of Proposition 4.8(iii), no generality is lost by assuming that k = 0. If λ1 is real then, by Lemma 4.3(i) and the irrationality of log |λ1|, the sequence (log |x⊤Any|) is u.d. mod 1. If λ1 is not real, then apply Proposition 5.33 with m = 1, ρ0 = log |λ1|, and ρ1 = 1 2π arg λ1. In either case, (x⊤Any) is Benford.
Case 2: |λ1| = . . . = |λl| > |λl+1| for some l ≤s.
Here several different eigenvalues of the same magnitude occur. Let k be the maximal degree of p1, p2, . . . pl and cj := limn→∞n−kpj(n) for j = 1, 2, . . ., l.
Note that if x⊤Any ̸= 0 infinitely often then at least one of the numbers c1, c2, . . . , cl is non-zero. As before, |x⊤Any| = |λ1|nnk ℜ n−kp1(n) λ1 |λ1| n + n−kp2(n) λ2 |λ1| n + . . .
+ n−kps(n) λs |λ1| n = |λ1|nnk|ℜ(c1eın arg λ1 + . . . + cleın arg λl + zn)| , where zn = Xl j=1(n−kpj(n) −cj) λj |λ1| n + Xs j=l+1 n−kpj(n) λj |λ1| n →0 .
Propositions 4.8(iii) and 5.33 with m = l and ρ0 = log |λ1|, ρ1 = 1 2π arg λ1, . . . , ρl = 1 2π arg λl imply that log |x⊤Any| = n log |λ1| + k log n + log |ℜ(c1eın arg λ1 + . . . + cleın arg λl + zn)| is u.d. mod 1, hence (x⊤Any) is Benford.
The assertion concerning (∥Anx∥) is proved in a completely analogous man-ner.
Example 5.34. According to Example 5.31, the matrix 0 1 1 1 is Benford regular. By Theorem 5.32, every solution of the difference equation xn+1 = 90 A. Berger and T.P. Hill xn + xn−1 is Benford, except for the trivial solution xn ≡0 resulting from x0 = x1 = 0. In particular, therefore, the sequences of Fibonacci and Lucas numbers, (Fn) = (1, 1, 2, 3, 5, . . .) and (Ln) = (−1, 2, 1, 3, 4, . . .), generated respectively from the initial values x0 x1 = 1 1 and x0 x1 = −1 2 , are Benford. For the former sequence, this has already been seen in Example 4.12.
Note that (F 2 n), for instance, is Benford as well by Corollary 4.7(i), see Fig 24.
Example 5.35. Recall from Example 5.30 that A = 1 −1 1 1 is not Benford regular. Hence Theorem 5.32 does not apply, and the sequence (x⊤Any) may, for some x, y ∈R2, be neither Benford nor terminating. Indeed, pick for example x = y = 1 0 ⊤and note that x⊤Any = 1 0 2n/2 " cos( 1 4πn) −sin( 1 4πn) sin( 1 4πn) cos( 1 4πn) # 1 0 = 2n/2 cos 1 4πn .
Hence (x⊤Any) is clearly not Benford as x⊤Any = 0 whenever n = 2 + 4l for some l ∈N0. It will be seen later (in Theorem 5.37) that in the case of a 2 × 2-matrix A, the Benford regularity of A is actually necessary for every sequence of the form (x⊤Any) to be either Benford or terminating. Note, however, that this does of course not rule out the possibility that some sequences derived from iterating A may be Benford nevertheless. For a concrete example, fix any x ̸= 0 and, for each n ∈N, denote by En the area of the triangle with vertices at Anx, An−1x, and the origin. Then En = 1 2 det(Anx, An−1x) = 2n−2∥x∥2 , n = 1, 2, . . . , so (En) is Benford, see Fig 24.
Remark. According to Theorem 5.32, Benford regularity of a matrix A is a simple condition guaranteeing the widespread generation of Benford sequences of the form (x⊤Any). Most d×d-matrices are Benford regular, under a topological as well as a measure-theoretic perspective. To put this more formally, let Bd := {A ∈Rd×d : A is Benford regular } .
While the complement of Bd is dense in Rd×d, it is a topologically small set: Rd×d\Bd is of first category, i.e. a countable union of nowhere dense sets. A (topologically) typical (“generic”) d × d-matrix therefore belongs to Bd, i.e. is Benford regular. Similarly, if A is an Rd×d-valued random variable, that is, a random matrix, whose distribution is a.c. with respect to the d2-dimensional Lebesgue measure on Rd×d, then P(A ∈Bd) = 1, i.e., A is Benford regular with probability one. Similar statements hold for instance within the family of stochastic matrices, see [BHKR].
♣ While Benford regularity of A is a property sufficient for all sequences (x⊤ Any) to be either Benford or terminating, the following example shows that this property is not in general necessary.
A basic theory of Benford’s Law 91 1 1 2 3 4 4 5 8 8 8 16 16 A = 1 −1 1 1 A9 1 0 A8 1 0 F 2 6 F 2 5 F 2 4 F 2 3 E9 E8 E7 E6 Fig 24. Two Benford sequences derived from linear 2-dimensional systems, see Examples 5.34 and 5.35. Note that the matrix A associated with (En) is not Benford regular.
Example 5.36. Consider the 4 × 4-matrix A = 10 √ 2 cos(2π √ 3) −sin(2π √ 3) 0 0 sin(2π √ 3) cos(2π √ 3) 0 0 0 0 cos(4π √ 3) −sin(4π √ 3) 0 0 sin(4π √ 3) cos(4π √ 3) , for which σ(A)+ = {10 √ 2e−2πı √ 3, 10 √ 2e4πı √ 3} =: {λ1, λ2}. Since 2 arg λ1 + arg λ2 = 0, the matrix A is not Benford regular. It will now be shown that nevertheless for any x, y ∈R4 the sequence (x⊤Any) is either Benford or termi-nating. Indeed, with x⊤= x1 x2 x3 x4 and y = y1 y2 y3 y4 ⊤, a straightforward calculation confirms that x⊤Any = 10n √ 2ℜ (x1+ıx2)(y1−ıy2)e−2πın √ 3+(x3+ıx4)(y3−ıy4)e−4πın √ 3 .
Unless (x2 1 + x2 2)(y2 1 + y2 2) + (x2 3 + x2 4)(y2 3 + y2 4) = 0, therefore, (x⊤Any) is not terminating, and log |x⊤Any| = n √ 2 + f(n √ 3) , with the function f : [0, 1) →R given by f(s) = log ℜ (x1 + ıx2)(y1 −ıy2)e−2πıs + (x3 + ıx4)(y3 −ıy4)e−4πıs .
92 A. Berger and T.P. Hill Note that f has at most finitely many discontinuities. Moreover, 1, √ 2, √ 3 are Q-independent, and hence [Ber2, Cor.2.6] implies that (x⊤Any) is Benford.
The dimension d = 4 in Example 5.36 is smallest possible. Indeed, as the fol-lowing result shows, Benford regularity is (not only sufficient but also) necessary in Theorem 5.32 whenever d < 4.
Theorem 5.37. Assume d < 4, and let A ∈Rd×d be invertible. Then the following statements are equivalent: (i) A is Benford regular.
(ii) For every x, y ∈Rd the sequence (x⊤Any) is either Benford or terminat-ing.
Proof. As demonstrated by Theorem 5.32, assumption (i) implies (ii) even with-out any restrictions on d.
Conversely, assume that (ii) holds. Notice that whenever A has a real eigen-value λ ̸= 0, with a corresponding eigenvector eλ ̸= 0, then choosing x = y = eλ results in x⊤Any = λn∥eλ∥2. Hence log |λ| must be irrational. For d = 1, this shows that A is Benford regular.
Next let d = 2. In this case, two different eigenvalues of the same modulus can occur either in the form ±λ with λ > 0, i.e. as non-zero eigenvalues of opposite sign, or in the form λ = |λ|e±2πıρ with |λ| > 0 and 0 < ρ < 1 2, i.e. as a pair of conjugate non-real eigenvalues. In the former case, let e−and e+ be normalized eigenvectors corresponding to −λ and λ, respectively. Note that 1 + e⊤ +e−> 0, by the Cauchy–Schwarz inequality. Then (e+ + e−)⊤An(e+ + e−) = 2λn(1 + e⊤ +e−) if n is even , 0 if n is odd , showing that (x⊤Any) is not Benford for x = y = e+ + e−. Assuming (ii), therefore, implies that A does not have real eigenvalues of opposite sign. On the other hand, if σ(A)+ = {|λ|e2πıρ} then there exists a regular matrix P ∈R2×2 such that P −1AP = |λ| cos(2πρ) −sin(2πρ) sin(2πρ) cos(2πρ) .
Specifically choosing x⊤= 0 1 P −1 and y = P 1 0 ⊤yields x⊤Any = |λ|n sin(2πnρ) , n = 0, 1, . . . .
(5.7) If log |λ| is rational, say log |λ| = p q , then the sequence ⟨q log |x⊤Any|⟩= ⟨q log | sin(2πnρ)|⟩ is either periodic (if ρ is rational) or else distributed according to T∗λ0,1, with T : [0, 1) →[0, 1) given by T (s) = ⟨q log | sin(2πs)|⟩. As in Example 5.27(iii), it can be shown that T∗λ0,1 ̸= λ0,1. Thus, as before, rationality of log |λ| is ruled out by assumption (ii). If ρ is rational then x⊤Any = 0 holds for infinitely many A basic theory of Benford’s Law 93 but not all n, and hence (x⊤Any) is neither Benford nor terminating. Again, this possibility is ruled out by assumption (ii). To conclude the case d = 2, assume that log |λ| and ρ are both irrational, yet 1, log |λ| and ρ are rationally dependent, i.e., there exist integers k1, k2, k3 with k2k3 ̸= 0 such that k1 + k2 log |λ| + k3ρ = 0 .
Without loss of generality, assume k3 > 0. For every j ∈{1, 2, . . ., k3} and n ∈N0 therefore log |x⊤Ank3+jy| = (nk3+j) log |λ|+log sin 2πj k1 k3 + 2π j k2 k3 + nk2 log |λ| , so (log |x⊤Ank3+jy|) is distributed according to Tj∗λ0,1, with Tj : [0, 1) →[0, 1) given by Tj(s) = k3s + j log |λ| + log sin 2πj k1 k3 + 2π k2 k3 j log |λ| + 2πk2s , and (⟨log |x⊤Any|⟩) is distributed according to 1 k3 Pk3 j=1 Tj∗λ0,1. Again it can be shown that the latter probability measure on [0, 1), B[0, 1) does not equal λ0,1.
Overall, therefore, for d = 2 and σ(A)+ = {|λ|e2πıρ}, assumption (ii) implies that 1, log |λ|, and 1 2π arg λ are rationally independent. In other words, A is Benford regular.
Finally, consider the case d = 3. The only eigenvalue configuration not cov-ered by the preceding arguments is that of three different eigenvalues with the same modulus, i.e. with |λ| > 0 and 0 < ρ < 1 2 either σ(A)+ = {|λ|, |λ|e2πıρ} or σ(A)+ = {−|λ|, |λ|e2πıρ}. In both cases, there exists a regular matrix P ∈R3×3 such that P −1AP = |λ| ±1 0 0 0 cos(2πρ) −sin(2πρ) 0 sin(2πρ) cos(2πρ) , and choosing x⊤= 0 0 1 P −1 and y = P 0 1 0 ⊤again yields (5.7).
As before, assumption (i) implies that 1, log |λ|, and ρ are rationally indepen-dent.
Finally, it is worth noting that even if A is not Benford regular, many or even most sequences of the form (x⊤Any) may nevertheless be Benford.
Example 5.38. Recall from Example 5.30 that A = 1 −1 1 1 is not Benford regular because σ(A)+ = { √ 2eıπ/4} is resonant. However, a short calculation with x⊤= x1 x2 , y = y1 y2 ⊤confirms that x⊤Any = 2n/2∥x∥∥y∥cos 1 4πn + ψ , n = 0, 1, . . . ; here ψ ∈[−π, π) is the angle of a counter-clockwise rotation moving x/∥x∥ into y/∥y∥. (Note that ψ is unique unless ∥x∥∥y∥= 0 in which case x⊤Any ≡0 94 A. Berger and T.P. Hill anyway.) By virtue of Proposition 4.8(ii), if ψ ̸∈[−π, π)∩1 4πZ then (log |x⊤Any|) is u.d. mod 1. Thus, if ψ is not an integer multiple of 1 4π, or equivalently if (x2 1 −x2 2)y1y2 −x1x2(y2 1 −y2 2) (x2 1 −x2 2)(y2 1 −y2 2) + 4x1x2y1y2 ̸= 0 , then (x⊤Any) is Benford.
The present section closes with two examples of non-linear systems. The sole purpose of these examples is to hint at possible extensions of the results pre-sented earlier; for more details the interested reader is referred to the references provided.
Example 5.39. Consider the non-linear map T : R2 →R2 given by T : x1 x2 7→ 2 0 0 2 x1 x2 + f(x1) f(x2) , with the bounded continuous function f(t) = 3 2|t + 2| −3|t + 1| + 3|t −1| −3 2|t −2| = 0 if |t| ≥2 , 3t + 6 if −2 < t < −1 , −3t if −1 ≤t < 1 , 3t −6 if 1 ≤t < 2 .
Sufficiently far away from the x1- and x2-axes, i.e. for min{|x1|, |x2|} sufficiently large, the dynamics of T is governed by the matrix 2 0 0 2 , and since the latter is Benford regular, one may reasonably expect that x⊤T n(y) should be Benford.
It can be shown that this is indeed the case. More precisely, by means of a multi-dimensional shadowing argument, the following statement can be proved, see [Ber2, Thm.4.1]: Let T : Rd →Rd be of the form T (x) = Ax + f(x) with A ∈Rd×d and a bounded continuous f : Rd →Rd. If A is Benford regular and has no eigenvalues inside the unit disc, that is, |λ| > 1 holds for every eigenvalue λ of A, then the sequence x⊤T n(y) is Benford whenever it is unbounded. Notice that the provision concerning boundedness is already needed in the present simple example: For instance, if |ξ| ≤3 2 and x⊤= ξ 0 then T n(x) is eventually 2-periodic and hence x⊤T n(x) is not Benford.
Example 5.40. Consider the non-linear map T : R2 →R2 defined as T : x1 x2 7→ " 3x3 1x2 2 + 4x1 5x2 1x4 2 −2x2 2 + 1 # .
Unlike in the previous example, the map T is now genuinely non-linear and cannot be considered a perturbation of a linear map. Rather, T may be thought of as a 2-dimensional analogue of the polynomial map x 7→1 + x2. Clearly, if |x1| or |x2| is small, then the behavior of T n(x) may be complicated. For instance, on the x2-axis, i.e. for x1 = 0, the map T reduces to x2 7→1 −2x2 2 which, up to a change of coordinates, is nothing else but the chaotic map Q4 A basic theory of Benford’s Law 95 studied in Example 5.26. If, however, |x1| and |x2| are sufficiently large then a two-dimensional version of Theorem 5.12 asserts that, for (Lebesgue) almost every x, each component of OT (x) is Benford, see [BS, Thm.16]; at the same time, there is also an abundance of exceptional points [BS, Cor.17].
5.3. Differential equations By presenting a few results on, and examples of differential equations, i.e. de-terministic continuous-time processes, this short section aims at convincing the reader that the emergence of BL is not at all restricted to discrete-time dynam-ics. Rather, solutions of ordinary or partial differential equations often turn out to be Benford as well. Recall that a (Borel measurable) function f : [0, +∞) →R is Benford if and only if log |f| is u.d. mod 1.
Consider the initial value problem (IVP) ˙ x = F(x) , x(0) = x0 , (5.8) where F : R →R is continuously differentiable with F(0) = 0, and x0 ∈R. In the simplest case, F(x) ≡αx with some α ∈R. In this case, the unique solution of (5.8) is x(t) = x0eαt. Unless αx0 = 0, therefore, every solution of (5.8) is Benford, by Example 4.5(i). As in the discrete-time setting, this feature persists for arbitrary C2-functions F with F ′(0) < 0. The direct analog of Theorem 5.8 is Theorem 5.41 ([BBH]). Let F : R →R be C2 with F(0) = 0. Assume that F ′(0) < 0. Then, for every x0 ̸= 0 sufficiently close to 0, the unique solution of (5.8) is Benford.
Proof. Pick δ > 0 so small that xF(x) < 0 for all 0 < |x| ≤δ. As F is C2, the IVP (5.8) has a unique local solution whenever |x0| ≤δ, see [Wa]. Since the interval [−δ, δ] is forward invariant, this solution exists for all t ≥0. Fix any x0 with 0 < |x0| ≤δ and denote the unique solution of (5.8) as x = x(t).
Clearly, limt→+∞x(t) = 0. With y : [0, +∞) →R defined as y = x−1 therefore y(0) = x−1 0 =: y0 and limt→+∞|y(t)| = +∞. Let α := −F ′(0) > 0 and note that there exists a continuous function g : R →R such that F(x) = −αx + x2g(x).
From ˙ y = −˙ x x2 = αy −g(y−1) , it follows via the variation of constants formula that, for all t ≥0, y(t) = eαty0 − Z t 0 eα(t−τ)g y(τ)−1 dτ .
As α > 0 and g is continuous, the number y0 := y0 − Z +∞ 0 e−ατg y(τ)−1 dτ 96 A. Berger and T.P. Hill is well defined. (Note that y0 is simply the continuous-time analogue of the auxiliary point x in Lemma 5.5.) Moreover, for all t > 0, y(t) −eαty0 = Z +∞ t eα(t−τ)g y(τ)−1 dτ ≤ Z +∞ 0 e−ατ g y(t + τ)−1 dτ ≤∥g∥∞ α , where ∥g∥∞= max|x|≤δ |g(x)|, and Lemma 5.7(ii) shows that y is Benford if and only if t 7→eαty0 is. An application of Corollary 4.7(ii), together with Example 4.5(i) therefore completes the proof.
Example 5.42. (i) The function F(x) = −x+x4e−x2 satisfies the assumptions of Theorem 5.41. Thus except for the trivial x = 0, every solution of ˙ x = −x + x4e−x2 is Benford.
(ii) The function F(x) = −x3 + x4e−x2 is also smooth with xF(x) < 0 for all x ̸= 0. Hence for every x0 ∈R, the IVP (5.8) has a unique solution with limt→+∞x(t) = 0. However, F ′(0) = 0, and as will be shown now, this prevents x from being Benford. To see this, fix x0 ̸= 0 and integrate −˙ x x3 = 1 −xe−x2 from 0 to t to obtain the implicit representation x2(t) = x2 0 1 + 2tx2 0 −2x2 0 Z t 0 x(τ)e−x(τ)2dτ .
(5.9) Note that limt→+∞x(t) = 0 implies limt→+∞1 t R t 0 x(τ)e−x(τ)2dτ = 0. Hence it follows from (5.9) that limt→+∞2tx(t)2 = 1. Consequently, t 7→| log x|/ log t is bounded as t →+∞, and (the continuous-time version of) Proposition 4.8(iv) shows that x is not Benford.
Informally, the fact that F ′(0) = 0 causes the solutions of ˙ x = F(x) to approach the equilibrium x = 0 too slowly in order to be Benford. It is not hard to see that this is true in general: If F is C2 and xF(x) < 0 for all x ̸= 0 in a neighborhood of 0, and hence F(0) = 0, yet F ′(0) = 0 then, for all |x0| sufficiently small the solution of (5.8) is not Benford.
(iii) As the previous example showed, for solutions of (5.8) with F(0) = F ′(0) = 0 to be Benford for all x0 ̸= 0 sufficiently close to 0, it is necessary that F not be C2. (In fact, F must not even be C1+ε for any ε > 0, see [BBH, Thm.6.7].) For an example of this type, consider F(x) = − x p 1 + (log x)4 , x ̸= 0 .
A basic theory of Benford’s Law 97 With F(0) := 0, the function F is C1 with F ′(0) = 0, and every non-trivial solution of ˙ x = F(x) is Benford. To see this, fix x0 ̸= 0 and let y = −log x.
Then ˙ y = 1 ln 10 p 1 + y4 , from which it is straightforward to deduce that |y(t) − 3 p 3t/ ln 10| →0 as t →+∞, which in turn shows that y is u.d. mod 1, i.e., x is Benford.
(iv) Theorem 5.41 applies to the smooth function F(x) = −x+x log(1+x2).
In this case, ˙ x = F(x) has three equilibria, namely x = 0 and x = ±3, and consequently the solution of (5.8) is Benford whenever 0 < |x0| < 3.
To analyze the behavior of solutions outside of [−3, 3], fix x0 > 3 and let y := log x −1 2. Then ˙ y = 2y ln 10 + log(1 + 10−1−2y) ln 10 , and hence, for all t ≥0, y(t) = e2t/ ln 10y0 + Z t 0 e2(t−τ)/ ln 10 log(1 + 10−1−2y(τ)) ln 10 dτ .
With the usual auxiliary point y0 := y0 + R +∞ 0 e−2τ)/ ln 10 log(1+10−1−2y(τ)) ln 10 dτ, y(t) −e2t/ ln 10y0 = Z +∞ t e2(t−τ)/ ln 10 log(1 + 10−1−2y(τ)) ln 10 dτ ≤ Z +∞ 0 e−2τ/ ln 10 log(1 + 10−1−2y(t+τ)) ln 10 dτ ≤log p 1 + 10−1−2y(t) →0 as t →+∞.
By the same reasoning as in Example 4.5(iii), the function y is u.d. mod 1. Thus by Theorem 4.2, x is Benford for |x0| > 3 as well. Note that |x| goes to +∞ faster than exponentially in this case, i.e. limt→+∞|x(t)e−αt| = +∞for every α > 0.
Also, note that the case |x0| > 3 could be rephrased in the setting of Theorem 5.41 as well. Indeed, with z := x−1 one finds ˙ z = z log(z2) + z −z log(1 + z2) =: e F(z) .
With e F(0) := 0, the function e F : R →R is continuous but not C1, as limz→0 e F(z)/z = −∞. Thus Theorem 5.41 does not apply. The lack of smooth-ness of e F corresponds to the fact that solutions of the IVP ˙ z = e F(z), z(0) = z0, though still unique and globally defined, approach z = 0 faster than exponen-tially whenever |z0| < 1 3. For a result in the spirit of Theorem 5.41 that does apply to ˙ z = e F(z) directly, see [BBH, Thm.6.9].
98 A. Berger and T.P. Hill Just as their discrete-time counterparts, linear differential equations in higher dimensions are also a rich source of Benford behavior. Consider for instance the IVP ¨ x −x = 0 , x(0) = x0, ˙ x(0) = v0 , (5.10) with given numbers x0, v0 ∈R. The unique solution of (5.10) is x(t) = x0 + v0 2 et + x0 −v0 2 e−t which clearly is Benford unless x0 = v0 = 0. Using matrix-vector notation, (5.10) can be written as d dt x ˙ x = 0 1 1 0 x ˙ x , x ˙ x t=0 = x0 v0 .
Much more generally, therefore, consider the linear d-dimensional ordinary dif-ferential equation ˙ x = Ax , (5.11) where A is a real d × d-matrix. Recall that every solution of (5.11) is given by x : t 7→etAx0 for some x0 ∈Rd, in fact x0 = x(0), with the matrix exponential etA defined as etA = X∞ l=0 tl l!Al .
To ensure that every component of x, or that, more generally, for any x, y ∈Rd the function t 7→x⊤etAy is either Benford or trivial, a condition reminiscent of Benford regularity has to be imposed on A.
Definition 5.43. A matrix A ∈Rd×d is exponentially Benford regular (base 10) if eτA is Benford regular for some τ > 0.
Note that in the simplest case, i.e. for d = 1, the matrix A = [a] is ex-ponentially Benford regular if and only if a ̸= 0. Moreover, every exponen-tially Benford regular matrix is regular. It is easily checked that a matrix A fails to be exponentially Benford regular exactly if there exist λ1, λ2, . . . , λl in σ(A)+ with ℜλ1 = ℜλ2 = . . . = ℜλl such that ℜλ1/ ln 10 and the elements of { 1 2πℑλ1, 1 2πℑλ2, . . . , 1 2πℑλl}\ 1 2Z are rationally dependent. Also, it is not hard to see that if A is exponentially Benford regular then the set {t ∈R : etA is not Benford regular } actually is at most countable, i.e. finite (possibly empty) or countable. With this, the continuous-time analog of Theorem 5.32 is Theorem 5.44. Assume that A ∈Rd×d is exponentially Benford regular. Then, for every x, y ∈Rd, the function t 7→x⊤etAy is either Benford or identically equal zero. Also, t 7→∥etAx∥is Benford for every x ̸= 0.
A basic theory of Benford’s Law 99 Proof. Given x, y ∈Rd, define f : R →R according to f(t) := x⊤etAy. As observed above, for almost every h > 0 the matrix ehA is Benford regular and, by Theorem 5.32, the sequence (x⊤(ehA)ny) = f(nh) is either terminating or Benford. In the former case, f = 0 due to the fact that f is real-analytic. In the latter case, log |f(nh)| is u.d. mod 1 for almost all h > 0, and [KN, Thm.9.6] shows that log |f| is u.d. mod 1, i.e., f is Benford. The function t 7→∥etAx∥is dealt with similarly.
Example 5.45. (i) The matrix A = 0 1 1 0 associated with (5.10) is expo-nentially Benford regular, as σ(A)+ = {−1, 1}, and hence, as seen earlier, the solution of (5.10) is Benford unless x0 = v0 = 0.
(ii) For A = 0 1 1 1 recall from Example 5.31 that σ(A)+ = {−ϕ−1, ϕ} with ϕ = 1 2(1 + √ 5). Hence A is exponentially Benford regular, and every function of the form t 7→x⊤etAy is either Benford or vanishes identically. This is also evident from the explicit formula etA = etϕ 2 + ϕ 1 ϕ ϕ 1 + ϕ + e−tϕ−1 2 + ϕ ϕ + 1 −ϕ −ϕ 1 , which shows that the latter is the case if and only if x and y are proportional to 1 ϕ ⊤and −ϕ 1 ⊤(or vice versa), i.e. to the two perpendicular eigendi-rections of A.
(iii) Consider A = 1 −π/ ln 10 π/ ln 10 1 with σ(A)+ = {1 + ıπ/ ln 10}. In this case, A fails to be exponentially Benford regular because, with λ = 1+ıπ/ ln 10, ℜλ ln 10 −2ℑλ 2π = 0 .
As a matter of fact, no function t 7→x⊤etAy is Benford. Indeed, etA = et cos(πt/ ln 10) −sin(πt/ ln 10) sin(πt/ ln 10) cos(πt/ ln 10) , and picking for instance x⊤= 0 1 and y = 1 0 ⊤yields log |x⊤etAy| = log et sin πt ln 10 = t ln 10 + log sin πt ln 10 = g t ln 10 , where g(s) = s + log | sin(πs)|. As in Example 5.27(iii), it can be shown that g is not u.d. mod 1.
This example suggests that exponential Benford regularity of A may (not only be sufficient but) also be necessary in Theorem 5.44. In analogy to Example 5.36 and Theorem 5.37, one can show that this is indeed true if d < 4, but generally false otherwise; details are left to the interested reader.
Finally, it should be mentioned that at present little seems to be known about the Benford property for solutions of partial differential equations or more 100 A. Berger and T.P. Hill general functional equations such as e.g. delay or integro-differential equations.
Quite likely, it will be very hard to decide in any generality whether many, or even most solutions of such systems exhibit the Benford property in one form or another.
Example 5.46. A fundamental example of a partial differential equation is the so-called one-dimensional heat (or diffusion) equation ∂u ∂t = ∂2u ∂x2 , (5.12) a linear second-order equation for u = u(t, x). Physically, (5.12) describes e.g.
the diffusion over time of heat in a homogeneous one-dimensional medium. With-out further conditions, (5.12) has many solutions of which for instance u(t, x) = cx2 + 2ct , with any constant c ̸= 0, is neither Benford in t (“time”) nor in x (“space”), whereas u(t, x) = e−c2t sin(cx) is Benford (or identically zero) in t but not in x, and u(t, x) = ec2t+cx is Benford in both t and x. Usually, to specify a unique solution an equation like (5.12) has to be supplemented with initial and/or boundary conditions. A prototypical example of an Initial-boundary Value Problem (IBVP) consists of (5.12) together with u(0, x) = u0(x) for all 0 < x < 1 , u(t, 0) = u(t, 1) = 0 for all t > 0 .
(5.13) Physically, the conditions (5.13) may be interpreted as the ends of the medium, at x = 0 and x = 1, being kept at a reference temperature u = 0 while the initial distribution of heat (or temperature etc.) is given by the function u0 : [0, 1] →R.
It turns out that, under very mild assumptions on u0, the IBVP consisting of (5.12) and (5.13) has a unique solution which, for any t > 0, can be written as u(t, x) = X∞ n=1 une−π2n2t sin(πnx) , where un = 2 R 1 0 u0(s) sin(πns) ds. From this it is easy to see that, for every 0 ≤x ≤1, the function t 7→u(t, x) either vanishes identically or else is Benford.
Another possible set of initial and boundary data is u(0, x) = u0(x) for all x > 0 , u(t, 0) = 0 for all t > 0 , (5.14) A basic theory of Benford’s Law 101 corresponding to a semi-infinite one-dimensional medium kept at zero temper-ature at its left end x = 0, with an initial heat distribution given by the (in-tegrable) function u0 : [0, +∞) →R. Again, (5.12) together with (5.14) has a unique solution, for any t > 0 given by u(t, x) = 1 2 √ πt Z +∞ 0 u0(y) e−(x−y)2/(4t) −e−(x+y)2/(4t) dy .
Assuming R +∞ 0 y|u0(y)| dy < +∞, it is not hard to see that, for every x ≥0, limt→+∞t3/2u(t, x) = x 2√π Z +∞ 0 yu0(y) dy , and hence, for any fixed x ≥0, the function u is not Benford in time. On the other hand, if for example u0(x) = xe−x then a short calculation confirms that, for every t > 0, limx→+∞ exu(t, x) x = et , showing that u is Benford in space. Similarly, if u0(x) = 1[0,1)(x) then limx→+∞xe(x−1)2/(4t)u(t, x) = r t π holds for every t > 0, and again u is Benford in space.
6. Benford’s Law for random processes The purpose of this chapter is to show how BL arises naturally in a variety of stochastic settings, including products of independent random variables, mix-tures of random samples from different distributions, and iterations of random maps. Perhaps not surprisingly, BL arises in many other important fields of stochastics as well, such as geometric Brownian motion, order statistics, ran-dom matrices, L´ evy processes, and Bayesian models. The present chapter may also serve as a preparation for the specialized literature on these advanced topics [EL, LSE, MN, Sch¨ u1].
6.1. Independent random variables The analysis of sequences of random variables, notably the special case of (sums or products of) independent, identically distributed (i.i.d.) sequences of ran-dom variables, constitutes an important classical topic in probability theory.
Within this context, the present section studies general scenarios that lead to BL emerging as an “attracting” distribution. The results nicely complement the observations made in previous chapters.
Recall from Chapter 3 that a (real-valued) random variable X by definition is Benford if P(S(X) ≤t) = log t for all t ∈[1, 10). Also, recall that a sequence 102 A. Berger and T.P. Hill (Xn) of random variables converges in distribution to a random variable X, symbolically Xn D →X, if limn→∞P(Xn ≤t) = P(X ≤t) holds for every t ∈R for which P(X = t) = 0. By a slight abuse of terminology, say that (Xn) converges in distribution to BL if S(Xn) D →S(X), where X is a Benford random variable, or equivalently if limn→∞P(S(Xn) ≤t) = log t for all t ∈[1, 10) .
Another important concept is almost sure convergence. Specifically, the se-quence (Xn) converges to X almost surely (a.s.), in symbols Xn a.s.
→X, if P(limn→∞Xn = X) = 1. It is easy to check that Xn a.s.
→1 implies Xn D →X.
The reverse implication does not hold in general, as is evident from any i.i.d.
sequence (Xn) for which X1 is not constant: In this case, all Xn have the same distribution, so trivially Xn D →X1, yet P( limn→∞Xn exists ) = 0.
An especially simple way of generating a sequence of random variables is this: Fix a random variable X, and set Xn := Xn for every n ∈N. While the sequence (Xn) thus generated is clearly not i.i.d. unless X = 0 a.s. or X = 1 a.s., Theorems 4.10 and 4.17 imply Theorem 6.1. Assume that the random variable X has a density. Then: (i) (Xn) converges in distribution to BL.
(ii) With probability one, (Xn) is Benford.
Proof. To prove (i), note that the random variable log |X| has a density as well.
Hence, by Theorem 4.17 P(S(Xn) ≤t) = P(⟨log |Xn|⟩≤log t) = P(⟨n log |X|⟩≤log t) →log t as n →∞ holds for all t ∈[1, 10), i.e. (Xn) converges in distribution to BL.
To see (ii), simply note that log |X| is irrational with probability one. By Theorem 4.10, therefore, P (Xn) is Benford = 1.
Example 6.2. (i) Let X be uniformly distributed on [0, 1). For every n ∈N, FS(Xn)(t) = t1/n −1 101/n −1 , 1 ≤t < 10 , and a short calculation, together with the elementary estimate et−1−t et−1 < t 2 for all t > 0 shows that FS(Xn)(t) −log t ≤101/n −1 −ln 10 n 101/n −1 < ln 10 2n →0 as n →∞, and hence (Xn) converges in distribution to BL. Since P( log X is rational ) = 0, the sequence (Xn) is Benford with probability one.
A basic theory of Benford’s Law 103 (ii) Assume that X = 2 a.s. Thus PX = δ2, and X does not have a density.
For every n, S(Xn) = 10⟨n log 2⟩with probability one, so (Xn) does not converge in distribution to BL. On the other hand, (Xn) is Benford a.s.
Remarks. (i) In the spirit of Theorem 6.1, several results from Chapter 5 can be extended to a stochastic context. For a prototypical result, consider the map T : x 7→1 + x2 from Example 5.13(iii). If X has a density, then so has Y := log |X|. Recall from the proof of Theorem 5.12 that log |T n(X)| −2nY a.s.
→0 as n →∞, with a uniquely defined Y = h(Y ), and a close inspection of the map h shows that Y has a density as well. Hence by Theorems 4.2 and 4.17, OT (X) = T n−1(X) converges in distribution to BL, whereas Theorem 5.12 implies that P(OT (X) is Benford ) = 1.
(ii) For any random variable, it is not hard to see that assertion (ii) in The-orem 6.1 holds whenever (i) does. In the case of an i.i.d. sequence (Xn), the convergence of (Xn) in distribution to BL also implies that (Xn) is Benford for all n, so by independence it is easy to see that (Xn) is Benford with proba-bility one. In general, however, these two properties are independent. For one implication see Example 6.2(ii). For the other implication consider the constant sequence (X, X, X, . . .) which is evidently not Benford, but if X is a Benford random variable then (X) trivially converges in distribution to BL.
♣ The sequence of random variables considered in Theorem 6.1 is very special in that Xn is the product of n quantities that are identical, and hence dependent in extremis. Note that Xn is Benford for all n if and only if X is Benford. This invariance property of BL persists if, unlike the case in Theorem 6.1, products of independent factors are considered.
Theorem 6.3. Let X, Y be two independent random variables with P(XY = 0) = 0. Then: (i) If X is Benford then so is XY .
(ii) If S(X) and S(XY ) have the same distribution, then either log S(Y ) is rational with probability one, or X is Benford.
Proof. As in the proof of Theorem 4.13, the argument becomes short and trans-parent through the usage of Fourier coefficients. Note first that log S(XY ) = ⟨log S(X) + log S(Y )⟩and, since the random variables X0 := log S(X) and Y0 := log S(Y ) are independent, \ Plog S(XY ) = \ P⟨X0+Y0⟩= d PX0 · d PY0 .
(6.1) To prove (i), simply recall that X being Benford is equivalent to PX0 = λ0,1, and hence d PX0(k) = 0 for every integer k ̸= 0. Consequently, \ Plog S(XY )(k) = 0 as well, i.e., XY is Benford.
104 A. Berger and T.P. Hill To see (ii), assume that S(X) and S(XY ) have the same distribution. In this case, (6.1) implies that d PX0(k) 1 −d PY0(k) = 0 for all k ∈Z .
If d PY0(k) ̸= 1 for all non-zero k, then d PX0 = d λ0,1, i.e., X is Benford. Alterna-tively, if d PY0(k0) = 1 for some k0 ̸= 0 then, as seen in the proof of Theorem 4.13(iii), PY0( 1 |k0|Z) = 1, hence |k0|Y0 = |k0| log S(Y ) is an integer with proba-bility one.
Example 6.4. Let V , W be independent random variables distributed accord-ing to U(0, 1). Then X := 10V and Y := W are independent and, by Theorem 6.3(i), XY is Benford even though Y is not. If, on the other hand, X := 10V and Y := 101−V then X and Y are both Benford, yet XY is not. Hence the independence of X and Y is crucial in Theorem 6.3(i). It is essential in asser-tion (ii) as well, as can be seen by letting X equal either 10 √ 2−1 or 102− √ 2 with probability 1 2 each, and choosing Y := X−2. Then S(X) and S(XY ) = S(X−1) have the same distribution, but neither X is Benford nor log S(Y ) is rational with probability one.
Corollary 6.5. Let X be a random variable with P(X = 0) = 0, and let α be an irrational number. If S(X) and S(αX) have the same distribution, i.e., if X and αX have the same distribution of significant digits, then X is Benford.
Now consider a sequence (Xn) of independent random variables. From The-orem 6.3 it is clear that if the product Qn j=1 Xj is Benford for all sufficiently large n then one of the factors Xj is necessarily Benford. Clearly, this is a very restrictive assumption, especially in the i.i.d. case, where all Xj would have to be Benford. Much greater applicability is achieved by requiring Qn j=1 Xj to conform to BL only asymptotically. As the following theorem, an i.i.d. counter-part of Theorem 6.1, shows, convergence of Qn j=1 Xj in distribution to BL is a very widespread phenomenon. The result may help explain why BL often ap-pears in mathematical models that describe e.g. the evolution of stock prices by regarding the future price as a product of the current price times a large number of successive, multiplicative changes in price, with the latter being modeled as independent continuous random variables.
Theorem 6.6. Let (Xn) be an i.i.d. sequence of random variables that are not purely atomic, i.e. P(X1 ∈C) < 1 for every countable set C ⊂R. Then: (i) Qn j=1 Xj converges in distribution to BL.
(ii) With probability one, Qn j=1 Xj is Benford.
Proof. Let Yn = log |Xn|. Then (Yn) is an i.i.d. sequence of random variables that are not purely atomic. By Theorem 4.13(iii), the sequence of Pn j=1 Yj = log | Qn j=1 Xj| converges in distribution to U(0, 1). Thus Qn j=1 Xj converges in distribution to BL.
A basic theory of Benford’s Law 105 To prove (ii), let Y0 be u.d. mod 1 and independent of (Yn)n∈N, and define Sj := ⟨Y0 + Y1 + . . . + Yj⟩, j ∈N0 .
Recall from Theorem 4.13(i) that Sj is uniform on [0, 1) for every j ≥0. Also note that, by definition, the random variables Yj+1, Yj+2, . . . are independent of Sj. The following argument is most transparent when formulated in ergodic theory terminology. (For an alternative approach see e.g. [Ro].) To this end, endow T∞:= [0, 1)N0 = {(xj)j∈N0 : xj ∈[0, 1) for all j } with the σ-algebra B∞:= O j∈N0 B[0, 1) := σ {B0×B1×. . .×Bj×[0, 1)×[0, 1)×. . . : j ∈N0, B0, B1, . . . , Bj ∈B[0, 1)} .
A probability measure P∞is uniquely defined on (T∞, B∞) by setting P∞(B0×B1×. . .×Bj×[0, 1)×[0, 1)×. . .) = P(S0 ∈B0, S1 ∈B1, . . . , Sj ∈Bj) for all j ∈N0 and B0, B1, . . . , Bj ∈B[0, 1). The map σ∞: T∞→T∞with σ∞ (xj) = (xj+1), often referred to as the (one-sided) left shift on T∞(cf.
Example 5.25), is clearly measurable, i.e. σ−1 ∞(A) ∈B∞for every A ∈B∞. As a consequence, (σ∞)∗P∞is a well-defined probability measure on (T∞, B∞). In fact, since S1 is u.d. mod 1 and (Yn) is an i.i.d. sequence, (σ∞)∗P∞(B0×B1×. . .×Bj×[0, 1)×[0, 1)×. . .) = P∞([0, 1)×B0×B1×. . .×Bj×[0, 1)×[0, 1)×. . .) = P(S1 ∈B0, S2 ∈B1, . . . , Sj+1 ∈Bj) = P(S0 ∈B0, S1 ∈B1, . . . , Sj ∈Bj) = P∞(B0×B1×. . .×Bj×[0, 1)×[0, 1)×. . .) , showing that (σ∞)∗P∞= P∞, i.e., σ∞is P∞-preserving. (In probabilistic terms, this is equivalent to saying that the random process (Sj)j∈N0 is stationary, see [Sh, Def.V.1.1].) It will now be shown that σ∞is even ergodic with respect to P∞. Recall that this simply means that every invariant set A ∈B∞has measure zero or one, or, more formally, that P∞(σ−1 ∞(A)∆A) = 0 implies P∞(A) ∈{0, 1}; here the symbol ∆denotes the symmetric difference of two sets, i.e. A∆B = A\B ∪B\A. Assume, therefore, that P∞(σ−1 ∞(A)∆A) = 0 for some A ∈B∞.
Given ε > 0, there exists a number N ∈N and sets B0, B1, . . . , BN ∈B[0, 1) such that P∞ A ∆(B0×B1×. . .×BN ×[0, 1)×[0, 1)×. . .) < ε .
For notational convenience, let Aε := B0×B1×. . .×BN×[0, 1)×[0, 1)×. . . ∈B∞, and note that P∞ σ−j ∞(A)∆σ−j ∞(Aε) < ε for all j ∈N0. Recall now from 106 A. Berger and T.P. Hill Theorem 4.13(iii) that, given S0, S1, . . . , SN, the random variables Sn converge in distribution to U(0, 1). Thus, for all sufficiently large M, P∞ Ac ε ∩σ−M ∞(Aε) −P∞(Ac ε)P∞(Aε) < ε , (6.2) and similarly P∞ Aε ∩σ−M ∞(Ac ε) −P∞(Aε)P∞(Ac ε) < ε. (Note that (6.2) may not hold if X1, and hence also Y1, is purely atomic, see for instance Example 4.14(ii).) Overall, therefore, 2P∞(Aε) 1 −P∞(Aε) ≤2ε + P∞ Aε∆σ−M ∞(Aε) ≤2ε + P∞(Aε∆A) + P∞ A∆σ−M ∞(A) + P∞ σ−M ∞(A)∆σ−M ∞(Aε) < 4ε , and consequently P∞(A) 1 −P∞(A) < 3ε + ε2. Since ε > 0 was arbitrary, P∞(A) ∈{0, 1}, which in turn shows that σ∞is ergodic. (Again, this is equiv-alent to saying, in probabilistic parlance, that the random process (Sj)j∈N0 is ergodic, see [Sh, Def.V.3.2].) By the BirkhoffErgodic Theorem, for every (mea-surable) function f : [0, 1) →C with R 1 0 |f(x)| dx < +∞, 1 n Xn j=0 f(xj) → Z 1 0 f(x) dx as n →∞ holds for all (xj)j∈N0 ∈T∞, with the possible exception of a set of P∞-measure zero. In probabilistic terms, this means that limn→∞ 1 n Xn j=0 f(Sj) = Z 1 0 f(x) dx a.s.
(6.3) Assume from now on that f is actually continuous with limx↑1 f(x) = f(0), e.g.
f(x) = e2πıx. For any such f, as well as any t ∈[0, 1) and m ∈N, denote the set ω∈Ω: lim supn→∞ 1 n Xn j=1f ⟨t + Y1(ω) + . . . + Yj(ω)⟩ − Z 1 0 f(x) dx < 1 m simply by Ωf,t,m. According to (6.3), 1 = R 1 0 P(Ωf,t,m) dt, and hence P(Ωf,t,m) = 1 for a.e. t ∈[0, 1). Since f is uniformly continuous, for every m ≥2 there exists tm > 0 such that P(Ωf,tm,m) = 1 and Ωf,tm,m ⊂Ωf,0,⌊m/2⌋. From 1 = P \ m≥2 Ωf,tm,m ≤P \ m≥2 Ωf,0,⌊m/2⌋ ≤1 , it is clear that limn→∞ 1 n Xn j=1 f ⟨Y1 + . . . + Yj⟩ = Z 1 0 f(x) dx a.s.
(6.4) A basic theory of Benford’s Law 107 As the intersection of countably many sets of full measure has itself full measure, choosing f(x) = e2πıkx, k ∈Z in (6.4) shows that, with probability one, limn→∞ 1 n Xn j=1 e2πık(Y1+...+Yj) = Z 1 0 e2πıkxdx = 0 for all k ∈Z, k ̸= 0 .
(6.5) By Weyl’s criterion [KN, Thm.2.1], (6.5) is equivalent to P Xn j=1 Yj is u.d. mod 1 = 1 .
In other words, (Qn j=1 Xj) is Benford with probability one.
Remarks. (i) As has essentially been observed already in Example 4.14(ii), for Theorem 6.6(i) to hold it is necessary and sufficient that P(log |X1| ∈a + 1 mZ) < 1 for all a ∈R, m ∈N .
(6.6) On the other hand, it is not hard to see that (ii) holds if and only if P(log |X1| ∈1 mZ) < 1 for all m ∈N , (6.7) which is a strictly weaker assumption than (6.6). Clearly, if X1 is not purely atomic then (6.6), and hence also (6.7) hold. However, if e.g. X1 = 2 with probability one then (6.6) does not hold, and correspondingly Qn j=1 Xj = (2n) does not converge in distribution to BL, whereas (6.7) does hold, and Qn j=1 Xj is Benford with probability one, cf. Example 6.2(ii).
(ii) For more general results in the spirit of Theorem 6.6 the reader is referred to [Sch¨ u1, Sch¨ u2].
♣ Example 6.7. (i) Let (Xn) be an i.i.d. sequence with X1 distributed according to U(0, a), the uniform distribution on [0, a) with a > 0. The k-th Fourier coefficient of P⟨log X1⟩is \ P⟨log X1⟩(k) = e2πık log a ln 10 ln 10 + 2πık , k ∈Z , so that, for every k ̸= 0, \ P⟨log X1⟩(k) = ln 10 p (ln 10)2 + 4π2k2 < 1 .
As seen in the proof of Theorem 4.13(iii), this implies that Qn j=1 Xj converges in distribution to BL, a fact apparently first recorded in [AS]. Note also that E log X1 = log a e . Thus with probability one, Qn j=1 Xj converges to 0 or +∞, depending on whether a < e or a > e. In fact, by the Strong Law of Large Numbers [CT], n rYn j=1 Xj a.s.
→a e 108 A. Berger and T.P. Hill holds for every a > 0. If a = e then P lim infn→∞ Yn j=1 Xj = 0 and lim supn→∞ Yn j=1 Xj = +∞ = 1 , showing that in this case the product Qn j=1 Xj does not converge but rather attains, with probability one, arbitrarily small as well as arbitrarily large positive values. By Theorem 6.6(ii), the sequence Qn j=1 Xj is a.s. Benford, regardless of the value of a.
(ii) Consider an i.i.d. sequence (Xn) with X1 distributed according to a log-normal distribution such that log X1 is standard normal. Denote by fn the density of log Qn j=1 Xj . Since log Qn j=1 Xj = Pn j=1 log Xj is normal with mean zero and variance n, fn(s) = 1 √ 2πn X k∈Z e−(k+s)2/(2n) , 0 ≤s < 1 , from which it is straightforward to deduce that limn→∞fn(s) = 1 , uniformly in 0 ≤s < 1 .
Consequently, for all t ∈[1, 10), P S Yn j=1 Xj ≤t = P D log Yn j=1 Xj E ≤log t = Z log t 0 fn(s) ds → Z log t 0 1 ds = log t , i.e., Qn j=1 Xj converges in distribution to BL. By Theorem 6.6(ii) also P Yn j=1 Xj is Benford = 1 , even though E log Qn j=1 Xj = Pn j=1 E log Xj = 0, and hence, as in the previous example, the sequence Qn j=1 Xj a.s. oscillates forever between 0 and +∞.
Having seen Theorem 6.6, the reader may wonder whether there is an anal-ogous result for sums of i.i.d. random variables. After all, the focus in classical probability theory is on sums much more than on products. Unfortunately, the statistical behavior of the significands is much more complex for sums than for products. The main basic reason is that the significand of the sum of two or more numbers depends not only on the significand of each each number (as in the case of products), but also on their exponents. For example, observe that S 3 · 103 + 2 · 102 = 3.2 ̸= 5 = S 3 · 102 + 2 · 102 , while clearly S 3 · 103 × 2 · 102 = 6 = S 3 · 102 × 2 · 102 .
A basic theory of Benford’s Law 109 Practically, this difficulty is reflected in the fact that for positive real numbers u, v, the value of log(u + v), relevant for conformance with BL via Theorem 4.2, is not easily expressed in terms of log u and log v, whereas log(uv) = log u + log v.
In view of these difficulties, it is perhaps not surprising that the analog of Theorem 6.6 for sums arrives at a radically different conclusion.
Theorem 6.8. Let (Xn) be an i.i.d. sequence of random variables with finite variance, that is EX2 1 < +∞. Then: (i) Not even a subsequence of Pn j=1 Xj converges in distribution to BL.
(ii) With probability one, Pn j=1 Xj is not Benford.
Proof. Assume first that EX1 ̸= 0. By the Strong Law of Large Numbers, 1 n Pn j=1 Xj converges a.s., and hence also in distribution, to the constant |EX1|. Since log S Xn j=1 Xj = D log Xn j=1 Xj E = log 1 n Xn j=1 Xj + log n , any subsequence of S 1 n Pn j=1 Xj either does not converge in distribution at all or else converges to a constant; in neither case, therefore, is the limit a Benford random variable. Since, with probability one, Pn j=1 Xj →+∞, it follows from log Xn j=1 Xj −log n = log 1 n Xn j=1 Xj a.s.
→|EX1| , together with Lemma 4.3(i) and Proposition 4.8(iii) that Pn j=1 Xj) is, with probability one, not Benford.
It remains to consider the case EX1 = 0. Without loss of generality, it can be assumed that EX2 1 = 1. By the Central Limit Theorem 1 √n Pn j=1 Xj con-verges in distribution to the standard normal distribution. Thus for sufficiently large n, and up to a rotation (i.e. an addition mod 1) of [0, 1), the distri-bution of ⟨log | Pn j=1 Xj|⟩differs by arbitrarily little from the distribution of Y := ⟨log |Z|⟩, where Z is standard normal. Intuitively, it is clear that PY ̸= λ0,1, i.e., Y is not uniform on [0, 1). To see this more formally, note that FY (s) = 2 X k∈Z Φ 10s+k −Φ 10k , 0 ≤s < 1 , (6.8) with Φ (= FZ) denoting the standard normal distribution function, see Example 4.16(ii). Thus |FY (s) −s| ≥FY (s) −s > 2 Φ (10s) −Φ (1) −s =: R(s) , 0 ≤s < 1 , and since R is concave on [0, 1) with R(0) = 0 and R′(0) = 2 ln 10 √ 2πe −1 = 0.1143 . . . > 1 9, it follows that max0≤s<1 |FY (s) −s| > max0≤s<1 R(s) > 0 , 110 A. Berger and T.P. Hill showing that indeed PY ̸= λ0,1, and hence Pn j=1 Xj does not converge in distribution to BL.
The verification of (ii) in the case EX1 = 0 uses an almost sure version of the Central Limit Theorem, see [LP]. With the random variables Xn de-fined on some (abstract) probability space (Ω, A, P), let Ω1 := ω ∈Ω: Pn j=1 Xj(ω) is Benford . By Theorem 4.2 and Proposition 4.8(iii), the se-quence xn(ω) with xn(ω) = log 1 √n Xn j=1 Xj(ω) , n ∈N , is u.d. mod 1 for all ω ∈Ω1. For every interval [a, b) ⊂[0, 1), therefore, 1 ln N XN n=1 1[a,b) xn(ω) n →b −a as N →∞.
(Recall the remark on p.18.) However, as a consequence of [LP, Thm.2], for every [a, b) ⊂[0, 1), 1 ln N XN n=1 1[a,b) xn n a.s.
→FY (b) −FY (a) , with FY given by (6.8). As shown above, FY (s) ̸≡s, and therefore P(Ω1) = 0.
In other words, P (Pn j=1 Xj) is Benford = 0.
Example 6.9. (i) Let (Xn) be an i.i.d. sequence with P(X1 = 0) = P(X1 = 1) = 1 2. Then EX1 = EX2 1 = 1 2, and by Theorem 6.8(i) neither Pn j=1 Xj nor any of its subsequences converges in distribution to BL. Note that Pn j=1 Xj is binomial with parameters n and 1 2, i.e. for all n ∈N, P Xn j=1 Xj = l = 2−n n l , l = 0, 1, . . ., n .
The Law of the Iterated Logarithm [CT] asserts that Xn j=1 Xj = n 2 + Yn √ n ln ln n for all n ≥3 , (6.9) where the sequence (Yn) of random variables is bounded, in fact |Yn| ≤1 a.s. for all n. From (6.9) it is clear that, with probability one, the sequence Pn j=1 Xj is not Benford.
(ii) Let (Xn) be an i.i.d. sequence of Cauchy random variables. As E|X1| is even infinite, Theorem 6.8 does not apply. However, recall from Example 4.14(i) that 1 n Pn j=1 Xj is again Cauchy, and hence the distribution of ⟨log |(Pn j=1 Xj)|⟩ is but a rotated version of P⟨log |X1|⟩, the density of which is given by f⟨log |X1|⟩(s) = ln 10 π X k∈Z 1 cosh (s + k) ln 10 , 0 ≤s < 1.
A basic theory of Benford’s Law 111 The density f⟨log |X1|⟩is a smooth function, and f⟨log |X1|⟩(0) = ln 10 π X k∈Z 2 10k + 10−k > ln 10 π 1 + 40 101 > 1 + 2 101 , showing that log | Pn j=1 Xj| is not u.d. mod 1. Hence the sequence Pn j=1 Xj does not converge in distribution to BL, and nor does any of its subsequences.
This example shows that the conclusions of Theorem 6.8 may hold, at least in parts, even if the Xn do not have finite first, let alone finite second moments.
Remark. Recall from the remark on p.18 that a sequence failing to be Benford may conform to a weaker form of BL. As seen above, under mild conditions the stochastic sequence Pn j=1 Xj is not Benford. Under the appropriate assump-tions, however, it does obey a weaker form of BL, see [Scha2].
♣ 6.2. Mixtures of distributions The characterizations of the Benford distribution via scale-, base- and sum-invariance, given in Chapter 4, although perhaps mathematically satisfying, hardly help explain the appearance of BL empirically in real-life data. Applica-tion of those theorems requires explaining why the underlying data is scale- or base-invariant in the first place. BL nevertheless does appear in many real-life datasets. Thus the question arises: What do the population data of three thou-sand U.S. counties according to the 1990 census have in common with the usage of logarithm tables during the 1880s, numerical data from newspaper articles of the 1930’s collected by Benford, or universal physical constants examined by Knuth in the 1960’s? Why should these data exhibit a logarithmically dis-tributed significand or equivalently, why should they be scale- or base-invariant?
As a matter of fact, most data-sets do not follow BL closely. Benford already observed that while some of his tables conformed to BL reasonably well, many others did not. But, as Raimi [Ra1] points out, “what came closest of all, how-ever, was the union of all his tables.” Combine the molecular weight tables with baseball statistics and drainage areas of rivers, and then there is a very good fit.
Many of the previous explanations of BL have first hypothesized some universal table of constants, such as Raimi’s [Ra1] “stock of tabular data in the world’s libraries”, or Knuth’s [Kn] “imagined set of real numbers”, and then tried to prove why certain specific sets of real observations were representative of either this mysterious universal table or the set of all real numbers. What seems more natural though is to think of data as coming from many different distributions.
This was clearly the case in Benford’s original study. After all, he had made an effort “to collect data from as many fields as possible and to include a wide variety of types”, noting that “the range of subjects studied and tabulated was as wide as time and energy permitted”.
The main goal of this section is to provide a statistical derivation of BL, in the form of a central-limit-like theorem that says that if random samples are taken from different distributions, and the results combined, then — provided 112 A. Berger and T.P. Hill the sampling is “unbiased” as to scale or base — the resulting combined samples will converge to the Benford distribution.
Denote by M the set of all probability measures on (R, B). Recall that a (real Borel) random probability measure, abbreviated henceforth as r.p.m., is a function P : Ω→M, defined on some underlying probability space (Ω, A, P), such that for every B ∈B the function ω 7→P(ω)(B) is a random variable.
Thus, for every ω ∈Ω, P(ω) is a probability measure on (R, B), and, given any real numbers a, b and any Borel set B, {ω : a ≤P(ω)(B) ≤b} ∈A , see e.g. [Ka] for an authoritative account on random probability measures. In more abstract conceptual terms, an r.p.m. can be interpreted as follows: When endowed with the topology of convergence in distribution, the set M becomes a complete and separable metrizable space. Denote by BM its Borel σ-algebra, defined as the smallest σ-algebra containing all open subsets of M. Then P∗P simply is a probability measure on (M, BM).
Example 6.10. (i) Let P be an r.p.m. that is U(0, 1) with probability 1 2, and otherwise is exp(1), i.e. exponential with mean 1, hence P(X > t) = min(1, e−t) for all t ∈R, see Example 3.10(i,ii). Thus, for every ω ∈Ω, the probability mea-sure P is either U(0, 1) or exp(1), and P P =U(0, 1) = P P =exp(1) = 1 2. For a practical realization of P simply flip a fair coin — if it comes up heads, P(ω) is a U(0, 1)-distribution, and if it comes up tails, then P is an exp(1)-distribution.
(ii) Let X be distributed according to exp(1), and let P be an r.p.m. where, for each ω ∈Ω, P(ω) is the normal distribution with mean X(ω) and variance 1. In contrast to the example in (i), here P is continuous, i.e., P(P = Q) = 0 for each probability measure Q ∈M.
The following example of an r.p.m. is a variant of a classical construction due to L. Dubins and D. Freedman which, as will be seen below, is an r.p.m. leading to BL.
Example 6.11. Let P be the r.p.m. with support on [1, 10), i.e. P [1, 10) = 1 with probability one, defined by its (random) cumulative distribution function FP , i.e.
FP (t) := FP (ω)(t) = P(ω) [1, t] , 1 ≤t < 10 , as follows: Set FP (1) = 0 and FP (10) = 1. Next pick FP (101/2) according to the uniform distribution on [0, 1). Then pick FP (101/4) and FP (103,4) in-dependently, uniformly on 0, FP (101/2) and FP (101/2), 1 , respectively, and continue in this manner. This construction is known to generate an r.p.m. a.s.
[DF, Lem.9.28], and as can easily be seen, is dense in the set of all probability measures on [1, 10), B[1, 10) , i.e., it generates probability measures that are arbitrarily close to any Borel probability measure on [1, 10).
The next definition formalizes the notion of combining data from different distributions. Essentially, it mimics what Benford did in combining baseball A basic theory of Benford’s Law 113 statistics with square-root tables and numbers taken from newspapers, etc. This definition is key to everything that follows. It rests upon using an r.p.m. to generate a random sequence of probability distributions, and then successively selecting random samples from each of those distributions.
Definition 6.12. Let m be a positive integer and P an r.p.m. A sequence of P-random m-samples is a sequence (Xn) of random variables on (Ω, A, P) such that, for all j ∈N and some i.i.d. sequence (Pn) of r.p.m.s with P1 = P, the following two properties hold: Given Pj = Q, the random variables X(j−1)m+1, X(j−1)m+2, . . . , Xjm (6.10) are i.i.d. with distribution Q ; The variables X(j−1)m+1, X(j−1)m+2, . . . , Xjm are independent of (6.11) Pi, X(i−1)m+1, X(i−1)m+2, . . . , Xim for every i ̸= j .
Thus for any sequence (Xn) of P-random m-samples, for each ω ∈Ωin the underlying probability space, the first m random variables are a random sample (i.e., i.i.d.) from P1(ω), a random probability distribution chosen according to the r.p.m. P; the second m-tuple of random variables is a random sample from P2(ω) and so on. Note the two levels of randomness here: First a probability is selected at random, and then a random sample is drawn from this distribution, and this two-tiered process is continued.
Example 6.13. Let P be the r.p.m. in Example 6.10(i), and let m = 3. Then a sequence of P-random 3-samples is a sequence (Xn) of random variables such that with probability 1 2, X1, X2, X3, are i.i.d. and distributed according to U(0, 1), and otherwise they are i.i.d. but distributed according to exp(1); the random variables X4, X5, X6 are again equally likely to be i.i.d. U(0, 1) or exp(1), and they are independent of X1, X2, X3, etc. Clearly the (Xn) are all identically distributed as they are all generated by exactly the same process.
Note, however, that for instance X1 and X2 are dependent: Given that X1 > 1, for example, the random variable X2 is exp(1)-distributed with probability one, whereas the unconditional probability that X2 is exp(1)-distributed is only 1 2.
Remark. If (Xn) is a sequence of P-random m-samples for some m and some r.p.m. P, then the Xn are a.s. identically distributed according to the distri-bution that is the average (expected) distribution of P (see Proposition 6.15 below), but they are not in general independent (see Example 6.13). On the other hand, given (P1, P2, . . .), the (Xn) are a.s. independent, but clearly are not in general identically distributed.
♣ Although sequences of P-random m-samples have a fairly simple structure, they do not fit into any of the familiar categories of sequences of random vari-ables. For example, they are not in general independent, exchangeable, Markov, martingale, or stationary sequences.
Example 6.14. Assume that the r.p.m. P is, with equal probability, the Dirac measure concentrated at 1 and the probability measure 1 2(δ1 + δ2), respectively, 114 A. Berger and T.P. Hill i.e. P(P = δ1) = P P = 1 2(δ1 + δ2) = 1 2. Let (Xn) be a sequence of P-random 3-samples. Then the random variables X1, X2, . . . are not independent as P(X2 = 2) = 1 4 but P(X2 = 2|X1 = 2) = 1 2 ; not exchangeable as P (X1, X2, X3, X4)=(1, 1, 1, 2) = 9 64 ̸= 3 64 =P (X1, X2, X3, X4)=(2, 2, 2, 1) ; not Markov as P(X3 = 1|X1 = X2 = 1) = 9 10 ̸= 5 6 = P(X3 = 1|X2 = 1) ; not martingale as E(X2|X1 = 2) = 3 2 but EX2 = 5 4 ; not stationary as P (X1, X2, X3) = (1, 1, 1) = 9 16 ̸= 15 32 = P (X2, X3, X4) = (1, 1, 1) .
Recall that, given an r.p.m. P and any Borel set B, the quantity P(B) is a random variable with values between 0 and 1. The following property of the expectation of P(B), as a function of B, is easy to check.
Proposition 6.15. Let P be an r.p.m. Then EP, defined as (EP)(B) := EP(B) = Z Ω P(ω)(B) dP(ω) for all B ∈B , is a probability measure on (R, B).
Example 6.16. (i) Let P be the r.p.m. of Example 6.10(i). Then EP is the Borel probability measure with density fEP (t) = 0 if t < 0 , 1 2 + 1 2e−t if 0 ≤t < 1 , 1 2e−t if t ≥1 , = 1 21[0,1)(t)+ 1 2e−t1[0,+∞) , t ∈R .
(ii) Consider the r.p.m. P of Example 6.10(ii), that is, P(ω) is normal with mean X(ω) and variance 1, where X is distributed according to exp(1). In this case, EP is also a.c., with density fEP (t) = 1 √ 2π Z +∞ 0 e−1 2 (t−τ)2e−τdτ = e 1 2 −t 1 −Φ(1 −t) , t ∈R .
A basic theory of Benford’s Law 115 (iii) Even if P is a.c. only with probability zero, it is possible for EP to be a.c. As a simple example, let X be exp(1)-distributed and P = 1 2(δ−X + δX).
Then P(P is purely atomic ) = 1, yet EP is the standard Laplace (or double-exponential) distribution; i.e., EP is a.c. with density fEP (t) = e−|t| 2 , t ∈R .
The next lemma shows that the limiting proportion of times that a sequence of P-random m-sample falls in a (Borel) set B is, with probability one, the average P-value of the set B, i.e., the limiting proportion equals EP(B). Note that this is not simply a direct corollary of the classical Strong Law of Large Numbers as the random variables in the sequence are not in general independent (see Examples 6.13 and 6.14).
Lemma 6.17. Let P be an r.p.m., and let (Xn) be a sequence of P-random m-samples for some m ∈N. Then, for every B ∈B, #{1 ≤n ≤N : Xn ∈B} N a.s.
→EP(B) as N →∞.
Proof. Fix B ∈B and j ∈N, and let Yj = #{1 ≤i ≤m : X(j−1)m+i ∈B}. It is clear that limN→∞ #{1 ≤n ≤N : Xn ∈B} N = 1 m limn→∞ 1 n Xn j=1 Yj , (6.12) whenever the limit on the right exists. By (6.10), given Pj, the random variable Yj is binomially distributed with parameters m and E Pj(B) , hence a.s.
EYj = E E(Yj|Pj) = E mPj(B) = mEP(B) (6.13) since Pj has the same distribution as P. By (6.11), the Yj are independent. They are also uniformly bounded, as 0 ≤Yj ≤m for all j, hence P∞ j=1 EY 2 j /j2 < +∞.
Moreover, by (6.13) all Yj have the same mean value mEP(B). Thus by [CT, Cor.5.1] 1 n Xn j=1 Yj a.s.
→mEP(B) as n →∞, (6.14) and the conclusion follows by (6.12) and (6.14).
Remark. The assumption that each Pj is sampled exactly m times is not essen-tial: The above argument can easily be modified to show that the same con-clusion holds if the j-th r.p.m. is sampled Mj times where (Mj) is a sequence of independent, uniformly bounded N-valued random variables which are also independent of the rest of the process.
♣ The stage is now set to give a statistical limit law (Theorem 6.20 below) that is a central-limit-like theorem for significant digits mentioned above. Roughly speaking, this law says that if probability distributions are selected at random, 116 A. Berger and T.P. Hill and random samples are then taken from each of these distributions in such a way that the overall process is scale- or base-neutral, then the significant digit frequencies of the combined sample will converge to the logarithmic distribution.
This theorem may help explain and predict the appearance of BL in significant digits in mixtures of tabulated data such as the combined data from Benford’s individual datasets, and also his individual dataset of numbers gleaned from newspapers.
In order to draw any conclusions concerning BL for the process of sampling from different distributions, clearly there must be some restriction on the un-derlying r.p.m. that generates the sampling procedure. Otherwise, if the r.p.m.
is, say, U(0, 1) with probability one, for example, then any resulting sequence of P-random m-samples will be i.i.d. U(0, 1), and hence a.s. not Benford, by Example 3.10(i). Similarly, it can easily be checked that sequences of P-random m-samples from the r.p.m.s in Example 6.10 (i) and (ii) will not generate Benford sequences. A natural assumption to make concerning an r.p.m. in this context is that on average the r.p.m. is unbiased (i.e. invariant) with respect to changes in scale or base.
Definition 6.18. An r.p.m. P has scale-unbiased (decimal) significant digits if, for every significand event A, i.e. for every A ∈S, the expected value of P(A) is the same as the expected value P(αA) for every α > 0, that is, if E P(αA) = E P(A) for all α > 0, A ∈S .
Equivalently, the Borel probability measure EP has scale-invariant significant digits.
Similarly, P has base-unbiased significant (decimal) digits if, for every A ∈S the expected value of P(A) is the same as the expected value of P(A1/n) for every n ∈N, that is, if E P(A1/n) = E P(A) for all n ∈N, A ∈S , i.e., if EP has base-invariant significant digits.
An immediate consequence of Theorems 4.20 and 4.30 is Proposition 6.19. Let P be an r.p.m. with EP({0}) = 0. Then the following statements are equivalent: (i) P has scale-unbiased significant digits.
(ii) P({±10k : k ∈Z}) = 0, or equivalently S∗P({1}) = 0 holds with probabil-ity one, and P has base-unbiased significant digits.
(iii) EP(A) = B(A) for all A ∈S, i.e., EP is Benford.
Random probability measures with scale- or base-unbiased significant digits are easy to construct mathematically (see Example 6.22 below). In real-life examples, however, scale- or base-unbiased significant digits should not be taken for granted. For instance, picking beverage-producing companies in Europe at A basic theory of Benford’s Law 117 random, and looking at the metric volumes of samples of m products from each company, is not likely to produce data with scale-unbiased significant digits, since the volumes in this case are probably closely related to liters. Conversion of the data to another unit such as gallons would likely yield a radically different set of first-digit frequencies. On the other hand, if species of mammals in Europe are selected at random and their metric volumes sampled, it seems more likely that the latter process is unrelated to the choice of human units.
The question of base-unbiasedness of significant digits is most interesting when the units in question are universally agreed upon, such as the numbers of things, as opposed to sizes. For example, picking cities at random and looking at the number of leaves of m-samples of trees from those cities is certainly less base-dependent than looking at the number of fingers of m-samples of people from those cities.
As will be seen in the next theorem, scale- or base-unbiasedness of an r.p.m.
imply that sequence of P-random samples are Benford a.s. A crucial point in the definition of an r.p.m. P with scale- or base-unbiased significant digits is that it does not require individual realizations of P to have scale- or base-invariant significant digits. In fact, it is often the case (see Benford’s original data in [Ben] and Example 6.22 below) that a.s. none of the random probabilities has either of these properties, and it is only on average that the sampling process does not favor one scale or base over another. Recall from the notation introduced above that S∗P({1}) = 0 is the event {ω ∈Ω: P(ω)(S = 1) = 0}.
Theorem 6.20 ([Hi2]). Let P be an r.p.m. Assume P either has scale-unbiased significant digits, or else has base-unbiased significant digits and S∗P({1}) = 0 with probability one. Then, for every m ∈N, every sequence (Xn) of P-random m-samples is Benford with probability one, that is, for all t ∈[1, 10), #{1 ≤n ≤N : S(Xn) < t} N a.s.
→log t as N →∞.
Proof. Assume first that P has scale-unbiased significant digits, i.e., the proba-bility measure EP has scale-invariant significant digits. According to Theorem 4.20, EP is Benford. Consequently, Lemma 6.17 implies that for every sequence (Xn) of P-random m-samples and every t ∈[1, 10), #{1 ≤n≤N :S(Xn) < t} N = # 1 ≤n≤N :Xn ∈S k∈Z 10k (−t, −1] ∪[1, t) N a.s.
→EP [ k∈Z 10k (−t, −1] ∪[1, t) = log t as N →∞.
Assume in turn that S∗P({1}) = 0 with probability one, and that P has base-unbiased significant digits. Then S∗EP({1}) = EP S−1({1}) = Z Ω S∗P(ω)({1}) dP(ω) = 0 .
Hence q = 0 holds in (4.8) with P replaced by EP, proving that EP is Benford, and the remaining argument is the same as before.
118 A. Berger and T.P. Hill Corollary 6.21. If an r.p.m. P has scale-unbiased significant digits, then for every m ∈N, every sequence (Xn) of P-random m-samples, and every d ∈ {1, 2, . . ., 9}, #{1 ≤n ≤N : D1(Xn) = d} N a.s.
→log(1 + d−1) as N →∞.
A main point of Theorem 6.20 is that there are many natural sampling pro-cedures that lead to the same logarithmic distribution. This helps explain how the different empirical evidence of Newcomb, Benford, Knuth and Nigrini all led to the same law. It may also help explain why sampling the numbers from newspaper front pages or almanacs [Ben], or accumulating extensive accounting data [Ni], often tends toward BL, since in each of these cases various distribu-tions are being sampled in a presumably unbiased way. In a newspaper, perhaps the first article contains statistics about population growth, the second arti-cle about stock prices, the third about forest acreage. None of these individual distributions itself may be unbiased, but the mixture may well be.
Justification of the hypothesis of scale- or base-unbiasedness of significant digits in practice is akin to justification of the hypothesis of independence (and identical distribution) when applying the Strong Law of Large Numbers or the Central Limit Theorem to real-life processes: Neither hypothesis can be formally proved, yet in many real-life sampling procedures, they appear to be reasonable assumptions.
Many standard constructions of r.p.m. automatically have scale- and base-unbiased significant digits, and thus satisfy BL in the sense of Theorem 6.20.
Example 6.22. Recall the classical Dubins–Freedman construction of an r.p.m.
P described in Example 6.11. It follows from [DF, Lem.9.28] that EP is Benford.
Hence P has scale- and base-unbiased significant digits. Note, however, that with probability one P will not have scale- or base-invariant significant digits. It is only on average that these properties hold but, as demonstrated by Theorem 6.20, this is enough to guarantee that random sampling from P will generate Benford sequences a.s.
In the Dubins–Freedman construction, the fact that FP (101/2), FP (101/4), FP (103/4), etc. are chosen uniformly from the appropriate intervals is not cru-cial: If Q is any probability measure on (0, 1), and the values of FP (101/2) etc.
are chosen independently according to an appropriately scaled version on Q, then, for the r.p.m. thus generated, EP will still be Benford, provided that Q is symmetric about 1 2, see [DF, Thm.9.29]. As a matter of fact, real-world pro-cesses often exhibit this symmetry in a natural way: Many data may be equally well recorded using certain units or their reciprocals, e.g. in miles per gallon vs.
gallons per mile, or Benford’s “candles per watt” vs. “watts per candle”. This suggests that the distribution of log S should be symmetric about 1 2.
Data having scale- or base-unbiased significant digits may be produced in many ways other than through random samples. If such data are combined with unbiased random m-samples then the result will again conform to BL in the sense of Theorem 6.20. (Presumably, this is what Benford did when combining A basic theory of Benford’s Law 119 mathematical tables with data from newspaper statistics.) For example, consider the sequence (2n) which may be thought of as the result of a periodic sampling from a (deterministic) geometric process. As (2n) is Benford, any mixture of this sequence with a sequence of unbiased random m-samples will again be Benford.
Finally, it is important to note that many r.p.m. and sampling processes do not conform to BL, and hence necessarily are scale- and base-biased.
Example 6.23. (i) Let P be the constant r.p.m. P ≡δ1. Since EP = δ1 has base-invariant significant digits, P has base-unbiased significant digits. Never-theless, for every sequence (Xn) of P-random m-samples, the sequence of first significant digits is constant, namely D1(Xn) ≡1.
Similarly, if P = λ0,1 with probability one, then EP = λ0,1 does not have scale- or base-invariant significant digits. Consequently, every sequence of P-random m-samples is an i.i.d. U(0, 1)-sequence and hence not Benford, by Ex-ample 3.10(i).
(ii) The r.p.m. considered in Example 6.10 do not have scale- or base-unbiased significant digits, simply because EP is not Benford.
(iii) As a another variant of the classical construction in [DF], consider the following way of generating an r.p.m. on [1, 10): First let X1/2 be uniformly distributed on [1, 10) and set FP (X1/2) = 1 2. Next let X1/4 and X3/4 be inde-pendent and uniformly distributed on [1, X1/2) and [X1/2, 10), respectively, and set FP (X1/4) = 1 4 and FP (X3/4) = 3 4, etc. It follows from [DF, Thm.9.21] that FEP (t) = 2 π arcsin log t , 1 ≤t < 10 , and hence EP is not Benford. The r.p.m. P thus constructed, therefore, has scale- and base-biased significant digits.
6.3. Random maps The purpose of this brief concluding section is to illustrate and prove one simple basic theorem that combines the deterministic aspects of BL studied in Chapter 5 with the stochastic considerations of the present chapter. Specifically, it is shown how applying randomly selected maps successively may generate Benford sequences with probability one. Random maps constitute a wide and intensely studied field, and for stronger results than the one discussed here the interested reader is referred e.g. to [Ber3].
For a simple example, first consider the map T : R →R with T (x) = p |x|.
Since T n(x) = |x|2−n →1 as n →∞whenever x ̸= 0, the orbit OT (x0) is not Benford for any x0. More generally, consider the randomized map T (x) = ( p |x| with probability p , x3 with probability 1 −p , (6.15) 120 A. Berger and T.P. Hill and assume that, at each step, the iteration of T is independent of the entire past process. If p = 1, this is simply the map studied before, and hence for every x0 ∈R, the orbit OT (x0) is not Benford. On the other hand, if p = 0 then Theorem 5.12 implies that, for almost every x0 ∈R, OT (x0) is Benford. It is plausible to expect that the latter situation persists for small p > 0. As the following theorem shows, this is indeed that case even when the non-Benford map p |x| occurs more than half of the time: If p < log 3 log 2 + log 3 = 0.6131 . . . , (6.16) then, for a.e. x0 ∈R, the (random) orbit OT (x0) is Benford with probability one.
To concisely formulate this result, recall that for any (deterministic or random) sequence (Tn) of maps mapping R or parts thereof into itself, the orbit OT (x0) of x0 ∈R simply denotes the sequence Tn−1 ◦. . . ◦T1(x0) n∈N.
Theorem 6.24 ([Ber3]). Let (βn) be an i.i.d. sequence of positive random vari-ables, and assume that log β1 has finite variance, i.e. E(log β1)2 < +∞. For the sequence (Tn) of random maps given by Tn : x 7→xβn and a.e. x0 ∈R, the orbit OT (x0) is Benford with probability one or zero, depending on whether E log β1 > 0 or E log β1 ≤0.
Proof. For every x ∈R and n ∈N, log Tn ◦. . . ◦T1(x) = Yn j=1 βj log |x| = 10Bn log |x| , where Bn = Pn j=1 log βj. Assume first that E log β1 > 0. In this case, Bn n a.s.
→ log β1 as n →∞, and it can be deduced from [KN, Thm.4.2] that, with proba-bility one, the sequence (10Bny) is u.d. for a.e. y ∈R. Since x 7→log |x| maps the family of (Lebesgue) nullsets into itself, with probability one OT (x0) is Benford for a.e. x0 ∈R. More formally, with (Ω, A, P) denoting the underlying probabil-ity space, there exists Ω1 ∈A with P(Ω1) = 1 such that for every ω ∈Ω1 the sequence OT (x0) is Benford for all x0 ∈R\Bω, where Bω ∈B with λ(Bω) = 0.
Denote by N ⊂R × Ωthe set of all (x0, ω) for which OT (x0) is not Benford, and let Nx = {ω ∈Ω: (x, ω) ∈N} , x ∈R , N ω = {x ∈R : (x, ω) ∈N} , ω ∈Ω.
Then Nx ∈A and N ω ∈B for all x ∈R and ω ∈Ω, respectively, and λ(N ω) = 0 for all ω ∈Ω1. By Fubini’s Theorem, 0 = Z Ω λ(N ω) dP(ω) = Z R×Ω 1N d(λ × P) = Z R P(Nx) dλ(x) , showing that P(Nx) = 0 for a.e. x ∈R. Equivalently P(OT (x0) is Benford ) = 1 holds for a.e. x0 ∈R.
A basic theory of Benford’s Law 121 Next assume that E log β1 < 0. Then Tn ◦. . .◦T1(x) a.s.
→1 as n →∞for every x ̸= 0, and hence OT (x) is not Benford. (Note, however, that (Tn◦. . .◦T1(x)−1) may be Benford in this case.) Finally, it remains to consider the case E log β1 = 0. It follows from the Law of the Iterated Logarithm that, for every t ∈R, lim supN→∞ #{1 ≤n ≤N : Bn ≤t} N ≥1 2 with probability one .
Clearly, this implies P(OT (x0) is Benford ) = 0 for every x0 ∈R.
Example 6.25. (i) For the random map given by (6.15), P β = 1 2 = p = 1 −P(β = 3) , and the condition E log β = −p log 2 + (1 −p) log 3 > 0 is equivalent to (6.16).
Note that E log β > 0 is not generally equivalent to the equally plausible (yet incorrect) condition Eβ > 1. In the present example, the latter reduces to p < 4 5.
(ii) Consider the sequence (Tn) of random maps Tn : x 7→|x|102n+γn where (γn) is an i.i.d. sequence of Cauchy random variables. Since E|γ1| = +∞, Theo-rem 6.24 does not apply. However, Bn = n(n+1)+Pn j=1 γj, and [CT, Thm.5.22] shows that Bn n2 a.s.
→1 as n →∞. The latter is enough to deduce from [KN, Thm.4.2] that (10Bny) is u.d. mod 1 for a.e. y ∈R. The same argument as in the above proof shows that P(OT (x0) is Benford ) = 1 for a.e. x0 ∈R. Thus the conclusions of Theorem 6.24 may hold under weaker assumptions.
(iii) Statements in the spirit of Theorem 6.24 are true also for more general random maps, not just monomials [Ber3].
List of symbols N, N0, Z, Q, set of positive integer, non-negative integer, integer, rational, R+, R, C positive real, real, complex numbers (Fn) sequence of Fibonacci numbers, (Fn) = (1, 1, 2, 3, 5, 8, 13, . . .) (pn) sequence of prime numbers, (pn) = (2, 3, 5, 7, 11, 13, 17, . . .) ⌊x⌋ largest integer not larger than x ∈R ⟨x⟩ fractional part of x ∈R, i.e. ⟨x⟩= x −⌊x⌋ ℜz, ℑz real, imaginary part of z ∈C z, |z| conjugate, absolute value (modulus) of z ∈C Cl set of all l times continuously differentiable functions, l ∈N0 C∞ set of all smooth (i.e. infinitely differentiable) functions, i.e.
C∞= T l≥0 Cl 122 A. Berger and T.P. Hill S significand function (Definition 2.3) D1, D2, D3 etc.
first, second, third etc. significant decimal digit (Definition 2.1) D(b) m m-th significant digit base b log x base 10 logarithm of x ∈R+ ln x natural logarithm of x ∈R+ #A cardinality (number of elements) of the finite set A O order symbol; an = O(bn) as n →∞provided that |an| ≤c|bn| for some c > 0 and all n (Ω, A, P) probability space Ac complement of A in some ambient space Ωclear from the con-text, i.e. Ac = {ω ∈Ω: ω ̸∈A} A\B set of elements of A not in B, i.e. A\B = A ∩Bc A∆B symmetric difference of A and B, i.e. A∆B = A\B ∪B\A σ(f) σ-algebra generated by the function f : Ω→R f∗P probability measure on R induced by P and the measurable function f : Ω→R, via f∗P(•) := P f −1(•) δa Dirac probability measure concentrated at a ∈Ω B Borel σ-algebra on R or parts thereof λ Lebesgue measure on (R, B) or parts thereof S significand σ-algebra (Definition 2.7) 1A indicator function of the set A λa,b normalized Lebesgue measure (uniform distribution) on [a, b), B[a, b) i.i.d.
independent, identically distributed (sequence or family of ran-dom variables) a.e.
(Lebesgue) almost every a.s.
almost surely, i.e. with probability one u.d. mod 1 uniformly distributed modulo one (Definition 4.1) X, Y, . . .
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3094 | https://www.youtube.com/watch?v=sSMRlIsDy0A | Use Geogebra to Construct Reflection Over Intersecting Lines and Create a Rotation
MrCaryMath
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Posted: 15 May 2024
This video shows how to create a dynamic visualization of a rotation using Geogebra. The video details the construction of two reflections over intersecting lines to create the isometry known as a rotation. The video also explores the relationship between the angle between the intersecting lines and the angle of rotation.
Transcript:
in this video we're going to explore some Transformations using geogebra so you need to go to uh geogebra.org that should take you to this page and then you hit uh calculators and we want to go to Geometry you might want to be signed in I'm signed in over here so you can save your work okay so we're going to start with the first one that we did by hand the other day which is um reflecting a triangle over two intersecting lines so the first thing we want to do is draw our intersecting lines and we want them to be at a 25 degree angle so I'm going to click line and do two points and then I'll do I'll make a my my Center here okay so go back over here and click that to choose something else now what we want to do is we want to measure this angle in here and make sure it's at 25 degrees so I'm going to go to my measure angle pick three points there we go click that again move this so I kind see okay so that's at 30 so I'm going to take this and I'm G to move this into 25 perfect now I think we called this Center o right here we can click on that and rename that o there we go capital O let's do that okay I'm having in troubles here let's try this again okay capital o oh no what did I do you'd think this was the hardest part of what we're doing here huh all right there we go so and we were going to use c and b for some other things so let's let's change these to other stuff maybe um let's call this point L because remember we were going to talk about line L versus line M I'm not sure we can name our lines we could try it but maybe if we just name these points little L and Little M then we can use those as references for this is going to be line L and this is going to be line M oh it might let me name the line let's try that is it naming it I don't see it anywhere here yeah I don't know okay so let's put our triangle in there now so we're going to we're going to choose a polygon polygon and I'm going to pick a triangle and like we talked about yesterday I I want to pick a triangle that's not equilateral and it's labeling it ABC which is perfect because that's what I want okay so remember yesterday the first thing we did we used the mirror to reflect this over line L now we could move this wherever we want but you know we can set it right there is fine if you want to move it around before you reflect it that's fine and actually you're going to be able to move it around after you reflect it which is really cool so let's go down and find transform okay so notice here Transformations ah we see some things we've been working on translate by a vector rotate around a point reflect over a line these are three of the isometries that we've learned three out of the four the fourth one being a Glide reflection which a Glide reflection can be obtained by a translation followed by a reflection or vice versa so can use a combination of translation and reflection to accomplish a Glide we're not doing a Glide we're just doing a reflection here right so I'm going to choose my triangle and then I'm going to come here I'm going to go reflect about a line it says select object to reflect and then the line of reflection so I'm going to select the object to reflect and then the line ah look at that and it even labeled it a double prime bble prime cble prime for me wonderful okay so now I want to reflect this triangle over line M so I'm going to hit reflect again I'm gonna say reflect this triangle over line M beautiful now notice because line M went through a prime B Prime C Prime that um a prime B Prime C Prime overlapped with a double Prime and it did all the labeling for us pretty cool okay now watch this I went and chose my um you know just my pointer again I'm going to grab this triangle you ready this going be amazing ooh look at that see it's it's still if I move it over here that's the reflection over L reflection over m to the double Prime and I could oh I oh I can even change the triangle huh and it still maintains that reflection oh that wasn't what I was trying to do but that's kind of cool too so mess around with this a little bit that's kind of neat pretty cool okay so after you're messing around with that make that stagnant just pick something now remember the next thing we did was we measured the angle of the rotation itself so the transformation that we're looking at is triangle ABC over to Triangle a prime a double Prime B Prime C Prime this one right here the prime one is sort of an intermediate step of the translation so I'm going to click on this and I'm going to color this a different color maybe a lighter color like orange and maybe I'll even can I make it like uh less transparent Maybe like that and then this one maybe I'll make that even um a little bit darker how do I do that yeah here I'm going to choose that to be a little bit darker like that you you mess around with what you like but what I'm trying to do here is really highlight the fact that the transformation is from this blue triangle to this blue triangle that's the transformation I'm looking at that's the rotation I'm measuring not anything to do with this orange one okay so now we wanted to measure angle a o a prime so I am going to draw um I can draw two rays I may be able to draw an angle and then I'll measure it uh we know an angle is two rays or two line segments we could do either do segment or Ray oh let's just do segment to keep it from going off the page so I'm going to go segment a to O and then segment o to a double Prime I'm going to pick my pointer here I may even change the color of this again to highlight it maybe we'll change it to Red change this to Red okay so the angle I'm measuring is the angle between these two red lines here so now I'm going to go to measure angle a o able Prime choose my pointer I got to go down here and grab this angle there it is right there there's the 50 maybe I'll move this 25 maybe I'll even change the color of this 50 to Red that's nice so that really highlights the fact that this 50 corresponds to this red angle right here where the 25 remember is the angle between the two lines so again I can grab this triangle and move it around oh man that's cool now the 50 is not see that red angle is still 50° this little indicator is not moving with me but isn't that neat that's that's pretty neat pretty Dynamic now you could do the same thing with um b o b double prime or C C double Prime maybe do one more of those it can certainly get a little messy if we do too much but let's do c o and then o c double Prime let's make those a different color so it's not messy how about green we'll make this one green that's lovely and then we're going to measure the angle from C to O to couble Prime choose my pointer grab that it's still 50 maybe I'll make that green to again highlight the fact that's the measure of the green angle pretty cool all right no matter where I start here those angles are still 50 okay now now we had this hypothesis that the angle of rotation was twice the angle between the lines certainly looks to be true for 25 degrees we could test that hypothesis Now by actually changing the angle between the lines so if I pick this line this point L right here I'm going to pick this point L and I'm going to drag it and change the angle between the lines to something bigger than 25 and we can watch how that affects this angle of rotation you ready here we go oh that's exciting let's just move it to 60° and take a look at what we've got okay so now notice the construction still stands how we built it this initial triangle ABC is reflected over L over here and then this orange triangle is reflected over M down here and we have the angle of rotation is twice as much as the angle between the lines 60.7 in this case um one more thing I want to highlight here about the reflections themselves we've talked about how if I create a segment with the reflection that that segment is bi seed by the reflecting line so let's put those segments in there so I'm going to do segment a nope cancel not I want I don't want segment with a given length just a segment a to Able prime or excuse me a prime and then a prime to a double Prime okay and I'm going to make those a different color because it's getting messy let's see what colors do we have left um orange let's do orange and then this line orange oh we can add a color oh look at that we have a whole palette over there okay I don't want to get distracted This Is Where It Gets In The Weeds right let's just choose orange so okay I want you to really I want you to look here let's just look up here because this one's easy to see notice this is the perpendicular bis sector and then this is the perpendicular bis sector here m is the perpendicular bis sector of a prime a double Prime let's actually measure that okay so I'm going to put a point I'm going to put a point at this this intersection right here that's the intersection of a prime a prime and line M and then I am going to move these angles out of the way there we go I'm going to show that m is the perpendicular bis sector of a prime a dou prime so to do that I first need to show that these are perpendicular so I'm going to measure that angle measure angle m d and then a double Prime 90° awesome okay how do I show it's a bis sector to show it's a bis sector I need to measure the length of a prime to D and then D to Able Prime measure where's measure length angle angle distance or length there we go a prime to D and D to a double Prime look they're both 7.3 it's bisecting and we can go back we can pick this move thing we can move this whole business around wow move it around stop at it and stare at it for a minute you can even take one of the points of your triangle and like move it out of the way to see things better if you want and you could do this angle b or uh perpendicular bis sector bit between a a and a prime if you want measuring the angle and the distances like we did from Able Prime to a prime as well okay so hopefully this convinces you of a couple things first of all that the angle of rotation is indeed twice the angle between the lines and that when you're dealing with a reflection that if you create a segment between the pre-image and the image that the reflecting line is perpendicular to and will bict the segment connecting the pre-image and the image all right so before we do anything else let's um save this thing let's see how do we save that's okay so we hit this button up here hit save let's call this um rotation over L then M save okay now if you wanted to um copy this and paste it into your submission for the lesson today that will be your credit for this particular activity okay all right so we could do the same thing reflecting over the lines in the reverse order I'm not going to have you build that in geogebra right now but we are going to do something over parallel lines so I'm going to help you set up the parallel lines one and then have you um kind of finish that on your own so I always get a little bit scared when I start over that I'm going to lose something so make sure that you save this either save it online if you logged in you can actually save it to your computer as a file I usually save my stuff okay it makes me nervous and then I'm going to go clear all which makes me so nervous but if I go back here and click this and then go open hopefully it's going to be there it is right there okay it saved it for me all right so the next one we're going to do is the uh parallel lines how do I make this go away I want that to go away oh there's a settings button just outside you got you can't see it but I think if you click that settings button uh it might go away oh there's a little x button out here there we go okay so now we need need parallel lines so we're going to construct parallel lines if I click on this it says select parallel line and a point so I have to start with a line and a point and then I can construct the parallel lines so I'm going to go up here I'm going to make a line very good and then I'm going to pick a point not on the line anywhere you want and then I can do my parallel line and I click the line and the point and it creates those parallel lines now from here you could drag C closer notice it keeps the lines parallel because that's how you built it you could move any of these points now again we were going to label our triangle ABC right so maybe we rename these points again we could um call this little l oh I didn't want to do that edit undo we could call this little L to signify that that's line L and we could call this little M to signify line M and then I don't care what we call this let's call it o all right so we're going to build our triangle again just like we did be oh no we're doing a rectang or a quadrilateral this time right so we're gon to oh I don't want a regular polygon just a polygon one two three four close it you can go back and reshape it if you want into some other kind of shape and then you're going to do the reflecting over the lines again and then then you're going to measure uh how far that it moved and how far it translated so I want you to do the reflections okay I I don't I know in the Transformations down here you can do rotations and translations but I don't want you to use these I want you to do the two Reflections just like we did by hand and just like we did in the last activity okay we can go back and mess around with these later but we're doing it by doing a composite or a series of two Reflections and then you're going to measure how far that it it translated and you're going to measure the distance between the parallel lines you can do that here uh we can select these two lines does it measure the dist oh no to measure the distance between the lines uh we actually have to make a perpendicular to measure the distance between the lines so let me show you how to do that we can do um perpendicular line so we can do a line perpendicular to this and then we can call this point right here e and then actually we could uh measure the distance measure distance between here and here and there's the distance between our lines I kind of don't like having this perpendicular in here it's like kind of distracting to make it a little less conspicuous maybe we could could like color it white we could color that line white and then it sort of goes away but we still have this distance between these lines here right but now it's not in the way that's kind of interesting okay so I want you to complete this uh activity in geogebra activity from the worksheet that you did yesterday and um you will save that and submit that as well |
3095 | https://www.khanacademy.org/math/statistics-probability/summarizing-quantitative-data/variance-standard-deviation-sample/a/population-and-sample-standard-deviation-review | Use of cookies
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3096 | https://www.holycross.edu/document/sobczak2005aslobulletinv14p53 | LINDEMAN’S TROPHIC-DYNAMIC ASPECT OF ECOLOGY: “WILL YOU STILL NEED ME WHEN I’M 64?”
William V . Sobczak , Biology Department, Holy Cross College, Worcester, MA 01610 USA; wsobczak@holycross.edu
In the fall of 1941, only six years after the word “ecosystem” was coined, Raymond Lindeman submitted “The trophic-dynamic aspect of ecology” (the final chapter from his Ph.D. thesis) to the journal Ecology (Cook 1977). Lindeman’s thesis work at the University of Minnesota was ambitious in that it explored the mechanisms underlying bog succession. Specifically, Lindeman studied the succession of Cedar Creek Bog in Northern Minnesota, and provided a novel conceptual template for studying not only succession, but also energy flow in aquatic ecosystems. Lindeman’s approach to studying aquatic ecosystems was initially supported by G. Evelyn Hutchinson, and later embraced by Eugene Odum and a generation of aquatic ecologists. Lindeman’s Ecology
manuscript is now considered foundational work in the histories of ecology and limnology (Real and Brown 1991; Wetzel 2001), and seminal work in ecosystem ecology (Golley 1993). This work was one of the first papers to implement Arthur Tansley’s newly proposed ecosystem concept in a quantitative manner and was groundbreaking in several ways (Cook 1977; Golley 1993). Lindeman’s paper argued that trophic structure affects community patterns and succession; hence, a quantitative food web is essential to understanding temporal change in aquatic ecosystems. Remarkably, the paper recognized the central role of bacteria and detritus in ecosystem energy flow placing “ooze” as the hub of all trophic interactions (Lindeman 1942; Golley 1993; Figure 1). The work’s most enduring contribution is that it provided a common currency (organic matter or energy flow) for studying interactions among trophic levels and for comparing disparate aquatic ecosystems, thus establishing a theoretical orientation in ecology (Cook 1977). Lindeman was fortunate that his Ecology manuscript was reviewed by arguably two of the leading limnologists in the world: Chancey Juday at the University of Wisconsin and Paul Welch at the University of Michigan (Cook 1977). Unfortunately, neither Juday nor Welch recognized the merits of Lindeman’s work and both rejected the manuscript for publication in Ecology (Cook 1977). Juday wrote: “The chances are that the author’s beliefs and imaginary lakes would be very different entities if he had a background of observations on fifty or a hundred of the 10,000 lakes claimed by the state of Minnesota, instead of only one, and that a special type.” Welsh wrote: “Limnology is not yet ready for generalizations of this kind. The basic background data for such a paper is far too
Figure 1 . Generalized food-cycle relationships for aquatic ecosystems showing energy flow among four trophic levels and highlighting the central role of bacteria (Lindeman 1942). Reprinted with permission from Ecological Society of America.
The Limnology and Oceanography Bulletin
The American Society of Limnology and Oceanography is a membership-driven scientific society (501(c)(3)) that promotes the interests of limnology (the study of inland waters), oceanography and related aquatic science disciplines by fostering the exchange of information and furthering investigations through research and education. ASLO also strives to link knowledge in the aquatic sciences to the identification and solution of problems generated by human interactions with the environment.
Editor: Greg Cutter, Old Dominion University, Norfolk, VA 23529-0276 USA, Tel: 757-683-4929, gcutter@odu.edu
Associate Editor: Fred C. Dobbs, Old Dominion University, Norfolk, VA 23529-0276 USA, Tel: 757-683-5329, fdobbs@odu.edu
ASLO Business Manager:
Helen Schneider Lemay, ASLO Business Office, Waco, TX 76710 USA, Tel: 254-399-9635 or 800-929-2756, Fax: 254-776-3767, business@aslo.org
Advertising: Helen Schneider Lemay, ASLO Business Manager, Tel: 254-399-9635 or 800-929-2756; business@aslo.org © 2005 American Society of Limnology and Oceanography. Material in this issue may be photocopied by individual scientists for research or classroom use. Permission is also granted to use short quotes, figures, and tables for publication in scientific books and journals. For permission for any other uses, contact the ASLO Business Office. The L&O Bulletin is published quarterly by the American Society of Limnology and Oceanography, 5400 Bosque Blvd., Suite 680, Waco, TX 76710 USA. Postage paid at Waco, Texas. POSTMASTER: Send address changes to ASLO Business Office, 5400 Bosque Blvd., Suite 680, Waco, TX 76710 USA. Subscription price to regular members is included in annual dues. Information on institutional subscriptions is available upon request from the ASLO Business Office. Views expressed in this publication do not necessarily reflect official positions of the American Society of Limnology and Oceanography unless expressly stated. The L&O Bulletin publishes brief, peer-reviewed articles of broad interest to the ASLO membership, Letters to the
Bulletin (typically responses to articles), as well as ASLO News on a quarterly basis. Information on the preparation and submission of articles and letters can be found on the ASLO web site (www.aslo.org). It is recommended that you contact the editor before preparing an article or letter.
fragmentary. If Dr. Lindeman could put this paper aside for ten years, then bring it out and see how it looks in the light of what we hope will be the added accumulation of limnological information, he might possibly congratulate himself that he deferred its publication.” Lindeman received a rejection letter from the editor of Ecology , Thomas Park, in November 1941. G. Evelyn Hutchinson, Lindeman’s post-doctoral advisor at Yale University, recognized that Lindeman’s approach to studying aquatic ecosystems as described in the “Trophic-dynamic aspect of ecology” was powerful and would allow comparative analyses among aquatic ecosystems (Golley 1993). Hutchinson was discouraged by the reviews and promptly wrote a passionate letter to Park in defense of Lindeman’s work (Cook 1977). Hutchinson wrote: “I feel that a number of far-reaching hypotheses that can be tested by actual data and which, if confirmed, would become significant generalizations, are far more valuable than an unending number of marks on paper indicating that a quantity of rather unrelated observations has been made.” Swayed by Hutchinson’s letter, Park ultimately accepted a revised version of Lindeman’s manuscript recognizing that “time is a great sifter in these matters and it alone will judge the question.” Ray Lindeman died in June 1942 at age twenty-seven prior to the paper’s publication. In an obituary attached to the end of the publication, Hutchinson wrote: “Knowing that one man’s life at best is too short for intensive studies of more than a few localities, and before the manuscript was completed, that he might never return again to the field, he wanted others to think in the same terms as he found so stimulating, and for them to collect material that would confirm, extend, or correct his theoretical conclusions.” In 2002, “Trophic-dynamic aspect of ecology” was cited by members of the American Institute of Biological Sciences as the third most influential journal article with respect to their career training (Barrett and Mabry 2002). The fields of ecology and limnology were apparently not ready for the generalizations proposed by Lindeman after all, as the paper and its approach were not widely embraced (outside of Yale) during the decade following its publication. Finally, in 1953, with the publication of Eugene Odum’s Fundamentals of Ecology , Lindeman’s model for studying energy flow in aquatic ecosystems was promoted and widely distributed, thus greatly influencing an emerging body of research on energy and nutrient cycling in aquatic and terrestrial ecosystems (Golley 1993). Odum’s writings and those of his numerous colleagues and students rapidly made the study of energy flow the principal approach for studying diverse biological processes and comparing diverse ecosystems (Golley 1993). John Teal’s classic study of “Energy flow of the salt marsh ecosystem of Georgia” (1962) embodies the effectiveness of Lindeman’s approach 20 years after the original publication. Using Lindeman’s work on Cedar Creek Bog as a template, Teal (1962) quantified a salt marsh’s allochthonous and autochthonous organic matter production and assessed the fate of these pools of potential energy for marsh heterotrophs. Teal found that much of the organic matter produced within and delivered to salt marshes could not be accounted for in secondary production within the salt marsh ecosystem and hypothesized that as much as 45% of the organic matter entering the ecosystem might be transported to adjacent ecosystems. Lindeman’s approach to studying energy flow in a closed bog ecosystem was now being used to study open lotic ecosystems that required quantification of organic matter inputs and outputs. Odum’s “outwelling hypothesis” that terrestrial organic matter inputs were subsidizing secondary production in estuaries and coastal waters was a logical extension of the findings of Teal and others. It promoted much debate in the literature, thus extending Lindeman’s bog-ecosystem model to include all aquatic ecosystems (e.g., springs, swamps, salt marshes, streams, and sloughs). Four decades after the publication of “Trophic-dynamic aspect of ecology” and several years after the inception of the “microbial loop” (Pomeroy 1974), Lindeman’s model of energy flow had become central to the study of inland waters (see preface of Wetzel 1983) and was at the base of one of stream ecology’s most influential conceptual models: the River Continuum Concept (RCC; Vannote et al. 1980). Although not cited nearly as much as the RCC (now approaching 1600 citations!), Lindeman’s seminal work on a lentic-bog ecosystem provided the template for examining the energy flow of stream ecosystems at multiple scales. The RCC’s classic conceptual figure builds on Lindeman’s food web depiction (Fig. 1) by placing microbes at the center of feeding relationship diagrams. Vannote et al.’s RCC (1980) and Wetzel’s leading limnology textbook (1983) inspired a tremendous body of research on energy flow in inland waters throughout the 1980s and 1990s (see Wetzel 2001). By the end of the 20 th Century, Lindeman’s blueprint for studying energy flow was being used at a wide range of spatial scales across freshwater and estuarine ecosystems. Experimental studies of energy flow in lake ecosystems close to Cedar Creek Bog have verified the power and longevity of Lindeman’s approach and further quantified the important role of terrestrial energy subsidies to aquatic biota (Pace et al. 2004 and references within). Experimental manipulations of allochthonous energy flow in low-order, forested streams in the southeast U.S. are grounded in Lindeman’s approach (Hall et al. 2000), and have greatly advanced it (Eggert and Wallace 2003 and references within). A growing body of food web studies using stable isotope analyses has provided new, and potentially powerful tools for examining energy flow from lower to higher trophic levels (e.g., Bastviken et al. 2003). We now know that the world’s large rivers and estuaries are routinely net heterotrophic confirming the importance of terrestrial-derived energy inputs to downstream-ecosystem metabolism (Frankignoulle et al. 1998; Cole and Caraco 2001). However, net ecosystem production in higher-trophic levels remains uncertain as spatial scale increases in aquatic ecosystems (Thorp and Delong 2002). This net heterotrophy paradox has roots that extend decades into the past (e.g., Teal 1962; Vannote et al. 1980), but has received renewed attention after the documentation of community changes and declines in higher trophic levels in many riverine, estuarine, and marine ecosystems (Pauly et al. 2002; Thorp and Delong 2002; Jacobs et al. 2003). Lindeman’s approach can be incorporated into efforts to understand the mechanisms underlying these food web changes and, in some cases, efforts to restore food webs. San Francisco Bay’s freshwater Delta provides one such case study for examining energy flow within a large-scale aquatic ecosystem that has experienced significant food web alterations. Recent research here helps address basic questions regarding energy flow in riverine and estuarine ecosystems, and, more specifically, questions regarding the multiple mechanisms underlying declines in native fish and shifts in forage food (Jassby et al. 2002; Lopez et al. 2005; Sobczak et al. 2005). The San Francisco Bay’s Delta is a complex mosaic of tidal, freshwater and estuarine habitats that consists of 1127 km of leveed waterways and 57 human-made islands at the confluence of the Sacramento and San Joaquin Rivers, which drain a 1.6 x 10 5 km 2 catchment (Jacobs et al. 2003). California’s Delta is at the heart of the state’s water conveyance projects that provide drinking water for twenty million people and irrigation water for much of the Central Valley’s agriculture. The Delta’s engineered and highly-disturbed waterways are also a complex aquatic ecosystem supporting 130 species of fish, including many that are threatened (e.g., winter-run Chinook salmon) or endangered (e.g., Sacramento splittail; Jacobs et al. 2003; Lopez et al. 2005). Historical declines in higher trophic levels including native fishes and zooplankton may be the result of multiple stressors. Did alterations in food resources at the base of the Delta’s food web contribute to declines in higher trophic levels? The Delta’s large spatial scale and complex hydrology provide inherent problems to addressing this seemingly basic ecological question. Further, the Delta’s migratory and spatially variable fish populations complicate an inventory of fish populations. Large-scale ecosystem restoration plans (comparable to the Florida Everglades’ restoration in scope) aim to stabilize or increase the Delta’s declining native fish populations (Jacobs et al. 2003). Lindeman’s blueprint for examining aquatic food web structure provided a tractable solution to this problem: examine the energetic base of the Delta’s food web and determine if it is sufficient to support higher trophic levels. The strategy to study the base of the Delta’s food web required a team of researchers willing to share multiple backgrounds and scientific approaches to studying energy flow (see Likens 1998 on ecosystem ecology- team building). The outcome of this collaboration was a group of complementary papers that characterized the base of the Delta’s food web and energy flow to higher trophic levels. Allochthonous detritus dominates organic matter supply to the Delta and supports the majority of ecosystem respiration; however, allochthonous detritus is not strongly coupled to the Delta’s metazoan food web (Sobczak et al. 2002). Autochthonous phytoplankton production, a small component of the Delta’s organic matter budget, is the dominant energy source for metazoans and zooplankton (Müller-Solger et al. 2002; Sobczak et al. 2002). Zooplankton growth and reproduction are routinely limited by low phytoplankton biomass (Müller-Solger et al. 2002) even in many of the Delta’s shallow water habitats (Sobczak et al. 2005). Phytoplankton production in the Delta has declined over 40% from 1975 to 1995, suggesting the long-term loss of an important food resource (Jassby et al. 2002). Phytoplankton biomass is highly variable in time and space, and appears to be related to turbidity and low growth rates, water transport and residence time, and benthic grazing by the invasive clam Corbicula fluminea (Jassby et al. 2002; Lopez et al. 2005). Our collective findings have provided multiple lines of evidence (and thus strong inference) regarding a linkage between declines in phytoplankton production and declines in higher trophic levels. These findings contribute to an evolving conceptual model of riverine and estuarine food webs (e.g., Thorp and Delong 2002) and aid ecosystem restoration efforts. Restoration actions geared at augmenting higher trophic levels need to consider disparities in phytoplankton biomass and bioavailable organic matter among Delta habitats (Lopez et al. 2005; Sobczak et al. 2005). In addition, restoration efforts aimed at promoting the energetic base of the Delta’s food web must critically examine the role of multiple stressors to the Delta’s native fish populations, such as predation and freshwater flow modification (Sommer et al. 2004). Why is Lindeman’s model paper for studying aquatic ecosystems still important 64 years after its publication? While limnological methods and analytical tools have changed dramatically in the 64 years since Lindeman’s work was submitted to Ecology , Lindeman’s approach to studying aquatic ecosystems is clearly central to many of today’s most important limnological research efforts (Bastviken et al. 2003; Eggert and Wallace 2003; Pace et al. 2004; Sommer et al. 2004). Further, limnologists and oceanographers are now being asked to contribute to global change models that require new knowledge regarding energy flow: what is the fate of terrestrial organic matter inputs into the world’s aquatic ecosystems (Findlay and Sinsabaugh 2003); what is the age and source of the terrestrial-world’s detrital subsidies to coastal ecosystems (Raymond and Bauer 2001; Mayorga et al. 2005); what is the role of humans in altering energy flow to the world’s coastal ecosystems (Cloern 2001)? These current challenges facing limnologists and oceanographers occur over large portions of the globe and a shifting mixture of physical, chemical, and biological factors likely control their underlying processes. Lindeman’s insight on energy flow helps researchers aggregate and organize some of the complexity of aquatic ecosystems, thus greatly increasing the likelihood that we will successfully answer these pressing questions. Further, limnologists and oceanographers will be increasingly asked to contribute to ecosystem restoration projects that aim to sustain or augment biodiversity and important native populations (Wetzel 2001; Pauly et al. 2002). Lindeman’s approach for studying aquatic ecosystems may be an essential organizational component of ecosystem restoration programs designed to manage higher trophic levels. Responding to Lindeman’s critical reviewers in fall 1941, Hutchinson wrote: “In field ecology, it is necessary to have data collected over many months or years, and for comparative purposes studies are needed on localities very widely separated in space. To obtain the kind of data required takes two or three years’ work on a single locality; to suggest that any one individual should wait until he has completed investigations on fifty or one hundred lakes is ironical rather than practical. It is therefore most important that all ecologists should have the opportunity to acquaint themselves with the theoretical possibilities that may guide them in their collection of data.”
ACKNOWLEDGMENTS
Jim Cloern, Jon Cole, Greg Cutter, Stuart Findlay, Alan Jassby, Gene Likens, and two anonymous reviewers made helpful comments on an earlier version of the manuscript.
REFERENCES
Barrett, G. W., and K. E. Mabry. 2002. Twentieth-century classic books and benchmark publications in Biology. BioScience 52: 282-285. Bastviken, D., J. Ejlertsson, I. Sundh, and L. Tranvik. 2003. Methane as a source of carbon and energy for lake pelagic food webs. Ecology 84: 969-981. Cloern, J. E. 2001. Our evolving conceptual model of the coastal eutrophication problem. Mar. Ecol. Prog. Ser. 210: 223-253. Cole, J. J., and N. A. Caraco. 2001. Carbon in catchments: connecting terrestrial carbon losses with aquatic metabolism. J. Mar. Freshw. Res. 52: 101-110. Cook, R. E. 1977. Raymond Lindeman and the trophic-dynamic concept in Ecology. Science 198: 22-26. Eggert, S. L. and J. B. Wallace. 2003. Litter breakdown and invertebrate detritivores in a resource-depleted Appalachian stream. Arch. Hydrobiol. 156: 315-338. Findlay, S.E.G., and R. L. Sinsabaugh. 2003. Aquatic Ecosystems: Interactivity of Dissolved Organic Matter. Academic Press. Frankignoulle, M., G. Abril, A. Borges, I. Bourge, C. Canon, B. Delille, E. Libert, and J. Theate. 1998. Carbon dioxide emission from European estuaries. Nature 282: 434-436. Golley, F. B. 1993. A History of the Ecosystem Concept in Ecology. Yale University Press. Hall, R. O., J. B. Wallace, and S. L. Eggert. 2000. Organic matter flow in stream food webs with reduced detrital resource base. Ecology 81: 3445-3463. Lopez, C. B., J. E. Cloern, T. S. Schraga, A. J. Little, L. V. Lucas, J. K. Thompson, and J. R. Burau. 2005. Ecological values of shallow-water habitats: implications for restoration of disturbed ecosystems. Ecosystems (in press). Jacobs, K. L., S. N. Luoma, and K. A. Taylor. 2003. CALFED: An experiment in science and decision making. Environment 45: 30-41. Jassby, A. D., J. E. Cloern, and B. E. Cole. 2002. Annual primary production: Patterns and mechanisms of change in a nutrient-rich tidal ecosystem. Limnol. Oceanogr. 47: 698-712. Lindeman, R. E. 1942. Trophic-dynamic aspect of ecology. Ecology 23: 399-418. Likens, G. E. 1998. Limitations to intellectual progress in ecosystem science. In: M. L. Pace, and P. M. Groffman (Eds.). Successes, Limitations, and Frontiers in Ecosystem Science. Springer. Mayorga, E., A. K. Aufdenkampe, C. A. Masiello, A. V. Krusche, J. I. Hedges, P. D. Quay, J. E. Richey, and T. A. Brown. 2005. Young organic matter as a source of carbon dioxide outgassing from Amazonian Rivers. Nature 436: 538-541. Müller-Solger, A. B., A. D. Jassby, and D. C. Müller-Navarra. 2002. Nutritional quality of food resources for zooplankton (Daphnia) in a tidal freshwater system (Sacramento-San Joaquin River Delta, USA). Limnol Oceanogr 47: 1468-1476. Pace, M. L., J. J. Cole, S. R. Carpenter, J. F. Kitchell, J. R. Hodgson, M. C. Van de Bogert, D. L. Bade, E. S. Kritzberg, and D. Bastviken. 2004. Whole-lake carbon-13 additions reveal terrestrial support of aquatic food webs. Nature 427: 240-243. Pomeroy, L. R. 1974. The ocean’s food web: A changing paradigm. BioScience 24: 499-504. Pauly, D., V. Christensen, S. Guenette, T. J. Pitcher, U. R. Sumaila, C. J. Walters, R. Watson, and D. Zeller. 2002. Towards sustainability in world fisheries. Nature 418: 689-695. Raymond, R. A., and J. E. Bauer. 2001. Riverine export of aged terrestrial organic matter to the North Atlantic Ocean. Nature 409: 497-500. Real, L. A., and J. H. Brown. 1991. Foundations of Ecology: Classic Papers with Commentaries. The University of Chicago Press. Sobczak, W. V., J. E. Cloern, A. D. Jassby, and A. B. Müller-Solger. 2002. Bioavailability of organic matter in a highly disturbed estuary: The role of detrital and algal resources. Proceedings of the National Academy of Sciences 99: 8101-8105. Sobczak, W.V., J. E. Cloern, A. D. Jassby, B. E. Cole, T. S. Schraga, and A. Arnsberg. 2005. Detritus fuels metabolism but not metazoan food webs in San Francisco Estuary’s freshwater delta. Estuaries 28: 124-137. Sommer, T. R., W. C. Harwell, A. B. Mueller-Solger, 2004. Effects of flow variation on channel and floodplain biota and habitats of the Sacramento River, California, USA. Aquatic Conservation of Marine and Freshwater Ecosystems 14: 247-261. Teal, J. M. 1962. Energy flow in the salt marsh ecosystem of Georgia. Ecology 23: 614-624. Thorp, J. H., and M. D. Delong. 2002. Dominance of autochthonous autotrophic carbon in food webs of heterotrophic rivers? Oikos 96: 543-550. Vannote, R. L., G. W. Minshall, K. W. Cummins, J. R. Sedell, and C. E. Cushing. 1980. The river continuum concept. Can. J. Fish. Aquat. Sci. 37: 130-137. Wetzel, R. G. 1983. Limnology. Saunders Press. Wetzel, R. G. 2001. Limnology: Lake and River Ecosystems. Academic Press. |
3097 | https://referencecenter.dynamicgeometry.com/gsp-reference-center/objects_loci_anatomy.htm | Anatomy of a Locus
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| Navigation: Objects > Loci > Anatomy of a Locus | Previous pageReturn to chapter overviewNext page |
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Navigation:
Objects > Loci >
Anatomy of a Locus
Recall the definition of a locus: A locus is a set of positions of a driven object as a driver varies over its domain.
Possible driven objects include points, straight objects, circles, arcs, interiors, pictures, function plots, and loci themselves.
Possible drivers are points on paths and parameters.
The domain for a point driver is normally its path, which might be a straight object, a circle, an arc, a polygon or other interior, a function plot, or another point locus. If the point driver is not constructed on a path, you must explicitly select a path to serve as the domain.
The domain for a parameter driver is a numeric domain that you specify when you construct the locus.
Examples of Loci
Locus of a point, driven by another point: The illustration below left shows segment CD with end point C attached to circle AB; the illustration below right shows the locus of point E on the segment.
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| Driver: Point C Domain: Circle AB Driven Object: Point E on Segment | Locus of Point E |
Driver: Point C
Domain: Circle AB
Driven Object: Point E on Segment
Locus of Point E
Locus of a point, driven by another point: In the illustration below left, point P is the intersection of line j (the perpendicular bisector of segment CF) and line k (the perpendicular to d through C). This construction guarantees that point P is equally distant from point F and segment d. The illustration below right shows a parabola: the locus of point P as point C moves along segment d.
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| Driver: Point C Domain: Segment d Driven Object: Intersection Point P | Locus of Intersection P |
Driver: Point C
Domain: Segment d
Driven Object: Intersection Point P
Locus of Intersection P
Locus of a picture, driven by a parameter: In the illustration below left, parameter Elevation has been used to rotate point P to P', and a picture has been attached to P'. On the right is the locus of the picture as the parameter varies from 0° to 180°.
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| Driver: Parameter Elevation Domain: 0° ≤ Elevation ≤ 180° Driven Object: Picture Attached to P' | Locus of Picture |
Driver: Parameter Elevation
Domain: 0° ≤ Elevation ≤ 180°
Driven Object: Picture Attached to P'
Locus of Picture
Locus of a segment, driven by a point: In the illustration below left, segment k is constructed to have a slope determined by a calculation involving its x and y values. As driver point A moves along segment CD, the calculation's value changes, and so does the slope of segment k. The locus below right shows the locus of the slope segment for various positions of point A on its domain.
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| Driver: Point A Domain: Segment CD Driven Object: Slope Segment k | Locus of Slope Segment |
Driver: Point A
Domain: Segment CD
Driven Object: Slope Segment k
Locus of Slope Segment
Locus of a circle, driven by a point: In the illustration below left, circle AB was constructed first, and then circle CB was constructed with point C defined on circle AB. Point C is the driver, circle CB is the driven object, and circle AB is the domain. (As before, there’s no need to select the domain, because point C is constructed on it.) The illustration below right shows the resulting locus of circle CB.
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| Driver: Point C Domain: Circle AB Driven Object: Circle CB | Locus of Circle CB |
Driver: Point C
Domain: Circle AB
Driven Object: Circle CB
Locus of Circle CB
Locus of a segment, driven by an independent point using an arc domain: In the illustrations below, the driver is independent of the domain, so the selections include all three objects: the driver (the left endpoint of the segment), the driven object (the segment), and the domain (the arc). The illustration below right shows the resulting locus of the segment as its endpoint moves along the arc.
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| Driver: Independent Point Domain: Arc Driven object: Segment | Locus of the Segment |
Driver: Independent Point
Domain: Arc
Driven object: Segment
Locus of the Segment
Family of functions (locus of a function plot), driven by a parameter: These illustrations show a family of functions. In this example the driver is parameter a. With the driver and the function plot selected, the command becomes Construct | Family of Functions. The illustration below right shows the resulting family of functions as parameter a varies across its domain. (You can do this same construction using a slider rather than a parameter. When using a slider, the driver would be the adjustable point of the slider.)
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| Driver: Parameter a Domain: –2 ≤ a ≤ 2 Driven Object: Function Plot | Family of Functions |
Driver: Parameter a
Domain: –2 ≤ a ≤ 2
Driven Object: Function Plot
Family of Functions
Family of curves (locus of a point locus), driven by another point: These illustrations show a family of point loci. In this example the driver is point F, the focus of the parabola locus. With the driver and the locus selected, the command becomes Construct | Family of Curves. The illustration below right shows the resulting family of parabolas as focus F moves along its segment.
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| Driver: Point F Domain: Segment Driven Object: Parabola (Point Locus) | Family of Curves |
Driver: Point F
Domain: Segment
Driven Object: Parabola (Point Locus)
Family of Curves
Family of loci (locus of a non-point locus), driven by a point: These illustrations show a family of loci. Driver point B can move along the vertical segment, determining the vertical position of the red slope segment and of the locus of the slope segment. With the driver and the segment locus selected, the command becomes Construct | Family of Loci. The illustration below right shows the resulting family of loci: the slope field of the differential equation used to determine the slope of the red segment.
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| Driver: Point B Domain: Segment on x-axis Driven Object: Slope Segment Locus | Family of Loci |
Driver: Point B
Domain: Segment on x-axis
Driven Object: Slope Segment Locus
Family of Loci
Two metaphors may help you better understand how the various parts of a locus relate to each other. One way of thinking about a locus with a point driver is as an abstract mathematical function, a function in which the elements of the ordered pairs are not necessarily numbers. For a locus driven by a parameter, the first element of the ordered pair is a specific value of the parameter (a number) and the second element is the corresponding position of the driven geometric object. A locus driven by a point is similar, but the first element is the position of the driver point. Sketchpad moves or varies the independent variable (the driver) along its domain (its path), while keeping track of the position of the dependent variable (the driven object). Each sample of the locus represents one value of the function, and the entire locus is an approximation of the range of the function. (It’s an approximation because Sketchpad uses only a finite number of ordered pairs — or samples — in constructing the locus.) The abstract mathematical function in this analogy is the construction by which the independent variable (the driver) determines the position of the dependent variable (the driven object).
A second way of thinking about a locus is as a durable form of a traced animation. An animating point or parameter (the driver) moves or varies along its domain and defines the position of some object (the driven object). If you trace that object, you eventually see all of its positions (its locus) for the animated positions or values of the driver. (Where animation and tracing require time and motion to trace out the locus of an object, a construction of its locus gives you the entire result instantaneously, allowing you to use time and motion to then further explore or manipulate that result.)
Choose Edit | Properties | Plot to set the number of samples and to determine whether the locus is displayed in discrete or continuous form.
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| The difficulty of naming these dynamic concepts has a long history: When Johan De Witt and Sir Isaac Newton studied locus constructions of the conics in the 17th century, they used the term directrix to refer to what we call the driver. Today, when discussing the same type of locus, mathematicians use directrix to refer to the domain instead! | |
The difficulty of naming these dynamic concepts has a long history: When Johan De Witt and Sir Isaac Newton studied locus constructions of the conics in the 17th century, they used the term directrix to refer to what we call the driver. Today, when discussing the same type of locus, mathematicians use directrix to refer to the domain instead! |
3098 | https://www.chm.uri.edu/weuler/chm401/studyaids/redoxpracticeproblems_answers.html | Oxidation/Reduction Practice Problems Answers
Chemistry 401
Intermediate Inorganic Chemistry
University of Rhode Island
Practice Problems
Oxidation & Reduction
Some anaerobic bacteria utilize oxidizing agents other than O 2 as an energy source; for example, SO 4 2–, NO 3–, and Fe 3+. One half-reaction is FeO(OH)(s) + HCO 3–(aq) + 2 H+(aq) + e– → FeCO 3(s) + 2 H 2 O(l), for which E° = +1.67 V. What mass of iron gives the same standard reaction Gibbs energy as 1.00 g of oxygen?
The standard half-reaction for the reduction of oxygen under acid conditions is:
O 2(g) + 4 H+(aq) + 4 e– → 2 H 2 O(l)
for which E° = +1.23 V. 1.00 g of oxygen = 0.03125 moles, which would give 0.125 moles of electrons. Thus, the Gibbs energy is:
ΔG° = –nFE° = –(0.125)(96485)(1.23) = 14.8 kJ
To obtain the same free energy from the iron reaction would require n = ΔG°/FE° = (14800)/(96485)(1.67) = 0.0919 moles of electrons. Since the mole ratio between iron and electrons is 1, this is also the number of moles of iron required, or 5.13 g.
Using the following aqueous acid solution reduction potentials E° (Pd 2+, Pd) = +0.915 V and E° ([PdCl 4]2–, Pd) = +0.600 V, calculate the equilibrium constant for Pd 2+(aq) + 4 Cl–(aq) →← [PdCl 4]2–(aq) in 1 M HCl(aq).
Break the desired reaction into two components that use the available potential information:
Pd 2+(aq) + 2 e– → Pd(s)E° = +0.915 V
Pd(s) + 4 Cl–(aq) → [PdCl 4]2–(aq) + 2 e–E° = +0.600 V
Summing the two half-reactions gives the target reaction:
Pd 2+(aq) + 4 Cl–(aq) →← [PdCl 4]2–(aq)
and the 'potential' for the desired reaction is given by E° = +0.915 – 0.600 = +0.315 V. The equilibrium constant is found from the relationship E° = (RT/nF)lnK. Solving gives:
K = exp[nFE°/RT] = exp[(2)(96500)(0.315)/(8.314)(298)] = 4.5×10 10
Many of the tabulated data for standard potentials have been determined from thermochemical data rather than direct electrochemical measurements of cell potentials. Carry out a calculation to illustrate this approach for the half-reaction Sc 2 O 3(s) + 3 H 2 O(l) + 6 e– → 2 Sc(s) + 6 OH–(aq).
Sc 3+(aq)OH–(aq)H 2 O(l)Sc 2 O 3(s)Sc(s)
ΔH f° (kJ/mol)–614.2–230.0–285.8–1908.7 0
S m°(J K–1 mol–1)–255.2–10.75+69.91+77.0+34.76
Direct calculation of the reduction potential from the half-cell is not possible because the heat of formation and absolute entropy for the electron are not known. Thus, an oxidation reaction with known potential must be used:
Sc(s) → Sc 3+(aq) + 3e–E° = 2.03 V
Multiplying by 2 and adding this to the unknown reaction gives a net reaction of
Sc 2 O 3(s) + 3 H 2 O(l) → 2 Sc 3+(aq) + 6 OH–(aq)
The heat of reaction for the net reaction is:
ΔH° = [2Δ f H°(Sc 3+(aq)) + 6Δ f H°(OH–(aq))] – [Δ f H°(Sc 2 O 3(s)) + 3Δ f H°(H 2 O(l))]
= [2(–614.2) + 6(–230.0)] – [(–1908.7) + 3(–285.8)] = +157.7 kJ mol–1
The change in entropy for the net reaction is:
ΔS° = [2 S m°(Sc 3+(aq)) + 6 S m°(OH–(aq))] – [S m°(Sc 2 O 3(s)) + 3 S m°(H 2 O(l))]
= [2(–255.2) + 6(–10.75)] – [(+77.0) + 3(+69.91)] = –861.6 J K–1 mol–1
At 298 K:
ΔG° = ΔH° – TΔS° = +157.7 – (298)(–0.8616) = +414.5 kJ mol–1
ΔG° = ΔG°reduction + ΔG°oxidation = ΔG°reduction + –nFE°
414500 J mol–1 = ΔG°reduction – (6)(96485)(2.03)
ΔG°reduction = 414500 + 1175000 = 1590000 J mol–1
E°reduction = –1590000/(6)(96485) = –2.75 V
Calculate the reduction potential at 25 °C for the conversion of MnO 4–(aq) to MnO 2(s) in aqueous solution at pH = 9.00 and 1 M MnO 4–(aq) given that E°(MnO 4–, MnO 2) = +1.69 V.
The balanced half-reaction under basic conditions is:
MnO 4–(aq) + 2 H 2 O(l) + 3 e– → MnO 2(s) + 4 OH–(aq)
To find the potential at pH = 9.00 ([OH–] = 1.0×10–5 M) the Nernst equation must be used.
Balance the following redox reaction in acid solution: MnO 4–(aq) + H 2 SO 3(aq) → Mn 2+(aq) + HSO 4–(aq). Predict the qualitative pH dependence on the net potential for this reaction (i.e. increases, decreases, remains the same).
Split the reaction into the two half-reactions and balance each separately:
MnO 4–(aq) + 8 H+(aq) + 5 e– → Mn 2+(aq) + 4 H 2 O(l)
H 2 SO 3(aq) + H 2 O(l) → HSO 4–(aq) + 3 H+(aq) + 2 e–
so the net reaction is:
2 MnO 4–(aq) + 5 H 2 SO 3(aq) + H+(aq) → 2 Mn 2+(aq) + 5 HSO 4–(aq) + 3 H 2 O(l)
Since hydrogen ion is a reactant, an increase in pH lowers the hydrogen ion concentration, driving the reaction towards reactants, thereby decreasing the net potential.
This can be thought of in terms of the Nernst equation, as well:
That is, increasing the pH subtracts from the standard potential so the net potential decreases with increasing pH.
From the following Latimer diagram, calculate the net E° value for the reaction 2 HO 2(aq) → O 2(g) + H 2 O 2(aq)
Given your value of E°, comment on the thermodynamic tendency of HO 2 to undergo disproportionation.
For the disproportionation reaction E° = +0.125 + 1.510 = + 1.635 V. This is a large positive value so the disproportionation is strongly favored thermodynamically.
Comment on the likelihood that the following reactions occur by a simple out-sphere electron transfer, simple atom transfer, or a multistep mechanism:
(a) HIO(aq) + I–(aq) → I 2(aq) + OH–(aq)
(b) [Co(phen)3]3+(aq) + [Cr(bipy)3]2+(aq) → [Co(phen)3]2+(aq) + [Cr(bipy)3]3+(aq)
(c) IO 3–(aq) + 8 I–(aq) + 6 H+(aq) → 3 I 3–(aq) + 3 H 2 O(l)
(a) HIO(aq) + I–(aq) → I 2(aq) + OH–(aq)
This reaction is probably a simple atom transfer. The iodine atom in HIO bonds to the iodide ion to form the iodine molecule, leaving hydroxide ion behind.
(b) [Co(phen)3]3+(aq) + [Cr(bipy)3]2+(aq) → [Co(phen)3]2+(aq) + [Cr(bipy)3]3+(aq)
This reaction is almost certainly a simple outer-sphere electron transfer. Both reactants have relatively small metal ions from the first transition series bound with bulky Lewis bases, which prevents any inner-sphere bond formation for steric reasons. Further, the phen and bipy Lewis bases do not have any available lone pairs to bond to a neighboring reactant to form an inner-sphere transition state.
(c) IO 3–(aq) + 8 I–(aq) + 6 H+(aq) → 3 I 3–(aq) + 3 H 2 O(l)
This is a multistep mechanism. Given the stoichiometry, it is unlikely that a single electron transfer step, either outer-sphere or inner-sphere, could transfer the required atoms and electrons from one transition state.
Complete and balance the following oxidation-reduction reactions.
a) H 2 O 2(aq) → O 2(g)(pH = 0)
b) MnO 4–(aq) + C 2 O 4 2–(aq) → CO 2(g) + Mn 2+(aq)(pH = 0)
c) MnO 4–(aq) + CH 3 CH 2 OH(aq) → CH 3 CHO(aq) + MnO 2(s)(pH = 14)
a) H 2 O 2(aq) → O 2(g)(pH = 0)
Reduction half-reaction:H 2 O 2(aq) + 2 H+(aq) + 2 e– → 2 H 2 O(l)
Oxidation half-reaction:H 2 O 2(aq) → O 2(g) + 2 H+(aq) + 2 e–
Net reaction:2 H 2 O 2(aq → O 2(g) + 2 H 2 O(l)
b) MnO 4–(aq) + C 2 O 4 2–(aq) → CO 2(g) + Mn 2+(aq)(pH = 0)
Reduction half-reaction:MnO 4–(aq) + 8 H+(aq) + 5 e– → Mn 2+(aq) + 4 H 2 O(l)
Oxidation half-reaction:C 2 O 4 2–(aq) → 2 CO 2(g) + 2 e–
Net reaction:2 MnO 4–(aq) + 5 C 2 O 4 2–(aq) + 16 H+(aq) → 2 Mn 2+(aq) + 10 CO 2(g) + 8 H 2 O(l)
c) MnO 4–(aq) + CH 3 CH 2 OH(aq) → CH 3 CHO(aq) + MnO 2(s)(pH = 14)
Reduction half-reaction:MnO 4–(aq) + 4 H+(aq) + 3 e– → MnO 2(s) + 2 H 2 O(l)
Oxidation half-reaction:CH 3 CH 2 OH(aq) → CH 3 CHO(aq) + 2 H+(aq) + 2 e–
Net reaction (in acid):2 MnO 4–(aq) + 3 CH 3 CH 2 OH(aq) + 2 H+(aq) → 2 MnO 2(s) + 3 CH 3 CHO(aq) + 4 H 2 O(l)
Acid/base neutralization:2 H 2 O(l) → 2 H+(aq) + 2 OH–(aq)
Net reaction (in base):2 MnO 4–(aq) + 3 CH 3 CH 2 OH(aq) → 2 MnO 2(s) + 3 CH 3 CHO(aq) + 2 H 2 O(l) + 2 OH–(aq) |
3099 | https://en.wikipedia.org/?title=Granulosa_cell_tumor&redirect=no | Granulosa cell tumor - Wikipedia
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